SOURCES IN RECREATIONAL
MATHEMATICS
AN ANNOTATED BIBLIOGRAPHY
EIGHTH PRELIMINARY EDITION
DAVID SINGMASTER
Contact via the Puzzle Museum
http://puzzlemuseum.com
Last
updated on 19 March 2004.
This is a copy of the current version
from my source files. I had intended to
reorganise the material before producing a Word version, but have decided to
produce this version for G4G6 and to renumber it as the Eighth Preliminary
Edition.
If I have perchance omitted anything
more or less proper or necessary, I beg indulgence, since there is no one who
is blameless and utterly provident in all things. [Fibonacci, translated by Grimm.])
DIACRITICAL MARKS AND NOTATION
SOME OTHER RECURRING REFERENCES
1. BIOGRAPHICAL
MATERIAL -- in chronological order
2. GENERAL
PUZZLE COLLECTIONS AND SURVEYS
3. GENERAL
HISTORICAL AND BIBLIOGRAPHICAL MATERIAL
3.A. GENERAL
HISTORICAL MATERIAL
4.A. GENERAL
THEORY AND NIM‑LIKE GAMES
4.B.1. TIC‑TAC‑TOE =
NOUGHTS AND CROSSES
4.B.5. OVID'S GAME AND NINE MEN'S MORRIS
4.B.12. RITHMOMACHIA
= THE PHILOSOPHERS' GAME
5.A.2. THREE DIMENSIONAL VERSIONS
5.B.1. LOWERING FROM TOWER PROBLEM
5.B.2. CROSSING A BRIDGE WITH A TORCH
5.C. FALSE
COINS WITH A BALANCE
5.C.1RANKING COINS WITH A BALANCE
5.D.2. RULER WITH MINIMAL NUMBER OF MARKS
5.D.3 FALSE COINS WITH A WEIGHING SCALE
5.D.4. TIMING WITH HOURGLASSES
5.E.2. MEMORY WHEELS =
CHAIN CODES
5.F.1. KNIGHT'S TOURS AND PATHS
5.F.2. OTHER HAMILTONIAN CIRCUITS
5.F.3. KNIGHT'S TOURS IN HIGHER DIMENSIONS
5.F.4. OTHER CIRCUITS IN AND ON A CUBE
5.G.1. GAS, WATER AND ELECTRICITY
5.H. COLOURED
SQUARES AND CUBES, ETC.
5.H.1. INSTANT INSANITY =
THE TANTALIZER
5.I. LATIN
SQUARES AND EULER SQUARES
5.I.2. COLOURING CHESSBOARD WITH NO REPEATS
IN A LINE
5.J.3. TILING A SQUARE OF SIDE 70 WITH
SQUARES OF SIDES 1, 2, ..., 24
5.K.1. DERANGED BOXES OF A,
B AND A & B
5.K.2. OTHER LOGIC PUZZLES BASED ON
DERANGEMENTS
5.M. SIX
PEOPLE AT A PARTY -- RAMSEY THEORY
5.N. JEEP OR
EXPLORER'S PROBLEM
5.O. TAIT'S
COUNTER PUZZLE: BBBBWWWW TO
WBWBWBWB
5.P. GENERAL
MOVING PIECE PUZZLES
5.Q. NUMBER
OF REGIONS DETERMINED BY N LINES OR PLANES
5.Q.1. NUMBER
OF INTERSECTIONS DETERMINED BY N LINES
5.R.2. FROGS AND TOADS: BBB_WWW
TO WWW_BBB
5.R.3. FORE AND AFT -- 3
BY 3 SQUARES MEETING AT A CORNER
5.R.4. REVERSING FROGS AND TOADS: _12...n
TO _n...21 , ETC.
5.S. CHAIN
CUTTING AND REJOINING
5.S.1. USING CHAIN LINKS TO PAY FOR A ROOM
5.W. MAKING
THREE PIECES OF TOAST
5.X. COUNTING
FIGURES IN A PATTERN
5.X.2. COUNTING RECTANGLES OR SQUARES
5.Y. NUMBER
OF ROUTES IN A LATTICE
5.Z. CHESSBOARD
PLACING PROBLEMS
5.AB. FOLDING
A STRIP OF STAMPS
5.AC. PROPERTIES OF THE SEVEN BAR DIGITAL DISPLAY
5.AD. STACKING
A DECK TO PRODUCE A SPECIAL EFFECT
5.AG. RUBIK'S
CUBE AND SIMILAR PUZZLES
6.F.1. OTHER CHESSBOARD DISSECTIONS
6.F.2. COVERING DELETED CHESSBOARD WITH
DOMINOES
6.F.3. DISSECTING A CROSS INTO Zs
AND Ls
6.F.4. QUADRISECT AN L‑TROMINO, ETC.
6.F.5. OTHER DISSECTIONS INTO POLYOMINOES
6.G.2. DISSECTION OF 63 INTO 33, 43 AND 53, ETC.
6.G.3. DISSECTION OF A DIE INTO NINE 1 x 1 x 3
6.G.4. USE OF OTHER POLYHEDRAL PIECES
6.I. SYLVESTER'S
PROBLEM OF COLLINEAR POINTS
6.J. FOUR BUGS
AND OTHER PURSUIT PROBLEMS
6.K. DUDENEY'S
SQUARE TO TRIANGLE DISSECTION
6.N. DISSECTION
OF A 1 x 1 x 2 BLOCK TO A CUBE
6.O. PASSING
A CUBE THROUGH AN EQUAL OR SMALLER CUBE. PRINCE RUPERT'S PROBLEM
6.P.1. PARADOXICAL DISSECTIONS OF THE
CHESSBOARD BASED ON FIBONACCI NUMBERS 286
6.Q. KNOTTING
A STRIP TO MAKE A REGULAR PENTAGON
6.R.1. EVERY TRIANGLE IS ISOSCELES
6.R.2. A RIGHT ANGLE IS OBTUSE
6.R.3. LINES APPROACHING BUT NOT MEETING
6.T. NO THREE
IN A LINE PROBLEM
6.U.2. PACKING BRICKS IN BOXES
6.V. SILHOUETTE
AND VIEWING PUZZLES
6.W.2. SIX PIECE BURR =
CHINESE CROSS
6.W.3. THREE PIECE BURR WITH IDENTICAL
PIECES
6.W.4. DIAGONAL SIX PIECE BURR =
TRICK STAR
6.W.5. SIX PIECE BURR WITH IDENTICAL
PIECES
6.X. ROTATING
RINGS OF POLYHEDRA
6.Z. LANGLEY'S
ADVENTITIOUS ANGLES
6.AD.1. LARGEST PARCEL ONE CAN POST
6.AE. 6" HOLE THROUGH SPHERE LEAVES CONSTANT VOLUME
6.AF. WHAT
COLOUR WAS THE BEAR?
6.AJ.2. TRIBAR AND IMPOSSIBLE STAIRCASE
6.AK. POLYGONAL
PATH COVERING N x N LATTICE OF POINTS,
QUEEN'S TOURS, ETC.
6.AN. VOLUME OF
THE INTERSECTION OF TWO CYLINDERS
6.AO.1. PLACE FOUR POINTS EQUIDISTANTLY =
MAKE FOUR TRIANGLES WITH SIX MATCHSTICKS
6.AO.2. PLACE AN EVEN NUMBER ON EACH LINE
6.AP .
DISSECTIONS OF A TETRAHEDRON
6.AQ. DISSECTIONS
OF A CROSS, T OR H
6.AR. QUADRISECTED
SQUARE PUZZLE
6.AS. DISSECTION
OF SQUARES INTO A SQUARE
6.AS.1. TWENTY 1, 2, Ö5
TRIANGLES MAKE A SQUARE OR FIVE EQUAL SQUARES TO A SQUARE
6.AS.1.a. GREEK CROSS TO A SQUARE
6.AS.1.b. OTHER GREEK CROSS DISSECTIONS
6.AS.2. TWO (ADJACENT) SQUARES TO A SQUARE
6.AS.2.a. TWO EQUAL SQUARES TO A SQUARE
6.AS.3. THREE EQUAL SQUARES TO A SQUARE
6.AS.3.a. THREE EQUAL 'SQUARES' TO A HEXAGON
6.AS.4. EIGHT EQUAL SQUARES TO A SQUARE
6.AS.5. RECTANGLE TO A SQUARE OR OTHER
RECTANGLE
6.AT. POLYHEDRA
AND TESSELLATIONS
6.AT.2 STAR AND STELLATED POLYHEDRA
6.AT.5. REGULAR‑FACED POLYHEDRA
6.AT.6.a. TESSELLATING WITH CONGRUENT FIGURES
6.AU. THREE
RABBITS, DEAD DOGS AND TRICK MULES
MODERN VERSIONS OF THE THREE RABBITS PUZZLE
6.AV. CUTTING
UP IN FEWEST CUTS
6.AW. DIVISION
INTO CONGRUENT PIECES
6.AW.3. DIVIDING A SQUARE INTO CONGRUENT
PARTS
6.AW.4. DIVIDING AN L-TROMINO INTO CONGRUENT PARTS
6.AY. DISSECT 3A x 2B
TO MAKE 2A x 3B, ETC.
6.BA. CUTTING
A CARD SO ONE CAN PASS THROUGH IT
6.BB. DOUBLING
AN AREA WITHOUT CHANGING ITS HEIGHT OR WIDTH
6.BD. BRIDGE A MOAT WITH PLANKS
6.BE. REVERSE A
TRIANGULAR ARRAY OF TEN CIRCLES
6.BF.2. SLIDING SPEAR =
LEANING REED
6.BF.3. WELL BETWEEN TWO TOWERS
6.BF.5. TRAVELLING ON SIDES OF A RIGHT
TRIANGLE.
6.BG. QUADRISECT
A PAPER SQUARE WITH ONE CUT
6.BI. VENN DIAGRAMS FOR N SETS
6.BL. TAN-1
⅓ + TAN-1 ½ = TAN-1 1, ETC.
6.BM. DISSECT
CIRCLE INTO TWO HOLLOW OVALS
6.BN. ROUND PEG
IN SQUARE HOLE OR VICE VERSA
6.BP. EARLY MATCHSTICK
PUZZLES
6.BQ. COVERING A
DISC WITH DISCS
6.BR. WHAT IS A
GENERAL TRIANGLE?
6.BS. FORM SIX
COINS INTO A HEXAGON
6.BT. PLACING
OBJECTS IN CONTACT
6.BW. DISTANCES
TO CORNERS OF A SQUARE
7. ARITHMETIC &
NUMBER‑THEORETIC RECREATIONS
7.B. JOSEPHUS
OR SURVIVOR PROBLEM
7.E. MONKEY
AND COCONUTS PROBLEMS
7.E.1. VERSIONS WITH ALL GETTING THE SAME
7.F. ILLEGAL
OPERATIONS GIVING CORRECT RESULT
7.G.1. HALF + THIRD + NINTH, ETC.
7.H. DIVISION
AND SHARING PROBLEMS -- CISTERN PROBLEMS
7.H.1. WITH GROWTH -- NEWTON'S CATTLE
PROBLEM
7.H.3. SHARING UNEQUAL RESOURCES --
PROBLEM OF THE PANDECTS
7.H.4. EACH DOUBLES OTHERS' MONEY TO MAKE
ALL EQUAL, ETC.
7.H.5. SHARING COST OF STAIRS, ETC.
7.H.7. DIGGING PART OF A WELL.
7.I.1. LARGEST NUMBER USING FOUR ONES, ETC.
7.L.1. 1 + 7 + 49 + ... &
ST. IVES
7.L.2.b. HORSESHOE NAILS PROBLEM
7.L.2.c. USE OF 1, 2, 4, ... AS WEIGHTS, ETC.
7.L.3. 1 + 3 + 9 + ... AND OTHER SYSTEMS OF
WEIGHTS
7.M. BINARY
SYSTEM AND BINARY RECREATIONS
7.M.2.a. TOWER OF HANOI WITH MORE PEGS
7.M.4.b. OTHER DIVINATIONS USING BINARY OR
TERNARY
7.M.5. LOONY
LOOP =
GORDIAN KNOT
7.N.3. ANTI‑MAGIC SQUARES AND
TRIANGLES
7.P.1. HUNDRED FOWLS AND OTHER LINEAR
PROBLEMS
7.P.2. CHINESE REMAINDER THEOREM
7.P.3. ARCHIMEDES' CATTLE PROBLEM
7.P.5 . SELLING DIFFERENT AMOUNTS 'AT SAME
PRICES' YIELDING THE SAME
7.P.6. CONJUNCTION OF PLANETS, ETC.
7.Q.1. REARRANGEMENT ON A CROSS
7.Q.2. REARRANGE A CROSS OF SIX TO MAKE TWO
LINES OF FOUR, ETC.
7.R. "IF
I HAD ONE FROM YOU, I'D HAVE TWICE YOU"
7.R.1. MEN FIND A PURSE AND 'BLOOM OF
THYMARIDAS'
7.R.2. "IF I HAD 1/3
OF YOUR MONEY, I COULD BUY THE HORSE"
7.R.4. "IF I SOLD YOUR EGGS AT MY
PRICE, I'D GET ...."
7.S. DILUTION
AND MIXING PROBLEMS
7.S.1. DISHONEST BUTLER DRINKING SOME AND
REPLACING WITH WATER
7.S.2. WATER IN WINE VERSUS WINE IN WATER
7.V. XY = YX AND ITERATED EXPONENTIALS
7.X. HOW OLD
IS ANN? AND OTHER AGE PROBLEMS
7.Y. COMBINING
AMOUNTS AND PRICES INCOHERENTLY
7.Y.1. REVERSAL OF AVERAGES PARADOX
7.Z. MISSING
DOLLAR AND OTHER ERRONEOUS ACCOUNTING
7.AC. CRYPTARITHMS,
ALPHAMETICS AND SKELETON ARITHMETIC
7.AC.1. CRYPTARITHMS: SEND + MORE
= MONEY, ETC.
7.AC.2. SKELETON ARITHMETIC: SOLITARY SEVEN, ETC.
7.AC.3.a INSERTION OF SIGNS TO MAKE 100,
ETC.
7.AC.6. OTHER PAN‑DIGITAL AND SIMILAR
PROBLEMS
7.AC.7. SELF-DESCRIPTIVE NUMBERS, PANGRAMS,
ETC.
7.AD. SELLING, BUYING AND SELLING
SAME ITEM
7.AE. USE OF
COUNTERFEIT BILL OR FORGED CHEQUE
7.AH. MULTIPLYING
BY REVERSING
7.AH.1. OTHER REVERSAL PROBLEMS
7.AI.
IMPOSSIBLE EXCHANGE RATES
7.AJ.1. MULTIPLYING BY APPENDING DIGITS
7.AN. THREE
ODDS MAKE AN EVEN, ETC.
7.AO. DIVINATION
OF A PERMUTATION
7.AP. KNOWING SUM
VS KNOWING PRODUCT
7.AQ. NUMBERS IN
ALPHABETIC ORDER
7.AU. NUMBER OF
CUTS TO MAKE N PIECES
7.AV. HOW LONG
TO STRIKE TWELVE?
7.AZ.
DIVINATION OF A PAIR OF CARDS FROM ITS ROWS
7.BA. CYCLE OF
NUMBERS WITH EACH CLOSER TO TEN THAN THE PREVIOUS
7.BB. ITERATED
FUNCTIONS OF INTEGERS
7.BC. UNUSUAL DIFFICULTY
IN GIVING CHANGE
8.D. ATTEMPTS
TO MODIFY BOY‑GIRL RATIO
8.G. PROBABILITY
THAT THREE LENGTHS FORM A TRIANGLE
8.H.2. BERTRAND'S CHORD PARADOX
8.J. CLOCK
PATIENCE OR SOLITAIRE
9.A. ALL
CRETANS ARE LIARS, ETC.
9.B. SMITH
-- JONES -- ROBINSON PROBLEM
GENERAL STUDIES OF KINSHIP RELATIONS
9.E.1. THAT MAN'S FATHER IS MY FATHER'S SON,
ETC.
9.E.2. IDENTICAL SIBLINGS WHO ARE NOT TWINS
10.A. OVERTAKING
AND MEETING PROBLEMS
10.A.3. TIMES FROM MEETING TO FINISH GIVEN
10.A.6. DOUBLE CROSSING PROBLEMS
10.C. LEWIS
CARROLL'S MONKEY PROBLEM
10.D.1 MIRROR REVERSAL PARADOX
10.E.1. ARISTOTLE'S WHEEL PARADOX
10.E.2. ONE WHEEL ROLLING AROUND ANOTHER
10.E.4. RAILWAY WHEELS PARADOX
10.H. SNAIL
CLIMBING OUT OF WELL
10.I. LIMITED
MEANS OF TRANSPORT -- TWO MEN AND A BIKE, ETC.
10.K. PROBLEM
OF THE DATE LINE
10.L. FALLING
DOWN A HOLE THROUGH THE EARTH
10.N. SHIP'S
LADDER IN RISING TIDE
10.O. ERRONEOUS
AVERAGING OF VELOCITIES
10.U. SHORTEST
ROUTE VIA A WALL, ETC.
10.V. PICK UP
PUZZLES = PLUCK IT
10.X. HOW FAR
DOES A PHONOGRAPH NEEDLE TRAVEL?
10.Y. DOUBLE
CONE ROLLING UPHILL
10.AB. BICYCLE
TRACK PROBLEMS.
11.B. TWO
PEOPLE JOINED BY ROPES AT WRISTS
11.C. TWO BALLS
ON STRING THROUGH LEATHER HOLE AND STRAP
= CHERRIES PUZZLE
11.H. REMOVING
WAISTCOAT WITHOUT REMOVING COAT
11.H.1. REMOVING LOOP FROM ARM
11.I. HEART
AND BALL PUZZLE AND OTHER LOOP PUZZLES
11.K.1. RING AND SPRING PUZZLE
11.K.2. STRING AND SPRING PUZZLE
11.K.3. MAGIC CHAIN =
TUMBLE RINGS
11.K.6. INTERLOCKED NAILS, HOOKS, HORNS,
ETC.
11.L. JACOB'S
LADDER AND OTHER HINGING DEVICES
11.N. THREE
KNIVES MAKE A SUPPORT
11.Q. TURNING AN
INNER TUBE INSIDE OUT
Recreational mathematics is as old as
mathematics itself. Recreational
problems already occur in the oldest extant sources -- the Rhind Papyrus and
Old Babylonian tablets. The Rhind
Papyrus has an example of a purely recreational problem -- Problem 79 is like
the "As I was going to St. Ives" nursery rhyme. The Babylonians give fairly standard
practical problems with a recreational context -- a man knows the area plus the
difference of the length and width of his field, a measurement which no
surveyor would ever make! There is even
some prehistoric mathematics which could not have been practical -- numerous
'carved stone balls' have been found in eastern Scotland, dating from the
Neolithic period and they include rounded forms of all the regular polyhedra
and some less regular ones. Since these
early times, recreations have been a feature of mathematics, both as pure
recreations and as pedagogic tools. In
this work, I use recreational in a fairly broad sense, but I tend to omit the
more straightforward problems and concentrate on those which 'stimulate the
curiosity' (as Montucla says).
In addition, recreational mathematics
is certainly as diffuse as mathematics.
Every main culture and many minor ones have contributed to the
history. A glance at the Common
References below, or at almost any topic in the text, will reveal the diversity
of sources which are relevant to this study.
Much information arises from material outside the purview of the
ordinary historian of mathematics -- e.g.
patents; articles in newspapers,
popular magazines and minor journals;
instruction leaflets; actual
artifacts and even oral tradition.
Consequently, it is very difficult to
determine the history of any recreational topic and the history given in
popular books is often extremely dubious or even simply fanciful. For example, Nim, Tangrams, and Magic
Squares are often traced back to China of about 2000 BC. The oldest
known reference to Nim is in America in 1903.
Tangrams appear in China and Europe at essentially the same time, about
1800, though there are related puzzles in 18C Japan and in the Hellenistic
world. Magic Squares seem to be
genuinely a Chinese invention, but go back to perhaps a few centuries BC and
are not clearly described until about 80AD.
Because of the lack of a history of the field, results are frequently
rediscovered.
When I began this bibliography in
1982, I had the the idea of producing a book (or books) of the original
sources, translated into English, so people could read the original
material. This bibliography began as
the table of contents of such a book. I
thought that this would be an easy project, but it has become increasingly
apparent that the history of most recreations is hardly known. I have recently realised that mathematical
recreations are really the folklore of mathematics and that the historical
problems are similar to those of folklore.
One might even say that mathematical recreations are the urban myths or
the jokes or the campfire stories of mathematics. Consequently I decided that an annotated bibliography was the
first necessity to make the history clearer.
This bibliography alone has grown into a book, something like Dickson's History
of the Theory of Numbers. Like that
work, the present work divides the subject into a number of topics and treats
them chronologically.
I have printed six preliminary
editions of this work, with slightly varying titles. The first version of 4 Jul 1986 had 224 topics and was spaced out
so entries would not be spread over two pages and to give room for page
numbers. This stretched the text from
110pp to 129pp and was printed for the Strens Memorial Conference at the Univ.
of Calgary in Jul/Aug 1986. I no longer
worry about page breaks. The following
editions had: 250 topics on 152 pages; 290 topics on 192 pages; 307 topics on 223 pages; 357 topics on 311 pages and 392 topics on 456 pages. The seventh edition was never printed, but
was a continually changing computer file.
It had about 419 topics (as of 20 Oct 95) and 587 pages, as of 20 Oct
1995. I then carried out the conversion
to proportional spacing and this reduced the total length from 587 to 488
pages, a reduction of 16.87% which is conveniently estimated as 1/6. This reduction was fairly consistent
throughout the conversion process.
This eighth edition is being prepared
for the Gathering for Gardner 6 in March 2004.
The text is 818 pages as of 18 Mar 2004. There are about 457 topics as of 18 Mar 2004.
A fuller description of this project
in 1984-1985 is given in my article Some early sources in recreational
mathematics, in: C. Hay et al., eds.; Mathematics from Manuscript to
Print; Oxford Univ. Press, 1988, pp. 195‑208. A more recent description is in my article: Recreational
mathematics; in: Encyclopedia of the History and Philosophy of the
Mathematical Sciences; ed. by I. Grattan-Guinness; Routledge & Kegan
Paul, 1993; pp. 1568-1575.
Below I compare this work with Dickson
and similar works and discuss the coverage of this work.
As already mentioned, the work which
the present most resembles is Dickson's History of the Theory of Numbers.
The history of science can
be made entirely impartial, and perhaps that is what it should be, by merely
recording who did what, and leaving all "evaluations" to those who
like them. To my knowledge there is
only one history of a scientific subject (Dickson's, of the Theory of Numbers)
which has been written in this coldblooded, scientific way. The complete success of that unique example
-- admitted by all who ever have occasion to use such a history in their work
-- seems to indicate that historians who draw morals should have their own
morals drawn.
E. T. Bell. The Search for Truth. George Allen & Unwin, London, 1935,
p. 131.
Dickson attempted to be exhaustive and
certainly is pretty much so. Since his time,
many older sources have been published, but their number-theoretic content is
limited and most of Dickson's topics do not go back that far, so it remains the
authoritative work in its field.
The best previous book covering the
history of recreational mathematics is the second edition of Wilhelm Ahrens's Mathematische
Unterhaltungen und Spiele in two volumes.
Although it is a book on recreations, it includes extensive histories of
most of the topics covered, far more than in any other recreational book. He also gives a good index and a
bibliography of 762 items, often with some bibliographical notes. I will indicate the appropriate pages at the
beginning of any topic that Ahrens covers.
This has been out of print for many years but Teubner has some plans to
reissue it.
Another similar book is the 4th
edition of J. Tropfke's Geschichte der Elementarmathematik, revised by
Vogel, Reich and Gericke. This is quite
exhaustive, but is concerned with older problems and sources. It presents the material on a topic as a
history with references to the sources, but it doesn't detail what is in each
of the sources. Sadly, only one volume,
on arithmetic and algebra, appeared before Vogel's death. A second volume, on geometry, is being
prepared. For any topic covered in
Tropfke, it should be consulted for further references to early material which
I have not seen, particularly material not available in any western
language. I cite the appropriate pages
of Tropfke at the beginning of any topic covered by Tropfke.
Another book in the field is W. L.
Schaaf's Bibliography of Recreational Mathematics, in four volumes. This is a quite exhaustive bibliography of
recent articles, but it is not chronological, is without annotation and is
somewhat less classified than the present work. Nonetheless it is a valuable guide to recent material.
Collecting books on magic has been
popular for many years and quite notable collections and bibliographies have
been made. Magic overlaps recreational
mathematics, particularly in older books, and I have now added references to
items listed in the bibliographies of Christopher, Clarke & Blind, Hall,
Heyl, Toole Stott and Volkmann & Tummers -- details of these works are
given in the list of Common References below.
There is a notable collection of Harry Price at Senate House, University
of London, and a catalogue was printed in 1929 & 1935 -- see HPL in Common
References.
Another related bibliography is
Santi's Bibliografia della Enigmistica, which is primarily about word
puzzles, riddles, etc., but has some overlap with recreational mathematics --
again see the entry in the list of Common References. I have not finished working through this.
Other relevant bibliographies are
listed in Section 3.B.
In selecting topics, I tend to avoid
classical number theory and classical geometry. These are both pretty well known. Dickson's History of the Theory of Numbers and Leveque's
and Guy's Reviews in Number Theory cover number theory quite well. I also tend to avoid simple exercises, e.g.
in the rule of three, in 'aha' or 'heap' problems, in the Pythagorean theorem
(though I have now included 6.BF) or in two linear equations in two unknowns,
though these often have fanciful settings which are intended to make them
amusing and some of these are included -- see
7.R, 7.X, 7.AX.
I also leave out most divination (or 'think of a number') techniques
(but a little is covered in 7.M.4.b) and most arithmetic fallacies. I also leave out Conway's approach to
mathematical games -- this is extensively covered by Winning Ways and
Frankel's Bibliography.
The classification of topics is still
ad‑hoc and will eventually get rationalised -- but it is hard to sort
things until you know what they are! At
present I have only grouped them under the general headings: Biography,
General, History & Bibliography, Games, Combinatorics, Geometry,
Arithmetic, Probability, Logic, Physics, Topology. Even the order of these should be amended. The General section should be subsumed under
the History & Bibliography.
Geometry and Arithmetic need to be subdivided.
I have recently realised that some
general topics are spread over several sections in different parts. E.g. fallacies are covered in 6.P, 6.R,
6.AD, 6.AW.1, 6.AY, 7.F, 7.Y, 7.Z, 7.AD, 7.AI, 7.AL, 7.AN, most of 8, 10.D,
10.E, 10.O. Perhaps I will produce an
index to such topics. I try to make
appropriate cross-references.
Some topics are so extensive that I
include introductory or classifactory material at the beginning. I often give a notation for the problems
being considered. I give brief
explanations of those problems which are not well known or are not described in
the notation or the early references.
There may be a section index. I
have started to include references to comprehensive surveys of a given topic --
these are sometimes given at the beginning.
Recreational problems are repeated so
often that it is impossible to include all their occurrences. I try to be exhaustive with early material,
but once a problem passes into mathematical and general circulation, I only
include references which show new aspects of the problem or show how the
problem is transmitted in time and/or space.
However, the point at which I start leaving out items may vary with time
and generally slowly increases as I learn more about a topic. I include numerous variants and developments
on problems, especially when the actual origin is obscure.
When I began, I made minimal
annotations, often nothing at all. In rereading
sections, particularly when adding more material, I have often added
annotations, but I have not done this for all the early entries yet.
Recently added topics often may exist
in standard sources that I have not reread recently, so the references for such
topics often have gaps -- I constantly discover that Loyd or Dudeney or Ahrens
or Lucas or Fibonacci has covered such a topic but I have forgotten this --
e.g. looking through Dudeney recently, I added about 15 entries. New sections are often so noted to indicate
that they may not be as complete as other sections.
Some of the sources cited are lengthy
and I originally added notes as to which parts might be usable in a book of
readings -- these notes have now been mostly deleted, but I may have missed a
few.
I would like to think that I am about
75% of the way through the relevant material.
However, I recently did a rough measurement of the material in my study
-- there is about 8 feet of read but unprocessed material and about 35 feet of
unread material, not counting several boxes of unread Rubik Cube material and
several feet of semi-read material on my desk and table. I recently bought two bookshelves just to
hold unread material. Perhaps half of
this material is relevant to this work.
In particular, the unread material
includes several works of Folkerts and Sesiano on medieval MSS, a substantial
amount of photocopies from Schott, Schwenter and Dudeney (400 columns), some
2000 pages of photocopies recently made at Keele, some 500 pages of photocopies
from Martin Gardner's files, as well as a number of letters. Marcel Gillen has made extracts of all US,
German and EURO patents and German registered designs on puzzles -- 26 volumes,
occupying about two feet on my shelves.
I have recently acquired an almost complete set of Scripta
Mathematica (but I have previously read about half of it),
Schwenter-Harsdörffer's Deliciæ Physico‑Mathematicae, Schott's Joco-Seriorum
and Murray's History of Board Games Other Than Chess. I have recently acquired the early issues of Eureka,
but there are later issues that I have not yet read and they persist in not
sending the current copies I have paid for!
I have not yet seen some of the
earlier 19C material which I have seen referred to and I suspect there is much
more to be found. I have examined some
18C & 19C arithmetic and algebra books looking for problem sections --
these are often given the pleasant name of Promiscuous Problems. There are so many of these that a reference
to one of them probably indicates that the problem appears in many other
similar books that I have not examined.
My examination is primarily based on those books which I happen to have
acquired. There are a few 15-17C books
which I have not yet examined, notably those included at the end of the last
paragraph.
In working on this material, it has
become clear that there were two particularly interesting and productive eras
in the 19C. In the fifteen years from
1857, there appeared about a dozen books in the US and the UK: The
Magician's Own Book (1857); Parlour
Pastime, by "Uncle George" (1857); The Sociable (1858);
The Boy's Own Toymaker, by Landells (1858); The Book of 500 Curious Puzzles
(1859); The Secret Out
(1859); Indoor and Outdoor Games for
Boys and Girls (c1859); The
Boy's Own Conjuring Book (1860); The
Illustrated Boy's Own Treasury (1860, but see below); The Parlor Magician (1863); The Art of Amusing, by Bellew
(1866); Parlour Pastimes
(1868); Hanky Panky (1872); Within Doors, by Elliott (1872); Magic No Mystery (1876), just to name
those that I know. Most of these are of
uncertain authorship and went through several editions and versions. The Magician's Own Book, The Book
of 500 Curious Puzzles, The Secret Out, The Sociable, The
Parlor Magician, Hanky Panky, and Magic No Mystery seem to be
by the same author(s). I have recently
had a chance to look at a number of previously unseen versions at Sotheby's and
at Edward Hordern's and I find that sometimes two editions of the same title are
essentially completely different! This
is particularly true for US and UK editions.
Many of the later UK editions say 'By the author of Magician's Own Book
etc., translated and edited by W. H. Cremer Jr.' From the TPs, it appears that they were written by Wiljalba
Frikell (1818‑1903) and then translated into English. However, BMC and NUC generally attribute the
earlier US editions to George Arnold (1834-1865), and some catalogue entries
explicitly say the Frikell versions are later editions, so it may be that
Frikell produced later editions in some other language (French or German ??)
and these were translated by Cremer. On
the other hand, the UK ed of The Secret Out says it is based on Le
Magicien des Salons. This is
probably Le Magicien des Salons ou le Diable Couleur de Rose, for which
I have several references, with different authors! -- J. M. Gassier, 1814; M.
[Louis Apollinarie Christien Emmanuel] Comte, 1829; Richard (pseud. of A. O. Delarue), 1857 and earlier. There was a German translation of this. Some of these are at HPL but ??NYS. Items with similar names are: Le Magicien de Société, Delarue,
Paris, c1860? and Le Manuel des Sorciers (various Paris
editions from 178?-1825, cf in Common References). It seems that this era was inspired by these earlier French
books. To add to the confusion, an
advertisement for the UK ed. of Magician's Own Book (1871?) says it is
translated from Le Magicien des Salons which has long been a standard in
France and Germany. Toole Stott opines
that Frikell had nothing to do with these books -- as a celebrated conjuror of
the times, his name was simply attached to the books. Toole Stott also doubts whether Le Magicien des Salons
exists -- but it now seems pretty clear that it does, though it may not have
been the direct source for any of these works, but see below.
Christopher 242 cites the following
article on this series.
Charles L. Rulfs. Origins of some conjuring works. Magicol 24 (May 1971) 3-5. He discusses the various books, saying that
Cremer essentially pirated the Dick & Fitzgerald productions. He says The Magician's Own Book draws
on Wyman's Handbook (1850, ??NYS), Endless Amusement, Parlour
Magic (by W. Clarke?, 1830s, ??NYS), Brewster's Natural Magic
(??NYS). He says The Secret Out
is largely taken, illustrations and all, from Blismon de Douai's Manuel du
Magicien (1849, ??NYS) and Richard & Delion's Magicien des salons ou
le diable couleur de rose (1857 and earlier, ??NYS).
Christopher 622 says Harold Adrian
Smith [Dick and Fitzgerald Publishers; Books at Brown 34 (1987) 108-114]
has studied this series and concludes that Williams was the author of Magician's
Own Book, assisted by Wyman.
Actually Smith simply asserts: "The book was undoubtedly [sic]
written by H. L. Williams, a "hack writer" of the period, assisted by
John Wyman in the technical details."
He gives no explanation for his assertion. He later says he doubts whether Cremer ever wrote anything. He suggests The Secret Out book is
taken from DeLion. He states that The
Boy's Own Conjuring Book is a London pirate edition.
Several of the other items are
anonymous and there was a tremendous amount of copying going on -- problems are
often reproduced verbatim with the same diagram or sometimes with minor
changes. In some cases, the same error
is repeated in five different books! I
have just discovered some earlier appearances of the same material in The
Family Friend, a periodical which ran in six series from 1849 to 1921 and
which I have not yet tracked down further.
However, vol. 1-3 of 1849‑1850 and the volume for Jul‑Dec
1859 contain a number of the problems which appear repeatedly and identically
in the above cited books. Toole Stott
407 is an edition of The Illustrated Boy's Own Treasury of c1847 but the
BM copy was destroyed in the war and the other two copies cited are in the
US. If this date is correct, then this
book is a forerunner of all the others and a major connection between Boy's
Own Book and Magician's Own Book.
I would be most grateful to anyone who can help sort out this material
-- e.g. with photocopies of these or similar books or magazines.
The other interesting era was about
1900. In English, this was largely
created or inspired by Sam Loyd and Henry Dudeney. Much of this material first appeared in magazines and
newspapers. I have seen much less than
half of Loyd's and Dudeney's work and very little of similar earlier material
(but see below). Consequently problems
due to Loyd or Dudeney may seem to first appear in the works of Ball (1892, et
seq.), Hoffmann (1893) and Pearson (1907).
Further examination of Loyd's and Dudeney's material will be needed to
clarify the origin and development of many problems. Though both started puzzle columns about 1896, they must have
been producing material for a decade or more previously which does not seem to
be known. I have just obtained
photocopies of 401 columns by Dudeney in the Weekly Dispatch of
1897-1903, but have not had time to study them. Will Shortz and Angela Newing have been studying Loyd and Dudeney
respectively and turning up their material.
The works of Lucas (1882‑1895),
Schubert (1890s) and Ahrens (1900‑1918) were the main items on the
Continent and they interacted with the English language writers. Ahrens was the most historical of these and
his book is one of the foundations of the present work. All of these also wrote in newspapers and
magazines and I have not seen all their material.
I would be happy to hear from anyone
with ideas or suggestions for this bibliography. I would be delighted to hear from anyone who can locate missing
information or who can provide copies of awkward material. I am particularly short of information about
recreations in the Arabic period. I
prepared a separate file, 'Queries and Problems in the History of Recreational
Mathematics', which is about 23 pages, and has recently been updated. I have also prepared three smaller letters
of queries about Middle Eastern, Oriental and Russian sources and these are
generally more up-to-date.
I have prepared a CD containing this
and much else of my material. I divided
Sources into four files when I used floppy discs as it was too big to
fit on one disc, and I have not yet changed this. The files are: 1:
Introductory material and list of abbreviations/references; 2: Sections 1 - 6; 3: Section 7; 4: Sections 8 - 11. It is
convenient to have the first file separate from the main material, but I might
combine the other three files. (I have
tried to send it by email in the past, but this document is very large
(currently c4.1MB and the Word version will be longer) and most people who
requested it by email found that it overflowed their mailbox and created chaos
in their system -- this situation has changed a bit with larger memories and
improved transmission speeds.)
This file started on a DEC-10, then
was transferred to a VAX. It is now on
my PC using Script Professional, the development of LocoScript on the
Amstrad. Even in its earliest forms, this
provided an easy and comprehensive set of diacritical marks, which are still
not all available nor easy to use in WordPerfect or Word (except perhaps by
using macros and/or overstriking??). It
also provides multiple cut and paste buffers and easy formatting, though I have
learned how to overcome these deficiencies in Word.
Script provides an ASCII output, but
this uses IBM extended ASCII which has 8-bit codes. Not all computers will accept or print such characters and
sometimes they are converted into printer control codes causing considerable
confusion. I have a program that
converts these codes to 7-bits -- e.g. accents and umlauts are removed. However, ASCII loses a great deal of the
information, such as sub- and superscripts, so this is not a terribly useful
format.
Script also provides WordStar and
"Revisable-Form-Text DCA" output, but neither of these seems to be
very successful (DCA is better than WordStar).
Script later added a WordPerfect exporting facility. This works well, though some (fairly rare)
characters and diacritical marks are lost and the output requires some
reformatting. (Nob Yoshigahara reports
that Japanese WordPerfect turns all the extended ASCII characters into Kanji
characters!)
Reading the WordPerfect output in Word
(you may need to install this facility) gives a good approximation to my text,
but in Courier 10pt. Selecting All and changing to Times New
Roman 12pt gives an better
approximation. (Some files use a
smaller font of 10pt and I may have done some into 9pt.) You have to change this in the Header
separately, using View Header and Footer. The page layout is awkward as my page numbering header gets put
into the text, leaving a large gap at the top.
I go into Page Setup and set the
Paper Size to A4 and the Top, Bottom and Header Margins to 15mm and the Left and Right Margins to
25mm. (It has taken me some time to
work this out and some earlier files may have other settings.) However, I find that lines are a bit too
close together and underlines and some diacritical marks are lost, so one needs
to also go into Format Paragraph
Spacing -- Line Spacing and
choose At least and
12pt (or 10pt). I use hanging indentation in most of the
main material and this feature is not preserved in this conversion. By selecting a relevant section and going
into Format Paragraph Indentation --
Special and selecting Hanging,
it should automatically select
10.6mm which corresponds to my
automatic spacing of five characters in 12pt.
Further, I use second level hanging indentation in quite a number of
places. You need to create a style
which is the basic style with the left hand margin at 10.6mm (or 10 or 11 mm).
When second level indenting is needed, select the desired section and
apply this style to it.
However, this still leaves some
problems. I use em dashes a bit,
i.e. –, which gets converted into an underline, _. In Word, this is
obtained by use of CTRL and the
- sign on the numeric
keypad. One can use the find and
replace feature, EXCEPT that a number of other characters are also converted
into underlines. In particular,
Cyrillic characters are all converted into underlines. This is not insuperable as I always(?) give
a transliteration of Cyrillic (using the current Mathematical Reviews
system) and one can reconstruct the original Cyrillic from it. I notice that the Cyrillic characters are
larger than roman characters and hence may overlap. One can amend this by selecting the Cyrillic text and going
into Format Font Character
Spacing Spacing and choosing Expanded By 2 pt (or thereabout). But a number of characters with unusual
diacritical marks are also converted to underlines or converted to the unmarked
character and not all of these are available in Word. E.g. ĭ, which is the transliteration of й
becomes just i. I am slowly forming a Word file containing
the Word versions of entries having the Cyrillic or other odd characters, and I
will include this file on my CD, named
CYRILLIC.DOC. For diacritical
marks not supported by Word, I use an approximation and/or an explanation.
It is very tedious to convert the
underlines back to em dashes, so I will convert every em dash to a double
hyphen --.
Finally, I have made a number of
diagrams by simple typing without proportional spacing and Word does not permit
changing font spacing in mid-line and ignores spaces before a right-alignment
instruction. The latter problem can be
overcome by using hard spaces and the former problem is less of a problem, and
I think it can be overcome.
Later versions of Script support
Hewlett-Packard DeskJets and I am now on my second generation of these, so the
7th and future editions will be better printed (if they ever are!). However, this required considerable
reformatting as the text looks best in proportional spacing (PS) and I found I
had to check every table and every mathematical formula and diagram. Also, to set off letters used as
mathematical symbols within text, I find PS requires two spaces on each side of
the letter -- i.e. I refer to x
rather than to x. (I find this
easier to do than to convert to italics.)
I also sometimes set off numbers with two spaces, though I wasn't
consistent in doing this at the beginning of my reformatting. The conversion to proportional spacing
reduced the total length from 587 to
488 pages, a reduction of 16.87%
which is conveniently estimated as
1/6. The percentage of reduction
was fairly consistent throughout the conversion process.
The printing of Greek characters went
amiss in the second part of the 6th Preliminary Edition, apparently due to the
printer setting having been changed without my noticing -- this happens if an
odd character gets sent to the printer, usually in DOS when trying to use or
print a corrupted file, and there is no indication of it. I was never able to reproduce the effect!
The conversion to (Loco)Script
provided many improved features compared to my earlier DEC versions. I am using an A4 page (8¼
by 11⅔ inches) rather than an 8½ by 11 inch page, which gives
60 lines of text per page, four
more or 7% more than when using the DEC or VAX.
[SIXTH
EDITION: 1: Fibonacci, 1: Montucla;
3.B; 4.A.1.a, 4.B.9, 4.B.10,
4.B.11, 4.B.12; 5.R.1.a, 5.W.1, 5.AA,
5.AB; 6.AS.1.b, 6.AS.2.a, 6.AS.5,
6.AW.4, 6.BP, 6.BQ, 6.BR; 7.I.1, 7.Y.2,
7.AY, 7.AZ; 7.BA; 8.I, 8.J; 9.E.2, 9.K;
10.A.4, 10.A.5, 10.U, 10.V, 10.W;
11.K.6, 11.K.7, 11.K.8.]
In
the last edition, I had 8.K instead of 8.J in the list of New Sections and in
the Contents.
1:
Pacioli, Carroll, Perelman; 4.B.13,
4.B.14, 4.B.15; 5.B.2, 5.H.3 (the
previous 5.H.3 has been renumbered 5.H.4), 5.K.3, 5.R.1.b, 5.X.4, 5.AC, 5.AD,
5.AE, 5.AF, 5.AG.1, 5.AG.2; 6.AJ.4,
6.AJ.5, 6.AS.3.a, 6.AT.8, 6.AT.9, 6.AY.2, 6.BF.4, 6.BF.5, 6.BS, 6.BT, 6.BU,
6.BV, 6.BW; 7.H.6, 7.H.7 (formerly part
of 7.H.5), 7.M.4.a, 7.M.4.b, 7.M.6, 7.R.4, 7.AC.3.a, 7.AC.7, 7.AH.1, 7.AJ.1,
7.BB, 7.BC; 8.K, 8.L; 10.A.6, 10.A.7, 10.A.8, 10.D has become
10.D.1, 10.D.2, 10.D.3, 10.E.4, 10.X, 10.Y, 10.Z, 10.AA, 10.AB, 10.AC, 10.AD,
10.AE; 11.N, 11.O, 11.P, 11.Q, 11.R,
11.S. (65 new sections)
I am immensely indebted to many
mathematicians, historians, puzzlers, bookdealers and others who have studied particular
topics, as will be apparent.
I have had assistance from so many
sources that I have probably forgotten some, but I would like to give thanks
here to the following, and beg forgiveness from anyone inadvertently omitted --
if you remind me, I will make amendment.
In some cases, I simply haven't got to your letter yet! Also I have had letters from people whose
only identification is an undecipherable signature and phone messages from
people whose name and phone number are unintelligible.
Sadly, a few of these have died since
I corresponded with them and I have indicated those known to me with †.
André
Allard, Eric J. Aiton†, Sue Andrew,
Hugh Ap Simon, Gino Arrighi, Marcia Ascher, Mohammad Bagheri,
Banca Commerciale Italiana,
Gerd Baron, Chris Base, Rainier [Ray] Bathke, John Beasley, Michael Behrend, Jörg
Bewersdorff, Norman L. Biggs, C. [Chris] J. Bouwkamp, Jean Brette, John Brillhart, Paul
J. Campbell,
Cassa di Risparmio di Firenze, Henry Cattan, Marianna Clark,
Stewart Coffin,
Alan & Philippa Collins,
John H. Conway,
H. S. M. Coxeter,
James Dalgety,
Ann E. L. Davis,
Yvonne Dold, Underwood Dudley, Anthony W. F. Edwards, John Ergatoudis, John Fauvel†, Sandro Ferace,
Judith V. Field,
Irving Finkel,
Graham Flegg,
Menso Folkerts,
David Fowler,
Aviezri S. Fraenkel,
Raffaella Franci,
Gregory N. Frederickson,
Michael Freude,
Walter W. Funkenbusch,
Nora Gädeke,
Martin Gardner,
Marcel Gillen, Leonard
J. Gordon, Ron Gow, Ivor Grattan‑Guinness, Christine Insley Green, Jennifer Greenleaves (Manco), Tom Greeves, H. [Rik] J. M. van Grol, Branko Grünbaum, Richard K. Guy, John Hadley, Peter Hajek, Diana
Hall, Joan Hammontree, Anton Hanegraaf†, Martin Hansen, Jacques Haubrich, Cynthia Hay,
Takao Hayashi,
Robert L. Helmbold,
Hanno Hentrich,
Richard I. Hess,
Christopher Holtom,
Edward Hordern†,
Peter Hosek,
Konrad Jacobs,
Anatoli Kalinin,
Bill Kalush, Michael
Keller,
Edward S. Kennedy,
Sarah Key (The Haunted Bookshop), Eberhard Knobloch, Don Knuth,
Bob Koeppel,
Joseph D. E. Konhauser,
David E. Kullman,
Mogens Esrom Larsen,
Jim Lavis (Doxa (Oxford)), John Leech†, Elisabeth Lefevre, C. Legel, Derrick [Dick] H. Lehmer†, Emma Lehmer, Leisure Dynamics,
Hendrik W. Lenstra,
Alan L. Mackay,
Andrzej Makowski,
John Malkevitch,
Giovanni Manco,
Tatiana Matveeva,
Ann Maury, Max Maven, Jim McArdle, Patricia McCulloch,
Peter McMullen,
Leroy F. Meyers†, D.
P. Miles, Marvin Miller, Nobuo Miura, William O. J. Moser, Barbara Moss,
Angela Newing,
Jennie Newman,
Tom and Greta O'Beirne††, Owen O'Shea,
Parker Brothers, Alan
Parr, Jean J. Pedersen, Luigi Pepe,
William Poundstone, Helen
Powlesland, Oliver Pretzel, Walter Purkert, Robert A. Rankin†, Eleanor Robson, David J. A. Ross, Lee Sallows, Christopher Sansbury,
Sol Saul, William L. Schaaf, Doris Schattschneider, Jaap Scherphuis, Heribert Schmitz,
Š. Schwabik,
Eileen Scott†, Al
Seckel, Jacques Sesiano, Claude E. Shannon†, John Sheehan, A. Sherratt,
Will Shortz,
Kripa Shankar Shukla,
George L. Sicherman, Deborah
Singmaster, Man‑Kit Siu,
Gerald [Jerry] K. Slocum, Cedric A. B. Smith† (and Sue Povey & Jim
Mallet at the Galton Laboratory for letting me have some of Cedric's
books), Jurgen Stigter, Arthur H. Stone, Mel Stover†, Michael Stueben,
Shigeo Takagi†, Michael Tanoff, Gary J. Tee, Andrew Topsfield, George Tyson†, Dario Uri, Warren Van Egmond,
Carlo Viola,
Kurt Vogel†,
Anthony Watkinson,
Chris Weeks,
Maurice Wilkes,
John Winterbottom,
John Withers,
Nob. Yoshigahara,
Claudia Zaslavsky.
I would also like to thank the
following libraries and museums which I have used:
University
of Aberdeen;
University of Bristol;
Buckleys Shop Museum, Battle, East Sussex; University of Calgary; University of Cambridge; Marsh's Library, Dublin;
FLORENCE:
Biblioteca Nazionale; Biblioteca Riccardiana;
University of Keele -- The Turner Collection(†)
and its librarian Martin Phillips;
Karl‑Marx‑Universität, Leipzig:
Universität Bibliothek and Sektion Mathematik Bibliothek,
especially Frau Letzel at the latter;
LONDON:
Birkbeck College; British Library (at Bloomsbury and
then at St. Pancras; also at Colindale);
The London Library; School of Oriental and African
Studies, especially Miss Y. Yasumara, the Art Librarian; Senate House, particularly the Harry Price
Library; South Bank
University;
Southwark Public Library;
University College London, especially the Graves
Collection and the Rare Book Librarians Jill Furlong, Susan Stead and their
staff; Warburg Institute;
MUNICH:
Deutsches Museum; Institut für
Geschichte der Naturwissenschaften;
NEW
YORK:
Brooklyn Public Library;
City College of New York; Columbia University;
Newark Public Library, Newark, New
Jersey;
University
of Newcastle upon Tyne -- The Wallis Collection and its librarian Lesley
Gordon;
OXFORD:
Ashmolean Museum; The Bodleian Library;
Museum of the History of Science, and its librarian
Tony Simcock;
University
of Reading; University of St. Andrews;
SIENA:
Biblioteca Comunale degli
Intronati;
Dipartimento di Matematica, Università di Siena;
University of Southampton; Mathematical Institute, Warsaw.
I would like especially to thank the
following.
Interlibrary Loans (especially Brenda
Spooner) at South Bank University and the British Library Lending Division for
obtaining many strange items for me.
Richard Guy, Bill Sands and the Strens
bequest for a most useful week at the Strens/Guy Collection at Calgary in early
1986 and for organizing the Strens Memorial Meeting in summer 1986 and for
printing the first preliminary edition of these Sources.
Gerd Lassner, Uwe Quasthoff and the
Naturwissenschaftlich‑Theoretisches Zentrum of the Karl‑Marx‑Universität,
for a very useful visit to Leipzig in 1988.
South Bank University Computer Centre
for the computer resources for the early stages of this project, and especially
Ann Keen for finding this file when it was lost.
My School for printing these preliminary
editions.
Martin Gardner for kindly allowing me
to excavate through his library and files.
James Dalgety, Edward Hordern, Bill
Kalush, Chris Lewin, Tom Rodgers and Will Shortz for allowing me to rummage
through their libraries.
John Beasley, Edward Hordern, Bill
Kalush, Will Shortz and Jerry Slocum for numerous photocopies and copies from
their collections.
Menso Folkerts, Richard Lorch, Michael
Segre and the Institut für Geschichte der Naturwissenschaft, Munich, for a most
useful visit in Sep 1994 and for producing a copy of Catel.
Raffaella Franci and the Dipartimento
di Matematica and the Centro Studi della Matematica Medioevale at Università di
Siena for a most useful visit in Sep 1994.
Takao Hayashi for much material from
Japan and India.
My wife for organizing a joint trip to
Newcastle in Sep 1997 where I made use of the Wallis Collection.
Finally, I would like to thank a large
number of publishers, distributors, bookdealers and even authors who have
provided copies of the books and documents upon which much of this work is
based. Bookdealers have often let me
examine books in their shops. Their
help is greatly appreciated. There are
too many of these to record here, but I would like to mention Fred Whitehart
(†1999), England's leading dealer in secondhand scientific books for many years
who had a real interest in mathematics.
DIACRITICAL MARKS AND NOTATION
Before
converting to LocoScript, I used various conventions, given below, to represent
diacritical marks. Each symbol
(except ') occurred after the letter it
referred to. I have now converted these
and all mathematical conventions into correct symbols, so far as possible, but
I may have missed some, so I am keeping this information for the present.
Common
entries using such marks are given later in this section and only the
abbreviated or simplified form is used later -- e.g. I use Problemes for
Bachet's work rather than Problèmes.
(Though this may change??)
Initially,
I did not record all diacritical marks, so some may be missing though I have
checked almost all items. I may omit
diacritical marks which are very peculiar.
Transliterations
of Arabic, Sanskrit, Chinese, etc. are often given in very different
forms. See Smith, History, vol. 1, pp. xvii-xxii
for a discussion of the problems. The
use of ^ and ˉ seems interchangeable and I have used ^
when different versions use both
^ and ˉ , except when quoting a title or passage when I use the
author's form. [Smith, following Suter,
uses ^ for Arabic, but
uses ˉ for Indian. Murray uses ˉ
for both. Wieber uses ˉ
for Arabic. Van der Linde
uses ´
for Arabic. Datta & Singh
use ^
for Indian.]
There
are two breathing marks in Arabic -- ayn
‘ and alif/hamzah ’ --
but originally I didn't have two forms easily available, so both were
represented by '. I have now converted almost all of these
to ‘
and ’. These don't seem to be as distinct in the printing as on my
screen.
French
practice in accenting capitals is variable and titles are often in capitals, so
expected marks may be missing. Also,
older printing may differ from modern usage -- e.g. I have seen: Liège and
Liége; Problèmes, Problêmes and
Problémes. When available, I have
transcribed the material as printed without trying to insert marks, but many
places insert the marks according to modern French spelling.
Greek
and Cyrillic titles are now given in the original with an English
transliteration (using the Amer. Math. Soc. transliteration for Cyrillic).
I
usually ignore the older usage of
v for u and i
for j, so that I give
mathematiqve as mathematique and xiij as
xiii.
I
used a1, a2, ..., ai, etc. for subscripted variables, though I
also sometimes used a(1), a(2),
..., a(i), etc. Superscripts or exponents were indicated by
use of ^, e.g. 2^3 is 8. These have been converted to ordinary sub-
and superscript usage, but ^ may be used when the superscript is
complicated -- e.g. for 2^ai or
9^(99).
Greek
letters were generally spelled out in capitals or marked with square brackets,
e.g. PI, [pi], PHI, but these have probably all been converted.
My
word processor does not produce binomial coefficients easily, so I use BC(n, k)
for n!/k!(n‑k)!
Many
problems have solutions which are sets of fractions with the same denominator
and I abbreviate a/z, b/z,
c/z as (a, b, c)/z. Notations
for particular problems are explained at the beginning of the topic.
Rather
than attempting to italicise letters used as symbols, I generally set them off
by double-spaces on each side -- see examples above. Other mathematical notations may be improvised as necessary and
should be obvious.
Recall
that the symbols below occurred after the letter they referred to, except
for ' .
" denoted umlaut or diaeresis in general, e.g.: ä, ë, ï, ö, ü.
/ was used after a letter for accent acute, ́, after l for ł
in Polish, and after o for
ø in Scandinavian.
\ denoted accent grave,
̀.
^ denoted the circumflex,
^, in Czech, etc.; the overbar
(macron) ˉ or
^ for a long vowel in Sanskrit,
Hindu, etc.; and the overbar used to indicate omission in medieval MSS.
@ denoted the cedilla (French
ç and Arabic ş)
and the ogonek or Polish hook (Polish
ą).
. denoted the underdot in
ḥ, ḳ, ṇ, ṛ,
ṣ, ṭ, in Sanskrit, Hindu, Arabic.
These are sometimes written with a following h -- e.g. k
may also be written kh and I may sometimes have used this. (It is difficult to search for ḥ. , etc., so not all of these
may be converted.) This mark vanishes
when converted to WordPerfect.
* denoted the overdot for
ġ, ṁ, ṅ, in Sanskrit, Hindu,
Arabic. This vanishes over m
and n in WordPerfect.
~ denoted the Spanish tilde
~ and the caron or hachek ˇ,
in ğ, š. The breve is a curved version, ˘,
of the same symbol and is essentially indistinguishable from the
caron. It occurs in Russian й,
which is translitereated as
ĭ.
_ denoted the underbar in
ḏ , j, ṯ (I cannot find a j
with an underbar in Arial). This mark vanishes
in WordPerfect.
' denotes breathing marks in Arabic, etc. There are actually two forms of this --
ayn ’
and alif/hamzah ‘ -- but I didn't have two forms easily
available and originally entered both as apostrophe ' . These normally occur between letters and I
placed the ' in the same space. I have
converted most of these.
Commonly
occurring words with diacritical marks are: Académie, arithmétique,
bibliothèque, Birkhäuser, café, carré, école, Erdös, für, géomètre, géométrie,
Göttingen, Hanoï -- in French only, ‑ième, littéraire, mathématique,
mémoire, ménage, misère, Möbius, moiré, numérique, Pétersbourg, probabilités,
problème (I have seen problêmes??), Rätsel, récréation, Sändig, siècle,
société, Thébault, théorie, über, umfüllung.
I
have used ?? to indicate uncertainty and points where further work needs to be
done. The following symbols after ??
indicate the action to be done.
* check for diacritical marks,
etc.
NX no Xerox or other copy
NYS not yet seen
NYR not yet read
o/o on order
SP check spelling
Other
comments may be given.
ABBREVIATIONS
OF JOURNALS AND SERIES.
See: AMM, CFF, CM,
CMJ, Family Friend, G&P,
G&PJ, HM, JRM,
MG, MiS, MM,
MS, MTg, MTr, M500,
OPM, RMM, SA,
SM, SSM in Common References below.
See: AMS, C&W, CUP,
Loeb Classical Library, MA, MAA,
NCTM, OUP in Common References below.
ABBREVIATIONS
OF MONTHS. All months are given by their first three letters in
English: Jan, Feb, ....
PUBLISHERS'
LOCATIONS. The following publisher's locations will not be cited each
time. Other examples may occur and can
be found in the file PUBLOC.
AMS (American Mathematical
Society), Providence, Rhode Island,
USA.
Chelsea Publishing, NY, USA.
CUP (Cambridge University
Press), Cambridge, UK.
Dover, NY, USA.
Freeman, San Francisco, then NY, USA.
Harvard University Press, Cambridge, Massachusetts, USA.
MA (Mathematical
Association), Leicester, UK.
MAA (Mathematical Association of
America), Washington, DC, USA.
NCTM (National Council of
Teachers of Mathematics), Reston,
Virginia, USA.
Nelson, London, UK.
OUP (Oxford University
Press), Oxford, UK (and also NY, USA).
Penguin, Harmondsworth, UK.
Simon & Schuster, NY, USA.
NOTES. When referring to items below, I will
usually include the earliest reasonable date, even though the citation may be
to a much later edition. For example, I
would say "Canterbury Puzzles, 1907", even though I am citing problem
numbers or pages from the 1958 Dover reprint of the 1919 edition. Sometimes the earlier editions are hard to
come by and I have sometimes found that the earlier edition has different
pagination -- in that case I will (eventually) make the necessary changes.
Edition
information in parentheses indicates items or editions that I have not seen,
though I don't always do this when the later version is a reprint or facsimile.
Abbaco. See:
Pseudo-dell'Abbaco.
Abbot Albert. Abbot Albert von Stade. Annales Stadenses. c1240.
Ed. by J. M. Lappenberg.
In: Monumenta Germaniae Historica, ed. G. H. Pertz, Scriptorum
t. XVI, Imp. Bibliopolii Aulici Hahniani, Hannover, 1859 (= Hiersemann,
Leipzig, 1925), pp. 271‑359.
(There are 13 recreational problems on pp. 332‑335.) [Vogel, on p. 22 of his edition of the
Columbia Algorism, dates this as 1179, but Tropfke gives 1240, which is more in
line with Lappenberg's notes on variants of the text. The material of interest, and several other miscellaneous
sections, is inserted at the year 1152 of the Annales, so perhaps Vogel was
misled by this.] I have prepared an
annotated translation of this: The problems of Abbot Albert (c1240). I have numbered the problems and will cite
this problem number.
Abraham. R. M. Abraham.
Diversions and Pastimes.
Constable, London, 1933
= Dover, 1964 (slightly amended and with different pagination,
later retitled: Tricks and Amusements with Coins, Cards, String, Paper and
Matches). I will cite the Constable
pages (and the Dover pages in parentheses).
Ackermann. Alfred S. E. Ackermann. Scientific Paradoxes and Problems and Their
Solutions. The Old Westminster Press,
London, 1925.
D. Adams. New Arithmetic. 1835.
Daniel
Adams (1773-1864). ADAMS NEW
ARITHMETIC. Arithmetic, in which the
principles of operating by numbers are analytically explained, and
synthetically applied; thus combining the advantages to be derived both from
the inductive and synthetic mode of instructing: The whole made familiar by a great variety of useful and
interesting examples, calculated at once to engage the pupil in the study, and
to give him a full knowledge of figures in their application to all the
practical purposes of life. Designed
for the use of schools and academies in the United States. J. Prentiss, Keene, New Hampshire, 1836,
boarded. 1-262 pp + 2pp publisher's
ads, apparently inserted backward.
[Halwas 1-6 lists 1st ed as 1835, then has 1837, 1838, 1839, 1842,
c1850.] This is a reworking of The
Scholar's Arithmetic of 1801.
D. Adams. Scholar's Arithmetic. 1801.
Daniel
Adams (1773-1864). The Scholar's
Arithmetic; or, Federal Accountant: Containing.
I. Common arithmetic, .... II.
Examples and Answers with Blank Spaces, ....
III. To each Rule, a Supplement, comprehending, 1. Questions .... 2. Exercises. IV. Federal Money, ....
V. Interest cast in Federal Money, ....
VI. Demonstration by engravings ....
VII. Forms of Notes, .... The
Whole in a Form and Method altogether New, for the Ease of the Master and the
greater Progress of the Scholar. Adams
& Wilder, Leominster, Massachusetts, 1801; 2nd ed, 1802. 3rd ed ??.
4th ed, by Prentiss, 1807; 6th ed, 1810; 10th ed, 1816; Stereotype
Edition, Revised and Corrected, with Additions, 1819, 1820, 1824; John Prentiss,
Keene, New Hampshire, 1825. [Halwas
8-14.] I have the 1825, whose Preface
is for the 10th ed of 1816, so is probably identical to that ed. The Preface says he has now made some
revisions. The only change of interest
to us is that he has added answers to some problems. So I will cite this as 1801 though I will be giving pages from
the 1825 ed. The book was thoroughly
reworked as Adams New Arithmetic, 1835.
M. Adams. Indoor Games. 1912.
Morley
Adams, ed. The Boy's Own Book of Indoor
Games and Recreations. "The Boy's
Own Paper" Office, London, 1912; 2nd ptg, The Religious Tract Society,
London (same address), 1913. [This is a
major revision of: G. A. Hutchison, ed.; Indoor Games and
Recreations; The Boy's Own Bookshelf; New ed., Religious Tract Society,
London, 1891 (possibly earlier) -- see 5.A.]
M. Adams. Puzzle Book. 1939.
Morley
Adams. The Morley Adams Puzzle
Book. Faber & Faber, London, 1939.
M. Adams. Puzzles That Everyone Can Do. 1931.
Morley
Adams. Puzzles that Everyone Can Do. Grant Richards, London, 1931, boarded.
AGM. Abhandlungen zur Geschichte der Mathematischen
Wissenschaften mit Einschluss ihrer Anwendungen. Begründet von Moritz Cantor.
Teubner, Leipzig. The first ten
volumes were Supplements to Zeitschrift für Math. u. Physik, had a slightly
different title and are often bound in with the journal volume.
Ahrens, Wilhelm Ernst Martin
Georg (1872-1927). See: A&N,
MUS, 3.B, 7.N.
al‑Karkhi. Aboû Beqr Mohammed Ben
Alhaçen Alkarkhî [= al‑Karagi
= al‑Karajī].
Untitled MS called Kitāb al-Fakhrī (or just Alfakhrî) (The
Book Dedicated to Fakhr al-Din).
c1010. MS 952, Supp. Arabe de la
Bibliothèque Impériale, Paris. Edited
into French by Franz Woepcke as: Extrait du Fakhrî. L'Imprimerie Impériale, Paris, 1853; reprinted by Georg Olms
Verlag, Hildesheim, 1982. My page
citations will be to Woepcke. Woepcke
often refers to Diophantos, but his numbering gets ahead of Heath's.
Alberti. 1747. Giuseppe
Antonio (or Giusepp-Antonio) Alberti (1715-1768). I Giochi Numerici Fatti Arcani Palesati da Giuseppe Antonio
Alberti. Bartolomeo Borghi, Bologna,
1747, 1749. Venice, 1780, 1788(?). 4th ed., adornata di figure, Giuseppe
Orlandelli for Francesco di Niccolo' Pezzana, Venice, 1795 (reprinted: Arnaud, Florence, 1979), 1813. Adornata di 16 figure, Michele Morelli,
Naples, 1814. As: Li Giuochi Numerici
Manifestati, Edizione adorna di Figure in rame, Giuseppe Molinari, Venice,
1815.
The
editions have almost identical content, but different paginations. I have compared several editions and seen
little difference. The 1747 ed. has a
dedication which is dropped in the 2nd ed. which also omits the last paragraph
of the Prefazione. I only saw one other
point where a few words were changed. I
will give pages of 1747 (followed by 1795 in parenthesis). Much of Alberti, including almost all the
material of interest to us and many of the plates, is translated from vol. 4 of
the 1723 ed. of Ozanam.
(Serge
Plantureux's 1993 catalogue describes a 1747-1749 ed. with Appendice al
Trattato de' Giochi Numerici (1749, 72 pp) & Osservazioni all'Appendice de'
Giochi Numerici (38 pp), ??NYS. The
copy in the Honeyman Collection had the Appendice. Christopher 3 has the Osservazioni. The Appendice is described by Riccardi as a severe criticism of
Alberti, attributed to Giovanni Antonio Andrea Castelvetri and published by
Lelio dall Volpe, Bologna, 1749. The
Osservazioni are Alberti's response.)
Alcuin (c735-804).
Propositiones
Alcuini doctoris Caroli Magni Imperatoris ad acuendos juvenes. 9C.
IN:
B. Flacci Albini seu Alcuini, Abbatis et Caroli Magni Imperatoris
Magistri. Opera Omnia: Operum pars
octava: Opera dubia. Ed. D. Frobenius,
Ratisbon, 1777, Tomus secundus, volumen secundum, pp. 440‑448. ??NYS.
Revised and republished by J.‑P. Migne as: Patrologiae Cursus
Completus: Patrologiae Latinae, Tomus 101, Paris, 1863, columns 1143‑1160.
A
different version appears in: Venerabilis Bedae, Anglo‑Saxonis
Presbyteri. Opera Omnia: Pars Prima,
Sectio II -- Dubia et Spuria: De Arithmeticus propositionibus. Tomus 1, Basel, 1563. (Rara, 131, says there were earlier
editions: Paris, 1521 (part), 1544‑1545 (all), 1554, all ??NYS.) Revised and republished by J.‑P. Migne
as: Patrologiae Cursus Completus: Patrologiae Latinae, Tomus 90, Paris, 1904,
columns 665‑672. Incipiunt aliae
propositiones ad acuendos juvenes is col. 667‑672. A version of this occurs in Ens'
Thaumaturgus Mathematicus of 1636 -- cf under Etten.
The
Alcuin has 53 numbered problems with answers.
The Bede has 3 extra problems, but the problems are not numbered, there
are only 31 1/2 answers and there are several transcription errors. The editor has used the Bede to rectify the
Alcuin.
There
is a recent critical edition of the text by Folkerts -- Die älteste mathematische
Aufgabensammlung in lateinischer Spräche: Die Alkuin zugeschriebenen
Propositiones ad Acuendos Iuvenes; Denkschriften der Österreichischen Akademie
der Wissenschaften, Mathematische‑naturwissenschaftliche Klasse 116:6
(1978) 13‑80. (Also separately
published by Springer, Vienna, 1978.
The critical part is somewhat revised as: Die Alkuin zugeschriebenen
"Propositiones ad Acuendos Iuvenes"; IN: Science in Western and
Eastern Civilization in Carolingian Times, ed. by P. L. Butzer & D.
Lohrmann; Birkhäuser, Basel, 1993, pp. 273-281.) He finds that the earliest text is late 9C and is quite close to
the first edition cited above. He uses
the same numbers for the problems as above and numbers the extra Bede problems
as 11a, 11b, 33a. I use Folkerts for
the numbering and the titles of problems.
John
Hadley kindly translated Alcuin for me some years ago and made some amendments
when Folkerts' edition appeared. I
annotated it and it appeared as: Problems to Sharpen the Young, MG 76 (No. 475)
(Mar 1992) 102-126. A slightly
corrected and updated edition, containing some material omitted from the MG
version, is available as Technical Report SBU-CISM-95-18, School of Computing,
Information Systems, and Mathematics, South Bank University, Oct 1995, 28pp.
Menso
Folkerts and Helmuth Gericke have produced a German edition: Die Alkuin
zugeschriebenen Propositiones ad Acuendos Juvenes (Aufgabe zur Schärfung des
Geistes der Jugend); IN: Science in Western and Eastern Civilization in
Carolingian Times, ed. by P. L. Butzer & D. Lohrmann; Birkhäuser, Basel,
1993, pp. 283-362.
See
also: David Singmaster. The history of some of Alcuin's Propositiones. IN: Charlemagne and his Heritage 1200 Years of Civilization and Science in
Europe: Vol. 2 Mathematical Arts; ed.
by P. L. Butzer, H. Th. Jongen & W. Oberschelp; Brepols, Turnhout, 1998,
pp. 11‑29.
AM. 1917. H.
E. Dudeney. Amusements in
Mathematics. Nelson, 1917. (There were reprintings in 1919, 1920, 1924,
1925, 1927, 1928, 1930, 1932, 1935, 1938, 1939, 1941, 1943, 1946, 1947, 1949,
1951, but it seems that the date wasn't given before 1941?) = Dover, 1958.
AMM. American Mathematical Monthly.
AMS. American Mathematical Society.
Les Amusemens. 1749.
Les
Amusemens Mathématiques Precedés Des Elémens d'Arithmétique, d'Algébre & de
Géométrie nécessaires pour l'intelligence des Problêmes. André‑Joseph Panckoucke, Lille,
1749. Often listed with Panckoucke as
author (e.g. by the NUC, the BNC and Poggendorff), but the book gives no such
indication. Sometimes spelled
Amusements. There were 1769 and 1799
editions.
Apianus. Kauffmanss Rechnung. 1527.
Petrus
Apianus (= Peter Apian or Bienewitz or Bennewitz) (1495‑1552). Eyn Newe Unnd wolgegründte underweysung
aller Kauffmanss Rechnung in dreyen Büchern / mit schönen Regeln uň
[NOTE: ň denotes an n with an overbar.] fragstucken
begriffen. Sunderlich was fortl unnd
behendigkait in der Welschē Practica uň Tolletn gebraucht wirdt / des
gleychen fürmalss wider in Teützscher noch in Welscher sprach nie
gedrückt. durch Petrum Apianū von
Leyssnick / d Astronomei zů Ingolstat Ordinariū / verfertiget. Georgius Apianus, Ingolstadt, (1527),
facsimile, with the TP of the 1544 ed. and 2pp of publication details added at
the end, Polygon-Verlag, Buxheim-Eichstätt, 1995, with 8pp commentary leaflet
by Wolfgang Kaunzner. (The TP of this
has the first known printed version of Pascal's Triangle.) Smith, Rara, pp. 155-157. (The
d is an odd symbol, a bit like
a δ or an 8, which is used regularly for der
both as a single word and as the ending of a word, e.g. and
for ander.) Smith notes that Apianus follows Rudolff
(1526) very closely.
AR. c1450. Frater
Friedrich Gerhart (attrib.). Latin
& German MSS, c1450, known as Algorismus Ratisbonensis. Transcribed and edited from 6 MSS by Kurt
Vogel as: Die Practica des Algorismus Ratisbonensis; C. H. Beck'sche
Verlagsbuchhandlung, Munich, 1954.
(Kindly sent by Prof. Vogel.)
Vogel classifies the problems and gives general comments on the
mathematics on pp. 155‑189. He
gives detailed historical notes on pp. 203‑232. When appropriate, I will cite these pages before the specific
problems. He says (on p. 206) that
almost all of Munich 14684 (see below) is included in AR.
Arnold, George. See:
Book of 500 Puzzles, Boy's Own
Conjuring Book, Hanky Panky.
Arrighi, Gino. See:
Benedetto da Firenze,
Calandri, Pseudo-Dell'Abbaco, della Francesca, Gherardi, Lucca
1754, P. M. Calandri.
Aryabhata. Āryabhata (I))
[NOTE: ţ denotes a t with a dot under it and ş
denotes an s with a dot under it.] (476-
). Āryabhatīya.
499. Critically edited and
translated into English by Kripa Shankar Shukla, with K. V. Sarma. Indian National Science Academy, New Delhi,
1976. (Volume 1 of a three volume series
-- the other two volumes are commentaries, of which Vol. 2 includes the
commentary Āryabhatīya-Bhāşya, written by Bhaskara I in
629. Aryabhata rarely gives numerical
examples, so Bhaskara I provided them and these were later used by other Indian
writers such as Chaturveda, 860. The
other commentaries are later and of less interest to us. Prof. Shukla has sent a photocopy of an
introductory booklet, which is an abbreviated version of the introductory
material of Vol. 1, with some extensions relating Aryabhata to other
writers.) The material is organized
into verses. There is an older
translation by Walter Eugene Clark as:
The Âryabhaţîya of
Âryabhaţa; Univ. of Chicago Press, Chicago, 1930. (There was an Aryabhata II, c950, but he
only occurs in 7.K.1.)
A&N. Wilhelm Ahrens. Altes und Neues aus der Unterhaltungsmathematik. Springer, Berlin, 1918.
Bachet, Claude‑Gaspar
(1581-1638). See: Problemes.
Bachet-Labosne. See:
Problemes.
Badcock. Philosophical Recreations, or, Winter
Amusements. [1820].
Philosophical
Recreations, or, Winter Amusements.
Thomas Hughes, London, nd [1820].
[BCB 18-19; OCB, pp. 180 & 197.
Heyl 22-23. Toole Stott
75-77. Christopher 54-56. Wallis 34 BAD, 35 BAD. These give dates of 1820, 1822, 1828.] HPL [Badcock] RBC has three versions with
slightly different imprints on the title pages, possibly the three dates
mentioned.
Wallis
34 BAD has this bound after the copy of:
John Badcock; Domestic Amusements, or Philosophical Recreations ...; T.
Hughes, London, nd [1823], and it is lacking its Frontispiece and TP -- cf in
6.BH. HPL [Badcock] has both books,
including the folding Frontispieces.
The earlier does not give an author, but its Preface is signed
J. B. and the later book does give his name and calls itself a sequel to
the earlier. Toole Stott 75-80 clearly
describes both works. Some of the
material is used in Endless Amusement II.
Baker. Well Spring of Sciences.
1562?
Humfrey
Baker (fl. 1557-1587). The Well Sprynge
of Sciences Which teacheth the perfect worke and practise of Arithmeticke both
in whole numbers and fractions, with such easye and compendious instruction
into the sayde arte, .... Rouland Hall
for James Rowbotham, London, 1562.
[Smith, Rara, p. 327, says it was written in 1562 but wasn't actually
printed until 1568, but a dealer says the 1st ed. was 1564 and there was a 4th
ed. in 1574, which I have examined.]
Apparently much revised and extended, (1580). Reprinted, with title: The Wel [sic] Spring of Sciences: Which
teacheth the perfect worke and practise of Arithmetike; Thomas Purfoote,
London, 1591. I have seen Thomas
Purfoot, London, 1612, which is essentially identical to 1591. I have also seen: Christopher Meredith, London, 1646; Christopher Meredith, London, 1650; R. & W. L. for Andrew Kemb, London, 1655; which are all the same, but differently
paged than the 1591. I have also seen
Baker's Arithmetick, ed. by Henry Phillippes, Edward Thomas, London, 1670,
which has different pagination and some additional problems compared to the
1646/1655 ed. [Smith, Rara, 327-330
& 537, says it was rewritten in 1580, but there is little difference
between the 1580 and the many later editions, so the 1591 ed. is probably close
to the 1580 ed. The copy of the 1562 in
the Graves collection ends on f. 160r, but an owner has written a query as
to whether the book is complete.
Neither Smith nor De Morgan seems to have seen a 1562 so they don't give
a number of pages for it. (STC records
no copies of the 1562, 1564, 1576, 1584, 1607 editions, but there was a 1576 by
[T. Purfoote], apparently the 5th ed., of c500pp, in the Honeyman
Collection.) Almost all the problems of
interest occur on ff. 189r-198r of the 1591 ed. and hence are not in the Graves
copy of the 1562 ed., but H&S 61 refers to one of these problems as being
in Baker, 1568. The 1574 ends at fol.
200 (misprinted as 19?, where the ? is an undecipherable blob) and Chapter 16,
which is headed: The 16 Chapter
treateth of sportes and pastime, done by number, is on ff. 189r-200v, and contains just a few recreations, as in
Recorde. So I will date the book as
1562?, but most of the later material as 1580?. The problems of 7.AF.1 and 10.A may be in Graves copy of the 1562
ed. -- ??check. I will cite the 1580?,
1646 and 1670 editions, e.g. 1580?:
ff. 192r 193r; 1646: pp. 302-304; 1670: pp. 344-345.] Bill Kalush has recently sent a CD with
1574, 1580, 1591, 1598, 1602, 1607, 1612, 1617, 1650, 1655 on it -- ??NYR.
Bakhshali MS. The
Bakhshālī Manuscript, c7C.
This MS was found in May 1881 near the village of Bakhshālī,
in the Yusufzāī district of the Peshawer division, then at the
northwestern frontier of India, but apparently now in Pakistan. This is discussed in several places, such as
the following, but a complete translation has only recently appeared. David Pingree says it is 10C, but his
student Hayashi opts for 7C which seems pretty reasonable and I will adopt c7C.
1. A. F. Rudolf Hoernle. Extract of his report in some journal of the
previous year. The Indian Antiquary 12 (Mar
1883) 89-90. A preliminary report,
saying it was found near Bakhshâlî in the Yusufzai District in the Panjâb.
2. A. F. Rudolf Hoernle. On the Bakhshālī Manuscript. Berichte des VII. Internationalen
Orientalisten‑Congresses, Wien, 1886.
Alfred Hölder, Vienna, 1889.
Arische Section, p. 127-147 plus three folding plates. Cf next item. I will cite this as Hoernle, 1886.
3. A. F. Rudolf Hoernle. The Bakhshali manuscript. The Indian Antiquary 17 (Feb 1888) 33‑48
& Plate I opp. p. 46; 275‑279
& Plates II & III opp. pp. 276 & 277. This is essentially a reprint of the
previous item, with a few changes or corrections, but considerable additional
material. He dates it c4C. I will cite this as Hoernle, 1888.
4. G. R. Kaye. The Bakhshālī Manuscript – A Study
in Medieval Mathematics. Archæeological
Survey of India – New Imperial Series XLIII: I-III, with parts I & II as
one volume, (1927‑1933).
(Facsimile reprint in two volumes, Cosmo Publications, New Delhi, 1981 –
this is a rather poor facsimile, but all the text is preserved. I have a letter detailing the changes
between the original and this 'facsimile'.)
I will only cite Part I – Introduction, which includes a discussion of
the text. Part II is a discussion of
the script, transliteration of the text and pictures of the entire MS. Part III apparently was intended to deal
with the language used, but Kaye died before completing this and the published
Part III consists of only a rearranged version of the MS with footnotes
explaining the mathematics. Gupta,
below, cites part III, as Kaye III and I will reproduce these citations. He dates it c12C.
5. B. Datta. The Bakhshâlî mathematics.
Bull. Calcutta Math. Soc. 21 (1929) 1‑60. This is largely devoted to dating of the MS and
of its contents. He asserts that the MS
is a copy of a commentary on some lost work of 4C or 5C (?).
6. R. C. Gupta. Some equalization problems from the
Bakhshālī manuscript. Indian
Journal of the History of Science 21 (1986) 51-61. Notes that Hoernle gave the MS to the Bodleian Library in 1902,
where it remains, with shelf mark MS. Sansk. d.14. He follows Datta in believing that this is a commentary on a
early work, though the MS is 9C, as stated by Hoernle. He gives many problems from Kaye III,
sometimes restoring them, and he discusses them in more detail than the
previous works.
7. Takao Hayashi. The Bakhshālī Manuscript An ancient Indian mathematical
treatise. Egbert Forsten, Groningen,
Netherlands, 1995. (Based on his PhD
Dissertation in History of Mathematics, Brown University, May 1985,
774pp.) A complete edition and
translation with extensive discussion of the context of the problems. He dates it as 7C.
Ball, Walter William Rouse
(1850-1925). See: Ball‑FitzPatrick; MRE.
Ball‑FitzPatrick.
French
translation of MRE by J. Fitz‑Patrick, published by Hermann, Paris.
1st
ed., Récréations et Problèmes Mathématiques des Temps Anciens &
Modernes. From the 3rd ed, 1896, of
MRE, 'Revue et augmentée par l'auteur'.
1898. The Note says 'M. Ball ...
a bien voulu apporter à la troisième édition anglaise des additions et des
modifications importantes.' 352pp.
2nd
ed., Récréations et Problèmes Mathématiques des Temps Anciens et Modernes. From the 4th ed, 1905, of MRE, 'et enrichie
de nombreuses additions'.
As three volumes, 1907‑09. [I have vol. 1, 1907, which is 356pp. Pp. 327‑355 is a note by A. Hermann,
Comptabilité d'une persone qui dépense plus que son revenu. I have not yet seen the other volumes to
compare with the 1926 reprint, but Strens's notes in his copy indicate that
they are identical.]
Reprinted in one vol., Gabay, Paris, 1992,
544pp.
Reprinted, 1926-1927. The only copies that I have seen are bound
as one volume, but with separate pagination.
My copy has left out the title pages of vols. 2 & 3. The copy in the Strens Collection has these
title pages, but its vol. II is 1908.
The 1926 vol. 1 says Nouvelle édition française, but the 1927 vol. 3
says Deuxième édition française.
[Vol.
1 is 326pp, omitting the note by Hermann.
Vol. 2 is 363pp (pp. 322‑355 is a historical note on the cubic,
based on Cossali (1797)). Vol. 3 is
363pp including: Notes diverses de M. Aubry, pp. 137‑206 (or 340? -- the
Table des Matières and the page set up do not make it clear if Aubry's Notes
end on p. 206); Note de M. Fitz‑Patrick,
La géométrie par le pliage et découpage du papier, pp. 341‑360; A. Margossian, De l'ordonnance des nombres
dans les carrés magiques impairs, pp. 1‑60 (pp. 61-64 is a Note on the
same subject, presumably part of Margossian's material); Capt. Reinhart, some geometric notes, pp.
130-136.]
Barnard. 50 Observer Brain-Twisters. 1962.
Douglas
St. Paul Barnard. Fifty Observer Brain‑Twisters A Book of Mathematical and Reasoning
Problems. Faber, 1962. US ed.:
A Book of Mathematical and Reasoning Problems: Fifty Brain
Twisters; Van Nostrand, 1962. The editions have identical pagination.
Bartl. c1920. János
Bartl. Preis-Verzeichnis von Bartl's
Akademie für moderne magische Kunst.
Hamburg, c1920. Reprinted by
Olms Verlag, Zürich, 1983, as: Zauberkatalog Bartl. References are to the section: Vexier- und Geduldspiele,
pp. 305‑312.
Bartoli. Memoriale.
c1420.
Francesco
Bartoli ( -1425). Memoriale (= Notebook) containing some 30
mathematical problems copied during 1400?-1425. Ms 1 F 54 of the Archives départementales du Vaucluse,
France. ??NYS -- described and quoted
in: Jacques Sesiano; Les problèmes
mathématiques du Memoriale de F. Bartoli; Physis 26:1 (1984) 129-150.
BC. Binomial Coefficient, i.e. BC(n, k) = n!/k!(n-k)!.
BCB. See: Hall, BCB.
BDM. See under DSB.
Bede, The Venerable
(c672-735). (Now St. Bede.) See:
Alcuin.
Benedetto da Firenze. c1465.
Benedetto
da Firenze. Trattato d'Abacho. c1465.
This was a popular treatise and Van Egmond's Catalog 356 lists 18 copies
under Benedetto. Six show B as
author, one has Benedetto, one has Benedetto da Firenze, one has Po Ma and one has Filipo Chalandri, so it seems
Benedetto is the most likely author.
The MSS date from c1465 to c1525 and contain 9 to 25 chapters.
The
version in Cod. Acq. e doni 154, Biblioteca Medicea Laurenziana, Florence,
c1480. has been transcribed and edited
by Gino Arrighi as: Pier Maria
Calandri; Tractato d'Abbacho; Domus Galilaeana, Pisa, 1974. The incipit names Po Ma
as author. Cf Van Egmond's
Catalog 96. This version has 23
chapters.
Benson. 1904. J.
K. Benson. The Book of Indoor Games for
Young People of All Ages. C. Arthur
Pearson, London, 1904. [This copies a lot
from Hoffmann (or a common ancestor?).]
Much
of the material of Indoor Games is repeated in: J. K. Benson, ed.; The Pearson
Puzzle Book; C. Arthur Pearson, London, nd [1921 -- BMC]. This is not in BMC or NUC under Benson --
but I have seen an ad listing this as by Mr. X and it is listed under Mr. X in
BMC. Puzzle Book pp. 1-96 = Indoor Games pp. 189-257; Puzzle Book pp. 109-114 =
Indoor Games pp. 258-262. The
only different material in Puzzle Book is pp. 97-108. Neither book refers to the other. Cf Mr. X in Section 4.A.1
Berkeley & Rowland. Card Tricks & Puzzles. 1892.
"Berkeley"
[Peel, Walter H.] & Rowland, T. B. Card Tricks and Puzzles.
The Club Series, George Bell
& Sons, London, 1892 -- according to BMC, but my copy is 1897. Card Puzzles, etc., pp. 1-74 is by Berkeley;
Arithmetical Puzzles, pp. 75-120 is by Rowland.
Berlekamp, Elwyn R. (1940- )
See: Winning Ways.
Bestelmeier. 1801-1803.
G.
H. [Georg Hieronimus] Bestelmeier.
Magazin von verschiedenen Kunst‑ und andern nützlichen Sachen
.... [Toy catalogues.] Nuremberg, 1801‑1803.
Eight
issues and cumulative classified index reprinted by Olms, Zurich, 1979. Issue VII is 1801; the others are 'neue
verbesserte Auflage', 1803. This
includes items numbered 1 through 1111.
Selections,
with English translations. Daniel S.
Jacoby, ed. The Amazing Catalogue of
the Esteemed Firm of George Hieronimus Bestelmeier. Selected Excerpts from Editions of 1793 and 1807. [A comment inside makes me wonder if
1793-1807 is meant??] Merrimack
Publishing Corp., NY, 1971, 82pp. The
numeration is the same as in the Olms edition, but the Jacoby continues to item
1321. Obviously these later items come
from the 1807 edition, but we cannot tell if they might date from 1805, say,
nor whether all the earlier items go back to 1793. Jerry Slocum uses Jacoby in his Compendium and has kindly
provided photocopies of Jacoby's pp. 70-82 containing all the items after 1111
and some examples of the earlier items.
Jacoby does not translate the texts, but just provides English labels
for each picture and these labels are sometimes unconnected with the text.
Many
of Bestelmeier's items are taken from Catel; Kunst-Cabinet; 1790. Sometimes the figure is identical (often
reversed) or is a poor copy. Texts are
often copied verbatim, or slightly modified, but often abbreviated. E.g. Catel often explains the puzzle and
this part is frequently omitted in Bestelmeier. Bestelmeier was the successor to Catel, qv. The booklet by Slocum & Gebhardt (qv
under Catel) gives precise datings for the various parts of these catalogues,
but I have not yet entered these details.
Bhaskara I. 629.
Bhāskara
I.
Āryabhaţīya-Bhāşya. [NOTE: ţ denotes a
t with a dot under it and ş
denotes an s with a dot under it.] 629.
Critically edited, including an English Appendix of the numerical
examples used, by Kripa Shankar Shukla.
Indian National Science Academy, New Delhi, 1976. (Vol. 2 of a three volume series devoted to
the Āryabhaţīya (499) of Aryabhata (476- ), qv.)
Bhaskara I repeats and exposits Aryabhata verse by verse, but
Aryabhata rarely gives numerical examples, so Bhaskara I provided them and
these were later used by other Indian writers such as Chaturveda, 860. His earlier Maha-Bhaskariya (Mahā‑Bhāskarīya)
of c629 is cited in 7.P.2. Shukla's
Appendix is sometimes brief, but sometimes very detailed, e.g. on the 26
examples of Chinese remainder problems.
Bhaskara II (1114-c1185).
Bhâskara
II (1114-c1185, see Colebrooke).
Biggs, Norman L. See:
BLW.
Bijaganita. Bîjaganita of Bhaskara II,
1150 (see Colebrooke).
The Bile Beans Puzzle Book. 1933.
Bile
Beans (C. E. Fulford, Ltd., Leeds, England).
The Bile Beans Puzzle Book.
1933.
Birtwistle. Math. Puzzles & Perplexities. 1971.
Claude
Birtwistle. Mathematical Puzzles and
Perplexities How to Make the Most of
Them. George Allen & Unwin, London,
1971.
Birtwistle. Calculator Puzzle Book. 1978.
Claude
Birtwistle. The Calculator Puzzle
Book. Paperfronts (Elliot Right Way
Books), Kingswood, Surrey, 1978. (There
is a US ed. by Bell, NY, 1978.)
BL(LD). British Library (Lending Division).
Blasius. 1513. Johannis
(or Joannes) Martinus Blasius (later denoted Sileceus or Sciliceus). Liber Arithmetice Practice Astrologis Phisicis
et Calculatioribus admodum utilis.
Thomas Kees for Joannis Parui & Joannis Lambert (in colophon; TP has
Jehanlambert), Paris, 1513. Facsimile
by Heffer Scientific Reprint, Cambridge, 1960.
See Smith, Rara, pp. 95-97.
The Glaisher article in 7.P.5 [Messenger of Mathematics 53 (1923-24) 1‑131]
discusses this book and says he only knows one example of it, which he has in
front of him, so I suspect this facsimile is from that copy. See Rara 95-97. The Honeyman Collection had a copy, saying it was printed for J.
Petit and J. Lambert and that copy had Petit's device on the TP while the TP
shown in Rara has Lambert's device, which is as in this facsimile. There was a reprinting in 1514 and extended
editions in 1519 (ed. by Oronce Finé) and 1526 (ed. by T. Rhaetus) [Honeyman
Collection, nos. 350-352].
BLC. British Library Catalogue, replacing BMC,
in progress since 1970s.
BLC-Ø Indicates that I could not find the item in
the BLC.
BLW. 1976. Norman L. Biggs, E. Keith Lloyd & Robin J.
Wilson. Graph Theory 1736‑1936. OUP, 1976.
Blyth. Match-Stick Magic. 1921.
Will
Blyth. Match-Stick Magic. C. Arthur Pearson, London, 1921, reprinted
1923, 1939.
BM(C). British Museum (Catalogue (of books) to
1955. c1963).
BMC65. Supplement to the above Catalogue for 1956‑1965. c1968.
BN(C). Bibliothèque National, Paris. (Catalogue, 1897-1981.)
Bodleian. The Bodleian Library, University of Oxford, or
its catalogue.
Bonnycastle. Algebra.
1782
John
Bonnycastle (??-1821). An Introduction
to Algebra, with Notes and Observations; designed for the Use of Schools, and
Other Places of Public Education.
1782. The first nine editions
appeared "without any material alterations". In 1815, he produced a 10th ed., "an
entire revision of the work" which "may be considered as a concise
abridgment" of his two volume Treatise on Algebra, 1813, (2nd ed. in
1820). The 1815 ed. had an Appendix: On
the application of Algebra to Geometry.
I have a copy of the 7th ed., 1805, printed for J. Johnson, London, and
it is identical to the 2nd ed. of 1788 except for a problem in the final
section of Miscellaneous Questions.
However, the 9th ed. of 1812 has page numbers advanced by 10 except
toward the end of the book. I also have
the 13th ed. of 1824, printed for J. Nunn and 11 other publishers, London,
1824. This version has an Addenda: A
New Method of resolving Numerical Equations, by his son Charles Bonnycastle
(1797-1840), but is otherwise identical to the 10th ed. of 1815. The earlier text was expanded by about 10%
in 1815, so a number of problems only occur in later editions. I will cite these later problems as 1815 and
will cite the earlier problems as 1782.
[Halwas 36-38 gives some US editions.]
Book of 500 Puzzles. 1859.
The
Book of 500 Curious Puzzles: Containing a Large Collection of Entertaining
Paradoxes, Perplexing Deceptions in Numbers, and Amusing Tricks in
Geometry. By the author of "The
Sociable," "The Secret Out," "The Magician's Own
Book," "Parlor Games," and " Parlor Theatricals,"
etc. Illustrated with a great Variety
of Engravings. Dick & Fitzgerald,
NY, 1859. Compiled from The Sociable
(qv) and Magician's Own Book. Pp. 1-2
are the TP and its reverse. Pp. 3‑36,
are identical to pp. 285-318 of The Sociable; pp. 37-54 are identical to
pp. 199-216 of Magician's Own Book and pp. 55-116 are identical to pp. 241-302
of Magician's Own Book. [Toole Stott
103 lists it as anonymous. NUC, under
Frikell, says to see title. NUC, under
Book, also has an 1882 ed, compiled by William B. Dick. Christopher 129. C&B lists it under Cremer.]
The
authorship of this and the other books cited -- The Sociable, The Secret Out,
The Magician's Own Book, Parlor Games, and Parlor Theatricals, etc. -- is
confused. BMC & NUC generally
assign them to George Arnold (1834-1865) or Wiljalba (or Gustave) Frikell (1818
(or 1816) - 1903), sometimes with Frikell as UK editor of Arnold's US version
-- but several UK versions say they are translated and edited by W. H. Cremer
Jr, and one even cites an earlier French book (though the given title may not
exist!, but cf Manuel des Sorciers, 1825) -- see the discussion under Status of
The Project, in the Introduction, above.
The names of Frank Cahill, Henry Llewellyn Williams and Gustave Frikell
(Jr.) are sometimes associated with versions of these as authors or
coauthors. The Preface of The Sociable
says that most of the Parlor Theatricals are by Frank Cahill and George Arnold
-- this may indicate they had little to do with the parts that interest
us. Toole Stott 640 opines that this
reference led Harry Price to ascribe these books to these authors.
A
publisher's ad in the back says: "The above five books are compiled from
the "Sociable" and "Magician's Own."", referring to:
The Parlor Magician [Toole Stott 543, 544]; Book of Riddles and Five Hundred
Home Amusements [Toole Stott 107, 951]; Book of Fireside Games [possibly Toole
Stott 300??]; Parlor Theatricals; The Book of 500 Curious Puzzles. However, [Toole Stott 951] is another version
of The Book of Riddles and Five Hundred Home Amusements "by the author of
"Fireside Games" [Toole Stott 300], "The Parlor Magic"
[perhaps Toole Stott 543, 544], "Parlor Tricks with Cards" [Toole
Stott 1056 lists this as by Frikell, "abridged from The Secret Out"
(see also 547, 1142)], ..."; Dick & Fitzgerald, 1986 [sic, but must
mean 1886??].
See
Magician's Own Book for more about the authorship.
See
also: Boy's Own Book, Boy's Own Conjuring Book, Illustrated Boy's Own Treasury, Indoor and Outdoor, Landells: Boy's Own Toy-Maker, The Secret Out, Hanky Panky, The
Sociable.
Book of Merry Riddles. 1629?
The
Book of Merry Riddles. London,
1629. [Santi 235.]
Several
reprints. Also known as Prettie
Riddles.
A
Booke of Merry Riddles; Robert Bird, London, 1631. [Mark Bryant; Dictionary of Riddles; Routledge, 1990, p. 100.]
Booke
of Merry Riddles; John Stafford & W. G., London, 1660.
Reprint
of the 1629 in: J. O. Halliwell; The literature of the sixteenth and
seventeenth centuries; London, 1851, pp. 67‑102. [Santi 235.]
Reprint
of the 1660 in: J. O. Halliwell; The Booke of Merry Riddles, together with
proper questions, and witty proverbs, to make pleasant pastime. Now first reprinted from the unique edition
printed at London in the year 1660. For
the author, London, 1866. This was a
printing of 25 copies. There is a copy
at UCL and a MS note at the end says 15 copies were destroyed on 9 Apr 1866,
signed: J. O. H., with Number 9 written below.
[Santi 307.] I have seen this,
but some of the riddles are quoted by other authors and I will date all items
as 1629? until I examine other material.
Reprint
of the 1629 in: Alois Brandl; Shakespeares Book of Merry Riddles und die
anderen Räthselbücher seiner Zeit; Jahrbuch der deutschen
Shakespeare-Gesellschaft 42 (1906) 1-64 (with the 1631 ed on pp. 53-63). ??NYR.
[Santi 235 & 237.]
Borghi. Arithmetica. 1484.
Pietro
Borghi = Piero Borgo or Borgi (?? - ³1494). Qui comenza la nobel opera de arithmethica
ne la qual se tracta tute cosse amercantia pertinente facta & compilata p
Piero borgi da veniesia. Erhard
Ratdolt, Venice, 1484. 2 + 116 numbered
ff. This is the second commercial
arithmetic printed in Italy and was reprinted many times. See Rara 16-22. This edition was reproduced in facsimile, with notes by Kurt
Elfering, as: Piero Borghi; Arithmetica
Venedig 1484; Graphos, Munich, 1964;
in: Veröffentlichungen des Forschungsinstituts des Deutschen Museums für
die Geschichte der Naturwissenschaften und der Technik, Reihe C -- Quellentexte
und Übersetzunge, Nr. 2, 1965.
The 3rd ed of 1491 had a title: Libro
dabacho. From the 4th ed of 1501, the
title was Libro de Abacho, so this is sometimes used as the title for the first
editions also. Rara indicates that the
printing was revised to 100 numbered ff by the 4th ed. of 1491. I have examined a 1509 ed. by Jacomo Pentio,
Venice, ??NX. This has 100 numbered ff,
but the last three ff contain additional material, though Rara doesn't mention
this until the 11th ed of 1540. H&S
discusses a problem and the folio in the 1540 ed is the same as in the 1509
ed. The locations of interest in the
1509 ed. are c18ff before the corresponding locations of the 1484. Van Egmond's Catalog 293-297 lists 13
Venetian editions from 1484 to 1567.
It
has been conjectured that this was a pseudonym of Luca Pacioli, but there is no
evidence for this [R. Emmett Taylor; No Royal Road Luca Pacioli and His Times; Univ. of North Carolina Press, Chapel
Hill, 1942, pp. 60 & 349].
See
also: D. E. Smith; The first great
commercial arithmetic; Isis 8 (1926) 41-49.
Bourdon. Algèbre.
7th ed., 1834.
Louis
Pierre Marie Bourdon (1779-1854).
Élémens d'Algèbre. 7th ed.,
Bachelier, Paris, 1834. (1st ed, 1817;
5th, 1828; 6th, 1831; 8th, 1837; 1840.
Undated preface in the 7th ed. describes many changes, so I will cite
this as 1834, though much of the material would have occurred earlier.)
Boy's Own Book. 1828.
William
Clarke, ed. The Boy's Own Book. The bibliography of this book is extremely
complex -- by 1880, it was described as having gone through scores of
editions. My The Bibliography of Some Recreational Mathematics Books has 11 pages listing 76 English (40 UK, 37
US, 1 Paris) versions and a Danish version, implying 88 English (50 UK, 37
US, 1 Paris) versions, and 10 (or 11) related versions, and giving a detailed
comparison of the versions that I have seen.
Because of the multiplicity of versions, I have cited it by title rather
than by the original editor's name, which is not in any of the books (except
the modern facsimile) though this attribution seems to be generally
accepted. I have examined the following
versions, sometimes in partial photocopies or imperfect copies.
Vizetelly,
Branston and Co., London, 1828, 448pp.;
2nd ed., 1828, 462pp.; 3rd ed.,
1829, 464pp (has an inserted advertisement sheet); 6th ed??, c1830, 462pp?? (my copy lacks TP, pp. 417-418, 431-436,
461-462); 9th ed., 1834, 462pp. Longman, Brown & Co., London, 24th ed.,
1846, 462pp. [The latter five are
identical, except for a bit in the Prelude (and the extra sheet in 3rd ed), so
I will just cite the first of these as 1828‑2. It seems that all editions from the 2nd of 1828 through the 29th
of 1848, 462pp. are actually identical except for a bit of the Prelude (and the
advertisement sheet in the 3rd ed.)]
First
American Edition. Munroe & Francis,
Boston & Charles S. Francis, NY, 1829, 316pp. Facsimile by Applewood Books, Bedford, Massachusetts, nd
[1998?]. This is essentially an
abridgement of the 2nd ed of 1828, copying the Prelude and adding "So far the London Preface. The American publishers have omitted a few
articles, entirely useless on this side of the Atlantic, ...." The type is reset, giving some reduction in
pages. A number of the woodcuts have
been omitted. The section title pages
are omitted. Singing Birds, Silkworms,
White Mice, Bantams, Magnetism, Aerostatics, Chess and Artificial Fireworks are
omitted. Angling, Rabbits, Pigeons,
Optics are reduced. Rosamond's Bower is
omitted from Paradoxes and Puzzles.
Surprisingly, The Riddler is increased in size. The 2pp Contents is omitted and an 8pp Index
is added.
Baudry's
European Library & Stassin & Xavier, Paris, 1843, 448pp. [The existence of a Paris edition was
previously unknown to the vendor and myself, but it is Heyl 354 and he cites
Library of Congress. It is very
different than the English and US editions, listing J. L. Williams as
author. Even when the topic is the
same, the text, and often the topic's name, are completely rewritten. See my
The Bibliography of Some Recreational Mathematics Books for details -- in it I have found it
generally necessary to treat this book separately from all other editions. I will cite it as 1843 (Paris). Much of this, including almost all of the
material of interest is copied exactly in
Anon: Boy's Treasury, 1844, qv,
and in translated form in
de Savigny, Livre des Écoliers, 1846, qv. The problem of
finding the number of permutations of the letters of the alphabet assumes 24
letters, which makes me wonder if these books are based on some earlier French
work. Heyl 355 is probably the same
book, with slight variations in the title, by Dean and Munday, London, c1845.]
David
Bogue, London, 1855, 611pp. [It seems
that this version first appears in 1849 and continues through about 1859, when
two sections were appended.]
[W.
Kent (late D. Bogue), London, 1859, 624pp??.
For almost all material of interest, this is identical to the 1855 ed,
so I will rarely (if ever?) cite it.]
[Lockwood
& Co., London, 1861, 624pp.
Identical to the 1859 ed., so I will not cite it.]
Lockwood
& Co., London, 1868, 696pp.
[Lockwood
& Co., London, 1870, 716pp.
Identical to 1868 with 20pp of Appendices, so page numbers for material
of interest are the same as in 1868, so I will not cite it.]
[Crosby
Lockwood & Co., London, 1880, 726pp.
Identical to 1870, but having the Appendices and 20 more pages
incorporated into a new section. For
almost all material of interest, the page numbers are 30 ahead of the 1868
& 1870 page numbers, so I will not cite it except when the page numbers are
not as expected.]
[5th
(US?) ed., Worthington, NY, 1881, 362pp.
For almost all material of interest, this is identical to the 1829 (US)
ed., so I will rarely (if ever?) cite it.]
I
will cite pages with edition dates and edition numbers or locations if needed
(e.g. 1828-2: 410 or
1829 (US): 216). See also: Book of 500 Puzzles, Boy's Own Conjuring Book, Illustrated Boy's Own Treasury.
Anonymous. The Riddler; A Collection of Puzzles,
Charades, Rebusses, Conundrums, Enigmas, Anagrams, &c. for the Amusement of
Little Folks. S. Babcock, New Haven,
Connecticut, 1835. 22pp. My copy has leaf 11/12 half missing and leaf
17/18 missing; NUC & Toole Stott 1392 say it should be 24pp, so presumably
leaf 23/24 is also missing here. [Toole
Stott 1392 has The Riddler: or, Fire-Side Recreations; a collection ..., 1838,
also listed in NUC.] Paradoxes and
Puzzles section consists of the introduction and 11 puzzles copied almost
exactly from the Paradoxes and Puzzles section of Boy's Own Book, 2nd ed. of
1828 and this material is all in the first American edition of 1829. Other material is charades, etc. and is all
in both these versions of Boy's Own Book.
Shortz states that this is the first American book with puzzles -- but
there were at least five American versions of Boy's Own Book before this and
all the material in The Riddler, except some woodcuts, is taken from Boy's Own
Book, so this pamphlet seems to be a pirate version. NUC also lists a 1838 version.
Boy's Own Conjuring Book. 1860.
The
Boy's Own Conjuring Book: Being a Complete Hand-book of Parlour Magic; and
Containing over One Thousand Optical, Chemical, Mechanical, Magnetical, and
Magical Experiments, Amusing Transmutations, Astonishing Sleights and Subtleties,
Celebrated Card Deceptions, Ingenious Tricks with Numbers, Curious and
Entertaining Puzzles, Charades, Enigmas, Rebuses, etc., etc., etc. Illustrated with nearly two hundred
engravings. Intended as a source of
amusement for one thousand and one evenings.
Dick and Fitzgerald, NY, 1860.
384pp. [Toole Stott 115,
corrected, lists this as (1859), and under 114, describes it as an extended
edition of The Magician's Own Book -- indeed the running head of the book is
The Magician's Own Book! -- but see below.
Toole Stott 481 cites a 1910 letter from Harris B. Dick, of the
publishers Dick & Fitzgerald. He
describes The Boy's Own Conjuring Book as a reprint of Magician's Own Book
"evidently gotten up and printed in London, but singularly enough it had
printed in the book on the title-page -- New York, Dick &
Fitzgerald." Indeed, all the
monetary terms are converted into British.
Harold Adrian Smith [Dick and Fitzgerald Publishers; Books at Brown 34
(1987) 108-114] states that this is a London pirate edition. BMC has 384pp, c1860. NUC has a 384pp version, nd. Christopher 145-149 are five versions from
1859 and 1860, though none has the blue cover of my copy. Christopher 145 says the 1859 versions were
printed by Milner & Sowerby, Halifax, and describes it as an extraction
from Magician's Own Book, but see below.
Christopher 148 cites Smith's article.]
I also have a slightly different version with identical contents except
omitting the date and frontispiece, but with a quite different binding, probably
Christopher 149. [NUC lists 334pp, nd;
416pp, nd and 416pp, 1860. Toole Stott
114 is a 416pp version, 1861. Toole
Stott 959 is a 534pp version, 1861.
C&B cite a New York, 1859 with 416pp, a New York, nd, 334pp and
London, c1850 (surely too early?).]
I
have now compared this with The Magician's Own Book of 1857 and it is
essentially a minor reworking of that book.
The Magician's Own Book has 17 chapters and an answers chapter and a
miscellaneous chapter of items which are almost all also listed in the Contents
under earlier sections. All together,
there are some 635 items. The Boy's Own
Conjuring Book copies about 455 of these items essentially directly, completely
omitting the chapters on Electricity, Galvanism, Magnetism, Geometry, Art, Secret
Writing and Strength, and almost completely omitting the chapter on
Acoustics. Of the 488 items in the
other chapters, 453 are copied into the Boy's Own Conjuring Book, and this has
in addition two of the acoustic problems, 125 new miscellaneous problems and
38pp of charades, riddles, etc. (The
later UK edition of Magician's Own Book is very different from the US
edition.) Many of the problems are
identical to the Boy's Own Book or the Illustrated Boy's Own Treasury. See also:
Book of 500 Puzzles, Boy's Own
Book, Illustrated Boy's Own
Treasury, Landells: Boy's Own Toy‑Maker.
Boy's Treasury. 1844.
Anonymous. The Boy's Treasury of Sports, Pastimes, and
Recreations. With four hundred
engravings. By Samuel Williams. [The phrasing on the TP could be read as
saying Williams is the author, but the NUC entry shows he was clearly listed as
the designer in later editions and his name appears on the Frontispiece.] D. Bogue, London, 1844. Despite the similarity of title, this is
quite different from Illustrated Boy's Own Treasury and the similar books of
c1860. [Toole Stott 116. Toole Stott 117 is another ed., 1847,
'considerably extended'. Toole Stott
gives US editions: 959; 960; 118; 199 & 961-965 are 1st, 1847; 2nd, 1847;
3rd, 1848; 6 versions of the 4th, 1850, 1848, 1849, 1852, 1854, 1848. Hall, BCB 37 is a US ed. of 1850 = Toole
Stott 119. Christopher 151 is a US
version of 1850? NUC lists 9 versions,
all included in Toole Stott. Toole
Stott cites some BM copies, but I haven't found this in the BMC. A section of this, with some additional
material, was reissued as Games of Skill and Conjuring: ..., in 1860, 1861,
1862, 1865, 1870 -- see Toole Stott 312-317.]
I
have now found that much of this, including all the material of interest, is
taken directly from the 1843 Paris edition of
Boy's Own Book, qv, by J. L.
Williams, including many of the illustrations - indeed they have the same Frontispiece,
with S. Williams' name on it.
BR. c1305. Greek
MS, c1305, Codex Par. Suppl. Gr. 387, fol. 118v‑140v. Transcribed, translated and annotated by
Kurt Vogel as: Ein Byzantinisches Rechenbuch des frühen 14.Jahrhunderts; Wiener
Byzantinistische Studien, Band VI; Hermann Böhlaus Nachf., Wien, 1968. I will cite problem numbers and pages from
this -- Vogel gives analysis of the methods on pp. 149‑153 and historical
comments on pp. 154‑160, but I will not cite these.
Brahmagupta, c628. See:
Brahma‑sphuta‑siddhanta;
Colebrooke.
Brahma‑sphuta‑siddhanta.
Bráhma‑sphuta‑siddhânta
of Brahmagupta, 628 (see Colebrooke).
He only states rules, which are sometimes obscure. It appears from Colebrooke, p. v, and Datta
(op. cit. under Bakhshali, p. 10), that almost all the illustrative examples
and all the solutions are due to Chaturveda Prthudakasvâmî in 860. Brahmagupta's rules are sometimes so general
that one would not recognise their relevance to these examples and I have often
not cited Brahmagupta. E.g. cistern
problems are given as examples to Brahmagupta's verse on how to add and
subtract fractions. (See also Datta
& Singh, I, p. 248.) Some of these
comments are taken from Bhaskara I in 629.
Brush. Hubert Phillips.
Brush Up Your Wits. Dent,
London, 1936.
BSHM. British Society for the History of
Mathematics. The produce a useful
Newsletter.
Buteo. Logistica. 1559.
Johannes
Buteo (= Jean Borrel, c1485-c1560 or c1492-1572). Ioan. Buteonis Logistica, quæ & Arithmetica vulgò dicitur in
libros quinque digesta: quorum index summatim habetur in tergo. Gulielmus Rovillius, Lyons, 1559. Most of the material is in books IV and
V. H&S cites some problems in the
1560 ed with the same pages as in the 1559 ed, so these editions are presumably
identical. See Rara 292-294.
c. circa, e.g. c1300. Also
c= means "approximately
equal", though @ will be used in mathematical contexts.
C. Century, e.g. 13C, -5C.
Calandri. Arimethrica. 1491.
Philippo
Calandri. Untitled. Frontispiece is labelled "Pictagoras
arithmetrice introductor". Text
begins: "Philippi Calandri ad nobilem et studiosus Julianum Laurentii
Medicē de arimethrica opusculū." Lorenzo de Morgiani & Giovanni Thedesco da Maganza, Florence,
1491. Van Egmond's Catalog
298-299. The Graves collection has two
copies dated 1491, one with the folio number
c iiii misprinted as b iiii - cf Van Egmond for other
differences in this unique variant.
There was a reprint by Bernardo Zucchetta, Florence, 1518 -- ??NYS but
mentioned: in a handwritten note in one
of the Graves copies of the 1491 (giving Bernardo Zucchecta, 1517); in Smith, Rara, p. 48 (giving Bernardo
Zuchetta, 1518); in Riccardi [I, col.
208-209] (giving Bernardo Zuchecta, 1515)
and in Van Egmond's Catalog 299.
"It is the first printed Italian arithmetic with illustrations
accompanying problems, ...." (Smith, Rara, pp. 46‑49). There are about 50 of these illustrations,
which appear to be woodcuts, but they are quite small, about 25mm (1")
square, and the same picture is sometimes repeated for a related but
inappropriate problem. Rara reproduces
some of these, slightly reduced.
Riccardi [I, col. 208-209] says there may have been a 1490 ed. by
Bernardo Zuchecta, but Van Egmond did not find any example.
Calandri. Aritmetica.
c1485.
Filippo
Calandri. Aritmetica. c1485 [according to Van Egmond's Catalog
158-159]. Italian MS in Codex 2669,
Biblioteca Riccardiana di Firenze.
Edited by Gino Arrighi, Edizioni Cassa di Risparmio di Firenze,
Florence, 1969. 2 vol.: colour
facsimile; transcription of the text.
Copies of the facsimile were exhausted about 1980 and repeated requests
to the Cassa di Risparmio have not produced a reprint, though they usually send
a copy of the text volume every time I write!
I have now (1996) acquired a example of the 2 vol. set and I find that
copies of the text volume which are not part of a set have 8 colour plates
inserted, but these are not in the copy in the set.
I
cite folios from the facsimile volume and pages from the text volume. These are in direct correspondence with the
original except for those pages with full page illustrations. The original begins with a blank side with a
Frontispiece verso, then 9 sheets (18 pp.) of full page tables, then two blank
sheets. The numbered folios then begin
and go through 110. Ff. 1r - 32r are
pp. 3 - 65 of the text. F. 32v is a full page calculation which is
not in the text. Then ff. 33r - 110r are pp. 66 - 220 of the text. F. 110v is a full page illustration omitted in the text. The first 80 folio numbers are in elaborate
Roman numerals centred at the head of the page. (These are sometimes unusually written -- e.g. XXIIIIII.) The later folios were not originally
numbered and were later numbered in the top right corner using Hindu-Arabic
numerals.
In
Sep 1994, I examined the original MS, though it is on restricted access. The original colours are rather more
luminous than in the facsimile, but the facsimile is a first class job. The history of this codex is obscure. It is said to have belonged to Piero di
Lorenzo dei Medici and it may be the book catalogued in the library of
Francesco Pandolfini, c1515, as 'uno libretto ... di Filippo Calandri in
arithmetica'. The Riccardi family
collected continuously from their rise in the mid 15C until the library was
acquired by the city in 1813. A number
of items from the Pandolfini catalogue can be identified as being in the
Riccardiana. Van Egmond's dating may be
early as some claim this was produced for Giuliano de' Medici, who was born in
1479.
Calandri. Raccolta.
c1495.
Filippo
Calandri. Una Raccolta di Ragioni. In: Cod. L.VI.45, Biblioteca Comunale di
Siena. Ed. by D. Santini. Quaderni del Centro Studi della Matematica
Medioevale, No. 4, Univ. di Siena, 1982.
Van Egmond's Catalog 193 identifies this as ff. 75r-111v of the codex,
titles it Ragone Varie and gives a date of c1495.
Calandri. See also:
Benedetto da Firenze, P. M.
Calandri.
Cardan. Ars Magna.
1545.
Jerome
Cardan = Girolamo Cardano = Hieronymus Cardanus (1501‑1576). Artis Magnae sive de Regulis
Algebraicis Liber Unus. Joh. Petreium, Nuremberg, 1545, ??NYS Included in Vol. IV of the Opera Omnia,
Joannis Antonius Huguetan & Marcus Antonius Ravaud, Lyon, 1663, and often
reprinted, e.g. in 1967. NEVER CITED??
Cardan. Practica Arithmetice. 1539.
Jerome
Cardan = Girolamo Cardano = Hieronymus Cardanus (1501‑1576). Practica Arithmetice, & Mensurandi
Singularis. (Or: Practica Arithmeticae
Generalis Omnium Copiosissima & Utilissima, in the 1663 ed.) Bernardini Calusci, Milan, 1539. Included in Vol. IV of the Opera Omnia,
1663, see above. Some of the section
numbers are omitted in the Opera Omnia and have to be intuited. I will give the folios from the 1539 ed.
followed by the pages of the 1663 ed., e.g. ff. T.iiii.v-T.v.r (p. 113).
Cardan. De Rerum Varietate. 1557.
Jerome
Cardan = Girolamo Cardano = Hieronymus Cardanus (1501‑1576). De Rerum Varietate. Henricus Petrus, Basel, 1557; 2nd ed., 1557;
5th ed., 1581, ??NYS. Included in Vol.
III of the Opera Omnia, 1663, see above.
Cardan. De Subtilitate. 1550.
Jerome
Cardan = Girolamo Cardano = Hieronymus Cardanus (1501‑1576). De Subtilitate Libri XXI. J. Petreium, Nuremberg, 1550; Basel, 1553;
6th ed., 1560; and five other 16C editions, part ??NYS. Included in Vol. III of the Opera Omnia,
1663, see above. French ed. by Richard
Leblanc, Paris, 1556, 1584, titled: Les Livres d'Hieronymus Cardanus: De la
Subtilité et subtiles Inventions, ensemble les causes occultes et les raisons
d'icelles; 9th ed., 1611. I have seen a
note that the 1582 ed. by Henricus Petrus, Basel, was augmented by a riposte to
attacks by Scaliger with further illustrations.
Carlile. Collection.
1793.
Richard
Carlile. A Collection of One Hundred
and Twenty Useful and Entertaining Arithmetical, Mathematical, Algebraical, and
Paradoxical Questions: With the Method of Working Each. Printed by T. Brice for the author, Exeter,
1793. Wallis 227 CAR, ??NX. Includes a number of straightforward
problems covered here, but I have only entered the more unusual examples.
Carroll-Collingwood. 1899.
The
Lewis Carroll Picture Book. Stuart
Dodgson Collingwood, ed. T. Fisher
Unwin, London, 1899. = Diversions
and Digressions of Lewis Carroll, Dover, 1961.
= The Unknown Lewis Carroll, Dover, 1961(?). Reprint, in reduced format, Collins, c1910. The pagination of the main text is the same
in the original and in both Dover reprints, but is quite different than the
Collins. I will indicate the Collins
pages separately. The later Dover has
42 additional photographs.
Carroll-Gardner. c1890?
or 1996
Martin
Gardner. The Universe in a
Handkerchief. Lewis Carroll's
Mathematical Recreations, Games, Puzzles and Word Plays. Copernicus (Springer, NY), 1996. As with Carroll-Wakeling, Carroll material
will be dated as 1890?, but there is much material by Gardner which is dated
1996.
Carroll-Wakeling. c1890?
Lewis
Carroll's Games and Puzzles. Newly
Compiled and Edited by Edward Wakeling.
Dover and the Lewis Carroll Birthplace Trust, 1992. This is mostly assembled from various
manuscript sheets of Carroll's containing problems which he intended to
assemble into a puzzle book. Wakeling
has examined a great deal of such material, including a mass of Carroll's notes
to Bartholomew Price (1818‑1898) who was Sedleian Professor of Natural
Philosophy at Oxford in 1853-1898.
Price was at Pembroke College, becoming the Master, adjacent to
Carroll's Christ Church. He had tutored
Carroll (1833‑1898) and they were close friends and in continual contact
until their deaths, both in 1898.
However, few of the papers are dated and they are simply loose sheets
with no indication of being in order, so there is no way to date the undated
sheets and I have given a fairly arbitrary date of c1890? for these, though
Carroll was more active before then rather than after. Some items are taken from Carroll's youthful
magazines or his correspondence and hence are more precisely dated. The correspondence is more fully given in
Carroll-Collingwood.
In
response to an inquiry, Wakeling wrote on 28 May 2003 and said that some of the
Carroll-Price notes were typewritten 'probably using Dodgson's Hammond
typewriter, purchased in 1888.' This
gives a somewhat more precise dating than my c1890? and I will give: 1888 to 1898 for such items, unless there is other evidence.
Carroll-Wakeling II. c1890?
Rediscovered
Lewis Carroll Puzzles. Newly Compiled
and Edited by Edward Wakeling. Dover,
1995. See the notes to
Carroll-Wakeling, above.
Cassell's. 1881.
Cassell's
Book of In‑Door Amusements, Card Games, and Fireside Fun. Cassell, Peter, Gilpin & Co., London,
1881; Cassell, London, 1973. 217pp
[probably + 1p + 6pp Index] (pp. 1-8 are preliminary matter). [There was a companion volume: Cassell's
Book of Sports and Pastimes. In 1887,
the two were combined, with the spine titled
Cassell's Book of Outdoor Sports and Indoor Amusements. The front cover says Out Door Sports, the
back cover says Indoor Amusements, while the title page says Cassell's Book of
Sports and Pastimes. It contains all
the main text of Book of In‑Door Amusements, ..., advanced by 744
pages. From at least 1896, Card Games
and Parlour Magic were completely revised and later there were a few other
small changes. The title varies
slightly. Manson (qv) is a 1911
revision and extension to 340pp of main text.]
Catel. Kunst-Cabinet. 1790.
Peter
Friedrich Catel. Mathematisches und
physikalisches Kunst-Cabinet, dem Unterrichte und der Belustigung der Jugend
gewidmet. Nebst einer zweckmässigen
Beschreibung der Stücke, und Anzeige der Preise, für welche sie beim Verfassser
dieses Werks P. F. Catel in Berlin zu bekommen sind. [I.e. this is a catalogue of items for sale by post!] Lagarde und Friedrich, Berlin & Libau,
1790. [MUS #113.] P. iv says he started his business in 1780.
There
is a smaller Vol. 2, with the same title, except 'beim Verfasser dieses Werkes
P. F. Catel' is replaced by 'in der P. F. Catelschen Handlung', and the
publisher is F. L. Lagarde, Berlin, 1793.
My
thanks to M. Folkerts for getting a copy of the example in the Deutsches Museum
made for me.
All
citations are to vol. 1 unless specified.
Many
of Bestelmeier's items are taken from Catel.
Sometimes the figure is identical (often reversed) or is a poor
copy. Texts are often copied verbatim,
or slightly modified, but usually abbreviated.
E.g. Catel often explains the puzzle and this part is frequently omitted
in Bestelmeier. Bestelmeier was the
successor to Catel. Dieter Gebhardt has
searched for the various editions and associated price lists of the Catel and
Bestelmeier catalogues in German libraries and he and Jerry Slocum have
published the details in: Jerry
Slocum & Dieter Gebhardt. Puzzles
from Catel's Cabinet and Bestelmeier's Magazine 1785 to 1823. English
translations of excerpts from the German Catel-Katalog and
Bestelmeier-Katalog. Intro. by David
Singmaster. History of Puzzles
Series. The Slocum Puzzle Foundation,
PO Box 1635, Beverly Hills, California, 90213, USA, 1997. I have not yet made detailed entries from
this which gives precise dates for the various parts of these catalogues.
CFF. Cubism for Fun. This is the Newsletter of the Nederlandse
Kubus Club (NKC) (Dutch Cubists Club) which has been in English since the mid
1980s.
Chambers -- see: Fireside
Amusements.
Charades, Enigmas, and Riddles. 1859.
Charades,
Enigmas, and Riddles. Collected by A
Cantab. [BLC gives no author. "A
Cantab." was a common pseudonym.
One such author of about the right time and nature was George
Haslehurst.] (Cambridge, 1859).
3rd ed.,
J. Hall and Son, Cambridge, 1860, HB.
Half-title, 6 + 96pp.
4th
ed., Bell & Daldy, London, 1862. 8
preliminaries (i = half-title; FP facing iii = TP; v-viii = Introduction;
errata slip; two facing plates illustrating a charade for Harrowgate [sic]
Waters), 1-160, 32pp publisher's ads, dated Jan 1863; (my copy is lacking pp.
63-64). The three plates are signed
J.R.J. This is a substantial expansion
of the 3rd ed.
I
also have photocopy of part of the 5th ed., Bell and Daldy, London, 1865, and
this shows it was even larger than the 4th ed, but most of the problems of
interest have the same or similar problem numbers in the three editions that I
have seen. I will cite them as in the
following example. 1860: prob. 28, pp.
59 & 63; 1862: prob. 29, pp. 135 & 141; 1865: prob. 573, pp. 107
& 154.
Chaturveda. Chaturveda Pŗthudakasvâmî [NOTE: ŗ
denotes an r with an underdot.]. Commentator on the Brahma‑sphuta‑siddhanta
(qv), 860. Some of these comments are
taken from Bhaskara I in 629.
Shukla calls him Pŗthūdaka, but Colebrooke cites him as Ch.
Chessics. Chessics.
The Journal of Generalised Chess.
Produced by G. P. Jelliss, 5 Biddulph Street, Leicester, LE2 1BH. No. 1 (Mar 1976) -- No. 29 & 30 (1987). Succeeded by G&PJ.
Child. Girl's Own Book.
Mrs.
L. Maria Child [= Mrs. Child = Lydia
Maria Francis, later Child]. The Girl's
Own Book. The bibliography of this book
is confused. According to the Opies
[The Singing Game, p. 481], the first edition was Boston, 1831 and there was a
London 4th ed. of 1832, based on the 2nd US ed. However the earliest edition in the BMC is a 6th ed. of
1833. I have examined and taken some notes
from the 3rd ed., Thomas Tegg, London, 1832 -- unfortunately I didn't have time
to go through the entire book so I may have missed some items of interest. I have also examined the following.
Clark
Austin & Co., NY, nd [back of original TP says it was copyrighted by
Carter, Hendee, & Babcock in Massachusetts in 1833]; facsimile by Applewood Books, Bedford,
Massachusetts, nd [new copy bought in 1998 indicates it is 4th ptg, so
c1990]. The facsimile is from a copy at
Old Sturbridge Village. The back of the
modern TP says the book was first published in 1834 and the Cataloguing-in-Publication
data says it was originally published by Carter, Hendee and Babcock in
1834. However, the earliest version in
the NUC is Clark, Austin, 1833. I am
confused but it seems likely that Carter, Hendee and Babcock was the original
publisher in Boston in 1831 and that that this facsimile is likely to be from
1833 or an 1834 reprint of the same.
The pagination is different than in the 1832 London edition I have seen.
The
Tenth Edition, with Great Additions. By
Mrs. Child. Embellished with 144 Wood
Cuts. Thomas Tegg, London (& three
other copublishers), 1839. 12 + 307 pp
+ 1p publisher's ad. Has Preface to the
Second Edition but no other prefaces.
This Preface is identical to that in the 1833 NY ed, except that it
omits the final P.S. of season's greetings.
The 1833 NY essentially has the same text, but they have different
settings and different illustrations with some consequent rearrangement of
sections. However the main difference
is that the NY ed omits 41pp of stories.
There are a number of minor differences which lead to the NY ed having 9
extra pages of material.
The
Eleventh Edition, with Great Additions.
By Mrs. Child. Embellished with
124 Wood Cuts. Thomas Tegg, London
(& three other copublishers), 1842.
12 + 363 pp + 1p publisher's ad.
The Preface is identical to that in the 10th ed, but omits 'to the
Second Edition' after Preface. 90 pp of
games and 40 pp of enigmas, charades, rebuses, etc. have been added; 56 pp of
stories have been dropped.
The
Girl's Own Book of Amusements, Studies and Employments. New Edition. Considerably enlarged and modernized by Mrs. L. Valentine, and
others. William Tegg, London,
1876. This differs considerably from
the previous editions.
I
will cite the above by the dates 1832,
1833, 1839, 1842, 1876.
Various
sources list: 13th ed., 1844 [BMC,
Toole Stott 831]; Clark Austin, NY,
1845 [NUC]; 16th ed., 1853 [BMC]; 17th ed. by Madame de Chatelain, 1856 [BMC,
NUC, Toole Stott 832]; 18th ed. by
Madame de Chatelain, 1858 [BMC, Toole Stott 833]; 1858 [Osborne Collection (at Univ. of Toronto)]; rev. by Mrs. R. Valentine, 1861 [BMC,
Osborne Collection]; rev. by Mrs. R. Valentine,
1862 [BMC, NUC]; rev. by Mrs. R.
Valentine, 1864 [BMC]; rev. by Mrs. R.
Valentine, 1867 [BMC]; enlarged by Mrs.
L. Valentine, 1868 [NUC]; enlarged by
Mrs. L. Valentine, 1869 [BMC]; enlarged
by Mrs. L. Valentine, 1873 [NUC];
enlarged by Mrs. L. Valentine, 1875 [NUC]; enlarged by Mrs. L. Valentine, 1876 [BMC];
Heyl
gives the following under the title The
Little Girl's Own Book: Carter, Hendee
and Co., Boston, 1834; American
Stationers Co, John B. Russell, Boston, 1837;
Edward Kearney, NY, 1847; NY,
1849.
I think there were at least 33 editions. See my
The Bibliography of Some Recreational Mathematics Books for more details. Cf Fireside Amusements, below, which is largely copied from
Child.
Chiu Chang Suan Ching. c-150?
Chiu
Chang Suan Ching (Nine Chapters on the Mathematical Art). (Also called Chiu Chang Suan Shu and
variously transliterated. The pinyin is
Jiŭ Zhāng Suàn Shù.) c‑150? German translation by K. Vogel; Neun
Bücher arithmetischer Technik; Vieweg, Braunschweig, 1968. My citations will be to chapter and problem,
and to the pages in Vogel. (Needham
said, in 1958, that Wang Ling was translating this, but it doesn't seem to have
happened.) Some of the material dates
from the early Han Dynasty or earlier, say c-200, but Chap. 4 & 9, the most
original of all, have no indication of so early a date. A text of c50 describes the contents of all
the chapters and Høyrup suggests that Chap. 4 & 9 and the final assembly of
the book should be dated to the [early] 1C.
Christopher. 1994.
Maurine
Brooks Christopher & George P. Hansen.
The Milbourne Christopher Library.
Magic, Mind Reading, Psychic Research, Spiritualism and the Occult 1589-1900.
Mike Coveney's Magic Words, Pasadena, 1994. 1118 entries. References
are to item numbers.
Christopher II. 1998.
Maurine
Brooks Christopher & George P. Hansen.
The Milbourne Christopher Library -- II. Magic, Mind Reading, Psychic Research, Spiritualism and the
Occult 1589-1900. Mike Coveney's Magic Words, Pasadena, 1998. 3067 entries. References are to item numbers.
Recently received, ??NYR.
Chuquet. 1484. Nicolas
Chuquet. Problèmes numériques faisant
suite et servant d'application au Triparty en la science des nombres de Nicolas
Chuquet Parisien. MS No. 1346 du Fonds
Français de la Bibliothèque Nationale, 1484, ff. 148r-210r. Published in an abbreviated version as:
Aristide Marre; Appendice au Triparty en la science des nombres de Nicolas
Chuquet Parisien; Bulletino di bibliografia e di storia delle scienze
matematiche e fisiche 14 (1881) 413‑460.
(The first part of the MS was published by Marre; ibid. 13 (1880)
593-814; ??NYS) Marre generally
transcribes the text of the problem, but just gives the answer without any of
the text of the solution. I will cite
problems by number. There are 166
problems. (Much of this was used in his
student's book: Estienne de la Roche; Larismethique novellement composee par
maistre Estienne de la roche dict Villefrāche; Lyons, 1520, ??NYS. (Rara 128‑130).)
FHM Graham Flegg, Cynthia Hay & Barbara Moss. Nicolas Chuquet, Renaissance Mathematician. A study with extensive translation of Chuquet's
mathematical manuscript completed in 1484.
Reidel, Dordrecht, 1985. This
studies the entire MS, of which the above Appendice is only the second quarter. It often gives a full English translation of
the text of the problem and the solution, but it may summarize or skip when
there are many similar problems. The
problems in the first part of the MS are not numbered in FHM. I will cite this as FHM xxx, where xxx is
the page number, with 'English in FHM xxx' when the problem is explicitly
translated.
Clark. Mental Nuts. 1897, 1904,
1916.
A book
of Old Time Catch or Trick Problems
Regular old Puzzlers that kept your Grandad up at night. Copyright, 1897, by S. E. Clark,
Philadelphia. Flood & Conklin
Co. Makers of Fine Varnishes, Newark,
N.J. 100 problems and answers. 32pp + covers.
A book
of 100 Catch or Trick Problems Their
simplicity invites attack, while their cunningly contrived relations call forth
our best thought and reasoning.
Copyright, 1897, by S. E. Clark, Philadelphia. Revised 1904 Edition.
Waltham Watches, Waltham, Massachusetts. This was an promotional item and jewellers would have their
address printed on the cover. My
example has: With the compliments
of J. H. Allen Jeweler [sic] Shelbina, Mo. Thanks to
Jerry Slocum for this. In fact there
are only 95 problems; numbers 68, 75, 76, 78, 84 are skipped. 32pp + covers.
Revised
Edition 1916, with no specific company mentioned. Enlarged PHOTOCOPY from Robert L. Helmbold. 100 numbered problems, but some figures
inserted after no. 75 are the solutions to a problem in the other editions and
I have counted this as a problem (no. 75A), making 101 problems. 28pp + covers.
The editions are
considerably different. Only 40
problems occur in all three editions. There
are 50 problems common to 1897 and 1904, 42 common to 1897 and 1916 and 71
common to 1904 and 1916, though this counting is a bit confused by the fact
that problems are sometimes combined or expanded or partly omitted, etc. Solutions are brief. It includes a number of early examples or
distinct variants, which is remarkable for a promotional item. I have entered 36 of the 1897 problems plus
13 of the 1904 problems not in 1897 and 7 of the 1916 problems not in 1897 or
1904. Many others are standard examples
of topics covered in this work, but are not sufficiently early to be worth
entering.
I originally had the 1904
ed and cited the 1904 problems as 1897 on the grounds that editions of this
period do not change much, but having now seen the 1897 and 1916 eds, I realise
that the editions are very different, so I will cite the actual dates. Since only the 1897 version is paginated, I
will just cite problem numbers; the solutions are at the back.
Clarke, William. See:
Boy's Own Book.
CM. Crux Mathematicorum (originally titled
Eureka until 4:3)
CMJ. The College Mathematics Journal. Before the early 1980s, this was the Two
Year College Mathematics Journal.
Colebrooke. 1817.
Henry
Thomas Colebrooke (1765-1837), trans.
Algebra, with Arithmetic and Mensuration from the Sanscrit of
Brahmegupta and Bháscara. John Murray,
London, 1817. Contains Lîlâvatî and
Bîjaganita of Bhâskara II (1150) and Chapters XII (Arithmetic) and XIII
(Algebra) of the Bráhma‑sphuta‑siddhânta of Brahmagupta (628). There have been several reprints, including
Sändig, Wiesbaden, 1973. (Edward
Strachey produced a version: Bija Ganita: or the Algebra of the Hindus; W.
Glendinning, London, 1813; by translating a Persian translation of 1634/5.)
Collins. Book of Puzzles. 1927.
A.
Frederick Collins. The Book of
Puzzles. D. Appleton and Co., NY,
1927.
Collins. Fun with Figures. 1928.
A.
Frederick Collins. Fun with
Figures. D. Appleton and Co., NY,
1928.
Columbia Algorism. c1350.
Anonymous
Italian MS, c1350 [according to Van Egmond's Catalog 253‑254], Columbia
X511 .A1 3. Transcribed and edited by
K. Vogel; Ein italienisches Rechenbuch aus dem 14.Jahrhundert;
Veröffentlichungen des Forschungsinstituts des Deutschen Museums für die
Geschichte der Naturwissenschaften und der Technik, Reihe C, Quellentexte und
Übersetzungen, Nr. 33, Munich, 1977. My
page references will be to this edition.
Van Egmond says it has a title in a later hand: Rascioni de Algorismo.
The Algorism is discussed at length in
Elizabeth B. Cowley; An Italian mathematical manuscript; Vassar Medieval
Studies, New Haven, 1923, pp. 379‑405.
Conway, John Horton. (1937- ).
See: Winning Ways.
Cowley, Elizabeth B. See:
Columbia Algorism.
CP. 1907. H.
E. Dudeney. Canterbury Puzzles. (1907);
2nd ed. "with some fuller solutions and additional notes",
Nelson, 1919; 4th ed. = Dover, 1958. (I have found no difference between the 2nd and 4th editions,
except Dover has added an extra note on British coins and stamps. I now have a 1st ed, which has different
page numbers, but I have not yet added them.)
CR Comptes Rendus des Séances de
l'Académie des Sciences, Paris.
Crambrook. 1843. W.
H. M. Crambrook. Crambrook's Catalogue
of Mathematical & Mechanical Puzzles Deceptions and Magical Curiosities,
contained in the Necromantic Tent, Royal Adelaide Gallery, West Strand,
London. ... To which is added, a Complete Exposé [of] the Baneful Arts by
which unwary Youth too often become the prey of professed gamesters. And ... an extract from The Anatomy of
Gambling. Second Edition, Corrected
& Enlarged. T. C. Savill, 107 St.
Martin's Lane, 1843. 23pp. Photocopy provided by Slocum. [According to: Edwin A. Dawes; The Great
Illusionists; Chartwell Books, Secaucus, New Jersey, 1979, p. 138, this is the
first known magical catalogue. It has a
list of about 100 puzzles on pp. 3-5, with the rest devoted to magic
tricks. Unfortunately there are no
pictures. Comparison with Hoffmann
helped identify some of the puzzles, but I can not identify many of them. I have marked almost all these entries with
?? or check??, but the only way one can check is if actual examples or an
illustrated catalogue turn up. Some of
the names are so distinctive that it seems certain that the item does fit where
I have cited it; others are rather speculative. There are several names which may turn up with more
investigation. Toole Stott 190 says
there should be 48pp, though the later pages may be the added material on
gambling.]
Cremer, William Henry, Jr. See under: Book of 500 Puzzles, Hanky
Panky, Magician's Own Book.
CUP. Cambridge University Press.
Cyclopedia. 1914.
Sam
Loyd's Cyclopedia of 5,000 Puzzles, Tricks and Conundrums (ed. by Sam Loyd
Jr). Lamb Publishing, 1914 = Pinnacle or Corwin, 1976. This is a reprint of Loyd's "Our Puzzle
Magazine", a quarterly which started in June 1907 and ran till 1908. See OPM for details.
C&B. 1920. Sidney
W. Clarke & Adolphe Blind. The Bibliography of Conjuring
And Kindred Deceptions. George
Johnson, London, 1920. Facsimile by
Martino Fine Books, Mansfield Centre, Connecticut, nd [obtained new in 1998].
C&W. Chatto & Windus, London.
Datta & Singh. Bibhutibhusan Datta & Avadhesh
Narayan Singh. History of Hindu
Mathematics. Combined edition of Parts I
(1935) and II (1938), Asia Publishing House, Bombay, 1962. NOTE: This book makes some contentious
assertions. Readers are referred to the
following reviews.
O.
Neugebauer. Quellen und Studien zur
Geschichte der Mathematik 3B (1936) 263-271.
S.
Gandz. Isis 25 (1936) 478-488.
Datta, B. See:
Bakhshali MS; Datta & Singh.
De Morgan (1806-1871). See:
Rara.
De Viribus. See:
Pacioli.
dell'Abbaco. See:
Pseudo-dell'Abbaco.
Depew. Cokesbury Game Book.
Arthur
M. Depew. The Cokesbury Game Book. Abingdon-Cokesbury Press, NY &
Nashville, 1939. [The back of the TP
says it is copyright by Whitmore & Smith -- ?? The Acknowledgements say material has been assembled from various
sources and colleagues who have been collecting and writing over the previous
thirty years.]
Dickson. Leonard Eugene Dickson (1874-1954). History of the Theory of Numbers, 3
vols. Carnegie Institution of
Washington, Publication 256, 1919-1923;
facsimile reprint by Chelsea, 1952.
Dilworth. Schoolmaster's Assistant.
Thomas
Dilworth. The Schoolmaster's
Assistant: Being a Compendium of
Arithmetic, both Practical and Theoretical.
(1743); 11th ed., Henry Kent,
London, 1762 (partly reproduced by Scott, Foresman, 1938.) 20th ed., Richard & Henry Causton,
London, 1780. De Morgan suggests the
1st ed. was 1744 or 1745, but the testimonials are dated as early as Jan 1743,
so I will assume 1743. Comparison of a
1762 ed. (Wallis 321 DIL) with my 1780 ed. shows the 1780 ed. is identical to
the 1762 ed., except the section on exchange is much expanded, so the page
numbers of all material of interest are increased by 12pp. I will cite the pages of the 1762 ed., but
give the date as 1743. [Wallis also has: 14th ed., 1767; 15th ed., 1768;
1783; 22nd ed., 1785; 1791;
24th ed., 1792; 1793; 33rd ed., 179-; 1799; 1800; 1804.]
[Halwas 149‑162 are some US editions.]
Diophantos. c250.
Diophantos. Arithmetica. c250. In: T. L. Heath;
Diophantos of Alexandria; 2nd ed., (OUP, 1910); Dover, 1964. Note: Bachet edited a Greek and Latin
version of Diophantos in 1620, which inserted 45 problems from the Greek
Anthology at the end of Book V. (It was
in Fermat's copy of this work that Fermat wrote the famous marginal note now
called his Last Theorem; Fermat's son published an edition with his father's
annotations in 1670, but the original copy was lost in a fire.)
DNB. Leslie Stephen, ed. The Dictionary of National Biography. Smith, Elder and Co., London, 1885‑1901
in 22 volumes. OUP took it over in
1917. Decennial Supplements were
added.
Compact
Edition, with Supplement amalgamating the six decennial supplements to 1960,
OUP, 1975. The Compact Ed. shows the
original volumes and pages so I will cite them in ( ), followed by the pages in
the Compact Ed.
Dodson. Math. Repository. (1747?); 1775.
James
Dodson. The Mathematical
Repository. Containing Analytical
Solutions of near Five Hundred Questions, mostly selected from Scarce and
Valuable Authors. Designed As Examples
to Mac-Laurin's and other Elementary Books of Algebra; And To conduct Beginners
to the more difficult Properties of Numbers.
2nd ed., J. Nourse, London, 1775, HB.
(I have now acquired vols. II & III (1753 & 1755), but these are
largely concerned with annuities, etc., except the beginning of vol. II has a
section on indeterminate equations, entered in 7.P.1. From references in these volumes, it seems that the 1775 ed. of
volume I is pretty close to the first ed. of c1747, but has been a little
rearranged, so I have redated the entries as above.)
Doubleday -- n. 1969, etc.
Eric
Doubleday. Test Your Wits, Vols. 1 -
5. Ace Publishing, NY, 1969; 1971;
1972; 1969[sic]; (1969), revised 1973.
[Vols. 1 - 3 are good collections, with a number of novel variations of
standard problems. Vols. 4 & 5 are
vol. 1 split into two parts and much padded by putting each answer on a
separate page! The books refer to
Doubleday as puzzle setter for a London newspaper and one of the best known
setters in the English speaking world.
However, none of the older puzzle setters/editors in England have ever
heard of him and there is no book by him in the British Library Catalogue. Surprisingly, there is also no book by him
in the Library of Congress Catalogue! I
am beginning to think the author is a deception, but the first three books are
better than scissors and paste hack work.]
DSB. Dictionary of Scientific Biography. Ed. by Charles C. Gillespie for the American
Council of Learned Societies.
Scribner's, NY, 1970-1977, in 18 volumes. I will give the volume and the pages.
The
mathematical material has been reprinted in four volumes as: Biographical Dictionary of
Mathematicians Reference Biographies
from the Dictionary of Scientific Biography. Scribner's, NY, 1990?
This has new pagination, continuous through the four volumes. If I don't have the DSB details, I will cite
this as BDM.
Dudeney, Henry Ernest
(1857-1930). See: AM,
CP, MP, PCP,
536. I also cite his columns or
contributions in The Captain, Cassell's Magazine, Daily Mail, London Magazine,
The Nineteenth Century, The Royal Magazine, Strand Magazine, Tit-Bits, The
Tribune, The Weekly Dispatch.
Eadon. Repository. 1794.
John
Eadon. The Arithmetical and
Mathematical Repository, Being a New Improved System of Practical Arithmetic,
in all its Branches; Designed for the Use of Schools, Academies,
Counting-Houses, and Also for the Benefit of private Persons who have not the
Assistance of a Teacher. In Four
Volumes. Volume 1. In Three Books. Printed for the author, and sold by G. G. and J. Robinson,
Paternoster Row, London, 1794.
EB Encyclopædia Britannica. I tend to use my 1971 ed.
Endless Amusement I. c1818.
Anonymous. Endless Amusement; A Collection of Nearly
400 Entertaining Experiments In various
Branches of Science; ..., All the Popular Tricks and Changes of the Cards,
.... 3rd ed.(?), Thomas Boys, London,
nd [1825]. Frontispiece & TP are
missing, but James Dalgety has inserted a photocopy of the TP of the 3rd
ed. [BMC lists 2nd ed., 1819?; 3rd ed., 1825? BMC65
lists 1st ed. by Thorp & Burch, c1818;
2nd ed., c1820. Toole Scott
255-267 lists 3rd ed., c1820; 4th ed., 1822; 5th ed., 1830; 6th ed.,
1834; 7th ed., 1839. Hall, BCB 116-123 are: 1st ed., c1815; 2nd ed., c1820; 3rd ed.,
c1820; 3rd ed., Philadelphia,
1822; 4th ed., c1825; 5th ed., c1830; 6th ed., 1834;
Philadelphia, 1847. Heyl
110-115, 121 are 1819; 2nd & 3rd ed., M. Carey & Sons,
Philadelphia, 1821 & 1822; 3rd ed,
C. Tilt, London, 1825; Borradaile, NY,
1831; Henry Washburne & Thomas
Tegg, London, 1839; Lea &
Blanchard, Philadelphia, 1847. Almost
all of these are listed as 2 + 216 pp, so the editions are probably all the
same as the 1st ed., except that Hall notes that the 1st ed. title is slightly
different: Endless Amusement; A Collection of Upwards of 400 Entertaining and
Astonishing Experiments. Among a
Variety of other Subjects, are Amusements in Arithmetic, Mechanics, Hydraulics
.... All the Popular Tricks and Changes
of the Cards, ..., and Heyl gives a
similar title for the 1825 ed. and the 1831 NY ed. has some variations. Christopher 330-338 are 2nd ed., Philadelphia, 1821; 3rd ed., c1820; 3rd ed., Philadelphia, 1822;
4th ed., c1822; 4th ed., c1822
(slightly different to preceding; 5th
ed., c1830; 6th ed., 1834; 7th ed., 1839; Philadelphia, 1847.
C&B list it under Thorp and Burch, the publishers, with no dates.] [There is a Recreations in Science, ..., by
the author of Endless Amusement, 1830.]
Endless Amusement II. 1826?
Anonymous. A Sequel to the Endless Amusement,
Containing Nearly Four Hundred Interesting Experiments, In various Branches of
Science, ..., to Which are Added, Recreations with cards, and a Collection of
Ingenious Problems. Thomas Boys,
London, nd [1826?]. Pp. 203-216 are
missing, but James Dalgety has inserted photocopies. [BMC lists one ed., 1826?
Hall, BCB 252 gives c1825. Heyl
says this refers to Thomas Boys ... and Thorp and Burch, London (1825). Toole Stott 623 gives 1825.]
=
Anonymous. The Endless Amusement. New Series Containing Nearly Four Hundred
Interesting Experiments, ... (as above).
Thomas Tegg & Sons, London, 1837.
Angela Newing has provided a photocopy of the interesting parts of this
and it is virtually identical to the 1826? ed., though it has been reset, resulting
in an extra word fitting on some lines, and it has rather poorer pictures. One problem has been replaced by
another. [Heyl 122.] 21 problems, including the replacement
problem, are copied in The New Sphinx.
=
Anonymous. A Companion to the Endless
Amusement. James Gilbert, London,
1831. [Toole Stott 172 says this is a
reprint of A Sequel ..., from the same type, with new TP, and this is clear
from examination of the example Wallis 187.5 COM. Heyl 66 dates it as c1820?]
Some
of the material is taken from Badcock.
van Etten/Leurechon. 1624.
Recreation
Mathematicque.
The
bibliography of this book is very complicated.
I have now made a separate bibliography of this, augmented by many
contributions from Voignier, which is now (Aug 2001) 19pp, listing 50 French
editions, 5 English editions, 4 Latin editions and 8 (or 9) Dutch editions -- a
total of 67 (or 68) editions from 1624 to 1706, though at least 10 of the
French editions may be 'ghosts'. This
is part of my The Bibliography of Some
Recreational Mathematics Books.
This
book has traditionally been attributed (since 1643) to Père Jean Leurechon, SJ
(c1591-1670), who was probably van Etten's university teacher, but the book
specifically names van Etten and there seems to be very little real evidence
for Leurechon's authorship. Trevor
Hall's booklet and chapter are a substantial study of this question and he
concludes that there is no real reason to doubt van Etten's authorship, though
he may well have had help or inspiration from his teacher. Hall has also shown that van Etten and his
uncle, the book's dedicatee, were real people.
(Toole Stott 429‑431 dismisses Hall's work as a result of
completely misunderstanding it!)
However, the book was amended, revised and translated many times, so
that versions may occur under the following names: Hendrik van Etten; Jean
Leurechon; D.H.P.E.M. = Denis (or Didier) Henrion, professeur en
mathématique (or Professeur ès Mathematiques), a pseudonym of Clément
Cyriaque de Mangin, who also called himself Pierre Hérigone; Claude Mydorge; Caspar (or Gaspar) Ens;
Wynant van Westen; William
Oughtred; William Leake; not to mention
Anonymous and versions of the title -- I have found it under Recreations or
Récréations and under Vermakelijkheden. E.g. Lucas, RM1, 239-240, has 7 entries for this book under five
different authors and twice under Récréations.
Jacques
Voignier; Who was the author of "Recreation Mathematique" (1624)?; The
Perennial Mystics #9 (1991) 5-48 (& 1-2 which are the cover and its
reverse). [This journal is edited and
published by James Hagy, 2373 Arbeleda Lane, Northbrook, Illinois, 60062,
USA.] This is the second serious study
of this book. He points out evidence
for Leurechon's connection with the book, which makes it seem more likely, but
definite evidence is still lacking, so I am suggesting that it may have been
some kind of joint production and I will change my references it to van Etten/Leurechon. The work of Hall and Voignier form the basis
of the following discussion, supplemented by the standard catalogues and
personal inspection of about a half of the French and English editions --
generally after 1630.
Henrik
van Etten. Recreation
Mathematicque. Composee de Plusieurs
Problemes Plaisants et Facetieux. En faict
d'Arithmeticque Geometrie, Mechanicque, Opticque, & autres parties de ces
belles sciences. Jean Appier Hanzelet,
Pont‑a‑Mousson, 1624 [taken from facsimile of the 1626 ed.]. 155pp., ??NYS.
2nd
ed., 1626, ibid. = recent facsimile
with no details, but with 'Pont à Mousson
13 ‑ 10 ‑ 54' written inside the back cover. [An apparently identical copy at the Museum
of the History of Science, Oxford, has a small insert saying it was reissued by
La Compagnie de Pont-à-Mousson, printed by l'Imprimerie Berger-Levrault,
nd.]. 91 problems on xiv + 144 =
158pp. [The extra pages include
questions V and VI of problem 91 -- these questions occur in no other edition,
except probably in the 1629 reissue in Pont-à-Mousson.]
After
the two Hanzelet editions, there were three editions in Paris in 1626, by Rolet
Boutonné (2nd ed.), by Antoine Robinot (2nd ed.) and by Jean Moreau &
Guillaume Loyson (3rd ed.). Boutonné
and Robinot were closely associated and their output was interchangeable. Their 2nd eds. appear to be essentially the
1624 ed. The Moreau & Loyson has
Notes added to the problems and 8pp. of Additions. This was the first to put the illustrations as woodcuts in the text
rather than using copperplates for five separate sheets of 8 figures. The Notes are signed D.A.L.G., but are due
to Claude Mydorge. (NUC indicates the
Robinot had further comments signed D.H.P.E.M., later identified as Denis (or
Didier) Henrion Professeur En Mathématique (though Henrion is a pseudonym of
Clément Cyriaque de Mangin!) -- but this seems to be a confusion.) In the next few years, editions appeared in
Paris, Rouen and Lyon. In 1627,
Boutonné issued a '4th ed.' with "Nottes sur les recreations mathematiques
... Par D. H. P. E. M." and the D.A.L.G. notes were omitted. In 1627, Claude Rigaud & Claude Obert,
Lyon, issued a version with 9pp of Additions as in the 1626 Moreau &
Loyson.
In
1628, Charles Osmont, Rouen, issued a version in three parts: Récréations
mathématiques ... 1re et 2de partie. La 3e partie contient un recueil
de plusieurs gentilles et récréatives inventions de feux d'artifice .... Part 1 was van Etten's 91 problems, with
questions V & VI of prob. 91 omitted, omitting the D.A.L.G. notes and the
Additions. Part 2 had 45 new problems,
often attributed to Mydorge and/or Henrion, but they had no connection with
this and the authorship of these problems is unknown, though Voignier suggests
the printer, Osmont. Part 3 is an
independent treatise on fireworks which Hall attributes to Hanzelet. This edition was reissued in Rouen by
various publishers in 1628, 1629, 1630, 1634, 1638 and in Lyon in 1642-1643,
1653, 1656, 1658, 1669, 1680.
In
1630, Boutonné and Robinot (their printing is indistinguishable and volumes
often have parts from both of them, indeed the Privilege is issued to them
jointly) issued an extended version in four parts, titled Examen du Livre des
Recreations Mathematiques, stated to be by Mydorge. Part 1 is van Etten's 91 problems, with parts V & VI of prob.
91 omitted, with many problems being followed by an Examen signed D.A.L.G. These are by Mydorge and are a revision of
his material in the 1626 Moreau & Loyson.
Part 2 has its own TP and had 45 new problems, taken from the 1628 Rouen
ed. Part 3 has its own TP, but doesn't
state the publisher, and is the independent treatise on fireworks, also taken
from the 1628 Rouen ed. Part 4 again
has its own TP and is Nottes [sic] sur les Recreation Mathematiques by
D.H.P.E.M. and are additions to 27 of van Etten's problems, taken or extended
from the Notes in the 1627 4th ed. The
book is also described as 3 parts with the Nottes, but Parts 2 and 3 are
consecutively paged, leading to some descriptions of the book as being in 3
parts. The parts were probably issued
separately as they sometimes are catalogued separately and different copies of
the whole work often have a mixture of the Boutonné and Robinot printings. This most extended form was reissued by
various publishers in Paris: 1634(??), 1638, 1639, and in Rouen: 1639 (two publishers), 1643 (two publishers ??),
1648?, 1649?
In
1659, Cardin Besonge, Paris, issued Les
Récréations mathématiques, avec l'examen de ses problèmes en arithmétique,
géométrie, .... Premièrement reveu par
D. Henrion, depuis par M. Mydorge, et tout nouvellement corrigé et
augmenté, 5e et dernière édition.
The Nottes are incorporated in the text (or perhaps omitted??). The entire text is consecutively
page-numbered. Reissued in Paris: 1660,
1661 and in Rouen as the 6th ed.: 1660?, 1663?, 1664, 1669 (seven publishers!).
The
1630 Paris ed. and the 1626 ed. have the same problem numbers for the first 91
problems, as do almost all French editions.
I will cite the problem number and the pages of the 1626 ed. I will add reference to the 1630 Paris ed.,
when the latter has additional information.
Only one of the additional problems in part 2 (prob. 7) is of any
interest to us, but several of Henrion's Nottes give corrections, extensions,
additional references and even additional problems. I didn't find any of the D.A.L.G. notes of any interest.
English editions.
Mathematicall
Recreations. Or a Collection of sundrie
[1653 has: many] Problemes, extracted out of the Ancient and Moderne
Philosophers, as secrets in nature, and experiments in Arithmeticke, Geometrie,
Cosmographie, Horologographie, Astronomie, Navigation, Musicke, Opticks,
Architecture, Staticke, Machanicks, Chimestrie, Waterworkes, Fireworks,
&c. Not vulgarly made manifest
untill this time: Fit for Schollers, Students, and Gentlemen, that desire to know
the Philosophicall cause of many admirable Conclusions. Usefull for others, to acuate and stirre
them up to the search of further knowledge; and serviceable to all for many
excellent things, both for pleasure and Recreation. Most of which were written first in Greeke and Latine, lately
compiled in French, by Henry Van Etten Gent.
And now delivered in the English tongue, with the Examinations,
Corrections, and Augmentations. Printed
by T. Cotes for Richard Hawkins, London, 1633.
328pp. (This ed. is 'excessively
rare'.)
Reissued
by William Leake, London: 2nd ed., 1653; (1667(??)); 1674. 344pp.
[Hall, OCB, says the 2nd ed. is similar to the 1633 edition, but with an
extra 16pp description (of 1636) of some dials by Oughtred (which led to the
book or the translation often being attributed to Oughtred). Hall also states that the English editions
are based on the Rouen ed. of 1628.
Sadly some interesting problems were omitted in the English, leading to
confusion in plate numbers. However, I
have just noticed that Prob. 63 is about two pages longer than the
corresponding Prob. 70 of the French editions.] The 1633, 1653 and 1674 editions are identical except for the
additional English material in the later editions. I will add citations to the English editions in parentheses. I now have an imperfect copy of the 1674 ed,
covers missing and lacking the Frontispiece and pp. 273-282 and later
material. Heyl 311 is a 1753 ed., which
must be an error for 1653.
W.
Leybourn, qv, takes several sections directly from the English editions.
Latin editions.
In
1628(??), Caspar (or Gaspar) Ens made a Latin translation but added some other
material, e.g. 49 problems from Alcuin.
I have only studied the 1636 ed. carefully.
Thaumaturgus
Mathematicus, Id est, Admirabilium Effectorum e Mathematicarum Disciplinarum
Fontibus Profluentium Sylloge. Casparo
Ens L. Collectore & Interprete.
1628. [Taken from 1636 TP. MUS #30 says this is only a translation of
van Etten. There is some doubt whether
the 1628 edition exists!]
Reissued
in 1636 and 1651. It has 89 of van
Etten's problems (omitting 38 & 46) and adding 25 new problems, with some
numbering errors so the last is numbered 113.
This is followed by 55 problems of Alcuin, using the Bede version of 56
problems, but omitting 18.
Thaumaturgus
mathematicus Gasparo Ens lectore collectore, & interprete, Nunc denuò
Correctus, & Auctus. Apollonius
Zambonus, Venice, 1706. 113 problems +
49 from Alcuin (check??). There are
some differences between this and the 1636 ed.
[MUS
#30 gives Köln, 1651, and further editions.]
Dutch editions.
In
1641, Wynant van Westen translated van Etten into Dutch. The title is: Het eerste [- derde] deel van
de Mathematische vermaecklyckheden. Te
samen ghevoeght van verscheyden ghenuchlijcke ende boertige werckstucken, soo
uyt arithmetica, geometria, astronomie, geographia, cosmographia, musica,
physica, optica, catoptrica, architectonica, sciotetica, als uyt andere
ongehoorde mysterien meer.
Ghetranslateert uyt het fransch in nederduytsche tale: ende verrijckt,
vermeerdert, ende verbetert met verscheyden observatien ende annotatien,
dienende tot onderrichtinge van eenige duystere questien, ende mis-slaghen in
den franschen druck. Door Wynant van
Westen .... op nieus oversien verbetert.
Jacob van Biesen, Arnhem, 1641.
3 parts with separate title pages and pagination, perhaps in 3 vols, but
later in 1 vol.
This
was reissued: Van Biesen, Arnhem, 1641,
??, 1644, 1662, 1671-72; Lootsman and Jacobsz, Amsterdam, 1673. I haven't examined any of these.
Euler. Algebra. 1770.
Leonard
Euler (1707-1783). Vollständig
Anleitung zur Algebra. Royal Academy of
Sciences at Petersburg, 1770. [A
Russian translation appeared in 1768.]
Translated into French by John III Bernoulli, with additions by
Bernoulli and La Grange (pp. 463-593 here), 1774. Translated from French into English as Element of Algebra, with
further notes, by Rev. John Hewlett, with a Memoir of Euler by Francis Horner
[Horner actually did the translation; Hewlett edited it.], (1797), 5th ed., Longman, Orme, and Co., London,
1840. Reprinted, with Introduction by
C. Truesdell (1972), omitting 4 pp of Horner, Springer, NY, 1984 [hidden on
back of title page]. I will cite part,
section, chapter, article and the pages from the Springer ed. (Part II has no sections.) Unfortunately these numbers seem to have
little connection with other editions.
[Though most of the recreational material in Euler is much older than
Euler, I have included it as a representative 18C text.] [Halwas 175-176 are some US editions -- the
1818 edition was the first example of a translated algebra in the US.]
Family Friend. The Family Friend. This was a magazine founded by Robert Kemp
Philp in 1849. The dating is awkward --
vol. 1 is dated 1850 on the cover, but the Preface is dated 15 Nov 1849 and
refers to the success of the past year, when it appeared monthly. It also says the magazine will henceforth
appear twice a month with two volumes per year, due on the first of June and
December. The Gardening section of vol.
1 goes from Jan to Dec. The Preface of
Vol. 2 is dated 10 Jun 1850 and its gardening section covers Jan - Jun. The Preface of Vol. 3 is dated 15 Dec 1850
and its Gardening section goes Jul - Dec.
BMC shows Philp left in 1852 and the magazine continued with two volumes
per year through a fifth series, ending in 1867, then restarted with one volume
a year from 1870 until 1921. I have
vols. 1 - 3 & the second half of 1858, which is dated 1858-9, but appears
to be Jul-Dec. None of the text is
signed. At the back of volumes are
included answers to correspondents. The
puzzles are often identical to those in The Magician's Own Book or The
Illustrated Boy's Own Treasury, etc., but are considerably earlier.
FHM. Graham Flegg, Cynthia Hay & Barbara Moss -- see under Chuquet.
Fibonacci. Leonardo Pisano, called
Fibonacci (c1170->1240). Liber
Abbaci. (1202); 2nd ed., 1228. In: Scritti di Leonardo Pisano; vol. I, ed.
and pub. by B. Boncompagni; Tipografia delle Scienze Matematiche e
Fisiche, Rome, 1857. The title pages
give 'abbaci', but Boncompagni's text begins 'Incipit liber Abaci ... Anno
MCCII.', while the c1275 MS starts 'Incipit abbacus'. Both forms are used, sometimes even in the same article -- e.g.
Loria's biographical article, see in Section 1.
Richard
E. Grimm was working on a critical edition of this and he kindly gave me some
details. There are 15 known MSS, all of
the 1228 2nd ed. Six of these consist
of 1½ to 3 chapters only; five of the others lack Chapter 10 and the second
half of Chapter 9; one lacks Chapter 10 and one lacks much of Chapter 15,
leaving two essentially complete texts.
The last four MSS mentioned are the most important: Siena L.IV.20, c1275, lacking much of Chap.
15, "the oldest and best";
Siena L.IV.21, 1463 [Grimm said c1465 -- there are dates up through 1464
in interest calculations, but the Incipit specifically says 1463], which
includes much other material from later writers, so it is at least double the
size of L.IV.20; Vatican Palatino
#1343, end of 13C, lacking Chap. 10;
Florence Bibl. Naz. Conventi Soppressi C. 1. 2616, early 14C, "handsome
but frequently badly faded" so "that a later hand found it necessary
to rewrite what he saw there."
When I examined it in Sep 1994, the black ink was indeed sometimes badly
faded but the numbers were in a clear red -- perhaps these are what was
rewritten?? L.IV.20 has the beginning
sentence ending "et correctus ab eodem a MCCXXVIII", but Grimm says
all the others are also of the 1228 ed even if they do not carry this addition
or the extra initial dedication. Sadly,
I heard in Aug 1998 that Grimm had Alzheimer's disease and was in a nursing
home. Inquiry has revealed no trace of
the photocopies of all the Liber Abbaci MSS which he said he had obtained and
in summer 2000 I heard he had died.
Boncompagni
used only one MS, then denoted Codex Magliabechiana, C. I, 2616, Badia
Fiorentina, no. 73, now Conventi Soppressi, C. I. 2616, the badly faded fourth
MS described above.
In
Sep 1994 and Mar 1998, I examined Siena L.IV.20 and 21 and Conv. Soppr.
C.1.2616. I have slides of the Incipit
& Fibonacci numbers from all of these and some other material.
The
dates of 1202 and 1228 are based on the Pisan calendar.
Fibonacci-Sigler.
Liber Abaci. Translated by
Laurence E. Sigler as: Fibonacci's
Liber Abaci A Translation into Modern
English of Leonardo Pisano's Book of Calculation. Springer, New York, 2002.
I have added page references to this, denoted S, after the Boncompagni
pages, e.g. pp. 397-398 (S:
543-544). I have given Sigler's English
wherever I previously had just quoted the Latin.
Fibonacci. Flos
and Epistola.
Leonardo
Pisano, called Fibonacci. MS of c1225
which begins "Incipit flos Leonardi bigolli pisani ...", Biblioteca
Ambrosiana, Milan, E. 75. In: Scritti
di Leonardo Pisano, vol. II, ed. and pub. by B. Boncompagni, Rome, 1862, pp.
227-252.
Part
of the MS has a separate heading: "Epistola suprascripsit Leonardi ad
Magistrum Theodorum phylosophum domini Imperatoris" and is sometimes
considered a separate work. It occupies
pp. 247-252 of the printed version. For
an English description, see: A. F.
Horadam; Fibonacci's mathematical letter to Master Theodorus; Fibonacci
Quarterly 29 (1991) 103-107.
Italian
translation (including the Epistola) and commentary: E. Picutti; Il 'Flos' di
Leonardo Pisano; Physis 25 (1983) 293-387.
Fireside Amusements. 1850.
Fireside
Amusements. Chambers's Library for
Young People. William and Robert
Chambers, Edinburgh, 1850, 188pp. The
BMC has this under Fireside Amusements and refers to Chambers for the Library,
which was 19 vols, 1848-1851. Pp. 187+
are missing in the copy I have seen, but it seems that just one page of
solutions is missing -- the NUC gives 188pp.
The NUC lists a 1870 reprint.
[The
BMC lists an 1880 ed with 159pp, part of Chambers's Juvenile Library, NYS.]
Fireside
Amusements A Book of Indoor Games. W. & R. Chambers, London and Edinburgh,
nd, 128pp. The BMC lists this as
1890[1889]. Though laid out entirely
differently, almost all the material is taken from the 1850 ed. I will cite both editions.
Much
of the material of interest is taken from Child: Girl's Own Book.
Folkerts. Aufgabensammlungen. 13-15C.
Menso
Folkerts. Mathematische
Aufgabensammlungen aus dem ausgehenden Mittelalter. Sudhoffs Archiv 55 (1971) 58-75.
He examines 33 anonymous Latin manuscript problem collections from 13-15
C in Oxford, London, Berlin, Munich, Vienna and Erfurt and catalogues the
problems therein. Of these, only Munich
14684 is published (cf below). He notes
that many more such sources exist. His
catalogue covers 14 of my topics. I
will not try to cite the individual MSS, since many of the topics occur in over
a dozen of them. I will simply say he
has n
sources, though some of the sources have several examples.
Folkerts, Menso. See:
Alcuin.
della Francesca. Trattato.
c1480.
Piero
della Francesca (1412-1492). Trattato d'Abaco. Italian MS in Codex Ashburnhamiano 359*
[291*] - 280 in the Biblioteca Mediceo-Laurenziana, Florence. c1480 [according to Van Egmond's Catalog 84,
based on watermarks in the paper which date from 1470 to 1500, but Davis,
below, p. 16, says c1450]. Transcribed
and annotated by Gino Arrighi, Testimonianze di Storia della Scienze 6, Domus
Galilæana, Pisa, 1970. Arrighi
uses c. (for carta) instead of
f. (for folio), but I will
use f.
for consistency with other usage, followed by the pages in Arrighi in (
). Arrighi reproduces many of the
diagrams, but he doesn't say anything about whether he has included all of
them. This MS appears to be that which
was in the possession of Piero's descendents until 1835 when it was reported as
having disappeared. Guglielmo Libri,
the noted historian of mathematics, who was also a shady bookdealer,
transcribed part of this MS in vol. 3 of his Histoire de la Mathématique en
Italie in 1840 as an anonymous work, then sold it to Lord Ashburnham in 1847
(recorded in his collection in 1881) whose collection was bought for the
Laurentian Library in 1884. There are
three different catalogue numbers - I use the format used in Van Egmond's
Catalog. The MS had passed out of
common knowledge until it was rediscovered in the Laurentian Library in 1917 by
Girolamo Mancini who recognised the handwriting as Piero's.
This
work and Piero's Libellus de Quinque Corporibus Regularibus are the subject of
a long standing plagiarism argument.
Giorgio Vasari [Le Vite de' più eccellenti pittori, scultori e
architetti; 1550; The Essential Vasari,
ed. by Betty Burroughs from the 1850 translation of Mrs. Jonathan Foster, Unwin
Books, London, 1962] states: "...
Piero della Francesca, who was a master of perspective and mathematics but who
first went blind and then died before his books were known to the public. Fra Luca di Borgo, who should have cherished
the memory of his master and teacher, Piero, did his best, on the contrary, to
obliterate his name, taking to himself all the honour by publishing as his own
work that of that good old man.
... Maestro Luca di Borgo caused
the works of his master, Piero della Francesca, to be printed as his own after
Piero died." The mathematical
works of Piero were unknown until they were rediscovered in 1850/1880 and
1917. Examination shows that Pacioli
certainly used 105 problems, many unusual, from Piero in the Summa. But he does praise Piero in the Epistola (f.
2r) of the Summa, as "the monarch of painting of our times". It has been suggested that Pacioli had a
large hand in the writing of Piero's works and hence was just reusing his own
material and he frequently expands on it.
However, there is no evidence that Pacioli was ever a student of Piero. Entire books have been written on the question,
so I will not try to say any more.
See: Margaret Daly Davis; Piero
della Francesca's Mathematical Treatises
The "Trattato d'abaco" and "Libellus de quinque
corporibus regularibus"; Longo Editore, Ravenna, 1977, for detailed comparisons and the work of R.
E. Taylor in Section 1: Pacioli. Davis
identifies 139 problems in the Libellus, of which 85
(= 61%) are taken from the
Trattato. Davis notes that Pacioli's
Summa, Part II, ff. 68v ‑ 73v, prob. 1-56, are essentially
identical to della Francesca's Trattato, ff. 105r ‑ 120r. See also section 6.AT.3 where the Libellus
and the Pacioli & da Vinci: De Divina Proportione are discussed.
The
work is discussed and 42 problems are given in English in: S. A. Jayawardene; The 'Trattato d'Abaco' of Piero della Francesca; IN: Cecil H. Clough, ed.; Cultural Aspects of the Italian
Renaissance Essays in Honour of Paul
Oskar Kristeller; Manchester Univ.
Press, Manchester, nd [1976?]; pp.
229-243. I will note 'English in
Jayawardene.' when relevant.
Frikell, Wiljalba (1818 (or
1816) - 1903). (The given name Gustave
sometimes occurs -- I thought Gustave might be a son of Wiljalba, but the son
was named Adalbert ( -1889) and his
name was pirated by a clumsy imposter in England.)
See
the discussion at: Book of 500
Puzzles, Boy's Own Conjuring Book, Hanky Panky, Magician's Own Book, The
Secret Out. Frikell was a noted
conjuror of the time and his name has been associated with the UK versions of
these books, but there is no evidence he had anything to do with them. The Art of Amusing, by Frank Bellew, Hotten,
London, 1866?, op. cit. in 5.E, has a note on the back of the TP saying The
Secret Out is a companion volume, just issued, by Hermann Frikell. C&B, under Williams, Henry Llewellyn ("W. Frikell") lists:
Hanky Panky; Magician's Own
Book, London & New York; (Magic No
Mystery); The Secret Out and says to also see Cremer.
Gamow & Stern. 1958.
George
Gamow & Marvin Stern. Puzzle‑Math. Macmillan, London, 1958.
Gardner. Martin Gardner (1914- ).
Many references are to both his SA column, cited by (month & year),
e.g. SA (Mar 1982), and to the appearance of the column as a chapter in one of
his books, abbreviated as shown below.
In general, I will only give the chapter reference as the various
editions and translations are differently paginated. Answers, comments and extensions appeared in succeeding issues of
SA, usually in Gardner's column, but sometimes in the Letters. All this material is collected in the book
chapter, sometimes by rewriting of the article, sometimes as notes or an
Addendum at the end of the chapter.
Since many years usually passed before the book version, the Addenda
often contain material that never appeared in SA, as well as references to work
done as a result of the SA article. I
have not tried to enter all of Gardner's references here, so anyone interested
in a topic that Gardner has considered should consult the book version of
Gardner's column. Currently some of the
earlier books are being reissued in new editions, with further extensions and
updating. See also the next entry.
For
years from at least 1950, SA appeared in two volumes per year, each of six
issues. In year 1950 + n,
vol. 182 + 2n covers Jan-Jun
and vol. 183 + 2n covers
Jul-Dec.
1st Book The Scientific American Book
of Mathematical Puzzles and Diversions.
Simon & Schuster, 1959.
UK version: Mathematical Puzzles and
Diversions from Scientific American.
Bell, London, 1961; Penguin (without the words 'from Scientific American'),
1965.
2nd Book The Second
Scientific American Book of Mathematical Puzzles and Diversions. Simon & Schuster, 1961.
UK version: More Mathematical Puzzles and
Diversions from Scientific American.
Bell, London, 1963. Penguin
(without the words 'from Scientific American'), 1966. (The UK versions omit Chapter 20: "The Mysterious Dr.
Matrix". The dust wrapper of the
HB has a sentence referring to this chapter which has been blacked out. ??)
New MD
Martin Gardner's New Mathematical Diversions from Scientific American. Simon & Schuster, 1966.
Unexpected The
Unexpected Hanging and Other Mathematical Diversions. Simon & Schuster, 1969.
UK version: Further Mathematical
Diversions. Allen & Unwin, London,
1970; Penguin, 1977.
6th Book Martin
Gardner's Sixth Book of Mathematical Games from Scientific American. Freeman, 1971.
Carnival Mathematical
Carnival. Knopf, NY, 1975; Penguin,
1978.
Magic Show Mathematical
Magic Show. Random House, NY, 1978.
Circus Mathematical
Circus. Knopf, NY, 1979.
Wheels Wheels,
Life and Other Mathematical Amusements.
Freeman, 1983.
Knotted Knotted
Doughnuts and Other Mathematical Entertainments. Freeman, 1986.
Time Travel Time
Travel and Other Mathematical Bewilderments.
Freeman, 1988.
Penrose Tiles Penrose
Tiles to Trapdoor Ciphers. Freeman,
1989.
Fractal Fractal
Music, Hypercards and More ....
Freeman, 1992.
Last The
Last Recreations Hydras, Eggs, and
Other Mathematical Mystifications.
Copernicus (Springer), NY, 1997.
??NYR.
Magic Numbers The
Magic Numbers of Dr. Matrix.
Prometheus, Buffalo, 1985.
Chaps. 1-18 previously appeared as: The Incredible Dr. Matrix;
Scribner's, NY, 1976. Chaps. 1‑7
& 9 previously appeared as: The Numerology of Dr. Matrix; Simon &
Schuster, NY, 1967. In contrast to his
other books above, the answers and comments occur at the end of this book
instead of following the original articles.
Workout A
Gardner's Workout Training the Mind and
Entertaining the Spirit. A. K.
Peters, Natick, Massachusetts, 2001.
This comprises 41 chapters of articles written after his retirement from
SA.
Gardner. MM&M.
1956.
Martin
Gardner. Mathematics, Magic and
Mystery. Dover, NY, 1956.
General Trattato. 1556.
Nicolo
Tartaglia (c1499-1557). (La Prima Parte
del) General Trattato di Numeri et Misure.
Curtio Troiano, Venice, 1556.
(Modern Italian spells his given name as Niccolò, but it appears as
Nicolo on the title page.) Six parts
actually appeared in 1556-1560. All
references are to Part 1. Unless
otherwise specified, reference is to Book 16 (of Part 1), but I also have
references to Books 12 and 17. CAUTION
-- the running head in Book 17 says Libro Decimosesto for several pages before
changing to Libro Decimosettimo. Since
it is hard to find the beginnings of books, this can cause confusion. See Rara 275-279; Van Egmond's Catalog
345-346.
In
1578, Guillaume Gosselin produced an annotated translation of parts 1 & 2
into French as: L'Arithmetique de Nicolas Tartaglia -- cf Van Egmond's Catalog
347.
Ghaligai. Practica D'Arithmetica. 1521.
Francesco
Ghaligai. Practica D'Arithmetica di
Francesco Ghaligai Fiorentino.
Nuovamente Rivista, & con somma Diligenza Ristampata. I Giunti, Florence, 1552. Smith, Rara, says that this is identical to
the first (Latin?) edition by Bernardo Zucchetta, Florence, 1521, except that
edition was titled Summa De Arithmetica, so I will date the entries as
1521. See Rara 132; Van Egmond's
Catalog 316-317.
Gherardi. Libro di ragioni and Liber habaci. 1328 & c1310.
Paolo
Gherardi. Two Italian MSS in Codici
Magliabechiani Classe XI, no. 87 & 88 in Bib. Naz. di Firenze. Van Egmond's Catalog 115-116. The first is dated 1327 (but see
below). The second is undated, but
clearly of a similar date which I originally denoted 1327? - see below. Transcribed by Gino Arrighi; Collana di
Storia della Scienza e della Tecnica, No. 2; Maria Pacini Fazzi, Lucca, 1987. See also:
Warren Van Egmond; The earliest vernacular treatment of algebra: the Libro
di ragioni of Paolo Gerardi (1328); Physis 20 (1978) 155-189. Van Egmond notes that the date of 30 Jan
1327 is in our year 1328 and uses this in his Catalog. He doubts whether Liber habaci is actually
by Gherardi and his Catalog assigns no author to it, so I will put Gherardi? as
author. He dates it to c1310. His paper is concerned with the quadratic
and cubic equations and hence of little interest to us.
Good, Arthur. See:
Tom Tit.
Gori. Libro di arimetricha.
1571.
Dionigi
Gori. Libro di arimetricha. 1571.
Italian MS in Biblioteca Comunale di Siena, L. IV. 23. ??NYS.
Extensively quoted and discussed in: R. Franci & L. Toti Rigatelli;
Introduzione all'Aritmetica Mercantile del Medioevo e del Rinascimento;
Istituto di Matematica dell'Università di Siena, nd [1980?]. (Later published by Quattroventi, Urbino,
1981.) (I will quote Gori's folios and
also give the pages of this Introduzione.)
Van Egmond's Catalog 191-192.
Graves. The Graves Collection of early mathematical
books at University College London (UCL).
Guy, Richard Kenneth (1916- ).
See: Winning Ways.
G4Gn Gathering for Gardner n, held in
Atlanta. 1: Jan 1993; 2: Jan 1996; 3: Jan 1998; 4: Feb
2000; 5: Apr 2002.
G&P. Games & Puzzles. The first version ran from 1972 through
1981. The second series started in Apr
1994 and finished with No. 16 in Jul 1995.
G&PJ. Games and Puzzles Journal. Successor to Chessics. Ran through 12 issues, Sep 1987 -- Dec 1989,
then restarted intermittently in May 1996.
Haldeman-Julius. 1937.
E.
Haldeman-Julius. Problems, Puzzles and
Brain-teasers. Haldeman-Julius
Publications, Girard, Kansas, 1937.
Facsimile (I believe) presented by Bob Koeppel at IPP13, 1993.
Hall. BCB. 1957.
Trevor
H. Hall. A Bibliography of Books on
Conjuring in English from 1580 to 1850.
(Carl Waring Jones, Minneapolis, 1957); Palmyra Press, Lepton,
W. Yorks., 1957. 323 entries. I will cite item numbers. A Supplement is in Hall, OCB. See Heyl for a list of items not in BCB.
Hall. OCB. 1972.
Trevor
H. Hall. Old Conjuring Books. Duckworth, London, 1972. This covers books in English up through 1850
and it includes a Supplement to his BCB and should be checked for further
information on items in BCB. This
contains 39 new items and additional notes to 36 previous items. New items are given interpolated item
numbers, e.g. 24.5. OCB also includes a
slightly revised version of his booklet on van Etten, see Section 1 below.
Halwas. Robin Halwas, Ltd. List XV. American
Mathematical Textbooks 1760-1850. Catalogue of 511 items being sold as a
collection. London, 1997, 144pp. Quite a number of English works and a few
French works had US editions which are detailed in this.
Hanky Panky. 1872.
Hanky
Panky A Book of Easy and Difficult
Conjuring Tricks Edited by W. H.
Cremer, Jun. (John Camden Hotten,
London, 1872 [BMC & Toole Stott 193, listed under Cremer, while C&B,
under Cremer, give London, 1872]; Hotten was succeeded by Chatto & Windus
c1873 and they produced several editions [NUC has 1872, Toole Stott 1017 &
Shortz have 1875].) My copy says: A new
edition with 250 practical illustrations.
John Grant, Edinburgh, nd [Toole Stott 1016 gives 1874; NUC gives
1875? Christopher 235, under Cremer, is
c1890. NUC says it is also attributed
to Henry Llewellyn Williams, but their entry under Williams says 'supposed
author'. C&B also list it under
Williams. (This has been attributed to
Frikell, but Toole Stott doubts that Frikell had anything to do with this. I may put this under Cremer.)
HB.XI.22. 1488.
Stuttgart
Landesbibliothek German MS HB.XI.22, 1488.
Brief description by E. Rath;
Über einen deutschen Algorismus aus dem Jahr 1488; Bibl. Math. (3) 14 (1913‑14)
244‑248.
Heath, Sir Thomas L. See:
Diophantos; HGM.
Heyl. 1963. Edgar
Heyl (1911-1993). A Contribution to
Conjuring Bibliography. English
Language 1580 to 1850. Edgar Heyl conjuring books, Baltimore,
1963. Facsimile edition of 100 copies
by Maurizio Martino Fine Books, PO Box 373, Mansfield Center, Connecticut,
06250, nd [1998?]. 360 entries +
appendix of 14 more, almost all not in Hall, BCB.
HGM. 1921. Sir Thomas L.
Heath. A History of Greek Mathematics,
2 vols. (OUP, 1921); corrected reprint,
Dover, 1981.
HM. Historia Mathematica.
Hoernle, A. F. Rudolf. See:
Bakhshali MS.
Hoffmann. 1893. Professor
Louis Hoffmann [pseudonym of Angelo John Lewis (1839‑1919)]. Puzzles Old and New. Warne, London, 1893. Reprinted with Foreword by
L. E. Hordern; Martin Breese, London, 1988.
In
1984, Hordern published a limited edition (15 copies) of "The Hordern
Collection of Hoffmann Puzzles 1850‑1920",
which gives colour photos of examples from his collection and the appropriate
text. I often cite these pictures as
they often differ from those in the following item, with the heading Hordern
Collection. Generally, the next item
gives more specific dating and/or older examples.
In
1993, Hordern produced a corrected edition of all of Hoffmann as: Hoffmann's
Puzzles Old & New; published by himself.
This has colour photos of all puzzles for which known examples
exist. I will cite this as
Hoffmann-Hordern. The 1893 edition
gives solutions for each chapter in a following chapter, but both of Hordern's
illustrated versions give each solution immediately after the problem, with
colour picture nearby. A small section
on Elementary Properties of Numbers is omitted from the 1993 edition.
See
also: Tom Tit.
Honeyman Collection.
The
Honeyman Collection of Scientific Books and Manuscripts. Sold by Sotheby's [Sotheby Parke Bernet],
1978-1981. Seven volumes -- details
given in Section 3.B.
Hordern, L. Edward
(1941-2000). See under Hoffmann and in
5.A.
HPL. The Harry Price Library, Senate House,
University of London OR its catalogues.
Harry
Price (1881-1948). Short-Title
Catalogue and Supplementary Catalogue of Works on Psychical Research,
Spiritualism, Magic, Psychology, Legerdemain and Other Methods of Deception,
Charlatanism, Witchcraft, and technical Works for the Scientific Investigation
of Alleged Abnormal Phenomena from Circa 1450 A.D. to 1935 A.D. Compiled by Harry Price. (The first part was originally ... to 1929 A. D.; Proc. National Lab. of
Psychical Research 1:2 (1929); National Laboratory of Psychical Research,
London, 1929. The second part was
originally: Short-Title Catalogue of the Research Library, for 1472 A.D. to the
Present Day; Bull. Univ. of London Council for Psychical Investigation 1
(1935); Univ. of London Council for Psychical Investigation, 1935.) New Introduction by R. W. Rieber and Andy
Whitehead With an appendix entitled
"The St. Louis Magnet" (which originally appeared in 1845) by T. J. McNair
and J. F. Slafter. Da Capo Press
(Plenum), NY, 1982. NOTE: The works
listed here are now in the Harry Price Library, though the editors have added
36 items in their Introduction.
Hummerston. Fun, Mirth & Mystery. 1924.
R.
A. Hummerston. The Book of Fun, Mirth
& Mystery A feast of delightful
entertainment, including games, tricks, puzzles and solutions, "how to
makes," and various other means of amusement. Pearson, London, 1924.
Hunt. 1631 & 1651.
Nich.
Hunt. Newe Recreations or The Mindes
release and solacing. Aug. Math. for
Luke Fawne, 1631.
Nich.
Hunt. New Recreations or A Rare and
Exquisite Invention. J. M. for Luke
Fawn, London, 1651. This edition
contains a few more pages and several problems of interest and is differently
paginated. I will give both page
numbers for problems in both editions.
Bill Kalush has sent both texts on a CD.
Hutton. A Course of Mathematics. 1798?
Charles
Hutton (1737-1823). A Course of
Mathematics. Composed for the Use of
the Royal Military Academy. (In 2 vols,
plus a third, 1798-1811.) A New
Edition, entirely Remodelled. By William
Ramsay, B. A., Trinity College, Cambridge.
T. T. & J. Tegg, London, and Richard Griffin & Co.,
Glasgow, 1833 (in one volume). 8 +
822 pp.
Hutton-Rutherford. A Course of Mathematics. 1841?
Charles
Hutton. A Course of Mathematics, Composed for the Use of The Royal Military
Academy. By Charles Hutton, LL.D.,
F.R.S., Late Professor of Mathematics in that Institution. A new and carefully corrected Edition,
Entirely Re-modelled, and Adapted to the Course of Instruction Now Pursued in
the Royal Military Academy. By William
Rutherford, F.R.A.S. Royal Military
Academy. William Tegg, London, 1857
[Preface dated Nov 1840, so probably identical or nearly identical to the 1841
ed]. 8 + 895 pp.
[Two
volume versions: 1798/1801; 3rd, 1800/1801; 4th, 1803/1804; 5th, [1810, NUC gives 1806- and 1807]; 6th, [1810-1811 -- NUC].
Three volume versions -- apparently the early forms were just the
earlier 2 volumes with an additional third vol: 6th, 1811; 1813; 7th, 1819-1820; 8th, 1824;
9th, 1827/1828. 10th ed by
Olinthus Gregory, in 3 vols., 1827-1831.
New ed by William Ramsay in one vol., 1833; 1838. 11th ed by Gregory in 2 vols,
1836-1837. 12th ed, revised by Thomas
Stephens Davies, 2 vols., 1841-1843.
Ed by William Rutherford in one vol, 1841; 1843; 1846; 1849; 1851; 1853;
1857; 1860.
This
is a pretty straightforward text, but it well illustrates the situation in
early 19C England, outside Oxford and Cambridge. Almost all the material of interest is in the first two sections:
Arithmetic and Algebra and is identical in these two editions. I imagine most of these problems appeared in
the first edition, so I will date this as 1798?, citing pages as 1833 and
1857. However the 1857 has an
additional three pages on Practical Questions in Arithmetic which has 44
problems, some of which are recreational, and another new problem. Assuming the 1857 is essentially the same as
the 1841, I will cite this as 1841?.]
See
also: Ozanam‑Hutton.
H&S. 1927. Vera
Sanford. The History and Significance of
Certain Standard Problems in Algebra.
(Teachers College, Columbia University, NY, Contributions to Education,
No. 251, 1927) = AMS Press, NY, 1972.
Illustrated Boy's Own
Treasury. c1847.
The
Illustrated Boy's Own Treasury of I. - Science, II. - Drawing, III. - Painting,
IV. - Constructive Wonders, V. - Rural Affairs, VI. - Wild and Domesticated
Animals, Outdoor Sports & Indoor Pastimes forming a Complete Repository of
Home Amusements & Healthful Recreations embellished with five hundred descriptive
engravings. (John & Robert Maxwell,
London, c1847 [Toole Stott 407]). Ward
and Lock, 1860 [Toole Stott 1091]. (3rd
ed, for the Proprietors, 1865? [Toole Stott 408].) [Toole Stott's descriptions make it seem that these editions are
identical.] See also: Boy's Own Book, Boy's Own Conjuring Book.
[Many of the problems are the same as in the other two books, but the
illustrations here occasionally omit some labels, so these must be errors in
copying from some earlier source. If
the c1847 date is correct, then this considerably changes the chronology of
these problems, with this book being the major known intermediate between Boy's
Own Book and Magician's Own Book. I
will hold off making these changes until I see the c1847 ed -- this may take
some time as the BM copy was lost in the war and the other two copies cited are
in the US. Hall, BCB 187 is: The
Illustrated Boys' Own Treasury of Indoor Pastimes; Robson, London, c1845. This may be related to this book.]
Indoor & Outdoor. c1859.
Indoor
and Outdoor Games for Boys and Girls: Comprising Parlour Pastimes, Charades,
Riddles, Fireside Games, Chess, Draughts, &c, &c. With a Great Variety of Athletic Sports,
Parlour Magic, Exercises for Ingenuity, and Much That is Curious, Entertaining,
and Instructive. James Blackwood,
London, nd [c1859]. This is a
combination of two earlier books, comprising two separately paginated
parts. The earlier books are Parlour
Pastime (1857 -- qv) and Games for All Seasons [Toole Stott 311 & BMC give
1858]. There is a later version of the
first part -- Parlour Pastimes, qv, which the BMC dates as 1868. Also there is another version of the
combined ed "with additions by Oliver Optic", as Sports and Pastimes
for Indoors and Out, G. W. Cottrell, Boston, 1863 [Toole Stott 1186, which
identifies Optic as William Taylor Adams].
Both BMC and NUC say Indoor & Outdoor is by George Frederick
Pardon. Hence the problems in the first
part will be cited as: Parlour Pastime, 1857
= Indoor & Outdoor, c1859, Part 1
= Parlour Pastimes, 1868. Many
of the problems are identical to Book of 500 Puzzles.
IPPn n-th International Puzzle Party. 10 = London, 1989; 13 = Amsterdam, 1993; 16 = Luxembourg, 1996;
19 = London, 1999;
20 = Los Angeles, 2000;
22 = Antwerp, 2002.
This are the ones I have attended, but some material has appeared at
other IPPs.
Jackson. Rational Amusement. 1821.
John
Jackson. Rational Amusement for Winter
Evenings; or, A Collection of above 200 Curious and Interesting Puzzles and
Paradoxes relating to Arithmetic, Geometry, Geography, &c. With Their Solutions, and Four Plates. Designed Chiefly for Young Persons. By John Jackson, Private Teacher of the
Mathematics. London: Sold by J. and A.
Arch, Cornhill; and by Barry & Son, High‑Street; and P. Rose; Bristol. 1821.
[Other copies, apparently otherwise identical, say: London: Sold by
Longman, Hurst, Rees, Orme, and Brown; G. and W. B. Whittaker; and Harvey and
Darton. And Barry and Son, High-Street,
Bristol. 1821. [Heyl 185.
Toole Stott 413.] Will Shortz
says this is the first English-language book devoted to non-word puzzles.]
JRM. Journal of Recreational Mathematics.
Kanchusen. Wakoku Chiekurabe. 1727.
Tagaya
Kanchusen [pseud. of Fuwa Senkuro].
Wakoku Chiekurabe [Japanese Wisdom Competition -- in Japanese]. 2 vols, 1727, 12 & 29 pp. PHOTOCOPY from Shigeo Takagi's copy sent by
Naoaki Takashima. Edited into modern
Japanese, with commentary on Kanchusen, by Shigeo Takagi, 1991, 42pp, present
from Takagi. Translated into English by
Hiroko Dean, 1999, 15pp plus annotations on the 42pp. Takagi and Takashima are working on a translation and annotation
into modern Japanese. We intend to
produce an English version from Dean's translation with commentary on the
puzzles. I will cite pages from
Takagi's edition. (Partly reproduced in
Akira Hirayama; Tôzai Sûgaku Monogatari [Mathematical Stories from East and
West]; (1973), 3rd ed., 1981, p. 208, ??NYS, from which it has been reproduced
in the exhibition Horizons Mathématiques at La Villette, Paris, and elsewhere.)
Kaye, George R. See:
Bakhshali MS.
King. Best 100. 1927
Tom
King. The Best 100 Puzzles. W. Foulsham, London, nd [1927, according to
BMC -- my copy says 'Wartime reprint'.]
A selection of these are reproduced in a booklet: Foulsham's Games and
Puzzles Book; W. Foulsham, London, nd [c1930].
I will indicate this by =
Foulsham's, no. & pp.
Knott, Cargill G. See under:
Tom Tit.
Labosne. See under:
Problemes.
Ladies' Diary. See under: T. Leybourn.
Landells. Boy's Own Toy-Maker. 1859.
E[benezer]
Landells. The Boy's Own Toy-Maker: A
Practical Illustrated Guide to the Useful Employment of Leisure Hours. Griffith & Farran, London, 1859(1858);
Shepard, Clark & Brown, Boston, 1859; Griffith & Farran, 3rd ed., 1860;
D. Appleton, NY, 1860; Griffith & Farran, 6th ed., 1863 [Toole Stott 1286‑1290]. [BMC has 1859(1858) and a longer 10th ed,
1881. NUC has the latter four of the
versions given by Toole Stott.] [The
Preface to the Second Edition, reproduced in the 3rd ed., says it appeared just
two months after the first edition.
Toole Stott indicates that all the versions he cites are identical. I have 3rd ed., 1860. Shortz has Appleton, 1860.] The date 1859(1858) indicates that the book
appeared in late 1858 to catch the Christmas trade, but was postdated 1859 to
seem current for the whole of 1859, so I will date this as 1858. The 2nd ed. must be 1859. Comparison shows that the section on
Practical Puzzles is essentially an exact subset of the material in Boy's Own
Conjuring Book.
Leeming. 1946. Joseph
Leeming. Fun with Puzzles. (Lippincott, Philadelphia, 1946); Comet
Books (Pocket Books), NY, 1949.
Lemon. 1890. "Don
Lemon" [= "The Sphinx" =
Eli Lemon Sheldon], selector.
Everybody's Illustrated Book of Puzzles. Saxon & Co., London, 1890 (with 1891 on the back cover) and
1892. The 1892 ed. omits the text on
the back cover and adds some pages of publisher's advertisements, but is
otherwise identical. 794 problems,
about 100 being mathematical, on 125pp.
This looks like a UK reprint of a US book, but the NUC only lists London
editions, so perhaps it is just selected from US publications. Some of the problems are attributed to
Golden Days, Good Housekeeping, St. Nicholas, etc.
I
also have an undated edition which says 'Selected by the Sphinx'. This has 744 problems on 122pp, about 45% of
which come from the other edition. The
NUC dates this as 1895. I will refer to
this edition as: Sphinx. 1895.
Leopold. At Ease!
1943.
Jules
Leopold. At Ease! Ill. by Warren King. Whittlesey House (McGraw‑Hill),
1943. [This appears to be largely drawn
from Yank, The Army Weekly, over the previous few years.]
Leske. Illustriertes Spielbuch für Mädchen. 1864?
Marie
Leske. Illustriertes Spielbuch für Mädchen Unterhaltende und anregende Belustigungen,
Spiele and Beschäftigungen für Körper und Geist, im Zimmer sowie im
Freien. (1864; 19th ed., 1904); 20th ed., Otto Spamer, Leipzig, 1907. There is no indication of any updating in
the Foreword to the 19th ed. which is included here, and it was common to
describe new printings as new editions, so I will date this as 1864? This book is jammed with material of all
sorts, including lots of rebuses, riddles and puzzles. It is a bit like Boy's Own Book. My copy is lacking pp. 159-160 and 211‑212.
Leurechon, Jean
(c1591-1670). See: van Etten.
T. Leybourn.
Thomas
Leybourn, ed. The Mathematical
Questions, proposed in the Ladies' Diary, and Their Original Answers, Together
with some New Solutions, from its commencement in the year 1704 to 1816. 4 vols., J. Mawman, London, and two
co-publishers, 1817. I have only
examined vols. I & II so far. The
problems are proposed each year with solutions in the following year. Leybourn puts the solutions just after the
problem and numbers almost all the problems consecutively, though I don't know
if these numbers are in the Ladies' Diary.
Problems do not have any names and sometimes have pseudonyms or vague
names, e.g. Mr. Deare. I will give the
names of the proposer and solver(s), followed by Ladies' Diary and the two
years involved, then = T. Leybourn, his
volume and pages and his question number.
E.g. Chr. Mason, proposer; Rob.
Fearnside, solver. Ladies' Diary,
1732-33 = T. Leybourn, I: 223, quest. 168.
W. Leybourn. Pleasure with Profit. 1694.
William
Leybourn. Pleasure with Profit:
Consisting of Recreations of Divers Kinds, viz. Numerical, Geometrical,
Mechanical, Statical, Astronomical, Horometrical, Cryptographical, Magnetical,
Automatical, Chymical, and Historical.
Published to Recreate Ingenious Spirits; and to induce them to make
farther scrutiny into these (and the like) Sublime Sciences. And
To divert them from following such Vices, to which Youth (in this age)
are so much Inclin'd. To this work is
also Annext, A Treatise of Algebra, ..., by R. Sault. Richard Baldwin and John Dunton, London, 1694. The text consists of several parts, labelled
Tract. I, Tract. II, ..., which are separately paginated. All material is from Tract. I unless otherwise
specified. Several sections are taken
from the English editions of van Etten.
[Santi 371.]
Li & Du. 1987. Li
Yan & Du Shiran. Chinese
Mathematics: A Concise History. (In
Chinese: Commercial Press, Hong Kong, c1965.)
English translation by John Crossley & Anthony W.‑C. Lun. OUP, 1987.
Libbrecht. 1973. Ulrich
Libbrecht. Chinese Mathematics in the
Thirteenth Century. MIT Press,
Cambridge, Mass., 1973.
Lilavati. 1150. Lîlâvatî
of Bhaskara II, 1150 (see Colebrooke).
Lloyd, E. Keith. See:
BLW.
Loeb Classical Library.
Published
by Harvard Univ. Press, or Putnam's, NY, & Heinemann, London.
Loyd, Sam (1841-1911) (&
Sam Loyd Jr. (1873-1934).
See: Cyclopedia, MPSL,
OPM, SLAHP.
Lucas, Édouard (1842-1891). See:
RM and the following.
Lucas. L'Arithmétique Amusante. 1895.
Édouard
Lucas. L'Arithmétique Amusante. Ed. by H. Delannoy, C.‑A. Laisant
& E. Lemoine. (Gauthier-Villars,
Paris, 1895.) = Blanchard, Paris, 1974.
Lucca 1754. c1330.
Scuola
Lucchese. Libro d'abaco. c1390.
Dal Codice 1754 (sec. XIV) della Biblioteca Statale di Lucca. Edited by Gino Arrighi. Cassa di Risparmio di Lucca, 1973. Arrighi gives folio numbers and I will cite
these and the pages of his edition.
Arrighi has c1390, but Van Egmond's Catalog 163-164 gives c1330.
MA. Mathematical Association (UK).
MAA. Mathematical Association of America.
Magician's Own Book. 1857.
The
Magician's Own Book, or The Whole Art of Conjuring. Being a Complete Hand-Book of Parlor Magic, and Containing over
One Thousand Optical, Chemical, Mechanical, Magnetical, and Magical
Experiments, Amusing Transmutations, Astonishing Sleights and Subtleties,
Celebrated Card Deceptions, Ingenious Tricks with Numbers, Curious and
Entertaining Puzzles, Together with All the Most Noted Tricks of Modern
Performers. The Whole Illustrated with
over 500 Wood Cuts, and Intended as a Source of Amusement for One Thousand and
One Evenings. Dick & Fitzgerald,
NY, ©1857. 12 + 362 pp. + 10 pp.
publisher's ads. My thanks to Jerry
Slocum for providing a copy of this.
[Toole Stott 481 lists this as anonymous and entirely different from the
UK ed. He cites a 1910 letter from
Harris B. Dick who says H. L. Williams may have edited it, but both Dick's
father and John Wyman may also have had a hand in it. Toole Stott 929, 930, 1378, 931 lists Dick & Fitzgerald,
1862, 1866, 1868, 1870, all apparently identical to the 1857. 929-930 are listed under Arnold and he there
cites Cushing's Anonyms as saying the book is by Arnold and Cahill. Christopher 622-625 are all Dick &
Fitzgerald; 622-623 are 1st ed., 624-625 are reprints of about the same time
and my copy seems most likely to be 625.
C&B, under Cremer, say "It is believed that they were all
written by H. L. Williams, a prolific hack writer of the period." Christopher 622 says Harold Adrian Smith
[Dick and Fitzgerald Publishers; Books at Brown 34 (1987) 108-114] has studied
this book and concludes that Williams was the author, assisted by Wyman. Actually Smith simply asserts: "The
book was undoubedly [sic] written by H. L. Williams, a "hack writer"
of the period, assisted by John Wyman in the technical details." He gives no explanation for his assertion,
but it may be based on C&B. NUC
lists this as by George Arnold (1834-1865) and Frank Cahill, under both Arnold
and Cahill. C&B list it under
Cremer, attributed to Arnold & Cahill, but they give a date of 1851, which
must be a transcription error. C&B
also list it under Magician's, from New York, 1857, and under Williams, but as
London, 1857.] See the discussion under
Status of the Project in the Introduction for the sources of the material.
Boy's
Own Conjuring Book, qv, appears to be a UK pirate edition largely drawn from
this. Book of 500 Puzzles copies about
80 pages of this. See the comments under
Book of 500 Puzzles. A fair number of
the problems are identical to or similar to the Boy's Own Book and a woodcut, a
poem and the introduction to a section are taken directly from Boy's Own Book. Otherwise I had thought that this book was
the source for the spate of puzzle books in the following 15 years, but I have
found that some of the identical puzzles appeared in The Family Friend
c1850.
Magician's Own Book (UK
version). 1871.
This
is quite different than the previous book.
The
Magician's Own Book. By the Author of
"The Secret Out," "The Modern Conjuror," &c. Edited by W. H. Cremer, Jun. Containing Ample Instructions for
Recreations in Chemistry, Acoustics, Pneumatics, Legerdemain, Prestidigitation,
Electricity (with and without apparatus).
(In the middle of the page is an illustration of a wizard in white on
red.) Performances with Cups and Balls,
Eggs, Hats, Flowers, Coin, Books, Cards, Keys, Rings, Birds, Boxes, Bottles,
Handkerchiefs, Glasses, Dice, Knives, &c., &c. With 200 Practical Illustrations. John Camden Hotten, nd [1871]. This has a two page list of Very Important
New Books at the beginning on pp. i-ii.
This lists Magician's Own Book as by the Author of "The Secret
Out" and The Secret Out as by the Author of the "Magician's Own
Book". But a further note says
"Under the title of "Le Magicien des Salons" the first has long
been a standard Magic Book with all French and German Professors of the
Art." -- see the discussion under Status of the Project in the
Introduction, above. This list is
followed by a half-title, p. iii, whose reverse (p. iv) has the printer's
colophon, then a blank page v, backed by a Frontispiece, p. vi, comprising
Figures 105 & 110 from the text.
The TP is p. vii,, backed by a blank p. viii. The Preliminary on pp. ix-x has the address Piccadilly
at the end and states this is "an Entirely New Edition" and is
by the same author as The Secret Out.
It refers to Cremer's display of Toys of the World at the recent
International Exhibition (possibly the 1862??) and to [Frank Bellew's]
"The Art of Amusing" (of 1866 and published by Hotten in 1870) and
Clara Bellew's "The Merry Circle".
Contents are given on pp. xi‑xii and then the text on pp.
13-326. [Toole Stott 194 lists this
under Cremer and says there are 30pp of publisher's catalogue at the end, dated
1872 -- I didn't record these details.
Christopher 239 lists this as "An entirely new edition" with
200 illustrations, 325 pp. + 30 pp. publisher's ads.]
I
have also seen a John Grant, Edinburgh, ed. which omits pp. i-vi, and has a
simplified title page, saying it is A New Edition, and drops the address Piccadilly.
Otherwise the text appears to be identical. It has no date but Toole Stott 1015 gives the date 1871. C&B, under Cremer, have London, 1871
with no indication of the New York ed.
C&B also list it under Williams.
I
have a Chatto & Windus ed., which omits pp. i-vi, and whose TP has only
slight changes from the Hotten TP, saying it is A New Edition, but the text
appears to be identical, except for dropping the address Piccadilly
from the end of the Preliminary.
It is dated 1890, with separately paginated publisher's catalogue of
32pp. dated Apr 1893.
Mahavira. 850. Mahāvīrā(cārya). Gaņita‑sāra‑sangraha
[NOTE: ņ denotes an n with a dot under it and ń
denotes an n with a dot over it.] (= Gaņita‑sāra‑samgraha
[The m
should have a dot over it.] = Ganitasar Samgrha). 850.
Translated by M. Rańgācārya. Government Press, Madras, 1912.
The sections in this are verses.
I will refer to the integral part of the first verse of the
problem. E.g. where he uses 121½ ‑ 123, I will use v. 121. [This work is described by David Eugene
Smith; The Ganita-Sara-Sangraha of Mahāvīrācārya;
Bibliotheca Mathematica (3) (1908/09) 106-110.
In: [G. R. Kaye; A brief
bibliography of Hindu mathematics; J.
Asiatic Society of Bengal (NS) 7:10 (Nov 1911) 679-686], this work is cited as
1908 with the note: "This is really an advance copy of a work not yet
actually published, kindly supplied to me by the author." See the entry under Pearson, 1907, in 7.E.]
Mair. 1765?.
John
Mair. Arithmetic, Rational and
Practical. Wherein The Properties of Numbers are clearly
pointed out, the Theory of the science deduced from first principles, the
methods of Operation demonstratively explained, and the whole reduced to
Practice in a great variety of useful Rules.
Consisting of Three Parts, viz.
I. Vulgar Arithmetic. II.
Decimal Arithmetic. III. Practical
Arithmetic. A. Kincaid & J. Bell,
Edinburgh, in three vols, 1765-1766 (Turner G1.14/1-3); 2nd ed., A. Kincaid & W. Creech, and J.
Bell, Edinburgh, 1772; 3rd ed, John
Bell and William Creech, Edinburgh, 1777.
I have the 3rd ed and have seen the 2nd ed. All the material of interest is in part 3, which first appeared
as vol. 3 in 1765 and books of this era often had little or no change between
editions, so I will date entries as 1765?
Manson. Indoor Amusements. 1911.
J.
A. Manson, compiler. Indoor
Amusements. Cassell & Co., London,
1911. FP + 8pp + 340pp + 8pp Index
(341-348). This is an extension of
Cassell's Book of In-Door Amusements ..., expanding the earlier 209pp of main
text to 340pp. This is partly due to
using larger type, getting 47 lines per page instead of 54. The material which was in Cassell's is
generally unchanged.
Manuel des Sorciers. 1825.
Manuel
des Sorciers ou Cours de Récréations Physiques,
Mathématiques, Tours de Cartes et de Gibecière, suive Des Petits Jeux de Société, et de Leurs Pénitences. (Conort, Paris, 178?; 2nd ed, Metier & Levacher, Paris, 1802
[Christopher 642, C&B]; 4th ed,
Ferra Jeune, Paris, 1815 [C&B]; 5th
ed, Ferra Jeune, 1820 [C&B]); 6th
ed, Augmentée d'une Notice sur la Magie noire, Ferra Jeune, Paris, 1825
[Christopher 643, C&B, HPL]. There
are a great many French books with similar titles from this era. They seem to be the predecessors of
Magician's Own Book, etc. -- cf the discussion under Status of the Project in
the Introduction and under Book of 500 Puzzles and Magician's Own Book. This is the first that I have found and
examined carefully, at HPL. It is
likely that most of the material dates back to the first ed of 178?, but until
I see some earlier editions, I'll date it as 1825, Gaidoz, in Section 7.B, cites the 2nd ed, but the material is on
a different page than in the 1825.
Marinoni, Augusto. See:
Pacioli. De Viribus. c1500.
McKay. At Home Tonight. 1940.
Herbert
McKay. At Home Tonight. OUP, 1940.
Section V: Puzzles and problems, pp. 63-88.
McKay. Party Night. 1940.
Herbert
McKay. Party Night. OUP, 1940.
Sections on Dinner-Table Tricks, pp. 134-171; Some Tricks in English, pp. 174-175; Arithmetical Catches and Puzzles, pp. 176-184.
Metrodorus. c510. In: The Greek Anthology, W. R. Paton,
trans. Loeb Classical Library, 1916‑1918. Vol. 5, Book 14. This contains 44 mathematical problems, most of which are
attributed to Metrodorus, though he is clearly simply a compiler and some may
be much older. I have cited the pages
of the English translation -- the Greek is on the previous page. Paton gives English answers, but they are
not in the Greek.
The
Greek Anthology is the modern name for a combination of two anthologies which
were based on earlier compilations dating back to the time of Alexander (-4C),
e.g. the Garlands of Meleager (c-95) and Philippus of Thessalonika (c40) and
the Cycle of Agathias (c570). In the
late 9C or early 10C, Konstantinos Kephalas assembled these into one
collection, but distributed them according to type and then added several other
collections -- this was somewhat revised c980.
In 1301, Maximus Planudes re-edited Kephalas' anthology, omitting much
and adding some (Paton believes that Planudes' source was missing a book). The Planudean version replaced Kephalas's
version and was printed in 1484.
However a copy of Kephalas's version of c980 was discovered at
Heidelberg in the Palatine Library (hence the anthology is sometimes described
as Palatine) in 1606 and modern versions now use this version as the first 15
books and put all of Planudes' additions as Book 16, the Planudean
Appendix. The Anthology comprises some
4000 poems. Modern scholars view Paton
as obsolete, but I don't know of any later versions of the Anthology which
include the mathematical problems -- e.g. they are not even mentioned in Peter
Jay; The Greek Anthology; Allen Lane, 1973; Penguin, 1981.
See
also: David Singmaster; Puzzles from
the Greek Anthology; Math. Spectrum
17:1 (1984/85) 11-15 for a survey of
these problems.
Meyer. Big Fun Book. 1940.
Jerome
S. Meyer. The Big Fun Book. Greenberg : Publisher, NY, 1940.
MG. Mathematical Gazette.
Mikami. 1913. Yoshio
Mikami. The Development of Mathematics
in China and Japan. Teubner, Leipzig,
1913; reprinted, Chelsea, 1961? See
also: Smith & Mikami.
Minguet. 1733.
Pablo
Minguet (or Minguét) è (or é or e or y) Yról (or Irol) ( -1801?).
Engaños à Ojos Vistas, y Diversion de Trabajos Mundanos, Fundada en
Licitos Juegos de Manos, que contiene todas las diferencias de los Cubiletes, y
otras habilidades muy curiosas, demostradas con diferentes Láminas, para que
los pueda hacer facilmente qualquier entretenido. Pedro Joseph Alonso y Padilla, Madrid, nd [1733]. Frontispiece + 12 + 218 pp. Imprimaturs or licenses dated 3 Nov 1733, 10
Nov 1733 & 12 Dec 1733. [NUC. Christopher 672 for a 18 + 110 pp
version. C&B gives versions with 18
+ 110 pp and with 12 + 218 pp. HPL has
4 versions -- the one I examined was 12 + 218 pp. The BM has a version, but it is not clear which.]
The
early history of this book is confused.
The first edition may have only had 18 + 110 pp (or 105 pp ??), then was
apparently frequently reprinted, without changing the dates, sometimes with
additions. Or it may be that both
versions appeared in 1733. However, my
copy is identical to one in HPL which is catalogued as 1733 and Palau (a list
of Spanish book sales, c1955) lists six sales of the 12 + 218 pp version, dated
1733, and only one sale of the 18 + 110 pp version, dated 1733. Christopher dates the 12 + 218 pp versions
to c1760.
I
now have discovered 26 editions of this work, and 2 and 9 editions of two
derivative works, but I have only seen a few versions. If anyone has access to a copy, I would like
a photocopy of the TP and other publishing details and of the Indice.
3rd
ed, Domingo Fernandez de Arrojo, Madrid, 1755, 18 + 157 + 3 pp. [This includes the same material as 1733,
but it is reset with smaller type and much rearranged and includes some new
material. It seems to be the same as
the 1766. Palau gives this as 18 + 150
+ 10 pp and says there was a cheap edition of 14 + 171 + 10 pp. NUC gives 14 + 171 +11 pp, with publisher
Domingo Fernandez. HPL has two
'entirely different' editions of 1755 -- the one I examined was 18 + 157 + 3
pp. BM has a 1755, but it is not clear
which.]
3rd
ed, Dionisio Hernández, Madrid, 1755?
[Palau, with nd.]
Antonio
del Valle, Madrid, 1756. 171 + 20 (or
40??) pp. [Palau.]
Revised. Pedro J. Alfonso y Padilla, Madrid, c1760,
14 + 218 pp. [Christopher 673 &
674.]
Idem. Añadido en esta quarta impresion, 18
enigmas, 6 quisicosas muy curiosas. 4th
ptg, D. Gabriel Ramirez, Madrid, 1766, 16 + 150 + 10 pp. [BM; Palau; Christopher 675.]
Palau
says he has seen a catalogue with a 1793 ed, but thinks this is a printing
error for either 1733 or 1893.
Palau
lists an 1804 reprint of the 1733, 14 + 218 pp.
Sierra
y Martí, Barcelona, 1820, 1 + 224 pp.
[Identical to 1733, 12 + 218 pp, except the text has been reset,
spelling modernized, the front matter updated and the text starts on p. 10,
with everything the same as the 1733, except page numbers are increased by 9,
and the index is reduced in size. HPL;
Palau; Christopher 676.]
Neuva
Edicion corregida y aumentada por D. J. M. de L. Juan Francisco Piferrer, Barcelona, 1822, 1 + 224 pp. [This is quite differently arranged than the
1733, 12 + 218 pp, and 1820 eds, and contains some extra material. Palau, with
1 + 240 pp; NUC; Shortz; Christopher 677.].
Retitled:
Juegos de Manos ó sea Arte de Hacer Diabluras, y Juegos de Prendas. Que contiene varias demonstraciones de
mágia, fantasmagoria, sombras y otros entretenimientos de diversion para
tertulias y sociedades caseras.
ilustrado con láminas Por D.
Pablo Minguet, y aumentado considerablemente en esta nueva edicion con infinidad
de juegos nuevos, y con laminas intercaladas en el texto. D. Manuel Saurí, Barcelona, 1847, 189 +
10 pp. [Palau; Christopher 678.]
Juegos
de manos; ó sea, Arte de hacer diabluras ...ilustrado con 60 grabados.... New ed.
Simon Blanquet, Mexico, nd [1856].
7 + 498 pp. [NUC.]
Palau
cites an 1857 reprint of the 1847, presumably the 2nd Saurí ed..
Title
varied: Juegos de Manos ó sea Arte de Hacer Diabluras. Contiene: juegos de prendas, de naipes,
varias demonstraciones de majia, fantasmagoria, sombras y otros
entretenimientos de diversion, para tertulias y sociedades caseras. Por D. Pablo Minguet. Tercera Edicion Aumentada con gran número de Juegos nuevos, y grabados
intercalados en el texto. Manuel Saurí,
Barcelona, 1864. 1 + 213 pp. [From TP of 1993 facsimile. This omits one problem and some discussion
that was in 1733 and adds 22 new problems, but I see some of these already
appeared in 1755 and 1822. Palau,
noting that this is the 3rd ed from Saurí and mentioning a recent facsimile; HPL.]
Palau
cites an 1875 Barcelona reprint as 185pp, but has no publisher's name, probably
the 4th Saurí ed.
Title
varied: Juegos de manos; ó sea, El arte de hacer diabluras, contiene 150 clases
de juegos, de prendas, de naipes, varias demonstraciones de mágia,
fantasmagoria, sombras y otros entretenimientos de diversion, para tertulias y
sociedades caseras. 5. ed., aumentada
con gran numero de juegos nuevos y 70 grabados intercalados en el texto. Manuel Saurí, Barcelona, 1876. 192pp.
[NUC.]
Palau
confusingly cites Sauri reprints of 192pp:
8th ed, 1885; 8th ed, 1888; 9th ed, 1888. [Christopher 679 & 680 are the latter two. Perhaps there is an error in the dates, e.g.
the 1885 might really be the 7th ed.]
Title
varied: Juegos de Manos o sea Arte de hacer Diabluras, contiene 150 clases de
juegos, de prendas de naipes, varias demonstraciones de magia, fantasmagorias,
sombras y otros entretenimientos de mucha diversion, para tertulias y
sociedades caseras. Décima edición
aumentada con juegos nuevos, y 70 grabados intercalados en el texto. Juan Tarroll y Cla., Barcelona, 1893. 192pp.
[Palau.]
Palau
lists 11th ed [from Sauri]. Sauri y
Sabater, Barcelona, 1896. 192pp. [NUC lists 2 copies of the 11th ed as Sauri
y Sabater, 1897, 192pp & 189pp.]
12th
ed [from Sauri]. Sauri, Barcelona,
1906. 190pp. [Christopher II 2102. HPL
with nd.]
Facsimile
of the 1864 ed, with Presentación por Joan Brossa: De la brujería blanca. Editorial Alta Fulla, Barcelona, 1981. 11pp new material + II-VIII + 9-213 pp of
facsimile (II is FP; III is TP; V-VIII is Al Lector; 9-12 is a poetic
introductory Relacion; 203-213 is Indice).
2nd
ed (i.e. printing) of the 1981 facsimile of the 1864 ed, with a new cover,
1993. [DBS.]
Pp.
1-25 is a fairly direct translation of the 1723/1725 Ozanam, vol. IV,
pp. 393‑406. A number of
other pictures and texts also are taken from Ozanam. I will give the date as 1733, though the expansion may not have
occurred until c1755. I will cite the
1755, 1822 and 1864 pages in parentheses, e.g.
Pp. 158-159 (1755: 114-115; 1822: 175-176; 1864: 151).
MiS. Mathematics in School.
Mittenzwey. 1880.
Louis
Mittenzwey. Mathematische
Kurzweil. Julius Klinkhardt,
Leipzig. 1880; 2nd ed., 1883; 3rd ed., 1895; 4th ed.,
1904; 5th ed., 1907; 6th ed., 1912; 7th ed., 1918. I will
give the date as 1880 or 1895? I now
have copies of the 1st, 4th, 5th and 7th eds.
In working through the other editions, I have seen many more items than
I had previously recorded and I now think this is one of the most important 19C
puzzle books.
[Ahrens,
MUS #363 lists the first ed. being 1879, apparently taking the date of the
Foreword. He also has 3rd ed., 1903,
but V&T (and another reference) cite 3rd ed., 1895. I think this is a misinterpretation of the
last Vorrede in the 4th ed., which is for the 3rd and 4th ed. and dated
1903. Trey Kazee has obtained a copy of
the 2nd ed. The first two editions have
the author's given name and have 300 problems; the 4th, 5th and 7th have
333. The Vorrede to the 2nd ed in the
2nd ed. says it has no major changes.
The Vorrede to the 3rd & 4th eds in the 4th ed says the 3rd ed was
extended and the 4th ed. has replaced some problems, so it seems that the 3rd
probably had 333 problems. Compared to
the 1st ed, the 4th drops 9 problems and adds 42. The Vorrede to the 3rd, 4th and 5th eds in the 5th ed also says
some problems have been replaced, but I have not discovered any differences
between the 4th and 5th eds except one minor amendment, some resetting which changes
the page breaks on pp. 1-10 and 36-37 and a few line breaks. I suspect that the 4th and 5th eds are very
similar to the 3rd. The 5th and 7th
eds. are very similar, but the 7th is reset with smaller type and occupies five
fewer pages. There are a few amendments
and reorderings and four problems (36, 92, 122, 137) have been replaced, but a
number of simple misprints persist through all editions. I will give pages in 1st, 3rd?, and 7th eds,
e.g.
Prob. 193 & 194, pp. 36 & 89; 1895?: 218 & 219, pp. 41 & 91; 1917: 218 & 219, pp. 37 & 87.]
MM. Mathematics Magazine.
Montucla, Jean Étienne
(1725-1799). See: Ozanam;
Ozanam‑Montucla.
MP. 1926. H.
E. Dudeney. Modern Puzzles. C. Arthur Pearson, London, 1926; new ed., nd
[1936?]. (Almost all of this is in
536.)
MPSL1. 1959 & MPSL2.
1960.
Mathematical
Puzzles of Sam Loyd, vol. 1 & 2, ed. by Martin Gardner, Dover, 1959,
1960. (These contain about 2/3 of the
mathematical problems in the Cyclopedia, often with additional material by
Gardner.)
MRE. W. W. Rouse Ball (1850-1925). Mathematical Recreations and Essays. (First three editions titled Mathematical
Recreations and Problems of Past and
Present Times.) I have compiled an 8
page detailed comparison of the contents of all editions as part of my The Bibliography of Some Recreational
Mathematics Books.
1st
ed., Feb 1892, 240pp.
2nd
ed., May 1892 ('No material changes').
3rd
ed., 1896, 276pp.
4th
ed., 1905, 388pp. [In the 4th ed., it
says 'First Edition, Feb. 1892. Reprinted, May 1892. Second Edition, 1896. Reprinted, 1905.' However, it calls itself the 4th ed. and the 3rd ed. calls itself
3rd and there are substantial changes in the 4th ed.]
5th
ed., 1911, 492pp.
6th
ed., 1914, 492pp.
7th
ed., 1917, 492pp.
8th
ed., 1919, 492pp.
9th
ed., 1920, 492pp.
10th
ed., 1922, 366pp.
11th
ed., 1939, revised by H. S. M. Coxeter, 418pp.
Macmillan
(for 1st to 11th ed.).
12th
ed., 1974, revised by H. S. M. Coxeter, 428pp, U. of Toronto Press.
13th
ed., 1987, revised by H. S. M. Coxeter, 428pp, Dover.
A
few of the editions are actually reprintings of the previous edition: 2 is essentially the
same
as 1, 6 as 5, 9 as 8, 13 as 12. Consequently I will rarely, if ever, cite
editions
2, 6, 9, 13. For each topic occurring
in Ball, I have examined all the
editions,
so the absence of a reference indicates the topic does not occur in that
edition
-- unless it is buried in some highly unlikely place.
See
also: Ball‑FitzPatrick.
MS. Mathematical Spectrum (Sheffield, UK).
MTg. Mathematics Teaching (UK).
MTr. Mathematics Teacher (US).
Munich 14684. 14C.
Munich
Codex Lat. 14684. 14C. Ff. 30‑33. Published by M. Curtze; Mathematisch‑historische Miscellan:
6 -- Arithmetische Scherzaufgaben aus dem 14 Jahrhundert; Bibliotheca Math. (2)
9 (1895) 77‑88. 34 problems. Curtze gives brief notes in German. Curtze says these problems also appear in
Codex Amplonianus Qu. 345, ff. 16‑16', c1325, ??NYS. Munich 14684 comes from the same monastery
(St. Emmeran) as AR and AR incorporates much of it. In a later paper, Curtze says this is 13C -- ??NYR. Cf Folkerts, Aufgabensammlungen.
Murray. 1913. Harold
James Ruthven Murray. A History of
Chess. (OUP, 1913); reprinted by
Benjamin Press, Northampton, Massachusetts, nd [c1986].
MUS. 1910 & 1918.
Wilhelm
Ernst Martin Georg Ahrens (1872-1927).
Mathematische Unterhaltungen und Spiele. 2nd ed., 2 vols., 1910, 1918, Teubner, Leipzig. [The first ed. was 1901 in one volume. There is a 3rd ed. of Vol 1, 1921, but it is
a reprint of the 2nd ed. with just 2 pages of extra notes and the typographical
corrections made.] I will tend to cite
this, as e.g. MUS 1 153-155. MUS #n denotes item n in the substantial
Literarischer Index, MUS 2 375-431.
[MUS
II vii lists 22 items which he had been unable to see and which he suspected
might not exist. I have seen the
following items from the list: 86 (=
Les Amusemens, above), 102 (Hooper,
listed in Some Other Recurring References, below; cf Section 4.A.1), 145 (Jackson, see above), 212 (author's name is Horatio Nelson
Robinson); item 32 exists as an English
edition of van Etten, but the citation gives the publisher as though he was the
author; I have seen a copy of item 82
advertised for sale. A number of items
(135, 152, 164, 221, 223, 277, 297, 317) are cited by Wölffing - op. cit. in
Section 3.B.]
Muscarello. 1478.
Pietro
Paolo Muscarello. Algorismus. MS of 1478.
Published in 2 vols.: I -- facsimile; II -- transcription with notes and
commentaries; Banca Commerciale Italiana, Milan, 1972. I will cite the folios (given in vol. I) and
the pages of the transcribed version in vol. II. Van Egmond's Catalog 275-276.
M500. This is the actual name of the journal of the M500 Society,
the Open University student mathematics society.
NCTM. National Council of Teachers of Mathematics.
nd. no date. Estimated dates may follow in
[ ].
Needham. 1958. Joseph
Needham (1900-1995). Science and
Civilization in China, Vol. 3.
CUP, 1958. (Occasional references
may be made to other volumes: Vol. 2, 1956; Vol. 5, Part IV, 1980.)
The New Sphinx. c1840.
Anonymous. The New Sphinx An elegant Collection of upwards of 500 Enigmas Charades Rebusses
Logogriphes Anagrams Conundrums
&c. &c. To which are
added, a Number of Ingenious Problems.
T. Tegg & Son, London, nd, HB with folding frontispiece. [Vendor suggests it is 1840s. Shortz has 4th ed, by Gye & Baine and
says it is c1840. He also has 'a new(?)
edition', by T. Tegg, and says it is c1843.
He says the chapter of geometrical problems and brainteasers was new in
the 4th ed. Heyl 238 is a 7th ed.,
London, 18??, referring to HPL, where I find it in the Supplement.] The chapter of problems has 27 problems, of
which 21 are copied from Endless Amusement II, 1837 ed., 20 of which come from
the 1826? ed.
E. P. Northrop. Riddles in Mathematics. 1944.
Eugene
P. Northrop. Riddles in
Mathematics. Van Nostrand, 1944; English Universities Press, 1945; revised ed., Penguin, 1961. The Van Nostrand ed has the main text on
pp. 1-262. The EUP ed. has it on
pp. 1-242. The Penguin ed. has it on
pp. 11-240. The revision seems to
consist of only a few additional notes.
I will cite the dates and pages, e.g.
1944: 209-211 & 239;
1945: 195‑197 & 222;
1961: 197‑198 & 222.
H. D. Northrop. Popular Pastimes. 1901.
Henry
Davenport Northrop. Popular
Pastimes for Amusement and Instruction
being a Standard Work on Games,
Plays, Magic and Natural Phenomena Suitable for All Occasions containing
Parlor Games; Charming Tableaux; Tricks of Magic; Charades and Conundrums; Curious Puzzles; Phrenology
and Mind Reading; Palmistry, or How to
Read the Hand; Humorous and Pathetic
Recitations, Dialogues, Etc., Etc. including
The Delightful Art of Entertaining
The Whole Forming a Charming
Treasury of Pastimes for the Home, Public Schools and Academies, Lodges, Social Gatherings, Sunday Schools, Etc., Etc. Frank S. Brant, Philadelphia, 1901. [Not in any of my bibliographies. Vendor says only two copies in NUC, none in
BL.]
NUC. National Union Catalogue Pre‑1956 Imprints. Library of Congress, USA. c1960.
??check details
Nuts to Crack. Nuts to Crack, Part nn. Or, Enigmatical Repository; containing near
mmm Hieroglyphics, Enigmas, Conundrums, Curious Puzzles, and Other Ingenious
Devices. R. Macdonald, 30 Great Sutton
Street, Clerkenwell, London. These are
single broadsheets. The publisher's
details are often trimmed from the bottom of the sheet. At least 25 annual parts appeared, from Part
I of 1832, but the year is not always given.
Answer books -- The Nutcrackers -- also appeared and the publishers kept
the old sheets available for some years.
This series is very rare -- Will Shortz and James Dalgety have the only
examples known to me. I have photocopy
of almost all from Dalgety and Shortz, ??NYR.
mmm is either 250 or 200 and the problems are individually numbered in
each part. I will cite problems as,
e.g. Nuts to Crack I (1832), no. 200.
NX. No copy. Usually prefixed by
?? as a flag for further action.
NYR. Not yet read -- i.e. I have a copy which I
have not yet studied. Usually prefixed
by ??
as a flag for further action.
NYS. Not yet seen. Usually prefixed by
?? as a flag for further action.
OCB. See: Hall, OCB.
OED. Oxford English Dictionary. (As: New English Dictionary, OUP, 1884‑1928),
reprinted with supplements, OUP, 1933 and in various formats since.
o/o. On order.
OPM. 1907-1908. Our Puzzle
Magazine. Produced by Sam Loyd. The pages were unnumbered. The Magazine was reprinted as the Cyclopedia
as follows, but some pages of the Magazine were omitted and the answers to each
magazine were normally in the next one.
Vol. 1, No. 1 (Jun 1907) =
Cyclopedia pp. 7‑70.
Vol. 1, No. 2 (Oct 1907) =
Cyclopedia pp. 71‑121.
Vol. 1, No. 3 (Jan 1908) =
Cyclopedia pp. 122‑178.
Vol. 1, No. 4 (Apr 1908) =
Cyclopedia pp. 179‑234.
Since the Cyclopedia goes
to p. 339, there appear to have been two further issues which have not been
seen by anyone??
(The above data were provided by
Jerry Slocum.)
OUP. Oxford University Press.
Ozanam. 1694.
The
bibliography of this book is a little complicated. I have prepared a more detailed 7 pp. version covering the 19 (or
20) French and 10 English editions, from 1694 to 1854, as well as 15 related
versions -- as part of my The
Bibliography of Some Recreational Mathematics Books.
Jacques
Ozanam (1640-1717). Recreations
Mathematiques et Physiques, qui contiennent Plusieurs Problémes [sic] utiles &
agreables, d'Arithmetique, de Geometrie, d'Optique, de Gnomonique, de
Cosmographie, de Mecanique, de Pyrotechnie, & de Physique. Avec un Traité nouveau des Horloges
Elementaires. 2 vols., Jombert,
Paris, 1694, ??NYS. [Title taken from
my 1696 ed.]
[BNC. NUC -- but NUC lists a 1693 3rd ed. from
Amsterdam, which appears to be a misreading for 1698. MUS II 380 says 1st ed. was Récréations Mathématiques, Paris,
1694, 2 vols. Serge Plantureux's 1993
catalogue describes a 1694 edition in 2 vols., by Jombert and notes that
it is the original edition, that the privilege is dated 11 Jan 1692, but
that it was not printed until 30 Apr 1694.
The dating of the privilege may account for some references to the first
edition being 1692 -- e.g. the Preface to the 1778 Ozanam-Montucla ed. says the
first ed. was 1692, but Hutton changes this to 1694. MRE, 1st ed, 1892, pp. 3‑4, says the the 1st ed. was 2
volumes, Amsterdam, 1696, but this was amended in his 4th ed., 1905.
This
first appeared in two volumes, but later versions were sometimes in one
volume. I have references to versions
in Paris: 1694, 1696, 1697, 1698, possibly 1700? and apparently 1720; and in
Amsterdam: 1696, 1697, 1698, 1700.
There is no indication of any textual changes in these, except that the
pages are numbered consecutively in later versions. Plantureaux describes a 1696 edition as: Paris, Jombert (mais
Hollande), so there seems to have been some piracy going on. I have the 1696 ed. I will assume that the 1696 is essentially
identical in content to the 1694, though in the second volume, the page numbers
may be different and there is some confusion of plate numbering.
The
Traité des Horologes élémentaires, which appears in the 1694 ed., is a
translation of Domenico Martinelli's Horologi elementari. NUC says this was separately paginated in
1694, but it occupies pp. 473-583 of the 1696 ed and pp. 301-482 of vol. 3 of
the 1725 ed.]
About
1723, the work was revised into 4 vols., sometimes described as 3 vols. and a
supplement. MUS #52 gives 1720, 1723,
1724, 1725 and says the dates vary in the literature. The 1725 has privilege dated 1720, but I haven't found any
catalogue entry for a 1720 ed. of this revision, so it may be a spurious date
based on the privilege.
Nouv. ed.
Recreations
Mathematiques et Physiques, qui contiennent Plusieurs Problêmes [sic]
d'Arithmétique, de Géométrie, de Musique, d'Optique, de Gnomonique, de
Cosmographie, de Mécanique, de Pyrotechnie, & de Physique. Avec un Traité des Horloges
Elementaires. Par feu [misprinted
Parfeu in vol. 1] M. Ozanam, de l'Académie Royale des Sciences, &
Professeur en Mathematique. Nouvelle
edition, Revûë, corrigée & augmentée.
Vol. 4 has different title page.
Recreations
Mathematiques et Physiques, ou l'on traite Des Phosphores Naturels &
Artificiels, & des Lampes Perpetuelles.
Dissertation Physique & Chimique.
Avec l'Explication des Tours de Gibeciere, de Gobelets, & autres
récréatifs & divertissans. Nouvelle
edition, Revûë, corrigée & augmentée.
Claude
Jombert, Paris, 1723. [Taken from my
1725 ed.]
[Ball
and Glaisher [op. cit. in 7.P.5, p. 119] both cite a 1723 ed. as though they
had seen it, but there is no BMC entry for this date -- perhaps there is a copy
at Cambridge??. I have seen one volume
in an exhibition which was 1723. MUS
#52 says it was edited by Grandin. NUC
-- "The editor is said to be one Grandin." I have a brief 1899 reference to this ed.
I
have 1725, which is apparently a reprint of the 1723. The privilege/approbation is dated 16 May 1722 in Vol. 3 and Vol.
4 and also 28 Apr and 15 May 1720 in Vol. 3.
NUC says this is 1723 with new title pages. BNC has: Nouvelle édition ... augmentée [par Grandin], 4 vols, Paris,
1725. It was reprinted in 1735, 1737?,
1741, 1750/1749, 1770.
The
text and plates of the 1725 and 1735 eds. seem identical, though some of the
accessory material -- lists of corrections and of plates -- has been omitted
and other has been rearranged. I have
seen two versions of the 1735 -- one has the plates inserted in the text, the
other has them at the end as ordinary pages, while my 1725 has them at the end
on folding pages. Most of the 1725
plates are identical to the 1696 plates, but there were a number of additions
and reorderings. The 1725 plates have
their 1696 plate numbers and 1725 page references at the top with new, more
sequential, plate numbers at the bottom.
The 1725 text sidenotes refer to the plate numbers at the top, while the
1735 and later sidenotes refer to the bottom numbers. (However some of the new illustrations in vol. 4 are not
described in the text and this makes me wonder if there was an earlier version
with these new plates??) I will give
the 1725 top plate numbers, followed by the bottom numbers in ( )
-- e.g. plate 12 (14). The 1741
and the 1750/1749 eds. are essentially identical to the 1735 ed.
Ball,
MRE, and MUS #52 say the 1750 and/or the 1770 ed. were revised by Montucla, but
all other sources say his revision was 1778.
Indeed Montucla was only 25 in 1750.
Inspection of 1750 copies in the Turner Collection and at the Institute
für Geschichte der Naturwissenschaft shows the 1750 is identical to my 1725 ed.
except for some accents and a new publisher.]
English versions
Recreations
Mathematical and Physical; Laying down, and Solving Many Profitable and
Delightful Problems of Arithmetick, Geometry, Opticks, Gnomonicks, Cosmography,
Mechanicks, Physicks, and Pyrotechny.
By Monsieur Ozanam, Professor of the Mathematicks at Paris. Done into English, and illustrated with very
Many cuts. R. Bonwick, et al., London,
1708.
[Pp.
130-191 are omitted, but there is no gap in the text and the Contents also
shows these pages are lacking. Ball,
MRE. BMC. NUC. Hall, OCB, p.
165. Hall, BCB 216. UCL.
Toole Stott 520, noting the gap.
Bodleian. This is a pretty
direct translation of the 1696 French ed. or an early simple revision. Prob. 18-20 of Cosmographie have been
omitted. C&B list this and say
there were three later editions, though they then list the 1756 and 1790
editions.]
2nd ed.
Recreations
for gentlemen and ladies: or, Ingenious Amusements. Being Curious and diverting sports and pastimes, natural and
artificial. With Many Inventions,
pleasant Tricks on the Cards and Dice, Experiments, artificial Fireworks, and other
Curiosities, affording variety of entertainments. James Hoey, Dublin, 1756.
[Taken from Toole Stott entry.]
[NUC. Hall, BCB 212. Hall, OCB, p. 165, but giving the 1790 title. Toole Stott 518, but he has Hoez, which
seems to be either a misreading or a miswriting??]
3rd ed.
Recreations
for Gentlemen and Ladies; being Ingenious Sports and Pastimes. Containing Many curious Inventions, pleasant
Tricks on Cards and Dice; Arithmetical Sports; new Games; Rules for Assuredly
winning at all Games, whether of Cards or Dice; Recreative Fire-works; Tricks
to promote Diversion in Company, and other Curiosities.... James Hoey, Dublin, 1759. [Taken from Hall, BCB 213.]
[Ball,
MRE. Hall, BCB 213. Hall, OCB, p. 165. Bodleian.]
4th ed.
Recreations
for gentlemen and ladies; being ingenious sports and pastimes: containing Many
curious Inventions, pleasant Tricks on Cards and Dice; Arithmetical Sports; new
Games; Rules for Assuredly winning at all Games, whether of Cards or Dice;
Recreative Fire-works; Tricks to promote Diversion in Company, and other
Curiosities.... The fourth
edition. Peter Hoey, Dublin, 1790. [Taken from Toole Stott entry.]
[BMC
calls it the 4th ed. [abridged].
NUC. Hall, BCB 214. Hall, OCB, p. 165. Toole Stott 519.
Bodleian.]
I
will cite these and the following editions by date, though the varying problem
numbers and volume numbers will make this a bit unwieldy. All references to the 4 volume versions are
to vol. I unless specified otherwise.
See the entry in 4.A.1 as an example.
Note that I give the problem number first because this is usually the
same in the 1790, 1803 and 1814 editions, and often in the 1840. Figure numbers will also been given with the
problem number, although the 1840 has the figures in the text with different
numbers. Additions will be entered at
the relevant point.
Ozanam‑Hutton. 1803.
Recreations
in Mathematics and Natural Philosophy: ...
first composed by M. Ozanam, of the Royal Academy of Sciences, &c.
Lately recomposed, and greatly enlarged, in a new Edition, by the celebrated M.
Montucla. And now translated into
English, and improved with many Additions and Observations, by Charles Hutton,
.... 4 vols., T. Davison for G.
Kearsley, London, 1803.
[This
is a pretty direct translation of the 1790 Ozanam-Montucla ed. with a few
changes, and some notes and extra material, generally in sections which do not
interest us. Only one or two plates
have been changed. Erroneous problem
numbers have been retained.
There
was an 1814 ed. by Longman, Hurst, Rees, Orme and Brown, London. The texts are basically identical, but the
1814 has been reset to occupy about 15% fewer pages, the problem numbers have
been corrected, a few corrections/additions have been made and the plates do
not fold out.]
Ozanam‑Montucla. 1778.
J.
E. Montucla (1725-1799)'s revision of Ozanam.
Récréations
Mathématiques et Physiques, Qui contiennent les Problêmes & les Questions
les plus remarquables, & les plus propres à piquer la curiosité, tant des
Mathématiques que de la Physique; le tout traité d'une maniere à la portée des
Lecteurs qui ont seulement quelques connoissances légeres de ces Sciences. Par feu M. Ozanam, de l'Académie royale des
Sciences, &c. Nouvelle Edition,
totalement refondue & considérablement augmentée par M. de C. G. F. Claude Antoine Jombert, fils aîné, Paris,
1778, 4 vols. Approbation &
privilege dated 5 Aug 1775 & 4 Sep 1775.
[BMC
says this is "par M. de C. G. F. [i.e. M. de Chanla, géomètre forézien,
pseudonym of J. E. Montucla.]"
BMC, under Montucla, says M. de Chanla is a pseudonym of Montucla. BNC, under Montucla, describes Montucla as
Éd. Lucas, RM1, p. 242 lists this under
Chaula "attribué à Montucla".
MUS #52 has Chaula. NUC has
Chanla. Montucla's connection with the
book was so little known that the 1778 version was sent to him in his role as
Mathematical Censor and he made some additions to it before approving it. Hutton says the 'last edition' (presumably
1790) bears Montucla's initials.
'Forézien' means from Feurs or the Forez region. Reprinted in 1785-1786 and 1790 (see below).
This
is a considerable reworking of the earlier versions. In particular, the interesting material on conjuring and
mechanical puzzles in Vol. IV has been omitted. There are occasional misnumberings of problems. I recall the plates are folding at the end
of each volume, but I didn't note this specifically.]
Nouv. ed.
"Nouvelle
édition, totalement refondue et considérablement augmentée par M. de M***"
[BMC adds: [i.e. Jean Étienne Montucla]].
Firmin Didot, Paris, 1790.
[NUC
lists this as a reissue of 1778 with new TPs.
I have this ed.]
Ozanam‑Riddle. 1840.
(Edward
Riddle (1788-1854)'s revision of Ozanam‑Hutton.)
Recreations
in mathematics and natural philosophy: translated from Montucla's edition of
Ozanam, by Charles Hutton, LL.D. F.R.S. &c. A new and revised edition, with numerous additions, and
illustrated with upwards of four hundred woodcuts, By Edward Riddle, Master of the Mathematical School, Royal
Hospital, Greenwich. Thomas Tegg,
London, 1840. [Taken from my copy. MUS #130 asserts this is by a C. Biddle, but
this must be due to be a misreading or misprint.]
Another ed.
Recreations
in science and natural philosophy: Dr. Hutton's translation of Montucla's
edition of Ozanam. The present new
edition of this celebrated work is revised By Edward Riddle, Master of the
Mathematical School, Royal Hospital, Greenwich, who has corrected it to the
present era, and made numerous additions.
This Edition is also Illustrated by upwards of Four Hundred
Woodcuts. Thomas Tegg, London, 1844.
Another ed.
Recreations
in Science and Natural Philosophy. Dr.
Hutton's translation of Montucla's edition of Ozanam. New Edition, revised and corrected, with numerous additions, by
Edward Riddle, Master of the Mathematical School, Royal Hospital,
Greenwich. Illustrated by upwards of
Four Hundred Woodcuts. William Tegg,
London, 1851.
[Reprinted by William Tegg, 1854.]
All
the Riddle printings seem to be from the same plates -- the date of the
Prefatory Note and the Erratum given on p. xiv are the same in 1840 and
1851. Marcia Ascher cites an 1844
edition from Nuttall & Hodgson, but this is the printer.
These
have the figures in the text, but otherwise seem to be little different than
Ozanam-Hutton. I will generally cite it
just as 1840.
P. M. Calandri. See:
Benedetto da Firenze.
Pacioli. De Viribus.
c1500.
Luca
Pacioli (or Paciuolo) (c1445-1517). De
Viribus Quantitatis. c1500. Italian MS in Codex 250, Biblioteca
Universitaria di Bologna. Santi 3 dates
it as 1498 and describes the later parts which deal with riddles, etc., but
include some magic, etc. However,
Pacioli petitioned for a privilege to print this in 1508 and Part 2, Chap.
CXXIX, ff 228r-228v, has a date of 1509, so he may have been working on the MS
for many years. Van Egmond's Catalog
55-56 is not as helpful as usual.
Part
1: Delle forze numerali cioe di Arithmetica is described in: A. Agostini;
Il "De viribus quantitatis" di Luca Pacioli; Periodico di Matematiche
(4) 4 (1924) 165‑192 (also separately published with pp. 1-28). Agostini's descriptions are sometimes quite
brief -- unless one knows the problem already, it is often difficult to figure
out what is intended. Further, he
sometimes gives only one case from Pacioli, while Pacioli does the general
situation and all the cases. All
references are to the problem numbers in this part, unless specified
otherwise. I will use problem numbers
and names as in the MS at the problem -- these often differ considerably from
the numbers and names given in the index at the beginning of the MS. There are 81 problems in part 1, but the
Index lists 120 -- in a few cases, the Index name clearly indicates a problem
similar to an actual problem and I will mention this. There are microfilms at the Warburg Institute (currently
misplaced) and at Munich and Siena. I
printed Part 1 and some other relevant material from the Warburg microfilm,
some 125pp. When I read this, I saw
that the material near the last pages I had copied would be of interest, but
when I went back to copy this material, the microfilm had been misplaced.
Dario
Uri has photographed the entire MS and enhanced the images and put them all on
a CD. This has 614 images, including
the insides of the covers. This is
often more legible than the microfilm, but the folio numbers are often faint,
sometimes illegible. It was not until
the transcription (below) became available that I could read the material just
after this point and see that there were more interesting problems, but I found
it difficult to read the Italian (many words are run together and/or archaic)
and the diagrams referred to are lacking.
Dario was able to carry on and found the Chinese Rings and about a dozen
other interesting items. He has put
some material up on his website:
http://digilander.libero.it/maior2000/.
This includes the indexes and a number of the most interesting items,
with his comments and diagrams of later examples of the puzzles. I have now gone through all of the text and
found a number of problems of interest for this bibliography. Nonetheless, quite a number of problems,
some clearly of interest, remain obscure.
Transcription
by Maria Garlaschi Peirani, with Preface and editing by Augusto Marinoni. Ente Raccolta Vinciana, Milano, 1997. (The publisher did not reply to a letter,
but Bill Kalush kindly obtained a copy for me.
Dario Uri says it can be bought from:
Libreria Pecorini, 48 foro Buonaparte, Milano; tel: 02 8646 0660;
fax: 02 7200 1462; web: www.pecorini.com.) I will cite the text as Peirani and any
references to Marinoni's work as Marinoni.
The transcription is not exactly literal in that Peirani has expanded
abbreviations and inserted punctuation, etc.
Also, Peirani seems to have worked from the microfilm or a poor copy as
she sometimes says the manuscript has an incorrect form which she corrects, but
Dario Uri's version clearly shows the MS has the correct form. Peirani uses the problem numbers and names
in the MS (see comment above about these differing from those in the Index),
but with some amendments. I will
probably give problem names as in the MS, with some of Peirani's amendments. I will give translations of the names, but
some of them are pretty uncertain and some have defeated me completely. Marinoni, pp. VIII‑IX indicates
the MS was written in about 1496-1509.
The
title is a bit cryptic, but I think the best English version is: On the Powers
of Numbers, but Pacioli [f. Ir] has 'forze numerali' and R. E. Taylor [op. cit.
in Section 1, pp. 307 & 339] follows Agostini and uses: On the Forces of
Quantity.
Pacioli. Summa.
1494.
Luca
Pacioli (or Paciuolo or Paccioli) (??-c1509).
Sūma de Arithmetica Geometria Proportioni &
Proportionalita. Paganino de Paganini,
Venice, 1494 -- cf Van Egmond's Catalog 325-326; facsimile printed by Istituto Poligrafico e Zecca dello Stato,
Rome, for Fondazione Piero della Francesca, Comune di Sansepolcro, 1994, with
descriptive booklet edited by Enrico Giusti.
There was a second ed., Paganino de Paganinis, Toscolano, 1523, but the
main text seems identical, except for corrections (and errors) and somewhat
different usage of initial letters and colour.
For extensive studies of this book, see the works by Narducci, Taylor,
Davis and Rankin given in Section 1.
Davis identifies material taken from Piero della Francesca's works. Taylor says 99 copies of the 1494 and 36
copies of the 1523 are known. The text
is in two parts -- the second part is geometry and is separately paged. All page references will be to the first
part unless specified as Part II.
Problems are numbered at the right hand edge of the last line of the
previous problem. See Rara 54-59.
Davis
notes that Pacioli's Summa, Part II, ff. 68v - 73v, prob. 1-56, are essentially
identical to della Francesca's Trattato, ff. 105r - 120r.
Panckoucke, André Joseph
(1700-1753). See: Les Amusemens.
Pardon, George Frederick
(1824-1884). See: Indoor & Outdoor; Parlour Pastime; Parlour Pastimes.
Parlour Pastime. 1857.
Parlour
Pastime for the Young: Consisting of Pantomime and Dialogue Charades, Fire-side
Games, Riddles, Enigmas, Charades, Conundrums, Arithmetical and Mechanical Puzzles,
Parlour Magic, etc. etc. Edited by
Uncle George [NUC says this is George Frederick Pardon]. James Blackwood, London, 1857 [Toole Stott
545; Christopher 724]. This was
combined with Games for All Seasons [Toole Stott 311; Christopher 723] into Indoor
& Outdoor, c1859, qv. There is a
later edition, Parlour Pastimes, 1868, qv.
Hence the problems will be cited as: Parlour Pastime, 1857 = Indoor & Outdoor, c1859, Part 1 = Parlour Pastimes, 1868. Many of the problems are identical to Book
of 500 Puzzles.
Parlour Pastimes. 1868?
Parlour
Pastimes: A Repertoire of Acting Charades, Fire-side Games Enigmas, Riddles,
Charades, Conundrums, Arithmetical and Mechanical Puzzles, Parlour Magic, etc.,
etc. James Blackwood, London, nd. [BMC, NUC and Toole Stott 1136 date this
1868 and say it is by George Frederick Pardon.
Toole Stott 1136 indicates that By G. F. P. is on the TP, but it is not
in my example. Toole Stott 1137 is
1870, a slightly smaller ed.] It is an
expanded version of Parlour Pastime, with the material of interest to us being
directly copied, though the page layout varies slightly. The running head of this is actually Parlour
Pastime. Hence the problems will be
cited as: Parlour Pastime, 1857 =
Indoor & Outdoor, c1859, Part 1 =
Parlour Pastimes, 1868. Many of the
problems are identical to Book of 500 Puzzles.
PCP. 1932. H. E.
Dudeney. Puzzles and Curious
Problems. Nelson, 1932; revised ed., nd
[1936?]. (Almost all of this is in
536.) There are almost no changes in
the revised ed., except that problem 175 and its solution have been corrected
-- it is a cross number puzzle and the text was for a different diagram. See: 7.AM.
Peano. Giochi. 1924.
Giuseppe
Peano (1858‑1932). Giochi di
Aritmetica e Problemi Interessanti. G. B.
Paravia, Torino, nd [1924 and later reprints].
(Thanks to Luigi Pepe for a photocopy of this.)
Pearson. 1907. A.
Cyril Pearson. The Twentieth Century
Standard Puzzle Book. Routledge,
London, nd [1907]. Three parts in one
volume, separately paginated. The parts
were also published separately. Each
part has several numbered sequences of problems.
Peck & Snyder. 1886.
Price
List of Out & Indoor Sports & Pastimes. Peck & Snyder, 126‑130 Nassau Street, N. Y., 1886. Reprinted, with some explanatory material,
in the American Historical Catalogue Collection, Pyne Press, Princeton,
1971. Unpaginated -- I have numbered
the pages starting with 1 as the original cover.
Peirani, Maria Garlaschi. See:
Pacioli. De Viribus. c1500.
Perelman. FFF.
1934.
Yakov
Isidorovich Perelman [Я. И. Перелман]
(1882-1942). Figures for Fun. Живая
Математика [Zhivaya
Matematika],
Наука [Nauka], Moscow. The books give no indication of the original
dates, but Tatiana Matveeva has kindly searched the Russian State Library and
found it was originally published by Гос.
техн.-теор.
издат., Leningrad-Moscow, 1934. Schaaf I 9 cites Recreational Arithmetic,
6th ed., Leningrad, 1935 and Sphinx 5 (1935) 96 reviews the 5th ed. of
L'Arithmétique Récréative, Leningrad, 1934 -- both presumably the same book ??
Translated
by G. Ivanov‑Mumjiev.
Foreign Languages Publishing House, Moscow, 1957, 120 sections.
(2nd
ed., 1973 -- used for MCBF, below, apparently the same as the 3rd ed.)
Translated
by G. Ivanov‑Mumjiev. 3rd ed.,
MIR, 1979, 123 sections. The 3rd ed.
drops 3 sections and adds 6 sections and has some amended English.
Perelman. FMP.
1984.
Yakov
Isidorovich Perelman [Я. И. Перелман]. Fun with Maths and Physics. MIR, Moscow, 1984. [This is a translation of Занимательныи
Задачи и Оыты
[Zanimatel'nye Zadachi i Oryty],
Детская
Литературы [Detskaya
Literatura], Moscow.) (There was a
Занимательныи
Задачи, with 4th ed. in 1935 – no
earlier version in Russian State Library.
This title originally published by
Деттиз, 1959.) Compiled
by I. I. Pruskov. Translated by
Alexander Repyev. This is a compilation
from several of his books from 1913 to c1942 (when he died). I have not yet seen the earlier books
(??NYS), but if this is mainly based on the earlier book of the similar title,
this would date the material to c1935? and I will use this date.
Perelman. MCBF.
c1980?
Yakov
Isidorovich Perelman [Я. И. Перелман]. Mathematics Can Be Fun. 3rd ed., MIR, 1985. This consists of the following, originally
separate, works, but with the second part having its page and problem numbers
continued from the first part. Both
these works exist in many other editions and translations.
Figures
for Fun, translated from Живая
Математика
[Zhivaya Matematika], Наука [Nauka], Moscow,
translated 1973 -- with 123 sections.
Translator not specified, but presumably G. Ivanov-Mumjiev, as in FFF
above.
Algebra
Can be Fun, translated from Занимательная
Алгебра
[Zanimatel'naya Algebra] (3rd ed. was published by ОНТИ,
Leningrad-Moscow, 1937 -- The Russian State Library apparently has no earlier
edition), edited and supplemented by V. Boltyansky, Наука [Nauka], Moscow, 1976,
translated by G. Yankovsky, 1976.
References
to this will be to material not in FFF, so will be dated 1937.
Phillips, Hubert
(1891-1964). See: Brush;
Week‑End.
Pike. Arithmetic. 1788.
Nicolas
Pike (1743-1819). A New and Complete
System of Arithmetic, composed for the Use of the Citizens of the United
States. John Mycall, Newbury-Port,
Massachusetts, 1788. (I have a 2nd ed.,
1797, [Halwas 318], ??NYR. This went
through at least 5 editions and then at least six variants, often abridged for
schools [Halwas 318-326].)
Poggendorff. J. C. Poggendorff. Biographisch-Literarisches Handwörterbuch zur Geschichte der
Exacten Wissenschaften enthaltend Nachweisungen über Lebensverhältnisse und
Leistungen von Mathematikern, Astronomen, Physikern, Chemikern, Mineralogen,
Geologen usw aller Völker und Zeiten.
Johann Ambrosius Barth, Leipzig.
Facsimile by Maurizio Martino, Storrs-Mansfield (later Mansfield
Center), Connecticut.
Vols.
I (A - L) and II (M - Z). (1863), nd
[bought in 1996].
Vol.
III (1858-1883), edited by B. W. Feddersen & A. J. von Oettingen;
Parts
I (A - L) and II (M - Z). (1898), nd
[bought in 1998].
Vol.
IV (1883-1904), edited by Arthur von Oettingen;
Parts
I (A - L) and II (M - Z). (1904), nd
[bought in 1998].
[Vol.
V covers 1904-1922. Vol. VI covers
1922-1949. One can get I-VI on
microfiche. Vol. VIIa covers 1932-1953
and apparently comprises 5 volumes.
There is also a VIIa Supplement which gives material supplementary to
vols. I-VI.]
Prévost. Clever and Pleasant Inventions. (1584), 1998.
J.
Prévost. (La Première Partie des
Subtiles et Plaisantes Inventions, Contenant Plusieurs Jeux de Récréation. Antoine Bastide, Lyons, 1584. ??NYS.)
Translated by Sharon King as: Clever
and Pleasant Inventions Part One Containing Numerous Games of Recreations and
Feats of Agility, by Which One May Discover the Trickery of Jugglers and Charlatans. Hermetic Press, Seattle, 1998. [No Second Part ever appeared. Hall, OCB, pp. 43, 100 & 113.] This is apparently the first book primarily
devoted to conjuring. Only five copies
of the original are known. There was a
facsimile in 1987. My thanks to Bill
Kalush for bringing this work to my attention.
Price, Harry (1881-1948). See: HPL.
Problemes. 1612.
Claude‑Gaspar
Bachet (c1587-1638). Problèmes plaisans
& délectables qui se font par les nombres.
(1st ed., 1612); 2nd ed., 1624, P.
Rigaud, Lyon (for 1 & 2 ed.);
revised by A. Labosne, (3rd ed., 1874; 4th ed., 1879); 5th ed., 1884,
Gauthier‑Villars, Paris (for 3, 4, 5 ed.).) 5th ed. reprinted by Blanchard, Paris, 1959 et al., with a
Frontispiece portrait and an introduction by J. Itard, based on the article by
Collet and Itard cited in 1 below.
I
have now obtained a photocopy of the 2nd ed. and have examined a 1st ed. I had believed that Bachet added 10 problems
in the 2nd ed., but the additional section of 10 problems, beginning "S'Ensuivent quelques autres ..."
is already in the 1612 1st ed. In the
1612, there are two problems V, but in
1624, these are made into two parts of prob. V. However, he does extend the initial section of 22 problems to 25
problems, inserting the new material as problems 3, 16 and 21. Prob. 16 (1612) = 18 (1624) has additional
material. Also, Bachet greatly expands
his preliminary material on the properties of numbers from 14 to 52 pages, but
Labosne drops this. Otherwise, the
material seems identical and the main text seems pretty much identical with the
fifth edition except that orthography is modernised -- e.g. plaisans becomes
plaisants, mesme becomes même, luy becomes lui, etc. I have now compared the 3rd ed with the 5th ed and I could find
no differences between them -- though I didn't check every word. Labosne adds a Supplement of 15 problems,
four Notes and a table of contents.
Labosne's Préface given in the 5th ed. is for the 3rd ed. I will cite problem numbers and pages from
the 1st ed., 1612; 2nd ed., 1624 and the 5th ed., 1884 (1959 reprint),
e.g. Prob. XIX, 1612, 99-103. Prob. XXII, 1624: 170-173; 1884: 115‑117. I will generally not give problem titles as
they usually run to several lines. I
will denote Labosne's supplementary problems as Bachet-Labosne, 1874.
I
have seen a 4th ed. by Gauthier‑Villars, 1905, no editor named,
containing only 37 of the 50 problems in the 5th ed. A contemporary review by E. Lampe (Fortschritte der Math. 36
(1905) 300‑301) was also mystified by this edition. C&B list this ed.
Pŗthudakasvâmî or
Pŗthūdaka. See: Chaturveda.
Pseudo-dell'Abbaco. c1440.
This
is attributed to Paolo dell'Abbaco (sometimes called Dagomari)
(c1281-1367). Trattato
d'Aritmetica. (c1370, according to
Arrighi, but see below). Codex
Magliabechiano XI, 86 at Biblioteca Nazionale di Firenze. Edited by Gino Arrighi, Domus Galilaeana,
Pisa, 1964. Arrighi gives some black
& white reproductions of illustrations.
I examined this MS in Sep 1994 and found the illustrations are often
lightly coloured and that Arrighi's illustrations were probably made from
poorish photocopies -- the writing on the opposite side shows through much more
in several of his illustrations than it does in the originals. I have colour slides of 11 pages.
Warren
Van Egmond [New light on Paolo dell'Abbaco; Annali dell'Istituto e Museo di
Storia della Scienza di Firenze 2:2 (1977) 3‑21 and Van Egmond's Catalog
114-115] asserts this MS is a c1440 compilation, based on watermark evidence,
and doubts that it is due to dell'Abbaco, giving the author as
pseudo-dell'Abbaco.
Smith,
Rara, 435‑440 describes a different MS at Columbia, headed 'Trattato
d'Abbaco, d'Astronomia e di segreti naturali e medicinali', which he dates
c1339. Van Egmond, above, gives the
title of this as 'Trattato di tutta l'arte dell'abacho', but Van Egmond's
Catalog 254-255 describes it as Plimpton 167, a codex containing two
works. The first is the dell'Abbaco:
Trattato di tutta l'arte dell'abacho; the second is Rinaldo da Villanova:
Medichamento Generale, which has the title Trattato d'Abbaco, d'Astronomia e di
segreti naturali e medicinali added in a later hand. The first includes the Regoluzze which is sometimes cited
separately. This is quite a different
book than the c1440 Trattato. There is
a c1513 version at Bologna, MS B 2433, which is dated 1339 -- Dario Uri has
sent a CD of images of it. See the
entry for dell'Abbaco in 7.E.
Putnam. Puzzle Fun.
1978.
Graham
R. Putnam, ed. Puzzle Fun. Fun Incorporated, np [Chicago?], 1978.
Rara. 1970. David
Eugene Smith. Rara Arithmetica. (1908; with some addenda, 1910; Addenda,
1939, published both separately and with the 1910 ed.); 4th ed., combining the
original with both Addenda and with De Morgan's Arithmetical Books of 1847 and
a new combined index, Chelsea, 1970.
References are to the main entry of this. Check the index for references to the Addenda and to De Morgan.
Rational Recreations. 1824.
Rational
Recreations. Midsummer MDCCCXXIV. Knight and Lacey, London. This is a six part work, but is bound
together -- perhaps the parts were issued monthly. The parts are consecutively paginated. [Toole Stott 590. Toole
Stott 591 is 2nd ed, 1825 and 592 is 3rd ed, 1825, copublished in Dublin. Hall BCB 235, 236 are 1824 and 1825. C&B.
HPL. Not in Christopher. I have examined the BL copy.]
Recorde. First Part.
1543.
Recorde. Second Part. 1552.
Recorde-Mellis. Third Part.
1582.
Recorde
(or Record), Robert (1510?-1558). The
Grounde of Artes Teaching the worke and practice of Arithmetike. ....
The dating of this book is uncertain.
Smith, Rara, p. 526 records a 1540 edition. An edition by Reynold
Wolff, London, at the Bodleian (Douce R.301) has generally been dated as 1542
and there is a facsimile by Theatrum
Orbis Terrarum, Amsterdam & Da Capo Press, NY, 1969, with the date 1542.
There were reprints in 1543 and 1549.
However, the DSB entry for Recorde ignores the Smith 1540 edition
(presumably because it has not been confirmed) and says the 1542 is now dated
as 1550?, making the 1543 the first edition. This edition only contained material on whole numbers.
In
1552, Recorde added a Second Part dealing with fractions, so the earlier
material will be called the First Part.
At
some stage, John Dee augmented it, but it appears he simply made some revisions
and additions to the existing text without adding new topics. The Dee material was added in 1590 (Smith)
or 1573 (De Morgan).
In
1582, John Mellis added a Third Part, mostly on rules of calculation,
published by J. Harison & H. Bynneman, London.
By
1640, the title was changed to:
Record's Arithmetick, or, The Ground of Arts; Teaching The perfect Work
and Practice of ARITHMETICK, both in whole Numbers and Fractions, after a more
easie and exact form then in former time hath been set forth. Afterwards augmented by Mr. JOHN DEE. And since enlarged with a third part of
RULES of PRACTICE, abridged into a briefer method then hitherto ..., by JOHN
MELLIS.
By
1648, more material was added by Robert Hartwell. I have a 1668 edition which has a little more material by Thomas
Willsford, but these latter two extensions are of no interest to us. It is clear that the text was increased by
accretion, with only minor revisions of Recorde's text, which is generally
preserved in gothic (= black-letter) type, and this is indicated by Smith. So the presence of a problem in Part One or
Part Two or Part Three of the 1668 ed almost certainly indicates its presence in
the first version of these parts. This
is certainly true for Part One, as I have the facsimile to compare, and it is
confirmed by brief examination of a 1582 ed, though at the time, I was looking
at Part Three and did not know of the material in Part Two. I will cite the pages from my 1662 ed and
the side notes which are titles of the problems, and pages of any earlier
editions that I have seen.
Riccardi. Pietro Riccardi. Biblioteca Matematica Italiana dalla Origine della Stampa ai
Primi Anni del Secolo XIX.
G. G. Görlich, Milan, 1952, 2 vols. This work appeared in several parts and supplements in the late
19C and early 20C, mostly published by the Società Tipografica Modense, Modena,
1878-1893. For details, see in Section
3.B.
Riddle, Edward (1788-1854). See:
Ozanam‑Riddle.
The Riddler. See under Boy's Own Book.
Riese. Coss. 1524.
Adam
Riese (c1489-1559). Die Coss. German MS of 1524 found at Marienberg in
1855. Described and abstracted in
Programm der Progymnasial‑ und Realschulanstalt zu Annaberg 1860. Reprinted in 1892. My reference to this comes from Johannes Lehmann; Rechnen und
Raten; Volk und Wissen, Berlin (DDR), 1987, pp. 7‑14, esp. p. 13. I have since seen the Glaisher paper, op.
cit. in 7.G.1 under Widman, esp. p. 37.
Glaisher and Lehmann cite: Bruno Berlet; Adam Riese, sein Leben und
seine Art zu rechnen; Die Coss von Adam Riese; Leipzig & Frankfurt,
1892. Glaisher notes that this was a
pamphlet. BLLD provided a copy from
Biblioth. Regia Berolinen. G., but it was lacking a title page. It seems to have the title: Zur Feier des
vierhundertsten Geburtsjahres von Adam Riese.
It was printed by Königl. Universitätsdruckerei von H. Stürtz in
Würzburg. The Vorwort is dated 1892, but
only signed 'Der Verfasser' and his name does not appear anywhere except on the
spine of the library's cover. The
booklet has two parts.
Adam
Riese, sein Leben, seine Rechenbücher und seine Art zu rechnen (from the
Programm for 1855), pp. 1‑26.
This is a discussion of Riese's Rechnung, but it also mentions some
material from his 'grosse Rechenbuch' titled Rechenung nach der lenge, auf den
Linihen und Feder, written in 1525 but not published until 1550. Glaisher, loc. cit., p. 43, says he sees no
authority for the date of 1525 and assumes it was written c1550. (I have recently obtained a 1976 reprint of
this work, ??NYR)
Die
Coss von Adam Riese (with Abdruck der Coss) (from the Programm for 1860), pp.
27‑62. This gives many numbered
problems -- I will cite the problem number and pages from this.
There
is a recent facsimile of the MS which I have just received -- ??NYR.
Riese. Rechnung. 1522.
Adam
Riese (c1489-1559). Rechnung auff der
Linien unnd Federn ... Erfurt,
1522. I have two reprinted editions. See Rara 138-143.
Christian
Egenolph, Frankfurt, 1544. (Riese's
text is dated 1525. There is a
supplement on gauging by Erhard Helm, dated 1544.) Facsimile by Th. Schäfer, Hannover, 1978.
Christian
Egenolff's Erben, Frankfurt, 1574.
(Riese's text is dated 1525 and appears to be the same text as above,
but reset. The Supplement has further
material.) Facsimile by Th. Schäfer,
Hannover, 1987.
Ripley's Puzzles and Games. 1966.
Ripley's
Believe It or Not! Puzzles and
Games. Essandess Special Edition (Simon
& Schuster), New York, 1966. Much
of the material occurred in the various Ripley's Believe it or Not! books. Most of the material is well known, but
there are a number of unusual variations and some interesting incomplete
assertions and mistakes!
RM. (François-) Édouard (-Anatole) Lucas
(1842-1891). Récréations
mathématiques. (Gauthier‑Villars,
Paris, 4 vols, 1882, 1883, 1893, 1894, 2nd eds. of vol. 1, 1891, vol. 2,
1893) = Blanchard, Paris, (1960), 1975‑1977,
using 2nd ed. of vol. 1 and 1st ed. of vol. 2 (however, there seem to be very
few differences in the editions). I
will cite the Blanchard reprint volumes as RM1, etc. (Dates are as given in Harkin, op. cit. in 1 below, and on the
books, but I have seen other dates cited.)
Lucas;
L'Arithmétique Amusante, 1895, Note IV, pp. 210-260 gives various fragments of
material for further volumes which were found after Lucas's untimely
death. His draft Tables of Contents for
volumes 5 and 6 are given on p. 210, but no material exists for most of the
chapters.
RMM. Recreational Mathematics Magazine. Nos. 1 - 14 (Feb 1961 - Jan/Feb 1964). Quite a bit of the material in this was
abstracted and sometimes extended in: Joseph S. Madachy; Mathematics on
Vacation; Scribner's, NY, 1966; somewhat corrected as: Madachy's Mathematical
Recreations; Dover, 1979.
Rocha. Libro Dabaco. 1541. SEE:
Tagliente. Libro de Abaco. (1515).
1541.
Rohrbough. Lynn Rohrbough edited a series of 20
booklets, called Handy Series, Kits A-J and M‑V, for Cooperative
Recreation Service, Delaware, Ohio, during at least 1925-1941. Most of these were reprinted and revised
several times. Two of these are of
especial interest to us and are listed below.
Several others are cited a few times.
Rohrbough. Brain Resters and Testers. c1935.
Lynn
Rohrbough, ed. Brain Resters and Testers. Handy Series, Kit M, Cooperative Recreation
Service, Delaware, Ohio, nd [c1935].
Rohrbough. Puzzle Craft. 1932.
Lynn
Rohrbough, ed. Puzzle Craft Plans for Making and Solving 40 Puzzles in
Wire, Wood and String. Handy Series,
Kit U, Cooperative Recreation Service, Delaware, Ohio, (1930), 1932.
I
have another version of this, unfortunately undated, but apparently later, so I
have dated it as 1940s? Jerry Slocum
located this by tracking down Rohrbough's successors. The 1932 version has 39 puzzles in its index, but 4 more that
were not indexed and some Notes which are listed in the index of this version,
making 44 items in all. 13 items are
omitted and replaced by 5 in the present version, giving 36 indexed items. The outside of the back cover of the 1932
version shows a number of puzzles, including several not described in either
version of the booklet.
Rudin. 1936. Jacob
Philip Rudin. So You Like Puzzles! Frederick A. Stokes Co., NY, 1936.
SA. Scientific American, usually Martin
Gardner's Mathematical Games column.
For years from at least 1950, SA appeared in two volumes per year, each
of six issues. In year 1950 + n,
vol. 182 + 2n covers Jan-Jun
and vol. 183 + 2n covers
Jul-Dec. See also under Gardner.
Sanford. 1930. Vera
Sanford. A Short History of
Mathematics. Houghton Mifflin, Boston,
1930 & 1958. See also: H&S.
Santi. 1952. Aldo Santi. Bibliografia della Enigmistica. Sansoni Antiquariato, Florence, 1952. 2541 entries, often citing several editions
and versions, arranged chronologically from 1479 onward. My thanks to Dario Uri for providing
this. I will site item numbers. I have only entered part of the information
so far.
de Savigny. Livre des Écoliers. 1846.
M.
l'Abbé de Savigny. Le Livre des
Écoliers Illustré de 400
vignettes. Jeux. -- Récréations. Exercises. -- Arts utiles et
d'agrément. Amusements de la
science. Gustave Havard, Paris, nd
[dealer has written in 1846], HB. [This
is almost entirely the same as Boy's
Own Book, 1843 (Paris). Each has a few
sections the other does not.
de Savigny's illustrations seem to have been copied by a new hand,
generally simplifying a little. Many of
the copies have been done in reverse and this leads to one erroneous
chessboard. However, there is one diagram
in Boy's Own Book, 1843 (Paris), which looks like it was badly copied, so it is
possible that both these books are based on an earlier book.]
Schaaf. 1955-1978.
William
L. Schaaf. A Bibliography of
Recreational Mathematics. Vol. 1,
(1955, 1958, 1963); 4th ed., 1970. Vol.
2, 1970. Vol. 3, 1973. Vol. 4, 1978. National Council of Teachers of Mathematics, Washington, DC. See Schaaf & Singmaster in Section 3.B
for a Supplement to these.
Schott. 1674. Gaspare
Schott. Cursus Mathematicus. Joannis Arnoldi Cholini, Frankfurt,
1674. The material of interest is in
Liber II, Caput VI: De Arithmetica Divinatoria, pp. 57-60, and in Liber XXVI:
Algebra, Pars III: De Exercitatione Algebraicam, Caput I, II, IV, VI, pp.
551-563, and Pars V: Exercitationes Algebraicae, pp. 570-571.
Schwenter. 1636.
Daniel
Schwenter (1585-1636). Deliciæ Physico‑Mathematicae. Oder Mathemat‑ und Philosophische
Erquickstunden, Darinnen Sechshundert Drey und Sechsig, Schöne, Liebliche und
Annehmliche Kunststücklein, Auffgaben und Fragen, auf; der Rechenkunst,
Landtmessen, Perspectiv, Naturkündigung und andern Wissenschafften
genomēn, begriffen seindt, Wiesolche uf der andern seiten dieses blats
ordentlich nacheinander verzeichnet worden: Allen Kunstliebenden zu Ehren,
Nutz, Ergössung des Gemüths und sonderbahren Wolgefallen am tag gegeben Durch
M. Danielem Schwenterum. Jeremiæ
Dümlers, Nuremberg, 1636. [Note: the
text is in elaborate Gothic type with additional curlicues so that it is not
always easy to tell what letter is intended!]
Probably edited for the press by Georg Philip Harsdörffer.
Extended
to three volumes by Harsdörffer in 1651 & 1653, with vol. 1 being a reprint
of the 1636 vol. Vol. 2 & 3 have
titles as follows.
Delitiæ
Mathematicæ et Physicæ Der Mathematischen
und Philosophischen Erquickstunden
Zweiter Theil: Bestehend in
Fünffhundert nutzlichen und lustigen Kunstfragen / nachsinnigen Aufgaben / und
derselben grundrichtigen Erklärungen / Auss Athanasio Kirchero Petro Bettino,
Marion Mersennio, Renato des Cartes, Orontio Fineo, Marino Gethaldo, Cornelio
Drebbelio, Alexandron Tassoni, Sanctorio Sanctorii, Marco Marco, und vielen
anderen Mathematicis und Physicis zusammen getragen durch Georg Philip
Harsdörffern. Jeremia Dümlern,
Nürnberg, 1651.
Delitiæ
Philosophicæ et Mathematicæ Der
Philosophischen und Mathematischen
Erquickstunden / Dritter Theil:
Bestehend in Fünffhundert nutzlichen und lustigen Kunstfragen / und
derselben gründlichen Erklärung: Mit vielen nothwendigen Figuren / so wol in
Kupffer als Holz / gezieret. Und Aus allen neuen berühmten Philosophis und
Mathematicis, mit grossem Fleiss zusammen getragen. Durch Georg Philip
Harsdörffern. Wolffgang dess Jüngern
und Joh. Andreas Endtern, Nürnberg, 1653.
This
3 vol. version was reprinted in 1677 and 1692. Modern facsimile of the 3 vol. version edited by Jörg Jochen
Berns, Keip Verlag, Frankfurt Am Main, 1991, HB. See also MUS II 325-326.
Schott described this as a German translation of van Etten/Leurechon,
but this is quite wrong. V&T, p.
152, say it is 'partially derived from' van Etten/Leurechon. C&B list just the 1636, under
Schwenterum. P. 549 of vol. 1 is
misprinted 249 which indicates that it was the first issue of the 1st ed.
I
have not yet entered all the items from this.
The Secret Out. 1859.
The
Secret Out; or, One Thousand Tricks with Cards, And Other Recreations. Illustrated with over three hundred
engravings. And containing Clear and comprehensive explanations how to
perform with ease, all the curious card deceptions, and slight of hand tricks
extant. With an endless variety of
entertaining experiments in drawing room or white magic, including the
celebrated science of second sight.
Together with a choice collection of intricate and puzzling questions,
amusements in chance, natural magic, etc., etc., etc. By the Author of "The Sociable, or, One Thousand and One
Home Amusements," "The Magician's Own Book," etc, etc. Dick & Fitzgerald, NY, 1859.
[Toole
Stott 191, listing all versions under Cremer.
C&B, under Frikell, have New York, 1859. H. A. Smith dates this as 1869.]
The Secret Out (UK). c1860.
The
Secret Out or, One Thousand Tricks in Drawing-room or White Magic, with an
Endless Variety of Entertaining Experiments.
By the author of "The Magician's Own Book." Translated and edited by W. H. Cremer,
Junr. With three hundred
illustrations.
I
have seen several editions. [Toole
Stott lists all versions under Cremer.]
C&W
(based on the John Camden Hotten, London, 1871?) (with ads from Sep 1886 at back
and inscription dated 12 Oct 1887 on flyleaf).
[NUC; Toole Stott 192; C&B, under Cremer, have London & New
York, 1871, and under Frikell, have London, 1870.]
C&W,
nd [1871? -- NUC lists several dates; Toole Stott 1013 is 1870; Christopher 242
is 1878?].
John
Grant, Edinburgh, nd [1872 -- Toole Stott 1014, no ads].
All
these copies have identical green covers with five magic tricks on the cover.
[Toole
Stott 192 discusses the authorship, saying that Wiljalba (or Gustave) Frikell
is named on the TP of some editions, but that most of the tricks are taken from
the US ed of The Magician's Own Book.
In the US ed, The Author acknowledges his indebtedness to The Sociable
and The Magician's Own Book 'and many other works of similar character and value',
but claims 'that the greater portion of it [i.e. the book] is entirely
original.' In the Preliminary to the UK
ed he says he is indebted to '"Le
Magicien des Salons," revised by references to Decremps, Servière, Leopold, Besson, Kircher, Hildebrandt,
Ozanam, &c., &c.' though an
1874 ad by C&W indicates that it is translated from Le Magicien des
Salons. (This may be Le Magicien de
Société, Delarue, Paris, c1860, but see Rulfs, below.) The back of the TP of Bellew's The Art of
Amusing, Hotten, 1866?, says The Secret Out is a companion volume, just issued,
by Hermann Frikell. BMC & Toole
Stott say it is also attributed to Henry L. Williams. Toole Stott 481 cites a 1910 letter from Harris B. Dick, of the
publishers Dick & Fitzgerald, who thinks their version of The Secret Out
"was a reprint of an English book by W. H. Cremer" -- but there seems
to be no record of a UK ed before the US one.
NUC says an 1871 ed. gives author as Gustave Frikell. Christopher 240-242 are two copies from Dick
& Fitzgerald, c1859, and a C&W, 1878?
He repeats most of the above comments from Toole Stott and 242 cites the
Rulfs article mentioned under Magician's Own Book, above. Rulfs says The Secret Out is largely taken,
illustrations and all, from Blismon de Douai's Manuel du Magicien (1849) and
Richard & Delion's Magicien des salons ou le diable couleur de rose (1857
and earlier). H. A. Smith [op. cit.
under Magician's Own Book] says the first US ed is 1869 (this must be a misprint
or misreading -- though the date is a little hard to read in my copy, it is
clearly 1859) and the UK eds are basically a condensed version with a few
additions. He suggests the book is
taken from DeLion. He doubts whether
Cremer ever wrote anything. C&B,
under Gustave Frikell, say it is a translation of Richard & Delion. C&B, under Herrman Frikell, list London,
1870. C&B, under Secret, list New
York, nd. C&B also list it under
Williams, as London, 1871.]
[Toole
Stott 1056 is [Frikell, Wiljalba]; Parlor Tricks with Cards, ...; By the Author
of Book of Riddles and 500 Home Amusements, etc.; Dick & Fitzgerald, 1860?;
which is described as "abridged from The Secret Out. Toole Stott 547 and 1142 are two versions of
1863, but without the description of the author and hence listed anonymously.]
The
US and UK editions are fairly different.
The US ed has 382 sections, of which 157 (41%) are used in the UK
ed. The UK ed has 323 sections, so 51%
of it is taken from the US ed. The US
ed seems like a magic book, with chapters on Scientific Amusements and
Miscellaneous Tricks. The UK ed has
much less magic and tricks, adding other general tricks and a lot more
scientific tricks. The illustrations
for the common sections are not quite identical -- one was probably copied from
the other. The amount taken from The
Magician's Own Book and The Sociable is fairly small, perhaps 10% from each, in
either edition.
Shortz. Will Shortz's library or his catalogues
thereof, called "Puzzleana".
The most recent I have is: May
1992, 88pp with 1175 entries in 26 categories, with indexes of authors and
anonymous titles. Some entries cover
multiple items. In Jan 1995, he
produced a 19pp Supplement extending to a total of 1451 entries.
SIHGM. 1939-1941.
Ivor
Bulmer Thomas. Selections Illustrating
the History of Greek Mathematics. Loeb
Classical Library, 1939‑1941, 2 vols.
I will give volume and pages as in
SIHGM I 308‑309.
Simpson. Algebra.
1745.
Thomas
Simpson (1710-1761). A Treatise of
Algebra; Wherein the Fundamental Principles Are fully and clearly demonstrated,
and applied to the Solution of a great Variety of Problems. To which is added, The Construction of a
great Number of Geometrical Problems; with the Method of resolving the same
Numerically. John Nourse, London,
1745.
I
also have the 7th ed., with the slightly different title: A Treatise of Algebra. Wherein the Principles Are Demonstrated, And
Applied In many useful and interesting Enquiries, and in the Resolution of a
great Variety of Problems of different Kinds.
To which is added, The Geometrical Construction of a great Number of
Linear and Plane Problems; with the Method of resolving the same
Numerically. The Seventh Edition,
carefully Revised. F. Wingrave,
Successor to Mr. Nourse, London, 1800.
I
have now seen a 6th ed., F. Wingrave, 1790 and it appears identical to the 7th
ed. Both have an Author's Preface to
the Second Edition.
I
will give the 1745 details with the 1790/1800 details in parenthesis like
(1790: ...).
SLAHP. 1928. Sam Loyd
Jr. (1873-1934) Sam Loyd and His
Puzzles. An Autobiographical
Review. Barse & Co., NY, 1928. (This contains somewhat more original
material than I expected, but he claims that he devised a lot of his father's
puzzles given in the Cyclopedia, OPM, and even earlier.)
Slocum. Compendium.
1977.
Jerry
(= G. K.) Slocum. Compendium of
Mechanical Puzzles from Catalogues.
Published by the author, Beverly Hills, 1977, 57pp. This is a compendium of illustrations and
descriptions from 30 catalogues. The
earliest ones are Bestelmeier, 1793, 1807 (Jacoby edition); The Youth's
Companion, 1875; Montgomery Ward, 1886, 1889, 1903, 1930; Peck & Snyder,
1886; Joseph Bland, c1890. Others of
some interest are: Gamage's, 1913, c1915, c1928; Johnson Smith, 1919, 1935,
1937, 1938, 1942.
Slocum & Gebhardt -- see
under Catel.
SM. Scripta Mathematica.
Smith, David Eugene (1860‑1944). See:
Rara and the following four
items.
Smith. History. 1923.
David
Eugene Smith. History of
Mathematics. Two vols. (Ginn, NY, 1923) = Dover, 1958.
Smith. Number Stories. 1919.
David
Eugene Smith. Number Stories of Long
Ago. NCTM, (1919), reprinted 1968? Chaps. IX and X.
Smith. Source Book. 1929.
David
Eugene Smith. A Source Book in
Mathematics. Two vols. (1929)
= Dover, 1959.
Smith & Mikami. 1914.
David
Eugene Smith & Yoshio Mikami. A History of Japanese Mathematics. Open Court, Chicago, 1914.
The Sociable. 1858.
The
Sociable; or, One Thousand and One Home Amusements. Containing Acting Proverbs; Dramatic Charades; Acting Charades,
or Drawing-room Pantomimes; Musical Burlesques; Tableaux Vivants; Parlor Games;
Games of Action; Forfeits; Science in Sport, and Parlor Magic; and a Choice
Collection of Curious Mental and Mechanical Puzzles; &c,&c. Illustrated with nearly three hundred
engravings and diagrams, the whole being a fund of never-ending
entertainment. By the author of
"The Magician's Own Book."
(Dick & Fitzgerald, NY, 1858 [Toole Stott 640 lists this as
anonymous; C&B list it under the title, with no author]); G. G. Evans,
Philadelphia, nd, but the back of the title gives ©1858 by Dick & Fitzgerald, NY, so this copy seems to be a
reprint of the 1858 book. Cf. Book of
500 Puzzles for discussion of possible authorship. The Preface here says that most of the Parlor Theatricals are by
Frank Cahill and George Arnold -- Toole Stott opines that this reference led
Harry Price to ascribe this and the related books to these authors. My thanks to Jerry Slocum for providing a
copy of this.
The
entire section Puzzles and Curious Paradoxes, pp. 285-318, is identical to the
same section in Book of 500 Puzzles, pp. 3-36.
Rulfs (see under Status of the Project in the Introduction) says this
draws on the same sources as Magician's Own Book, with more taken from Endless
Amusement and Parlour Magic.
See
also: Book of 500 Puzzles; Boy's Own Conjuring Book; Illustrated Boy's Own Treasury; Indoor and Outdoor; Landells: Boy's Own Toy-Maker; The Secret Out; Hanky Panky.
SP. Prefixed by ?? is a flag to check
spelling.
"The Sphinx". See:
Lemon.
Sridhara. c900. Śrīdharācārya. Patiganita (= Pâţîgaņita
[NOTE: ţ, ņ denote
t, n with underdot.]). c900.
Transcribed and translated by Kripa Shankar Shukla. Lucknow Univ., Lucknow, 1959. The text is divided into verses and
examples, separately numbered by the editor.
I will cite verse (v.) and example (ex.) and the page of the English
text. The editor has appended answers
on pp. 93‑96, some of which were given by an unknown commentator. (I have seen this dated 8C, which would put
it before Mahavira -- ??)
SSM. School Science and Mathematics.
Struik. Source Book. 1969.
D.
J. Struik (1894- ), ed. A Source Book in Mathematics 1200‑1800. Harvard Univ. Press, 1969.
Sullivan. Unusual.
1943 & 1947.
Orville
A. Sullivan. Problems involving unusual
situations. SM 9 (1943) 114‑118
& 13 (1947) 102‑104.
(Previously listed in Section 2 below.)
Suter. 1900-1902. Heinrich
Suter. Die Mathematiker und Astronomen
der Araber und Ihre Werke. (AGM 10
(1900) & 14 (1902)); reprinted by APA -- Academic Publishers Associated,
Amsterdam, 1981. See also: H. P. J. Renaud; Additions et corrections à
Suter "Die Mathematiker und Astronomen der Araber"; Isis 18 (1932)
166-183.
S&B. 1986. Jerry
(= G. K.) Slocum & Jack Botermans.
Puzzles Old & New -- How to Make and Solve Them. Univ. of Washington Press, Seattle,
1986. Slocum had produced a detailed
index for this and has extended it to a joint index with New Book of Puzzles.
Tabari. Miftāh al-mu‘āmalāt. c1075.
Mohammed
ibn Ayyūb Ţabarī [NOTE:
Ţ denotes a T
with a underdot.]. Miftāh
al-mu‘āmalāt. c1075. Ed. by Mohammed [the h
should have an underdot] Amin Riyāhi [the h should have an
underdot], Teheran, 1970. ??NYS --
frequently cited and sometimes quoted by Tropfke and others.
Tagliente. Libro de Abaco. (1515). 1541.
Girolamo
[& Giannantonio] Tagliente. Libro
de abaco che insegnia a fare ogni raxone marcadantile & apertegare le terre
con larte di la giometria & altre nobilissime raxone straordinarie cō
la tarifa come raspondeno li pexi & monete de molte terre del mondo con la
inclita citta de. Venetia. El qual
Libro se chiama Texauro universale ...; Venice, 1515. See Rara 114-116, 495, 511‑512, which seems to confuse this
with another work whose title starts:
Opera che insegna .... Riccardi
lists 27 editions of this and three editions of the other work. He says Boncompagni has studied this work
and found that Giovanni Rocha made some corrections and that several editions
have only his name, so it is sometimes catalogued under Rocha -- cf below. Smith mentions 25 editions under
Tagliente and says Riccardi mentions
11 others. Van Egmond's Catalog 334-344 lists 31
editions to 1586. It is clear that this
was a major book of its time. I have
briefly looked at a few examples and they seem to have the same material,
though the woodcuts were often changed.
I
have examined: Giovanne Rocha; Libro Dabaco che insegna a fare ogni ragione
mercadãtile: & a ptegare le terre cõ larte di la geometria: & altre
nobilissime ragione straordinarie cõ la Tariffa come respõdeno li pesi &
mone de molte terre del mõdo con la inclita di Vinegia. El qual Libro se chiama Thesauro universale. Venturino Rossinello, Venice, 1541. Smith, Rara, p. 529, only records a 1550 ed.
printed by Giovanni Padovano in Venice.
The Crawford Collection has 1544 & 1550. Not under Rocha in Riccardi.
I found this in the Turner Collection at Keele as A4.32, but it is not
under Rocha in Hill's Catalogue (listed in 3.B), but is under Tagliente. It has nice woodcut(?) illustrations in the
text -- see Rara 512 for an example.
Having first found this book on the shelf at the Turner Collection, I
originally thought it was by Rocha and hence an extraordinarily rare book. I am grateful to Bill Kalush for identifying
this as a version of Tagliente and for pointing out its importance. Cf Van Egmond's Catalogue 338, item 15.
Tartaglia, Nicolo (or Niccolò)
(c1506-1559). See: General Trattato.
Thomas, Ivor B. See:
SIHGM.
Tissandier. Récréations Scientifiques. 1880.
Gaston
Tissandier. Les Récréations
Scientifiques ou L'Enseignement par les Jeux.
G. Masson, Paris, (1880); 2nd
ed., 1881; 3rd ed., 1883; (4th ed., 1884); 5th ed., 1888; (6th ed.,
1893; 7th ed., 1894.) I have seen 3rd ed., 1883, and I have 2nd
ed., 1881, & 5th ed., 1888.
Tissandier was editor of La Nature and the articles often have fine
illustrations by L. Poyet and others, frequently copied elsewhere. See Tom Tit for more of Poyet's work. I have seen the date of the first two
editions as 1881 & 1882, but the Avertissement of the 2nd ed. says the 1st
ed. was Nov 1880 and the Avertissement is dated Apr 1881. [C&B only list 1881.]
Tissandier. Popular Scientific Recreations. 1881.
Translation
and enlargement of the above. Ward
Lock, London, ([1881, 1882]); New enlarged ed., nd [1890, 1891]. [Not clear which French edition the
translation is based on. The new
enlarged ed. contains a Supplement on pp. 775‑876 which includes material
which is in the 5th French ed. of 1888, but not in the 3rd French ed. of 1883,
so it seems the the main text is c1885 and the supplement is c1890. C&B list a London edition, nd,
780pp.] The index refers to a puzzle of
knots and cords on p. 775 which is not present. Most of the Supplement appeared as a series of Scientific
Amusements in (Beeton's) Boy's Own Magazine from 1889 -- I have vol. 3 (1889)
which has the first 10 articles, comprising 62 pages and 67 problems.
Some
of the material appeared in: Marvels of Invention and Scientific Puzzles. Being
A Popular Account of Many Useful and Interesting Inventions and
Discoveries. Ward, Lock, & Co., nd
[c1890]. My copy has no authors listed,
but Jerry Slocum has a copy with Tissandier and Firth on the TP, though it is
difficult to see what Firth could have done to warrant his inclusion. This consists of Chapters 56-60, pp.
726-774, and Chap. 32, pp. 448-465, of Popular Scientific Recreations, set on
smaller pages, plus a few extra items:
An economical mouse-trap (pp. 57-58);
Flying bridges (pp. 64-66);
Performing fleas (pp. 77-79);
Knots and cords and A Curious Toy (pp. 83-85). This last must be what was on pp. 775-776 of
the earlier edition of Popular Scientific Recreations and the other material
seems to have come from some edition of that work or from the articles in
(Beeton's) Boy's Own Magazine. I won't
bother to cite this version.
Tom Tit. 1890-1893?
Arthur
Good [= "Tom Tit"]. La
Science Amusante. 3 vols.,
Larousse, Paris, 1890, 1892, 1893? [I
have seen the dates given as 1889, 1891, 1893, but the Introductions are dated
as I have given, the last being Dec 1893.]
The material originally appeared in the magazine L'Illustration with
classic engravings by Poyet which have been often reproduced, e.g. in: Beeton's Boy's Own Magazine; The Boy's Own Paper; Kolumbus‑Eier (1890, 1976) --
translated as: Columbus' Egg (1978).
Arthur Good's name is clearly English and I have wondered if the
articles were written originally in French or if they were translated from his
English.
I
have the three volume set and five one volume selections/adaptations. I will give references to these by the
initials shown below.
C. François Caradec, ed. La Science Amusante -- 100 Experiences de Physique. Les Editions 1900 [no accent], Paris,
1989. [This consists of all of Vol. 1,
reordered, but otherwise little changed.
Caradec's Preface gives a little about the author. The illustrations are a bit dark.]
H. Magic at Home.
A Book of Amusing Science.
Annotated translation of La Science Amusante, vol. 1, by Prof.
Hoffmann. Cassell, 1891. Cf. VBM below.
VBM. The Victorian Book of Magic Illustrated or Professor
Hoffman's [sic] Curious & Innocent Diversions for Parlour & Refined
Gatherings. Selected [from the above]
& with a note to readers by C. Raymond Reynolds. (Stephen Greene Press, Japan, 1969); Hugh Evelyn, London, nd
[c1970]. [This has 26 of the items in
Vol. 1. Illustrations are small but
good.]
K. Tom Tit.
Scientific Amusements. Selected
and translated by Cargill G. Knott.
Nelson, nd [1918]. [This
contains 178 items, mostly from Vols. 2 & 3, but Knott has added a few
others, possibly taken from Tom Tit's later articles? Knott has also extended some items. Sadly, the illustrations were poorly redrawn for this edition.]
R&A. David Roberts & Cliff Andrew, eds. 100 Amazing Magic Tricks. Cape, London, 1977. [Selections from all three volumes. Although this refers to the original books
and L'Illustration, it avoids mentioning the original author's name! Illustrations are very good.]
[I
have also seen an Italian translation.
There was a US ed., trans. by Camden Curwen & Robert Waters; Magical
Experiments or Science in Play; (© Worthington, 1892); David McKay,
Philadelphia, 1894, 329 pp. [Christopher 398.]]
Todhunter. Algebra, 5th ed. 1870.
Isaac
Todhunter (1820-1884). Algebra For the Use of Colleges and Schools. With Numerous Examples. Macmillan, (1858; 5th ed, 1870); new
edition, 1879, HB. [1st ed was 1858;
2nd, 1861; 5th, 1870. The Preface in my
1879 copy is dated 1870 and says the work has been carefully revised, with two
chapters and 300 miscellaneous examples added, so it was quite different than
previous editions and I will date citations as 1870.]
Tonstall. De Arte Supputandi. 1522.
Cuthbert
Tonstall [often spelled Tunstall] (c1474-1559). De Arte Supputandi Libri Quatuor. ([With Quattuor] Richard Pynson, London, 1522 -- the first
arithmetic printed in England, with TP engraved by Holbein.) I have seen: Robert Stephan, Paris, 1538.
Though the TP and pagination are different, Smith, Rara, gives no
indication that the 1538 text is any different than the 1522, so I will cite
this as 1522. Most citations are to
Book III, whose problems are numbered.
See Rara 132-136.
Toole Stott. 1976-1978.
Raymond
Toole Stott. A Bibliography of English
Conjuring 1581‑1876. 2 vols., published by the author, Derby,
1976, 1978; distributed by Harpur & Sons, Derby. 1414 entries. References
are to the item number.
TP Title Page.
Tropfke. 1980. Johannes
Tropfke, revised by Kurt Vogel, Karin Reich and Helmuth Gericke. Geschichte der Elementarmathematik. 4th ed., Vol. 1: Arithmetik und Algebra. De Gruyter, Berlin, 1980. (Prof. Folkerts says (1994) that vol. 2 is
being edited.)
[The
1st ed. was De Gruyter?, Leipzig, 1902, 2 vols. 2nd ed., De Gruyter, Berlin & Leipzig, 1921-1924, 7
vols. 3rd ed., De Gruyter, Berlin &
Leipzig, 1930-1940, vols. 1-4 (the MSS of the remaining volumes were destroyed
in 1945).]
Tunstall, Cuthbert -- see: Tonstall, Cuthbert.
Turner. The Turner Collection, formerly at
University of Keele. Sadly this
collection was secretly sold by the University in 1998 and has now been
dispersed. A useful catalogue was prepared: Susan Hill; Catalogue of the Turner
Collection of the History of Mathematics Held in the Library of the University
of Keele; University Library, Keele, 1982.
UCL. University College London or its Library,
which includes the Graves Collection, cf Graves.
Uncle George. See:
Parlour Pastime.
Unger. Arithmetische Unterhaltungen.
1832.
Ephraim
Salomon Unger. Arithmetische
Unterhaltungen, bestehend in einer systematisch geordneten Sammlung von mehr
als 900 algebraischen Aufgaben, verbunden mit einer Anleitung, diese Aufgaben
mittelst der einfachsten Regeln der Arithmetick zu lösen. (Erfurt, 1832, 4 + 253 pp., ??NYS); 2nd ed., Keysersche Buchhandlung, Erfurt,
1838, 10 + 268pp. MUS 166 only mentions
the 1838 ed. but a copy of the 1st ed. was advertised by Sändig in Jun 1997. I haven't seen any other reference to this
work. In general, this book believes in
beating problems to death -- each type of problem is done several times. E.g. there are 10 problems of the Chinese
Remainder Theorem with two moduli.
Hence I will generally not describe all the problems.
van Etten. See:
Etten, above.
Van Egmond's Catalog.
Warren
Van Egmond. Practical Mathematics in
the Italian Renaissance: A Catalog of Italian Abbacus Manuscripts and Printed
Books to 1600. Supp. to Annali dell'Istituto
e Museo di Storia della Scienza (1980), fasc. 1. = Istituto e Museo di Storia della Scienza, Monografia N. 4. Florence, 1980. I have consulted this for (almost?) all MSS cited in these
Sources and have made a number of changes of dates and even authorship based on
it. Like any such catalog, there are
some omissions and errors, but it is by far the most authoritative listing of
the material and I have adopted his dates and attributions -- cf Benedetto da
Firenze, P. M. Calandri, dell'Abbaco, pseudo-dell'Abbaco.
Vinot. 1860.
Joseph
Vinot. Récréations Mathématiques Nouveau Recueil de Questions Curieuses et
Utiles Extraites des Auteurs Anciens et Modernes. Larousse & Boyer, Paris, 1860, (3rd ed., Larousse, Paris,
1893; 1898; 1902; 6th ed., nd
[1911]). I have found no difference
between the 1st and 6th eds -- indeed I have found a simple typographical error
repeated.
Vogel, Kurt (1888-1985). See:
AR; BR; Chiu Chang Suan Ching; Columbia Algorism; Tropfke; items in 7.K
under al‑Khwârizmî; items in
7.P.7 and 7.R.2 under Fibonacci.
Vyse. Tutor's Guide. 1771?
Charles
Vyse (fl. 1770-1815). The Tutor's
Guide, being a Complete System of Arithmetic; with Various Branches in the
Mathematics. In Six Parts, .... To which is added, An Appendix, Containing
Different Forms of Acquittances, Bills of Exchange, &c. &c. The whole being designed for the Use of
Schools, .... The Eighth Edition,
corrected and improved, with Additions.
G. G. J. and J. Robinson, London, 1793, HB. 12 + 324 pp. [1st ed. was
1770 or 1771; 2nd ed., 1772; 4th ed., 1779; 6th ed., 1785; 1790; 8th ed., 1793; 1799; 12th ed., 1804; 13th ed., 1807; 14th ed., 1810; 15th ed.,
1815; 1817; 16th ed., 1821.] Since
books of this nature rarely had major changes, I will date this as 1771? until
I see further editions.
Charles
Vyse (fl. 1770-1815). The Tutor's
Guide, being a Complete System of Arithmetic; with Various Branches in the
Mathematics. In Six Parts, .... To which is added, An Appendix, Containing
Different Forms of Acquittances, Bills of Exchange, &c. &c. The whole being designed for the Use of
Schools, .... 10th ed., ed. by J.
Warburton. S. Hamilton for G. G. and J.
Robinson, London, 1799. 12 + 335
pp. [Dedication removed; Preface by the
Editor and references to the Key added.
Though the text is reset with one or two more lines per page, the text
seems to be preserved, though the editor has added substantial footnotes in
places.]
The
Key to the Tutor's Guide; or the Arithmetician's Repository: containing
Solutions to the Questions, &c. in the Tutor's Guide, with references to
the pages where they stand. To which
are added Some Useful Rules, &c.
Likewise An Appendix; showing the Combination of Quantities; The
different Ways they may be varied; with the method of filling the magic
squares, &c. .... Eighth edition; Carefully revised,
corrected, and augmented. G. & J.
Robinson, London, 1802. 8 + 370 pp +
2pp publisher's ads. [1st ed.,
1773; 3rd ed., 1779; 4th ed., 1785; 7th ed., 1799; 8th ed.,
1802; 9th ed., 1807; 11th ed., 1818.]
Despite
the claim in the title of the Key, the references to the answers are in the
text of the 10th ed at the beginning of each set of exercises. There is no mention of the Key in the 8th ed
of the book. I have the 8th and 9th
eds. of the Key and have not seen any difference between them -- indeed I have
found a common misprint -- and they give the answers to the 10th ed. of the
book, on the pages cited in the book, so I assume they are essentially
identical to the 7th ed. of the Key.
V&T, 1952. Volkmann, Kurt &
Tummers, Louis. Bibliographie de
la Prestidigitation Tome I Allemagne et Autriche. Cercle Belge d'Illusionnisme, Bruxelles,
1952. With a 2 page list of
Bibliographies of Conjuring.
Walkingame. Tutor's Assistant.
Francis
Walkingame. The Tutor's Assistant;
Being a Compendium of Arithmetic, and a Complete Question-Book. ...
To which are added, A new and very short Method of extracting the Cube
Root, and a General Table for the ready calculating the Interest .... The Fifteenth Edition. Printed for the Author, London (Great
Russel-Street, Bloomsbury), 1777.
[First ed. was 1751. There were
about a hundred editions in all. See
Wallis's article on this book in MG 47 (No. 361) (1963) 199-208.]
The
Tutor's Assistant. Being a Compendium
of Arithmetic, and a Complete Question-Book.
... To which are added, A new
and very short Method of extracting the Cube Root, and a General Table for the
ready calculating the Interest .... The
Twentieth Edition. Printed for the
Author, London (Kensington), 1784.
Appears to be identical to the 15th ed., except for resetting and some
small changes, both corrections and errors, so I won't cite this separately.
The
Tutor's Assistant, being a Compendium of Arithmetic, and Complete
Question-Book; ...; to which is added, A Compendium of Book-keeping, by Single
Entry, by Isaac Fisher. Thomas
Richardson, Derby; Simpkin, Marshall & Co., London, 1835. [The first Derby edition by Fisher was
1826.]
The
Tutor's Companion; or, Complete Practical Arithmetic. To which is added A Complete Course of Mental Arithmetic, ..., by
Isaac Butler. Webb, Millington, and
Co., London, 1860.
The
editions are pretty similar, but the interesting collection of problems at the
end is much shortened in the 1835 ed. and a few are omitted in the 1860
ed. Consequently I will assume the
problems date from 1751, unless they vary in some important way. I also have a Key to Walkingame from 1840,
but it does not correlate with any of the editions I have!
Wallis. The Wallis Collection of early English
mathematics books. Gathered by Peter J.
Wallis and left to the University of Newcastle upon Tyne in 1992. A typical catalogue entry is Wallis 227 CAR and there are special sections for Newtoniana and Record(e).
Week‑End. 1932. Hubert
Phillips. The Week‑End Problems
Book. Nonesuch Press, London, 1932.
Wehman. New Book of 200 Puzzles. 1908.
Wehman
Bros.' New Book of 200 Puzzles. Wehman Bros., 126 Park Row, NY, 1908. Largely copied from various 19C works: Boy's Own Book, Magician's
Own Book, The Sociable, sometimes with typographical omissions. I only count 130 puzzles! There seems to have been a Johnson Smith
reprint at some time.
Wells. 1698. Edward Wells. Elementa Arithmeticæ Numerosæ et
Speciosæ. In Usum Juventutis
Academicæ. At the Sheldonian Theatre
(i.e. OUP), Oxford, 1698. (My copy was
previously in the Turner Collection, Turner D1.1.) All the material cited is in Appendix Posterior: Viz. Problemata
sive Quæstiones ad exercendas Regulas Arithmeticæ.
Western Puzzle Works. 1926.
Western
Puzzle Works, 979 Marshall Avenue, St. Paul, Minnesota. 1926 Puzzle Catalogue. Photocopy provided by Slocum. Unpaginated, 8pp.
Williams. Home Entertainments. 1914.
Archibald
& F. M. Williams. Home
Entertainments. The Hobby Books, ed. by
Archibald Williams. Nelson, London, nd
[1914 -- BMC].
Williams, Henry Llewellyn, Jr.
(1842-??). Books by Frikell?,
particularly Magician's Own Book, are often attributed to him. [C&B, under Williams, Henry Llewellyn ("W. Frikell") lists:
Hanky Panky; Magician's Own
Book, London & New York; (Magic No
Mystery); The Secret Out and says to also see Cremer.]
Williams, J. L. See:
Boy's Own Book, 1843 (Paris) edition, which lists him as author.
Wilson, Robin J. See:
BLW.
Wingate/Kersey. 1678?.
Edmund
Wingate (1596-1656). Mr. Wingate's
Arithmetick; Containing A Plain and Familiar Method For Attaining the Knowledge
and Practice of Common Arithmetick. (1629.) The Seventh(?) Edition, very much
Enlarged. First Composed by Edmund
Wingate, late of Gray's-Inn, Esquire.
Afterwards, upon Mr. Wingate's Request, Enlarged in his Life-time: Also
since his Decease carefully Revised, and much Improved; ... By John Kersey,
late Teacher of the Mathematicks. My
copy is lacking the TP and pp. 345+, but it appears to be identical to The
Tenth Edition; J. Philips, J. Taylor & J. Knapton, London, 1699 --
though reset, it has the same pagination throughout except for the Dedication. A Librarian's note suggests my earlier
version is the 7th ed. of 1678. This
would be the fourth and last of Kersey's versions. Kersey began editing from the second or third (1658) ed. and did
four versions, the last in 1678.
The
material of interest is almost all in Chapter 10 of Kersey's Appendix: A
Collection of subtil Questions to exercise all the parts of Vulgar Arithmetick;
to which also are added various practical Questions about the mensuration of
Superficial Figures and Solids, with the Gauging of Vessels, pp. 475-527, 75
questions. There are a few further
items in Chapter 11: Sports and Pastimes, pp. 528-544, 7 problems. Chapter 11 is not clearly marked as being by
Kersey in these copies, but is so marked in later editions and it is pretty
clear that the entire Appendix is due to Kersey. At the end, he says he has taken the problems of Chap. 11 from
Bachet's Problemes.
I
have seen a 14th ed. of 1720 which has the same text, reset and repaginated, with
some supplementary material by George Shelley.
I have also seen a 19th ed. of 1760 which has been considerably
reorganized by James Dodson. The two
chapters of puzzle problems have become Chapters XLIII and XLIV and the
material has been changed, generally omitting some problems of interest and
only adding two.
Winning Ways. 1982.
Elwyn
R. Berlekamp, John H. Conway & Richard K. Guy. Winning Ways for Your Mathematical Plays: Vol. 1: Games in
General; Vol. 2: Games in Particular. 2
vols, Academic Press, NY, 1982.
Witgeest. Het Natuurlyk Tover-Boek. 1686.
Simon
Witgeest. Het Natuurlyk Tover-Boek,
Of't Nieuw Speel-Toneel Der Konsten.
Verhandelende over de agt hondert natuurlijke Tover-Konsten. so uyt de
de Gogel-tas, als Kaartspelen, Mathematische Konsten, en meer andered
diergelijke aerdigheden, die tot vermaek, en tijtkorting verstrecken. Mitsgaders een Tractaet van alderley
Waterverwen, en verligteryen; Als oock
Een verhandelinge van veelderley Blanketsels.
Om verscheyde wel-ruykende Wateren, Poederen en Balsemen, als ook
kostelijke beeydselen, om het Aensicht, Hals en Handen, wit en sagt te maecken,
door Simon Witgeest. Jan ten Hoorn,
Amsterdam, 1686. ??NYS -- some
photocopies sent by Jerry Slocum.
'Boek' is 'Boeck' on the frontispiece and running heads. This is a much expanded and retitled 3rd
edition of Witgeest's 1679 work. The
new material is stated to already be in the 2nd ed. of 1682.
[There
were many later editions: 1695; 1698;
Ten Hoorn, 1701; G. de Groot Keur, Amsterdam, 1725 (10th ed.),
1739; 1749; 1760; Amsterdam,
1773; 1815 [Christopher 1098-1099, C&B, HPL]. It was translated as: Naturliches Zauber-Buch (or Zauber=Buch)
oder neuer Spiel-Platz der Künste; Hoffmanns sel Wittw. & Engelbert Streck,
Nürnberg, 1702. There were later
editions (all in Nürnberg?) of 1713 (in 1 or 2 vols); Hoffmann, Nürnberg, 1718;
1730; 1739; 1740; 1745; 1753; Johann Adam Stein, Nürnberg, 1755; 1760-1762;
1763; 1766; 1781; 1786; 1798 and a Lindau reprint of 1978 [Christopher 1092-1097, C&B, V&T],
all apparently based on the 1682 Dutch ed.]
Witgeest. Het Nieuw Toneel der Konsten. 1679.
Simon
Witgeest. Het Nieuw Toneel der Konsten,
Bestaande uyt Sesderley Stukken: het eerste, handelt van alderley aardige
Speeltjes en Klugjes: het tweede, van de Verligt-konst in 't Verwen en
Schilderen: het derde, van het Etzen en Plaat-shijden: het vierde, van de
Glas-konst: het vijfde, heest eenige aardige remedien tegen alderley Ziekten:
het sesde, is van de Vuur-werken. Uyt
verscheyde Autheuren by een vergadert, door S. Witgeest, Middel-borger. Jan ten Hoorn, Amsterdam, 1679; facsimile with epilogue by John Landwehr, A.
W. Sijthoff's Uitgeversmaatschappij N. V., Leiden, 1967 (present from Bill
Kalush).
There
were many later editions, but Nanco Bordewijk has examined these and discovered
that the 3rd ed. of 1686 (I can't recall if he saw the 1682 ed.) was so
extensively revised and extended as to constitute a new book, and it has the
different title given in the previous entry.
(Other sources indicate these revisions are already in the 2nd ed. of
1682.) Landwehr has written a
bibliographical article on this book -- ??NYR.
Wood. Oddities. Clement
Wood. A Book of Mathematical
Oddities. Little Blue Book 1210. Haldeman-Julius, Girard, Kansas, nd [1927].
Young Man's Book. 1839.
Anonymous. The Young Man's Book of Amusement. Containing the Most Interesting and
Instructive Experiments in Various Branches of Science. To Which is Added All the Popular Tricks and
Changes in Cards; and the Art of Making Fire Works. William Milner, Halifax, 1844, HB. 2 + 384 pp + folding plate (originally a frontispiece). My copy has a number of annotations as
though in preparation for another edition.
[Hall BCB 322. Toole Stott
751. BCB 320-323 are 1839, 1840, 1844, 1848. Heyl 358-360 are 1846, 1846 (Milner & Sowerby ??), 1850. Toole Stott 749-752, 1216 are 1839, 1840, 1844, 1846, 1850. Christopher 1111-1113 are 1839, 1846, 1859
(Milner & Sowerby). All these are
apparently the same except for the publisher's name change.]
Young World. c1960.
Young
World Productions. Tricks and
Teasers. 303 Gags Games Tongue Twisters Problems Tricks. Young World Productions, London, nd
[inscribed 1965 on first page, so probably c1960; BLC-Ø].
536. H. E. Dudeney. 536 Puzzles and Curious Problems. Ed. by M. Gardner. Scribner's, NY, 1967.
(This consists of almost all the puzzles from Modern Puzzles (MP) and
Puzzles and Curious Problems (PCP).)
[There is also a Fontana, London, 1970, edition in two volumes: Puzzles
and Curious Problems (258 problems); More Puzzles and Curious Problems (261
problems).]
?? indicates uncertainty and points
where further work needs to be done.
‑ BC, e.g. ‑330 is 330 BC
and ‑5C is
5th century BC.
¹ Inequality or
incongruence (mod m). (My word
processor does not have an incongruence sign.
I may change this in Word using an Arial character.)
Ø Nothing, used after catalogues,
etc., to indicate that I have looked in that catalogue and found no entry. E.g.
BLC-Ø.
SOME OTHER RECURRING REFERENCES
Details
of these items are given under the first reference in Sources. Later references often cite the first
reference. I tend to make entries below
when I use the item, but sometimes I have entered the entry long after the
first usage of the item, and I haven't searched the text for other (perhaps
entered long ago) appearances of these items.
For: See Sections:
Anon: Home Book ..., 1941 4.B.1, 4.B.3, 5.U, 6.AO, 6.BF.4,
7.B, 7.AT, 9.E.1
Anon: Treatise, 1850 7.H, 7.P.6, 7.S, 7.X,
10.A, 10.R
Allen, 1991 5.B,
5.I.1, 5.N, 6.L, 7.E, 7.I, 7.P.1, 7.R.3, 7.AC.6, 7.AH, 7.AL, 9.B, 10.A.4
Always: More Puzzles to Puzzle
You, 1967
6.BF.4,
Always: Puzzles for Puzzlers,
1971
5.D.2,
5.D.5, 7.G.1, 7.AC.1, 9.J, 10.I, 10.K
Always: Puzzles to Puzzle You,
1965
5.K.2,
5.W.1, 5.X.2, 7.AC.3, 7.AC.6, 7.AS,
Always: Puzzling You Again, 1969
5.C,
6.BD, 7.AH,
Ananias of Shirak, c640 7.E, 7.H, 10.A
André, 1876 7.H,
7.H.1, 7.S.1, 7.S.2, 7.AF, 7.AF.2
August, 1939 5.X.1,
6.BE, 7.I, 7.X, 7.Z, 7.AL, 7.AN, 7.AT, 7.AV, 10.H
Badcock, 1823 6.BH, 7.H.3,
7.P.5, 7.Q
Bagley: Paradox Pie, 1944 6.BN, 7.Z, 7.AI, 7.AW, 10.F,
10.Q, 10.S
Bagley: Puzzle Pie, 1944 5.D.5, 6.O, 6.P.1, 6.P.2,
6.R.1, 6.R.2, 6.Y, 6.AF, 6.AI, 7.AV, 10.L
Bath, 1959 5.C,
7.G.1, 7.I, 7.P.5, 7.AC.3, 7.AC.6, 7.AM
Bellew, 1866 5.E,
6.AO.1, 6.AQ
Berloquin, 1981 5.N, 7.H.5,
7.N.3, 10.R
Black, 1952 [1946?] 5.T, 6.F.2, 9.D, 9.F
Bourdon, 1834 7.E.1, 7.H,
7.P.1, 7.S, 7.X, 7.AF.1, 7.AK, 10.A, 10.R
Brandreth Puzzle Book [1895] 5.B, 5.B.1, 5.O, 6.AW.1, 6.AY.1,
7.B, 7.G.1
Bullen, 1789 7.G.1,
7.H, 7.H.5, 7.L.2.b, 7.S.1, 7.AF.1
Bullivant, 1910 5.S, 6.T, 6.AK
[Chambers], 1866? 7.H, 7.L.2.a,
7.L.2.b, 7.Y, 7.AF.2
Chang Chhiu‑Chien -- see
Zhang Qiujian
Colenso, Algebra, 1849 7.P.1, 10.G
Colenso, Arithmetic, 1853 7.H, 7.X, 10.G, 10.R
Devi, 1976 5.D.1,
5.X.1, 7.E, 7.P.1, 7.AC.3, 7.AC.6, 7.AE, 10.A.3; 10.K
Dresner, 1962 5.B.1, 5.C,
5.D.4, 5.K.1, 5.W
Dudeney: World's best puzzles,
1908
2,
5.P.1, 5.S, 6.P.1, 6.S, 6.AI, 6.AO, 6.AW
Elliott, 1872 6.V, 6.AQ,
6.AV, 6.AZ, 11.B, 11.C, 11.D
Filipiak, 1942 5.H.1,
6.W.1, 6.W.2, 6.AK
Fisher, 1968 6.E,
6.P.2, 7.M.4, 7.AI
Fisher, 1973 1, 5.E,
7.S.2, 10.L
Fourrey, Cur. Geom., 1907 6.S.1, 8.G (also 6.R.1)
Fourrey, Rec. Arith., 1899 4.A.1, 5.B, 5.P.1, 5.U, 7.N.1
Fuss, 1843 5.F.1
Goldston, nd [1910?] 6.AK, 11.E
Gomme, 1894/98 4.B.1, 5.R.5 (Also 4.A.3)
Greenblatt, 1965 6.U.2, 6.AE,
7.AG
Heald, 1941 7.Z,
10.E.3, 10.G, 10.0
Hooper, 1774 4.A.1, 5.AA,
6.F, 6.P.2, 7.B, 7.AO, 7.AZ
Hutton, 1804 7.G.2, 7.H,
7.X, 7.AF.1, 7.AK, 10.A, 10.R
Kamp, 1877 5.B,
5.D.1, 5.E, 5.R.7, 7.B, 7.L, 7.Q
Kraitchik, MJ, 1930 4.A.2, 5.J, 7.E,
7.G.1, 7.H.3, 7.AR, 10.B, 10.P
Kraitchik, MR, 1943 4.A.2, 5.J, 6.M,
7.H.2, 7.H.3, 9.D, 9.G, 10.P
Laisant, 1906 6.P.1, 7.AR,
10.A.2, 10.B, 10.H, 10.I
Larte de labbacho, 1478 See: Treviso Arithmetic.
Von der Lasa, 1897 5.F.1, 7.B (Also 4.B.1, 4.B.5)
Licks, 1917 5.A,
6.R.4, 6.AG, 7.P.3, 7.S.2, 7.AC, 7.AD
Van der Linde, 1874 5.F.1, 7.N (Also 4.B.1, 4.B.5)
Littlewood, 1953 5.C, 5.W, 6.J,
8.B, 9.C, 9.D
Lucas, Théorie, 1891 5.L, 5.Z.5, 5.AB
Madachy, 1966 5.O, 6.D, 6.X,
7.N.3, 7.N.4, 7.AC.3, 7.BB
Meyer, 1965 7.I,
7.AC.4, 7.AH, 7.AR, 7.AX
Milne, 1881 7.E, 7.H,
7.R, 7.X, 7.Y, 10.A, 10.A.3, 10.G, 10.R
W. O. J. Moser, 1981 6.I, 6.T
Nordmann, 1927 4.A.4, 5.G.1,
6.AR, 7.AC.3, 7.AR, 11.C, 11.E
Papyrus Rhind, c‑1650 7.C, 7.G.1, 7.L, 7.S.1
Phillips: Playtime Omnibus, 1933 6.AF, 7.S.2, 7.AC.1, 7.AD.1, 7.AE, 9.D
Phillips: Question Time, 1937 5.U, 7.E, 7.AG, 9.G
Ransom, 1955 6.M, 7.F,
7.X, 7.AC.2, 8.B, 10.A.1, 10.B, 10.I
Smith: Origin, 1917 3A, 7.G.2, 7.H, 10.A
Steinhaus: 1938,1950,1960,1969 5.C.1, 6.E, 6.G.1, 6.H, 6.AB
Strutt, 1791?, 1801 4.B.1, 5.R.1,
5.R.5 (Also 4.A.3)
Strutt-Cox, 1903 4.B.1, 5.R.1,
5.R.5 (Also 4.A.3)
Trenchant, 1566 7.L.2.a, 7.S.1
(also 5.B, 5.D.1, 7.E, 7.S.1, 7.AF.1)
Treviso Arithmetic, 1478 7.H, 7.K.1, 7.AL, 10.A
Trigg: Quickies, 1967 5.Q, 6.AE, 6.AN, 7.N.3,
7.W
Wagner, Rechenbuch, 1483 7.G.1, 7.G.2, 7.H, 7.AK, 10.A
Wecker, (1582), 1660 7.L.3, 7.AO, 10.P, 11.I,
11.N
A. C. White, 1913 1, 5.I.1, 6.T,
6.AK, 7.X, 11.E
Widman(n), 1489 7.G.1, 7.H, 7.L.2,
7.P.1, 7.P.5, (7.AL)
Williams & Savage, 1940 7.P.5, 7.X, 7.AC.2, 7.AM,
7.AP, 8.I, 10.E.2
Wolff, 1937 7.R.3,
7.S.2, 7.AC.1, 7.AE, 9.E, 9.E.1, 10.O
Workman, 1902 7.H.1, 7.H.4,
7.J, 7.S.2, 7.AC.2, 10.G
Wyatt, 1928 5.H.1,
6.V, 6.W.1, 6.W.2, 6.AI
Mr. X, 1903, 1911 4.A.1, 5.B, 5.P.1,
5.S, 6.AF, 6.AU, 7.H.3, 7.I, 7.J, 7.M.4, 9.E, 9.J, 10.H
Yang Hui, 1275 7.N, 7.P.1,
7.P.2, 10.A
Zhang Qiujian, 468 7.E, 7.L, 7.P.1, 7.P.6, 10.A
1. BIOGRAPHICAL MATERIAL -- in chronological order
ALCUIN (c735‑804)
Phillip Drennon Thomas. Alcuin of York. DSB I, 104‑105.
Robert Adamson. Alcuin, or Albinus. DNB, (I, 239‑240), 20.
Andrew Fleming West. Alcuin and the Rise of the Christian
Schools. (The Great Educators --
III.) Heinemann, 1893. The only book on Alcuin that I found which
deals with the Propositiones.
Stephen Allott. Alcuin of York c. A.D. 732 to 804
-- his life and letters. William Sessions, York, 1974.
FIBONACCI
[LEONARDO PISANO] (c1170->1240)
See
also the entries for Fibonacci in Common References.
Fibonacci. (1202 -- first paragraph); 1228 -- second paragraph, on p. 1. In this paragraph he narrates almost
everything we know about him. [In the
second ed., he inserted a dedication as the first paragraph.]
The
paragraph ends with the notable sentence which I have used as a motto for this
work. "Si quid forte minus aut
plus iusto vel necessario intermisi, mihi deprecor indulgeatur, cum nemo sit
qui vitio careat et in omnibus undique sit circumspectus." (If I have perchance omitted anything more
or less proper or necessary, I beg indulgence, since there is no one who is
blameless and utterly provident in all things.
[Grimm's translation.])
Richard E. Grimm. The autobiography of Leonardo Pisano. Fibonacci Quarterly 11:1 (Feb 1973)
99-104. He has collated six MSS of the
autobiographical paragraph and presents his critical version of it, with
English translation and notes. Sigler,
below, gives another translation. I
give Grimm's translation, omitting his notes.
After
my father's appointment by his homeland as state official in the customs house
of Bugia for the Pisan merchants who thronged to it, he took charge; and, in
view of its future usefulness and convenience, had me in my boyhood come to him
and there wanted me to devote myself to and be instructed in the study of
calculation for some days. There,
following my introduction, as a consequence of marvelous instruction in the
art, to the nine digits of the Hindus, the knowledge of the art very much
appealed to me before all others, and for it I realized that all its aspects
were studied in Egypt, Syria, Greece, Sicily, and Provence, with their varying
methods; and at these places thereafter, while on business, I pursued my study
in depth and learned the give-and-take of disputation. But all this even, and the algorism, as well
as the art of Pythagoras I considered as almost a mistake in respect to the
method of the Hindus. Therefore,
embracing more stringently that method of the Hindus, and taking stricter pains
in its study, while adding certain things from my own understanding and
inserting also certain things from the niceties of Euclid's geometric art, I
have striven to compose this book in its entirety as understandably as I could,
dividing it into fifteen chapters.
Almost everything which I have introduced I have displayed with exact
proof, in order that those further seeking this knowledge, with its pre-eminent
method, might be instructed, and further, in order that the Latin people might
not be discovered to be without it, as they have been up to now. If I have perchance omitted anything more or
less proper or necessary, I beg indulgence, since there is no one who is
blameless and utterly provident in all things.
F. Bonaini. Memoria unica sincrona di Leonardo Fibonacci
novamente scoperta. Giornale Storico
degli Archivi Toscani 1:4 (Oct-Dec 1857) 239-246. This reports the discovery of a 1241 memorial of the Comune of
Pisa, which I reproduce as it is not well known. This grants Leonardo an annual honorarium of 20 pounds. In 1867, a plaque bearing this inscription
and an appropriate heading was placed in the atrium of the Archivio di Stato in
Pisa.
"Considerantes
nostre civitatis et civium honorem atque profectum, qui eis, tam per doctrinam
quam per sedula obsequia discreti et sapientis viri magistri Leonardi Bigolli,
in abbacandis estimationibus et rationibus civitatis eiusque officialium et
aliis quoties expedit, conferunter; ut eidem Leonardo, merito dilectionis et
gratie, atque scientie sue prerogativa, in recompensationem laboris sui quem
substinet in audiendis et consolidandis estimationibus et rationibus
supradictis, a Comuni et camerariis publicis, de Comuni et pro Comuni, mercede
sive salario suo, annis singulis, libre xx denariorum et amisceria consueta
dari debeant (ipseque pisano Comuni et eius officialibus in abbacatione de cetero
more solito serviat), presenti constitutione firmamus."
A
translation follows, but it can probably be improved. My thanks to Steph Maury Gannon for many improvements over my
initial version.
Considering
the honour and progress of our city and its citizens that is brought to them
through both the knowledge and the diligent application of the discreet and
wise Maestro Leonardo Bigallo in the art of calculation for valuations and
accounts for the city and its officials and others, as often as necessary; we
declare by this present decree that there shall be given to the same Leonardo,
from the Comune and on behalf of the Comune, by reason of affection and
gratitude, and for his excellence in science, in recompense for the labour
which he has done in auditing and consolidating the above mentioned valuations
and accounts for the Comune and the public bodies, as his wages or salary, 20
pounds in money each year and his usual fees (the same Pisano shall continue to
render his usual services to the Comune and its officials in the art of
calculation etc.).
Bonaini
also quotes a 1506 reference to Lionardo Fibonacci.
Mario Lazzarini. Leonardo Fibonacci Le sue Opere e la sua Famiglia.
Bolletino di Bibliografia e Storia delle Scienze Matematiche 6 (1903) 98‑102 &
7 (1904) 1-7. Traces the family
to late 11C, saying Leonardo's father was Guglielmo and his grandfather was
probably Bonaccio. He estimates the
birth date as c1170. He describes a
contract of 28 Aug 1226 in which Leonardo Bigollo, his father, Guglielmo, and
his brother, Bonaccingo, buy a piece of land from a relative. This land included a tower and other
buildings, outside the city, near S. Pietro in Vincoli. [G. Milanesi; Documento inedito intorno a
Leonardo Fibonacci; Rome, 1867 -- ??NYS].
Says nothing is known of the 1202 ed of Liber Abbaci. Quotes the above memorial.
R. B. McClenon. Leonardo of Pisa and his liber
quadratorum. AMM 26:1 (Jan 1919) 1-8.
Gino Loria. Leonardo Fibonacci. Gli Scienziati Italiana dall'inizio del
medio evo ai nostri giorni. Ed. by Aldo
Mieli. (Dott. Attilio Nardecchia
Editore, Rome, 1921;) Casa Editrice
Leonardo da Vinci, Rome, 1923. Vol. 1,
pp. 4-12. This reproduces much of the
material in Lazzarini and the opening biographical paragraph of Liber
Abaci.
Ettore Bortolotti. Article on Fibonacci in: Enciclopedia Italiana. G. Treccani, Rome, 1949 (reprint of 1932
ed.).
Charles King. Leonardo Fibonacci. Fibonacci Quarterly 1:4 (Dec 1963) 15-19.
Gino Arrighi, ed. Leonardo Fibonacci: La Practica di Geometria
-- Volgarizzata da Cristofano di Gherardo di Dino, cittadino pisano. Dal Codice 2186 della Biblioteca Riccardiana
di Firenze. Domus Galilaeana, Pisa, 1966. The Frontispiece is the mythical portrait of
Fibonacci, taken from I Benefattori dell'Umanità, vol. VI; Ducci, Florence,
1850. (Smith, History II 214 says it is
a "Modern engraving. The portrait
is not based on authentic sources".)
P. 15 shows the plaque erected in the Archivio di Stato di Pisa in 1855
which reproduces the above memorial with an appropriate heading, but Arrighi
has no discussion of it. P. 19 is a
photo of the statue in Pisa and p. 16 describes its commissioning in 1859.
Joseph and Francis Gies. Leonard of Pisa and the New Mathematics of
the Middle Ages. Crowell, NY,
1969. This is a book for school
students and contains a number of dubious statements and several false
statements.
Kurt Vogel. Fibonacci, Leonardo, or Leonardo of
Pisa. DSB IV, 604-613.
A. F. Horadam. Eight hundred years young. Australian Mathematics Teacher 31 (1975) 123‑134. Good survey of Fibonacci's life &
work. Gives English of a few
problems. This is available on
Kimberling's website - see below.
Ettore Picutti. Leonardo Pisano. Le Scienze 164 (Apr 1982) ??NYS.
= Le Scienze, Quaderni; 1984, pp. 30-39. (Le Scienze is a magazine;
the Quaderni are collections of articles into books.) Mostly concerned with the Liber Quadratorum,
but surveys Fibonacci's life and work.
Says he was born around 1170.
Includes photo of the plaque in the Archivo di Stato di Pisa.
Leonardo Pisano Fibonacci. Liber quadratorum, 1225. Translated and edited by L. E. Sigler
as: The Book of Squares; Academic
Press, NY, 1987. Introduction: A brief
biography of Leonardo Pisano (Fibonacci) [1170 - post 1240], pp. xv-xx. This is the best recent biography,
summarizing Picutti's article. Says he
was born in 1170 and his father's name was Guilielmo -- cf Loria above. Gives another translation of the
biographical paragraph of the Liber Abbaci.
A. F. Horadam & J. Lahr. Letter to the Editor. Fibonacci Quarterly 28:1 (Feb 1990) 90. The authors volunteer to act as coordinators
for work on the life and work of Fibonacci.
Addresses: A. F. Horadam,
Mathematics etc., Univ. of New England, Armidale, New South Wales, 2351,
Australia; J. Lahr, 14 rue des Sept
Arpents, L‑1139 Luxembourg, Luxembourg.
Thomas Koshy. Fibonacci and Lucas Numbers with
Applications. Wiley-Interscience,
Wiley, 2001. Claims to be 'the first
attempt to compile a definitive history and authoritative analysis' of the
Fibonacci numbers, but the history is generally second-hand and marred with a
substantial number of errors, The
mathematical work is extensive, covering many topics not organised before, and
is better done, but there are more errors than one would like.
Laurence E. Sigler. Translation of Liber Abaci as: Fibonacci's Liber Abaci A Translation into Modern English of
Leonardo Pisano's Book of Calculation.
Springer, 2002.
Clark Kimberling's site web
includes biographical material on Fibonacci and other similar number
theorists.
http://cedar.evansville.edu/~ck6/bstud/fibo.html .
Ron Knott has a huge website on
Fibonacci numbers and their applications, with material on many related topics,
e.g. continued fractions, π, etc. with some history.
www.ee.surrey.ac.uk/personal/r.knott/fibonacci/fibnat.html .
Luca
PACIOLI (c1445-1517)
S. A. Jayawardene. Luca Pacioli. BDM 4, 1897-1900.
Bernardino Baldi (Catagallina)
(1553-1617). Vita di Pacioli. (1589, first published in his Cronica de Mathematici
of 1707.) Reprinted in: Bollettino di
bibliografia e di storia delle scienze matematiche e fisiche 12 (1879)
421-427. ??NYS -- cited by Taylor,
p. 338.
Enrico Narducci. Intorno a due edizioni della "Summa de
arithmetica" di Fra Luca Pacioli.
Rome, 1863. ??NYS -- cited by
Riccardi [Biblioteca Matematica Italiana, 1952]
D. Ivano Ricci. Luca Pacioli, l'uomo e lo scienziato. San Sepolcro, 1940. ??NYS -- cited in BDM.
R. Emmett Taylor. No Royal Road Luca Pacioli and His Times.
Univ. of North Carolina Press, Chapel Hill, 1942. BDM describes this as lively but unreliable.
Ettore Bortolotti. La Storia della Matematica nella Università
di Bologna. Nicola Zanichelli Editore,
Bologna, 1947. Chap. I, § V, pp. 27-33:
Luca Pacioli.
Margaret Daly Davis. Piero della Francesca's Mathematical
Treatises The "Trattato
d'abaco" and "Libellus de quinque corporibus regularibus". Longo Editore, Ravenna, 1977. This discusses Piero's reuse of his own
material and Pacioli's reuse of Piero's material.
Fenella K. C. Rankin. The Arithmetic and Algebra of Luca
Pacioli. PhD thesis, Univ. of London,
1992 (copy at the Warburg Institute), ??NYR.
Enrico Giusti, ed. Descriptive booklet accompanying the 1994
facsimile of the Summa -- qv in Common References.
Edward A. Fennell. Figures in Proportion: Art, Science and the
Business Renaissance. The contribution
of Luca Pacioli to culture and commerce in the High Renaissance. Catalogue for the exhibition, The Institute
of Chartered Accountants in England and Wales, London, 1994.
Claude-Gaspar
BACHET de Méziriac (1581‑1638)
C.‑G. Collet & J.
Itard. Un mathématicien humaniste --
Claude‑Gaspar Bachet de Méziriac (1581‑1638). Revue d'Histoire des Sciences et leurs
Applications 1 (1947) 26‑50.
J. Itard. Avant-propos. IN: Bachet; Problemes;
1959 reprint, pp. v‑viii. Based
on the previous article.
There is a Frontispiece portrait
in the reprint.
Underwood Dudley. The first recreational mathematics
book. JRM 3 (1970) 164‑169. On Bachet's Problemes.
William Schaaf. Bachet de Méziriac, Claude‑Gaspar. DSB I, 367‑368.
Jean
LEURECHON (c1591‑1670)
and Henrik VAN ETTEN
A. Deblaye. Étude sur la récréation mathématique du P.
Jean Leurechon, Jésuite. Mémoires de la
Société Philotechnique de Pont-à-Mousson 1 (1874) 171-183. [MUS #314.
Schaaf. Hall, OCB, pp. 86, 88
& 114, says the only known copy of this journal is at Harvard, which has
kindly supplied me with a photocopy of this article. Hall indicates the article is in vol. II and says it is 12 pages,
but only cites pp. 171 & 174.] This
simply assumes Leurechon is the author and gives a summary of his life. The essential content is described by Hall.
G. Eneström. Girard Desargues und D.A.L.G. Biblioteca Mathematica (3) 14 (1914) 253‑258. D.A.L.G. was an annotator of van Etten's
book in c1630. Although D.A.L.G. was
used by Mydorge on one of his other books, it had been conjectured that this
stood for Des Argues Lyonnais Girard
(or Géomètre). Eneström can find no
real evidence for this and feels that Mydorge is the most likely person.
Trevor H. Hall. Mathematicall Recreations. An Exercise in Seventeenth Century
Bibliography. Leeds Studies in
Bibliography and Textual Criticism, No. 1.
The Bibliography Room, School of English, University of Leeds, 1969,
38pp. Pp. 18‑38 discuss the
question of authorship and Hall feels that van Etten probably was the author
and that there is very little evidence for Leurechon being the author. Much of the mathematical content is in
Bachet's Problemes and may have been copied from it or some common source. [This booklet is reproduced as pp. 83-119 of
Hall, OCB, with the title page of the 1633 first English edition reproduced as
plate 5, opp. p. 112. Some changes have
been made in the form of references since OCB is a big book, but the only other
substantial change is that he spells the name of the dedicatee of the book as
Verreyken rather than Verreycken.]
William Schaaf. Leurechon, Jean. DSB VIII, 271‑272.
Jacques Voignier. Who was the author of "Recreation
Mathematique" (1624)? The
Perennial Mystics #9 (1991) 5-48 (& 1-2 which are the cover and its
reverse). [This journal is edited and
published by James Hagy, 2373 Arbeleda Lane, Northbrook, Illinois, 60062,
USA.] Presents some indirect evidence
for Leurechon's authorship.
Jacques
OZANAM (1640‑1717)
On the flyleaf of J. E.
Hofmann's copy of the 1696 edition of Ozanam's Recreations is a pencil portrait
labelled Ozanam -- the only one I know of.
This copy is at the Institut für Geschichte der Naturwissenschaft in
Munich. Hofmann published the picture
-- see below.
Charles Hutton. A Mathematical and Philosophical
Dictionary. 1795-1796. Vol. II, pp. 184-185. ??NYS
[Hall, OCB, p. 166.]
Charles Hutton. On the life and writings of Ozanam, the
first author of these Mathematical Recreations. Ozanam-Hutton. Vol.
I. 1803: xiii-xv; 1814: ix-xi.
William L. Schaaf. Jacques Ozanam on mathematics .... MTr 50 (1957) 385-389. Mostly based on Hutton. Includes a sketchy bibliography of Ozanam's
works, generally ignoring the Recreations.
Joseph Ehrenfried Hofmann. Leibniz und Ozanams Problem, drei Zahlen so
zu bestimmen, dass ihre Summe eine Quadratzahl und ihre Quadratsumme eine
Biquadratzahl ergibt. Studia Leibnitiana
1:2 (1969) 103-126. Outlines Ozanam's
life, gives a bibliography of his works and reproduces the above-mentioned
drawing as a plate opp. p. 124. (My
thanks to Menso Folkerts for this information and a copy of Hofmann's article.)
William L. Schaaf. Ozanam, Jacques. DSB X, 263‑265.
Jean
Étienne MONTUCLA (1725-1799)
Charles Hutton. Some account of the life and writings of
Montucla. Ozanam‑Hutton. Vol. I.
1803: viii-xii; 1814: iv-viii.
Charles Hutton. A Philosophical and Mathematical
Dictionary. 2nd ed. of the Dictionary
cited under Ozanam, 1815, Vol. II, pp.
63-64. ??NYS. According to Hall, OCB, p. 167, this is not in the 1795-1796 ed.
and is a reworking of the previous item.
Lewis
CARROLL (1832-1898)
Pseudonym
of Charles Lutwidge Dodgson. There is
so much written on Carroll that I will only give references to his specifically
recreational work and some basic references.
The Diaries of Lewis
Carroll. Edited by Roger Lancelyn
Green. (OUP, 1954); 2 vols, Greenwood Publishers, Westport,
Connecticut, 1971, HB.
Lewis Carroll's Diaries The private journals of Charles Lutwidge
Dodgson (Lewis Carroll) The first
complete version of the nine surviving volumes with notes and annotations by
Edward Wakeling. Introduction by Roger
Lancelyn-Green. The Lewis Carroll
Society, Publications Unit, Luton, Bedfordshire. [There were 13 journals, but 4 are lost.]
Vol.
1. Journal 2, Jan-Sep 1855. 1993, 158pp.
Vol.
2. Journal 4, Jan-Dec 1856. 1994, 158pp.
Vol.
3. Journal 5, Jan 1857 - Apr 1858. 1995, 199pp.
Vol.
4. Journal 8, May 1862 - Sep 1864 and a
reconstruction of the four missing
years,
1858-1862. 1997, 399pp.
Vol.
5. Journal 9, Sep 1864 - Jan 1868,
including the Russian Journal.
1999,
416pp.
Vol.
6. Journal 10, Apr 1868 - Dec
1876. 2001, 552pp.
Vol.
7. Journal 11, Jan 1877 - Jun
1883. 2003, 606pp.
The Letters of Lewis
Carroll. Edited by Morton N. Cohen with the assistance of Roger Lancelyn
Green. Volume One ca.1837 - 1885; Volume Two 1886 -
1898. Macmillan London, 1979.
Stuart Dodgson Collingwood. The Life and Letters of Lewis Carroll. T. Fisher Unwin, London, 1898.
Stuart Dodgson Collingwood,
ed. The Lewis Carroll Picture
Book. T. Fisher Unwin, London,
1899. = Diversions and
Digressions of Lewis Carroll, Dover, 1961.
= The Unknown Lewis Carroll, Dover, 1961(?). Reprint, in reduced format, Collins, c1910. The pagination of the main text is the same
in the 1899 and in both Dover reprints, but is quite different than the
Collins. Cited as: Carroll-Collingwood,
qv in Common References.
R. B. Braithwaite. Lewis Carroll as logician. MG 16 (No. 219) (Jul 1932) 174-178. He notes that Carroll assumed that a
universal statement implied the existence of an object satisfying the
antecedent, e.g. 'all unicorns are blue' would imply the existence of unicorns,
contrary to modern convention.
Derek Hudson. Lewis Carroll -- An Illustrated
Biography. Constable, 1954; new illustrated ed., 1976.
Warren Weaver. Lewis Carroll: Mathematician. SA 194:4 (Apr 1956) 116‑128. +
Letters and response. SA 194:6
(Jun 1956) 19-22.
Martin Gardner. The Annotated Alice. C. N. Potter, NY, 1960. Penguin, 1965; 2nd ed., 1971. Revised as: More Annotated Alice, 1990, qv.
Martin Gardner. The Annotated Snark. Bramhall House, 1962. Penguin, 1967; revised, 1973 & 1974.
John Fisher. The Magic of Lewis Carroll. Nelson, 1973. Penguin, 1975.
Morton N. Cohen, ed. The Selected Letters of Lewis Carroll. Papermac (Macmillan), 1982.
Martin Gardner. More Annotated Alice. [Extension of The Annotated Alice.] Random House, 1990.
Edward Wakeling. Lewis Carroll's Games and Puzzles. Dover and the Lewis Carroll Birthplace
Trust, 1992. Cited as Carroll-Wakeling,
qv in Common References.
Francine F. Abeles, ed. The Pamphlets of Lewis Carroll -- Vol. 2:
The Mathematical Pamphlets of Charles Lutwidge Dodgson and Related Pieces. Lewis Carroll Society of North America,
distributed by University Press of Virginia, Charlottesville, 1994.
Edward Wakeling. Rediscovered Lewis Carroll Puzzles. Dover, 1995. Cited as Carroll‑Wakeling II, qv in Common References.
Martin Gardner. The Universe in a Handkerchief. Lewis Carroll's Mathematical Recreations,
Games, Puzzles and Word Plays.
Copernicus (Springer, NY), 1996.
Cited as Carroll‑Gardner, qv in Common References.
Martin Gardner. The Annotated Alice: The Definitive
Edition. 1999. [A combined version of The Annotated Alice
and More Annotated Alice.]
Professor
Louis HOFFMANN (1839‑1919)
Pseudonym
of Angelo John Lewis.
Joseph Foster. Men-at-the-Bar: A biographical Hand-List of
the Members of the Various Inns of Court, including Her Majesty's Judges,
etc. 2nd ed, the author, 1885. P. 277 is the entry for Lewis. Born in London, eldest son of John
Lewis. Graduated from Wadham College,
Oxford. Entered Lincoln's Inn as a
student in 1858, called to the bar there in 1861. Married Mary Ann Avery in 1864.
Author of Manual of Indian Penal Code and Manual of Indian
Civil Procedure. Address: 12
Crescent Place, Mornington Crescent, London, NW. (My thanks to the Library of Lincoln's Inn for this information.)
Anonymous. Professor Hoffmann. Mahatma 4:1 (Jul 1900) 377-378. A brief note, with photograph, stating that
he is Mr. Angelo Lewis, M.A. and Barrister-at-Law.
Will Goldston. Will Goldston's Who's Who in Magic. My version is included in a compendium
called: Tricks that Mystify; Will Goldston, London, nd [1934-NUC]. Pp. 106-107. Says he was a barrister, retired to Hastings about 1903 and died
in 1917.
Who Was Who, 1916-1928, p.
627. This says he attended North London
Collegiate School and that he only practised law until 1876. He was on the staff of the Saturday Review
and a contributor to many journals. Won
the £100 prize offered by Youth's Companion (Boston) for best short story for
boys. Lists 36 books by him and 9 card
games he invented. Address:
Manningford, Upper Bolebrooke Road, Bexhill-on-Sea. (My thanks to the Library of Lincoln's Inn for this information.)
J. B. Findlay & Thomas A.
Sawyer. Professor Hoffmann: A
Study. Published by Thomas A. Sawyer,
Tustin, California, 1977. A short book,
12 + 67 pp, with two portraits (one from Mahatma) and 27pp of
bibliography. He was born at 3 Crescent
Place, Mornington Crescent, London. He
was a barrister and wrote two books on Indian law.
Charles Reynolds. Introduction -- to the reprint of
Hoffmann's Modern Magic, Dover, 1978, pp. v‑xiv. This says Lewis was a barrister, which is
mentioned in another reprint of a Hoffmann book and in S. H. Sharpe's
translation of Ponsin on Conjuring.
Edward Hordern. Foreword to this edition. In:
Hoffmann's Puzzles Old and New (see under Common References), 1988
reprint, pp. v‑vi. This says he
was the Reverend Lewis, but this is corrected in Hoffmann-Hordern to saying he
was a barrister.
Hoffmann-Hordern, p. viii, is a
version of the photograph in Mahatma.
Hall, OCB, p. 189, gives
Hoffmann's address as Ireton Lodge, Cromwell Ave., N. -- presumably the
Cromwell Ave. in Highgate.
Toole Stott 386 gives a little
information about Hoffmann and Modern Magic, including an address in Mornington
Crescent in 1877.
No DNB or DSB entry -- I have
suggested a DNB entry.
Sam
LOYD (1841‑1911) and
Sam LOYD JR. (1873‑1934)
[W. R. Henry.] Samuel Loyd. [Biography.] Dubuque
Chess Journal, No. 66 (Aug-Sep 1875) 361-365.
??NX -- o/o (11 Jul 91).
Loyd. US Design 4793 -- Design for Puzzle-Blocks. 11 April 1871. These are solid pieces, but unfortunately the drawing did not
come with this, so I am not clear what they are. ??Need drawing -- o/o (11 Jul 91).
Anonymous & Sam Loyd. Loyd's puzzles (Introductory column). Brooklyn Daily Eagle (22 Mar 1896) 23. Says he lives at 153 Halsey St., Brooklyn.
L. D. Broughton Jr. Samuel Loyd. [A Biography.] Lasker's
Chess Magazine 1:2 (Dec 1904) 83-85. About
his chess problems with a mention of some of his puzzles.
G. G. Bain. The prince of puzzle‑makers. An interview with Sam Loyd. Strand Magazine 34 (No. 204) (Dec 1907) 771‑777. Solutions of Sam Loyd's puzzles. Ibid. 35 (No. 205) (Jan 1908) 110.
Walter Prichard Eaton. My fifty years in puzzleland -- Sam Loyd and
his ten thousand brain‑teasers.
The Delineator (New York) (April 1911) 274 & 328. Drawn portrait of Loyd, age 69.
Anon. Puzzle inventor dead.
New-York Daily Tribune (12 Apr 1911) 7.
Says he died at his house, 153 Halsey St. "He declared no one had ever succeeded in solving [the
"Disappearing Chinaman"]."
Says he is survived by a son and two daughters (!! -- has anyone ever
tracked the daughters and their descendents??).
Anon. Sam Loyd, puzzle man, dies.
New York Times (12 Apr 1911) 13.
Says he was for some time editor of The Sanitary Engineer and a shrewd
operator on Wall Street.
Anon. Sam Loyd. SA (22 Apr
1911) 40-41?? Says he was for some
years chess editor of SA and was puzzle editor of Woman's Home Companion when
he died.
W. P. Eaton. Sam Loyd.
The American Magazine 72 (May 1911) 50, 51, 53. Abridged version of Eaton's earlier article. Photo of Loyd on p. 50.
P. J. Doyle. Letter to the Chess column. The Sunday Call [Newark, NJ] (21 May 1911),
section III, p. 10.
A. C. White. Sam Loyd and His Chess Problems. Whitehead and Miller, Leeds, UK, 1913; corrected, Dover, 1962.
Alain C. White. Supplement to Sam Loyd and His Chess
Problems. Good Companion Chess Problem
Club, Philadelphia, vol. I, nos. 11-12 (Aug 1914), 12pp. This is mostly corrections of the chess
problems, but adds a few family details with a picture of the Loyd Homestead
and Grist Mill in Moylan, Pennsylvania.
Alain C. White. Reminiscences of Sam Loyd's family. The Problem [Pittsburgh]
(28 Mar 1914) 2, 3, 6, 7.
Louis C. Karpinski. Loyd, Samuel. Dictionary of American Biography, Scribner's, NY, vol. XI,
1933, pp. 479‑480.
Loyd Jr. SLAHP.
1928. Preface gives some details
of his life, making little mention of his father, "who was a famous
mathematician and chess player".
He claims to have created over 10,000 puzzles. There are some vague
biographical details on pp. 1‑22, e.g. 'Father conducted a printing
establishment.' 'My "Missing Chinaman
Puzzle"'. (It may have been some
such assertion that led me to estimate his birthdate as 1865, but I now see it
is well known to be 1873.)
Anonymous. Sam Loyd dead; puzzle creator. New York Times (25 Feb 1934). Obituary of Sam Loyd Jr. Says he resided at 153 Halsey St.,
Brooklyn -- the same address as his father -- see the Brooklyn Daily Eagle
article of 1896, above. He worked from
a studio at 246 Fulton St., Brooklyn.
It says Jr. invented 'How Old is Ann?'.
Clark Kinnaird. Encyclopedia of Puzzles and Pastimes. Grosset & Dunlap, NY, 1946. Pp. 263‑267: Sam Loyd. Asserts that Loyd Jr. invented 'How Old is
Ann?'
Gardner. Sam Loyd: America's greatest puzzlist. SA (Aug 1957) c= First Book, Chap. 9.
Gardner. Advertising premiums. SA (Nov 1971) c= Wheels, chap. 12.
Will Shortz is working on a
biography.
No DSB entry.
François
Anatole Édouard LUCAS (1842‑1891)
Jeux Scientifiques de Ed.
Lucas. Advertisement by Chambon &
Baye (14 rue Etienne-Marcel, Paris) for the 1re Serie of six games. Cosmos.
Revue des Sciences et Leurs Applications 39 (NS No. 254) (7 Dec 1889) no
page number on my photocopy.
B. Bailly [name not given, but
supplied by Hinz]. Article on Lucas's
puzzles. Cosmos. Revue des Sciences et Leurs Applications. NS, 39 (No. 259) (11 Jan 1890) 156-159. NEED 156‑157.
Nécrologie: Édouard Lucas. La Nature 19 (1891) II, 302.
Obituary notice: "La
Nature announces the death of Prof. Edouard Lucas ...." Nature 44 (15 Oct 1891) 574-575.
Duncan Harkin. On the mathematical work of François‑Édouard‑Anatole
Lucas. L'Enseignement Math. (2) 3
(1957) 276‑288. Pp. 282‑288
is a bibliography of 184 items. I have
found many Lucas publication not listed here and have started a new
Bibliography -- see below.
P. J. Campbell. Lucas' solution to the non‑attacking
rooks problem. JRM 9 (1976/77) 195‑200. Gives life of Lucas.
A photo of Lucas is available
from Bibliothèque Nationale, Service Photographique, 58 rue Richelieu, F‑75084
Paris Cedex 02, France. Quote Cote du
Document Ln27 . 43345 and Cote du Cliche 83 A 51772. (??*)
I have obtained a copy, about 55 x 85 mm, with the photo in an oval
surround. It looks like a carte-de-visite,
but has Édouard LUCAS (1842-1891). --
Phot. Zagel. underneath. (Thanks to H. W. Lenstra for the information.)
Norman T. Gridgeman. Lucas, François‑Édouard‑Anatole. DSB VIII, 531‑532.
Susanna S. Epp. Discrete Mathematics with Applications. Wadsworth, Belmont, Calif., 1990, p. 477
gives a small photo of Lucas which looks nothing like the photo from the
BN. I have since received a note from
Epp via Paul Campbell that a wrong photo was used in the first edition, but
this was corrected in later editions.
Alain Zalmanski. Edouard Lucas Quand l'arithmétique devient amusante. Jouer Jeux Mathématiques 3 (Jul/Sep 1991) 5. Brief notice of his life and work.
Andreas M. Hinz. Pascal's triangle and the Tower of
Hanoi. AMM 99 (1992) 538-544. Sketches Lucas' life and work, giving
details that are not in the above items.
David Singmaster. The publications of Édouard Lucas. Draft version, 14pp, 1998. I discovered many items in Dickson's History
of the Theory of Numbers and elsewhere which are not given by Harkin (cf
above). This has 248 items, though many
of these are multiple items so the actual count is perhaps 275. However, Dickson does not give article
titles, and may not give the pages of the entire article, so the same article
may be cited more than once, at different pages. I hope to fill in the missing information at some time.
Hermann
Cäsar Hannibal SCHUBERT
(1848-1911)
Acta Mathematica 1882-1912.
Table Générale des Tomes 1-35.
1913. P. 169. Portrait of Schubert.
Werner Burau. Schubert, Hermann Cäsar Hannibal. DSB XII, 227‑229.
Walter
William Rouse BALL (1850‑1925)
Anon. Obituary: Mr. Rouse
Ball. The Times (6 Apr 1925) 16.
Anon. Funeral notice: Mr. W. W.
R. Ball. The Times (9 Apr 1925) 13.
(Lord) Phillimore. Letter:
Mr. Rouse Ball. The Times (9 Apr
1925) 15.
"An old pupil". The late Mr. Rouse Ball. The Times (13 Apr 1925) 12.
J. J. Thomson. W. W. Rouse Ball. The Cambridge Review (24 Apr 1925) 341-342.
Anon. Obituary of W. W. Rouse Ball.
Nature 115 (23 May 1925) 808‑809.
Anon. The late Mr. W. W. Rouse Ball.
The Trinity Magazine (Jun 1925) 53-54.
Anon. Entry in Who's Who, 1925, p. 127.
Anon. Wills and bequests: Mr.
Walter William Rouse Ball. The Times
(7 Sep 1925) 15.
E. T. Whittaker. Obituary.
W. W. Rouse Ball. Math. Gaz. 12
(No. 178) (Oct 1925) 449-454, with photo opp. p. 449.
F. Cajori. Walter William Rouse Ball. Isis 8 (1926) 321‑324. Photo on plate 15, opp. p. 321. Copy of Ball's 1924 Xmas card on p. 324.
J. A. Venn. Alumni Cantabrigienses. Part II:
From 1752 to 1900. Vol. I, p.
136. CUP, 1940.
David Singmaster. Walter William Rouse Ball (1850-1925). 6pp handout for 1st UK Meeting on the
History of Recreational Mathematics, 24 Oct 1992. Plus extended biographical (6pp) and bibliographical (8pp) notes
which repeat some of the material in the handout.
No DNB or DSB entry -- however I
have offered to write a DNB entry. I
have since seen the proposed list of names for the next edition and Ball is
already on it.
Henry
Ernest DUDENEY (1857‑1930)
Anon. & Dudeney. A chat with the puzzle king. The Captain 2 (Dec? 1899) 314‑320, with
photo. Partly an interview. Includes photos of Littlewick Meadow.
Anon. Solutions to "Sphinx's puzzles". The Captain 2:6 (Mar 1900) 598‑599 &
3:1 (Apr 1900) 89.
Anon. Master of the breakfast table problem. Daily Mail (1 Feb 1905) 7.
An interview with Dudeney in which he gives the better version of his
spider and fly problem.
Fenn Sherie. The Puzzle King: An Interview with Henry E.
Dudeney. Strand Magazine 71 (Apr 1926)
398‑4O4.
Alice Dudeney. Preface to PCP, dated Dec 1931, pp. vii‑x. The date of his death is erroneously given
as 1931.
Gardner. Henry Ernest Dudeney: England's greatest
puzzlist. SA (Jun 1958) c= Second Book, chap. 3.
Angela Newing. The Life and Work of H. E. Dudeney. MS 21 (1988/89) 37‑44.
Angela Newing is working on a biography.
No DNB or DSB entry. I have suggested a DNB entry.
Wilhelm
Ernst Martin Georg AHRENS (1872‑1927)
Wilhelm Lorey. Wilhelm Ahrens zum Gedächtnis. Archiv für Geschichte der Mathematik, der
Naturwissenschaften und der Technik 10 (1927/28) 328‑333. Photo on p. 328.
O. Staude. Dem Andenken an Dr. Wilhelm Ahrens. Jahresbericht DMV 37 (1928) 286-287.
No DSB entry.
Yakov
Isidorovich PERELMAN [Я. И. Перелман] (1882-1942)
Perelman. FMP.
1984. P. 2 (opp. TP) is a sketch
of his life and the history of the book.
There is a small drawing of Perelman at the top of the page.
Patricio Barros. Website -- Yakov I. Perelman [in Spanish]:
www.geocities.com/yakov_perelman/index.html.
This includes a four page biography, in collaboration with Antonio Bravo,
and two photos.
Hubert
PHILLIPS (1891-1964)
Hubert Phillips. Journey to Nowhere. A Discursive Autobiography. Macgibbon & Kee, London, 1960. ??NYR
No DNB entry -- I have suggested
one.
2. GENERAL PUZZLE COLLECTIONS AND SURVEYS
H. E. Dudeney. Great puzzle crazes. London Magazine 13?? (Nov 1904) 478‑482. Fifteen Puzzle. Pigs in Clover, Answers, Pick-me-up (spiral ramp) and other
dexterity puzzles. Get Off the Earth. Conjurer's Medal (ring maze). Chinese Rings. Chinese Cross (six piece burr).
Puzzle rings. Solitaire. The Mathematician's Puzzle (square, circle,
triangle). Imperial Scale. Heart and Balls.
H. E. Dudeney. Puzzles from games. Strand Magazine 35 (No. 207) (Mar 1908)
339‑344. Solutions. Ibid. 35 (No. 208) (Apr 1908) 455‑458.
H. E. Dudeney. Some much‑discussed puzzles. Strand Magazine 35 (No. 209) (May 1908)
580‑584. Solutions. Ibid. 35 (No. 210) (Jun 1908) 696.
H. E. Dudeney. The world's best puzzles. Strand Magazine 36 (No. 216) (Dec 1908) 779‑787. Solutions.
Ibid. 37 (No. 217) (Jan 1909) 113‑116.
H. E. Dudeney. The psychology of puzzle crazes. The Nineteenth Century 100:6 (Dec 1926) 868‑879. Repeats much of his 1904 article.
Sam Loyd Jr. Are you good at solving puzzles? The American Magazine (Sep 1931) 61‑63,
133‑137.
Orville A. Sullivan. Problems involving unusual situations. SM 9 (1943) 114‑118 &
13 (1947) 102‑104.
3. GENERAL HISTORICAL AND BIBLIOGRAPHICAL MATERIAL
I have tried to divide this material into historical and
bibliographical parts, but the two overlap considerably.
3.A. GENERAL HISTORICAL MATERIAL
Raffaella Franci. Giochi matematici in trattati d'abaco del
medioevo e del rinascimento. Atti del
Convegno Nazionale sui Giochi Creative, Siena, 11-14 Jun 1981. Tipografia Senese for GIOCREA (Società
Italiana Giochi Creativi), 1981. Pp.
18-43. Describes and quotes many
typical problems. 17 references, several
previously unknown to me.
Heinrich Hermelink. Arabische Unterhaltungsmathematik als
Spiegel Jahrtausendealter Kulturbeziehungen zwischen Ost und West. Janus 65 (1978) 105-117, with English
summary. An English translation
appeared as: Arabic recreational
mathematics as a mirror of age-old cultural relations between Eastern and
Western civilizations; in: Ahmad Y. Al-Hassan,
Ghada Karmi & Nizar Namnum, eds.; Proceedings of the First International
Symposium for the History of Arabic Science, April 1976 -- Vol. Two: Papers in
European Languages; Institute for the History of Arabic Science, Aleppo, 1978,
pp. 44-52. (There are a few translation
and typographical errors, which make it clear that the English version is a
translation of the German.)
D. E. Smith. On the origin of certain typical
problems. AMM 24 (1917) 64‑71. (This is mostly contained in his History,
vol. II, pp. 536‑548.)
Many of the items cited in the Common
References have extensive bibliographies.
In particular: BLC; BMC;
BNC; DNB; DSB;
Halwas; NUC; Schaaf;
Smith & De Morgan: Rara; Suter
are basic bibliographical sources.
Datta & Singh; Dickson; Heath: HGM;
Murray; Sanford: H&S &
Short History; Smith:
History & Source Book; Struik; Tropfke
are histories with extensive bibliographical references. AR;
BR are editions of early texts
with substantial bibliographical material.
Ahrens: MUS; Ball: MRE; Berlekamp, Conway & Guy: Winning Ways; Gardner;
Lucas: RM are recreational books
with some useful bibliographical material.
Of these, the material in Ahrens is by far the most useful. The magic bibliographies of Christopher,
Clarke & Blind, Hall, Heyl, Price (see HPL), Toole Stott and Volkmann &
Tummers have considerable overlap with the present material, particularly for
older books, though Hall, Heyl and Toole Stott restrict themselves to English
material, while Volkmann & Tummers only considers German. Santi is also very useful. Below I give some additional bibliographical
material which may be useful, arranged in author order.
Anonymous. Mathematical bibliography. SSM 48 (1948) 757‑760. Covers recreations.
Wilhelm Ahrens. Mathematische Spiele. Section I G 1 of Encyklopadie der Math.
Wiss., Vol. I, part 2, Teubner, Leipzig, 1900‑1904, pp. 1080‑1093.
Raymond Clare Archibald. Notes on some minor English mathematical serials. MG 14 (1928-29) 379-400.
Elliott M. Avedon &
Brian Sutton‑Smith. The
Study of Games. (Wiley, NY, 1971); Krieger, Huntington, NY, 1979.
Anthony S. M. Dickins. A Catalogue of Fairy Chess Books and
Opuscules Donated to Cambridge University Library, 1972‑1973, by Anthony
Dickins M.A. Third ed., Q Press,
Kew Gardens, UK, 1983.
Underwood Dudley. An annotated list of recreational
mathematics books. JRM 2:1
(Jan 1969) 13-20. 61 titles, in
English and in print at the time.
Aviezri S. Fraenkel. Selected Bibliography on Combinatorial Games
and Some Related Material. There have
been several versions with slightly varying titles. The most recent printed version is: 400 items, 28 pp., including 4 pp of text, Sep 1990. Technical Report CS90‑23, Weizmann
Institute of Science, Rehovot, Israel.
= Proc. Symp. Appl. Math. 43 (1991) 191-226. Fraenkel has since produced Update 1 to this which lists 430
items on 31pp, Aug 1992; and Update 2,
480 items on 33pp, with 5 pp of text, accidentally dated Aug 1992 at the top
but produced in Feb 1994. On 22 Nov
1994, it became a dynamic survey on the Electronic J. Combinatorics and can be
accessed from:
http://ejc.math.gatech.edu:8080/journal/surveys/index.html.
It
can also be accessed via anonymous ftp from
ftp.wisdom.weizmann.ac.il. After
logging in, do cd pub/fraenkel and then
get one of the following three
compressed files: games.tex.z; games.dvi.z; games.ps.z.
Martin P. Gaffney &
Lynn Arthur Steen. Annotated
Bibliography of Expository Writing in the Mathematical Sciences. MAA, 1976.
JoAnne S. Growney. Mathematics and the arts -- A
bibliography. Humanistic Mathematics
Network Journal 8 (1993) 22-36. General
references. Aesthetic standards for
mathematics and other arts.
Biographies/autobiographies of mathematicians. Mathematics and display of information (including
mapmaking). Mathematics and humor. Mathematics and literature (fiction and
fantasy). Mathematics and music. Mathematics and poetry. Mathematics and the visual arts.
JoAnne S. Growney. Mathematics in Literature and Poetry. Humanistic Mathematics Network Journal 10
(Aug 1994) 25-30. Short survey. 3 pages of annotated references to 29
authors, some of several books.
R. C. Gupta. A bibliography of selected book [sic] on history
of mathematics. The Mathematics
Education 23 (1989) 21-29.
Trevor H. Hall. Mathematicall Recreations. Op. cit. in 1. This is primarily concerned with the history of the book by van
Etten. [This booklet is revised as pp.
83-119 of Hall, OCB -- see Section 1.]
Catherine Perry Hargrave. A History of Playing Cards and a
Bibliography of Cards and Gaming.
(Houghton Mifflin, Boston, 1930);
Dover, 1966.
Susan Hill. Catalogue of the Turner Collection of the
History of Mathematics Held in the Library of the University of Keele. University Library, Keele, 1982. (Sadly this collection was secretly sold by
Keele University in 1998 and has now been dispersed.)
Honeyman Collection -- see:
Sotheby's.
Horblit Collection -- see:
Sotheby's and H. P. Kraus.
Else Høyrup. Books about Mathematics. Roskilde Univ. Center, PO Box 260, DK‑4000,
Roskilde, Denmark, 1979.
D. O. Koehler. Mathematics and literature. MM 55 (1982) 81-95. 64 references. See Utz for some further material.
H. P. Kraus (16 East 46th Street,
New York, 10017). The History of
Science including Navigation.
Catalogue
168. A First Selection of Books from
the Library of Harrison D. Horblit. Nd
[c1976].
Catalogue
169. A Further Selection of Books, 1641-1700
(Wing Period) from the Library of Harrison D. Horblit. Nd [c1976].
Catalogue
171. Another Selection of Books from
the Library of Harrison D. Horblit. Nd
[c1976].
These are the continuations of the catalogues
issued by Sotheby's, qv.
John S. Lew. Mathematical references in literature. Humanistic Mathematics Network Journal 7
(1992) 26-47.
Antonius van der Linde. Das erst Jartausend [sic] der
Schachlitteratur -- (850‑1880).
(1880); Facsimile reprint by
Caissa Limited Editions, Yorklyn, Delaware, 1979, HB.
Andy Liu. Appendix III: A selected bibliography on popular mathematics. Delta-k 27:3 (Apr 1989) --
Special issue: Mathematics for
Gifted Students, 55-83.
Édouard Lucas. Récréations mathématiques, vol 1 (i.e. RM1),
pp. 237-248 is an Index Bibliographique.
Felix Müller. Führer durch die mathematische Literature
mit besonderer Berücksichtigung der historisch wichtigen Schriften. Abhandlungen zur Geschichte der Mathematik
27 (1903).
Charles W. Newhall. "Recreations" in secondary mathematics. SSM 15 (1915) 277‑293.
Mathematical Association. 259 London Road, Leicester, LE2 3BE.
Catalogue of Books and Pamphlets
in the Library. No details, [c1912],
19pp, bound in at end of Mathematical Gazette, vol. 6 (1911‑1912).
A
First List of Books & Pamphlets in the Library of the Mathematical
Association -- Books and Pamphlets acquired before 1924. Bell, London, 1926.
A
Second List of Books & Pamphlets in the Library of the Mathematical
Association -- Books and Pamphlets acquired during 1924 and 1925. Bell, London, 1929.
A
Third List of Books & Pamphlets in the Library of the Mathematical
Association -- Books and Pamphlets added from 1926 to 1929. Bell, London, 1930.
A
Fourth List of Books & Pamphlets in the Library of the Mathematical
Association -- Books and Pamphlets added from 1930 to 1935. Bell, London, 1936.
Lists 1‑4 edited by E. H.
Neville.
Books
and Periodicals in the Library of the Mathematical Association. Ed. by R. L. Goodstein. MA, 1962.
Includes the four previous lists and additions through 1961.
SEE
ALSO: Riley; Rollett; F. R. Watson.
Stanley Rabinowitz. Index to Mathematical Problems 1980-1984.
MathPro Press, Westford, Massachusetts, 1992.
Cecil B. Read &
James K. Bidwell.
Selected
articles dealing with the history of elementary Mathematics. SSM 76 (1976) 477-483.
Periodical
articles dealing with the history of advanced mathematics -- Parts I &
II. SSM 76 (1976) 581-598 &
687-703.
Rudolf H. Rheinhardt. Bibliography on Whist and Playing
Cards. From: Whist Scores and Card-table Talk, Chicago, 1887. Reprinted by L. & P. Parris, Llandrindod
Wells, nd [1980s].
Pietro Riccardi. Biblioteca Matematica Italiana dalla Origine
della Stampa ai Primi Anni del Secolo XIX.
G. G. Görlich, Milan, 1952, 2 vols. This work appeared in several parts and supplements in the late
19C and early 20C, mostly published by the Società Tipografica Modense, Modena,
1878-1893. Because it appeared in
parts, the contents of early copies are variable and even the reprints may vary. The contents of this set are as follows.
I. 20pp prelims +
Col. 1 - 656 (Abaco - Kirchoffer).
[= original Vol. I.]
Col.
1 - 676 (La Cometa - Zuzzeri) + 2pp correzioni. [= original Vol. II.]
II. 4pp
titles and reverses. Correzioni ed
Aggiunte. [= original Appendice.]
Serie
I.a Col. 1 - 78 +
1½pp Continuazione delle
Correzioni (note that these
have Pag.
when they mean Col.).
Serie
II.a. Col. 81 - 156.
Serie
III.a. Col. 157 - 192 +
Aggiunte al Catalogo delle Opere di sovente citate,
col.
193-194 + 1p Continuazione delle Correzioni (note that these have
Pag. when they mean Col.).
Serie
IV.a. Col. 197 - 208 +
Seconda Aggiunta al Catalogo delle Opere più di
sovente
citate, col. 209 - 212 + Continuazione delle Correzioni in
col.
211-212.
Serie
V.a. Col. 1 - 180.
Serie
VI.a. Col. 179 - 200.
Serie
V & VI must have been published as one volume as Serie V ends
halfway
down a page and then Serie VI begins on the same page.
Serie
VII.a. 2pp introductory note
by Ettore Bortolotti in 1928 saying that this
material
was left as a manuscript by Riccardi and never previously
published +
Col. 1 - 106.
Indice
Alfabetico, of authors, covering the original material and all seven Series
of
Correzioni ed Aggiunte, in 34 unnumbered columns.
Parte
Seconda. Classificazione per materie
delle opere nella Parte I. 18pp
(including
a chronological table) + subject index, pp. 1 - 294.
Catalogo Delle opere più di sovente citate, col. 1 -
54.
[I
have seen an early version which had the following parts: Vol. I, 1893, col. 1‑656; Vol. II, 1873, col. 1-676; Appendice, 1878-1880-1893, col. 1-228. Appendice, nd, col. 1-212. Serie V, col. 1-228. Parte 2, Vol. 1, 1880, pp. 1-294. Renner Katalog 87 describes it as 5 in 2
vols.]
A. W. Riley. School Library Mathematics List --
Supplement No. 1. MA, 1973.
SEE
ALSO: Rollett.
Tom Rodgers. Catalog of his collection of books on
recreational mathematics, etc. The
author, Atlanta, May 1991, 40pp.
Leo F. Rogers. Finding Out in the History of
Mathematics. Produced by the author,
London, c1985, 52pp.
A. P. Rollett. School Library Mathematics List. Bell, London, for MA, 1966.
SEE
ALSO: Riley.
Charles L. Rulfs. Origins of some conjuring works. Magicol 24 (May 1971) 3-5.
José A. Sánchez Pérez. Las Matematicas en la Biblioteca del
Escorial. Imprenta de Estanislao
Maestre, Madrid, 1929.
William L. Schaaf.
List
of works on recreational mathematics.
SM 10 (1944) 193-200.
PLUS: A. Gloden; Additions to Schaaf's "List of works on
mathematical recreations"; SM 13 (1947) 127.
A
Bibliography of Recreational Mathematics.
Op. cit. in Common References, 4 vols., 1955-1978. In these volumes he gives several lists of
relevant books.
Books for the periods 1900-1925 and
1925-c1956 are given as Sections 1.1 (pp. 2-3) and 1.2 (pp. 4-12) in Vol.
1.
Chapter 9, pp. 144-148, of Vol. 1, is
a Supplement, generally covering c1954-c1962, but with some older items.
In Vol. 2, 1970, the Appendix, pp.
181-191, extends to c1969, including some older items and repeating a few from
the Supplement of Vol. 1.
Appendix A of Vol. 3, 1973, pp.
111-113, adds some more items up through 1972.
Appendix A, pp. 134-137, of Vol. 4,
1978, extends up through 1977.
The following VESTPOCKET BIBLIOGRAPHIES are
extensions of the material
in his Bibliographies.
No. 1: Pythagoras and rational triangles; Geoboards and lattices. JRM 16:2 (1983-84) 81-88.
No. 2: Combinatorics; Gambling and sports. JRM
16:3 (1983-84) 170-181.
No. 3: Tessellations and polyominoes; Art and music. JRM 16:4 (1983-84) 268‑280.
No. 4: Recreational miscellany. JRM 17:1 (1984-85) 22-31.
No. 5: Polyhedra;
Topology; Map coloring. JRM 17:2 (1984-85) 95-105.
No. 6: Sundry algebraic notes. JRM 17:3 (1984-85) 195-203.
No. 7: Sundry geometric notes. JRM 18:1 (1985-86) 36-44.
No. 8: Probability;
Gambling. JRM 18:2 (1985-86)
101-109.
No. 9: Games and puzzles. JRM 18:3 (1985-86) 161-167.
No.
10: Recreational
mathematics; Logical puzzles; Expository mathematics. JRM 18:4 (1985-86) 241-246.
No.
11: Logic, Artificial
intelligence, and Mathematical foundations.
JRM 19:1 (1987) 3-9.
No.
12: Magic squares and cubes; Latin squares; Mystic arrays and Number patterns. JRM 19:2 (1987) 81-86.
The
High School Mathematics Library. NCTM,
(1960, 1963, 1967, 1970, 1973); 6th
ed., 1976; 7th ed., 1982; 8th ed., 1987.
SEE ALSO:
Wheeler; Wheeler & Hardgrove.
Early
Books on Magic Squares. JRM 16:1
(1983-84) 1-6.
William L. Schaaf &
David Singmaster. Books on
Recreational Mathematics. A Supplement
to the Lists in William L. Schaaf's A Bibliography of Recreational
Mathematics. Collected by William L.
Schaaf; typed and annotated by David Singmaster. School of Computing, Information Systems and
Mathematics, South Bank University, London, SE1 0AA. 18pp, Dec 1992 and revised several times afterwards.
Peter Schreiber.
Mathematik
und belletristik [1.] & 2. Teil.
Mitteilungen der Mathematischen Gesellschaft der Deutschen
Demokratischer Republik. (1986), no. 4,
57-71 & (1988), no. 1-2, 55-61.
Good on German works relating mathematics and arts.
Mathematiker
als Memoirenschreiber. Alpha (Berlin)
(1991), no. 4, no page numbers on copy received from author. Extends previous work.
S. N. Sen. Scientific works in Sanskrit, translated
into foreign languages and vice‑versa in the 18th and 19th century
A.D. Indian J. History of Science 7
(1972) 44‑70.
Will Shortz. Puzzleana [catalogue of his puzzle
books]. Produced by the author. 14 editions have appeared. The latest is: May 1992, 88pp with 1175 entries in 26 categories, with indexes
of authors and anonymous titles. Some
entries cover multiple items. In Jan
1995, he produced a 19pp Supplement extending to a total of 1451 entries.
David Singmaster.
The
Bibliography of Some Recreational Mathematics Books. School of Computing, Information Systems
and Mathematics, South Bank Univ.
13 Nov 1994, 39pp. Technical Report SBU-CISM-94-09.
2nd ed., Aug 1995, 41pp. Technical Report SBU-CISM-95-08.
3rd ed., Jun 1996, 42pp. Technical Report SBU-CISM-96-12.
4th ed., Jun 1998, 44pp. Technical Report SBU-CISM-98-02.
(Current version is 61pp.)
Books
on Recreational Mathematics. School of Computing,
Information Systems and
Mathematics, South Bank Univ., until
1996.
21 Jan 1991.
Approx. 2951 items on 120pp, ringbound.
30 Jan 1992.
Approx. 3314 items on 138pp, ringbound.
10 Jan 1993.
Approx. 3606 items on 95pp, ringbound.
10 Dec 1994.
Approx. 4303 items plus 67 Old Books on 110pp. Technical
Report SBU‑CISM-94-11.
10 Oct 1996.
Approx. 4842 items plus 84 Old Books on 127pp. Technical
Report SBU-CISM-96-17.
24 May 1999.
Approx. 6015 items plus 133 Old Books on 166pp. Technical
Report SBU-CISM-99-14.
26 Feb 2002.
Approx. 7185 items plus 192 Old Books plus Supplement of
Calculating Devices, on 220pp.
thermal bound.
22 Nov 2003.
Approx. 7811 items plus 202 Old Books plus Supplement of
Calculating Devices, on 244pp.
thermal bound.
Index
to Martin Gardner's Columns and Cross Reference to His Books. (Oct 1993.)
Slightly revised as: Technical
Report SBU-CISM-95-09; School of Computing, Information Systems, and
Mathematics; South Bank University, London, Aug 1995, 22pp. (Current version is 23pp and Don Knuth has
sent 9pp of additional material and I will combine these at some time.)
Harold Adrian Smith. Dick and Fitzgerald Publishers. Books at Brown 34 (1987) 108-114.
Sotheby's [Sotheby Parke
Bernet].
Catalogue
of the J. B. Findlay Collection Books
and Periodicals on Conjuring and the Allied Arts. Part I: A-O 5-6 Jul 1979. Part II: P-Z plus: Mimeographed Books
and Instructions; Flick Books
Catalogues of Apparatus and Tricks
Autograph Letters, Manuscripts, and Typescripts 4-5 Oct 1979. Part III: Posters and Playbills
3-4 Jul 1980. Each with estimates
and results lists.
The
Celebrated Library of Harrison D. Horblit Esq.
Early Science Navigation &
Travel Including Americana with a few medical books. Part I
A - C 10/11 Jun 1974. Part II
D - G 11 Nov 1974. HB.
The sale was then cancelled and the library was sold to E. P. Kraus, qv,
who issued three further catalogues, c1976.
The
Honeyman Collection of Scientific Books and Manuscripts. Seven volumes, each
with estimates and results booklets.
Part I: Printed Books A-B, 30-31 Oct 1978.
Part II:
Printed Books C-E, 30 Apr - 1 May 1979.
Part III: Manuscripts and Autograph Letters of
the 12th to the 20th Centuries.
Part IV: Printed Books F-J, 5-6 Nov 1979.
Part V: Printed Books K-M, 12-13 May 1980.
Part VI: Printed Books N-Sa, 10-11 Nov 1980.
Part VII:
Printed Books Sc-Z and Addenda, 19-20 May 1981.
Lynn A. Steen, ed.
Library
Recommendations for Undergraduate Mathematics.
MAA Reports No. 4, 1992.
Two-Year
College Mathematics Library Recommendations.
MAA Reports No. 5, 1992.
Strens/Guy Collection. Author/Title Listing. Univ. of Calgary. Preliminary Catalogue, 319 pp., July 1986. [The original has a lot of blank space. I have a computer version which is reduced
to 67pp.]
Eva Germaine Rimington
Taylor. The Mathematical Practitioners
of Tudor & Stuart England
1485-1714. CUP for the Institute
of Navigation, 1970.
Eva Germaine Rimington
Taylor. The Mathematical Practitioners
of Hanoverian England 1714‑1840. CUP for the Institute of Navigation,
1966.
PLUS: Kate Bostock, Susan Hurt & Michael Hart;
An Index to the Mathematical Practitioners of Hanoverian England 1714-1840; Harriet Wynter Ltd., London,
1980.
W. R. Utz. Letter:
Mathematics in literature. MM 55
(1982) 249‑250. Utz has sent his
3pp original more detailed version along with 4pp of further citations. This extends Koehler's article.
George Walker. The Art of Chess-Play: A New Treatise on the
Game of Chess. 4th ed., Sherwood,
Gilbert & Piper, London, 1846.
Appendix: Bibliographical
Catalogue of the chief printed books, writers, and miscellaneous articles on
chess, up to the present time, pp. 339-375.
Frank R. [Joe] Watson, ed. Booklists.
MA.
Puzzles,
Problems, Games and Mathematical Recreations.
16pp, 1980.
Selections
from the Recommended Books. 18pp, 1980.
Full
List of Recommended Books. 105pp, 1984.
Margariete Montague
Wheeler. Mathematics Library --
Elementary and Junior High School. 5th
ed., NCTM, 1986.
SEE
ALSO: Schaaf; Wheeler & Hardgrove.
Margariete Montague Wheeler &
Clarence Ethel Hardgrove.
Mathematics Library -- Elementary and Junior High School. NCTM, (1960; 1968; 1973); 4th ed., 1978.
SEE
ALSO: Schaaf; Wheeler.
Ernst Wölffing. Mathematischer Bücherschatz. Systematisches Verzeichnis der wichtigsten
deutschen und ausländischen Lehrbücher und Monographien des 19. Jahrhunderts
auf dem Gebiete der mathematischen Wissenschaften. I: Reine Mathematik; (II:
Angewandte Mathematik never appeared).
AGM 16, part I (1903).
Aviezri S. Fraenkel. Selected Bibliography on Combinatorial Games
and Some Related Material. Op. cit. in
3.B.
4.A. GENERAL THEORY AND NIM‑LIKE GAMES
Conway's
extension of this theory is well described in Winning Ways and later work is
listed in Fraenkel's Bibliography -- see section 3.B & 4 -- so I will not
cover such material here.
See
MUS I 145-147.
(a,
b) denotes the game where one can
take 1, 2, ..., or
a away from one pile, starting
with b
in the pile, with the last player winning. The version (10,
100) is sometimes called Piquet des
Cavaliers or Piquet à Cheval, a name which initially perplexed me. Piquet is one of the older card games, being
well known to Rabelais (1534) and was known in the 16C as Cent (or Saunt or
Saint) because of its goal of 100 points.
See: David Parlett; (Oxford Guide
to Card Games, 1990 =) A History of Card Games; OUP, 1991, pp. 24
& 175-181. The connection with
horses undoubtedly indicates that (10,
100) was viewed as a game which could
be played without cards, while riding -- see Les Amusemens, Decremps.
( 3, 13) Dudeney, Stong
( 3, 15) Mittenzwey, Hoffmann, Mr. X,
Dudeney, Blyth,
( 3, 17) Fourrey,
( 3, 21) Blyth, Hummerston,
( 4, 15) Mittenzwey,
( 6, 30) Pacioli, Leske, Mittenzwey,
Ducret,
( 6, 31) Baker,
( 6, 50) Ball-FitzPatrick,
( 6, 52) Rational Recreations
( 6, 57) Hummerston,
( 7, 40) Mittenzwey,
( 7, 41) Sprague,
( 7, 45) Mittenzwey,
( 7, 50) Decremps,
( 7, 60) Fourrey,
( 8, 100) Bachet,
Carroll,
( 9, 100) Bachet,
Ozanam, Alberti
(10, 100) Bachet, Henrion, Ozanam, Alberti, Les Amusemens,
Hooper, Decremps,
Badcock,
Jackson, Rational Recreations, Manuel des Sorciers,
Boy's Own
Book, Nuts to Crack, Young Man's Book, Carroll,
Magician's Own
Book, Book of 500 Puzzles, Secret Out,
Boy's Own Conjuring
Book, Vinot, Riecke, Fourrey, Ducret, Devant,
(10, 120) Bachet,
(12, 134) Decremps,
General case: Bachet, Ozanam, Alberti, Decremps, Boy's Own
Book, Young Man's Book, Vinot, Mittenzwey, (others ?? check)
Versions
with limited numbers of each value or using a die -- see 4.A.1.a.
Version
where an odd number in total has to be taken:
Dudeney, Grossman & Kramer, Sprague.
Versions
with last player losing: Mittenzwey,
Pacioli. De Viribus.
c1500. Ff. 73v - 76v. XXXIIII effecto afinire qualunch' numero
na'ze al compagno anon prendere piu de un termi(n)ato .n. (34th effect to
finish whatever number is before the company, not taking more than a limiting
number) = Peirani 109‑112.
Phrases it as an addition problem.
Considers (6, 30) and the general problem.
David Parlett. (Originally: The Oxford Guide to Card Games; OUP, 1990); reissued as: A History of Card Games.
Penguin, 1991, pp. 174-175.
"Early references to 'les luettes', said to have been played by
Anne de Bretagne and Archduke Philip the Fair in 1503, and by Gargantua in
1534, seem to suggest a game of the Nim family (removing numbers of objects
from rows and columns)."
Cardan. Practica Arithmetice. 1539.
Chap. 61, section 18, ff. T.iiii.v - T.v.r (p. 113). "Ludi mentales". One has
1, 3, 6 and the other has 2, 4, 5;
or one has
1, 3, 5, 8, 9 and
the other has 2, 4, 6, 7, 10; one one wants to make 100.
"Sunt magnæ inventionis, & ego inveni æquitando & sine
aliquo auxilio cum socio potes ludere & memorium exercere ...."
Baker. Well Spring of Sciences.
1562? Prob. 5: To play at 31
with Numbers, 1670: pp. 353‑354.
??NX. (6, 31).
Bachet. Problemes.
1612. Prob. XIX: 1612,
99-103. Prob. XXII, 1624: 170-173; 1884: 115‑117. Phrases it as an addition problem. First considers (10, 100), then (10, 120), (8, 100), (9, 100), and the general case. Labosne omits the demonstration.
Dennis Henrion. Nottes to van Etten. 1630.
Pp. 19-20. (10, 100) as an addition problem, citing Bachet.
Ozanam. 1694.
Prob. 21, 1696: 71-72; 1708: 63‑64. Prob. 25, 1725: 182‑184. Prob. 14, 1778: 162-164; 1803: 163-164; 1814: 143-145. Prob. 13,
1840: 73-74. Phrases it as an addition
problem. Considers (10, 100)
and (9, 100) and remarks on the general case.
Alberti. 1747.
Due persone essendo convenuto ..., pp. 105‑108 (66‑67). This is a slight recasting of Ozanam.
Les Amusemens. 1749.
Prob. 10, p. 130: Le Piquet des Cavaliers. (10, 100) in additive
form. "Deux amis voyagent à
cheval, l'un propose à l'autre un cent de Piquet sans carte."
William Hooper. Rational Recreations, In which the
Principles of Numbers and Natural Philosophy Are clearly and copiously
elucidated, by a series of Easy, Entertaining, Interesting Experiments. Among which are All those commonly performed
with the cards. [Taken from my 2nd ed.]
4 vols., L. Davis et al., London, 1774;
2nd ed., corrected, L. Davis et al., London, 1783-1782 (vol. 1 says
1783, the others say 1782; BMC gives 1783-82);
3rd ed., corrected, 1787; 4th
ed., corrected, B. Law et al., London, 1794.
[Hall, BCB 180-184 & Toole Stott 389-392. Hall says the first four eds. have identical
pagination. I have not seen any
difference in the first four editions, except as noted in Section 6.P.2. Hall, OCB, p. 155. Heyl 177 notes the different datings of the 2nd ed, Hall, BCB 184 and Toole Stott 393 is a 2
vol. 4th ed., corrected, London, 1802.
Toole Stott 394 is a 2 vol. ed. from Perth, 1801. I have a note that there was an 1816 ed, but
I have no details. Since all relevant
material seems the same in all volumes, I will cite this as 1774.] Vol. 1, recreation VIII: The magical
century. (10, 100) in additive form. Mentions other versions and the general rule.
I
don't see any connection between this and Rational Recreations, 1824.
Henri Decremps. Codicile de Jérôme Sharp, Professeur de
Physique amusante; Où l'on trouve parmi
plusieurs Tours dont il n'est point parlé dans son Testament, diverses
récréations relatives aux Sciences & Beaux-Arts; Pour servir de troisième suite
À La Magie Blanche Dévoilée. Lesclapart, Paris, 1788.
Chap. XXVII, pp. 177-184: Principes mathématiques sur le piquet à
cheval, ou l'art de gagner son diner en se promenant. Does (10, 100) in
additive form, then discusses the general method, illustrating with (7, 50) and (12, 134).
Badcock. Philosophical Recreations, or, Winter
Amusements. [1820]. Pp. 33-34, no. 48: A curious recreation with
a hundred numbers, usually called the magical century. (10, 100) as an additive problem where each person starts with 50
counters. Discusses general
case, but doesn't notice that the limitation to 50 counters each
considerably changes the game!
Jackson. Rational Amusement. 1821.
Arithmetical Puzzles, no. 47, pp. 11 & 64. Additive form of (10,
100).
Rational Recreations. 1824. Exercise 12(?), pp. 57-58.
As in Badcock. Then says it can
be generalised and gives (6, 52).
Manuel des Sorciers. 1825.
Pp. 57-58, art. 30: Le piquet sans cartes. ??NX (10, 100) done subtractively.
The Boy's Own Book.
The
certain game. 1828: 177; 1828-2: 236;
1829 (US): 104; 1855: 386‑387; 1868: 427.
The
magical century. 1828: 180; 1828-2: 236‑237; 1829 (US): 104-105; 1855: 391‑392.
Both are additive phrasings of (10, 100).
The latter mentions using other numbers and how to win then.
Nuts to Crack V (1836), no.
70. An arithmetical problem. (10, 100).
Young Man's Book. 1839.
Pp. 294-295. A curious
Recreation with a Hundred Numbers, usually called the Magical Century. Almost identical to Boy's Own Book.
Lewis Carroll.
Diary
entry for 5 Feb 1856. In
Carroll-Gardner, pp. 42-43. (10,
100). Wakeling's note in the Diaries
indicates he is not familiar with this game.
Diary
entry for 24 Oct 1872. Says he has
written out the rules for Arithmetical Croquet, a game he recently invented. Roger Lancelyn Green's abridged version of
the Diaries, 1954, prints a MS version dated 22 Apr 1889. Carroll-Wakeling, prob. 38, pp. 52-53 and
Carroll-Gardner, pp. 39 & 42 reprint this, but Gardner has a misprinted
date of 1899. Basically (8, 100),
but passing the values 10, 20,
..., requires special moves and one may
have to go backward. Also, when a move
is made, some moves are then barred for the next player. Overall, the rules are typically
Carrollian-baroque.
Magician's Own Book. 1857.
The certain
game, p. 243. As in Boy's Own Book.
The
magical century, pp. 244-245. As in
Boy's Own Book.
Book of 500 Puzzles. 1859.
The
certain game, p. 57. As in Boy's Own
Book.
The
magical century, pp. 58-59. As in Boy's
Own Book.
The Secret Out. 1859.
Piquet on horseback, pp. 397-398 (UK: 130‑131) -- additive (10, 100)
unclearly explained.
Boy's Own Conjuring Book. 1860.
The
certain game, pp. 213‑214. As in
Boy's Own Book.
Magical
century, pp. 215. As in Boy's Own Book.
Vinot. 1860. Art. XI: Un cent de
piquet sans cartes, pp. 19-20. (10.
100). Says the idea can be generalised,
giving (7, 52) as an example.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 563-III, pp.
247: Wer von 30 Rechenpfennigen den letzen wegnimmt, hat gewonnen. (6, 30).
F. J. P. Riecke. Mathematische Unterhaltungen. 3 vols., Karl Aue, Stuttgart, 1867, 1868
& 1873; reprint in one vol.,
Sändig, Wiesbaden, 1973. Vol. 3, art
22.2, p. 44. Additive form of (10, 100).
Mittenzwey. 1880.
Probs. 286-287, pp. 52 & 101-102;
1895?: 315-317, pp. 56 & 103-104;
1917: 315-317, pp. 51 & 98.
(6,
30), last player wins.
(4,
15), last player loses, the solution discusses other cases: (7, 40), (7, 45)
and indicates the general solution.
(added
in 1895?) (3, 15), last player loses.
Hoffmann. 1893.
Chap VII, no. 19: The fifteen matches puzzle, pp. 292 & 300‑301
= Hoffmann-Hordern, p. 197. (3,
15). c= Benson, 1904, The fifteen match
puzzle, pp. 241‑242.
Ball-FitzPatrick. 1st ed., 1898. Deuxième exemple, pp. 29-30.
(6, 50).
E. Fourrey. Récréations Arithmétiques. (Nony, Paris, 1899; 2nd ed., 1901); 3rd ed., Vuibert & Nony, Paris, 1904; (4th ed., 1907); 8th ed., Librairie Vuibert, Paris, 1947. [The 3rd and 8th eds are identical except
for the title page, so presumably are identical to the 1st ed.] Sections 65‑66: Le jeu du piquet à
cheval, pp. 48‑49. Additive forms
of (10, 100) and (7, 60). Then gives subtractive form for a pile of
matches for (3, 17).
Étienne Ducret. Récréations Mathématiques. Garnier Frères, Paris, nd [not in BN, but a
similar book, nouv. ed., is 1892]. Pp.
102‑104: Le piquet à cheval.
Additive version of (10,
100) with some explanation of the use
of the term piquet. Discusses (6, 30).
Mr. X [possibly J. K. Benson --
see entry for Benson in Abbreviations].
His Pages. The Royal Magazine
9:3 (Jan 1903) 298-299. A good game for
two. (3, 15) as a subtraction game.
David Devant. Tricks for Everyone. Clever Conjuring with Everyday Objects. C. Arthur Pearson, London, 1910. A counting race, pp. 52-53. (10, 100).
Dudeney. AM.
1917. Prob. 392: The pebble
game, pp. 117 & 240. (3, 15) &
(3, 13) with the object being to
take an odd number in total. For 15,
first player wins; for 13, second
player wins. (Barnard (50 Telegraph
..., 1985) gives the case (3, 13).)
Blyth. Match-Stick Magic.
1921.
Fifteen
matchstick game, pp. 87-88. (3, 15).
Majority
matchstick game, p. 88. (3, 21).
Hummerston. Fun, Mirth & Mystery. 1924.
Two
second-sight tricks (no. 2), p. 84. (6,
57), last player losing.
A
match mystery, p. 99. (3, 21), last player losing.
H. D. Grossman & David
Kramer. A new match-game. AMM 52 (1945) 441‑443. Cites Dudeney and says Games Digest (April
1938) also gave a version, but without solution. Gives a general solution whether one wants to take an odd total
or an even total.
C. L. Stong. The Amateur Scientist. Ill. by Roger Hayward. S&S, 1960. How to design a "Pircuit" or Puzzle circuit, pp.
388-394. On pp. 388-391, Harry Rudloe
describes a relay circuit for playing the subtractive form of (3, 13), which he calls the "battle of
numbers" game.
Ronald Sprague. Unterhaltsame Mathematik. Vieweg, Braunschweig, 1961. Translated by T. H. O'Beirne as: Recreations in Mathematics, Blackie, London,
1963. Problem 24: "Ungerade"
gewinnt, pp. 16 & 44‑45. (=
'Odd' is the winner, pp. 18 & 53‑55.) (7, 41) with the winner
being the one who takes an odd number in total. Solves (7, b) and states the structure for (a, b).
I
also have some other recent references to this problem. Lewis (1983) gives a general solution which
seems to be wrong.
Numerical
variations: Badcock, Gibson, McKay.
Die
versions: Secret Out (UK), Loyd,
Mott-Smith, Murphy.
Baker. Well Spring of Sciences.
1562? Prob. 5: To play at 31
with Numbers, 1670: pp. 353‑354.
??NX. (6, 31). ??CHECK if this has the limited use of
numbers.
John Fisher. Never Give a Sucker an Even Break. (1976);
Sphere Books, London, 1978.
Thirty-one, pp. 102-104. (6,
31) additively, but played with just 4
of each value, the 24 cards of ranks 1
-- 6, and the first to exceed 31
loses. He says it is played extensively
in Australia and often referred to as "The Australian Gambling Game of
31". Cites the 19C gambling expert
Jonathan Harrington Green who says it was invented by Charles James Fox (1749‑1806). Gives some analysis.
Badcock. Philosophical Recreations, or, Winter
Amusements. [1820]. Pp. 33-34, no. 48: A curious recreation with
a hundred numbers, usually called the magical century. (10, 100) as an additive problem where each person starts with 50
counters. Discusses general
case, but doesn't notice that the limitation to 50 counters each
considerably changes the game!
Nuts to Crack V (1836), no.
71. (6, 31) additively, with four of
each value. "Set down on a slate,
four rows of figures, thus:-- ... You agree to rub out one figure alternately,
to see who shall first make the number thirty-one."
Magician's Own Book. 1857.
Art. 31: The trick of thirty‑one, pp. 70‑71. (6, 31)
additively, but played with just 4 of each value -- e.g. the 24 cards of
ranks 1 -- 6. The author advises you not to play it for money with
"sporting men" and says it it due to Mr. Fox. Cf Fisher.
= Boy's Own Conjuring Book; 1860; Art. 29: The trick of thirty‑one,
pp. 78‑79. = The Secret Out;
1859, pp. 65-66, which adds a footnote that the trick is taken from the book
One Hundred Gambler Tricks with Cards by J. H. Green, reformed gambler,
published by Dick & Fitzgerald.
The Secret Out (UK), c1860. To throw thirty‑one with a die before
your antagonist, p. 7. This is
incomprehensible, but is probably the version discussed by Mott-Smith.
Edward S. Sackett. US Patent 275,526 -- Game. Filed: 9 Dec 1882; patented: 10 Apr 1883. 1p
+ 1p diagrams. Frame of six rows
holding four blocks which can be slid from one side to the other to play the 31
game, though other numbers of rows, blocks and goal may be used. Gives an example of a play, but doesn't go
into the strategy at all.
Larry Freeman. Yesterday's Games. Taken from "an 1880 text" of games. (American edition by H. Chadwick.) Century House, Watkins Glen, NY, 1970. P. 107: Thirty-one. (6, 31) with 4 of each value -- as in Magician's Own Book.
Algernon Bray. Letter:
"31" game. Knowledge 3
(4 May 1883) 268, item 806. "...
has lately made its appearance in New York, ...." Seems to have no idea as how to win.
Loyd. Problem 38: The twenty‑five up puzzle. Tit‑Bits 32 (12 Jun &
3 Jul 1897) 193 & 258.
= Cyclopedia. 1914. The dice game, pp. 243 & 372. = SLAHP: How games originate, pp. 73 &
114. The first play is arbitrary. The second play is by throwing a die. Further values are obtained by rolling the
die by a quarter turn.
Ball-FitzPatrick. 1st ed., 1898. Généralization récente de cette question, pp. 30-31. (6, 50)
with each number usable at most 3 times. Some analysis.
Ball. MRE, 4th ed., 1905, p. 20.
Some analysis of (6, 50) where each player can play a value at most 3
times -- as in Ball-FitzPatrick, but with the additional sentence: "I have never seen this extension
described in print ...." He also
mentions playing with values limited to two times. In the 5th ed., 1911, pp. 19-21, he elaborates his analysis.
Dudeney. CP.
1907. Prob. 79: The thirty-one
game, pp. 125-127 & 224. Says it
used to be popular with card-sharpers at racecourses, etc. States the first player can win if he starts
with 1, 2 or 5, but the analysis of cases 1 and 2 is complicated. This occurs as No. 459: The thirty-one
puzzle, Weekly Dispatch (17 Aug 1902) 13 & (31 Aug 1902) 13, but he leaves
the case of opening move 2 to the reader, but I don't see the answer given in
the next few columns.
Devant. Tricks for Everyone. Op. cit. in 4.A.1. 1910. The thirty-one
trick, pp. 53-54. Says to get to 3, 10, 17, 24.
Hummerston. Fun, Mirth & Mystery. 1924.
Thirty-one -- a game of skill, pp. 95-96. This uses a layout of four copies of the numbers 1, 2, 3, 4, 5, 6 with one copy of 20 in a
5 x 5 square with the 20
in the centre. Says to get
to 3, 10, 17, 24, but that this will lose to an experienced
player.
Loyd Jr. SLAHP.
1928. The "31 Puzzle
Game", pp. 3 & 87. Loyd Jr
says that as a boy, he often had to play it against all comers with a $50 prize
to anyone who could beat 'Loyd's boy'.
This is the game that Loyd Sr called 'Blind Luck', but I haven't found
it in the Cyclopedia. States the first
player wins with 1, 2 or 5, but only sketches the case for opening with 5. I have seen an example of Blind Luck -- it
has four each of the numbers 1 - 6 arranged around a frame containing a
horseshoe with 13 in it.
McKay. Party Night. 1940. The 21 race, pp. 166. Using the numbers 1, 2, 3, 4, at most four
times, achieve 21. Says to get 1, 6, 11, 16. He doesn't realise that the sucker can be mislead into playing
first with a 1 and losing! Says that
with 1, ..., 5 at most four times, one wants to achieve 26
and that with 1, ..., 6 at most four times, one wants to achieve 31. Gives just the key numbers each time.
Geoffrey Mott-Smith. Mathematical Puzzles for Beginners and
Enthusiasts. (Blakiston, 1946); revised 2nd ed., Dover, 1954.
Prob.
179: The thirty-one game, pp. 117-119
& 231-232. As in Dudeney.
Prob.
180: Thirty-one with dice, p. 119
& 232-233. Throw a die, then make quarter turns to
produce a total of 31. Analysis based
on digital roots (i.e. remainders (mod 9)).
First player wins if the die comes up 4, otherwise the second player can
win. He doesn't treat any other totals.
"Willane". Willane's Wizardry. Academy of Recorded Crafts, Arts and
Sciences, Croydon, 1947. "Trente
et un", pp. 56-57. Says he doesn't
know any name for this. Get 31 using 4
each of the cards A, 2, ..., 6. Says first player loses easily if he starts
with 4, 5, 6 (not true according to Dudeney) and that gamblers dupe the sucker
by starting with 3 and winning enough that the sucker thinks he can win by
starting with 3. But if he starts with
a 1 or 2, then the second player must play low and hope for a break.
Walter B. Gibson. Fell's Guide to Papercraft Tricks, Games and
Puzzles. Frederick Fell, NY, 1963. Pp. 54-55: First to fifty. First describes (50, 6), but then adds a
version with slips of paper: eight
marked 1 and seven marked with 2, 3, 4,
5, 6 and you secretly extract a 6 slip
when the other player starts.
Harold Newman. The 31 Game. JRM 23:3 (1991) 205-209.
Extended analysis. Confirms
Dudeney. Only cites Dudeney &
Mott-Smith.
Bernard Murphy. The rotating die game. Plus 27 (Summer 1994) 14-16. Analyses the die version as described by
Mott-Smith and finds the set,
S(n), of winning moves for
achieving a count of n by the first player, is periodic with period
9 from n = 8, i.e. S(n+9) =
S(n) for n ³ 8. There is no first player winning move if and only if n is
a multiple of 9. [I have confirmed this
independently.]
Ken de Courcy. The Australian Gambling Game of 31. Supreme Magic Publication, Bideford, Devon,
nd [1980s?]. Brief description of the game
and some indications of how to win. He
then plays the game with face-down cards!
However, he insures that the cards by him are one of of each rank and he
knows where they are.
Loyd?? Problem 43: The daisy game.
Tit‑Bits 32 (17 Jul
& 7 Aug 1897) 291 &
349. (= Cyclopedia. 1914.
A daisy puzzle game, pp. 85 & 350.
c= MPSL2, prob. 57, pp. 40‑41 & 140. c= SLAHP: The daisy game, pp. 42 & 99.) Circular version of Kayles with 13
objects. Solution uses a symmetry
argument -- but the Tit‑Bits solution was written by Dudeney.
Dudeney. Problem 500: The cigar puzzle. Weekly Dispatch (7 Jun, 21 Jun, 5 Jul,
1903) all p. 16. (= AM, prob. 398,
pp. 119, 242.) Symmetry in placement
game, using cigars on a table.
Loyd. Cyclopedia. 1914. The great Columbus problem, pp. 169 &
361. (= MPSL1, prob. 65, pp. 62 &
144. = SLAHP: When men laid eggs,
pp. 75 & 115.) Placing eggs on
a table.
Maurice Kraitchik. La Mathématique des Jeux. Stevens, Bruxelles, 1930. Section XII, prob. 1, p. 296. (= Mathematical Recreations; Allen &
Unwin, London, 1943; Problem 1, pp. 13‑14.) Child plays black and white against two
chess players and guarantees to win one game.
[MJ cites L'Echiquier (1925) 84, 151.]
CAUTION. The 2nd edition of Math. des Jeux, 1953, is
a translation of Mathematical Recreations and hence omits much of the earlier
edition.
Leopold. At Ease!
1943. Chess wizardry in two
minutes, pp. 105‑106. Same as
Kraitchik.
This
has objects in a line or a circle and one can remove one object or two adjacent
objects (or more adjacent objects in a generalized version of the game). This derives from earlier games with an
array of pins at which one throws a ball or stick.
Murray 442 cites Act 17 of Edward IV,
c.3 (1477): "Diversez novelx
ymagines jeuez appellez Cloishe Kayles ..." This outlawed such games.
A 14C picture is given in [J. A. R. Pimlott; Recreations; Studio Vista,
1968, plate 9, from BM Royal MS 10 E IV f.99] showing a 3 x 3
array of pins. A version is
shown in Pieter Bruegel's painting "Children's Games" of 1560 with
balls being thrown at a row of pins by a wall, in the back right of the
scene. Versions of the game are given
in the works of Strutt and Gomme cited in 4.B.1. Gomme II 115‑116 discusses it under Roly‑poly, citing
Strutt and some other sources. Strutt
270‑271 (= Strutt-Cox 219-220) calls it "Kayles, written also cayles
and keiles, derived from the French word quilles". He has redrawings of two 14C engravings
(neither that in Pimlott) showing lines of pins at which one throws a stick (=
plate opp. 220 in Strutt-Cox). He also
says Closh or Cloish seems to be the same game and cites prohibitions of it in
c1478 et seq. Loggats was analogous and
was prohibited under Henry VIII and is mentioned in Hamlet.
14C MS in the British Museum,
Royal Library, No. 2, B. vii.
Reproduced in Strutt, p. 271.
Shows a monk(?) standing by a line of eight conical pins and another
monk(?) throwing a stick at the pins.
Anonymous. Games of the 16th Century. The Rockliff New Project Series. Devised by Arthur B. Allen. The Spacious Days of Queen Elizabeth. Background Book No. 5. Rockliff Publishing, London, ©1950, 4th
ptg. The Background Books seem to be
consecutively paginated as this booklet is paginated 129-152. Pp. 133-134 describes loggats, quoting
Hamlet and an unknown poet of 1611. P.
137 is a photograph of the above 14C illustration. The caption is "Skittles, or "Kayals", and
Throwing a Whirling Stick".
van Etten. 1624.
Prob. 72 (misnumbered 58) (65), pp 68‑69 (97‑98): Du jeu des
quilles (Of the play at Keyles or Nine-Pins).
Describes the game as a kind of ninepins.
Loyd. Problem 43: The daisy game.
Tit‑Bits 32 (17 Jul
& 7 Aug 1897) 291 &
349. (= Cyclopedia. 1914.
A daisy puzzle game, pp. 85 & 350.
c= MPSL2, prob. 57, pp. 40‑41 & 140. c= SLAHP: The daisy game, pp. 42 &
99.) Circular version of Kayles with 13
objects. See also 4.A.2.
Dudeney. Sharpshooters puzzle. Problem 430. Weekly Dispatch (26 Jan, 9 Feb, 1902) both p. 13. Simple version of Kayles.
Ball. MRE, 4th ed., 1905, pp. 19-20.
Cites Loyd in Tit‑Bits.
Gives the general version:
place p counters in a circle and one can take not
more than m adjacent ones.
Dudeney. CP.
1907. Prob. 73: The game of
Kayles, pp. 118‑119 & 220.
Kayles with 13 objects.
Loyd. Cyclopedia. 1914. Rip van Winkle puzzle, pp. 232 & 369‑370. (c= MPSL2, prob. 6, pp. 5 & 122.) Linear version with 13 pins and the second
knocked down. Gardner asserts that Dudeney
invented Kayles, but it seems to be an abstraction from the old form of the
game.
Rohrbough. Puzzle Craft, later version, 1940s?. Daisy Game, p. 22. Kayles with 13 petals of a daisy.
Philip Kaplan. More Posers. (Harper & Row, 1964);
Macfadden-Bartell Books, 1965.
Prob. 45, pp. 48 & 95.
Circular kayles with five objects.
Doubleday - 2. 1971.
Take your pick, pp. 63-65. This
is Kayles with a row of 10, but he says the first player can only take one.
Nim is the game with a number of piles
and a player can take any number from one of the piles. Normally the last one to play wins.
David Parlett. (Originally: The Oxford Guide to Card Games; OUP, 1990); reissued as: A History of Card Games.
Penguin, 1991. Pp. 174-175. "Early references to 'les luettes',
said to have been played by Anne de Bretagne and Archduke Philip the Fair in
1503, and by Gargantua in 1534, seem to suggest a game of the Nim family
(removing numbers of objects from rows and columns)."
Charles L. Bouton. Nim: a game with a complete mathematical
theory. Annals of Math. (2) 3 (1901/02)
35‑39. He says Nim is played at
American colleges and "has been called Fan‑Tan, but as it is not the
Chinese game of that name, the name in the title is proposed for it." He says Paul E. More showed him the misère
(= last player loses) version in 1899,
so it seems that Bouton did not actually invent the game himself.
Ahrens. "Nim", ein amerikanisches Spiel
mit mathematischer Theorie.
Naturwissenschaftliche Wochenschrift
17:22 (2 Mar 1902) 258‑260.
He says that Bouton has admitted that he had confused Nim and Fan‑Tan. Fan‑Tan is a Chinese game where you
bet on the number of counters (mod 4) in someone's hand. Parker, Ancient Ceylon, op. cit. in 4.B.1,
pp. 570-571, describes a similar game, based on odd and even, as popular in
Ceylon and "certainly one of the earliest of all games".
For
more about Fan-Tan, see the following.
Stewart Culin. Chess and playing cards. Catalogue of games and implements for
divination exhibited by the United States National Museum in connection with
the Department of Archæology and Paleontology of the University of Pennsylvania
at the Cotton States and International Exposition, Atlanta, Georgia, 1895. IN: Report of the U. S. National Museum,
year ending June 30, 1896. Government
Printing Office, Washington, 1898, HB, pp. 665-942. [There is a reprint by Ayer Co., Salem, Mass., c1990.] Fan-Tan (= Fán t‘án = repeatedly
spreading out) is described on pp. 891 & 896, with discussion of related
games on pp. 889-902.
Alan S. C. Ross. Note 2334:
The name of the game of Nim. MG
37 (No. 320) (May 1953) 119‑120.
Conjectures Bouton formed the word 'nim' from the German 'nimm'. Gives some discussion of Fan‑Tan and
quotes MUS I 72.
J. L. Walsh. Letter:
The name of the game of Nim. MG
37 (No. 322) (Dec 1953) 290.
Relates that Bouton said that he had chosen the word from the German
'nimm' and dropped one 'm'.
W. A. Wythoff. A modification of the game of Nim. Nieuw Archief voor Wiskunde (Groningen) (2)
7 (1907) 199‑202. He considers a
Nim game with two piles allows the extra move of taking the same amount from
both piles. [Is there a version with
more piles where one can take any number from one pile or equal amounts from
two piles?? See Barnard, below for a
three pile version.]
Ahrens. MUS I.
1910. III.3.VII: Nim, pp. 72‑88. Notes that Nim is not the same as Fan‑Tan,
has been known in Germany for decades and is played in China. Gives a thorough discussion of the theory of
Nim and of an equivalent game and of Wythoff's game.
E. H. Moore. A generalization of the game called
Nim. Annals of Math. (2) 11 (1910) 93‑94. He considers a Nim game with n
piles and one is allowed to take any number from at most k
piles.
Ball. MRE, 5th ed., 1911, p. 21.
Sketches the game of Nim and its theory.
A. B. Nordmann. One Hundred More Parlour Tricks and
Problems. Wells, Gardner, Darton &
Co., London, nd [1927 -- BMC]. No. 13:
The last match, pp. 10-11. Thirty
matches divided at random into three heaps.
Last player loses. Explanation
of how to win is rather cryptic:
"you must try and take away ... sufficient ... to leave the matches
in the two or three heaps remaining, paired in ones, twos, fours, etc., in
respect of each other."
Loyd Jr. SLAHP.
1928. A tricky game, pp. 47
& 102. Nim (3, 4, 8).
Emanuel Lasker. Brettspiele der Völker. 1931.
See comments in 4.A.5. Jörg
Bewersdorff [email of 6 Jun 1999] says that Lasker considered a three person
Nim and found an equilibrium for it -- see: Jörg Bewersdorff; Glück, Logik und
Bluff Mathematik im Spiel -- Methoden,
Ergebnisse und Grenzen; Vieweg, 1998, Section 2.3 Ein Spiel zu dritt, pp. 110-115.
Lynn Rohrbough, ed. Fun in Small Spaces. Handy Series, Kit Q, Cooperative Recreation
Service, Delaware, Ohio, nd [c1935].
Take Last, p. 10. Last player
loses Nim (3, 5, 7).
Rohrbough. Puzzle Craft. 1932.
Japanese
Corn Game, p. 6 (= p. 6 of 1940s?).
Last player loses Nim (1, 2, 3, 4, 5).
Japanese
Corn Game, p. 23. Last player loses
Nim (3, 5, 7).
René de Possel. Sur la Théorie Mathématique des Jeux de
Hasard et de Réflexion. Actualités
Scientifiques et Industrielles 436.
Hermann, Paris, 1936. Gives the
theory of Nim and also the misère version.
Depew. Cokesbury Game Book.
1939. Make him take it, pp.
187-188. Nim (3, 4, 5), last player loses.
Edward U. Condon, Gereld L.
Tawney & Willard A. Derr. US Patent
2,215,544 -- Machine to Play Game of Nim.
Filed: 26 Apr 1940; patented: 24
Sep 1940. 10pp + 11pp diagrams.
E. U. Condon. The Nimatron. AMM 49 (1942) 330‑332.
Has photo of the machine.
Benedict Nixon & Len
Johnson. Letters to the Notes &
Queries Column. The Guardian
(4 Dec 1989) 27. Reprinted in: Notes & Queries, Vol. 1; Fourth Estate,
London, 1990, pp. 14-15. These describe
the Ferranti Nimrod machine for playing Nim at the Festival of Britain,
1951. Johnson says it played Nim (3, 5, 6) with a maximum move of
3. The Catalogue of the
Exhibition of Science shows this as taking place in the Science Museum.
H. S. M. Coxeter. The golden section, phyllotaxis, and
Wythoff's game. SM 19 (1953) 135‑143. Sketches history and interconnections.
H. S. M. Coxeter. Introduction to Geometry. Wiley, 1961. Chap. 11: The golden section and phyllotaxis, pp. 160-172. Extends his 1953 material.
A. P. Domoryad. Mathematical Games and Pastimes. (Moscow, 1961). Translated by Halina Moss.
Pergamon, Oxford, 1963. Chap.
10: Games with piles of objects, pp. 61‑70. On p. 62, he asserts that Wythoff's game is 'the Chinese national
game tsyanshidzi ("picking stones")'. However M.‑K. Siu cannot recognise such a Chinese game,
unless it refers to a form of jacks, which has no obvious connection with
Wythoff's game or other Nim games. He
says there is a Chinese character, 'nian', which is pronounced 'nim' in
Cantonese and means to pick up or take things.
N. L. Haddock. Note 2973:
A note on the game of Nim. MG 45
(No. 353) (Oct 1961) 245‑246.
Wonders if the game of Nim is related to Mancala games.
T. H. O'Beirne. Puzzles and Paradoxes. OUP, 1965.
Section on misère version of Wythoff's game, p. 133. Richard Guy (letter of 27 Feb 1985) says
this is one of O'Beirne's few mistakes -- cf next entry.
Winning Ways. 1982.
P. 407 says Wythoff's game is also called Chinese Nim or Tsyan‑shizi. No reference given. See comment under Domoryad above. This says many authors have done this
incorrectly.
D. St. P. Barnard. 50 Daily Telegraph Brain‑Twisters. Javelin Books, Poole, Dorset, 1985. Prob. 30: All buttoned up, pp. 49‑50,
91 & 115. He suggests three pile
game where one can take any number from one pile or an equal number from any
two or all three piles. [See my note to
Wythoff, above.]
Matthias Mala. Schnelle Spiele. Hugendubel, Munich, 1988.
San Shan, p. 66. This describes
a nim-like game named San Shan and says it was played in ancient China.
Jagannath V. Badami. Musings on Arithmetical Numbers Plus Delightful Magic Squares. Published by the author, Bangalore, India,
nd [Preface dated 9 Sep 1999]. Section
4.16: The game of Nim, pp. 124-125.
This is a rather confused description of one pile games (21, 5) and (41,
5), but he refers to solving them by (mentally) dividing the pile into
piles. This makes me think of combining
the two games, i.e. playing Nim with several piles but with a limit on the
number one can take in a move.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. Ff. 134-144 are: Essay 10 Part 5. See 4.B.1 for more details.
At the top of f. 134.r, he has added a note: "This is probably my earliest Note on
Games of Skill. I do not recollect the
date. 3 March 1865". He then describes Tit Tat To and makes some
simple analysis, but he never uses a name for it.
Charles Babbage. Notebooks -- unpublished collection of MSS
in the BM as Add. MS 37205. ??NX. See 4.B.1 for more details. On f. 304, he starts on analysis of
games. Ff. 310‑383 are
almost entirely devoted to Tit-Tat-To, with some general discussions. F. 321.r, 10 Sep 1860, is the beginning
of a summary of his work on games of skill in general. F. 324-333, Oct 1844, studies "General
laws for all games of Skill between two players" and draws flow charts
showing the basic recursive analysis of a game tree (ff. 325.v &
325.r). On f. 332, he counts the
number of positions in Tit Tat To as
9! + 8! + ... + 1! = 409,113.
F. 333 has an idea of the tree structure of a game.
John M. Dubbey. The Mathematical Work of Charles
Babbage. CUP, 1978, pp. 96‑97
& 125‑130. See 4.B.1 for more
details. He discusses the above Babbage
material. On p. 127, Dubbey
has: "The basic problem is one
that appears not to have been previously considered in the history of
mathematics." Dubbey, on p. 129,
says: "This analysis ... must
count as the first recorded stochastic process in the history of
mathematics." However, it is
really a deterministic two-person game.
E. Zermelo. Über eine Anwendung der Mengenlehre auf die
Theorie des Schachspiels. Proc. 5th ICM
(1912), CUP, 1913, vol. II, 501‑504. Gives general idea of first and second
person games.
Ahrens. A&N.
1918. P. 154, note. Says that each particular Dots and Boxes
board, with rational play, has a definite outcome.
W. Rivier. Archives des Sciences Physiques et
Naturelles (Nov/Dec 1921). ??NYS --
cited by Rivier (1935) who says that the later article is a new and simpler
version of this one.
H. Steinhaus. Difinicje potrzebne do teorji gry i
pościgu (Definitions for a theory of games and pursuit). Myśl Akademicka (Lwów) 1:1 (Dec 1925)
13‑14 (in Polish). Translated,
with an introduction by Kuhn and a letter from Steinhaus in: Naval Research Logistics Quarterly 7 (1960)
105‑108.
Dénès König. Über eine Schlussweise aus dem Endlichen ins
Unendliche. Mitteilungen der Universitä
Szeged 3 (1927) 121-130. ??NYS -- cited
by Rivier (1935). Kalmár cites it to
the same Acta as his article.
László Kalmár. Zur Theorie der abstracten Spiele. Acta Litt. Sci. Regia Univ. Hungaricae
Francisco‑Josephine (Szeged) 4 (1927) 62‑85. Says there is a gap in Zermelo which has
been mended by König. Lengthy approach,
but clearly gets the idea of first and second person games.
Max Euwe. Proc. Koninklijke Akadamie van Wetenschappen
te Amsterdam 32:5 (1929). ??NYS --
cited by Rivier (1935).
Emanuel Lasker. Brettspiele der Völker. Rätsel‑ und mathematische Spiele. A. Scherl, Berlin, 1931, pp. 170‑203. Studies the one pile game (100, 5)
and the sum of two one‑pile games: (100, 5) + (50, 3). Discusses Nimm, "an old Chinese game according to Ahrens"
and says the solver is unknown. Gives
Lasker's Nim -- one can take any amount from a pile or split it in two -- and
several other variants. Notes that 2nd person + 2nd person is
2nd person while 2nd person + 1st person is
1st person. Gives the idea of
equivalent positions. Studies three
(and more) person games, assuming the pay‑offs are all different. Studies some probabilistic games. Jörg Bewersdorff [email of 6 Jun 1999]
observes that Lasker's analysis of his Nim got very close to the idea of the
Sprague-Grundy number. See: Jörg
Bewersdorff; Glück, Logik und Bluff
Mathematik im Spiel -- Methoden, Ergebnisse und Grenzen; Vieweg, 1998,
Section 2.5 Lasker-Nim: Gewinn auf
verborgenem Weg, pp. 118-124.
W. Rivier. Une theorie mathématique des jeux de
combinaisions. Comptes-Rendus du
Premier Congrès International de Récréation Mathématique, Bruxelles, 1935. Sphinx, Bruxelles, 1935, pp. 106‑113. A revised and simplified version of his 1921
article. He cites and briefly discusses
Zermelo, König and Euwe. He seems to be
classifying games as first player or second player.
René de Possel. Sur la Théorie Mathématique des Jeux de
Hasard et de Réflexion. Actualités
Scientifiques et Industrielles 436.
Hermann, Paris, 1936. Gives the
theory of Nim and also the misère version.
Shows that any combinatorial game is a win, loss or draw and describes
the nature of first and second person positions. He then goes on to consider games with chance and/or bluffing,
based on von Neumann's 1927 paper.
R. Sprague. Über mathematische Kampfspiele. Tôhoku Math. J. 41 (1935/36) 438‑444.
P. M. Grundy. Mathematics and games. Eureka 2 (1939) 6‑8. Reprinted, ibid. 27 (1964) 9‑11. These two papers develop the Sprague-Grundy
Number of a game.
D. W. Davies. A theory of chess and noughts and
crosses. Penguin Science News 16 (Jun
1950) 40-64. Sketches general ideas of
tree structure, Sprague-Grundy number, rational play, etc.
H. Steinhaus. Games, an informal talk. AMM 72 (1965) 457‑468. Discusses Zermelo and says he wasn't aware
of Zermelo in 1925. Gives Mycielski's
formulation and proof via de Morgan's laws.
Goes into pursuit and infinite games and their relation to the Axiom of
Choice.
H. Steinhaus. (Proof that a game without ties has a
strategy.) In: M. Kac; Hugo Steinhaus -- a
reminiscence and a tribute; AMM 81 (1974) 572‑581. Repeats idea of his 1965 talk.
See
5.M for Sim and 5.R.5 for Fox and Geese, etc.
Most
of the board games described here are classic and have been extensively
described and illustrated in the various standard books on board games,
particularly the works of Robert C. Bell, especially his Board and
Table Games from Many Civilizations;
OUP, vol. I, 1960, vol. II, 1969;
combined and revised ed., Dover, 1979 and the older work of Edward G.
Falkener; Games Ancient and Oriental and How to Play Them; Longmans, Green, 1892; Dover, 1961. The works by Culin (see 4.A.4, 4.B.5 and 4.B.9) are often
useful. Several general works on games
are cited in 4.B.1 and 4.B.5 -- I have read Murray's History of Board Games
Other than Chess, but not yet entered the material. Note that many of these works are more concerned with the game
than with its history and have a tendency to exaggerate the ages of games by
assuming, e.g. that a 3 x 3 board must have been used for
Tic-Tac-Toe. I will not try to
duplicate the descriptions by Bell, Falkener and others, but will try to
outline the earliest history, especially when it is at variance with common
belief. The most detailed mathematical
analyses are generally in Winning Ways.
4.B.1. TIC‑TAC‑TOE = NOUGHTS AND CROSSES
Popular
belief is that the game is ancient and universal -- e.g. see Brandreth,
1976. However the game appears to have
evolved from earlier three‑in‑a‑row games, e.g. Nine Holes or
Three Men's Morris, in the early 19C.
See also the historical material in 4.B.5. The game is not mentioned in Strutt nor most other 19C books on games,
not even in Kate Greenaway's Book of Games (1889), nor in Halliwell's section
on slate games (op. cit. in 7.L.1, 1849, pp. 103-104), but there may be an
1875 description in Strutt-Cox of 1903.
Babbage refers to it in his unpublished MSS of c1820 as a children's
game, but without giving it a name. In
1842, he calls it Tit Tat To and he uses slight variations on this name in his
extended studies of the game -- see below.
The OED's earliest references are:
1849 for Tip‑tap‑toe;
1855 for Tit‑tat‑toe;
1861 for Oughts and Crosses.
However, the first two entries may be referring to some other game --
e.g. the entries for Tick‑tack‑toe for 1884 & 1899 are clearly
to the game that Gomme calls Tit‑tat‑toe. Von der Lasa cites a 1838-39 Swedish book for Tripp, Trapp,
Trull. Van der Linde (1874, op. cit. in
5.F.1) gives Tik, Tak, Tol as the Dutch name.
Using the works of Strutt, Gomme, Strutt-Cox, Fiske, Murray, the OED and
some personal communications, I have compiled a separate index of 121
variant names which refer to
5 basic games, with a few
variants and a few unknown games. The
Murray and Parker material is given first, as it deals generally with the
ancient history. Then I list several
standard sources and then summarize their content. Other material follows that.
Fiske says that van der Linde and von der Lasa (see 5.F.1) mention early
appearances of Morris games, but rather briefly and I don't always have that
material.
The
usual # shape board will be so indicated. If one is setting down pieces, then the board is often drawn as a
'crossed square', i.e. a square with its horizontal and vertical midlines
drawn, and one plays on the intersections.
Fiske 127 says this form is common in Germany, but unknown in England
and the US. In addition, the diagonals
are often drawn, producing a 'doubly crossed square'. The squares are sometime drawn as circles giving a 'crossed
circle' and a 'doubly crossed circle', though it is hard to identify the
corners in a crossed circle. The 3 x 3
array of dots sometimes occurs.
The standard # pattern is sometimes surrounded by a square
producing a '3 x 3 chessboard'.
Fiske
129 says the English play with O and
+, while the Swedes play with O
and 1. My experience is that English and Americans play with O
and X. One English friend said that where she grew up, it was called 'Exeter's
Nose' as a deliberate corruption of 'Xs and Os'.
The
first clear references to the standard game of Noughts and Crosses are Babbage
(1820) and the items discussed under Tic-tac-toe below. Further clear references are: Cassell's, Berg, A wrangler ..., Dudeney,
White and everything entered below after White.
Misère
version: Gardner (1957); Scotts (1975);
Murray mentions Morris, which he
generally calls Merels, many times.
Besides the many specific references mentioned below and in 4.B.5, he shows,
on p. 614, under Nine Holes and Three Men's Morris, a number of 3 x 3
diagrams.
Kurna,
Egypt, (-14C) -- a double crossed square and a double crossed circle -- see
Parker below.
Ptolemaic
Egypt (in the BM, no. 14315) -- a square with
# drawn inside. See below where I describe this, from a
recent exhibition, as just a # board.
Ceylon
-- a doubly crossed square -- see Parker below.
Rome
and Pompeii -- doubly crossed circles.
Under
Nine Holes, he says a piece can be moved to any vacant point; under Three Men's Morris, he says a man can
only be moved along a marked line to an adjacent point, i.e. horizontally,
vertically or along a main diagonal.
Under
Nine Holes, he shows the # board for English Noughts and Crosses. He specifically notes that the pieces do not
move. His only other mention of this
board is for a Swedish game called Tripp, Trapp, Trull, but he does not state
that the pieces do not move. He gives
no other examples of the # board nor of non‑moving pieces.
He
also mentions Five (or Six) Men's Morris, of which little is known. On p. 133, he mentions a 3 x 3
"board of nine points used for a game essentially identical with
the 'three men's merels', which has existed in China from at least the time of
the Liang dynasty (A.D. 502‑557).
The 'Swei shu' (first half of the 7th c.) gives the names of twenty
books on this game."
H. Parker. Ancient Ceylon. ??, London, 1909; Asian
Educational Services, New Delhi, 1981.
Nerenchi keliya, pp. 577‑580 & 644. There is a crossed square with small holes at the intersections
at the Temple of Kurna, Upper Egypt, ‑14C. [Rohrbough, loc. cit. in 4.B.5, says this temple was started by
Ramses I and completed by Seti in -1336/-1333, citing J. Royal Asiatic Soc.
(1783) 17.] On p. 644, he shows 34
mason's diagrams from Kurna, which include
#, # in a circle, crossed square with small holes at the
intersections, doubly crossed square, doubly crossed circle. He cites Bell, Arch. Survey of Ceylon, Third
Progress Report, p. 5 note, for for a doubly crossed square in Ceylon, c1C, but
Noughts and Crosses is not found in the interior of Ceylon. The doubly crossed square was used in 18C
Ireland. On pp. 643-665, he discusses
appearances of the crossed square and doubly crossed circle as designs or
characters and claims they have mystic significance. On p. 662, he lists many early appearances of the #
pattern.
Murray 440, note 63, includes a
reference to Soutendam; Keurboek van Delft; Delft, c1425, f. 78 (or p. 78?);
who says games of subtlety are allowed, e.g. ... ticktacken. There is no indication if this may be our
game and the OED indicates that such names were used for backgammon back to
1558. The OED doesn't cite: W. Shakespeare; Measure for Measure,
c1604. Act I, scene ii, line 180 (or
196): "foolishly lost at a game of
ticktack". Later it was more
common as Tric-trac.
Murray 746 notes a Welsh game
Gwyddbwyll mentioned in the Mabinogion (14C).
The name is cognate with the Irish Fidchell and may be a Three Men's
Morris, but the game was already forgotten by the 15C.
STANDARD
SOURCES ON GAMES
Joseph Strutt. The Sports and Pastimes of the People of
England. (With title starting: Glig‑Gamena Angel-Ðeod., or the Sports
...; J. White, London, 1791, 1801, 1810).
A new edition, with a copious index, by William Hone. Tegg, London, 1830, 1831, 1833, 1834, 1838,
1841, 1850, 1855, 1875, 1876, 1891.
[The 1830 ed. has a preface, omitted in 1833, stating that the 1810 ed.
is the same as the 1801 ed. and that Hone has only changed it by adding the
Index and incorporating some footnotes into the text.] [Hall, BCB 263-266 are: 1801,
1810, 1830, 1831.
Toole Stott 647-656 are:
1791; 1801; 1810;
1828-1830 in 10 monthly parts with Index by Hone; 1830; 1830;
1833; 1838; 1841;
1876, an expanded ed, ed by Hone.
Heyl 300-302 gives 1830; 1838;
1850. Toole Stott 653 says the
sheets were remaindered to Hone, who omitted the first 8pp and issued it
in 1833, 1834, 1838, 1841.
I have seen an 1855 ed. C&B
list 1801, 1810, 1830, 1903. BMC has
1801, 1810, 1830, 1833, 1834, 1838, 1841, 1875, 1876, 1898.]
Strutt-Cox. The Sports and Pastimes of the People of
England. By Joseph Strutt. 1801.
A new edition, much enlarged and corrected by J. Charles Cox. Methuen, 1903. The Preface sketches Strutt's life and says this is based on the
'original' 1801 in quarto, with separate plates which were often hand coloured,
but not consistently, while the 1810 reissue had them all done in a terra‑cotta
shade. Hone reissued it in octavo in 1830
with the plates replaced by woodcuts in the text and this was reissued in 1837,
1841 and 1875. (From above we see that
there were other reissues.) "Mr.
Strutt has been left for the most part to speak in his own characteristic
fashion .... A few obvious mistakes and
rash conclusions have been corrected, ... certain unimportant omissions have
been made. ... Nearly a third of the book is new." Reprinted in 1969 and in the 1960s?
J. T. Micklethwaite. On the indoor games of school boys in the
middle ages. Archaeological Journal 49
(Dec 1892) 319-328. Describes
various 3 x 3 boards and games on them, including Nine Holes and "tick,
tack, toe; or oughts and crosses, which I suppose still survives wherever
slate and pencil are used as implements of education", Three Men's Morris
and also Nine Men's Morris, Fox and Geese, etc.
Alice B. Gomme. The Traditional Games of England, Scotland,
and Ireland. 2 vols., David Nutt,
London, 1894 & 1898. Reprinted in
one vol., Thames & Hudson, London, 1984.
Willard Fiske. Chess in Iceland and in Icelandic Literature
with Historical Notes on Other Table-Games.
The Florentine Typographical Society, Florence, 1905. Esp. pp. 97-156 of the Stray Notes. P. 122 lists a number of works on ancient
games.
These
and the OED have several entries on Noughts and Crosses and Tic‑tac‑toe
and many on related games, which are summarised below. Gomme often cites or quotes Strutt. The OED often gives the same quotes as
Gomme. Gomme's references are highly
abbreviated but full details of the sources can usually be found in the OED.
(Nine
Men's) Morris,
where Morris is spelled about 30 different ways, e.g. Marl, Merelles, Mill,
Miracles, Morals, and Nine Men's may be given as, e.g. Nine‑peg, Nine
Penny, Nine Pin. Also known as Peg
Morris and Shepherd's Mill. Gomme I 80
& 414‑419 and Strutt 317‑318 (c= Strutt-Cox 256-258 & plate
opp. 246, which adds reference to Micklethwaite) are the main entries. See 4.B.5 for material more specifically on
this game.
Nine
Holes, also
known as Bubble‑justice, Bumble‑puppy, Crates, and possibly Troll‑madam,
Troule‑in‑Madame. Gomme I
413‑414 and Strutt 274‑275 & 384 (c= Strutt‑Cox
222-223 & 304) are the main entries.
Twelve Holes is similar [Gomme II 321 gives a quote from 1611]. There seem to be cases where Nine Men's
Morris was used in referring to Nine Holes [Gomme I 414‑419]. There are two forms of the game: one form has holes in an upright board that
one must roll a ball or marble through;
the other form has holes in the ground, usually in a 3 x 3 array, that one must roll balls into. Unfortunately, none of the references
implies that one has to get three in a row -- see Every Little Boys Book for a
version where this is certainly not the case.
There are references going back to 1572 for Crates (but mentioning
eleven holes) [Gomme I 81 & II 309] and 1573 [OED] for Nine Holes. Botermans et al.; The World of Games; op.
cit. in 4.B.5; 1989; p. 213, shows a 17C engraving by Ménian showing Le Jeu de
Troumadame as having a board with holes in it, held vertically on a table and
one must roll marbles through the holes.
They say it is nowadays known as 'bridge'.
Three
Men's Morris. This is less common, but occurs in several
variant spellings corresponding to the variants of Nine Men's Morris,
including, e.g. Three‑penny Morris, Tremerel. The game is played on a 3
x 3 board and each player has three men. After making three plays each, consisting of setting men on the
cells, further play consists of picking up one of your own men and placing it
on a vacant cell, with the object of getting three in a row. There are several versions of this game,
depending on which cells one may play to, but the descriptions given rarely
make this clear. [Gomme I 414‑419]
quotes from F. Douce; Illustrations of Shakespeare and of Ancient Manners;
1807, i.184. "In the French
merelles each party had three counters only, which were to be placed in a line
to win the game. It appears to have
been the tremerel mentioned in an old fabliau.
See Le Grand, Fabliaux et Contes, ii.208. Dr. Hyde thinks the morris, or merrils, was known during the time
that the Normans continued in possession of England, and that the name was
afterwards corrupted into three men's morals, or nine men's morals." [Hyde.
Hist. Nederluddi [sic], p. 202.]
In practice, the board is often or usually drawn as a crossed
square. If one can move along all
winning lines, then it would be natural to draw a doubly crossed square. See under Alfonso MS (1283) in 4.B.5 for
versions called marro, tres en raya and riga di tre. Again, much of the material on this game is in 4.B.5.
Five‑penny
Morris. None of the references make it clear, but
this seems to be (a form of) Three Men's Morris. Gomme I 122 and the OED [under Morrell] quote: W. Hawkins; Apollo Shroving (a play of
1627), act III, scene iv, pp. 48-49.
"...,
Ovid hath honour'd my exercises.
He describes in verse our boyes play.
Twise
three stones, set in a crossed square where he wins the game
That
can set his three along in a row,
And
that is fippeny morrell I trow."
Most of the references (and
myself) are perplexed by the reference to five, though the fact that one has at
most five moves in Tic‑tac‑toe might have something to do with
it?? Since Three Men's Morris is less
well known, some writers have assumed Five‑penny Morris was Nine Men's
Morris and others have called all such games by the same name. A few lines later, Hawkins has: "I
challenge him at all games from blowpoint upward to football, and so on to
mumchance, and ticketacke. ... rather
than sit out, I will give Apollo three of the nine at Ticketacke,
..."
Corsicrown [Gomme I 80] seems to be a
version of Three Men's Morris, but using seven of the nine cells, omitting two
opposite side cells. Gomme quotes from
J. Mactaggart; The Scottish Gallovidian Encyclopedia; (1871 or possibly
1824?): "each has three men ....
there are seven points for these men to move about on, six on the edges of the
square and one at the centre."
Tic‑tac‑toe. The earliest clearly described versions are given in Babbage
(with no name given), c1820, and Gomme I 311, under Kit‑cat‑cannio,
where she quotes from: Edward Moor;
Suffolk Words and Phrases; 1823 (This word does not occur in the OED). Gomme also gives entries for Noughts and
Crosses [I 420‑421] and Tip‑tap‑toe [II 295‑296] with
variants Tick‑tack‑toe and Tit‑tat‑toe. In 1842-1865, Babbage uses Tit Tat To and
slight variants. Under Tip‑tap‑toe,
Gomme says the players make squares and crosses and that a tie game is a score
for Old Nick or Old Tom. (When I was
young, we called it Cat's Game, and this is an old Scottish term [James T. R.
Ritchie; The Singing Street Scottish
Children's Games, Rhymes and Sayings; (O&B, 1964); Mercat Press,
Edinburgh, 2000, p. 61].) She quotes
regional glossaries for Tip‑tap‑toe (1877), Tit‑tat‑toe
(1866 & 1888), Tick‑tack‑toe (1892). The OED entry for Oughts and Crosses seems to be this game and
gives an 1861 quote. Von der Lasa cites
a 1838-39 Swedish book for Tripp, Trapp, Trull. Van der Linde (1874, op. cit. in 5.F.1) gives Tik, Tak, Tol as
the Dutch name.
Tit‑tat‑toe [Gomme II 296‑298]. This is a game using a slate marked with a
circle and numbered sectors. The player
closes his eyes and taps three times with a pencil and tries to land on a good
sector. Gomme gives the verse:
Tit,
tat, toe, my first go,
Three
jolly butcher boys all in a row
Stick
one up, stick one down,
Stick
one in the old man's ground.
But cf Games and Sports for
Young Boys, 1859, below.
The
OED entries under Tick‑tack, Tip‑tap and Tit give a number of
variant spellings and several quotations, which are often clearly to this game,
but are sometimes unclear. Also some
forms seem to refer to backgammon.
In
her 'Memoir on the study of children's games' [Gomme II 472‑473], Gomme
gives a somewhat Victorian explanation of the origin of Old Nick as the winner
of a tie game as stemming from "the primitive custom of assigning a
certain proportion of the crops or pieces of land to the devil, or other earth
spirit."
Franco Agostini & Nicola
Alberto De Carlo. Intelligence
Games. (As: Giochi della Intelligenza; Mondadori, Milan, 1985.) Simon & Schuster, NY, 1987. P. 81 says examples of boards were
discovered in the lowest level of Troy and in the Bronze Age tombs in Co.
Wicklow, Ireland. Their description is
a bit vague but indicates that the Italian version of Tic-tac-toe is actually
Three Men's Morris.
Anonymous. Play the game. Guardian Education section (21 Sep 1993) 18-19. Shows a stone board with the #
incised on it 'from Bet Shamesh, Israel, 2000 BC'. This might be the same as the first board
below??
A small exhibition of board
games organized by Irving Finkel at the British Museum, 1991, displayed the
following.
Stone
slab with the usual # Tick-Tac-Toe board incised on it, but
really a 4 x 3 board. With nine stone men. From Giza, >-850. BM items EA 14315 & 14309, donated by W.
M. Flinders Petrie. Now on display in
Room 63, Case C.
Stone
Nine Holes board from the Temple of Artemis, Ephesus, 2C-4C. Item BM GR 1873.5.5.150. This is a
3 x 3 array of
depressions. Now on display in Room 69,
Case 9.
Robbie Bell & Michael
Cornelius. Board Games Round the
World. CUP, 1988. P. 6 states that the crossed square board
has been found at Kurna (c-1400) and at the Ptolemaic temple at Komombo
(c-300). They state that Three Men's
Morris is the game mentioned by Ovid in Ars Amatoria. They say that it was known to the Chinese at the time of
Confucius (c-500) under the name of Yih, but is now known as Luk tsut k'i. They also say the game is also known as Nine Holes -- which seems
wrong to me.
The Spanish Treatise on
Chess-Play written by order of King Alfonso the Sage in the year 1283. [= Libro de Acedrex, Dados e Tablas of
Alfonso El Sabio, generally known as the Alfonso MS.] MS in Royal Library of the Escorial (j.T.6. fol). Complete reproduction in 194 Phototypic
Plates. 2 vols., Karl W. Hirsemann, Leipzig, 1913. (There was also an edition by Arnald
Steiger, Geneva, 1941.) See 4.B.5 for
more details of this work. Vol. 2, f.
93v, p. CLXXXVI, shows a doubly crossed square board. ??NX -- need to study text.
Pieter Bruegel (the Elder). Children's Games. Painting dated 1560 at the Kunsthistorisches Museum, Vienna. In the right background, children are
playing a game involving throwing balls into holes in the ground, but the holes
appear to be in a straight line.
Anonymous. Games of the 16th Century. 1950.
Op. cit. in 4.A.3. P. 134
describes nine-holes, quoting an unknown poet of 1611: "To play at
loggats, Nine-holes, or Ten-pinnes".
The author doesn't specify what positions the balls are to be rolled
into. P. 152 describes Troll-my-dames
or Troule-in-madame: "they may have in the end of a bench eleven holes
made, into which to troll pummets, or bowls of lead, ...."
William Wordsworth. The Prelude, Book 1. Completed 1805, published 1850. Lines 509‑513.
At
evening, when with pencil, and smooth slate
In
square divisions parcelled out and all
With
crosses and with cyphers scribbled o'er,
We
schemed and puzzled, head opposed to head
In
strife too humble to be named in verse.
It
is not clear if this is referring to Noughts and Crosses.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. F. 4r is part of
the Table of Contents. It shows Noughts
and Crosses games played on the # board and on a 4 x 4 board adjacent to
entry 4: The Mill. Ff. 124-146
are: Essay 10 -- Of questions requiring
the invention of new modes of analysis.
On f. 128.r, he refers to a game in which "the relative
positions of three of the marks is the object of inquiry." Though the reference is incomplete, a Noughts
and Crosses game is drawn on the facing page, f. 127.v. Ff. 134-144 are: Essay 10 Part 5. At the
top of f. 134.r, he has added a note:
"This is probably my earliest Note on Games of Skill. I do not recollect the date. 3 March 1865". The Essay begins: "Amongst the simplest of those games requiring any degree of
skill which amuse our early years is one which is played at in the following
manner." He then describes the
game in detail and makes some simple analysis, but he never uses a name for it.
Charles Babbage. Notebooks -- unpublished collection of MSS
in the BM as Add. MS 37205. ??NX. On f. 304, he starts on analysis of games. Ff. 310-383 are almost entirely devoted to
Tit-Tat-To, with some general discussions.
Most of this material comprises a few sheets of working, carefully
dated, sometimes amended and with the date of the amendment. A number of sheets describe parts of the
automaton that he was planning to build which would play the game, but no such
machine was built until 1949. The
sheets are not always in strict chronological order.
F.
310.r is the first discussion of the game, called Tit Tat To, dated 17 Sep
1842. On F. 312.r, 20 Sep 1843, he says
he has "Reduced the 3024 cases D to 199 which include many Duplicates by
Symmetry." F. 321.r, 10 Sep 1860,
is the beginning of a summary of his work on games of skill in general. He refers to Tit-tat-too. F. 322.r continues and he says: "I have found no game of skill more
simple that that which children often play and which they call Tit‑tat-to." F. 324-333, Oct 1844, studies "General
laws for all games of Skill between two players" and draws flow charts
showing the basic recursive analysis of a game tree (ff. 325.v & 325.r). On f. 332, he counts the number of
positions as 9! + 8! + ... + 1!
= 409,113. F. 333 has an idea of the tree structure of a game. On ff. 337-338, 8 Sep 1848, he has Tit-tat
too. On ff. 347.r-347.v, he suggests
Nine Men's Morris boards in triangular and pentagonal shapes and does various
counting on the different shapes. On
ff. 348-349, 26 Oct 1859, he uses Tit-Tat-To.
John M. Dubbey. The Mathematical Work of Charles
Babbage. CUP, 1978, pp. 96‑97
& 125‑130. He discusses the
above Babbage material. On p. 127,
Dubbey has: "After a surprisingly
lengthy explanation of the rules, he attempts a mathematical formulation. The basic problem is one that appears not to
have been previously considered in the history of mathematics." Babbage represents the game using roots of
unity. Dubbey, on p. 129, says: "This analysis ... must count as the
first recorded stochastic process in the history of mathematics." However, it is really a deterministic
two-person game.
Games and Sports for Young
Boys. Routledge, nd [1859 - BLC]. P. 70, under Rhymes and Calls: "In the game of Tit-tat-toe, which is
played by very young boys with slate and pencil, this jingle is used:--
Tit,
tat, toe, my first go:
Three
jolly butcher boys all in a row;
Stick
one up, stick one down.
Stick
one on the old man's crown."
Baron Tassilo von Heydebrand und
von der Lasa. Ueber die griechischen
und römischen Spiele, welche einige ähnlichkeit mit dem Schach hatten. Deutsche Schachzeitung (1863) 162-172, 198-199,
225-234, 257-264. ??NYS -- described on
Fiske 121-122 & 137, who says van der Linde I 40-47 copies much of it. Von der Lasa asserts that the Parva Tabella
of Ovid is Kleine Mühle (Three Men's Morris).
He says the game is called Tripp, Trapp, Trull in the Swedish book
Hand-Bibliothek för Sällkapsnöjen, of 1839, vol. II, p. 65 (or 57) --
??NYS. Van der Linde says that the
Dutch name is Tik, Tak, Tol. Fiske
notes that both of these refer to Noughts and Crosses, but it is unclear if von
der Lasa or van der Linde recognised the difference between Three Men's Morris
and Noughts and Crosses.
C. Babbage. Passages from the Life of a
Philosopher. 1864. Chapter XXXIV -- section on Games of Skill,
pp. 465‑471. (= pp. 152‑156
in: Charles Babbage and His Calculating
Engines, Dover, 1961.) Partial
analysis. He calls it tit‑tat‑to.
The Play Room: or, In-door Games
for Boys and Girls. Dick &
Fitzgerald(?), 1866. [Reprinted as: How
to Amuse an Evening Party. Dick &
Fitzgerald, NY, 1869.] ??NX -- the 1869
was seen at Shortz's. P. 22:
Tit-tat-to. Uses O
and +. "This is a game that small boys enjoy, and some big ones who
won't own it."
Anonymous. Every Little Boy's Book A Complete Cyclopædia of in and outdoor
games with and without toys, domestic pets, conjuring, shows, riddles, etc. With two hundred and fifty
illustrations. Routledge, London,
nd. HPL gives c1850, but the text is
clearly derived from Every Boy's Book, whose first edition was 1856. But the main part of the text considered
here is not in the 1856 ed of Every Boy's Book (with J. G. Wood as unnamed
editor), but is in the 8th ed of 1868 (published for Christmas 1867), which was
the first seriously revised edition, with Edmund Routledge as editor. So this may be c1868. This is the first published use of the term
Noughts and Crosses found so far -- the OED's 1861 quote is to Oughts and
Crosses..
Pp.
46-47: Slate games: Noughts and crosses.
"This is a capital game, and one which every school-boy truly
enjoys." Though the example shown
is a draw, there is no mention of the fact that the game should always be a tie.
Pp.
85-86: Nine-holes. This has nine holes
in a row and each player has a hole.
The ball is rolled to them and the person in whose hole it lands must
run and pick up the ball and try to hit one of the others who are running
away. So this has nothing to do with
our games or other forms of Nine Holes.
P.
106: Nine-holes or Bridge-board. This
has nine holes in an upright board and the object is get one's marbles through
the holes. (This material is in the
1856 ed. of Every Boy's Book.)
Correspondent to Notes and
Queries (1875) ??NYS -- quoted by
Strutt-Cox 257. Describes a game called
Three Mans' Marriage [sic] in Derbyshire which seems to be Noughts and Crosses
played on a crossed square board.
Pieces are not described as moving, but in the next description of a
Nine Men's Morris, they are specifically described as moving. However, the use of a crossed square board
may indicate that diagonals were not considered.
Cassell's. 1881.
Slate Games: Noughts and Crosses, or Tit‑Tat‑To, p. 84,
with cross reference under Tit-Tat-To, p. 87.
= Manson, 1911, pp. 202-203 & 208.
Albert Norman. Ungdomens Bok [Book for Youth] (in
Swedish). 2nd ed., Stockholm,
1883. Vol. I, p. 162++. ??NYS -- quoted and described in Fiske
134-136. Description of Tripp, Trapp,
Trull, with winning cry: "Tripp,
trapp, trull, min qvarn är full."
(Qvarn = mill.)
Lucas. RM2, 1883. Pp.
73-99. Analysis of Three Men's Morris,
on a board with the main diagonals drawn, with moves of only one square along a
winning line. He shows this is a first
person game. If the first player is not
permitted to play in the centre, then it is a tie game. No mention of Tic-Tac-Toe.
Albert Ellery Berg, ed. The Universal Self‑Instructor. Thomas Kelly, NY, 1883. Tit‑tat‑to, p. 379. Brief description.
Mark Twain. The Adventures of Huckleberry Finn. 1884.
Chap. XXXIV, about half-way through.
"It's as simple as tit-tat-toe, three-in-a-row, ..., Huck
Finn."
"A wrangler and late master
at Harrow school." The science of
naughts and crosses. Boy's Own Paper
10: (No. 498) (28 Jul 1888) 702‑703; (No. 499) (4 Aug 1888) 717; (No. 500) (11 Aug 1888) 735; (No. 501) (18 Aug 1888) 743. Exhaustive analysis, including odds of
second player making a correct response to each opening. For first move in: middle, side, corner, the odds of a correct response are: 1/2,
1/2, 1/8. He implies that the analysis is not widely
known.
"Tom Wilson". Illustred Spelbok (in Swedish). Nd [late 1880s??]. ??NYS -- described by Fiske 136-137. This gives Tripp, Trapp, Trull as a Three Men's Morris game on
the crossed square, with moves "according to one way of playing, to
whatever points they please, but according to another, only to the nearest
point along the lines on which the pieces stand. This last method is always employed when the board has, in
addition to the right lines, or lines joining the middles of the exterior
lines, also diagonals connecting the angles". He then describes a drawn version using the #
board and 0 and
+ (or 1 and 2 in
the North) which seems to be genuinely Noughts and Crosses. Fiske says the book seems to be based on an
early edition of the Encyclopédie des Jeux or a similar book, so it is
uncertain how much the above represents the current Swedish game. Fiske was unable to determine the author's
real name, though he was still living in Stockholm at the time.
Il Libro del Giuochi. Florence, 1894. ??NYS -- described in Fiske, pp. 109-110. Gives doubly crossed square board and
mentions a Three Men's Morris game.
T. de Moulidars. Grande Encyclopédie des Jeux. Montgredien or Librairie Illustree, Paris,
nd. ??NYS -- Fiske 115 (in 1905) says
it appeared 'not very long ago' and that Gelli seems to be based on it. Fiske quotes the clear description of Three
Men's Morris as Marelle Simple, using a doubly crossed square, saying that
pieces move to adjoining cells, following a line, and that the first player
should win if he plays in the centre.
Fiske notes that Noughts and Crosses is not mentioned.
J. Gelli. Come Posso Divertirmi? Milan, 1900. ??NYS -- described in Fiske 107.
Fiske quotes the description of Three Men's Morris as Mulinello
Semplice, essentially a translation from Moulidars.
Dudeney. CP.
1907. Prob. 109: Noughts and
crosses, pp. 156 & 248.
(c= MP, prob. 202: Noughts and crosses, pp. 89 & 175‑176. = 536, prob. 471: Tic tac toe, pp. 185 &
390‑392. Asserts the game is a
tie, but gives only a sketchy analysis.
MP gives a reasonably exhaustive analysis. Looks at Ovid's game.
A. C. White. Tit‑tat‑toe. British Chess Magazine (Jul 1919) 217‑220. Attempt at a complete analysis, but has a
mistake. See Gardner, SA (Mar
1957) = 1st Book, chap. 4.
D'Arcy Wentworth Thompson. Science and the Classics. OUP, 1940.
Section V Games and Playthings,
pp. 148-165. On p. 160, he quotes Ovid
and says it is Noughts and Crosses, or in Ireland, Tip-top-castle.
The Home Book of Quizzes, Games
and Jokes. Blue Ribbon Books, NY,
1941. This is excerpted from several
books -- this material is most likely taken from: Clement Wood & Gloria Goddard; Complete Book of Games; same
publisher, nd [late 1930s]. P. 150:
Tit-tat-toe, noughts and crosses. Brief
description of the game on the # board.
"To win requires great ingenuity."
G. E. Felton &
R. H. Macmillan. Noughts and
Crosses. Eureka 11 (Jan 1949) 5-9. Mentions Dudeney's work on the 3 x 3
board and asks for generalizations.
Mentions pegotty = go-bang. Then
studies the 4 x 4 x 4 game -- see 4.B.1.a. Adds some remarks on pegotty, citing
Falkener, Lucas and Tarry.
Stanley Byard. Robots which play games. Penguin Science News 16 (Jun 1950)
65-77. On p. 73, he says D. W.
Davies, of the National Physical Laboratory, has built, and exhibited to the
Royal Society in May 1949, an electro-mechanical noughts and crosses machine. A photo of the machine is plate 16. He also mentions Babbage's interest in such
a machine and an 1874 paper to the US National Academy by a Dr. Rogers --
??NYS.
P. C. Parks. Building a noughts and crosses machine. Eureka 14 (Oct 1951) 15-17. Cites Babbage, Rogers, Davies, Byard. Parks built a simple machine with wire and
tin cans in 1950 at a cost of about £6.
Says G. Eastell of Thetford, Norfolk, built a machine using sixty valves
for the Festival of Britain.
Gardner. Ticktacktoe. SA (Mar 1957) c= 1st
Book, chap. 4. Quotes Wordsworth,
discusses Three Men's Morris (citing Ovid) and its variants (including versions
on 4 x 4 and 5 x 5 boards), the misère version (the person who
makes three in a row loses), three and
n dimensional forms (giving L.
Moser's result on the number of winning lines on a kn board),
go-moku, Babbage's proposed machine, A. C. White's article. Addendum mentions the Opies' assertion that
the name comes from the rhyme starting
"Tit, tat, toe, My first
go,".
C. L. Stong. The Amateur Scientist. Ill. by Roger Hayward. S&S, 1960. A ticktacktoe machine, pp. 384-385. Noel Elliott gives a brief description of a relay logic machine
to play the game.
Donald Michie. Trial and error. Penguin Science Survey 2 (1961) 129-145. ??NYS.
Describes his matchbox and bead learning machine, MENACE (Matchbox
educable noughts and crosses engine), for the game.
Gardner. A matchbox game-learning machine. SA (Mar 1962) c= Unexpected, chap. 8.
Describes Michie's MENACE. Says it
used 300 matchboxes. Gardner adapts it
to Hexapawn, which is much simpler, requiring only 24 matchboxes. Discusses other games playable by
'computers'. Addendum discusses results
sent in by readers including other games.
Barnard. 50 Observer Brain-Twisters. 1962.
Prob. 34: Noughts and crosses, pp. 39‑40, 64 & 93‑94. Asserts there are 1884 final winning
positions. He doesn't consider
equivalence by symmetry and he allows either player to start.
Donald Michie &
R. A. Chambers. Boxes: an experiment
in adaptive control. Machine
Intelligence 2 (1968) 136-152.
Discusses MENACE, with photo of the pile of boxes. Says there are 288 boxes, but doesn't
explain exactly how he found them.
Chambers has implemented MENACE as a general game-learning computer
program using adaptive control techniques designed by Michie. Results for various games are given.
S. Sivasankaranarayana
Pillai. A pastime common among South
Indian school children. In: P. K. Srinivasan, ed.; Ramanujan Memorial
Volumes: 1: Ramanujan -- Letters and
Reminiscences; 2: Ramanujan -- An
Inspiration; Muthialpet High School,
Number Friends Society, Old Boys' Committee, Madras, 1968. Vol. 2, pp. 81-85. [Taken from Mathematics Student, but no date or details given --
??] Shows ordinary tic-tac-toe is a
draw and considers trying to get t in a row on an n x n board. Shows that
n = t ³ 3 is a draw and
that if t ³ n
+ 1 - Ö(n/6), then the game
is a draw.
L. A. Graham. The Surprise Attack in Mathematical
Problems. Dover, 1968. Tic-tac-toe for gamblers, prob. 8, pp.
19-22. Proposed by F. E. Clark,
solutions by Robert A. Harrington & Robert E. Corby. Find the probability of the first player
winning if the game is played at random.
Two detailed analyses shows that the probabilities for first player, second player, tie are
(737, 363, 160)/1260.
[Henry] Joseph & Lenore
Scott. Quiz Bizz. Puzzles for Everyone -- Vol. 6. Ace Books (Charter Communications), NY,
1975. P. 143: Ha-ho-ha. Misère version of noughts and crosses proposed. No discussion.
Gyles Brandreth. Pencil and Paper Games and Puzzles. Carousel, 1976. Noughts and Crosses, pp. 11-12. Asserts "It's been played all around the world for hundreds,
if not thousands, of years ...."
I've included it as a typical example of popular belief about the
game. = Pencil & Paper Puzzle
Games; Watermill Press, Mahwah, New Jersey, 1989, Tic-Tac-Toe, pp. 11‑12.
Winning Ways. 1982.
Pp. 667-680. Complete and
careful analysis, including various uncommon traps. Several equivalent games.
Discusses extensions of board size and dimension.
Sheila Anne Barry. The World's Best Travel Games. Sterling, NY, 1987. Tic-tac-toe squared, pp. 88-89. Get
3 in a row on the 4 x 4
board. Also considers
Tic-tac-toe-toe -- get 4 in a row on
5 x 5 board.
George Markovsky. Numerical tic-tac-toe -- I. JRM 22:2 (1990) 114-123. Describes and studies two versions where the
moves are numbered 1, 2, .... One
is due to Ron Graham, the other to P. H. Nygaard and Markowsky sketches the histories.
Ira Rosenholtz. Solving some variations on a variant of
tic-tac-toe using invariant subsets.
JRM 25:2 (1993) 128-135. The
basic variant is to avoid making three in a row on a 4 x 4
board. By playing symmetrically,
the second player avoids losing and 252 of the 256 centrally symmetric
positions give a win for the second player.
Extends analysis to 2n x
2n, 5 x 5, 4 x 4 x 4, etc.
Bernhard Wiezorke. Sliding caution. CFF 32 (Aug 1993) 24-25
& 33 (Feb 1994) 32. This describes a sliding piece puzzle on the
doubly crossed square board -- see under 5.A.
See: Yuri I. Averbakh; Board
games and real events; 1995; in 5.R.5, for a possible connection.
C. Planck. Four‑fold magics. Part 2 of chap. XIV, pp. 363‑375, of
W. S. Andrews, et al.; Magic Squares and Cubes; 2nd ed., Open Court, 1917; Dover, 1960. On p. 370, he notes that the number of m‑dimensional directions through a cell of the n‑dimensional board is the m‑th term of the binomial expansion
of ½(1+2)n.
Maurice Wilkes says he
played 3-D noughts and crosses at
Cambridge in the late 1930s, but the game was to get the most lines on a 3 x 3 x 3
board. I recall seeing a
commercial version, called Plato?, of this in 1970.
Cedric Smith says he played 3-D
and 4-D versions at Cambridge in the early 1940s.
Arthur Stone (letter to me of 9
Aug 1985) says '3 and 4 dimensional forms of tic-tac-toe produced by Brooks,
Smith, Tutte and myself', but it's not quite clear if they invented these. Tutte became expert on the 43
board and thought it was a first person game.
They only played the 54 game once - it took a long time.
Funkenbusch & Eagle, National
Mathematics Mag. (1944) ??NYR.
G. E. Felton &
R. H. Macmillan. Noughts and
crosses. Eureka 11 (1949) 5‑9. They say they first met the 4 x 4 x 4
game at Cambridge in 1940 and they give some analysis of it, with
tactics and problems.
William Funkenbusch & Edwin
Eagle. Hyper‑spacial tit‑tat‑toe
or tit‑tat‑toe in four dimensions.
National Mathematics Magazine 19:3 (Dec 1944) 119‑122. ??NYR
A. L. Rubinoff, proposer; L. Moser, solver. Problem E773 -- Noughts and crosses. AMM 54 (1947) 281
& 55 (1948) 99. Number of winning lines on a kn board is {(k+2)n ‑ kn}/2. Putting
k = 1 gives Planck's result.
L. Buxton. Four dimensions for the fourth form. MG 26 (1964) 38‑39. 3 x 3 x 3
and
3 x 3 x 3 x 3
games are obviously first person, but he proposes playing for most lines
and with the centre blocked on the 3 x
3 x 3 x 3 board. Suggests
3n and 4 x 4 x 4
games.
Anon. Puzzle page: Noughts and crosses. MTg 33 (1965) 35. Says
practice shows that the 4 x 4 x
4 game is a draw. [I only ever had one drawn game!] Conjectures
nn is first player
and (n+1)n is a draw.
Roland Silver. The group of automorphisms of the game of 3‑dimensional
ticktacktoe. AMM 74 (1967) 247‑254. Finds the group of permutations of cells
that preserve winning lines is generated by the rigid motions of the cube and
certain 'eviscerations'. [It is
believed that this is true for the kn board, but I don't know of a simple proof.]
Ross Honsberger. Mathematical Morsels. MAA, 1978.
Prob. 13: X's and O's, p. 26.
Obtains L. Moser's result.
Kathleen Ollerenshaw. Presidential Address: The magic of
mathematics. Bull. Inst. Math. Appl.
15:1 (Jan 1979) 2-12. P. 6 discusses my
rediscovery of L. Moser's 1948 result.
Paul Taylor. Counting lines and planes in generalised
noughts and crosses. MG 63 (No. 424)
(Jun 1979) 77-82. Determines the number pr(k) of r-sections of a kn board by means of a recurrence
pr(k) = [pr-1(k+2) - pr-1(k)]/2r which generalises L. Moser's 1948
result. He then gets an explicit sum
for it. Studies some other
relationships. This work was done while
he was a sixth form student.
Oren Patashnik. Qubic:
4 x 4 x 4 tic‑tac‑toe. MM 53 (1980) 202‑216. Computer assisted proof that 4 x 4 x 4
game is a first player win.
Winning Ways. 1982.
Pp. 673-679, esp. 678-679.
Discusses getting k in a row on a n x n board. Discusses
43 game
(Tic-Toc-Tac-Toe) and kn game.
Victor Serebriakoff. A Mensa Puzzle Book. Muller, London, 1982. (Later combined with A Second Mensa Puzzle
Book, 1985, Muller, London, as: The Mensa Puzzle Book, Treasure Press, London,
1991.) Chapter 7: Conceptual conflict
in multi-dimensional space, pp. 80-94 (1991: 98-112) & answers on pp. 99,
100, 106 & 131 (1991: 115, 116, 122 & 147). He considers various higher dimensional noughts and crosses on
the 33, 34 and 35 boards.
He finds that there are 49 winning lines on the 33 and he finds how to determine the number of d-facets on an n-cube as the
coefficients in the expansion of (2x +
1)n. He also considers games
where one has to complete a 3 x 3 plane to win and gives a problem: OXO three hypercube planes, p. 91 (1991:
109) & Answer 29, p. 106 (1991: 122)
which asks for the number of planes in the hypercube 34. The answer says there are
123 of them, but in 1985 I
found 154 and the general formula for the number of d-sections of a kn board. When I wrote to Serebriakoff, he responded
that he could not follow the mathematics and that "I arrived at the
figures ... from a simple formula published in one of Art [sic] Gardner's books
which checked out as far as I could take it.
Several other mathematicians have looked through it and not
disagreed." I wrote for a
reference to Gardner but never had a response.
I presented my work to the British Mathematical Colloquium at Cambridge
on 2 Apr 1985 and discovered that the results were known -- I had found the
explicit sum given by Taylor above, but not the recurrence.
David Fielker sent some pages
from a Danish book on games, but the TP is not present in his copies, so we
don't have details. This says that Hein
introduced the game in a lecture to students at the Institute for Theoretical
Physics (now the Niels Bohr Institute) in Copenhagen in 1942. After its appearance in Politiken, specially
printed pads for playing the game were sold, and a game board was marketed in
the US as Hex in 1952.
Piet Hein. Article or column in Politiken (Copenhagen)
(26 Dec 1942). ??NYR, but the diagrams
show a board of hexagons.
Gardner (1957) and others have
related that the game was independently invented by John Nash at Princeton in
1948-1949. Gardner had considerable
correspondence after his article which I have examined. The key point is that one of Niels Bohr's
sons, who had known the game in Copenhagen, was a visitor at the Institute for
Advanced Study at the time and showed it to friends. I concluded that it was likely that some idea of the game had
permeated to Nash who had forgotten this and later recalled and extensively
developed the idea, thinking it was new to him. I met Harold Kuhn in 1998, who was a student with Nash at the
time and he has no doubt that Nash invented the idea. In particular, Nash started with the triangular lattice, i.e. the
dual of Hein's board, for some time before realising the convenience of the
hexagonal lattice. Nash came to
Princeton as a graduate student in autumn 1948 and had invented the game by the
spring of 1949. Kuhn says he observed
Nash developing the ideas and recognising the connections with the Jordan Curve
Theorem, etc. Kuhn also says that there
was not much connection between students at Princeton and at the Institute and
relates that von Neumann saw the game at Princeton and asked what it was,
indicating that it was not well known at the Institute. In view of this, it seems most likely that
Nash's invention was independent, but I know from my own experience that it can
be difficult to remember the sources of one's ideas -- a casual remark about a
hexagonal game could have re-emerged weeks or months later when Nash was
studying games, as the idea of looking at hexagonal boards in some form, from
which the game would be re-invented.
Sylvester was notorious for publishing ideas which he had actually
refereed or edited some years earlier, but had completely forgotten the earlier
sources. In situations like Hex, we
will never know exactly what happened -- even if we were present at the time,
it is difficult to know what is going on in the mind of the protagonist and the
protagonist himself may not know what subconscious connections his mind is
making. Even if we could discover that
Nash had been told something about a hexagonal game, we cannot tell how his
mind dealt with this information and we cannot assume this was what inspired
his work. In other words, even a time
machine will not settle such historical questions -- we need something that
displays the conscious and the unconscious workings of a person's mind.
Parker Brothers. Literature on Hex, 1952. ??NYS or NYR.
Claude E. Shannon. Computers and automata. Proc. Institute of Radio Engineers 41
(Oct 1953) 1234‑1241. Describes
his Hex machine on p. 1237.
M. Gardner. The game of Hex. SA (Jul 1957) = 1st Book, chap. 8. Description of Shannon's 8 by 7 'Hoax' machine, pp. 81‑82,
and its second person strategy, p. 79.
Anatole Beck, Michael N.
Bleicher & Donald W. Crowe. Excursions
into Mathematics. Worth Publishers, NY,
1969. Chap. 5: Games (by Beck), Section
3: The game of Hex, pp. 327-339 (with photo of Hein on p. 328). Says it has been attributed to Hein and
Nash. At Yale in 1952, they played on
a 14 x 14 board. Shows it is a
first player win, invoking the Jordan Curve Theorem
David Gale. The game of Hex and the Brouwer fixed-point
theorem. AMM 86:10 (Dec 1979)
818-827. Shows that the non-existence
of ties (Hex Theorem) is equivalent to the Brouwer Fixed-Point Theorem in two
and in n dimensions. Says the use
of the Jordan Curve Theorem is unnecessary.
Winning Ways. 1982.
Pp. 679-680 sketches the game and the strategy stealing argument which
is attributed to Nash.
C. E. Shannon. Photo of his Hoax machine sent to me in
1983.
Cameron Browne. Hex Strategy: Making the Right
Connections. A. K. Peters, Natick,
Massachusetts, 2000.
Lucas. Le jeu de l'École Polytechnique.
RM2, 1883, pp. 90‑91. He
gives a brief description, starting: "Depuis quelques années, les élèves
de l'École Polytechnique ont imaginé un nouveaux jeu de combinaison assez
original." He clearly describes
drawing the edges of the game board and that the completer of a box gets to go
again. He concludes: "Ce jeu nous
a paru assez curieux pour en donner ici la description; mais, jusqu'a présent,
nous ne connaissons pas encore d'observations ni de remarques assez importantes
pour en dire davantage."
Lucas. Nouveaux jeux scientifiques de M. Édouard Lucas. La Nature 17 (1889) 301‑303. Clearly describes a game version of La
Pipopipette on p. 302, picture on p. 301, "... un nouveau jeu ... dédié
aux élèves de l'école Polytechnique."
This is dots and boxes with the outer edges already drawn in.
Lucas. L'Arithmétique Amusante.
1895. Note III: Les jeux scientifiques de Lucas,
pp. 203‑209 -- includes his booklet: La Pipopipette, Nouveau jeu de combinaisons, Dédié aux élèves de
l'École Polytechnique, Par un Antique de la promotion de 1861, (1889), on pp.
204‑208. On p. 207, he says
the game was devised by several of his former pupils at the École
Polytechnique. On p. 37, he remarks
that "Pipo est la désignation abrégée de Polytechnique, par les
élèves de l'X, ...."
Robert Marquard & Georg
Frieckert. German Patent 108,830 -- Gesellschaftsspiel. Patented: 15 Jun 1899. 1p + 1p diagrams. 8 x 8 array of boxes on a
board with slots for inserting edges.
No indication that the player who completes a box gets to play
again. They have some squares with
values but also allow all squares to have equal value.
C. Ganse. The dot game. Ladies' Home Journal (Jun 1903) 41. Describes the game and states that one who makes a box gets to go
again.
Loyd. The boxer's puzzle.
Cyclopedia, 1914, pp. 104 & 352.
= MPSL1, prob. 91, pp. 88‑89 & 152‑153. c= SLAHP: Oriental tit‑tat‑toe,
pp. 28 & 92‑93. Loyd
doesn't start with the boundaries drawn.
He asserts it is 'from the East'.
Ahrens. A&N.
1918. Chap. XIV: Pipopipette,
pp. 147‑155, describes it in more detail than Lucas does. He says the game appeared recently.
Blyth. Match-Stick Magic.
1921. Boxes, pp. 84-85. "The above game is familiar to most boys
and girls ...." No indication that
the completer of a box gets to play again.
Heinrich Voggenreiter. Deutsches Spielbuch Sechster Teil: Heimspiele. Ludwig Voggenreiter, Potsdam, 1930. Pp. 84-85: Die Käsekiste. Describes a version for two or more players. The first player must start at a corner and
players must always connect to previously drawn lines. A player who completes a box gets to play
again.
Meyer. Big Fun Book. 1940. Boxes, p. 661. Brief description, somewhat vaguely stating that a player who
completes a box can play again.
The Home Book of Quizzes, Games
and Jokes. Op. cit. in 4.B.1,
1941. P. 151: Dots and squares. Clearly says the completer gets to play
again. "The game calls for great
ingenuity."
"Zodiastar". Fun with Matches and Match Boxes. (Cover says: Match Tricks From the 1880s
to the 1940s.) Universal Publications,
London, nd [late 1940s?]. The game of
boxes, pp. 48-49. Starts by laying out
four matches in a square and players put down matches which must touch the
previous matches. Completing a box
gives another play. No indication that
matches must be on lattice lines, but perhaps this is intended.
Readers' Research
Department. RMM 2 (Apr 1961) 38‑41,
3 (Jun 1961) 51‑52, 4 (Aug 1961) 52‑55. On pp. 40‑41 of No. 2, it says that Martin Gardner suggests
seeking the best strategy. Editor notes
there are two versions of the rules -- where the one who makes a box gets an
extra turn, and where he doesn't -- and that the game can be played on other
arrays. On p. 51 of No. 3, there is a
symmetry analysis of the no‑extra‑turn game on a board with an odd
number of squares. On pp. 52‑54
of No. 4, there is some analysis of the extra‑turn case on a board with
an odd number of boxes.
Everett V. Jackson. Dots and cubes. JRM 6:4 (Fall 1973) 273‑279. Studies 3‑dimensional game where a play is a square in the
cubical lattice.
Gyles Brandreth. Pencil and Paper Games and Puzzles. Carousel, 1976. Worm, pp. 18-19. This is
a sort of 'anti-boxes' -- one draws segments on the lattice forming a path without
any cycles -- last player wins. =
Pencil & Paper Puzzle Games; Watermill Press, Mahwah, New Jersey, 1989, pp.
18-19.
Winning Ways. 1982.
Chap. 16: Dots-and-Boxes, pp. 507-550
David B. Lewis. Eureka!
Perigee (Putnam), NY, 1983. Pp.
44‑45 suggests playing on the triangular lattice.
Sheila Anne Barry. The World's Best Travel Games. Sterling, NY, 1987.
Eternal
triangles, pp. 80-81. Gives the game on
the triangular lattice.
Snakes,
pp. 81-82. Same as Brandreth's
Worm. I think 'snake' would be a better
title as only one path is drawn.
M. Gardner. SA (Jul
1967) = Carnival, chap. 1. Describes Michael
Stewart Paterson and John Horton Conway's invention of the game on 21 Feb 1967
at tea time in the Department common room at Cambridge. The idea of adding a spot was due to
Paterson and they agreed the credit for the game should be 60% Paterson to 40%
Conway.
Gyles Brandreth. Pencil and Paper Games and Puzzles. Carousel, 1976. Sprouts, p. 13. "...
actually born in Cambridge about ten years ago." c= Pencil & Paper Puzzle Games; Watermill Press, Mahwah, New
Jersey, 1989, p. 13: "... was invented about ten years ago."
Winning Ways. 1982.
Sprouts, pp. 564-570 & 573.
Says the game was "introduced by M. S. Paterson and J. H. Conway
some time ago". Also describes
Brussels Sprouts and Stars-and-Stripes.
An answer for Brussels Sprouts and some references are on p. 573.
Sheila Anne Barry. The World's Best Travel Games. Sterling, NY, 1987. Sprouts, pp. 95-97.
Karl-Heinz Koch. Pencil & Paper Games. (As: Spiele mit Papier und Bleistift, no
details); translated by Elisabeth E. Reinersmann. Sterling, NY, 1992.
Sprouts, pp. 36-37, says it was invented by J. H. Conway & M. S.
Paterson on 21 Feb 1976 [sic -- misprint of 1967] during their five o'clock tea
hour.
4.B.5. OVID'S GAME AND NINE MEN'S MORRIS
See
also 4.B.1 for historical material.
The
classic Nine Men's Morris board consists of three concentric squares with their
midpoints joined by four lines. The
corners are sometimes also joined by another four diagonal lines, but this
seems to be used with twelve men per side and is sometimes called Twelve Men's
Morris -- see 1891 below. Fiske 108
says this is common in America but infrequent in Europe, though on 127 he says
both forms were known in England before 1600, and both were carried to the US,
though the Nine form is probably older.
Murray 615 discusses Nine Men's
Morris. He cites Kurna, Egypt (‑14C),
medieval Spain (Alquerque de Nueve), the Gokstad ship and the steps of the
Acropolis of Athens. He says the board
sometimes has diagonals added and then is played with 9, 11 or 12 pieces.
Dudeney. AM. 1917.
Introduction to Moving Counter Problems, pp. 58-59. This gives a brief survey, mentioning a
number of details that I have not seen elsewhere, e.g. its occurrence in Poland
and on the Amazon. Says the board was
found on a Roman tile at Silchester and on the steps of the Acropolis in Athens
among other sites.
J. A. Cuddon. The Macmillan Dictionary of Sports and
Games. Macmillan, London, 1980. Pp. 563‑564. Discusses the history.
Says there is a c‑1400 board cut in stone at Kurna, Egypt and
similar boards were made in years 9 to 21 at Mihintale, Ceylon. Says Ars Amatoria may be describing Three
Men's Morris and Tristia may be describing a kind of Tic‑tac‑toe. Cites numerous medieval descriptions and
variants.
Claudia Zaslavsky. Tic Tac Toe and Other Three‑in‑a‑Row
Games from Ancient Egypt to the Modern Computer. Crowell, NY, 1982. This
is really a book for children and there are no references for the historical
statements. I have found most of them
elsewhere, and the author has kindly
send me a list of source books, but I have not yet tracked down the following
items -- ??.
There
is an English court record of 1699 of punishment for playing Nine Holes in
church.
There
is a Nine Men's Morris board on a stone on the temple of Seti I (presumably
this is at Kurna). There is a picture
in the 13C Spanish 'Book of Games' (presumably the Alfonso MS -- see below) of
children playing Alquerque de Tres (c= Three Men's Morris). A 14C inventory of the Duc de Berry lists
tables for Mérelles (=? Nine Men's Morris) (see Fiske 113-115 below) and a book
by Petrarch shows two apes playing the game.
H. Parker. Ancient Ceylon. Loc. cit. in 4.B.1. Nine
Men's Morris board in the Temple of Kurna, Egypt, ‑14C. [Rohrbough, below, says this temple was
started by Ramses I and completed by Seti in -1336/-1333, citing J. Royal
Asiatic Soc. (1783) 17.] Two diagrams
for Nine Men's Morris are cut into the great flight of steps at Mihintale,
Ceylon and these are dated c1C. He
cites Bell; Arch. Survey of Ceylon, Third Progress Report, p. 5 note, for
another diagram of similar age.
Jack Botermans, Tony Burrett,
Pieter van Delft & Carla van Spluntern.
The World of Games. (In Dutch,
1987); Facts on File, NY, 1989.
P.
35 describes Yih, a form of Three Men's Morris, played on a doubly crossed
square with a man moving "one step along any line". A note adds that only the French have a rule
forbidding the first player to play in the centre, which makes the game more
challenging and is recommended.
Pp.
103-107 is the beginning of a section:
Games of alignment and configuration and discusses various games, but
rather vaguely and without references.
They mention Al-Qurna, Mihintale, Gokstad and some other early
sites. They say Yih was described by
Confucius, was played c-500 and is "the game, that we now know as
tic-tac-toe, or three men's morris."
They describe Noughts and Crosses in the usual way. They then distinguish Tic-Tac-Toe, saying
"In Britain it is generally known as three men's morris ...." and say
it is the same as Yih, "which was known in ancient Egypt". They say "Ovid mentions
tic-tac-toe" in Ars Amatoria, that several Roman boards have survived and
that it was very popular in 14C England with several boards for this and Nine
Men's Morris cut into cloister seats.
They then describe Three-in-a-Row, which allows pieces to move one step
in any direction, as a game played in Egypt.
They then describe Five or Six Men's Morris, Nine Men's Morris, Twelve
Men's Morris and Nine Men's Morris with Dice, with nice 13C & 15C
illustration of Nine Men's Morris.
Bell & Cornelius. Board Games Round the World. Op. cit. in 4.B.1. 1988. Pp. 6-8. They discuss the crossed square board -- see
4.B.1 -- and describe Three Men's Morris with moves only along the lines to an
adjacent vacant point. They then
describe Achi, from Ghana, on the doubly crossed square with the same
rules. They then describe Six Men's
Morris which was apparently popular in medieval Europe but became obsolete by
c1600.
Ovid. Ars Amatoria. -1. II, 203-208
& III, 353-366. Translated by J. H. Mozley; Loeb Classical
Library, 1929, pp. 80-81 & 142-145.
Translated by B. P. Moore, 1935, used in A. D. Melville; Ovid The Love Poems; OUP, 1990, pp. 113, 137, 229
& 241.
II,
203-208 are three couplets apparently referring to three games: two dice games
and Ludus Latrunculorum. Mozley's prose
translation is:
"If she be gaming, and throwing
with her hand the ivory dice, do you throw amiss and move your throws amiss; or
if is the large dice you are throwing, let no forfeit follow if she lose; see
that the ruinous dogs often fall to you; or if the piece be marching under the
semblance of a robbers' band, let your warrior fall before his glassy
foe."
'Dogs'
is the worst throw in Roman dice games.
Moore's
verse translation of 207-208 is:
"And when the raiding chessmen
take the field, Your champion to his
crystal foe must yield."
Melville's
note says the original has 'bandits' and says the game is Ludus Latrunculorum.
III,
357-360 is probably a reference to the same game since 'robbers' occurs again,
though translated as brigands by Mozley, and again it immediately follows a
reference to throwing dice. Mozley's
translation of 353-366 is:
"I am ashamed to advise in
little things, that she should know the throws of the dice, and thy powers, O
flung counter. Now let her throw three
dice, and now reflect which side she may fitly join in her cunning, and which challenge, Let her cautiously and not foolishly play
the battle of the brigands, when one piece falls before his double foe and the
warrior caught without his mate fights on, and the enemy retraces many a time
the path he has begun. And let smooth
balls be flung into the open net, nor must any ball be moved save that which
you will take out. There is a sort of
game confined by subtle method into as many lines as the slippery year has
months: a small board has three counters on either side, whereon to join your
pieces together is to conquer."
Moore's
translation of 357-360 is:
"To guide with wary skill the
chessmen's fight, When foemen twain
o'erpower the single knight, And caught
without his queen the king must face
The foe and oft his eager steps retrace".
This
is clearly not a morris game -- Mozley's note above and the next entry make it
clear it is Ludus Latrunculorum, which had a number of forms. Mozley's note on pp. 142-143 refers to
Tristia II, 478 and cites a number of other references for Ludus Latrunculorum.
Moore's
translation of 363-366 is:
"A game there is marked out in
slender zones As many as the fleeting
year has moons; A smaller board with
three a side is manned, And victory's
his who first aligns his band."
Mozley's
notes and Melville's notes say the first two lines refer to the Roman game of
Ludus Duodecim Scriptorum -- the Twelve Line Game -- which is the ancestor of
Backgammon. Mozley says the game in the
latter two lines is mentioned in Tristia, "but we have no information
about it." Melville says it is "a
'position' game, something like Nine Men's Morris" and cites R. C. Bell's
article on 'Board and tile games' in the Encyclopaedia Britannica, 15th ed.,
Macropaedia ii.1152‑1153, ??NYS.
Ovid. Tristia. c10. II, 471‑484. Translated by A. L. Wheeler.
Loeb Classical Library, 1945, pp. 88‑91. This mentions several games and the text parallels that of Ars
Amatoria III.
"Others have written of the arts
of playing at dice -- this was no light sin in the eyes of our ancestors --
what is the value of the tali, with what throw one can make the highest
point, avoiding the ruinous dogs; how the tessera is counted, and when
the opponent is challenged, how it is fitting to throw, how to move according
to the throws; how the variegated soldier steals to the attack along the
straight path when the piece between two enemies is lost, and how he
understands warfare by pursuit and how to recall the man before him and to
retreat in safety not without escort; how a small board is provided with three
men on a side and victory lies in keeping one's men abreast; and the other
games -- I will not describe them all -- which are wont to waste that precious
thing, our time."
A
note says some see a reference to Ludus Duodecim Scriptorum at the beginning of
this. The next note says the next text
refers to Ludus Latrunculorum, a game on a squared board with 30 men on a side,
with at least two kinds of men. The
note for the last game says "This game seems to have resembled a game of draughts
played with few men." and refers to Ars Amatoria and the German
Mühlespiel, which he describes as 'a sort of draughts', but which is Nine Men's
Morris.
R. G. Austin. Roman board games -- I & II. Greece and Rome 4 (No. 10) (Oct 1934) 24‑34 & 4 (No. 11) (Feb
1935) 76-82. Claims the Ovid references
are to Ludus Latrunculorum (a kind of Draughts?), Ludus Duodecim Scriptorum
(later Tabula, an ancestor of Backgammon) and (Ars Amatoria.iii.365-366) a kind
of Three Men's Morris. In the last, he
shows a doubly crossed 3 x 3 board, but it is not clear which rule he
adopts for the later movement of pieces, but he says: "the first player is
always able to force a win if he places his first man on the centre point, and
this suggests that the dice may have been used to determine priority of play,
although there is no evidence of this."
He says no Roman name for this game has survived. He discusses various known artifacts for all
the game, citing several Roman 8 x
8 boards found in Britain. He gives an informal bibliography with
comments as to the value of the works.
D'Arcy Wentworth Thompson. Science and the Classics. OUP, 1940.
Section V Games and Playthings,
pp. 148-165. On p. 160, he quotes Ovid,
Ars Amatoria.iii.365-366 and says it is Noughts and Crosses, or in Ireland,
Tip-top-castle.
The British Museum has a Nine
Men's Morris board from the Temple of Artemis, Ephesus, 2C-4C. Item BM GR 1872,8-3,44. This was in a small exhibition of board
games in 1990. I didn't see it on
display in late 1996.
Murray, p. 189. There was an Arabic game called Qirq, which
Murray identifies with Morris.
"Fourteen was a game played with small stones on a wooden board
which had three rows of holes (al‑Qâbûnî)." Abû‑Hanîfa [the H
should have a dot under it], c750, held that Fourteen was illegal and
Qirq was held illegal by writers soon afterward. On p. 194, Murray gives a 10C passage mentioning Qirq being
played at Mecca.
Fiske 255 cites the Kitāb
al Aghāni, c960, for a reference to qirkat, i.e. morris boards.
Paul B. Du Chaillu. The Viking Age. Two vols., John Murray, London, 1889. Vol. II, p.168, fig. 992 -- Fragments of wood from Gokstad ship. Shows a partial board for Nine Men's Morris
found in the Gokstad ship burial. There
is no description of this illustration and there is only a vague indication
that this is 10C, but other sources say it is c900.
Gutorm Gjessing. The Viking Ship Finds. Revised ed., Universitets Oldsaksamling,
Oslo, 1957. P. 8: "... there are two boards which were
used for two kinds of games; on one side figures appear for use in a game which
is frequently played even now (known as "Mølle")."
Thorlief Sjøvold. The Viking Ships in Oslo. Universitets Oldsaksamling, Oslo, 1979. P. 54:
"... a gaming board with one antler gaming piece, ...."
In medieval Europe, the game is
called Ludus Marellorum or Merellorum or just Marelli or Merelli or Merels,
meaning the game of counters. Murray
399 says the connection with Qirq is unclear.
However, medieval Spain played various games called Alquerque, which is
obviously derived from Qirq. Alquerque
de Nueve seems to be Nine Men's Morris.
However, in Italy and in medieval France, Marelle or Merels could mean
Alquerque (de Doze), a draughts‑like game with 12 men on a side played on
a 5 x 5 board (Murray 615). Also
Marro, Marella can refer to Draughts which seems to originate in Europe
somewhat before 1400.
Stewart Culin. Korean Games, with Notes on the
Corresponding Games of China and Japan.
University of Pennsylvania, Philadelphia, 1895. Reprinted as: Games of the Orient; Tuttle, Rutland, Vermont, 1958. Reprinted under the original title, Dover
and The Brooklyn Museum, 1991. P. 102,
section 80: Kon-tjil -- merrells. This
is the usual Nine Men's Morris. The
Chinese name is Sám-k'i (Three Chess).
"I am told by a Chinese merchant that this game was invented by
Chao Kw'ang-yin (917-975), founder of the Sung dynasty." This is the only indication of an oriental
source that I have seen.
Gerhard Leopold. Skulptierte Werkstücke in der Krypta der
Wipertikirche zu Quedlinburg. IN:
Friedrich Möbius & Ernst Schubert, eds.; Skulptur des Mittelalters; Hermann
Böhlaus Nachfolger, Weimar, 1987, pp. 27-43; esp. pp. 37 & 43. Describes and gives photos of several
Nine-Men's-Morris boards carved on a pillar of the crypt of the Wipertikirche,
Quedlinburg, Sachsen-Anhalt, probably from the 10/11 C.
Richard de Fournivall. De Vetula.
13C. This describes various
games, including Merels. Indeed the
French title is: Ci parle du gieu des
Merelles .... ??NYS -- cited by Murray,
pp. 439, 507, 520, 628. Murray 620
cites several MSS and publications of the text.
"Bonus Socius"
[Nicolas de Nicolaï?]. This is a
collection of chess problems, compiled c1275, which exists in many manuscript
forms and languages. See 5.F.1 for more
details of these MSS. See Murray 618‑642. On pp. 619‑624 & 627, he mentions
several MSS which include 23, 24, 25 or 28 Merels problems. On p. 621, he cites "Merelles a
Neuf" from 14C. Fiske 104 &
110-111 discusses some MSS of this collection.
The Spanish Treatise on
Chess-Play written by order of King Alfonso the Sage in the year 1283. [= Libro de Acedrex, Dados e Tablas of
Alfonso El Sabio, generally known as the Alfonso MS.] MS in Royal Library of the Escorial (j.T.6. fol). Complete reproduction in 194 Phototropic
Plates. 2 vols., Karl W. Horseman, Leipzig, 1913. (See in 4.A.1 for another ed.) This is a collection of chess problems
produced for Alfonso X, the Wise, King of Castile (Castilla). Vol. 2, ff. 92v‑93r,
pp. CLXXXIV‑CLXXXV, shows Nine Men's Morris boards. ??NX -- need to study text. See:
Murray 568‑573; van der
Linde I 137 & 279 ??NYS & Quellenstudien 73 & 277‑278, ??NYS
(both cited by Fiske 98); van der Lasa
116, ??NYS (cited by Fiske 99).
Fiske
98-99 says that the MS also mentions Alquerque, Cercar de Liebre and Alquerque
de Neuve (with 12 men against one).
Fiske 253-255 gives a more detailed study of the MS based on a
transcript. He also quotes a
communication citing al Querque or al Kirk in Kazirmirski's Arabic dictionary
and in the Kitāb al Aghāni, c960.
José
Brunet y Bellet. El Ajedrez. Barcelona, 1890. ??NYS -- described by Fiske 98.
This has a chapter on the Alfonso MS and refers to Alquerque de Doce,
saying that it is known as Tres en Raya in Castilian and Marro in Catalan
(Fiske 102 says this word is no longer used in Spanish). Brunet notes that there are five miniatures
pertaining to alquerque. Fiske says
that all this information leaves us uncertain as to what the games were. Fiske says Brunet's chapter has an appendix
dealing with Carrera's 1617 discussion of 'line games' and describing Riga di
Tre as the same as Marro or Tres en Raya as a form of Three Men's Morris
Murray gives many brief
references to the game, which I will note here simply by his page number and
the date of the item.
438‑439
(12C); 446 (14C);
449
(c1400 -- 'un marrelier', i.e. a Merels board);
431
(c1430); 447 (1491); 446 (1538).
Anon. Romance of Alexander.
1338. (Bodleian Library, Mss
Bodl. 264). ??NYS. Nice illustration clearly showing Nine Men's
Morris board. I. Disraeli (Amenities of
Literature, vol. I, p. 86) also cites British Museum, Bib. Reg. 15, E.6 as a
prose MS version with illustrations.
Prof. D. J. A. Ross tells me there is nothing in the text corresponding
to the illustrations and that the Bodleian text was edited by M. R. James,
c1920, ??NYS. Illustration reproduced
in: A. C. Horth; 101 Games to Make and
Play; Batsford, London, (1943; 2nd ed.,
1944); 3rd ed., 1946; plate VI facing
p. 44, in B&W. Also in: Pia Hsiao et al.; Games You Make and Play;
Macdonald and Jane's, London, 1975, p. 7, in colour.
Fiske 113-115 gives a number of
quotations from medieval French sources as far back as mid 14C, including an
inventory of the Duc de Berry in 1416 listing two boards. Fiske notes that the game has given rise to
several French phrases. He quotes a
1412 source calling it Ludus Sanct Mederici or Jeu Saint Marry and also
mentions references in city statutes of 1404 and 1414.
MS, Montpellier, Faculty of
Medicine, H279 (Fonts de Boulier, E.93).
14C. This is a version of the
Bonus Socius collection. Described in
Murray 623-624, denoted M, and in van der Linde I 301, denoted K. Lucas, RM2, 1883, pp. 98-99 mentions it and
RM4, 1894, Quatrième Récréation: Le jeu des mérelles au XIIIe
siècle, pp. 67-85 discusses it extensively.
This includes 28 Merels problems which are given and analysed by
Lucas. Lucas dates the MS to the 13C.
Household accounts of Edward IV,
c1470. ??NYS -- see Murray 617. Record of purchase of "two foxis and 46
hounds" to form two sets of "marelles".
Civis Bononiae [Citizen of
Bologna]. This is a collection of chess
problems compiled c1475, which exists in several MSS. See Murray 643‑703.
It has 48 or 53 merels problems.
On p. 644, 'merelleorum' is quoted.
A Hundred Sons. Chinese scroll of Ming period
(1368-1644). 18C copy in BM. ??NYS -- extensively reproduced and
described in: Marguerite Fawdry;
Chinese Childhood; Pollock's Toy Theatres, London, 1977. On p. 12 of Fawdry is a scene, apparently
from the scroll, in which some children appear to be playing on a Twelve Men's
Morris board.
Elaborate boards from Germany
(c1530) and Venice (16C) survive in the National Museum, Munich and in South
Kensington (Murray 757‑758).
Murray shows the first in B&W facing p. 757.
William Shakespeare. A Midsummer Night's Dream. c1610.
Act II, scene I, lines 98-100:
"The nine men's morris is fill'd up with mud, And the quaint mazes in the wanton
green For lack of tread are
indistinguishable." Fiske 126
opines that the latter two lines may indicate that the board was made in the
turf, though he admits that they may refer just to dancers' tracks, but to me
it clearly refers to turf mazes.
J. C. Bulenger. De Ludis Privatis ac Domesticus
Veterum. Lyons, 1627. ??NYS
Fiske 115 & 119 quote his description of and philological note on
Madrellas (Three Men's Morris).
Paul Fleming (1609-1640). In one of his lyrics, he has Mühlen. ??NYS -- quoted by Fiske 132, who says this is
the first German mention of Morris.
Fiske 133 gives the earliest
Russian reference to Morris as 1675.
Thomas Hyde. Historia Nerdiludii, hoc est dicere,
Trunculorum; .... (= Vol. 2 of De Ludis
Orientalibus, see 7.B for vol. 1.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Historia Triodii, pp. 202-214, is on morris games. (Described in Fiske 118-124, who says there
is further material in the Elenchus at the end of the volume -- ??NYS) Hyde asserts that the game was well known to
the Romans, though he cannot find a Roman name for it! He cites and discusses Bulenger, but
disagrees with his philology. Gives
lots of names for the game, ranging as far as Russian and Armenian. He gives both the Nine and Twelve Men's
Morris boards on p. 210, but he has not found the Twelve board in Eastern
works. On p. 211, he gives the doubly
crossed square board with a title in Chinese characters, pronounced 'Che-lo',
meaning 'six places', and having three white and three black men already placed
along two sides. He says the Irish name
is Cashlan Gherra (Short Castle) and that the name Copped Crown is common in
Cumberland and Westmorland. He then
describes playing the Twelve Man and Nine Man games, and then he considers the
game on the doubly crossed square board.
He seems to say there are different rules as to how one can move. ??need to study the Latin in detail. This is said to throw light on the Ovid
passages. Hyde believes the game was
well known to the Romans and hence must be much older. Fiske remarks that this is history by
guesswork.
Murray 383 describes Russian
chess. He says Amelung identifies the
Russian game "saki with Hölzchenspiel (?merels)". Saki is mentioned on this page as being
played at the Tsar's court, c1675.
Archiv der Spiele. 3 volumes, Berlin, 1819-1821. Vol. 2 (1820) 21-27. ??NYS
Described and quoted by Fiske 129-132.
This only describes the crossed square and the Nine Men's Morris
boards. It says that the Three Men's
Morris on the crossed square board is a tie, i.e. continues without end, but it
is not clear how the pieces are allowed to move. Fiske says this gives the most complete explanation he knows of
the rules for Nine Men's Morris.
Charles Babbage. Notebooks -- unpublished collection of MSS
in the BM as Add. MS 37205. ??NX. For more details, see 4.B.1. On ff. 347.r-347.v, 8 Sep 1848, he suggests
Nine Men's Morris boards in triangular and pentagonal shapes and does various
counting on the different shapes.
The Family Friend (1856)
57. Puzzle 17. -- Two and a Bushel. Shows the standard # board. "This very simple and amusing games, --
which we do not remember to have seen described in any book of games, -- is
played, like draughts, by two persons with counters. Each player must have three, ...
and the game is won when one of the players succeeds in placing his
three men in a row; ...." There is
no specification of how the men move.
The word 'bushel' occurs in some old descriptions of Three Men's Morris
and Nine Men's Morris as the name of the central area.
The Sociable. 1858.
Merelles: or, nine men's morris, pp. 279-280. Brief description, notable for the use of Merelles in an English
book.
Von der Lasa. Ueber die griechischen und römischen Spiele,
welche einige ähnlichkeit mit dem Schach hatten. Deutsche Schachzeitung (1863) 162-172, 198-199, 225-234, 257‑264. ??NYS -- described on Fiske 121-122 &
137, who says van der Linde I 40-47 copies much of it. He asserts that the Parva Tabella of Ovid is
Kleine Mühle (Three Men's Morris). Von
der Lasa says the game is called Tripp, Trapp, Trull in the Swedish book
Hand-Bibliothek för Sällkapsnöjen, of 1839, vol. II, p. 65 (or 57??). Van der Linde says that the Dutch name is
Tik, Tak, Tol. Fiske notes that both of
these refer to Noughts and Crosses, but it is unclear if von der Lasa or van
der Linde recognised the difference between Three Men's Morris and Noughts and
Crosses.
Albert Norman. Ungdomens Bok [Book for Youth] (in
Swedish). 2nd ed., Stockholm,
1883. Vol. I, p. 162++. ??NYS -- quoted and described in Fiske
134-136. Plays Nine Men's Morris on a
Twelve Men's Morris board.
Webster's Dictionary. 1891.
??NYS -- Fiske 118 quotes a definition (not clear which) which includes
"twelve men's morris". Fiske
says: "Here we have almost the
only, and certainly the first mention of the game by its most common New
England name, "twelve men's morris," and also the only hint we have
found in print that the more complicated of the morris boards -- with the
diagonal lines ... -- is used with twelve men, instead of nine, on each
side." Fiske 127 says the name
only appears in American dictionaries.
Dudeney. CP.
1907. Prob. 110: Ovid's game,
pp. 156‑157 & 248. Says the
game "is distinctly mentioned in the works of Ovid." He gives Three Men's Morris, with moves to
adjacent cells horizontally or diagonally, and says it is a first player win.
Blyth. Match-Stick Magic.
1921. Black versus white, pp.
79-80. 4 x 4 board with four men each.
But the men must be initially placed
WBWB in the first row and BWBW
in the last row. They can move
one square "in any direction" and the object is to get four in a row
of your colour.
Games and Tricks -- to make the
Party "Go". Supplement to
"Pearson's Weekly", Nov. 7th, no year indicated [1930s??]. A matchstick game, p. 11. On a
4 x 4 board, place eight
men, WBWB on the top row and
BWBW on the bottom row. Players alternately move one of their men by
one square in any direction -- the object is to make four in a line.
Lynn Rohrbough, ed. Ancient Games. Handy Series, Kit N, Cooperative Recreation Service, Delaware,
Ohio, (1938), 1939.
Morris
was Player [sic] 3,300 Years Ago, p. 27.
Says the temple of Kurna was started by Ramses I and completed by Seti
in -1336/-1333, citing J. Royal Asiatic Soc. (1783) 17.
Three
Men's Morris, p. 27. After placing
their three men, players 'then move trying to get three men in a row.' Contributor says he played it in Cardiff
more than 50 years ago.
Winning Ways. 1982.
Pp. 672-673. Says Ovid's Game is
conjectured to be Three Men's Morris.
The current version allows moves by one square orthogonally and is a
first person win if the first person plays in the centre. If the first player cannot play in the
centre, it is a draw. They use Three
Men's Morris for the case with one step moves along winning lines, i.e.
orthogonally or along main diagonals.
An American Indian game, Hopscotch, permits one step moves orthogonally
or diagonally (along any diagonal). A
French game, Les Pendus, allows any move to a vacant cell. All of these are draws, even allowing the
first player to play in the centre.
They briefly describe Six and Nine Men's Morris.
Ralph Gasser &
J. Nievergelt. Es ist
entscheiden: Das Muehle-Spiel ist unentscheiden. Informatik Spektrum 17 (1994) 314-317. ??NYS -- cited by Jörg Bewersdorff [email of 6 Jun 1999].
L. V. Allis. Beating the World Champion -- The state of
the art in computer game playing. IN:
Alexander J. de Voogt, ed.; New Approaches to Board Games Research: Asian
Origins and Future Perspectives; International Institute for Asian Studies,
Leiden, 1995; pp. 155-175. On p. 163,
he states that Ralph Gasser showed that Nine Men's Morris is a draw in Oct
1993, but the only reference is to a letter from Gasser.
Ralph Gasser. Solving Nine Men's Morris. IN: Games of No Chance; ed. by Richard
Nowakowski; CUP, 1996, pp. 101-113.
??NYS -- cited by Bewersdorff [loc. cit.] and described in William Hartston; What mathematicians get up
to; The Independent Long Weekend (29 Mar 1997) 2. Demonstrates that Nine Men's Morris is a draw. Gasser's abstract: "We describe the
combination of two search methods used to solve Nine Men's Morris. An improved analysis algorithm computes
endgame databases comprising about 1010 states. An 18-ply alpha-beta search the used these databases
to prove that the value of the initial position is a draw. Nine Men's Morris is the first non-trivial
game to be solved that does not seem to benefit from knowledge-based
methods." I'm not sure about the
last statement -- 4 x 4 x 4 noughts and crosses (see 4.B.1.a) and
Connect-4 were solved in 1980 and 1988, though the first was a computer aided
proof and the original brute force solution of Connect-4 by James Allen in Sep
1988 was improved to a knowledge-based approach by L. V. Allis by Aug 1989. The five-in-a-row version of Connect-4 was
shown to be a first person win in 1993.
Bewersdorff [email of 11 Jun 1999] clarifies this by observing that draw
here means a game that continues forever -- one cannot come to a stalemate
where neither side can move.
Winning Ways. 1982.
Philosopher's football, pp. 688‑691. In 1985, Guy said this was the only published description of the
game.
This is best viewed as played on
a n x n array of squares.
The n(n+1) vertical edges belong to one player, say
red, while the n(n+1) horizontal edges belong to black. Players alternate marking a square with a
line of their colour between edges of their colour. A square cannot be marked twice.
The object is to complete a path across the board. In practice, the edges are replaced by
coloured dots which are joined by lines.
As with Hex, there can be no ties and there must be a first person strategy.
M. Gardner. SA (Oct 1958) c= 2nd Book, Chap. 7. Introduces David Gales's game, later called
Bridg‑it. Addendum in the book
notes that it is identical to Shannon's 'Bird Cage' game of 1951 and that it
was marketed as Bridg‑it in 1960.
M. Gardner. SA (Jul 1961) c= New MD, Chap. 18. Describes Oliver Gross's simple strategy for
the first player to win. Addendum in
the book gives references to other solutions and mentions.
M. Gardner. SA (Jan 1973) c= Knotted, Chap. 9. Article says Bridg‑it was still on the
market.
Winning Ways. 1982.
Pp. 680-682. Covers Bridg-it and
Shannon Switching Game.
In Oct 2000, I bought a
second-hand copy of a 5 x 5 version called Connections, attributed to
Tom McNamara, but with no date.
Fred Schuh. Spel van delers (Game of divisors). Nieuw Tijdschrift vor Wiskunde 39 (1951‑52)
299‑304. ??NYS -- cited by Gardner, below.
M. Gardner. SA (Jan 1973) c= Knotted, Chap. 9. Gives David Gale's description of his game
and results on it. Addendum in Knotted
points out that it is equivalent to Schuh's game and gives other references.
David Gale. A curious Nim-type game. AMM 81 (1974) 876-879. Describes the game and the basic
results. Wonders if the winning move is
unique. Considers three dimensional and
infinite forms. A note added in proof
refers to Gardner's article, says two programmers have consequently found that
the 8 x 10 game has two winning first moves and mentions Schuh's game.
Winning Ways. 1982.
Pp. 598-600. Brief description
with extensive table of good moves.
Cites an earlier paper of Gale and Stewart which does not deal with this
game.
I
have included this because it has an interesting history and because I found a
nice way to express it as a kind of Markov process or random walk, and this
gives an expression for the average time the game lasts. I then found that the paper by Daykin et al.
gives most of these ideas.
The
game has two or three rules for finishing.
A. One finishes by going exactly to the last
square, or beyond it.
B. One finishes by going exactly to the last
square. If one throws too much, then
one stands still.
C. One finishes by going exactly to the last
square. If one throws too much, one
must count back from the last square.
E.g., if there are 100 squares and one is at 98 and one throws 6, then
one counts: 99, 100, 99, 98, 97,
96 and winds up on 96. (I learned this from a neighbour's child but
have only seen it in one place -- in the first Culin item below.)
Games
of this generic form are often called promotion games. If one considers the game with no snakes or
ladders, then it is a straightforward race game, and these date back to
Egyptian and Babylonian times, if not earlier.
In fact, the same theory applies to
random walks of various sorts, e.g. random walks of pieces on a chessboard,
where the ending is arrival exactly at the desired square.
In the British Museum, Room 52,
Case 24 has a Babylonian ceramic board (WA 1991-7.20,I) for a kind of snakes
and ladders from c-1000. The label says
this game was popular during the second and first millennia BC.
Sheng-kuan t'u [The game of
promotion]. 7C. Chinese game. This is described in:
Nagao Tatsuzo; Shina Minzoku-shi [Manners and Customs of the Chinese];
Tokyo, 1940-1942, perhaps vol. 2, p. 707, ??NYS This is cited in:
Marguerite Fawdry; Chinese Childhood; Pollock's Toy Theatres, London,
1977, p. 183, where the game is described as "played on a board or plan
representing an official career from the lowest to the highest grade, according
to the imperial examination system. It
is a kind of Snakes and Ladders, played with four dice; the object of each
player being to secure promotion over the others."
Thomas Hyde. Historia Nerdiludii, hoc est dicere,
Trunculorum; .... (= Vol. 2 of De Ludis
Orientalibus, see 7.B for vol. 1.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De ludo promotionis mandarinorum, pp. 70-101 -- ??NX. This
is a long description of Shing quon
tu, a game on a board of 98 spaces,
each of which has a specific description which Hyde gives. There is a folding plate showing the Chinese
board, but the copy in the Graves collection is too fragile to photocopy. I did not see any date given for the game.
Stewart Culin. Chinese Games with Dice and Dominoes. From the Report of the U. S. National Museum
for 1893, pp. 489‑537. Pp.
502-507 describes several versions of the Japanese Sugoroku (Double Sixes)
which is a generic name for games using dice to determine moves, including
backgammon and simple race games, as well as Snakes and Ladders games. One version has ending in the form C. Then says
Shing Kún T’ò (The Game of the
Promotion of Officials) is described by
Hyde as The Game of the Promotion of the Mandarins and gives an extended
description of it. There is a legend
that the game was invented when the Emperor Kienlung (1736-1796) heard a
candidate playing dice and the candidate was summoned to explain. He made up a story about the game, saying
that it was a way for him and his friends to learn the different ranks of the
civil service. He was sent off to bring
back the game and then made up a board overnight. However Hyde had described the game a century before this
date. It seems that this is not really
a Snakes and Ladders game as the moves are determined by the throw of the dice
and the position -- there are no interconnections between cells. But Culin notes that the game is complicated
by being played for money or counters which permit bribery and rewards, etc.
Culin. Chess and Playing Cards.
Op. cit. in 4.A.4. 1898.
Pp.
820-822 & plates 24 & 25 between 821 & 822. Says
Shing Kún T’ò (The Game of the
Promotion of Officials) is described by
Hyde as The Game of the Promotion of
the Mandarins and refers to the above
for an extended description. Describes
the Korean version:
Tjyong-Kyeng-To (The Game of Dignitaries) and several others from Korea and Tibet,
with 108, 144, 169 and 64 squares.
Pp.
840-842 & plate 28, opp. p. 841 describes
Chong ün Ch’au (Game of the
Chief of the Literati) as 'in many
respects analogous' to Shing Kún
T’ò and the Japanese game Sugoroku
(Double Sixes) -- in several versions.
Then mentions modern western versions -- Jeu de L'Oie, Giuoco
dell'Oca, Juego de la Oca, Snake Game.
Pp. 843-848 is a table listing 122 versions of the game in the
University of Pennsylvania Museum of Archaeology and Paleontology. These are in 11 languages, varying from 22
to 409 squares.
Bell & Cornelius. Board Games Round the World. Op. cit. in 4.B.1. 1988. Snakes and Ladders
and the Chinese Promotion Game, pp. 65‑67. They describe the Hindu version of Snakes and Ladders,
called Moksha-patamu. Then they discuss Shing Kun t'o (Promotion
of the Mandarins), which was played in
the Ming (1368-1616) with four or more players racing on a board with 98 spaces
and throwing 6 dice to see how many equal faces appeared. They describe numerous modern variants.
Deepak Shimkhada. A preliminary study of the game of Karma in
India, Nepal, and Tibet. Artibus Asiae
44 (1983) 4. ??NYS - cited in Belloli
et al, p. 68.
Andrew Topsfield. The Indian game of snakes and ladders. Artibus Asiae 46:3 (1985) 203‑214 + 14
figures. Basically a catalogue of
extant Indian boards. He says the game
is called Gyān caupad [the d
should have an underdot] or Gyān chaupar in Hindi. He states
that Moksha-patamu sounds like it is Telugu and that this name
appeared in Grunfield's Games of the World (1975) with no reference to a source
and that Bell has repeated this. Game
boards were drawn or painted on paper or cloth and hence were perishable. The oldest extant version is believed to be
an 84 square board of 1735, in the Museum of Indology, Jaipur. There were Hindu, Jain, Muslim and Tibetan
versions representing a kind of Pilgrim's Progress, finally arriving at God or
Heaven or Nirvana. The number of squares
varies from 72 to 360.
He
gives many references and further details.
An Indian version of the game was described by F. E. Pargiter; An Indian
game: Heaven or Hell; J. Royal Asiatic Soc. (1916) 539-542, ??NYS. He cites the version by Ayres (and Love's
reproduction of it -- see below) as the first English version. He cites several other late 19C versions.
F. H. Ayres. [Snakes and ladders game.] No. 200682 Regd. Example in the Bethnal Green Museum, Misc. 8 - 1974. Reproduced in: Brian Love; Play The Game; Michael Joseph, London, 1978; Snakes
& Ladders 1, pp. 22-23. This is the
earliest known English version of the game, with 100 cells in a spiral
and 5
snakes and 5 ladders.
N. W. Bazely &
P. J. Davis. Accuracy of Monte
Carlo methods in computing finite Markov chains. J. of Res. of the Nat. Bureau of Standards -- Mathematics and
Mathematical Physics 64B:4 (Oct-Dec 1960) 211-215. ??NYS -- cited by Davis & Chinn and Bewersdorff. Bewersdorff [email of 6 Jun 1999] brought
these items to my attention and says it is an analysis based on absorbing
Markov chains.
D. E. Daykin, J. E. Jeacocke
& D. G. Neal. Markov chains and
snakes and ladders. MG 51 (No. 378)
(Dec 1967) 313-317. Shows that the game
can be modelled as a Markov process and works out the expected length of play
for one player (47.98 moves) or two players (27.44
moves) on a particular board with finishing rule A.
Philip J. Davis &
William G. Chinn. 3.1416 and All
That. S&S, 1969, ??NYS; 2nd ed, Birkhäuser, 1985, chap. 23 (by
Davis): "Mr. Milton, Mr. Bradley -- meet Andrey Andreyevich Markov",
pp. 164-171. Simply describes how to
set up the Markov chain transition matrix for a game with 100
cells and ending B. Doesn't give
any results.
Lewis Carroll. Board game for one. In: Lewis Carroll's Bedside Book; ed. by
Gyles Brandreth (under the pseud. Edgar Cuthwellis); Methuen, 1979, pp.
19-21. ??look for source; not in
Carroll-Wakeling, Carroll-Wakeling II or Carroll-Gardner. Board of 27 cells with pictures in the odd cells. If you land on any odd cell, except the last
one, you have to return to square 1.
"Sleep is almost certain to have overwhelmed the player before he
reaches the final square." Ending
A is probably intended. (The average
duration of this game should be computable.)
S. C. Althoen, L. King & K.
Schilling. How long is a game of snakes
and ladders? MG 77 (No. 478) (Mar 1993)
71-76. Similar analysis to Daykin,
Jeacocke & Neal, using finishing rule B, getting 39.2 moves. They also use a simulation to find the
number of moves is about 39.1.
David Singmaster. Letter [on Snakes and ladders]. MG 79 (No. 485) (Jul 1995) 396-397. In response to Althoen et al. Discusses history, other ending rules and
wonders how the length depends on the number of snakes and ladders.
Irving L. Finkel. Notes on two Tibetan dice games. IN: Alexander J. de Voogt, ed.; New
Approaches to Board Games Research: Asian Origins and Future Perspectives;
International Institute for Asian Studies, Leiden, 1995; pp. 24-47. Part II: The Tibetan 'Game of Liberation',
pp. 34-47, discusses promotion games with many references to the literature and
describes a particular game.
Jörg Bewersdorff. Glück, Logik und Bluff Mathematik im Spiel -- Methoden, Ergebnisse
und Grenzen. Vieweg, 1998. Das Leiterspiel, pp. 67-68 &
Das Leiterspiel als Markow‑Kette.
Discusses setting up the Markov chain, citing Bazley & Davis, with
the same board as in Davis & Chinn, then states that the average duration
is 39.224 moves.
Jay Belloli, ed. The Universe A Convergence of Art, Music, and Science. [Catalogue for a group of exhibitions and
concerts in Pasadena and San Marino, Sep 2000 - Jun 2001.] Armory Center for the Arts, Pasadena,
2001. P. 68 has a discussion of the Jain
versions of the game, called 'gyanbazi', with a colour plate of a 19C example
with a 9 x 9 board with three extra cells.
This
is a Maori game which can be found in several books on board games. I have included it because it has been
completely analysed. There are eight
(or 2n) points around a central area.
Each player has four (or n) markers, originally placed on consecutive
points. One can move from a point to an
adjacent point or to the centre, or one can move from the centre to a point,
provided the position moved to is empty.
The first player who cannot move is the loser. To prevent the game becoming trivial, it is necessary to require
that the first two (or one) moves of each player involve his end pieces, though
other restrictions are sometimes given.
Marcia Ascher. Mu Torere: An analysis of a Maori game. MM 60 (1987) 90-100. Analyses the game with 2n
points. For n = 1,
there are 6 inequivalent positions (where equivalence is by rotation or
reflection of the board) and play is trivially cyclic. For
n = 2, there are 12
inequivalent positions, but there are no winning positions. For
n = 3, there are 30
inequivalent positions, some of which are wins, but the game is a
tie. Obtains the number of positions
for general n. For the traditional version with n = 4, there are 92 inequivalent positions, some of which are
wins, but the game is a tie, though this is not at all obvious to an
inexperienced player. In 1856, it was
reported that no foreigner could win against a Maori. For n = 5, there are
272 inequivalent positions, but
the game is a easy win for the first player -- the constraints on first moves
need to be revised. Ascher gives
references to the ethnographic literature for descriptions of the game.
Marcia Ascher. Ethnomathematics. Brooks/Cole Publishing, Pacific Grove, California, 1991. Sections 4.4-4.7, pp. 95-109 &
Notes 4-7, pp. 118-119.
Amplified version of her MM article.
There
were a number of earlier guessing games of the Mastermind type before the
popular version devised by Marco Meirovitz in 1973 -- see: Reddi.
One of these was the English Bulls and Cows, but I haven't seen anything
written on this and it doesn't appear in Bell, Falkener or Gomme. Since 1975 there have been several books on
the game and a number of papers on optimal strategies. I include a few of the latter.
NOTATION. If there are h holes and c
choices at each hole, then I abbreviate this as ch.
A. K. Austin. How do You play 'Master Mind'. MTg 71 (Jun 1975) 46-47. How to state the rules correctly.
S. S. Reddi. A game of permutations. JRM 8:1 (1975) 8-11. Mastermind type guessing of a permutation of 1,2,3,4
can win in 5 guesses.
Donald E. Knuth. The computer as Master Mind. JRM 9:1 (1976-77) 1-6. 64 can be won in 5
guesses.
Robert W. Irving. Towards an optimum Mastermind strategy, JRM 11:2 (1978-79) 81-87. Knuth's algorithm takes an average of 5804/1296 = 4.478 guesses. The author
presents a better strategy that takes an average of 5662/1296 = 4.369
guesses, but requires six guesses in one case. A simple adaptation eliminates this, but increases the average
number of guesses to 5664/1296 =
4.370. An intelligent setter will
choose a pattern with a single repetition, for which the average number of
guesses is 3151/720 = 4.376.
A. K. Austin. Strategies for Mastermind. G&P 71 (Winter 1978) 14-16. Presents Knuth's results and some other
work.
Merrill M. Flood. Mastermind strategy. JRM 18:3 (1985-86) 194-202. Cites five earlier papers on strategy,
including Knuth and Irving. He
considers it as a two-person game and considers the setter's strategy. He has several further papers in JRM
developing his ideas.
Antonio M. Lopez, Jr. A PROLOG Mastermind program. JRM 23:2 (1991) 81-93. Cites Knuth, Irving, Flood and two other
papers on strategy.
Kenji Koyama and Tony W.
Lai. An optimal Mastermind
strategy. JRM 25:4 (1994) 251‑256. Using exhaustive search, they find the
strategy that minimizes the expected number of guesses, getting expected
number 5625/1296 = 4.340. However, the worst case in this problem
requires 6 guesses. By a slight
adjustment, they find the optimal strategy with worst case requiring 5 guesses
and its expected number of guesses is
5626/1296 = 4.341. 10 references
to previous work, not including all of the above.
Jörg Bewersdorff. Glück, Logik und Bluff Mathematik im Spiel -- Methoden, Ergebnisse
und Grenzen. Vieweg, 1998. Section 2.15 Mastermind: Auf Nummer sicher, pp. 227-234 &
Section 3.13 Mastermind:
Farbcodes und Minimax, pp. 316-319. Surveys
the work on finding optimal strategies.
Then studies Mastermind as a two-person game. Finds the minimax strategy for the 32 game and
describes Flood's approach.
4.B.12. RITHMOMACHIA = THE PHILOSOPHERS' GAME
I have generally not tried to include board games in any
comprehensive manner, but I have recently seen some general material on this
which will be useful to anyone interested in the game. The game is one of the older and more
mathematical of board games, dating from c1000, but generally abandoned about
the end of the 16C along with the Neo-Pythagorean number theory of Boethius on
which the game was based.
Arno Borst. Das mittelalterliche Zahlenkampfspiel. Sitzungsberichten der Heidelberger Akademie
der Wissenschaften, Philosophisch-historische Klasse 5 (1986) Supplemente. Available separately: Carl Winter, Heidelberg, 1986. Edits the surviving manuscripts on the
game. ??NYS -- cited by Stigter &
Folkerts.
Detlef Illmer, Nora Gädeke,
Elisabeth Henge, Helen Pfeiffer & Monika Spicker-Beck. Rhythmomachia. Hugendubel, Munich, 1987.
Jurgen Stigter. Emanuel Lasker: A Bibliography AND
Rithmomachia, the Philosophers' Game: a reference list. Corrected, 1988 with annotations to 1989, 1
+ 15 + 16pp preprint available from the author, Molslaan 168, NL‑2611 CZ
Delft, Netherlands. Bibliography of the
game.
Jurgen Stigter. The history and rules of Rithmomachia, the
Philosophers' Game. 14pp preprint available
from the author, as above.
Menso Folkerts. 'Rithmimachia'. In: Die deutsche
Litteratur des Mittelalters: Verfasserlexikon; 2nd ed., De Gruyter, Berlin,
1990; vol. 8, pp. 86-94. Sketches
history and describes the 10 oldest texts.
Menso Folkerts. Die Rithmachia des Werinher von
Tegernsee. In: Vestigia Mathematica, ed. by M. Folkerts
& J. P. Hogendijk, Rodopi, Amsterdam, 1993, pp. 107-142. Discusses Werinher's work (12C), preserved
in one MS of c1200, and gives an edition of it.
This
is a very broad field and I will only mention a few early items. Four row mancala games are played in south
and east Africa. Three row games are
played in Ethiopia and adjacent parts of Somaliland. Two row games are played everywhere else in Africa, the Middle
East and south and south-east Asia. See
the standard books by R. C. Bell and Falkener for many examples. Many general books mention the game, but I
only know a few specific books on the game -- these are listed first below.
One
article says that game boards have been found in the pyramids of Khamit (-1580)
and there are numerous old boards carved in rocks in several parts of Africa.
An anonymous article, by a member of the Oware Society in
London, [Wanted: skill, speed, strategy; West Africa (16-22 Sep 1996)
1486-1487] lists the following names for variants of the game: Aditoe (Volta
region of Ghana), Awaoley (Côte d'Ivoire), Ayo (Nigeria), Chongkak (Johore),
Choro (Sudan), Congclak (Indonesia), Dakon (Philippines), Guitihi (Kenya),
Kiarabu (Zanzibar), Madji (Benin), Mancala (Egypt), Mankaleh (Syria), Mbau
(Angola), Mongola (Congo), Naranji (Sri Lanka), Qai (Haiti), Ware (Burkina
Faso), Wari (Timbuktu), Warri (Antigua),
Stewart Culin. Mancala, The National Game of Africa. IN: US National Museum Annual Report 1894,
Washington, 1896, pp. 595-607.
Chief A. O. Odeleye. Ayo
A Popular Yoruba Game.
University Press Ltd., Ibadan, Nigeria, 1979. No history.
Laurence Russ. Mancala Games. Reference Publications, Algonac, Michigan, 1984. Photocopy from Russ, 1995.
Kofi Tall. Oware
The Abapa Version. Kofi Tall
Enterprise, Kumasi, Ghana, 1991.
Salimata Doumbia &
J. C. Pil. Les Jeux de
Cauris. Institut de Recherches
Mathématiques, 08 BP 2030, Abidjan 08, Côte d'Ivoire, 1992.
Pascal Reysset &
François Pingaud. L'Awélé. Le jeu des semailles africaines. 2nd ed., Chiron, Paris, 1995 (bought in Dec
1994). Not much history.
François Pingaud. L'awélé
jeu de strategie africain.
Bornemann, 1996.
Alexander J. de Voogt. Mancala
Board Games. British Museum
Press, 1997. ??NYR.
Larry (= Laurence) Russ. The Complete Mancala Games Book How to Play the World's Oldest Board
Games. Foreword by Alex de Voogt. Marlowe & Co., NY, 2000. His 1984
book is described as an earlier edition of this.
William Flinders Petrie. Objects of Daily Use. (1929);
Aris & Phillips, London??, 1974.
P. 55 & plate XLVII.
??NYS -- described with plate reproduced in Bell, below. Shows and describes a 3 x 14
board from Memphis, ancient Egypt, but with no date given, but Bell
indicates that the context implies it is probably earlier than ‑1500. Petrie calls it 'The game of forty-two and
pool' because of the 42 holes and a large hole on the side, apparently for
storing pieces either during play or between games.
R. C. Bell. Games to Play. Michael Joseph (Penguin), 1988.
Chap. 4, pp. 54-61, Mancala games.
On pp. 54-55, he shows the ancient Egyptian board from Petrie and his
own photo of a 3 x 6 board cut into the roof of a temple at
Deir-el-Medina, probably about ‑87.
Thomas Hyde. Historia Nerdiludii, hoc est dicere,
Trunculorum; .... (= Vol. 2 of De Ludis
Orientalibus, see 7.B for vol. 1.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De Ludo Mancala, pp. 226-232.
Have X of part of this.
R. H. Macmillan. Wari.
Eureka 13 (Oct 1950) 12. 2 x
6 board with each cup having four to
start. Says it is played on the Gold
Coast.
Vernon A. Eagle. On some newly described mancala games from
Yunnan province, China, and the definition of a genus in the family of mancala
games. IN: Alexander J. de Voogt, ed.;
New Approaches to Board Games Research: Asian Origins and Future Perspectives;
International Institute for Asian Studies, Leiden, 1995; pp. 48-62. Discusses the game in general, with many references. Attempts a classification in general. Describes six forms found in Yunnan.
Ulrich Schädler. Mancala in Roman Asia Minor? Board Games Studies International Journal for the Study of Board
Games 1 (1998) 10-25. Notes that mancala could have been played on
a flat board of two parallel rows of squares, i.e. something like a 2 x n
chessboard, but that archaeologists have tended to view such patterns as
boards for race games, etc. Describes
52 examples from Asia Minor. Some
general discussion of Greek and Roman games.
John Romein &
Henri E. Bal (Vrije Universiteit, Amsterdam). New computer cluster solves 3500-year old game. Posted on
www.alphagalileo.org on 29 Aug
2002. They show that Awari is a tie
game. They determined all 889,063,398,406 possible positions and stored them in a 778 GByte
database. They then used a 144
processor cluster to analyse the graph, which 'only' took 51 hours.
R. C. Bell. Games to Play. 1988. Op. cit. in
4.B.13. P. 136 gives some history. The Académie Français adopted the word for
both the pieces and the game in 1790 and it was generally thought that they
were an 18C invention. However, a
domino was found on the Mary Rose, which sank in 1545, and a record of
Henry VIII (reigned 1509-1547) losing £450 at dominoes has been found.
Bell, p. 131, describes the
modern variant Tri-Ominos which are triangular pieces with values at the
corners. They were marketed c1970 and
marked © Pressman Toy Corporation, NY.
Hexadoms are hexagonal pieces
with numbers on the edges -- opposite edges have the same numbers. These were also marketed in the early 1970s
-- I have a set made by Louis Marx, Swansea, but there is no date on it.
Anonymous [R. S. &
J. M. B[rew ?]]. Svoyi kosiri is
an easy game. Eureka 16 (Oct 1953) 8‑12. This is an intriguing game of pure strategy
commonly played in Russia and introduced to Cambridge by Besicovitch. It translates roughly as 'One's own trumps'. There are two players and the hands are
exposed, with one's spades and clubs being the same as the other's hearts and
diamonds. At Cambridge, the cards below
6 are removed, leaving 36 cards in the deck.
The article doesn't explain how trumps are chosen, but if one has spades
as trumps, then the other has hearts as trumps! Players alternate playing to a central discard pile. A player can take the pile and start a new
pile with any card, or he can 'cover' the top card and then play any card on
that. 'Covering' is done by playing a
higher card of the same suit or one of the player's own trumps -- if this
cannot be done, e.g. if the ace of the player's own trumps has been played, the
player has to take the pile. The object
is to get rid of all one's cards.
7.AZ
is actually combinatorial rather than arithmetical and I may shift it.
Pictorial versions: The Premier (1880), Lemon (1890), Stein (1898), King
(1927).
Double-sided versions: The Premier (1880), Brown (1891).
Relation to Magic Squares: Loyd (1896), Cremer (1880), Tissandier
(1880 & 1880?), Cassell's (1881), Hutchison (1891).
Making a magic square with the
Fifteen Puzzle: Dudeney (1898),
Anon & Dudeney (1899), Loyd (1914), Dudeney
(1917), Gordon (1988). See also: Ollerenshaw & Bondi
in 7.N.
Peter Hajek. 1995 report of his 1992 visit to the Museum
of Money, Montevideo, Uruguay, with later pictures by Jaime Poniachik. In this Museum is a metal chest made in England
in 1870 for the National State Bank of Uruguay. The front has a 7 x
7 array of metal squares with bolt
heads. These have to be slid in a 12
move sequence to reveal the three keyholes for opening the chest. This opens up a whole new possible background
for the 15 Puzzle -- can anyone provide details of other such sliding devices?
S&B, pp. 126‑129,
shows several versions of the puzzle.
L. Edward Hordern. Sliding Piece Puzzles. OUP, 1986.
Chap. 2: History of the sliding block puzzle, pp. 18‑30. This is the most extensive survey of the
history. He concludes that Loyd did not
invent the general puzzle where the 15 pieces are placed at random, which
became popular in 1879(?). Loyd may
have invented the 14‑15 version or he may have offered the $1000 prize
for it, but there is no evidence of when (1881??) or where. However, see the entries for Loyd's Tit‑Bits
article and Dudeney's 1904 article which seem to add weight to Loyd's
claims. Most of the puzzles considered
here are described by Hordern and have code numbers beginning with a letter,
e.g. E23, which I will give.
I
contributed a note about computer techniques of solving such puzzles and hoping
that programmers would attack them as computer power increased.
In 1993-1995, I produced four
Sliding Block Puzzle Circulars, totalling 24 pages (since reformatted to 21),
largely devoted to reporting on computer solutions of puzzles in Hordern. Since then, a large number of solution
programs have appeared and many more puzzles have appeared. The best place to look is on Nick Baxter's
Sliding Block Home Page:
http://www.johnrausch.com/slidingblockpuzzles/index.html .
Embossing Co. Puzzle labelled "No. 2 Patent Embossed
puzzle of Fifteen and Magic Sixteen.
Manufactured by the Embossing Co.
Patented Oct 24 1865".
Illustrated in S&B, p. 127.
Examples are in the collections of Slocum and Hordern. Hordern, p. 25, says that searching has not
turned up such a patent.
Edward F. [but drawing gives E.]
Gilbert. US Patent 91,737 --
Alphabetical Instruction Puzzle.
Patented 22 Jun 1869. 1p + 1p
diagrams. Described by Hordern, p. 26. This is not really a puzzle -- it has the
sliding block concept, but along several tracks and with many blank
spaces. I recall a similar toy from
c1950.
Ernest U. Kinsey. US Patent 207,124 -- Puzzle-Blocks. Applied: 22 Nov 1877; patented: 20 Aug 1878. 2pp + 1p diagrams. Described by Hordern, p. 27.
6 x 6 square sliding
block puzzle with one vacant space and tongue & grooving to prevent falling
out. Has letters to spell words. He suggests use of triangular and diamond‑shaped
pieces. This seems to be the most
likely origin of the Fifteen Puzzle craze.
Montgomery Ward & Co. Catalogue.
1889. Reproduced in: Joseph J.
Schroeder, Jr.; The Wonderful World of Toys, Games & Dolls 1860··1930; DBI Books, Northfield, Illinois,
1977?, p. 34. Spelling Boards. Like Gilbert's idea, but a more compact
layout.
Loyd prize puzzle: One hundred pounds. Tit-Bits (14 Oct 1893) 25 &
(18 Nov 1893) 111. Loyd is
described as "author of "Fifteen Puzzle," ...."
Loyd. Tit‑Bits 31 (24 Oct 1896) 57. Loyd asserts he developed the 15 puzzle from a 4 x 4
magic square. "[The fifteen
block puzzle] had such a phenomenal run some twenty years ago. ... There was one of the periodical revivals of
the ancient Hindu "magic square" problem, and it occurred to me to
utilize a set of movable blocks, numbered consecutively from 1 to 16, the conditions
being to remove one of them and slide the others around until a magic square
was formed. The "Fifteen Block
Puzzle" was at once developed and became a craze.
I
give it as originally promulgated in 1872 ..." and he shows it with the 15
and 14 interchanged. "The puzzle
was never patented" so someone used round blocks instead of square
ones. He says he would solve such
puzzles by turning over the 6 and the 9.
"Sphinx" [= Dudeney] says he well remembers the sensation and
hopes "Mr. Loyd is duly penitent."
Dudeney. Great puzzle crazes. Op. cit. in 2. 1904. "... the
"Fifteen Puzzle" that in 1872 and 1873 was sold by millions,
.... When this puzzle was brought out
by its inventor, Mr. Sam Loyd, ... he thought so little of it that he did not
even take any steps to protect his idea, and never derived a penny profit from
it.... We have recently tried all over
the metropolis to obtain a single example of the puzzle, without success." Dudeney says the puzzle came with 16 pieces
and you removed the 16. He also says he
recently could not find a single example in London.
Loyd. The 14‑15 puzzle in puzzleland. Cyclopedia, 1914, pp. 235 & 371 (= MPSL1, prob. 21, pp.
19‑20 & 128). He says he
introduced it 'in the early seventies'.
One problem asks to move from the wrong position to a magic square with
sum = 30 (i.e. the blank is counted
as 0).
This is c= SLAHP, pp. 17‑18 & 89.
G. G. Bain. Op. cit. in 1, 1907. Story of Loyd being unable to patent it.
Anonymous & Sam Loyd. Loyd's puzzles, op. cit. in 1, 1896. Loyd "owns up to the great sin of
having invented the "15 block puzzle"", but doesn't refer to the
patent story or the date.
W. P. Eaton. Loc. cit. in 1, 1911. Loyd refers to it as the 'Fifteen block'
puzzle, but doesn't say he couldn't patent it.
Loyd Jr. SLAHP.
1928. Pp. 1‑3 &
87. "It was in the early 80's, ...
that the world‑disturbing "14‑15 Puzzle" flashed across
the horizon, and the Loyds were among its earliest victims." He gives many of the stories in the
Cyclopedia and two of the same problems.
He doesn't mention the patent story.
W. W. Johnson. Notes on the 15‑Puzzle -- I. Amer. J. Math. 2 (1879) 397‑399.
W. E. Story. Notes on the 15‑Puzzle -- II. Ibid., 399‑404.
J. J. Sylvester. Editorial comment. Ibid., 404.
(This
issue may have been delayed to early 1880??
Johnson & Story are not terribly readable, but Sylvester is
interesting, asserting that this is the first time that the parity of a
permutation has become a popular concept.)
Anonymous. Untitled editorial. New York Times (23 Feb 1880) 4. "... just now the chief amusement of
the New York mind, ... a mental epidemic ....
In a month from now, the whole population of North America will be at
it, and when the 15 puzzle crosses the seas, it is sure to become an English
mania."
Anonymous. EUREKA!
The Popular but Perplexing Problem Solved at Last. "THIRTEEN -- FOURTEEN -- FIFTEEN" New York Herald (28 Feb 1880) 8. ""Fifteen" is a puzzle of
seeming simplicity, but is constructed with diabolical cunning. At first sight the victim feels little or no
interest; but if he stops for a single moment to try it, or to look at any one
else who is trying it, the mania strikes him.
... As to the last two numbers,
it depends entirely upon the way in which the blocks happen to fall in the
first place .... Two or three
enterprising gamblers took up the puzzle and for a time made an excellent
living.... The subject was brought up
in the Academy of Sciences by the veteran scientist Dr. P. H. Vander
Weyde", who showed it could not be solved. The Herald reporter discovered that the problem is solvable if
one turns the board 90o, i.e. runs the numbers down instead of
across, and Vander Weyde was impressed.
The article implies the puzzle had already been widely known for some
time.
Mary T. Foote. US Patent 227,159 -- Game apparatus. Filed: 4 Mar 1880; patented: 4 May 1880. 1p
+ 1p diagrams. The patent is for a box
with sliding numbered blocks for teaching the multiplication tables. Lines 57-63: "I am aware that it is not novel to produce a game apparatus
in which blocks are to be mixed and then replaced by a series of moves; also,
that it is not novel to number such blocks, as in the "game of 15,"
so called, where the fifteen numbers are first mixed and then moved into
place."
Persifor Frazer Jr. Three methods and forty‑eight
solutions of the Fifteen Problem. Proc.
Amer. Philos. Soc. 18 (1878‑1880) 505‑510. Meeting of 5 Mar 1880. Rather cryptic presentation of some possible
patterns. Asserts his 26 Feb article in
the Bulletin (??NYS -- ??where -- Philadelphia??) was the first "solution
for the 13, 15, 14 case".
J. A. Wales. 15 - 14 - 13 -- The Great Presidential
Puzzle. Puck 7 (No. 158) (17 Mar 1880)
back cover.
Anonymous. Editorial:
"Fifteen". New York
Times (22 Mar 1880) 4. "No
pestilence has ever visited this or any other country which has spread with the
awful celerity of what is popularly called the "Fifteen Puzzle." It is only a few months ago that it made its
appearance in Boston, and it has now spread over the entire country." Asserts that an unregenerate Southern
sympathiser has introduced it into the White House and thereby disrupted a
meeting of President Hayes' cabinet.
Sch. [H. Schubert]. The Boss Puzzle. Hamburgischer Correspondent (= Staats- und Gelehrte Zeitung des
Hamburgischen unpartheyeischen Correspondent) No. 82 (6 Apr 1880) 11, with
response on 87 (11 Apr 1880) 12
(Sprechsaal). Gives a fairly careful description
of odd and even permutations and shows the puzzle is solvable if and only if it
is in an even permutation. The response
is signed X and says that when the problem is insoluble, just turn the box by
90o to see another side of the problem!
Gebr. Spiro, Hofliefer (Court
supplier), Jungfernsteig 3(?--hard to read), Hamburg. Hamburgischer Correspondent (= Staats- und Gelehrte Zeitung des
Hamburgischen unpartheyeischen Correspondent) No. 88 (13 Apr 1880) 7. Advertises Boss Puzzles: "Kaiser-Spiel 50Pf. Bismarck-Spiel 50 Pf. Spiel der 15 u.
16, 50 Pf. Spiel der 16 separat, 15
Pf. System und Lösung, 20 Pf."
G. W. Warren. Letter:
Clew to the Fifteen Puzzle. The
Nation 30 (No. 774) (29 Apr 1880) 326.
Anon. Shavings. The London
Figaro (1 May 1880) 12. "The
"15 Puzzle," which has for some months past been making a sensation in
New York equal to that aroused by "H. M. S. Pinafore" last
year, has at length reached this country, and bids fair to become the rage here
also." (Complete item!)
George Augustus Sala. Echoes of the Week. Illustrated London News 76 (No. 2138) (22
May 1880) 491.
Mary T. Foote. US Patent 227,159 -- Game Apparatus. Applied: 4 Mar 1880; patented: 4 May 1880. 1p + 1p diagrams. Described in Hordern, p. 27.
3 x 12 puzzles based on
multiplication tables. Refers to the
"game of 15" and Kinsey.
Arthur Black. ??
Brighton Herald (22 May 1880).
??NYS -- mentioned by Black in a letter to Knowledge 1 (2 Dec 1881) 100.
Anonymous. Our latest gift to England. From the London Figaro. New York Times (11 Jun 1880) 2(?). ??page
The Premier. First (?) double‑sided version, with
pictures of Gladstone and Beaconsfield, apparently produced for the 1880 UK
election. Described in Hordern, pp. 32‑33
& plate I.
Ahrens. MUS II 227.
1918. Story of Reichstag being
distracted in 1880.
P. G. Tait. Note on the Theory of the "15
Puzzle". Proc. Roy. Soc. Edin. 10
(1880) 664‑665. Brief but valid
analysis. Mentions Johnson &
Story. First mention of the possibility
of a 3D version.
T. P. Kirkman. Question 6489 and Note on the solution
of the 15‑puzzle in question 6489.
Mathematical Questions with their Solutions from the Educational Times
34 (1880) 113‑114 & 35 (1881) 29‑30. The question considers the n x n
problem. The note is rather
cryptic. (No use??)
Messrs. Cremer (210 Regent St.
and 27 New Bond St., London). Brilliant
Melancholia. Albrecht Durer's Game of
the Thirty Four and "Boss" Game of the Fifteen. 1880.
Small booklet, 16pp + covers, apparently instructions to fit in a box
with pieces numbered 1 to 16 to be used for making magic squares as well as for
the 15 puzzle. Explains that only half
the positions of the 15 puzzle are obtainable and describes them by
examples. (Photo in The Hordern
Collection of Hoffmann Puzzles, p. 74, and in Hordern, op. cit. above, plate
IV.) Possibly written by "Cavendish"
(Henry Jones).
H. Schubert. Theoretische Entscheidung über das Boss‑Puzzle
Spiel. 2nd ed., Hamburg, 1880. ??NYS
(MUS, II, p. 227)
Gaston Tissandier. Les carrés magiques -- à propos du
"Taquin," jeu mathématique.
La Nature 8 (No. 371) (10 Jul 1880) 81‑82. Simple description of the puzzle called
'Taquin' which came from America and has had a very great success for several
weeks. Says it had 16 squares and was
usable as a sliding piece puzzle or a magic square puzzle. Cites Frénicle's 880 magic squares of order
4.
Anon. & C. Henry. Gaz. Anecdotique Littéraire, Artistique et
Bibliographique. (Pub. by
G. d'Heylli, Paris) Year 5, t. II,
1880, pp. 58‑59 & 87‑92.
??NYS
Piarron de Mondésir. Le dernier mot du taquin. La Nature 8 (No. 382) (25 Sep 1880) 284‑285. Simple description of parity decision for
the 15 puzzle. Says 'la Presse
illustrée' offered 500 francs for achieving the standard pattern from a random
pattern, but it was impossible, or rather it was possible in only half the
cases.
Jasper W. Snowdon. The "Fifteen" Puzzle. Leisure Hour 29 (1880) 493‑495.
Gwen White. Antique Toys. Batsford, London, 1971;
reprinted by Chancellor Press, London, nd [1982?]. On p. 118, she says: "The French game of Taquin was played
in 1880, in which 15 pieces had to be moved into 16 compartments in as few
moves as possible; the word 'taquin' means 'a teaser'." She gives no references.
Tissandier. Récréations Scientifiques. 1880?
2nd
ed., 1881 -- unlabelled section, pp. 143-153.
As: Le taquin et les carrés magiques;
seen in 1883 ed., ??NX; 1888:
pp. 208-215. Adapted from the 1880 La
Nature articles of Tissandier and de Mondésir.
1881 says it came from America -- 'récemment une nouvelle apparition',
but this is dropped in 1888 -- otherwise the two versions are the same.
Translated
in Popular Scientific Recreations, nd [c1890], pp. 731‑735. Text says "Mathematical games, ...,
have recently obtained a new addition ....
... from America, ...." The
references to contemporary reactions are deleted and the translation is
confused. E.g. the newspaper is now
just "a French paper" and the English says the problem is impossible
in nine cases out of ten!
Lucas. Récréations scientifiques sur l'arithmétique et sur la géométrie
de situation. Sixième récréation: Sur
le jeu du taquin ou du casse‑tête américain. Revue scientifique de France et de l'étranger (3) 27 (1881) 783‑788. c= Le jeu du taquin, RM1, 1882, pp. 189‑211. Revue says that Sylvester told him that it
was invented 18 months ago by an American deaf‑mute. RM1 says "vers la fin de
1878". Cf Schubert, 1895.
Cassell's. 1881.
Pp. 96‑97: American puzzles "15" and
"34". = Manson, pp.
246-248. Says "articles ... have
appeared in many periodicals, but no one has ... publish[ed] a solution."
Then sketches the parity concept and its application.
Richard A. Proctor. The fifteen puzzle. Gentlemen's Magazine 250 (No. 1801) (1881)
30‑45.
"Boss". Letter:
The fifteen puzzle. Knowledge 1
(11 Nov 1881) 37-38, item 13. This magazine
was edited by Proctor. The letter
starts: "I am told that in a
magazine article which appeared some time ago, you have attempted to show that
there are positions in the Fifteen Puzzle from which the won position can never
be obtained." I suspect the letter
was produced by Proctor. The response
is signed Ed. and begins: "I
thought the Fifteen Puzzle was dead, and hoped I had had some share in killing
the time-absorbing monster." Notes
that many people get to the position starting
blank, 1, 2, 3 and view this as
a win. Sketches parity argument and
suggests "Boss" work on the
3 x 3 or 3 x 2 or even the 2 x 2 version.
Editorial comment. The fifteen puzzle. Knowledge 1 (25 Nov 1881) 79. "I supposed every one knew the Fifteen
Puzzle." Proceeds to explain, obviously
in response to readers who didn't know it.
Arthur Black. Letter:
The fifteen puzzle. Knowledge 1
(2 Dec 1881) 100, item 80. Sketches a
proof which he says he published in the Brighton Herald of 22 May 1880.
"Yawnups". Letter:
The fifteen puzzle. Knowledge 1
(30 Dec 1881) 185. Solution from the
15-14 position obtained by turning the box.
Editorial comment says the solution uses 102 moves and the editor gets
an easy solution in 57 moves. Adds that
a 60 move solution has been received.
Arthur Black. Letter:
The fifteen puzzle. Knowledge 1
(13 Jan 1882) 230. Finds a solution
from the 15-14 position in 39 moves by turning the box and asserts no shorter
solution is possible. Says he also gave
this in the Brighton Herald in May 1880.
An addition says J. Watson has provided a similar solution, which takes
38 moves??
A. B. Letter: The fifteen
puzzle. Knowledge 2 (20 Oct 1882) 345,
item 598. Finds a box-turning solution
in 39 moves.
C. J. Malmsten. Göteborg Handl 1882, p. 75. ??NYS -- cited by Ahrens in his Encyklopadie
article, op. cit. in 3.B, 1904.
Anonymous. Enquire Within upon Everything. Houlston and Sons, London. This was a popular book with editions almost
every year -- I don't know when the following material was added. Section 2591: Boss; or the Fifteen Puzzle,
p. 363. Place the pieces
'indifferently' in the box. Half the
positions are unsolvable. Cites
Cavendish for the solution by turning the box 90o but notes this
only works with round pieces. Goes on
to The thirty-four puzzle, citing Dürer.
I found this material in the 66th ed., 862nd thousand, of 1883, but I
didn't find the material in the 86th ed of 1892.
Letters received and short
answers. Knowledge 4 (16 Nov 1883)
310. 'Impossible'.
P. G. Tait. Listing's Topologie. Philosophical Mag. (5) 17 (No. 103) (Jan
1884) 30‑46 & plate opp. p. 80. Section 11, p. 39. Simple
but cryptic solution.
Letters received and short
answers. Letter from W. S. B. asks how
to solve the problem when the last row has 13, 14, 15 [sic!]; Answer by Ed. points out the misprint and
says the easiest solution is to remove the 15 and put it after the 14, or to
invert the 6 and 9. Knowledge
6 (No. 159) (14 Nov 1884) 412
& 6 (No. 160) (21 Nov 1884)
429.
Don Lemon. Everybody's Pocket Cyclopedia .... Saxon & Co., London, (1888), revised 8th
ed., 1890. P. 137: The fifteen
puzzle. Brief description, with pieces
placed randomly in the box -- "to get the last three into order is often a
puzzle indeed".
John D. Champlin & Arthur E. Bostwick. The
Young Folk's Cyclopedia of Games and Sports.
1890. ??NYS Cited in Rohrbough; Brain Resters and
Testers; c1935; Fifteen Puzzle, p. 20.
Describes idea of parity of number of exchanges. [Another reference provided more details of
Champlin & Bostwick.]
Lemon. 1890. A trick puzzle, no.
202, pp. 31 & 105 (= Sphinx, no. 422, pp. 60 & 112). 15 puzzle with lines on the pieces to arrange
as "a representation of a president with only one eye". The solution is a spelling of the word 'president'. Attributed to Golden Days -- ??. After The Premier puzzle of c1880, this is
the second suggestion of using a picture and the first publication of the idea
that I have seen.
G. A. Hutchison, ed. Indoor Games and Recreations. The Boy's Own Bookshelf. (1888);
New ed., Religious Tract Society, London, 1891. (See M. Adams; Indoor Games for a much
revised version, but which doesn't contain this material.) Chap. 19: The American Puzzles., pp. 240‑241. "These puzzles, known as the 'Thirty‑four
Game' and the 'Fifteen Game,' on their introduction amongst us some years ago
...." "The '15' puzzle would
appear to have been, on its coming to England a few years ago, strictly a new
introduction ...." He sketches the
parity concept. [NOTE. I have seen a reference to the editor as
Hutchinson, but the book definitely omits the first n.]
Daniel V. Brown. US Patent 471,941 -- Puzzle. Applied: 23 Apr 1891; patented: 29 Mar 1892, 2pp + 1p diagrams. Double-sided 16 block puzzle to spell George
Washington on one side and Benjamin Harrison on the other. No sliding involved.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
American fifteen puzzle, pp. 105-107.
"The Fifteen Puzzle was introduced by a shrewd American some ten
years ago, ...." Refers to Tait's
1880 paper. Says half the positions are
impossible, but solves them by turning the box 90o or by inverting
the 6 and the 9.
Hoffmann. 1893.
Chap IV, no. 69: The "Fifteen" or "Boss" puzzles,
pp. 161‑162 & 217‑218 = Hoffmann-Hordern, pp. 142-144, with
photo of five early examples, two or three of which also are thirty-four
puzzles. (Hordern Collection, p. 74,
has a photo of a version by Cremer, cf above.)
"This, like a good many of the best puzzles, hails from America,
where, some years ago, it had an extraordinary vogue, which a little later
spread to this country, the British public growing nearly as excited over the
mystic "Fifteen" as they did at a later date over the less innocent
"Missing Word" competitions."
He distinguishes between the ordinary Fifteen where one puts the pieces
in at random, and the Boss or Master puzzle which has the 14 and 15
reversed. "Notwithstanding the
enormous amount of energy that has been expended over the "Fifteen"
Puzzle, no absolute rule for its solution has yet been discovered and it
appears to be now generally agreed by mathematicians that out of the vast
number of haphazard positions ... about half admit [of solution]. To test whether ... the following rule has
been suggested." He then says to
count the parity of the number of transpositions.
Hoffmann. 1893.
Chap. IV, no. 70: The peg‑away puzzle, pp. 163 & 218
= Hoffmann‑Hordern, p. 145.
This is a 3 x 3 version of the Fifteen puzzle, made by Perry
& Co. Start with a random pattern
and get to standard form. "The
possibility of success in solving this puzzle appears to be governed by
precisely the same rule as the "Fifteen" Puzzle." Hoffmann-Hordern has no photo of this -- do
any examples exist??
H. Schubert. Zwölf Geduldspiele. Dümmler, Berlin, 1895. [Taken from his columns in
Naturwissenschaftlichen Wochenschrift, 1891-1894.] Chap. VII: Boss-Puzzle oder Fünfzehner-Spiel, pp. 75-94?? Pp. 75-77 sketches the history, saying it
was called "Jeu du Taquin" (Neck-Spiel) in France and was popular in
1879-1880 in Germany. Cites Johnson
& Story and his own 1880 booklet.
Gives the story of a deaf and dumb American inventing it in Dec 1878,
saying "Sylvester communicated this at the annual meeting of the
Association Française pour l'Avancement des Sciences at Reims". Cf Lucas, 1881. [There is a second edition, Teubner??,
Leipzig, 1899, ??NYS. However this
material is almost identical to the beginning of Chap. 15 in Schubert's
Mathematische Mussestunden, 3rd ed., Göschen, Leipzig, 1909, vol. 2. The later version omits only some of the
Hamburg details of 1879-1880. Hence the
2nd ed. of Zwölf Geduldspiele is probably very close to these versions.]
Dudeney. Problem 49: The Victoria Cross puzzle. Tit‑Bits 32 (4 &
25 Sep 1897) 421 & 475.
= AM, 1917, prob. 218, pp. 60 & 194. B7. 3 x 3 board with letters Victoria going clockwise
around the edges, leaving the middle empty, and starting with V in
a corner. Slide to get Victoria
starting at an edge cell, in the fewest moves.
Does it in 18 moves, by interchanging the i's and says there are 6 such solutions.
Dudeney. Problem 65: The Spanish dungeon. Tit‑Bits 33 (1 Jan &
5 Feb 1898) 257 & 355.
= AM, 1917, prob. 403, pp. 122-123 & 244. B14.
Convert 15 Puzzle, with pieces in correct order, into a magic
square. Does it in 37 moves.
Conrad F. Stein. US Design 29,649 -- Design for a
Game-Board. Applied: 29 Sep 1898; patented: 8 Nov 1898 as No. 692,242. 1p + 1p diagrams. This
appears to be a 3 x 4 puzzle with a picture of a city with a Spanish
flag on a tower. Apparently the object
is to move an American flag to the tower.
Anon. & Dudeney. A chat with the Puzzle King. The Captain 2 (Dec? 1899) 314-320; 2:6 (Mar 1900) 598-599 &
3:1 (Apr 1900) 89. The eight fat
boys. 3 x 3 square with pieces: 1 2
3; 4 X 5; 6 7 8 to be shifted into
a magic square. Two solutions in 19
moves. Cf Dudeney, 1917.
Addison Coe. US Patent 785,665 -- Puzzle or Game
Apparatus. Applied: 17 Nov 1904; patented: 1 Mar 1905. 4pp + 3pp diagrams. Gives a
3 x 5 flat version and a 3‑dimensional
version -- cf 5.A.2.
Dudeney. AM.
1917.
Prob.
401: Eight jolly gaol birds, pp. 122 & 243. E23. Same as 'The eight
fat boys' (see Anon. & Dudeney, 1899) with the additional condition that
one person refuses to move, which occurs in one of the two previous solutions.
Prob.
403: The Spanish dungeon, pp. 122-123 & 244. = Tit-Bits prob. 65 (1898).
B14.
Prob.
404: The Siberian dungeons, pp. 123 & 244.
B16. 2 x 8 array with prisoners 1, 2, ..., 8 in top row and 9, 10,
..., 16 in bottom row. Two extra rows of 4 above the right hand end
(i.e. above 5, 6, 7, 8) are empty.
Slide the prisoners into a magic square. Gives a solution in 14 moves, due to G. Wotherspoon, which they
feel is minimal. This allows long moves
-- e.g. the first move moves 8 up two and left 3.
"H. E. Licks" [pseud.
of Mansfield Merriman]. Recreations in
Mathematics. Van Nostrand, NY,
1917. Art. 28, pp. 20‑21. 'About the year 1880 ... invented in 1878 by
a deaf and dumb man....'
[From
sometime in the 1980s, I suspected the author's name was a pseudonym. On pp. 132-138, he discusses the Diaphote
Hoax, from a Pennsylvania daily newspaper of 10 Feb 1880, which features the
following people: H. E. Licks, M. E. Kannick, A. D. A. Biatic, L. M.
Niscate. The diaphote was essentially a
television. He says this report was
picked up by the New York Times and the New York World. An email from Col. George L. Sicherman on 5
Jun 2000 agrees that the name is false and suggested that the author was
"the eminent statistician Mansfield Merriman" who wrote the article
on The Cattle Problem of Archimedes in Popular Science Monthly (Nov
1905), which is abridged on pp. 33-39 of the book, but omitting the author's
name. Sichermann added that Merriman was
one of the authors of Pillsbury's List.
William Hartston says this was an extraordinary list of some 30 words
which Pillsbury, who did memory feats, was able to commit to memory quite
rapidly. Sicherman continued to
investigate Merriman and got Prof. Andri Lange interested and Lange
corresponded with a James A. McLennan, author of a history of the physics
department at Lehigh University where Merriman had been. McLennan found Merriman's obituary from the
American Society of Civil Engineers which states that Merriman used H. E. Licks
as a pseudonym. [Email from Sicherman
on 25 Feb 2002.]]
Stephen Leacock. Model Memoirs and Other Sketches from Simple
to Serious. John Lane, The Bodley Head,
1939, p. 300. "But this puzzle
stuff, as I say, is as old as human thought.
As soon as mankind began to have brains they must have loved to exercise
them for exercise' sake. The 'jig-saw'
puzzles come from China where they had them four thousand years ago. So did the famous 'sixteen puzzle' (fifteen
movable squares and one empty space) over which we racked our brains in the
middle eighties."
G. Kowalewski. Boss‑Puzzle und verwandte Spiele. K. F. Kohler Verlag, Leipzig, 1921
(reprinted 1939). Gives solution of
general polygonal versions, i.e. on a graph with a Hamilton circuit and one or
more diagonals.
Hummerston. Fun, Mirth & Mystery. 1924.
1 2
9 Push,
pp. 22 & 25. This is played on the
board
3 10 11
4 shown at the left
with its orthogonal lines, like
12 13 3,
10, 11, 4, and its diagonal lines, like
5 14 15
6 1, 9, 11, 13,
6. 10, 15 and 11, 14 are not
16 connected,
so this is an octagram. Take 16
7 8 numbered counters and place
them at random on
the
board and remove counter 16. Move the
pieces
to
their correct locations. He asserts
that 'unlike the original ["Sixteen" Puzzle],
no
position can be set up in "Push" that cannot be solved'.
The six bulls puzzle, Puzzle no.
34, pp. 90 & 177. This uses
the 2 x 3 + 1 0
board
shown at the right, where the 0 is the blank space. Exchange 1 2 3
3
and 6 and 4 and 5. He does it
in 20
moves. [This is Hordern's 4 5 6
B3,
first known from 1977 under the name Bull Pen, but is a variant of
Hordern's B2, first known from 1973.]
Q. E. D. -- The sergeant's
problem, Puzzle no. 40, pp. 106 & 178.
Take a 2 x 3 board, with the centre of one long side
blank. Interchange the men along one
short side. He does this in 17
moves, but the blank is not in its initial position nor are the other
men. [This is Hordern's B1, first known
from Loyd's Cyclopedia, 1914.]
King. Best 100. 1927. No. 26, p. 15. = Foulsham's, no. 9, pp. 7 & 10. "An entertaining variation ... is to draw, and colour, if
you like, a small picture; then cut it into sixteen squares and discard the
lower right hand square."
G. Kowalewski. Alte und neue mathematische Spiele. Teubner, Leipzig, 1930, pp. 61‑81. Gives solution of general polygonal
versions.
Dudeney. PCP.
1932. The Angelica puzzle, prob.
253, pp. 76 & 167. = 435, prob.
378, pp. 136 & 340. B8. 3 x 3
problem -- convert: A C I L E G
N A X to A N G
E L I C A X. Requires interchanging the As.
Solution in 36 moves.
In the answer in 435, Gardner notes that it can be done in 30
moves.
H. V. Mallison. Note 1454:
An array of squares. MG 24 (No.
259) (May 1940) 119‑121.
Discusses 15 Puzzle and says any legal position can be achieved in at
most about 150 moves. But if one fixes
cells 6, 7, 11, then a simple problem requires about 900
moves.
McKay. At Home Tonight.
1940. Prob. 44: Changing the
square, pp. 73 & 88. In the usual
formation, colour the pieces alternately blue and red, as on a chessboard, with
the blank at the lower right position 16 being a missing red, so there are 7
reds. Move so the colours are still
alternating but the blank is at the lower left, i.e. position 13. Takes 15 moves.
Sherley Ellis Stotts. US Patent 3,208,753 -- Shiftable Block
Puzzle Game. Filed: 7 Oct 1963; patented: 28 Sep 1965. 4pp + 2pp diagrams. Described in Hordern, pp. 152-153, F10‑12. Rectangular pieces of different sizes. One can also turn a piece.
Gardner. SA (Feb 1964) = 6th Book, chap. 7.
Surveys sliding-block puzzles with non-square pieces and notes there is
no theory for them. Describes a number
of early versions and the minimum number of moves for solution, generally done
by hand and then confirmed by computer.
Pennant Puzzle, C19; L'Âne
Rouge, C27d; Line Up the Quinties,
C4; Ma's Puzzle, D1; a form of Stotts' Baby Tiger Puzzle, F10.
Gardner. SA (Mar & Jun 1965) c= 6th Book, chap. 20. Prob. 9: The eight-block puzzle. B5.
3 x 3 problem -- convert: 8 7 6
5 4 3 2 1 X to
1 2 3 4 5 6 7 8 X.
Compares it with Dudeney's Angelica puzzle (1932, B8) but says it can be
done if fewer than 36 moves.
Many readers found solutions in
30 moves; two even found all 10 minimal solutions by
hand! Says Schofield (see next entry)
has been working on this and gives the results below, but this did not quite
resolve Gardner's problem. William F.
Dempster, at Lawrence Radiation Laboratory, programmed a IBM 7094 to find all
solutions, getting 10 solutions in 30 moves; 112
in 32 moves and 512
in 34 moves. Notes it is
unknown if any problem with the blank in a side or corner requires more
than 30 moves. (The description
of Schofield's work seems a bit incorrect in the SA solution, and is changed in
the book.)
Peter D. A. Schofield. Complete solution of the 'Eight‑Puzzle'. Machine Intelligence 1 (1967) 125‑133. This is the
3 x 3 version of the 15 Puzzle,
with the blank space in the centre.
Works with the corner twists which take the blank around a 2 x 2
corner in four moves. Shows that
the 5-puzzle, which is the 3 x 2 version, has every position reachable in at
most 20 moves, from which he shows that an upper bound for the 8-puzzle
is 48
moves. Since the blank is in the
middle, the 8!/2 = 20160 possible positions fall into 2572
equivalence classes. He also
considers having inverse permutations being equivalent, which reduces to 1439
classes, but this was too awkward to implement. An ATLAS program found that the maximum
number of moves required was 30 and
60 positions of 12
classes required this maximum number, but no example is given -- but see
previous entry.
A. L. Davies. Rotating the fifteen puzzle. MG 54 (No. 389) (Oct 1970) 237‑240. Studies versions where the numbers are printed
diagonally so one can make a 90o turn of the puzzle. Then any pattern can be brought to one of
two 'natural' patterns. He then asks
when this is true for an m x n board and obtains a complicated
solution. For an n x n
board, n must be divisible by 4.
R. M. Wilson. Graph puzzles, homotopy and the alternating
group. J. Combinatorial Thy., Ser. B,
16 (1974) 86‑96. Shows that a
sliding block puzzle, on any graph of n
+ 1 points which is non‑separable
and not a cycle, has at least An as its group -- except for one case on 7
points.
Alan G. & Dagmar R.
Henney. Systematic solutions of the
famous 15‑14 puzzles. Pi Mu
Epsilon J. 6 (1976) 197‑201. They
develop a test‑value which significantly prunes the search tree. Kraitchik gave a problem which took him 114
moves -- the authors show the best solution has 58
moves!
David Levy. Computer Gamesmanship. Century Publishing, London, 1983. [Most of the material appeared in Personal
Computer World, 1980‑1981.] Pp.
16‑29 discusses 8‑puzzle and uses the Henney's test‑value as
an evaluation function. Cites
Schofield.
Nigel Landon & Charles
Snape. A Way with Maths. CUP, 1984.
Cube moving, pp. 23 & 46. Consider
a 9-puzzle in the usual arrangement: 1
2 3, 4 5 6, 7 8 x. Move the 1 to
the blank position in the minimal number of moves, ignoring what happens to the
other pieces. Generalise. Their answer only says 13
is minimal for the 3 x 3 board.
My
student Tom Henley asked me the m x
n problem in 1993 and gave a
conjectural minimum, which I have corrected to: if m = n, then it can be done in 8m ‑ 11 moves;
but if n < m, then it can be done in 6m + 2n - 13 moves, using a straightforward method. However, I don't see how to show this is minimal, though it seems
pretty clear that it must be. I call
this a one-piece problem. See also
Ransom, 1993.
Len Gordon. Sliding the 15‑1 [sic, but 15‑14
must have been meant] puzzle to magic squares.
G&PJ 4 (Mar 1988) 56.
Reports on computer search to find minimal moves from either ordinary or
15‑14 forms to a magic square.
However, he starts with the blank before the 1, i.e. as a 0
rather than a 16.
Leonard J. Gordon. The 16‑15 puzzle or trapezeloyd. G&PJ 10 (1989) 164. Introduces his puzzle which has a
trapezoidal shape with a triangular wedge in the 2nd and 3rd row so the last
row can hold 5 pieces, while the other rows hold four pieces. Reversing the last two pieces can be done
in 85
moves, but this may not be minimal.
George T. Gilbert & Loren C.
Larson. A sliding block problem. CMJ 23:4 (Sep 1992) 315‑319. Essentially the same results as obtained by
R. Wilson (1974). Guy points this out
in 24:4 (Sep 1993) 355-356.
P. H. R. [Peter H. Ransom]. Adam's move. Mathematical Pie 128 (Spring 1993) 1017 & Notes, p. 3. Considers the one piece problem of Langdon
& Snape, 1984. Solution says the
minimal solution on a n x n board is
8n - 11, but doesn't give the
answer for the m x n board.
Bernhard Wiezorke. Sliding caution. CFF 32 (Aug 1993) 24-25
& 33 (Feb 1994) 32. In 1986, the German games company ASS
(Altenburg Stralsunder Spielkarten AG) produced a game called Vorsicht
(= Caution). Basically this is a 3 x 3
board considered as a doubly crossed square. It has pieces marked with
+ or x. The +
pieces can only move orthogonally;
the x pieces can only move diagonally.
The pieces are coloured and eight are placed on the board to be played
as a sliding piece puzzle from given starts to given ends. The diagonal moves are awkward to make and
Wiezorke suggests the board be spread out enough for diagonal moves to be
made. A note at the end says he has
received two similar games made by Y. A. D. Games in Israel.
Bala Ravikumar. The Missing Link and the Top-Spin. Report TR94-228, Department of Computer
Science and Statistics, University of Rhode Island, Jan 1994. The Missing Link is a cylindrical form of
the Fifteen Puzzle, with four layers and four pieces in each layer. The middle two layers are rigidly joined,
but that makes little difference in solving the puzzle. After outlining the relevant group theory
and solving the Fifteen Puzzle, he shows the state space of the Missing Link
is S15.
Richard E. Korf &
Ariel Felner. Disjoint pattern
database heuristics. Artificial
Intelligence 134 (2002) 9-22. Discusses
heuristic methods of solving the Fifteen Puzzle, Rubik's Cube, etc. The authors applied their method to 1000
random positions of the Fifteen Puzzle.
The optimal solution length averaged 52.522 and the average time
required was 27 msec. They also did 50
random positions of the Twenty-Four Puzzle and found an average optimal
solution length of 100.78, with average time being two days on a 440MHz
machine.
S&B, pp. 130‑133, show
many versions.
See Kinsey, 1878, above, for
mention of triangular and diamond‑shaped pieces.
Henry Walton. US Patent 516,035 -- Puzzle. Applied: 14 Mar 1893; patented: 6 Mar 1894. 1p + 1p diagrams. Described in Hordern, pp. 27 & 68‑69,
C1. 4 x 4 area with five
1 x 2 & two
2 x 1 pieces.
Lorman P. Shriver. US Patent 526,544 -- Puzzle. Applied: 28 Jun 1894; patented: 25 Sep 1894, 2pp + 1p diagrams. Described in Hordern, p. 27. 4 x 5
area with two 2 x 1 &
15 1 x 1
pieces. Because there is only
one vacant space, the rectangles can only move lengthwise and so this is a dull
puzzle.
Frank E. Moss. US Patent 668,386 -- Puzzle. Applied: 8 Jun 1900; patented: 19 Feb 1901. 2pp + 1p diagrams. Described in Hordern, pp. 27‑28
& 75, C14. 4 x 4 area with six 1 x 1, two 1 x 2
&
two 2 x 1
pieces, allowing sideways movement of the rectangles.
William H. E. Wehner. US Patent 771,514 -- Game Apparatus. Applied: 15 Feb 1904; patented: 4 Oct 1904. 2pp + 1p diagrams. First to use L-shaped
pieces. Described in Hordern, pp. 28
& 107, D5.
Lewis W. Hardy. US Patent 1,017,752 -- Puzzle. Applied: 14 Dec 1907; patented: 20 Feb 1912. 3pp + 1p diagrams. Described in Hordern, pp. 29 & 89‑90, C43-45. 4 x 5
area with one 2 x 2, two
1 x 2, three 2 x 1
& four 1 x 1
pieces.
L. W. Hardy. Pennant Puzzle. Copyright 1909. Made by
OK Novelty Co., Chicago. No known
patent. Described in Gardner, SA (Feb
1964) = 6th Book, chap. 7 and in Hordern, pp. 28‑29 & 78-79,
C19. 4 x 5 area with one 2 x 2,
two 1 x 2, four 2 x 1, two
1 x 1 pieces.
Nob Yoshigahara designed Rush
Hour in the late 1970s and it was produced in Japan as Tokyo Parking Lot. Binary Arts introduced it to the US in 1996
and it became very popular.
Winning Ways. 1982.
Pp. 769-777: A trio of sliding block puzzles. This covers Dad's Puzzler (c19, with piece 8 moved two places to
the right), The Donkey (C27d, with all the central pieces moved down one
position) and The Century (C42), showing how one can examine partial problems
which allow one to consider many positions the same and much reduce the number
of positions to be studied. This allows
the graph to be written on a large sheet and solutions to be readily found.
Andrew N. Walker. Checkmate and other sliding-block
puzzles. Mathematics Preprint Series,
University of Nottingham, no. 95-32, 1995, 8pp + covers. Describes a version by W. G. H. [Wil]
Strijbos made by Pussycat. 4 x 4 with an extra position below the left
column. Pieces are alternately black
and white and have a black king, a white king and a white rook on them and the
object is to produce checkmate, but all positions must be legal in chess,
except that the black and white markings do not have to be correct in the
intermediate positions. However, one
soon finds that one tile is fixed in place and two other tiles are joined
together. He discusses general computer
solving techniques and finds there are five optimal solutions in 68 moves. He then discusses other problems, citing
Winning Ways, Hordern and my Sliding Block Puzzle Circulars. He gives the UNIX shell scripts that he
used.
Ivars Peterson. Simple puzzles can give computers an
unexpectedly strenuous workout. Science
News 162:7 (17 Aug 2002) 6pp PO from their website, http:''sciencenews.org . Reports on recent work by Gary W. Flake
& Eric B. Baum that Nob Yoshigahara's Rush Hour puzzle is PSPACE complete,
but is not polynomial time. Robert A.
Hearn and Erik D. Demaine have verified and extended this, showing other
sliding block puzzles are PSPACE complete, including the case where all pieces
are dominoes and can slide sideways as well as front and back.
5.A.2. THREE DIMENSIONAL VERSIONS
See Hordern, pp. 27, 156‑160
& plates IX & X.
P. G. Tait. Note on the Theory of the "15
Puzzle". Proc. Roy. Soc. Edin. 10
(1880) 664‑665. "...
conceivable, but scarcely realisable ..."
Charles I. Rice. US Patent 416,344 -- Puzzle. Applied: 9 Sep 1889; patented: 3 Dec 1889. 2pp + 1p diagrams. Described in Hordern, pp. 27 & 157‑158,
G2. 2 x 2 x 2 version with peepholes in the faces.
Ball. MRE, 1st ed., 1892, p. 78.
Mentions possibility.
Hoffmann. 1893.
Chap. X, No. 1: The John Bull political puzzle, pp. 331 &
357-358 = Hoffmann-Hordern, pp.
215-216. A 3 x 3 board in the form of
a cylinder, with an extra cell attached to one bottom cell. Pieces can move back and forth around each
level, but the connections from one level to the next are all parallel to one
of the diagonals -- though this isn't really a complication compared to having
vertical connections. The pieces have
two markings: three colours and three letters.
When they are randomly placed on the board, you have to move them so
they form a pair of orthogonal 3 x
3 Latin squares. Fortunately there are such arrangements
which differ by an odd permutation, so the puzzle can be solved from any random
starting point. Two examples done. Says the game is produced by Jaques &
Son.
Addison Coe. US Patent 785,665 -- Puzzle or Game
Apparatus. Applied: 17 Nov 1904; patented: 1 Mar 1905. 4pp + 3pp diagrams. Mentioned in Hordern, pp. 158‑159,
G3. Gives a 3 x 5 flat version and
a 3 x 3 x 3 cubical version with 3 x
3 arrays of holes in the six faces (in
order to push the pieces) and a
3 x 5 cylindrical version.
Burren Loughlin &
L. L. Flood. Bright-Wits Prince of Mogador. H. M. Caldwell Co., NY, 1909.
The nine disks, pp. 29-34 & 60.
Same as Hoffmann except pieces have colour and shape.
Guy thinks Hein patented
Bloxbox, but I have not found any US patent of it -- ??CHECK.
Gardner. SA (Feb 1973). First mention of Hein's Bloxbox.
Daniel Kosarek. US Patent 3,845,959 -- Three-Dimensional
Block Puzzle. Filed: 14 Nov 1973; patented: 5 Nov 1974. 3pp + 1p diagrams (+ 1p abstract). Mentioned in Hordern, pp. 158‑159,
G3. 3 x 3 x 3 box with 3 x 3 array of portholes on each face. Mentions
4 x 4 x 4 and
larger versions.
Gabriel Nagorny. US Patent 4,428,581 -- Tri-dimensional
Puzzle. Filed: 16 Jun 1981; patented: 31 Jan 1984. Cover page + 3pp + 3pp diagrams. Three dimensional sliding cube puzzles with
central pieces joined together. A 3 x 3 x 3
version was made in Hungary and marketed as a Varikon Box. Inventor's address is in France and he cites
earlier French applications of 19 Jun 1980 and 19 Nov 1980. He also describes a 3 x 4 x 4 version with the central areas of each face
joined to a 1 x 2 x 2 block in the middle.
Here one has a set of solid pieces in
a tray and one tilts or rolls a piece into the blank space.
Thomas Henry Ward.
UK Patent 2,870 -- Apparatus for
Playing Puzzle or Educational Games.
Provisional: 8 Jun 1883;
Complete as: An Improved Apparatus to be Employed in Playing Puzzle or
Educational Games, 6 Dec 1883. 3pp + 1p
diagrams.
US Patent 287,352 -- Game
Apparatus. Applied: 13 Sep 1883; patented: 23 Oct 1883. 1p + 1p diagrams. Hexagonal board of 19 triangles with 18
tetrahedra to tilt.
George Mitchell &
George Springfield. UK Patent
6867 -- A novel puzzle, and improvements in the construction of apparatus
therefor. Applied: 16 Mar 1897; accepted: 5 Jun 1897. 2pp + 1p diagrams. Rolling cubes puzzle, where the cube faces are hollowed and fit
onto domes in the tray. Basic form has
four cubes in a row with two extra spaces above the middle cubes, but other
forms are shown.
Sven Bergling invented the
rolling ball labyrinth puzzle/game and they began to be produced in 1946. [Kenneth Wells; Wooden Puzzles and Games;
David & Charles, Newton Abbot, 1983, p. 114.]
Ronald Sprague. Unterhaltsame Mathematik. Vieweg, Braunschweig, 1961. Translated by T. H. O'Beirne as: Recreations in Mathematics, Blackie, London,
1963. Problem 3: Schwere Kiste, pp. 3-4
& 22-23 (= Heavy boxes, pp. 4-5
& 25-26). Three problems with 5
boxes some of which are so heavy that one has to tilt or roll them.
Gardner. SA (Dec 1963). = Sixth Book, chap. 8.
Gives Sprague's first problem.
Gardner. SA (Nov 1965). c= Carnival, chap. 9.
Prob. 1: The red-faced cube. Two
problems of John Harris involving one cube with one red face rolling on a
chessboard. Gardner says that the field
is new and that only Harris has made any investigations of the problem. The book chapter cites Harris's 1974 article,
below, and a 1971 board game called Relate with each player having four coloured
cubes on a 4 x 4 board.
Charles W. Trigg. Tetrahedron rolled onto a plane. JRM 3:2 (Apr 1970) 82-87. A tetrahedron rolled on the plane forms the
triangular lattice with each cell corresponding to a face of the
tetrahedron. He also considers rolling
on a mirror image tetrahedron and rolling octahedra.
John Harris. Single vacancy rolling cube problems. JRM 7:3 (1974) 220-224. This seems to be the first appearance of the
problem with one vacant space. He
considers cubes rolling on a chessboard.
Any even permutation of the pieces with the blank left in place is
easily obtained. From the simple
observation that each roll is an odd permutation of the pieces and an odd
rotation of the faces of a cube, he shows that the parity of the rotation of a
cube is the same as the parity of the number of spaces it has moved. He shows that any such rotation can be
achieved on a 2 x 3 board.
Rotating one cube 120o about a diagonal takes 32 moves. If the blank is allowed to move, the the
parity of the permutation of the pieces is the parity of the number of spaces
the blank moves, but each cube still has to have the parity of its rotation the
same as the parity of the number of spaces it has moved. If the identical pieces are treated as
indistinguishable, the parity of the permutation is only shown by the location
of the blank space. He suggests the use
of ridges on the board so that the cube will roll automatically -- this was
later used in commercial versions. He
gives a number of problems with different colourings of the cubes.
Gardner. SA (Mar 1975). = Time Travel, chap. 9.
Prob. 8: Rolling cubes. This is
the first of Harris's problems.
Computer analysis has found that it can be done in fewer moves than
Harris had. Gardner also reports on the
last of Harris's problems, which has also been resolved by computer.
A 3 x 3 array with 8
coloured cubes was available from Taiwan in the early 1980s. It was called Color Cube Mental Game -- I
called it 'Rolling Cubes'. The cubes
had thick faces, producing grooved edges which fit into ridges in the bottom of
the plastic frame, causing automatic rolling quite nicely. I wonder if this was inspired by Harris's
article.
John Ewing & Czes
Kośniowski. Puzzle it Out --
Cubes, Groups and Puzzles. CUP,
1982. The 8 Cubes Puzzle, pp.
58-59. Analysis of the Rolling Cubes
puzzle. The authors show how to rotate
a single cube about a diagonal in 36 moves.
Invented by Toshio Akanuma
(??SP). Manufactured by Tricks Co.,
Japan, in 1983. Described in Hordern,
pp. 144-145 & 220, E35, and in S&B, p. 135. This looks like a Tower of Hanoi (cf 7.M.2) with two differently
coloured piles of 10 pieces on the outside two tracks of three tracks of height
12 joined like a letter E. This is made
as a sliding block puzzle, but with blockages -- a piece cannot slide down a
track further than its original position.
Mark Manasse, Danny Sleator & Victor K. Wei. Some Results on the Panex Puzzle. Preprint sent by Jerry Slocum, 23pp, nd
[1983, but S&B gives 1985]. For
piles of size n, the minimum number of moves, T(n),
to move one pile to the centre track is determined by means of a 2nd
order, non-homogeneous recurrence which has different forms for odd and
even n. Compensating for this leads to a 2nd order non‑homogeneous
recurrence, giving T(10) =
4875 and T(n) ~ C(1 + Ö2)n. This solution doesn't ever move the other
pile. The minimum number of moves, X(n),
to exchange the piles is bounded above and below and determined exactly
for n £ 6 by computer search. X(5) = 343,
compared to bounds of 320 and
343. X(6) = 881, compared to the bounds of 796
and 881. For
n = 7, the bounds are 1944
and 2189, For
n = 10, the bounds are 27,564
and 31,537. The larger bounds are considered as probably
correct.
Christoph Hausammann. US Patent 5,261,668 -- Logic Game. Filed: 6 Aug 1992; patented: 16 Nov 1993. 1p
abstract + 2pp text + 3pp diagrams.
Essentially identical to Panex.
Vladimir Dubrovsky. Nesting Puzzles -- Part I: Moving oriental
towers. Quantum 6:3 (Jan/Feb 1996)
53-59 & 49-51. Says Panex was
produced by the Japanese Magic Company in the early 1980s. Discusses it and cites S&B for the
bounds given above. Sketches a number
of standard configurations and problems, leading to "Problem 9. Write out a complete solution to the Panex
puzzle." He says his method is
about 1700 moves longer than the upper bound given above.
Nick Baxter. Recent results for the Panex Puzzle. 4pp handout at G4G5, 2002. Describes the puzzle and its history. David Bagley wrote a program to implement
the Manasse, Sleator & Wei methods.
On 7 Feb 2002, this confirmed the conjecture that X(7) = 2189. On 26 Mar 2002, it obtained X(8) = 5359, compared to bounds of
4716 and 5359.
It is estimated that the cases
n = 9 and 10 will take 10
and 1200 years! If Moore's Law on the
increase of computing power continues for another 20 years, the latter answer
may be available by then. He gives a
simplified version of the algorithm for the upper bound, which gets 31,544
for n = 10. He has a Panex page: www.baxterweb.com/puzzles/panex/ and will be publishing an edited and
annotated version of the Manasse, Sleator & Wei paper on it.
See MUS I 1-13, Tropfke 658 and
also 5.N.
Wolf, goat and cabbages: Alcuin,
Abbot Albert, Columbia
Algorism, Munich 14684, Folkerts,
Chuquet, Pacioli, Tartaglia,
van Etten, Merry Riddles, Ozanam,
Dilworth, Wingate/Dodson, Jackson,
Endless Amusement II,
Boy's Own Book, Nuts to Crack, Taylor; The Riddler, Child,
Fireside Amusements,
Magician's Own Book,
Book of 500 Puzzles,
Boy's Own Conjuring Book, Secret Out (UK),
Mittenzwey, Carroll 1873, Kamp,
Carroll 1878, Berg, Lemon,
Hoffmann, Brandreth Puzzle
Book, Carroll 1899, King,
Voggenreiter, Stein, Stong,
Zaslavsky, Ascher, Weismantel (a film),
Verse version: Taylor,
Version with only one pair of
incompatibles: Voggenreiter
Extension to four items: Gori,
Phillips, M. Adams, Gibbs,
Ascher
Adults and children: Alcuin,
Kamp, Hoffmann, Parker?,
Voggenreiter, Gibbs
Three jealous husbands: Alcuin,
Abbot Albert, Columbia
Algorism, Munich 14684, Folkerts,
Rara, Chuquet, Pacioli,
Cardan, Tartaglia, H&S ‑ Trenchant, Gori,
Bachet, van Etten, Wingate/Kersey, Ozanam, Minguét, Dilworth,
Les Amusemens,
Wingate/Dodson, Jackson, Endless Amusement II, Nuts to Crack, Young Man's Book,
Family Friend,
Magician's Own Book,
The Sociable,
Book of 500 Puzzles,
Boy's Own Conjuring Book, Vinot, Secret Out
(UK), Lemon, Hoffmann, Fourrey, H. D. Northrop, Mr. X, Loyd, Williams,
Clark, Goodstein, O'Beirne,
Doubleday, Allen,
Verse mnemonic: Abbot
Albert, Munich 14684,
Verse solution:
Ozanam, Vinot,
Four or more jealous
husbands: Pacioli, Filicaia,
Tartaglia, Bachet, Delannoy,
Ball, Carroll-Collingwood, Dudeney,
O'Beirne
Jealous husbands, with island in
river: De Fontenay, Dudeney,
Ball, Loyd, Dudeney,
Pressman & Singmaster
Missionaries and cannibals: Jackson,
Mittenzwey, Cassell's, Lemon,
Pocock, Hoffmann, Brandreth Puzzle Book, H. D. Northrop, Schubert, Arbiter, H&S,
Abraham, Bile Beans,
Goodstein, Beyer, O'Beirne,
Pressman & Singmaster.
With only one cannibal who can row: Brandreth Puzzle Book,
Abraham, Beyer.
Bigger boats: Pacioli,
Filicaia?,
Bachet(-Labosne), Delannoy, Ball,
Dudeney, Abraham?, Goodstein,
Kaplan, O'Beirne,
Alcuin. 9C.
Prob.
17: Propositio de tribus fratribus singulas habentibus sorores. 3 couples, rather earthily expressed.
Prob.
18: Propositio de lupo et capra et fasciculo cauli. Wolf, goat, cabbages.
Prob.
19: Propositio de viro et muliere ponderantibus plaustrum. Man, wife and two small children.
Prob.
20: Propositio de ericiis. Rewording of
Prob. 19.
Ahrens. MUS II 315‑318, cites many sources,
mostly from folklore and riddle collections, with one from the 12C and several
from the 14C. ??NYS.
Abbot Albert. c1240.
Prob.
5, p. 333. Wolf, goat & cabbages.
Prob.
6, p. 334. 3 couples, with verse
mnemonic.
Columbia Algorism. c1350.
No.
122, pp. 130‑131 & 191: wolf, goat, bundle of greens. See also Cowley 402 & plate
opposite. P. 191 and the Cowley plate
are reproductions of the text with a crude but delightful illustration. P. 130 gives a small sketch of the
illustration. I have a colour slide
from the MS.
No.
124, p. 132: 3 couples. See also Cowley
403 & plate opposite. The plate
shows another crude but delightful illustration. I have a colour slide from the MS.
Munich 14684. 14C.
Prob.
XXVI, pp. 82‑83: 3 couples, with verse mnemonic.
Prob.
XXVII, p. 83: wolf, goat, cabbage.
Folkerts. Aufgabensammlungen. 13-15C.
11 sources with wolf, goat, cabbage.
12 sources with three jealous couples.
Rara, 459‑465, cites two
Florentine MSS of c1460 which include 'the jealous husbands'. ??NYS.
Chuquet. 1484.
Prob.
163: wolf, goat & cabbages. FHM 233
says that a 12C MS claims that every boy of five knows this problem.
Prob.
164: 3 couples. FHM 233.
Pacioli. De Viribus.
c1500.
Ff.
103v - 105v. LXI. C(apitolo). de .3.
mariti et .3. mogli gelosi (About 3 husbands and 3 wives). = Peirani 146-148. 3 couples. Says that
4 or 5 couples requires a 3
person boat.
F.
IIIv. = Peirani 6. The Index lists the above as Problem 66 and
lists a Problem 65: Del modo a salvare la capra el capriolo dal lupo al passar
de un fiume ch' non siano devorati (How to save the goat and the kid from the
wolf in crossing a river so they are not eaten).
Piero di Nicolao d'Antonio da
Filicaia. Libro dicto giuochi
mathematici. Early 16C -- ??NYS,
mentioned in Franci, op. cit. in 3.A.
Franci, p. 23, says Pacioli and Filicaia deal with the case of four or
five couples and that Pacioli considers bigger boats, but I'm not clear if
Filicaia also does so.
Cardan. Practica Arithmetice. 1539.
Chap. 66, section 73, f. FF.v.v (p. 157). (The 73 is not printed in the Opera Omnia). Three jealous husbands.
Tartaglia. General Trattato, 1556, art. 141‑143,
p. 257r‑ 257v.
Art.
141: wolf, goat and cabbages.
Art.
142: three couples.
Art.
143: four couples -- erroneously -- see Bachet.
H&S 51 says 3 couples occurs
in Trenchant (1566), ??NYS.
Gori. Libro di arimetricha.
1571.
Ff.
71r‑71v (p. 77). 3 couples.
F. 80v
(p. 77). Dog, wolf, sheep, horse to cross
river in boat which holds 2, but each cannot abide his neighbours in the
given list, so each cannot be alone with such a neighbour.
Bachet. Problemes.
1612. Addl. prob. IV: Trois
maris jaloux ..., 1612: 140-142; 1624: 212‑215; 1884: 148‑153. Three couples; four couples -- notes that Tartaglia is wrong by showing that one
can never get five persons on the far side.
Labosne gives a solution with a
3 person boat and does n
couples with an n‑1 person boat.
van Etten. 1624.
Prob.
14: Des trois maistres & trois valets, p. 14. 3 men and 3
valets. (The men hate the other
valets and will beat them if given a chance.)
(Not in English editions.)
Prob.
15: Du loup, de la chevre & du chou, pp. 14‑15. Wolf, goat & cabbages. (Not in English editions.)
Book of Merry Riddles. 1629
72 Riddle, pp. 43-44. "Over
a water I must passe, and I must carry a lamb, a woolfe, and a bottle of hay if
I carry any more than one at once my boat will sink." Tony Augarde; The Oxford Guide to Word
Games; OUP, 1984; p. 6 says wolf, goat, cabbage appears in the 1629 ed.
Wingate/Kersey. 1678?.
Prob. 6., p. 543. Three jealous
couples. Cf 1760 ed.
Ozanam. 1725.
Prob.
2, 1725: 3‑4. Prob. 18, 1778:
171; 1803: 171; 1814: 150.
Prob. 17, 1840: 77. Wolf, goat
and cabbage.
Prob.
3, 1725: 4‑5. Prob. 19, 1778:
171-172; 1803: 171-172; 1814: 150-151. Prob. 18, 1840: 77.
Jealous husbands. Latin verse
solution. He also discusses three
masters and valets: "none of the
the masters can endure the valets of the other two; so that if any one of them
were left with any of the other two valets, in the absence of his master, he
would infallibly cane him."
Minguet. 1733.
Pp. 158-159 (1755: 114-115; 1822: 175-176; 1864: 151). Three jealous couples.
Dilworth. Schoolmaster's Assistant. 1743.
Part IV: Questions: A short Collection of pleasant and diverting
Questions, p. 168.
Problem
6: Fox, goose and peck of corn. = D.
Adams; Scholar's Arithmetic; 1801, p. 200, no. 8.
Problem
7: Three jealous husbands. (Dilworth cites
Wingate for this -- but this is in Kersey's additions -- cf Wingate/Kersey,
1678? above.) = D. Adams; Scholar's
Arithmetic; 1801, p. 200, no. 9.
Les Amusemens. 1749.
Prob. 14, p. 136: Les Maris jaloux.
Solution is incorrect and has been corrected by hand in my copy.
Edmund Wingate (1596-1656). A Plain and Familiar Method for Attaining
the Knowledge and Practice of Common Arithmetic. .... 19th ed., previous
ed. by John Kersey (1616-1677) and George Shell(e)y, now by James Dodson. C. Hitch and L. Hawes, et al., 1760.
Art.
749. Prob. VI. P. 379.
Three jealous husbands. As in
1678? ed.
Art.
750. Prob. VII. P. 379.
Fox, goose and corn.
Jackson. Rational Amusement. 1821.
Arithmetical Puzzles.
No.
7, pp. 2 & 52. Fox, goose and corn. One solution.
No.
13, pp. 4 & 54. Three jealous
husbands.
No.
21, pp. 5 & 56. Three masters and
servants, where the servants will murder the masters if they outnumber them --
i.e. missionaries and cannibals. First
appearance of this type.
Endless Amusement II. 1826?
Prob.
17, pp. 198-199. Wolf, goat and
cabbage.
Prob.
25, pp. 201-202. Three jealous
husbands.
Boy's Own Book. The wolf, the goat and the cabbages. 1828: 418‑419; 1828-2: 423; 1829 (US): 214;
1855: 570; 1868: 670.
Nuts to Crack III (1834).
No.
209. Fox, goose and peck of corn.
No.
214. Three jealous husbands.
The Riddler. 1835.
The wolf, the goat and the cabbages, pp. 5-6. Identical to Boy's Own Book.
Young Man's Book. 1839.
Pp. 39-40. Three jealous
Husbands ..., identical to Wingate/Kersey.
Child. Girl's Own Book. 1842:
Enigma 49, pp. 237-238; 1876: Enigma
40, p. 200. Fox, goose and corn. Says it takes four trips instead of three --
but the solution has 7 crossings.
Walter Taylor. The Indian Juvenile Arithmetic, or Mental
Calculator; to which is added an appendix, containing arithmetical recreations
and amusements for leisure hours ....
For the author at the American Press, Bombay, 1849. [Quaritch catalogue 1224, Jun 1996, says
their copy has a note in French that Ramanujan learned arithmetic from this and
that it is not in BMC nor NUC. Graves
14.c.35.] P. 211, No. 8. Wolf, goat and cabbage in verse! No solution.
Upon
a river's brink I stand, it is both deep and wide;
With
a wolf, a goat, and cabbage, to take to the other side.
Tho'
only one each time can find, room in my little boat;
I
must not leave the goat and wolf, not the cabbage and the goat.
Lest
one should eat the other up, -- now how can it be done --
How
can I take them safe across without the loss of one?
Fireside Amusements. 1850: No. 24, pp. 111 & 181; 1890: No. 24, p. 100. Fox, goose and basket of corn.
Family Friend 3 (1850) 344 &
351. Enigmas, charades, etc. -- No. 17:
The three jealous husbands.
Magician's Own Book. 1857.
The
three jealous husbands, p. 251.
The
fox, goose, and corn, p. 253.
The Sociable. 1858.
Prob. 33: The three gentlemen and their servants, pp. 296 &
314-315. "None of the gentlemen shall
be left in company with any of the servants, except when his own servant is
present" -- so this is like the Jealous Husbands. = Book of 500 Puzzles, 1859, prob. 33, pp.
14 & 32-33. = Illustrated Boy's Own
Treasury, 1860, prob. 11, pp. 427-428 & 431.
Book of 500 Puzzles. 1859.
Prob.
33: The three gentlemen and their servants, pp. 14 & 32-33. As in The Sociable.
The
three jealous husbands, p. 65.
The
fox, goose and corn, p. 67.
Both identical to Magician's Own Book.
Boy's Own Conjuring Book. 1860.
The
three jealous husbands, pp. 222‑223.
The
fox, goose, and corn, pp. 225.
Both identical to Magician's Own Book.
Vinot. 1860. Art. XXXVII: Les
trois maris jaloux, pp. 56-57. Three
jealous husbands, with verse solution taken from Ozanam.
The Secret Out (UK). c1860.
A comical dilemma, p. 27.
Wolf, goat and cabbage. Varies
it as fox, goose and corn and then as gentlemen and servants, which is jealous
husbands, rather than the same problem.
Lewis Carroll. Letter of 15 Mar 1873 to Helen Feilden. Pp. 212-215 (Collins: 154-155). Fox, goose and bag of corn. "I rashly proposed to her to try the
puzzle (I daresay you know it) of "the fox, and goose, and bag of
corn."" Cf
Carroll-Collingwood, pp. 212-215 (Collins: 154-155); Carroll-Wakeling, prob. 28,
pp. 36-37 and Carroll-Gardner, p. 51. Cf Carroll, 1878.
Wakeling writes that this does not appear elsewhere in Carroll.
Bachet-Labosne. 1874.
For details, see Bachet, 1612.
Jens Kamp. Danske Folkeminder, Aeventyr, Folksagen,
Gaader, Rim og Folketro, Samlede fra Folkemende. R. Neilsen, Odense, 1877.
Marcia Ascher has kindly sent me a photocopy of the relevant material
with a translation by Viggo Andressen.
No.
18, pp. 326‑327: Fox, lamb and cabbage.
No.
19, p. 327: Husband, wife and two half‑size sons.
Lewis Carroll. Letter of 22 Jan 1878 to Jessie
Sinclair. Fox, goose and bag of
corn. Cf Carroll-Collingwood, pp.
205-207 (Collins: 150); Carroll-Wakeling, prob. 26: The fox, the goose and the
bag of corn, pp. 34 & 72. Cf
Carroll. 1872.
Mittenzwey. 1880.
Prob. 227-228, pp. 42 & 92;
1895?: 254-255, pp. 46 & 94;
1917: 254‑255, pp. 42 & 90. Bear, goat and cabbage, mentioning second
solution; three kings and three
servants, where the servants will rob the kings if they outnumber them, i.e.
like missionaries and cannibals.
Cassell's. 1881.
P. 105: The dishonest servants.
The servants are rogues who will murder masters if they outnumber them,
so this is equivalent to the missionaries and cannibals version.
Lucas. RM1. 1882. Pp. 1-18 is a general discussion of the
problem.
De Fontenay. Unknown source and date -- 1882?? Described in RM1, 1882, pp. 15‑18
(check 1st ed.??). n > 3 couples,
2 person boat, island in river,
can be done in 8n ‑ 8 passages.
Lucas says this was suggested at the Congrès de l'Association française
pour l'avancement des sciences at Montpellier in 1879, ??NYS. (De Fontenay is unclear -- sometimes he
permits bank to bank crossings, other times he only permits bank to island crossings. His argument really gives 8n - 6
if bank to bank crossings are prohibited. See Pressman & Singmaster, below, for clarification.)
Albert Ellery Berg, ed. Op. cit. in 4.B.1. 1883. P. 377: Fox, goose
& peck of corn.
Lemon. 1890.
Gentlemen
and their servants, no. 101, pp. 17‑18 & 101. This is the same as missionaries and
cannibals.
The
three jealous husbands, no. 151, pp. 24 & 103 (= Sphinx, no. 478, pp. 66
& 114.) The solution mentions
Alcuin.
Crossing
the river, no. 450, pp. 59 & 114.
English travellers and native servants
= missionaries and cannibals.
Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 136, no. 14. Fox,
goose and corn. No solution.
Herbert Llewelyn Pocock. UK Patent 15,358 -- Improvements in Toy
Puzzles. Applied: 29 Sep 1890; complete specification: 29 Jun 1891; accepted: 22 Aug 1891. 2pp + 1p diagrams. Three whites and three blacks and the blacks must never outnumber
the whites, i.e. same as missionaries and cannibals. He describes the puzzle as "well known".
Delannoy. Described in RM1, 1891, Note 1: Sur le jeu des traversées, pp. 221‑222. ??check 1882 ed. Shows n couples can cross in an x
person boat in N trips, for
n, x, N = 2, 2, 5; 3, 2, 11;
4, 3, 9; 5, 3, 11; n > 5, 4, 2n ‑ 3. (He has
2n ‑ 1 by mistake. Simple modification shows we also have 5, 4, 7;
6, 5, 9; 7, 6, 5; 8, 7, 7;
n > 8, n ‑ 1, 5.)
Ball. MRE, 1st ed., 1892, pp. 45‑47, says Lucas posed the problem
of minimizing x for a given
n and quotes the Delannoy
solution (with erroneous 2n ‑
1) and also gives De Fontenay's version
and solution. (He spells it De Fonteney
as does his French translator, though Ahrens gives De Fontenay and the famous
abbey in Burgundy is Fontenay -- ??)
The Ballybunnion and Listowel
Railway in County Kerry, Ireland, was a late 19C railway using the Lartigue
monorail system. This had a single
rail, about three feet off the ground, with a carriage hanging over both sides
of the rail. The principle job of the
conductor/guard to make sure the passengers and goods were equally distributed
on both sides. Kerry legend asserts
that a piano had to be sent on this railway and there were not enough
passengers or goods to balance it. So a
cow was sent on the other side. At the
far end, the piano was unloaded and replaced with two large calves and the
carriage sent back. The cow was then
unloaded and one calf moved to the other side, so the carriage could be sent
back to the far end and everyone was happy.
Hoffmann. 1893.
Chap. IV, pp. 157‑158 & 211‑213 = Hoffmann-Hordern, pp.
136-138, with photos.
No.
56: The three travellers. Masters and
servants, equivalent to missionaries and cannibals. Solution says Jaques & Son make a puzzle version with six
figures, three white and three black.
Photos in Hoffmann-Hordern, pp. 136 & 137 -- the latter shows Caught
in the Rain, 1880-1905, where Preacher, Deacon, Janitor and their wives have to
get somewhere using one umbrella.
No.
57: The wolf, the goat, and the cabbages.
Photo on p. 136 of La Chevre et le Chou. with box, by Watilliaux,
1874-1895. Hordern Collection, p. 72,
and S&B, p. 134, show the same puzzle.)
No.
58: The three jealous husbands.
No.
59: The captain and his company. This
is Alcuin's prop. 19 with many adults.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co.,
NY), nd [1895].
P. 7:
The wolf, the goat and the cabbages.
Identical to Hoffmann No. 57, with nice colour picture. No solution.
P. 9:
The missionaries' and cannibals' puzzle.
Usual form, with nice colour picture, but only one cannibal can
row. No solution. This seems to be the first to use the
context of missionaries and cannibals and the first to restrict the number of
rowers.
Lucas. L'Arithmétique Amusante.
1895. Les vilains maris jaloux,
pp. 125-144 & Note II, pp. 198-202.
Prob.
XXXVI: La traversée des trois ménages, pp. 125-130. 3 couples. Gives Bachet's 1624 reasoning for the
essentially unique solution -- but attributes it to 1613.
Prob.
XXXVII: La traversée des quatre ménages, pp. 130-132. 4 couples in a 3
person boat done in 9 crossings.
L'erreur
de Tartaglia, pp. 133-134. Discusses
Tartaglia's error and Bachet's notice of it and gives an easy proof that 4
couples cannot be done with a
2 person boat.
Prob.
XXXVIII: La station dans une île, pp. 135-140.
4 couples, 2
person boat, with an island.
Gives De Fontenay's solution in 24 crossings.
Prob.
XXXIX: La traversée des cinq ménages, pp. 141-143. 5 couples, 3
person boat in 11 crossings.
Énoncé
général du problème des traversées, pp. 143-144. n couples, x
person boat, can be done in N crossings as given by Delannoy above. He corrects
2n - 1 to 2n - 3
here.
Note
II: Sur les traversées, pp.
198-202. Gives Tarry's version with an
island and with n men having harems of size m, where the women are obviously unable to row. He gives solutions in various cases. For the ordinary case, i.e. m = 1, he finds a solution for
4 couples in 21
moves, using the basic ferrying technique that Pressman and Singmaster
found to be optimal, but the beginning and end take longer because the women
cannot row. He says this gives a
solution for n couples in
4n + 5 crossings. He then considers the case of n - 1
couples and a ménage with m wives and finds a solution in 8n + 2m + 7
crossings. I now see that this
solution has the same defects as those in Pressman & Singmaster, qv.
Ball. MRE, 3rd ed., 1896, pp. 61‑64, repeats 1st ed., but adds
that Tarry has suggested the problem for harems -- see above.
Dudeney. Problem 68: Two rural puzzles. Tit‑Bits 33 (5 Feb &
5 Mar 1898) 355 & 432.
Three men with sacks of treasure and a boat that will hold just two men
or a man and a sack, with additional restrictions on who can be trusted with
how much. Solution in 13
crossings.
Carroll-Collingwood. 1899.
P. 317 (Collins: 231 or 232 (missing in my copy)) Cf Carroll-Wakeling II, prob. 10:
Crossing the river, pp. 17 & 66.
Four couples -- only posed, no solution. Wakeling gives a solution, but this is incorrect. After one wife is taken across, he has
another couple coming across and from Bachet onward, this is considered
improper as the man could get out of the boat and attack the first, undefended,
wife.
E. Fourrey. Op. cit. in 4.A.1, 1899. Section 211: Les trois maîtres et les trois valets. Says a master cannot leave his valet with
the other masters for fear that they will intimidate him into revealing the
master's secrets. Hence this is the
same as the jealous couples.
H. D. Northrop. Popular Pastimes. 1901.
No. 5:
The three gentlemen and their servants, pp. 67 & 72. = The Sociable.
No.
12: The dishonest servants, pp. 68 & 73.
"... the servants on either side of the river should not outnumber
the masters", so this is the same as missionaries and cannibals.
Mr. X [cf 4.A.1]. His Pages.
The Royal Magazine 10:2 (Jun 1903) 140-141. A matrimonial difficulty.
Three couples. No answer given.
Dudeney. Problem 523. Weekly Dispatch (15
& 29 Nov 1903), both p. 10, (= AM, prob. 375, pp. 113 & 236‑237). 5
couples in a 3 person boat.
Johannes Bolte. Der Mann mit der Ziege, dem Wolf und dem
Kohle. Zeitschrift des Vereins für
Volkskunde 13 (1903) 95-96 & 311.
The first part is unaware of Alcuin and Albert. He gives a 12C Latin
solution: It capra, fertur olus, redit
hec, lupus it, capra transit [from Wattenbach; Neuen Archiv für ältere deutsche
Geschichtskunde 2 (1877) 402, from Vorauer MS 111, ??NYS] and a 14C solution: O natat, L sequitur, redit O, C navigat
ultra, / Nauta recurrit ad O, bisque natavit ovis (= ovis, lupus, ovis, caulis, ovis) [from Mone; Anzeiger für
Kunde der deutschen Vorzeit 45 (No. 105) (1838), from Reims MS 743,
??NYS]. Cites Kamp and several other
versions, some using a fox, a sheep, or a lamb. The addendum cites and quotes Alcuin and Albert as well as
relatively recent French and Italian versions.
H. Parker. Ancient Ceylon. Op. cit. in 4.B.1.
1909. Crossing the river, p.
623.
A
King, a Queen, a washerman and a washerwoman have to cross a river in a boat
that holds two. However the King and
Queen cannot be left on a bank with the low caste persons, though they can be
rowed by the washerperson of the same sex.
Solution in 7 crossings.
Ferry-man
must transport three leopards and three goats in a boat which holds himself and
two others. If leopards ever outnumber
goats, then the goats get eaten. So
this is like missionaries and cannibals, but with a ferry-man. Solution in 9 crossings.
H. Schubert. Mathematische Mussestunde. Vol. 2, 3rd ed., Göschen, Leipzig,
1909. Pp. 160‑162: Der drei
Herren und der drei Sklaven. (Same as
missionaries and cannibals.)
Arbiter Co. (Philadelphia). 1910.
Capital and Labor Puzzle. Shown
in S&B, p. 134. Equivalent to
missionaries and cannibals.
Ball. MRE, 5th ed., 1911, pp. 71-73, repeats 3rd ed., but omits the
details of De Fonteney's solution in
8(n-1) crossings.
Loyd. Cyclopedia, 1914.
Summer
tourists, pp. 207 & 366. 3 couples,
2 person boat, with additional
complications -- the women cannot row and there have been some arguments. Solution in
17 crossings.
The
four elopements, pp. 266 & 375.
4 couples, 2
person boat, with an island and the stronger constraint that no man is
to get into the boat alone if there is a girl alone on either the island or the
other shore. "The [problem]
presents so many complications that the best or shortest answer seems to have
been overlooked by mathematicians and writers on the subject." "Contrary to published answers, ... the
feat can be performed in 17 trips, instead of 24."
Ball. MRE, 6th ed., 1914, pp. 71-73, repeats 5th ed., but adds
that 6n ‑ 7 trips suffices for n couples with an island,
though he gives no reference.
Williams. Home Entertainments. 1914.
Alcuin's riddle, pp. 125-126.
"This will be recognized as perhaps the most ancient British riddle
in existence, though there are several others conceived on the same
lines." Three jealous
couples.
Clark. Mental Nuts. 1916, no.
67. The men and their wives. "... no man shall be left alone with
another's wife."
Dudeney. AM.
1917. Prob. 376: The four
elopements, pp. 113 & 237. 4 couples,
2 person boat, with island, can
be done in 17 trips and that this cannot be improved. This is the same solution as given by Loyd. (See Pressman and Singmaster, below.)
Ball. MRE, 8th ed., 1919, pp. 71-73 repeats 6th ed. and adds a citation
to Dudeney's AM prob. 376 for the solution in
6n ‑ 7 trips for n
couples.
Hummerston. Fun, Mirth & Mystery. 1924.
Crossing the river puzzles, Puzzle no, 52, pp. 128 & 180. 'Puzzles of this type ... interested people
who lived more than a thousand years ago'.
No. 1:
The eight travellers. Six men and two
boys who weigh half as much.
No. 2:
White and black. = Missionaries and
cannibals.
No. 3:
The fox, the goose, and the corn.
No. 4:
the jealous husbands.
H&S, 1927, p. 51 says
missionaries and cannibals is 'a modern variant'.
King. Best 100. 1927. No. 10, pp. 10 & 40. Dog, goose and corn.
Heinrich Voggenreiter. Deutsches Spielbuch Sechster Teil: Heimspiele. Ludwig Voggenreiter, Potsdam, 1930.
P.
106: Der Wolf, die Ziege und der Kohlkopf.
Usual wolf, goat, cabbage.
Pp.
106-107: Die 100 Pfund-Familie. Parents
weigh 100 pounds; the two children weigh 100 pounds together.
P.
107: Der Landjäger and die Strolche [The policeman and the vagabonds]. Two of the vagabonds hate each other so much
that they cannot be left together. As
far as I recall, this formulation is novel and I was surprised to realise that
it is essentially equivalent to the wolf, goat and cabbage version.
Phillips. Week‑End. 1932. Time tests of
intelligence, no. 41, pp. 22 & 194.
Rowing explorer with 4 natives:
A, B, C, D, who cannot abide
their neighbours in this list. A can row.
They get across in seven trips.
Abraham. 1933.
Prob. 54 -- The missionaries at the ferry, pp. 18 & 54 (14 &
115). 3 missionaries and 3 cannibals.
Doesn't specify boat size, but says 'only one cannibal can row'. 1933 solution says 'eight double journeys',
1964 says 'seven crossings'. This seems
to assume the boat holds 3. (For a
2 man boat, it takes 11
crossings with one missionary and two cannibals who can row or 13
crossings with one missionary and one cannibal who can row.)
The Bile Beans Puzzle Book. 1933.
No. 34: Missionaries & cannibals.
Three of each but only one of each can row. Done in 13 crossings.
Phillips. Brush.
1936. Prob. L.2: Crossing the
Limpopo, pp. 39‑40 & 98. Same
as in Week‑End, 1932.
M. Adams. Puzzle Book. 1939. Prob. C.63: Going
to the dance, pp. 139 & 178. Same
as Week‑End, 1932, phrased as travelling to a dance on a motorcycle which
carries one passenger.
R. L. Goodstein. Note 1778:
Ferry puzzle. MG 28 (No. 282)
(1944) 202‑204. Gives a graphical
way of representing such problems and considers m soldiers and m
cannibals with an n person boat, 3 jealous husbands and
how many rowers are required.
David Stein. Party and Indoor Games. P. M. Productions, London, nd [c1950?]. P. 98, prob. 5: Man with cat, parrot
and bag of seeds.
C. L. Stong. The Amateur Scientist. Ill. by Roger Hayward. S&S, 1960.
A
puzzle-solving machine, pp. 377-384.
Describes how Paul Bezold made a logic machine from relays to solve the
fox, goose, corn problem.
How to
design a "Pircuit" or Puzzle circuit, pp. 388-394. On pp. 391-394, Harry Rudloe describes relay
circuits for solving the three jealous couples problem, which he attributes to
Tartaglia, and the missionaries and cannibals problem.
Nathan Altshiller Court. Mathematics in Fun and in Earnest. Mentor (New American Library), NY,
1961. [John Fauvel sent some pages from
a different printing which has much different page numbers than my copy.] "River crossing" problems,
pp. 168‑171. Discusses
various forms of the problem and adds a problem with two parents weighing 160,
two children weighing 80 and a dog weighing 12, with a boat
holding 160.
E. A. Beyer, proposer; editorial solution. River‑crossing dilemma. RMM 4 (Aug 1961) 46 &
5 (Oct 1961) 59. Explorers and
natives (= missionaries and cannibals), with all the explorers and one native
who can row. Solves in 13
crossings, but doesn't note that only one rowing explorer is
needed. (See note at Abraham, 1933,
above.)
Philip Kaplan. Posers.
(Harper & Row, 1963);
Macfadden Books, 1964. Prob. 36,
pp. 41 & 91. 5 men and a
3 person boat on one side, 5
women on the other side. One man
and one woman can row. Men are not
allowed to outnumber women on either side nor in the boat. Exchange the men and the women in 7
crossings.
T. H. O'Beirne. Puzzles and Paradoxes, 1965, op. cit. in
4.A.4, chap. 1, One more river to cross, pp. 1‑19. Shows
2n ‑ 1 couples (or 2n ‑ 1 each of missionaries and cannibals ?) can cross in a n
person boat in 11 trips.
2n ‑ 2 can cross in 9
trips. He also considers
variants on Gori's second version.
Doubleday - 2. 1971.
Family outing, pp. 49-50. Three
couples, but one man has quarrelled with the other men and his wife has
quarrelled with the other women, so this man and wife cannot go in the boat nor
be left on a bank with others of their sex.
Further men cannot be outnumbered by women on either bank. Gives a solution in 9
crossings, but I find the conditions unworkable -- e.g. the initial position
is prohibited!
Claudia Zaslavsky. Africa Counts. Prindle, Weber & Schmidt, Boston, 1973. Pp. 109‑110 says that leopard, goat
and pile of cassava leaves is popular with the Kpelle children of Liberia. However, Ascher's Ethnomathematics (see
below), p. 120, notes that this is based on an ambiguous description and that
an earlier report of a Kpelle version has the form described below.
Ball. MRE, 12th ed., 1974, p. 119, corrects Delannoy's 2n ‑ 1 to 2n ‑ 3 and corrects De Fontenay's 8n ‑ 8 to 8n ‑ 6, but still gives the solution for n = 4
with 24 crossings.
W. Gibbs. Pebble Puzzles -- A Source Book of Simple
Puzzles and Problems. Curriculum
Development Unit, Solomon Islands, 1982.
??NYS, o/o??. Excerpted in: Norman K. Lowe, ed.; Games and Toys in the
Teaching of Science and Technology; Science and Technology Education, Document
Series No. 29, UNESCO, Paris, 1988, pp. 54‑57. On pp. 56‑57 is a series of river crossing problems. E.g. get people of weights 1, 2, 3
across with a boat that holds a weight of at most 3.
Also people numbered 1, 2, 3, 4,
5 such that no two consecutive people
can be in the boat or left together.
In about 1986, James Dalgety
designed interactive puzzles for Techniquest in Cardiff. Their version has a Welshman with a dragon,
a sheep and a leek!
Ian Pressman & David
Singmaster. Solutions of two river
crossing problems: The jealous husbands and the missionaries and the
cannibals. Extended Preprint, April
1988, 14pp. MG 73 (No. 464) (Jun 1989)
73‑81. (The preprint contains
historical and other detail omitted from the article as well as some further
information.) Observes that De Fontenay
seems to be excluding bank to bank crossings and that Lucas' presentation is
cryptic. Shows that De Fontenay's
method should be 8n ‑ 6 crossings for n > 3 and that this is
minimal. If bank to bank crossings are
permitted, as by Loyd and Dudeney, a computer search revealed a solution
with 16 crossings for n = 4, using an ingenious move that Dudeney could
well have ignored. For n > 4,
there is a simple solution in 4n
+ 1 crossings, and these numbers are
minimal. [When this was written, I had
forgotten that Loyd had done the problem for
4 couples in 17
moves, which changes the history somewhat. However, I now see that Loyd was copying from Dudeney's Weekly
Dispatch problem 270 of 23 Apr 1899 & 11 Jun 1899. Loyd states what appears to be a stronger
constraint but all the methods in our article do obey the stronger
constraint. However, one could make the
constraints stronger -- e.g. our solutions have a husband taking the boat from
bank to bank while his wife and another wife are on the island -- the solution
of Loyd & Dudeney avoids this and may be minimal in this case --??.]
For
the missionaries and cannibals problem, the
16 crossing solution reduces
to 15
and gives a general solution in
4n ‑ 1 crossings, which is
shown to be minimal. If bank to bank
crossings are not permitted, then De Fontenay's amended 8n ‑ 6 solution is still optimal.
Marcia Ascher. A river‑crossing problem in cultural
perspective. MM 63 (1990) 26‑28. Describes many appearances in folklore of
many cultures. Discusses African
variants of the wolf, goat and cabbage problem in which the man can take two of
the items in the boat. This is much easier,
requiring only three crossings, but some versions say that the man cannot
control the items in the boat, so he cannot have the wolf and goat or the goat
and cabbage in the boat with him. This
still only takes three crossings.
Various forms of these problems are mentioned: fox, fowl and corn;
tiger, sheep and reeds; jackal,
goat and hay; caged cheetah, fowl and
rice; leopard, goat and leaves -- see
below for more details.
She
also discusses an Ila (Zambia) version with leopard, goat, rat and corn which
is unsolvable!
Marcia Ascher. Ethnomathematics. Op. cit. in 4.B.10.
1991. Section 4.8, pp.
109-116 & Note 8, pp. 119-121. Good
survey of the problem and numerous references to the folklore and ethnographic
literature. Amplifies the above
article. A version like the Wolf, goat
and cabbage is found in the Cape Verde Islands, in Cameroon and in
Ethiopia. The African version is found
as far apart as Algeria and Zanzibar, but with some variations. An Algerian version with jackal, goat and hay
allows one to carry any two in the boat, but an inefficient solution is presented
first. A Kpelle (Liberia) version with
cheetah, fowl and rice adds that the man cannot keep control while rowing so he
cannot take the fowl with either the cheetah or the rice in the boat. A Zanzibar version with leopard, goat and
leaves adds instead that no two items can be left on either bank together. (A similar version occurs among
African-Americans on the Sea Islands of South Carolina.) Ascher notes that Zaslavsky's description is
based on an ambiguous report of the Kpelle version and probably should be like
the Algerian or Kpelle version just described.
Liz Allen. Brain Sharpeners. New English Library (Hodder & Stoughton), London, 1991. Crossing the river, pp. 62 & 125. Three mothers and three sons. The sons are unwilling to be left with
strange mothers, so this is a rephrasing of the jealous husbands.
Yuri B. Chernyak & Robert S.
Rose. The Chicken from Minsk. BasicBooks, NY, 1995. Chap. 1, probs. 4-6: The knights and the
pages; More knights and pages; Yet more knights and pages: no man is an island,
pp. 4-5 & 100-102. Equivalent to
the jealous couples. Prob. 4 is three
couples, solved in 11 crossings. Prob.
5 is four couples -- "There is no solution unless one of the four pages is
sacrificed. (In medieval times, this
was not a problem.)" Prob. 6 is
four couples with an island in the river, solved in general by moving all pages
to the island, then having the pages go back and accompany his knight to other
side, then return to the island. After
the last knight is moved, the pages then move from the island to the other
side. This takes 7n - 6
steps in general. It satisfies the
jealousy conditions used by Pressman & Singmaster, but not those of Loyd
& Dudeney.
John P. Ashley. Arithmetickle. Arithmetic Curiosities, Challenges, Games and Groaners for all
Ages. Keystone Agencies, Radnor, Ohio,
1997. P. 16: The missionaries and the
pirates. Politically correct rephrasing
of the missionaries and the cannibals version.
All the missionaries, but only one pirate, can row. Solves in 13 crossings.
Prof. Dr. Robert Weismantel,
Otto-von-Guericke-Universität Magdeburg, Fakultät für Mathematik, PSF 3120,
D-39016 Magdeburg, Germany; tel: 0391/67-18745; email:
weismantel@imo.math.uni-magdeburg.de; has produced a 45 min. film: "Der
Wolf, die Ziege und Kohlköpfe
Transportprobleme von Karl dem Grossen bis heute", suitable for the
final years of school.
5.B.1. LOWERING FROM TOWER PROBLEM
The problem is for a collection of
people (and objects or animals) to lower themselves from a window using a rope
over a pulley, with baskets at each end.
The complication is that the baskets cannot contain very different
weights, i.e. there is a maximum difference in the weights, otherwise they go
too fast. This is often attributed to
Carroll.
Carroll-Collingwood. 1899.
P. 318 (Collins: 232-233 (232 is lacking in my copy)). = Carroll-Wakeling II, prob. 4: The
captive queen, pp. 8 & 65-66.
3 people of weights 195, 165, 90 and a weight of 75, with difference at most 15.
He also gives a more complex form.
No solutions. Although the text
clearly says 165, the prevalence of the
exact same problem with 165 replaced by
105 makes me wonder if this was
a misprint?? Wakeling says there is no
explicit evidence that Carroll invented this, and neither book assigns a date,
but Carroll seems a more original source than the following and he was more
active before 1890 than after.
An
addition is given in both books: add
three animals, weighing 60, 45,
30.
Lemon. 1890. The prisoners in
the tower, no. 497, pp. 65 & 116.
c= Sphinx, The escape, no. 113, pp. 19 & 100‑101. Three people of weights 195, 105, 90 with a weight of 75. The difference in weights cannot be more
than 15.
Hoffmann. 1893.
Chap. IV, no. 28: The captives in the tower, pp. 150 & 196
= Hoffmann‑Hordern, p. 123.
Same as Lemon.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co.,
NY), nd [1895]. P. 3: The captives in
the tower. Same as Lemon. Identical to Hoffmann. With colour picture. No solution.
Loyd. The fire escape puzzle.
Cyclopedia, 1914, pp. 71 & 348.
c= MPSL2, prob. 140, pp. 98‑99 & 165. = SLAHP: Saving the family, pp. 59 &
108. Simplified form of Carroll's
problem. Man, wife, baby & dog,
weighing a total of 390.
Williams. Home Entertainments. 1914.
The escaping prisoners, pp. 126-127.
Same as Lemon.
Rudin. 1936. No. 92, pp. 31-32
& 94. Same as Lemon.
Haldeman-Julius. 1937.
No. 150: Fairy tale, pp. 17 & 28.
Same as Lemon, except the largest weight is printed as 196, possibly an
error.
Kinnaird. Op. cit. in 1 -- Loyd. 1946.
Pp. 388‑389 & 394.
Same as Lemon.
Simon Dresner. Science World Book of Brain Teasers. Scholastic Book Services, NY, 1962. Prob. 61: Escape from the tower, pp. 29
& 99‑100. Same as Lemon.
Robert Harbin [pseud. of Ned
Williams]. Party Lines. Oldbourne, London, 1963. Escape, p. 29. As in Lemon.
Howard P. Dinesman. Superior Mathematical Puzzles. Allen & Unwin, London, 1968. No. 60: The tower escape, pp. 78 &
118. Same as Carroll. Answer in
15 stages. He cites Carroll, noting that Carroll did
not give a solution and he asks if a shorter solution can be found.
F. Geoffrey Hartswick. In:
H. O. Ripley & F. G. Hartswick; Detectograms and Other Puzzles;
Scholastic Book Services, NY, 1969. No.
15: Stolen treasure puzzle, pp. 54‑55 & 87. Same as Lemon.
5.B.2. CROSSING A BRIDGE WITH A TORCH
New
section.
Four
people have to get across a bridge which is dark and needs to be lit with the
torch. The torch can serve for at most
two people and the gap is too wide to throw the torch across, so the torch has
to be carried back and forth. The
various people are of different ages and require 5, 10, 20, 25 minutes to
cross and when two cross, they have to go at the speed of the slower. But the torch (= flashlight) battery will
only last an hour. Can it be done? I heard this about 1997, when it was claimed
to be used by Microsoft in interviewing candidates. I never found any history of it, until I recently found a
discussion on Torsten Sillke's site: Crossing the bridge in an hour
(www.mathematik.uni‑bielefeld.de/~sillke/PUZZLES/crossing-bridge),
starting in Jun 1997 and last updated in Sep 2001. This cites the 1981 source and the other references below. Denote the problem with speeds a, b, c, d
and total time t by
(a,
b, c, d; t), etc. t is
sometimes given, sometimes not.
Saul X. Levmore &
Elizabeth Early Cook. Super
Strategies for Puzzles and Games.
Doubleday, 1981, p. 3 -- ??NYS.
(5, 10, 20, 25; 60), as in the
introduction to this section..
Heinrich Hemme. Das Problem des Zwölf-Elfs. Vandenhoeck & Ruprecht, 1998. Prob. 81: Die Flucht, pp. 40 & 105-106,
citing a web posting by Gunther Lientschnig on 4 Dec 1996. (2, 4, 8, 10; t).
Dick Hess. Puzzles from Around the World. Apr 1997.
Prob. 107: The Bridge.
(1,
2, 5, 10; 17). Poses versions with more
people: (1, 3, 4, 6, 8, 9; 31) and, with a three-person bridge, (1, 2, 6, 7, 8, 19, 10; 25).
Quantum (May/Jun 1997) 13. Brainteaser B 205: Family planning. Problem (1, 3, 8, 10; 20).
Karen Lingel. Email of 17 Sep 1997 to rec.puzzles. Careful analysis, showing that the 'trick' solution is better
than the 'direct' solution if and only if
a + c > 2b.
[Indeed, a + c - 2b is the time saved by the 'trick' solution.] She cites
(2, 3, 5, 8; 19) and (2, 2, 3, 3; 11) to Sillke and (1, 3, 6,
8, 12; 30), from an undated
website. Expressing the solution for
more people seems to remain an open question.
5.C. FALSE COINS WITH A BALANCE
See
5.D.3 for use of a weighing scale.
There
are several related forms of this problem.
Almost all of the items below deal with 12 coins with one false, either
heavy or light, and its generalizations, but some other forms occur, including
the following.
8 coins, £1 light: Schell, Dresner
26
coins, £1 light: Schell
8 coins,
1 light: Bath (1959)
9 coins,
1 light: Karapetoff, Meyer
(1946), Meyer (1948), M. Adams, Rice
I have been sent an article by Jack Sieburg; Problem
Solving by Computer Logic; Data Processing Magazine, but the date is cut off --
??
E. D. Schell, proposer; M. Dernham, solver. Problem E651 -- Weighed and found
wanting. AMM 52:1 (Jan 1945) 42 &
7 (Aug/Sep 1945) 397. 8 coins,
at most one light -- determine the light one in two weighings.
Benjamin L. Schwartz. Letter:
Truth about false coins. MM 51
(1978) 254. States that Schell told
Michael Goldberg in 1945 that he had originated the problem.
Emil D. Schell. Letter of 17 Jul 1978 to Paul J.
Campbell. Says he did NOT originate the
problem, nor did he submit the version published. He first heard of it from Walter W. Jacobs about
Thanksgiving 1944 in the form of finding at most one light coin among 26 good
coins in three weighings. He submitted
this to the AMM, with a note disclaiming originality. The AMM problem editor published the simpler version described
above, under Schell's name. Schell says
he has heard Eilenberg describe the puzzle as being earlier than Sep 1939. Campbell wrote Eilenberg, but had no
response.
Schell's
letter is making it appear that the problem derives from the use of 1, 3, 9, ... as weights.
This usage leads one to discover that a light coin can be found in 3n coins using n weighings.
This is the problem mentioned by Karapetoff. If there is at most one light coin, then n
weighings will determine it among
3n ‑ 1 coins,
which is the form described by Schell.
The problem seems to have been almost immediately converted into the
case with one false coin, either heavy or light.
Walter W. Jacobs. Letter of 15 Aug 1978 to Paul J.
Campbell. Says he heard of the problem
in 1943 (not 1944) and will try to contact the two people who might have told
it to him. However, Campbell has had no
further word.
V. Karapetoff. The nine coin problem and the mathematics of
sorting. SM 11 (1945) 186‑187. Discusses 9 coins, one light, and asks for a
mathematical approach to the general problem.
(?? -- Cites AMM 52, p. 314, but I cannot find anything relevant in the
whole volume, except the Schell problem.
Try again??)
Dwight A. Stewart,
proposer; D. B. Parkinson & Lester
H. Green, solvers. The counterfeit
coin. In: L. A. Graham, ed.; Ingenious Mathematical Problems and Methods;
Dover, 1959; pp. 37‑38 & 196‑198. 12 coins. First appeared
in Oct 1945. Original only asks for the
counterfeit, but second solver shows how to tell if it is heavy or light.
R. L. Goodstein. Note 1845:
Find the penny. MG 29 (No. 287)
(Dec 1945) 227‑229. Non‑optimal
solution of general problem.
Editorial Note. Note 1930:
Addenda to Note 1845. Ibid. 30
(No. 291) (Oct 1946) 231. Comments on how to extend to optimal solution.
Howard D. Grossman. The twelve‑coin problem. SM 11:3/4 (Sep/Dec 1945) 360‑361. Finds counterfeit and extends to 36 coins.
Lothrop Withington, Jr. Another solution of the 12‑coin
problem. Ibid., 361‑362. Finds also whether heavy or light.
Donald Eves, proposer; E. D. Schell & Joseph Rosenbaum,
solvers. Problem E712 -- The extended
coin problem. AMM 53:3 (Mar 1946) 156 &
54:1 (Jan 1947) 46‑48.
12 coins.
Jerome S. Meyer. Puzzle Paradise. Crown, NY, 1946. Prob.
132: The nine pearls, pp. 94 & 132.
Nine pearls, one light, in two weighings.
N. J. Fine, proposer &
solver. Problem 4203 -- The generalized
coin problem. AMM 53:5 (May 1946)
278 & 54:8 (Oct 1947) 489‑491.
General problem.
H. D. Grossman. Generalization of the twelve‑coin
problem. SM 12 (1946) 291‑292. Discusses Goodstein's results.
F. J. Dyson. Note 1931:
The Problem of the Pennies. MG
30 (No. 291) (Oct 1946) 231‑234.
General solution.
C. A. B. Smith. The Counterfeit Coin Problem. MG 31 (No. 293) (Feb 1947) 31‑39.
C. W. Raine. Another approach to the twelve‑coin
problem. SM 14 (1948) 66‑67. 12 coins only.
K. Itkin. A generalization of the twelve‑coin
problem. SM 14 (1948) 67‑68. General solution.
Howard D. Grossman. Ternary epitaph on coin problems. SM 14 (1948) 69‑71. Ternary solution of Dyson & Smith.
Jerome S. Meyer. Fun-to-do.
A Book of Home Entertainment.
Dutton, NY, 1948. Prob. 40: Nine
pearls, pp. 41 & 188. Nine pearls,
one light, in two weighings.
Blanche Descartes [pseud. of
Cedric A. B. Smith]. The twelve coin
problem. Eureka 13 (Oct 1950) 7 &
20. Proposal and solution in verse.
J. S. Robertson. Those twelve coins again. SM 16 (1950) 111‑115. Article indicates there will be a
continuation, but Schaaf I 32 doesn't cite it and I haven't found it yet.
E. V. Newberry. Note 2342:
The penny problem. MG 37 (No.
320) (May 1953) 130. Says he has made a
rug showing the 120 coins problems and makes comments similar to Littlewood's,
below.
J. E. Littlewood. A Mathematician's Miscellany. Methuen, London, 1953; reprinted with minor corrections, 1957
(& 1960). [All the material cited
is also in the later version:
Littlewood's Miscellany, ed. by B. Bollobás, CUP, 1986, but on different
pages. Since the 1953 ed. is scarce, I
will also cite the 1986 pages in ( ).] Pp. 9 & 135 (31 & 114).
"It was said that the 'weighing‑pennies' problem wasted
10,000 scientist‑hours of war‑work, and that there was a proposal
to drop it over Germany."
John Paul Adams. We Dare You to Solve This! Berkley Publishing, NY, nd [1957?]. [This is apparently a collection of problems
used in newspapers. The copyright is
given as 1955, 1956, 1957.] Prob. 18:
Weighty problem, pp. 13 & 46. 9
equal diamonds but one is light, to be found in 2 weighings.
Hubert Phillips. Something to Think About. Revised ed., Max Parrish, London, 1958. Foreword, p. 6 & prob. 115: Twelve
coins, pp. 81 & 127‑128. Foreword says prob. 115 has been added to this edition and
"was in oral circulation during the war.
So far as I know, it has only appeared in print in the Law Journal,
where I published both the problem and its solution." This may be an early appearance, so I should
try and track this down. ??NYS
Dan Pedoe. The Gentle Art of Mathematics. (English Universities Press, 1958); Pelican (Penguin), 1963. P. 30:
"We now come to a problem which is said to have been planted over
here during the war by enemy agents, since Operational Research spent so many
man‑hours on its solution."
Philip E. Bath. Fun with Figures. The Epworth Press, London, 1959.
No. 7: No weights -- no guessing, pp. 8 & 40. 8
balls, including one light, to be determined in two weighings. Method actually works for £ 1 light.
M. R. Boothroyd &
J. H. Conway. Problems drive,
1959. Eureka 22 (Oct 1959) 15-17 &
22-23. No. 9. Five boxes of sugar, but some has been taken from one box and put
in another. Determine which in least
number of weighings. Does by weighing
each division of A, B, C, D into two pairs.
Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. The "False
Coin" problem, pp. 178-182.
Sketches history and solution.
Simon Dresner. Science World Book of Brain Teasers. 1962.
Op. cit. in 5.B.1. Prob. 46: Dud
reckoning, pp. 21 & 94. Find one
light among eight in two weighings.
Philip Kaplan. More Posers. (Harper & Row, 1964);
Macfadden-Bartell Books, 1965.
Prob. 55, pp. 57 & 98.
Six identical appearing coins, three of which are identically
heavy. In two weighings, identify two
of the heavy coins.
Charlie Rice. Challenge!
Hallmark Editions, Kansas City, Missouri, 1968. Prob. 7, pp. 22 & 54-55. 9 pearls, one light.
Jonathan Always. Puzzling You Again. Tandem, London, 1969. Prob. 86: Light‑weight contest, pp. 51‑52
& 106‑107. 27 weights of
sizes 1, 2, ..., 27, except one is light.
Find it in 3 weighings. He
divides into 9 sets of three having equal weights. Using two weighings, one locates the light weight in a set of
three and then weighing two of these with good weights reveals the light
one. [3 weights 1, 2, 3 cannot be done
in one weighing, but 9 weights 1, 2, ..., 9 can be done in two weighings.]
Robert H. Thouless. The 12‑balls problem as an
illustration of the application of information theory. MG 54 (No. 389) (Oct 1970) 246‑249. Uses information theory to show that the
solution process is essentially determined.
Ron Denyer. Letter.
G&P, No. 37 (Jun 1975) 23. Asks
for a mnemonic for the 12 coins puzzles.
He notes that one can use three predetermined weighings and find the
coin from the three answers.
Basil Mager & E. Asher. Letters:
Coining a mnemonic. G&P, No.
40 (Sep 1975) 26. One mnemonic for a
variable method, another for a predetermined method.
N. J. Maclean. Letter:
The twelve coins. G&P, No.
45 (Feb 1976) 28-29. Exposits a ternary
method for predetermined weighings for
(3n-3)/2 in n
weighings. Each weighing
determines one ternary digit and the resulting ternary number gives both the
coin and whether it is heavy or light.
Tim Sole. The Ticket to Heaven and Other Superior
Puzzles. Penguin, 1988. Weighty problems -- (iii), pp. 124 &
147. Nine equal pies, except someone
has removed some filling from one and inserted it in a pie, possibly the same
one. Determine which, if any, are the
heavy and light ones in 4 balancings.
Calvin T. Long. Magic in base 3. MG 76 (No. 477) (Nov 1992) 371-376. Good exposition of the base 3 method for 12 coins.
Ed Barbeau. After Math.
Wall & Emerson, Toronto, 1995.
Problems for an equal-arm balance, pp. 137-141.
1. Six balls, two of each of three
colours. One of each colour is lighter
than normal and all light weights are equal.
Determine the light balls in three weighings.
2. Five balls, three normal, one heavy, one
light, with the differences being equal, i.e. the heavy and the light weigh as
much as two normals. Determine the
heavy and light in three weighings.
3. Same problem with nine balls and seven
normals, done in four weighings.
5.C.1RANKING COINS WITH A BALANCE
If
one weighs only one coin against another, this is the problem of sorting except
that we don't actually put the objects in order. If one weighs pairs, etc., this is a more complex problem.
J. Schreier. Mathesis Polska 7 (1932) 154‑160. ??NYS -- cited by Steinhaus.
Hugo Steinhaus. Mathematical Snapshots. Not in Stechert, NY, 1938, ed. OUP, NY:
1950: pp. 36‑40 & 258;
1960: pp. 51‑55 & 322;
1969 (1983): pp. 53‑56 & 300. Shows n objects can be ranked in M(n) = 1 + kn ‑ 2k steps where
k = 1 + [log2 n].
Gets M(5) = 8.
Lester R. Ford Jr. & Selmer
M. Johnson. A tournament problem. AMM 66:5 (May 1959) 387‑389. Note that
élog2 n!ù = L(n) is a lower bound from information
theory. Obtain a better upper bound
than Steinhaus, denoted U(n), which is too complex to state here. For convenience, I give the table of these
values here.
n 1 2
3 4 5 6 7
8 9 10 11 12
13
M(n) 0
1 3 5 8 11
14 17 21 25 29
33 37
U(n) 0
1 3 5 7 10
13 16 19 22 26
30 34
L(n) 0 1 3 5
7 10 13 16 19
22 26 29 33
U(n)
= L(n) also holds at n = 20 and 21.
Roland Sprague. Unterhaltsame Mathematik. Op. cit. in 4.A.1. 1961. Prob. 22: Ein noch
ungelöstes Problem, pp. 16 & 42‑43.
(= A still unsolved problem, pp. 17 & 48‑49.) Sketches Steinhaus's method, then does 5
objects in 7 steps. Gives the lower
bound L(n) and says the case n =
12 is still unsolved.
Kobon Fujimura, proposer; editorial comment. Another balance scale problem.
RMM 10 (Aug 1962) 34 & 11 (Oct 1962) 42. Eight coins of different weights and a balance. How many weighings are needed to rank the
coins? In No. 11, it says the solution
will appear in No. 13, but it doesn't appear there or in the last issue,
No. 14. It also doesn't appear in the
proposer's Tokyo Puzzles.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 6: In the
balance, pp. 18 & 85-86. Rank five
balls in order in seven weighings.
John Cameron. Establishing a pecking order. MG 55 (No. 394) (Dec 1971) 391‑395. Reduces Steinhaus's M(n)
by 1 for n ³
5, but this is not as good as Ford
& Johnson.
W. Antony Broomhead. Letter:
Progress in congress? MG 56 (No.
398) (Dec 1972) 331. Comments on
Cameron's article and says Cameron can be improved. States the values
U(9) and U(10),
but says he doesn't know how to do
9 in 19 steps. Cites Sprague for numerical values, but
these don't appear in Sprague -- so Broomhead presumably computed L(9)
and L(10). He gets
10 in 23 steps, which is better
than Cameron.
Stanley Collings. Letter:
More progress in congress. MG 57
(No. 401) (Oct 1973) 212‑213. Notes
the ambiguity in Broomhead's reference to Sprague. Improves Cameron by
1 (or more??) for n ³ 10, but still not as good as Ford & Johnson.
L. J. Upton, proposer; Leroy J. Myers, solver. Problem 1138. CM 12 (1986) 79 & 13 (1987) 230‑231. Rank coins weighing 1, 2, 3, 4
with a balance in four weighings.
See
MUS I 105-124, Tropfke 659.
NOTATION: I-(a, b, c)
means we have three jugs of sizes
a, b, c with a
full and we want to divide
a in half using b
and c. We normally assume a ³ b ³
c and
GCD(a, b, c) = 1.
Halving a is clearly impossible if GCD(b, c)
does not divide a/2 or if
b+c < a/2, unless
one has a further jug or one can drink some.
If a ³
b+c ³ a/2 and GCD(b, c) divides a/2, then the problem is solvable.
More
generally, the question is to determine what amounts can be produced, i.e.
given a, b, c as above, can one measure out an amount d? We denote this by II-(a, b, c; d). Since this also produces
a-d, we can assume that d £ a/2. Then we must have d £ b+c for a solution. When a ³ b+c ³
d, the condition GCD(b, c) ½ d guarantees that d can be produced. This also holds for a =
b+c‑1 and a = b+c‑2. The simplest impossible cases are I‑(4, 4, 3) = II-(4, 4, 3; 2) and
II‑(5, 5, 3; 1). Case I‑(a,
b, c) is the same as II-(a, b, c; a/2).
If a is
a large source, e.g. a stream or a big barrel, we have the problem of
measuring d using b and
c without any constraint on a
and we denote this II-(¥,
b, c; d). However, the solution may not
use the infiniteness of the source and such a problem may be the same as II‑(b+c, b, c; d).
The
general situation when a < b+c is more complex and really requires us to
consider the most general three jug problem:
III‑(A; a, b, c; d) means
we have three jugs of sizes
a, b, c, containing a
total amount of liquid A (in some initial configuration) and we wish
to measure out d. In our previous problems, we had A = a.
Clearly we must have a+b+c ³ A. Again, producing d also produces A-d,
so we can assume d £
A/2. By considering the amounts of
empty space in the containers, the problem
III-(A; a, b, c; d) is
isomorphic to III‑(a+b+c‑A; a, b, c; d') for several possible d'.
NOTES. I have been re-examining this problem and I
am not sure if I have reached a final interpretation and formulation. Also, I have recently changed to the above
notation and I may have made some errors in so doing. I have long had the problem in my list of projects for students,
but no one looked at it until 1995-1996 when Nahid Erfani chose it. She has examined many cases and we have have
discovered a number of properties which I do not recall seeing. E.g. in case I-(a,b,c) with a ³ b ³
c and
GCD(b,c) = 1, there are two ways
to obtain a/2. If we start by pouring into b,
it takes b + c - 1 pourings; if we start by pouring into c,
it takes b + c pourings; so it is always best to start
pouring into the larger jug. A number
of situations II-(a,b,c;d) are solvable for all values of d,
except a/2. E.g.
II‑(a,b,c;a/2) with b+c > a
and c > a/2 is unsolvable.
From about the mid 19C, I have not
recorded simple problems.
I-( 8, 5, 3): almost
all the entries below
I-(10, 6, 4): Pacioli, Court
I-(10, 7, 3): Yoshida
I-(12, 7, 5): Pacioli, van Etten/Henrion, Ozanam, Bestelmeier, Jackson,
Manuel des Sorciers,
Boy's Own Conjuring Book
I-(12, 8, 4): Pacioli
I-(12, 8, 5): Bachet, Arago
I-(16, 9, 7): Bachet-Labosne
I-(16,11, 6): Bachet-Labosne
I-(16,12, 7): Bachet-Labosne
I-(20,13, 9): Bachet-Labosne
I-(42,27,12): Bachet-Labosne
II-(10,3,2;6) Leacock
= II(10,3,2;4)
II-(11,4,3;9): McKay
= II(11,4,3;2)
II-( ¥,5,3;1): Wood, Serebriakoff, Diagram Group
II-( ¥,5,3;4): Chuquet, Wood,
Fireside Amusements,
II-( ¥,7,4;5): Meyer, Stein,
Brandes
II-( ¥,8,5;11): Young World,
III-(20;19,13,7;10): Devi
General problem, usually form I,
sometimes form II: Bachet‑Labosne, Schubert,
Ahrens, Cowley, Tweedie,
Grossman, Buker, Goodstein,
Browne, Scott, Currie,
Sawyer, Court, O'Beirne,
Lawrence, McDiarmid & Alfonsin.
Versions with 4 or more
jugs: Tartaglia, Anon: Problems drive (1958), Anon (1961), O'Beirne.
Impossible versions: Pacioli,
Bachet, Anon: Problems drive
(1958).
Abbot Albert. c1240.
Prob. 4, p. 333. I-(8,5,3) -- one solution.
Columbia Algorism. c1350.
Chap. 123: I-(8,5,3). Cowley 402‑403 & plate opposite
403. The plate shows the text and three
jars. I have a colour slide of the
three jars from the MS.
Munich 14684. 14C.
Prob. XVIII & XXIX, pp. 80 & 83. I-(8,5,3).
Folkerts. Aufgabensammlungen. 13-15C.
16 sources with I-(8,5,3).
Pseudo-dell'Abbaco. c1440.
Prob. 66, p.62. I-(8,5,3) -- one solution. "This problem is of little utility ...." I have a colour slide of this.
Chuquet. 1484.
Prob. 165. Measure 4 from a cask
using 5 and 3. You can pour back into
the cask, i.e. this is II-(¥,5,3;4). FHM 233 calls this the tavern-keeper's
problem.
HB.XI.22. 1488.
P. 55 (= Rath 248). Same as
Abbot Albert.
Pacioli. De Viribus.
c1500.
Ff.
97r - 97v. LIII. C(apitolo). apartire
una botte de vino fra doi (To divide a bottle of wine between two). = Peirani 137-138. I-(8,5,3). One solution.
Ff.
97v - 98v. LIIII. C(apitolo). a partire
unaltra botte fra doi (to divide another bottle between two). = Peirani 138-139. I-(12,7,5). Dario Uri
points out that the solution is confused and he repeats himself so it takes
him 18
pourings instead of the usual
11. He then says one can
divide 18 among three brothers who have containers of sizes 5, 6, 7,
which he does by filling the
6 and then the problem is
reduced to the previous problem. [He
could do it rather more easily by pouring the
6 into the 7
and then refilling the 6!]
Ff.
98v - 99r. LV. (Capitolo) de doi altri
sotili divisioni. de botti co'me se dira (Of two other subtle divisions of
bottles as described). = Peirani
139-140. I‑(10,6,4) and
I-(12,8,4). Pacioli suggests
giving these to idiots.
Ghaligai. Practica D'Arithmetica. 1521.
Prob. 20, ff. 64v-65r. I‑(8,5,3). One solution.
Cardan. Practica Arithmetice. 1539.
Chap. 66, section 33, f. DD.iiii.v (p. 145). I-(8,5,3). Gives one
solution and says one can go the other way.
H&S 51 says I-(8,5,3)
case is also in Trenchant (1566).
??NYS
Tartaglia. General Trattato, 1556, art. 132 & 133,
p. 255v‑256r.
Art.
132: I-(8,5,3).
Art.
133: divide 24 in thirds, using 5, 11, 13.
Buteo. Logistica. 1559. Prob. 73, pp. 282-283. I-(8,5,3).
Gori. Libro di arimetricha.
1571. Ff. 71r‑71v (p.
76). I-(8,5,3).
Bachet. Problemes.
1612. Addl. prob. III: Deux bons
compagnons ont 8 pintes de vin à partager entre eux également, ..., 1612: 134-139; 1624: 206-211; 1884: 138‑147. I‑(8,5,3) -- both solutions; I-(12,8,5)
(omitted by Labosne). Labosne
adds I‑(16,9,7); I‑(16,11,6); I‑(42,27,12);
I-(20,13,9); I-(16,12,7) (an impossible case!) and discusses general case. (This seems to be the first discussion of
the general case.)
van Etten. 1624.
Prob. 9 (9), pp. 11 & fig. opp. p. 1 (pp. 22‑23). I‑(8,5,3) -- one solution.
Henrion's Nottes, 1630, pp. 11‑13, gives the second solution and
poses and solves I‑(12,7,5).
Hunt. 1631 (1651). P. 270
(262). I-(8,5,3). One solution.
Yoshida (Shichibei) Kōyū (= Mitsuyoshi Yoshida)
(1598-1672). Jinkō‑ki. 2nd ed., 1634 or 1641??. ??NYS
The recreational problems are discussed in Kazuo Shimodaira; Recreative
Problems on "Jingōki", a 15 pp booklet sent by Shigeo
Takagi. [This has no details, but Takagi
says it is a paper that Shimodaira read at the 15th International Conference
for the History of Science, Edinburgh, Aug 1977 and that it appeared in
Japanese Studies in the History of Science 16 (1977) 95-103. I suspect this is a copy of a preprint.] This gives both Jingōki and
Jinkōki as English versions of the title and says the recreational
problems did not appear in the first edition, 4 vols., 1627, but did appear in
the second edition of 5 vols. (which may be the first use of coloured wood cuts
in Japan), with the recreational problems occurring in vol. 5. He doesn't give a date, but Mikami, p. 179,
indicates that it is 1634, with further editions in 1641, 1675, though an
earlier work by Mikami (1910) says 2nd ed. is 1641. Yoshida (or Suminokura) is the family name. Shimodaira refers to the current year as the
350th anniversary of the edition and says copies of it were published
then. I have a recent transcription of
some of Yoshida into modern Japanese and a more recent translation into English,
??NYR, but I don't know if it is the work mentioned by Shimodaira.
Shimodaira
discusses a jug problem on p. 14:
I-(10,7,3) -- solution in 10
moves. Shimodaira thinks Yoshida heard
about such puzzles from European contacts, but without numerical values, then
made up the numbers. I certainly can
see no other example of these numbers.
The recent transcription includes this material as prob. 7 on pp. 69-70.
Wingate/Kersey. 1678?.
Prob. 7, pp. 543-544.
I-(8,5,3). Says there is a
second way to do it.
Witgeest. Het Natuurlyk Tover-Boek. 1686.
Prob. 38, p. 308. I-(8,5,3).
Ozanam. 1694.
Prob.
36, 1696: 91-92; 1708: 82‑83. Prob. 42, 1725: 238‑240. Prob. 21, 1778: 175‑177; 1803: 174-176; 1814: 153-154. Prob. 20,
1840: 79. I-(8,5,3) -- both solutions.
Prob.
43, 1725: 240‑241. Prob. 22,
1778: 177-178; 1803: 176-177; 1814: 154-155. Prob. 21, 1840: 79‑80.
I-(12,7,5) -- one solution.
Dilworth. Schoolmaster's Assistant. 1743.
Part IV: Questions: A short Collection of pleasant and diverting
Questions, p. 168. Problem 8. I-(8,5,3).
(Dilworth cites Wingate for this -- cf in 5.B.) = D. Adams; Scholar's Arithmetic; 1801, p.
200, no. 10.
Les Amusemens. 1749.
Prob. 17, p. 139: Partages égaux avec des Vases inégaux. I-(8,5,3)
-- both solutions.
Bestelmeier. 1801.
Item 416: Die 3 Maas‑Gefäss.
I-(12,7,5).
Badcock. Philosophical Recreations, or, Winter
Amusements. [1820]. Pp. 48-49, no. 75: How to part an eight
gallon bottle of wine, equally between two persons, using only two other bottles,
one of five gallons, and the other of three.
Gives both solutions.
Jackson. Rational Amusement. 1821.
Arithmetical Puzzles.
No.
14, pp. 4 & 54. I-( 8,5,3).
One solution.
No.
52, pp. 12 & 67. I-(12,7,5). One solution.
Rational Recreations. 1824.
Exer. 10, p. 55. I-(8,5,3) one way.
Manuel des Sorciers. 1825.
??NX
Pp.
55-56, art. 27-28. I-(8,5,3) two ways.
P. 56,
art. 29. I-(12,7,5).
Endless Amusement II. 1826?
Prob. 7, pp. 193-194.
I-(8,5,3). One solution. = New Sphinx, c1840, p. 133.
Nuts to Crack III (1834), no.
212. I-(8,5,3). 8 gallons of spirits.
Young Man's Book. 1839.
Pp. 43-44. I-(8,5,3). Identical to Wingate/Kersey.
The New Sphinx. c1840.
P. 133. I-(8,5,3). One solution.
Boy's Own Book. 1843 (Paris): 436 & 441, no. 7. The can of ale: 1855: 395; 1868:
432. I‑(8,5,3). One solution. The 1843 (Paris) reads as though the owners of the 3 and 5 kegs
both want to get 4, which would be a problem for the owner of the 3. = Boy's Treasury, 1844, pp. 425 & 429.
Fireside Amusements. 1850.
Prob. 9, pp. 132 & 184. II-(¥,5,3;4). One solution.
Arago. [Biographie de] Poisson (16 Dec 1850). Oeuvres, Gide & Baudry, Paris, vol. 2, 1854, pp. 593‑??? P. 596 gives the story of Poisson's being
fascinated by the problem I‑(12,8,5). "Poisson résolut à l'instant cette
question et d'autres dont on lui donna l'énoncé. Il venait de trouver sa véritable vocation." No solution given by Arago.
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Arithmetical puzzles, no. 8, pp. 174-175
(1868: 185-186). I-(8,5,3). Milkmaid with eight quarts of milk.
Magician's Own Book. 1857.
P.
223-224: Dividing the beer: I-(8,5,3).
P.
224: The difficult case of wine:
I-(12,7,5).
Pp.
235-236: The two travellers:
I-(8,5,3) posed in verse.
Each problem gives just one solution.
Boy's Own Conjuring Book. 1860.
P.
193: Dividing the beer: I-(8,5,3).
P.
194: The difficult case of wine:
I-(12,7,5).
Pp.
202‑203: The two travellers:
I-(8,5,3) posed in verse.
Each problem gives just one solution.
Illustrated Boy's Own
Treasury. 1860. Prob. 21, pp. 428-429 & 433. I‑(8,5,3). "A man coming from the Lochrin distillery with an 8-pint jar
full of spirits, ...."
Vinot. 1860. Art. XXXVIII: Les
cadeaux difficiles, pp. 57-58.
I-(8,5,3). Two solutions.
The Secret Out (UK). c1860.
To divide equally eight pints of wine ..., pp. 12-13.
Bachet-Labosne. 1874.
For details, see Bachet, 1612.
Labosne adds a consideration of the general case which seems to be the
first such.
Kamp. Op. cit. in 5.B.
1877. No. 17, p. 326: I-(8,5,3).
Mittenzwey. 1880.
Prob. 106, pp. 22 & 73-74;
1895?: 123, pp. 26 & 75-76;
1917: 123, pp. 24 & 73-74.
I-(8,5,3). One solution.
Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 135, no. 1.
I-(8,5,3). No solution.
Loyd. Problem 11: "Two thieves of Damascus". Tit‑Bits 31 (19 Dec 1896 &
16 Jan 1897) 211 & 287.
Thieves found with 2 &
2 quarts in pails of size 3 & 5.
They claim the merchant measured the amounts out from a fresh
hogshead. Solution is that this could
only be done if the merchant drained the hogshead, which is unreasonable!
Loyd. Problem 13: The Oriental problem. Tit‑Bits 31 (19 Jan,
30 Jan & 6 Feb 1897) 269, 325 & 343.
= Cyclopedia, 1914, pp. 188 & 364: The merchant of Bagdad. Complex problem with hogshead of water,
barrel of honey, three 10 gallon jugs to be filled with 3 gallons of water, of
honey and of half and half honey & water.
There are a 2 and a 4 gallon measure and also 13 camels to receive 3
gallons of water each. Solution
takes 521 steps. 6 Feb reports
solutions in 516 and
513 steps. Cyclopedia gives solution in 506
steps.
Dudeney. The host's puzzle. London Magazine 8 (No. 46) (May 1902) 370 &
8 (No. 47) (Jun 1902) 481‑482 (= CP, prob. 6, pp. 28‑29
& 166‑167). Use 5 and 3
to obtain 1 and 1 from a cask. One must drink some!
H. Schubert. Mathematische Mussestunden, 3rd ed.,
Göschen, Leipzig, 1907. Vol. 1, chap.
6, Umfüllungs‑Aufgaben, pp. 48‑56.
Studies general case and obtains some results. (The material appeared earlier in Zwölf Geduldspiele, 1895, op.
cit. in 5.A, Chap. IX, pp. 110-119.
The 13th ed. (De Gruyter, Berlin, 1967), Chap. 9, pp. 62‑70, seems
to be a bit more general (??re-read).)
Ahrens. MUS I, 1910, chap. 4, Umfüllungsaufgaben,
pp. 105‑124. Pp. 106‑107 is
Arago's story of Poisson and this problem.
He also extends and corrects Schubert's work.
Dudeney. Perplexities: No. 141: New measuring
puzzle. Strand Magazine 45 (Jun 1913)
710 & 46 (Jul 1913) 110. (= AM,
prob. 365, pp. 110 & 235.) Two 10
quart vessels of wine with 5 and 4 quart measures. He wants 3 quarts in each measure. (Dudeney gives numerous other versions in AM.)
Loyd. Cyclopedia. 1914. Milkman's puzzle, pp. 52 & 345. (= MPSL2, prob. 23, pp. 17 & 127‑128 = SLAHP: Honest John, the milkman, pp. 21
& 90.) Milkman has two full 40
quart containers and two customers with 5 and 4 quart pails, but both want 2
quarts. (Loyd Jr. says "I first published [this] in
1900...")
Williams. Home Entertainments. 1914.
The measures puzzle, p. 125.
I-(8,5,3).
Hummerston. Fun, Mirth & Mystery. 1924.
A shortage of milk, Puzzle no. 75, pp. 164 & 183. I-(8,5,3),
one solution.
Elizabeth B. Cowley. Note on a linear diophantine equation. AMM 33 (1926) 379‑381. Presents a technique for resolving I-(a,b,c),
which gives the result when a =
b+c. If a < b+c, she only
seems to determine whether the method gets to a point with A
empty and neither B nor
C full and it is not clear to me
that this implies impossibility. She
mentions a graphical method of Laisant (Assoc. Franç. Avance. Sci, 1887, pp.
218-235) ??NYS.
Wood. Oddities. 1927.
Prob.
15: A problem in pints, pp. 16-17.
Small cask and measures of size 5 and 3, measure out 1 in each measure. Starts by filling the 5 and the 3 and then emptying the cask, so
this becomes a variant of II-(¥,5,3;1).
Prob.
26: The water-boy's problem, pp. 28-29.
II-(¥;,5,3;4).
Ernest K. Chapin. Scientific Problems and Puzzles. In:
S. Loyd Jr.; Tricks and Puzzles,
Vol. 1 (only volume to appear);
Experimenter Publishing Co., NY, nd [1927] and Answers to Sam Loyd's
Tricks and Puzzles, nd [1927]. [This
book is a selection of pages from the Cyclopedia, supplemented with about 20
pages by Chapin and some other material.]
P. 89 & Answers p. 8. You
have a tablet that has to be dissolved in
7½ quarts of water, though you
only need 5 quarts of the resulting mixture.
You have 3 and 5 quart measures and a tap.
Stephen Leacock. Model Memoirs and Other Sketches from Simple
to Serious. John Lane, The Bodley Head,
1939, p. 298. "He's trying to
think how a farmer with a ten-gallon can and a three-gallon can and a
two-gallon can, manages to measure out six gallons of milk." II-(10,3,2;6) = II-(10,3,2;4).
M. C. K. Tweedie. A graphical method of solving Tartaglian
measuring puzzles. MG 23 (1939) 278‑282. The elegant solution method using triangular
coordinates.
H. D. Grossman. A generalization of the water‑fetching
puzzle. AMM 47 (1940) 374‑375. Shows
II-(¥,b,c;d) with GCD(b,c) = 1 is solvable.
McKay. Party Night. 1940.
No.
18, p. 179. II-(11,4,3;9).
No.
19, pp. 179-180. I-(8,5,3).
Meyer. Big Fun Book. 1940. No. 10, pp. 165 & 753. II-(¥,7,4,5).
W. E. Buker, proposer. Problem E451. AMM 48 (1941) 65.
??NX. General problem of what
amounts are obtainable using three jugs, one full to start with, i.e. I-(a,b,c).
See Browne, Scott, Currie below.
Eric Goodstein. Note 153:
The measuring problem. MG 25
(No. 263) (Feb 1941) 49‑51.
Shows II-(¥,b,c;d) with
GCD(b,c) = 1 is solvable.
D. H. Browne & Editors. Partial solution of Problem E451. AMM 49 (1942) 125‑127.
W. Scott. Partial solution of E451 -- The generalized
water‑fetching puzzle. AMM 51
(1944) 592. Counterexample to
conjecture in previous entry.
J. C. Currie. Partial solution of Problem E451. AMM 53 (1946) 36‑40. Technical and not complete.
W. W. Sawyer. On a well known puzzle. SM 16 (1950) 107‑110. Shows that
I-(b+c,b,c) is solvable if b & c
are relatively prime.
David Stein. Party and Indoor Games. Op. cit. in 5.B. c1950. Prob. 13, pp. 79‑80. Obtain 5 from a spring using measures 7 and
4, i.e. II-(¥,7,4,5).
Anonymous. Problems drive, 1958. Eureka 21 (Oct 1958) 14-16 & 30. No. 8.
Given an infinite source, use:
6, 10, 15 to obtain 1, 6, 7
simultaneously; 4, 6, 9, 12 to obtain
1, 2, 3, 4 simultaneously; 6, 9, 12, 15, 21 to obtain 1, 3, 6, 8,
9 simultaneously. Answer simply says the first two are possible
(the second being easy) and the third is impossible.
Young World. c1960.
P. 58: The 11 pint problem. II-(¥,8,5;11). This is the same as II‑(13,8,5;11) or
II-(13,8,5,2).
Anonymous. Moonshine sharing. RMM 2 (Apr 1961) 31
& 3 (Jun 1961) 46. Divide
24 in thirds using cylindrical
containers holding 10, 11, 13. Solution in No. 3 uses the cylindricity of a
container to get it half full.
Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. "Pouring"
problems -- The "robot" method.
General description of the problem.
Attributes Tweedie's triangular 'bouncing ball' method to Perelman, with
no reference. Does I‑(8,5,3) two ways, also
I-(12,7,5) and I-(16,9,7),
then considers type II questions.
Considers the problem with
II-(10,6,4;d) and extends
to II-(a,6,4;d) for
a > 10, leaving it to
the reader to "try to formulate some rule about the results." He then considers II‑(7,6,4;d),
noting that the parallelogram has a corner trimmed off. Then considers II-(12,9,7;d) and II-(9,6,3;d).
Lloyd Jim Steiger. Letter.
RMM 4 (Aug 1961) 62. Solves the
RMM 2 problem by putting the 10 inside the 13 to measure 3.
Irving & Peggy Adler. The Adler Book of Puzzles and Riddles. Or Sam Loyd Up-To-Date. John Day, NY, 1962. Pp. 32 & 46. Farmer has two full 10-gallon cans. Girls come with 5-quart and 4-quart cans and each wants 2 quarts.
Philip Kaplan. More Posers. (Harper & Row, 1964);
Macfadden-Bartell Books, 1965.
Prob. 80, pp. 81 & 109.
Tavern has a barrel with 15 pints of beer. Two customers, with 3 pint and 5 pint jugs appear and ask for 1
pint in each jug. Bartender finds it
necessary to drink the other 13 pints!
T. H. O'Beirne. Puzzles and Paradoxes. OUP, 1965.
Chap. 4: Jug and bottle department, pp. 49‑75. This gives an extensive discussion of
Tweedie's method and various extensions to four containers, a barrel of unknown
size, etc.
P. M. Lawrence. An algebraic approach to some pouring
problems. MG 56 (No. 395) (Feb 1972) 13‑14. Shows
II-(¥,b,c,d) with d £ b+c and GCD(b,c) = 1 is possible and extends to more jugs.
Louis Grant Brandes. The Math. Wizard. revised ed., J. Weston Walch, Portland, Maine, 1975. Prob. 5: Getting five gallons of water: II‑(¥,7,4,5).
Shakuntala Devi. Puzzles to Puzzle You. Orient Paperbacks (Vision Press), Delhi,
1976.
Prob.
53: The three containers, pp. 57 & 110.
III-(20;19,13,7;10). Solution in
15 steps. Looking at the triangular
coordinates diagram of this, one sees that it is actually isomorphic to II-(19,13,7;10) and this can be seen by considering the amounts of empty space in
the containers.
Prob.
132: Mr. Portchester's problem, pp. 82 & 132. Same as Dudeney (1913).
Victor Serebriakoff. A Mensa Puzzle Book. Muller, London, 1982. (Later combined with A Second Mensa Puzzle
Book, 1985, Muller, London, as: The Mensa Puzzle Book, Treasure Press, London,
1991.) Problem T.16: Pouring puttonos,
part b, pp. 19-20 (1991: 37-38) & Answer 19, pp. 102-103 (1991:
118-119). II-( ¥,5,3;1).
The Diagram Group. The Family Book of Puzzles. The Leisure Circle Ltd., Wembley, Middlesex,
1984. Problem 161, with Solution at the
back of the book. II-(¥,5,3;1),
which can be done as II-(8,5,3;1).
D. St. P. Barnard. 50 Daily Telegraph Brain Twisters. 1985.
Op. cit. in 4.A.4. Prob. 4:
Measure for measure, pp. 15, 79‑80, 103.
Given 10 pints of milk, an 8 pint bowl, a jug and a flask. He describes how he divides the milk in
halves and you must deduce the size of the jug and the flask.
Colin J. H. McDiarmid &
Jorge Ramirez Alfonsin. Sharing
jugs of wine. Discrete Mathematics 125
(1994) 279-287. Solves I-(b+c,b,c)
and discusses the problem of getting from one state of the problem to
another in a given number of steps, showing that GCD(b,c) = 1 guarantees the
graph is connected. indeed essentially cyclic.
Considers GCD(b,c) ¹
1. Notes that the work done easily
extends to a > b + c. Says the second author's PhD at Oxford,
1993, deals with more cases.
John P. Ashley. Arithmetickle. Arithmetic Curiosities, Challenges, Games and Groaners for all
Ages. Keystone Agencies, Radnor, Ohio,
1997. P. 11: The spoon and the
bottle. Given a 160 ml bottle and a 30
ml spoon, measure 230 ml into a bucket.
5.D.2. RULER WITH MINIMAL NUMBER OF MARKS
Dudeney. Problem 518: The damaged measure. Strand Mag. (Sep 1920) ??NX.
Wants a minimal ruler for 33 inches total length. (=? MP 180)
Dudeney. Problem 530: The six cottagers. Strand Mag. (Jan 1921) ??NX.
Wants 6 points on a circle to give all arc distances 1, 2, ..., 20. (=? MP 181)
Percy Alexander MacMahon. The prime numbers of measurement on a
scale. Proc. Camb. Philos. Soc. 21
(1922‑23) 651‑654. He
considers the infinite case, i.e. a(0)
= 0,
a(i+1) = a(i) + least integer which is not yet
measurable. This gives:
0, 1, 3, 7, 12, 20, 30, 44, ....
Dudeney. MP.
1926.
Prob.
180: The damaged measure, pp. 77 & 167.
(= 536, prob. 453, pp. 173, 383‑384.) Mark a ruler of length 33 with 8 (internal)
marks. Gives 16 solutions.
Prob.
181: The six cottagers, pp. 77‑78 & 167. = 536, prob. 454, pp. 174 & 384.
A. Brauer. A problem of additive number theory and its
application in electrical engineering.
J. Elisha Mitchell Sci. Soc. 61 (1945) 55‑56. Problem arises in designing a resistance box.
Л.
Редеи
& А. Реньи [L. Redei & A. Ren'i
(Rényi)]. О
представленин
чисел
1, 2, ..., N
лосредством
разностей [O
predstavlenin chisel 1, 2, ... , N losredstvom raznosteĭ (On the
representation of 1, 2, ..., N by differences)]. Мат.
Сборник [Mat. Sbornik] 66
(NS 24) (1949) 385‑389.
Anonymous. An unsolved problem. Eureka 11 (Jan 1949) 11 & 30. Place as few marks as possible to permit
measuring integers up to n. For
n = 13, an example is: 0, 1, 2, 6, 10, 13. Mentions some general results for a circle.
John Leech. On the representation of 1, 2, ..., n by differences. J. London
Math. Soc. 31 (1956) 160‑169.
Improves Redei & Rényi's results.
Gives best examples for small n.
Anon. Puzzle column: What's your potential? MTg 19 (1962) 35
& 20 (1962) 43. Problem posed in terms of transformer
outputs -- can we arrange 6 outputs to give every integral voltage up
through 15? Problem also asks for the general case. Solution asserts, without real proof, that the optimum occurs
with 0, 1, 4, 7, 10, ..., n‑11, n‑8,
n‑5, n‑2 or its complement.
T. H. O'Beirne. Puzzles and Paradoxes. OUP, 1965.
Chap. 6 discusses several versions of the problem.
Gardner. SA (Jan 1965) c= Magic Numbers, chap.
6. Describes 1, 2, 6, 10 on a ruler 13 long. Says 3
marks are sufficient on 9 and
4 marks on 12
and asks for proof of the latter and for the maximum number of distances
that 3
marks on 12 can produce. How can you mark a ruler
36 long? Says Dudeney, MP prob. 180, believed that 9
marks were needed for a ruler longer than 33, but Leech managed to
show 8
was sufficient up to 36.
C. J. Cooke. Differences. MTg 47 (1969) 16. Says
the problem in MTg 19 (1962) appears in H. L. Dorwart's The Geometry of
Incidence (1966) related to perfect difference sets but with an erroneous
definition which is corrected by references to H. J. Ryser's Combinatorial
Mathematics. However, this doesn't
prove the assertions made in MTg 20.
Jonathan Always. Puzzles for Puzzlers. Tandem, 1971. Prob. 22: Starting and stopping, pp. 18 & 66. Circular track, 1900 yards around. How can one place marker posts so every
multiple of 100 yards up to 1900 can be run.
Answer: at 0, 1, 3, 9, 15.
Gardner. SA (Mar 1972) = Wheels, Chap. 15.
5.D.3 FALSE COINS WITH A WEIGHING SCALE
H. S. Shapiro, proposer; N. J. Fine, solver. Problem E1399 -- Counterfeit coins. AMM 67 (1960) 82 & 697‑698. Genuines weigh 10, counterfeits
weigh 9. Given 5 coins and a scale, how many weighings are
needed to find the counterfeits? Answer
is 4.
Fine conjectures that the ratio of weighings to coins decreases to 0.
Kobon Fujimura & J. A. H.
Hunter, proposers; editorial
solution. There's always a way. RMM 6 (Dec 1961) 47 &
7 (Feb 1962) 53.
(c= Fujimura's The Tokyo Puzzles (Muller, London, 1979), prob. 29:
Pachinko balls, pp. 35 & 131.) Six
coins, one false. Determine which is
false and whether it is heavy or light in three weighings on a scale. In fact one also finds the actual weights.
K. Fujimura, proposer; editorial solution. The 15‑coin puzzle. RMM 9 (Jun 1962) & 10 (Aug 1962) 40‑41. Same problem with fifteen coins and four
weighings.
5.D.4. TIMING WITH HOURGLASSES
I
have just started these and they are undoubtedly older than the examples here. I don't recall ever seeing a general approach
to these problems.
Simon Dresner. Science World Book of Brain Teasers. 1962.
Op. cit. in 5.B.1. Prob. 17: Two‑minute
eggs, pp. 9 & 87. Time 2
minutes with 3 & 5 minute timers.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 21: The sands
of time, pp. 35 & 93. Time 9
minutes with 4 & 7 minute timers.
David B. Lewis. Eureka!
Perigee (Putnam), NY, 1983. Pp.
73‑74. Time 9
minutes with 4 & 7 minute timers.
Yuri B. Chernyak & Robert S.
Rose. The Chicken from Minsk. BasicBooks, NY, 1995. Chap. 1, prob. 8: Grandfather's breakfast,
pp. 6 & 102. Time 15
minutes with 7 & 11 minute timers.
I
have just started this and there must be much older examples.
Benson. 1904.
The water‑glass puzzle, p. 254.
Dudeney. AM.
1917. Prob. 364: The barrel
puzzle, pp. 109-110 & 235.
King. Best 100. 1927. No. 1, pp. 7 & 38.
Collins. Fun with Figures. 1928. The dairymaid's
problem, pp. 29-30.
William A. Bagley. Puzzle Pie.
Vawser & Wiles, London, nd [BMC gives 1944]. [There is a revised edition, but it only
affects material on angle trisection.]
No. 14: 'Arf an' 'arf, p. 15.
Anon. The Little Puzzle Book.
Peter Pauper Press, Mount Vernon, NY, 1955. P. 52: The cider barrel.
Jonathan Always. Puzzles for Puzzlers. Tandem, London, 1971. Prob. 87: But me no butts, pp. 42 & 88.
Richard I. Hess. Email Christmas message to NOBNET, 24 Nov
2000. Solution sent by Nick Baxter on
the same day. You have aquaria (assumed
cuboidal) which hold 7 and 12 gallons and a water supply. The 12 gallon aquarium has dots accurately
placed in the centre of each side face.
How many steps are required to get 8 gallons into the 12 gallon
aquarium? Fill the 12 gallon aquarium
and tilt it on one corner so the water level passes through the centres of the
two opposite faces. This leaves 8
gallons! Nick says this is two steps.
Euler
circuits have been used in primitive art, often as symbols of the passage of
the soul to the land of the dead. [MTg
110 (Mar 1985) 55] shows examples from Angola and New Hebrides. See Ascher (1988 & 1991) for many other
examples from other cultures.
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Above
is the 'five‑brick pattern'.
See: Clausen, Listing,
Kamp, White, Dudeney,
Loyd Jr, Ripley, Meyer,
Leeming, Adams, Anon.,
Ascher. Prior to Loyd Jr, the
problem asked for the edges to be drawn in three paths, but about 1920 the
problem changed to drawing a path across every wall.
Trick
solutions: Tom Tit, Dudeney (1913), Houdini, Loyd Jr, Ripley,
Meyer, Leeming, Adams,
Gibson, Anon. (1986).
Non-crossing
Euler circuits: Endless Amusement
II, Bellew, Carroll 1869, Mittenzwey, Bile Beans,
Meyer, Gardner (1964), Willson,
Scott, Singmaster.
Kn denotes the complete graph on n
vertices.
Matthäus Merian the Elder. Engraved map of Königsberg. Bernhard Wiezorke has sent me a coloured
reproduction of this, dated as 1641. He
used an B&W version in his article: Puzzles und Brainteasers; OR News,
Ausgabe 13 (Nov 2001) 52-54. BLW use a
B&W version on their dust jacket and on p. 2 which they attribute to M.
Zeiller; Topographia Prussiae et Pomerelliae; Frankfurt, c1650. I have seen this in a facsimile of the
Cosmographica due to Merian in the volume on Brandenburg and Pomerania, but it
was not coloured. There seem to be at
least two versions of this picture --??CHECK.
L. Euler. Solutio problematis ad geometriam situs
pertinentis. (Comm. Acad. Sci.
Petropol. 8 (1736(1741)) 128‑140.)
= Opera Omnia (1) 7 (1923) 1‑10.
English version: Seven Bridges
of Königsberg is in: BLW, 3‑8; SA 189 (Jul 1953) 66‑70;
World of Mathematics, vol. 1, 573‑580; Struik, Source Book, 183‑187.
My late colleague Jeremy Wyndham
became interested in the seven bridges problem and made inquiries which turned
up several maps of Königsberg and a list of all the bridges and their dates of
construction (though there is some ambiguity about one bridge). The first bridge was built in 1286 and until
the seventh bridge of 1542, an Euler path was always possible. No further bridge was built until a railway
bridge in 1865 which led to Saalschütz's 1876 paper -- see below. In 1905 and later, several more bridges were
added, reaching a maximum of ten bridges in 1926 (with 4512 paths from the
island), then one was removed in 1933.
Then a road bridge was added, but it is so far out that it does not show
on any map I have seen. Bombing and
fighting in 1944-1945 apparently destroyed all the bridges and the Russians
have rebuilt six or seven of them. I
have computed the number of paths in each case -- from 1865 until 1935 or 1944,
there were always Euler paths.
L. Poinsot. Sur les polygones et les polyèdres. J. École Polytech. 4 (Cah. 10) (1810) 16‑48. Pp. 28‑33 give Euler paths on K2n+1 and Euler's criterion.
Discusses square with diagonals.
Endless Amusement II. 1826?
Prob. 34, p. 211. Pattern of two
overlapping squares has a non-crossing Euler circuit.
Th. Clausen. De linearum tertii ordinis
propietatibus. Astronomische
Nachrichten 21 (No. 494) (1844), col. 209‑216. At the very end, he gives the five‑brick pattern and says
that its edges cannot be drawn in three paths.
J. B. Listing. Vorstudien zur Topologie. Göttinger Studien 1 (1847) 811‑875. ??NYR.
Gives five brick pattern as in Clausen.
?? Nouv. Ann. Math. 8
(1849?) 74. ??NYS. Lucas says this poses the problem of finding
the number of linear arrangements of a set of dominoes. [For a double N set, N = 2n,
this is (2n+1)(n+1) times the number of circular arrangements,
which is n2n+1 times the number of Euler circuits on K2n+1.]
É. Coupy. Solution d'un problème appartenant a la
géométrie de situation, par Euler.
Nouv. Ann. Math. 10 (1851) 106‑119. Translation of Euler.
Translator's note on p. 119 applies it to the bridges of Paris.
The Sociable. 1858.
Prob. 7: Puzzle pleasure garden, pp. 288 & 303. Large maze-like garden and one is to pass
over every path just once -- phrased in verse.
= Book of 500 Puzzles, 1859, prob. 7, pp. 6 & 21. = Illustrated Boy's Own Treasury, 1860,
prob. 49, pp. 405 & 443. In fact,
if one goes straight across every intersection, one finds the path, so this is
really almost a unicursal problem.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 587, pp. 297
& 410: Ariadnerätsel. Three
diagrams to trace with single lines. No
attempt to avoid crossings.
Frank Bellew. The Art of Amusing. Carleton, NY (& Sampson Low & Co., London), 1866
[C&B list a 1871]; John Camden
Hotten, London, nd [BMC & NUC say 1870] and John Grant, Edinburgh, nd
[c1870 or 1866?], with slightly different pagination. 1866: pp. 269-270;
1870: p. 266. Two
overlapping squares have a non-crossing Euler circuit.
Lewis Carroll. Letter of 22 Aug 1869 to Isabel
Standen. Taken from: Stuart Dodgson
Collingwood; The Life and Letters of Lewis Carroll; T. Fisher Unwin, London,
(Dec 1898), 2nd ed., Jan 1899, p. 370: "Have you succeeded in drawing the
three squares?" On pp. 369-370,
the recipient is identified as Isabel Standen and she is writing Collingwood,
apparently sending him the letter.
Collingwood interpolates:
"This puzzle was, by the way, a great favourite of his; the problem
is to draw three interlaced squares without going over the same lines twice, or
taking the pen off the paper". But
no diagram is given.
Dudeney;
Some much‑discussed puzzles; op. cit. in 2; 1908, quotes Collingwood,
gives the diagram and continues: "This is sometimes ascribed to him [i.e.
Carroll] as its originator, but I have found it in a little book published in
1835." This was probably a
printing of Endless Amusement II, qv above and in Common References, though
this has two interlaced squares. John
Fisher; The Magic of Lewis Carroll; op. cit. in 1; pp. 58‑59, says
Carroll would ask for a non‑crossing Euler circuit, but this is not
clearly stated in Collingwood. Cf
Carroll-Wakeling, prob. 29: The three squares, pp. 38 & 72, which
clearly states that a non-crossing circuit is wanted and notes that there is
more than one solution. Cf Gardner
(1964). Carroll-Gardner, pp. 52-53.
Mittenzwey. 1880.
Prob. 269-279, pp. 47-48 & 98-100;
1895?: 298-308, pp. 51-52 & 100‑102; 1917: 298-308, pp. 46-48 & 95-97. Straightforward unicursal patterns. The first is K5, but one of the diagonals was missing in my
copy of the 1st ed. -- the path is not to use two consecutive outer edges. The third is the 'envelope' pattern. The fourth is three overlapping squares,
where the two outer squares just touch in the middle. The last is a simple maze with no dead ends and the path is not
to cross itself. See also the entry for
Mittenzwey in 5.E.1, below.
M. Reiss. Évaluation du nombre de combinaisons
desquelles les 28 dés d'un jeu de dominos sont susceptibles d'après la règle de
ce jeu. Annali di Matematica Pura ed
Applicata (2) 5 (1871) 63‑120.
Determines the number of linear arrangements of a double‑6 set of
dominoes, which gives the number of Euler circuits on K7.
L. Saalschütz. [Report of a lecture.] Schriften der Physikalisch‑Ökonomischen
Gesellschaft zu Königsberg 16 (1876) 23‑24. Sketches Euler's work, listing the seven bridges. Says that a recent railway bridge, of 1865,
connecting regions B and
C on Euler's diagram, can be
considered within the walkable region.
He shows there are 48 x 2 x
4 = 384 possible paths -- the
48 are the lists of regions
visited starting with A; the
2 corresponds to reversing these
lists; the 4 (= 2 x 2) corresponds
to taking each of the two pairs of bridges connecting the same regions in
either order, He lists the 48
sequences of regions which start at
A. I wrote a program to compute
Euler paths and I tested it on this situation.
I find that Saalschütz has omitted two cases, leading to four sequences
or 16
paths starting at A or
32 paths considering both
directions. That is, his 48
should be 52 and his
384 should be 416.
Kamp. Op. cit. in 5.B. 1877. Pp. 322‑327 show several unicursal
problems.
No. 8
is the five‑brick pattern as in Clausen.
No. 10
is two overlapping squares.
No. 11
is a diagram from which one must remove some lines to leave an Eulerian figure.
C. Hierholzer. Ueber die Möglichkeit, einen Linienzug ohne
Wiederholung und ohne Unterbrechung zu umfahren. Math. Annalen 6 (1873) 30‑32. (English is in BLW, 11‑12.)
G. Tarry. Géométrie de situation: Nombre de manières
distinctes de parcourir en une seule course toutes les allées d'un labyrinthe
rentrant, en ne passant qu'une seule fois par chacune des allées. Comptes Rendus Assoc. Franç. Avance. Sci.
15, part 2 (1886) 49‑53 & Plates I & II. General technique for the number of Euler
circuits.
Lucas. RM2. 1883. Le jeu de dominos -- Dispositions
rectilignes, pp. 63‑77
& Note 1: Sur le jeu de dominos, p. 229.
RM4. 1894.
La géométrie des réseaux et le problème des dominos, pp. 123‑151.
Cites
Reiss's work and says (in RM4) that it has been confirmed by Jolivald. The note in RM2 is expanded in RM4 to
explain the connection between dominoes and
K2n+1. There are
obviously 2 Euler circuits on K3. He sketches Tarry's method and uses it to
compute that K5 has
88 Euler circuits and K7 has 1299 76320. [This gives
28 42582 11840 domino rings for
the double-6 set.] He says Tarry has
found that K9 has
911 52005 70212 35200.
Tom Tit, vol. 3. 1893.
Le rectangle et ses diagonales, pp. 155-156. = K, no. 16: The rectangle and its diagonals, pp. 46‑48. = R&A, The secret of the rectangle, p.
100. Trick solutions by folding the
paper and making an arc on the back.
Hoffmann. 1893.
Chap. X, no. 9: Single‑stroke figures, pp. 338 & 375 =
Hoffmann-Hordern, pp. 230-231. Three
figures, including the double crescent 'Seal of Mahomet'. Answer states Euler's condition.
Dudeney. The shipman's puzzle. London Mag. 9 (No. 49) (Aug 1902) 88‑89 &
9 (No. 50) (Sep 1902) 219 (= CP, prob. 18, pp. 40‑41 &
173). Number of Euler circuits on K5.
Benson. 1904.
A geometrical problem, p. 255. Seal
of Mahomet.
William F. White. A Scrap‑Book of Elementary
Mathematics. Open Court, 1908. [The 4th ed., 1942, is identical in content
and pagination, omitting only the Frontispiece and the publisher's catalogue.] Bridges and isles, figure tracing, unicursal
signatures, labyrinths, pp. 170‑179.
On p. 174, he gives the five‑brick puzzle, asking for a route
along its edges.
Dudeney. Perplexities: No. 147: An old three‑line
puzzle. Strand Magazine 46 (Jul 1913)
110 & (Aug 1913) 221. c= AM,
prob. 239: A juvenile puzzle, pp. 68‑69 & 197. Five‑brick form to be drawn or rubbed
out on a board in three strokes. Either
way requires doing two lines at once, either by folding the paper as you draw
or using two fingers to rub out two lines at once. "I believe Houdin, the conjurer, was fond of showing this to
his child friends, but it was invented before his time -- perhaps in the Stone
Age."
Loyd. Problem of the bridges.
Cyclopedia, 1914, pp. 155 & 359‑360. = MPSL1, prob. 28, pp. 26‑27 & 130‑131. Eight bridges. Asks for number of routes.
Loyd. Puzzle of the letter carrier's route. Cyclopedia, 1914, pp 243 & 372. Asks for a circuit on a 3
x 4 array with a minimal length of
repeated path.
Dudeney. AM.
1917.
Prob.
242: The tube inspector's puzzle, pp. 69 & 198. Minimal route on a 3 x
4 array.
Prob.
261: The monk and the bridges, pp. 75-76 & 202-203. River with one island. Four bridges from island, two to each side
of the river, and another bridge over the river. How many Euler paths from a given side of the river to the
other? Answer: 16.
Collins. Book of Puzzles. 1927. The fly on the
octahedron, pp. 105-108. Asserts there
are 1488 Euler circuits on the edges of an octahedron. He counts the reverse as a separate circuit.
Harry Houdini [pseud. of Ehrich
Weiss] Houdini's Book of Magic. 1927 (??NYS); Pinnacle Books, NY, 1976, p.
19: Can you draw this? Take a square
inscribed in a circle and draw both diagonals.
"The idea is to draw the figure without taking your pencil off the
paper and without retracing or crossing a line. There is a trick to it, but it can be done. The trick in drawing the figure is to fold
the paper once and draw a straight line between the folded halves; then, not
removing your pencil, unfold the paper.
You will find that you have drawn two straight lines with one
stroke. The rest is simple." This perplexed me for some time, but I
believe the idea is that holding the pencil between the two parts of the folded
sheet and moving the pencil parallel to the fold, one can draw a line, parallel
to the fold, on each part.
Loyd Jr. SLAHP.
1928. Pp. 7‑8. Discusses what he calls the "Five‑brick
puzzle", the common pattern of five rectangles in a rectangle. He says that the object was to draw the
lines in four strokes -- which is easily done -- but that it was commonly
misprinted as three strokes, which he managed to do by folding the paper. He says "a similar puzzle ... some ten
or fifteen years ago" asked for a path crossing each of the 16 walls once,
which is also impossible.
The Bile Beans Puzzle Book. 1933.
No.
32. Draw the triangular array of three
on an edge without crossing.
No.
36. Draw the five-brick pattern in
three lines. Folds paper and draws two
lines at once.
R. Ripley. Believe It Or Not! Book 2. (Simon &
Schuster, 1931); Pocket Books, NY,
1948, pp. 70‑71. = Omnibus
Believe It Or Not! Stanley Paul, London, nd [c1935?], p. 270. Gives the five‑brick problem of
drawing a path crossing each wall once, with the trick solution having the path
going along a wall. Asserts "This
unicursal problem was solved thus by the great Euler himself." and cites
the Euler paper above!!
Meyer. Big Fun Book. 1940.
Tryangle,
pp. 98 & 731. Triangle subdivided
into triangles, with three small triangles along each edge. Draw an Euler circuit without crossings.
Cutting
the walls, pp. 637 & 794.
Five-brick problem. Solution has
line crossing through a vertex.
Ern Shaw. The Pocket Brains Trust - No. 2. W. H. Allen, London, nd but inscribed
1944. Prob. 29: Five bricks teaser, pp.
10 & 39.
Leeming. 1946.
Chap. 6, prob. 2: Through the walls, pp. 70 & 184. Five‑brick puzzle, with trick solution
having the path go through an intersection.
John Paul Adams. We Dare You to Solve This!. Op. cit. in 5.C. 1957? Prob. 49: In just
one line, pp. 30 & 48-49.
Five-brick puzzle, with answer having the path going along a wall, as in
Ripley. Asserts Euler invented this
solution.
Gibson. Op. cit. in 4.A.1.a. 1963.
Pp. 70 & 75: The "impossible" diagram. Same as Tom Tit.
Gardner. SA (Apr 1964) = 6th Book, chap. 10. Says Carroll knew that a planar Eulerian
graph could be drawn without crossings.
Gives a method of O'Beirne for doing this -- two colour the regions and
then make a path which separates the colours into simply connected regions.
Ripley's Puzzles and Games. 1966.
P. 39. Euler paths on the
'envelope', i.e. a rectangle with its diagonals drawn and an extra connection
between the top corners, looking like an unfolded envelope. Asserts the envelope has 50 solutions, but
it is not clear if the central crossing is a further vertex. I did this by hand but did not get 50, so I
wrote a program to count Euler paths.
If the central crossing is not a vertex, then I find 44 paths from one
of the odd vertices to the other, and of course 44 going the other way -- and I
had found this number by hand. However,
if the central crossing is a vertex, then my hand solution omitted some cases
and the computer found 120 paths from one odd vertex to the other.
Pp.
40-43 give many problems of drawing non-crossing Euler paths or circuits.
W. Wynne Willson. How to abolish cross‑roads. MTg 42 (Spring 1968) 56‑59. Euler circuit of a planar graph can be made
without crossings.
[Henry] Joseph & Lenore
Scott. Master Mind Brain Teasers. Tempo (Grosset & Dunlap), NY, 1973
(& 1978?? -- both dates are given -- I'm presuming the 1978 is a 2nd ptg or
a reissue under a different imprint??).
One line/no crossing, pp. 85-86.
Non-crossing Euler circuits on the triangular array of side 3 and
non-crossing Euler paths on the 'envelope' -- cf under Ripley's, above. Asserts the envelope has 50 solutions. I adapted the program mentioned above to
count the number of non-crossing Euler paths -- one must rearrange the first case
as a planar graph -- and there are
16 in the first case and 26
in the second case. Taking the
reversals doubles these numbers so it is possible that the Scotts meant the
second case and missed one path and its reversal.
David Singmaster, proposer; Jerrold W. Grossman & E. M. Reingold,
solvers. Problem E2897 -- An Eulerian
circuit with no crossings. AMM 88:7
(Aug 1981) 537-538 & 90:4 (Apr 1983) 287-288. A planar Eulerian graph can be drawn with no
crossings. Solution cites some previous
work.
Anon. [probably Will Shortz ??check with Shortz]. The impossible file. No. 2: In just one line. Games (Apr 1986) 34 & 64 & (Jul 1986) 64. Five brick pattern -- draw a line crossing
each wall once. Says it appeared in a
1921 newspaper [perhaps by Loyd Jr??].
Gives the 1921 solution where the path crosses a corner, hence two walls
at once. Also gives a solution with the
path going along a wall. In the July
issue, Mark Kantrowitz gives a solution by folding over a corner and also a
solution on a torus.
Marcia Ascher. Graphs in cultures: A study in
ethnomathematics. HM 15 (1988) 201‑227. Discusses the history of Eulerian circuits
and non-crossing versions and then exposits many forms of the idea in many
cultures.
Marcia Ascher. Ethnomathematics. Op. cit. in 4.B.10.
1991. Chapter Two: Tracing
graphs in the sand, pp. 30-65. Sketches
the history of Eulerian graphs with some interesting references -- ??NYS. Describes graph tracing in three cultures:
the Bushoong and the Tshokwe of central Africa and the Malekula of Vanuatu
(ex-New Hebrides). Extensive references
to the ethnographic literature.
This
section is mainly concerned with the theory.
The history of mazes is sketched first, with references to more detailed
sources. There is even a journal,
Caerdroia (53 Thundersley Grove, Thundersley, Essex, SS7 3EB, England), devoted
to mazes and labyrinths, mostly concentrating on the history. It is an annual, began in 1980 and issue 31
appeared in 2000.
Mazes
are considered under Euler Circuits, since the method of Euler Circuits is
often used to find an algorithm.
However, some mazes are better treated as Hamiltonian Circuits -- see
5.F.2.
A
maze can be considered as a graph formed by the nodes and paths -- the path
graph. For the usual planar maze, one
can also look at the graph formed by the walls -- the wall graph, which is a
kind of dual to the path graph. In
later mazes, the walls do not form a connected whole, and an isolated part of the
wall appears as a region or 'face' in the path graph. Such isolated bits of walling are sometimes called islands, but
they are the same as the components of the wall graph, with the outer wall
being one component, so the number of components is one more than the number of
islands. The 'hand-on-wall' method will
solve a maze if and only if the goals are adjacent to walls in the component of
the outer wall.
A
'ring maze' is a plate with holes and raised areas with an open ring which must
be removed by moving it from hole to hole.
I have put these in 11.K.5 as they are a kind of mechanical or
topological puzzle, though there are versions with a simple two legged spacer.
HISTORICAL
SOURCES
W. H. Matthews. Mazes & Labyrinths: A General Account of Their History and
Developments. Longmans, Green and Co.,
London, 1922. = Mazes and
Labyrinths: Their History and
Development. Dover, 1970. (21 pages of references.) [For more about the book and the author,
see: Zeta Estes; My Father, W. H.
Matthews; Caerdroia (1990) 6-8.]
Walter Shepherd. For Amazement Only. Penguin, 1942; Let's go amazing, pp. 5-12.
Revised as: Mazes and Labyrinths
-- A Book of Puzzles. Dover, 1961; Let's go a‑mazing, pp. v‑xi. (Only a few minor changes are made in the
text.) Sketch of the history.
Sven Bergling invented the
rolling ball labyrinth puzzle/game and they began being produced in 1946. [Kenneth Wells; Wooden Puzzles and Games;
David & Charles, Newton Abbot, 1983, p. 114.]
Walter Shepherd. Big Book of Mazes and Labyrinths. Dover, 1973, More amazement,
pp. vii-x. Extends the historical
sketch in his previous book, arguing that mazes with multiple choices perhaps
derive from Iron Age hill forts whose entrances were designed to confuse an
enemy.
Janet Bord. Mazes and Labyrinths of the World. Latimer, London, 1976. (Extensively illustrated.)
Nigel Pennick. Mazes and Labyrinths. Robert Hale, London, 1990.
Adrian Fisher [& Georg
Gerster (photographer)]. The Art of the
Maze. Weidenfeld and Nicolson, London,
1990. (Also as: Labyrinth; Solving the Riddle of the Maze;
Harmony (Crown Publishers), NY, 1990.)
Origins and History occupies pp. 11-56, but he also describes many
recent developments and innovations. He
has convenient tables of early examples.
Adrian Fisher & Diana Kingham. Mazes.
Shire Album 264. Shire,
Aylesbury, 1991.
Adrian Fisher & Jeff
Saward. The British Maze Guide. Minotaur Designs, St. Alban's, 1991.
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HISTORICAL
SKETCH
Up
to about the 16C, all mazes were unicursal, i.e. with no decision points. The word labyrinth is sometimes used to
distinguish unicursal mazes from others, but this distinction is not made consistently. Until about 1000, all mazes were of the
classical 'Cretan' seven-ring type shown above. (However, see Shepherd's point in his 1973 book, above.) The oldest examples are rock carvings, the
earliest being perhaps that in the Tomba del Labirinto at Luzzanas, Sardinia, c‑2000
[Fisher, pp. 12, 25, 26, with photo on p. 12].
(In fact, Luzzanas is a local name for an uninhabited area of fields, so
does not appear on any ordinary map. It
is near Benetutti. See my A
Mathematical Gazetteer or Mazing in Sardinia (Caerdroia 30 (1999) 17-21). Jeff Saward writes that current
archaeological feeling is that the maze is Roman, though the cave is probably
c-2000.) On pottery, there are
labyrinths on fragments, c‑1300, from Tell Rif'at, Syria [the first photos
of this appeared in [Caerdroia 30 (2000) 54-55]), and on tablets, c‑1200,
from Pylos. Fisher [p. 26] lists
the early examples. Staffen Lundán; The
labyrinth in the Mediterranean; Caerdroia 27 (1996) 28-54, catalogues all known
'Cretan' labyrinths from prehistory to the end of antiquity, c250, excluding
the Roman 'spoked' form. All these
probably had some mystical significance about the difficulty of reaching a
goal, often with substantial mythology -- e.g. Theseus in the Labyrinth or,
later, the Route to Jerusalem.
Roman
mosaics were unicursal but essentially used the Cretan form four times over in
the four corners. Lundán, above, calls
these 'spoked'. Most of the extant
examples are 2C‑4C, but some BC examples are known -- the earliest seems
to be c-110 at Selinunte, Sicily.
Fisher [pp. 36-37] lists all surviving examples. Saward says the earliest Roman example is at
Pompeii, so £ 79.
In
the medieval period, the Christians developed a quite different unicursal
maze. See Fisher [pp. 60-67] for
detailed comparison of this form with the Roman and Cretan forms. The earliest large Christian example is the
Chemin de Jerusalem of 1235 on the floor of Chartres Cathedral. Fisher [pp. 41 & 48] lists early and
later Christian examples.
The
legendary Rosamund's Bower was located in Woodstock Park, Oxfordshire, and its
purported site is marked by a well and fountain. It was some sort of maze to conceal Rosamund Clifford, the
mistress of Henry II (1133‑1189), from the Queen, Eleanor of
Aquitaine. Legend says that about 1176,
Eleanor managed to solve the maze and confronted Rosamund with the choice of a
dagger or poison -- she drank the poison and Henry never smiled again. [Fisher, p. 105]. Historically, Henry had imprisoned Eleanor for fomenting rebellion
by her sons and Rosamund was his acknowledged mistress. Rosamund probably spent her last days at a
nunnery in Godstow, near Oxford. The
legend of the bower dates from the 14C and her murder is a later addition
[Collins, Book of Puzzles, 1927, p. 121.]
In the 19C, many puzzle collections had a maze called Rosamund's Bower.
The
earliest record of a hedge maze is of one destroyed in a siege of Paris in
1431.
Non-unicursal
mazes and islands in the wall graph start to appear in the late 16C. Matthews [p. 96] says that: "A simple
"interrupted-circle" type of labyrinth was adopted as a heraldic
device by Gonzalo Perez, a Spanish ecclesiastic ... and published ... in 1566
..." in his translation of the Odyssey.
Matthews doesn't show this, but he then [pp. 96-97] describes and
illustrates a simple maze used as a device by Bois-dofin de Laval, Archbishop
of Embrum. He copies it from Claude
Paradin; Devises Héroiques et Emblèmes of the early 17C. It has four entrances and possibly three
goals, with walls having 8 components, two being part of the outer wall. The central goals is accessible from two of
the entrances, but the two minor goals are each accessible from just one of the
other entrances. Presumably this sort
of thing is what Matthews meant as an "interrupted circle".
However,
Saward has found a mid 15C anonymous English poem, The Assembly of Ladies,
which describes the efforts of a group of ladies to reach the centre of a maze,
which, as he observes, implies there must be some choices involved.
[Matthews,
p. 114] has three examples from a book by Androuet du Cerceau; Les Plus
Excellents Bastiments de France of 1576.
Fig. 82 was in the gardens at Charleval and has four entrances, only one
of which goes to the central goal.
There are four minor goals. The
N entrance connects to the NE and SE goals, with several dead ends. The E entrance is a dead end. The S entrance goes to the SW goal. The W entrance goes to the central goal, but
the NW goal is on an island, though 'left-hand-on-wall' goes past it. Figs. 83 and 84 are essentially identical
and seem to be corruptions of unicursal examples so that most of the maze is
bypassed. In fig. 84, one has to walk
around to the back of the maze to find the correct entrance to get to the
central goal, which is an interesting idea.
A small internal change in both cases and moving the entrances converts
them to a standard unicursal pattern.
Matthews'
Chap. XIII [pp. 100-109] is on floral mazes and reproduces some from Jan
Vredeman De Vries; Hortorum Viridariorumque Formae; Antwerp, 1583. Fig. 74 is one of these and has two
components and a short dead-end, but the 'hand-on-wall' rule solves it. Fig. 73 is another of De Vries's, but
it is not all shown. It appears to have
two entrances and there is certainly a decision point by the far gate, but one
route goes to the apparent exit at the bottom of the page. There is a small dead end near the central
goal. Fig. 78 shows a maze from a 17C
manuscript book in the Harley Manuscripts at the BL, identified on p. 224 as
BM Harl. 5308 (71, a, 12). This
has two components with the central goal in the inner component, so the
'hand-on-wall' rule fails, but it brings you within sight of the centre and
Matthews describes it as unicursal!
Fig. 79 is from Adam Islip; The Orchard and the Garden, compiled from
continental sources and published in 1602.
It has 5 components, but four of these are small enclosures which could
be considered as minor goals, especially if they had seats in them. The 'hand-on-wall' rule gets to the central
goal. There is a lengthy dead end which
goes to two of the inner islands. Fig.
80 is from a Dutch book: J. Commelyn; Nederlantze Hesperides of 1676. It has two components, a central goal and
four minor goals. The 'hand-on-wall'
gets you to the centre and passes two minor goals. One minor goal is on a dead end so 'left-hand-on-wall' gets to
it, but 'right-hand-on-wall' does not.
The fourth minor goal is on the island.
At
Versailles, c1675, André Le Nôtre built a Garden Maze, but the objective was to
visit, in correct order, 40 fountains based on Aesop's Fables. Each node of the maze had at least one
fountain. Some fountains were not at
path junctions, but one can consider these as nodes of degree two. This is an early example of a Hamiltonian
problem, except that one fountain was located at the end of a short dead
end. [Fisher, pp. 49, 79, 130 &
144-145, with contemporary map on p. 144.
Fisher says there are 39 fountains, and the map has 40. Close examination shows that the map counts
two statues at the entrance but omits to count a fountain between numbers 37
and 38. Matthews, pp. 117‑121,
says it was built by J. Hardouin-Mansart and his map has 39 fountains.] It has a main entrance and exit but there is
another exit, so the perimeter wall already has three components, and there are
14 other components. Sadly, it was
destroyed in 1775.
Several
other mazes, of increasing complexity, occur in the second half of the 17C
[Matthews, figs. 93-109, opp. p. 120 - p. 127]. Several of these could be from 20C maze books. Fig. 94, designed for Chantilly by Le Nôtre,
is surprisingly modern in that there are eight paths spiralling to the
centre. The entrance path takes you
directly to the centre, so the real problem is getting back out! One of the mazes presently at Longleat has
this same feature.
The
Hampton Court Maze, planted c1690, is the oldest extant hedge maze and one of
the earliest puzzle mazes.
([Christopher Turner; Hampton Court, Richmond and Kew Step by Step;
(As part of: Outer London Step by Step, Faber, 1986); Revised and published in
sections, Faber, 1987, p. 16] says the present shape was laid out in 1714,
replacing an earlier circular shape, but I haven't seen this stated
elsewhere.) Matthews [p. 128] says it
probably replaced an older maze. It has
dead ends and one island, i.e. the graph has two components, though the 'hand
on wall' rule will solve it.
The
second Earl Stanhope (1714-1786) is believed to be the first to design mazes
with the goal (at the centre) surrounded by an island, so that the 'hand on
wall' rule will not solve it. It has
seven components and only a few short dead ends.. The fourth Earl planted one of these at Chevening, Kent, in c1820
and it is extant though not open to the public. [Fisher, p. 71, with photo on p. 72 and diagram on p. 73.] However, investigation in Matthews revealed
the earlier examples above. Further
Bernhard Wiezorke (below at 2001) has found a hedge maze in Germany, dating
from c1730, which is not solved by the 'hand on wall' rule. This maze has 12 components.
In
1973, Stuart Landsborough, an Englishman settled at Wanaka, South Island, New
Zealand, began building his Great Maze.
This was the first of the board mazes designed by Landsborough which
were immensely popular in Japan. Over
200 were built in 1984-1987, with 20 designed by Landsborough. Many of these were three dimensional -- see
below. About 60 have been demolished
since then. [Fisher, pp. 78‑79
& 118-121 has 6 colour photos, pp. 156-157 lists Landsborough's designs.]
If Minos' labyrinth ever really
existed, it may have been three dimensional and there may have been garden
examples with overbridges, but I don't know of any evidence for such early
three dimensional mazes. Lewis Carroll
drew mazes which had paths that crossed over others making a simple three
dimensional maze, in his Mischmasch of c1860, see below. John Fisher [The Magic of Lewis Carroll;
(Nelson, 1973), Penguin, 1975, pp. 19-20] gives this and another example. Are there earlier examples? Boothroyd & Conway, 1959, seems to be
the earliest cubical maze. Much more
complex versions were developed by Larry Evans from about 1970 and published in
a series of books, starting with 3-Dimensional Mazes (Troubador Press, San
Francisco, 1976). His 3‑Dimensional
Maze Art (Troubador, 1980) sketches some general history of the maze and
describes his development of pictures of three dimensional mazes. The first actual three dimensional maze
seems to be Greg Bright's 1978 maze at Longleat House, Warminster. [Fisher, pp. 74, 76, 94-95 & 152-153,
with colour photos on pp. 94-95.]
Since then, Greg Bright, Adrian Fisher, Randoll Coate, Stuart
Landsborough and others have made many innovations. Bright seems to have originated the use of colour in mazes c1980
and Fisher has extensively developed the idea.
[Fisher, pp. 73-79.]
Abu‘l-Rayhan Al‑Biruni (=
’Abû-alraihân [the h should have an underdot] Muhammad ibn ’Ahmad
[the h
should have an underdot] Albêrûnî).
India. c1030. Chapter XXX. IN: Al‑Beruni's India, trans. by E. C. Sachau, 2 vols.,
London, 1888, vol. 1, pp. 306-307
(= p. 158 of the Arabic ed., ??NYS). In
describing a story from the fifth and sixth books of the Ramayana, he says that
the demon Ravana made a labyrinthine fortress, which in Muslim countries
"is called Yâvana-koti, which has been frequently explained as
Rome." He then gives "the
plan of the labyrinthine fortress", which is the classical Cretan
seven-ring form. Sachau's notes do not
indicate whether this plan is actually in the Ramayana, which dates from
perhaps -300.
Pliny. Natural History.
c77. Book 36, chap. 19. This gives a brief description of boys
playing on a pavement where a thousand steps are contained in a small
space. This has generally been
interpreted as referring to a maze, but it is obviously pretty vague. See: Michael Behrend; Julian and Troy names;
Caerdroia 27 (1996) 18-22, esp. note 5 on p. 22.
Pacioli. De Viribus.
c1500. Part II: Cap.
(C)XVII. Do(cumento). de saper fare
illa berinto con diligentia secondo Vergilio, f. 223v = Peirani 307-308. A
sheet (or page) of the MS has been lost.
Cites Vergil, Æneid, part six, for the story of Pasiphæ and the
Minotaur, but the rest is then lost.
Sebastiano Serlio. Architettura, 5 books, 1537-1547. The separate books had several editions
before they were first published together in 1584. The material of interest is in Book IV which shows two unicursal
mazes for gardens. I have seen the
following.
Tutte
l'Opere d'Architetture et Prospetiva, ....
Giacomo de'Franceschi, Venice, 1619;
facsimile by Gregg Press, Ridgewood, New Jersey, 1964. F. 199r shows the designs and f. 197v has
some text, partly illegible in my photocopy.
[Cf Caerdroia 30 (1999) 15.]
Sebastiano
Serlio on Architecture Volume One Books I-V of 'Tutte l'Opere d'Architettura
et Prospetiva'. Translated and edited
by Vaughan Hart and Peter Hicks. Yale
Univ. Press, New Haven, 1996. P. 388
shows the designs and p. 389 has the text, saying these 'are for the
compartition of gardens'. The sidenotes
state that these pages are ff. LXXVr and LXXIIIIr of the 3rd ed. of 1544 and
ff. 198v-199r and 197v-198r of the 1618/19 ed.
William Shakespeare. A Midsummer Night's Dream. c1610.
Act II, scene I, lines 98-100:
"The nine men's morris is fill'd up with mud, And the quaint mazes in the wanton
green For lack of tread are
undistinguishable." Fiske 126
opines that the latter two lines may indicate that the board was made in the
turf, though he admits that they may refer just to dancers' tracks, but to me
it clearly refers to turf mazes.
John Cooke. Greene's Tu Quoque; or the Cittie Gallant; a
Play of Much Humour. 1614. ??NYS -- quoted by Matthews, p. 135. A challenge to a duel is given by Spendall
to Staines.
Staines. I accept it ; the meeting place?
Spendall. Beyond the maze in Tuttle.
This
refers to a maze in Tothill Fields, close to Westminster Abbey.
Lewis Carroll. Untitled maze. In: Mischmasch, the last of his youthful MS magazines, with
entries from 1855 to 1862. Transcribed
version in: The Rectory Umbrella and
Mischmasch; Cassell, 1932; Dover, 1971; p. 165 of the Dover ed. John Fisher [The Magic of Lewis Carroll;
(Nelson, 1973), Penguin, 1975, pp. 19-20] gives this and another example. Cf Carroll-Wakeling, prob. 35: An amazing
maze, pp. 46-47 & 75 and Carroll-Gardner, pp. 80-81 for the Mischmasch
example. I don't find the other example
elsewhere, but it was for Georgina "Ina" Watson, so probably c1870.
Mittenzwey. 1880.
Prob. 281, pp. 50 & 100;
1895?: 310, pp. 53-54 & 102;
1917: 310, pp. 49 & 97.
The garden of a French place has a maze with 31 points to see. Find a path past all of them with no
repeated edges and no crossings. The
pattern is clearly based on the Versailles maze of c1675 mentioned in the Historical
Sketch above, but I don't recall the additional feature of no crossings
occurring before.
C. Wiener. Ueber eine Aufgabe aus der Geometria
situs. Math. Annalen 6 (1873) 29‑30. An algorithm for solving a maze. BLW asserts this is very complicated, but it
doesn't look too bad.
M. Trémaux. Algorithm.
Described in Lucas, RM1, 1891, pp. 47‑51. ??check 1882 ed. BLW assert Lucas' description is faulty. Also described in MRE, 1st ed., 1892, pp. 130‑131; 3rd ed., 1896, pp. 155-156; 4th ed., 1905, pp. 175-176 is vague; 5th-10th ed., 1911‑1922, 183; 11th ed., 1939, pp. 255‑256 (taken
from Lucas); (12th ed. describes
Tarry's algorithm instead) and in Dudeney, AM, p. 135 (= Mazes, and how to
thread them, Strand Mag. 37 (No. 220) (Apr 1909) 442‑448, esp. 446‑447).
G. Tarry. Le problème des labyrinthes. Nouv. Annales de Math. (3) 4 (1895) 187‑190. ??NYR
Collins. Book of Puzzles. 1927. How to thread any
maze, pp. 122-124. Discusses right hand
rule and its failure, then Trémaux's method.
M. R. Boothroyd &
J. H. Conway. Problems drive,
1959. Eureka 22 (Oct 1959) 15-17 &
22-23. No. 2. 5 x 5 x 5 cubical
maze. Get from a corner to an antipodal
corner in a minimal number of steps.
Anneke Treep. Mazes... How to get out! (part I).
CFF 37 (Jun 1995) 18-21. Based
on her MSc thesis at Univ. of Twente.
Notes that there has been very little systematic study. Surveys the algorithms of Tarry, Trémaux,
Rosenstiehl. Rosenstiehl is greedy on
new edges, Trémaux is greedy on new nodes and Trémaux is a hybrid of
these. ??-oops-check. Studies probabilities of various routes and
the expected traversal time. When the
maze graph is a tree, the methods are equivalent and the expected traversal
time is the number of edges.
Bernhard Wiezorke. Puzzles und Brainteasers. OR News, Ausgabe 13 (Nov 2001) 52-54. This reports his discovery of a hedge maze
in Germany -- the first he knew of. It
is in Altjessnitz, near Dessau in Sachsen-Anhalt. (My atlas doesn't show such a place, but Jessnitz is about 10km
south of Dessau.) This maze dates from
1720 and has 12 components, with the goal completely separated from the outside
so that the 'hand on wall' rule does not solve it. Torsten Silke later told Wiezorke of two other hedge mazes in
Germany. One, in Probststeierhagen,
Schleswig-Holstein, about 12km NE of Kiel, is in the grounds of the restaurant
Zum Irrgarten (At the Labyrinth) and is an early 20C copy of the Altjessnitz
example. The other, in Kleinwelka,
Sachsen, about 50km NE of Dresden, was made in 1992 and is private. Though it has 17 components, the 'hand on
wall' method will solve it. He gives
plans of both mazes. He discusses the
Seven Bridges of Königsberg, giving a B&W print of the 1641 plan of the
city mentioned at the beginning of Section 5.E -- he has sent me a colour
version of it. He also describes
Tremaux's solution method.
5.E.2. MEMORY WHEELS = CHAIN CODES
These
are cycles of 2n 0s
and 1s such that each n‑tuple
of 0s
and 1s appears just once. They
are sometimes called De Bruijn sequences, but they have now been traced back to
the late 19C. An example for n = 3
is 00010111.
Émile Baudot. 1884.
Used the code for 25 in telegraphy. ??NYS -- mentioned by Stein.
A. de Rivière, proposer; C. Flye Sainte-Marie, solver. Question no. 58. L'Intermédiare des Mathématiciens 1 (1894) 19-20 & 107-110. ??NYS -- described in Ralston and
Fredricksen (but he gives no. 48 at one point). Deals with the general problem of a cycle of kn symbols such that every n‑tuple
of the k basic symbols occurs just once.
Gives the graphical method and shows that such cycles always exist and
there are k!g(n)/ kn of them,
where g(n) = kn‑1. This work was unknown to the following authors until about 1975.
N. G. de Bruijn. A combinatorial problem. Nederl. Akad. Wetensch. Proc. 49 (1946) 758‑764. ??NYS -- described in Ralston and
Fredricksen. Gives the graphical method
for finding examples and finds there are
2f(n) solutions,
where f(n) = 2n-1
- n.
I. J. Good. Normal recurring decimals. J. London Math. Soc. 21 (1946) 167-169. ??NYS -- described in Ralston and
Fredricksen. Shows there are solutions
but doesn't get the number.
R. L. Goodstein. Note 2590:
A permutation problem. MG 40
(No. 331) (Feb 1956) 46‑47.
Obtains a kind of recurrence for consecutive n‑tuples.
Sherman K. Stein. Mathematics: The Man‑made Universe.
Freeman, 1963. Chap. 9: Memory
wheels. c= The mathematician as
explorer, SA (May 1961) 149‑158.
Surveys the topic. Cites the
c1000 Sanskrit word: yamátárájabhánasalagám used as the mnemonic for 01110100(01) giving all triples of short and long beats in Sanskrit poetry and
music. Describes the many reinventions,
including Baudot (1882), ??NYS, and the work of Good (1946), ??NYS, and de
Bruijn (1946), ??NYS. 15 references.
R. L. Goodstein. A generalized permutation problem. MG 54 (No. 389) (Oct 1970) 266‑267. Extends his 1956 note to find a cycle
of an symbols such that the n‑tuples are distinct.
Anthony Ralston. De Bruijn sequences -- A model example of
the interaction of discrete mathematics and computer science. MM 55 (1982) 131‑143 & cover. Deals with the general problem of cycles
of kn symbols such that every n‑tuple of the k
basic symbols occurs just once.
Discusses the history and various proofs and algorithms which show that
such cycles always exist. 27
references.
Harold Fredricksen. A survey of full length nonlinear shift
register cycle algorithms. SIAM Review
24:2 (Apr 1982) 195-221. Mostly about
their properties and their generation, but includes a discussion of the door
lock connection, a mention of using the
23 case as a switch
for three lights, and gives a good history.
The door lock connection is that certain push button door locks will
open when a four digit code is entered, but they open if the last four buttons
pressed are the correct code, so using a chain code reduces the number of
button pushes required by a burglar to
1/4 of the number required if he
tries all four digit combinations. 58
references.
At G4G2, 1996, Persi Diaconis
spoke about applications of the chain code in magic and mentioned uses in
repeated measurement designs, random number generators, robot location, door
locks, DNA comparison.
They
were first used in card tricks by Charles T. Jordan in 1910. Diaconis' example had a deck of cards which
were cut and then five consecutive cards were dealt to five people in a
row. He then said he would determine
what cards they had, but first he needed some help so he asked those with red
cards to step forward. The position of
the red cards gives the location of the five cards in a cycle of 32
(which was the size of the deck)!
Further, there are simple recurrences for the sequence so it is fairly
easy to determine the location. One can
code the binary quintuples to give the suit and value of the first card and
then use the succeeding quintuples for the succeeding cards.
Long
versions of the chain code are printed on factory floors so that a robot can
read it and locate itself.
In Jan 2000, I discussed the
Sanskrit chain code with a Sanskrit scholar, Dominik Wujastyk, who said that
there is no known Sanskrit source for it.
He has asked numerous pandits who did not know of it and he said there
is is a forthcoming paper on it, but that it did not locate any Sanskrit
source.
Haubrich's
1995-1996 surveys, op. cit. in 5.H.4, include this.
B. Astle. Pantactic squares. MG 49 (No. 368) (May 1965) 144‑152. This is a two‑dimensional version of
the memory wheel. Take a 5 x 5
array of cells marked 0 or
1 (or Black or White). There are
16 ways to take a 2 x 2
subarray from the 5 x 5 array.
If these give all 16 2 x 2
binary patterns, the array is called pantactic. The author shows a number of properties and
some types of such squares.
C. J. Bouwkamp, P. Janssen &
A. Koene. Note on pantactic
squares. MG 54 (No. 390) (Dec 1970) 348‑351. They find 800 such squares, forming 50
classes of 16 forms.
[Surprisingly, neither paper
considers a 4 x 4 array viewed toroidally, which is the
natural generalization of the memory wheel.
Precisely two of the fifty classes, namely nos. 25 & 41, give such a
solution and these are the same pattern on the torus. One can also look at the
4 x 4 subarrays of a 131 x 131
or a 128 x 128 array, etc., as well as 3 and higher
dimensional arrays. I submitted the
question of the existence and numbers of these as a problem for CM, but it was
considered too technical.]
Ivan Moscovich. US Patent 3,677,549 -- Board Game
Apparatus. Applied: 14 Jun 1971; patented: 18 Jul 1972. Front page, 1p diagrams, 2pp text. Reproduced in Haubrich, About ..., 1996, op.
cit. in 5.H.4. 2pp + 2pp diagrams. This uses the 16 2 x 2 binary patterns as game pieces. He allows the pieces to be rotated, scoring
different values according to the orientation.
No mention of reversing pieces or of the use of the pieces as a puzzle.
John Humphries. Review of Q-Bits. G&P 54 (Nov 1976) 28.
This is Moscovich's game idea, produced by Orda. Though he mentions changing the rules to
having non-matching, there is no mention of two-sidedness.
Pieter van Delft & Jack
Botermans. Creative Puzzles of the
World. (As: Puzzels uit de hele wereld; Spectrum Hobby, 1978); Harry N. Abrams, NY, 1978. The colormatch square, p. 165. See Haubrich,1994, for description.
Jacques Haubrich. Pantactic patterns and puzzles. CFF 34 (Oct 1994) 19-21. Notes the toroidal property just mentioned. Says Bouwkamp had the idea of making the 16
basic squares in coloured card and using them as a MacMahon-type puzzle, with
the pieces double-sided and such that when one side had MacMahon matching, the
other side had non-matching. There are
two different bijections between matching patterns and non-matching patterns,
so there are also 800 solutions in 50 classes for the non-matching
problem. Bouwkamp's puzzle appeared in
van Delft & Botermans, though they did not know about and hence did not
mention the double-sidedness. [In an
email of 22 Aug 2000, Haubrich says he believes Bouwkamp did tell van Delft and
Botermans about this, but somehow it did not get into their book.] The idea was copied by two manufacturers
(Set Squares by Peter Pan Playthings and Regev Magnetics) who did not
understand Bouwkamp's ideas -- i.e. they permitted pieces to rotate. Describes Verbakel's puzzle of 5.H.2.
Jacques Haubrich. Letter: Pantactic Puzzles = Q-Bits. CFF 37 (Jun 1995) 4. Says that Ivan Moscovich has responded that
he invented the version called "Q-Bits" in 1960-1964, having the same
tiles as Bouwkamp's (but only one-sided [clarified by Haubrich in above
mentioned email]). His US Patent
3,677,549 (see above) is for a game version of he idea. The version produced by Orda Ltd. was
reviewed in G&P 54 (Nov 1976) (above).
So it seems clear that Moscovich had the idea of the pieces before
Bouwkamp's version was published, but Moscovich's application was to use them
in a game where the orientations could be varied.
For
queen's, bishop's and rook's tours, see 6.AK.
A
tour is a closed path or circuit.
A
path has end points and is sometimes called an open tour.
5.F.1. KNIGHT'S TOURS AND PATHS
GENERAL
REFERENCES
Antonius van der Linde. Geschichte und Literatur des
Schachspiels. (2 vols., Springer, Berlin,
1874); one vol. reprint, Olms, Zürich,
1981. [There are two other van der
Linde books: Quellenstudien zur
Geschichte des Schachspiels, Berlin, 1881, ??NYS; and Das Erste Jartausend
[sic] der Schachlitteratur (850‑1880),
(Berlin, 1880); reprinted with some notes and corrections, Caissa Limited
Editions, Delaware, 1979, which is basically a bibliography of little use
here.]
Baron Tassilo von Heydebrand und
von der Lasa. Zur Geschichte und
Literatur des Schachspiels.
Forschungen. Leipzig, 1897. ??NYS.
Ahrens. MUS I.
1910. Pp. 319-398.
Harold James Ruthven
Murray. A History of Chess. OUP, 1913;
reprinted by Benjamin Press, Northampton, Massachusetts, nd [c1986]. This has many references to the problem,
which are detailed below.
Reinhard Wieber. Das Schachspiel in der arabischen Literatur
von den Anfängen bis zur zweiten Hälfte des 16.Jahrhunderts. Verlag für Orientkunde Dr. H. Vorndran,
Walldorf‑Hessen, 1972.
George P. Jelliss.
Special
Issue: Notes on the Knight's Tour. Chessics 22 (Summer 1985) 61‑72.
Further
notes on the knight's tour. Chessics 25
(Spring 1986) 106‑107.
Notes
on Chessics 22 continued. Chessics 29
& 30 (1987) 160.
This is a progress report on his
forthcoming book on the knight's tour.
I will record some of his comments at the appropriate points below. He also studies the 3 x n
board extensively.
Two
problems with knights on a 3 x 3 board are generally treated here, but cf
5.R.6.
The
4 knights problem has two W and two B knights at the corners (same colours at
adjacent corners) and the problem is to exchange them in 16 moves. The graph of knight's connections is an
8-cycle with the pieces at alternate nodes.
[Putting same colours at opposite corners allows a solution in 8 moves.]
The
7 knights problem is to place 7 knights on a
3 x 3 board in the 4 corners and
3 of the sides so each is a knight's move from the previously placed one. This is equivalent to the octagram puzzle of
5.R.6.
4
knights problem -- see: at‑Tilimsâni,
1446; Civis Bononiae, c1475;
7
knights problem -- see: King's Library
MS.13, A.xviii, c1275; "Bonus
Socius", c1275; at‑Tilimsâni,
1446;
Al‑Adli (c840) and as‑Suli
(c880‑946) are the first two great Arabic chess players. Although none of their works survive, they
are referred to by many later writers who claim to have used their material.
Rudraţa:
Kāvyālaʼnkāra [NOTE:
ţ denotes a t
with an underdot and ʼn denotes an
n with an overdot.]. c900.
??NYS -- described in Murray 53‑55, from an 1896 paper by Jacobi,
??NYS. The poet speaks of verses which
have the shapes of "wheel, sword, club, bow, spear, trident, and plough,
which are to be read according to the chessboard squares of the chariot [=
rook], horse [= knight], elephant [c= bishop], &c." According to Jacobi, the poet placed
syllables in the cells of a half chessboard so that it reads the same straight
across as when following a piece's path.
With help from the commentator Nami, of 1069, the rook's and knight's
path's are reconstructed, and are given on Murray 54. Both are readily extended to full board paths, but not
tours. The elephant's path is confused.
Kitâb ash‑shatranj mimma’l‑lafahu’l‑‘Adli
waş‑Şûlî wa ghair‑huma [Book of the Chess; extracts from
the works of al‑'Adlî, aş‑Şûlî and others]. [NOTE:
ş, Ş denote
s, S with underdot.] Copied by
Abû Ishâq [the h should have an underdot] Ibrâhîm ibn al‑Mubârak
ibn ‘Alî al‑Mudhahhab al‑Baghdâdî.
Murray 171‑172 says it is MS ‘Abd‑al‑Hamid
[the H
should have an underdot] I, no. 560, of 1140, and denotes it AH. Wieber 12‑15 says it is now MS Lala
Ismail Efendi 560, dates it July‑August 1141, and denotes it L. Both cite van der Linde, Quellenstudien, no.
xviii, p. 331+, ??NYS. The author is
unknown. This MS was discovered in
1880. Catalogues in Istanbul listed it
as Risâla fi’sh‑shaţranj by Abû’l‑‘Abbâs Ahmad [the h
should have an underdot] al‑‘Adlî. It is sometimes attributed to al‑Lajlâj who wrote one short
section of this book. Murray, van der
Linde and Wieber (p. 41) cite another version:
MS Khedivial Lib., Cairo, Mustafa Pasha, no. 8201, copied c1370, which
Murray denotes as C and Wieber lists as unseen.
Murray
336 gives two distinct tours: AH91 & AH92.
The solution of AH91 is a numbered diagram, but AH92 is 'solved' four times
by acrostic poems, where the initial letters of the lines give the tour in an
algebraic notation. Wieber 479‑480
gives 2 tours from ff. 74a‑75b: L74a = AH91 and L74b = reflection of
AH92. [Since the 'solutions' of AH92
are poetic, it is not unreasonable to consider the reflection as
different.] Also AH94 = L75b is a
knight/bishop tour, where moves of the two types alternate. These tours may be due to as‑Suli. AH196 is a knight/queen tour.
Arabic MS Atif Efendi 2234
(formerly Vefa (‘Atîq Efendî) 2234), Eyyub, Istanbul. Copied by Muhammad [the
h should have an underdot] ibn
Hawâ (or Rahwâr -- the MS is obscure) ibn ‘Othmân al‑Mu’addib in
1221. Murray 174‑175 describes it
as mostly taken from the above book and denotes it V. A tour is shown on p. 336 as V93 = AH92. Wieber 20‑24 denotes it A. On p. 479, he shows the tour from f. 68b
which is the same as L74b, the reflection of AH92.
King's Library MS.13, A.xviii,
British Museum, in French, c1275. Described in van der Linde I 305‑306. Described and transcribed in Murray 579‑582
& 588‑600, where it is denoted as K.
Van der Linde discusses the knight's path on I 295, with diagram no. 244
on p. 245. Murray 589 gives the text
and a numbered diagram of a knight's path as K1. The path splits into two half board paths: a1 to d1 and e3 to h1, so the first half and
the whole are corner to corner. The
first half is also shown as diagram K2 with the half board covered with pieces
and the path described by taking of pieces.
K3 is the 7 knights problem
"Bonus Socius"
[perhaps Nicolas de Nicolaï]. This is
the common name of a collection of chess problems, assembled c1275, which was
copied and translated many times. See
Murray 618‑642 for about 11 MSS.
Some of these are given below.
Fiske 104 & 110‑111 discusses some MSS of this collection.
MS
Lat. 10286, Nat. Lib., Paris.
c1350. Van der Linde I 293‑295
describes this but gives the number as 10287 (formerly 7390). Murray 621 describes it and denotes it
PL. Van der Linde describes a half board
knight's path, with a diagram no. 243 shown on p. 245. The description indicates a gap in the path
which can only be filled in one way.
This is a path from a8 to h8 which cannot be extended to the full
board. Murray 641 says that PL275
is the same as problems in two similar MSS and as CB244, diagrammed on p.
674. However, this is not the same as
van der Linde's no. 243, though cells 1‑19 and 31‑32 are the same
in both paths, so this is also an a8 to h8 path which does not extend to a full
board.
Murray
620 mentions a path in a late Italian MS version of c1530 (Florence, Nat. Lib.
XIX.7.51, which he denotes It) which may be the MS described by van der Linde
I 284 as no. 4 and the half board path described on I 295 with diagram no.
245 on I 245. Fiske 210-211 describes
this and says von der Lasa 163-165 (??NYS) describes it as early 16C, but
Murray does not mention von der Lasa.
Fiske says it contains a tour on f. 28b, which von der Lasa claims
is "das älteste beispiel eines vollkommenen rösselsprunges", but
Murray does not detail the problems so I cannot compare these citations. Fiske also says it also contains the 7
knights problem.
Dresden MS 0/59, in French,
c1400. Murray describes this on pp. 607‑613
and denotes it D. On p. 609, Murray
describes D57 which asks for a knight's path on a 4 x 4 board. No solution is given -- indeed this is
impossible, cf Persian MS 211 in the RAS.
Ibid. is D62 which asks for a half board tour, but no answer is
provided.
Persian MS 211 in Royal Asiatic
Society. Early 15C. ??NYS.
Extensively
described as MS 250 bequeathed by Major David Price in: N. Bland; On the Persian game of chess; J.
Royal Asiatic Soc. 13 (1852) 1‑70.
He dates it as 'at least 500 years old' and doesn't mention the knight's
tour.
Described,
as MS No. 260, and partially translated in Duncan Forbes; The History of Chess;
Wm. H. Allen, London, 1860. Forbes says
Bland's description is "very detailed but unsatisfactory". On p. 82 is the end of the translation of
the preface: '"Finally I will show
you how to move a Knight from any individual square on the board, so that he
may cover each of the remaining squares in as many moves and finally come to
rest on that square whence he started.
I will also show how the same thing may be done by limiting yourself
only to one half, or even to one quarter (1) of the board." -- Here the
preface abruptly terminates, the following leaf being lost.' Forbes's footnote (1) correctly doubts that
a knight's tour (or even a knight's path) is possible on the 4 x 4
board.
Murray
177 cites it as MS no. 211 and denotes it RAS.
He says that it has been suggested that this MS may be the work of
‘Alâ'addîn Tabrîzî = ‘Alî ash‑Shatranjî, late 14C, described on Murray
171. Murray mentions the knight's tour
passage on p. 335. This may be in van
der Linde, ??NX. Wieber 45 mentions the
MS.
Abû Zakarîyâ Yahya [the h
should have an underdot] ibn Ibrâhîm al‑Hakîm[the H
should have an underdot]. Nuzhat
al‑arbâb al‑‘aqûl fî’sh‑shaţranj [NOTE: ţ
denotes a t with an underdot] al‑manqûl (The
delight of the intelligent, a description of chess). Arabic MS 766, John Rylands Library, Manchester.
Bland,
loc. cit., pp. 27‑28, describes this as no. 146 of Dr. Lee's catalogue
and no. 76 of the new catalogue.
Forbes, loc. cit., says that Dr. Lee had loaned his two MSS to someone
who had not yet returned them, so Forbes copies Bland's descriptions (on
pp. 27‑31) as his Appendix C, with some clarifying notes. (The other of Dr. Lee's MSS is described
below.) Van der Linde I 107ff (??NX)
seems to copy Bland & Forbes.
Murray
175‑176 describes it as Arab. 59 at John Rylands Library and denotes it
H. He says it was Bland who had
borrowed the MSS from Dr. Lee and Murray traces their route to Dr. Lee and to
Manchester. Murray says it is late 15C,
is based on al‑Adli and as‑Suli and he also describes a later
version, denoted Z, late 18C. Wieber 32‑35
cites it as MS 766(86) at John Rylands, dates it 1430 and denotes it Y1.
Murray
336 gives three paths. H73 = H75 are
the same tour, but with different keys, one poetic as in Rudraţa
[NOTE: ţ denotes a t with an underdot.], one numeric. H74 is a path attributed to Ali Mani with
similar poetic solution. Wieber 480
shows two diagrams. Y1‑39a, Y1‑39b,
Y1‑41b are the same tour as H73, but with different descriptions, the
latter two being attributed to al‑Adli.
Y1‑39a (second diagram) = H74 is attributed to ‘Ali ibn Mani.
Shihâbaddîn Abû’l‑‘Abbâs Ahmad [the h should have an
underdot] ibn Yahya [the h should have an underdot] ibn Abî Hajala
[the H
should have an underdot] at‑Tilimsâni alH‑anbalî [the H
should have an underdot]. Kitâb
’anmûdhaj al‑qitâl fi la‘b ash‑shaţranj [NOTE: ţ
denotes a t with an underdot] (Book of the examples of
warfare in the game of chess). Copied
by Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî in 1446.
Bland,
loc. cit., pp. 28‑31, describes this as the second of Dr. Lee's MSS, old
no. 147, new no. 77. Forbes copies
this and adds notes. Van der Linde I
105‑107 seems to copy from Bland and Forbes. Murray 176‑177 says the author died in 1375, so this might
be c1370. He says it is Dr. Lee's on
175‑176, that it is MS Arab. 93 at the John Rylands Library and denotes
it Man. Wieber 29‑32 cites it as
MS 767(59) at the Rylands Library and denotes it H. On p. 481, he shows a half‑board path which cannot be
extended to the full board.
This
MS also gives the 4 knights and 7 knights problems. Murray 337, 673 (CB236) & 690 and Wieber 481 show these
problems.
Risâlahĭ Shatranj. Persian poem of unknown date and
authorship. A copy was sent to Bland by
Dr. Sprenger of Delhi. See Bland, loc.
cit., pp. 43‑44. [Bland uses á
for â.] Bland says it has the problem
of the knight's tour or path. [I think
this is the poem mentioned on Murray 182-183 and hence on Wieber 42.]
Şifat mal ‘ûb al‑faras
fî gamî abyât aš‑šaţranğ [NOTE: Ş, ţ. denote
S, t with underdot.] MS Gotha 10, Teil 6; ar. 366; Stz. Hal. 408. Date unknown. Wieber 37 & 480 describes this and gives a path from h8 to e4
which occurs on ff. 70 & 68.
Civis Bononiae [Citizen of
Bologna]. Like Bonus Socius, this is a
collection of chess problems, from c1475, which exists in several MSS and
printings. All are in Latin, from Italy,
and give essentially the same 288 problems.
See Murray 643‑703 for description of about 10 texts and
transcription of the problems. Many of
the texts are not in van der Linde.
Murray 643 cites MS Lasa, in the library of Baron von der Lasa, c1475,
as the most accurate and complete of the texts. Two other well known versions are described below.
Paulo
Guarino (di Forli) (= Paulus Guarinus).
No real title, but the end has 'Explicit liber de partitis scacorum'
with the writer's name and the date 4 Jan 1512. This MS was in the Franz Collection and is now (1913) in the John
G. White Collection in Cleveland, Ohio.
This version only contains 76 problems.
Van der Linde I 295‑297 describes the MS and on p. 294 he
describes a half board path and says Guarino's 74 is a reflection of his no.
243. Murray 645 describes the MS but
doesn't list the individual problems.
He implies that CB244, on p. 674, is the tour that appears in all of the
Civis Bononiae texts, but this is not the same as van der Linde's no. 243. CB236, pp. 673 & 690, is the 4 knights
problem, which is Guarino's 42 [according to Lucas, RM4, p. 207], but I don't
have a copy of van der Linde's no. 215 to check this, ??NX.
Anon.
Sensuit Jeux Partis des eschez: composez nouvellement Pour recreer tous nobles
cueurs et pour eviter oysivete a ceulx qui ont voulente: desir et affection de
le scavoir et apprendre et est appelle ce Livre le jeu des princes et
damoiselles. Published by Denis Janot,
Paris, c1535, 12 ff. ??NYS. (This is the item described by von der Lasa
as 'bei Janot gedrucktes Quartbändchen' (MUS #195).) This a late text of 21 problems, mostly taken from Civis
Bononiae. Only one copy is known, now
(1913) in Vienna. See van der Linde I
306‑307 and Murray 707‑708 which identify no. 18 as van der Linde's
no. 243 and with CB244, as with the Guarino work. I can't tell but van der Linde may identify no. 11 as the 4
knights problem (??NX).
Murray
730 gives another half board path, C92, of c1500 which goes from a8 to g5. Murray 732 notes that a small rearrangement
makes it extendable to the whole board.
Horatio Gianutio della
Mantia. Libro nel quale si tratta della
Maniera di giuocar' à Scacchi, Con alcuni sottilissimi Partiti. Antonio de' Bianchi, Torino, 1597. ??NYS.
Gives half board tours which can be assembled into to a full tour. (Not in the English translation: The Works of Gianutio and Gustavus Selenus,
on the game of Chess, Translated and arranged by J. H. Sarratt; J. Ebers,
London, 1817, vol. 1. -- though the copy I saw didn't say vol. 1. Van der Linde, Erste Jartausend ... says
there are two volumes.)
Bhaţţa
Nīlakaņţha. [NOTE: ţ,
ņ denote t,
n with underdot.] Bhagavantabhāskara. 17C.
End of 5th book. ??NYS,
described by Murray 63‑66. The
author gives three tours, in the poetic form of Rudraţa [NOTE: ţ
denotes a t with an underdot.], which are the same tour
starting at different points. The tour
has 180 degree rotational symmetry.
Ozanam. 1725.
Prob. 52, 1725: 260‑269.
Gives solutions due to Pierre Rémond de Montmort, Abraham de Moivre,
Jean‑Jacques d'Ortous de Mairan (1678-1771). Surprisingly, these are all distinct and different from the
earlier examples. Ozanam says he had
the problem and the solution from de Mairan in 1722. Says the de Moivre is the simplest. Kraitchik, Math. des Jeux, op. cit. in 4.A.2, p. 359, dates the
de Montmort as 1708 and the de Moivre as 1722, but gives no source for
these. Montmort died in 1719. Ozanam died in 1717 and this edition was
edited by Grandin. Van der Linde and
Ahrens say they can find no trace of these solutions prior to Ozanam
(1725). See Ozanam-Montucla, 1778.
Ball,
MRE, 1st ed., 1892, p. 139, says the earliest examples he knows are the De
Montmort & De Moivre of the late 17C, but he only cites them from
Ozanam-Hutton, 1803, & Ozanam-Riddle, 1840. In the 5th ed., 1911, p. 123, he adds that
"They were sent by their authors to Brook Taylor who seems to have
previously suggested the problem."
He gives no reference for the connection to Taylor and I have not seen
it mentioned elsewhere. This note is
never changed and may be the source of the common misconception that knight's
tours originated c1700!
Les Amusemens. 1749.
Prob. 181, p. 354. Gives de
Moivre's tour. Says one can imagine
other methods, but this is the simplest and most interesting.
L. Euler. Letter to C. Goldbach, 26 Apr 1757. In:
P.‑H. Fuss, ed.; Correspondance Mathématique et Physique de
Quelques Célèbres Géomètres du XVIIIème Siècle; (Acad. Imp. des Sciences, St.
Pétersbourg, 1843) = Johnson Reprint,
NY, 1968, vol. 1, pp. 654‑655.
Gives a 180o symmetric tour.
L. Euler. Solution d'une question curieuse qui ne
paroit soumise à aucune analyse. (Mém.
de l'Académie des Sciences de Berlin, 15 (1759 (1766)), 310‑337.) = Opera Omnia (1) 7 (1923) 26‑56. (= Comm. Arithm. Coll., 1849, vol. 1, pp.
337‑355.) Produces many
solutions; studies 180o
symmetry, two halves, and other size boards.
[Petronio dalla Volpe]. Corsa del Cavallo per tutt'i scacchi dello
scacchiere. Lelio della Volpe, Bologna,
1766. 12pp, of which 2 and 12 are
blanks. [Lelio della Volpe is sometimes
given as the author, but he died c1749 and was succeeded by his son
Petronio.] Photographed and printed by
Dario Uri from the example in the Libreria Comunale Archiginnasio di Bologna,
no. 17 CAPS XVI 13. The booklet is
briefly described in: Adriano Chicco;
Note bibliografiche su gli studi di matematica applicata agli scacchi,
publicati in Italia; Atti del Convegno Nazionale sui Giochi Creative, Siena, 11‑14
Jun 1981, ed. by Roberto Magari; Tipografia Senese for GIOCREA (Società
Italiana Giochi Creativi), 1981; p. 155.
The
Introduction by the publisher cites Ozanam as the originator of this 'most
ingenious' idea and says he gives examples due to Montmort, Moivre and
Mairan. He also says this material has
'come to hand' but doesn't give any source, so it is generally thought he was
the author. He gives ten paths,
starting from each of the 10 essentially distinct cells. He then gives the three cited paths from
Ozanam. He then gives six tours. Each path is given as a numbered board and a
line diagram of the path, which led Chicco to say there were 38 paths. The line drawing of the first tour is also
reproduced on the cover/title page.
Ozanam-Montucla. 1778.
Prob. 23, 1778: 178-182; 1803:
177-180; 1814: 155-157. Prob. 22, 1840: 80‑81. Drops the reference to de Mairan as the
source of the problem and adds a fourth tour due to "M. de W***, capitaine
au régiment de Kinski". All of
these have a misprint of 22 for 42 in the right hand column of De Moivre's
solution.
H. C. von Warnsdorff. Des Rösselsprunges einfachste und
allgemeinste Lösung. Th. G. Fr.
Varnhagenschen Buchhandlung, Schmalkalden, 1823, 68pp. ??NYS -- details from Walker. Rule to make the next move to the cell with
the fewest remaining neighbours. Lucas,
L'Arithmétique Amusante, p. 241, gives the place of publication as Berlin.
Boy's Own Book. Not in 1828. 1828-2: 318 states a knight's tour can be made.
George Walker. The Art of Chess-Play: A New Treatise on the
Game of Chess. (1832, 80pp. 2nd ed., Sherwood & Co, London, 1833,
160pp. 3rd ed., Sherwood & Co.,
London, 1841, 300pp. All ??NYS --
details from 4th ed.) 4th ed., Sherwood,
Gilbert & Piper, London, 1846, 375pp.
Chap. V -- section: On the knight, p. 37. "The problem respecting the Knight's covering each square of
the board consecutively, has attracted, in all ages, the attention of the first
mathematicians." States
Warnsdorff's rule, without credit, but gives the book in his bibliography on p.
375, and asserts the rule will always give a tour. No diagram.
Family Friend 2 (1850) 88 &
119, with note on 209. Practical Puzzle
-- No. III. Find a knight's path. Gives one answer. Note says it has been studied since 'an early period' and cites
Hutton, who copies some from Montucla, an article by Walker in Frasers Magazine
(??NYS) which gives Warnsdorff's rule and an article by Roget in Philosophical
Magazine (??NYS) which shows one can start and end on any two squares of
opposite colours. Describes using a
pegged board and a string to make pretty patterns.
Boy's Own Book. Moving the knight over all the squares
alternately. 1855: 511-512; 1868: 573; 1881 (NY): 346-347. 1855
says the problem interested Euler, Ozanam, De Montmart [sic], De Moivre, De
Majron [sic] and then gives Warnsdorff's rule, citing George Walker's 'Treatise
on Chess' for it -- presumably 'A New Treatise', London, 1832, with 2nd ed.,
1833 & 3rd ed., 1841, ??NYS.
Walker also wrote On Moving the Knight, London, 1840, ??NYS. 1868 drops all the names, but the NY ed. of
1881 is the same as the 1855. Gives a
circuit due to Euler.
Magician's Own Book. 1857.
Art. 46: Moving the knight over all the squares alternately,
pp. 283-287. Identical to Boy's
Own Book, 1855, but adds Another Method.
= Book of 500 Puzzles; 1859, art. 46, pp. 97-101. = Boy's Own Conjuring Book, 1860, prob. 45,
pp. 246‑251.
Landells. Boy's Own Toy-Maker. 1858.
Moving the knight over all the squares alternately, p. 143. This is the Another Method of Magician's Own
Book, 1857. Cf Illustrated Boy's Own
Treasury, 1860.
Illustrated Boy's Own
Treasury. 1860. Prob. 47: Practical chess puzzle, pp. 404
& 443. Knight's tour. This is the Another Method of Magician's Own
Book.
C. F. de Jaenisch. Traité des Applications de l'Analyse
Mathématiques au Jeu des Échecs.
3 vols., no publisher, Saint-Pétersbourg. 1862-1863. Vol. 1: Livre I: Section III: De la marche
du cavalier, pp. 186-259 & Plate III.
Vol. 2: Livre II: Problème du Cavalier, pp. 1-296 & 31 plates
(some parts ??NYS). Vol. 3: Addition au
Livre II, pp. 239-243 (This Addition ??NYS).
This contains a vast amount of miscellaneous material and I have not yet
read it carefully. ??NYR
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 323, pp.
153-154 & 393: Rösselsprung-Aufgaben.
Three arrays of syllables and one must find a poetic riddle by following
a knight's tour. Arrays are 8 x 8,
8 x 8, 6 x 4.
C. Flye Sainte‑Marie. Bull. Soc. Math. de France (1876) 144‑150. ??NYS -- described by Jelliss. Shows there is no tour on a 4 x n
board and describes what a path must look like.
Mittenzwey. 1880.
Prob. 222-223, pp. 40 & 91;
1895?: 247-248, pp. 44 & 93;
1917: 247‑248, pp. 40-41 & 89. First is a knight's path.
Second is a board with word fragments and one has to make a poem, which
uses the same path as in the first problem.
Paul de Hijo [= Abbé
Jolivald]. Le Problème du Cavalier des
Échecs. Metz, 1882. ??NYS -- described by Jelliss and quoted by
Lucas. Jelliss notes the BL copy of de
Hijo was destroyed in the war, but he has since told me there are copies in The
Hague and Nijmegen. First determination
of the five 6 x 6 tours with 4-fold rotational symmetry, the 150
ways to cover the 8 x 8 with two circuits of length 32 giving a
pattern with 2‑fold rotational symmetry, the 378 ways giving reflectional
symmetry in a median, the 140 ways with four circuits giving 4-fold rotational
symmetry and the 301 ways giving symmetry in both medians (quoted in Lucas,
L'Arithmétique Amusante, pp. 238-241).
Lucas. Nouveaux jeux scientifiques ..., 1889, op. cit. in 4.B.3. (Described on p. 302, figure on p.
301.) 'La Fasioulette' is an 8 x 8
board with 64 links of length Ö5 to form knight's tours.
Knight's move puzzles. The Boy's Own Paper 11 (Nos. 557 &
558) (14 & 21 Sep 1889) 799 & 814. Four Shakespearean quotations concealed as
knight's tours on a 8 x 8 board.
Beginnings not indicated!
Hoffmann. 1893.
Chap. X, no. 6: The knight's tour, pp. 335-336 & 367-373
= Hoffmann‑Hordern, pp. 225-229.
Gives knight's paths due to Euler and Du Malabare, a knight's tour due
to Monneron, and four other unattributed tours. Gives Warnsdorff's rule, citing Walker's A New Treatise on Chess,
1832.
Ahrens. Mathematische Spiele. Encyklopadie article, op. cit. in 3.B. 1904, pp. 1080‑1093. Pp. 1084‑1086 gives many references to
19C work, including estimates of the number of tours and results on 'semi‑magic
tours'.
C. Planck. Chess Amateur (Dec 1908) 83. ??NYS -- described by Jelliss. Shows there
are 1728 paths on the 5 x 5 board.
Jelliss notes that this counts each path in both directions and there
are only 112 inequivalent tours.
Ahrens. 1910.
MUS I 325. Use of knight's tours
as a secret code.
Dudeney. AM.
1917. Prob. 339: The four
knight's tours, pp. 103 & 229.
Quadrisect the board into four congruent pieces such that there is a
knight's tour on the piece. Jelliss
asserts that the solution is unique and says this may be what Persian MS 260
(i.e. 211) intended. He notes that the
four tours can be joined to give a tour with four fold rotational symmetry.
W. H. Cozens. Cyclically symmetric knight's tours. MG 24 (No. 262) (Dec 1940) 315‑323. Finds symmetric tours on various odd‑shaped
boards.
H. J. R. Murray. The Knight's Tour. ??NYS. MS of 1942
described by G. P. Jelliss, G&PJ 2 (No. 17) (Oct 1999) 315. Observes that a knight can move from
the (0, 0) cell to the (2, 1) and
(1, 2) cella and that the angle
between these lines is the smaller angle of a
3, 4, 5 triangle. One can see this by extending the lines
to (8, 4) and (5, 10) and seeing these points form a 3, 4, 5
triangle with (0, 0).
W. H. Cozens. Note 2761:
On note 2592. MG 42 (No. 340)
(May 1958) 124‑125. Note 2592
tried to find the cyclically symmetric tours on the 6 x 6 board and found
4. Cozens notes two are reflections of
the other two and that three such tours were omitted. He found all these in his 1940 paper.
R. C. Read. Constructing open knight's tours
blindfold! Eureka 22 (Oct 1959)
5-9. Describes how to construct easily
a tour between given cells of opposite colours, correcting a method of Roget
described by Ball (MRE 11th ed, p. 181).
Says he can do it blindfold.
W. H. Cozens. Note 2884:
On note 2592. MG 44 (No. 348)
(May 1960) 117. Estimates there are 200,000
cyclically symmetric tours on the
10 x 10 board.
Roger F. Wheeler. Note 3059:
The KNIGHT's tour on 42 and other boards. MG 47 (No. 360) (May 1963) 136‑141. KNIGHT means a knight on a toroidal
board. He finds 2688 tours of 19 types
on the 42 toroid.
(Cf Tylor, 1982??)
J. J. Duby. Un algorithme graphique trouvant tous les
circuits Hamiltoniens d'un graphe.
Etude No. 8, IBM France, Paris, 22 Oct 1964. [In English with French title and summary.] Finds there are 9862 knight's tours on
the 6 x 6 board, where the tours all start at a fixed corner and then go to
a fixed one of the two cells reachable from the corner. He also finds 75,000 tours on the 8 x 8 board which have the same first 35 moves. He believes there may be over a million
tours.
Karl Fabel. Wanderungen von Schachfiguren. IN:
Eero Bonsdorff, Karl Fabel & Olavi Riihimaa; Schach und Zahl; Walter
Rau Verlag, Düsseldorf, 1966, pp. 40-50.
On p. 50, he says that there are
122,802,512 tours where the
knight does two joined half-board paths.
He also says there are upper bounds, determined by several authors, and
he gives 1.5 x 1026 as an example.
Gardner. SA (Oct 1967) = Magic Show, chap. 14. Surveys results of which boards have tours
or paths.
D. J. W. Stone. On the Knight's Tour Problem and Its Solution
by Graph‑Theoretic and Other Methods.
M.Sc. Thesis, Dept. of Computing Science, Univ. of Glasgow, Jan.
1969. Confirms Duby's 9862 tours on
the 6 x 6 board.
David Singmaster. Enumerating unlabelled Hamiltonian
circuits. International Series on
Numerical Mathematics, No. 29.
Birkhäuser, Basel, 1975, pp. 117‑130. Discusses the work of Duby and Stone and gives an estimate, which
Stone endorses, that there are 1023±3 tours on the 8 x 8 board.
C. M. B. Tylor. 2‑by‑2 tours. Chessics 14 (Jul‑Dec 1982) 14. Says there are 17 knight's tours on a 2 x 2
torus and gives them. Doesn't
mention Wheeler, 1963.
Robert Cannon & Stan
Dolan. The knight's tour. MG 70 (No. 452) (Jun 1986) 91‑100. A rectangular board is tourable if it has a
knight's path between any two cells of opposite colours. They prove that m x n is tourable if
and only if mn is even and
m ³ 6, n ³ 6. They also prove that m x
n has a knight's tour if and only
if mn is even and [(m ³ 5, n ³ 5)
or (m = 3, n ³ 10)] and that when
mn is even, m x n has a knight's path if and only if m ³ 3, n ³ 3, except for the 3 x 6 and 4 x 4
boards. (These later results are
well known -- see Gardner. The authors
only cite Ball's MRE.)
George Jelliss. Figured tours. MS 25:1 (1992/93) 16-20.
Exposition of paths and tours where certain stages of the path form an
interesting geometric figure. E.g.
Euler's first paper has a path on the 5
x 5 such that the points on one
diagonal are in arithmetic progression:
1, 7, 13, 19, 25.
Martin Loebbing & Ingo
Wegener. The number of knight's tours
equals 33,439,123,484,294 ‑--
Counting with binary decision diagrams.
Electronic Journal of Combinatorics 3 (1996) article R5. A somewhat vague description of a method for
counting knight's tours -- they speak of directed knight's tours, but it is not
clear if they have properly accounted for the symmetries of a tour or of the
board. Several people immediately
pointed out that the number is incorrect because it has to be divisible by
four. Two comments have appeared,
ibid. On 15 May 1996, the authors
admitted this and said they would redo the problem, but they have submitted no
further comment as of Jan 2001. On 18 Feb
1997, Brendan McKay announced that he had done the computation another way and
found 13,267,364,410,532.
In
view of the difference between this and my 1975 estimate of 1023±3 tours, it might be worth explaining my reasoning. In 1964, Duby found 75,000
tours with the same first 35 moves.
The average valence for a knight on an
8 x 8 board is 5.25, but one
cannot exit from a cell in the same direction as one entered, so we might
estimate the number of ways that the first 35 moves can be made as 4.2535 = 9.9 x 1021. Multiplying by 75,000 then gives 7.4 x 1026. I think I assumed that some of the first
moves had already been made, e.g. we only allow one move from the starting
cell, and factored by 8 for the symmetries of the square, to get 2.2 x 1025. I can't find my original calculations, and I
find the estimate 1025 in later papers, so I suppose I tried to
reduce the effect of the 4.2535 some more.
In retrospect, I had no knowledge of how many of these had already been
tried. If about half of all moves from
a cell had already been tried before any circuit was found, then the estimate
would be more like 2.2534 x 75,000 =
7.1 x 1016. If we
divide the given number of circuits by
75,000 and take the 34th root,
we get an average valence of 1.78 remaining, far less than I would have
guessed.
I
am grateful to Don Knuth for this reference.
Neither he nor I expected to ever see this number calculated!
5.F.2. OTHER HAMILTONIAN CIRCUITS
For
circuits on the n‑cube, see also 5.F.4 and 7.M.1,2,3.
For
circuits on the chessboard, see also 6.AK.
Le Nôtre. Le Labyrinte de Versailles, c1675. This was a hedge or garden maze, but the
objective was to visit, in correct order, 40 fountains based on Aesop's Fables. Each node of the maze had at least one
fountain. Some fountains were not at
path junctions, but one can consider these as nodes of degree two. This is an early example of a Hamiltonian
problem, except that one fountain was located at the end of a short dead
end. [Fisher, op. cit. in 5.E.1, pp.
49, 79, 130 & 144-145, with contemporary diagram on p. 144. He says there are 39 fountains, but the
diagram has 40.]
T. P. Kirkman. On the partitions of the R‑pyramid,
being the first class of R‑gonous X‑edra. Philos. Trans. Roy. Soc. 148 (1858) 145‑161.
W. R. Hamilton. The Icosian Game. 4pp instructions for the board game. J. Jaques and Son, London, 1859.
(Reproduced in BLW, pp. 32-35, with frontispiece photo of the board at
the Royal Irish Academy.)
For a long time, the only known
example of the game, produced by Jaques, was at the Royal Irish Academy in
Dublin. This example is inscribed on
the back as a present from Hamilton to his friend, J. T. Graves. It is complete, with pegs and instructions. None of the obvious museums have an
example. Diligent searching in the
antique trade failed to turn up an example in twenty years, but in Feb 1996,
James Dalgety found and acquired an example of the board -- sadly the pegs and
instructions were lacking. Dalgety
obtained another board in 1998, again without the pegs and instructions, but in
1999 he obtained another example, with the pegs.
Mittenzwey. 1880.
Prob. 281, pp. 50 & 100;
1895?: 310, pp. 53-54 & 102;
1917: 310, pp. 49 & 97.
The garden of a French palace has a maze with 31 points to see. Find a path past all of them with no repeated
edges and no crossings. The pattern is
clearly based on the Versailles maze of c1675 mentioned above, but I don't
recall the additional feature of no crossings occurring before.
T. P. Kirkman. Solution of problem 6610, proposed by
himself in verse. Math. Quest. Educ.
Times 35 (1881) 112‑116. On p.
115, he says Hamilton told him, upon occasion of Hamilton presenting him 'with
his handsomest copy of the puzzle', that Hamilton got the idea for the Icosian
Game from p. 160 of Kirkman's 1858 article,
Lucas. RM2, 1883, pp. 208‑210.
First? mention of the solid version.
The 2nd ed., 1893, has a footnote referring to Kirkman, 1858.
John Jaques & Son. The Traveller's Dodecahedron; or,
A Voyage Round the World. A New
Puzzle. "This amusing puzzle, exercising
considerable skill in its solution, forms a popular illustration of Sir William
Hamilton's Icosian Game. A wood
dodecahedron with the base pentagon stretched so that when it sits on the base,
all vertices are visible. With ivory?
pegs at the vertices, a handle that screws into the base, a string with rings
at the ends and one page of instructions, all in a box. No date.
The only known example was obtained by James Dalgety in 2002.
Pearson. 1907.
Part III, no. 60: The open door, pp. 60 & 130. Prisoner in one corner of an 8 x 8
array is allowed to exit from from the other corner provided he visits
every cell once. This requires him to
enter and leave a cell by the same door.
Ahrens. Mathematische Spiele. 2nd ed., Teubner, Leipzig, 1911. P. 44, note, says that a Dodekaederspiel is
available from Firma Paul Joschkowitz -- Magdeburg for .65
mark. This is not in the 1st ed.
of 1907 and the whole Chapter is dropped in the 3rd ed. of 1916 and the later
editions.
Anonymous. The problems drive. Eureka 12 (Oct 1949) 7-8 & 15. No. 3.
How many Hamiltonian circuits are there on a cube, starting from a given
point? Reflections and reversals count
as different tours. Answer is 12, but
this assumes also that rotations are different. See Singmaster, 1975, for careful definitions of how to
count. There are 96 labelled circuits,
of which 12 start at a given vertex.
But if one takes all the 48 symmetries of the cube as equivalences (six
of which fix the given vertex), there are just 2 circuits from a given starting
point. However, these are actually the
same circuit started at different points.
Presumably Kirkman and Hamilton knew of this.
C. W. Ceram. Gods, Graves and Scholars. Knopf, New York, 1956, pp. 26-29. 2nd ed., Gollancz, London, 1971, pp.
24-25. Roman knobbed dodecahedra -- an
ancient solid version??
R. E. Ingram. Appendix 2: The Icosian Calculus. In:
The Mathematical Papers of Sir William Rowan Hamilton. Vol. III: Algebra. Ed. by H. Halberstam & R. E. Ingram. CUP, 1967, pp. 645‑647. [Halberstam told me that this Appendix is
due to Ingram.] Discusses the method
and asserts that the tetrahedron, cube and dodecahedron have only one
unlabelled circuit, the octahedron has two and the icosahedron has 17.
David Singmaster. Hamiltonian circuits on the regular
polyhedra. Notices Amer. Math. Soc. 20
(1973) A‑476, no. 73T‑A199.
Confirms Ingram's results and gives the number of labelled circuits.
David Singmaster. Op. cit. in 5.F.1. 1975. Carefully defines
labelled and unlabelled circuits.
Discusses results on regular polyhedra in 3 and higher dimensions.
David Singmaster. Hamiltonian circuits on the n‑dimensional octahedron. J. Combinatorial Theory (B) 18 (1975) 1‑4. Obtains an explicit formula for the number
of labelled circuits on the n‑dimensional
octahedron and shows it is @
(2n)!/e. Gives numbers for n £ 8. In unpublished work, it is shown that the number of unlabelled
circuits is asymptotic to the above divided by
n!2n×4n.
Angus Lavery. The Puzzle Box. G&P 2 (May 1994) 34-35.
Alternative solitaire, p. 34.
Asks for a knight's tour on the 33-hole solitaire board. Says he hasn't been able to do it and offers
a prize for a solution. In Solutions,
G&P 3 (Jun 1994) 44, he says it cannot be done and the proof will be given
in a future issue, but I never saw it.
5.F.3. KNIGHT'S TOURS IN HIGHER DIMENSIONS
A.‑T. Vandermonde. Remarques sur les problèmes de
situation. Hist. de l'Acad. des Sci.
avec les Mémoires (Paris) (1771 (1774)) Mémoires: pp. 566‑574 & Plates
I & II. ??NYS. First? mention of cubical problem. (Not
given in BLW excerpt.)
F. Maack. Mitt. über Raumschak. 1909, No. 2, p. 31. ??NYS -- cited by Gibbins, below. Knight's tour on 4 x 4 x 4 board.
Dudeney. AM.
1917. Prob. 340: The cubic knight's
tour, pp. 103 & 229. Says
Vandermonde asked for a tour on the faces of a
8 x 8 x 8 cube. He gives it as a problem with a solution.
N. M. Gibbins. Chess in three and four dimensions. MG 28 (No. 279) (1944) 46‑50. Gives knight's tour on 3 x 3 x 4
board -- an unpublished result due to E. Hubar‑Stockar of
Geneva. This is the smallest 3‑D
board with a tour. Gives Maack's tour
on 4 x 4 x 4 board.
Ian Stewart. Solid knight's tours. JRM 4:1 (Jan 1971) 1. Cites Dudeney. Gives a tour through the entire
8 x 8 x 8 cube by stacking 8
knight's paths.
T. W. Marlow. Closed knight tour of a 4 x 4 x 4
board. Chessics 29 & 30
(1987) 162. Inspired by Stewart.
5.F.4. OTHER CIRCUITS IN AND ON A CUBE
The number of Hamiltonian Circuits on
the n-dimensional cube is the same as
the number of Gray codes (see 7.M.3) and has been the subject of considerable
research. I will not try to cover this
in detail.
D. W. Crowe. The
n‑dimensional cube and the Tower of Hanoi. AMM 63:1 (Jan 1956) 29‑30.
E. N. Gilbert. Gray codes and paths on the n-cube.
Bell System Technical Journal 37 (1958) 815-826. Shows there are 9 inequivalent circuits on
the 4-cube and 1 on the n-cube for n =
1, 2, 3. The latter cases are
sufficiently easy that they may have been known before this.
Allen F. Dreyer. US Patent 3,222,072 -- Block Puzzle. Filed: 11 Jun 1965; patented: 7 Dec 1965. 4pp + 2pp diagrams. 27 cubes on an elastic. The holes are straight or diagonal so that
three consecutive cubes are either in a line or form a right angle. A solution is a Hamiltonian path through the
27 cells. Such puzzles were made in
Germany and I was given one about 1980 (see Singmaster and Haubrich &
Bordewijk below). Dreyer gives two
forms.
Gardner. The binary Gray code. SA (Aug 1972) c= Knotted, chap. 2.
Notes that the number of circuits on the n-cube, n > 4, is not known. SA (Apr 1973) reports that three (or four) groups had found the
number of circuits on the 4-cube -- this material is included in the Addendum
in Knotted, chap. 2, but none of the groups ever seem to have published their
results elsewhere. Unfortunately, none
of these found the number of inequivalent circuits since they failed to take
all the equivalences into account -- e.g. for
n = 1, 2, 3, 4, 5, their enumerations give: 2, 8, 96, 43008, 5 80189 28640 for the numbers of labelled circuits. Gardner's Addendum describes some further work
including some statistical work which estimates the number on the 6-cube is
about 2.4 x 1025.
David Singmaster. A cubical path puzzle. Written in 1980 and submitted to JRM, but
never published. For the 3 x 3 x 3
problem, the number, S, of straight through pieces (ignoring the
ends) satisfies 2 £ S £
11.
Mel A. Scott. Computer output, Jun 1986, 66pp. Determines there are 3599 circuits through
the 3 x 3 x 3 cube such that the resulting string of 27 cubes can be made into
a cube in just one way. But cf the next
article which gives a different number??
Jacques Haubrich & Nanco
Bordewijk. Cube chains. CFF 34 (Oct 1994) 12‑15. Erratum, CFF 35 (Dec 1994) 29. Says Dreyer is the first known reference to
the idea and that they were sold 'from about 1970' Reproduces the first page of diagrams from Dreyer's patent. Says his first version has a unique
solution, but the second has 38 solutions.
They have redone previous work and get new numbers. First, they consider all possible strings of
27 cubes with at most three in a line (i.e. with at most a single 'straight'
piece between two 'bend' pieces and they find there are 98,515
of these. Only 11,487
of these can be folded into a 3
x 3 x 3 cube. Of these, 3654 can be folded up in only one way. The chain with the most solutions had 142
different solutions. They refer to Mel
Scott's tables and indicate that the results correspond -- perhaps I miscounted
Scott's solutions??
5.G.1. GAS, WATER AND ELECTRICITY
Dudeney. Problem 146 -- Water, gas, and
electricity. Strand Mag. 46 (No. 271)
(Jul 1913) 110 & (No. 272) (Aug 1913) 221 (c= AM, prob. 251,
pp. 73 & 200‑201). Earlier
version is slightly more interesting, saying the problem 'that I have called
"Water, Gas, and Electricity" ... is as old as the hills'. Gives trick solution with pipe under one
house.
A. B. Nordmann. One Hundred More Parlour Tricks and
Problems. Wells, Gardner, Darton &
Co., London, nd [1927 -- BMC]. No. 96:
The "three houses" problem, pp. 89-90 & 114. "Were all the houses connected up with
all three supplies or not?" Answer
is no -- one connection cannot be made.
Loyd, Jr. SLAHP.
1928. The three houses and three
wells, pp. 6 & 87‑88. "A
puzzle ... which I first brought out in 1900 ..." The drawing is much less polished than
Dudeney's. Trick solution with a pipe
under one house, a bit differently laid out than Dudeney.
The Bile Beans Puzzle Book. 1933.
No. 46: Water, gas & electric light. Trick solution almost identical to Dudeney.
Philip Franklin. The four color problem. In:
Galois Lectures; Scripta Mathematica Library No. 5; Scripta Mathematica,
Yeshiva College, NY, 1941, pp. 49-85.
On p. 74, he refers to the graph as "the basis of a familiar
puzzle, to join each of three houses with each of three wells (or in a modern
version to a gas, water, and electricity plant)".
Leeming. 1946.
Chap. 6, prob. 4: Water, gas and electricity, pp. 71 & 185. Dudeney's trick solution.
H. ApSimon. Note 2312:
All modern conveniences. MG 36
(No. 318) (Dec 1952) 287‑288.
Given m houses and
n utilities, the maximum number
of non‑crossing connections is 2(m+n‑2) and this
occurs when all the resulting regions are 4‑sided. He extends to p‑partite graphs in general and a special case.
John Paul Adams. We Dare You to Solve This! Op. cit. in 5.C. 1957? Prob. 50: Another enduring
favorite appears below, pp. 30 & 49.
Electricity, gas, water.
Dudeney's trick solution.
Young World. c1960.
P. 4: Crossed lines.
Electricity, TV and public address lines. Trick solution with a line passing under a house.
T. H. O'Beirne. For boys, men and heroes. New Scientist 12 (No. 266) (21 Dec 1961) 751‑753. Shows you can join 4 utilities to 4 houses
on a torus without crossing.
5.H. COLOURED SQUARES AND CUBES, ETC.
5.H.1. INSTANT INSANITY = THE TANTALIZER
Note.
Often the diagrams do not show all sides of the pieces so I cannot tell
if one version is the same as another.
Frederick A. Schossow. US Patent 646,463 -- Puzzle. Applied: 19 May 1899; patented: 3 Apr 1900. 1p + 1p diagrams. Described in S&B, p. 38, which also says it is described in
O'Beirne, but I don't find it there??
Four cubes with suit patterns.
The net of each cube is shown.
The fourth cube has three clubs.
George Duncan Moffat. UK Patent 9810 -- Improvements in or
relating to Puzzle-apparatus. Applied:
28 May 1900; accepted: 30 Jun
1900. 2pp + 1p diagrams. For a six cube version with "letters R, K, B,
W, F and B-P, the initials of the names of General
Officers of the South African Field Force."
Joseph Meek. UK Patent 2775 -- Improved Puzzle Game. Applied: 5 Feb 1909; complete specification: 16 Jun 1909; accepted: 3 Feb 1910. 2pp + 1p diagrams. A four cube version with suit patterns. His discussion seems to describe the pieces drawn by Schossow.
Slocum. Compendium.
Shows: The Great Four Ace Puzzle
(Gamage's, 1913); Allies Flag Puzzle
(Gamage's, c1915); Katzenjammer Puzzle
(Johnson Smith, 1919).
Edwin F. Silkman. US Patent 2,024,541 -- Puzzle. Applied: 9 Sep 1932; patented: 17 Dec 1935. 2pp + 1 p diagrams. Four cubes marked with suits. The net of each cube is shown. The third cube has three hearts. This is just a relabelling of Schossow's
pattern, though two cubes have to be reflected which makes no difference to the
solution process.
E. M. Wyatt. The bewitching cubes. Puzzles in Wood. (Bruce Publishing, Co., Milwaukee, 1928) = Woodcraft Supply Corp., Woburn, Mass.,
1980, p. 13. A six cube, six way
version.
Abraham. 1933.
Prob. 303 -- The four cubes, p. 141 (100). 4 cube version "sold ... in 1932".
A. S. Filipiak. Four ace cube puzzle. 100 Puzzles, How do Make and How to Solve
Them. A. S. Barnes, NY, (1942) = Mathematical Puzzles, and Other Brain
Twisters; A. S. Barnes, NY, 1966;
Bell, NY, 1978; p. 108.
Leeming. 1946.
Chap. 10, prob. 9: The six cube puzzle, pp. 128‑129 &
212. Identical to Wyatt.
F. de Carteblanche [pseud. of
Cedric A. B. Smith]. The coloured cubes
problem. Eureka 9 (1947) 9‑11. General graphical solution method, now the
standard method.
T. H. O'Beirne. Note 2736:
Coloured cubes: A new "Tantalizer". MG 41 (No. 338) (Dec 1957) 292-293. Cites Carteblanche, but says the current
version is different. Gives a nicer
version.
T. H. O'Beirne. Note 2787:
Coloured cubes: a correction to Note 2736. MG 42 (No. 342) (Dec 1958) 284.
Finds more solutions than he had previously stated.
Norman T. Gridgeman. The 23 colored cubes. MM 44:5 (Nov 1971) 243-252. The
23 colored cubes are the
equivalence classes of ways of coloring the faces with 1
to 6 colors. He cites and
describes some later methods for attacking Instant Insanity problems.
Jozsef Bognár. UK Patent Application 2,076,663 A -- Spatial
Logical Puzzle. Filed 28 May 1981; published 9 Dec 1981. Cover page + 8pp + 3pp diagrams. Not clear if the patent was ever
granted. Describes Bognár's Planets,
which is a four piece instant insanity where the pieces are spherical and held
in a plastic tube. This was called
Bolygok in Hungarian and there is a reference to an earlier Hungarian patent. Also describes his version with eight pieces
held at the corners of a plastic cube.
Haubrich's
1995-1996 surveys, op. cit. in 5.H.4, include MacMahon puzzles as one class.
I
have just added the Carroll result that there are 30 six-coloured cubes, but
this must be older??
Frank H. Richards. US Patent 331,652 -- Domino. Applied: 13 Jun 1885; patented: 1 Dec 1885. 2pp + 2pp diagrams. Cited by Gardner in Magic Show, but with
date 1895. Reproduced in Haubrich,
About ..., 1996, op. cit. in 5.H.4. For
triangular matching games, specifically showing the MacMahon 5-coloured
triangles, but considering reflections as equivalences, so he has 35
pieces. [One of the colours is blank
and hence Gardner said it was a 4-colouring.]
Carroll-Wakeling. c1890?
Prob. 15: Painting cubes, pp. 18-19 & 67. This is one of the problems on undated sheets of paper that
Carroll sent to Bartholomew Price. How
many ways can one six-colour a cube?
Wakeling gives a solution, but this apparently is not on Carroll's MS.
Percy Alexander MacMahon &
Julian Robert John Jocelyn. UK
Patent 3927 A.D. 1892 -- Appliances to
be used in Playing a New Class of Games.
Applied: 29 Feb 1892; Complete
Specification Left: 28 Nov 1892;
Accepted: 28 Jan 1893. 5pp + 2pp
diagrams. Reproduced in Haubrich, About
..., 1996, op. cit. in 5.H.4. Describes
the 24 triangles with four types of edge and mentions other numbers of edge
types. Describes various games and
puzzles.
Percy Alexander MacMahon &
Julian Robert John Jocelyn. UK
Patent 8275 A.D. 1892 -- Appliances for
New Games of Puzzles. Applied: 2 May
1892; Complete Specification Left: 31
Jan 1893; Accepted: 4 Mar 1893. 2pp.
27 cubes with three colours, opposite faces having the same colour. Similar sets of 8, 27, 64, etc. cubes. Various matching games suggested. Using six colours and all six on each cube
gives 30 cubes -- the MacMahon Cubes.
Gives a complex matching problem of making two 2 x 2 x 2 cubes. Paul Garcia (email of 15 Nov 2002)
commented: "8275 describes 2
different sets of blocks, using either three colours or six colours. The three colour blocks form a set of 27
that can be assembled into a large cube with single coloured faces and internal
contact faces matching. For the six
colour cubes, the puzzle suggested is to pick out two associated cubes, and
find the sixteen cubes that can be assembled to make a copy of each. Not quite Mayblox, although using the same
colouring system."
James Dalgety. R. Journet & Company A Brief History of the Company & its
Puzzles. Published by the author, North
Barrow, Somerset, 1989. On p. 13, he
says Mayblox was patented in 1892. In
an email on 12 Nov 2002, he cited UK
Patent 8275.
Anon. Report:
"Mathematical Society, February 9". Nature 47 (No. 1217) (23 Feb 1893) 406. Report of MacMahon's talk:
The group of thirty cubes composed by six differently coloured squares.
See: Au Bon Marché, 1907, in 5.P.2, for a puzzle of hexagons with
matching edges.
Manson. 1911.
Likoh, pp. 171-172. MacMahon's
24 four-coloured isosceles right triangles, attributed to MacMahon.
"Toymaker". The Cubes of Mahomet Puzzle. Work, No. 1447 (9 Dec 1916) 168. 8 six-coloured cubes to be assembled into a
cube with singly-coloured faces and internal faces to have matching colours.
P. A. MacMahon. New Mathematical Pastimes. CUP, 1921.
The whole book deals with variations of the problem and calculates the
numbers of pieces of various types. In
particular, he describes the 24 4-coloured triangles, the 24
3-coloured squares, the MacMahon cubes, some right-triangular and
hexagonal sets and various subsets of these.
With n colours, there are n(n2+2)/3 triangles,
n(n+1)(n2‑n+2)/4
squares and n(n+1)(n4-n3+n2+2)/6 hexagons.
[If one allows reflectional equivalence, one gets n(n+1)(n+2)/6 triangles, n(n+1)(n2+n+2)/8 squares
and n(n+1)(n4-n3+4n2+2)/12 hexagons.
Problem -- is there an easy proof that the number of triangles is BC(n+2, 3)?] On p. 44, he says that Col. Julian R. Jocelyn told him some years
ago that one could duplicate any cube with 8 other cubes such that the internal
faces matched.
Slocum. Compendium.
Shows Mayblox made by R. Journet from Will Goldston's 1928 catalogue.
F. Winter. Das Spiel der 30 bunten Würfel MacMahon's Problem. Teubner, Leipzig, 1934, 128pp. ??NYR.
Clifford Montrose. Games to play by Yourself. Universal Publications, London, nd
[1930s?]. The coloured squares, pp.
78-80. Makes 16 squares with
four-coloured edges, using five colours, but there is no pattern to the
choice. Uses them to make a 4 x 4
array with matching edges, but seems to require the orientations to be
fixed.
M. R. Boothroyd &
J. H. Conway. Problems drive,
1959. Eureka 22 (Oct 1959) 15-17 &
22-23. No. 6. There are twelve ways to colour the edges of a pentagon, when
rotations and reflections are considered as equivalences. Can you colour the edges of a dodecahedron
so each of these pentagonal colourings occurs once? [If one uses tiles, one has to have reversible tiles.] Solution says there are three distinct
solutions and describes them by describing contacts between 10 pentagons
forming a ring around the equator.
Richard K. Guy. Some mathematical recreations I &
II. Nabla [= Bull. Malayan Math.
Soc.] 7 (Oct & Dec 1960) 97-106 & 144-153. Pp. 101-104 discusses MacMahon triangles,
squares and hexagons.
T. H. O'Beirne. Puzzles and paradoxes 5: MacMahon's
three-colour set of squares. New
Scientist 9 (No. 220) (2 Feb 1961) 288-289.
Finds 18 of the 20 possible monochrome border patterns.
Gardner. SA (Mar 1961) = New MD, Chap. 16. MacMahon's 3-coloured squares and his
cubes. Addendum in New MD cites
Feldman, below.
Gary Feldman. Documentation of the MacMahon Squares
Problem. Stanford Artificial
Intelligence Project Memo No. 12, Stanford Computation Center, 16 Jan 1964. ??NYS
Finds 12,261 solutions for the 6 x 4 rectangle with
monochrome border -- but see Philpott, 1982, for 13,328 solutions!!
Gardner. SA (Oct 1968) = Magic Show, Chap. 16. MacMahon's four-coloured triangles and
numerous variants.
Wade E. Philpott. MacMahon's three-color squares. JRM 2:2 (1969) 67-78. Surveys the topic and repeats Feldman's
result.
N. T. Gridgeman, loc. cit. in
5.H.1, 1971, covers some ideas on the MacMahon cubes.
J. J. M. Verbakel. Digitale tegels (Digital tiles). Niet piekeren maar puzzelen (name of a
puzzle column). Trouw (a Dutch
newspaper) (1 Feb 1975). ??NYS --
described by Jacques Haubrich; Pantactic patterns and puzzles; CFF 34 (Oct
1994) 19-21. There are 16 ways to 2‑colour
the edges of a square if one does not allow them to rotate. Assemble these into a 4 x 4
square with matching edges.
There are 2,765,440 solutions in 172,840 classes of
16. One can add further constraints to
yield fewer solutions -- e.g. assume the
4 x 4 square is on a torus and
make all internal lines have a single colour.
Gardner. Puzzling over a problem‑solving
matrix, cubes of many colours and three‑dimensional dominoes. SA 239:3 (Sep 1978) 20‑30 &
242 c= Fractal, chap. 11. Good review of MacMahon (photo) and his
coloured cubes. Bibliography cites
recent work on Mayblox, etc.
Wade E. Philpott. Instructions for Multimatch. Kadon Enterprises, Pasadena, Maryland,
1982. Multimatch is just the 24
MacMahon 3-coloured squares. This
surveys the history, citing several articles ??NYS, up to the determination of
the 13,328 solutions for the 6 x
4 rectangle with monochrome border, by
Hilario Fernández Long (1977) and John W. Harris (1978).
Torsten Sillke. Three
3 x 3 matching puzzles. CFF 34 (Oct 1994) 22-23. He has wanted an interesting 9 element
subset of the MacMahon pieces and finds that of the 24 MacMahon 3-coloured
squares, just 9 of them contain all three colours. He considers both the corner and the edge versions. The editor notes that a 3 x 3
puzzle has 36 x 32/2 =
576 possible edge contacts and that the
number of these which match is a measure of the difficulty of the puzzle, with
most 3 x 3 puzzles having 60 to
80 matches. The corner version of Sillke's puzzle
has 78
matches and one solution. The
edge version has 189 matches and many solutions, hence Sillke
proposes various further conditions.
Here we have a set of pieces and one
has to join them so that some path is formed.
This is often due to a chain or a snake, etc. New section. Again,
Haubrich's 1995-1996 surveys, op. cit. in 5.H.4, include this as one class.
Hoffmann. 1893.
Chap. III, No. 18: The endless chain, pp. 99-100 & 131
= Hoffmann‑Hordern, pp. 91-92, with photo. 18 pieces, some with
parts of a chain, to make into an 8 x
8 array with the chain going
through 34 of the cells. All the
pieces are rectangles of width one.
Photo shows The Endless Chain, by The Reason Manufacturing Co.,
1880-1895. Hordern Collection, p. 62,
shows the same and La Chaine sans fin, 1880-1905.
Loyd. Cyclopedia. 1914. Sam Loyd's endless chain puzzle, pp. 280
& 377. Chain through all 64 cells
of a chessboard, cut into 13 pieces.
The chessboard dissection is of type:
13: 02213 131.
Hummerston. Fun, Mirth & Mystery. 1924.
The dissected serpent, p. 131.
Same pieces as Hoffmann, and almost the same pattern.
Collins. Book of Puzzles. 1927. The dissected snake
puzzle, pp. 126-127. 17 pieces forming
an 8 x 8 square. All the piece are
rectangular pieces of width one except for one L‑hexomino -- if this were
cut into straight tetromino and domino, the pieces would be identical to
Hoffmann. The pattern is identical to
Hummerston.
See Haubrich in 5.H.4.
These
all have coloured edges unless specified.
See S&B, p. 36, for examples.
Edwin L[ajette] Thurston. US Patent 487,797 -- Puzzle. Applied: 30 Sep 1890; patented: 13 Dec 1892. 3pp + 3pp diagrams. Reproduced in Haubrich, About ..., 1996, op.
cit. below. 4 x 4 puzzles with 6-coloured corners or edges,
but assuming no colour is repeated on a piece -- indeed he uses the 15 = BC(6,2) ways of choosing 4 out of 6 colours once only and then has a
sixteenth with the same colours as another, but in different order. Also a star-shaped puzzle of six
parallelograms.
Edwin L. Thurston. US Patent 487,798 -- Puzzle. Applied: 30 Sep 1890; patented: 13 Dec 1892. 2pp + 1p diagrams. Reproduced in Haubrich, About ..., 1996, op. cit. below. As far as I can see, this is the same as the 4 x 4
puzzle with 6‑coloured edges given above, but he seems to be
emphasising the 15 pieces.
Edwin L. Thurston. US Patent 490,689 -- Puzzle. Applied: 30 Sep 1890; patented: 31 Jun 1893. 2pp + 1p diagrams. Reproduced in Haubrich, About ..., 1996, op. cit. below. The patent is for 3 x 3 puzzles with 4‑coloured
corners or edges, but with pieces having no repeated colours and in a fixed
orientation. He selects some 8 of these
pieces for reasons not made clear and mentions moving them "after the manner of the old 13, 14, 15
puzzle." S&B, p. 36, describes
the Calumet Puzzle, Calumet Baking Powder Co., Chicago, which is a 3 x 3
head to tail puzzle, claimed to be covered by this patent.
Le Berger Malin. France, c1900. 3 x 3 head to tail
puzzle, but the edges are numbered and the matching edges must add to 10. ??NYS -- described by K. Takizawa, N.
Takashima & N. Yoshigahara; Vess Puzzle and Its Family -- A Compendium
of 3 by 3 Card Puzzles; published by the authors, Tokyo, 1983. Slocum has this in two different boxes and
dates it to c1900 -- I had c1915 previously.
Haubrich has one version, Produced by GB&O N.K. Atlas.
Angus K. Rankin. US Patent 1,006,878 -- Puzzle. Applied: 3 Feb 1911; patented: 24 Oct 1911. 2pp + 1p diagrams. Reproduced in Haubrich, About ..., 1996, op. cit. below. Described in S&B, p. 36. Grandpa's Wonder Puzzle. 3 x 3
square puzzle. Each piece has
corners coloured, using four colours, and the colours meeting at a corner must
differ. The patent doesn't show the
advertiser's name -- Grandpa's Wonder Soap -- but is otherwise identical to
S&B's photo.
Daily Mail World Record Net Sale
puzzle. 1920‑1921. Instructions and picture of the pieces. Letter from Whitehouse to me describing its
invention. 19 6-coloured hexagons without repeated colours. Daily Mail articles as follows. There may be others that I missed and
sometimes the page number is a bit unclear.
Note that 5 Dec was a Sunday.
9 Nov 1920, p. 5. "Daily Mail" puzzle.
To be issued on 7 Dec.
13 Nov
1920, p. 4. Hexagon mystery.
17 Nov
1920, p. 5. New mystery puzzle. Asserts the inventor does not know the
solution -- i.e. the solution has been locked up in a safe.
20 Nov
1920, p. 4. What is it?
23 Nov
1920, p. 5. Fascinating puzzle. The most fascinating puzzle since "Pigs
in Clover".
25 Nov
1920, p. 5. Can you do it?
29 Nov
1920, p. 5. £250 puzzle.
1 Dec 1920, p. 4. Mystery puzzle clues.
2 Dec 1920, p. 5. £250 puzzle race.
3 Dec 1920, p. 5. The puzzle.
4 Dec 1920, p. 4. The puzzle. Amplifies on
the inventor not knowing the solution -- after the idea was approved, a new
pattern was created by someone else and locked up.
6 Dec 1920, unnumbered back page. Photo with caption: £250 for solving this.
7 Dec 1920, p. 7. "Daily Mail" Puzzle. Released today. £100 for
getting the locked up solution. £100
for the first alternative solution and £50 for the next alternative solution. "It is believed that more than one
solution is possible."
8 Dec 1920, p. 5. "Daily Mail" puzzle.
9 Dec 1920, p. 5. Can you do it?
10 Dec
1920, p. 4. It can be done.
13 Dec
1920, p. 9. Most popular pastime. "More than 500,000 Daily Mail
Puzzles have been sold."
15 Dec
1920, p. 4. Puzzle king & the 19
hexagons. Dudeney says he does not
think it can be solved "except by trial."
16 Dec
1920, p. 4. Tantalising 19 hexagons.
16 Dec
1920, unnumbered back page. Banner at
top has: "The Daily Mail"
puzzle. Middle of page has a cartoon of
sailors trying to solve it.
17 Dec
1920, p. 5? The Xmas game.
18 Dec
1920, p. 7. Puzzle Xmas 'card'.
20 Dec
1920, p. 7. Hexagon fun.
22 Dec
1920, p. 3. 3,000,000 fascinated. It is assumed that about 5 people try each
example and so this indicates that nearly 600,000 have been sold.
23 Dec
1920, p. 3. Too many cooks.
23 Dec
1920, unnumbered back page.
Cartoon: The hexagonal dawn!
28 Dec
1920, p. 3? Puzzled millions. "On Christmas Eve the sales exceeded
600,000 ...."
29 Dec
1920, p. 3? "I will do
it."
30 Dec
1920, p. 8. Puzzle fun.
3 Jan 1921, p. 3. The Daily Mail Puzzle. C.
Lewis, aged 21, a postal clerk solved it within two hours of purchase and
submitted his solution on 7 Dec.
Hundreds of identical solutions were submitted, but no alternative
solutions have yet appeared. There are
two pairs of identical pieces: 1 &
12, 4 & 10.
3 Jan 1921, p. 10 = unnumbered back
page. Hexagon Puzzle Solved, with photo
of C. Lewis and diagram of solution.
10 Jan
1921, p. 4. Hexagon puzzle. Since no alternative hexagonal solutions
were received, the other £150 is awarded to those who submitted the most
ingenious other solution -- this was judged to be a butterfly shape, submitted
by 11 persons, who shared the £150.
Horace Hydes & Francis
Reginald Beaman Whitehouse. UK Patent
173,588 -- Improvements in Dominoes.
Applied: 29 Sep 1920; complete
application: 29 Jun 1921; accepted:
29 Dec 1921. Reproduced in
Haubrich, About ..., 1996, op. cit. below.
3pp + 1p diagrams. This is the
patent for the above puzzle, corresponding to provisional patent 27599/20
on the package. The illustration
shows a solved puzzle based on 'A
stitch in time saves nine'.
George Henry Haswell. US Patent 1,558,165 -- Puzzle. Applied: 3 Jul 1924; patented: 11 Sep 1925. Reproduced in Haubrich, About ..., 1996, op.
cit. below. 2pp + 1p diagrams. For edge-matching hexagons with further
internal markings which have to be aligned.
[E.g. one could draw a diagonal and require all diagonals to be vertical
-- this greatly simplifies the puzzle!]
If one numbers the vertices 1,
2, ..., 6, he gives an example formed
by drawing the diagonals 13, 15, 42,
46 which produces six triangles along
the edges and an internal rhombus.
C. Dudley Langford. Note 2829:
Dominoes numbered in the corners.
MG 43 (No. 344) (May 1959) 120‑122. Considers triangles, squares and hexagons with numbers at the
corners. There are the same number of
pieces as with numbers on the edges, but corner numbering gives many more kinds
of edges. E.g. with four numbers, there
are 24 triangles, but these have 16 edge patterns instead of 4. The editor (R. L. Goodstein) tells Langford
that he has made cubical dominoes "presumably with faces
numbered". Langford suggests cubes
with numbers at the corners. [I find 23
cubes with two corner numbers and 333 with three corner numbers. ??check]
Piet Hein. US Patent 4,005,868 -- Puzzle. Applied: 23 Jun 1975; patented: 1 Feb 1977. Front page + 8pp diagrams + 5pp text. Basically non-matching puzzles using marks
at the corners of faces of the regular polyhedra. He devises boards so the problems can be treated as planar.
Kiyoshi Takizawa; Naoaki Takashima & Nob.
Yoshigahara. Vess Puzzle and Its Family
-- A Compendium of 3 by
3 Card Puzzles. Published by the authors, Tokyo, Japan,
1983. Studies 32 types (in 48
versions) of 3 x 3 'head to tail' matching puzzles and 4
related types (in 4 versions).
All solutions are shown and most puzzles are illustrated with colour
photographs of one solution. (Haubrich
counts 51 versions -- check??)
Melford D. Clark. US Patent 4,410,180 -- Puzzle. Applied: 16 Nov 1981; patented: 18 Oct 1983. Reproduced in Haubrich, About ..., 1996, op.
cit. in 5.H.4. 2pp + 2pp diagrams. Corner matching squares, but with the pieces
marked 1, 2, ..., so that the pieces marked 1 form a
1 x 1 square, the pieces
marked 2 allow this to be extended to a
2 x 2 square, etc. There are
n2 - (n-1)2
pieces marked n.
Jacques Haubrich. Compendium of Card Matching Puzzles. Printed by the author, Aeneaslaan 21,
NL-5631 LA Eindhoven, Netherlands, 1995.
2 vol., 325pp. describing over
1050 puzzles. He classifies them by the nine most common
matching rules: Heads and Tails; Edge Matching (i.e. MacMahon); Path Matching; Corner Matching; Corner
Dismatching; Jig-Saw-Like; Continuous Path; Edge Dismatching;
Hybrid. He does not include
Jig-Saw-Like puzzles here. Using the
number of cards and their shape, then the matching rules, he has 136
types. 31 different numbers of cards occur: 4, 6-16, 18-21, 23-25, 28, 30, 36, 40, 45, 48, 56, 64, 70, 80,
85, 100. There is an index of 961
puzzle names. He says Hoffmann is the
earliest published example. He notes
that most path puzzles have a global criterion that the result have a single
circuit which slightly removes them from his matching criterion and he does not
treat them as thoroughly. He has
developed computer programs to solve each type of puzzle and has checked them
all.
Jacques Haubrich. About, Beyond and Behind Card Matching
Puzzles. [= Vol. 3 of above]. Ibid, Apr 1996, 87pp. This is a general discussion of the
different kinds of puzzles, how to solve them and their history, reproducing
ten patents and two obituaries.
5.I. LATIN SQUARES AND EULER SQUARES
This topic ties in with certain
tournament problems but I have not covered them. See also Hoffmann and Loughlin & Flood in 5.A.2 for examples
of two orthogonal 3 x 3 Latin squares.
The derangement problems in 5.K.2 give Latin rectangles.
Ahrens-1 & Ahrens-2. Opp. cit. in 7.N. 1917 & 1922. Ahrens-1
discusses and cites early examples of Latin squares, going back to medieval
Islam (c1200), where they were used on amulets. Ahrens-2 particularly discusses work of al‑Buni -- see
below.
(Ahmed [the h
should have an underdot] ibn ‘Alî ibn Jûsuf) el‑Bûni, (Abû'l‑‘Abbâs,
el‑Qoresî.) = Abu‑l‘Abbas
al‑Buni. (??= Muhyi'l‑Dîn
Abû’l-‘Abbâs al‑Bûnî -- can't
relocate my source of this form.) Sams
al‑ma‘ârif = Shams al‑ma‘ârif
al‑kubrâ = Šams
al-ma‘ārif. c1200. ??NYS.
Ahrens-1 describes this briefly and incorrectly. He expands and corrects this work in
Ahrens-2. See 7.N for more
details. Ahrens notes that a 4 x 4
magic square can be based on the pattern of two orthogonal Latin squares
of order 4, and Al-Buni's work indicates knowledge of such a pattern,
exemplified by the square
8, 11, 14,
1; 13, 2,
7, 12; 3, 16, 9,
6; 10, 5,
4, 15 considered (mod 4).
He also has Latin squares of order
4 using letters from a name of
God. He goes on to show 7
Latin squares of order 7, using the same 7 letters each time --
though four are corrupted. (Throughout,
the Latin squares also have 'Latin' diagonals, i.e. the diagonals contain all
the values.) These are arranged so each
has a different letter in the first place.
It is conjectured that these are associated with the days of the week or
the planets.
Tagliente. Libro de Abaco. (1515). 1541. F. 18v.
7 x 7 Latin square with
entries 1, 13, 2, 14, 3, 10,
4 cyclically shifted forward -- i.e.
the second row starts 13, 2, .... This is an elaborate plate which notes that
the sum of each file is 47 and has a motto: Sola Virtu la Fama Volla, but I
could find no text or other reason for its appearance!
Inscription on memorial to
Hannibal Bassett, d. 1708, in Meneage parish church, St. Mawgan, Cornwall. I first heard of this from Chris Abbess, who
reported it in some newsletter in c1993.
However, [Peter Haining; The Graveyard Wit; Frank Graham, Newcastle,
1973, p. 133] cites this as being at Cunwallow, near Helstone, Cornwall. [W. H. Howe; Everybody's Book of Epitaphs
Being for the Most Part What the Living Think of the Dead; Saxon & Co.,
London, nd [c1895] (facsimile by Pryor Publications, Whitstable, 1995); p. 173]
says it is in Gunwallow Churchyard.
Spelling and punctuation vary a bit.
The following gives a detailed account.
Alfred Hayman Cummings. The Churches and Antiquities of Cury &
Gunwalloe, in the Lizard District, including Local Traditions. E. Marlborough & Co., London & Truro,
1875, pp. 130-131. ??NX. "It has been said that there once
existed ... the curious epitaph --" and gives a considerable rearrangement
of the inscription below. He continues
"But this is in all probability a mistake, as repeated search has been
made for it, not only by the writer, but by a former Vicar of Gunwalloe, and it
could nowhere be found, while there is a plate with an inscription in
the church at Mawgan, the next parish, which might be very easily the one
referred to." He gives the
following inscription, saying it is to Hannibal Basset, d. 1708-9. Chris Weeks was kind enough to actually go
to the church of St. Winwaloe, Gunwalloe, where he found nothing, and to St.
Mawgan in Meneage, a few miles away.
Chris Weeks sent pictures of Gunwallowe -- the church is close to the
cliff edge and it looks like there could once have been more churchyard on the
other side of the church where the cliff has fallen away. In the church at St. Mawgan is the brass
plate with 'the Acrostic Brass Inscription', but it is not clearly associated
with a grave and I wonder if it may have been moved from Gunwallowe when a
grave was eroded by the sea. It is on
the left of the arch by the pulpit. I
reproduce Chris Weeks' copy of the text.
He has sent a photograph, but it was dark and the photo is not very
clear, but one can make out the Latin square part.
Hanniball
Baòòet here Inter'd doth lye
Who
dying lives to all Eternitye
hee
departed this life the 17th of Ian
1709/8
in the 22th year of his age ~
A
lover of learning
Shall wee all dye
Wee shall dye all
all dye shall wee
dye all wee shall
The òò are old style long esses.
The superscript th is actually over the numeral. The
9 is over the 8 in
the year and there is no stroke. This
is because it was before England adopted the Gregorian calendar and so the year
began on 25 Mar and was a year behind the continent between 1 Jan and 25
Mar. Correspondence of the time
commonly would show 1708/9 at this time, and I have used this form for
typographic convenience, but with the
9 over the 8 as
on the tomb.
A
word game book points out that this inscription is also palindromic!!
Richard Breen. Funny Endings. Penny Publishing, UK, 1999, p. 35. Gives the following form:
Shall we all die? / We shall die all. / All die shall we? /
Die all we shall and notes that
it is a word palindrome and says it comes from Gunwallam [sic], near Helstone.
Joseph Sauveur. Construction générale des quarrés
magiques. Mémoires de l'Académie Royale
des Sciences 1710(1711) 92‑138.
??NYS -- described in Cammann‑4, p. 297, (see 7.N for details of
Cammann) which says Sauveur invented Latin squares and describes some of his
work.
Ozanam. 1725.
1725: vol. IV, prob. 29, p. 434
& fig. 35, plate 10 (12). Two 4 x 4
orthogonal squares, using A, K,
Q, J of the 4 suits, but it looks
like:
J¨, A©, K§, Qª; Qª, K§, A¨, J©; A§, Jª, Q©, K¨; K©, Q¨, J§, Aª; but the
§ and ª look very similar. From later versions of the same diagram, it is clear that the
first row should have its § and ª reversed. Note the
diagonals also contain all four ranks and suits. (I have a reference for this to the 1723 edition.)
Minguet. 1733.
Pp. 146-148 (1864: 142-143; not noticed in other editions). Two
4 x 4 orthogonal squares,
using A, K, Q, J (= As, Rey, Caballo (knight), Sota (knave))
of the 4 suits, but the Spanish suits, in descending order, are: Espadas,
Bastos, Oros, Copas. The result is
described but not drawn, as:
RO,
AE, CC, SB; SC, CB, AO, RE; AB, RC, SE, CO; CE, SO, RB, AC;
which
would translate into the more usual cards as:
K¨, A§, Qª, J©; J§, Q©, A¨, Kª; A©, K§, Jª, Q¨; Qª, J¨, K©, A§.
However,
I'm not sure of the order of the Caballo and Sota; if they were reversed, which
would interchange Q and J in the latter pattern, then both Ozanam and Minguet
would have the property that each row is a cyclic shift or reversal of A, K, Q, J.
Alberti. 1747.
Art. 29, p. 203 (108) & fig. 36, plate IX, opp. p. 204
(108). Two 4 x 4 orthogonal squares,
figure simplified from the correct form of Ozanam, 1725.
L. Euler. Recherches sur une nouvelle espèce de
Quarrés Magiques. (Verhandelingen
uitgegeven door het zeeuwsch Genootschap der Wetenschappen te Vlissingen
(= Flessingue) 9 (1782) 85‑239.)
= Opera Omnia (1) 7 (1923) 291‑392. (= Comm. Arithm. 2 (1849) 302‑361.)
Manuel des Sorciers. 1825.
Pp. 78-79, art. 39. ??NX Correct form of Ozanam.
The Secret Out. 1859.
How to Arrange the Twelve Picture Cards and the four Aces of a Pack in
four Rows, so that there will be in Neither Row two Cards of the same Value nor
two of the same Suit, whether counted Horizontally or Perpendicularly, pp.
90-92. Two 4 x 4
orthogonal Latin squares, not the same as in Ozanam.
Bachet-Labosne. Problemes.
3rd ed., 1874. Supp. prob. XI,
1884: 200‑202. Two 4 x 4
orthogonal squares.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
Card Puzzles, No. XVI, pp. 17-18.
Similar to Ozanam.
Hoffmann. 1893.
Chap. X, no. 14: Another card puzzle, pp. 342 & 378-379
= Hoffmann‑Hordern, pp. 234 & 236. Two orthogonal Latin squares, but the diagonals do not contain
all the suits and ranks.
Aª, J©, Q¨, K§; J¨, A§, Kª, Q©; Q§, K¨, A©, Jª; K©, Qª, J§, A¨.
G. Tarry. Le probleme de 36 officiers. Comptes Rendus de l'Association Française pour
l'Avancement de Science Naturel 1 (1900) 122‑123 &
2 (1901) 170‑203. ??NYS
Dudeney. Problem 521. Weekly Dispatch (1 Nov, 15 Nov, 1903) both p. 10.
H. A. Thurston. Latin squares. Eureka 9 (Apr 1947) 19-21.
Survey of current knowledge.
T. G. Room. Note 2569:
A new type of magic square. MG
39 (No. 330) (Dec 1955) 307. Introduces
'Room Squares'. Take the 2n(2n‑1)/2 combinations from 2n symbols and insert them in a 2n‑1 x 2n‑1 grid so that each row and column contains
all 2n
symbols. There are n
entries and n‑1 blanks in each row and column. There is an easy solution for n = 1.
n = 2 and n = 3
are impossible. Gives a solution
for n = 4. This is a design for a round‑robin tournament with the
additional constraint of 2n‑1 sites such that each player plays once at
each site.
Parker shows there are two
orthogonal Latin squares of order 10 in 1959.
R. C. Bose &
S. S. Shrikande. On the falsity
of Euler's conjecture about the nonexistence of two orthogonal Latin squares of
order 4t+2. Proc. Nat. Acad. Sci. (USA) 45: 5 (1959) 734‑737.
Gardner. SA (Nov 1959) c= New MD, chap. 14.
Describes Bose & Shrikande's work.
SA cover shows a 10 x 10 counterexample in colour. Kara Lynn and David Klarner actually made a quilt
of this, thereby producing a counterpane counterexample! They told me that the hardest part of the
task was finding ten sufficiently contrasting colours of material.
H. Howard Frisinger. Note:
The solution of a famous two-centuries-old problem: the Leonhard Euler-Latin square
conjecture. HM 8 (1981) 56-60. Good survey of the history.
Jacques Bouteloup. Carrés Magiques, Carrés Latins et
Eulériens. Éditions du Choix, Bréançon,
1991. Nice systematic survey of this
field, analysing many classic methods.
An Eulerian square is essentially two orthogonal Latin squares.
See MUS I 210-284. S&B 37 shows examples. See also 5.Z. See also 6.T for examples where no three are in a row.
Ahrens. Mathematische Spiele. Encyklopadie article, op. cit. in 3.B. 1904.
Pp. 1082‑1084 discusses history and results for the n
queens problem, with many references.
Paul J. Campbell. Gauss and the eight queens problem. HM 4:4 (Nov 1977) 397‑404. Detailed history. Demonstrates that Gauss did not obtain a complete solution and
traces how this misconception originated and spread.
"Schachfreund" (Max
Bezzel). Berliner Schachzeitung 3 (Sep
1848) 363. ??NYS
Solutions. Ibid. 4 (Jan 1849) 40. ??NYS
(Ahrens says this only gives two solutions. A. C. White says two or three. Jaenisch says a total of 5
solutions were published here and in 1854.)
Franz Nauck. Eine in das Gebiet der Mathematik fallende
Aufgabe von Herrn Dr. Nauck in Schleusingen.
Illustrirte Zeitung (Leipzig) 14 (No. 361) (1 Jun 1850) 352. Reposes problem. [The papers do not give a first name or initial. The only Nauck in the first six volumes of
Poggendorff is Ernst Friedrich (1819-1875), a geologist. Ahrens gives no initial. Campbell gives Franz.]
Franz Nauck. Briefwechseln mit Allen für Alle. Illustrirte Zeitung (Leipzig) 15 (No. 377)
(21 Sep 1850) 182. Complete
solution.
Editorial comments:
Briefwechsel. Illustrirte Zeitung
(Leipzig) 15 (No. 378) (28 Sep 1850) 207.
Thanks 6 correspondents for the complete solution and says Nauck reports
that a blind person has also found all
92 solutions.
Gauss read the Illustrirte
Zeitung and worked on the problem, corresponding with his friend Schumacher
starting on 1 Sep 1850. Campbell
discusses the content of the letters, which were published in: C. A. F. Peters, ed; Briefwechsel zwischen
C. F. Gauss und H. C. Schumacher; vol. 6, Altona, 1865, ??NYS. John Brillhart writes that there is some
material in Gauss' Werke, vol. XII: Varia kleine Notizen verschiednen Inhalts
... 5, pp. 19-28, ??NYS -- not
cited by Campbell.
F. J. E. Lionnet. Question 251. Nouvelles Annales de Mathématiques 11 (1852) 114‑115. Reposes problem and gives an abstract
version.
Giusto Bellavitis. Terza rivista di alcuni articoli dei Comptes
Rendus dell'Accademia delle Scienze di Francia e di alcuni questioni des
Nouvelles Annales des mathématiques.
Atti dell'I. R. Istituto Veneto di Scienze, Lettere ed Arti (3) 6 [=
vol. 19] (1860/61) 376-392 & 411‑436 (as part of Adunanza del Giorno
17 Marzo 1861 on pp. 347‑436).
The material of interest is: Q. 251.
Disposizione sullo scacchiere di otto regine, on pp. 434‑435. Gives the
12 essentially different
solutions. Lucas (1895) says Bellavitis
was the first to find all solutions, but see above. However this may be the first appearance of the 12
essentially different solutions.
C. F. de Jaenisch. Op. cit. in 5.F.1. 1862. Vol. 1, pp.
122-135. Gives the 12 basic solutions and
shows they produce 92. Notes that in
every solution, 4 queens are on white squares and 4 are on black.
A. C. Cretaine. Études sur le Problème de la Marche du
Cavalier au Jeu des Échecs et Solution du Problème des Huit Dames. A. Cretaine, Paris, 1865. ??NYS -- cited by Lucas (1895). Shows it is possible to solve the eight
queens problem after placing one queen arbitrarily.
G. Bellavitis. Algebra N. 72 Lionnet. Atti dell'Istituto Veneto (3) 15 (1869/70)
844‑845.
Siegmund Günther. Zur mathematische Theorie des
Schachbretts. Grunert's Archiv der
Mathematik und Physik 56 (1874) 281-292.
??NYS. Sketches history of the
problem -- see Campbell. He gives a
theoretical, but not very practical, approach via determinants which he carries
out for 4 x 4 and 5 x 5.
J. W. L. Glaisher. On the problem of the eight queens. Philosophical Magazine (4) 48 (1874)
457-467. Gives a sketch of Günther's
history which creates several errors, in particular attributing the solution to
Gauss -- see Campbell, who suggests Glaisher could not read German well. (However, in 1921 & 1923, Glaisher
published two long articles involving the history of 15-16C German mathematics,
showing great familiarity with the language.)
Simplifies and extends Günther's approach and does 6 x 6,
7 x 7, 8 x 8 boards.
Lucas. RM2, 1883. Note V: Additions du Tome premier. Pp. 238-240. Gives the solutions on the
9 x 9 board, due to P. H.
Schoute, in a series of articles titled Wiskundige Verpoozingen in Eigen
Haard. Gives the solutions on the 10 x 10
board, found by M. Delannoy.
S&B, p. 37, show an 1886
puzzle version of the six queens problem.
A. Pein. Aufstellung von n Königinnen auf einem
Schachbrett von n2 Feldern.
Leipzig. ??NYS -- cited by Ball,
MRE, 4th ed., 1905 as giving the 92 inequivalent solutions on the 10 x 10.
Ball. MRE, 1st ed., 1892. The
eight queens problem, pp. 85-88. Cites
Günther and Glaisher and repeats the historical errors. Sketches Günther's approach, but only cites
Glaisher's extension of it. He gives
the numbers of solutions and of inequivalent solutions up through 10 x 10
-- see Dudeney below for these numbers, but the two values in ( )
are not given by Dudeney. He
states results for the 9 x 9 and
10 x 10, citing Lucas. Says that a
6 x 6 version "is sold in
the streets of London for a penny".
Hoffmann. 1893.
Chap. VI, pp. 272‑273 & 286 = Hoffmann-Hordern,
pp. 187-189, with photo.
No.
24: No two in a row. Eight queens. Photo on p. 188 shows Jeu des Sentinelles,
by Watilliaux, dated 1874-1895.
No.
25: The "Simple" Puzzle. Nine
queens. Says a version was sold by
Messrs. Feltham, with a notched board but the pieces were allowed to move over
the gaps, so it was really a 9 x 9 board.
Chap. X, No. 18: The Treasure at Medinet, pp. 343‑344
& 381 = Hoffmann-Hordern, pp. 237-239. This is a solution of the eight queens problem, cut into four
quadrants and jumbled. The goal is to
reconstruct the solution. Photo on p.
239 shows Jeu des Manifestants, with box.
Hordern
Collection, p. 94, and S&B, p. 37, show a version of this with same box,
but which divides the board into eight
2 x 4 rectangles.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co.,
NY), nd [1895]. P. 1: The famous
Italian pin puzzle. 6 queens puzzle. No solution.
Lucas. L'Arithmétique Amusante.
1895. Note IV: Section I: Les huit dames, pp. 210-220. Asserts Bellavitis was the first to find all
solutions. Discusses symmetries and
shows the 12 basic solutions. Correctly
describes Jaenisch as obscure. Gives an
easy solution of Cretaine's problem which can be remembered as a trick. Shows there are six solutions which can be
superimposed with no overlap, i.e. six solutions using disjoint sets of cells.
C. D. Locock, conductor. Chess Column. Knowledge 19 (Jan 1896)
23-24; (Feb 1896) 47‑48; (May 1896) 119; (Jul 1896) 167-168. This
series begins by saying most players know there is a solution, "but,
possibly, some may be surprised to learn that there are ninety-two ways of performing
the feat, ...." He then enumerates
them. Second article studies various properties
of the solutions, particularly looking for examples where one solution shifts
to produce another one. Third article
notes some readers' comments. Fourth
article is a long communication from W. J. Ashdown about the number of distinct
solutions, which he gets as 24 rather than the usual 12.
T. B. Sprague. Proc. Edinburgh Math. Soc. 17 (1898-9)
43-68. ??NYS -- cited by Ball, MRE, 4th
ed., 1905, as giving the 341 inequivalent solutions on the 11 x 11.
Benson. 1904.
Pins and dots puzzle, p. 253. 6
queens problem, one solution.
Ball. MRE, 4th ed., 1905. The
eight queens problem, pp. 114-120.
Corrects some history by citing MUS, 1st ed., 1901. Gives one instance of Glaisher's method --
going from 4 x 4 to 5
x 5 and its results going up to 8 x 8.
Says the 92 inequivalent solutions on the 10 x 10 were given
by Pein and the 341 inequivalent solutions on the 11 x 11 were given by
Sprague. The 5th ed., pp. 113-119 calls
it "One of the classical problems connected with a chess-board" and
adds examples of solutions up to 21 x
21 due to Mr. Derington.
Pearson. 1907.
Part III, no. 59: Stray dots, pp. 59 & 130. Same as Hoffmann's Treasure at Medinet.
Burren Loughlin &
L. L. Flood. Bright-Wits Prince of Mogador. H. M. Caldwell Co., NY, 1909.
The eight provinces, pp. 14-15 & 65. Same as Hoffmann's Treasure at Medinet.
A. C. White. Sam Loyd and His Chess Problems. 1913.
Op. cit. in 1. P. 101 says Loyd
discovered that all solutions have a piece at
d1 or equivalent.
Williams. Home Entertainments. 1914.
A draughtboard puzzle, p. 115.
"Arrange eight men on a draughtboard in such a way that no two are
upon the same line in any direction." This is not well stated!!
Gives one solution:
52468317 and says "Work out other solutions for
yourself."
Dudeney. AM.
1917. The guarded chessboard,
pp. 95‑96. Gives the number of
ways of placing n queens and the number of inequivalent ways. The values in ( ) are given by Ball,
but not by Dudeney.
n 4 5 6 7 8 9 10 11 12 13
ways 2 10 4 40 92 (352) (724) ‑ - -
inequivalent ways 1 2 1 6 12 46 92 341 (1766) (1346)
Ball. MRE, 9th ed., 1920. The
eight queens problem, pp. 113-119.
Omits references to Pein and Sprague and adds the number of inequivalent
solutions for the 12 x 12 and
13 x 13.
Blyth. Match-Stick Magic.
1921. No pairs allowed, p.
74. 6 queens problem.
Hummerston. Fun, Mirth & Mystery. 1924.
No two in a line, p. 48.
Chessboard. Place 'so that no
two are upon the same line in any direction along straight or diagonal
lines?' Gives one solution: 47531682,
'but there are hundreds of other ways'.
You can let someone place the first piece.
Rohrbough. Puzzle Craft. 1932. Houdini Puzzle, p.
17. 6 x 6 case.
Rohrbough. Brain Resters and Testers. c1935.
Houdini Puzzle, p. 25. 6 x
6 problem. "-- From New York World some years ago, credited to
Harry Houdini." I have never seen
this attribution elsewhere.
Pál Révész. Mathematik auf dem Schachbrett. In:
Endre Hódi, ed. Mathematisches Mosaik.
(As: Matematikai Érdekességek;
Gondolat, Budapest, 1969.) Translated
by Günther Eisenreich. Urania‑Verlag,
Leipzig, 1977. Pp. 20‑27. On p. 24, he says that all solutions have 4
queens on white and 4 on black. He says
that one can place at most 5 non‑attacking queens on one colour.
Doubleday - 2. 1971.
Too easy?, pp. 97-98. The two
solutions on the 4 x 4 board are disjoint.
Dean S. Clark & Oved
Shisha. Proof without words: Inductive
construction of an infinite chessboard with maximal placement of nonattacking
queens. MM 61:2 (1988) 98. Consider a
5 x 5 board with queens in cells (1,1), (2,4), (3,2), (4,5), (5,3). 5 such boards can be similarly placed within
a 25 x 25 board viewed as a 5 x
5 array of 5 x 5 boards and this has
no queens attacking. Repeating the
inflationary process gives a solution on the board of edge 53, then the board of edge 54, ....
They cite their paper:
Invulnerable queens on an infinite chessboard; Annals of the NY Acad. of
Sci.: Third Intern. Conf. on Comb. Math.; to appear. ??NYS.
Liz Allen. Brain Sharpeners. Op. cit. in 5.B.
1991. Squares before your eyes,
pp. 21 & 106. Asks for solutions of
the eight queens problem with no piece on either main diagonal. Two of the 12 basic solutions have this, but
one of these is the symmetric case, so there are 12 solutions of this problem.
Donald E. Knuth. Dancing links. 25pp preprint of a talk given at Oxford in Sep 1999, sent by the
author. See the discussion in 6.F. He finds the following numbers of solutions
for placing n queens, n = 1, 2, ...,
18.
1, 0,
0, 2, 10, 4, 40,
92, 352, 724,
2680, 14200, 73712,
3 65596, 22 79184, 147 72512,
958 15104, 6660 90624.
5.I.2. COLOURING CHESSBOARD WITH NO REPEATS IN A LINE
New
section. I know there is a general
result that an n x n board can be n‑coloured if
n satisfies some condition
like n º 1 or 5
(mod 6), but I don't recall any other
old examples of the problem.
Dudeney. Problem 50: A problem in mosaics. Tit‑Bits 32 (11 Sep 1897) 439 &
33 (2 Oct 1897) 3.
An 8 x 8 board with two adjacent corners omitted can
be 8‑coloured with no two in a row, column or diagonal. = Anon. & Dudeney; A chat with the
Puzzle King; The Captain 2 (Dec? 1899) 314-320; 2:6 (Mar 1900) 598-599
& 3:1 (Apr 1900) 89.
Dudeney. AM.
1917. Prob. 302: A problem in
mosaics, pp. 90 & 215-216. The
solution to the previous problem is given and then it is asked to relay the
tiles so that the omitted squares are the
(3,3) and (3,6)
cells.
Hummerston. Fun, Mirth & Mystery. 1924.
Q.E.D. -- The office boy problem, Puzzle no. 30, pp. 82 & 176. Wants to mark the cells of a 4 x 4
board with no two the same in any 'straight line ..., either
horizontally, vertically, or diagonally.'
His answer is:
ABCD, CDEA,
EABC, BCDA, which has no two the same on any short
diagonal. The problem uses coins of
values: A, B, C, D, E =
12, 30, 120, 24, 6 and the
object is to maximize the total value of the arrangement. In fact, there are only two ways to 5-colour
the board and they are mirror images.
Four colours are used three times and one is used four times -- setting
the value 120 on the latter cells gives the maximum value of 696.
NOTE.
Perfect means no two squares are the same size. Compound means there is a squared
subrectangle. Simple means not
compound.
Dudeney. Puzzling Times at Solvamhall Castle: Lady
Isabel's casket. London Mag. 7
(No. 42) (Jan 1902) 584 & 8 (No. 43) (Feb 1902) 56. = CP, prob. 40, pp. 67 & 191‑193. Square into
12 unequal squares and a
rectangle.
Max Dehn. Über die Zerlegung von Rechtecken in
Rechtecke. Math. Annalen 57 (1903) 314‑332. Long and technical. No examples. Shows sides must be parallel and commensurable.
Loyd. The patch quilt puzzle.
Cyclopedia, 1914, pp. 39 & 344.
= MPSL1, prob. 76, pp. 73 & 147‑148. c= SLAHP: Building a patchquilt, pp. 30
& 92. 13 x 13 into
11 squares, not simple nor
perfect. (Gardner, in 536, says this
appeared in Loyd's "Our Puzzle Magazine", issue 1 (1907), ??NYS.)
Loyd. The darktown patch quilt party.
Cyclopedia, 1914, pp. 65 & 347.
12 x 12 into 11
squares, not simple nor perfect, in two ways.
P. J. Federico. Squaring rectangles and squares -- A
historical review with annotated bibliography.
In: Graph Theory and Related
Topics; ed. by J. A. Bondy & U. S. R. Murty; Academic Press, NY, 1979, pp.
173‑196. Pp. 189‑190 give
the background to Moroń's work.
Moroń later found the first example of Sprague but did not publish
it.
Z. Moroń. O rozkładach prostokątów na
kwadraty (In Polish) (On the dissection of a rectangle into squares). Przegląd Matematyczno‑Fizyczny
(Warsaw) 3 (1925) 152‑153.
Decomposes rectangles into 9 and 10 unequal squares. (Translation provided by A. Mąkowski,
1p. Translation also available from M.
Goldberg, ??NYS.)
M. Kraitchik. La Mathématique des Jeux, 1930, op. cit. in
4.A.2, p. 272. Gives Loyd's "Patch
quilt puzzle" solution and Lusin's opinion that there is no perfect solution.
A. Schoenflies. Einführung in der analytische Geometrie der
Ebene und des Raumes. 2nd ed., revised
and extended by M. Dehn, Springer, Berlin, 1931. Appendix VI: Ungelöste Probleme der Analytischen Geometrie,
pp. 402‑411. Same results as
in Dehn's 1903 paper.
Michio Abe. On the problem to cover simply and without
gap the inside of a square with a finite number of squares which are all
different from one another (in Japanese).
Proc. Phys.‑Math. Soc. Japan 4 (1931) 359‑366. ??NYS
Michio Abe. Same title (in English). Ibid. (3) 14 (1932) 385‑387. Gives
191 x 195 rectangle into 11
squares. Shows there are squared
rectangles arbitrarily close to squares.
Alfred Stöhr. Über Zerlegung von Rechtecken in
inkongruente Quadrate. Schr. Math.
Inst. und Inst. angew. Math. Univ. Berlin 4:5 (1939), Teubner, Leipzig, pp. 119‑140. ??NYR.
(This was his dissertation at the Univ. of Berlin.)
S. Chowla. The division of a rectangle into unequal
squares. Math. Student 7 (1939) 69‑70. Reconstructs Moroń's 9
square decomposition.
Minutes of the 203rd Meeting of
the Trinity Mathematical Society (Cambridge) (13 Mar 1939). Minute Books, vol. III, pp. 244‑246. Minutes of A. Stone's lecture: "Squaring the Square". Announces Brooks's example with 39
elements, side 4639, but containing a perfect subrectangle.
Minutes of the 204th Meeting of
the Trinity Mathematical Society (Cambridge) (24 Apr 1939). Minute Books, vol. III, p. 248. Announcement by C. A. B. Smith that Tutte
had found a perfect squared square with no perfect subrectangle.
R. Sprague. Recreation in Mathematics. Op. cit. in 4.A.1. 1963. The expanded
foreword of the English edition adds comments on Dudeney's "Lady Isabel's
Casket", which led to the following paper.
R. Sprague. Beispiel einer Zerlegung des Quadrats in
lauter verschiedene Quadrate. Math.
Zeitschr. 45 (1939) 607‑608.
First perfect squared square --
55 elements, side 4205.
R. Sprague. Zur Abschätzung der Mindestzahl
inkongruenter Quadrate, die ein gegebenes Rechteck ausfüllen. Math. Zeitschrift 46 (1940) 460‑471. Tutte's 1979 commentary says this shows
every rectangle with commensurable sides can be dissected into unequal squares.
A. H. Stone, proposer; M. Goldberg & W. T. Tutte, solvers. Problem E401. AMM 47:1 (Jan 1940) 48
& AMM 47:8 (Oct 1940) 570‑572. Perfect squared square -- 28
elements, side 1015.
R. L. Brooks, C. A. B. Smith, A. H. Stone & W. T. Tutte. The dissection of rectangles into squares. Duke Math. J. 7 (1940) 312‑340. = Selected Papers of W. T. Tutte; Charles
Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 10-38, with
commentary by Tutte on pp. 1-9. Tutte's
1979 commentary says Smith was perplexed by the solution of Dudeney's
"Lady Isabel's Casket" -- see also his 1958 article.
A. H. Stone, proposer; Michael Goldberg, solver. Problem E476. AMM 48 (1941) 405 ??NYS
& 49 (1942) 198-199. An isosceles right triangle can be dissected
into 6
similar figures, all of different sizes. Editorial notes say that Douglas and Starke found a different
solution and that one can replace
6 by any larger number, but it
is not known if 6 is the least such. Stone asks if there is any solution where the smaller triangles
have no common sides.
M. Kraitchik. Mathematical Recreations, op. cit. in 4.A.2,
1943. P. 198. Shows the compound perfect squared square with 26
elements and side 608 from Brooks, et al.
C. J. Bouwkamp. On the construction of simple perfect
squared squares. Konink. Neder. Akad.
van Wetensch. Proc. 50 (1947) 72-78 = Indag. Math. 9 (1947) 57-63. This criticised the method of Brooks, Smith,
Stone & Tutte, but was later retracted.
Brooks, Smith, Stone &
Tutte. A simple perfect square. Konink. Neder. Akad. van Wetensch. Proc. 50
(1947) 1300‑1301. = Selected
Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre,
Manitoba, 1979; pp. 99-100, with commentary by Tutte on p. 98. Bouwkamp had published several notes and was
unable to make the authors' 1940 method work.
Here they clarify the situation and give an example. One writer said they give details of Sprague's
first example, but the example is not described as being the same as in
Sprague.
W. T. Tutte. The dissection of equilateral triangles into
equilateral triangles. Proc. Camb. Phil
Soc. 44 (1948) 464‑482. =
Selected Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre,
Manitoba, 1979; pp. 106-125, with commentary by Tutte on pp. 101-105.
T. H. Willcocks, proposer and
solver. Problem 7795. Fairy Chess Review 7:1 (Aug 1948) 97 &
106 (misnumberings for 5 & 14). Refers
to prob. 7523 -- ??NYS. Finds compound
perfect squares of orders 27, 27, 28
and 24.
T. H. Willcocks. A note on some perfect squares. Canadian J. Math. 3 (1951) 304‑308. Describes the result in Fairy Chess Review
prob. 7795.
T. H. Willcocks. Fairy Chess Review (Feb & Jun
1951). Prob. 8972. ??NYS -- cited and described by G. P.
Jelliss; Prob. 44 -- A double squaring, G&PJ 2 (No. 17) (Oct 1999) 318-319. Squares of edges 3, 5, 9, 11, 14, 19, 20, 24, 31, 33, 36, 39, 42 can be formed into a 75 x 112
rectangle in two different ways.
{These are reproduced, without attribution, as Fig. 21, p. 33 of Joseph
S. Madachy; Madachy's Mathematical Recreations; Dover, 1979 (this is a
corrected reprint of Mathematics on Vacation, 1966, ??NYS). The 1979 ed. has an errata slip inserted for
p. 33 as the description of Fig. 21 was omitted in the text, but the erratum
doesn't cite a source for the result.}
The G&PJ problem then poses a new problem from Willcocks involving 21
squares to be made into a rectangle in two different ways -- it is not
clear if these have to be the same shape.
M. Goldberg. The squaring of developable surfaces. SM 18 (1952) 17‑24. Squares cylinder, Möbius strip, cone.
W. T. Tutte. Squaring the square. Guest column for SA (Nov 1958). c= Gardner's 2nd Book, pp. 186‑209. The latter
= Selected Papers of W. T.
Tutte; Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp.
244-266, with a note by Tutte on p. 244, but the references have been
omitted. Historical account -- cites
Dudeney as the original inspiration of Smith.
R. L. Hutchings &
J. D. Blake. Problems drive
1962. Eureka 25 (Oct 1962) 20-21 &
34-35. Prob. G. Assemble squares of sides 2, 5, 7, 9, 16, 25, 28, 33, 36 into a rectangle. The rectangle is 69 x
61 and is not either of Moroń's
examples.
W. T. Tutte. The quest of the perfect square. AMM 72:2, part II (Feb 1965) 29-35. = Selected Papers of W. T. Tutte;
Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 432-438,
with brief commentary by Tutte on p. 431.
General survey, updating his 1958 survey.
Blanche Descartes [pseud. of
Cedric A. B. Smith]. Division of a
square into rectangles. Eureka 34
(1971) 31-35. Surveys some history and
Stone's dissection of an isosceles right triangle into 6
others of different sizes (see above).
Tutte has a dissection of an equilateral triangle into 15
equilateral triangles -- but some of the pieces must have the same area
so we consider up and down pointing triangles as + and - areas and then all the
areas are different. Author then
considers dissecting a square into incongruent but equiareal rectangles. He finds it can be done in n
pieces for any n ³ 7.
A. J. W. Duijvestijn. Simple perfect squared square of lowest
order. J. Combinatorial Thy. B 25
(1978) 240‑243. Finds a perfect
square of minimal order 21.
A. J. W. Duijvestin, P. J.
Federico & P. Leeuw. Compound
perfect squares. AMM 89 (1982) 15‑32. Shows Willcocks' example has the smallest
order for a compound perfect square and is the only example of its order, 24.
This is the problem of cutting a
square into smaller squares.
Loyd. Cyclopedia, 1914, pp. 248 & 372, 307 & 380. Cut 3 x 3
into 6 squares: 2 x 2 and
5 1 x 1.
Dudeney. AM.
1917. Prob. 173: Mrs Perkins's
quilt, pp. 47 & 180. Same as Loyd's
"Patch quilt puzzle" in 5.J.
Dudeney. PCP.
1932. Prob. 117: Square of
Squares, pp. 53 & 148‑149.
= 536, prob. 343, pp. 120 & 324‑325. c= "Mrs Perkins's quilt".
N. J. Fine & I. Niven,
proposers; F. Herzog, solver. Problem E724 -- Admissible Numbers. AMM 53 (1946) 271 & 54 (1947) 41‑42. Cubical version.
J. H. Conway. Mrs Perkins's quilt. Proc. Camb. Phil. Soc. 60 (1964) 363‑368.
G. B. Trustrum. Mrs Perkins's quilt. Ibid. 61 (1965) 7‑11.
Ripley's Puzzles and Games. 1966.
Pp. 16-17, item 7. "Can you
divide a square into 6 perfect squares?"
Answer as in Loyd.
Nick Lord. Note 72.11:
Subdividing hypercubes. MG 72
(No. 459) (Mar 1988) 47‑48. Gives
an upper bound for impossible numbers in
d dimensions.
David Tall. To prove or not to prove. Mathematics Review 1:3 (Jan 1991)
29-32. Tall regularly uses the question
as an exercise in problem solving.
About ten years earlier, a 14 year old girl pointed out that the problem
doesn't clearly rule out rejoining pieces.
E.g. by cutting along the diagonals and rejoining, one can make two
squares.
S. Chowla. Problem
1779. Math. Student 7 (1939) 80. (Solution given in Brooks, et al., Duke
Math. J., op. cit. in 5.J, section 10.4, but they give no reference to a
solution in Math. Student.)
5.J.3. TILING A SQUARE OF SIDE 70 WITH SQUARES OF SIDES 1, 2, ..., 24
J. R. Bitner. Use of Macros in Backtrack Programming. M.Sc. Thesis, ref. UIUCDCS‑R‑74‑687,
Univ. of Illinois, Urbana‑Champaign, 1974, ??NYS. Shows such a tiling is impossible.
Let D(n) = the number of derangements of n
things, i.e. permutations
leaving no point fixed.
Eberhard Knobloch. Euler and the history of a problem in
probability theory.
Gaņita-Bhāratī [NOTE:
ņ denotes an n
with an underdot] (Bull. Ind. Soc. Hist. Math.) 6 (1984) 1‑12. Discusses the history, noting that many 19C
authors were unaware of Euler's work.
There is some ambiguity in his descriptions due to early confusion
of n
as the number of cards and
n as the number of the card on
which a match first occurs. Describes
numerous others who worked on the problem up to about 1900: De Moivre, Waring, Lambert, Laplace, Cantor,
etc.
Pierre Rémond de Montmort. Essai d'analyse sur les jeux de
hazards. (1708); Seconde edition revue & augmentee de
plusieurs lettres, (Quillau, Paris,
1713 (reprinted by Chelsea, NY, 1980));
2nd issue, Jombert & Quillau, 1714.
Problèmes divers sur le jeu du trieze, pp. 54‑64. In the original game, one has a deck of 52
cards and counts 1, 2, ...,
13 as one turns over the cards. If a card of rank i occurs at the i-th count, then the player wins. In general, one simplifies by assuming there
are n
distinct cards numbered 1, ...,
n and one counts 1, ..., n.
One can ask for the probability of winning at some time and of winning
at the k-th draw. In 1708, Montmort already gives tables of
the number of permutations of n cards such that one wins on the k-th draw, for n = 1, ..., 6.
He gives various recurrences and the series expression for the
probability and (more or less) finds its limit. In the 2nd ed., he gives a proof of the series expression, due to
Nicholas Bernoulli, and John Bernoulli says he has found it also. Nicholas' solution covers the general case
with repeated cards. [See: F. N. David; Games, Gods and Gambling; Griffin,
London, 1962, pp. 144‑146 & 157.]
(Comtet and David say it is in the 1708 ed. I have seen it on pp. 54-64 of an edition which is uncertain, but
probably 1708, ??NX. Knobloch cites
1713, pp. 130-143, but adds that Montmort gave the results without proofs in
the 1708 ed. and includes several letters from and to John I and Nicholas I
Bernoulli in the 1713 ed., pp. 290-324, and mentions the problem in his Preface
-- ??NYS.)
Abraham de Moivre. The Doctrine of Chances: or, A Method of
Calculating the Probability of Events in Play.
W. Pearson for the Author, London, 1718. Prob. XXV, pp. 59-63.
(= 2nd ed, H. Woodfall for the Author, London, 1738. Prob. XXXIV, pp. 95-98.) States and demonstrates the formula for
finding the probability of p items to be correct and q
items to be incorrect out of
n items. One of his examples is the probability of
six items being deranged being 53/144.
L. Euler. Calcul de la probabilité dans le jeu de
rencontre. Mémoires de l'Académie des
Sciences de Berlin (7) (1751(1753)) 255‑270. = Opera Omnia (1) 7 (1923) 11‑25. Obtains the series for the probability and
notes it approaches 1/e.
L. Euler. Fragmenta ex Adversariis Mathematicis
Deprompta. MS of 1750‑1755. Pp. 287‑288: Problema de
permutationibus. First published in
Opera Omnia (1) 7 (1923) 542‑545.
Obtains alternating series for
D(n).
Ozanam-Montucla. 1778.
Prob. 5, 1778: 125-126; 1803:
123-124; 1814: 108-109; 1840: omitted. Describes Jeu du Treize, where a person
takes a whole deck and turns up the cards, counting 1, 2, ..., 13 as he
goes. He wins if a card of rank i
appears at the i‑th
count. Montucla's description is brief
and indicates there are several variations of the game. Hutton gives a lengthier description of one
version. Cites Montmort for the
probability of winning as .632..
L. Euler. Solutio quaestionis curiosae ex doctrina
combinationum. (Mem. Acad. Sci. St.
Pétersbourg 3 (1809/10(1811)) 57‑64.)
= Opera Omnia (1) 7 (1923) 435‑440. (This was presented to the Acad. on 18 Oct 1779.) Shows
D(n) = (n‑1) [D(n‑1) + D(n‑2)] and
D(n) = nD(n‑1) + (‑1)n.
Ball. MRE. 1st ed., 1892. Pp. 106-107: The mousetrap and Treize. In the first, one puts out n
cards in a circle and counts out.
If the count k occurs on the k-th card, the card is removed and one starts again. Says Cayley and Steen have studied
this. It looks a bit like a derangement
question.
Bill Severn. Packs of Fun. 101 Unusual Things to Do with Playing Cards and To Know about
Them. David McKay, NY, 1967. P. 24: Games for One: Up and down. Using a deck of 52 cards, count through 1, 2, ..., 13 four times. You lose if a
card of rank i appears when you count i, i.e. you win if the cards are a
generalized derangement. Though a
natural extension of the problem, I can't recall seeing it treated, perhaps
because it seems to get very messy.
However, a quick investigation reveals that the probability of such a
generalized derangement should approach
e-4.
Brian R. Stonebridge. Derangements of a multiset. Bull. Inst. Math. Appl. 28:3 (Mar 1992)
47-49. Gets a reasonable extension to
multisets, i.e. sets with repeated elements.
5.K.1. DERANGED BOXES OF A, B AND A & B
Three boxes contain A
or B or A & B, but they have been shifted about so each is
in one of the other boxes. You can look
at one item from one box to determine what is in all of them. This is just added and is certainly older
than the examples below.
Simon Dresner. Science World Book of Brain Teasers. 1962.
Op. cit. in 5.B.1. Prob. 84:
Marble garble, pp. 40 & 110. Black
and white marbles.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 26: Mexican
jumping beans, pp. 40-41 & 96. Red
and black beans in matchboxes. The
problem continues with a Bertrand box paradox -- see 8.H.1.
Doubleday - 3. 1972.
Open the box, pp. 147-148. Black
and white marbles.
5.K.2. OTHER LOGIC PUZZLES BASED ON DERANGEMENTS
These
typically involve a butcher, a baker and a brewer whose surnames are Butcher,
Baker and Brewer, but no one has the profession of his name. I generally only state the beginning of the
problem.
New
section -- there must be older examples.
Gardner, in an article: My ten
favorite brainteasers in Games (collected in Games Big Book of Games, 1984, pp.
130-131) says this is one of his favorite problems. ??locate
I
now see these lead to Latin rectangles, cf Section 5.I.
R. Turner, proposer: The sons of
the dons; Eureka 2 (May 1939) 9-10. K.
Tweedie, solver: On the problem of the sons of the dons. Eureka 4 (May 1940) 21-23. Six dons, in analysis, geometry, algebra,
dynamics, physics and astronomy, each have a son who studies one of these
subjects, but none studies the subject of his father. Several further restrictions, e.g., there are no two students who
each study the subject of the other's father.
M. Adams. Puzzle Book. 1939. Prob. B.91: Easter
bonnet, pp. 80 & 107. Women named
Green, Black, Brown and White with 4 colours of hats and 4 colours of dresses,
but name, hat and dress are always distinct.
J. B. Parker. Round the table. Eureka 5 (Jan 1941)
20-21 & 6 (May 1941) 11. Seven
men, whose names are colours, with ties, socks and cars, being coloured with
three of the names of other men and all colours used for each item, sitting at
a table with eight places.
Anonymous. The umbrella problem. Eureka
9 (Apr 1947) 22 & 10 (Mar 1948) 25. Six men 'of negligible honesty' each go away with another's
umbrella.
Jonathan Always. Puzzles to Puzzle You. Tandem, London, 1965. No. 30: Something about ties, pp. 16 &
74-75. Black, Green and Brown are
wearing ties, but none has the colour of his name, remarked the green tie
wearer to Mr. Black.
David Singmaster. The deranged secretary. If a secretary puts n
letters all in wrong envelopes, how many envelopes must one open before
one knows what is each of the unopened envelopes?
Problem
proposal and solution 71.B. MG 71 (No.
455) (Mar 1987) 65 & 71 (No. 457) (Oct 1987) 238-239.
Open
question. The Weekend Telegraph (11 Jun
1988) XV & (18 Jun 1988) XV.
This is a solitaire (= patience) game
developed by Cayley, based on Treize.
Take a deck of cards, numbered
1, 2, ..., n, and shuffle them. Count through them. If a card does not match its count, put it
on bottom and continue. If it matches,
set it aside and start counting again from 1. One wins if all cards are set aside. In this case, pick up the deck and start a
new game.
T. W. O. Richards,
proposer; Richard I. Hess, solver. Prob. 1828.
CM 19 (1993) 78 & 20 (1994) 77-78. Asks whether there is any arrangement which allows three or more
consecutive wins. No theoretical
solution. Searching finds one solution
for n = 6 and n = 8 and
8 solutions for n = 9.
How
many ways can n couples be seated, alternating sexes, with
no couples adjacent?
A. Cayley. On a problem of arrangements. Proc. Roy. Soc. Edin. 9 (1878) 338‑342. Problem raised by Tait. Uses inclusion/exclusion to get a closed
sum.
T. Muir. On Professor Tait's problem of
arrangements. Ibid., 382‑387. Uses determinants to get a simple n‑term recurrence.
A. Cayley. Note on Mr. Muir's solution of a problem of
arrangement. Ibid., 388‑391. Uses generating function to simplify to a
usable form.
T. Muir. Additional note on a problem of
arrangement. Ibid., 11 (1882) 187‑190. Obtains Laisant's 2nd order and 4th order
recurrences.
É. Lucas. Théorie des Nombres. Gauthier‑Villars, Paris, 1891; reprinted by Blanchard, Paris, 1958. Section 123, example II, p. 215 &
Note III, pp. 491‑495.
Lucas appears not to have known of the work of Cayley and Muir. He describes Laisant's results. The 2nd order, non‑homogeneous
recurrence, on pp. 494‑495, is attributed to Moreau.
C. Laisant. Sur deux problèmes de permutations. Bull. Soc. Math. de France 19 (1890‑91)
105‑108. General approach to
problems of restricted occupancy. His
work yields a 2nd order non-homogeneous recurrence and homogeneous 3rd and 4th
order recurrences. He cites Lucas, but
says Moreau's work is unpublished.
H. M. Taylor. A problem on arrangements. Messenger of Math. 32 (1903) 60‑63. Gets almost to Muir & Laisant's 4th
order recurrence.
J. Touchard. Sur un problème de permutations. C. R. Acad. Sci. Paris 198 (1934) 631‑633. Solution in terms of a complicated
integral. States the explicit
summation.
I. Kaplansky. Solution of the "problème des
ménages". Bull. Amer. Math. Soc.
49 (1943) 784‑785. Obtains the
now usual explicit summation.
I. Kaplansky & J.
Riordan. The problème de ménages. SM 12 (1946) 113‑124. Gives the history and a uniform approach.
J. Touchard. Permutations discordant with two given
permutations. SM 19 (1953) 109‑119. Says he prepared a 65pp MS developing the
results announced in 1934 and rediscovered in Kaplansky and in Kaplansky &
Riordan. Proves Kaplansky's lemma on
selections by finding the generating functions which involve Chebyshev
polynomials. Obtains the explicit
summation, as done by Kaplansky.
Extends to more general problems.
M. Wyman & L. Moser. On the 'problème des ménages'. Canadian J. Math. 10 (1958) 468‑480. Analytic study. Updates the history -- 26 references. Gives table of values for
n = 0 (1) 65.
Jacques Dutka. On the 'Problème des ménages'. Math. Intell. 8:3 (1986) 18‑25 &
33. Thorough survey & history -- 25
references.
Kenneth P. Bogart & Peter G.
Doyle. Non‑sexist solution of the
ménage problem. AMM 93 (1986) 514‑518. 14 references.
5.M. SIX PEOPLE AT A PARTY -- RAMSEY THEORY
In a group of six people, there is a
triple who all know each other or there is a triple who are all strangers. I.e., the Ramsey number R(3,3) = 6.
I will not go into the more complex aspects of this -- see Graham &
Spencer for a survey.
P. Erdös & G. Szekeres. A combinatorial problem in geometry. Compositio Math. 2 (1935) 463‑470. [= Paul Erdös; The Art of Counting; Ed. by
Joel Spencer, MIT Press, 1973, pp. 5‑12.] They prove that if n ³
BC(a+b-2, a-1), then any two‑colouring
of Ka contains a monochromatic Ka or Kb.
William Lowell Putnam Examination,
1953, part I, problem 2. In: L. E. Bush; The William Lowell Putnam
Mathematical Competition; AMM 60 (1953) 539-542. Reprinted in: A. M.
Gleason, R. E. Greenwood & L. M. Kelly; The William Lowell Putnam
Mathematical Competition Problems and Solutions -- 1938‑1964; MAA, 1980;
pp. 38 & 365‑366. The classic
six people at a party problem.
R. E. Greenwood & A. M.
Gleason. Combinatorial relations and
chromatic graphs. Canadian J. Math. 7
(1955) 1-7. Considers n = n(a,b,...) such that a two colouring of
Kn contains a Ka of the first colour or a
Kb of the second
colour or .... Thus n(3,3) = 6.
They find the bound and many other results of Erdös & Szekeres.
C. W. Bostwick, proposer; John Rainwater & J. D. Baum,
solvers. Problem E1321 -- A gathering
of six people. AMM 65 (1958) 446 &
66 (1959) 141‑142.
Gamow & Stern. 1958.
Diagonal strings. Pp. 93‑95.
G. J. Simmons. The Game of Sim. JRM 2 (1969) 66.
M. Gardner. SA (Jan 1973) c= Knotted, chap. 9. Exposits Sim. Reports Simmons' result that it is second person (determined
after his 1969 article above). The
Addendum in Knotted reports that several people have shown that Sim on five
points is a draw. Numerous references.
Ronald L. Graham & Joel H.
Spencer. Ramsey theory. SA 263:1 (Jul 1990) 80‑85. Popular survey of Ramsey theory beginning
from Ramsey and Erdös & Szekeres.
5.N. JEEP OR EXPLORER'S PROBLEM
See
Ball for some general discussion and notation.
Alcuin. 9C.
Prob. 52: Propositio de homine patrefamilias. Wants to get 90 measures over a distance of 30 leagues. He is trying to get the most to the other
side, so this is different than the 20C versions. Solution is confusing, but Folkerts rectifies a misprint and this
makes it less confusing. Alcuin's
camels only eat when loaded!! (Or else
they perish when their carrying is done??)
The camel take a load to a point 20 leagues away and leaves 10 there,
then returns. This results in getting
20 to the destination.
The
optimum solution is for the camel to make two return trips and a single trip
to 10
leucas, so he will have consumed
30 measures and he has 60
measures to carry on. He now
makes one return and a single trip of another
15 leucas, so he will
have consumed another 30 measures, leaving 30 to carry on the
last 5
leucas, so he reaches home with
25 measures.
Pacioli. De Viribus.
c1500. Probs. 49‑52. Agostini only describes Prob. 49 in some
detail.
Ff.
94r - 95v. XLIX. (Capitolo) de doi
aportare pome ch' piu navanza (Of two ways to transport as many apples as
possible). = Peirani 134‑135. One has 90 apples to transport 30 miles from
Borgo [San Sepolcro] to Perosia [Perugia], but one eats one apple per mile and
one can carry at most 30 apples. He
carries 30 apples 20 miles and leaves 10 there and returns, without eating on
the return trip! (So this = Alcuin.) Pacioli continues and gives the optimum solution!
F.
95v. L. C(apitolo). de .3. navi per
.30. gabelle 90. mesure (Of three ships
holding 90 measures, passing 30 customs points). Each ship has to pay one measure at each customs point. Mathematically the same as the
previous.
F.
96r. LI. C(apitolo). de portar .100.
perle .10. miglia lontano 10. per volta
et ogni miglio lascia 1a (To carry 100 pearls 10 miles, 10 at a time,
leaving one every mile). = Peirani
136-137. Takes them 2 miles in ten
trips, giving 80 there. Then takes them
to the destination in 8 trips, getting 16 to the destination.
Ff.
96v - 97r. LII. C(apitolo). el medesimo
con piu avanzo per altro modo (The same with more carried by another
method). Continues the previous problem
and takes them 5 miles in ten trips, giving 50 there. Then takes them to the destination in 5 trips, getting 25 to the
destination.
[This is optimal for a single stop --
if one makes the stop at distance
a, then one gets a(10-a)
to the destination. One can make
more stops, but this is restricted by the fact that pearls cannot be
divided. Assuming that the amount of
pearls accumulated at each depot is a multiple of ten, one can get 28 to the
destination by using depots at 2 and
7 or
5 and 7. One can get 27 to the destination
with depots at 4 and 9 or 5
and 9. These are all the ways one can
put in two depots with integral multiples of 10 at each depot and none of these
can be extended to three such depots.
If the material being transported was a continuous material like grain,
then I think the optimal method is to first move 1 mile to get 90 there, then
move another 10/9 to get 80 there, then another 10/8
to get 70 there, ..., continuing until we get 40 at 8.4563...,
and then make four trips to the destination. This gets 33.8254 to the destination. Is this the best method??]
Cardan. Practica Arithmetice. 1539.
Chap. 66, section 57, ff. EE.vi.v - EE.vii.v (pp. 152‑153). Complicated problem involving carrying food
and material up the Tower of Babel!
Tower is assumed 36 miles high and seems to require 15625 porters.
Mittenzwey. 1880.
Prob. 135, pp. 28-29; 1895?:
153, p. 32; 1917: 153, pp. 29. If eight porters can carry eight full loads
from A
to B in an hour, how long will it take four porters? The obvious answer is two hours, but he
observes that the porters have to return from
B to A and it will take three
hours. [Probably a little less as they
should return in less time than they go.]
Pearson. 1907.
Part II, pp. 139 & 216. Two
explorers who can carry food for 12 days.
(No depots, i.e. form A of Ball, below.)
Loyd. A dash for the South Pole.
Ladies' Home Journal (15 Dec 1910).
??NYS -- source?? -- WS??
Ball. MRE, 5th ed., 1911.
Exploration problems, pp. 23‑24.
He distinguishes two forms of the problem, with n
explorers who can carry food for
d days.
A. Without depots, they can get one man nd/(n+1)
days into the desert and back.
B. With depots permitted, they can get a
man
d/2 (1/1 + 1/2 + ... + 1/n) into the desert and back. This is the more common form.
Dudeney. Problem 744: Exploring the desert. Strand Mag. (1925). ??NX. (??= MP 49)
Dudeney. MP.
1926. Prob. 49: Exploring the
desert, pp. 21 & 111 (= 536, prob. 76, pp. 22 & 240). A version of Ball's form A,
with n = 9, d = 10,
but replacing days by stages of length 40 miles.
Abraham. 1933.
Prob. 34 -- The explorers, pp. 13 & 25 (9‑10 & 112). 4 explorers, each carrying food for 5
days. Mentions general case. This is Ball's form A.
Haldeman-Julius. 1937.
No. 10: The four explorers, pp. 4 & 21. Ball's form A, with n =
4, d = 5.
Olaf Helmer. Problem in logistics: The Jeep problem. Project Rand Report RA‑15015
(1 Dec 1946) 7pp.
N. J. Fine. The jeep problem. AMM 54 (1947) 24‑31.
C. G. Phipps. The jeep problem: a more general
solution. AMM 54 (1947) 458‑462.
G. G. Alway. Note 2707:
Crossing the desert. MG 41 (No.
337) (1957) 209. If a jeep can carry
enough fuel to get halfway across, how much fuel is needed to get across? For a desert of width 2,
this leads to the series 1 + 1/3
+ 1/5 + 1/7 + .... See Lehmann and Pyle
below.
G. C. S[hephard, ed.] The problems drive. Eureka 11 (Jan 1949) 10-11 & 30. No. 2.
Four explorers, starting from a supply base. Each can carry food for 100 miles and goes 25 miles per day. Two men do the returning to base and
bringing out more supplies. the third
man does ferrying to the fourth man. How
far can the fourth man get into the desert and return? Answer is 100 miles. Ball's form B would give 104 1/6.
Gamow & Stern. 1958.
Refueling. Pp. 114‑115.
Pyle, I. C. The explorer's problem. Eureka 21 (Oct 1958) 5-7. Considers a lorry whose load of fuel takes
it a distance which we assume as the unit.
What is the widest desert one can cross? And how do you do it?
This is similar to Alway, above.
He starts at the far side and sees you have to have a load at distance 1
from the far side, then two loads at distance
1 + 1/3 from the far side, then three loads at 1 + 1/3 + 1/5, .... This diverges, so
any width can be crossed. Does examples
with given widths of 2, 3 and 4 units.
Editor notes that he is not convinced the method is optimal.
Martin Gardner, SA (May &
June 1959) c= 2nd Book, chap. 14, prob. 1. (The book gives extensive references which were not in SA.)
R. L. Goodstein. Letter: Explorer's problem. Eureka 22 (Oct 1959) 23. Says Alway shows that Pyle's method is
optimal. Editor notes Gardner's article
and that Eureka was cited in the solutions in Jun.
David Gale. The jeep once more or jeeper by the
dozen & Correction to "The Jeep once more or jeeper by the
dozen". AMM 77:5 (May 1970)
493-501 & 78:6 (Jun-Jul 1971) 644-645.
Gives an elaborate approach via a formula of Banach for path lengths in
one dimension. This formally proves
that the various methods used are actually optimal and that a continuous string
of depots cannot help, etc. Notes that
the cost for a round trip is only slightly more than for a one-way trip -- but
the Correction points out that this is wrong and indeed the round trip is
nearly four times as expensive as a one-way trip. Considers sending several jeeps.
Says he hasn't been able to do the round trip problem when there is fuel
on both sides of the desert. Comments
on use of dynamic programming, noting that R. E. Bellman [Dynamic Programming;
Princeton Univ. Press, 1955, p. 103, ex. 54-55] gives the problem as exercises
without solution and that he cannot see how to do it!
Birtwistle. Math. Puzzles & Perplexities. 1971.
The
expedition, pp. 124-125, 183 & 194.
Ball's form A, first with n =
5, d = 6, then in general.
Second
expedition, pp. 125-126. Ball's form B,
done in general.
Third
expedition, pp. 126, 183-184 & 194.
Three men want to cross a 180 mile wide desert. They can travel 20 miles per day and can
carry food for six days, which can be stored at depots. Minimize the total distance travelled. Solution seems erroneous to me.
A. K. Austin. Jeep trips and card stacks. MTg 58 (1972) 24‑25. There are
n flags located at
distances a1, a1 + a2, a1 + a2 + a3, ....
Jeep has to begin at the origin, go to the first flag, return to the
origin, go to the second flag, return, ....
He can unload and load fuel at the flags. Can he do this with
F fuel? Author shows this is equivalent to
successfully stacking cards over a cliff with successive overhangs being a1, a2, a3, ....
Doubleday - 3. 1972.
Traveller's Tale, pp. 63-64. d =
8 and we want one man to get across the
desert of width 12. How many porters, who return to base, are
needed? The solution implies that no
depots are used. Reasoning as in Ball's
case A, we see that n men can support one man crossing a desert of
width 2nd/(n+1). If depots are permitted, this is essentially
the jeep problem and n men can support a man getting across a
desert of width d [1 + 1/3 + 1/5 + ...
+ 1/(2n-1)]
Johannes Lehmann. Kurzweil durch Mathe. Urania Verlag, Leipzig, 1980. No. 13, pp. 27 & 129. d = 4
and we want to get a man across a desert of width 6.
Similar to Doubleday - 3.
Pierre Berloquin. [Le Jardin du Sphinx. Dunod, Paris, 1981.] Translated by Charles Scribner Jr as: The Garden of the Sphinx. Scribner's, NY, 1985.
Prob.
1: Water in the desert, pp. 3 & 85.
Prob.
40: Less water in the desert, pp. 26 & 111.
Prob.
80: Beyond thirst, pp. 48 & 140.
Prob.
141: The barrier of thirst, pp. 79 & 181.
Prob.
150: No holds barred, pp. 82 & 150.
In
all of these, d = 5 and we want to get a man across a desert of
width 4, and sometimes back, which is slightly different than the problem
of getting to the maximum distance and back.
Prob.
1 is Ball's form A, with
n = 4 men, using 20
days' water.
Prob.
40 is Ball's form B, but using only whole day trips, using 14
days' water.
Prob.
80 is Ball's form B, optimized for width 4, using
11½ days' water.
Prob.
141 uses depots and bearers who don't return, as in Alcuin?? You can get one man, who is the only one to
return, a distance d (1/2 + 1/3 + ... +
1/(n+1)) into the desert this way. He gives the optimum form for width 4,
using 9½ days' water.
Prob.
150 is like prob. 141, except that no one returns! You can get him d (1 + 1/2
+ ... + 1/n) into the desert this
way. The optimum here uses 4
days' water.
D. R. Westbrook. Note 74.7:
The desert fox, a variation of the jeep problem. MG 74 (No. 467) (1990) 49‑50. A more complex version, posed by A. K.
Dewdney in SA (Jan 1987), is solved here.
Liz Allen. Brain Sharpeners. Op. cit. in 5.B.
1991. Round-trip, pp. 96-97
& 140. Plane wants to circle the
earth, but can only carry fuel to go half-way.
Other planes can accompany and transfer fuel, but must return to base.
Dylan Gow. Flyaway.
MS 25:3 (1992/3) 84-86.
Considers the standard problem without return as in Alway, Pyle and
Lehmann -- but finds a non-optimal solution.
Wolfram Hinderer. Optimal crossing of a desert. MS 26:4 (1993/4) 100-102. Finds optimal solutions for Gow's problem
and for the case with return -- i.e. Ball's
B. Also considers use of extra
jeeps that do not return, i.e. Berloquin's 141 & 150. Notes that extra jeeps that must return to
base do not change the distance that one jeep can reach. [But it changes the time required.]
Harold Boas. Letter:
Crossing deserts. MS 26:4
(1993/4) 122. Notes the problem has a
long history and cites Fine, Phipps, Gale (and correction), Alway.
David Singmaster. Letter:
Crossing deserts. MS 27:3
(1994/5) 63. Points out that the
history is far older and sketches the history given above.
Günter Rote & Guochuan
Zhang. Optimal Logistics for
Expeditions: The Jeep Problem with Complete Refilling. Karl-Franzens-Universität Graz &
Technische Universität Graz. Bericht 71
(24 Jun 1996). This deals with a
variant. "We have n
cans of fuel on the edge of a desert and a jeep with an empty tank whose
capacity is just one can. The jeep can
carry one can in addition to the fuel in its tank. Moreover, when a can is opened, the fuel must immediately be
filled into the jeep's tank. The goal
is to find the farthest point in the desert which the jeep can reach by
consuming the n cans of fuel. Derick Wood [1984] treated this problem similarly to the
classical problem and gave the first solution.
Ute and Wilfried Brauer [1989] presented a new strategy and got a better
solution than Wood's. They also
conjectured that their solution was optimal for infinitely many values of n.
We give an algorithm which produces a better solution than Brauers' for
all n > 6, and we use a linear
programming formulation to derive an upper bound which shows that our solution
is optimal." 14 references,
several not given above.
5.O. TAIT'S COUNTER PUZZLE: BBBBWWWW TO WBWBWBWB
See
S&B 125.
The
rules are that one can move two counters as an ordered pair, e.g. from BBBBWWWW
to BBB..WWWBW, but not to
BBB..WWWWB -- except in Lucas
(1895) and AM prob. 237, where such reversal must be done. Also, moving to BBB..WWW.BW is sometimes
explicitly prohibited, but it is not always clear just where one can move
to. It is also not always specified
where the blank spaces are at the beginning and end positions.
Gardner,
1961, requires that the two counters must be
BW or WB.
Barbeau, 1995, notes that moving
to BWBWBWBW is a different problem, requiring an extra move. I had not noticed this difference before --
indeed I previously had it the wrong way round in the heading of this
section. I must check to see if this
occurs earlier. See Achugbue &
Chin, 1979-80, for this version.
Genjun Nakane (= Hōjiku
Nakane). Kanja‑otogi‑soshi
(Book of amusing problems for the entertainment of thinkers). 1743.
??NYS. (See: T. Hayashi; Tait's problem with counters in
the Japanese mathematics; Bibl. Mathem. (3) 6 (1905) 323, for this and other
Japanese references of 1844 and 1879, ??NYS.)
P. G. Tait. Listing's Topologie. Philosophical Mag. (Ser. 5) 17 (No. 103)
(Jan 1884) 30‑46 & plate opp. p. 80.
Section 12, pp. 39‑40. He
says he recently saw it being played on a train.
George Hope Verney (= Lloyd‑Verney). Chess eccentricities. Longmans, 1885. P. 193: The pawn puzzle.
??NX With 4 & 4.
Lucas. Amusements par les jetons.
La Nature 15 (1887, 2nd sem.) 10-11.
??NYS -- cited by Ahrens, title obtained from Harkin. Probably c= the material in RM3, below.
Ball. MRE, 1st ed., 1892, pp. 48‑49.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
Card Puzzles No. XIV: The eight-card puzzle, pp. 14-15. Uses cards:
BRBRBRBR and asks to bring the
colours together, explicitly requiring the moved cards to be placed in contact
with the unmoved cards.
Hoffmann. 1893.
Chap. VI, pp. 270‑271 & 284‑285 = Hoffmann-Hordern,
pp. 184-186, with photo.
No.
19: The "Four and Four" puzzle.
Photo on p. 184 shows a version named Monkey Puzzle advertising Brooke's
Soap to go from BBBBBWWWW.. to
..WBWBWBWB .
No.
20: The "Five and Five" puzzle.
No.
21: The "Six and Six" puzzle.
Lucas. RM3. 1893. Amusements par les jetons, pp. 145‑151. He gives Delannoy's general solution
for n
of each colour in n moves.
Remarks that one can reverse the moved pair.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co.,
NY), nd [1895]. P. 11: The Egyptian
disc puzzle. 4 & 4. "Two discs adjoining each other to be
moved at a time; no gaps to be left in the line." -- this seems to prevent one from making any
moves at all!! No solution.
Lucas. L'Arithmétique Amusante.
1895. Pp. 84-108.
Prob.
XXI - XXIV and Méthode générale, pp. 84-97.
Gives solution for 4, 5, 6,
7 and the general solution for n & n
in n moves due to Delannoy.
Rouges
et noires, avec interversion, prob. XXV - XXVIII and Méthode générale, pp. 97‑108. Interversion means that the two pieces being
moved are reversed or turned over, e.g. from
BBBBWWWW to BBB..WWWWB,
but not to BBB..WWWBW. Gives solutions for 4, 5, 6, 7, 8 pairs and in general in n moves, but he ends with a gap, e.g. ....BB..BB
and it takes an extra move to close up the gap.
Ball. MRE, 3rd ed., 1896, pp. 65‑66. Cites Delannoy's solution as being in La Nature (Jun 1887)
10. ??NYS.
Ahrens. MUS I.
1910. Pp. 14-15 &
19-25. Cites Tait and gives Delannoy's
general solution, from Lucas.
Ball. MRE, 5th ed., 1911, pp. 75-77.
Adds a citation to Hayashi, but incorrectly gives the date as 1896.
Loyd. Cyclopedia. 1914. After dinner tricks, pp. 41 & 344. 4 & 4.
Williams. Home Entertainments. 1914.
The eight counters puzzle, pp. 116-117.
Standard version, but with black and white reversed, in four moves. Says the moved counters must be placed in
line with and touching the others.
Dudeney. AM.
1917.
Prob.
236: The hat puzzle, pp. 67 & 196-197.
BWBWBWBWBW.. to have the Bs
and Ws together and two blanks at an end. Uses 5 moves to get to ..WWWWWBBBBB.
Prob.
237: Boys and girls, pp. 67-68 & 197.
..BWBWBWBW to have the Bs
and Ws together with two blanks at an end, but pairs must be reversed as
they are moved. Solution in 5 moves to WWWWBBBB...
= Putnam, no. 2. Cf Lucas, 1895.
Blyth. Match-Stick Magic.
1921. Transferring in twos, pp.
80-81. WBWBWBWB.. to
..BBBBWWWW in four moves.
King. Best 100. 1927. No. 66, pp. 27 & 55. = Foulsham's, no. 9, pp. 9 & 13. BWBWBWBW..
to ..WWWWBBBB, specifically prescribed.
Rohrbough. Brain Resters and Testers. c1935.
Alternate in Four Moves, p. 4.
..BBBBWWWW to WBWBWBWB.. , but he doesn't specify the
blanks, showing all stages as closed up to 8 spaces, except the first two
stages have a gap in the middle.
McKay. At Home Tonight. 1940.
Prob.
43: Arranging counters, pp. 73 & 87-88.
RBRBRB.... to ....BBBRRR
in three moves. Sketches general
solution.
Prob.
45: Triplets, pp. 74 & 88.
YRBYRBYRB.. to BBBYYYRRR..
in 5 moves.
McKay. Party Night. 1940. Heads and tails again, p. 151. RBRBR..
to ..BBRRR in three moves. RBRBRB.. to ..BBBRRR in four moves. RBRBRBRB.. to ..BBBBRRRR
in four moves. Notes that the
first move takes coins 2 & 3 to the end and thereafter one is always
filling the spaces just vacated.
Gardner. SA (Jun & Jul 1961) =
New MD, chap. 19, no. 1: Collating the coins. BWBWB to BBBWW,
moving pairs of BW or
WB only, but the final position
may be shifted. Gardner thanks H. S.
Percival for the idea. Solution in 4
moves, using gaps and with the solution shifted by six spaces to the
right. Thanks to Heinrich Hemme for
this reference.
Joseph S. Madachy. Mathematics on Vacation. (Scribners, NY, 1966, ??NYS); c= Madachy's Mathematical Recreations. Dover, 1979. Prob. 3: Nine-coin move, pp. 115 & 128-129 (where the
solution is headed Eight-coin move).
This uses three types of coin, which I will denote by B, R, W.
BRWBRWBRW ® WWWRRRBBB by moving two adjacent unlike coins
at a time and not placing the two coins away from the rest. Eight move solution leaves the coins in the
same places, but uses two extra cells at each end. From the discussion of Bergerson's problem, see below, it is
clear that the earlier book omitted the word unlike and had a nine move
solution, which has been replaced by Bergerson's eight move solution.
Yeong‑Wen Hwang. An interlacing transformation problem. AMM 67 (1967) 974‑976. Shows the problem with 2n
pieces, n > 2, can be solved in n moves and this is
minimal.
Doubleday - 1. 1969.
Prob. 70: Oranges and lemons, pp. 86 & 170. = Doubleday - 4, pp. 95‑96. BWBWBWBWBW.. considered as a cycle.
There are two solutions in five moves:
to ..WWWWWBBBBB, which never
uses the cycle; and to: BBWWWWWW..BBB.
Howard W. Bergerson,
proposer; Editorial discussion; D. Dobrev, further solver; R. H. Jones, further solver. JRM
2:2 (Apr 1969) 97; 3:1 (Jan
1970) 47-48; 3:4 (Oct 1970)
233-234; 6:2 (Spring 1973) 158. Gives Madachy's 1966 problem and says there
is a shorter solution. The editor points
out that Madachy's book and Bergerson have omitted unlike. Bergerson has an eight move solution of the intended
problem, using two extra cells at each end, and Leigh James gives a six move
solution of the stated problem, also using two extra cells at each end. Dobrev gives solutions in six and five
steps, using only two extra cells at the right. Jones notes that the problem does not state that the coins have
to be adjacent and produces a four move solution of the stated problem, going
from ....BRWBRWBRW.... to
WWW..R..R..R..BBB.
Jan M. Gombert. Coin strings. MM 42:5 (Nov 1969) 244-247.
Notes that BWBWB...... ®
......BBBWW can be done in four
moves. In general, BWB...BWB,
with n Ws and n+1
Bs alternating can be
transformed to BB...BWW...W in n2 moves and this is minimal. This requires shifting the whole string n(n+1)
to the right and a move can go to places separated from the rest of the
pieces. By symmetry, ......BWBWB ® WWBBB...... in the same number of moves.
Doubleday - 2. 1971.
Two by two, pp. 107-108.
..BWBWBWBW to WWWWBBBB...
He doesn't specify where the extra spaces are, but says the first two
must move to the end of the row, then two more into the space, and so on. The solution always has two moving into an internal
space after the first move.
Wayne A. Wickelgren. How to Solve Problems. Freeman, 1974. Checker-rearrangement problem, pp. 144‑146. BWBWB
to BBBWW by moving two adjacent checkers, of
different colours, at a time. Solves in
four moves, but the pattern moves six places to the left.
Putnam. Puzzle Fun.
1978.
No. 1:
Nickles [sic] & dimes, pp. 1 & 25.
Usual version with 8 coins.
Solution has blanks at the opposite end to where they began.
No. 2:
Nickles [sic] & dimes variation, pp. 1 & 25. Same, except the order of each pair must be reversed as it
moves. Solution in five moves with
blanks at opposite end to where they started.
= AM 237. Cf Lucas, 1895.
James O. Achugbue &
Francis Y. Chin. Some new
results on a shuffling problem. JRM 12:2
(1979-80) 125-129. They demonstrate
that any pattern of n & n occupying
2n consecutive cells can be
transformed into any other pattern in the same cells, using only two extra
cells at the right, except for the case
n = 3 where 10
cells are used. They then find
an optimal solution for BB...BW...WW ®
BWBW...BW in n+1 moves using two extra
cells. They seem to leave open the
question of whether the number of moves could be shortened by using more cells.
Walter Gibson. Big Book of Magic for All Ages. Kaye & Ward, Kingswood, Surrey,
1982.
Six
cents at a time, p. 117. Uses pennies
and nickels. .....PNPNP to
NNPPP..... in four moves.
Tricky
turnover, p. 137. HTHTHT to
HHHTTT in two moves. This requires turning over one of the two
coins on each move.
Ed Barbeau. After Math.
Wall & Emerson, Toronto, 1995.
Pp. 117, 119 & 123-126. He
asks to move BBBWWW to
WBWBWB and to BWBWBW
and notes that the latter takes an extra move. He sketches the general solutions.
5.P. GENERAL MOVING PIECE PUZZLES
See
also under 5.A.
See Hordern, op. cit. in 5.A, pp. 167‑177,
for a survey of these puzzles. The
Chifu‑Chemulpo (or Russo‑Jap Railway) Puzzle of 1903 is actually
not of this type since all the pieces can move by themselves -- Hordern, pp.
124‑125 & plate VIII.
See
S&B 124‑125.
A
'spur' is a dead‑end line. A
'side‑line' is a line or siding joined to another at both ends.
Mittenzwey. 1880.
Prob. 219-221, pp. 39-40 & 91;
1895?: 244-246, pp. 43-44 & 93;
1917: 244-246, pp. 40 & 89.
First two have a canal too narrow to permit boats to pass, with a
'bight', or widening, big enough to hold one boat while another passes. First problem has two boats meeting one
boat; second problem has two boats meeting two boats. The third problem has a single track railway with a side-line big
enough to hold an engine and 16 wagons on the side-line or on the main line
between the switches. Two trains
consisting of an engine and 20 wagons meet.
Lucas. RM2, 1883, pp. 131‑133.
Passing with a spur and with a side‑line.
Alexander Henry Reed. UK Patent 15,051 -- Improvements in
Puzzles. Complete specification: 8 Dec
1885. 4pp + 1p diagrams. Reverse a train using a small turntable on
the line. This has forms with one line
and with two crossing lines. One object
is to spell 'Humpty Dumptie'. He also
has a circular line with three turntables (equivalent to the recent Top-Spin
Puzzle of F. Lammertinck).
Pryse Protheroe. US Patent 332,211 -- Puzzle. Applied: 18 Sep 1885; patented: 8 Dec 1885. 3pp + 1p diagrams. Described in Hordern, p. 167.
Identical to the Reed patent above!
Both Reed and Protheroe are described as residents of suburban
London. The Reed patent says it was
communicated from abroad by an Israel J. Merritt Jr of New York and it doesn't
assert that Reed is the inventor, so perhaps Reed and Merritt were agents for
Protheroe.
Jeffrey & Son (Syracuse,
NY). Great Railroad Puzzle. Postcard puzzle produced in 1888. ??NYS.
Described in Hordern, pp. 175‑176. Passing with a turntable that holds two wagons.
Arthur G. Farwell. US Patent 437,186 -- Toy or Puzzle. Applied: 20 May 1889; patented: 30 Sep 1890. 1p + 1p diagrams. Described in Hordern, pp.
167‑169. Great Northern Puzzle. This requires interchanging two cars on the
legs of a 'delta' switch which is too short to allow the engine through, but
will let the cars through. Hordern
lists 6 later patents on the same basic idea.
Ball. MRE, 1st ed., 1892, pp. 43‑44. Great Northern Puzzle "which I bought some eight or nine
years ago." (Hordern, p. 167,
erroneously attributes this quote to Ahrens.)
Loyd. Problem 28: A railway puzzle.
Tit‑Bits 32 (10 Apr
& 1 May 1897) 23 &
79. Engine and 3 cars need to
pass 4 cars by means of a 'delta' switch whose branches and tail hold only one
car. Solution with 28 reversals.
Loyd. Problem 31: The turn‑table puzzle. Tit‑Bits 32 (1 &
22 May 1897) 79 & 135.
Reverse an engine and 9 cars with an 8 track turntable whose lines hold
3 cars. The turntable is a double
curved connection which connects, e.g. track 1 to tracks 4 or 6.
E. Fourrey. Récréations Arithmétiques. Op. cit. in 4.A.1. 1899. Art. 239: Problèmes
de Chemin de fer, pp. 184-189.
I. Three parallel tracks with two switched
crossing tracks. Train of 21 wagons on
the first track must leave wagons 9 & 12 on third track.
II.
Delta shape with a turntable at the point of the delta, which can only hold the
wagons and not the engine, so this is isomorphic to Farwell.
III. This
is a more complex railway problem involving timetables on a circular line.
J. W. B. Shunting!
c1900. ??NYS. Described in Hordern, pp. 176‑177
& plate XII. Reversing a train with
a turntable that holds three wagons.
Orril L. Hubbard. US Patent 753,266 -- Puzzle. Applied: 21 Apr 1902; patented: 1 Mar 1904. 3pp + 1p diagrams. Great Railroad Puzzle, described in Hordern, pp. 175‑176. Improved version of the Jeffrey & Son
puzzle of 1888. Engine & 2 cars to
pass engine & 3 cars, using a turntable that holds two cars, preserving
order of each train.
Harry Lionel Hook &
George Frederick White. UK
Patent 26,645 -- An Improved Puzzle or Game.
Applied: 3 Dec 1902; accepted:
11 Jun 1903. 2pp + 1p diagrams. This is very cryptic, but appears to be a
kind of sliding piece Puzzle using turntables.
Mr. X [cf 4.A.1]. His Pages.
The Royal Magazine 10:1 (May 1903) 50-51 & 10:2 (Jun 1903)
140-141 & 10:4 (Aug 1903)
336-337. A railway puzzle. One north-south line with a spur heading
north which is holding 7 trucks, but cannot hold the engine as well, so the
engine is on the main line heading south.
An engine pulling seven trucks arrives from the north and wants to get
past. First solution uses 17 stages;
second uses 12 stages.
Mr. X [cf 4.A.1]. His Pages.
The Royal Magazine 10:5 (Sep 1903) 426-427 & 10:6
(Oct 1903) 530‑531. A
shunting problem. Same as Fourrey - II,
hence isomorphic to Farwell. Solution
in 17 stages.
Celluloid Starch Puzzle. c1905.
Described in Hordern, pp. 169‑170. Cars on the three parts of a 'delta' switch with an engine
approaching. Reverse the engine,
leaving all cars on their original places.
More complexly, suppose the tail of the 'delta' only holds one car or
the engine.
Livingston B. Pennell. US Patent 783,589 -- Game Apparatus. Applied: 20 Mar 1902; patented: 28 Feb 1905. 3pp + 1p diagrams. Described in Hordern, p. 173.
Passing with a side line -- engine & 3 cars to pass engine & 3
cars using a siding which already contains 3 cars, without couplings, so these
three can only be pushed. Also the
engines can move at most three cars at a time.
William Rich & Harry
Pritchard. UK Patent 7647 -- Railway
Game and Puzzle. Applied: 11 Apr
1905; complete specification: 11 Oct
1905; accepted: 14 Dec 1905. 2pp + 1p diagrams. Main line with two short and two long spurs.
Ball. MRE, 4th ed., 1905, pp. 61-63, adds a problem with a side-line,
"on sale in the streets in 1905“.
The 5th ed., 1911, pp. 69-71 & 82, adds the name "Chifu-Chemulpo
Puzzle" and that the minimum number of moves is 26, in more than one
way. P. 82 gives solutions of both
problems.
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. Great Northern
Puzzle. He says the "Railway
puzzle" was very popular "about twenty years ago".
Ahrens. MUS I.
1910. Pp. 3-4. Great Northern. Says it is apparently modern and cites Fourrey for other
examples.
Anon. Prob. 6. Hobbies 32 (No. 814) (20 May 1911) 145 &
(No. 817) (10 Jun 1911) 208.
Great Northern Puzzle. Solution
asks if readers know any other railway puzzles.
Loyd. The switch problem
& Primitive railroading
problem. Cyclopedia, 1914, pp. 167
& 361; 89 & 350 (= MPSL2, prob.
24, pp. 18‑19; MPSL1, prob. 95,
pp. 92 & 155). Passing with a
'delta' switch & passing with a spur. The first is like Tit-Bits Problem 28, but
the engine and 3 cars have to pass 5 cars.
Solution in 32 moves. See
Hordern, pp. 170‑171.
Hummerston. Fun, Mirth & Mystery. 1924.
The Chinese railways, pp. 103 & 188. Imagine a line of positions:
ABCEHGJLMN with single
positions D, I, F, K attached to
positions C, H, G, L. You have eight engines at ABCD
and KLMN and the object is to exchange them,
preserving the order. He does it in 18
moves, where a move can be of any length.
King. Best 100. 1927. No. 14, pp. 12 & 41. Side‑line with a bridge over it too
low for the engine. Must interchange
two wagons on the side‑line which are on opposite sides of the bridge.
B. M. Fairbanks. Railroad switching problems. IN:
S. Loyd Jr., ed.; Tricks and Puzzles; op. cit. in 5.D.1 under
Chapin; 1927. P. 85 & Answers p.
7. Three realistic problems with
several spurs and sidelines.
Loyd Jr. SLAHP.
1928. Switching cars, pp. 54
& 106. Great Northern puzzle. See Hordern, pp. 168‑169.
Doubleday - 2. 1971.
Traffic jam, pp. 85-86. Version
with cars in a narrow lane and a lay-by.
Two cars going each way. Though
the lay-by is three cars wide and just over a car long, he restricts its use so
that it acts like it is two cars wide.
Jacques
Haubrich has kindly enlightened me that 'taquin' simply means 'teaser'. So these items should be re-categorised.
Lucas. RM3. 1893. 3ème Récréation -- Le jeu du caméléon et le
jeu des jonctions de points, pp. 89‑103.
Pp. 91‑97 -- Le taquin de neuf cases avec un seul port. I thought that taquin was the French generic
term for such puzzles, but I find no other usage than that below, except in
referring to the 15 Puzzle -- see references to taquin in 5.A.
Au Bon Marché (the Paris
department store). Catalogue of 1907,
p. 13. Reproduced in Mary Hillier;
Automata and Mechanical Toys; An Illustrated History; Jupiter Books, London,
1976, p. 179. This shows Le Taquin Japonais Jeu de Patience
Casse-tete. This comprises 16
hexagonal pieces, looking like a corner view of a die, so each has three
rhombic parts containing a pattern of pips.
They are to be placed as the corners of four interlocked hexagons with
the numbers on adjacent rhombi matching.
5.Q. NUMBER OF REGIONS DETERMINED BY N LINES OR PLANES
Mittenzwey. 1880.
Prob. 200, pp. 37 & 89;
1895?: 225, pp. 41 & 91;
1917: 225, pp. 38 & 88. Family
of 4 adults and 4 children. With three
cuts, divide a cake so the adults and the children get equal pieces. He makes two perpendicular diametrical cuts
and then a circular cut around the middle.
He seems to mean the adults get equal pieces and the children get equal
pieces, not necessarily the same. But
if the circular cut is at Ö2/2 of the radius, then the areas are all
equal. Not clear where this should go
-- also entered in 5.T.
Jakob Steiner. Einige Gesetze über die Theilung der Ebene
und des Raumes. (J. reine u. angew.
Math. 1 (1826) 349‑364) = Gesam.
Werke, 1881, vol. 1, pp. 77‑94.
Says the plane problem has been raised before, even in a Pestalozzi school
book, but believes he is first to consider 3‑space. Considers division by lines and circles
(planes and spheres) and allows parallel families, but no three coincident.
Richard A. Proctor. Some puzzles; Knowledge 9 (Aug 1886) 305-306
& Three puzzles; Knowledge 9 (Sep 1886) 336-337. "3.
A man marks 6 straight lines on a field in such a way as to enclose 10
spaces. How does he manage
this?" Solution begins: "III.
To inclose ten spaces by six ropes fastened to nine pegs." Take
(0,0), (1,0), ..., (n,0), (0,n), ..., (0,1), as 2n+1 points, using n+2 ropes from (0,0) to (n,0) and to (0,n) and from
(i,0) to (0,n+1-i) to enclose n(n+1)/2
areas.
Richard A. Proctor. Our puzzles. Knowledge 10 (Nov 1886)
9 & (Dec 1886) 39-40.
Describes several ways of solving previous problem and asks for a
symmetric version.
G.
Chrystal. Algebra -- An Elementary
Text-Book. Vol. 2, A. & C. Black,
Edinburgh, 1889. [Note -- the 1889
version of vol. 1 is a 2nd ed.] Chap.
23, Exercises IV, p. 34. Several
similar problems and the following.
No. 7
-- find number of interior and of exterior intersections of the diagonals of a
convex n-gon.
No. 8
-- n
points in general position in space, draw planes through every three and
find number of lines and of points of intersection.
L. Schläfli. Theorie der vielfachen Kontinuität. Neue Denkschriften der allgemeinen
schweizerischen Gesellschaft für die Naturwissenschaften 38:IV, Zürich, 1901,
239 pp. = Ges. Math. Abh., Birkhäuser,
Basel, 1950‑1956, vol. 1, pp. 167‑392. (Pp. 388‑392 are a Nachwort by J. J. Burckhardt.) Material of interest is Art. 16: Über die
Zahl der Teile, ..., pp. 209‑212.
Obtains formula for k hyperplanes in n space.
Loyd, Dudeney, Pearson &
Loyd Jr. give various puzzles based on this topic.
Howard D. Grossman. Plane- and space-dissection. SM 11 (1945) 189-190. Notes Schläfli's result and observes that
the number of regions determined by
k+1 hyperspheres in n
space is twice the number of regions determined by k
hyperplanes and gives a two to one correspondence for the case n = 2.
Leo Moser, solver. MM 26 (Mar 1953) 226. ??NYS.
Given in: Charles W. Trigg;
Mathematical Quickies; (McGraw‑Hill, NY, 1967); corrected ed., Dover, 1985.
Quickie 32: Triangles in a circle, pp. 11 & 90‑91. N
points on a circle with all diagonals drawn. Assume no three diagonals are concurrent. How many triangles are formed whose vertices
are internal intersections?
Timothy Murphy. The dissection of a circle by chords. MG 56 (No. 396) (May 1972) 113‑115 +
Correction (No. 397) (Oct 1972) 235‑236. N points on a circle, in
a plane or on a sphere; or N
lines in a plane or on a sphere, all simply done, using Euler's formula.
Rowan Barnes-Murphy. Monstrous Mysteries. Piccolo, 1982. Slicing cakes, pp. 33 & 61.
Cut a circular cake into 12 equal pieces with 4 cuts. [From this, we see that N
full cuts can yield either
2N or 4(N-1) equal pieces. Further, if we make k
circular cuts producing k+1 regions of equal area and then make N-k
diametric cuts equally spaced, we get
2(k+1)(N-k) pieces of the same
size.]
Looking at this problem, I see
that one can obtain any number of pieces from
N+1 up through the maximum.
5.Q.1. NUMBER OF INTERSECTIONS DETERMINED BY N LINES
Chrystal. Text Book of Algebra. 2nd ed., vol. 2, 1889, p. 34, ex. 7. See above.
Loyd Jr. SLAHP.
1928. When drummers meet, pp. 74
& 115. Six straight railroads can
meet in 15 points.
Paul Erdös, proposer; Norbert Kaufman & R. H. Koch and Arthur
Rosenthal, solvers. Problem E750. AMM 53 (1946) 591 & 54 (1947) 344. The first solution is given in Trigg, op.
cit. in 5.Q, Quickie 191: Intersections of diagonals, pp. 53 & 166‑167. In a convex
n‑gon, how many intersections of diagonals are there? This counts a triple intersection as three
ordinary (i.e. double) intersections or assumes no three diagonals are
concurrent. Editorial notes add some
extra results and cite Chrystal.
See
also 5.O. Some of these are puzzles, but
some are games and are described in the standard works on games -- see the
beginning of 4.B.
See
MUS I 182-210.
Ahrens, MUS I 182‑183,
gives legend associating this with American Indians. Bergholt, below, and Beasley, below, find this legend in the 1799
Encyclopédie Méthodique: Dictionnaire des Jeux Mathématique (??*), ??NYS. Ahrens also cites some early 19C material
which has not been located. Bergholt
says some maintain the game comes from China.
Thomas Hyde. Historia Nerdiludii, hoc est dicere,
Trunculorum; .... (= Vol. 2 of De Ludis
Orientalibus, see 7.B for vol. 1.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De Ludo dicto Ufuba wa Hulana, p. 233. This has a
5 x 5 board with each
side having 12 men, but the description is extremely
brief. It seems to have two players,
but this may simply refer to the two types of piece. I'm not clear whether it's played like solitaire (with the jumped
pieces being removed) or like frogs & toads. I would be grateful if someone could read the Latin
carefully. The name of the puzzle is
clearly Arabic and Hyde cites an Arabic source, Hanzoanitas (not further
identified on the pages I have) -- I would be grateful to anyone who can track
down and translate Arabic sources.
G. W. Leibniz. Le Jeu du Solitaire. Unpublished MS LH XXXV 3 A 10 f. 1-2,
of c1678. Transcribed in: S. de Mora-Charles; Quelques jeux de hazard
selon Leibniz; HM 19 (1992) 125-157.
Text is on pp. 152-154. 37
hole board. Says the Germans call it
'Die Melancholy' and that it is now the mode at the French court.
Claude‑Auguste Berey. Engraving:
Madame la Princesse de Soubize jouant au Jeu de Solitaire. 1697(?).
Beasley (below) discovered and added this while his book was in
proof. It shows the 37‑hole
French board. Reproduced in: Pieter van Delft & Jack Botermans;
Creative Puzzles of the World; op. cit. in 5.E.2.a, p. 170.
G. W. Leibniz. Jeu des Productions. Unpublished MS LH XXXV 8,30 f. 4, of
1698. Transcribed in: de Mora-Charles, loc. cit. above. Text is on pp. 154-155. 37 hole board. Considers the game in reverse.
Trouvain. Engraving:
Dame de Qualité Jouant au Solitaire.
1698(?).
Claude‑Auguste Berey. Engraving:
Nouveau Jeu de Solitaire.
Undated, but Berey was active c1690‑c1730. Reproduced in: R. C. Bell; The Board Game Book; Marshall Cavendish, London,
1979, pp. 54‑55 and in: Jasia
Reichardt, ed.; Play Orbit [catalogue of an exhibition at the ICA, London, and
elsewhere in 1969-1970]; Studio International, 1969, p. 38. Beasley's additional notes point out that
this engraving is well known, but he had not realised its date until the
earlier Berey engraving was discovered.
This engraving includes the legend associating the game with the American
indians -- "son origine vient de l'amerique ou les Peuples vont seuls à la
chasse, et au retour plantent leurs flèches en des trous de leur cases, ce qui
donna idée a un françois de composer ce jeu ...." Reichardt says the original is in the Bibliothèque
Nationale.
The three engravings above are
reproduced in: Henri d'Allemagne; Musée
rétrospectif de la classe 100, Jeux, à l'exposition universelle international
de 1900 à Paris, Tome II, pp. 152‑158. D'Allemagne says the originals are in the Bibliothèque Nationale,
Paris. He (and de Mora-Charles) also
cites Rémond de Montmort, 2nd ed., 1713 -- see below.
G. W. Leibniz. Annotatio de quibusdam Ludis; inprimis de
Ludo quodam Sinico, differentiaque Scachici et Latrunculorum & novo genere
Ludi Navalis. Misc. Berolinensia (= Misc.
Soc. Reg., Berlin) 1 (1710) 24. Last
para. on p. 24 relates to solitaire.
(English translation on p. xii of Beasley, below.)
Pierre Rémond de Montmort. Essai d'analyse sur les jeux de
hazards. (1708); Seconde edition revue & augmentee de
plusieurs lettres, (Quillau, Paris,
1713 (reprinted by Chelsea, NY, 1980));
2nd issue, Jombert & Quillau, 1714.
Avertissement (to the 2nd ed.), xli-xl.
"J'ai trouvé dans le premier volume de l'Academie Royale de Berlin,
...; il propose ensuite des Problèmes sur un jeu qui a été à la mode en France
il y a douze ou quinze ans, qui se nomme Le Solitaire."
Edward Hordern's collection has
a wooden 37 hole board on the back of which is inscribed "Invented by Lord
Derwentwater when Imprisoned in the Tower". The writing is old, at least 19C, possibly earlier. However the Encyclopedia Britannica article
on Derwentwater and the DNB article on Radcliffe, James, shows that the
relevant Lord was most likely to have been James Radcliffe (1689-1716), the 3rd
Earl from 1705, who joined the Stuart rising in 1715, was captured at Preston,
was imprisoned in the Tower and was beheaded on 24 Feb 1716, so the
implied date of invention is 1715 or 1716.
The third Earl became a figure of romance and many stories and books
appeared about him, so the invention of solitaire could well have been
attributed to him.
Though
the title was attainted and hence legally extinct, it was claimed by
relatives. Both James's brother Charles
(1693‑1746), the claimed 5th Earl from 1731, and Charles's son James Bartholomew
(1725-1786), the claimed 6th Earl from 1746, spent time in prison for their
Stuart sympathies. Charles escaped from
Newgate Prison after the 1715 rising, but both were captured on their way to
the 1745 rising and taken to the Tower where Charles was beheaded. If either of these is the Lord Derwentwater
referred to, then the date must be 1745 or 1746. A guide book to Northumberland, where the family lived at
Dyvelston (or Dilston) Castle, near Hexham, asserts the last Derwentwater was
executed in 1745, while the [Blue Guide] says the last was executed for
his part in the 1715 uprising.
In
any case, the claim seems unlikely.
G. W. Leibniz. Letter to de Montmort (17 Jan 1716). In:
C. J. Gerhardt, ed.; Die Philosophischen Schriften von Gottfried
Wilhelm Leibniz; (Berlin, 1887) = Olms,
Hildesheim, 1960; Vol. 3, pp. 667‑669.
Relevant passage is on pp. 668‑669. (Poinsot, op. cit. in 5.E, p. 17, quotes this as letter VIII
in Leibn. Opera philologica.)
J. C. Wiegleb. Unterricht in der natürlichen Magie. Nicolai, Berlin & Stettin, 1779. Anhang von dreyen Solitärspielen, pp. 413‑416,
??NYS -- cited by Beasley. First known
diagram of the 33‑hole board.
Catel. Kunst-Cabinet. 1790. Das Grillenspiel (Solitaire), p. 50 &
fig. 167 on plate VI. 33 hole
board. (Das Schaaf- und Wolfspiel, p.
52 & fig. 169 on plate VI, is a game on the 33-hole board.)
Bestelmeier. 1801.
Item 511: Ein Solitair, oder Nonnenspiel. 33 hole board.
Strutt. Op. cit. in 4.B.1. The Solitary Game. (1801:
Book IV, p. 238. ??NYS -- cited by
Beasley -- may be actually 1791??)
1833: Book IV, chap. II, art. XV, p. 319. c= Strutt-Cox, p. 259.
Beasley says this is the first attribution to a prisoner in the
Bastille. The description is vague:
"fifty or sixty" holes and "a certain number of pegs". Strutt-Cox adds a note that "The game
of Solitaire, reimported from France, ..., came again into Fashion in England
in the late" 1850s and early 1860s.
Ada Lovelace. Letter of 16 Feb 1840 to Charles
Babbage. BM MSS 37191, f. 331. ??NYS -- reproduced in Teri Perl; Math
Equals; Addison-Wesley, Menlo Park, California, 1978, pp. 109-110. Discusses the 37 hole board and wonders if
there is a mathematical formula for it.
M. Reiss. Beiträge zur Theorie des Solitär‑Spiels. J. reine angew. Math. 54 (1857) 376‑379.
St. v. Kosiński & Louis
Wolfsberg. German Patent 42919 --
Geduldspiel. Patented:
25 Sep 1877. 1p + 1p
diagrams. 33 hole version.
The Sociable. 1858.
The game of solitaire, pp. 282-284.
37 hole board. "It is
supposed to have been invented in America, by a Frenchman, to beguile the
wearisomeness attendant upon forest life, and for the amusement of the Indians,
who pass much of their time alone at the chase, ...."
Anonymous. Enquire Within upon Everything. 66th ed., 862nd thousand, Houlston and Sons,
London, 1883, HB. Section 135:
Solitaire, p. 49. Mentions a 37
hole board but shows a 33 hole board.
This material presumably goes back some time before this edition. It later shows Fox and Geese on the 33 hole
board.
Hoffmann. 1893.
Chap. X, no. 11: Solitaire problems, pp. 339-340 & 376-377
= Hoffmann‑Hordern, pp. 232-233, with photo on p. 235. Three problems. Photo on p. 235 shows a 33-hole board in a square frame,
1820-1840, and a 37-hole board with a holding handle, 1840-1890.
Ernest Bergholt. Complete Handbook to the Game of Solitaire
on the English Board of Thirty-three Holes.
Routledge, London, nd [Preface dated Nov 1920] -- facsimile produced by
Naoaki Takashima, 1993. This is the
best general survey of the game prior to Beasley.
King. Best 100. 1927. No. 68, pp. 28 & 55. = Foulsham's no. 24, pp. 9 & 13. 3 x 3
array of men in the middle of a
5 x 5 board. Men can jump diagonally as well as
orthogonally. Object is to leave one
man in the centre.
Rohrbough. Puzzle Craft. 1932. Note on
Solitaire & French Solitaire, pp. 14-15
(= pp. 6-7 of 1940s?). 33
hole board, despite being called French.
B. M. Stewart. Solitaire on a checkerboard. AMM 48 (1941) 228-233. This surveys the history and then considers
the game on the 32 cell board comprising the squares of one colour on a
chessboard. He tilts this by 45o to get a board with 7 rows, having 2, 4, 6, 8, 6, 4, 2 cells in each row. He
shows that each beginning-ending problem which is permitted by the parity rules
is actually solvable, but he gives examples to show this need not happen on
other boards.
Gardner. SA (Jun 1962). Much amended as: Unexpected, chap. 11, citing results of Beasley,
Conway, et al. Cites Leibniz and
mentions Bastille story.
J. D. Beasley. Some notes on solitaire. Eureka 25 (Oct 1962) 13-18. No history of the game.
Jeanine Cabrera & René
Houot. Traité Pratique du
Solitaire. Librairie Saint‑Germain,
Paris, 1977. On p. 2, they give the
story that it was invented by a prisoner in the Bastille, late 18C, and they
even give the name of the reputed inventor:
"Comte"(?) Pellisson.
They say that a Paul Pellisson‑Fontanier was in the Bastille in
1661‑1666 and was a man of some note, but history records no connection
between him and the game.
The Diagram Group. Baffle Puzzles -- 3: Practical Puzzles. Sphere, 1983. No. 12. On the 33-hole
board place 16 markers:
1 in row 2; 3 in row 3; 5 in row 4;
7 in row 5; making a triangle
centred on the mid-line. Can you remove
all the men, except for one in the central square? Gives a solution in 15 jumps.
J. D. Beasley. The Ins and Outs of Peg Solitaire. OUP, 1985.
History, pp. 3‑7; Selected
Bibliography, pp. 253‑261.
PLUS Additional notes, from the
author, 1p, Aug 1985. 57 references and
5 patents, including everything known before 1850.
Franco Agostini & Nicola
Alberto De Carlo. Intelligence
Games. (As: Giochi della Intelligenza; Mondadori, Milan, 1985.) Simon & Schuster, NY, 1987. This gives the legend of the nobleman in the
Bastille. Then says that "it would
appear that a very similar game" is mentioned by Ovid "and again, it
was widely played in ancient China -- hence its still frequent alternative name,
"Chinese checkers."" I
have included this as an excellent example of how unreferenced statements are
made in popular literature. I have
never seen either of these latter statements made elsewhere. The connection with Ovid is pretty tenuous
-- he mentions a game involving three in a row and otherwise is pretty cryptic
and I haven't seen anyone else claiming Ovid is referring to a solitaire game
-- cf 4.B.5. The connection with
Chinese checkers is so far off that I wonder if there is a translation problem
-- i.e. does the Italian name refer to some game other than what is known as
Chinese checkers in English??
Nob Yoshigahara. Puzzlart.
Tokyo, 1992. Coin solitaire, pp.
5 & 90. Four problems on a 4 x 4
board.
Marc Wellens, et al. Speelgoed Museum Vlaanderen -- Musée du
Jouet Flandre -- Spielzeug Museum Flandern -- Flanders Toy Museum. Speelgoedmuseum Mechelen, Belgium, 1996, p.
90 (in English), asserts 'It was
invented by the French nobleman Palissen, who had been imprisoned in the
Bastille by Louis XIV' in the early 18C.
The triangular version of the game has
only recently been investigated. The
triangular board is generally numbered as below.
1
2 3
4 5
6
7 8 9 10
11 12 13 14
15
Herbert M. Smith. US Patent 462,170 -- Puzzle. Filed: 13 Mar 1891; issued: 27 Oct 1891. 2pp + 1p diagrams. A board based on a triangular lattice.
Rohrbough. Puzzle Craft. 1932. Triangle Puzzle, p.
5 (= p. 6 in 1940s?). Remove peg 13 and
leave last peg in hole 13.
Maxey Brooke. (Fun for the Money, Scribner's, 1963); reprinted as: Coin Games and Puzzles; Dover, 1973. All the following are on the 15 hole board.
Prob.
1: Triangular jump, pp. 10-11 & 75.
Remove one man and jump to leave one man on the board. Says Wesley Edwards asserts there are just
six solutions. He removes the middle
man of an edge and leaves the last man there.
Prob.
2: Triangular jump, Ltd., pp. 12-13 & 75.
Removes some of the possible jumps.
Prob.
3: Headless triangle, pp. 14 & 75.
Remove a corner man and leave last man there.
M. Gardner. SA (Feb 1966) c= Carnival, 1975, chap. 2.
Says a 15 hole version has been on sale as Ke Puzzle Game by S. S. Adams
for some years. Addendum cites Brooke
and Hentzel and says much unpublished work has been done.
Irvin Roy Hentzel. Triangular puzzle peg. JRM 6:4 (1973) 280-283. Gives basic theory for the triangular
version. Cites Gardner.
[Henry] Joseph & Lenore
Scott. Quiz Bizz. Puzzles for Everyone -- Vol. 6. Ace Books (Charter Communications), NY,
1975. Pennies for your thoughts, pp.
179-182. Remove a coin and solve. Hint says to remove the coin at 13
and that you should be able to have the last coin at 13.
The solution has this property.
Alan G. Henney & Dagmar R.
Henney. Computer oriented
solutions. CM 4:8 (1978) 212‑216. Considers the 'Canadian I. Q. Problem',
which is the 15 hole board, but they also permit such jumps as 1 to 13,
removing 5. They find solutions from each initial
removal by random trial and error on a computer.
Putnam. Puzzle Fun.
1978. No. 15: Jumping coins, pp.
5 & 28. 15 hole version, remove peg 1
and leave last man there.
Benjamin L. Schwarz & Hayo
Ahlburg. Triangular peg solitaire -- A
new result. JRM 16:2 (1983-84)
97-101. General study of the 15 hole
board showing that starting and ending with
5 is impossible.
J. D. Beasley. The Ins and Outs of Peg Solitaire. Op. cit. above, 1985. Pp. 229-232 discusses the triangular
version, citing Smith, Gardner and Hentzel, saying that little has been
published on it.
Irvin Roy Hentzel & Robert
Roy Hentzel. Triangular puzzle
peg. JRM 18:4 (1985-86) 253‑256. Develops theory.
John Duncan & Donald
Hayes. Triangular solitaire. JRM 23:1 (1991) 26-37. Extended analysis. Studies army advancement problem.
William A. Miller. Triangular peg solitaire on a
microcomputer. JRM 23:2 (1991)
109-115 & 24:1 (1992) 11.
Summarises and extends previous work.
On the 10 hole triangular board, the classic problem has essentially a
unique solution -- the removed man must be an edge man (e.g. 2) and the last
man must be on the adjacent edge and a neighbour of the starting hole (i.e. 3
if one starts with 2). On the 15 hole
board, the removed man can be anywhere and there are many solutions in each
case.
Remove
man from hole: 1 2 4 5
Number
of solutions: 29760 14880 85258 1550
Considers
the 'tree' formed by the first four rows and hole 13.
New section. See also King and Stewart in 5.R.1 for some forms based on a
square board.
A
B C D E
F G
H
I J
Putnam. Puzzle Fun.
1978. No. 53: Checker star, pp.
10 & 34. Use the 10 points of a
pentagram, as above, and leave one of the inner points empty. Reduce to one man. [Parity shows the one man must be at an outer point and any outer
point can be achieved. If one leaves an
outer point empty, then the last man must be on an inner point and any of these
can be achieved.]
Hummerston. Fun, Mirth & Mystery. 1924.
Perplexity,
pp. 22-23. Using the octagram board
shown in 5.A, place 15 markers on it, leaving cell 16 empty. It is possible to remove all but one man. [I can't see how to apply parity to this
board.]
Solplex,
p. 25. In playing his Perplexity,
specify where you will leave the last man?
Leap
frog, Puzzle no. 22, pp. 64 & 175.
Take a 4 x 3 board with the long edge extended by one
more cell at the upper left and lower right.
Put white counters on the 4 x
3 area, put a black counter in one of
the extra cells and leave the other extra cell empty. Remove all but the black man.
Counting multiple jumps of the same man as a single move, he does it in
eight moves, getting the black man back to its starting point.
5.R.2. FROGS AND TOADS: BBB_WWW TO WWW_BBB
In
the simplest version, one has n black men at the left and n
white men at the right of a strip of
2n+1 cells, e.g. BBB_WWW.
One can slide a piece forward (i.e. blacks go left and whites go right)
into an adjacent place or one can jump forward over one man of the other colour
into an empty place. The object is to
reverse the colours, i.e. to get
WWW_BBB. S&B 121 & 125,
shows versions.
One
finds that the solution never has a man moving backward nor a man jumping
another man of the same colour. Some
authors have considered relaxing these restrictions, particularly if one has
more blank spaces, when these unusual moves permit shorter solutions. Perhaps the most general form of the
one-dimensional problem would be the following. Suppose we have m men at the left of the board, n men
at the right and b blank spaces in the middle. The usual case has b = 1, but when b > 1,
the kinds of move permitted do change the number of moves in a minimal
solution. First, considering slides,
can a piece slide backward? Can a piece
slide more than one space? If so, is
there a maximum distance, s, that it is allowed to slide? (The usual problem has s = 1.)
Of course s £
b. Second, considering jumps, can a
piece jump backward? Can a piece jump
over a piece (or pieces) of its own colour and/or a blank space (or spaces)
and/or a mixture of these? If so, is
there a maximum number of pieces,
p, that it can jump over? (The usual problem has p = 1.)
It is not hard to construct simple examples with s > 1
such that shorter solutions exist when unusual moves are permitted. Are there situations where one can show that
backward moves are not needed?
The
game is sometimes played on a 2-dimensional board, where one colour can move
down or right and the other can move up or left. See: Hyde ??; Lucas (1883); Ball; Hoffmann and 5.R.3.
Chinese checkers is a later variation of this same idea. On these more complex boards, one is usually
allowed to make multiple jumps and the object is usually to minimize the number
of moves to accomplish the interchange of pieces.
There
is a trick version to convert full and empty glasses: FFFEEE to FEFEFE
in one move, which is done by pouring.
I've just noted this in a 1992 book and I'll look for earlier examples.
Thomas Hyde. Historia Nerdiludii, hoc est dicere,
Trunculorum; .... (= Vol. 2 of De Ludis
Orientalibus, see 7.B for vol. 1.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De Ludo dicto Ufuba wa Hulana, p. 233. This has a
5 x 5 board with each
side having 12 men, but the description is extremely brief. It seems to have two players, but this may
simply refer to the two types of piece.
I'm not clear whether it's played like solitaire (with the jumped pieces
being removed) or like frogs & toads.
I would be grateful if someone could read the Latin carefully. The name of the puzzle is clearly Arabic and
Hyde cites an Arabic source, Hanzoanitas (not further identified on the pages I
have) -- I would be grateful to anyone who can track down and translate Arabic
sources.
American Agriculturist (Jun
1867). Spanish Puzzle. ??NYR -- copy sent by Will Shortz.
Anonymous. Every Little Boy's Book A Complete Cyclopædia of in and outdoor games
with and without toys, domestic pets, conjuring, shows, riddles, etc. With two hundred and fifty
illustrations. Routledge, London,
nd. HPL gives c1850, but the material
is clearly derived from Every Boy's Book, whose first edition was 1856. But the text considered here is not in the
1856 ed of Every Boy's Book (with J. G. Wood as unnamed editor), nor in the 8th
ed of 1868 (published for Christmas 1867), which was the first seriously
revised edition, with Edmund Routledge as editor, nor in the 13th ed. of
1878. So this material is hard to date,
though in 4.A.1, I've guessed this book may be c1868.
P.
12: Frogs and toads. "A new and
fascinating game of skill for two players; played on a leather board with
twelve reptiles; the toads crawling, and the frogs hopping, according to
certain laws laid down in the rules.
The game occupies but a few minutes, but in playing it there is scarcely
any limit to the skill that can be exhibited, thus forming a lasting amusement. (Published by Jaques, Hatton Garden.)" This does not sound like our puzzle, but
perhaps it is related. Unfortunately
Jaques' records were destroyed in WW2, so it is unlikely they can shed any
light on what the game was. Does anyone
know what it was?
Hanky Panky. 1872.
Checker puzzle, p. 124.
Three and three, with solution.
Mittenzwey. 1880.
Prob. 239, pp. 44 & 94;
1895?: 267-268, pp. 48 & 96;
1917: 267-268, pp. 44 & 91‑92. Problem with 3 & 3 brown and white
horses in stalls. 1895? adds a version
with 4 & 4.
Bazemore Bros. (Chattanooga,
Tennessee). The Great "13"
Puzzle! Copyright No. 1033 ‑ O ‑ 1883. Hammond & Jones Printers.
Advertising puzzle consisting of two
3 and 3 versions arranged in
an X
pattern.
Lucas. RM2. 1883.
Pp.
141‑143. Finds number of moves
for n and n.
Pp.
144‑145. Considers game on 5 x 5,
7 x 7, ..., boards and gives number of moves.
Edward Hordern's collection has
an example called Sphinxes and
Pyramids from the 1880s.
Sophus Tromholt. Streichholzspiele. (1889; 5th ed,
1892.) Revised from 14th ed. of 1909 by
R. Thiele; Zentralantiquariat der DDR, Leipzig, 1986. Prob. 11, 41, 81 are the game for 4 & 4, 2 & 2, 3 & 3.
Ball. MRE, 1st ed., 1892, pp. 49‑51. 3 & 3 case, citing
Lucas, with generalization to n &
n; 7 x 7 board, citing Lucas, with generalization to 2n+1 x 2n+1.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
German counter puzzle, p. 112. 3
& 3 case.
Hoffmann. 1893.
Chap. VI, pp. 269‑270 & 282‑284 = Hoffmann-Hordern,
pp. 182-185, with photo.
No.
17: The "Right and Left" puzzle.
Three and three. Photo on p. 184
shows: a cartoon from Punch (18 Dec 1880): The Irish Frog Puzzle -- with a Deal
of Croaking; and an example of a handsome carved board with square pieces with
black and white frogs on the tops, registered 1880. Hordern Collection, p. 77, shows the latter board and two
further versions: Combat Sino-Japonais (1894‑1895) and Anglais &
Boers (1899-1902).
No.
18. Extends to a 7 x 5
board.
Puzzles with draughtsmen. The Boy's Own Paper 17 or 18?? (1894??)
751. 3 and 3.
Lucas. L'Arithmétique Amusante.
1895. Prob. XXXV: Le bal des
crapauds et des grenouilles, pp. 117-124.
Does 2 and 2, 3 and 3,
4 and 4 and the general case of n and n,
showing it can be done in
n(n+2) moves -- n2 jumps and
2n steps. The general solution is attributed to M. Van
den Berg. M. Schoute notes that each
move should make as little change as possible from the previous with respect to
the two aspects of changing type of piece and changing type of move.
Clark. Mental Nuts. 1904, no.
72; 1916, no. 62. A good study. 3 and 3.
Burren Loughlin &
L. L. Flood. Bright-Wits Prince of Mogador. H. M. Caldwell Co., NY, 1909.
Doola's Game, pp. 42-43 & 61-62.
3 and 3.
Anon. Prob. 47: The monkey's dilemma.
Hobbies 30 (No. 762) (21 May
1910) 168 & 182 & (No. 765) (11 Jun 1910) 228. Basically
3 & 3, but there are eight
posts for crossing a river, with the monkeys on 1,2,3 and 6,7,8.
The monkeys can jump onto the bank and we want the monkeys to all get to
the bank they are headed for, so this is not the same as BBB..WWW
to WWW..BBB. The solution doesn't spell out all the
steps, so it's not clear what the minimum number of moves is -- could we have a
monkey jumping another of the same colour?
Ahrens. MUS I.
1910. Pp. 17-19. Basically repeats some of Lucas's work from
1883 & 1895.
Williams. Home Entertainments. 1914.
The cross-over puzzle, pp. 119-120.
3 and 3 with red and white
counters. Doesn't say how many moves
are required.
Dudeney. AM.
1917. Prob. 216: The educated
frogs, pp. 59-60 & 194.
_WWWBBB to BBBWWW_
with frogs able to jump either way over one or two men of either
colour. Solution in 10 jumps.
Ball. MRE, 9th ed., 1920, pp. 77-79, considers the m & n
case, giving the number of steps in the solution.
Blyth. Match-Stick Magic.
1921. Matchstick circle
transfer, pp. 81‑82. 3 and 3 in 15 moves.
Hummerston. Fun, Mirth & Mystery. 1924.
The frolicsome frogs, Puzzle no. 2, pp. 17 & 172. Two
3 & 3 problems with the
boards crossing at the centre cell. He
notes that the easiest solution is to solve the boards one at a a time. He says: "It is not good play to jump a
counter over another of the same colour."
Lynn Rohrbough, ed. Socializers. Handy Series, Kit G, Cooperative Recreation Service, Delaware,
Ohio, 1925. Six Frogs, p. 5. Dudeney's 1917 problem done in 11 moves.
Botermans et al. The World of Games. Op. cit. in 4.B.5. 1989. P. 235 describes
this as The Sphinx Puzzle, "very popular around the turn of the century,
particularly in the United States and France" and they show an example of
the period labelled The Sphinx and Pyramid Puzzle -- An Egyptian Novelty.
Haldeman-Julius. 1937.
No. 162: Checker problem, pp. 18 & 29. 3 & 3.
See Harbin in 5.R.4 for a 1963
example.
Doubleday - 1. 1969.
Prob.
77: Square dance, pp. 93 & 171. =
Doubleday - 5, pp. 103-104. Start
with _WWWBBB. He says they must change places, with a piece able to move into
the vacant space by sliding (either way) or by jumping one or two pieces of any
colour. Asks for a solution in 10
moves. His solution gets to BBBWWW_,
which does not seem to be 'changing places' to me.
Prob.
79: All change, pp. 95 & 171. =
Doubleday - 5, pp. 105-106. BB_WW
Start with the pattern at the right and change
the whites and BB_WW
blacks in 10 moves, where a piece can slide one
place into an
adjacent vacant square or jump one or two
pieces into a vacant square. However,
the solution simply does each row separately.
Katharina Zechlin. (Dekorative Spiele zum Selbermachen; Verlag
Frech, WWWWW
Stuttgart-Botnang,
1973.) Translated as: Making Games in Wood Games BWWWW
you
can build yourself. Sterling, 1975, pp.
24-27: The chess knight game. BBOWW
5
x 5 board with 12 knights of each of
two colours, arranged as at the right. BBBBW
The
object is to reverse them by knight's moves.
Says it can be done within BBBBB
50
moves and 'is almost impossible to do it in less than 45'.
Wickelgren. How to Solve Problems. Op. cit. in 5.O. 1974. Discrimination
reversal problem, pp. 78‑81.
_WWWBBB to BBBWWW
with the extra place not specified in the goal, with pieces allowed to move
into the vacant space by sliding or by hopping over one or two pieces. Gets to
BBBWW_W in 9 moves. [I find it takes 10 moves to get to BBBWWW_ .]
Joe Celko. Jumping frogs and the Dutch national
flag. Abacus 4:1 (Fall 1986)
71-73. Same as Wickelgren. Celko attributes this to Dudeney. Gives a solution to BBBWWW_
in 10 moves and asks for results for higher numbers.
Johnston Anderson. Seeing induction at work. MG 75 (No. 474) (Dec 1991) 406-414. Example 2: Frogs, pp. 408-411. Careful proof that BB...BB_WW...WW to WW...WW_BB...BB, with n counters of each colour, requires n2 + 2n moves.
5.R.3. FORE AND AFT -- 3 BY 3 SQUARES MEETING AT A CORNER
This is Frogs and Toads on part of
the 5 x 5 board consisting of two 3
x 3 subarrays at diagonally opposite
corners. They overlap in the central
square. One square has 8 black men and
the other has 8 white men, with the centre left vacant.
Ball. MRE, 1st ed., 1892, pp. 51‑52. 51 move solution. In the
third ed., 1896, pp. 69‑70, he says he believes he was the first to
publish the puzzle but "that it has been since widely distributed in
connexion with an advertisement and probably now is well known". He gives a 48 move solution.
Hoffmann. 1893.
Chap. VI, no. 26: The "English Sixteen" puzzle, pp. 273‑274
& 287 = Hoffmann-Hordern, pp. 188-189, with photo. Mentions that it is produced by Messrs
Heywood, as below. Solution in 52
moves, which he believes is minimal.
Hordern notes that the minimum is 46.
Photo on p. 188 of the Heywood version, see next entry.
John Heywood, Manchester,
produced a version called 'The English Sixteen Puzzle', undated, but by 1893 as
Hoffmann cites it. Photo in
Hoffmann-Hordern, p. 188, dated 1880‑1895.
Charles A. Emerson. US Patent 522,250 -- Puzzle. Applied: 3 Nov 1893; patented: 3 July 1894. 2pp + 1p diagrams. The Fore and Aft Puzzle.
Says it can be done in 48, 49,
50, 51 or 52 moves.
Dudeney. Problem 66: The sixteen puzzle. Tit‑Bits 33 (1 Jan &
5 Feb 1898) 257 & 355.
"It was produced, I believe, in America, many years ago, and has
since been issued over here in the form of an advertisement by a prominent
commercial house." Solution in 46
moves. He says published solutions
assert the minimum number of moves is
53, 52 or 50. The 46 move
solution is given in Ball, MRE, 5th ed., 1911, 79‑80.
Ball. MRE, 5th ed., 1911, pp. 79-80.
Drops his historical claims and includes a 46 move solution due to
Dudeney.
Loyd. Fore and aft puzzle.
Cyclopedia, 1914, pp. 108 & 353 (solution misprinted, but claimed to
be 47 moves in contrast to 52 move solutions 'in the puzzle books'.) (c= MPSL1, prob. 4, pp. 3‑4 &
121 (only referring to Dudeney's 46 move solution)).
Loyd Jr. SLAHP.
1928. A joke on granddad, pp. 29
& 93. Says 'our granddaddies, who
used to play this puzzle game 75 years ago, when it was universally popular. The old‑time books explain how the
solution is accomplished in 52 moves, "the shortest possible
method."' He then asks for and
gives a 46 move solution.
M. Adams. Puzzles That Everyone Can Do. 1931.
Prob. 24, pp. 17 & 132: "General post". Gives a solution which takes 46 moves, but
gives no discussion of it.
Rohrbough. Puzzle Craft. 1932. Migration (or Fore
and Aft), p. 12 (= p. 15 of 1940s?).
Says it was popular 75 years ago and it has recently been shown that it
can be done in 46 moves, then gives a solution which stops at 42 moves!
M. Gardner. SA (Sep 1959) = 2nd Book, pp. 210‑219. Discusses the puzzle. On pp. 218‑219, he gives
Dudeney's 46 move solution and says 48 different solutions and several proofs
that 46 is minimal were sent to him.
Uwe Schult. Das Seemanns‑Spiel: Mathematisch
erledigt. Reported in Das Mathematische
Kabinett column, Bild der Wissenschaft 19:11 (Nov 1982) 181-184. (A version is given in Neues aus dem
Mathematischen Kabinett, ed. by Thiagar Devendran, Hugendubel, Munich, 1985,
pp. 102‑103.) There are 218,790
possible patterns of the pieces.
Reversing black and white takes
46 moves and there are 1026
different halfway positions that can occur in a 46
move solution. There are two
patterns which require 47 moves, namely, after reversing black and
white, put one of the far corner pieces in the centre.
Nob Yoshigahara, postcard to me
on 18 Aug 1994, announces he has found the worst solution -- in 58 moves.
5.R.4. REVERSING FROGS AND TOADS: _12...n TO _n...21 , ETC.
A
piece can slide into the empty cell or jump another piece into the empty cell.
Dudeney. AM.
1917.
Prob.
214: The six frogs, pp. 59 & 193.
Case of n = 6, solved in
21 moves, which he says is
minimal. In general, the minimal solution
takes n(n+1)/2 moves, including n steps, when n is
even and (n2+3n-8)/2
moves, including 2n-4 steps, when
n is odd. "This complete general solution is
published here for the first time."
Prob.
215: The grasshopper puzzle, pp. 59 & 193-194. Problem for a circular arrangement. Example has n = 12. Says he invented it in 1900. Solvable in
44 moves. General solution is complex -- he says that
for n > 4, it can be done in (n2+4n-16)/4 moves when
n is even and in (n2+6n-31)/4 moves when
n is odd.
Rohrbough. Puzzle Craft. 1932. The Reversible
Frogs, p. 22 (= The Jumping Frogs, pp. 20‑21 of 1940s?). n = 8, citing Dudeney, AM.
Robert Harbin. Party Lines. Op. cit. in 5.B.1.
1963. Hopover, p. 89. First gives
3 and 3 Frogs and Toads, then
asks for complete reversal from
123_456 to 654_321.
[Henry] Joseph and Lenore
Scott. Master Mind Pencil Puzzles. Tempo Books (Grosset & Dunlap), NY, 1973
(& 1978?? -- both dates are given -- I'm presuming the 1978 is a 2nd ptg or
a reissue under a different imprint??).
Reverse the numbers, pp. 117‑118. Give the problem for n =
6 and a solution in 21 moves. For
n even, the method gives a
solution in n(n+1)/2, it is not shown that this is optimal, nor is
a general method given for odd n.
[Henry] Joseph & Lenore
Scott. Master Mind Brain Teasers. 1973.
Op. cit. in 5.E. 13-hour clock,
pp. 43-44. Case n = 12 considered in a circle can be done in 44 moves.
Joe Celko. Jumping frogs and the Dutch national
flag. Abacus 4:1 (Fall 1986)
71-73. Cites Dudeney and gives the
results.
Jim Howson. The Computer Weekly Book of Puzzlers. Computer Weekly Publications, Sutton,
Surrey, 1988, unpaginated. [The
material comes from his column which started in 1966, so an item may go back to
then.] Prob. 54 -- same as the Scotts
in Master Mind Pencil Puzzles.
There
are a number of similar games on different boards -- too many to describe
completely here, so I will generally just cite extensive descriptions. See any of the main books on games mentioned
at the beginning of 4.B, such as Bell or Falkener. The key feature is that one side has more, but weaker, pieces. These are sometimes called hunt games. The standard Fox and Geese is played on a 33
hole Solitaire board, with diagonal moves allowed. I have recently acquired but not yet read Murray's History of
Board Games other than Chess which should have lots of material.
Gretti's Saga, late 12C. Mention of Fox and Geese. Also in Edward IV's accounts. ??NYS -- cited by Botermans et al, below.
Shackerley (or Schackerley or
Shakerley) Marmion. A Fine Companion (a
play). 1633. IN: The Dramatic Works of Shackerley Marmion; William Paterson,
Edinburgh & H. Sotheran & Co., London, 1875.
II, v, pp. 140-141. "...,
let him sit in the shop ..., and play at fox and geese with the foreman,
....." Earliest English occurrence
of fox-and-geese. Quoted by OED and
cited by Fiske, below.
Richard Lovelace. To His Honoured Friend On His Game of
Chesse-Play or To Dr. F. B. on his Book of Chesse. 1656?, published in his Posthume Poems,
1659. Lines 1-4. My edition of Lovelace notes that F. B. was
Francis Beale, author of 'Royall Game of Chesse Play,' 1656. Lovelace died in 1658.
Sir,
now unravell'd is the Golden Fleece,
Men
that could onely fool at Fox and Geese,
Are
new-made Polititians by the Book,
And
can both judge and conquer with a look.
Henry Brooke. Fool of Quality. [A novel.]
1766-1768. Vol. I, p. 367. ??NYS -- quoted by Fiske, below. "Can you play at no kind of game,
Master Harry?" "A little at
fox‑and‑geese, madam."
Catel. Kunst-Cabinet. 1790.
Das
Fuchs- und Hühnerspiel, pp. 51-52 & fig. 168 on plate VI. 11 chickens against one fox on a 4 x 4
board with all diagonals drawn, giving
16 + 9 playing points.
Das
Schaaf- und Wolfspiel, p. 52 & fig. 169 on plate VI, is the same game on
the 33-hole solitaire board with 11 sheep and one wolf, no diagonals
Bestelmeier. 1801.
Item
83: Das Schaaf- und Wolfspiel. Same
diagram and game as Catel, p. 52.
Item
833: Ein Belagerungspiel. 33 hole board
with a fortress on one arm, with diagonals drawn.
Strutt. Op. cit. in 4.B.1. Fox and Geese. 1833: Book
IV, chap. II, art. XIV, pp. 318‑319.
= Strutt-Cox, p. 258 & plate opp. p. 246. Fig. 107 (= plate opp. p. 246) shows the 33
hole board with its diagonals drawn.
Gomme. Op. cit. in 4.B.1. I 141‑142
refers to Strutt and Micklethwaite.
Illustrated Boy's Own
Treasury. 1860. Fox and Geese, pp. 406‑407. 33 hole Solitaire board with diagonals
drawn.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 320, p. 152:
Fuchs und Gänse. Shows 33 hole
solitaire board with diagonals drawn.
Stewart Culin. Chinese Games with Dice and Dominoes. From the Report of the U. S. National Museum
for 1893, pp. 489‑537. Pp.
874-877 describes: the Japanese game
of Juroku Musashi (Sixteen Soldiers) with 16 men versus a general;
the Chinese game of Shap luk kon
tséung kwan (The sixteen pursue the
commander); another Chinese game
of Yeung luk sz' kon tséung kwan with 27 men against a commander (described
by Hyde -- ?? I didn't see this); the
Malayan game of Dam Hariman (Tiger Game), identical to the Hindu game of
Mogol Putt'han (= Mogul Pathan
(Mogul against Pathan)), similar to a
Peruvian game of Solitario and the Mexican game of Coyote; the Siamese game of Sua ghin gnua (Tiger and Oxen) and the
similar Burmese game of Lay gwet
kyah, with three big tigers versus 11
or 12 little tigers; the Samoan game of Moo;
the Hawaiian game of Konane; a
similar Madagascarian game; the Hindu
game of Pulijudam (Tiger Game) with three tigers versus 15
lambs.
Fiske. Op. cit. in 4.B.1.
1905. Fox-and-Geese, pp. 146-156
& 359, discusses the history of the game, especially as to whether it is
identical to the old Norse game of Hnefatafl.
On p. 359, he says that John of Salisbury (c1150) used 'vulpes' as
the name of a game, but there is no indication of what it was. He says "the fox-and-geese board, in
comparatively modern times, has begun to be used for games more or less
different in their nature, especially for one called in England solitaire and
in France "English solitaire", and for another, known in Spain and
Italy as asalto (assalto), in French as assaut, in Danish as
belejringsspel." He then surveys
the various sources that he treated under Mérelles -- see 4.B.1 and 4.B.5 for
details. He is not sure that Brunet is
really describing the game in the Alfonso MS (op. cit. in 4.B.5 and
below). He cites an 1855 Italian usage
as Jeu de Renard or Giuoco della Volpe.
In Come Posso Divertirmi? (Milan, 1901, pp. 231-233), it is said that
the game is usually played with 17 geese rather than 13 -- Fiske notes that
this assertion is of "some historical value, if it be true." Moulidars calls it Marelle Quintuple, quotes
Maison des Jeux Académiques (Paris, 1668) for a story that it was invented by
the Lydians and gives the game with 13 or 17 geese. Asalto has 2 men against 24.
Fiske quotes Shackley Marmion, above, for the oldest English occurrence
of fox-and-geese and then Henry Brooke, above.
Fiske follows with German, Swedish and Icelandic (with 13 geese)
references.
H. Parker. Ancient Ceylon. Op. cit. in 4.B.1, 1909.
Pp. 580-583 & 585 describe four forms of The Leopards Game, with one
tiger against seven leopards, three leopards against 15 dogs, two leopards
against 24 cattle and one leopard against six cattle on a 12 x 12
board. The first two are played
on a triangular board.
Robert Kanigel. The Man Who Knew Infinity. A Life of the Genius Ramanujan. (Scribner's, NY, 1991); Abacus (Little, Brown & Co. (UK)),
London, 1992. Pp. 18 & 377: Ramanujan and his mother used to play the
game with three tigers and fifteen goats on a kind of triangular board.
The Spanish Treatise on
Chess-Play written by order of King Alfonso the Sage in the year 1283. [= Libro de Acedrex, Dados e Tablas of
Alfonso El Sabio, generally known as the Alfonso MS.] MS in Royal Library of the Escorial (j.T.6. fol). Complete reproduction in 194 Phototypic
Plates. 2 vols., Karl W. Hirsemann, Leipzig, 1913. See 4.B.5 for more details of this work. See below.
Botermans et al. The World of Games. Op. cit. in 4.B.5. 1989. P. 147 says De
Cercar La Liebre (Catch the Hare) occurs in the Alfonso MS and is the earliest
example of a hunt game in European literature, but undoubtedly derived from an
Arabic game of the Alquerque type -- I didn't see this when I briefly looked at
the facsimile -- ??NYS. They say Murray
has noted that hunt games are popular in Asia, but not in Africa, leading to
the conjecture that they originated in Asia.
They describe it on a 5 x 5 array of points with verticals and
horizontals and some diagonals drawn, with one hare against 12 hunters.
Botermans
et al. continue on pp. 148-155 to describe the following.
Shap
Luk Kon Tseung Kwan (Sixteen Pursue the General) played on a 5 x 5
board like Catch the Hare with an extra triangle on one side and
capturing by interception.
Yeung
Luk Sz'Kon Tseung Kwan, seen in Nanking by Hyde and described by him in 1694,
somewhat similar to the above, but with 26 rebels against a general. (??NYS)
Fox
and Geese, mentioned in Gretti's Saga of late 12C and in Edward IV's
accounts. They give a version called
Lupo e Pecore from a 16C Venetian book, using a Solitaire board extended by
three points on each arm, giving 45 points.
They give a 1664 engraving showing Le Jeu du Roi which they say is a
rather complex form of fox and geese, but looks like a four-handed game on a
cross-shaped board with 7 x 5 arms on a
7 x 7 central square and 4
groups of 7 x 4 men.
Leopard
games, from Southeast Asia, with a kind of triangular board. Len Choa, from Thailand, has a tiger against
six leopards. Hat Diviyan Keliya, from
Sri Lanka, has a tiger against seven leopards.
Tiger
games, also from Southeast Asia, are similar to leopard games, but use an
extended Alquerque board (as in Catch the Hare). Rimau (Tiger), from Malaysia, has 24 men versus a tiger and
Rimau-Rimau (Tigers) is a version with two tigers versus 22 men.
Murray. 1913.
P. 347
cites a 1901 Indian book for 2 lions against 32 goats on a chessboard.
P. 371
cites a Soyat (North Asia) example (19C?) of Bouge‑Shodra (Boar's Chess)
with 2 boars against 24 calves on a chessboard.
Pp.
569 & 616‑617 cite the Alfonso MS of 1283 for 'De cercar la liebre',
played on a 5 x 5 board with
10, 11 or 12 men against a hare.
P. 585
shows Cott. 6 (c1275) of 8 pawns against a king on a chessboard.
Pp.
587 & 590 give Cott. 11 = K6: Le Guy de Alfins with king and 4 bishops
against a king on a chessboard.
Pp.
589-590 shows K4 = CB249: Le Guy de Dames and No. 5 = K5: Le Guy de
Damoyselles, which have 16 pawns against a king on a chessboard.
P. 617
discusses Fox and Geese, with 13, 15 or 17 geese against a fox on the solitaire
board. Edward IV, c1470, bought
"two foxis and 46 hounds".
Murray says more elaborate forms exist and refers to Hyde and Fiske (see
4.B.1 and 5.F.1 for more on these), ??NYS.
Pp.
675 & 692 show CB258: Partitum regis Francorum with king and four pawns
against king on the chessboard. It says
the first side wins.
P. 758
describes a 16C Venetian board (then) at South Kensington (V&A??) with the
Solitaire board for Fox and Geese and an enlarged board for Fox and Geese.
P. 857
mentions Fox and Geese in Iceland.
Family Friend 2 (1850) 59. Fox and geese. 4 geese against 1 fox on a chess board.
The Sociable. 1858.
Fox and geese, p. 281. 17 geese
against a fox on the solitaire board.
Four men versus a king on the draughts board, saying the first side wins
even allowing the king to be placed anywhere against the men who start on one
side.
Stewart Culin. Korean Games, op. cit. in 4.B.5, 1895. Pp. 76-77 describes some games of this type,
in particular a Japanese game called Yasasukari Musashi with 16 soldiers versus
a general on a 5 x 5 board, taken from a 1714 (or 1712) Japanese
book: Wa Kan san sai dzu e
"Japanese, Chinese, Three Powers picture collection", published in
Osaka.
Anonymous. Enquire Within upon Everything. 66th ed., 862nd thousand, Houlston and Sons,
London, 1883, HB. Section 2593: Fox and
Geese, p. 364. 33 hole Solitaire board
with 17 geese against a fox. 4 geese
against a fox on the chessboard. Says
the geese should win in both cases.
Slocum. Compendium.
Shows Solitaire and Solitaire & Tactic Board from Gamage's 1913
catalogue. Like Bestelmeier's 833, but
without diagonals.
Bell & Cornelius. Board Games Round the World. Op. cit. in 4.B.1. 1988. Games involving unequal
forces, pp. 43-52. Discusses the
following.
The
Maharajah and the Sepoys. 1 against 16
on a chessboard.
Fox
and Geese. Cites an Icelandic work of
c1300 (probably Gretti's Saga?). 1
against 13 or 17 on a Solitaire board.
Lambs
and Tigers, from India. 3 against 15.
Cows
and Leopards, from SE Asia. 2 against
24.
Vultures
and Crows, also called Kaooa, from India.
1 against 7 on a pentagram board.
The
New Military Game of German Tactics, c1870.
2 against 24 on a Solitaire Board with a fortress, as in Bestelmeier.
Yuri I. Averbakh. Board games and real events. IN: Alexander J. de Voogt, ed.; New
Approaches to Board Games Research: Asian Origins and Future Perspectives;
International Institute for Asian Studies, Leiden, 1995; pp. 17-23. Notes that Murray believes hunt games
evolved from war games, but he feels the opposite is true. He describes a Nepalese game of Baghachal
with four tigers versus 20 goats -- this is Murray's 5.6.22. He corrects some of Murray's assertions
about Boar Chess and describes other Tuvinian hunt games: Bull's Chess and
Calves' Chess, probably borrowed from the Mongols. The latter has a three-in-a-row pattern and he wonders if there
is some connection with morris or noughts and crosses (which he says is
"played everywhere"). He
mentions Cercar la Liebre from the Alfonso MS.
Fox and Geese type games are mentioned in the Icelandic sagas as 'the
fox game'. He describes several forms.
One
has an octagram and seven men. One has to
place a man on a vacant point and then slide him to an adjacent vacant point,
then do the same with the next man, ...,
so as to cover seven of the points.
The diagram is just an 8-cycle and is the same as the knight's
connections on the 3 x 3 board, so the octagram puzzle is equivalent
to the 7 knights problem mentioned in 5.F.1.
Further, the 4 knights problem of 5.F.1 has the same 8-cycle, with men
at alternate points of it.
Versions
with different numbers of points.
5 points:
Rohrbough.
7 points: Mittenzwey;
Meyer.
9 points: Dudeney.
10 points: Bell & Cornelius; Hoffmann; Cohen;
Williams; Toymaker; Rohrbough; Putnam.
13 points: Berkeley & Rowland.
Bell & Cornelius. Board Games Round the World. Op. cit. in 4.B.1. 1988. Pentalpha, p.
15. Says that a pentagram board occurs
at Kurna, Egypt, c-1400 and that the solitaire game of Pentalpha is played in
Crete. This has 9 men to be placed on
the vertices and the intersections of the pentagram. Each man must be placed on a vacant point, then slid ahead two
positions along one straight line. The
intermediate point may be occupied, but the ending point must be
unoccupied. Unfortunately we don't know
if the Egyptian board was used for this game.
Pacioli. De Viribus.
c1500.
Ff.
112r - 113v. .C(apitolo). LXVIII. D(e).
cita ch' a .8. porti ch' cosa convi(e)ne arepararli (Chap. 68. Of a city with 8 gates which admits of
division ??). = Peirani 158-160. Octagram puzzle with a complex story about a
city with 8 gates and 7 disputing factions to be placed at the gates.
F.
IVv. = Peirani 8. The Index gives the above as Problem
83. Problem 82: De .8. donne ch' sonno
aun ballo et de .7. giovini quali con loro sa con pagnano (Of 8 ladies who are
at a ball and of 7 youths who accompany them).
Schwenter. 1636.
Part 2, exercise 36, pp. 149-150.
Octagram.
Witgeest. Het Natuurlyk Tover-Boek. 1686.
Prob. 4, pp. 224-225. Octagram,
taken from Schwenter.
Les Amusemens. 1749.
P. xxxiii.
Catel. Kunst-Cabinet. 1790. Das Achteck, pp. 12-13 & fig. 36 on
plate II. The rules are not clearly
described.
Bestelmeier. 1801.
Item 290: Das Achteckspiel. Text
copies part of Catel.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. Ff. 131-133 are an analysis of
the heptagram puzzle.
Rational Recreations. 1824.
Feat 34, pp. 161-164. Octagram.
Endless Amusement II. 1826?
Prob. 28, pp. 203-204. = New
Sphinx, c1840, pp. 137-138.
Nuts to Crack IV (1835), no. 194
-- part of a long section called Tricks upon Travellers.
Family Friend (Dec 1858)
359. Practical puzzles -- 1. I don't have the answer.
The Boy's Own Magazine 3 (1857)
159 & 192. Puzzle of the points.
Illustrated Boy's Own
Treasury. 1860. Practical Puzzles No. 6, pp. 396 & 436.
The Secret Out. 1859.
To Place Seven Counters upon an Eight-Pointed Star, pp. 373-374.
J. J. Cohen, New York. Star puzzle. Advertising card for Star Soap, Schultz & Co., Zanesville,
Ohio, Copyright May 1887. Reproduced
in: Bert Hochberg; As advertised
Puzzles from the collection of Will Shortz; Games Magazine 17:1 (No.
113) 10-13, on p. 11. Identical to
pentalpha - see Bell & Cornelius above.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
Card Puzzles No. IX: The reversi puzzle, pp. 8-10. Version with 13 cards in a circle and one
can move ahead by any number of steps.
If there are x cards and one moves ahead s
steps, then x and
s must have no common factor.
Hoffmann. 1893.
Chap. VI, pp. 267-268 & 280-281 = Hoffmann-Hordern,
pp. 180-181, with photos.
No.
13. No name. Basic octagram puzzle.
Photo on p. 181 shows: The Seven Puzzles, by W. & T. Darton, dated
1806-1811; a Tunbridge ware version dated 1825-1840; and Jeu de Zig‑Zag,
by M. D., Paris, 1891-1900.
No.
14. The "Okto" puzzle, pp.
268 & 281. Here the counters and
points are coloured. Photo on p. 181 of
The "Okto" Puzzle by McGaw, Stevenson & Orr, Ltd. for John
Stewart, dated 1880-1895.
Chap. X, no. 8: Crossette, pp. 337 & 374-375 =
Hoffmann-Hordern, pp. 229-230, with photo.
10 counters in a circle. Start
anywhere and move ahead three. Photo on
p. 230 shows The Mystic Seven, a seven counter version, by the Lord Roberts
Workshops, 1914-1920.
Mittenzwey. 1895?
Prob. 329, pp. 58 & 106;
1917: 329, pp. 52 & 101.
Heptagram.
Dudeney. Problem 58: A wreath puzzle. Tit‑Bits 33 (6 &
27 Nov 1897) 99 & 153.
Complex nonagram puzzle involving moves in either direction and
producing the original word again.
Clark. Mental Nuts. 1897, no.
54; 1904, no. 80; 1916, no. 69. A little puzzle. Usual
octagram.
Benson. 1904.
The eight points puzzle, pp. 250‑251. c= Hoffmann, no. 13.
Slocum. Compendium.
Shows the "Octo" Star Puzzle from Gamage's 1913 catalogue.
Williams. Home Entertainments. 1914.
Crossette,
pp. 115-116. Ten points, advancing
three places.
Eight
points puzzle, pp. 120-121. Usual
octagram.
"Toymaker". Top in Hole Puzzle. Work (23 Dec 1916) 200. 10 holes and one has to move to the third
position and reverse the top in that hole.
Blyth. Match-Stick Magic.
1921. Crossing the points, pp.
83-84.
Hummerston. Fun, Mirth & Mystery. 1924.
The
sacred seven, Puzzle no. 5, pp. 26 & 173.
Octagram puzzle on the outer points of the diagram shown in 5.A.
The
four rabbits, Puzzle no. 6, pp. 26 & 173.
Using the octagram shown in 5.A, put black counters on locations 1 and 2
and white counters on 7 and
8. The object is to interchange the colours. This is like the 4 knights problem except
the corresponding 8-cycle has men at positions
1, 2, 5, 6. He counts a sequence
of steps by the same man as a move and hence solves it in 6 moves (comprising
16 steps).
Will Blyth. Money Magic. C. Arthur Pearson, London, 1926.
Turning the tails, pp. 66-69. 8
coins in a circle, tails up. Count from
a tail four ahead and reverse that coin.
Get 7 heads up. Counting four
ahead means that if you start at 1, you count
1, 2, 3, 4 and reverse 4.
King. Best 100. 1927. No. 64, pp. 26-27 & 54.
Rohrbough. Puzzle Craft. 1932.
Count
4, p. 6. 10 points on a circle, moving
ahead 3. (= Rohrbough; Brain Resters
and Testers; c1935, p. 21.)
Star
Puzzle, p. 8 (= p. 10 of 1940s?).
Consider the pentagram with its internal vertices. First puzzle is Pentalpha. Second is to place a counter and move ahead
three positions. The object is to get four
counters on the points, which is the same as the pentagram puzzle, moving one
position.
Jerome S. Meyer. Fun-to-do.
Op. cit. in 5.C. 1948. Prob. 18:
Odd man out, pp. 27 & 184. Version
with 7 positions in a circle and 6 men where one must place a man and then move
him three places ahead.
Putnam. Puzzle Fun.
1978. No. 63: Ten card turnover,
pp. 11 & 35. Ten face down cards in
a circle. Mark a card, count ahead
three and turnover.
The
usual version is to have 8 counters in a row which must be converted to 4 piles
of two, but each move must pass a counter over two others. Martin Gardner pointed out to me that the
problem for 10, 12, 14, ... counters is easily reduced to that for
8. The problem is impossible for 2, 4, 6.
There are many later appearances of the problem than given here. In describing solutions, 4/1
means move the 4th piece on top of the 1st piece.
There
are trick solutions where a counter moves to a vacated space or even lands
between two spaces. See: Mittenzwey; Haldeman-Julius; Hemme.
Berkeley & Rowland give a problem
where each move must pass a counter over two piles. This makes the problem easier and it is solvable for any even
number of counters ³ 6,
but it gives more solutions.
See: Berkeley & Rowland; Wood; Indoor Tricks & Games; Putnam;
Doubleday - 1.
One
could also permit passing over one pile, which is solvable for any even
number ³ 4.
Mittenzwey,
Double Five Puzzle, Hummerston, and Singmaster & Abbott deal with the
problem in a circle and with piles to be left in specific locations.
Mittenzwey,
Lucas and Putnam consider making piles of three by passing over 3, etc.
Kanchusen. Wakoku Chiekurabe. 1727. Pp. 38-39. Jukkoku-futatsu-koshi (Ten stones jumping
over two). Ten counters, one solution.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. F. 4r is "Analysis of the
Essay of Games". F. 4v has
"The question of the shillings passing at each time over two or a certain
number 8 is the least number Any number being given and any law of transit Dr Roget" The layout suggests that Roget had posed the
general version. Adjacent is a diagram
with a row of 10 counters and the first move 1 to 4 shown, but with some unclear
later moves.
Endless Amusement II. 1826?
Prob. 10, p. 195. 10 halfpence. One solution: 4/1 7/3 5/9 2/6 8/10. = New Sphinx, c1840, pp. 135-135.
Nuts to Crack II (1833), no.
122. 10 counters, identical to Endless
Amusement II.
Nuts to Crack V (1836), no.
68. Trick of the eight sovereigns. Usual form.
Young Man's Book. 1839.
P. 234. Ingenious Problem. 10 halfpence. Identical to Endless Amusement II.
Family Friend 2 (1850) 178 &
209. Practical Puzzle, No. VI. Usual form with eight counters or
coins. One solution.
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles, no. 2, p. 176 (1868:
187). Passing over coins. Gives two symmetric solutions.
Magician's Own Book. 1857.
Prob. 34: The counter puzzle, pp. 277 & 300. Identical to Book of 500 Puzzles, prob. 34.
The Sociable. 1858.
Prob. 16: Problem of money, pp. 291-292 & 308. Start with 10 half‑dimes, says to pass
over one, but solution has passing over two.
One solution. = Book of 500
Puzzles, 1859, prob. 16, pp. 9-10 & 26.
Book of 500 Puzzles. 1859.
Prob. 16:
Problem of money, pp. 9-10 & 26. As
in The Sociable.
Prob.
34: The counter puzzle, pp. 91 & 114.
Eight counters, two solutions given.
Identical to Magician's Own Book.
The Secret Out. 1859.
The Crowning Puzzle, p. 386.
'Crowning' is here derived from the idea of crowning in draughts or
checkers. One solution: 4/1 6/9 8/3 2/5 7/10.
Boy's Own Conjuring Book. 1860.
Prob.
33: The counter puzzle, pp. 240 & 264.
Identical to Magician's Own Book, prob. 34.
The
puzzling halfpence, p. 342. Almost identical
to The Sociable, prob. 16, with half-dimes replaced by halfpence.
Illustrated Boy's Own
Treasury. 1860. Prob. 17, pp. 398 & 438. Same as prob. 34 in Magician's Own Book but
only gives one solution.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 593, part 6,
pp. 299-411: Sechs Knacknüsse. 10
counters, one solution.
Hanky Panky. 1872.
Counter puzzle, p. 132. Gives
two solutions for 8 counters and one for 10 counters.
Kamp. Op. cit. in 5.B.
1877. No. 12, p. 325.
Mittenzwey. 1880.
Prob. 235-238, pp. 43-44 & 93-94;
1895?: 262-266, pp. 47-48 & 95-96;
1917: 262-266, pp. 43-44 & 91.
235
(262). Usual problem, with 10
counters. Two solutions.
---
(263). Added in 1895? Same with 8 counters. Two solutions.
236
(264). 12 numbered counters in a
circle. Pass over two to leave six
piles of two on the first six positions.
Solution is misprinted in all editions!
237
(265). 12 counters in a circle. Pass over three to leave six piles of two,
except the last move goes over six. The
solution allows landing counters on vacated locations!
238
(266). 15 counters in a row. Pass over 3 to leave five piles of
three. The solution allows landing a
counter on a vacated location and landing a counter between two locations!!
Lucas. RM2. 1883. Les huit pions, pp. 139-140. Solves for
8, 10, 12, ... counters. Says Delannoy has generalized to the problem
of mp
counters to be formed into
m piles (m ³ 4) of p by passing over p counters.
[More
generally, using one of Berkeley & Rowland's variations (see below), one
can ask when the following problem is solvable: form a line of n =
kp counters into k
piles of p by passing over q [counters or
piles]. Does q have to be £ p?]
Double Five Puzzle. c1890.
??NYS -- described by Slocum from his example. 10 counters in a circle,
but the final piles must alternate with gaps, e.g. the final piles are at the
even positions. This is also solvable
for 4, 8, 12, 16, ..., and I conjectured it was only solvable
for 4n
or 10 counters -- it is easy to see there is no solution for 2
or 6 counters and my computer gave no solutions for 14
or 18. For 4, 8, 10 or 12 counters, one can also leave the final piles
in consecutive locations, but there is no such solution for 6, 14, 16 or 18 counters. See Singmaster
& Abbott, 1992/93, for the resolution of these conjectures.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
Card Puzzles.
No.
VII: The halma puzzle, pp. 6-7. Arrange
the first ten cards of a suit in a row so that passing over two cards leaves
five piles whose cards total 11 and are in the odd places. Arrangement is 7,6,3,4,5,2,1,8,9,10.
Move 2 to 9, 4 to 7,
8 to 3, 6 to 5, 10 to 1.
No.
VIII: Another version, p. 7. With the
cards in order and passing over two piles, leave five piles of two. But this is so easy, he adds that one wants
to leave as low a total as possible on the tops of the piles. He moves
7 to 10, 6 to 3, 4 to 9,
1 to 5, 2 to 8, leaving a total of 20.
[However, this is not minimal --
there are six solutions leaving 18 exposed, e.g. 1 to 4, 3 to 6, 7 to 10,
5 to 9, 2 to 8. For 6 cards, the minimum is 6, achieved
once; for 8 cards, the minimum is 11,
achieved 3 times; for 12 cards, the
minimum is 27, achieved 10 times. For
the more usual case of passing over two cards, the minimum for 8 cards is 15,
achieved twice; for 10 cards, the
minimum is 22, achieved 4 times, e.g. by
7 to 10, 5 to 2, 3 to 8,
1 to 4, 6 to 9; for 12 cards, the minimum is 31, achieved 6
times; for 14 cards, the minimum is 42,
achieved 8 times. For passing over one
pile, the minimum for 4 cards is 3, achieved once; the minimum for 6 cards is 7, achieved twice; the minimum for 8 cards is 13, achieved 3 times; for 10 cards, the minimum is 21, achieved 4
times; the minimum for 12 cards is 31,
achieved 5 times. Maxima are obtained
by taking mirror images of the minimal solutions.]
Puzzles with draughtsmen. The Boy's Own Paper 17 or 18?? (1894??)
751. 8 men, passing over two men each
time. Notes that it can be extended to
any even number of counters.
Clark. Mental Nuts. 1897, no.
63: Toothpicks; 1904, no. 83 &
1916, no. 70: Place 8 toothpicks
in a row. One solution.
Parlour Games for
Everybody. John Leng, Dundee &
London, nd [1903 -- BLC], p. 32: The five pairs. 10 counter version, one solution.
Wehman. New Book of 200 Puzzles. 1908.
P. 15: The counter puzzle and Problem of money. 8 and 10 counter versions, the latter using pennies. Two and one solutions.
Ahrens. MUS I.
1910. Pp. 15-17. Essentially repeats Lucas.
Manson. 1911.
Decimal game, pp. 253-254. Ten
rings on pegs. "Children are
frequently seen playing the game out of doors with pebbles or other convenient
articles."
Blyth. Match-Stick Magic. 1921. Straights and
crosses, pp. 85-87. 10 matchsticks, one
must pass over two of them. Two
solutions, both starting with 4 to 1.
Hummerston. Fun, Mirth & Mystery. 1924.
The
pairing puzzle, Puzzle no. 8, pp. 27 & 173. Essential 8 counters in a circle, with four in a row being white,
the other four being black. Moving only
the whites, and passing over two, form four piles of two.
Pairing
the pennies, Puzzle no. 39, pp. 102 & 178.
Ten pennies, one solution.
Will Blyth. Money Magic. C. Arthur Pearson, London, 1926.
Marrying the coins, pp. 113‑115. Ten coins or eight coins, passing over 2. Gives two solutions for 10, not noting that
the case of 10 is immediately reduced to 8.
Says there are several solutions for 8 and gives two.
Wood. Oddities. 1927. Prob. 45: Fish in the basket, pp.
39-40. 12 fish in baskets in a circle. Move a fish over two baskets, continuing
moving in the same direction, to get get two fish in each of six baskets, in
the fewest number of circuits.
Rudin. 1936. No. 121, pp. 43
& 103. 10 matches. Two solutions.
J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Straights and crosses, pp. 105-106. As in Blyth, 1926.
Indoor Tricks and Games. Success Publishing, London, nd
[1930s??].
How to
pair the pennies, p. 4. 8 pennies, one
solution.
The
ten rings. p. 4. 10 rings, passing over
two piles, one solution.
Haldeman-Julius. 1937.
No.
91: Jumping pennies, pp. 11 & 25.
Six pennies to be formed into two piles of three by jumping over three
pennies each time. Solution has a trick
move. Jump 1 to 5, 6 to 3 and 2 to 1/5,
which gives the position: 6/3 4
2/1/5. He then says: "No. 4
jumps over 5, 1 and 2 -- then jumps back over 5, 1 and 2 and lands upon 3 and
6, ...." Since the rules are not
clear about where a jumping piece can land, the trick move can be viewed as a
legitimate jump to the vacant 6 position, then a legitimate move over the 2/1/5
pile and the now vacant 4 position onto the 6/3 pile. If the pennies are considered as a cycle, this trick is not
needed.
No.
148: Half dimes, pp. 16 & 143. 10
half dimes, passing over one dime (i.e. two counters).
Sullivan. Unusual.
1947. Prob. 39: On the
line. Ten pennies.
Doubleday - 1. 1969.
Prob. 75: Money moves, pp. 91 & 170. Ten pennies. Jump over
two piles. Says there are several
solutions and gives one, which sometimes jumps over three or four pennies.
Putnam. Puzzle Fun.
1978.
No.
26: Pile up the coins, pp. 7 & 31.
12 in a row. Make four piles of
three, passing over three coins each time.
No.
27: Pile 'em up again, pp. 7 & 31.
16 in a row. Make four piles of
four, passing over four or fewer each time.
No.
60: Coin assembly, pp. 11 & 35. Ten
in a row, passing over two each time.
No.
61: Alternative coin assembly, pp. 11 & 35. Ten in a row, passing over two piles each time.
David Singmaster, proposer; H. L. Abbott, solver. Problem 1767. CM 18:7 (1992) 207 & 19:6 (1993). Solves the general version of the Double Five Puzzle, which the
proposer had not solved. One can leave
the counters on even numbered locations if and only if the number of counters
is a multiple of 4 or a multiple of 10.
One can leave the counters in consecutive locations if and only if the
number of counters is 4, 8, 10 or 12.
Heinrich Hemme. Email of 25 Feb 1999. Points out that the rules in the usual
version should say that the counter must land on a pile of a single coin. This would eliminate the trick solutions
given by Mittenzwey and Haldeman-Julius.
Hemme says that without this rule, the problem is easy and can be solved
for 4 and 6 counters!
5.S. CHAIN CUTTING AND REJOINING
The basic problem is to minimise the
cost or effort of reforming a chain from some fragments.
Loyd. Problem 25: A brace of puzzles -- No. 25: The chain puzzle. Tit‑Bits 31 (27 Mar 1897) &
32 (17 Apr 1897) 41. 13
lengths: 5, 6, 7, 7, 7, 7, 8, 8, 8, 8,
8, 9, 12. (Not in the Cyclopedia.)
Loyd. Problem 42: The blacksmith puzzle. Tit‑Bits 32 (10
& 31 Jul &
21 Aug 1897) 273, 327 &
385. Complex problem involving
10 pieces of lengths from 3 to 23 to be joined.
Clark. Mental Nuts. 1897, no.
7 & 1904, no. 14: The chain question; 1916, no. 59: The chain puzzle.
5 pieces of 3 links to make into a single length.
Mr. X [cf 4.A.1]. His Pages.
The Royal Magazine 9:4 (Feb 1903) 390-391 & 9:5 (Mar 1903) 490-491.
The five chains. 5 pieces of 3
links to make into a single length.
Pearson. 1907.
Part II, no. 67, pp. 128 & 205.
5 pieces of 3 links to make into a single length.
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. He attributes such
puzzles to Loyd (Tit‑Bits prob. 25) and gives that problem.
Cecil H. Bullivant. Home Fun.
T. C. & E. C. Jack, London, 1910.
Part VI, Chap IV, No. 9: The broken chain, pp. 518 & 522. 5 3‑link pieces into an open chain.
Loyd. The missing link.
Cyclopedia, 1914, pp. 222 (no solution) (c= MPSL2, prob. 25,
pp. 19 & 129). 6 5‑link
pieces into a loop.
Loyd. The necklace puzzle.
Cyclopedia, 1914, pp. 48 & 345 (= MPSL1, prob. 47, pp. 45‑46
& 138). 12 pieces, with large and
small links which must alternate.
D. E. Smith. Number Stories. 1919. Pp. 119 & 143‑144. 5 pieces of 3 links to make into one length.
Hummerston. Fun, Mirth & Mystery. 1924.
Q.E.D. -- The broken chain, Puzzle no. 38, pp. 99 & 178. Pieces of lengths 2, 2, 3, 3, 4, 4, 6 to
make into a closed loop.
Ackermann. 1925.
Pp. 85‑86. Identical to
the Loyd example cited by Dudeney.
Dudeney. MP.
1926. Prob. 212: A chain puzzle,
pp. 96 & 181 (= 536, prob. 513, pp. 211‑212 & 408). 13 pieces, with large and small links which
must alternate.
King. Best 100. 1927. No. 7, pp. 9 & 40. 5 pieces of three links to make into one
length.
William P. Keasby. The Big Trick and Puzzle Book. Whitman Publishing, Racine, Wisconsin,
1929. A linking problem, pp. 161 &
207. 6 pieces comprising 2, 4, 4, 5, 5, 6 links to be made into one length.
5.S.1. USING CHAIN LINKS TO PAY FOR A ROOM
The
landlord agrees to accept one link per day and the owner wants to minimise the
number of links he has to cut. The
solution depends on whether the chain is closed in a cycle or open at the
ends. Some weighing problems in 7.L.2.c
and 7.L.3 are phrased in terms of making daily payment, but these are like having
the chain already in pieces. See the
Fibonacci in 7.L.2.c.
New
section. I recall that there are older
versions.
Rupert T. Gould. The Stargazer Talks. Geoffrey Bles, London, 1944. A Few Puzzles -- write up of a BBC talk on
10 Jan 1939, pp. 106-113. 63 link chain with three cuts. On p. 106, he says he believes it is
quite modern -- he first heard it in 1935.
On p. 113, he adds a postscript that he now believes it first appeared
in John O'London's Weekly (16 Mar 1935) ??NYS.
Anonymous. Problems drive, 1958. Eureka 21 (Oct 1958) 14-16 & 30. No. 3.
Man has closed chain of 182 links and wants to stay 182 days. What is the minimum number of links to be
opened?
Birtwistle. Math. Puzzles & Perplexities. 1971.
Pp. 13-16. Begins with seven
link open-ended bracelet. Then how big
a bracelet can be dealt with using only two cuts? Gets 23. Then does general case, getting n + (n+1)(2n+1 - 1).
Angela Fox Dunn. Second Book of Mathematical Bafflers. Dover, 1983. Selected from Litton's Problematical Recreations, which appeared
in 1959‑1971. Prob. 26, pp. 28
& 176. 23 link case.
Howson. Op. cit. in 5.R.4. 1988. Prob. 30. Says a
23 link chain need only be cut
twice, giving lengths 1, 1, 3, 6,
12, which make all values up to 23.
Asks for three cuts in a 63 link chain and the maximum length chain one
can deal with in n cuts.
Mittenzwey. 1880.
Prob. 200, pp. 37 & 89;
1895?: 225, pp. 41 & 91;
1917: 225, pp. 38 & 88.
Family of 4 adults and 4 children.
With three cuts, divide a cake so the adults and the children get equal
pieces. He makes two perpendicular
diametrical cuts and then a circular cut around the middle. He seems to mean the adults get equal pieces
and the children get equal pieces, not necessarily the same. But if the circular cut is at Ö2/2 of the radius, then the areas are all equal. Not clear where this should go -- also
entered in 5.Q.
B. Knaster. Sur le problème du partage pragmatique de H.
Steinhaus. Annales de la Société Polonaise
de Mathématique 19 (1946) 228‑230.
Says Steinhaus proposed the problem in a 1944 letter to Knaster. Outlines the Banach & Knaster method of
one cutting 1/n and each being allowed to diminish it --
last diminisher takes the piece. Also
shows that if the valuations are different, then everyone can get > 1/n
in his measure. Gives Banach's
abstract formulations.
H. Steinhaus. Remarques sur le partage pragmatique. Ibid., 230‑231. Says the problem isn't solved for irrational
people and that Banach & Knaster's method can form a game.
H. Steinhaus. The problem of fair division. Econometrica 16:1 (Jan 1948) 101‑104. This is a report of a paper given on 17
Sep. Gives Banach & Knaster's
method.
H. Steinhaus. Sur la division pragmatique. (With English summary) Econometrica 17
(Supplement) (1949) 315‑319.
Gives Banach & Knaster's method.
Max Black. Critical Thinking. Prentice‑Hall, Englewood Cliffs, (1946, ??NYS), 2nd ed.,
1952. Prob. 12, pp. 12 & 432. Raises the question but only suggests combining
two persons.
van Etten. 1624.
Prob. 89, part II, pp. 131‑132 (not in English editions). Two men have same number of hairs. Also:
birds & feathers, fish &
scales, trees & leaves, flowers or
fruit, pages & words -- if there
are more pages than words on any page.
E. Fourrey. Op. cit. in 4.A.1, 1899, section 213: Le
nombre de cheveux, p. 165. Two
Frenchmen have the same number of hairs.
"Cette question fut posée et expliquée par Nicole, un des auteurs
de la Logique de Port‑Royal, à la duchesse de Longueville." [This would be c1660.]
The same story is given in a
review by T. A. A. Broadbent in MG 25 (No. 264) (May 1941) 128. He refers to MG 11 (Dec 1922) 193,
??NYS. This might be the item
reproduced as MG 32 (No. 300) (Jul 1948) 159.
The question whether two trees
in a large forest have the same number of leaves is said to have been posed to
Emmanuel Kant (1724-1804) when he was a boy.
[W. Lietzmann; Riesen und Zwerge im Zahlbereich; 4th ed., Teubner,
Leipzig, 1951, pp. 23-24.] Lietzmann
says that an oak has about two million leaves and a pine has about ten million
needles.
Jackson. Rational Amusement. 1821.
Arithmetical Puzzles, no. 9, pp. 2-3 & 53. Two people in the world have the same number of hairs on their
head.
Manuel des Sorciers. 1825.
Pp. 84-85. ??NX Two men have the same number of hairs, etc.
Gustave Peter Lejeune
Dirichlet. Recherches sur les formes
quadratiques à coefficients et à indéterminées complexes. (J. reine u. angew. Math. (24 (1842) 291‑371) = Math. Werke, (1889‑1897), reprinted
by Chelsea, 1969, vol. I, pp. 533‑618.
On pp. 579‑580, he uses the principle to find good rational
approximations. He doesn't give it a
name. In later works he called it the
"Schubfach Prinzip".
Illustrated Boy's Own
Treasury. 1860. Arithmetical and Geometrical Problems, No.
34, pp. 430 & 434. Hairs on
head.
Pearson. 1907.
Part II, no. 51, pp. 123 & 201.
"If the population of Bristol exceeds by two hundred and thirty‑seven
the number of hairs on the head of any one of its inhabitants, how many of them
at least, if none are bald, must have the same number of hairs on their
heads?" Solution says 474!
Dudeney. The Paradox Party. Strand Mag. 38 (No. 228) (Dec 1909) 670‑676 (= AM,
pp. 137‑141). Two people
have same number of hairs.
Ahrens. A&N, 1918, p. 94. Two Berliners have same number of hairs.
Abraham. 1933.
Prob. 43 -- The library, pp. 16 & 25 (12 & 113). All books have different numbers of words
and there are more books than words in the largest book. (My copy of the 1933 ed. is a presentation
copy inscribed 'For the Athenaeum Library No 43 p 16 R M Abraham Sept 19th 1933'.)
Perelman. FMP.
c1935? Socks and gloves. Pp. 277 & 283‑284. = FFF, 1957: prob. 25, pp. 41 &
43; 1977, prob. 27, pp. 53‑54
& 56. = MCBF, prob. 27, pp. 51
& 54. Picking socks and gloves to
get pairs from 10 pairs of brown and 10 pairs of black socks and gloves.
P. Erdös & G. Szekeres. Op. cit. in 5.M. 1935. Any permutation of
the first n2 ‑ 1 integers contains an increasing or a
decreasing subsequence of length >
n.
P. Erdös, proposer; M. Wachsberger & E. Weiszfeld, M.
Charosh, solvers. Problem 3739. AMM 42 (1935) 396 & 44 (1937) 120. n+1
integers from first 2n have one dividing another.
H. Phillips. Question Time. Dent, London, 1937. Prob.
13: Marbles, pp. 7 & 179. 12
black, 8 red &
6 white marbles -- choose enough
to get three of the same colour.
The Home Book of Quizzes, Games
and Jokes. Op. cit. in 4.B.1,
1941. Pp. 148‑149, prob. 6. Blind maid bringing stockings from a drawer
of white and black stockings.
I am surprised that the context
of picking items does not occur before Perelman, Phillips and Home Book.
Sullivan. Unusual.
1943. Prob. 18: In a dark
room. Picking shoes and socks to get
pairs.
H. Phillips. News Chronicle "Quiz" No. 3:
Natural History. News Chronicle,
London, 1946. Pp. 22 & 43. 12
blue, 9 red and
6 green marbles in a bag. Choose enough to have three of one colour
and two of another colour.
H. Phillips. News Chronicle "Quiz" No. 4:
Current Affairs. News Chronicle,
London, 1946. Pp. 17 & 40. 6
yellow, 5 blue and
2 red marbles in a bag. Choose enough to have three of the same
colour.
L. Moser, proposer; D. J. Newman, solver. Problem 4300 -- The identity as a product of
successive elements. AMM 55 (1948)
369 & 57 (1950) 47. n elements from a group of order n
have a a subinterval with product
= 1.
Doubleday - 2. 1971.
In the
dark, pp. 145-146. How many socks do you
have to pick from a drawer of white and black socks to get two pairs (possibly
different)?
Lucky
dip, pp. 147-148. How many socks do you
have to pick from a drawer of with many white and black socks to get nine pairs
(possibly different)? Gives the general
answer 2n+1 for n pairs.
[Many means that the drawer contains more than n pairs.]
Doubleday - 3. 1972.
In the dark, pp. 35-36. Four
sweaters and 5, 12, 4, 9 socks of the same colours as the
sweaters. Lights go out. He can only find two of the sweaters. How many socks must he bring down into the
light to be sure of having a pair matching one of the sweaters?
I
managed to acquire one of these without instructions or packaging some years
ago. Michael Keller provided an example
complete with instructions and packaging.
I have recently seen Dockhorn's article on variations of the idea. This is related to Binary Recreations, 7.M.
The
device was produced by E.S.R., Inc. The
box or instructions give an address of 34 Label St., Montclair, New
Jersey, 07042, USA, but the company has long been closed. In Feb 2000, Jim McArdle wrote that he
believed that this became the well known Edmund Scientific Co. (101 East
Gloucester Pike, Barrington, New Jersey, 08007, USA; tel: 609‑547 3488;
email: scientifics@edsci.com; web: http://www.edmundscientific.com). But he later wrote that investigation of the
manuals of DifiComp, one of their other products, reveals that there appears to
be no connection. E.S.R. = Education
Science Research. The inventors of
DigiComp, as listed in the patent for it, are: Irving J. Lieberman, William H,
Duerig and Charles D. Hogan, all of Montclair, and they were the founders of
the company. The DigiComp manuals say
Think‑A‑Dot was later invented by John Weisbacker. There is a website devoted to DigiComp which
contains this material and/or pointers to related sites and has a DigiComp
emulator: http://members.aol.com/digicomp1/DigiComp.html . www.yahoo.com has a Yahoo club called Friends of DigiComp. There is another website with the DigiComp
manual: http://galena.tj.edu.inter.net/digicomp/ .
E.S.R. Instructions, 8pp, nd -- but box says ©1965. No patent number anywhere but leaflet says
the name Think-A-Dot is trademarked.
E.S.R., Inc. Corporation. US trademark registration no. 822,770. Filed: 8 Dec 1965; registered: 24 Jan 1967.
First used 23 Aug 1965.
Expired. The US Patent and
Trademark Office website entry says the owner is the company and gives no
information about the inventor(s). The
name has been registered for a computer game on 23 Jul 2002.
Benjamin L. Schwartz. Mathematical theory of Think‑A‑Dot. MM 40:4 (Sep 1967) 187‑193. Shows there are two classes of patterns and
that one can transform any pattern into any other pattern in the same class in
at most 15 drops.
Ray Hemmings. Apparatus Review: Think‑a‑Dot.
MTg 40 (1967) 45.
Sidney Kravitz. Additional mathematical theory of Think‑A‑Dot. JRM 1:4 (Oct 1968) 247‑250. Considers problems of making ball emerge
from one side and of viewing only the back of the game.
Owen Storer. A think about Think‑a‑dot. MTg 45 (Winter 1968) 50‑55. Gives an exercise to show that any possible
transformation can be achieved in at most 9 drops.
T. H. O'Beirne. Letter:
Think‑a‑dot. MTg 48
(Autumn 1969) 13. Proves Steiner's
(Storer?? - check) assertion about 9 drops and gives an optimal algorithm.
John A. Beidler. Think-A-Dot revisited. MM 46:3 (May 1973) 128-136. Answers a question of Schwarz by use of
automata theory. Characterizes all minimal
sequences. Suggests some generalized
versions of the puzzle.
Hans Dockhorn. Bob's binary boxes. CFF 32 (Aug 1993) 4-6. Bob Kootstra makes boxes with the same sort
of T-shaped switch present in Think-A-Dot, but with just one entrance. One switch with two exits is the simplest
case. Kootstra makes a box with three
switches and four exits along the bottom, and the successive balls come out of
the exits in cyclic sequence. Using a
reset connection between switches, he also makes a two switch, three exit, box.
Boob Kootstra. Box seven.
CFF 32 (Aug 1993) 7. Says he has
managed to design and make boxes with
5, 6, 7, 8 exits, again with
successive balls coming out the exits in cyclic order, but he cannot see any
general method nor a way to obtain solutions with a minimal number of movable
parts (switches and reset levers).
Further his design for 7 exits is awkward and the design of an optimal
box for seven is posed as a contest problem.
5.W. MAKING THREE PIECES OF TOAST
This involves an old‑fashioned
toaster which does one side of two pieces at a time. An alternative version is frying steaks or hamburgers on a grill
which holds two objects, assuming each side has to be cooked the same length of
time. The problem is probably older
than these examples.
Sullivan. Unusual.
1943. Prob. 7: For the busy
housewife.
J. E. Littlewood. A Mathematician's Miscellany. Op. cit. in 5.C. 1953. P. 4 (26). Mentions problem and solution.
Simon Dresner. Science World Book of Brain Teasers. 1962.
Op. cit. in 5.B.1. Prob. 40:
Minute toast, pp. 18 & 93.
D. St. P. Barnard. 50 Daily Telegraph Brain‑Twisters. 1985.
Op. cit. in 4.A.4. Prob. 5: Well
done, pp. 16, 80, 103‑104.
Grilling three steaks on a grill which only holds two. He complicates the problem in two ways: a)
each side takes a minute to season before cooking; b)
the steaks want to be cooked 4,
3, 2 minutes per side.
Edward Sitarski. When do we eat? CM 27:2 (Mar 2001) 133-135.
Hamburgers which require time T per side.
After showing that three hamburgers take 3T, he asks how long it
will take to cook H hamburgers.
Easily shows that it can be done in
HT, except for H = 1,
which takes 2T. Then remarks that this is an easy version of
a scheduling problem -- in reality, the hamburgers would have different numbers
of sides, there would be several grills and each hamburger would have different
parts requiring different grills, but in a particular order!
New
section. These are essentially parodies
of the Cistern Problem, 7.H.
McKay. Party Night. 1940. No. 28, p. 182. "An egg takes
3½ minutes to boil. How long should 12 eggs take?"
Jonathan Always. Puzzles to Puzzle You. Op. cit. in 5.K.2. 1965. No. 88: A
boiling problem, pp. 29 & 82.
"If it takes 3½ minutes to boil 2 eggs, how long will it
take to boil 4 eggs?"
John King, ed. John King
1795 Arithmetical Book. Published by the editor, who is the
great-great-grandson of the 1795 writer, Twickenham, 1995. P. 161, the editor mentions "If a girl
on a hilltop can see two miles, how far would two girls be able to see?"
5.X. COUNTING FIGURES IN A PATTERN
New
section -- there must be older examples.
There are two forms of such problems depending on whether one must use
the lattice lines or just the lattice points.
For
counting several shapes, see: Young
World (c1960); Gooding (1994) in 5.X.1.
Counting
triangles in a pattern is always fraught with difficulties, so I have written a
program to do this, but I haven't checked all the examples here.
Pearson. 1907.
Part II.
No.
74: A triangle of triangles, p. 74.
Triangular array with four on a side, but with all the altitudes also
drawn. Gets 653 triangles of various
shapes.
No. 75:
Pharaoh's seal, pp. 75 & 174.
Isosceles right triangles in a square pattern with some diagonals.
Anon. Prob. 76. Hobbies 31 (No. 791) (10 Dec 1910) 256 &
(No. 794) (31 Dec 1910) 318.
Make as many triangles as possible with six matches. From the solution, it seems that the
tetrahedron was expected with four triangles, but many submitted the figure of
a triangle with its altitudes drawn, but only one solver noted that this figure
contains 16 triangles! However, if the
altitudes are displaced to give an interior triangle, I find 17 triangles!!
Loyd. Cyclopedia. 1914. King Solomon's seal, pp. 284 & 378. = MPSL2, No. 142, pp. 100 & 165 c= SLAHP: Various triangles, pp. 25 &
91. How many triangles in the
triangular pattern with 4 on a side?
Loyd Sr. has this embedded in a larger triangle.
Collins. Book of Puzzles. 1927. The swarm of
triangles, pp. 97-98. Same as Pearson
No. 74. He says there are 653
triangles and that starting with 5 on a side gives 1196
and 10,000 on a side gives 6,992,965,420,382. When I
gave August's problem in the Weekend Telegraph, F. R. Gill wrote that this
puzzle with 5 on a side was given out as a competition problem by a furniture
shop in north Lancashire in the late 1930s, with a three piece suite as a prize
for the first correct solution.
Evelyn August. The Black-Out Book. Harrap, London, 1939. The eternal triangle, pp. 64 & 213. Take a triangle, ABC, with midpoints a, b, c,
opposite A, B, C. Take a point d between a
and B. Draw Aa, ab, bc, ca, bd,
cd. How many triangles? Answer is given as 24, but I (and my
program) find 27 and others have confirmed this.
Anon. Test your eyes.
Mathematical Pie 7 (Oct 1952) 51.
Reproduced in: Bernard Atkin,
ed.; Slices of Mathematical Pie; Math. Assoc., Leicester, 1991, pp. 15 & 71
(not paginated - I count the TP as p. 1).
Triangular pattern with 2 triangles on a side, with the three
altitudes drawn. Answer is 47 'obtained
by systematic counting'. This is
correct. Cf Hancox, 1978.
W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. How many triangles, pp. 43 & 130. Take a pentagon and draw the pentagram
inside it. In the interior pentagon,
draw another pentagram. How many
triangles are there? Answer is 85.
Young World. c1960.
P. 57: One for Pythagoras.
Consider a L-tromino. Draw all
the midlines to form 12 unit squares. Or take a 4 x 4 square array and remove a 2 x 2
array from a corner. Now draw
the two main diagonals of the 4 x
4 square - except half of one diagonal
would be outside our figure. How many
triangles and how many squares are present?
Gives correct answers of 26 &
17.
J. Halsall. An interesting series. MG 46 (No. 355) (Feb 1962) 55‑56. Larsen (below) says he seems to be the first
to count the triangles in the triangular pattern with n on a side, but he does
not give any proof.
Although there are few
references before this point, the puzzle idea was pretty well known and occurs
regularly. E.g. in the children's
puzzle books of Norman Pulsford which start c1965, he gives various irregular
patterns and asks for the number of triangles or squares.
J. E. Brider. A mathematical adventure. MTg (1966) 17‑21. Correct derivation for the number of
triangles in a triangle. This seems to
be the first paper after Halsall but is not in Larsen.
G. A. Briggs. Puzzle and Humour Book. Published by the author, Ilkley, 1966. Prob. 2/12, pp. 23 & 75. Consider an isosceles right triangle with
legs along the axes from (0,0) to
(4,0) and (0,4).
Draw the horizontals and verticals through the integer lattice points,
except that the lines through
(1,1) only go from the legs to
this point and stop. Draw the diagonals
through even-integral lattice points, e.g. from (2,0) to (0,2).
How many triangles. Says he
found 27, but his secretary then found
29. I find 29.
Ripley's Puzzles and Games. 1966.
Pp. 72-73 have several problems of counting triangles.
Item
3. Consider a Star of David with the
diameters of its inner hexagon drawn. How
many triangles are in it? Answer: 20,
which I agree with.
Item
4. Consider a 3 x 3 array of squares
with their diagonals drawn. How many
triangles are there? Answer: 150,
however, there are only 124.
Item
5. Consider five squares, with their
midlines and diagonals drawn, formed into a Greek cross. How many triangles are there? Answer: 104, but there are 120.
Doubleday - 2. 1971.
Count down, pp. 127-128. How
many triangles in the pentagram (i.e. a pentagon with all its diagonals)? He says
35.
Gyles Brandreth. Brandreth's Bedroom Book. Eyre Methuen, London, 1973. Triangular, pp. 27 & 63. Count triangles in an irregular pattern.
[Henry] Joseph & Lenore
Scott. Master Mind Brain Teasers. 1973.
Op. cit. in 5.E. An unusual
star, pp. 49-50. Consider a pentagram
and draw lines from each star point through the centre to the opposite crossing
point. How many triangles? They say 110.
[Henry] Joseph and Lenore
Scott. Master Mind Pencil Puzzles. 1973.
Op. cit. in 5.R.4.
Diamonds
are forever, pp. 35-36. Hexagon with
Star of David inside and another Star of David in the centre of that one. How many triangles? Answer is 76.
Count
the triangles, pp. 55-56. Ordinary
Greek cross of five squares, with all the diagonals and midlines of the five
squares drawn. How many
triangles> Answer is 104.
C. P. Chalmers. Note 3353:
More triangles. MG 58 (No. 403)
(Mar 1974) 52‑54. How many
triangles are determined by N points lying on M lines? (Not in Larsen.)
Nicola Davies. The 2nd Target Book of Fun and Games. Target (Universal-Tandem), London,
1974. Squares and triangles, pp. 18
& 119. Consider a chessboard of 4 x 4
cells. Draw all the diagonals, except
the two main ones. How many squares and
how many triangles?
Shakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 136: The
triangles, pp. 85 & 133. How many
triangles in a Star of David made of 12 equilateral triangles?
Michael Holt. Figure It Out -- Book Two. Granada, London, 1978. Prob. 67, unpaginated. How many triangles in a Star of David made
of 12 equilateral triangles?
Putnam. Puzzle Fun.
1978. No. 91: Counting
triangles, pp. 12 & 37. Same as
Doubleday - 2.
D. J. Hancox, D. J. Number Puzzles For all The Family.
Stanley Thornes, London, 1978.
Puzzle
8, pp. 2 & 47. Draw a line with
five points on it, say A, B, C, D,
E, making four segments. Connect all these points to a point F on
one side of the line and to a point
G on the other side of the line,
with FCG collinear. How many
triangles are there? Answer is 24,
which is correct.
Puzzle
53, pp. 24 & 54. Same as Anon.;
Test Your Eyes, 1952. Answer is 36, but
there are 47.
The Diagram Group. The Family Book of Puzzles. The Leisure Circle Ltd., Wembley, Middlesex,
1984.
Problem
40, with Solution at the back of the book.
Same as Doubleday - 2.
Problem
116, with Solution at the back of the book.
Count the triangles in a 'butterfly' pattern.
Sue Macy. Mad Math.
The Best of DynaMath Puzzles.
Scholastic, 1987. (Taken from
Scholastic's DynaMath magazine.)
Shape Up, pp. 5 & 56.
Take
a triangle, trisect one edge and join the points of trisection to the opposite
vertex. How many triangles? [More generally, if one has n
points on a line and joins them all to a vertex, there are 1 + 2 + ... + n-1 = n(n-1)/2
triangles.]
Take
a triangle, join up the midpoints of the edges, giving four smaller triangles,
and draw one altitude of the original triangle. How many triangles?
1980 Celebration of Chinese New
Year Contest Problem No. 5; solution by
Leroy F. Meyers. CM 17 (1991) 2 &
18 (1992) 272-273. n x n array of squares with all diagonals
drawn. Find the number of isosceles
right triangles. [Has this also been
done in half the diagram? That is, how
many isosceles right triangles are in the isosceles right triangle with legs
going from (0,0) to
(n,0) and (0,n)
with all verticals, horizontals and diagonals through integral points
drawn?]
Mogens Esrom Larsen. The eternal triangle -- a history of a
counting problem. Preprint, 1988. Surveys the history from Halsall on. The problem was proposed at least five times
from 1962 and solved at least ten times.
I have sent him the earlier references.
Marjorie Newman. The Christmas Puzzle Book. Hippo (Scholastic Publications), London,
1990. Star time, pp. 26 & 117. Consider a Star of David formed from 12
triangles, but each of the six inner triangles is subdivided into 4
triangles. How many triangles in
this pattern? Answer is 'at least 50'.
I find 58.
Erick Gooding. Polygon counting. Mathematical Pie No. 131 (Spring 1994) 1038 &
Notes, pp. 1-2. Consider the
pentagram, i.e. the pentagon with its diagonals drawn. How many triangles, quadrilaterals and
pentagons are there? Gets 35, 25, 92,
with some uncertainty whether the last number is correct.
When F. R. Gill (See Pearson and
Collins above) mentioned the problem of counting the triangles in the figure
with all the altitudes drawn, I decided to try to count them myself for the figure
with N
intervals on each side. The
theoretical counting soon gets really messy and I adapted my program for
counting triangles in a figure (developed to verify the number found for
August's problem). However, the number
of points involved soon got larger than my simple Basic could handle and I
rewrote the program for this special case, getting the answers of 653 and 1196
and continuing to N = 22. I expected
the answers to be like those for the simpler triangle counting problem so that
there would be separate polynomials for the odd and even cases, or perhaps for
different cases (mod 3 or 4 or 6 or 12 or ??).
However, no such pattern appeared for moduli 2, 3, 4 and I did not get
enough data to check modulus 6 or higher.
I communicated this to Torsten Sillke and Mogens Esrom Larsen. Sillke has replied with a detailed answer
showing that the relevant modulus is 60!
I haven't checked through his work yet to see if this is an empirical
result or he has done the theoretical counting.
Heather Dickson, Heather,
ed. Mind-Bending Challenging Optical
Puzzles. Lagoon Books, London, 1999,
pp. 40 & 91. Gives the version m = n = 4
of the following. I have seen
other versions of this elsewhere, but I found the general solution on 4 Jul
2001 and am submitting it as a problem to AMM.
Consider
a triangle ABC. Subdivide the side AB into m
parts by inserting m‑1 additional points. Connect these points to
C. Subdivide the side AC
into n parts by inserting n-1
additional points and connect them to
B. How many triangles are in
this pattern? The number is [m2n + mn2]/2. When
m = n, we get n3, but I cannot see any simple geometric interpretation for this.
5.X.2. COUNTING RECTANGLES OR SQUARES
I
have just seen M. Adams. There are probably
earlier examples of these types of problems.
Anon. Prob. 63. Hobbies 30 (No. 778) (10 Sep 1910) 488 &
31 (No. 781) (1 Oct 1910) 2. How
many rectangles on a 4 x 4 chessboard?
Solution says 100, which is correct, but then says they are of 17
different types -- I can only get
16 types.
Blyth. Match-Stick Magic.
1921. Counting the squares, p.
47. Count the squares on a 4 x 4
chessboard made of matches with an extra unit square around the central
point. The extra unit square gives 5
additional squares beyond the usual 1 +
4 + 9 + 16.
King. Best 100. 1927. No. 9, pp. 10 & 40. = Foulsham's, no. 5, pp. 6 & 10. 4 x 4
board with some diagonals yielding one extra square.
Loyd Jr. SLAHP.
1928. How many rectangles?, pp.
80 & 117. Asks for the number of
squares and rectangles on a 4 x 4 board (i.e. a 5 x 5 lattice of
points). Says answers are 1 + 4 + 9 + 16 and
(1 + 2 + 3 + 4)2 and that these generalise to any size of
board.
M. Adams. Puzzles That Everyone Can Do. 1931. o o
Prob.
37, pp. 22 & 134: 20 counter problem.
Given the pattern of o o
20 counters at the tight, 'how many perfect
squares are o
o o o o o
contained in the figure.' This means having their vertices o o o o o o
at counters.
There are surprisingly more than I expected. o o
Taking the basic spacing as one, one can have
squares of o o
edge
1, Ö2, Ö5, Ö8, Ö13,
giving 21 squares in all.
He then asks how many counters need
to be removed in order to destroy all the squares? He gives a solution deleting six counters.
Prob.
217, pp. 83 & 162: Match squares.
He gives 10 matches making a row of three equal squares and asks you to
add 14 matches to form 14 squares. The
answer is to make a 3 x 3 array of squares and count all of the
squares in it.
J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Counting the squares, pp. 84-85. As in Blyth.
Indoor Tricks and Games. Success Publishing, London, nd
[1930s??]. Square puzzle, p. 62. Start with a square and draw its diagonals
and midlines. Join the midpoints of the
sides to form a second level square inscribed in the first level original
square. Repeat this until the 9th
level. How many squares are there? Given answer is 16, but in my copy
someone has crossed this out and written
45, which seems correct to me.
Meyer. Big Fun Book. 1940. No. 9, pp. 162 & 752. Draw four equidistant horizontal lines and
then four equidistant verticals. How
many squares are formed? This gives
a 3 x 3 array of squares, but he counts all sizes of squares,
getting 9 + 4 + 1 = 14. (Also in 7.AU.)
Foulsham's New Party Book. Foulsham, London, nd [1950s?]. P. 103: How many squares? 4 x 4
board with some extra diagonals giving one extra square.
Although there are few
references before this point, the puzzle idea was pretty well known and occurs
regularly in the children's puzzle books of Norman Pulsford which start
c1965. He gives various irregular
patterns and asks for the number of triangles or squares.
Jonathan Always. Puzzles to Puzzle You. Op. cit. in 5.K.2. 1965. No. 140: A
surprising answer, pp. 43 & 90. 4 x
4 chessboard with four corner cells
deleted. How many rectangles are there?
Anon. Puzzle page: Strictly for squares. MTg 30 (1965) 48
& 31 (1965) 39 &
32 (1965) 39. How many squares
on a chessboard? First solution gets
S(8) = 1 + 4 + 9 + ... + 64 =
204. Second solution observes that
there are skew squares if one thinks of the board as a lattice of points and
this gives
S(1) + S(2) + ... + S(8) = 540 squares.
G. A. Briggs. Puzzle and Humour Book. Published by the author, Ilkley, 1966.
Prob.
2/11, pp. 22 & 74. 4 x 4 array of squares bordered on two sides by
bricks 1 x 2, 1 x 3, 2 x 1, 2 x 1. Count the squares and the rectangles. Gets 35 and 90.
Prob.
2/14, pp. 23 & 75. Pattern of
squares making the shape of a person -- how many squares in it?
Ripley's Puzzles and Games. 1966.
Pp. 72-73 have several problems of counting squares.
Item
4. Consider a 3 x 3 array of squares
with their diagonals drawn. The
solution says this has 30 squares. I
get 31, but perhaps they weren't counting the whole figure. I have computed the total number of squares for
an n x n array and get (2n3
+ n2)/2 squares for n
even and (2n3 + n2
-1)/2 squares for n
odd.
Unnumbered
item at lower right of p. 73. 4 x
4 array of squares with their diagonals
drawn, except that the four corner squares have only one diagonal -- the one
not pointing to an opposite corner -- and this reduces the number of squares by
eight, agreeing with the given answer of
64.
Doubleday - 2. 1971.
Bed of nails, pp. 129-130. 20
points in the form of a Greek cross with double-length arms (so that the axes
are five times the width of the central square, or the shape is a
9-omino). How many squares can be
located on these points? He finds 21.
W. Antony Broomhead. Note 3315:
Two unsolved problems. MG 55
(No. 394) (Dec 1971) 438. Find the
number of squares on an n x n array of dots, i.e. the second problem in
MTg (1965) above, and another problem.
W. Antony Broomhead. Note 3328:
Squares in a square lattice. MG
56 (No. 396) (May 1972) 129. Finds
there are n2(n2 ‑
1)/12 squares and gives a proof due to
John Dawes. Editorial note says the
problem appears in: M. T. L. Bizley;
Probability: An Intermediate Textbook; CUP, 1957, ??NYS. A. J. Finch asks the question for cubes.
Gyles Brandreth. Brandreth's Bedroom Book. Eyre Methuen, London, 1973. Squares, pp. 26 & 63. Same as Briggs.
Nicola Davies. The 2nd Target Book of Fun and Games. 1974.
See entry in 5.X.1.
Putnam. Puzzle Fun.
1978.
No.
107: Square the coins, pp. 17 & 40.
20 points in the form of a Greek cross made from five 2 x 2
arrays of points. How many
squares -- including skew ones? Gets
21.
No.
108: Unsquaring the coins, pp. 17 & 40.
How many points must be removed from the previous pattern in order to
leave no squares? Gets 6.
M. Adams. Puzzle Book. 1939. Prob. C.157: Making
hexagons, pp. 163 & 190. The
hexagon on the triangular lattice which is two units along each edge contains 8
hexagons. [It is known that the hexagon
of side n contains n3 hexagons.
I recently discovered this but have found that it is known, though I
don't know who discovered it first.]
The Diagram Group. The Family Book of Puzzles. The Leisure Circle Ltd., Wembley, Middlesex,
1984. Problem 32, with Solution at the
back of the book. Count the hexagons in
the hexagon of side three on the triangular lattice. They get 27.
G. A. Briggs. Puzzle and Humour Book. Published by the author, Ilkley, 1966. Prob. 2/13, pp. 23 & 75. Pattern with hexagonal symmetry and lots of overlapping
circles, some incomplete.
5.Y. NUMBER OF ROUTES IN A LATTICE
The
common earlier form was to have the route spell a word or phrase from the
centre to the boundaries of a diamond.
I will call this a word diamond.
Sometimes the phrase is a palindrome and one reads to the centre and
then back to the edge. See Dudeney, CP,
for analysis of the most common cases.
I have seen such problems on the surface of a 3 x 3 x 3 cube. The problems of counting Euler or
Hamiltonian paths are related questions, but dealt with under 5.E and 5.F.
New
section -- in view of the complexity of the examples below, there must be
older, easier, versions, but I have only found the few listed below. The first entry gives some ancient lattices,
but there is no indication that the number of paths was sought in ancient
times.
Roger Millington. The Strange World of the Crossword. M. & J. Hobbs, Walton‑on‑Thames,
UK, 1974. (This seems to have been
retitled: Crossword Puzzles: Their History and Cult for a US ed from Nelson,
NY.)
On
pp. 38-39 & 162, he gives the cabalistic triangle shown below and says it
is thought to have been constructed from the opening letters of the Hebrew
words Ab (Father), Ben (Son),
Ruach Acadash (Holy Spirit). He
then asks how many ways one can read
ABRACADABRA in it, though there
is no indication that the ancients did this.
His answer is 1024 which is correct.
A
B R A C A D A B R A
A B R A C A D A B R
A B R A C A D A B
A B R A C A D A
A B R A C A D
A B R A C A
A B R A
A B R
A B
A
On
pp. 39-40 he describes and illustrates an inscription on the Stele of Moschion
from Egypt, c300. This is a 39 x 39
square with a Greek text from the middle to the corner, e.g. like the
example in the following entry. The
text reads: ΟΥIΡIΔIΜΟΥΧIΩΝΥΓIΑΥΘΕIΥΤΟΝΠΟΔΑIΑΤΠΕIΑIΥ
which means: Moschion to Osiris, for the treatment which
cured his foot. Millington does not ask
for the number of ways to read the inscription, which is 4 BC(38,19) = 14 13810 55200.
Curiosities for the Ingenious
selected from The most authentic Treasures of E
D C D E
Nature,
Science and Art, Biography, History, and General Literature. D C B C D
(1821);
2nd ed., Thomas Boys, London, 1822.
Remarkable epitaph, C
B A B C
p.
97. Word diamond extended to a square,
based on 'Silo Princeps Fecit', D
C B C D
with
the ts
at the corners. An example based
on 'ABCDEF' is shown E
D C D E
at
the right. Says this occurs on the tomb
of a prince named Silo at the
entrance
of the church of San Salvador in Oviedo, Spain. Says the epitaph can be read in 270 ways. I find there are 4 BC(16, 8) = 51490 ways.
In the churchyard of St. Mary's,
Monmouth, is the gravestone of John Rennie, died 31 May 1832, aged 33
years. This has the inscription shown
below. Further down the stone it gives
his son's name as James Rennie.
Apparently an N has been dropped to get a message with an
odd number of letters. I have good
photos. Nothing asks for the number of
ways of reading the inscription. I
get 4 BC(16,9) =
45760 ways.
eineRnhoJsJohnRenie
ineRnhoJsesJohnReni
neRnhoJseiesJohnRen
eRnhoJseiliesJohnRe
RnhoJseileliesJohnR
nhoJseilereliesJohn
hoJseilerereliesJoh
oJseilereHereliesJo
hoJseilerereliesJoh
nhoJseilereliesJohn
RnhoJseileliesJohnR
eRnhoJseiliesJohnRe
neRnhoJseiesJohnRen
ineRnhoJsesJohnReni
eineRnhoJsJohnRenie
Nuts to Crack I (1832), no.
200. The example from Curiosities for
the Ingenious with 'SiloPrincepsFecit', but no indication of what is wanted --
perhaps it is just an amusing picture.
W. Staniforth. Letter.
Knowledge 16 (Apr 1893) 74-75.
Considers 1 2
3 4 5 6
"figure
squares" as at the right. "In
how many different ways may 2 3
4 5 6 7
the
figures in the square be read from
1 to 11 consecutively?" 3
4 5 6 7 8
He
computes the answers for the n x n case for the first few 4 5 6
7 8 9
cases
and finds a recurrence. "Has such
a series of numbers any 5 6
7 8 9 10
mathematical
designation?" The editor notes
that he doesn't 6 7
8 9 10 11
know.
J. J. Alexander. Letter.
Knowledge 16 (May 1893) 89. Says
Staniforth's numbers are the sums of the squares of the binomial
coefficients BC(n, k), the formula for which is BC(2n, n). Editor say he has received more than one note pointing this out
and cites a paper on such figure squares by T. B. Sprague in the Transactions
of the Royal Society of Edinburgh -- ??NYS, no more details provided.
Loyd. Problem 12: The temperance puzzle. Tit‑Bits 31 (2
& 23 Jan 1897) 251 &
307. Red rum & murder. = Cyclopedia, 1914, The little brown jug,
pp. 122 & 355. c= MPSL2, no. 61,
pp. 44 & 141. Word diamond based on
'red rum & murder', i.e. the central line is redrum&murder. He
allows a diagonal move from an E back to an inner R and this gives 372
paths from centre to edge, making
3722 = 138,384 in
total.
Dudeney. Problem 57: The commercial traveller's
puzzle. Tit‑Bits 33 (30 Oct &
20 Nov 1897) 82 & 140.
Number of routes down and right on a
10 x 12 board. Gives a general solution for any board.
Dudeney. A batch of puzzles. The Royal Magazine 1:3 (Jan 1899)
269-274 & 1:4 (Feb 1899) 368-372.
A "Reviver" puzzle.
Complicated pattern based on 'reviver'.
544 solutions.
Dudeney. Puzzling times at Solvamhall Castle. London Magazine 7 (No. 42) (Jan 1902) 580‑584 &
8 (No. 43) (Feb 1902) 53-56. The
amulet. 'Abracadabra' in a triangle
with A at top, two B's
below, three R's
below that, etc. Answer:
1024. = CP, 1907, No. 38,
pp. 64-65 & 190. CF Millington at
beginning of this section.
Dudeney. CP. 1907.
Prob.
30: The puzzle of the canon's yeoman, pp. 55-56 & 181-182. Word diamond based on 'was it a rat I
saw'. Answer is 63504
ways. Solution observes that for
a diamond of side n+1, with no diagonal moves, the number of routes
from the centre to an edge is 4(2n-1) and the number of ways to spell the phrase
is this number squared. Analyses four
types with the following central lines:
A ‑ 'yoboy'; B -
'level'; C - 'noonoon'; D ‑ 'levelevel'.
In
A, one wants to spell 'boy', so there are 4(2n-1) solutions.
In B, one wants to spell
'level' and there are [4(2n-1)]2 solutions.
In
C, one wants to spell 'noon' and there are 8(2n-1)
solutions.
In
D, one wants to spell 'level' and there are complications as one can
start and finish at the edge. He
obtains a general formula for the number of ways. Cf Loyd, 1914.
Prob.
38: The amulet, pp. 64-65 & 190.
See: Dudeney, 1902.
Pearson. 1907.
Part II: A magic cocoon, p. 147.
Word diamond based on 'cocoon', so the central line is noocococoon. Because one can start at the non‑central Cs,
and can go in as well as out, I get
948 paths. He says
756.
Loyd. Cyclopedia. 1914. Alice in Wonderland, pp. 164 & 360. = MPSL1, no. 109, pp. 107 & 161‑162. Word diamond based on 'was it a cat I
saw'. Cf Dudeney, 1907.
Dudeney. AM.
1917.
Prob.
256: The diamond puzzle, pp. 74 & 202.
Word diamond based on 'dnomaidiamond'.
This is type A of his discussion in CP and he states the
general formula. 252 solutions.
Prob.
257: The deified puzzle, pp. 74-75 & 202.
Word diamond based on 'deifiedeified'.
This is type D in CP and has 1992 solutions. He says 'madamadam' gives 400
and 'nunun' gives 64, while 'noonoon' gives 56.
Prob.
258: The voter's puzzle, pp. 75 & 202.
Word diamond built on 'rise to vote sir'. Cites CP, no. 30, for the result, 63504, and the general
formula.
Prob.
259: Hannah's puzzle, pp. 75 & 202.
6 x 6 word square based on
'Hannah' with Hs on the outside, As adjacent to the Hs
and four Ns in the middle. Diagonal moves allowed.
3468 ways.
Wood. Oddities. 1927. Prob. 44: The amulet problem, p. 39. Like the original ABRACADABRA triangle, but
with the letters in reverse order.
Collins. Book of Puzzles. 1927. The magic cocoon
puzzle, pp. 169-170. As in Pearson.
Loyd Jr. SLAHP.
1928. A strolling pedagogue, pp.
38 & 97. Number of routes to
opposite corner of a 5 x 5 array of points.
D. F. Lawden. On the solution of linear difference
equations. MG 36 (No. 317) (Sep 1952)
193-196. Develops use of integral
transforms and applies it to find that the number of king's paths going down or
right or down‑right from (0,
0) to
(n, n) is Pn(3) where Pn(x) is
the Legendre polynomial.
Leo Moser. King paths on a chessboard. MG 39 (No. 327) (Feb 1955) 54. Cites Lawden and gives a simpler proof of
his result Pn(3).
Anon. Puzzle Page: Check this.
MTg 36 (1964) 61 & 27 (1964) 65. Find the number of king's routes from corner to corner when he
can only move right, down or right‑down.
Gets 48,639 routes on
8 x 8 board.
Ripley's Puzzles and Games. 1966.
P. 32. Word diamond laid out
differently so A A
A
one
has to read from one side to the opposite side. Rotating by 45o, one gets B B
the
pattern at the right for edge three.
One wants the number of ways to C
C C
read ABCDEF.
In general, when the first line of
As has n positions, D D
the
total number of ways to reach the first row is
n. For each successive E E E
row,
the total number is alternately twice the number for the previous row less F F
twice
the end term of that row or just twice the the number for the previous
row. In our example with n = 3,
the number of ways to reach the second row is 4 = 2x3 - 2x1. The number of ways to reach the third row is 8 = 2x4.
The number of ways to reach the fourth row is 12 = 2x8 - 2x2, then we
get 24 = 2 x 12; 36 = 2x24 ‑ 2x6. It happens that the first n
end terms are the central binomial coefficients BC(2k,k),
so this is easy to calculate. I
find the total number of routes, for n
= 2, 3, ..., 7, is 4,
18, 232, 1300,
6744, 33320, the last being the desired and given answer
for the given problem.
Pál Révész. Op. cit. in 5.I.1. 1969. On p. 27, he gives
the number of routes for a king moving forward on a chessboard and a man moving
forward on a draughtsboard.
Putnam. Puzzle Fun.
1978. No. 8: Level - level, pp.
3 & 26. Form a wheel of 16 points
labelled LEVELEVELEVELEVE. Place 4
Es inside, joined to two
consecutive Vs and the intervening L.
Then place a V in the middle, joined to these four Es.
How many ways to spell LEVEL? He
gets 80, which seems right.
5.Z. CHESSBOARD PLACING PROBLEMS
See
MUS I 285-318, some parts of the previous chapter and the Appendix in II
351-360. See also 5.I.1, 6.T.
There
are three kinds of domination problems.
In
strong domination, a piece dominates the square it is on.
In
weak domination, it does not, hence more pieces may be needed to dominate the board.
Non‑attacking
domination is strong domination with no piece attacking another. Graph theorists say the pieces are
independent. This also may require more
pieces than strong domination, but it may require more or fewer pieces than
weak domination.
The
words 'guarded' or 'protected' are used for weak domination, but 'unguarded' or
'unprotected' may mean either strong or non‑attacking domination.
Though
these results seem like they must be old, the ideas seem to have originated
with the eight queens problem, c1850, (cf 5.I.1) and to have been first really
been attacked in the late 19C. There
are many variations on these problems, e.g. see Ball, and I will not attempt to
be complete on the later variations. In
recent years, this has become a popular subject in graph theory, where the
domination number, γ(G), is the size of the smallest strongly
dominating set on the graph G and the independent domination number, i(G),
is the size of the smallest non-attacking (= independent) dominating set.
Mario Velucchi has a web site devoted
to the non-dominating queens problem and related sites for similar
problems. See: http://www.bigfoot.com/~velucchi/papers.html and
http://www.bigfoot.com/~velucchi/biblio.html.
Ball. MRE, 3rd ed., 1896, pp. 109-110: Other problems with queens. Says:
"Captain Turton has called my attention to two other problems of a
somewhat analogous character, neither of which, as far as I know, has been
hitherto published, ...." These
ask for ways to place queens so as to attack as few or as many cells as
possible -- see 5.Z.2.
Ball. MRE, 4th ed., 1905, pp. 119-120: Other problems with queens; Extension to other chess pieces. Repeats above quote, but replaces 'hitherto
published' by 'published elsewhere', extends the previous text and adds the new
section.
Ball. MRE, 5th ed., 1911.
Maximum pieces problem; Minimum
pieces problem, pp. 119‑122.
[6th ed., 1914 adds that Dudeney has written on these problems in The
Weekly Dispatch, but this is dropped in the 11th ed. of 1939.] Considerably generalizes the problems. On the
8 x 8 board, the maximum number
of non-attacking kings is 16, queens is 8, bishops is 14 [6th ed., 1914, adds there are 256 solutions], knights is 32 with 2 solutions and
rooks is 8 with 88 solutions [sic, but changed to 8! in the 6th ed.]. The minimum number of pieces to strongly
dominate the board is 9 kings, 5 queens with 91 inequivalent solutions [the
91 is omitted in the 6th ed., since it is stated later], 8 bishops,
12 knights, 8 rooks. The minimum number of pieces to weakly
dominate the board is 5 queens, 10 bishops,
14 knights, 8 rooks.
Dudeney. AM.
1917. The guarded chessboard,
pp. 95‑96. Discusses different
ways pieces can weakly or non‑attackingly dominate n x n
boards.
G. P. Jelliss. Multiple unguard arrangements. Chessics 13 (Jan/Jun 1982) 8‑9. One can have 16 kings, 8 queens, 14 bishops,
32 knights or 8 rooks
non‑attackingly placed on a
8 x 8 board. He considers mixtures of pieces -- e.g. one
can have 10 kings and 4 queens non‑attacking. He tries to maximize the product of the numbers of each type in a
mixture -- e.g. scoring 40 for the example.
Ball. MRE, 4th ed., 1905. Other
problems with queens; Extension to
other chess pieces, pp. 119-120.
Says problems have been proposed for
k kings on an n x n,
citing L'Inter. des math. 8 (1901) 140, ??NYS.
Gilbert Obermair. Denkspiele auf dem Schachbrett. Hugendubel, Munich, 1984. Prob. 27, pp. 29 & 58. 9
kings strongly, and 12 kings weakly, dominate an 8 x 8
board.
Here
the graph is denoted Qn, but I will denote γ(Qn)
by γ(n) and
i(Qn) by i(n).
Murray. Pp. 674 & 691. CB249 (c1475) shows
16 queens weakly dominating an 8 x 8
board, but the context is unclear to me.
de Jaenisch. Op. cit. in 5.F.1. Vol. 3, 1863. Appendice,
pp. 244-271. Most of this is due to "un
de nos anciens amis, Mr de R***". Finds and describes the 91 ways of placing 5 queens so as to
non-attackingly dominate the 8 x 8 board.
Then considers the n x n board for
n = 2, ..., 7 with strong and
non-attacking domination. Up through 5,
he gives the number of pieces being attacked in each solution which allows one
to determine the weak solutions.
For n < 6, he gets the answers in the table below, but
for n = 6, he gets 21 non-attacking solutions instead of 17?.
Ball. MRE, 3rd ed., 1896. Other
problems with queens, pp. 109-110.
"Captain [W. H.] Turton has called my attention to two other problems
of a somewhat analogous character, neither of which, as far as I know, has been
hitherto published, or solved otherwise than empirically." The first is to place 8 queens so as to
strongly dominate the fewest squares.
The minimum he can find is 53.
(Cf Gardner, 1999.) The second
is to place m queens, m £
5, so as to strongly dominate as many
cells as possible. With 4 queens, the
most he can find is 62.
Dudeney. Problem 54: The hat‑peg puzzle. Tit‑Bits 33 (9 &
30 Oct 1897) 21 & 82. Problem
involves several examples of strong domination by 5 queens on an 8 x 8
board leading to a non‑attacking domination. He says there are just 728 such. This
= 8 x 91. = Anon. &
Dudeney; A chat with the Puzzle King; The Captain 2 (Dec? 1899) 314-320; 2:6
(Mar 1900) 598-599 & 3:1 (Apr 1900) 89. = AM; 1917; pp. 93-94 & 221.
Ball. MRE, 4th ed., 1905, loc. cit. in 5.Z.1. Extends 3rd ed. by asking for the minimum number of queens to
strongly dominate a whole n x n board.
Says there seem to be 91 ways of having 5 non-attacking queens on
the 8 x 8, citing L'Inter. des Math. 8 (1901) 88, ??NYS.
Ball. MRE, 5th ed., 1911, loc. cit. in 5.Z. On pp. 120-122, he considers queens and states the minimum
numbers of queens required to strongly dominate the board and the numbers of
inequivalent solutions for 2 x 2, 3 x 3,
..., 7 x 7, citing the article cited in the 4th ed. and
Jaenisch, 1862, without a volume number.
For n = 7, he gives the same unique solution for
strongly dominating as for non-attacking dominating. [In the 6th ed., this is corrected and he says it is a
solution.] He says Jaenisch also posed
the question of the minimum number of non-attacking queens to dominate the
board and gives the numbers and the number of inequivalent ways for the 4 x 4,
.., 8 x 8, except that he follows Jaenisch in stating
that there are 21 solutions on the 6 x
6. [This is changed to 17 in the 6th
ed.]
Dudeney. AM.
1917. Loc. cit. in 5.Z. He uses 'protected' for 'weakly', but he
seems to copy the values for 'strongly' from Jaenisch or Ball. His 'not protected' seems to mean
'non-attacking'. However, some values are different and I consequently
am very uncertain as to the correct values??
Pál Révész. Op. cit. in 5.I.1. 1969. On pp. 24‑25,
he shows 5 queens are sufficient to strongly dominate the board and says this
is minimal.
Below, min. denotes the minimum
number of queens to dominate and no. is the number of inequivalent ways to do
so.
STRONG WEAK NON-ATTACKING
n min. no. min.
no. min. no.
1 1 1 0 0 1 1
2 1 1 2 2 1 1
3 1 1 2 5 1 1
4 2 3 2 3 3 2
5 3 37 3 15 3 2
6 3 1 4 ³2 4
17?
7 4 4 5? 4 1
8 5 ³150 5 ³41 5
91
Rodolfo Marcelo Kurchan,
proposer; Henry Ibstedt & proposer,
solver. Prob. 1738 -- Queens in
space. JRM 21:3 (1989) 220 &
22:3 (1989) 237. How many queens
are needed to strongly dominate an n x
n x n cubical board? For
n = 3, 4, ..., 9, the best known
numbers are: 1, 4, 6, 8, 14, 20,
24. The solution is not clear if these
are minimal, but it seems to imply this.
Martin Gardner. Chess queens and maximum unattacked
cells. Math Horizons (Nov 1999). Reprinted in Workout, chap. 34. Considers the problem of Turton described in
Ball, 3rd ed, above: place 8
queens on an 8 x 8 board so as to strongly dominate the fewest
squares. That is, leave the maximum
number of unattacked squares. More
generally, place k queens on an n x n board to leave the
maximum number of unattacked squares.
He describes a simple problem by Dudeney (AM, prob. 316) and recent work
on the general problem. He cites Velucchi,
cf below, who provides the following table of maximum numbers of unattacked
cells and number of solutions for the maximum.
I'm not sure if some of these are still only conjectured.
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Max 0 0 0 1 3 5 7 11 18 22 30 36 47 56 72 82 97
Sols 0 0 0 25 1 3 38 7 1 1 2 7 1 4 3 1
Mario Velucchi has a web site
devoted to the non-dominating queens problem and related sites for similar
problems. See: http://www.bigfoot.com/~velucchi/papers.html and
http://www.bigfoot.com/~velucchi/biblio.html.
A. P. Burger & C. M.
Mynhardt. Symmetry and domination in
queens graphs. Bull. Inst.
Combinatorics Appl. 29 (May 2000) 11-24.
Extends results to n = 30, 45,
69, 77. Summarizes the field, with 14
references, several being earlier surveys.
The table below gives all known values.
It will be seen that the case n
= 4k + 1 seems easiest to deal
with. The values separated by
strokes, /, indicate cases where the value is one of the two given values,
but it is not known which.
n 4 5 6 7 8 9 10 11 12 13 14 15 16 17
γ(n) 2 3 3 4 5 5 5 5 6 7 7/8 8/9 8/9 9
i(n) 3 3 4 4 5 5 5 5 7 7 8 9 9 9
n 18 19 21 25 29 30 31 33 37 41 45 49 53 57
γ(n) 9 10 11 13 15 15 16 17 19 21 23 25 27 29
i(n) 11 13 17 23
n 61 69 77
γ(n) 31 35 39
Dudeney. AM.
1917. Prob. 299: Bishops in
convocation, pp. 89 & 215. There
are 2n ways to place 2n‑2 bishops non‑attackingly
on an n x n board. At loc. cit. in
5.Z, he says that for n = 2, ..., 8, there are
1, 2, 3, 6, 10, 20, 36
inequivalent placings.
Pál Révész. Op. cit. in 5.I.1. 1969. On pp. 25‑26,
he shows the maximum number of non‑attacking bishops on one colour is 7
and there are 16 ways to place them.
Obermair. Op. cit. in 5.Z.1. 1984. Prob. 17, pp. 23
& 50. 8 bishops strongly, and 10
bishops weakly, dominate the 8 x 8 board.
Ball. MRE, 4th ed., 1905. Loc.
cit. in 5.Z.1. Says questions as to the
maximum number of non-attacking knights and minimum number to strongly dominate
have been considered, citing L'Inter. des math. 3 (1896) 58, 4 (1897) 15-17 & 254, 5 (1898) 87
[5th ed. adds 230‑231], ??NYS.
Dudeney. AM.
1917. Loc. cit. in 5.Z. Notes that if n is odd, one can
have (n2+1)/2 non‑attacking knights in one way,
while if n is even, one can have n2/2 in two equivalent ways.
Irving Newman, proposer; Robert Patenaude, Ralph Greenberg and Irving
Newman, solvers. Problem E1585 --
Nonattacking knights on a chessboard.
AMM 70 (1963) 438 & 71 (1964) 210-211. Three easy proofs that the maximum number of non-attacking
knights is 32. Editorial note cites
Dudeney, AM, and Ball, MRE, 1926, p. 171 -- but the material is on p. 171 only
in the 11th ed., 1939.
Gardner. SA (Oct 1967, Nov 1967 & Jan 1968) c= Magic Show, chap. 14. Gives Dudeney's results for the 8 x 8.
Golomb has noted that Greenberg's solution of E1585 via a knight's tour
proves that there are only two solutions.
For the k x k board,
k = 3, 4, ..., 10, the minimal number of knights to strongly dominate is: 4, 4, 5, 8, 10, 12,
14, 16. He says the table may
continue: 21, 24, 28, 32, 37. Gives numerous examples.
Obermair. Op. cit. in 5.Z.1. 1984. Prob. 16, pp. 21
& 47. 14 knights are necessary for
weak domination of the 8x8 board.
E. O. Hare & S. T.
Hedetniemi. A linear algorithm for
computing the knight's domination number of a
k x n chessboard. Technical report 87‑May‑1, Dept.
of Computer Science, Clemson University.
1987?? Pp. 1‑2 gives the
history from 1896 and Table 2 on p. 13 gives their optimal results for strong
domination on k x n boards,
4 £ k £
9, k £ n £ 12 and also for k = n = 10. For the k x k
board, k = 3, ..., 10, they confirm the results in Gardner.
Anderson H. Jackson & Roy P.
Pargas. Solutions to the N x N
knight's cover problem. JRM 23:4
(1991) 255-267. Finds number of knights
to strongly dominate by a heuristic method, which finds all solutions up
through N = 10. Improves the value given by Gardner for N = 15
to 36 and finds solutions for N
= 16, ..., 20 with 42, 48, 54, 60, 65 knights.
É. Lucas. Théorie des Nombres. Gauthier‑Villars, Paris, 1891; reprinted by Blanchard, Paris, 1958. Section 128, pp. 220‑223. Determines the number of inequivalent
placings of n nonattacking rooks on an n x n
board in general and gives values for
n £ 12. For n = 1, ..., 8, there are
1, 1, 2, 7, 23, 115, 694, 5282
inequivalent ways.
Dudeney. AM.
1917. Loc. cit. at 5.Z. Notes there are n! ways to place n
non‑attacking rooks and asks how many of these are
inequivalent. Gives values for n = 1, ..., 5. AM prob. 296, pp. 88 & 214, is the case n = 4.
D. F. Holt. Rooks inviolate. MG 58 (No. 404) (Jun 1974) 131‑134. Uses Burnside's lemma to determine the
number of inequivalent solutions in general, getting Lucas' result in a more
modern form.
Ball. MRE, 5th ed., 1911. Loc.
cit. in 5.Z. P. 122: "There are endless similar questions in
which combinations of pieces are involved." 4 queens and king or queen or bishop or knight or rook or pawn
can strongly dominate 8 x 8.
King. Best 100. 1927.
No.
77, pp. 30 & 57. 4 queens and a
rook strongly dominate 8 x 8.
No.
78, pp. 30 & 57. 4 queens and a
bishop strongly dominate 8 x 8.
New
section. I have been meaning to add
this sometime, but I have just come across an expository article, so I am now
starting. The mathematics of this gets
quite formidable. See 5.AD for a
somewhat related topic.
A
faro, weave, dovetail or perfect (riffle) shuffle starts by cutting the deck in
half and then interleaving the two halves.
When the deck has an even number of cards, there are two ways this can
happen -- the original top card can remain on top (an out shuffle) or it
can become the second card of the shuffled deck (an in shuffle). E.g. if our deck is 123456,
then the out shuffle yields
142536 and the in shuffle
yields 415263. Note that removing the first and last cards
converts an out shuffle on 2n cards to an in shuffle on 2n-2
cards. When the deck has an odd
number of cards, say 2n+1, we cut above or below the middle card and
shuffle so the top of the larger pile is on top, i.e. the larger pile straddles
the smaller. If the cut is below the
middle card, we have piles of n+1 and
n and the top card remains on
top, while cutting above the middle card leaves the bottom card on bottom. Removing the top or bottom card leaves an in
shuffle on 2n cards.
Monge's
shuffle takes the first card and then alternates the next cards over and under
the resulting pile, so 12345678 becomes
86421357.
At G4G2, 1996, Max Maven gave a talk
on some magic tricks based on card shuffling and gave a short outline of the
history. The following is an attempt to
summarise his material. The faro
shuffle, done by inserting part of the deck endwise into the other part, but
not done perfectly, began to be used in the early 18C and a case of cheating
using this is recorded in 1726. The
riffle shuffle, which is the common American shuffle, depends on mass produced
cards of good quality and began to be used in the mid 19C. However, magicians did not become aware of
the possibilities of the perfect shuffle until the mid 20C, despite the early
work of Stanyans C. O. Williams and Charles T. Jordan in the 1910s.
Hooper. Rational Recreations. Op. cit. in 4.A.1. 1774. Vol. 1, pp. 78-85:
Of the combinations of the cards. This
describes a shuffle, where one takes the top two cards, then puts the next two
cards on top, then the next three cards underneath, then the next two on top,
then the next three underneath. For ten
cards 1234567890, it produces
8934125670, a permutation of
order 7. Tables of the first few
repetitions are given for 10, 24, 27
and 32 cards, having orders 7, 30, 30, 156.
The Secret Out. 1859.
Permutation table, pp. 394-395 (UK: 128-129). Describes Hooper's shuffle for ten cards.
Bachet-Labosne. Problemes.
3rd ed., 1874. Supp. prob. XV,
1884: 214-222. Discusses Monge's shuffle
and its period.
John Nevil Maskelyne. Sharps and Flats. 1894. ??NYS -- cited by
Gardner in the Addendum of Carnival.
"One of the earliest mentions". Called the "faro dealer's shuffle".
Ahrens. MUS I.
1910. Ein Kartenkunststück
Monges, pp. 152-145. Expresses the general
form of Monge's shuffle and finds its order for n = 1, 2, ..., 10.
Mentions the general question of finding the order of a shuffle.
Charles T. Jordan. Thirty Card Mysteries. The author, Penngrove, California, 1919
(??NYS), 2nd ed., 1920 (?? I have copy of part of this). Cited by Gardner in the Addendum to
Carnival. First magician to apply the
shuffle, but it was not until late 1950s that magicians began to seriously use
and study it. The part I have (pp.
7-10) just describes the idea, without showing how to perform it. The text clearly continues to some
applications of the idea. This material
was reprinted in The Bat (1948-1949).
Frederick Charles Boon. Shuffling a pack of cards and the theory of
numbers. MG 15 (1930) 17-20. Considers the Out shuffle and sees that it
relates to the order of 2 (mod 2n+1)
and gives some number theoretic observations on this. Also considers odd decks.
J. V. Uspensky & M. A.
Heaslet. Elementary Number Theory. McGraw-Hill, NY, 1939. Chap. VIII: Appendix: On card shuffling, pp.
244-248. Shows that an In shuffle of a
deck of 2n cards takes the card in position
i to position 2i (mod 2n+1), so the order of the
permutation is the exponent or order of
2 (mod 2n+1), which is 52 when
n = 26. [Though not discussed,
this shows that the order of the Out shuffle is the order of 2 (mod 2n-1), which is only 8
for n = 26. And the order of a shuffle of 2n+1
cards is the order of 2 (mod
2n+1).] Monge's shuffle is more
complex, but leads to congruences (mod 4n+1) and has order equal to the
smallest exponent e such that
2e º ±1 (mod 4n+1), which is
12 for n = 26.
T. H. R. Skyrme. A shuffling problem. Eureka 7 (Mar 1942) 17-18. Describes Monge's shuffle with the second
card going under or over the first.
Observes that in the under shuffle for an even number of cards, the last
card remains fixed, while the over shuffle for an odd number of cards also
leaves the last card fixed. By
appropriate choice, one always has the
n-th card becoming the first.
Finds the order of the shuffle essentially as in Uspensky &
Heaslet. Makes some further
observations.
N. S. Mendelsohn, proposer and
solver. Problem E792 -- Shuffling
cards. AMM 54 (1947) 545 ??NYS &
55 (1948) 430-431. Shows the
period of the out shuffle is at most
2n-2. Editorial notes cite
Uspensky & Heaslet and MG 15 (1930) 17-20 ??NYS.
Charles T. Jordan. Trailing the dovetail shuffle to its
lair. The Bat (Nov, Dec 1948; Jan, Feb, Mar, 1949). ??NYS -- cited by Gardner. I have
No. 59 (Nov 1948) cover & 431-432, which reprints some of the
material from his book.
Paul B. Johnson. Congruences and card shuffling. AMM 63 (1956) 718-719. ??NYS -- cited by Gardner.
Alexander Elmsley. Work in Progress. Ibidem 11 (Sep 1957) 222.
He had previously coined the terms 'in' and 'out' and represented them
by I
and O. He discovers and shows that to put the top card into the k‑th position, one writes k-1
in binary and reads off the sequence of
1s and 0s, from the most
significant bit, as I and
O shuffles. He asks but does not solve the question of
how to move the k-th card to the top --
see Bonfeld and Morris.
Alexander Elmsley. The mathematics of the weave shuffle, The Pentagram 11:9 (Jun 1957) 70‑71;
11:10 (Jul 1957) 77-79; 11:11 (Aug
1957) 85; 12 (May 1958) 62. ??NYR -- cited by Gardner in the
bibliography of Carnival, but he doesn't give the Ibidem reference in the
bibliography, so there may be some confusion here?? Morris only cites Pentagram.
Solomon W. Golomb. Permutations by cutting and shuffling. SIAM Review 3 (1961) 293‑297. ??NYS -- cited by Gardner. Shows that cuts and the two shuffles
generate all permutations of an even deck.
However, for an odd deck of
n cards, the two kinds of
shuffles can be intermixed and this only changes the cyclic order of the
result. Since cutting also only changes
the cyclic order, the number of possible permutations is n
times the order of the shuffle.
Gardner. SA (Oct 1966) = Carnival, chap. 10. Defines the in and out shuffles as above and
gives the relation to the order of 2.
Notes that it is easier to do the inverse operations, which consist of
extracting every other card. Describes
Elmsley's method. Addendum says no easy
method is known to determine shuffles to bring the k‑th card to the top.
Murray Bonfeld. A solution to Elmsley's problem. Genii 37 (May 1973) 195-196. Solves Elmsley's 1957 problem by use of an
asymmetric in-shuffle where the top part of the deck has 25 cards, so the first
top card becomes second and the last two cards remain in place. (If one ignores the bottom two cards this is
an in-shuffle of a 50 card deck.)
S. Brent Morris. The basic mathematics of the faro
shuffle. Pi Mu Epsilon Journal 6 (1975)
86-92. Obtains basic results, getting
up to Elmsley's work. His reference to
Gardner gives the wrong year.
Israel N. Herstein & Irving
Kaplansky. Matters Mathematical. 1974;
slightly revised 2nd ed., Chelsea, NY, 1978. Chap. 3, section 4: The interlacing shuffle, pp. 118-121. Studies the permutation of the in shuffle,
getting same results as Uspensky & Heaslet.
S. Brent Morris. Faro shuffling and card placement. JRM 8:1 (1975) 1-7. Shows how to do the faro shuffle. Gives Elmsley's and Bonfeld's results.
Persi Diaconis, Ronald L. Graham
& William M. Kantor. The
mathematics of perfect shuffles. Adv.
Appl. Math. 4 (1983) 175-196. ??NYS.
Steve Medvedoff & Kent
Morrison. Groups of perfect
shuffles. MM 60:1 (1987) 3-14. Several further references to check.
Walter Scott. Mathematics of card sharping. M500 125 (Dec 1991) 1-7. Sketches Elmsley's results. States a peculiar method for computing the
order of 2 (mod 2n+1) based on adding translates of the binary
expansion of 2n+1 until one obtains a binary number of
all 1s. The number of ones is the order
a and the method is thus
producing the smallest a such that
2a-1 is a multiple
of 2n+1.
John H. Conway & Richard K.
Guy. The Book of Numbers. Copernicus (Springer-Verlag), NY, 1996. Pp. 163-165 gives a brief discussion of
perfect shuffles and Monge's shuffle.
5.AB. FOLDING A STRIP OF STAMPS
É. Lucas. Théorie des Nombres. Gauthier‑Villars, Paris, 1891; reprinted by Blanchard, Paris, 1958. P. 120.
Exemple
II -- La bande de timbres-poste. -- De combien de manières peut-on replier, sur un
seul, une bande de p timbres-poste?
Exemple
III -- La feuille de timbres-poste. -- De combien de manières peut-on replir, sur un
seul, une feuille rectangulaire de pq timbres-poste?
"Nous
ne connaissons aucune solution de ces deux problèmes difficiles proposés par M.
Em. Lemoine."
M. A. Sainte-Laguë. Les Réseaux (ou Graphes). Mémorial des Sciences Mathématiques, fasc.
XVIII. Gauthier-Villars, Paris, 1926. Section 62: Problème des timbres-poste,
pp. 39‑41. Gets some basic
results and finds the numbers for a strip of
n,
n = 1, 2, ..., 10 as: 1, 2, 6, 16, 50, 144,
448, 7472, 17676, 41600.
Jacques Devisme. Contribution a l'étude du problème des
timbres-poste. Comptes-Rendus du
Deuxième Congrès International de Récréation Mathématique, Paris, 1937. Librairie du "Sphinx", Bruxelles,
1937, pp. 55-56. Cites Lucas (but in
the wrong book!) and Sainte-Laguë.
Studies the number of different forms of the result, getting
numbers: 1, 2, 3, 8, 18,
44, 115, 294, 783.
5.AC. PROPERTIES OF THE SEVEN BAR DIGITAL DISPLAY
┌─┐ 2
The
seven bar display, in the form of a figure
8, as at the right, is │ │ 1
3
now the standard form for
displaying digits on calculators, clocks, etc. ├─┤ 4
This lends itself to numerous
problems of a combinatorial/numerical │
│ 5 7
New
Section. └─┘ 6
For reference, we number the seven bars in the reverse-S pattern
shown. We can then refer to a pattern by its binary 7-tuple or its decimal equivalent. E.g. the number one is displayed by having
bars 3
and 7 on, which gives a binary pattern
1000100 corresponding to
decimal 68. NOTE that there is some ambiguity with the 6 / 9. Most versions use the upper / lower bar for
these, i.e. 1101111 / 1111011, but the bar is sometimes omitted,
giving 1001111 / 1111001. I will assume the first case unless
specified.
I have been interested in these for some time for several
reasons. First, my wife has such a
clock on her side of the bed and she often has a glass of water in front of it,
causing patterns to be reversed. At
other times the clock has been on the floor upside down, causing a different
reversal of patterns. Second, segments
often fail or get stuck on and I have tried to analyse which would be the worst
segment to fail or get stuck. As an
example, the clock in my previous car went from 16:59 to 15:00. Third, I
have analysed which segment(s) in a clock are used most/least often.
Birtwistle. Calculator Puzzle Book. 1978.
Prob. 35: New numbers, pp. 26-27 & 83. Asks for the number of new digits one can make, subject to their
being connected and full height. Says
it is difficult to determine when these are distinct -- e.g. calculators differ
as to the form of their 6s and 9s -- so he is not sure how to count, but he
gives 22 examples. I find there are 55
connected, full-height patterns.
Gordon Alabaster, proposer &
Robert Hill, solver. Problem
134.3 -- Clock watching. M500 134 (Aug
1993) 17 & 135 (Oct 1993) 14-15. Proposer notes that one segment of the units
digit of the seconds on his station clock was stuck on, but that the sequence
of symbols produced were all proper digits.
Which segment was stuck? Asks if
there are answers for 2, ..., 6
segments stuck on. Solver gives
systematic tables and discusses problems of how to determine which segment(s)
are stuck and whether one can deduce the correct time when the stuck segments
are known.
Martin Watson. Email to NOBNET, 17 Apr 2000 08:17:32 PDT
[NOBNET 2334]. Observes that the 10
digits have a total of 49 segments and asks if they can be placed on a 4 x 5
square grid. He calls these
forms 'digigrams'. He had been unable
to find a solution but Leonard Campbell has found 5 distinct solutions, though
they do no differ greatly. He has the
pieces and some discussion on his website:
http://martnal.tripod.com/puzzles.html .
Dario Uri [22 Apr 2000 14:44:35 +0200] found two extra solutions, but
Rick Eason [22 Apr 2000 09:37: -0400] also found these, but points out that
these have an error due to misreading the lattice which gives the two bars of
the 1 being parallel instead of end to end.
Eason's program also found the 5 solutions.
5.AD. STACKING A DECK TO PRODUCE A SPECIAL EFFECT
New section. This refers to the process of arranging a deck of cards or a
stack of coins so that dealing it by some rule produces a special effect. In many cases, this is just inverting the
permutation given by the rule and the Josephus problem (7.B) is a special
case. Other cases involve spelling out
the names of cards, etc.
Will Blyth. Money Magic. C. Arthur Pearson, London, 1926.
Alternate heads, pp. 61-63.
Stack of eight coins. Place one
on the table and the next on the bottom of the stack. The sequence of placed coins is to alternate heads and
tails. How do you arrange the
stack? Answer is HHTHHTTT.
This is the same process as counting out by 2s -- see 7.B.
Doubleday - 2. 1971.
Heads and tails, pp. 105-106.
Same as Blyth, but with six coins and solution HTTTHH.
New
section. There are several versions of
this and they usually involve parity.
The basic move is to reverse two of the cups. The classic problem seems to be to start with UDU
and produce DDD in three moves. A trick version is to demonstrate this several times to someone
and then leave him to start from
DUD. Another easy problem is to
leave three cups as they were after three moves. This is equivalent to a 3
x 3 array with an even number in each
row and column -- see 6.AO.2. These
problems must be much older than I have, but the following are the only
examples I have yet noted.
Anonymous. Social Entertainer and Tricks (thus on
spine, but running title inside is New Book of Tricks). Apparently a compilation with advertisements
for Johnson Smith (Detroit, Michigan) products, c1890?. P. 38a: Bottoms up. Given
UDU, produce DDD
in three moves.
Young World. c1960.
P. 39: Water switch. Full and
empty glasses: FFFEEE. Make them alternately full and empty in one
move.
Putnam. Puzzle Fun.
1978.
No. 3:
Tea for three, pp. 1 & 25. Cups
given as UDU. Produce DDD in three moves.
No.
16: Glass alignment, pp. 5 & 28.
Six cups arranged UUUDDD. Produce an alternating row. He gets
UDUDUD in three moves. I can get DUDUDU in four moves.
New
section. In the early 1980s, I asked
Richard Guy what was the 'standard' configuration for a die and later asked Ray
Bathke if he used a standard pattern. Assuming
opposite sides add to seven there are two handednesses. But also the spot pattern of the two, three
and six has two orientations, giving 16 different patterns of die. Ray said that when he furnished dice with
games, some customers had sent them back because they weren't the same. Within about three years, I had obtained
examples of all sixteen patterns!
Indeed, I often found several patterns in a single batch from one
manufacturer. Ray Bathke also pointed
out that the small dice that come from the oriental games have the two arranged
either horizontally or vertically rather than diagonally, giving another 16
patterns. I have only obtained five of
these, but with both handednesses included.
I used this idea in one of my Brain Twisters, cf below.
Since
the 2, 3 and 6 faces all meet at a corner, one has just to describe this
corner. The 2, 3, 6 can be clockwise
around the corner or anti-clockwise.
Note that 236 is clockwise if and only if 132
is clockwise. The position of
the 2 and 3 can be described by saying whether the pattern points toward or
away from the corner. If we place the 2
upward, then 6 will be a vertical face and we can describe it by saying whether
the lines of three spots are vertical or horizontal. Guy told me a system for describing a die, but it's not in
Winning Ways and I've forgotten it, so I'll invent my own.
We
write the sequence 236 if
236 is arranged clockwise at
the 236 corner and we write
263 otherwise. When looked at cornerwise, with the 2 on
top, the pattern of the 2 may appear vertical or horizontal. We write
2 when it is vertical and 2
when it is horizontal. (For
oriental dice, the 2 will appear on a diagonal and can be
indicated by 2 or
2. If we now rotate the cube to
bring the 3 on top, its pattern will appear either vertical or horizontal and
we write 3 or 3. Putting the
2 back on top, the 6
face will be upright and the lines of three spots will be either
vertical or horizontal, which we denote by
6 or 6.
David Singmaster. Dicing around. Weekend Telegraph (16 Dec 1989).
= Games & Puzzles No. 15 (Jun 1995) 22-23 & 16 (Jul 1995)
43-44. How many dice are there? Describes the normal 16 and mentions the
other 16.
Ian Stewart. The lore and lure of dice. SA (Nov 1997) ??. He asserts that the standard pattern has 132
going clockwise at a corner, except that the Japanese use the
mirror-image version in playing mah-jongg.
His picture has both 2 and 3 toward the 236 corner and the 6 being
vertical, i.e. in pattern 236. He
discusses crooked dice of various sorts and that the only way to make all
values from 1 to 12 equally likely is to have 123456 on one die and 000666 on
the other.
Ricky Jay. The story of dice. The New Yorker (11 Dec 2000) 90-95.
5.AG. RUBIK'S CUBE AND SIMILAR PUZZLES
I
have previously avoided this as being too recent to be covered in a historical
work, but it is now old enough that it needs to be covered, and there are some
older references. Much of the history
is given in my Notes on Rubik's Cube and my Cubic Circular. Jaap Scherphuis has sent me a file of puzzle
patents and several dozen of them could be entered here, but I will only enter
older or novel items. Scherphuis's file
has about a dozen patents for the 4 x 4
x 4 and 5 x 5 x 5 cubes! See Section 5.A for predecessors of the
idea. However, this Section will mostly
deal with puzzles where pieces are permuted without having any empty places, so
these are generally permutation puzzle.
New
section. Much to be added.
Richard E. Korf. Finding optimal solutions to Rubik's Cube
using pattern databases. Proc. Nat.
Conf. on Artificial Intelligence (AAAI-97), Providence, Rhode Island, Jul 1997,
pp. 700-705. Studies heuristic
methods of finding optimal solutions of the Cube. Claims to be the first to find optimal solutions for random
positions of the Cube -- but I think others such as Kociemba and Reid were
doing it up to a decade earlier. For
ten random examples, he found optimal solutions took 16 moves in one case, 17
moves in three cases, 18 moves in six cases, from which he asserts the median
optimal solution length seems to be 18.
He uses the idea of axial moves and obtains the lower bound of 18 for
God's Algorithm, as done in my Notes in 1980.
Cites various earlier work in the field, but only one reference to the
Cube literature.
Richard E. Korf &
Ariel Felner. Disjoint pattern
database heuristics. Artificial
Intelligence 134 (2002) 9-22. Discusses
heuristic methods of solving the Fifteen Puzzle, Rubik's Cube, etc. Asserts the median optimal solution length
for the Cube is only 18. Seems to say
one of the problems in the earlier paper took a couple of weeks running time,
but improved methods of Kociemba and Reid can find optimal solutions in about an
hour.
New
section. Much to be added.
William Churchill. US Patent 507,215 -- Puzzle. Applied: 28 May 1891; patented: 24 Oct 1893. 1p + 1p diagrams. Two rings of 22 balls, intersecting six spaces apart.
Hiester Azarus Bowers. US Patent 636,109 -- Puzzle. Filed: 16 Aug 1899; patented 31 Oct 1899. 2pp + 1p diagrams. 4 rotating discs which overlap in simple lenses.
Ivan Moscovich. US Patent 4,509,756 -- Puzzle with Elements
Transferable Between Closed‑loop Paths.
Filed: 18 Dec 1981; patented: 9
Apr 1985. Cover page + 3pp + 2pp
diagrams. Two rings of 18 balls, each
stretched to have two straight sections with semicircular ends. The rings cross in four places, at the ends
of the straight sections, so adjacent crossing points are separated by two
balls. I'm not sure this was ever
produced. Mentions three circular rings
version, but there each pair of rings only overlaps in two places so this is a
direct generalization of the Hungarian Rings.
David Singmaster. Hungarian Rings groups. Bull. Inst. Math. Appl. 20:9/10 (Sep/Oct
1984) 137-139. [The results were stated
in Cubic Circular 5 & 6 (Autumn & Winter 1982) 9‑10.] An article by Philippe Paclet [Des anneaux
et des groupes; Jeux et Stratégie 16 (Aug/Sep 1982) 30-32] claimed that all
puzzles of two rings have groups either the symmetric or the alternating group
on the number of balls. This article
shows this is false and determines the group in all cases. If we have rings of size m, n
and the intersections are distances
a, b apart on the two
rings. Then the group, G(m, n, a, b) is the symmetric group on
m+n-2 if mn
is even and is the alternating group if
mn is odd; except
that G(4, 4, 1, 1) is the exceptional group described in R. M. Wilson's
1974 paper: Graph puzzles, homotopy and the alternating group -- cited in
Section 5.A under The Fifteen Puzzle -- and is also the group generated by two
adjacent faces on the Rubik Cube acting on the six corners on those faces; and
except that G(2a, 2b, a, b) keeps antipodal pairs at antipodes and hence
is a subgroup of the wreath product Z2 wr Sa+b‑1,
with three cases depending on the parities of
a and b.
Bala Ravikumar. The Missing Link and the Top-Spin. Report TR94-228, Department of Computer
Science and Statistics, University of Rhode Island, Jan 1994. Top-Spin has a cycle of 20 pieces and a
small turntable which permits inverting a section of four pieces. After developing the group theory and doing
the Fifteen Puzzle and the Missing Link, he shows the state space of Top-Spin
is S20.
This is too big a topic to cover
completely. The first items should be
consulted for older material and the general history. Then I include material of particular interest. See also 6.BL which has some formulae which
are used to compute π. I have compiled a separate file on the
history of π.
Augustus De Morgan. A Budget of Paradoxes. (1872);
2nd ed., edited by D. E. Smith, (1915), Books for Libraries Press,
Freeport, NY, 1967.
J. W. Wrench Jr. The evolution of extended decimal
approximations to π. MTr 53 (Dec 1960) 644‑650. Good survey with 55 references, including
original sources.
Petr Beckmann. A History of π. The Golem Press,
Boulder, Colorado, (1970); 2nd ed.,
1971.
Lam Lay-Yong & Ang
Tian-Se. Circle measurements in ancient
China. HM 13 (1986) 325‑340. Good survey of the calculation of π
in China.
Dario Castellanos. The ubiquitous π. MM 61 (1988)
67-98 & 148-163. Good survey of
methods of computing π.
Joel Chan. As easy as pi. Math Horizons 1 (Winter 1993) 18-19. Outlines some recent work on calculating π and gives several
of the formulae used.
David Singmaster. A history of π. M500 168 (Jun 1999) 1-16. A chronology. (Thanks to Tony Forbes and Eddie Kent for carefully proofreading
and amending my file.)
Aristophanes. The Birds.
‑414. Lines 1001‑1005. In:
SIHGM I 308‑309. Refers to
'circle-squarers', possibly referring to the geometer/astronomer Meton.
E. J. Goodwin. Quadrature of the circle. AMM 1 (1894) 246‑247.
House Bill No. 246, Indiana
Legislature, 1897. "A bill for an
act introducing a new mathematical truth ..." In Edington's paper (below), p. 207, and in several of the
newspaper reports.
(Indianapolis)
Journal (19 Jan 1897) 3. Mentions the
Bill in the list of bills introduced.
Die
Quadratur des Zirkels. Täglicher
Telegraph (Indianapolis) (20 Jan 1897) ??.
Surveys attempts since -2000 and notes that Lindemann and Weierstrass
have shown that the problem is impossible, like perpetual motion.
A man
of 'genius'. (Indianapolis) Sun (6 Feb
1897) ??. An interview with Goodwin,
who says: "The astronomers have all been wrong. There's about 40,000,000 square miles on the surface of this earth
that isn't here." He says his
results are revelations and gives several rules for the circle and the sphere.
Mathematical
Bill passed. (Indianapolis) Journal (6
Feb 1897) 5. "This is the
strangest bill that has ever passed an Indiana Assembly." Gives whole text of the Bill.
Dr.
Goodwin's theaorem (sic) Resolution
adopted by the House of Representatives.
(Indianapolis) News (6 Feb 1897) 4.
Gives whole text of the Bill.
The
Mathematical Bill Fun-making in the
Senate yesterday afternoon -- other action.
(Indianapolis) News (13 Feb 1897) 11.
"The Senators made bad puns about it, ...." The Bill was indefinitely postponed.
House
Bills in the Senate. (Indianapolis)
Sentinel (13 Feb 1897) 2. Reports the
Bill was killed.
(No
heading??) (Indianapolis) Journal (13
Feb 1897) 3, col. 4. "...
indefinitely postponed, as not being a subject fit for legislation."
Squaring
the circle. (Indianapolis) Sunday
Journal (21 Feb 1897) 9. Says Goodwin
has solved all three classical impossible problems. Says π = 3.2, using
the fact that Ö2 = 10/7,
giving diagrams and a number of rules.
My
thanks to Underwood Dudley for locating and copying the above newspaper items.
C. A. Waldo. What might have been. Proc. Indiana Acad. Science 26 (1916) 445‑446.
W. E. Edington. House Bill No. 246, Indiana State
Legislature, 1897. Ibid. 45 (1935) 206‑210.
A. T. Hallerberg. House Bill No. 246 revisited. Ibid. 84 (1975) 374‑399.
Manuel H. Greenblatt. The 'legal' value of pi, and some related
mathematical anomalies. American
Scientist 53 (Dec 1965) 427A‑434A.
On p. 427A he tries to interpret the bill and obtains three different
values for π.
David Singmaster. The legal values of pi. Math. Intell. 7:2 (1985) 69‑72. Analyses Goodwin's article, Bill and other
assertions to find 23 interpretable statements giving 9 different values
of π !
Underwood Dudley. Mathematical Cranks. MAA, 1992.
Legislating pi, pp. 192-197.
C. T. Heisel. The Circle Squared Beyond Refutation. Published by the author, 657 Bolivar Rd.,
Cleveland, Ohio, 1st ed., 1931, printed by S. J. Monck, Cleveland; 2nd ed., 1934, printed by Lezius‑Hiles
Co., Cleveland, ??NX + Supplement: "Fundamental Truth",
1936, ??NX, distributed by the author
from 2142 Euclid Ave., Cleveland. This
is probably the most ambitious publication of a circle-squarer -- Heisel
distributed copies all around the world.
Underwood Dudley. πt: 1832-1879.
MM 35 (1962) 153-154. He plots
45 values of π as a function of time over the period
1832-1879 and finds the least-squares straight line which fits the data,
finding that πt =
3.14281 + .0000056060 t, for t
measured in years AD. Deduces
that the Biblical value of 3 was a good approximation for the time and that
Creation must have occurred when πt
= 0, which was in -560,615.
Underwood Dudley. πt. JRM 9 (1976-77) 178 & 180. Extends his previous work to 50 values
of π over 1826-1885, obtaining
πt = 4.59183 - .000773 t. The fact that πt
is decreasing is worrying -- when
πt = 1, all
circles will collapse into straight lines and this will certainly be the end of
the world, which is expected in 4646 on 9 Aug at 20:55:33 -- though this is
only the expected time and there is considerable variation in this
prediction. [Actually, I get that this
should be on 11 Aug. However, it seems
to me that circles will collapse once
πt = 2, as then
the circumference corresponds to going back and forth along the diameter. This will occur when t = 3352.949547, i.e. in 3352, on 13 Dec at 14:01:54 -- much earlier than Dudley's
prediction, so start getting ready now!]
See
Yates for a good survey of the field.
James Watt. UK Patent 1432 -- Certain New Improvements upon
Fire and Steam Engines, and upon Machines worked or moved by the same. Granted: 28 Apr 1784; complete specification: 24 Aug 1784. 14pp + 1 plate. Pp. 4-6 & Figures 7‑12 describe Watt's parallel
motion. Yates, below, p. 170 quotes one
of Watt's letters: "... though I
am not over anxious after fame, yet I am more proud of the parallel motion than
of any other invention I have ever made."
P. F. Sarrus. Note sur la transformation des mouvements
rectilignes alternatifs, en mouvements circulaires; et reciproquement. C. R. Acad. Sci. Paris 36 (1853) 1036‑1038. 6 plate linkage. The name should be Sarrus, but it is printed Sarrut on this and
the following paper.
Poncelet. Rapport sur une transformation nouvelle des
mouvements rectilignes alternatifs en mouvements circulaires et reciproquement,
par Sarrut. Ibid., 36 (1853) 1125‑1127.
A. Peaucellier. Lettre au rédacteur. Nouvelles Annales de Math. (2) 3 (1864) 414‑415. Poses the problem.
A. Mannheim. Proces‑Verbaux des sceances des 20 et
27 Juillet 1867. Bull. Soc.
Philomathique de Paris (1867) 124‑126.
??NYS. Reports Peaucellier's
invention.
Lippman Lipkin. Fortschritte der Physik (1871) 40‑?? ??NYS
L. Lipkin. Über eine genaue Gelenk‑Geradführung. Bull. Acad. St. Pétersbourg [=? Akad. Nauk,
St. Petersburg, Bull.] 16 (1871) 57‑60.
??NYS
L. Lipkin. Dispositif articulé pour la transformation
rigoureuse du mouvement circulaire en mouvement rectiligne. Revue Univers. des Mines et de la
Métallurgie de Liége 30:4 (1871) 149‑150. ??NYS. (Now spelled
Liège.)
A. Peaucellier. Note sur un balancier articulé a mouvement
rectiligne. Journal de Physique 2
(1873) 388‑390. (Partial English
translation in Smith, Source Book, vol. 2, pp. 324‑325.) Says he
communicated it to Soc. Philomath. in 1867 and that Lipkin has since also found
it. There is also an article in Nouv.
Annales de Math. (2) 12 (1873) 71‑78 (or 73?), ??NYS.
E. Lemoine. Note sur le losange articulé du Commandant
du Génie Peaucellier, destiné a remplacer le parallélogramme de Watt. J. de Physique 2 (1873) 130‑134. Confirms that Mannheim presented
Peaucellier's cell to Soc. Philomath. on 20 Jul 1867. Develops the inversive geometry of the cell.
[J. J. Sylvester.] Report of the Annual General Meeting of the
London Math. Soc. on 13 Nov 1873. Proc.
London Math. Soc. 5 (1873) 4 & 141.
On p. 4 is: "Mr. Sylvester
then gave a description of a new instrument for converting circular into
general rectilinear motion, and into motion in conics and higher plane curves,
and was warmly applauded at the close of his address." On p. 141 is an appendix saying that
Sylvester spoke "On recent discoveries in mechanical conversion of
motion" to a Friday Evening's Discourse at the Royal Institution on 23 Jan
1874. It refers to a paper 20 pages
long but is not clear if or where it was published.
H. Hart. On certain conversions of motion. Cambridge Messenger of Mathematics 4 (1874)
82‑88 and 116‑120 & Plate I.
Hart's 5 bar linkage. Obtains
some higher curves.
A. B. Kempe. On some new linkages. Messenger of Mathematics 4 (1875) 121‑124
& Plate I. Kempe's linkages for
reciprocating linear motion.
H. Hart. On two models of parallel motions. Proc. Camb. Phil. Soc. 3 (1876‑1880)
315‑318. Hart's parallelogram (a
5 bar linkage) and a 6 bar one.
V. Liguine. Liste des travaux sur les systèms
articulés. Bull. d. Sci. Math. 18 (or
(2) 7) (1883) 145‑160. ??NYS ‑
cited by Kanayama. Archibald; Outline
of the History of Mathematics, p. 99, says Linguine is entirely included in
Kanayama.
Gardner D. Hiscox. Mechanical Appliances Mechanical Movements and
Novelties of Construction. A
second volume to accompany his previous Mechanical Movements, Powers and
Devices. Norman W. Henley Publishing
Co, NY, (1904), 2nd ed., 1910. This is
filled with many types of mechanisms.
Pp. 245-247 show five straight-line linkages and some related
mechanisms.
(R. Kanayama). (Bibliography on linkages. Text in Japanese, but references in roman
type.) Tôhoku Math. J. 37 (1933) 294‑319.
R. C. Archibald. Bibliography of the theory of linkages. SM 2 (1933‑34) 293‑294. Supplement to Kanayama.
Robert C. Yates. Geometrical Tools. (As: Tools; Baton Rouge,
1941); revised ed., Educational Publishers,
St. Louis, 1949. Pp. 82-101 &
168-191. Gets up to outlining Kempe's
proof that any algebraic curve can be drawn by a linkage.
R. H. Macmillan. The freedom of linkages. MG 34 (No. 307) (Feb 1960) 26‑37. Good survey of the general theory of
linkages.
Michael Goldberg. Classroom Note 312: A six‑plate linkage in three
dimensions. MG 58 (No. 406) (Dec 1974)
287‑289.
Such
curves play an essential role in some ways to drill a square hole, etc.
L. Euler. Introductio in Analysin Infinitorum. Bousquet, Lausanne, 1748. Vol. 2, chap. XV, esp. § 355, p. 190 &
Tab. XVII, fig. 71. = Introduction to
the Analysis of the Infinite; trans. by John D. Blanton; Springer, NY,
1988-1990; Book II, chap. XV: Concerning curves with one or several diameters,
pp. 212-225, esp. § 355, p. 221 & fig. 71, p. 481. This doesn't refer to constant width, but
fig. 71 looks very like a Reuleaux triangle.
L. Euler. De curvis triangularibus. (Acta Acad. Petropol. 2 (1778(1781)) 3‑30) = Opera Omnia (1) 28 (1955) 298‑321. Discusses triangular versions.
M. E. Barbier. Note sur le problème de l'aiguille et jeu du
joint couvert. J. Math. pures appl. (2)
5 (1860) 273‑286. Mentions
that perimeter = π * width.
F. Reuleaux. Theoretische Kinematik; Vieweg,
Braunschweig, 1875. Translated: The Kinematics of Machinery. Macmillan, 1876; Dover, 1964. Pp. 129‑147.
Gardner D. Hiscox. Mechanical Appliances Mechanical Movements and
Novelties of Construction. A
second volume to accompany his previous Mechanical Movements, Powers and
Devices. Norman W. Henley Publishing
Co, NY, (1904), 2nd ed., 1910.
Item
642: Turning a square by circular motion, p. 247. Plain face, with four pins forming a centred square, is turned by
the lathe. A triangular follower is
against the face, so it is moved in and out as a pin moves against it. This motion is conveyed by levers to the
tool which moves in and out against the work which is driven by the same lathe.
Item
681: Geometrical boring and routing chuck, pp. 257-258. Shows it can make rectangles, triangles,
stars, etc. No explanation of how it
works.
Item
903A: Auger for boring square holes, pp. 353-354. Uses two parallel rotating cutting wheels.
H. J. Watts. US Patents 1,241,175‑7 -- Floating
tool‑chuck; Drill or boring
member; Floating tool‑chuck. Applied: 30 Nov 1915; 1 Nov 1916;
22 Nov 1916; all patented:
25 Sep 1917. 2 + 1, 2 + 1,
4 + 1 pp + pp diagrams. Devices for drilling square holes based on
the Reuleaux triangle.
T. Bonnesen & W.
Fenchel. Theorie der konvexen
Körper. Berlin, 1934; reprinted by Chelsea, 1971. Chap. 15: Körper konstanter Breite, pp. 127‑141. Surveys such curves with references to the
source material.
G. D. Chakerian & H.
Groemer. Convex bodies of constant
width. In: Convexity and Its Applications; ed. by Peter M. Gruber & Jörg
M. Wills; Birkhäuser, Boston, 1983.
Pp. 49‑96. (??NYS --
cited in MM 60:3 (1987) 139.)
Bibliography of some 250 items since 1930.
These
were discovered by Arthur H. Stone, an English graduate student at Princeton in
1939. American paper was a bit wider
than English and would not fit into his notebooks, so he trimmed the edge off
and had a pile of long paper strips which he played with and discovered the
basic flexagon. Fellow graduate
students Richard P. Feynman, Bryant Tuckerman and John W. Tukey joined in the
investigation and developed a considerable theory. One of their fathers was a patent attorney and they planned to
patent the idea and began to draw up an application, but the exigencies of the
1940s led to its being put aside, though knowledge of it spread as mathematical
folklore. E.g. Tuckerman's father,
Louis B. Tuckerman, lectured on it at the Westinghouse Science Talent Search in
the mid 1950s.
S&B, pp. 148‑149, show
several versions. Most square versions
(tetraflexagons or magic books) don't fold very far and are really just
extended versions of the Jacob's Ladder -- see 11.L
Martin Gardner. Cherchez la Femme [magic trick]. Montandon Magic Co., Tulsa, Okla.,
1946. Reproduced in: Martin Gardner Presents; Richard Kaufman and
Alan Greenberg, 1993, pp. 361-363.
[In: Martin Gardner Presents, p.
404, this is attributed to Gardner, but Gardner told me that Roger Montandon
had the copyright -- ?? I have learned
a little more about Gardner's early life -- he supported himself by inventing
and selling magic tricks about this time, so it may be that Gardner devised the
idea and sold it to Montandon.]. A
hexatetraflexagon.
"Willane". Willane's Wizardry. Academy of Recorded Crafts, Arts and
Sciences, Croydon, 1947. A trick book,
pp. 42-43. Same hexatetraflexagon.
Sidney Melmore. A single‑sided doubly collapsible
tessellation. MG 31 (No. 294) (1947)
106. Forms a Möbius strip of three
triangles and three rhombi. He sees it
has two distinct forms, but doesn't see the flexing property!!
Margaret Joseph. Hexahexaflexagrams. MTr 44 (Apr 1951) 247‑248. No history.
William R. Ransom. A six‑sided hexagon. SSM 52 (1952) 94. Shows how to number the 6 faces.
No history.
F. G. Maunsell. Note 2449:
The flexagon and the hexahexaflexagram.
MG 38 (No. 325) (Sep 1954) 213‑214. States that Joseph is first article in the field and that this is
first description of the flexagon.
Gives inventors' names, but with Tulsey for Tukey.
R. E. Rogers & Leonard L.
D'Andrea. US Patent 2,883,195 -- Changeable
Amusement Devices and the Like.
Applied: 11 Feb 1955; patented:
21 Apr 1959. 2pp + 1p correction + 2pp
diagrams. Clearly shows the 9 and 18
triangle cases and notes that one can trim the triangles into hexagons so the
resulting object looks like six small hexagons in a ring.
M. Gardner. Hexa‑hexa‑flexagon and
Cherchez la femme. Hugard's
MAGIC Monthly 13:9 (Feb 1956) 391.
Reproduced in his: Encyclopedia
of Impromptu Magic; Magic Inc., Chicago, 1978, pp. 439-442. Describes hexahexa and the hexatetra of
Gardner/Montandon & Willane.
M. Gardner. SA (Dec 1956) = 1st Book, chap. 1. His first article in SA!!
Joan Crampin. Note 2672:
On note 2449. MG 41 (No. 335)
(Feb 1957) 55‑56. Extends to a
general case having 9n triangles of 3n colours.
C. O. Oakley & R. J.
Wisner. Flexagons. AMM 64:3 (Mar 1957) 143‑154.
Donovan A. Johnson. Paper Folding for the Mathematics
Class. NCTM, 1957, section 61,
pp. 24-25: Hexaflexagons.
Describes the simplest case, citing Joseph.
Roger F. Wheeler. The flexagon family. MG 42 (No. 339) (Feb 1958) 1‑6. Improved methods of folding and colouring.
M. Gardner. SA (May 1958) = 2nd Book, chap. 2. Tetraflexagons and flexatube.
P. B. Chapman. Square flexagons. MG 45 (1961) 192‑194.
Tetraflexagons.
Anthony S. Conrad &
Daniel K. Hartline.
Flexagons. TR 62-11, RIAS, (7212
Bellona Avenue, Baltimore 12, Maryland,) 1962, 376pp. This began as a Science Fair project in 1956 and was then
expanded into a long report. The
authors were students of Harold V. McIntosh who kindly sent me one of the
remaining copies in 1996. They discover
how to make any chain of polygons into a flexagon, provided certain relations
among angles are satisfied. The
bibliography includes almost all the preceding items and adds the references to
the Rogers & D'Andrea patent, some other patents (??NYS) and a number of
ephemeral items: Conrad produced an
earlier RIAS report, TR 60-24, in 1960;
Allan Phillips wrote a mimeographed paper on hexaflexagons; McIntosh wrote an unpublished paper on flexagons; Mike Schlesinger wrote an unpublished paper
on Tuckerman tree theory.
Sidney H. Scott. How to construct hexaflexagons. RMM 12 (Dec 1962) 43‑49.
William R. Ransom. Protean shapes with flexagons. RMM 13 (Feb 1963) 35‑37. Describes 3‑D shapes that can be
formed. c= Madachy, below.
Robert Harbin. Party Lines. Op. cit. in 5.B.1.
1963. The magic book, pp.
124-125. As in Gardner's Cherchez la
Femme and Willane.
Pamela Liebeck. The construction of flexagons. MG 48 (No. 366) (Dec 1964) 397‑402.
Joseph S. Madachy. Mathematics on Vacation. Op. cit. in 5.O, (1966), 1979. Other flexagon diversions, pp. 76‑81. Describes 3‑D shapes that one can
form. Based on Ransom, RMM 13.
Lorraine Mottershead. Investigations in Mathematics. Blackwell, Oxford, 1985. Pp. 66-75.
Describes various tetra- and hexa-flexagons.
Douglas A. Engel. Hexaflexagon + HFG = slipagon! JRM 25:3 (1993) 161-166. Describes his slipagons, which are linked
flexagons.
Robert E. Neale (154 Prospect
Parkway, Burlington, Vermont, 05401, USA).
Self-designing tetraflexagons.
12pp document received in 1996 describing several ways of making
tetraflexagons without having to tape or paste. He starts with a creased square sheet, then makes some internal
tears or cuts and then folds things through to miraculously obtain a
flexagon! A slightly rearranged version
appeared in: Elwyn R. Berlekamp
& Tom Rodgers, eds.; The Mathemagician and Pied Puzzler A Collection in Tribute to Martin
Gardner; A. K. Peters, Natick,
Massachusetts, 1999, pp. 117-126.
Jose R. Matos. US Patent 5,735,520 -- Fold-Through Picture
Puzzle. Applied: 7 Feb 1997; patented: 7 Apr 1998. Front page + 6pp diagrams + 13pp text. Robert Byrnes sent an example of the
puzzle. This is a square in thin
plastic, 100mm on an edge. Imagine
a 2 x 2 array of squares with their diagonals drawn. Fold along all the diagonals and between the
squares. This gives an array of 16
isosceles right triangles. Now cut from
the centres of the four squares to the centre of the whole array. This produces an X cut in the middle. This object can now be folded through itself
in various ways to produce a double thickness square of half the area with
various logos. The example is 100mm
along the edge of the large square and has four logos advertising Beanoland (at
Chessington, 3 versions) and Strip Cheese.
The patent is assigned to Lulirama International, but Byrnes says it has
not been a commercial success as it is too complicated. The patent cites 19 earlier patents, back to
1881, and discusses the history of such puzzles. It also says the puzzle can form three dimensional objects.
This
is the square cylindrical tube that can be inverted by folding. It was also invented by Arthur H. Stone,
c1939, cf 6.D.
J. Leech. A deformation puzzle. MG 39 (No. 330) (Dec 1955) 307. Doesn't know source. Says there are three solutions.
M. Gardner. Flexa-tube puzzle. Ibidem 7 (Sep 1956) 129.
Cites the inventors of the flexagons and the articles of Maunsell and
Leech (but he doesn't have its details).
(I have a note that this came with attached sample, but the copy I have
doesn't indicate such.)
T. S. Ransom. Flexa-tube solution. Ibidem 9 (Mar 1957) 174.
M. Gardner. SA (May 1958) = 2nd Book, chap. 2. Says Stone invented it and shows Ransom's
solution.
H. Steinhaus. Mathematical Snapshots. Not in the Stechert, NY, 1938, ed. nor the
OUP, NY, 1950 ed. OUP, NY: 1960: pp. 189‑193 & 326; 1969 (1983): pp. 177-181 & 303. Erroneous attribution to the Dowkers. Shows a different solution than Ransom's.
John Fisher. John Fisher's Magic Book. Muller, London, 1968. Homage to Houdini, pp. 152‑155. Detailed diagrams of the solution, but no
history.
Highland Games (2 Harpers Court,
Dingwall, Ross-Shire, IV15 9HT) makes a version called Table Teaser, made in a
strip with end pieces magnetic. Pieces
are coloured so to produce several folding and inverting problems other than
the usual one. Bought in 1995.
See
S&B, pp. 15‑18. See 6.F.1,
6.F.3, 6.F.4 & 6.F.5 for early occurrences of polyominoes. See Lammertink, 1996 & 1997 for many
examples in two and three dimensions.
NOTATION. Each of the types of puzzle considered has a
basic unit and pieces are formed from a number of these units joined edge to
edge. The notation N: n1, n2, .... denotes a puzzle with N
pieces, of which ni pieces consist of i basic units. If ni
are single digit numbers, the intervening commas and spaces will be omitted,
but the digits will be grouped by fives, e.g. 15: 00382 11.
Polyiamonds: Scrutchin;
John Bull; Daily Sketch; Daily Mirror; B. T.s Zig-Zag;
Daily Mail;
Miller (1960); Guy (1960); Reeve & Tyrrell;
O'Beirne (2 & 9 Nov 1961); Gardner (Dec 1964 & Jul 1965); Torbijn;
Meeus; Gardner (Aug 1975); Guy (1996, 1999); Knuth,
Polycubes: Rawlings (1939); Editor (1948); Niemann
(1948); French (1948); Editor (1948); Niemann (1948);
Gardner (1958); Besley
(1962); Gardner (1972)
Solid Pentominoes: Nixon (1948); Niemann (1948); Gardner
(1958); Miller (1960); Bouwkamp (1967, 1969, 1978); Nelson (2002);
Cylindrical Pentominoes: Yoshigahara (1992);
Polyaboloes: Hooper (1774); Book of 500 Puzzles (1859);
J. M. Lester (1919); O'Beirne
(21 Dec 61 &
18 Jan 62)
Polyhexes: Gardner (1967); Te Riele & Winter
Polysticks: Benjamin;
Barwell; General
Symmetrics; Wiezorke & Haubrich; Knuth;
Jelliss;
Polyrhombs or
Rhombiominoes: Lancaster (1918); Jones (1992).
Polylambdas: Roothart.
Polyspheres -- see Section 6.AZ.
GENERAL
REFERENCES
G. P. Jelliss. Special Issue on Chessboard
Dissections. Chessics 28 (Winter 1986)
137‑152. Discusses many problems
and early work in Fairy Chess Review.
Branko Grünbaum & Geoffrey
C. Shephard. Tilings and Patterns. Freeman, 1987. Section 9.4: Polyiamonds, polyominoes and polyhexes,
pp. 497-511. Good outline of the
field with a number of references otherwise unknown.
Michael Keller. A polyform timeline. World Game Review 9 (Dec 1989)
4-5. This outlines the history of
polyominoes and other polyshapes.
Keller and others refer to polyaboloes as polytans.
Rodolfo Marcelo Kurchan (Parana
960 5 "A", 1017 Buenos Aires, Argentina). Puzzle Fun, starting with No. 1 (Oct 1994). This is a magazine entirely devoted to
polyomino and other polyform puzzles.
Many of the classic problems are extended in many ways here. In No. 6 (Aug 1995) he presents a labelling
of the 12 hexiamonds by the letters A,
C, H, I, J, M, O, P, S, V, X, Y, which
he obtained from Anton Hanegraaf. I
have never seen this before.
Hooper. Rational Recreations. Op. cit. in 4.A.1. 1774. Vol. 1, recreation
23, pp. 64-66. Considers figures formed
of isosceles right triangles. He has
eight of these, coloured with eight colours, and uses some of them to form
"chequers or regular four-sided figures, different either in form or
colour".
Book of 500 Puzzles. 1859.
Triangular problem, pp. 74-75.
Identical to Hooper, dropping the last sentence.
Dudeney. CP.
1907. Prob. 74: The broken
chessboard, pp. 119‑121 & 220‑221. The 12 pentominoes and a
2 x 2.
A. Aubry. Prob. 3224.
Interméd. Math 14 (1907) 122-124.
??NYS -- cited by Grünbaum & Shephard who say Aubry has something of
the idea or the term polyominoes.
G. Quijano. Prob. 3430.
Interméd. Math 15 (1908) 195. ??NYS
-- cited by Grünbaum & Shephard, who say he first asked for the number
of n‑ominoes.
Thomas Scrutchin. US Patent 895,114 -- Puzzle. Applied: 20 Feb 1908; patented: 4 Aug 1908. 2pp + 1p diagrams. Mentioned in S&B, p. 18.
A polyiamond puzzle -- triangle of side 8, hence with 64 triangles,
apparently cut into 10 pieces (my copy is rather faint -- replace??).
Thomas W. Lancaster. US Patent 1,264,944 -- Puzzle. Filed: 7 May 1917; patented: 7 May 1918. 2pp
+ 1p diagrams. For a general polyrhomb
puzzle making a rhombus. His diagram
shows an 11 x 11 rhombus filled with 19
pieces formed from 4 to
10 rhombuses.
John Milner Lester. US Patent 1,290,761 -- Game Apparatus. Filed: 6 Feb 1918; patented: 7 Jan 1919.
2pp + 3pp diagrams. Fairly general
assembly puzzle claims. He specifically
illustrates a polyomino puzzle and a polyabolo puzzle. The first has a Greek cross of edge 3
(hence containing 45 unit cells) to be filled with polyominoes
-- 11: 01154. The second has an 8-pointed star formed by superimposing two 4 x 4
squares. This has area 20
and hence contains 40 isosceles right triangles of edge 1, which is the basic unit of this type of
puzzle. There are 11: 0128 pieces.
Blyth. Match-Stick Magic.
1921. Spots and squares, pp.
68-73. He uses matchsticks broken in
thirds, so it is easier to describe with units of one-third. 6 units,
4 doubles and
2 triples. Some of the
pieces have black bands or spots.
Object is to form polyomino shapes without pieces crossing, but every
intersection must have a black spot.
19 polyomino shapes are given to
construct, including 7 of the pentominoes, though some of the
shapes are only connected at corners.
"John Bull" Star of
Fortune Prize Puzzle. 1922. This is a puzzle with 20
pieces, coloured red on one side, containing 6 through 13
triangles to be assembled into a star of David with 4
triangles along each edge (hence
12 x 16 = 192 triangles). Made by Chad Valley. Prize of £250 for a red star matching the
key solution deposited at a bank; £150
for solution closest to the key; £100
for a solution with 10 red and
10 grey pieces, or as nearly as
possible. Closing date of competition
is 27 Dec 1922. Puzzle made by Chad
Valley Co. as a promotional item for John Bull magazine, published by
Odhams Press. A copy is in the toy shop
of the Buckleys Shop Museum, Battle, East Sussex, to whom I am indebted for the
chance to examine the puzzle and a photocopy of the puzzle, box and solution.
Daily Sketch Jig-Saw
Puzzle. By Chad Valley. Card polyiamonds. 39: 0,0,1,5,6, 12,9,6,
with a path printed on one side, to assemble into a shape of 16
rows of 15 with four corners removed and so the printed
sides form a continuous circuit. In box
with shaped bottom. Instructions on
inside cover and loose sheet to submit solution. No dates given, but appears to be 1920s, though it is somewhat
similar to the Daily Mail Crown Puzzle of 1953 -- cf below -- so it might be
much later.
Daily Mirror Zig-Zag £500 Prize
Puzzle. By Chad Valley. Card polyiamonds. 29: 0,0,1,1,4, 5,6,3,1,4, 1,2,1.
One-sided pieces to fit into frame in card box. Three pieces are duplicated and one is
triplicated. Solution and claim
instructions appeared in Daily Mirror (17 Jan 1930) 1-2. See: Tom Tyler & Felicity Whiteley; Chad
Valley Promotional Jig-Saw Puzzles; Magic Fairy Publishing, Petersfield,
Hampshire, 1990, p. 55.
B. T.s Zig-Zag. B.T. is a Copenhagen newspaper. Polyiamond puzzle. 33: 0,0,1,2,5, 6,7,2,2,4, 1,1,1, Some repetitions, so I only see 20 different
shapes. To be fit into an irregular
frame. Solution given on 23 Nov 1931,
pp. 1-2. (I have a photocopy of the form
to fill in; an undated set of rules, apparently from the paper, saying the
solutions must be received by 21 Nov; and the pages giving the solution;
provided by Jan de Geus.)
Herbert D. Benjamin. Problem 1597: A big cutting-out design --
and a prize offer. Problemist Fairy
Chess Supplement (later called Fairy Chess Review) 2:9 (Dec 1934) 92. Finds the
35 hexominoes and asks if they
form a 14 x 15 rectangle.
Cites Dudeney (Tribune (20 Dec 1906)); Loyd (OPM (Apr-Jul 1908)) (see 6.F.1); Dudeney (CP, no. 74) (see above)
and some other chessboard dissections.
Jelliss says this is the first dissection problem in this journal.
F. Kadner. Solution 1597. Problemist Fairy Chess Supplement (later called Fairy Chess
Review) 2:10 (Feb 1935) 104-105. Shows
the 35
hexominoes cannot tile a rectangle by two arguments, both essentially
based on two colouring. Gives some other
results and some problems are given as 1679-1681 -- ??NYS.
William E. Lester. Correction to 1597. Problemist Fairy Chess Supplement (later
called Fairy Chess Review) 2:11 (Apr 1935) 121. Corrects an error in Kadner.
Finds a number of near-solutions.
Editor says Kadner insists the editor should take credit for the
two-colouring form of the previous proof.
Frans Hansson, proposer &
solver?. Problem 1844. Problemist Fairy Chess Supplement (later
called Fairy Chess Review) 2:12 (Jun 1935) 128
& 2:13 (Aug 1935) 135. Finds both
3 x 20 pentomino
rectangles.
W. E. Lester & B. Zastrow,
proposers. Problem 1923. Problemist Fairy Chess Supplement (later
called Fairy Chess Review) 2:13 (Aug 1935) 138. Take an 8 x 8 board and remove its corners. Fill this with the 12 pentominoes.
H. D. Benjamin, proposer. Problem 1924. Problemist Fairy Chess Supplement (later called Fairy Chess
Review) 2:13 (Aug 1935) 138. Dissect
an 8 x 8 into the 12 pentominoes and the I-tetromino. Need solution -- ??NYS.
Thomas Rayner Dawson &
William E. Lester. A notation for
dissection problems. Fairy Chess Review
3:5 (Apr 1937) 46‑47. Gives
all n‑ominoes up to n = 6.
Describes the row at a time notation.
Shows the pentominoes and a 2 x
2 cover the chessboard with the 2 x 2
in any position. Asserts there
are 108 7‑ominoes and
368 8‑ominoes -- citing F.
Douglas & W. E. L[ester] for the hexominoes and J. Niemann for the
heptominoes.
H. D. Benjamin, proposer. Problem 3228. Fairy Chess Review 3:12 (Jun 1938) 129. Dissect a 5 x 5 into the five tetrominoes and a pentomino so
that the pentomino touches all the tetrominoes along an edge. Asserts the solution is unique. Refers to problems 3026‑3030 -- ??NYS.
H. D. Benjamin, proposer. Problem 3229. Fairy Chess Review 3:12 (Jun 1938) 129. Dissect an 8 x 8 into the
12 pentominoes and a tetromino
so that all pieces touch the edge of the board. Asserts only one tetromino works.
T. R. D[awson], proposer. Problems 3230-1. Fairy Chess Review 3:12 (Jun 1938) 129. Extends prob. 3229 to ask for solutions with 12
pieces on the edge, using two other tetrominoes. Thinks it cannot be done with the remaining
two tetrominoes.
Editorial note: The colossal count. Fairy Chess Review 3:12 (Jun 1938) 131. Describes progress on enumerating 8-ominoes (four people get 368
but Niemann gets 369) and
9‑ominoes (numbers vary from
1237 to 1285).
All workers are classifying them by the size of the smallest containing
rectangle.
W. H. Rawlings, proposer. Problem 3930. Fairy Chess Review 4:3 (Nov 1939) 28. How many pentacubes are there?
Ibid. 4:4 (Feb 1940) 75, reports
that both 25 or 26 are claimed, but the editor has only seen 24.
Ibid. 4:5 (Apr 1940) 85, reports that R. J. F[rench] has
clearly shown there are 23 -- but this considers reflections as equal
-- cf the 1948 editorial note.
R. J. French, proposer and
solver. Problem 4149. Fairy Chess Review 4:3 (Nov 1939) 43 &
4:6 (Jun 1940) 93. Asks for
arrangement of the pentominoes with the largest hole and gives one with 127
squares in the hole. (See: G. P. Jelliss; Comment on Problem 1277; JRM
22:1 (1990) 69. This reviews various
earlier solutions and comments on Problem 1277.)
J. Niemann. Item 4154: "The colossal
count". Fairy Chess Review 4:3
(Nov 1939) 44-45. Announces that there
are 369 8-ominoes, 1285 9-ominoes and 4654 10-ominoes, but
Keller and Jelliss note that he missed a
10‑omino which was not corrected until 1966.
H. D. Benjamin. Unpublished notes. ??NYS -- cited and briefly described in G. P. Jelliss; Prob. 48
-- Aztec tetrasticks; G&PJ 2 (No. 17) (Oct 1999) 320. Jelliss says Benjamin studied polysticks,
which he called 'lattice dissections' around 1946-1948 and that some results by
him and T. R. Dawson were entered in W. Stead's notebooks but nothing is known
to have been published. For orders 1, 2, 3, 4,
there are 1, 2, 5, 16 polysticks.
Benjamin formed these into a 6 x
6 lattice square. Jelliss then mentions Barwell's rediscovery
of them and goes on to a new problem -- see Knuth, 1999.
D. Nixon, proposer and
solver. Problem 7560. Fairy Chess Review 6:16 (Feb 1948) 12 &
6:17 (Apr 1948) 131.
Constructs 3 x 4 x 5 from solid pentominoes.
Editorial discussion: Space
dissection. Fairy Chess Review 6:18 (Jun
1948) 141-142. Says that several people
have verified the 23 pentacubes but that 6 of
them have mirror images, making 29 if these are considered distinct. Says F. Hansson has found 77 6‑cubes
(these exclude mirror images and the
35 solid 6-ominoes).
Gives many problems using
n-cubes and/or solid polyominoes, which he calls flat n-cubes -- some are corrected in 7:2 (Oct 1948)
16 (erroneously printed as 108).
J. Niemann. The dissection count. Item 7803.
Fairy Chess Review 7:1 (Aug 1948) 8 (erroneously printed as 100). Reports on counting n‑cubes. Gets the following.
n = 4 5 6 7
flat
pieces 5 12 35 108
non-flats 2 11 77 499
TOTAL 7 23 112 607
mirror
images 1 6 55 416
GRAND TOTAL 8 29 167 1023
R. J. French. Space dissections. Fairy Chess Review 7:2 (Oct 1948) 16 (erroneously printed as
108). French writes that he and A. W.
Baillie have corrected the number of
6-cubes to
35 + 77 + 54 = 166. Baillie notes that every 6-cube lies in two layers -- i.e. has some
width £ 2 -- and asks for the result for
n‑cubes as prob. 7879. [I
suspect the answer is that n £
3k implies that an n-cube has some width £ k.] Editor adds some corrections to the discussion in 6:18.
Editorial note. Fairy Chess Review 7:3 (Dec 1948) 23. Niemann and Hansson confirm the number 166
given in 7:2.
Daily Mail Crown Puzzle. Made by Chad Valley Co. 1953.
26 pieces, coloured on one side, to be fit into a crown shape. 11 are border pieces and easily placed. The other 15 are polyiamonds: 15: 00112 24012 11. Prize of £100 for solution plus best slogan,
entries due on 8 Jun 1953.
S. W. Golomb. Checkerboards and polyominoes. AMM 61 (1954) 675‑682. Mostly concerned with covering the 8 x 8
board with copies of polyominoes. Shows one covering with the
12 pentominoes and the square
tetromino. Mentions that the idea can
be extended to hexagons. S&B, p.
18, and Gardner (Dec 1964) say he mentions triangles, but he doesn't.
Walter S. Stead. Dissection.
Fairy Chess Review 9:1 (Dec 1954) 2‑4. Gives many pentomino and hexomino patterns -- e.g. one of each
pattern of 8 x 8 with a
2 x 2 square deleted. "The possibilities of the 12
fives are not infinite but they will provide years of
amusement." Includes 3 x 20,
4 x 15, 5 x 12 and
6 x 10 rectangles. No reference to Golomb. In 1955, Stead uses the 108
heptominoes to make a 28 x
28 square with a symmetric hole of
size 28 in the centre -- first printed as cover of Chessics 28 (1986).
Jules Pestieau. US Patent 2,900,190 -- Scientific
Puzzle. Filed: 2 Jul 1956; patented: 18 Aug 1959. 2pp + 1p diagrams. For the 12 pentominoes! Diagram shows the 6 x
10 solution with two 5 x 6
rectangles and shows the two-piece non-symmetric equivalence of the N
and F pieces. Pieces have
markings on one side which may be used -- i.e. pieces may not be turned
over. Mentions possibility of using n-ominoes.
Gardner. SA (Dec 1957) = 1st Book, chap. 13. Exposits Golomb and Stead. Gives number of n-ominoes for n = 1, ...,
7. 1st Book describes Scott's
work. Says a pentomino set called
'Hexed' was marketed in 1957. (John
Brillhart gave me and my housemates an example in 1960 -- it took us two weeks
to find our first solution.)
Dana Scott. Programming a Combinatorial Puzzle. Technical Report No. 1, Dept. of Elec. Eng.,
Princeton Univ., 1958, 20pp. Uses
MANIAC to find 65 solutions for pentominoes on an 8 x 8
board with square 2 x 2 in the centre. Notes that the 3 x
20 pentomino rectangle has just two
solutions. In 1999, Knuth notes that
the total number of solutions with the
2 x 2 being anywhere does not
seem to have ever been published and he finds
16146.
M. Gardner. SA (Sep 1958) c= 2nd Book, chap. 6. First general mention of solid pentominoes,
pentacubes, tetracubes. In the Addendum
in 2nd Book, he says Theodore Katsanis of Seattle suggested the eight
tetracubes and the 29 pentacubes in a letter to Gardner on
23 Sep 1957. He also says that
Julia Robinson and Charles W. Stephenson both suggested the solid pentominoes.
C. Dudley Langford. Note 2793:
A conundrum for form VI. MG 42
(No. 342) (Dec 1958) 287. 4 each of the
L, N, and T (= Y) tetrominoes make a 7 x 7 square with the
centre missing. Also nine pieces make a 6 x 6
square but this requires an even number of Ts.
M. R. Boothroyd &
J. H. Conway. Problems drive,
1959. Eureka 22 (Oct 1959) 15-17 &
22-23. No. 8. Use the pentominoes to make two
5 x 5 squares at the same time. Solution just says there are several ways to
do so.
J. C. P. Miller. Pentominoes. Eureka 23 (Oct 1960) 13‑16. Gives the Haselgroves' number of
2339 solutions for the 6 x 10
and says there are 2 solutions for the 3 x 20. Says Lehmer
suggests assembling 12 solid pentominoes into a 3 x 4 x 5
and van der Poel suggests assembling the 12 hexiamonds into a
rhombus.
C. B. & Jenifer
Haselgrove. A computer program for
pentominoes. Ibid., 16‑18. Outlines program which found the 2339
solutions for the 6 x 10. It is usually said that they also found all
solutions of the 3 x 20, 4 x 15
and 5 x 12, but I don't see it mentioned here and in JRM
7:3 (1974) 257, it is reported that Jenifer (Haselgrove) Leech stated that only
the 6 x 10 and 3 x 20 were done
in 1960, but that she did the 5
x 12 and 4 x 15 with a
new program in c1966. See Fairbairn,
c1962, and Meeus, 1973.
Richard K. Guy. Some mathematical recreations I &
II. Nabla [= Bull. Malayan Math.
Soc.] 7 (Oct & Dec 1960) 97-106 & 144-153. Considers handed polyominoes, i.e. polyominoes
when reflections are not considered equivalent. Notes that neither the
5 plain nor the 7
handed tetrominoes can form a rectangle. The 10 chequered handed tetrominoes form 4 x 10
and 5 x 8 rectangles and he has several solutions of
each. There is no 2 x 20
rectangle. Discusses MacMahon
pieces -- cf 5.H.2 -- and polyiamonds.
He uses the word 'hexiamond', but not 'polyiamond' -- in an email of 8
Apt 2000, Guy says that O'Beirne invented all the terms. He considers making a 'hexagon' from the 19
hexiamonds. Part II considers solid
problems and uses the term 'solid pentominoes'.
Solomon W. Golomb. The general theory of polyominoes: part 2 --
Patterns and polyominoes. RMM 5 (Oct
1961) 3-14. ??NYR.
J. E. Reeve & J. A.
Tyrrell. Maestro puzzles. MG 45 (No. 353) (Oct 1961) 97‑99. Discusses hexiamond puzzles, using the 12
reversible pieces. [The puzzle
was marketed under the name 'Maestro' in the UK.]
T. H. O'Beirne. Pell's equation in two popular
problems. New Scientist 12 (No. 258)
(26 Oct 1961) 260‑261.
T. H. O'Beirne. Pentominoes and hexiamonds. New Scientist 12 (No. 259) (2 Nov 1961) 316‑317. This is the first use of the word
'polyiamond'. He considers the 19
one‑sided pieces. He says
he devised the pieces and R. K. Guy has already published many solutions in
Nabla. He asks for the number of ways
the 18 one-sided pentominoes can fill a
9 x 10. In 1999, Knuth found
this would take several months.
T. H. O'Beirne. Some hexiamond solutions: and an introduction to a set of 25
remarkable points. New Scientist
12 (No. 260) (9 Nov 1961) 378‑379.
Maurice J. Povah. Letter.
MG 45 (No. 354) (Dec 1961) 342.
States Scott's result of 65 and the Haselgroves' result of 2339
(computed at Manchester). Says
he has over 7000 solutions for the 8 x 8 board using a 2 x 2.
T. H. O'Beirne. For boys, men and heroes. New Scientist 12 (No. 266) (21 Dec 1961) 751‑752.
T. H. O'Beirne. Some tetrabolic difficulties. New Scientist 13 (No. 270) (18 Jan 1962) 158‑159. These two columns are the first mention of
tetraboloes, so named by S. J. Collins.
R. A. Fairbairn. Pentomino Problems: The 6 x 10,
5 x 12, 4 x 15, and
3 x 20 Rectangles -- The
Complete Drawings. Unpublished MS,
undated, but c1962, based on the Haselgroves' work of 1960. ??NYS -- cited by various authors, e.g. Madachy (1969), Torbijn (1969), Meeus
(1973). Madachy says Fairbairn is from
Willowdale, Ontario, and takes some examples from his drawings. However, the dating is at variance with
Jenifer Haselgrove's 1973 statement - cf Haselgrove, 1960. Perhaps this MS is somewhat later?? Does anyone know where this MS is now? Cf Meeus, 1973.
Serena Sutton Besley. US Patent 3,065,970 -- Three Dimensional
Puzzle. Filed: 6 Jul 1960; issued: 27 Nov 1962. 2pp + 4pp diagrams. For the
29 pentacubes, with one piece
duplicated giving a set of 30. Klarner had already considered omitting
the 1 x 1 x 5 and found that he could make two separate 2 x 5 x 7s. Besley says the following can be made: 5 x 5 x 6, 3 x 5 x
10, 2 x 5 x 15, 2 x 3 x 25;
3 x 5 x 6, 3 x 3 x 10, 2 x 5 x 9,
2 x 3 x 15; 3 x 4 x 5, 2 x 5 x 6, 2 x 3 x 10
(where the latter three are made with the 12 solid pentominoes and
the previous four are made with the
18 non-planar pentacubes) but
detailed solutions are only given for the
5 x 5 x 6, 3 x 5 x 6, 3 x 4 x 5.
Mentions possibility of
n-cubes.
M. Gardner. Polyiamonds. SA (Dec 1964) = 6th Book, chap. 18. Exposits basic ideas and results for the 12 double sided
hexiamonds. Poses several problems
which are answered by readers. The
six-pointed star using 8 pieces has a unique solution. John G. Fletcher and Jenifer (Haselgrove)
Leech both showed the 3 x 12 rhombus is impossible. Fletcher found the 3 x 11 rhombus has 24
solutions, all omitting the 'bat'.
Leech found 155 solutions for the 6 x 6 rhombus and 74
solutions for the 4 x 9. Mentions there are 160 9-iamonds, one with a
hole.
John G. Fletcher. A program to solve the pentomino problem by
the recursive use of macros. Comm. ACM
8 (1965) 621-623. ??NYS -- described by
Knuth in 1999 who says that Fletcher found the 2339 solutions for the 6 x 10
in 10 minutes on an IBM 7094 and that the program remains the fastest
known method for problems of placing the 12 pentominoes.
M. Gardner. Op art.
SA (Jul 1965) = 6th Book, chap. 24.
Shows the 24 heptiamonds and discusses which will tile the plane.
Solomon W. Golomb. Tiling with polyominoes. J. Combinatorial Theory 1 (1966)
280-296. ??NYS. Extended by his 1970 paper.
T. R. Parkin. 1966.
??NYS -- cited by Keller.
Finds 4655 10-ominoes.
M. Gardner. SA (Jun 1967) = Magic Show, chap. 11. First mention of polyhexes.
C. J. Bouwkamp. Catalogue of Solutions of the
Rectangular 3 x 4 x 5 Solid Pentomino Problem. Dept. of Math., Technische Hogeschool
Eindhoven, July 1967, reprinted 1981, 310pp.
C. J. Bouwkamp. Packing a rectangular box with the twelve
solid pentominoes.
J. Combinatorial Thy. 7 (1969) 278‑280. He gives the numbers of solutions for
rectangles as 'known'.
2 x 3 x 10 can be packed in
12 ways, which are given.
2 x 5 x
6 can be packed in 264
ways.
3 x 4 x
5 can be packed in 3940
ways. (See his 1967 report.)
T. R. Parkin, L. J. Lander & D. R. Parkin. Polyomino enumeration results. Paper presented at the SIAM Fall Meeting,
Santa Barbara, 1 Dec 1967. ??NYS --
described by Madachy, 1969. Gives
numbers of n-ominoes, with and without
holes, up to n = 15, done two independent ways.
Joseph S. Madachy. Pentominoes -- Some solved and unsolved
problems. JRM 2:3 (Jul 1969)
181-188. Gives the numbers of Parkin,
Lander & Parkin. Shows various
examples where a rectangle splits into two congruent halves. Discusses various other problems, including
Bouwkamp's 3 x 4 x 5 solid pentomino problem. Bouwkamp reports that the final total of
3940 was completed on 16 Mar 1967 after about three years work using three
different computers, but that a colleague's program would now do the whole
search in about three hours.
P. J. Torbijn. Polyiamonds. JRM 2:4 (Oct 1969) 216-227.
Uses the double sided hexiamonds and heptiamonds. A few years before, he found, by hand, that
there are 156 ways to cover the 6 x 6 rhombus with the 12 hexiamonds and 74
ways for the 4 x 9, but could find no way to cover the 3 x 12.
The previous year, John G. Fletcher confirmed these results with a
computer and he displays all of these -- but this contradicts Gardner (Dec 64)
-- ?? He gives several other problems
and results, including using the 24 heptiamonds to form 7 x 12,
6 x 14, 4 x 21 and
3 x 28 rhombuses.
Solomon W. Golomb. Tlling with sets of polyominoes. J. Combinatorial Theory 9 (1970) 60‑71. ??NYS.
Extends his 1966 paper. Asks
which heptominoes tile rectangles and says there are two undecided cases -- cf
Marlow, 1985. Gardner (Aug 75) says
Golomb shows that the problem of determining whether a given finite set of
polyominoes will tile the plane is undecidable.
C. J. Boukamp & D. A.
Klarner. Packing a box with Y-pentacubes. JRM 3:1 (1970) 10-26.
Substantial discussion of packings with
Y‑pentominoes and
Y-pentacubes. Smallest boxes
are 5 x 10 and 2 x 5 x 6 and
3 x 4 x 5.
Fred Lunnon. Counting polyominoes. IN:
Computers in Number Theory, ed. by A. O. L. Atkin &
B. J. Birch; Academic Press, 1971, pp. 347-372. He gets up through 18‑ominoes, but the larger ones can have
included holes. The numbers for n = 1, 2, ..., are as follows: 1, 1, 2,
5, 12, 35, 108, 369, 1285, 4655, 17073, 63600, 238591, 901971, 3426576, 13079255, 50107911, 192622052. These values have been quoted numerous
times.
Fred Lunnon. Counting hexagonal and triangular
polyominoes. IN: Graph Theory and Computing, ed. by R. C.
Read; Academic Press, 1972, pp. 87-100.
??NYS -- cited by Grünbaum & Shephard.
M. Gardner. SA (Sep 1972). c= Knotted, chap. 3. Says
the 8
tetracubes were made by E. S. Lowe Co. in Hong Kong and
marketed as "Wit's End". Says
an MIT group found 1390 solutions for the 2 x 4 x 4 box packed with
tetracubes. He reports that several
people found that there are 1023 heptacubes -- but see Niemann, 1948, above. Klarner reports that the heptacubes fill
a 2 x 6 x 83.
Jean Meeus. Some polyomino and polyamond [sic]
problems. JRM 6:3 (1973) 215-220. (Corrections in 7:3 (1974) 257.) Considers ways to pack a 5 x n
rectangle with some n pentominoes. A. Mank found the number of ways for n = 2, 3, ..., 11 as
follows, and the number for n = 12 was already known:
0,
7, 50, 107, 541,
1387, 3377, 5865, 6814,
4103, 1010.
Says
he drew out all the solutions for the area 60 rectangles in 1972 (cf Fairbairn,
c1962). Finds that 520
of the 6 x 10 rectangles can be divided into two congruent
halves, sometimes in two different ways.
For 5 x 12, there are
380; for 4 x 15,
there are 94. Gives some
hexomino rectangles by either deleting a piece or duplicating one, and an
'almost 11 x 19'. Says there are 46 solutions to the 3 x 30
with the 18 one-sided pentominoes and attributes this to Mrs
(Haselgrove) Leech, but the correction indicates this was found by A. Mank.
Jenifer Haselgrove. Packing a square with Y-pentominoes. JRM 7:3 (1974) 229. She finds and shows a way to pack 45
Y-pentominoes into a 15 x 15, but is unsure if there are more
solutions. In 1999, Knuth found 212
solutions. She also reports the
impossibility of using the Y-pentominoes to fill various other rectangles.
S. W. Golomb. Trademark for 'PENTOMINOES'. US trademark 1,008,964 issued
15 Apr 1975; published 21 Jan
1975 as SN 435,448. (First use: November 1953.) [These appear in the Official Gazette of the United States Patent
Office (later Patent and Trademark Office) in the Trademarks section.]
M. Gardner. Tiling with polyominoes, polyiamonds and
polyhexes. SA (Aug 75) (with slightly
different title) = Time Travel, chap. 14.
Gives a tiling criterion of Conway.
Describes Golomb's 1966 & 1970 results.
C. J. Bouwkamp. Catalogue of
solutions of the rectangular 2 x 5 x
6 solid pentomino problem. Proc. Koninklijke Nederlandse Akad. van
Wetenschappen A81:2 (1978) 177‑186.
Presents the 264 solutions which were first found in Sep
1967.
H. Redelmeier. Discrete Math.
36 (1981) 191‑203. ??NYS --
described by Jelliss. Obtains number of
n‑ominoes for n £ 24.
Karl Scherer. Problem 1045: Heptomino tessellations. JRM 14:1 (1981‑82) 64. XX
Says he has found that the heptomino at the right fills
a 26 x 42 rectangle. XXXXX
See
Dahlke below.
David Ellard. Poly-iamond enumeration. MG 66 (No. 438) (Dec 1982) 310‑314. For
n = 1, ..., 12,
he gets 1, 1, 1, 3, 4, 12, 24, 66, 160, 448, 1186, 3342
n-iamonds. One of the 8-iamonds has a hole and there are many
later cases with holes.
Anon. 31: Polyominoes. QARCH
1:8 (June 1984) 11‑13. [This is
an occasional publication of The Archimedeans, the student maths society at
Cambridge.] Good survey of counting and
asymptotics for the numbers of polyominoes, up to n = 24, polycubes,
etc. 10 references.
T. W. Marlow. Grid dissections. Chessics 23 (Autumn, 1985) 78‑79.
X XX
Shows XXXXX
fills a 23 x 24 and
XXXXX fills a 19 x 28.
Herman J. J. te Riele & D.
T. Winter. The tetrahexes puzzle. CWI Newsletter [Amsterdam] 10 (Mar 1986) 33‑39. Says there are: 7 tetrahexes, 22
pentahexes, 82 hexahexes,
333 heptahexes, 1448 octahexes.
Studies patterns of 28 hexagons.
Shows the triangle cannot be constructed from the 7
tetrahexes and gives 48 symmetric patterns that can be made.
Karl A. Dahlke. Science News 132:20 (14 Nov 1987) 310. (??NYS -- cited in JRM 21:3 and
XX
22:1
and by Marlow below.) Shows XXXXX
fills a 21 x 26 rectangle.
The
results of Scherer and Dahlke are printed in JRM 21:3 (1989) 221‑223 and
Dahlke's solution is given by Marlow below.
Karl A. Dahlke. J. Combinatorial Theory A51 (1989) 127‑128. ??NYS -- cited in JRM 22:1. Announces a
19 x 28 solution for the above heptomino
problem, but the earlier
21 x 26 solution is
printed by error. The 19 x 28
solution is printed in JRM 22:1 (1990) 68‑69.
Tom Marlow. Grid dissections. G&PJ 12 (Sep/Dec 1989) 185.
Prints Dahlke's result.
Brian R. Barwell. Polysticks.
JRM 22:3 (1990) 165-175.
Polysticks are formed of unit lengths on the square lattice. There are:
1, 2, 5, 16, 55 polysticks
formed with 1, 2, 3, 4,
5 unit lengths. He forms
5 x 5 squares with one 4-stick omitted, but he permits pieces to
cross. He doesn't consider the
triangular or hexagonal cases. See also
Blyth, 1921, for a related puzzle. Cf
Benjamin, above, and Wiezorke & Haubrich, below.
General Symmetrics (Douglas
Engel) produced a version of polysticks, ©1991, with 4 3‑sticks and 3
4-sticks to make a 3 x 3 square array with no crossing of pieces.
Nob Yoshigahara. Puzzlart.
Tokyo, 1992. Ip-pineapple
(pineapple delight), pp. 78-81. Imagine
a cylindrical solution of the 6 x
10 pentomino rectangle and wrap it
around a cylinder, giving each cell a depth of one along the radius. Hence each cell is part of an annulus. He reduces the dimensions along the short
side to make the cells look like tenths of a slice of pineapple. Nob constructed and example for Toyo Glass's
puzzle series and it was later found to have a unique solution.
Kate Jones, proposer; P. J. Torbijn, Jacques Haubrich,
solvers. Problem 1961 --
Rhombiominoes. JRM 24:2 (1992) 144-146 &
25:3 (1993) 223‑225. A
rhombiomino or polyrhomb is a polyomino formed using rhombi instead of
squares. There are 20
pentarhombs. Fit them into a 10 x 10
rhombus. Various other
questions. Haubrich found many
solutions. See Lancaster, 1918.
Bernard Wiezorke & Jacques
Haubrich. Dr. Dragon's polycons. CFF 33 (Feb 1994) 6-7. Polycons (for connections) are the same as
the polysticks described by Barwell in 1990, above. Authors describe a Taiwanese version on sale in late 1993,
using 10 of the 4‑sticks
suitably shortened so they fit into the grooves of a 4 x 4 board -- so crossings
are not permitted. (An n x n
board has n+1 lines of
n edges in each direction.) They fit
15 of the 4-sticks onto a 5 x 5 board and determine
all solutions.
CFF
35 (Dec 1994) 4 gives a number of responses to the article. Brain Barwell wrote that he devised them as
a student at Oxford, c1970, but did not publish until 1990. He expected someone to say it had been done
before, but no one has done so. He also
considered using the triangular and hexagonal lattice. He had just completed a program to consider
fitting 15 of the 4-sticks onto a 5
x 5 board and found over 180,000
solutions, with slightly under half having no crossings, confirming the
results of Wiezorke & Haubrich.
Dario
Uri also wrote that he had invented the idea in 1984 and called them polilati
(polyedges). Giovanni Ravesi wrote
about them in Contromossa (Nov 1984) 23 -- a defunct magazine.
Chris Roothart. Polylambdas. CFF 34 (Oct 1994) 26-28.
A lambda is a 30o-60o-90o triangle.
These may be joined along corresponding legs, but not along
hypotenuses. For n = 1, 2, 3, 4, 5, there are 1, 4, 4, 11, 12 n-lambdas.
He gives some problems using various sets of these pieces.
Richard Guy. Letters of 29 May and 13 Jun 1996. He is interested in using the 19
one-sided hexiamonds. Hexagonal
rings of hexagons contain 1, 6, 12 hexagons, so the hexagon with three hexagons
on a side has 19 hexagons. If these
hexagons are considered to comprise six equilateral triangles, we have a board
with 19 x 6 triangles. O'Beirne asked
for the number of ways to fill this board with the one-sided hexiamonds. Guy has collected over 4200
solutions. A program by Marc
Paulhus found 907 solutions in eight hours, from which it
initially estimated that there are about
30,000 solutions. The second letter gives the final results --
there are 124,518 solutions.
This is modulo the 12 symmetries of the hexagon. In 1999, Knuth found 124,519
and Paulhus has rerun his program and found this number.
Ferdinand Lammertink. Polyshapes.
Parts 1 and 2. The author,
Hengelo, Netherlands, 1996 & 1997.
Part 1 deals with two dimensional puzzles. Good survey of the standard polyform shapes and many others.
Hilarie Korman. Pentominoes: A first player win. IN: Games of No Chance; ed. by Richard
Nowakowski; CUP, 1997??, ??NYS - described in
William Hartston; What mathematicians get up to; The Independent Long
Weekend (29 Mar 1997) 2. This studies
the game proposed by Golomb -- players alternately place one of the pentominoes
on the chess board, aligned with the squares and not overlapping the previous
pieces, with the last one able to play being the winner. She used a Sun IPC Sparcstation for five
days, examining about 22 x 109 positions to show the game is a first player
win.
Nob Yoshigahara found in 1994
that the smallest box which can be packed with W-pentacubes is 5 x 6 x 6.
In 1997, Yoshya (Wolf) Shindo found that one can pack the 6 x 10 x 10
with Z-pentacubes, but it is not known if this is the smallest such
box. These were the last unsolved
problems as to whether a box could be packed with a planar pentacube
(= solid pentomino).
Marcel Gillen &
Georges Philippe. Twinform 462 Puzzles in one. Solutions for Gillen's puzzle exchange at
IPP17, 1997, 32pp + covers. Take 6 of
the pentominoes and place them in a 7 x
5 rectangle, then place the other six
to make the same shape on top of the first shape. There are 462 (= BC(12,6)/2) possible puzzles and all of them have solutions. Taking
F, T, U, W, X, Z for the first layer,
there is just one solution; all other cases have multiple solutions,
totalling 22,873 solutions, but only one solution for each
case is given here.)
Richard K. Guy. O'Beirne's hexiamond. In:
The Mathemagican and Pied Puzzler; ed. by Elwyn Berlekamp & Tom
Rodgers, A. K. Peters, Natick, Massachusetts, 1999, pp. 85‑96. He relates that O'Beirne discovered the 19
one-sided hexiamonds in c1959 and found they would fill a hexagonal shape in
Nov 1959 and in Jan 1960 he found a solution with the hexagonal piece in the
centre. He gives Paulhus's results (see
Guy's letters of 1996), broken down in various ways. He gives the number of double-sided (i.e. one can turn them over)
and single-sided n-iamonds for n = 1, ..., 7. Cf Ellard, 1982, for many more values for the double-sided case.
n 1 2
3 4 5 6 7
double 1
1 1 3 4 12
24
single 1
1 1 4 6 19
44
In
1963, Conway and Mike Guy considered looking for 'symmetric' solutions for
filling the hexagonal shape with the 19 one-sided hexiamonds. A number of these are described.
Donald E. Knuth. Dancing links. 25pp preprint of a talk given at Oxford in Sep 1999, sent by the
author. Available as:
http://www-cs-faculty.stanford.edu/~knuth/preprints.html . In this he introduces a new technique for
backtrack programming which runs faster (although it takes more storage) and is
fairly easy to adapt to different problems.
In this approach, there is a symmetry between pieces and cells. He applies it to several polyshape problems,
obtaining new, or at least unknown, results.
He extends Scott's 1958 results to get
16146 ways to pack the 8 x 8
with the 12 pentominoes and the
2 x 2. He describes
Fletcher's 1965 work. He extends
Haselgrove's 1974 work and finds 212 ways to fit 15 Y-pentominoes in a 15 x 15.
Describes Torbijn's 1969 work and Paulhus' 1996 work on hexiamonds,
correcting the latter's number to
124,519. He then looks for the
most symmetric solutions for filling the hexagonal shape with the 19 one-sided
hexiamonds, in the sense discussed by Guy (1999). He then considers the 18 one-sided pentominoes (cf Meeus (1973))
and tries the 9 x 10, but finds it would take a few months on his
computer (a 500 MHz Pentium III), so he's abandoned it for now. He then considers polysticks, citing an
actual puzzle version that I've not seen.
He adapts his program to them.
He considers the 'welded tetrasticks' which have internal junction
points. There are six of these and ten
if they are taken as one-sided. The ten
can be placed in a 4 x 4 grid. There are
15 unwelded, one-sided,
tetrasticks, but they do not form a square, nor indeed any nice shape. He considers all 25 one-sided tetrasticks
and asks if they can be fit into what he calls an Aztec Diamond, which is the
shape looking like a square tilted 45o on the square lattice. The rows contain 1, 3, 5, 7, 9, 7, 5, 3, 1
cells. He thinks an exhaustive
search is beyond present computing power.
G. P. Jelliss. Prob. 48 -- Aztec tetrasticks. G&PJ 2 (No. 17) (Oct 1999) 320. Jelliss first discusses Benjamin's work on
polysticks (see at 1946-1948 above) and Barwell's rediscovery of them (see
above). He then describes Knuth's Dancing
Links and gives the Aztec Diamond problem.
Jelliss has managed to get all but one of the polysticks into the shape,
but feels it is impossible to get them all in.
Harry L. Nelson. Solid pentomino storage, Question and
answer. 1p HO at G4G5, 2002. 1: Can one put all the solid
pentominoes into a cube of edge
4.5? What is the smallest cube
into which they can all be placed? He
gives 2 solutions to 1 and a solution due to Wei-Hwa Huang for a cube of edge
4.405889..., which is conjectured to be minimal. In fact, one edge of the packing is actually 4, so the volume is
less than (4.405889...)3. This leads me to ask what is the smallest
volume of a cuboid, with edges less than 5, that contains all the solid
pentominoes. In Summer 2002, Harry gave
me a set of solid pentominoes in a box with a list of various rectangles and
boxes to fit them into: 3 x 22; 3 x 21;
3 x 20; 4 x 16; 4 x 15;
5 x 13; 5 x 12; 6 x 11; 6 x 10; 7 x 9; 8 x 8;
2 x 4 x 8; 2
x 5 x 7; 2 x 5 x 6; 2 x 6 x 6;
3 x 4 x 6; 3 x 4 x
5; 3 x 5 x 5; 4 x 4 x 5; the given
box: 4.4 x 4.4 x 4.9.)
6.F.1. OTHER CHESSBOARD DISSECTIONS
See
S&B, pp. 12‑14. See also
6.F.5 for dissections of uncoloured boards.
Jerry Slocum. Compendium of Checkerboard Puzzles. Published by the author, 1983. Outlines the history and shows all manufactured
versions known then to him: 33 types in
61 versions. The first number in
Slocum's numbers is the number of pieces.
Jerry Slocum &
Jacques Haubrich. Compendium of
Checkerboard Puzzles. 2nd ed.,
published by Slocum, 1993. 90 types in
161 versions, with a table of which pieces are in which puzzles, making it much
easier to see if a given puzzle is in the list or not. This gives many more pictures of the puzzle
boxes and also gives the number of solutions for each puzzle and sometimes prints
all of them. The Slocum numbers are
revised in the 2nd ed. and I use the 2nd ed. numbers below. (There was a 3rd ed. in 1997, with new
numbering of 217 types in 376 versions.
NYR. Haubrich is working on an
extended version with Les Barton providing information.)
Henry Luers. US Patent 231,963 -- Game Apparatus or
Sectional Checker Board. Applied:
7 Aug 1880; patented: 7 Sep 1880. 1p + 1p diagrams. 15: 01329. Slocum
15.5.1. Manufactured as: Sectional
Checker Board Puzzle, by Selchow & Righter. Colour photo of the puzzle box cover is on the front cover of the
1st ed. of Slocum's booklet. B&W
photo is on p. 14 of S&B.
?? UK patent application 16,810.
1892. Not granted, so never
published. I have spoken to the UK
Patent Office and they say the paperwork for ungranted applications is
destroyed after about three to five years.
(Edward Hordern's collection has an example with this number on it, by
Feltham & Co. In the 2nd ed., the
cover is reproduced and it looks like the number may be 16,310, but that number
is for a locomotive vehicle.)
14: 00149. Slocum
14.20.1. Manufactured as: The Chequers
Puzzle, by Feltham & Co.
Hoffmann. 1893.
Chap. III, no. 16: The chequers puzzle, pp. 97‑98 & 129‑130
= Hoffmann‑Hordern, pp. 88-89, with photos. 14: 00149.
Slocum 14.20.1. Says it is made
by Messrs. Feltham, who state it has over 50 solutions. He gives two solutions. Photo on p. 89 of a example by Feltham &
Co., dated 1880-1895.
At
the end of the solution, he says Jacques & Son are producing a series of
three "Peel" puzzles, which have coloured squares which have to be
arranged so the same colour is not repeated in any row or column. Photo on p. 89 shows an example, 9: 023,
with the trominoes all being L-trominoes. This makes a 5 x 5 square, but the colours have almost faded
into indistinguishability.
Montgomery Ward & Co. Catalog No 57, Spring &
Summer, 1895. Facsimile by Dover, 1969,
??NX. P. 237 describes item 25470: "The "Wonder" Puzzle. The object is to place 18 pieces of 81
squares together, so as to form a square, with the colors running
alternately. It can be done in several
different ways."
Dudeney. Problem 517 -- Make a chessboard. Weekly Dispatch (4 & 18 Oct 1903), both
p. 10. 8: 00010 12111 001. Slocum 8.3.1.
Benson. 1904.
The chequers puzzle, pp. 202‑203.
As in Hoffmann, with only one solution.
Dudeney. The Tribune (20 & 24 Dec 1906) both
p. 1. ??NX Dissecting a chessboard.
Dissect into maximum number of different pieces. Gets 18: 2,1,4,10,0, 0,0,1. Slocum 18.1, citing later(?) Loyd versions.
Loyd. Sam Loyd's Puzzle Magazine (Apr-Jul 1908) -- ??NYS, reproduced
in: A. C. White; Sam Loyd and His Chess
Problems; 1913, op. cit. in 1; no. 58, p. 52.
= Cyclopedia, 1914, pp. 221 & 368, 250 & 373. = MPSL2, prob. 71, pp. 51 &
145. = SLAHP: Dissecting the
chessboard, pp. 19 & 87. Cut into
maximum number of different pieces -- as in Dudeney, 1906.
Burren Loughlin &
L. L. Flood. Bright-Wits Prince of Mogador. H. M. Caldwell Co., NY, 1909.
The rug, pp. 7-13 & 65. 14:
00149. Not in Slocum.
Loyd. A battle royal.
Cyclopedia, 1914, pp. 97 & 351 (= MPSL1, prob. 51, pp. 49 &
139). Same as Dudeney's prob. 517 of
1903.
Dudeney. AM. 1917.
Prob. 293: The Chinese chessboard, pp. 87 & 213‑214. Same as Loyd, p. 221.
Western Puzzle Works, 1926
Catalogue. No. 79: "Checker Board
Puzzle, in 16 pieces", but the picture only shows 14 pieces. 14: 00149.
Picture doesn't show any colours, but assuming the standard colouring of
a chess board, this is the same as Slocum 14.15.
John Edward Fransen. US Patent 1,752,248 -- Educational
Puzzle. Applied: 19 Apr 1929; patented: 25 Mar 1930. 1p + 1p diagrams. 'Cut thy life.'
11: 10101 43001.
Slocum 11.3.1.
Emil Huber-Stockar. Patience de l'echiquier. Comptes-Rendus du Premier Congrès
International de Récréation Mathématique, Bruxelles, 1935. Sphinx, Bruxelles, 1935, pp. 93-94. 15: 01329. Slocum 15.5. Says there
must certainly be more than 1000 solutions.
Emil Huber-Stockar. L'echiquier du diable. Comptes-Rendus du Deuxième Congrès
International de Récréation Mathématique, Paris, 1937. Librairie du "Sphinx", Bruxelles,
1937, pp. 64-68. Discusses how one
solution can lead to many others by partial symmetries. Shows several solutions containing about 40
altogether. Note at end says he has now
got 5275 solutions. This article
is reproduced in Sphinx 8 (1938) 36-41, but without the extra pages of
diagrams. At the end, a note says he
has 5330 solutions. Ibid., pp.
75-76 says he has got 5362 solutions and ibid. 91-92 says he has 5365.
By use of Bayes' theorem on the frequency of new solutions, he
estimates c5500 solutions.
Haubrich has found 6013.
Huber-Stocker intended to produce a book of solutions, but he died in
May 1939 [Sphinx 9 (1939) 97].
F. Hansson. Sam Loyd's 18-piece dissection -- Art. 48
& probs. 4152‑4153. Fairy
Chess Review 4:3 (Nov 1939) 44. Cites
Loyd's Puzzles Magazine. Asserts there
are many millions of solutions! He
determines the number of chequered handed
n-ominoes for
n = 1, 2, ..., 8
is 2, 1, 4, 10, 36, 110, 392,
1371. The first 17 pieces total 56
squares. Considers 8 ways to dissect
the board into 18 different pieces.
Problems ask for the number of ways to choose the pieces in each of
these ways and for symmetrical solutions.
Solution in 4:6 (Jun 1940) 93-94 (??NX of p. 94) says there are a total
of 3,309,579 ways to make the choices.
C. Dudley Langford. Note 2864:
A chess‑board puzzle. MG
43 (No. 345) (Oct 1959) 200. 15:
01248. Not in Slocum. Two diagrams followed by the following
text. "The pieces shown in the
diagrams can be arranged to form a square with either side uppermost. If the squares of the underlying grid are
coloured black and white alternately, with each white square on the back of a
black square, then there is at least one more way of arranging them as a
chess-board by turning some of the pieces over." I thought this meant that the pieces were double-sided with the
underside having the colours being the reverse of the top and the two diagrams
were two solutions for this set of pieces.
Jacques Haubrich has noted that the text is confusing and that the
second diagram is NOT using the set of double-sided pieces which are implied by
the first diagram. We are not sure if
the phrasing is saying there are two different sets of pieces and hence two
problems or if we are misinterpreting the description of the colouring.
B. D. Josephson. EDSAC to the rescue. Eureka 24 (Oct 1961) 10‑12 &
32. Uses the EDSAC computer to find two
solutions of a 12 piece chessboard dissection.
12: 00025 41. Slocum 12.9.
Leonard J. Gordon. Broken chessboards with unique
solutions. G&PJ 10 (1989) 152‑153. Shows Dudeney's problem has four
solutions. Finds other colourings which
give only one solution. Notes some
equivalences in Slocum.
6.F.2. COVERING DELETED CHESSBOARD WITH DOMINOES
See
also 6.U.2.
There
is nothing on this in Murray.
Pál Révész. Op. cit. in 5.I.1. 1969. On p. 22, he says
this problem comes from John [von] Neumann, but gives no details.
Max Black. Critical Thinking, op. cit. in 5.T. 1946 ed., pp. 142 & 394, ??NYS. 2nd ed., 1952, pp. 157 & 433. He simply gives it as a problem, with no
indication that he invented it.
H. D. Grossman. Fun with lattice points: 14 -- A chessboard
puzzle. SM 14 (1948) 160. (The problem is described with 'his clever
solution' from M. Black, Critical Thinking, pp. 142 & 394.)
S. Golomb. 1954.
Op. cit. in 6.F.
M. Gardner. The mutilated chessboard. SA (Feb 1957) = 1st Book, pp. 24 & 28.
Gamow & Stern. 1958.
Domino game. Pp. 87‑90.
Robert S. Raven, proposer; Walter P. Targoff, solver. Problem 85 -- Deleted checkerboard. In: L.
A. Graham; Ingenious Mathematical Problems and Methods; Dover, 1959, pp. 52
& 227.
R. E. Gomory. (Solution for deletion of any two squares of
opposite colour.) In: M. Gardner, SA (Nov 1962) = Unexpected, pp.
186‑187. Solution based on a
rook's tour. (I don't know if this was
ever published elsewhere.)
Michael Holt. What is the New Maths? Anthony Blond, London, 1967. Pp. 68 & 97. Gives the
4 x 4 case as a
problem, but doesn't mention that it works on other boards. (I include this as I haven't seen earlier
examples in the educational literature.)
David Singmaster. Covering deleted chessboards with
dominoes. MM 48 (1975) 59‑66. Optimum extension to n‑dimensions. For an
n-dimensional board, each dimension must be ³ 2. If the board has an even number of cells, then one can delete
any n-1 white cells and any n-1 black cells and still cover the board with dominoes
(i.e.
2 x 1 x 1 x ... x 1 blocks).
If the board has an odd number of cells, then let the corner cells be
coloured black. One can then delete
any n
black cells and any n-1 white cells and still cover the board with
dominoes.
I-Ping Chu & Richard
Johnsonbaugh. Tiling deficient boards
with trominoes. MM 59:1 (1986)
34-40. (3,n) = 1 and
n ¹ 5 imply that
an n x n board with one cell deleted can be covered with L
trominoes. Some 5 x 5
boards with one cell deleted can be tiled, but not all can.
6.F.3. DISSECTING A CROSS INTO Zs AND Ls
The L
pieces are not always drawn carefully, and in some cases the unit pieces
are not all square. I have enlarged and
measured those which are not clear and approximated them as n-ominoes.
Minguet. 1733.
Pp. 119-121 (1755: 85-86; 1822: 138-139; 1864: 116-117). The problem has two parts. The first is a cross into 5 pieces:
L-tetromino, 2 Z-pentominoes, L‑hexomino, Z-hexomino. The two hexominoes are like the
corresponding pentominoes lengthened by one unit. Similar to Les Amusemens, but one Z is longer and one L is
shorter. The diagram shows 8
L and Z shaped pieces formed
from squares, but it is not clear what the second part of the problem is doing
-- either a piece or a label is erroneous or missing. Says one can make different figures with the pieces.
Les Amusemens. 1749.
P. xxxi. Cross into 3 Z
pentominoes and 2 L pieces.
Like Minguet, but the Ls are much lengthened and are approximately a
L-heptomino and an L‑octomino.
Catel. Kunst-Cabinet. 1790. Das mathematische Kreuz, p. 10 & fig. 27
on plate I. As in Les Amusemens, but
the Ls are approximately a 9-omino and a 10-omino.
Bestelmeier. 1801.
Item 274 -- Das mathematische Kreuz.
Cross into 6 pieces, but the picture has an erroneous extra line. It should be the reversal of the picture in
Catel.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. F. 4r is "Analysis of the
Essay of Games". F. 4v has the
dissection of the cross into 3 Z pentominoes and two L
pieces. I don't have a copy of
this, but my sketch looks like the Ls are a tetromino and a pentomino, or
possibly a pentomino and a hexomino.
Manuel des Sorciers. 1825.
Pp. 204-205, art. 21. ??NX. Dissect a cross into three Zs
and two Ls. My notes don't indicate the size of the Ls.
Boy's Own Book. 1843 (Paris): 435 & 440, no. 3. As in Les Amusemens, but with the Ls
apparently intended to be a pentomino and a hexomino. = Boy's Treasury, 1844, pp. 424-425 & 428. = de Savigny, 1846, pp. 353 & 357, no.
2, except the solution has been redrawn with some slight changes and so the
proportions are less clear.
Family Friend 3 (1850) 330 &
351. Practical puzzle, No. XXI. As in Les Amusemens.
Magician's Own Book. 1857.
Prob. 31: Another cross puzzle, pp. 276 & 299. As in Les Amusemens.
Landells. Boy's Own Toy-Maker. 1858.
P. 152. As in Les Amusemens.
Book of 500 Puzzles. 1859.
Prob. 31: Another cross puzzle, pp. 90 & 113. As in Les Amusemens. = Magician's Own Book.
Indoor & Outdoor. c1859.
Part II, p. 127, prob. 5: The puzzle of the cross. As in Les Amusemens.
Illustrated Boy's Own
Treasury. 1860. Practical Puzzles, No. 24, pp. 399 &
439. Identical to Magician's Own Book.
Boy's Own Conjuring book. 1860.
Prob. 30: Another cross puzzle, pp. 239 & 263. = Magician's Own Book, 1857.
Leske. Illustriertes Spielbuch für Mädchen. 1864?
Prob.
584-2, pp. 286 & 404.
4 Z pentominoes to
make a (Greek) cross. (Also entered in
6.F.5.)
Prob.
584-8, pp. 287 & 405. 3 Z
pentominoes, L tetromino and L pentomino to make a
Greek cross. Despite specifically
asking for a Greek cross, the answer is a standard Latin cross with
height : width = 4 : 3.
Mittenzwey. 1880.
Prob. 173-174, pp. 33 & 85;
1895?: 198-199, pp. 38 & 87;
1917: 198‑199, pp. 35 & 84.
The first is 3 Z pentominoes, L tetromino and L
pentomino to make a cross. The
second is 4 Z pentominoes to
make a (Greek) cross. (Also entered in
6.F.5.)
Cassell's. 1881.
P. 93: The magic cross. =
Manson, 1911, p. 139. Same pattern as
Les Amusemens, but one end of the Zs is decidedly longer than the other and the
middle 'square' of the Zs is decidedly not square. The Ls are approximately a pentomino and a heptomino, But the middle 'square' of the Zs is almost
a domino and that makes the Zs into heptominoes, with the Ls being a hexomino
and a nonomino.
S&B, p. 20, shows a 7 piece
cross dissection, Jeu de La Croix, into 3
Zs, 2 Ls and 2 straights, from
c1890. The Zs are pentominoes, with the
centre 'square' lengthened a bit. The
Ls appear to be a heptomino and an octomino and the straights appear to be a
hexomino and a tetromino. Cf
Hoffmann-Hordern for a version without the straight pieces.
Handy Book for Boys and
Girls. Showing How to Build and
Construct All Kinds of Useful Things of Life.
Worthington, NY, 1892. Pp. 320-321:
The cross puzzle. As in Cassell's.
Hoffmann. 1893.
Chap. III, no. 29: Another cross puzzle, pp. 103 & 136
= Hoffmann‑Hordern, pp. 100-101, with photo. States that the two Ls are the same shape,
but the solution is as in Les Amusemens, with the Ls approximately a hexomino
and a heptomino. Hordern has corrected
the problem statement. Photo on
p. 100 shows an ivory version, dated 1850-1900, of the same
proportions. Hordern Collection, p. 65,
shows two wood versions, La Croix Brisée and Jeu de la Croix, dated 1880-1905,
both with Ls being approximately a heptomino and an octomino.
Benson. 1904.
The Latin cross puzzle, p. 200.
As in Hoffmann, but the solution is longer, as in Les Amusemens.
Wehman. New Book of 200 Puzzles. 1908.
Another cross puzzle, p. 32. As
in Les Amusemens, with the Ls being a pentomino and a hexomino.
S. Szabo. US Patent 1,263,960 -- Puzzle. Filed: 20 Oct 1917; patented: 23 Apr 1918. 1p + 1p diagrams. As in Les Amusemens, with even longer Ls, approximately a 10-omino and an 11-omino.
6.F.4. QUADRISECT AN L‑TROMINO, ETC.
See
also 6.AW.1 & 4.
Mittenzwey
and Collins quadrisect a hollow square obtained by removing a 2 x 2
from the centre of a 4 x 4.
Bile
Beans quadrisects a 5 x 5 after deleting corners and centre.
Minguet. 1733.
Pp. 114-115 (1755: 80; 1822: 133-134; 1864: 111-112). Quadrisect
L‑tromino.
Alberti. 1747.
Art. 30: Modo di dividere uno squadro di carta e di legno in quattro
squadri equali, p. ?? (131) & fig. 56, plate XVI, opp. p. 130.
Les Amusemens. 1749.
P. xxx. L-tromino
("gnomon") into 4 congruent pieces.
Vyse. Tutor's Guide. 1771? Prob. 9, 1793: p. 305, 1799: p. 317 &
Key p. 358. Refers to the land as a
parallelogram though it is drawn rectangular.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. F. 4r is "Analysis of the
Essay of Games". F. 4v has an
entry "8½ a Prob of figure" followed by the L‑tromino. 8½ b is
the same with a mitre and there are other dissection problems adjacent -- see
6.F.3, 6.AQ, 6.AW.1, 6.AY, so it seems clear that he knew this problem.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles, no. 3, pp. 23 & 83 & plate I, fig. 2.
Manuel des Sorciers. 1825.
Pp. 203-204, art. 20. ??NX. Quadrisect L-tromino.
Family Friend 2 (1850) 118 &
149. Practical Puzzle -- No. IV. Quadrisect L-tromino of land with four
trees.
Family Friend 3 (1850) 150 &
181. Practical puzzle, No. XV. 15/16
of a square with 10 trees to be divided equally. One tree is placed very close to another, cf
Magician's Own Book and Hoffmann, below.
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles, no. 8, p. 179 (1868:
190). Land in the shape of an L-tromino to be cut into four congruent
parts, each with a cherry tree.
Magician's Own Book. 1857.
Prob.
3: The divided garden, pp. 267 & 292.
15/16 of a square to be divided
into five (congruent) parts, each with two trees. The missing 1/16 is in the middle. One tree is placed very close to another, cf Family Friend 3,
above, and Hoffmann below.
Prob.
22: Puzzle of the four tenants, pp. 273 & 296. Same as Parlour Pastime, but with apple trees. (= Illustrated Boy's Own Treasury, 1860, No.
10, pp. 397 & 437.)
Prob.
28: Puzzle of the two fathers, pp. 275-276 & 298. Each father wants to divide
3/4 of a square. One has
L‑tromino, other has the mitre shape. See 6.AW.1.
Landells. Boy's Own Toy-Maker. 1858.
P.
144. = Magician's Own Book, prob. 3.
Pp.
148-149. = Magician's Own Book, prob.
27.
Book of 500 Puzzles. 1859.
Prob.
3: The divided garden, pp. 81 & 106.
Identical to Magician's Own Book.
Prob.
22: Puzzle of the four tenants, pp. 87 & 110. Identical to Magician's Own Book.
Prob.
28: Puzzle of the two fathers, pp. 89-90 & 112. Identical to Magician's Own Book. See also 6.AW.1.
Charades, Enigmas, and
Riddles. 1860: prob. 28, pp. 59 &
63; 1862: prob. 29, pp. 135 &
141; 1865: prob. 573, pp. 107 & 154. Quadrisect
L-tromino, attributed to Sir F. Thesiger.
Boy's Own Conjuring book. 1860.
Prob.
3: The divided garden, pp. 229 & 255.
Identical to Magician's Own Book.
Prob.
21: Puzzle of the four tenants, pp. 235 & 260. Identical to Magician's Own Book.
Prob.
27: Puzzle of the two fathers, pp. 237‑238 & 262. Identical to Magician's Own Book.
Illustrated Boy's Own
Treasury. 1860. Prob. 21, pp. 399 & 439. 15/16
of a square to be divided into five (congruent) parts, each with two
trees. c= Magician's Own Book,
prob. 3.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 175, p.
88. L-tromino into four congruent
pieces, each with two trees. The
problem is given in terms of the original square to be divided into five parts,
where the father gets a quarter of the whole in the form of a square and the
four sons get congruent pieces.
Hanky Panky. 1872. The divided orchards, p. 130.
L‑tromino into 4 congruent pieces, each with two trees.
Boy's Own Book. The divided garden. 1868: 675.
= Magician's Own Book, prob. 3.
Mittenzwey. 1880.
Prob.
192, pp. 36 & 89; 1895?: 217, pp.
40 & 91; 1917: 217, pp. 37 &
87. Cut 1 x 1 out of the centre
of a 4 x 4. Divide the rest into five parts of equal area with four being
congruent. He cuts a 2 x 2
out of the centre, which has a 1
x 1 hole in it, then divides the rest
into four L-trominoes.
Prob.
213, pp. 38 & 90; 1895?: 238, pp.
42 & 92; 1917: 238, pp. 39 &
88. Usual quadrisection of an
L-tromino.
Prob.
214, pp. 38 & 90; 1895?: 239, pp.
42 & 92; 1917: 239, pp. 39 &
88. Square garden with mother receiving
1/4 and the rest being divided into four congruent parts.
Cassell's. 1881.
P. 90: The divided farm. =
Manson, 1911, pp. 136-137. = Magician's
Own Book, prob. 3.
Lemon. 1890.
The
divided garden, no. 259, pp. 38 & 107.
= Magician's Own Book, prob. 3.
Geometrical
puzzle, no. 413, pp. 55 & 113 (= Sphinx, no. 556, pp. 76 & 116). Quadrisect
L-tromino.
Hoffmann. 1893.
Chap. X, no. 41: The divided farm, pp. 352‑353 & 391
= Hoffmann‑Hordern, p. 250.
= Magician's Own Book, prob. 3.
[One of the trees is invisible in the original problem, but Hoffmann-Hordern
has added it, in a more symmetric pattern than in Magician's Own Book.]
Loyd. Origin of a famous puzzle -- No. 18: An ancient puzzle. Tit‑Bits 31 (13 Feb &
6 Mar 1897) 363
& 419. Nearly 50 years ago he was told of the
quadrisection of 3/4 of a square, but drew the mitre shape
instead of the L‑tromino. See 6.AW.1.
Clark. Mental Nuts. 1897, no.
73; 1904, no. 31. Dividing the land. Quadrisect an L‑tromino. 1904 also has the mitre -- see 6.AW.1.
Benson. 1904.
The farmer's puzzle, p. 196.
Quadrisect an L‑tromino.
Wehman. New Book of 200 Puzzles. 1908.
The
divided garden, p. 17. = Magician's Own
Book, prob. 3
Puzzle
of the two fathers, p. 43. = Magician's
Own Book, prob. 28.
Puzzle
of the four tenants, p. 46. =
Magician's Own Book, prob. 22.
Dudeney. Some much‑discussed puzzles. Op. cit. in 2. 1908. Land in shape of
an L‑tromino to be
quadrisected. He says this is supposed
to have been invented by Lord Chelmsford (Sir F. Thesiger), who died in 1878 --
see Charades, Enigmas, and Riddles (1860).
But cf Les Amusemens.
M. Adams. Indoor Games. 1912. The clever farmer,
pp. 23‑25. Dissect L‑tromino into four congruent pieces.
Blyth. Match-Stick Magic.
1921. Dividing the inheritance,
pp. 20-21. Usual quadrisection of L-tromino set out with matchsticks.
Collins. Book of Puzzles. 1927. The surveyor's
puzzle, pp. 2-3. Quadrisect 3/4
of a square, except the deleted
1/4 is in the centre, so we are
quadrisecting a hollow square -- cf Mittenzwey,
The Bile Beans Puzzle Book. 1933.
No.
22: Paper squares. Quadrisect a
P-pentomino into P-pentominoes. One
solution given, I find another. Are
there more? How about quadrisecting
into congruent pentominoes? Which
pentominoes can be quadrisected into four copies of themself?
No.
41: Five lines. Consider a 5 x 5
square and delete the corners and centre. Quadrisect into congruent pentominoes. One solution given. I
find three more. Are there more? One can extend this to consider
quadrisecting the 5 x 5 with just the centre removed into congruent
hexominoes. I find seven ways.
Depew. Cokesbury Game Book.
1939. A plot of ground, p.
227. 3/4 of XX
a
square to be quadrisected, but the shape is as shown at the right. XXX
X XX
XXXX
Ripley's Puzzles and Games. 1966.
Pp. 18 & 19, item 8. Divide
an L-tromino into eight congruent pieces.
F. Göbel. Problem 1771: The L‑shape dissection
problem. JRM 22:1 (1990) 64‑65. The
L‑tromino can be dissected into
2, 3, or 4 congruent parts. Can it be divided into 5 congruent parts?
Rowan Barnes-Murphy. Monstrous Mysteries. Piccolo, 1982. Apple-eating monsters, pp. 40 & 63. Trisect into equal parts, the shape consisting of a 2 x 4
rectangle with a 1 x 1 square attached to one of the central
squares of the long side. [Actually,
this can be done with the square attached to any of the squares, though if it
is attached to the end of the long side, the resulting pieces are straight
trominoes.]
6.F.5. OTHER DISSECTIONS INTO POLYOMINOES
Catel. Kunst-Cabinet. 1790.
Das
Zakk- und Hakenspiel, p. 10 & fig. 11 on plate 1. 4 Z‑pentominoes and
4 L‑tetrominoes make a
6 x 6 square.
Die
zwolf Winkelhaken, p. 11 & fig. 26 on plate 1. 8 L‑pentominoes and
4 L‑hexominoes make a
8 x 8 square.
Bestelmeier. 1801.
Item 61 -- Das Zakken und Hakkenspiel.
As in Catel, p. 10, but not as regularly drawn. Text copies some of Catel.
Manuel des Sorciers. 1825.
Pp. 203-204, art. 20. ??NX Use four L-trominoes to make a 3 x 4 rectangle or a 4 x 4 square with four corners deleted.
Family Friend 3 (1850) 90 &
121. Practical puzzle -- No. XIII. 4 x 4
square, with 12 trees in the corners, centres of sides and four at the
centre of the square, to be divided into 4 congruent parts each with 3 trees. Solution uses 4 L-tetrominoes. The same problem is repeated as Puzzle 17 --
Twelve-hole puzzle in (1855) 339 with solution in (1856) 28.
Magician's Own Book. 1857.
Prob. 14: The square and circle puzzle, pp. 270 & 295. Same as Family Friend. = Book of 500 Puzzles, 1859, prob. 14, pp.
84 & 109. = Boy's Own Conjuring
book, 1860, prob. 13, pp. 231-232 & 257.
c= Illustrated Boy's Own Treasury, 1860, prob. 8, pp. 396 &
437. c= Hanky Panky, 1872, A square of
four pieces, p. 117.
Landells. Boy's Own Toy-Maker. 1858.
Pp. 146-147. Identical to Family
Friend.
Leske. Illustriertes Spielbuch für Mädchen. 1864?
Prob.
584-2, pp. 286 & 404.
4 Z‑pentominoes to make a Greek cross. (Also entered in 6.F.3.)
Prob.
584-3, pp. 286 & 404. 4 L-tetrominoes to make a square.
Prob.
584-5, pp. 286 & 404.
8 L‑pentominoes and 4 L‑hexominoes
make a 8 x 8 square. Same as Catel,
but diagram is inverted.
Prob.
584-7, pp. 287 & 405. 4 Z‑pentominoes and 4 L‑tetrominoes
make a 6 x 6 square. Same as Catel, but
diagram is inverted.
Mittenzwey. 1880.
Prob.
174, pp. 33 & 85; 1895?: 199, pp.
38 & 87; 1917: 199, pp. 35 &
84. 4 Z pentominoes to make a (Greek) cross. (Also entered in 6.F.5.)
Prob.
186, pp. 35 & 88; 1895?: 211, pp.
40 & 90; 1917: 211, pp. 36 &
87. 4 x 4 square into 4
L-tetrominoes.
Prob.
187, pp. 35 & 88; 1895?: 212, pp.
40 & 90; 1917: 212, pp. 36 &
87. 6 x 6 square into 4 Z‑pentominoes
and 4 L‑tetrominoes, as in Catel, p. 10.
Prob.
215, pp. 38 & 90; 1895?: 240, pp.
42 & 92; 1917: 240, pp. 39 &
88. Square garden with 12 trees
quadrisected into four L-tetrominoes.
S&B, p. 20, shows a 7 piece
cross dissection into 3 Zs,
2 Ls and 2 straights, from c1890.
Hoffmann. 1893.
Chap. X, no. 37: The orchard puzzle, pp. 350 & 390 = Hoffmann-Hordern, pp. 247, with
photo. Same as Family Friend 3. Photo on p. 247 shows St. Nicholas Puzzle
Card, © 1892 in the USA.
Tom Tit, vol 3. 1893.
Les quatre Z et des quatre L, pp. 181-182. = K, No. 27: The four Z's
and the four L's, pp. 70‑71. = R&A, Squaring the
L's and Z's,
p. 102. 6 x 6 square as in Catel, p. 10.
Sphinx. 1895.
The Maltese cross, no. 181, pp. 28 & 103. Make a Maltese cross (actually a Greek cross of five equal
squares) from 4 P-pentominoes. Also:
quadrisect a P‑pentomino.
Wehman. New Book of 200 Puzzles. 1908.
The square and circle puzzle, p. 5.
= Family Friend.
Burren Loughlin &
L. L. Flood. Bright-Wits Prince of Mogador. H. M. Caldwell Co., NY, 1909.
The Zoltan's orchard, pp. 24-28 & 64. = Family Friend.
Anon. Prob. 84. Hobbies 31 (No.
799) (4 Feb 1911) 443. Use at least one
each of: domino; L-tetromino; P and X pentominoes to
make the smallest possible square Due
to ending of this puzzle series, no solution ever appeared. I find numerous solutions for 5 x 5,
6 x 6, 8 x 8, of which the first is easily seen to be the
smallest possibility.
A. Neely Hall. Carpentry & Mechanics for Boys. Lothrop, Lee & Shepard, Boston, nd
[1918]. The square puzzle, pp. 20‑21. 7 x 7
square cut into 1 straight tromino, 1 L‑tetromino and
7 L‑hexominoes.
Collins. Book of Puzzles. 1927. The surveyor's
puzzle, pp. 2-3. Quadrisect 3/4
of a square, except the deleted
1/4 is in the centre, so we are
quadrisecting a hollow square.
Arthur Mee's Children's
Encyclopedia 'Wonder Box'. The
Children's Encyclopedia appeared in 1908, but versions continued until the
1950s. This looks like 1930s?? 4 Z‑pentominoes and
4 L‑tetrominoes make a
6 x 6 square and a 4 x 9
rectangle.
W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. All square, pp. 42 & 129. Make a
6 x 6 square from the staircase
hexomino, 2 Y-pentominoes, an N‑tetromino, an L-tetromino
and 3 T-tetrominoes. None of the pieces is turned over in the solution,
though this restriction is not stated.
Piet Hein invented the Soma Cube
in 1936. (S&B, pp. 40‑41.) ??Is there any patent??
M. Gardner. SA (Sep 1958) = 2nd Book, Chap 6.
Richard K. Guy. Loc. cit. in 5.H.2, 1960. Pp. 150-151 discusses cubical solutions
-- 234
found so far. He proposes the
'bath' shape -- a 5 x 3 x 2 cuboid with a 3 x 1 x 1 hole in the top
layer. In a 1985 letter, he said that
O'Beirne had introduced the Soma to him and his family. in 1959 and they found
234 solutions before Mike Guy went to Cambridge -- see below.
P. Hein, et al. Soma booklet. Parker Bros., 1969, 56pp.
Asserts there are 240 simple solutions and 1,105,920
total solutions, found by J. H. Conway & M. J. T. Guy with a a
computer (but cf Gardner, below) and by several others. [There seem to be several versions of this
booklet, of various sizes.]
Thomas V. Atwater, ed. Soma Addict. 4 issues, 1970‑1971, produced by Parker Brothers. (Gardner, below, says only three issues
appeared.) ??NYS -- can anyone provide
a set or photocopies??
M. Gardner. SA (Sep 1972) c= Knotted, chap. 3. States there are 240 solutions for the
cube, obtained by many programs, but first found by J. H. Conway & M. J. T.
Guy in 1962, who did not use a computer, but did it by hand "one wet
afternoon". Richard Guy's 1985
letter notes that Mike Guy had a copy of the Guy family's 234 solutions with
him.
SOMAP ??NYS -- ??details.
(Schaaf III 52)
Winning Ways, 1982, II, 802‑803
gives the SOMAP.
Jon Brunvall et al. The computerized Soma Cube. Comp. & Maths. with Appl. 12B:1/2
(1986 [Special issues 1/2
& 3/4 were separately printed as: I. Hargittai, ed.; Symmetry -- Unifying
Human Understanding; Pergamon, 1986.] 113‑121. They cite Gardner's 2nd Book which says the number of solutions
is unknown and they use a computer to find them.
See
also 6.N, 6.U.2, 6.AY.1 and 6.BJ. The
predecessors of these puzzles seem to be the binomial and trinomial cubes
showing (a+b)3 and
(a+b+c)3. I have an
example of the latter from the late 19C.
Here I will consider only cuts parallel to the cube faces -- cubes with
cuts at angles to the faces are in 6.BJ.
Most of the problems here involve several types of piece -- see 6.U.2
for packing with one kind of piece.
Catel. Kunst-Cabinet. 1790. Der algebraische Würfel, p. 6 & fig 50
on plate II. Shows a binomial
cube: (a + b)3 = a3 + 3a2b
+ 3ab2 + b3.
Bestelmeier. 1801.
Item 309 is a binomial cube, as in Catel. "Ein zerschnittener Würfel, mit welchem die Entstehung eines
Cubus, dessen Seiten in 2 ungleiche Theile a + b getheilet ist,
gezeigt ist."
Hoffmann. 1893.
Chap. III, no. 39: The diabolical cube, pp. 108 & 142
= Hoffmann-Hordern, pp. 108-109, with photos. 6: 0, 1, 1, 1, 1, 1, 1,
i.e. six pieces of volumes 2, 3,
4, 5, 6, 7. Photos on p. 108 shows Cube
Diabolique and its box, by Watilliaux, dated 1874-1895.
J. G.‑Mikusiński. French patent. ??NYS -- cited by Steinhaus.
H. Steinhaus. Mikusiński's
Cube. Mathematical Snapshots. Not in Stechert, 1938, ed. OUP, NY:
1950: pp. 140‑142 & 263;
1960, pp. 179‑181 & 326;
1969 (1983): pp. 168-169 & 303.
John Conway. In an email of 7 Apr 2000, he says he
developed the dissection of the 3 x 3 x
3 into
3 1 x 1 x 1 and
6 1 x 2 x 2 in c1960 and then adapted it to the 5 x 5 x 5 into
3 1 x 1 x 3, 1 2 x 2 x 2,
1 1 x 2 x 2 and
13 1 x 2 x 4 and the
5 x 5 x 5 into 3 1 x 1 x 3 and
29 1 x 2 x 2. He says his first publication of it was in
Winning Ways, 1982 (cf below).
Jan Slothouber &
William Graatsma. Cubics. Octopus Press, Deventer, Holland, 1970. ??NYS.
3 x 3 x 3 into 3 1
x 1 x 1 and 6 1 x 2 x 2. [Jan de Geus has sent a photocopy of some of
this but it does not cover this topic.]
M. Gardner. SA (Sep 1972) c= Knotted, chap. 3. Discusses Hoffmann's Diabolical Cube and
Mikusiński's cube. Says he has 8
solutions for the first and that there are just 2
for the second. The Addendum
reports that Wade E. Philpott showed there are just 13 solutions of the
Diabolical Cube. Conway has confirmed
this. Gardner briefly describes the
solutions. Gardner also shows the Lesk
Cube, designed by Lesk Kokay (Mathematical Digest [New Zealand] 58 (1978)
??NYS), which has at least 3 solutions.
D. A. Klarner. Brick‑packing puzzles. JRM 6 (1973) 112‑117. Discusses 3 x 3 x 3 into
3 1 x 1 x 1
and 6 1 x 2 x 2 attributed to
Slothouber‑Graatsma;
Conway's 5 x 5 x 5 into
3 1 x 1 x 3
and 29 1 x 2 x 2; Conway's 5 x 5 x 5 into
3 1 x 1 x 3, 1 2
x 2 x 2, 1 1 x 2 x
2 and
13 1 x 2 x 4. Because of the attribution to Slothouber
& Graatsma and not knowing the date of Conway's work, I had generally
attributed the 3 x 3 x 3 puzzle to them and Stewart Coffin followed
this in his book. However, it now seems
that it really is Conway's invention and I must apologize for misleading
people.
Leisure Dynamics, the US
distributor of Impuzzables, a series of
6 3 x 3 x 3 cube dissections identified by colours,
writes that they were invented by Robert Beck, Custom Concepts Inc.,
Minneapolis. However, the Addendum to
Gardner, above, says they were designed by Gerard D'Arcey.
Winning Ways. 1982.
Vol. 2, pp. 736-737 & 801.
Gives the 3 x 3 x 3 into
3 1 x 1 x 1 and
6 1 x 2 x 2 and the
5 x 5 x 5
into 3 1 x 1 x
3, 1
2 x 2 x 2, 1 1 x 2 x 2
and 13 1 x 2 x
4, which is called
Blocks-in-a-Box. No mention of the
other 5 x 5 x 5. Mentions Foregger & Mather, cf in 6.U.2.
Michael Keller. Polycube update. World Game Review 4 (Feb 1985) 13. Reports results of computer searches for solutions. Hoffmann's Diabolical Cube has 13;
Mikusinski's Cube has 2; Soma Cube has 240; Impuzzables: White -- 1;
Red -- 1; Green -- 16; Blue -- 8; Orange -- 30; Yellow --
1142.
Michael Keller. Polyform update. World Game Review 7 (Oct 1987) 10‑13. Says that Nob Yoshigahara has solved a
problem posed by O'Beirne: How many
ways can 9 L‑trominoes
make a cube? Answer is 111.
Gardner, Knotted, chap. 3, mentioned this. Says there are solutions with
n L‑trominoes and 9‑n
straight trominoes for n ¹ 1
and there are 4 solutions for n = 0. Says the Lesk Cube
has 4
solutions. Says Naef's Gemini
Puzzle was designed by Toshiaki Betsumiya.
It consists of the 10 ways to join two 1 x 2 x 2 blocks.
H. J. M. van Grol. Rik's Cube Kit -- Solid Block Puzzles. Analysis of all 3 x 3 x 3 unit solid
block puzzles with non‑planar 4‑unit
and 5‑unit shapes. Published by the author, The Hague, 1989,
16pp. There are 3
non‑planar tetracubes and
17 non‑planar
pentacubes. A 3 x 3 x 3 cube will
require the 3 non‑planar tetracubes and
3 of the non‑planar
pentacubes -- assuming no repeated pieces.
He finds 190 subsets which can form cubes, in 1
to 10 different ways.
Nob Yoshigahara. (Title in Japanese: (Puzzle in Wood)). H. Tokuda, Sowa Shuppan, Japan, 1987. Pp. 68-69 is a 3^3 designed by Nob
-- 6: 01005.
6.G.2. DISSECTION OF 63 INTO 33, 43 AND 53, ETC.
H. W. Richmond. Note 1672:
A geometrical problem. MG 27 (No.
275) (Jul 1943) 142. AND Note 1704:
Solution of a geometrical problem (Note 1672). MG 28 (No. 278) (Feb 1944) 31‑32. Poses the problem of making such a dissection, then gives a
solution in 12 pieces: three 1 x 3 x 3;
4 x 4 x 4; four 1 x 5 x 5; 1 x 4 x 4;
two 1 x 1 x 2 and a
V‑pentacube.
Anon. [= John Leech, according
to Gardner, below]. Two dissection
problems, no. 2. Eureka 13 (Oct 1950) 6 & 14 (Oct 1951)
23. Asks for such a dissection using at
most 10 pieces. Gives an 8 piece solution
due to R. F. Wheeler. [Cundy &
Rollett; Mathematical Models; 2nd ed., pp. 203‑205, say Eureka is the
first appearance they know of this problem.
See Gardner, below, for the identity of Leech.]
Richard K. Guy. Loc. cit. in 5.H.2, 1960. Mentions the 8 piece solution.
J. H. Cadwell. Some dissection problems involving sums of
cubes. MG 48 (No. 366) (Dec 1964)
391‑396. Notes an error in Cundy
& Rollett's account of the Eureka problem.
Finds examples for 123
+ 13 = 103 + 93 with 9 pieces and 93 = 83 + 63 + 13 with 9 pieces.
J. H. Cadwell. Note 3278:
A three‑way dissection based on Ramanujan's number. MG 54 (No. 390) (Dec 1970) 385‑387. 7 x 13 x 19
to 103 + 93 and
123 + 13
using 12 pieces.
M. Gardner. SA (Oct 1973) c= Knotted, chap. 16. He says that the problem was posed by John
Leech. He gives Wheeler's initials as
E. H. ?? He says that J. H.
Thewlis found a simpler 8‑piece solution, further simplified by T. H.
O'Beirne, which keeps the
4 x 4 x 4
cube intact. This is shown in Gardner. Gardner also shows an 8‑piece solution
which keeps the 5 x 5 x 5 intact, due to E. J. Duffy, 1970. O'Beirne showed that an 8‑piece
dissection into blocks is impossible and found a 9‑block solution in
1971, also shown in Gardner.
Harry Lindgren. Geometric Dissections. Van Nostrand, Princeton, 1984. Section 24.1, pp. 118‑120 gives
Wheeler's solution and admires it.
Richard K. Guy, proposer; editors & Charles H. Jepson [should be
Jepsen], partial solvers. Problem
1122. CM 12 (1987) 50 &
13 (1987) 197‑198. Asks
for such dissections under various conditions, of which (b) is the form given
in Eureka. Eight pieces is minimal in
one case and seems minimal in two other cases.
Eleven pieces is best known for the first case, where the pieces must be
blocks, but this appears to be the problem solved by O'Beirne in 1971, reported
in Gardner, above.
Charles H. Jepsen. Additional comment on Problem 1122. CM 14 (1988) 204‑206. Gives a ten piece solution of the first
case.
Chris Pile. Cube dissection. M500 134 (Aug 1993) 2-3.
He feels the 1 x 1 x 2 piece occurring in Cundy & Rollett is too
small and he provides another solution with 8 pieces, the smallest of which
contains 8 unit cubes. Asks how uniform
the piece sizes can be.
6.G.3. DISSECTION OF A DIE INTO NINE 1 x 1 x 3
Hoffmann. 1893.
Chap. III, no. 17: The "Spots" puzzle, pp. 98‑99 &
130‑131 = Hoffmann‑Hordern, pp. 90-91, with photo. Says it is made by Wolff & Son. Photo on p. 91 shows an example made by E.
Wolff & Son, London.
Benson. 1904.
The spots puzzle, pp. 203‑204.
As in Hoffmann.
Collins. Book of Puzzles. 1927. Pp. 131‑134:
The dissected die puzzle. The solution
is different than Hoffmann's.
Rohrbough. Puzzle Craft. 1932. P. 21 shows a
dissected die, but with no text. The
picture is the same as in Hoffmann's solution.
Slocum. Compendium.
Shows Diabolical Dice from Johnson Smith catalogue, 1935.
Harold Cataquet. The Spots puzzle revisited. CFF 33 (Feb 1994) 20-21. Brief discussion of two versions.
David Singmaster. Comment on the "Spots"
puzzle. 29 Sep 1994, 2pp. Letter in response to the above. I note that there is no standard pattern for
a die other than the opposite sides adding to seven. There are 23 =
8 ways to orient the spots forming 2, 3, and 6. There are two handednesses, so there are 16 dice altogether. (This was pointed out to me perhaps 10 years
before by Richard Guy and Ray Bathke. I
have since collected examples of all 16 dice.)
However, Ray Bathke showed me Oriental dice with the two spots of the 2
placed horizontal or vertically rather than diagonally, giving another 16 dice
(I have 5 types), making 32 dice in all.
A die can be dissected into 9 1 x 1 x 3
pieces in 6 ways if the layers have to alternate in direction, or in 21
ways in general. I then pose a number
of questions about such dissections.
6.G.4. USE OF OTHER POLYHEDRAL PIECES
S&B. 1986.
P. 42 shows Stewart Coffin's 'Pyramid Puzzle' using pieces made from
truncated octahedra and his 'Setting Hen' using pieces made from rhombic
dodecahedra. Coffin probably devised
these in the 1960s -- perhaps his book has some details of the origins of these
ideas. ??check.
Mark Owen & Matthew
Richards. A song of six splats. Eureka 47 (1987) 53‑58. There are six ways to join three truncated
octahedra. For reasons unknown, these
are called 3‑splats. They give
various shapes which can and which cannot be constructed from the six 3‑splats.
Georg Pick. Geometrisches zur Zahlenlehre. Sitzungsberichte des deutschen
naturwissenschaftlich‑medicinischen Vereines für Böhmen "Lotos"
in Prag (NS) 19 (1899) 311‑319.
Pp. 311‑314 gives the proof, for an oblique lattice. Pp. 318‑319 gives the extension to
multiply connected and separated regions.
Rest relates to number theory.
[I have made a translation of the material on Pick's Theorem.]
Charles Howard Hinton. The Fourth Dimension. Swan Sonnenschein & Co., London,
1906. Metageometry, pp. 46-60. [This material is in Speculations on the Fourth
Dimension, ed. by R. v. B. Rucker; Dover, 1980, pp. 130-141. Rucker says the book was published in 1904,
so my copy may be a reprint??] In the
beginning of this section, he draws quadrilateral shapes on the square lattice
and determines the area by counting points, but he counts I + E/2 + C/4, which works for quadrilaterals but is not valid in general.
H. Steinhaus. O mierzeniu pól płaskich. Przegląd Matematyczno‑Fizyczny 2
(1924) 24‑29. Gives a version of
Pick's theorem, but doesn't cite Pick. (My
thanks to A. Mąkowski for an English summary of this.)
H. Steinhaus. Mathematical Snapshots. Stechert, NY, 1938, pp. 16-17 &
132. OUP, NY: 1950: pp. 76‑77 & 260 (note 77); 1960: pp. 99‑100 & 324 (note
95); 1969 (1983): pp. 96‑97 &
301 (note 107). In 1938 he simply notes
the theorem and gives one example. In
1950, he outlines Pick's argument.
He refers to Pick's paper, but
in "Ztschr. d. Vereins 'Lotos' in Prag". Steinhaus also cites his own paper, above.
J. F. Reeve. On the volume of lattice polyhedra. Proc. London Math. Soc. 7 (1957) 378‑395. Deals with the failure of the obvious form
of Pick's theorem in 3‑D and finds a valid generalization.
Ivan Niven & H. S.
Zuckerman. Lattice points and polygonal
area. AMM 74 (1967) 1195‑1200. Straightforward proof. Mention failure for tetrahedra.
D. W. De Temple & J. M.
Robertson. The equivalence of Euler's
and Pick's theorems. MTr 67 (1974)
222‑226. ??NYS.
W. W. Funkenbusch. From Euler's formula to Pick's formula using
an edge theorem. AMM 81 (1974) 647‑648. Easy proof though it could be easier.
R. W. Gaskell, M. S. Klamkin
& P. Watson. Triangulations and
Pick's theorem. MM 49 (1976) 35‑37. A bit roundabout.
Richard A. Gibbs. Pick iff Euler. MM 49 (1976) 158. Cites
DeTemple & Robertson and observes that both Pick and Euler can be proven
from a result on triangulations.
John Reay. Areas of hex-lattice polygons, with short
sides. Abstracts Amer. Math. Soc. 8:2
(1987) 174, #832-51-55. Gives a formula
for the area in terms of the boundary and interior points and the
characteristic of the boundary, but it is an open question to determine when
this formula gives the actual area.
6.I. SYLVESTER'S PROBLEM OF COLLINEAR POINTS
If a set of non‑collinear points
in the plane is such that the line through any two points of the set contains a
third point of the set, then the set is infinite.
J. J. Sylvester. Question 11851. The Educational Times 46 (NS, No. 383) (1 Mar 1893) 156.
H. J. Woodall & editorial
comment. Solution to Question
11851. Ibid. (No. 385)
(1 May 1893) 231. A very
spurious solution.
(The above two items appear
together in Math. Quest. with their Sol. Educ. Times 59 (1893) 98‑99.)
E. Melchior. Über Vielseite der projecktiven Ebene. Deutsche Math. 5 (1940) 461‑475. Solution, but in a dual form.
P. Erdös, proposer; R. Steinberg, solver & editorial comment
giving solution of T. Grünwald (later = T. Gallai). Problem 4065. AMM 50
(1943) 65 & 51 (1944) 169‑171.
L. M. Kelly. (Solution.)
In: H. S. M. Coxeter; A problem
of collinear points; AMM 55 (1948) 26‑28. Kelly's solution is on p. 28.
G. A. Dirac. Note 2271:
On a property of circles. MG 36
(No. 315) (Feb 1952) 53‑54.
Replace 'line' by 'circle' in the problem. He shows this is true by inversion. He asks for an independent proof of the result, even for the case
when two, three are replaced by three, four.
D. W. Lang. Note 2577:
The dual of a well‑known theorem.
MG 39 (No. 330) (Dec 1955) 314.
Proves the dual easily.
H. S. M. Coxeter. Introduction to Geometry. Wiley, 1961. Section 4.7: Sylvester's problem of collinear points, pp.
65-66. Sketches history and gives
Kelly's proof.
W. O. J. Moser. Sylvester's problem, generalizations and
relatives. In his: Research Problems in Discrete Geometry 1981,
McGill University, Montreal, 1981.
Section 27, pp. 27‑1 -- 27‑14. Survey with 73 references.
(This problem is not in Part 1 of the 1984 ed. nor in the 1986 ed.)
6.J. FOUR BUGS AND OTHER PURSUIT PROBLEMS
The
general problem becomes too technical to remain recreational, so I will not try
to be exhaustive here.
Arthur Bernhart.
Curves
of pursuit. SM 20 (1954) 125‑141.
Curves
of pursuit -- II. SM 23 (1957) 49‑65.
Polygons
of pursuit. SM 24 (1959) 23‑50.
Curves
of general pursuit. SM 24 (1959) 189‑206.
Extensive history and analysis. First article covers one dimensional
pursuit, then two dimensional linear pursuit.
Second article deals with circular pursuit. Third article is the 'four bugs' problem -- analysis of
equilateral triangle, square, scalene triangle, general polygon, Brocard
points, etc. Last article includes such
variants as variable speed, the tractrix, miscellaneous curves, etc.
Mr. Ash, proposer; editorial
note saying there is no solver. Ladies'
Diary, 1748-47 = T. Leybourn, II: 15-17, quest. 310, with
'Solution by ΦIΛΟΠΟΝΟΣ, taken from
Turner's Exercises, where this question was afterwards proposed and answered
...' A fly is constrained to move on
the periphery of a circle. Spider
starts 30o away from the fly, but walks across the circle, always
aiming at the fly. If she catches the
fly 180o from her starting point, find the ratio of their
speeds.
ΦIΛΟΠΟΝΟΣ solves the more
general problem of finding the curve when the spider starts anywhere.
Carlile. Collection.
1793. Prob. CV, p. 62. A dog and a duck are in a circular pond of
radius 40 and they swim at the same speed.
The duck is at the edge and swims around the circumference. The dog starts at the centre and always
swims toward the duck, so the dog and the duck are always on a radius. How far does the dog swim in catching the
duck? He simply gives the result
as 20π. Letting R be the radius of the pond and V be
the common speed, I find the radius of the dog, r, is given by r = R sin Vt/R. Since the angle,
θ, of both the duck and the
dog is given by θ = Vt/R, the polar equation of the dog's path is r = R sin θ and the path is a semicircle whose diameter is the appropriate
radius perpendicular to the radius to the duck's initial position.
Cambridge Math. Tripos examination,
5 Jan 1871, 9 to 12. Problem 16, set by
R. K. Miller. Three bugs in general
position, but with velocities adjusted to make paths similar and keep the
triangle similar to the original.
Lucas. (Problem of three dogs.)
Nouvelle Correspondance Mathématique 3 (1877) 175‑176. ??NYS -- English in Arc., AMM 28 (1921) 184‑185
& Bernhart.
H. Brocard. (Solution of Lucas' problem.) Nouv. Corr. Math. 3 (1877) 280. ??NYS -- English in Bernhart.
Pearson. 1907.
Part II, no. 66: A duck hunt, pp. 66 & 172. Duck swims around edge of pond; spaniel starts for it from the centre at the
same speed.
A. S. Hathaway, proposer and
solver. Problem 2801. AMM 27 (1920) 31 & 28 (1921) 93‑97. Pursuit of a prey moving on a circle. Morley's and other solutions fail to deal
with the case when the velocities are equal.
Hathaway resolves this and shows the prey is then not caught.
F. V. Morley. A curve of pursuit. AMM 28 (1921) 54-61. Graphical solution of Hathaway's problem.
R. C. Archibald [Arc.] & H. P.
Manning. Remarks and historical notes
on problems 19 [1894], 160 [1902], 273 [1909] & 2801 [1920]. AMM 28 (1921) 91-93.
W. W. Rouse Ball. Problems -- Notes: 17: Curves of
pursuit. AMM 28 (1921) 278‑279.
A. H. Wilson. Note 19: A curve of pursuit. AMM 28 (1921) 327.
Editor's note to Prob. 2
(proposed by T. A. Bickerstaff), National Mathematics Magazine (1937/38) 417
cites Morley and Archibald and adds that some authors credit the problem to
Leonardo da Vinci -- e.g. MG (1930-31) 436 -- ??NYS
Nelson F. Beeler & Franklyn
M. Branley. Experiments in Optical
Illusion. Ill. by Fred H. Lyon. Crowell, 1951, An illusion doodle, pp. 68-71,
describes the pattern formed by four bugs starting at the corners of a square,
drawing the lines of sight at (approximately) regular intervals. Putting several of the squares together,
usually with alternating directions of motion, gives a pleasant pattern which
is now fairly common. They call this
'Huddy's Doodle', but give no source.
J. E. Littlewood. A Mathematician's Miscellany. Op. cit. in 5.C. 1953. 'Lion and man',
pp. 135‑136 (114‑117).
The 1986 ed. adds three diagrams and revises the text somewhat. I quote from it. "A lion and a man in a closed circular arena have equal
maximum speeds. What tactics should the
lion employ to be sure of his meal?"
This was "invented by R. Rado in the late thirties" and
"swept the country 25 years later".
[The 1953 ed., says Rado didn't publish it.] The correct solution "was discovered by Professor
A. S. Besicovitch in 1952".
[The 1953 ed. says "This has just been discovered ...; here is the
first (and only) version in print."]
C. C. Puckette. The curve of pursuit. MG 37 (No. 322) (Dec 1953) 256‑260. Gives the history from Bouguer in 1732. Solves a variant of the problem.
R. H. Macmillan. Curves of pursuit. MG 40 (No. 331) (Feb 1956) 1‑4. Fighter pursuing bomber flying in a straight line. Discusses firing lead and acceleration
problems.
Gamow & Stern. 1958.
Homing missiles. Pp. 112‑114.
Howard D. Grossman, proposer; unspecified solver. Problem 66 -- The walk around. In:
L. A. Graham; Ingenious Mathematical Problems and Methods; Dover,
1959, pp. 40 & 203‑205. Four
bugs -- asserts Grossman originated the problem.
I. J. Good. Pursuit curves and mathematical art. MG 43 (No. 343) (Feb 1959) 34‑35. Draws tangent to the pursuit curves in an
equilateral triangle and constructs various patterns with them. Says a similar but much simpler pattern was
given by G. B. Robison; Doodles; AMM 61 (1954) 381-386, but Robison's doodles
are not related to pursuit curves, though they may have inspired Good to use
the pursuit curves.
J. Charles Clapham. Playful mice. RMM 10 (Aug 1962) 6‑7.
Easy derivation of the distance travelled for n bugs at corners of a
regular n‑gon. [I don't see this result in Bernhart.]
C. G. Paradine. Note 3108:
Pursuit curves. MG 48 (No. 366)
(Dec 1964) 437‑439. Says Good
makes an error in Note 3079. He shows
the length of the pursuit curve in the equilateral triangle is ⅔
of the side and describes the curve as an equiangular spiral. Gives a simple proof that the length of the
pursuit curve in the regular n‑gon
is the side divided by (1 ‑ cos
2π/n).
M. S. Klamkin & D. J.
Newman. Cyclic pursuit or "The
three bugs problem". AMM 78 (1971)
631‑639. General treatment. Cites Bernhart's four SM papers and some of
the history therein.
P. K. Arvind. A symmetrical pursuit problem on the sphere
and the hyperbolic plane. MG 78 (No.
481) (Mar 1994) 30-36. Treats the n
bugs problems on the surfaces named.
Barry Lewis. A mathematical pursuit. M500 170 (Oct 1999) 1-8. Starts with equilateral triangular case,
giving QBASIC programs to draw the curves as well as explicit solutions. Then considers regular n-gons.
Then considers simple pursuit, one beast pursuing another while the
other moves along some given path.
Considers the path as a straight line or a circle. For the circle, he asserts that the analytic
solution was not determined until 1926, but gives no reference.
6.K. DUDENEY'S SQUARE TO TRIANGLE DISSECTION
Dudeney. Weekly Dispatch (6 Apr, 20 Apr, 4 May, 1902)
all p. 13.
Dudeney. The haberdasher's puzzle. London Mag. 11 (No. 64) (Nov 1903) 441 &
443. (Issue with solution not found.)
Dudeney. Daily Mail (1 & 8 Feb 1905) both p.
7.
Dudeney. CP.
1907. Prob. 25: The
haberdasher's puzzle, pp. 49‑50 & 178‑180.
Western Puzzle Works, 1926
Catalogue. No. 1712 -- unnamed, but
shows both the square and the triangle.
Apparently a four piece puzzle.
M. Adams. Puzzle Book. 1939. Prob. C.153:
Squaring a triangle, pp. 162 & 189.
Asserts that Dudeney's method works for any triangle, but his example is
close to equilateral and I recall that this has been studied and only certain
shapes will work??
Robert C. Yates. Geometrical Tools. (As: Tools; Baton Rouge,
1941); revised ed., Educational
Publishers, St. Louis, 1949. Pp.
40-41. Extends to dissecting a
quadrilateral to a specified triangle and gives a number of related problems.
Two
ladders are placed across a street, each reaching from the base of the house on
one side to the house on the other side.
The
simple problem gives the heights
a, b that the ladders reach on the walls. If the height of the crossing is
c, we easily get 1/c = 1/a + 1/b. NOTATION -- this problem will be denoted by (a, b).
The
more common and more complex problem is where the ladders have lengths a
and b, the height of their crossing is
c and one wants the width d of
the street. If the heights of the
ladder ends are x, y,
this situation gives x2
‑ y2 = a2 ‑ b2 and
1/x + 1/y = 1/c which leads to a
quartic and there seems to be no simple solution. NOTATION -- this will be denoted
(a, b, c).
Mahavira. 850.
Chap. VII, v. 180-183, pp. 243-244.
Gives the simple version with a modification -- each ladder reaches from
the top of a pillar beyond the foot of the other pillar. The ladder from the top of pillar Y
(of height y) extends by
m beyond the foot of pillar X
and the ladder from the top of pillar
X (of height x)
reaches n beyond the foot of pillar Y.
The pillars are d apart.
Similar triangles then yield:
(d+m+n)/c = (d+n)/x + (d+m)/y and one can compute the various distances along the ground. He first does problems with m = n = 0,
which are the simple version of the problem, but since d is
given, he also asks for the distances on the ground.
v.
181. (16, 16) with d = 16.
v.
182. (36, 20) with d = 12.
v.
183. x, y, d, m, n =
12, 15, 4, 1, 4.
Bhaskara II. Lilavati.
1150. Chap. VI, v. 160. In Colebrooke, pp. 68‑69. (10, 15).
(= Bijaganita, ibid., chap. IV, v. 127, pp. 205‑206.)
Fibonacci. 1202.
Pp. 397‑398 (S: 543-544) looks like a crossed ladders problem but
is a simple right triangle problem.
Pacioli. Summa.
1494. Part II.
F.
56r, prob. 48. (4, 6).
F.
60r, prob. 64. (10, 15).
Hutton. A Course of Mathematics. 1798?
Prob. VIII, 1833: 430; 1857: 508.
A ladder 40 long in a roadway can reach 33
up one side and, from the same point, can reach 21
up the other side. How wide is
the street? This is actually a simple
right triangle problem.
Victor Katz reports that
Hutton's problem, with values 60; 37,
23 appears in a notebook of Benjamin
Banneker (1731-1806).
Loyd. Problem 48: A jubilee problem.
Tit‑Bits 32 (21 Aug,
11 & 25 Sep 1897) 385, 439 & 475.
Given heights of the ladder ends above ground and the width of the
street, find the height of the intersection.
However one wall is tilted and the drawing has it covered in decoration
so one may interpret the tilt in the wrong way.
Jno. A. Hodge, proposer; G. B. M. Zerr, solver. Problem 131. SSM 8 (1908) 786 & 9 (1909) 174‑175. (100, 80, 10).
W. V. N. Garretson,
proposer; H. S. Uhler, solver. Problem 2836. AMM 27 (1920) & 29 (1922) 181. (40, 25, 15).
C. C. Camp, proposer; W. J. Patterson & O. Dunkel,
solvers. Problem 3173. AMM 33 (1926) 104 & 34 (1927) 50‑51. General solution.
Morris Savage, proposer; W. E. Batzler, solver. Problem 1194. SSM 31 (1931) 1000
& 32 (1932) 212. (100, 80, 10).
S. A. Anderson, proposer; Simon Vatriquant, solver. Problem E210. AMM 43 (1936) 242 & 642‑643. General solution in integers.
C. R. Green, proposer; C. W. Trigg, solver. Problem 1498. SSM 37 (1937) 484 & 860‑861. (40, 30, 15). Trigg cites
Vatriquant for smallest integral case.
A. A. Bennett, proposer; W. E. Buker, solver. Problem E433. AMM 47 (1940) 487 & 48 (1941) 268‑269. General solution in integers using four
parameters.
J. S. Cromelin, proposer; Murray Barbour, solver. Problem E616 -- The three ladders. AMM 51 (1944) 231 & 592. Ladders of length 60 & 77
from one side. A ladder from the
other side crosses them at heights
17 & 19.
How long is the third ladder and how wide is the street?
Geoffrey Mott-Smith. Mathematical Puzzles for Beginners and
Enthusiasts. (Blakiston, 1946); revised ed., Dover, 1954. Prob. 103: The extension ladder, pp. 58-59
& 176‑178. Complex problem
with three ladders.
Arthur Labbe, proposer; various solvers. Problem 25 -- The two ladders.
Sep 1947 [date given in Graham's second book, cited at 1961]. In:
L. A. Graham; Ingenious Mathematical Problems and Methods; Dover, 1959,
pp. 18 & 116‑118. (20, 30,
8).
M. Y. Woodbridge, proposer and
solver. Problem 2116. SSM 48 (1948) 749 & 49 (1949) 244‑245. (60, 40, 15). Asks for a trigonometric solution. Trigg provides a list of early references.
Robert C. Yates. The ladder problem. SSM 51 (1951) 400‑401. Gives a graphical solution using hyperbolas.
G. A. Clarkson. Note 2522:
The ladder problem. MG 39 (No.
328) (May 1955) 147‑148. (20, 30,
10). Let A = Ö(a2 ‑ b2) and set
x = A sec t, y = A tan t. Then
cos t + cot t = A and he gets a trigonometrical solution. Another method leads to factoring the
quartic in terms of a constant k whose square satisfies a cubic.
L. A. Graham. The Surprise Attack in Mathematical
Problems. Dover, 1968. Problem 6: Searchlight on crossed ladders,
pp. 16-18. Says they reposed Labbe's
Sep 1947 problem in Jun 1961. Solution
by William M. Dennis which is the same trigonometric method as Clarkson.
H. E. Tester. Note 3036:
The ladder problem. A solution
in integers. MG 46 (No. 358) (Dec 1962)
313‑314. A four parameter,
incomplete, solution. He finds the
example (119, 70, 30).
A. Sutcliffe. Complete solution of the ladder problem in
integers. MG 47 (No. 360) (May 1963)
133‑136. Three parameter
solution. First few examples are: (119, 70, 30); (116, 100, 35);
(105, 87, 35). Simpler than
Anderson and Bennett/Buker.
Alan Sutcliffe, proposer; Gerald J. Janusz, solver. Problem 5323 -- Integral solutions of the
ladder problem. AMM 72 (1965) 914 &
73 (1966) 1125-1127. Can the
distance f between the tops of the ladders be integral? (80342, 74226, 18837) has
x = 44758,
y = 32526, d =
66720, f = 67832. This is not known to be the smallest
example.
Anon. A window cleaner's problem.
Mathematical Pie 51 (May 1967) 399.
From a point in the road, a ladder can reach 30 ft up on one side
and 40
ft up on the other side. If the
two ladder positions are at right angles, how wide is the road?
J. W. Gubby. Note 60.3:
Two chestnuts (re-roasted). MG
60 (No. 411) (Mar 1976) 64-65. 1. Given
heights of ladders as a, b, what is the height c of their intersection? Solution:
1/c = 1/a + 1/b or c = ab/(a+b). 2.
The usual ladder problem -- he finds a quartic.
J. Jabłkowski. Note 61:11:
The ladder problem solved by construction. MG 61 (No. 416) (Jun 1977) 138.
Gives a 'neusis' construction.
Cites Gubby.
Birtwistle. Calculator Puzzle Book. 1978.
Prob. 83, A second ladder problem, pp. 58-59 & 115-118. (15, 20, 6). Uses xy as a variable to simplify the quartic for
numerical solution and eventually gets
11.61.
See: Gardner, Circus, p. 266 & Schaaf for more references. ??follow up.
Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. The tangled ladders, pp. 43-44 & 116. (30, 20, 10). Gives answer 12.311857... with no explanation.
A
ladder of length L is placed to just clear a box of width w
and height h at the base of a wall. How high does the ladder reach? Denote this by (w, h, L). Letting x be
the horizontal distance of the foot and
y be the vertical distance of
the top of the ladder, measured from the foot of the wall, we get x2 + y2 = L2 and
(x‑w)(y‑h) = wh,
which gives a quartic in general.
But if w = h, then use of
x + y as a variable reduces the
quartic to a quadratic. In this case,
the idea is old -- see e.g. Simpson.
The
question of determining shortest ladder which can fit over a box of width w
and height h is the same as determining the longest
ladder which will pass from a corridor of width w into another corridor
of width h. See Huntington below and section 6.AG.
Simpson. Algebra.
1745. Section XVIII, prob. XV,
p. 250 (1790: prob. XIX, pp. 272-273).
"The Side of the inscribed Square
BEDF, and the Hypotenuse AC
of a right-angled Triangle
ABC being given; to determine
the other two Sides of the Triangle AB and
BC." Solves "by
considering x + y as one Quantity".
Pearson. 1907.
Part II, no. 102: Clearing the wall, p. 103. For (15, 12, 52), the ladder reaches 48.
D. John Baylis. The box and ladder problem. MTg 54 (1971) 24. (2, 2, 10). Finds the
quartic which he solves by symmetry.
Editorial note in MTg 57 (1971) 13 says several people wrote to say that
use of similar triangles avoids the quartic.
Birtwistle. Math. Puzzles & Perplexities. 1971.
The ladder and the box problem, pp. 44-45. = Birtwistle; Calculator Puzzle Book; 1978; Prob. 53: A ladder
problem, pp. 37 & 96‑98.
(3, 3, 10). Solves by
using x + y - 6 as a variable.
Monte Zerger. The "ladder problem". MM 60:4 (1987) 239‑242. (4, 4, 16).
Gives a trigonometric solution and a solution via two quadratics.
Oliver D. Anderson. Letter.
MM 61:1 (1988) 63. In response
to Zerger's article, he gives a simpler derivation.
Tom Heyes. The old box and ladder problem --
revisited. MiS 19:2 (Mar 1990) 42‑43. Uses a graphic calculator to find roots
graphically and then by iteration.
A. A. Huntington. More on ladders. M500 145 (Jul 1995) 2-5.
Does usual problem, getting a quartic.
Then finds the shortest ladder.
[This turns out to be the same as the longest ladder one can get around
a corner from corridors of widths
w and h, so this problem is
connected to 6.AG.]
David Singmaster. Integral solutions of the ladder over box
problem. In preparation. Easily constructs all the primitive integral
examples from primitive Pythagorean triples.
E.g. for the case of a square box, i.e.
w = h, if X, Y, Z
is a primitive Pythagorean triple, then the corresponding primitive
solution has w = h = XY, x = X (X + Y), y = Y (X + Y), L = Z (X + Y), and
remarkably, x - h = X2, y - w = Y2.
These involve finding the shortest
distance over the surface of a cube or cylinder. I've just added the cylindrical form -- see Dudeney (1926),
Perelman and Singmaster. The shortest
route from a corner of a cube or cuboid to a diagonally opposite corner must
date back several centuries, but I haven't seen any version before 1937! I don't know if other shapes have been done
-- the regular (and other) polyhedra and the cone could be considered.
Two-dimensional
problems are in 10.U.
Loyd. The Inquirer (May 1900).
Gives the Cyclopedia problem.
??NYS -- stated in a letter from Shortz.
Dudeney. Problem 501 -- The spider and the fly. Weekly Dispatch (14 &
28 Jun 1903) both p. 16. 4
side version.
Dudeney. Breakfast table problems, No. 320 -- The
spider and the fly. Daily Mail (18 &
21 Jan 1905) both p. 7. Same
as the above problem.
Dudeney. Master of the breakfast table problem. Daily Mail (1 & 8 Feb 1905) both p.
7. Interview with Dudeney in which he gives
the 5 side version.
Ball. MRE, 4th ed., 1905, p. 66.
Gives the 5 side version, citing the Daily Mail of
1 Feb 1905. He says he heard
a similar problem in 1903 -- presumably Dudeney's first version. In the 5th ed., 1911, p. 73, he attributes
the problem to Dudeney.
Dudeney. CP.
1907. Prob. 75: The spider and
the fly, pp. 121‑122 & 221‑222. 5 side version with discussion of various generalizations.
Dudeney. The world's best problems. 1908.
Op. cit. in 2. P. 786 gives the
five side version.
Sidney J. Miller. Some novel picture puzzles -- No. 6. Strand Mag. 41 (No. 243) (Mar 1911) 372 &
41 (No. 244) (Apr 1911) 506.
Contest between two snails.
Better method uses four sides, similar to Dudeney's version, but with
different numbers.
Loyd. The electrical problem.
Cyclopedia, 1914, pp. 219 & 368 (= MPSL2, prob. 149, pp. 106
& 169 = SLAHP: Wiring the hall, pp.
72 & 114). Same as Dudeney's first,
four side, version. (In MPSL2, Gardner
says Loyd has simplified Dudeney's 5 side problem. More likely(?) Loyd had only seen Dudeney's earlier 4 side
problem.)
Dudeney. MP.
1926. Prob. 162: The fly and the
honey, pp. 67 & 157. (= 536, prob.
325, pp. 112 & 313.)
Cylindrical problem.
Perelman. FFF.
1934. The way of the fly. 1957: Prob. 68, pp. 111‑112 & 117‑118; 1979: Prob. 72, pp. 136 & 142‑144. MCBF: Prob. 72, pp. 134 & 141-142. Cylindrical form, but with different numbers
and arrangement than Dudeney's MP problem.
Haldeman-Julius. 1937.
No. 34: The louse problem, pp. 6 & 22. Room 40 x 20 x 10 with louse at a corner wanting to go to a
diagonally opposite corner. Problem
sent in by J. R. Reed of Emmett, Idaho.
Answer is 50!
M. Kraitchik. Mathematical Recreations, 1943, op. cit. in
4.A.2, chap. 1, prob. 7, pp. 17‑21.
Room with 8 equal routes from spider to fly. (Not in his Math. des Jeux.)
Sullivan. Unusual.
1943. Prob. 10: Why not
fly? Find shortest route from a corner
of a cube to the diagonally opposite corner.
William R. Ransom. One Hundred Mathematical Curiosities. J. Weston Walch, Portland, Maine, 1955. The spider problem, pp. 144‑146. There are three types of path, covering 3, 4 and 5
sides. He determines their
relative sizes as functions of the room dimensions.
Birtwistle. Math. Puzzles & Perplexities. 1971.
Round
the cone, pp. 144 & 195. What is
the shortest distance from a point
P around a cone and back to P?
Answer is "An ellipse", which doesn't seem to answer the
question. If the cone has height H,
radius R and
P is l from the apex, then the slant height L
is Ö(R2 + H2), the angle of the opened out cone is θ = 2πR/L and the required distance is 2l sin θ/2.
Spider
circuit, pp. 144 & 198. Spider is
at the midpoint of an edge of a cube.
He wants to walk on each of the faces and return. What is his shortest route? Answer is "A regular hexagon. (This may be demonstrated by putting a
rubber band around a cube.)"
David Singmaster. The spider spied her. Problem used as: More than one way to catch a fly, The Weekend Telegraph (2 Apr
1984). Spider inside a glass tube, open
at both ends, goes directly toward a fly on the outside. When are there two equally short paths? Can there be more than two shortest routes?
Yoshiyuki Kotani has posed the
following general and difficult problem.
On an a x b x c cuboid, which two points are furthest apart,
as measured by an ant on the surface?
Dick Hess has done some work on this, but I believe that even the case
of square cross-section is not fully resolved.
6.N. DISSECTION OF A 1 x 1 x 2 BLOCK TO A CUBE
W. F. Cheney, Jr.,
proposer; W. R. Ransom; A. H. Wheeler,
solvers. Problem E4. AMM 39 (1932) 489; 40 (1933) 113-114
& 42 (1934) 509-510. Ransom finds a solution in 8
pieces; Wheeler in 7.
Harry Lindgren. Geometric Dissections. Van Nostrand, Princeton, 1964. Section 24.2, p. 120 gives a variant of
Wheeler's solution.
Michael Goldberg. A duplication of the cube by dissection and
a hinged linkage. MG 50 (No. 373)
(Oct 1966) 304‑305. Shows that a
hinged version exists with 10 pieces. Hanegraaf, below, notes that there are actually 12 pieces here.
Anton Hanegraaf. The Delian Altar Dissection. Polyhedral Dissections, Elst, Netherlands,
1989. Surveys the problem, gives a 6
piece solution and a 7 piece hinged solution.
6.O. PASSING A CUBE THROUGH AN EQUAL OR SMALLER CUBE. PRINCE RUPERT'S PROBLEM
The
projection of a unit cube along a space diagonal is a regular hexagon of
side Ö2/Ö3. The largest square inscribable in this hexagon has edge Ö6 - Ö2 =
1.03527618. By passing the larger cube
on a slant to the space diagonal, one can get the larger cube having edge 3Ö2/4 = 1.06066172.
There
are two early attributions of this.
Wallis attributes it to Prince Rupert, but Hennessy says Philip Ronayne
of Cork invented it. I have discovered
a possible connection. Prince Rupert of
the Rhine (1619-1682), nephew of Charles I, was a major military figure of his
time, becoming commander-in-chief of Charles I's armies in the 1640s. In 1648-1649, he was admiral of the King's fleet
and was blockaded with 16 ships in Kinsale Harbor for 20 months. Kinsale is about 20km south of Cork.
Ronayne
wrote an Algebra, of which only a second edition of 1727 is in the BL. Schrek has investigated the family histories
and says Ronayne lived in the early 18C.
This would seem to make him too young to have met Rupert. Perhaps Rupert invented the problem while in
Kinsale and this was conveyed to Ronayne some years later. Does anyone know the dates of Ronayne or of
the 1st ed (Schrek only located the BL example of the 2nd ed)? I cannot find anything on him in Wallis,
May, Poggendorff, DNB, but Google has turned up a reference to a 1917 history
of the family which Schrek cites, but I have not yet tried to find this.
Hennessy's
article says a little about Daniel Voster and details are in Wallis's . His father, Elias (1682 ‑ >1728)
wrote an Arithmetic, of which Wallis lists 30 editions. The BL lists one as late as 1829. The son, Daniel (1705 ‑ >1760) was
a schoolmaster and instrument maker who edited later versions of his father's
arithmetic. The 1750 History of Cork
quoted by Hennessy says the author had seen the cubes with Daniel. Hennessy conjectures that his example was
made specially, perhaps under the direction of a mathematician. It seems likely that Daniel knew Ronayne and
made this example for him.
John Wallis. Perforatio cubi, alterum ipsi aequalem
recipiens. (De Algebra Tractatus; 1685;
Chap. 109) = Opera Mathematica,
vol. II, Oxford, 1693, pp. 470‑471, ??NYS. Cites Rupert as the source of the equal cube version. (Latin and English in Schrek.) Scriba, below, found an errata slip in
Wallis's copy of his Algebra in the Bodleian.
This corrects the calculations, but was published in the Opera, p. 695.
Ozanam‑Montucla. 1778.
Percer un cube d'une ouverture, par laquelle peut passer un autre cube
égal au premier. Prob. 30 & fig.
53, plate 7, 1778: 319-320; 1803:
315-316; 1814: 268-269. Prob. 29, 1840: 137. Equal cubes with diagonal movement.
J. H. van Swinden. Grondbeginsels der Meetkunde. 2nd ed., Amsterdam, 1816, pp. 512‑513,
??NYS. German edition by C. F. A.
Jacobi, as: Elemente der Geometrie,
Friedrich Frommann, Jena, 1834, pp. 394-395. Cites Rupert and Wallis and gives a simple construction, saying
Nieuwland has found the largest cube which can pass through a cube.
Peter Nieuwland. (Finding of maximum cube which passes
through another). In: van Swinden, op. cit., pp. 608‑610; van Swinden‑Jacobi, op. cit. above,
pp. 542-544, gives Nieuwland's proof.
Cundy and Rollett, p. 158, give
references to Zacharias (see below) and to Cantor, but Cantor only cites
Hennessy.
H. Hennessy. Ronayne's cubes. Phil. Mag. (5) 39 (Jan‑Jun 1895) 183‑187. Quotes, from Gibson's 'History of Cork', a
passage taken from Smith's 'History of Cork', 1st ed., 1750, vol. 1, p. 172,
saying that Philip Ronayne had invented this and that a Daniel Voster had made
an example, which may be the example owned by Hennessy. He gives no reference to Rupert. He finds the dimensions.
F. Koch & I. Reisacher. Die Aufgabe, einen Würfel durch einen andern
durchzuschieben. Archiv Math. Physik
(3) 10 (1906) 335‑336. Brief
solution of Nieuwland's problem.
M. Zacharias. Elementargeometrie und elementare
nicht-Euklidische Geometrie in synthetischer Behandlung. Encyklopädie der Mathematischen
Wissenschaften. Band III, Teil 1,
2te Hälfte. Teubner, Leipzig,
1914-1931. Abt. 28: Maxima und
Minima. Die isoperimetrische
Aufgabe. Pp. 1133-1134. Attributes it to Prince Rupert, following
van Swinden. Cites Wallis &
Ronayne, via Cantor, and Nieuwland, via van Swinden.
U. Graf. Die Durchbohrung eines Würfels mit einem
Würfel. Zeitschrift math. naturwiss.
Unterricht 72 (1941) 117. Nice photos
of a model made at the Technische Hochschule Danzig. Larger and better versions of the same photos can be found in: W.
Lietzmann & U. Graf; Mathematik in Erziehung und Unterricht; Quelle &
Meyer, Leipzig, 1941, vol. 2, plate 3, opp. p. 168, but I can't find any
associated text for it.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. No. 12: Curios [sic] cubes, p. 14. First says it can be done with equal cubes
and then a larger can pass through a smaller.
Claims that the larger cube can be about 1.1, but this is due to
an error -- he thinks the hexagon has the same diameter as the cube itself.
H. D. Grossman, proposer; C. S. Ogilvy & F. Bagemihl,
solvers. Problem E888 -- Passing a cube
through a cube of same size. AMM 56 (1949)
632 ??NYS & 57 (1950) 339. Only considers cubes of the same size, though Bagemihl's solution
permits a slightly larger cube. No
references.
D. J. E. Schrek. Prince Rupert's problem and its extension by
Pieter Nieuwland. SM 16 (1950) 73‑80
& 261‑267. Historical survey,
discussing Rupert, Wallis, Ronayne, van Swinden & Nieuwland. Says Ronayne is early 18C.
M. Gardner. SA (Nov 1966) = Carnival, pp. 41‑54. The largest square inscribable in a cube is
the cross section of the maximal hole through which another cube can pass.
Christoph J. Scriba. Das Problem des Prinzen Ruprecht von der
Pfalz. Praxis der Mathematik 10 (1968)
241-246. ??NYS -- described by Scriba
in an email to HM Mailing List, 20 Aug 1999.
Describes the correction to Wallis's work and considers the problem for
the tetrahedron and octahedron.
Gardner. MM&M.
1956. Chap. 7 & 8:
Geometrical Vanishing -- Parts I & II, pp. 114‑155. Best extensive discussion of the subject and
its history.
Gardner. SA (Jan 1963) c= Magic Numbers, chap.
3. Discusses application to making an
extra bill and Magic Numbers adds citations to several examples of people
trying it and going to jail.
Gardner. Advertising premiums to beguile the mind:
classics by Sam Loyd, master puzzle‑poser. SA (Nov 1971) = Wheels, Chap. 12. Discusses several forms.
S&B, p. 144, shows several
versions.
6.P.1. PARADOXICAL DISSECTIONS OF THE CHESSBOARD BASED ON FIBONACCI NUMBERS
Area
63 version: AWGL, Dexter, Escott,
White, Loyd, Ahrens, Loyd Jr., Ransom.
(W. Leybourn. Pleasure with Profit. 1694.
?? I cannot recall the source of
this reference and think it may be an error.
I have examined the book and find nothing relevant in it.)
Loyd. Cyclopedia, 1914, pp. 288 & 378. 8 x 8 to 5 x 13
and to an area of 63. Asserts Loyd presented the first of these in
1858. Cf Loyd Jr, below.
O. Schlömilch. Ein geometrisches Paradoxon. Z. Math. Phys., 13 (1868) 162. 8 x 8
to 5 x 13. (This article is only signed Schl. Weaver, below, says this is Schlömilch, and
this seems right as he was a co‑editor at the time. Coxeter (SM 19 (1953) 135‑143) says it
is V. Schlegel, apparently confusing it with the article below.) Doesn't give any explanation, leaving it as
a student exercise.
F. J. Riecke. Op. cit. in 4.A.1. Vol. 3, 1873. Art. 16:
Ein geometrisches Paradoxon. Quotes Schlömilch
and explains the paradox.
G. H. Darwin. Messenger of Mathematics 6 (1877) 87. 8 x 8
to 5 x 13 and generalizations.
V. Schlegel. Verallgemeinerung eines geometrischen
Paradoxons. Z. Math. Phys. 24
(1879) 123‑128 & Plate I. 8 x
8 to
5 x 13 and generalizations.
Mittenzwey. 1880.
Prob. 299, pp. 54 & 105;
1895?: 332, pp. 58 & 106-107;
1917: 332, pp. 53 & 101. 8 x
8 to
5 x 13. Clear explanation.
The Boy's Own Paper. No. 109, vol. III (12 Feb 1881) 327. A puzzle.
8 x 8 to 5 x 13
without answer.
Richard A. Proctor. Some puzzles. Knowledge 9 (Aug 1886) 305-306.
"We suppose all the readers ... know this old puzzle." Describes and explains 8 x 8
to 5 x 13. Gives a different method of cutting so that
each rectangle has half the error -- several typographical errors.
Richard A. Proctor. The sixty-four sixty-five puzzle. Knowledge 9 (Oct 1886) 360-361. Corrects the above and explains it in more
detail.
Will Shortz has a puzzle trade
card with the 8 x 8 to 5
x 13 version, c1889.
Ball. MRE, 1st ed., 1892, pp. 34‑36. 8 x 8 to 5 x 13
and generalizations. Cites
Darwin and describes the examples in Ozanam-Hutton (see Ozanam-Montucla in
6.P.2). In the 5th ed., 1911, p. 53, he
changes the Darwin reference to Schlömilch.
In the 7th ed., 1917, he only cites the Ozanam-Hutton examples.
Clark. Mental Nuts. 1897, no.
33; 1904, no. 41; 1916, no. 43. Four peculiar drawings. 8
x 8 to
5 x 13.
Carroll-Collingwood. 1899.
Pp. 316-317 (Collins: 231 and/or 232 (lacking in my copy)) = Carroll-Wakeling II, prob. 7: A
geometrical paradox, pp. 12 & 7.
8 x 8 to 5 x 13.
Carroll may have stated this as early as 1888. Wakeling says the papers among which this was found on Carroll's
death are now in the Parrish Collection at Princeton University and suggests
Schlömilch as the earliest version.
AWGL (Paris). L'Echiquier Fantastique. c1900.
Wooden puzzle of 8 x 8 to 5
x 13 and to area 63.
??NYS -- described in S&B, p. 144.
Walter Dexter. Some postcard puzzles. Boy's Own Paper (14 Dec 1901) 174‑175. 8 x 8
to 5 x 13 and to area
63.
C. A. Laisant. Initiation Mathématique. Georg, Geneva & Hachette, Paris,
1906. Chap. 63: Un paradoxe: 64 = 65,
pp. 150-152.
Wm. F. White. In the mazes of mathematics. A series of perplexing puzzles. III.
Geometric puzzles. The Open
Court 21 (1907) 241‑244.
Shows 8 x 8 to 5
x 13 and a two‑piece 11 x 13
to area 145.
E. B. Escott. Geometric puzzles. The Open Court 21 (1907) 502‑505. Shows 8 x 8 to area
63 and discusses the connection
with Fibonacci numbers.
William F. White. Op. cit. in 5.E. 1908. Geometric puzzles,
pp. 109‑117. Partly based on
above two articles. Gives 8 x 8
to 5 x 13 and to area
63. Gives an extension which
turns 12 x 12 into 8 x 18 and into area 144, but turns 23 x 23
into 16 x 33 and into area 145. Shows a puzzle of
Loyd: three‑piece 8 x 8
into 7 x 9.
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. 5 x 5 into four pieces that make a 3 x 8.
M. Adams. Indoor Games. 1912. Is 64
equal to 65? Pp. 345-346 with fig. on p. 344.
Loyd. Cyclopedia. 1914. See entry at 1858.
W. Ahrens. Mathematische Spiele. Teubner, Leipzig. 3rd ed., 1916, pp. 94‑95 & 111‑112. The 4th ed., 1919, and 5th ed., 1927, are
identical with the 3rd ed., but on different pages: pp. 101‑102
& pp. 118‑119. Art. X.
65 = 64 = 63 gives 8 x 8
to 5 x 13 and to area
63. The area 63
case does not appear in the 2nd ed., 1911, which has Art. V. 64 = 65, pp. 107 & 118‑119 and this material is not in the 1st
ed. of 1907.
Tom Tit?? In Knott, 1918, but I can't find it in Tom
Tit. No. 3: The square and the
rectangle: 64 = 65!, pp. 15-16.
Clearly explained.
Hummerston. Fun, Mirth & Mystery. 1924.
A puzzling paradox, pp. 44 & 185.
Usual 8 x 8 to
5 x 13, but he erases
the chessboard lines except for the cells the cuts pass through, so one way has
16 cells, the other way has 17 cells.
Reasonable explanation.
Collins. Book of Puzzles. 1927. A paradoxical
puzzle, pp. 4-5. 8 x 8 to 5
x 13. Shades the unit cells that the
lines pass through and sees that one way has 16 cells, the other way has 17
cells, but gives only a vague explanation.
Loyd Jr. SLAHP.
1928. A paradoxical puzzle, pp.
19‑20 & 90. Gives 8 x 8
to 5 x 13. "I have discovered a companion piece
..." and gives the 8 x 8 to area
63 version. But cf AWGL, Dexter, etc. above.
W. Weaver. Lewis Carroll and a geometrical
paradox. AMM 45 (1938) 234‑236. Describes unpublished work in which Carroll
obtained (in some way) the generalizations of the 8 x 8 to 5 x 13
in about 1890‑1893. Weaver
fills in the elementary missing arguments.
W. R. Ransom, proposer; H. W. Eves, solver. Problem E468. AMM 48 (1941) 266 & 49 (1942) 122‑123. Generalization of the 8 x 8
to area 63 version.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. No. 23: Summat for nowt?, pp. 27-28. 8 x 8
to 5 x 13, clearly drawn.
Warren Weaver. Lewis Carroll: Mathematician. Op. cit. in 1. 1956. Brief mention of 8 x 8
to 5 x 13. John B. Irwin's letter gives generalizations
to other consecutive triples of Fibonacci numbers (though he doesn't call them
that). Weaver's response cites his 1938
article, above.
In
several early examples, the authors appear unaware that area has vanished!
Pacioli. De Viribus.
c1500. Ff. 189v - 191r. Part 2.
LXXIX. Do(cumento). un tetragono saper lo longare con restregnerlo
elargarlo con scortarlo (a tetragon knows lengthening and contraction,
enlarging with shortening ??) = Peirani
250-252. Convert a 4 x 24
rectangle to a 3 x 32 using one cut into two pieces. Pacioli's
description
is cryptic but seems to have two cuts, making d c
three
pieces. There is a diagram at the bottom
of f. 190v, badly k f
e
redrawn
on Peirani 458. Below this is a
inserted note which Peirani
252
simply mentions as difficult to read, but can make sense. The g
points
are as laid out at the right. abcd is the original 4 x 24 h a
o b
rectangle. g is
one unit up from a and
e is one unit down from c.
Cut
from c
to g and from e parallel to the base, meeting cg
at f. Then move cdg to
fkh and move fec
to hag. Careful rereading of Pacioli seems to show
he is using a trick! He cuts from e
to f to g. then turns over the upper piece and slides
it along so that he can continue his cut from
g to h, which is where f
to c is now. This gives three
pieces from a single cut! Pacioli clearly
notes that the area is conserved.
Although
not really in this topic, I have put it here as it seems to be a predecessor of
this topic and of 6.AY.
Sebastiano Serlio. Libro Primo d'Architettura. 1545.
This is the first part of his Architettura, 5 books, 1537-1547, first
published together in 1584. I have seen
the following editions.
With
French translation by Jehan Martin, no publisher shown, Paris, 1545,
f. 22.r. ??NX
1559. F. 15.v.
Francesco
Senese & Zuane(?) Krugher, Venice, 1566, f. 16.r. ??NX
Jacomo
de'Franceschi, Venice, 1619, f. 16.r.
Translated
into Dutch by Pieter Coecke van Aelst as:
Den eerstē vijfsten boeck van architecturē; Amsterdam,
1606. This was translated into English
as: The Five Books of Architecture;
Simon Stafford, London, 1611 = Dover,
1982. The first Booke, f. 12v.
3
x 10 board is cut on a diagonal and
slid to form a 4 x 7 table with
3 x 1 left over, but
he doesn't actually put the two leftover pieces together nor notice the area
change!
Pietro Cataneo. L'Architettura di Pietro Cataneo
Senese. Aldus, Venice, 1567. ??NX.
Libro Settimo.
P.
164, prop. XXVIIII: Come si possa accresciere una stravagante larghezza. Gives a correct version of Serlio's process.
P.
165, prop. XXX: Falsa solutione del Serlio.
Cites p. xxii of Serlio.
Carefully explains the error in Serlio and says his method is
"insolubile, & mal pensata".
Schwenter. 1636.
Part 15, ex. 14, p. 541: Mit einem länglichten schmahlen Brett /
für ein bräites Fenster einen Laden zu
machen. Cites Gualtherus Rivius,
Architectur. Discusses Serlio's
dissection as a way of making a 4 x
7 from a 3 x 10 but doesn't notice
the area change.
Gaspar Schott. Magia Universalis. Joh. Martin Schönwetter, Bamberg, Vol. 3, 1677. Pp. 704-708 describes Serlio's error in
detail, citing Serlio. ??NX of plates.
I have a vague reference to the
1723 ed. of Ozanam, but I have not seen it in the 1725 ed. -- this may be an
error for the 1778 ed. below.
Minguet. 1755.
Pp. not noted -- ??check (1822: 145-146; 1864: 127-128). Same as Hooper. Not in 1733 ed.
Vyse. Tutor's Guide. 1771? Prob. 8, 1793: p. 304, 1799: p. 317 &
Key p. 358. Lady has a table 27
square and a board 12 x 48. She cuts the board into two 12 x 24
rectangles and cuts each rectangle along a diagonal. By placing the diagonals of these pieces on
the sides of her table, she makes a table
36 square. Note that
362 = 1296 and 272 + 12 x 48 =
1305. Vyse is clearly unaware
that area has been created. By dividing
all lengths by 3, one gets a version where one unit of area is
lost. Note that 4, 8, 9
is almost a Pythagorean triple.
William Hooper. Rational Recreations. 1774.
Op. cit. in 4.A.1. Vol. 4, pp.
286‑287: Recreation CVI -- The geometric money. 3 x 10 cut into four pieces
which make a 2 x 6 and a
4 x 5. (The diagram is shown in
Gardner, MM&M, pp. 131‑132.)
(I recently saw that an edition erroneously has a 3 x 6
instead of a 2 x 6 rectangle.
This must be the 1st ed. of 1774, as it is correct in my 2nd ed. of 1782.)
Ozanam-Montucla. 1778.
Transposition de laquelle semble résulter que le tout peut être égal à
la partie. Prob. 21 & fig. 127,
plate 16, 1778: 302-303 & 363;
1803: 298-299 & 361; 1814:
256 & 306; 1840: omitted. 3 x 11
to 2 x 7 and
4 x 5. Remarks that M. Ligier
probably made some such mistake in showing
172 = 2 x 122
and this is discussed further on the later page.
E. C. Guyot. Nouvelles Récréations Physiques et
Mathématiques. Nouvelle éd. La Librairie, Rue S. André‑des‑Arcs[sic],
Paris, Year 7 [1799]. Vol. 2, Deuxième
récréation: Or géométrique -- construction, pp. 41‑42 & plate 6, opp.
p. 37. Same as Hooper.
Manuel des Sorciers. 1825.
Pp. 202-203, art. 19. ??NX Same as Hooper.
The Boy's Own Book. The geometrical money. 1828: 413;
1828-2: 419; 1829 (US):
212; 1855: 566‑567; 1868: 669.
Same as Hooper.
Magician's Own Book. 1857.
Deceptive vision, pp. 258-259.
Same as Hooper. = Book of 500
Puzzles, 1859, pp. 72-73.
Illustrated Boy's Own
Treasury. 1860. Optics: Deceptive vision, p. 445. Same as Hooper. Identical to Book of 500 Puzzles.
Wemple & Company (New
York). The Magic Egg Puzzle. ©1880.
S&B, p. 144. Advertising card,
the size of a small postcard, but with ads for Rogers Peet on the back. Starts with 9 eggs. Cut into four rectangles and reassemble to
make 6, 7, 8, 10, 11, 12 eggs.
R. March & Co. (St. James's
Walk, Clerkenwell). 'The Magical Egg
Puzzle', nd [c1890]. (I have a
photocopy.) Four rectangles which
produce 6, 7, ..., 12 eggs.
Identical to the Wemple version, but with Wemple's name removed. I only have a photocopy of the front of this
and I don't know what's on the back. I also have a photocopy of the
instructions.
Loyd. US Patent 563,778 -- Transformation Picture. Applied: 11 Mar 1896; patented: 14 Jul 1896. 1p + 1p diagrams. Simple rotating version using 8 to 7 objects.
Loyd. Get Off the Earth. Puzzle
notices in the Brooklyn Daily Eagle (26 Apr ‑ 3 May 1896), printing
individual Chinamen. Presenting all of
these at an office of the newspaper gets you an example of the puzzle. Loyd ran discussions on it in his Sunday
columns until 3 Jan 1897 and he also sold many versions as advertising
promotions. S&B, p. 144, shows
several versions.
Loyd. Problem 17: Ye castle donjon.
Tit‑Bits 31 (6 & 27 Feb
& 6 &
20 Mar 1897) 343, 401, 419
& 455. = Cyclopedia, 1914, The architect's puzzle,
pp. 241 & 372. 5 x 25 to area 124.
Dudeney. Great puzzle crazes. Op. cit. in 2. 1904. Discusses and shows
Get Off the Earth.
Ball. MRE, 4th ed., 1905, pp. 50-51: Turton's seventy-seven
puzzle. Additional section describing
Captain Turton's 7 x 11 to 7
x 11 with one projecting square, using
bevelled cuts. This is dropped from the
7th ed., 1917.
William F. White. 1907 & 1908. See entries in 6.P.1.
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. Gives "Get Off
the Earth" on p. 785.
Loyd. Teddy and the Lions.
Gardner, MM&M, p. 123, says he has seen only one example, made as a
promotional item for the Eden Musee in Manhattan. This has a round disc, but two sets of figures -- 7 natives and 7
lions which become 6 natives and 8 lions.
Dudeney. A chessboard fallacy. The Paradox Party. Strand Mag. 38 (No. 228) (Dec 1909) 676 (= AM, prob. 413,
pp. 141 & 247). (There is a solution
in Strand Mag. 39 (No. 229) (Jan 1910) ??NYS.) 8 x 8 into 3 pieces which
make a 9 x 7.
Fun's Great Baseball
Puzzle. Will Shortz gave this out at
IPP10, 1989, as a colour photocopy, 433 x 280 mm (approx. A3). ©1912 by the Press Publishing Co (The New
York World). I don't know if Fun was
their Sunday colour comic section or what.
One has to cut it diagonally and slide one part along to change from 8
to 9 boys.
Loyd. The gold brick puzzle.
Cyclopedia, 1914, pp. 32 & 342 (= MPSL1, prob. 24, pp. 22 & 129). 24 x 24
to 23 x 25.
Loyd. Cyclopedia. 1914. "Get off the earth", p. 323. Says over 10 million were sold. Offers prizes for best answers received in
1909.
Loyd Jr. SLAHP.
1928. "Get off the
Earth" puzzle, pp. 5‑6. Says
'My "Missing Chinaman Puzzle"' of 1896. Gives a simple and clear explanation.
John Barnard. The Handy Boy's Book. Ward, Lock & Co., London, nd
[c1930?]. Some interesting optical
illusions, pp. 310-311. Shows a card
with 11 matches and a diagonal cut so that sliding it one place makes 10
matches.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. No. 24: A chessboard fallacy, pp. 28-29. 8 x 8
cut with a diagonal of a 8 x
7 region, then pieces slid and a
triangle cut off and moved to the other end to make a 9 x 7. Clear
illustration.
Mel Stover. From 1951, he devised a number of variations
of both Get off the Earth (perhaps the best is his Vanishing Leprechaun) and of
Teddy and the Lions (6 men and 4 glasses of beer become 5 men and 5 glasses). I have examples of some of these from Stover
and I have looked at his notebooks, which are now with Mark Setteducati. See
Gardner, MM&M, pp. 125-128.
Gardner. SA (May 1961) c= NMD, chap. 11.
Mentioned in Workout, chap. 27.
Describes his adaptation of a principle of Paul Curry to produce The
Disappearing Square puzzle, where 16 or 17 pieces seem to make the same
square. The central part of the 17
piece version consists of five equal squares in the form of a Greek cross. The central part of the 16 piece version has
four of the squares in the shape of a square.
This has since been produced in several places.
Ripley's Puzzles and Games. 1966.
P. 60. Asserts that when you cut
a 2½ x 4½ board into six right triangles with legs 1½ and 2½,
then they assemble into an equilateral triangle of edge 5. This has an area loss of about 4%.
John Fisher. John Fisher's Magic Book. Muller, London, 1968.
Financial
Wizardry, pp. 18-19. 7 x 8 region with
£ signs marking the area. A line cuts off a triangle of width 7 and height
2 at the top. The rest of the area is
divided by a vertical into strips of widths 4 and 3, with a small rectangle 3
by 1 cut from the bottom of the width 3 strip.
When the strips are exchanged, one unit of area is lost and one £
sign has vanished.
Try-Angle,
pp. 126-127. This is one of Curry's
triangles -- see Gardner, MM&M, p. 147.
Alco-Frolic!,
pp. 148-149. This is a form of
Stover's 6 & 4 to 5
& 5 version.
D. E. Knuth. Disappearances. In: The Mathematical
Gardner; ed. by David Klarner; Prindle, Weber & Schmidt/Wadsworth,
1981. P. 264. An eight line poem which rearranges to a seven line poem.
Dean Clark. A centennial tribute to Sam Loyd. CMJ 23:5 (Nov 1992) 402‑404. Gives an easy circular version with 11 & 12
astronauts around the earth and a
15 & 16 face version with
three pieces, a bit like the Vanishing Leprechaun.
6.Q. KNOTTING A STRIP TO MAKE A REGULAR PENTAGON
Urbano d'Aviso.
Trattato della Sfera e Pratiche per Uso di Essa. Col modeo di fare la figura celeste, opera cavata
dalli manoscritti del. P. Bonaventura Cavalieri. Rome, 1682. ??NYS cited
by Lucas (1895) and Fourrey.
Dictionary of Representative
Crests. Nihon Seishi Monshō
Sōran (A Comprehensive Survey of Names and Crests in Japan), Special issue of Rekishi Dokuhon (Readings in History), Shin Jinbutsu Oraisha, Tokyo, 1989, pp. 271-484. Photocopies of relevant pages kindly sent by
Takao Hayashi.
Crests
3504 and 3506 clearly show a strip knotted to make a pentagon. 3507 has two such knots and 3508 has five. I don't know the dates, but most of these
crests are several centuries old.
Lucas. RM2, 1883, pp. 202‑203.
Tom Tit.
Vol.
2, 1892. L'Étoile à cinq branches, pp.
153-154. = K, no. 5: The pentagon and
the five pointed star, pp. 20‑21.
He adds that folding over the free end and holding the knot up to the
light shows the pentagram.
Vol.
3, 1893. Construire d'un coup de poing
un hexagone régulier, pp. 159-161.
= K, no. 17: To construct a hexagon by finger pressure, pp. 49‑51. Pressing an appropriate size Möbius strip
flat gives a regular hexagon.
Vol.
3, 1893. Les sept pentagones, pp.
165-166. = K, no. 19: The seven
pentagons, pp. 54‑55. By
tying five pentagons in a strip, one gets a larger pentagon with a pentagonal
hole in the middle.
Somerville Gibney. So simple!
The hexagon, the enlarged ring, and the handcuffs. The Boy's Own Paper 20 (No. 1012) (4 Jun
1898) 573-574. As in Tom Tit, vol. 3,
pp. 159-161.
Lucas. L'Arithmétique Amusante.
1895. Note IV: Section II: Les Jeux de Ruban, Nos. 1 &
2: Le nœud de cravate &
Le nœud marin, pp. 220-222.
Cites d'Aviso and says he does both the pentagonal and hexagonal knots,
but Lucas only shows the pentagonal one.
E. Fourrey. Procédés Originaux de Constructions
Géométriques. Vuibert, Paris, 1924. Pp. 113 & 135‑139. Cites Lucas and cites d'Aviso as Traité de
la Sphère and says he gives the pentagonal and hexagonal knots. Fourrey shows and describes both, also
giving the pictures on his title page.
F. V. Morley. A note on knots. AMM 31 (1924) 237-239.
Cites Knott's translation of Tom Tit.
Says the process generalizes to
(2n+3)‑gons by using
n loops. Gets even-gons by using two strips. Discusses using twisted strips.
Robert C. Yates. Geometrical Tools. (As: Tools; Baton Rouge,
1941); revised ed., Educational
Publishers, St. Louis, 1949. Pp. 64-65
gives square (a bit trivial), pentagon, hexagon, heptagon and octagon. Even case need two strips.
Donovan A. Johnson. Paper Folding for the Mathematics
Class. NCTM, 1957, pp. 16-17: Polygons
constructed by tying paper knots. Shows
how to tie square, pentagon, hexagon, heptagon and octagon.
James K. Brunton. Polygonal knots. MG 45 (No. 354) (Dec 1961) 299‑302. All regular
n‑gons, n > 4, can be obtained, except n = 6
which needs two strips.
Discusses which can be made without central holes.
Marius Cleyet-Michaud. Le Nombre d'Or. Presses Universitaires de France, Paris, 1973. On pp. 47-48, he calls this the 'golden
knot' (Le "nœud doré") and describes how to make it.
General
surveys of such fallacies can be found in the following. See also:
6.P, 10.A.1.
These
fallacies are actually quite profound as the first two point out some major
gaps in Euclid's axioms -- the idea of a point being inside a triangle really
requires notions of order of points on a line and even the idea of continuity,
i.e. the idea of real numbers.
Ball. MRE. 1st ed., 1892, pp.
31‑34, two examples, discussed below.
3rd ed., 1896, pp. 39‑46
= 4th ed., 1905, pp. 41-48,
seven examples. 5th ed., 1911, pp.
44-52 = 11th ed., 1939, pp. 76-84, nine example.
Walther Lietzmann. Wo steckt der Fehler? Teubner, Stuttgart, (1950), 3rd ed.,
1953. (Strens/Guy has 3rd ed.,
1963(?).) (There are 2nd ed, 1952??; 5th ed, 1969; 6th ed,
1972. MG 54 (1970) 182 says the 5th ed
appears to be unchanged from the 3rd ed.)
Chap. B: V, pp. 87-99 has 18 examples.
(An
earlier version of the book, by Lietzmann & Trier, appeared in 1913, with
2nd ed. in 1917. The 3rd ed. of 1923
was divided into two books: Wo Steckt
der Fehler? and Trugschlüsse. There was a 4th ed. in 1937.
The relevant material would be in Trugschlüsse, but I have not seen any
of the relevant books, though E. P. Northrop cites Lietzmann, 1923, three times
-- ??NYS.)
E. P. Northrop. Riddles in Mathematics. 1944.
Chap. 6, 1944: 97-116, 232-236
& 249-250; 1945: 91-109, 215-219
& 230-231; 1961: 98-115, 216-219
& 229. Cites Ball, Lietzmann
(1923), and some other individual items.
V. M. Bradis, V. L. Minkovskii & A. K.
Kharcheva. Lapses in Mathematical
Reasoning. (As: [Oshibki v
Matematicheskikh Rassuzhdeniyakh], 2nd ed, Uchpedgiz, Moscow, 1959.) Translated by J. J. Schorr-Kon, ed. by E. A.
Maxwell. Pergamon & Macmillan, NY,
1963. Chap. IV, pp. 123-176. 20 examples plus six discussions of other
examples.
Edwin Arthur Maxwell. Fallacies in Mathematics. CUP, (1959), 3rd ptg., 1969. Chaps. II-V, pp. 13-36, are on geometric
fallacies.
Ya. S. Dubnov. Mistakes in Geometric Proofs. (2nd ed., Moscow?, 1955). Translated by Alfred K. Henn & Olga A.
Titelbaum. Heath, 1963. Chap 1-2, pp. 5-33. 10 examples.
А. Г.
Конфорович. [A. G.
Konforovich].
(Математичні
Софізми і
Парадокси
[Matematichnī Sofīzmi ī Paradoksi] (In Ukrainian). Радянська
Школа [Radyans'ka Shkola], Kiev, 1983.) Translated into German by Galina &
Holger Stephan as: Konforowitsch, Andrej Grigorjewitsch; Logischen Katastrophen
auf der Spur – Mathematische Sophismen und Paradoxa; Fachbuchverlag, Leipzig,
1990. Chap. 4: Geometrie,
pp. 102-189 has 68 examples, ranging from the type considered here up
through fractals and pathological curves.
S. L. Tabachnikov. Errors in geometrical proofs. Quantum 9:2 (Nov/Dec 1998) 37-39 &
49. Shows: every triangle is isosceles (6.R.1); the sum of the angles of a triangle is 180o without
use of the parallel postulate; a
rectangle inscribed in a square is a square;
certain approaching lines never meet (6.R.3); all circles have the same circumference (cf Aristotle's Wheel
Paradox in 10.A.1); the circumference
of a wheel is twice its radius; the
area of a sphere of radius R is
π2R2.
6.R.1. EVERY TRIANGLE IS ISOSCELES
This is sometimes claimed to
have been in Euclid's lost Pseudaria (Fallacies).
Ball. MRE, 1st ed., 1892, pp. 33‑34. On p. 32, Ball refers to Euclid's lost Fallacies and presents
this fallacy and the one in 6.R.2:
"I do not know whether either of them has been published
previously." In the 3rd ed., 1896,
pp. 42-43, he adds the heading: To
prove that every triangle is isosceles.
In the 5th ed., 1911, p. 45, he adds a note that he believes these two
were first published in his 1st ed. and notes that Carroll was fascinated by
them and they appear in The Lewis Carroll Picture Book (= Carroll-Collingwood)
-- see below.
Mathesis (1893). ??NYS.
[Cited by Fourrey, Curiosities Geometriques, p. 145. Possibly Mathesis (2) 3 (Oct 1893) 224,
cited by Ball in MRE, 3rd ed, 1896, pp. 44-45, cf in Section 6.R.4.]
Carroll-Collingwood. 1899.
Pp. 264-265 (Collins: 190-191).
= Carroll-Wakeling II, prob. 27: Every triangle has a pair of
equal sides!, pp. 43 & 27. Every
triangle is isosceles. Carroll may have
stated this as early as 1888.
Wakeling's solution just suggests making an accurate drawing. Carroll-Gardner, p. 65, mentions this and
says it was not original with Carroll.
Ahrens. Mathematische Spiele. Teubner.
Alle Dreiecke sind gleichschenklige.
2nd ed., 1911, chap. X, art. VI, pp. 108 & 119‑120. 3rd ed., 1916, chap. IX, art. IX, pp. 92-93
& 109-111. 4th ed., 1919 & 5th ed., 1927, chap IX, art. IX, pp. 99‑101 & 116‑118.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. Call Mr. Euclid -- No. 15: To prove all triangles
are equilateral, pp. 16-17. Clear
exposition of the fallacy.
See Read in 6.R.4 for a
different proof of this fallacy.
6.R.2. A RIGHT ANGLE IS OBTUSE
Ball. MRE, 1st ed.,
1892, pp. 32‑33. See 6.R.1. In the 3rd ed., 1896, pp. 40-41, he adds the
heading: To prove that a right angle is
equal to an angle which is greater than a right angle.
Mittenzwey. 1895?.
Prob. 331, pp. 58 & 106;
1917: 331, pp. 53 & 101.
Carroll-Collingwood. 1899.
Pp. 266‑267 (Collins 191-192).
An obtuse angle is sometimes equal to a right angle. Carroll-Gardner, p. 65, mentions this and
says it was not original with Carroll.
H. E. Licks. 1917.
Op. cit. in 5.A. Art. 82, p. 56.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. Call Mr. Euclid -- No. 16: To prove one right
angle greater than another right angle, pp. 18-19. "Here again, if you take the trouble to draw an accurate
diagram, you will find that the "construction" used for the alleged
proof is impossible."
E. A. Maxwell. Note 2121:
That every angle is a right angle.
MG 34 (No. 307) (Feb 1950) 56‑57.
Detailed demonstration of the error.
6.R.3. LINES APPROACHING BUT NOT MEETING
Proclus. 5C.
A Commentary on the First Book of Euclid's Elements. Translated by Glenn R. Morrow. Princeton Univ. Press, 1970. Pp. 289-291. Gives the argument and tries to refute it.
van Etten/Henrion/Mydorge. 1630.
Part 2, prob. 7: Mener une ligne laquelle aura inclination à une autre
ligne, & ne concurrera jamais contre l'Axiome des paralelles, pp. 13‑14.
Schwenter. 1636.
To be added.
Ozanam-Montucla. 1778.
Paradoxe géométrique des lignes ....
Prob. 70 & fig. 116-117, plate 13, 1778: 405-407; 1803: 411-413; 1814: 348-350. Prob. 69,
1840: 180-181. Notes that these
arguments really produce a hyperbola and a conchoid. Hutton adds that a great many other examples might be found.
E. P. Northrop. Riddles in Mathematics. 1944.
1944: 209-211 & 239;
1945: 195‑197 & 222;
1961: 197‑198 & 222.
Gives the 'proof' and its fallacy, with a footnote on p. 253
(1945: 234; 1961: 233) saying the argument "has been attributed to
Proclus."
Jeremy Gray. Ideas of Space. OUP, 1979. Pp. 37-39
discusses Proclus' arguments in the context of attempts to prove the parallel
postulate.
Ball. MRE, 3rd ed, 1896, pp. 44-45.
To prove that, if two opposite sides of a quadrilateral are equal, the
other two sides must be parallel. Cites
Mathesis (2) 3 (Oct 1893) 224 -- ??NYS
Cecil B. Read. Mathematical fallacies &
More mathematical fallacies. SSM
33 (1933) 575‑589 & 977-983.
There are two perpendiculars from a point to a line. Part of a line is equal to the whole
line. Every triangle is isosceles (uses
trigonometry). Angle trisection (uses a
marked straightedge).
P. Halsey. Class Room Note 40: The ambiguous case. MG 43 (No. 345) (Oct 1959) 204‑205. Quadrilateral ABCD with angle A = angle C and AB = CD. Is this a parallelogram?
Hoffmann. 1893.
Chap III, pp. 74‑90, 96-97, 111-124 & 128
= Hoffmann-Hordern, pp. 62‑79 & 86-87 with several
photos. Describes Tangrams and Richter
puzzles at some length. Lots of photos
in Hordern. Photos on pp. 67, 71, 75,
87 show Richter's: Anchor (1890‑1900, = Tangram), Tormentor (1898),
Pythagoras (1892), Cross Puzzle (1892), Circular Puzzle (1891), Star Puzzle
(1899), Caricature (1890-1900, = Tangram) and four non-Richter Tangrams in
Tunbridge ware, ivory, mother-of-pearl and tortoise shell. Hordern Collection, pp. 45-57 & 60,
(photos on pp. 46, 49, 50, 52, 54, 56, 60) shows different Richter versions of
Tormentor (1880-1900), Pythagoras (1880-1900), Circular Puzzle (1880-1900),
Star Puzzle (1880‑1900) and has a wood non-Richter version instead of the
ivory version in the last photo.
Ronald C. Read. Tangrams -- 330 Puzzles. Dover, 1965. The Introduction, pp. 1-6, is a sketch of the history. Will Shortz says this is the first serious
attempt to counteract the mythology created by Loyd and passed on by
Dudeney. Read cannot get back before
the early 1800s and notes that most of the Loyd myth is historically
unreasonable. However, Read does not
pursue the early 1800s history in depth and I consider van der Waals to be the
first really serious attempt at a history of the subject.
Peter van Note. Introduction. IN: Sam Loyd; The Eighth Book of Tan; (Loyd & Co., 1903);
Dover, 1968, pp. v-viii. Brief
debunking of the Loyd myth.
Jan van der Waals. History
& Bibliography. In:
Joost Elffers; Tangram; (1973), Penguin, 1976. Pp. 9‑27 & 29‑31. Says the Chinese term "ch'i ch'ae" dates from the Chu
era (‑740/‑330), but the earliest known Chinese book is 1813. The History reproduces many pages from early
works. The Bibliography cites 8
versions of 4 Chinese books (with locations!) from 1813 to 1826 and 18 Western
books from 1805 to c1850. The 1805, and
several other references, now seem to be errors.
S&B. 1986.
Pp. 22‑33 discusses loculus of Archimedes, Chie no Ita, Tangrams
and Richter puzzles.
Alberto Milano. Due giochi di società dell'inizio
dell'800. Rassegna di Studi e di
Notizie 23 (1999) 131-177. [This is a
publication by four museums in the Castello Sforzesco, Milan: Raccolta delle Stampe Achille Bertarelli;
Archivio Fotografico; Raccolte d'Arte Applicata; Museo degli Strumenti
Musicali. Photocopy from Jerry Slocum.] This surveys early books on tangrams, some
related puzzles and the game of bell and hammer, with many reproductions of TPs
and problems.
Jerry Slocum. The Tangram Book. (With Jack Botermans, Dieter Gebhardt, Monica Ma, Xiaohe Ma,
Harold Raizer, Dic Sonneveld and Carla van Spluntern.) ©2001 (but the first publisher collapsed),
Sterling, 2003. This is the long
awaited definitive history of the subject!
It will take me sometime to digest and summarize this, but a brief
inspection shows that much of the material below needs revision!
Recent
research by Jerry Slocum, backed up by The Admired Chinese Puzzle, indicates
that the introduction of tangrams into Europe was done by a person or persons
in Lord Amherst's 1815-1817 embassy to China, which visited Napoleon on St.
Helena on its return voyage. If so, then
the conjectural dating of several items below needs to be amended. I have amended my discussion accordingly and
marked such dates with ??. Although
watermarking of paper with the correct date was a legal requirement at the
time, paper might have been stored for some time before it was printed on, so
watermark dates only give a lower bound for the date of printing. I have seen several further items dated
1817, but it is conceivable that some material may have been sent back to
Europe or the US a few years earlier -- cf Lee.
On
2 Nov 2003, I did the following brief summary of Slocum's work in a letter to
an editor. I've made a few corrections
and added a citation to the following literature.
Tangrams. The history of this has now been definitely
established in Jerry Slocum's new book: The Tangram Book; ©2001 (but the first publisher collapsed),
Sterling, 2003. This history has been
extremely difficult to unravel because Sam Loyd deliberately obfuscated it in 1903,
claiming the puzzle went back to 2000 BC, because the only previous attempt at
a history had many errors, and because much of the material doesn't survive, or
only a few examples survive. The
history covers a wide range in both time and location, as evidenced by the
presence of seven co-authors from several countries.
Briefly,
the puzzle, in the standard form, dates from about 1800, in China. It is attributed to Yang-cho-chü-shih, but
this is a pseudonym, meaning 'dim-witted recluse', and no copies of his work
are known. The oldest known example of
the game is one dated 1802 in a museum near Philadelphia -- see Lee,
below. The oldest known book on the
puzzle had a preface by Sang-hsia-ko [guest under the mulberry tree] dated June
1813 and a postscript by Pi-wu-chü-shih dated July 1813. This is only known from a Japanese facsimile
of it made in 1839. This book was
republished, with a book of solutions, in two editions in 1815 -- one with
about four problems per page, the other with about eleven. The latter version was the ancestor of many
19C books, both in China and the west.
Another 2 volume version appeared later in 1815. Sang-hsia-ko explicitly says "The
origin of the Tangram lies within the Pythagorean theorem".
In
1816, several ships brought copies of the eleven problems per page books to the
US, England and Europe. The first
western publication of the puzzle is in early 1817 when J. Leuchars of 47
Piccadilly registered a copyright and advertised sets for sale. But the craze was really set off by the
publication of The Fashionable Chinese Puzzle and its Key by John
and Edward Wallis and John Wallis Jr in March 1817. This included a poem with a note that the game was "the
favourite amusement of Ex-Emperor Napoleon". This went through many printings, with some (possibly the first)
versions having nicely coloured illustrations.
By the end of the year, there were many other books, including examples
in France, Italy and the USA.
Dic
Sonneveld, one of the co-authors of Slocum's book, managed to locate the
tangram and books that had belonged to Napoleon in the Château de Malmaison,
outside Paris, but there is no evidence that Napoleon spent much time playing
with it. St. Helena was a regular stop
for ships in the China trade. Napoleon
is recorded as having bought a chess set from one ship and several notables are
recorded as having presented Napoleon with gifts of Chinese objects. A diplomatic letter of Jan 1817 records
sending an example of the game from St. Helena to Prince Metternich, but this
example has not been traced.
The
first American book was Chinese Philosophical and Mathematical Trangram
by James Coxe, appearing in Philadelphia in August 1817. The word 'trangram' meaning 'an odd,
intricately contrived thing' according to Johnson's Dictionary, was
essentially obsolete by 1817, but was still in some use in the US. The earliest known use of the word 'tangram'
is in Thomas Hill's Geometrical Puzzles for the Young, Boston,
1848. One suspects that he was
influenced by Coxe's book, but he may have known that 'T'ang' is the Cantonese
word for 'Chinese'. Hill later became
President of Harvard University and was an active promoter and inventor of
games for classroom use. In 1864, the
word was in Webster's Dictionary.
However,
the above is the story of the seven-piece tangram that we know today. There is a long background to this, dating
back to the 3rd century BC, when Archimedes wrote a letter to Eratosthenes
describing a fourteen piece puzzle, known as the Stomachion or Loculus of
Archimedes. The few surviving texts are
not very clear and there are two interpretations -- in one the standard
arrangement of the pieces is a square and in the other it is a rectangle twice
as wide as high. There are six (at
least) references to the puzzle in the classical world, the last being in the
6th century. The puzzle was used to
make a monstrous elephant, a brutal boar, a ship, a sword, etc., etc. The puzzle then disappears, and no form of
it appears in the Arabic world, which has always surprised me, given the Arabic
interest in patterns.
Further,
several eastern predecessors of the tangrams are known. The earliest is a Japanese version of 1742
by Ganriken (or Granreiken) which has seven pieces, attributed (as were many
things) to Sei Shonagon, a 10th century courtesan famous for her
ingenuity. By the end of the 18th century,
three other dissection/arrangement puzzles appeared in Japan, with 15, 19 and
19 pieces, including some semi-circles.
An 1804 print by Utamaro shows courtesans playing with some version of
the puzzle -- only two copies of this print have been located.
But
the basic puzzle idea has its roots in Chinese approaches to the Theorem of
Pythagoras and similar geometric proofs by dissection and rearrangement which
date back to the 3rd century (and perhaps earlier). But the tangram did not develop directly from these ideas. From the 12th century, there was a Chinese
tradition of making "Banquet Tables" in the form of several pieces
that could be arranged in several ways.
The first known Chinese book on furniture, by Huang Po-ssu in 1194,
describes a Banquet Table formed of seven rectangular pieces: two long, two
medium and three short. In 1617, Ko
Shan described 'Butterfly Wing" tables with 13 pieces, including isosceles
right triangles, right trapeziums and isosceles trapeziums. In 1856, a Chinese scholar noted the
resemblance of these tables with the tangram and a modern Chinese historian of
mathematics has observed that half of the butterfly arrangement can be easily
transformed into the tangrams. No
examples of these tables have survived, but tables (and serving dishes) in the
tangram pattern exist and are probably still being made in China.
Kanchusen. Wakoku Chiekurabe. 1727. Pp. 9 & 28-29:
a simple dissection puzzle with 8 pieces.
The problem appears to consist of a mitre comprising ¾ of
a unit square; 4 isosceles right
triangles of hypotenuse 1 and 3 isosceles right triangles of side ½,
but the solution shows that all the triangles are the same size, say
having hypotenuse 1, and the mitre shape is actually formed from
a rectangle of size 1 x Ö2.
"Ganriken" [pseud.,
possibly of Fan Chu Sen]. Sei
Shōnagon Chie-no-Ita (The Ingenious Pieces by Sei Shōnagon.) (In
Japanese). Kyoto Shobo, Aug 1742, 18pp,
42 problems and solutions. Reproduced
in a booklet, ed. by Kazuo Hanasaki, Tokyo, 1984, as pp. 19‑36. Also reproduced in a booklet, transcribed
into modern Japanese, with English pattern names and an English abstract, by
Shigeo Takagi, 1989. This uses a set of
seven pieces different than the Tangram.
S&B, p. 22, shows these pieces.
Sei Shōnagon (c965-c1010) was a famous courtier, author of The
Pillow Book and renowned for her intelligence.
The Introduction is signed Ganriken.
S&B say this is probably Fan Chu Sen, but Takagi says the author's
real name is unknown.
Utamaro. Interior of an Edo house, from the picture‑book: The Edo Sparrows (or Chattering Guide),
1786. Reproduced in B&W in: J. Hillier; Utamaro -- Colour Prints
and Paintings; Phaidon Press, Oxford, (1961), 2nd ed., 1979, p. 27, fig.
15. I found this while hunting for the
next item. This shows two women
contemplating some pieces but it is hard to tell if it is a tangram‑type
puzzle, or if perhaps they are cakes.
Hiroko and Mike Dean tell me that they are indeed cooking cakes.
Utamaro. Woodcut.
1792. Shows two courtesans
working on a tangram puzzle. Van der
Waals dated this as 1780, but Slocum has finally located it, though he has only
been able to find two copies of it! The
courtesans are clearly doing a tangram-like puzzle with 12(?) pieces -- the
pieces are a bit piled up and one must note that one of the courtesans is
holding a piece. They are looking at a
sheet with 10 problem figures on it.
Early 19C books from China --
??NYS -- cited by Needham, p. 111.
Jean Gordon Lee. Philadelphians and the China Trade 1784-1844.
Philadelphia Museum of Art, 1984, pp. 122-124. (Photocopy from Jerry Slocum.)
P. 124, item 102, is an ivory tangram in a cardboard box, inscribed on
the bottom of the box: F. Waln April 4th 1802. Robert Waln was a noted trader with China
and this may have been a present for his third son Francis (1799‑1822). This item is in the Ryerss Museum, a city
museum in Philadelphia in the country house called Burholme which was built by
one of Robert Waln's sons-in-law.
A New Invented Chinese Puzzle,
Consisting of Seven Pieces of Ivory or Wood, Viz. 5 Triangles, 1 Rhomboid,
& 1 Square, which may be so placed as to form the Figures represented in
the plate. Paine & Simpson,
Boro'. Undated, but the paper is
watermarked 1806. This consists of two
'volumes' of 8 pages each, comprising 159 problems with no solutions. At the end are bound in a few more pages
with additional problems drawn in -- these are direct copies of plates 21, 26,
22, 24, and 28 (with two repeats from plate 22) of The New and Fashionable
Chinese Puzzle, 1817. Bound in plain
covers. This is in Edward Hordern's
collection and he provided a photocopy.
Dalgety also has a copy.
Ch'i Ch'iao t'u ho‑pi (=
Qiqiao tu hebi) (Harmoniously combined book of tangram problems OR Seven clever
pieces). 1813. (Bibliothek Leiden 6891, with an 1815
edition at British Library 15257 d 13.)
van der Waals says it has 323 examples.
The 1813 seems to be the earliest Chinese tangram book of problems, with
the 1815 being the solutions. Slocum
says there was a solution book in 1815 and that the problem book had a preface
by Sang‑hsia K'o (= Sang-xia-ke), which was repeated in the solution book with the same date. Milano mentions this, citing Read and van
der Waals/Elffers, and says an example is on the BL. A version of this appears to have been the book given to Napoleon
and to have started the tangram craze in Europe. I have now received a photocopy from Peter Rasmussen & Wei
Zhang which is copied from van der Waals' copy from BL 15257 d 13. It has a cover, 6 preliminary pages and 28
plates with 318 problems. The pages are
larger than the photocopies of 1813/1815 versions in the BL that Slocum gave
me, which have 334 problems on 86 pages, but I see these are from 15257 d 5 and
14. I have a version of the smaller
page format from c1820s which has 334 problems on 84pp, apparently lacking its
first sheet. The problems are not
numbered, but given Chinese names. They
are identical to those appearing in Wallis's Fashionable Chinese Puzzle, below,
except the pages are in different order, two pages are inverted, Wallis
replaces Chinese names by western numbers and draws the figures a bit more
accurately. Wallis skips one number and
adds four new problems to get 323 problems - van der Waals seems to have taken
323 from Wallis.
Shichi‑kou‑zu
Gappeki [The Collection of Seven‑Piece Clever Figures]. Hobunkoku Publishing, Tokyo, 1881. This is a Japanese translation of an 1813
Chinese book "recognized as the earliest of existing Tangram book",
apparently the previous item. [The book
says 1803, but Jerry Slocum reports this is an error for 1813!] Reprinted, with English annotations by Y.
Katagiri, from N. Takashima's copy, 1989.
129 problems (but he counts 128 because he omits one after no. 124), all
included in my version of the previous item, no solutions.
Anonymous. A Grand Eastern Puzzle. C. Davenporte & Co. Registered on 24 Feb 1817, hence the second
oldest English (and European?) tangram book [Slocum, p. 71.] It is identical to Ch'i Ch'iao t'u ho‑pi,
1815, above, except that plates 25 and 27 have been interchanged. It appears to be made by using Chinese pages
and putting a board cover on it. On the
front cover is the only English text:
A
Grand
Eastern Puzzle
----------
THE
following Chineze Puzzle is recommended
to
the Nobility, Gentry, and others, being superior to
any
hitherto invented for the Amusement of the Juvenile
World,
to whom it will afford unceasing recreation and
information;
being formed on Geometrical principles, it
may
not be considered as trifling to those of mature
years,
exciting interest, because difficult and instructive,
imperceptibly
leading the mind on to invention and per-
severence.
-- The Puzzle consists of five triangles, a
square,
and a rhomboid, which may be placed in upwards
of
THREE HUNDRED and THIRTY Characters, greatly re-
sembling
MEN, BEASTS, BIRDS, BOATS, BOTTLES, GLASS-
ES,
URNS, &c. The whole being the
unwearied exertion
of
many years study and application of one of the Lite-
rati
of China, and is now offered to the Public for their
patronage
and support.
ENTERED
AT STATIONERS HALL
----
Published
and sold by
C.
DAVENPORTE and Co.
No.
20, Grafton Street, East Euston Square.
The Fashionable Chinese
Puzzle. Published by J. & E.
Wallis, 42, Skinner Street and J. Wallis Junr, Marine Library,
Sidmouth, nd [Mar 1817]. Photocopy from
Jerry Slocum. This has an illustrated
cover, apparently a slip pasted onto the physical cover. This shows a Chinese gentleman holding a
scroll with the title. There is a
pagoda in the background, a bird hovering over the scroll and a small person in
the foreground examining the scroll.
Slocum's copy has paper watermarked 1816.
PLUS
A Key to the New and Fashionable
Chinese Puzzle, Published by J. and E. Wallis, 42, Skinner Street, London,
Wherein is explained the method of forming every Figure contained in That
Pleasing Amusement. Nd [Mar 1817]. Photocopy from the Bodleian Library, Oxford,
catalogue number Jessel e.1176. TP
seems to made by pasting in the cover slip and has been bound in as a left hand
page. ALSO a photocopy from Jerry
Slocum. In the latter copy, the
apparent TP appears to be a paste down on the cover. The latter copy does not have the Stanzas mentioned below. Slocum's copy has paper watermarked 1815; I
didn't check this at the Bodleian.
NOTE. This is quite a different book than The New
and Fashionable Chinese Puzzle published by Goodrich in New York, 1817.
Bound
in at the beginning of the Fashionable Chinese Puzzle and the Bodleian copy of
the Key is: Stanzas, Addressed to
Messrs. Wallis, on the Ingenious Chinese Puzzle, Sold by them at the Juvenile
Repository, 42, Skinner Street. In the
Key, this is on different paper than the rest of the booklet. The Stanzas has a footnote referring to the
ex-Emperor Napoleon as being in a debilitated state. (Napoleon died in 1821, which probably led to the Bodleian
catalogue's date of c1820 for the entire booklet - but see below. Then follow 28 plates with 323 numbered
figures (but number 204 is skipped), solved in the Key. In the Bodleian copy of the Key, these are
printed on stiff paper, on one side of each sheet, but arranged as facing
pairs, like Chinese booklets.
[Philip
A. H. Brown; London Publishers and Printers
c. 1800-1870; British Library, 1982, p. 212] says the Wallis firm is
only known to have published under the imprint J. & E. Wallis during 1813
and Ruth Wallis showed me another source giving 1813?-1814. This led me to believe that the booklets
originally appeared in 1813 or 1814, but that later issues or some owner
inserted the c1820 sheet of Stanzas, which was later bound in and led the
Bodleian to date the whole booklet as c1820.
Ruth Wallis showed me a source that states that John Wallis (Jun.) set
up independently of his father at 186 Strand in 1806 and later moved to
Sidmouth. Finding when he moved to
Sidmouth might help date this publication more precisely, but it may be a later
reissue. However, Slocum has now found
the book advertised in the London Times in Mar 1817 and says this is the
earliest Western publication of tangrams, based on the 1813/1815 Chinese
work. Wallis also produced a second
book of problems of his own invention and some copies seem to be coloured.
In
AM, p. 43, Dudeney says he acquired the copy of The Fashionable Chinese Puzzle
which had belonged to Lewis Carroll.
He says it was "Published by J. and E. Wallis, 42 Skinner Street,
and J. Wallis, Jun., Marine Library, Sidmouth" and quotes the Napoleon
footnote, so this copy had the Stanzas included. This copy is not in the Strens Collection at Calgary which has
some of Dudeney's papers.
Van
der Waals cites two other works titled
The Fashionable Chinese Puzzle.
An 1818 edition from A. T. Goodridge [sic], NY, is in the American
Antiquarian Society Library (see below) and another, with no details given, is
in the New York Public Library. Could
the latter be the Carroll/Dudeney copy?
Toole
Stott 823 is a copy with the same title and imprint as the Carroll/Dudeney
copy, but he dates it c1840. This
version is in two parts. Part I has 1
leaf text + 26 col. plates -- it seems clear that col. means coloured, a
feature that is not mentioned in any other description of this book -- perhaps
these were hand-coloured by an owner.
Unfortunately, he doesn't give the number of puzzles. I wonder if the last two plates are missing
from this?? Part II has 1 leaf text +
32 col. plates, giving 252 additional figures.
The only copy cited was in the library of J. B. Findlay -- I have
recently bought a copy of the Findlay sale catalogue, ??NYR.
Toole
Stott 1309 is listed with the title: Stanzas, .... J. & F. [sic] Wallis ... and Marine Library, Sidmouth, nd
[c1815]. This has 1 leaf text and 28
plates of puzzles, so it appears that the Stanzas have been bound in and the
original cover title slip is lost or was not recognised by Toole Stott. The date of c1815 is clearly derived from
the Napoleon footnote but 1817 would have been more reasonable, though this may
be a later reissue. Again only one copy
is cited, in the library of Leslie Robert Cole.
Plates
1-28 are identical to plates 1-28 of The Admired Chinese Puzzle, but in
different order. The presence of the
Chinese text in The Admired Chinese Puzzle made me think the Wallis version was
later than it.
Comparison
of the Bodleian booklet with the first 27 plates of Giuoco Cinese, 1818?,
reveals strong similarities. 5 plates
are essentially identical, 17 plates are identical except for one, two or three
changes and 3 plates are about 50% identical.
I find that 264 of the 322 figures in the Wallis booklet occur in Giuoco Cinese, which is about 82%.
However, even when the plates are essentially identical, there are often
small changes in the drawings or the layout.
Some
of the plates were copied by hand into the Hordern Collection's copy of A New Invented Chinese Puzzle, c1806??.
The Admired Chinese Puzzle A New & Correct Edition From the Genuine
Chinese Copy. C. Taylor, Chester,
nd [1817]. Paper is clearly watermarked
1812, but the Prologue refers to the book being brought from China by someone
in Lord Amherst's embassy to China, which took place in 1815-1817 and which visited
Napoleon on St. Helena on its return.
Slocum dates this to after 17 Aug 1817, when Amherst's mission returned
to England and this seems to be the second western book on tangrams. Not in Christopher, Hall, Heyl or Toole
Stott -- Slocum says there is only one copy known in England! It originally had a cover with an
illustration of two Chinese, titled The
Chinese Puzzle, and one of the men
holds a scroll saying To amuse and
instruct. The Chinese text gives the
title Ch'i ch'iao t'u ho pi
(Harmoniously combined book of tangram problems). I have a photocopy of the cover from Slocum. Prologue facing TP; TP; two pp in Chinese,
printed upside down, showing the pieces;
32pp of plates numbered at the upper left (sometimes with reversed
numbers), with problems labelled in Chinese, but most of the characters are
upside down! The plates are printed
with two facing plates alternating with two facing blank pages. Plate 1 has 12 problems, with solution lines
lightly indicated. Plates 2 - 28
contain 310 problems. Plates 29-32
contain 18 additional "caricature Designs" probably intended to be
artistic versions of some of the abstract tangram figures. The Prologue shows faint guide lines for the
lettering, but these appear to be printed, so perhaps it was a quickly done
copperplate. The text of the Prologue
is as follows.
This
ingenious geometrical Puzzle was introduced into this Kingdom from China.
The
following sheets are a correct Copy from the Chinese Publication, brought to
England by a Gentleman of high Rank in the suit [sic] of Lord Amherst's late
Embassy. To which are added caricature
Designs as an illustration, every figure being emblematical of some Being or
Article known to the Chinese.
The
plates are identical to the plates in The Fashionable Chinese Puzzle above, but
in different order and plate 4 is inverted and this version is clearly upside
down.
Sy Hall. A New Chinese Puzzle, The Above Consists of Seven Pieces of Ivory
or Wood, viz. 5 Triangles, 1 Rhomboid, and 1 Square, which will form the 292
Characters, contained in this Book; Observing the Seven pieces must be used to
form each Character. NB. This Edition has been corrected in all its
angles, with great care and attention.
Engraved by Sy Hall, 14 Bury Street, Bloomsbury. 31 plates with 292 problems. Slocum, the Hordern Collection and BL have
copies. I have a photocopy from a
version from Slocum which has no date but is watermarked 1815. Slocum's recent book [The Tangram Book, pp.
74-75] shows a version of the book with the publisher's name as James Izzard
and a date of 1817. Sy
probably is an abbreviation of Sydney (or possibly Stanley?).
(The
BL copy is watermarked IVY MILL
1815 and is bound with a large folding
Plate 2 by Hall, which has 83 tinted examples with solution lines drawn in (by
hand??), possibly one of four sheets giving all the problems in the book. However there is no relationship between the
Plate and the book -- problems are randomly placed and often drawn in different
orientation. I have a photocopy of the
plate on two A3 sheets and a copy of a different plate with 72 problems,
watermarked J. Green 1816.)
A New Chinese Puzzle. Third Edition: Universally allowed to be the
most correct that has been published.
1817. Dalgety has a copy.
A New Chinese Puzzle Consisting
of Seven Pieces of Ivory or Wood, The Whole of which must be used, and will
form each of the CHARACTERS. J.
Buckland, 23 Brook Street, Holborn, London.
Paper watermarked 1816. (Dalgety
has a copy, ??NYS.)
Miss D. Lowry. A Key to the Only Correct Chinese Puzzle
Which has Yet Been Published, with above a Hundred New Figures. No. 1.
Drawn and engraved by Miss Lowry.
Printed by J. Barfield, London, 1817.
The initial D. is given on the next page. Edward Hordern's collection has a copy.
W. Williams. New Mathematical Demonstrations of Euclid,
rendered clear and familiar to the minds of youth, with no other mathematical
instruments than the triangular pieces commonly called the Chinese Puzzle. Invented by Mr. W. Williams, High Beech Collegiate
School, Essex. Published by the author,
London, 1817. [Seen at BL.]
Enigmes Chinoises. Grossin, Paris, 1817. ??NYS -- described and partly reproduced in
Milano. Frontispiece facing the TP
shows an oriental holding a banner which has the pieces and a few problems on
it. This is a small book, with five or
six figures per page. The figures seem
to be copied from the Fashionable Chinese Puzzle, but some figures are not in
that work. Milano says this is cited as
the first French usage of the term 'tangram', but this does not appear in
Milano's photos and it is generally considered that Loyd introduced the word in
the 1850s. Milano's phrasing might be
interpreted as saying this is the first French work on tangrams.
Chinesische-Raethsel. Produced by Daniel Sprenger with designs by
Matthaeus Loder, Vienna, c1818. ??NYS
-- mentioned by Milano.
Chinesisches Rätsel. Enigmes chinoises. Heinrich Friedrich Muller (or Mueller), Vienna, c1810??. ??NYS (van der Waals). This is probably a German edition of the
above and should be dated 1817 or 1818.
However, Milano mentions a box in the Historisches Museum der Stadt
Wien, labelled Grosse Chinesische
Raethsel, produced by Mueller and dated
1815-1820.
Passe-temps Mathématique, ou
Récréation à l'ile Sainte-Hélène. Ce
jeu qui occupé à qu'on prétend, les loisirs du fameux exilé à St.-Hélène. Briquet, Geneva, 1817. 21pp.
[Copy advertised by Interlibrum, Vaduz, in 1990.]
The New and Fashionable Chinese
Puzzle. A. T. Goodrich & Co., New
York, 1817. TP, 1p of Stanzas (seems
like there should be a second page??), 32pp with 346 problems. Slocum has a copy.
[Key] to the Chinese
Philosophical Amusements. A. T.
Goodrich & Co., New York, 1817. TP,
2pp of stanzas (the second page has the Napoleon footnote and a comment which
indicates it is identical to the material in the problem book), Index to the
Key to the Chinese Puzzle, 80pp of solutions as black shapes with white
spacing. Slocum has a copy.
NOTE. This is quite a different book than The
Fashionable Chinese Puzzle published in London by Wallis in 1817.
Slocum
writes: "Although the Goodrich problem book has the same title as the
British book by Wallis and Goodrich has the "Stanzas" poem (except
for the first 2 paragraphs which he deleted) the problem books have completely
different layouts and Goodrich's solution book largely copies Chinese
books."
Il Nuovo e Dilettevole Giuoco
Chinese. Bardi, Florence, 1817. ??NYS -- mentioned by Milano.
Buonapartes Geliefkoosste
Vermaack op St. Helena, op Chineesch Raadsel.
1er Rotterdam by J. Harcke.
Prijs 1 - 4 ??. 2e
Druck te(?) Rotterdam. Ter
Steendrukkery van F. Scheffers & Co.
Nanco Bordewijk has recently acquired this and Slocum has said it is a
translation of one of the English items in c1818. I have just a copy of the cover, and it uses many fancy letters
which I don't guarantee to have read correctly.
Recueil des plus jolis Jeux de
Sociéte, dans lequel on trouve les
gravures d'un grand nombre d'énigmes chinoises, et l'explication de ce nouveau
jeu. Chez Audot, Librairie, Paris,
1818. Pp. 158-162: Le jeu des énigmes
chinoises. This is a short
introduction, saying that the English merchants in Japan have sent it back to
their compatriots and it has come from England to France. This is followed by 11 plates. The first three are numbered. The first shows the pieces formed into a
rectangle. The others have 99 problems,
with 7 shown solved (all six of those on plate 2 and one (the square) on the
10th plate.)
Das grosse chinesische
Rätselspiel für die elegante Welt.
Magazin für Industrie (Leipzig) (1818).
??NYS (van der Waals). Jerry
Slocum informs me that 'Magazin' here denotes a store, not a periodical, and
that this is actually a game version with a packet of 50 cards of problems,
occurring in several languages, from 1818.
I have acquired a set of the cards which lacks one card (no. 17), in a
card box with labels in French and Dutch pasted on. One side has: Nouvelles /
ENIGMES / Chinoises / en Figures et en Paysages with a dancing Chinaman below.
The other side has: Chineesch /
Raadselspel, / voor / de Geleerde Waereld / in / 50 Beelaachlige /
Figuren. with two birds below. Both labels are printed in red, with the
dancing Chinaman having some black lines.
The cards are
82 x 55 mm and are beautifully printed with coloured
pictures of architectonic, anthropomorphic and zoomorphic designs in
appropriate backgrounds. The first card
has four shapes, three of which show the solution with dotted lines. All other cards have just one problem
shape. The reverses have a simple
design. Slocum says the only complete
set he has seen is in the British Library.
I have scanned the cards and the labels.
Gioco cinese chiamato il
rompicapo. Milan, 1818. ??NYS (van der Waals). Fratelli Bettali, Milan, nd, of which
Dalgety has a copy.
Al Gioco Cinese Chiamato Il
Rompicapo Appendice di Figure Rappresentanti ... Preceduta da un Discorso sul
Rompicapo e sulla Cina intitolato Passatempo Preliminare scritto dall'Autore
Firenze All'Insegna dell'Ancora 1818. 64pp
+ covers. The cover or TP has an almond
shape with the seven shapes inside. Pp.
3-43 are text -- the Passatempo Preliminare and an errata page. 12 plates.
The first is headed Alfabeto in fancy Gothic. Plates 1-3 give the alphabet (J and W are omitted). Plate 4 has the positive digits. Plates 5-12 have facing pages giving the
names of the figures (rather orientalized) and contain 100 problems. Hence a total of 133 problems, no solutions. The Hordern collection has a copy and I have
a photocopy from it. This has some
similarities to Giuoco Cinese.
Described and partly reproduced in Milano.
Al Gioco Cinese chiamato il
Rompicapo Appendice. Pietro &
Giuseppe Vallardi, Milan, 1818.
Possibly another printing of the item above. ??NYS -- described in Milano, who reproduces plates 1 & 2, which
are identical to the above item, but with a simpler heading. Milano says the plates are identical to
those in the above item.
Nuove e Dilettevole Giuoco
Chinese. Milano presso li Frat.
Bettalli Cont. del Cappello N. 4031.
Dalgety has a copy. It is
described and two pages are reproduced in Milano from an example in the
Raccolta Bertarelli. Milano dates it as
1818. Cover illustration is the same as
The Fashionable Chinese Puzzle, with the text changed. But it is followed by some more text: Questa ingegnosa invenzione è fondata sopra
principi Geometrici, e consiste in 7 pezzi cioè 5. triangoli, un quadrato ed un
paralellogrammo i quali possono essere combinati in modo da formare piu di 300
figure curiose. The second photo shows
a double page identical to pp. 3-4 of The Fashionable Chinese Puzzle, except
that the page number on p. 4 was omitted in printing and has been written
in. (Quaritch's catalogue 646 (1947)
item 698 lists this as Nuovo e
dilettevole Giuoco Chinese, from Milan,
[1820?])
Nuove e Dilettevole Giuoco
Chinese. Bologna Stamperia in pietra di
Bertinazzi e Compag. ??NYS -- described
and partly reproduced in Milano from an example in the Raccolta Bertarelli. Identical to the above item except that it
is produced lithographically, the text under the cover illustration has been
redrawn, the page borders, the page numbers and the figure numbers are a little
different. Milano's note 5 says the
dating of this is very controversial.
Apparently the publisher changed name in 1813, and one author claims the
book must be 1810. Milano opts for
1813? but feels this is not consistent with the above item. From Slocum's work and the examples above,
it seems clear it must be 1818?
Supplemento al nuovo giuoco
cinese. Fratelli Bettalli, Milan,
1818. ??NYS -- described in Milano, who
says it has six plates and the same letters and digits as Al Gioco Cinese
Chiamato Il Rompicapo Appendice.
Giuoco Cinese Ossia
Raccolta di 364. Figure Geometrica [last letter is blurred] formate con
un Quadrato diviso in 7. pezzi, colli quali si ponno formare infinite Figure
diversi, come Vuomini[sic], Bestie, Ucelli[sic], Case, Cocchi, Barche, Urne,
Vasi, ed altre suppelletili domestiche: Aggiuntovi l'Alfabeto, e li Numeri
Arabi, ed altre nuove Figure. Agapito Franzetti
alle Convertite, Rome, nd [but 1818 is written in by hand]. Copy at the Warburg Institute, shelf mark
FMH 4050. TP & 30 plates. It has alternate openings blank, apparently
to allow you to draw in your solutions, as an owner has done in a few cases. The first plate shows the solutions with
dotted lines, otherwise there are no solutions. There is no other text than on the TP, except for a florid
heading Alfabeto on plate XXVIII. The diagrams have no numbers or names. The upper part of the TP is a plate of three men, intended to be
Orientals, in a tent? The one on the
left is standing and cutting a card marked with the pieces. The man on the right is sitting at a low
table and playing with the pieces. He
is seated on a box labelled ROMPI CAPO. A third man is seated behind the table and
watching the other seated man. On the
ground are a ruler, dividers and right angle.
The Warburg does not know who put the date 1818 in the book, but the
book has a purchase note showing it was bought in 1913. James Dalgety has the only other copy
known. Sotheby's told him that
Franzetti was most active about 1790, but Slocum finds Sotheby's is no longer
very definite about this. I thought it
possible that a page was missing at the beginning which gave a different form
of the title, but Dalgety's copy is identical to this one. Mario Velucchi says it is not listed in a
catalogue of Italian books published in 1800-1900. The letters and numbers are quite different to those shown in
Elffers and the other early works that I have seen, but there are great
similarities to The New and Fashionable Chinese Puzzle, 1817 (check which??),
and some similarities to Al Gioco Cinese above. I haven't counted the figures to verify the 364. Mentioned in Milano, based on the copy I
sent to Dario Uri.
Jeu du Casse Tete Russe. 1817?
??NYS -- described and partly reproduced in Milano from an example in
the Raccolta Bertarelli but which has only four cards. Here the figures are given anthropomorphic
or architectonic shapes. There are four
cards on one coloured sheet and each card has a circle of three figures at the
top with three more figures along the bottom.
Each card has the name of the game at the top of the circle and
"les secrets des Chinois dévoliés" and "casse tête russe"
inside and outside the bottom of the circle.
The figures are quite different than in the following item.
Nuovo Giuoco Russo. Milano presso li Frat. Bettalli Cont. del
Cappello. [Frat. is an abbreviation of
Fratelli (Brothers) and Cont. is an abbreviation of Contrade (road).] Box, without pieces, but with 16 cards of
problems (one being examples) and instruction sheet (or leaflet). ??NYS - described by Milano with reproductions
of the box cover and four of the cards.
This example is in the Raccolta Bertarelli. Box shows a Turkish(?) man handing a box to another. On the first card is given the title and
publisher in French: Le Casse-Tête Russe
Milan, chez les Fr. Bettalli, Rue du Chapeau. The instruction sheet says that the Giuoco Chinese has had such
success in the principal cities of Europe that a Parisian publisher has
conceived another game called the Casse Tête Russe and that the Brothers
Bettalli have hurried to produce it.
Each card has four problems where the figures are greatly elaborated
into architectonic forms, very like those in Metamorfosi, below. Undated, but Milano first gives 1815‑1820,
and feels this is closely related to Metamorfosi and similar items, so he
concludes that it is 1818 or 1819, and this seems to be as correct as present
knowledge permits. The figures are
quite different than in the French version above.
Metamorfosi del Giuoco detto
l'Enimma Chinese. Firenze 1818
Presso Gius. Landi Libraio sul Canto di Via de Servi. Frontispiece shows an angel drawing a pattern
on a board which has the seven pieces at the top. The board leans against a plinth with the solution for making a
square shown on it. Under the drawing
is A. G. inv. Milano reproduces this
plate. One page of introduction,
headed Idea della Metamorfosi Imaginata dell'Enimma Chinese. 100 shapes, some solved, then with elegant
architectonic drawings in the same shapes, signed Gherardesce inv: et inc: Milano identifies the artist as Alessandro
Gherardesca (1779-1852), a Pisan architect.
See S&B, pp. 24‑25.
Grand Jeu du Casse Tête Français
en X. Pieces. ??NYS -- described and
partly reproduced in Milano, who says it comes from Paris and dates it
1818? The figures are anthropomorphic
and are most similar to those in Jeu du Casse Tete Russe.
Grande Giuocho del Rompicapo
Francese. Milano presso Pietro e
Giuseppe Vallardi Contrada di S. Margherita No 401(? my copy is
small and faint). ??NYS -- described
and partly reproduced in Milano, who dates it as 1818-1820. Identical problems as in the previous item,
but the figures have been redrawn rather than copied exactly.
Ch'i Ch'iao pan. c1820.
(Bibliothek Leiden 6891; Antiquariat Israel, Amsterdam.) ??NYS (van der Waals).
Le Veritable Casse‑tete,
ou Enigmes chinoises. Canu Graveur,
Paris, c1820. BL. ??NYS (van der Waals).
L'unico vero Enimma Chinese Tradotto dall'originale, pubblicato a
Londra, da J. Barfield. Florence,
[1820?]. [Listed in Quaritch's catalogue
646 (1947) item 699.)
A tangram appears in Pirnaisches
Wochenblatt of 16 Dec 1820. ??NYS --
described in Slocum, p. 60.
Ch'i Ch'iao ch'u pien ho‑pi. After 1820.
(Bibliothek Leiden 6891.) ??NYS
(van der Waals). 476 examples.
Nouveau Casse‑Tête
Français. c1820 (according to van der
Waals). Reproduced in van der Waals,
but it's not clear how the pages are assembled. Milano dates it a c1815 and indicates it is 16 cards, but van der
Waals looks like it may have been a booklet of 16 pp with TP, example page and
end page. The 16 pp have 80
problems.
Jerry
Slocum has sent 2 large pages with 58 figurative shapes which are clearly the
same pictures. The instructions are
essentially the same, but are followed by rules for a Jeu de Patience on the
second page and there is a 6 x 6 table of words on the first page headed
"Morales trouvées dans les ruines de la célébres Ville de Persépolis
..." which one has to assemble into moral proverbs. It looks like these are copies of folding
plates in some book of games.
Chinese Puzzle Georgina.
A. & S. Josh Myers, & Co 144, Leadenhall Street, London. Ganton Litho. 81 examples on 8 plates with elegant TP. Pages are one-sided sheets, sewn in the
middle, but some are upside down. Seen
at BL (1578/4938).
Bestelmeier, 1823. Item 1278: Chinese Squares. It is not in the 1812 catalogue.
Slocum. Compendium.
Shows the above Bestelmeier entry.
Anonymous. Ch'i ch'iao t'u ho pi (Harmoniously combined
book of Tangram problems) and Ch'i ch'iao t'u chieh (Tangram solutions). Two volumes of tangrams and solutions with
no title page, Chinese labels of the puzzles, in Chinese format (i.e. printed
as long sheets on thin paper, accordion folded and stitched with ribbon. Nd [c1820s??], stiff card covers with
flyleaves of a different paper, undoubtedly added later. 84 pages in each volume, containing 334 problems
and solutions. With ownership stamp of
a cartouche enclosing EWSHING, probably a Mr. E. W. Shing. Slocum says this is a c1820s reprint of the
earliest Chinese tangram book which appeared in 1813 & 1815. This version omits the TP and opening text. I have a photocopy of the opening material
from Slocum. The original problem book
had a preface by Sang‑hsia
K'o, which was repeated in the solution
book with the same date. Includes all
the problems of Shichi-kou-zu Gappeki, qv.
New Series of Ch'i ch'iau
puzzles. Printed by Lou Chen‑wan,
Ch'uen Liang, January 1826. ??NYS. (Copy at Dept. of Oriental Studies, Durham
Univ., cited in R. C. Bell; Tangram Teasers.)
Neues chinesisches Rätselspiel
für Kinder, in 24 bildlichen und alphabetischen Darstellungen. Friese, Pirna. Van der Waals, copying Santi, gives c1805, but Slocum, p. 60,
reports that it first appears in Pirnaisches Wochenblatt of 19 Dec 1829, though
there is another tangram in the issue of 16 Dec 1820. ??NYS.
Child. Girl's Own Book. 1833:
85; 1839: 72; 1842: 156. "Chinese
Puzzles -- These consist of pieces of wood in the form of squares, triangles,
&c. The object is to arrange them
so as to form various mathematical figures."
Anon. Edo Chiekata (How to Learn It??) (In Japanese). Jan 1837, 19pp, 306 problems. (Unclear if this uses the Tangram
pieces.) Reprinted in the same booklet
as Sei Shōnagon, on pp. 37‑55.
A Grand Eastern Puzzle. C. Davenport & Co., London. Nd.
??NYS (van der Waals). (Dalgety
has a copy and gives C. Davenporte (??SP) and Co., No. 20, Grafton Street, East
Euston Square. Chinese pages dated 1813
in European binding with label bearing the above information.)
Augustus De Morgan. On the foundations of algebra, No. 1. Transactions of the Cambridge Philosophical
Society 7 (1842) 287-300. ??NX. On pp. 289, he says "the well-known toy
called the Chinese Puzzle, in which a prescribed number of forms are given, and
a large number of different arrangements, of which the outlines only are drawn,
are to be produced."
Crambrook. 1843.
P. 4, no. 4: Chinese Puzzle.
Chinese Books, thirteen numbers.
Though not illustrated, this seems likely to be the Tangrams -- ??
Boy's Own Book. 1843 (Paris): 439.
No.
19: The Chinese Puzzle. Instructions
give five shapes and say to make one copy of some and two copies of the
others. As written, this has two medium
sized triangles instead of two large ones, though it is intended to be the
tangrams. 11 problem shapes given, no answers. Most of the shapes occur in earlier tangram collections,
particularly in A New Invented Chinese
Puzzle. "The puzzle may be
purchased, ..., at Mr. Wallis's, Skinner street, Snow hill, where numerous
books, containing figures for this ingenious toy may also be
obtained." = Boy's Treasury, 1844,
pp. 426-427, no. 16. It is also
reproduced, complete with the error, but without the reference to Wallis, as: de Savigny, 1846, pp. 355-356, no. 14: Le
casse-tête chinois; Magician's Own
Book, 1857, prob. 49, pp. 289-290;
Landells, Boy's Own Toy-Maker, 1858, pp. 139-140; Book of 500 Puzzles, 1859, pp. 103‑104; Boy's Own Conjuring Book, 1860, pp.
251-252; Wehman, New Book of 200
Puzzles, 1908, pp. 34‑35.
No.
20: The Circassian puzzle. "This
is decidedly the most interesting puzzle ever invented; it is on the same
principle, but composed of many more pieces than the Chinese puzzle, and may
consequently be arranged in more intricate figures. ..." No pieces or
problems are shown. In the next
problem, it says: "This and the Circassian puzzle are published by Mr.
Wallis, Skinner-street, Snow-hill."
= Boy's Treasury, 1844, p. 427, no. 17. = de Savigny, 1846, p. 356, no. 15: Le problème circassien, but
the next problem omits the reference to Wallis.
Although
I haven't recorded a Circassian puzzle yet -- cf in 6.S.2 -- I have just seen
that the puzzle succeeding The Chinese Puzzle in Wehman, New Book of 200
Puzzles, 1908, pp. 35-36, is called The Puzzle of Fourteen which might be the
Circassian puzzle. Taking a convenient
size, this has two equilateral triangles of edge 1 and four each of the
following: a 30o-60o-90o triangle with edges 2, 1, Ö3; a
parallelogram with angles 60o
and 120o with edges 1 and
2; a trapezium with base angles 60o and 60o, with lower and upper base edges 2 and 1,
height Ö3/4 and slant edges 1/2 and Ö3/2.
All 14 pieces make a rectangle 2Ö3 by 4.
Leske. Illustriertes Spielbuch für Mädchen. 1864?
Prob.
584-11, pp. 288 & 405: Chinesisches Verwandlungsspiel. Make a square with the tangram pieces. Shows just five of the pieces, but correctly
states which two to make two copies of.
Prob.
584-16, pp. 289 & 406. Make an
isosceles right triangle with the tangram pieces.
Prob.
584-18/25, pp. 289-291 & 407: Hieroglyphenspiele. Form various figures from various sets of pieces, mostly
tangrams, but the given shapes have bits of writing on them so the assembled
figure gives a word. Only one of the
shapes is as in Boy's Own Book.
Prob.
588, pp. 298 & 410: Etliche Knackmandeln.
Another tangram problem like the preceding, not equal to any in Boy's
Own Book.
Adams & Co., Boston. Advertisement in The Holiday Journal of
Parlor Plays and Pastimes, Fall 1868.
Details?? -- photocopy sent by Slocum.
P. 6: Chinese Puzzle. The
celebrated Puzzle with which a hundred or more symmetrical forms can be made,
with book showing the designs. Though
not illustrated, this seems likely to be the Tangrams -- ??
Mittenzwey. 1880.
Prob. 243-252, pp. 45 & 95-96;
1895?: 272-281, pp. 49 & 97-98;
1917: 272-282, pp. 45 & 92-93.
Make a funnel, kitchen knife, hammer, hat with brim being horizontal or
hanging down or turned up, church, saw, dovecote, hatchet, square, two equal
squares.
J. Murray (editor of the
OED). Two letters to H. E. Dudeney (9
Jun 1910 & 1 Oct 1910). The first inquires about the word 'tangram', following on
Dudeney's mention of it in his "World's best puzzles" (op. cit. in
2). The second says that 'tan' has no
Chinese origin; is apparently mid 19C,
probably of American origin; and the
word 'tangram' first appears in Webster's Dictionary of 1864. Dudeney, AM, 1917, p. 44, excerpts these
letters.
F. T. Wang & C.‑S.
Hsiung. A theorem on the tangram. AMM 49 (1942) 596‑599. They determine the 20 convex regions which
16 isosceles right triangles can form and hence the 13 ones which the Tangram
pieces can form.
Mitsumasa Anno. Anno's Math Games. (Translation of: Hajimete deau sugaku no ehon; Fufkuinkan Shoten,
Tokyo, 1982.) Philomel Books, NY,
1987. Pp. 38-43 & 95-96 show a
simplified 5-piece tangram-like puzzle
which I have not seen before. The
pieces are: a square of side 1; three isosceles right triangles of side 1; a
right trapezium with bases 1 and
2, altitude 1
and slant side Ö2. The trapezium can be viewed as putting
together the square with a triangle. 19
problems are set, with solutions at the back.
James Dalgety. Latest news on oldest puzzles. Lecture to Second Meeting on the History of
Recreational Mathematics, 1 Jun 1996.
10pp. In 1998, he extracted the
two sections on tangrams and added a list of tangram books in his collection
as: The origins of Tangram; © 1996/98; 10pp.
(He lists about 30 books, eight up to 1850.) In 1993, he was buying tangrams in Hong Kong and asked what they
called it. He thought they said
'tangram' but a slower repetition came out 'ta hau ban' and they wrote down the
characters and said it translates as 'seven lucky tiles'. He has since found the characters in 19C
Chinese tangram books. It is quite
possible that Sam Loyd (qv under Murray, above) was told this name and wrote
down 'tangram', perhaps adjusted a bit after thinking up Tan as the inventor.
At the International Congress on
Mathematical Education, Seville, 1996, the Mathematical Association gave
out The 3, 4, 5 Tangram, a cut card tangram, but in a 6 x 8
rectangular shape, so that the medium sized triangle was a 3-4-5 triangle. I modified this in Nov 1999, by stretching along a diagonal to
form a rhombus with angles double the angles of a 3-4-5 triangle, so that four
of the triangles are similar to 3-4-5 triangles. Making the small triangles be actually 3-4-5, all edges are
integral. I made up 35 problems with
these pieces. I later saw that Hans
Wiezorke has mentioned this dissection in CFF, but with no problems. I distributed this as my present at G4G4,
2000.
See S&B 22. I recall there is some dispute as to whether
the basic diagram should be a square or a double square.
E. J. Dijksterhuis. Archimedes.
Munksgaard, Copenhagen, 1956;
reprinted by Princeton Univ. Press, 1987. Pp. 408‑412 is the best discussion of this topic and
supplies most of the classical references.
Archimedes. Letter to Eratosthenes, c-250?. Greek palimpsest, c975, on MS no. 355, from
the Cloister of Saint Sabba (= Mar Saba), Jerusalem, then at Metochion of the
Holy Sepulchre, Constantinopole. [This
MS disappeared in the confusion in Asia Minor in the 1920s but reappeared in
1998 when it was auctioned by Christie's in New York for c2M$. Hopefully, modern technology will allow a
facsimile and an improved transcription in the near future.] Described by J. L. Heiberg (& H. G.
Zeuthen); Eine neue Schrift des Archimedes; Bibliotheca Math. (3) 7 (1906‑1907)
321‑322. Heiberg describes the
MS, but only mentions the loculus. The
text is in Heiberg's edition of Archimedes; Opera; 2nd ed., Teubner, Leipzig,
1913, vol. II, pp. 415‑424, where it has been restored using the Suter
MSS below. Heath only mentions the
problem in passing. Heiberg quotes
Marius Victorinus, Atilius Fortunatianus and cites Ausonius and Ennodius.
H. Suter. Der Loculus Archimedius oder das
Syntemachion des Archimedes. Zeitschr.
für Math. u. Phys. 44 (1899) Supplement
= AGM 9 (1899) 491‑499.
This is a collation from two 17C Arabic MSS which describe the
construction of the loculus. They are
different than the above MS. The German
is included in Archimedes Opera II, 2nd ed., 1913, pp. 420‑424.
Dijksterhuis discusses both of
the above and says that they are insufficient to determine what was
intended. The Greek seems to indicate
that Archimedes was studying the mathematics of a known puzzle. The Arabic shows the construction by cutting
a square, but the rest of the text doesn't say much.
Lucretius. De Rerum Natura. c‑70. ii, 778‑783. Quoted and discussed in H. J. Rose;
Lucretius ii. 778‑83; Classical Review (NS) 6 (1956) 6‑7. Brief reference to assembling pieces into a
square or rectangle.
Decimus Magnus Ausonius. c370.
Works. Edited & translated
by H. G. Evelyn White. Loeb Classical
Library, ??date. Vol. I, Book XVII:
Cento Nuptialis (A Nuptial Cento), pp. 370-393 (particularly the Preface,
pp. 374-375) and Appendix, pp. 395-397.
Refers to 14 little pieces of bone which form a monstrous elephant, a
brutal boar, etc. The Appendix gives
the construction from the Arabic version, via Heiberg, and forms the monstrous
elephant.
Marius Victorinus. 4C.
VI, p. 100 in the edition of Keil, ??NYS. Given in Archimedes Opera II, 2nd ed., 1913, p. 417. Calls it 'loculus Archimedes' and says it
had 14 pieces which make a ship, sword, etc.
Ennodius. Carmina: De ostomachio eburneo. c500.
In: Magni Felicis Ennodii Opera;
ed. by F. Vogel, p. 340. In: Monumenta Germaniae Historica, VII (1885)
249. ??NYS. Refers to ivory pieces to be assembled.
Atilius Fortunatianus. 6C.
??NYS Given in Archimedes Opera
II, p. 417. Same comment as for Marius
Victorinus.
E. Fourrey. Curiositiés Géométriques. (1st ed., Vuibert & Nony, Paris,
1907); 4th ed., Vuibert, Paris,
1938. Pp. 106‑109. Cites Suter, Ausonius, Marius Victorinus,
Atilius Fortunatianus.
Collins. Book of Puzzles. 1927. The loculus of
Archimedes, pp. 7-11. Pieces made from
a double square.
See Hoffmann
& S&B, cited at the beginning of 6.S, for general surveys.
See
Bailey in 6.AS.1 for an 1858 puzzle with 10 pieces and The Sociable and Book of
500 Puzzles, prob. 10, in 6.AS.1 for an 11 piece puzzle.
There
are many versions of this idea available and some are occasionally given in
JRM.
The Richter Anchor Stone puzzles and
building blocks were inspired by Friedrich Froebel (or Fröbel) (1782‑1852),
the educational innovator. He was the
inventor of Kindergartens, advocated children's play blocks and inspired both
the Richter Anchor Stone Puzzles and Milton Bradley. The stone material was invented by Otto Lilienthal (1848‑1896)
(possibly with his brother Gustav) better known as an aviation pioneer -- they
sold the patent and their machines to F. Adolph Richter for 1000 marks. The material might better be described as a
kind of fine brick which could be precisely moulded. Richter improved the stone and began production at Rudolstadt,
Thüringen, in 1882; the plant closed in 1964.
Anchor was the company's trademark.
He made at least 36 puzzles and perhaps a dozen sets of building blocks
which were popular with children, architects, engineers, etc. The Deutsches Museum in Munich has a whole
room devoted to various types of building blocks and materials, including the
Anchor blocks. The Speelgoed Museum 'Op
Stelten' (Sint Vincentiusstraat 86, NL-4902 (or 4901) GL Oosterhout,
Noord-Brabant, The Netherlands; tel: 0262 452 825; fax: 0262 452
413) has a room of Richter blocks and some puzzles. There was an Anker Museum in the Netherlands (Stichting Ankerhaus
(= Anker Museum); Opaalstraat 2‑4 (or Postf. 1061), NL-2400 BB Alphen aan
den Rijn, The Netherlands; tel: 01720‑41188) which produced
replacement parts for Anker stone puzzles.
Modern facsimiles of the building sets are being produced at Rudolstadt.
In 1996 I noticed the ceiling of
the room to the south of the Salon of the Ambassadors in the Alcazar of
Seville. This 15C? ceiling was built by
workmen influenced by the Moorish tradition and has 112 square wooden panels in
a wide variety of rectilineal patterns.
One panel has some diagonal lines and looks like it could be used as a
10 piece tangram-like puzzle. Consider
a 4 x 4 square. Draw both
diagonal lines, then at two adjacent corners, draw two lines making a unit
square at these corners. At the other
two corners draw one of these two lines, namely the one perpendicular to their
common side. This gives six isosceles
right triangles of edge 1; two pentagons with three right angles and
sides 1, 2, 1, Ö2, Ö2; two quadrilaterals with two right angles and
sides 2, 1, Ö2, 2Ö2. Since geometric patterns and panelling are
common features of Arabic art, I wonder if there are any instances of such
patterns being used for a tangram-like puzzle?
Grand Jeu du Casse Tête Français
en X. Pieces. ??NYS -- described and
partly reproduced in Milano, who says it comes from Paris and dates it
1818? The figures are anthropomorphic
and are most similar to those in Jeu du Casse Tete Russe.
Grande Giuocho del Rompicapo
Francese. Milano presso Pietro e
Giuseppe Vallardi Contrada di S. Margherita No 401(? my copy is
small and faint). ??NYS -- described
and partly reproduced in Milano, who dates it as 1818-1820. Identical problems as in the previous item,
but the figures have been redrawn rather than copied exactly.
Allizeau. Les Métamorphoses ou Amusemens
Géometriques Dédiée aux Amateurs Par Allizeau. A Paris chex Allizeau
Quai Malaquais, No 15.
??NYS -- described and partly reproduced in Milano. This uses 15 pieces and the problems tend to
be architectural forms, like towers.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles, nos. 20-27, pp. 27-29 & 88-89 & plate II,
figs. 15-22. This is a set of 20 pieces
of 8 shapes used to make a square, a right triangle, three squares, etc.
Crambrook. 1843.
P. 4, no. 1: Pythagorean Puzzle, with Book. Though not illustrated, this is probably(??) the puzzle described
in Hoffmann, below, which was a Richter Anchor puzzle No. 12 of the same name
and is still occasionally seen. See
S&B 28.
Edward Hordern's collection has
a Circassian Puzzle, c1870, with many pieces, but I didn't record the shapes --
cf Boy's Own Book, 1843 (Paris), in section 6.S.
Mittenzwey. 1880.
Prob.
177-179, pp. 34 & 86; 1895?:
202-204, pp. 38-39 & 88; 1917:
202-204, pp. 35 & 84-85. Consider
the ten piece version of dissecting 5 squares to one (6.AS.1). Use the pieces to make:
a squat octagon, a house gable-end, a
church (no solution), etc.;
two dissimilar rectangles;
three dissimilar parallelograms, two
dissimilar trapezoids. Solution says
one can make many other shapes with these pieces, e.g. a trapezoid with
parallel sides in the proportion 9 :
11.
Prob.
181-184, pp. 34-35 & 87-88; 1895?:
206-209, pp. 39 & 89-90; 1917:
206-209, pp. 36 & 85-86. Take six
equilateral triangles of edge 2. Cut an
equilateral triangle of edge 1 from the corner of each of them, giving 12
pieces. Make a hexagon in eight
different ways [there are many more -- how many??] and three tangram-like
shapes.
Prob.
195-196, pp. 36 & 89; 1895?:
220-221, pp. 41 & 91; 1917:
220-221, pp. 37 & 87. Use four
isosceles right triangles, say of leg 1, to make a square, a 1 x 4
rectangle and an isosceles right triangle.
Nicholas Mason. US Patent 232,140 - Geometrical
Puzzle-Block. Applied: 13 May 1880;
patented 14 Sep 1880. 1p plus 2pp
diagrams. Five squares, six units
square, each cut into four pieces in the same way. Start at the midpoint of a side and cut to an opposite
corner. (This is the same cut used to
produce the ten piece 'Five Squares to One' puzzle.) Cut again in the triangle just formed, from the same midpoint to
a point one unit from the right angle corner of the piece just made. This gives a right triangle of sides 3, 1, Ö10 and a triangle of sides 5, Ö10, 3Ö45. Cut again from the same midpoint across the
trapezoidal piece made by the first cut, to a point five units from the corner
previously cut to. This gives a
triangle of sides 5, 3Ö5, 2Ö10 and a right trapezoid with sides 2,Ö10, 1, 6, 3. This was produced as Hill's American
Geometrical Prize Puzzle in England ("Price, One Shilling.") in
1882. Harold Raizer produced a
facsimile version, with facsimile box label and instructions for IPP22. The instructions have 20 problems to solve
and the solutions have to be submitted by 1 May 1882.
Hoffmann. 1893.
Chap. III, no. 3: The Pythagoras Puzzle, pp. 83-85 & 117-118
= Hoffmann‑Hordern, pp. 69-72.
This has 7 pieces and is quite like the Tangram -- see comment under
Crambrook. Photo on p. 71, with
different version in Hordern Collection, p. 50.
C. Dudley Langford. Note 1538:
Tangrams and incommensurables.
MG 25 (No. 266) (Oct 1941) 233‑235. Gives alternate dissections of the square and some hexagonal
dissections.
C. Dudley Langford. Note 2861:
A curious dissection of the square.
MG 43 (No. 345) (Oct 1959) 198.
There are 5 triangles whose angles are multiples of π/8 = 22½o. He uses these to make a square.
See items at the end of 6.S.
6.T. NO THREE IN A LINE PROBLEM
See
also section 6.AO.2.
Loyd. Problem 14: A crow puzzle.
Tit‑Bits 31 (16 Jan
& 6 Feb 1897) 287 &
343. = Cyclopedia, 1914,
Crows in the corn, pp. 110 & 353.
= MPSL1, prob. 114, pp. 113 & 163‑164. 8 queens with no two attacking and no three
in any line.
Dudeney. The Tribune (7 Nov 1906) 1. ??NX.
= AM, prob. 317, pp. 94 & 222.
Asks for a solution with two men in the centre 2 x 2 square.
Loyd. Sam Loyd's Puzzle Magazine, January 1908. ??NYS.
(Given in A. C. White; Sam Loyd and His Chess Problems; 1913, op.
cit. in 1; p. 100, where it is described as the only solution with 2 pieces in
the 4 central squares.)
Ahrens, MUS I 227, 1910, says he
first had this in a letter from E. B. Escott dated 1 Apr 1909.
(W. Moser, below, refers this to the 1st ed., 1900, but this must be due
to his not having seen it.)
C. H. Bullivant. Home Fun, 1910, op. cit. in 5.S. Part VI, Chap. IV: No. 2: Another draught
puzzle, pp. 515 & 520. The problem
says "no three men shall be in a line, either horizontally or
perpendicularly". The solution
says "no three are in a line in any direction" and the diagram shows
this is indeed true.
Loyd. Picket posts. Cyclopedia,
1914, pp. 105 & 352. = MPSL2, prob.
48, pp. 34 & 136. 2 pieces
initially placed in the 4 central squares.
Blyth. Match-Stick Magic.
1921. Matchstick board game, p.
73. 6 x 6 version phrased as putting "only two in any one line:
horizontal, perpendicular, or diagonal."
However, his symmetric solution has three in a row on lines of slope 2.
King. Best 100. 1927. No. 69, pp. 28 & 55. Problem on the 6 x 6 board -- gives a
symmetric solution. Says "there
are two coins on every row" including "diagonally across it",
but he has three in a row on lines of slope 2.
Loyd Jr. SLAHP.
1928. Checkers in rows, pp. 40
& 98. Different solution than in
Cyclopedia.
M. Adams. Puzzle Book. 1939. Prob. C.83: Stars
in their courses, pp. 144 & 181.
Same solution as King, but he says "two stars in each vertical row,
two in each horizontal row, and two in each of the the two diagonals .... There must not be more than two stars in the
same straight line", but he has three in a row on lines of slope 2.
W. O. J. Moser & J.
Pach. No‑three‑in‑line
problem. In: 100 Research Problems in Discrete Geometry 1986; McGill Univ.,
1986. Problem 23, pp. 23.1 -- 23.4. Survey with 25 references. Solutions are known on the n x n
board for n £
16 and for even n £ 26. Solutions with the symmetries of the square are only known
for n = 2, 4, 10.
R. Penrose. The role of aesthetics in pure and applied
mathematical research. Bull. Inst.
Math. Appl. 10 (1974) 266‑272.
M. Gardner. SA (Jan 1977). Extensively rewritten as Penrose Tiles, Chaps. 1 & 2.
R. Penrose. Pentaplexity. Eureka 39 (1978) 16‑22.
= Math. Intell. 2 (1979) 32‑37.
D. Shechtman, I. Blech, D.
Gratias & J. W. Cohn. Metallic
phase with long‑range orientational order and no translational
symmetry. Physical Rev. Letters 53:20
(12 Nov 1984) 1951‑1953.
Describes discovery of 'quasicrystals' having the symmetry of a Penrose‑like
tiling with icosahedra.
David R. Nelson. Quasicrystals. SA 255:2 (Aug 1986) 32‑41 & 112. Exposits the discovery of
quasicrystals. First form is now called
'Shechtmanite'.
Kimberly-Clark Corporation has
taken out two patents on the use of the Penrose pattern for quilted toilet
paper as the non-repetition prevents the tissue from 'nesting' on the
roll. In Apr 1997, Penrose issued a
writ against Kimberly Clark Ltd. asserting his copyright on the pattern and
demanding damages, etc.
John Kay. Top prof goes potty at loo roll
'rip-off'. The Sun (11 Apr 1997) 7.
Patrick McGowan. It could end in tears as maths boffin sues
Kleenex over design. The Evening
Standard (11 Apr 1997) 5.
Kleenex art that ended in
tears. The Independent (12 Apr 1997) 2.
For a knight on the tiles. Independent on Sunday (13 Apr 1997) 24. Says they exclusively reported Penrose's
discovery of the toilet paper on sale in Dec 1996.
D. Trull. Toilet paper plagiarism. Parascope, 1997 -- available on
www.noveltynet.org/content/paranormal/www.parascope.com/arti...
6.U.2. PACKING BRICKS IN BOXES
In
two dimensions, it is not hard to show that a x b packs A x B
if and only if
a divides either A or
B; b
divides either A or B; A and B
are both linear combinations of
a and b. E.g. 2 x 3
bricks pack a 5 x 6 box.
See
also 6.G.1.
Anon. Prob. 52. Hobbies 30 (No. 767) (25 Jun 1910) 268 & 283 &
(No. 770) (16 Jul 1910) 328. Use
at least one of each of 5 x 7, 5 x 10,
6 x 10 to make the smallest
possible square. Solution says to use 4, 4, 1,
but doesn't show how. There are
lots of ways to make the assembly.
Manuel H. Greenblatt ( -1972, see JRM 6:1 (Winter 1973) 69). Mathematical Entertainments. Crowell, NY, 1965. Construction of a cube, pp. 80‑81. Can
1 x 2 x 4 fill 6 x 6 x 6?
He asserts this was invented by R. Milburn of Tufts Univ.
N. G. de Bruijn. Filling boxes with bricks. AMM 76 (1969) 37‑40. If a1
x ... x an fills A1 x ... x An and
b divides k of
the ai, then
b divides at least k of
the Ai. He previously presented the results, in
Hungarian, as problems in Mat. Lapok 12, pp. 110‑112, prob. 109 and
13, pp. 314‑317, prob. 119.
??NYS.
D. A. Klarner. Brick‑packing puzzles. JRM 6 (1973) 112‑117. General survey. In this he mentions a result that I gave him -- that 2 x 3 x 7
fills a 8 x 11 x 21, but that the box cannot be divided into two
packable boxes. However, I gave him the
case 1 x 3 x 4 in 5
x 5 x 12 which is the smallest example
of this type. Tom Lensch makes fine
examples of these packing puzzles.
T. H. Foregger, proposer; Michael Mather, solver. Problem E2524 -- A brick packing
problem. AMM 82:3 (Mar 1975) 300
& 83:9 (Nov 1976) 741-742. Pack
41 1 x 2 x 4 bricks in a 7 x 7 x 7 box.
One cannot get 42 such bricks into the box.
6.V. SILHOUETTE AND VIEWING PUZZLES
Viewing
problems must be common among draughtsmen and engineers, but I haven't seen
many examples. I'd be pleased to see
further examples.
2 silhouettes.
Circle &
triangle -- van Etten,
Ozanam, Guyot, Magician's Own Book (UK version)
Circle &
square -- van Etten
Circle &
rhombus -- van Etten,
Ozanam
Rectangle
with inner rectangle & rectangle with notch --
Diagram Group.
3 silhouettes.
Circle, circle,
circle -- Madachy
Circle, cross,
square -- Shortz collection (c1884), Wyatt,
Perelman
Circle, oval,
rectangle -- van Etten,
Ozanam, Guyot,
Magician's Own Book (UK version)
Circle, oval,
square -- van Etten,
Tradescant, Ozanam, Ozanam‑Montucla, Badcock,
Jackson, Rational
Recreations,
Endless Amusement II,
Young Man's Book
Circle, rhombus,
rectangle -- Ozanam,
Alberti
Circle, square,
triangle -- Catel,
Bestelmeier, Jackson, Boy's Own Book, Crambrook,
Family Friend, Magician's Own
Book,
Book of 500 Puzzles,
Boy's Own Conjuring Book,
Illustrated Boy's Own Treasury, Riecke, Elliott, Mittenzwey,
Tom Tit, Handy Book, Hoffmann,
Williams, Wyatt, Perelman,
Madachy. But see Note below.
Square, tee,
triangle -- Perelman
4 silhouettes.
Circle, square,
triangle, rectangle with curved
ends -- Williams
2 views.
Antilog, Ripley's,
Diagram Group;
3 views.
Madachy, Ranucci,
For
the classic Circle, Square, Triangle, version, the triangle cannot be not
equilateral. Consider a circle,
rectangle, triangle version. If D is
the diameter of the circle and H is the height of the plug, then the rectangle
has dimensions D x H and the triangle has base D
and side S, so S
= Ö(H2 + D2/4). Making the rectangle a square, i.e. H = D,
makes S = DÖ5/2, while making the triangle equilateral,
i.e. S = D, makes H = DÖ3/2.
van Etten. 1624.
Prob.
22 (misnumbered 15 in 1626) (Prob. 20), pp. 19‑20 & figs. opp.
p. 16 (pp. 35‑36): 2 silhouettes --
one circular, the other triangular, rhomboidal or square. (English ed. omits last case.) The 1630 Examen says the author could have
done better and suggests: isosceles
triangle, several scalene triangles, oval or circle, which he says can be done
with an elliptically cut cone and a scalene cone. I am not sure I believe these.
It seems that the authors are allowing the object to fill the hole and
to pass through the hole moving at an angle to the board rather than
perpendicularly as usually understood.
In the English edition the Examination is combined with that of the next
problem.
Prob.
23 (21), pp. 20‑21 & figs. opp. p. 16 (pp. 37‑38): 3
silhouettes -- circle, oval and square or rectangle. The 1630 Examen suggests:
square, circle, several parallelograms and several ellipses, which he
says can be done with an elliptic cylinder of height equal to the major diameter
of the base. The English Examination
says "a solid colume ... cut Ecliptick-wise" -- ??
John II Tradescant
(1608-1662). Musæum Tradescantianum:
Or, A Collection of Rarities Preserved at South-Lambeth neer London By John
Tradescant. Nathaniel Brooke, London,
1656. [Facsimile reprint, omitting the
Garden List, Old Ashmolean Reprints I, edited by R. T. Gunther, on the occasion
of the opening of the Old Ashmolean Museum as what has now become the Museum of
the History of Science, Oxford. OUP,
1925.] John I & II Tradescant were
gardeners to nobility and then royalty and used their connections to request
naval captains to bring back new plants, curiosities and "Any thing that
Is strang". These were accumulated
at his house and garden in south Lambeth, becoming known as Tradescant's Ark,
eventually being acquired by Elias Ashmole and becoming the foundation of the
Ashmolean Museum in Oxford. This
catalogue was prepared by Elias Ashmole and his friend Thomas Wharton, but they
are not named anywhere in the book. It
was the world's first museum catalogue.
P.
37, last entry: "A Hollow cut in wood, that will fit a round, square and
ovall figure."
Dudeney. Great puzzle crazes. Op. cit. in 2. 1904. He says square, circle and triangle is in a book in front of him dated
1674. I suspect this must be the 1674
English edition of van Etten, but I don't find the problem in the English
editions I have examined. Perhaps
Dudeney just meant that the idea was given in the 1674 book, though he is
specifically referring to the square, circle, triangle version.
Ozanam. 1725.
Vol. II, prob. 58 & 59, pp. 455‑458 & plate 25* (53 (note
there is a second plate with the same number)). Circle and triangle;
circle and rhombus; circle,
oval, rectangle; circle, oval,
square. Figures are very like van
Etten. See Ozanam-Montucla, 1778.
Ozanam. 1725.
Vol. IV. No text, but shown as
an unnumbered figure on plate 15 (17).
3 silhouettes: circle,
rhombus, rectangle.
Simpson. Algebra.
1745. Section XVIII, prob. XXIX,
pp. 279-281. (1790: prob. XXXVII, pp.
306-307. Computes the volume of an
ungula obtained by cutting a cone with a plane. Cf Riecke, 1867.
Alberti. 1747.
No text, but shown as an unnumbered figure on plate XIIII, opp. p. 218
(112), copied from Ozanam, 1725, vol IV.
3 silhouettes: circle, rhombus,
rectangle.
Ozanam-Montucla. 1778.
Faire passer un même corps par un trou quarré, rond &
elliptique. Prob. 46, 1778: 347-348; 1803: 345-346; 1814: 293. Prob. 45,
1840: 149-150. Circle, ellipse, square.
Catel. Kunst-Cabinet. 1790. Die mathematischen Löcher, p. 16 & fig.
42 on plate II. Circle, square,
triangle.
E. C. Guyot. Nouvelles Récréations Physiques et
Mathématiques. Op. cit. in 6.P.2.
1799. Vol. 2, Quatrième
récréation, p. 45 & figs. 1‑4, plate 7, opp. p. 45. 2 silhouettes: circle & triangle; 3
silhouettes: circle, oval, rectangle.
Bestelmeier.
1801. Item 536: Die 3 mathematischen Löcher. (See also the picture of Item 275, but that
text is for another item.) Square,
triangle and circle.
1807. Item 1126: Tricks includes the square, triangle and circle.
Badcock. Philosophical Recreations, or, Winter
Amusements. [1820]. P. 14, no. 23: How to make a Peg that will
exactly fit three different kinds of Holes.
"Let one of the holes be circular, the other square, and the third
an oval; ...." Solution is a
cylinder whose height equals its diameter.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles.
No.
16, pp. 26 & 86. Circle, square,
triangle, with discussion of the dimensions: "a wedge, except that its base must be
a circle".
No.
29, pp. 30 & 89-90. Circle, oval,
square.
Rational Recreations. 1824.
Feat 19, p. 66. Circle, oval,
square.
Endless Amusement II. 1826?
P. 62: "To make a Peg that
will exactly fit three different kinds of Holes." Circle, oval, square. c= Badcock.
The Boy's Own Book. The triple accommodation. 1828: 419;
1828-2: 424; 1829 (US):
215; 1855: 570; 1868: 677. Circle, square and triangle.
Young Man's Book. 1839.
Pp. 294-295. Circle, oval,
square. Identical to Badcock.
Crambrook. 1843.
P. 5, no. 16: The Mathematical Paradox -- the Circle, Triangle, and
Square. Check??
Family Friend 3 (1850) 60 &
91. Practical puzzle -- No. XII. Circle, square, triangle. This is repeated as Puzzle 16 -- Cylinder
puzzle in (1855) 339 with solution in (1856) 28.
Magician's Own Book. 1857.
Prob. 21: The cylinder puzzle, pp. 273 & 296. Circle, square, triangle. = Book of 500 Puzzles, 1859, prob. 21, pp.
87 & 110. = Boy's Own Conjuring
Book, 1860, prob. 20, pp. 235 & 260.
Illustrated Boy's Own
Treasury. 1860. Practical Puzzles, No. 42, pp. 403 &
442. Identical to Magician's Own Book,
with diagram inverted.
F. J. P. Riecke. Op. cit. in 4.A.1, vol. 1, 1867. Art. 33: Die Ungula, pp. 58‑61. Take a cylinder with equal height and
diameter. A cut from the diameter of
one base which just touches the other base cuts off a piece called an ungula
(Latin for claw). He computes the
volume as 4r3/3. He then makes the symmetric cut to produce
the circle, square, triangle shape, which thus has volume (2π ‑ 8/3) r3. Says he has seen such a shape and a board
with the three holes as a child's toy.
Cf Simpson, 1745.
Magician's Own Book (UK
version). 1871. The round peg in the square hole: To pass a cylinder through three different
holes, yet to fill them entirely, pp. 49-50.
Circle, oval, rectangle; circle
& (isosceles) triangle.
Alfred Elliott. Within‑Doors. A Book of Games and Pastimes for the Drawing
Room. Nelson, 1872. [Toole Stott 251. Toole Stott 1030 is a 1873 ed.]
No. 4: The cylinder puzzle, pp. 27‑28 & 30‑31. Circle, square, triangle.
Mittenzwey. 1880.
Prob. 257, pp. 46 & 97;
1895?: 286, pp. 50 & 99-100;
1917: 286, pp. 45 & 94-95.
Circle, square, triangle.
Will Shortz has a puzzle trade
card with the circle, cross, square problem, c1884.
Tom Tit, vol. 2. 1892.
La cheville universelle, pp. 161-162.
= K, no. 28: The universal plug, pp. 72‑73. = R&A, A versatile peg, p. 106. Circle, square, triangle.
Handy Book for Boys and Girls. Op. cit. in 6.F.3. 1892. Pp. 238-242:
Captain S's peg puzzle. Circle, square,
triangle.
Hoffmann. 1893.
Chap. X, no. 20: One peg to fit three holes, pp. 344 & 381‑382
= Hoffmann-Hordern, pp. 238-239, with photo. Circle, square, triangle.
Photo on p. 239 shows two examples: one simply a wood board and
pieces; the other labelled The Holes and Peg Puzzle, from Clark's Cabinet of
Puzzles, 1880-1900, but this seems to be just a card box with the holes.
Williams. Home Entertainments. 1914.
The plug puzzle, pp. 103-104.
Circle, square, triangle and rectangle with curved ends. This is the only example of this four-fold
form that I have seen. Nice drawing of
a board with the plug shown in each hole, except the curve on the sloping faces
is not always drawn down to the bottom.
E. M. Wyatt. Puzzles in Wood, 1928, op. cit. in
5.H.1.
The
"cross" plug puzzle, p. 17.
Square, circle and cross.
The
"wedge" plug puzzle, p. 18.
Square, circle and triangle.
Perelman. FMP.
c1935? One plug for three
holes; Further "plug"
puzzles, pp. 339‑340 & 346. 6
simple versions; 3 harder
versions: square, triangle,
circle; circle, square, cross; triangle, square, tee. The three harder versions are also in FFF,
1957: probs. 69-71, pp. 112 & 118-119; 1979: probs. 73‑75, pp. 137 & 144 = MCBF: probs. 73-75, pp. 134-135 &
142-143.
Anonymous [Antilog]. An elevation puzzle. Eureka 19 (Mar 1957) 11 & 19. Front and top views are a square with a
square inside it. What is the side
view? Gives two solutions.
Anonymous. An elevation puzzle. Eureka 21 (Oct 1958) 7 & 29. Front is the lower half of a circle. Plan (= top view) is a circle. What is the side view? Solution is a V shape, but it ought to
be the other way up! Nowadays, one can
buy potato crisps (= potato chips) in this shape.
Joseph S. Madachy. 3‑D in 2‑D. RMM 2 (Apr 1961) 51‑53 &
3 (Jun 1961) 47. Discusses 3
view and 3 silhouette problems.
3
circular silhouettes, but not a sphere.
Square,
circle, triangle.
Ernest R. Ranucci. Non‑unique orthographic
projections. RMM 14 (Jan‑Feb
1964) 50. 3 views such that there
are 10 different objects with these views.
Ripley's Puzzles and Games. 1966.
Pp. 18-19, item 1. Same problem
as Antilog, 1957. Gives one solution.
Cedric A. B. Smith. Simple projections. MG 62 (No. 419) (Mar 1978) 19-25. This is about how different projections
affect one's recognition of what an object is.
He starts with an example with two views and the isometric projection
which is very difficult to interpret.
He gives three other views, each of which is easily interpreted.
The Diagram Group. The Family Book of Puzzles. The Leisure Circle Ltd., Wembley, Middlesex,
1984. Problem 114, with Solution at the
back of the book. Front view is a
rectangle with an interior rectangle. Side
view is a rectangle with a rectangular notch on front side. Solution is a short cylinder with a straight
notch in it. This is a fairly classic
problem for engineers but I haven't seen it in print elsewhere.
Marek Penszko. Polish your wits -- 3: Loop the loop. Games 11:2 (Feb/Mar 1987) 28 & 58. Draw lines on a glass cube to produce three
given projections. Problem asks for all
three projections to be the same.
When
assembled, a burr looks like three sticks crossing orthogonally, forming a
'star' with six points at the vertices of an octahedron. Slocum says Wyatt [Puzzles in Wood, 1928,
op. cit. in 5.H.1] is the first to use the word 'burr'. Collins, Book of Puzzles, 1927, p. 135,
calls them "Cluster, Parisian or Gordian Knot Puzzles" and states:
"it is believed that they were first made in Paris, if, indeed, they were
not invented there." Since about
1990, there has been considerable development in new types of burr which use
plates or boards rather than sticks, or whose central volume is subdivided more
(cf in 6.W.1).
See
S&B, pp. 62‑85.
See
also 6.BJ.
Most
of these have three pieces which are rectangular in cross-section (1 x 3 x 5) with
slots of the same size and some of the pieces have notches from the slot to the
outside. When one piece is pushed, it
slides, revealing its notch. When
placed properly, this allows a second piece to slide off and out.
In
the 1990s, a more elaborate type of three piece burr appeared. These have three 3 x 3 x 5
pieces which intersect in a central
3 x 3 x 3 region. Within this region, some of the unit cubes
are not present, which allows sliding of the pieces. Some versions of the puzzle permit twisting of pieces though this
usually requires a bit of rounding of edges and the actual examples tend to
break, so these are not as acceptable.
Crambrook. 1843.
P. 5, no. 4: Puzzling Cross 3 pieces.
This seems likely to be a three piece burr, but perhaps is in 6.W.3 --
?? It is followed by "Maltese Cross 6 pieces".
Edward Hordern's collection has
examples in ivory from 1850-1900.
Hoffmann. 1893.
Chap. III, no. 35: The cross‑keys or three‑piece puzzle, pp.
106 & 139 = Hoffmann-Hordern, pp. 104-105, with photo. One piece has an extra small notch which
does not appear in other versions where the dimensions are better chosen. I have recently acquired an example which appears
identical to the illustrations but does not have the extra notch - this came
from a Jaques puzzle box, c1900, and Dalgety has several examples of such boxes
with the solution, where the puzzle is named The Cross Keys Puzzle (cf
discussion at the beginning of Section 11).
The photo on p. 105 is an assembled version, with verbal instructions,
by Jaques & Son, 1880-1895 (but Jaques was producing them up to at least
c1910). Hordern Collection, p. 67,
shows Le Noeud Mystérieux, 1880‑1905, with a pictorial solution and this
does not have the extra notch.
Benson. 1904.
The cross keys puzzle, pp. 205‑206.
Pearson. 1907.
Part III, no. 56: The cross‑keys, pp. 56 & 127‑128.
Anon. A puzzle in wood. Hobbies
31 (No. 795) (7 Jan 1911) 345. Three
piece burr with small extra notch as in Hoffmann.
Anon. Woodwork Joints. Evans,
London, (1918), 2nd ed., 1919. [I have
also seen a 4th ed., 1925, which is identical to the 2nd ed., except for
advertising pages at the end.]
A mortising puzzle, pp. 197‑199.
Collins. Book of Puzzles. 1927. Pp. 136-137: The
cross‑keys puzzle.
E. M. Wyatt. Three piece cross. Puzzles in Wood, 1928, op. cit. in 5.H.1, pp. 24‑25.
Arthur Mee's Children's
Encyclopedia 'Wonder Box'. The
Children's Encyclopedia appeared in 1908, but versions continued until the
1950s. This looks like 1930s?? 3-Piece Mortise with thin pieces.
A. S. Filipiak. Burr puzzle. Mathematical Puzzles, 1942, op. cit. in 5.H.1, p. 101.
Dic Sonneveld seems to be the
first to begin designing three piece burrs of the more elaborate style, perhaps
about 1985. Trevor Wood has made
several examples for sale.
Bill Cutler. Email announcement to NOBNET on 27 Jan
1999. He has begun analysing the newer
style of three piece burr, excluding twist moves. His first stage has examined cases where the centre cube of the
central region is occupied and the piece this central cube belongs to has no
symmetry. He finds 202 x 109 assemblies (I'm not sure if this is an exact
figure) and there are 33 level-8 examples (i.e. where it takes 8
moves to remove the first piece);
6674 level-7 examples; 73362 level-6 examples. He
thinks this is about 70% of the total and it is already about six
times the number of cases considered for the six piece burr (see 6.W.2).
Bill Cutler. Christmas letter of 4 Dec 1999. Says he has completed the above analysis and
found 25 x 1010 possibilities, which took 225 days on a
workstation. The most elaborate
examples require 8 moves to get a piece out and there are 80 of these. He used one for his IPP19 puzzle. He has a website with many of his results on
burrs, etc.: www.billcutlerpuzzles.com
.
6.W.2. SIX PIECE BURR = CHINESE CROSS
The
usual form of these has six sticks, 2 x 2 x 6 (or 8), which have various
notches in them. In the 1990s, new
forms were introduced, using plates or boards.
One version makes an open frame shape, something like a 3 x 3 x 3
chessboard. In the other, 1 x 4
x 6 boards are paired side by side and the result looks like a classic
six-piece burr with the end rectangle divided lengthwise rather than
crosswise. See also 6.W.7.
Jurgis Baltrušaitis. Anamorphoses ou magie artificielle des effets merveilleux. Olivier Perrin Éditeur, Paris, 1969. On pp. 110-116 & 184 is a discussion of
a 1698 engraving "L'Académie des Sciences et des Beaux Arts" by
Sébastien Leclerc (or Le Clerc). In the
right foreground is an object looking like a six piece burr. James Dalgety discusses this in his Latest
news on oldest puzzles; Lecture to Second Meeting on the History of
Recreational Mathematics, 1 Jun 1996.
This image also exists in a large painted version (950 x 480 mm) which
is more precise and more legible in many details, so it is supposed that the
engraving was done in conjunction with the painting. Though it was normal for a notable painting to be turned into an
engraving, the opposite sometimes happened and Leclerc was a famous
engraver. The painter is unknown. The divisions between the pairs of pieces of
the 'burr' are pretty clear in the engraving, but two of them are not visible
in the painting. The 'burr' is also not
quite correctly drawn, but all in all, it seems pretty convincing. James Dalgety was the first to discover this
picture and he has a copy of the engraving, but has not been able to locate the
painting, though it was in the Bernard Monnier Collection exhibited at the
Musée des Arts Decoratifs in Paris in 1975/76.
Camille Frémontier. Sébastien Leclerc and the British
Encyclopeaedists. Sphæra [Newsletter of
the Museum of the History of Science, Oxford] 6 (Aut 1997) 6-7. Discusses the Leclerc engraving which was
used as the frontispiece to several encyclopedias, the earliest being Chambers
Cyclopaedia of 1728.
Minguet. 1733.
Pp. 103-105 (1755: 51-52; 1822: 122-124; 1864: 103-104). Pieces diagrammed. One plain key piece.
Catel. Kunst-Cabinet. 1790. Die kleine Teufelsklaue, p. 10 & fig. 16
on plate I. Figure shows it assembled
and fails to draw one of the divisions between pieces. Description says it is 6 pieces, 2 inches
long, from plum wood and costs 3 groschen (worth about an English penny of the
time). (See also pp. 9-10, fig. 20 on
plate I for Die grosse Teufelsklaue -- the 'squirrelcage'.)
Bestelmeier. 1801.
Item 147: Die kleine Teufelsklaue.
(Note -- there is another item 147 on the next plate.) Only shows it assembled. Brief text may be copying part of
Catel. See also the picture for item
1099 which looks like a six‑piece burr included in a set of puzzles. (See also Item 142: Die grosse
Teufelsklaue.)
Edward Hordern's collection has
examples, called The Oak of Old England, from c1840.
Crambrook. 1843.
P. 5, no. 5: Maltese Cross 6 [pieces], three sorts. Not clear if these might be here or in 6.W.4
or 6.W.5 -- ??
Magician's Own Book. 1857.
Prob. 1: The Chinese cross, pp. 266-267 & 291. One plain key piece. Not the same as in Minguét.
Landells. Boy's Own Toy-Maker. 1858.
Pp. 137-139. Identical to
Magician's Own Book.
Book of 500 Puzzles. 1859.
1: The Chinese cross, pp. 80-81 & 105. Identical to Magician's Own Book.
A. F. Bogesen (1792‑1876). In the Danish Technical Museum, Helsingør (=
Elsinore) are a number of wooden puzzles made by him, including a 6 piece burr,
a 12 piece burr, an Imperial Scale? and a complex (trick??) joint.
Illustrated Boy's Own
Treasury. 1860. Practical Puzzles, No. 23: The Chinese
Cross, pp. 399 & 439. Identical to
Magician's Own Book, except one diagram in the solution omits two labels.
Boy's Own Conjuring Book. 1860.
Prob. 1: The Chinese cross, pp. 228 & 254. Identical to Magician's Own Book.
Hoffmann. 1893.
Chap. III, no. 36: The nut (or six‑piece) puzzle, pp. 106 &
139‑140 = Hoffmann-Hordern, pp. 104-106. Different pieces than in Minguét and Magician's Own Book.
Dudeney. Prob. 473 -- Chinese cross. Weekly Dispatch (23 Nov &
7 Dec 1902), both p. 13.
"There is considerable variety in the manner of cutting out the
pieces, and though the puzzle has been given in some of the old books, I have
purposely presented it in a form that has not, I believe, been published."
Dudeney. Great puzzle crazes. Op. cit. in 2. 1904. "... the
"Chinese Cross," a puzzle of undoubted Oriental origin that was
formerly brought from China by travellers as a curiosity, but for a long time
has had a steady sale in this country."
Wehman. New Book of 200 Puzzles. 1908.
The Chinese cross, pp. 40-41. =
Magician's Own Book.
Dudeney. The world's best puzzles. 1908.
Op. cit. in 2. P. 779 shows a
'"Chinese Cross" which ... is of great antiquity.'
Oscar W. Brown. US Patent 1,225,760 -- Puzzle. Applied: 27 Jun 1916; patented: 15 May 1917. 3pp + 1p diagrams. Coffin says this is the earliest US patent, with several others
following soon after.
Anon. Woodwork Joints, 1918, op. cit. in 6.W.1. Eastern joint puzzle, pp. 196‑197:
Two versions using different pieces.
Six‑piece joint puzzle, pp. 199‑200. Another version.
Western Puzzle Works, 1926
Catalogue. No. 86: 6 piece Wood
Block. Several other possible versions
-- see 6.W.7.
E. M. Wyatt. Six‑piece burr. Puzzles in Wood, 1928, op. cit. in 5.H.1,
pp. 27‑28. Describes 17 versions
from 13 types of piece.
A. S. Filipiak. Mathematical Puzzles, 1942, op. cit. in
5.H.1, pp. 79‑87. 73 versions
from 38 types of piece.
William H. [Bill] Cutler. The six‑piece burr. JRM 10 (1977‑78) 241‑250. Complete, computer assisted, analysis, with
help from T. H. O'Beirne and A. C. Cross.
Pieces are considered as 'notchable' if they can be made by a sequence
of notches, which are produced by two saw cuts and then chiselling out the
space between them. Otherwise viewed,
notches are what could be produced by a wide cutter or router. There are
25 of these which can occur in
solutions. (In 1994, he states that
there are a total of 59 notchable pieces and diagrams all of
them.) One can also have more general
pieces with 'right-angle notches' which would require four chisel cuts -- e.g.
to cut a single 1 x 1 x
1 piece out of a 2 x 2 x 8
rod. Alternatively, one can glue
cubes into notches. There are 369
which can occur in solutions.
(In 1994, he states that there are
837 pieces which produce 2225
different oriented pieces, and he lists them all.) He only considers solid solutions -- i.e.
ones where there are no internal holes.
He finds and lists the 314 'notchable' solutions. There are
119,979 general solutions.
C. Arthur Cross. The Chinese Cross. Pentangle, Over Wallop, Hants., UK, 1979. Brief description of the solutions in the
general case, as found by Cutler and Cross.
S&B, p. 83, describes holey
burrs.
W. H. [Bill] Cutler. Christmas letter, 1987. Sketches results of his (and other's) search
for holey burrs with notchable pieces.
Bill Cutler. Holey 6‑Piece Burr! Published by the author, Palatine,
Illinois. (1986); with addendum, 1988, 48pp. He is now permitting internal holes. Describes holey burrs with notchable pieces,
particularly those with multiple moves to release the first piece.
Bill Cutler. A Computer Analysis of All 6-Piece
Burrs. Published by the author, ibid.,
1994. 86pp. Sketches complete history of the project. (I have included a few details in the
description of his 1977/78 article, above.)
In 1987, he computed all the notchable holey solutions, using about 2
months of PC AT time, finding
13,354,991 assemblies giving 7.4 million solutions. Two of these were level 10 -- i.e. they
require 10 moves to remove the first piece (or pieces), but the highest level
occurring for a unique solution was 5.
After that he started on the general holey burrs and estimated it would
take 400 years of PC AT time -- running at 8 MHz. After some development, the actual time used was about 62.5 PC AT
years, but a lot of this was done on by Harry L. Nelson during idle time on the
Crays at Lawrence Livermore Laboratories, and faster PCs became available, so
the whole project only took about 2½ years, being completed in Aug 1990 and
finding 35,657,131,235 assemblies.
He hasn't checked if all assemblies come apart fully, but he estimates
there are 5.75 billion solutions. He
estimates the project used 45 times the computing power used in the proof of
the Four Color Theorem and that the project would only take two weeks on the
eight RS6000 workstations he now supervises.
Some 70,000 high-level solutions were specifically saved and can be
obtained on disc from him. The highest
level found was 12 and the highest level for a unique solution was 10. See 6.W.1 for a continuation of this
work. He has a website with many of his
results on burrs, etc.:
www.billcutlerpuzzles.com .
Bill Cutler & Frans de
Vreugd. Information leaflet
accompanying their separate IPP22 puzzles, 2002. In 2001, they did an analysis of six-board burrs, of the type
where the boards are paired side by side.
There are 4096 possible such boards, but only 219 usable boards
occur. They looked at all combinations
of six of these and found 14,563,061,989 assemblies. Of these, the highest level found was 13.
6.W.3. THREE PIECE BURR WITH IDENTICAL PIECES
See
S&B, p. 66.
Crambrook. 1843.
P. 5, no. 4: Puzzling Cross 3 pieces.
This seems likely to be a three piece burr, but perhaps is in 6.W.1 --
?? It is followed by "Maltese Cross 6 pieces".
Wilhelm Segerblom. Trick wood joining. SA (1 Apr 1899) 196.
6.W.4. DIAGONAL SIX PIECE BURR = TRICK STAR
This
version often looks like a stellated rhombic dodecahedron. It has two basic forms, one with a key
piece; the other with all pieces identical,
which assembles as two groups of three.
See
S&B, p. 78.
Crambrook. 1843.
P. 5, no. 5: Maltese Cross 6 [pieces], three sorts. Not clear if these belong here or in 6.W.2
or 6.W.5 -- ??
The Youth's Companion. 1875.
[Mail order catalogue.] Reproduced in: Joseph J. Schroeder, Jr.; The Wonderful World of
Toys, Games & Dolls 1860··1930; DBI
Books, Northfield, Illinois, 1977?, p. 19.
Star Puzzle. The picture does
not show which form it is. Slocum's
Compendium also shows this.
Samuel P. Chandler. US Patent 393,816 -- Puzzle. Applied: 9 Mar 1888; patented: 23 Apr 1888. 1p + 1p diagrams. Coffin says this is the earliest version, but it is more complex
than usual, with 12 pieces, and has a key piece.
John S. Pinnell. US Patent 774,197 -- Puzzle. Applied: 9 Oct 1902; patented: 8 Nov 1904. 2pp + 2pp diagrams. Coffin notes that this extends the idea
to 102
pieces!
William E. Hoy. US Patent 766,444 -- Puzzle‑Ball. Applied: 16 Oct 1902; patented: 2 Aug 1904. 2pp + 2pp diagrams. Spherical version with a key piece.
George R. Ford. US Patent 779,121 -- Puzzle. Applied: 16 May 1904; patented: 3 Jan 1905. 1p + 1p diagrams. With square rods, all identical.
He shows assembly by inserting a last piece rather than joining two
groups of three.
Anon. Simple wood puzzle.
Hobbies 31 (No. 786) (5 Nov 1910) 127.
With key piece.
E. M. Wyatt. Woodwork puzzles. Industrial Arts Magazine 12 (1923) 326‑327. Version with a key piece and square rods.
Collins. Book of Puzzles. 1927. The bonbon or nut
puzzle, pp. 137-139.
Iffland Frères (Lausanne). Swiss Patent 245,402 --
Zusammensetzspiel. Received:
19 Nov 1945; granted: 15 Nov
1946; published: 1 Jul 1947. 2pp + 1p diagrams. Stellated rhombic dodecahedral version with a key piece. (Coffin says this is the first to use this
shape, although Slocum has a version c1875.)
6.W.5. SIX PIECE BURR WITH IDENTICAL PIECES
One form has six identical pieces and
all move outward or inward together.
Another form with flat notched pieces has one piece with an extra notch
or an extended notch which allows it to fit in last, either by sliding or
twisting, but this is not initially obvious.
This form is sometimes made with equal pieces so that it can only be assembled
by force, perhaps after steaming, and it then makes an unopenable money
box. This might be considered under
11.M.
Edward Hordern's collection has
a version with one piece a little smaller than the rest from c1800.
Crambrook. 1843.
P. 5, no. 5: Maltese Cross 6 [pieces], three sorts. Not clear if these belong here or in 6.W.2
or 6.W.4 -- ??
C. Baudenbecher catalogue,
c1850s. Op. cit. in 6.W.7. This has an example of the six equal flat
pieces making an unopenable(?) money box.
F. Chasemore. Some mechanical puzzles. In:
Hutchison; op. cit. in 5.A; 1891, chap. 70, part 1, pp. 571‑572. Item 5: The puzzle box, p. 572. Six U pieces make a uniformly expanding
cubical box.
Hoffmann. 1893.
Chap. III, no.33: The bonbon nut puzzle, pp. 104 & 138
= Hoffmann‑Hordern, pp. 102-103, with photo. One piece has an extra notch to simplify the
assembly. Photo on p. 103 shows an
example, almost certainly by Jaques & Son, 1860-1895.
Burnett Fallow. How to make a puzzle money-box. The Boy's Own Paper 15 (No. 755)
(1 Jul 1893) 638. Equal flat
notched pieces forced together to make an unopenable box.
Burnett Fallow. How to make a puzzle picture-frame. The Boy's Own Paper 16 (No. 815)
(25 Aug 1894) 749. Each corner has
the same basic forced construction as used in the puzzle money-box.
Benson. 1904.
The bonbon nut puzzle, p. 204.
Bartl. c1920. Several versions
on p. 306.
Western Puzzle Works, 1926
Catalogue. Last page shows 20 Chinese
Wood Block Puzzles, High Grade. Some of
these are of the present type.
Collins. Book of Puzzles. 1927. The bonbon or nut
puzzle, pp. 137-139. As in Hoffmann.
Iona & Robert Opie and Brian
Alderson. Treasures of Childhood. Pavilion (Michael Joseph), London,
1989. P. 158 shows a "cluster
puzzle which Professor Hoffman [sic] names the 'Nut (or Six‑piece) Puzzle',
but which is usually called 'The Maltese Puzzle'."
William Altekruse. US Patent 430,502 -- Block-Puzzle. Applied: 3 Apr 1890; patented: 17 Jun 1890. 1p + 1p diagrams. Described in S&B, p. 72.
The standard version has 12 pieces, but variations discovered by Coffin
have 14, 36 & 38 pieces.
Western Puzzle Works, 1926
Catalogue. No. 112: 12 piece Wood
Block. Possibly Altekruse.
See
also 6.BJ for other 3D dissections. I
have avoided repeating items, so 6.BJ should also be consulted if you are
reading this section.
Catel. Kunst-Cabinet. 1790. Die grosse Teufelsklaue, pp. 9-10 & fig.
20 on plate I. 24 piece 'squirrel
cage'. Cost 16 groschen.
Bestelmeier. 1801.
Item 142: Die grosse Teufelsklaue.
The 'squirrelcage', identical to Catel, with same drawing, but
reversed. Text may be copying some of
Catel.
C. Baudenbecher, toy
manufacturer in Nuremberg. Sample book
or catalogue from c1850s. Baudenbecher
was taken over by J. W. Spear & Sons in 1919 and the catalogue is now in
the Spear's Game Archive, Ware, Hertfordshire.
It comprises folio and double folio sheets with finely painted
illustrations of the firm's products.
One whole folio page shows about 20 types of wooden interlocking puzzles,
including most of the types mentioned elsewhere in this section and in 6.W.5
and 6.BJ. Until I get a picture, I
can't be more specific.
The Youth's Companion. 1875.
[Mail order catalogue.]
Reproduced in: Joseph J. Schroeder, Jr.; The Wonderful World of Toys,
Games & Dolls 1860··1930; DBI
Books, Northfield, Illinois, 1977?, p. 19.
Shows a 'woodchuck' type puzzle, called White Wood Block Puzzle, from
The Youth's Companion, 1875. I can't
see how many pieces it has: 12 or
18?? Slocum's Compendium also shows
this.
Slocum. Compendium.
Shows: "Mystery", Magic "Champion Puzzle" and
"Puzzle of Puzzles" from Bland's Catalogue, c1890.
The
first looks like a 6 piece burr with circular segments added to make it look
like a ball. So it may be a 6 piece
burr in disguise. See also Hoffmann,
Chap. III, no. 38, pp. 107‑108 & 141‑142
= Hoffmann-Hordern, pp. 106-108 =
Benson, p. 205.
The
second is a six piece puzzle, but the pieces are flattish and it may be of the
type described in 6.W.5.
The
third is complex, with perhaps 18 pieces.
Bartl. c1920. Several versions
on pp. 306-307, including some that are in 6.W.5 and some 'Chinese block
puzzles'.
Western Puzzle Works, 1926
Catalogue. Shows a number of burrs and
similar puzzles.
No.
86: 6 piece Wood Block.
No.
112: 12 piece Wood Block. Possibly
Altekruse.
No.
212: 11 piece Wood Block
The
last page shows 20 Chinese Wood Block Puzzles, High Grade. Some of these are burrs.
Collins. Book of Puzzles. 1927. Other cluster
puzzles, pp. 139-142. Describes and
illustrates: The cluster; The cluster of clusters; The gun cluster; The point cluster; The
flat cluster; The cluster (or secret)
table; The barrel; The Ball;
The football. All of these have
a key piece.
Jan van de Craats. Das unmögliche Escher-puzzle. (Taken from: De onmogelijke Escher-puzzle; Pythagoras (Amsterdam)
(1988).) Alpha 6 (or: Mathematik Lehren / Heft 55 -- ??)
(1992) 12-13. Two Penrose tribars made
into an impossible 5-piece burr.
6.X. ROTATING RINGS OF POLYHEDRA
Generally,
these have edge to edge joints.
'Jacob's ladder' joints are used by Engel -- see 11.L for other forms of
this joint.
I am told these may appear in
Fedorov (??NYS).
Max Brückner. Vielecke und Vielfläche. Teubner, Leipzig, 1900. Section 162, pp. 215‑216 and Tafel
VIII, fig. 4. Describes rings of 2n
tetrahedra joined edge to edge, called stephanoids of the second order. The figure shows the case n = 5.
Paul Schatz. UK Patent 406,680 -- Improvements in or
relating to Boxes or Containers.
Convention date (Germany): 10 Dec 1931;
application date (in UK): 19 Jul 1932;
accepted: 19 Feb 1934. 6pp + 6pp
diagrams. Six and four piece rings of
prisms which fold into a box.
Paul Schatz. UK Patent 411,125 -- Improvements in
Linkwork comprising Jointed Rods or the like.
Convention Date (Germany): 31 Aug 1931;
application Date (in UK): 31 Aug 1932; accepted: 31 May 1934. 3p + 6pp diagrams. Rotating rings of six tetrahedra and linkwork versions of the
same idea, similar to Flowerday's Hexyflex.
Ralph M. Stalker. US Patent 1,997,022 -- Advertising Medium or
Toy. Applied: 27 Apr 1933; patented: 9 Apr 1935. 3pp + 2pp diagrams. "... a plurality of tetrahedron members
or bodies flexibly connected together."
Shows six tetrahedra in a ring and an unfolded pattern for such
objects. Shows a linear form with 14
tetrahedra of decreasing sizes.
Sidney Melmore. A single‑sided doubly collapsible
tessellation. MG 31 (No. 294) (1947)
106. Forms a Möbius strip of three
triangles and three rhombi, which is basically a flexagon (cf 6.D). He sees it has two distinct forms, but
doesn't see the flexing property!! He
describes how to extend these hexagons into a tessellation which has some
resemblance to other items in this section.
Alexander M. Shemet. US Patent 2,688,820 -- Changeable Display
Amusement Device. Applied: 25 Jul
1950; patented: 14 Sep 1954. 2pp + 2pp diagrams. Basically a rotating ring of six tetrahedra,
but says 'at least six'. Gives an
unfolded version or net for making it and a mechanism for flexing it
continually. Cites Stalker.
Wallace G. Walker invented his
"IsoAxis" ® in 1958 while a student at Cranbrook Academy of Art,
Michigan. This is approximately a ring
of ten tetrahedra. He obtained a US
Patent for it in 1967 -- see below. In
1973(?) he sent an example to Doris Schattschneider who soon realised that the
basic idea was a ring of tetrahedra and that Escher tessellations could be
adapted to it. They developed the idea
into "M. C. Escher Kaleidocycles", published by Ballantine in 1977
and reprinted several times since.
Douglas Engel. Flexahedrons. RMM 11 (Oct 1962) 3‑5.
These have 'Jacob's ladder' hinges, not edge‑to‑edge
hinges. He says he invented these in
Fall, 1961. He formed rings of 4, 6, 7, 8
tetrahedra and used a diagonal joining to make rings of 4 and 6 cubes.
Wallace G. Walker. US Patent 3,302,321 -- Foldable
Structure. Filed: 16 Aug 1963; issued: 7 Feb 1967. 2pp + 6pp diagrams.
Joseph S. Madachy. Mathematics on Vacation. Op. cit. in 5.O, (1966), 1979. Solid Flexagons, pp. 81‑84. Based on Engel, but only gives the ring of 6
tetrahedra.
D. Engel. Flexing rings of regular tetrahedra. Pentagon 26 (Spring 1967) 106‑108. ??NYS -- cited in Schaaf II 89 -- write
Engel.
Paul Bethell. More Mathematical Puzzles. Encyclopædia Britannica International,
London, 1967. The magic ring, pp.
12-13. Gives diagram for a
ten-tetrahedra ring, all tetrahedra being regular.
Jan Slothouber &
William Graatsma. Cubics. Octopus Press, Deventer, Holland, 1970. ??NYS. Presents versions of the flexing cubes and the 'Shinsei
Mystery'. [Jan de Geus has sent a
photocopy of some of this but it does not cover this topic.]
Jan Slothouber. Flexicubes -- reversible cubic shapes. JRM 6 (1973) 39‑46. As above.
Frederick George Flowerday. US Patent 3,916,559 -- Vortex Linkages. Filed: 12 Aug 1974 (23 Aug 1973 in UK); issued: 4 Nov 1975. Abstract + 2pp + 3pp diagrams. Mostly shows his Hexyflex, essentially a six
piece ring of tetrahedra, but with just four edges of each tetrahedron
present. He also shows his Octyflex
which has eight pieces. Text refers to
any even number ³ 6.
Naoki Yoshimoto. Two stars in a cube (= Shinsei
Mystery). Described in Japanese
in: Itsuo Sakane; A Museum of Fun;
Asahi Shimbun, Tokyo, 1977, pp. 208‑210.
Shown and pictured as Exhibit V‑1 with date 1972 in: The Expanding Visual World -- A Museum of
Fun; Exhibition Catalogue, Asahi Shimbun, Tokyo, 1979, pp. 102 & 170‑171. (In Japanese). ??get translated??
Lorraine Mottershead. Investigations in Mathematics. Blackwell, Oxford, 1985. Pp. 63-66.
Describes Walkers IsoAxis and rotating rings of six and eight tetrahedra.
The
first few examples illustrate what must be the origin of the idea in more straightforward
situations.
Lucca 1754. c1330.
F. 8r, pp. 31‑32. This
mentions the fact that a circumference increases by 44/7 times the increase
in the radius.
Muscarello. 1478.
Ff.
932-93v, p. 220. A circular garden has
outer circumference 150 and the wall is 3½ thick. What is the inner circumference? Takes
π as 22/7.
F.
95r, p. 222. The internal circumference
of a tower is 20 and its wall is 3 thick. What is the outer circumference? Again takes
π as 22/7.
Pacioli. Summa.
1494. Part II, f. 55r, prob.
33. Florence is 5 miles around the
inside. The wall is 3½
braccia wide and the ditch is
14 braccia wide -- how far is it
around the outside? Several other
similar problems.
William Whiston. Edition of Euclid, 1702. Book 3, Prop. 37, Schol. (3.). ??NYS -- cited by "A Lover" and
Jackson, below.
"A Lover of the
Mathematics." A Mathematical
Miscellany in Four Parts. 2nd ed., S.
Fuller, Dublin, 1735. The First Part
is: An Essay towards the Probable
Solution of the Forty five Surprising PARADOXES, in GORDON's Geography,
so the following must have appeared in Gordon.
Part I, no. 73, p. 56.
"'Tis certainly Matter of Fact, that three certain Travellers went
a Journey, in which, Tho' their Heads travelled full twelve Yards more than
their Feet, yet they all return'd alive, with their Heads on."
Carlile. Collection.
1793. Prob. XXV, p. 17. Two men travel, one upright, the other
standing on his head. Who "sails
farthest"? Basically he compares
the distance travelled by the head and the feet of the first man. He notes that this argument also applies to
a horse working a mill by walking in a circle; the outside of the horse travels
about six times the thickness of the horse further than the inside on each
turn.
Jackson. Rational Amusement. 1821.
Geographical Paradoxes, no. 54, pp. 46 & 115-116. "It is a matter of fact, that three
certain travellers went on a journey, in which their heads travelled full
twelve yards more than their feet; and yet, they all returned alive with their
heads on." Solution says this is
discussed in Whiston's Euclid, Book 3, Prop. 37, Schol. (3.). [This first appeared in 1702.]
K. S. Viwanatha Sastri. Reminiscences of my esteemed tutor. In:
P. K. Srinivasan, ed.; Ramanujan Memorial Volumes: 1: Ramanujan -- Letters and
Reminiscences; 2: Ramanujan -- An
Inspiration; Muthialpet High School,
Number Friends Society, Old Boys' Committee, Madras, 1968. Vol. 1, pp. 89-93. On p. 93, he relates that this was a favourite problem of his
tutor, Srinivasan Ramanujan. Though not
clearly dated, this seems likely to be c1908-1910, but may have been up to
1914. "Suppose we prepare a belt
round the equator of the earth, the belt being
2π feet longer, and if we
put the belt round the earth, how high will it stand? The belt will stand
1 foot high, a substantial
height."
Dudeney. The paradox party. Strand Mag. 38 (No. 228) (Dec 1909) 673‑674 (= AM, p. 139).
Anon. Prob. 58. Hobbies 30 (No. 773) (6 Aug 1910) 405 &
(No. 776) (27 Aug 1910) 448.
Double track circular railway, five miles long. Move all rails outward one foot. How much more material is needed? Solution notes the answer is independent of
the length.
Ludwig Wittgenstein was
fascinated by the problem and used to pose it to students. Most students felt that adding a yard to the
rope would raise it from the earth by a negligible amount -- which it is, in
relation to the size of the earth, but not in relation to the yard. See:
John Lenihan; Science in
Focus; Blackie, 1975, p. 39.
Ernest K. Chapin. Loc. cit. in 5.D.1. 1927.
Prob. 5, p. 87 & Answers p. 7.
A yard is added to a band around the earth. Can you raise it 5 inches?
Answer notes the size of the earth is immaterial.
Collins. Book of Puzzles. 1927. The globetrotter's puzzle,
pp. 68‑69. If you walk around the
equator, how much farther does your head go?
Abraham. 1933.
Prob. 33 -- A ring round the earth, pp. 12 & 24 (9 & 112).
Perelman. FMP.
c1935?? Along the equator, pp.
342 & 349. Same as Collins.
Sullivan. Unusual.
1943.
Prob.
20: A global readjustment. Take a wire
around the earth and insert an extra 40 ft into it -- how high up will it be?
Prob.
23: Getting ahead. If you walk around
the earth, how much further does your head go than your feet?
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. Things are seldom what they seem --
No. 42a, 43, 44, pp. 50-51. 42a
and 43 ask how much the radius increases for a yard gain of circumference. No. 44 asks if we add a yard to a rope
around the earth and then tauten it by pulling outward at one point, how far
will that point be above the earth's surface?
Richard I Hess. Puzzles from Around the World. The author, 1997. (This is a collection of 117 puzzles which he published in
Logigram, the newsletter of Logicon, in 1984-1994, drawn from many
sources. With solutions.) Prob. 28.
Consider a building 125 ft wide and a rubber band stretched around the
earth. If the rubber band has to
stretch an extra 10 cm to fit over the building, how tall is the building? He takes the earth's radius as 20,902,851 ft. He gets three trigonometric equations and uses iteration to
obtain 85.763515... ft.
Erwin Brecher &
Mike Gerrard. Challenging
Science Puzzles. Sterling, 1997. [Reprinted by Goodwill Publishing House, New
Delhi, India, nd [bought in early 2000]].
Pp. 38-39 & 77. The M25 is a
large ring road around London. A man
commutes from the south to the north and finds the distance is the same if goes
by the east or the west, so he normally goes to the east in the morning and to
the west in the evening. Recalling that
the English drive on the left, he realised that his right wheels were on the
outside in both journeys and he worried that they were wear out sooner. So he changed and drove both ways by the
east. But he then worried whether the
wear on the tires was the same since the evening trip was on the outer lanes of
the Motorway.
6.Z. LANGLEY'S ADVENTITIOUS ANGLES
Let
ABC be an isosceles triangle with Ð B
= Ð C = 80o.
Draw BD and
CE, making angles 50o and 60o with the base. Then Ð
CED = 20o.
JRM 15 (1982‑83) 150 cites
Math. Quest. Educ. Times 17 (1910) 75.
??NYS
Peterhouse and Sidney Entrance
Scholarship Examination. Jan 1916. ??NYS.
E. M. Langley. Note 644:
A Problem. MG 11 (No. 160) (Oct
1922) 173.
Thirteen solvers, including
Langley. Solutions to Note 644. MG 11 (No. 164) (May 1923) 321‑323.
Gerrit Bol. Beantwoording van prijsvraag No. 17. Nieuw Archief voor Wiskunde (2) 18 (1936) 14‑66. ??NYS.
Coxeter (CM 3 (1977) 40) and Rigby (below) describe this. The prize question was to completely
determine the concurrent diagonals of regular polygons. The
18‑gon is the key to Langley's problem. However Bol's work was not geometrical.
Birtwistle. Math. Puzzles & Perplexities. 1971.
Find the angle, pp. 86-87. Short
solution using law of sines and other simple trigonometric relations.
Colin Tripp. Adventitious angles. MG 59 (No. 408) (Jun 1975) 98‑106. Studies when Ð CED can be determined and all angles are an
integral number of degrees. Computer
search indicates that there are at most
53 cases.
CM 3 (1977) 12 gives 1939 & 1950 reappearances of the
problem and a 1974 variation.
D. A. Q. [Douglas A.
Quadling]. The adventitious angles
problem: a progress report. MG 61 (No. 415)
(Mar 1977) 55-58. Reports on a number
of contributions resolving the cases which Tripp could not prove. All the work is complicated trigonometry --
no further cases have been demonstrated geometrically.
CM 4 (1978) 52‑53 gives
more references.
D. A. Q. [Douglas A.
Quadling]. Last words on adventitious
angles. MG 62 (No. 421) (Oct 1978)
174-183. Reviews the history, reports
on geometric proofs for all cases and various generalizations.
J[ohn]. F. Rigby. Adventitious quadrangles: a geometrical approach. MG 62 (No. 421) (Oct 1978) 183-191. Gives geometrical proofs for almost all
cases. Cites Bol and a long paper of
his own to appear in Geom. Dedicata (??NYS).
He drops the condition that
ABC be isosceles. His adventitious quadrangles correspond to
Bol's triple intersections of diagonals of a regular n-gon.
MS 27:3 (1994/5) 65 has two straightforward letters on the
problem, which was mentioned in ibid. 27:1 (1994/5) 7. One letter cites 1938 and 1955 appearances. P. 66 gives another solution of the
problem. See next item.
Douglas Quadling. Letter: Langley's adventitious angles. MS 27:3 (1994/5) 65‑66. He was editor of MG when Tripp's article
appeared. He gives some history of the
problem and some life of Langley (d. 1933).
Edward Langley was a teacher at Bedford Modern School and the founding
editor of the MG in 1894-1895. E. T.
Bell was a student of Langley's and contributed an obituary in the MG (Oct
1933) saying that Langley was the finest expositor he ever heard -- ??NYS. Langley also had botanical interests and a
blackberry variety is named for him.
Albrecht Dürer. Underweysung der messung mit dem zirckel
uň [NOTE: ň denotes an
n with an overbar.] richtscheyt,
in Linien ebnen unnd gantzen corporen.
Nürnberg, 1525, revised 1538.
Facsimile of the 1525 edition by Verlag Dr. Alfons Uhl, Nördlingen,
1983. German facsimile with English
translation of the 1525 edition, with notes about the 1538 edition: The
Painter's Manual; trans. by Walter L. Strauss; Abaris Books, NY,
1977. Figures 29‑43 (erroneously
printed 34) (pp. 316-347 in The Painter's Manual, Dürer's 1525 ff. M-iii-v -
N-v-r) show nets and pictures of the regular polyhedra, an approximate sphere
(16 sectors by 8 zones), truncated tetrahedron, truncated cube,
cubo-octahedron, truncated octahedron, rhombi‑cubo-octahedron, snub cube,
great rhombi-cubo-octahedron, truncated cubo‑octahedron (having a pattern
of four triangles replacing each triangle of the cubo‑octahedron -- not
an Archimedean solid) and an elongated hexagonal bipyramid (not even regular
faced). (See 6.AT.3 for more
details.) (Panofsky's biography of
Dürer asserts that Dürer invented the concept of a net -- this is excerpted in
The World of Mathematics I 618‑619.)
In the revised version of 1538, figure 43 is replaced by the
icosi-dodecahedron and great rhombi-cubo-octahedron (figures 43 & 43a,
pp. 414‑419 of The Painter's Manual) to make 9 of
the Archimedean polyhedra.
Albrecht Dürer. Elementorum Geometricorum (?) -- the copy of
this that I saw at the Turner Collection, Keele, has the title page missing,
but Elementorum Geometricorum is the heading of the first text page and appears
to be the book's title. This is a Latin
translation of Unterweysung der Messung ....
Christianus Wechelus, Paris, 1532.
This has the same figures as the 1525 edition, but also has page
numbers. Liber quartus,
fig. 29-43, pp. 145-158 shows the same material as in the 1525
edition.
Cardan. De Rerum Varietate. 1557, ??NYS
= Opera Omnia, vol. III, pp. 246-247.
Liber XIII. Corpora, qua
regularia diei solent, quomodo in plano formentur. Shows nets of the regular solids, except the two halves of the
dodecahedron have been separated to fit into one column of the text.
Barbaro, Daniele. La Practica della Perspectiva. Camillo & Rutilio Borgominieri, Venice,
(1569); facsimile by Arnaldo Forni,
1980, HB. [The facsimile's TP doesn't
have the publication details, but they are given in the colophon. Various catalogues say there are several
versions with dates on the TP and colophon varying independently between 1568
and 1569. A version has both dates
being 1568, so this is presumed to be the first appearance. Another version has an undated title in an
elaborate border and this facsimile must be from that version.] Pp. 45-104 give nets and drawings of the
regular polyhedra and 11 of the 13 Archimedean polyhedra -- he omits the two
snub solids.
E. Welper. Elementa geometrica, in usum geometriae
studiosorum ex variis Authoribus collecta.
J. Reppius, Strassburg, 1620.
??NYS -- cited, with an illustration of the nets of the octahedron,
icosahedron and dodecahedron, in Lange & Springer Katalog 163 -- Mathematik
& Informatik, Oct 1994, item 1350 & illustration on back cover, but the
entry gives Trassburg.
Athanasius Kircher. Ars Magna, Lucis et Umbrae. Rome, 1646.
??NX. Has net of a
rhombi-cuboctahedron.
Pike. Arithmetic. 1788. Pp. 458-459. "As the figures of some of these bodies would give but a
confused idea of them, I have omitted them; but the following figures, cut out
in pasteboard, and the lines cut half through, will fold up into the several
bodies." Gives the regular
polyhedra.
Dudeney. MP.
1926. Prob. 146: The cardboard
box, pp. 58 & 149 (= 536, prob. 316, pp. 109 & 310). All
11 nets of a cube.
Perelman. FMP.
c1935? To develop a cube, pp.
179 & 182‑183. Asserts there
are 10
nets and draws them, but two "can be turned upside down and this
will add two more ...." One shape
is missing. Of the two marked as
reversible, one is symmetric, hence equal to its reverse, but the other isn't.
C. Hope. The nets of the regular star‑faced and
star‑pointed polyhedra. MG 35
(1951) 8‑11. Rather technical.
H. Steinhaus. One Hundred Problems in Elementary
Mathematics. (As: Sto Zadań, PWN -- Polish Scientific
Publishers, Warsaw, 1958.) Pergamon
Press, 1963. With a Foreword by M.
Gardner; Basic Books, NY, 1964. Problem
34: Diagrams of the cube, pp. 20 & 95‑96. (Gives all 11 nets.)
Gardner (pp. 5‑6) refers to Dudeney and suggests the four dimensional
version of the problem should be easy.
M. Gardner. SA (Nov 1966) c= Carnival, pp. 41‑54. Discusses the nets of the cube and the
Answers show all 11 of them.
He asks what shapes these
11 hexominoes will form -- they
cannot form any rectangles. He poses
the four dimensional problem; the
Addendum says he got several answers, no two agreeing.
Charles J. Cooke. Nets of the regular polyhedra. MTg 40 (Aut 1967) 48‑52. Erroneously finds 13 nets of the
octahedron.
Joyce E. Harris. Nets of the regular polyhedra. MTg 41 (Winter 1967) 29. Corrects Cooke's number to 11.
A. Sanders &
D. V. Smith. Nets of the
octahedron and the cube. MTg 42 (Spring 1968) 60‑63. Finds
11 nets for the octahedron and
shows a duality with the cube.
Peter Turney. Unfolding the tesseract. JRM 17 (1984‑85) 1‑16. Finds
261 nets of the 4‑cube. (I don't believe this has ever been confirmed.)
Peter Light &
David Singmaster. The nets of
the regular polyhedra. Presented at New
York Acad. Sci. Graph Theory Day X, 213 Nov 1985. In Notes from New York Graph Theory Day X, 23 Nov 1985;
ed. by J. W. Kennedy & L. V. Quintas; New York Acad. Sci., 1986,
p. 26. Based on Light's BSc
project in 1984-1984 under my supervision.
Shows there are 43,380 nets for the dodecahedron and
icosahedron. I may organize this into a
paper, but several others have since verified the result.
H. Steinhaus. Mathematical Snapshots. Stechert, NY, 1938. (= Kalejdoskop Matematyczny. Książnica‑Atlas, Lwów and
Warsaw, 1938, ??NX.) Pp. 74-75
describes the dodecahedron and says to see the model in the pocket at the end,
but makes no special observation of the self-rising property. Described in detail with photographs in OUP,
NY, eds: 1950: pp. 161-164; 1960: pp. 209‑212; 1969 (1983): pp. 196-198.
Donovan A. Johnson. Paper Folding for the Mathematics
Class. NCTM, 1957, p. 29, section 66:
Pop-up dodecahedron.
M. Kac. Hugo Steinhaus -- a reminiscence and a
tribute. AMM 81 (1974) 572‑581. Material is on pp. 580‑581, with
picture on p. 581.
A pop‑up octahedron was
used by Waddington's as an advertising insert in a trade journal at the London
Toy Fair about 1981. Pop-up cubes have
also been used.
There
is now a web page devoted to Life run by Bob Wainwright -- address is:
http://members.aol.com/life1ine/life/lifepage.htm
[sic!].
M. Gardner. Solitaire game of "Life". SA (Oct 1970). On cellular automata, self‑reproduction, the Garden‑of‑Eden
and the game of "Life". SA
(Feb 1971). c= Wheels, chap.
20-22. In the Oct 1970 issue, Conway
offered a $50 prize for a configuration which became infinitely large -- Bill
Gosper found the glider gun a month later.
At G4G2, 1996, Bob Wainwright showed a picture of Gosper's telegram to
Gardner on 4 Nov 1970 giving the coordinates of the glider gun. I wasn't clear if Wainwright has this or
Gardner still has it.
Robert T. Wainwright, ed. (12
Longue Vue Avenue, New Rochelle, NY, 10804, USA). Lifeline (a newsletter on Life), 11 issues, Mar 1971 -- Sep
1973. ??NYR.
John Barry. The game of Life: is it just a game? Sunday Times (London) (13 Jun 1971). ??NYS -- cited by Gardner.
Anon. The game of Life. Time
(21 Jan 1974). ??NYS -- cited by
Gardner.
Carter Bays. The Game of Three‑dimensional
Life. Dept. of Computer Science, Univ.
of South Carolina, Columbia, South Carolina, 29208, USA, 1986. 48pp.
A. K. Dewdney. The game Life acquires some successors in
three dimensions. SA 256:2
(Feb 1987) 8‑13. Describes
Bays' work.
Bays has started a quarterly 3‑D
Life Newsletter, but I have only seen one (or two?) issues. ??get??
Alan Parr. It's Life -- but not as we know it. MiS 21:3 (May 1992) 12-15. Life on a hexagonal lattice.
There is quite a bit of classical
history which I have not yet entered.
Magician's Own Book notes there is a connection between the Dido version
of the problem and Cutting a card so one can pass through it, Section
6.BA. There are several relatively modern
surveys of the subject from a mathematical viewpoint -- I will cite a few of
them.
Virgil. Aeneid.
‑19. Book 1, lines 360‑370. (p. 38 of the Penguin edition, translated by
W. F. Jackson Knight, 1956.) Dido came
to a spot in Tunisia and the local chiefs promised her as much land as she
could enclose in the hide of a bull.
She cut it into a long strip and used it to cut off a peninsula and
founded Carthage. This story was later
adapted to other city foundations. John
Timbs; Curiosities of History; With New Lights; David Bogue, London, 1857,
devotes a section to Artifice of the thong in founding cities, pp. 49-50,
relating that in 1100, Hengist, the first Saxon King of Kent, similarly
purchased a site called Castle of the Thong and gives references to Indian,
Persian and American versions of the story as well as several other English
versions.
Pappus. c290.
Synagoge [Collection]. Book V,
Preface, para. 1‑3, on the sagacity of bees. Greek and English in SIHGM II 588‑593. A different, abridged, English version is in
HGM II 389‑390.
The Friday Night Book (A Jewish Miscellany). Soncino Press, London, 1933. Mathematical Problems in the Talmud:
Arithmetical Problems, no. 2, pp. 135-136.
A Roman Emperor demanded the Jews pay him a tax of as much wheat as
would cover a space 40 x 40 cubits.
Rabbi Huna suggested that they request to pay in two instalments of 20 x 20
and the Emperor granted this.
[The Talmud was compiled in the period -300 to 500. This source says he is one of the few
mathematicians mentioned in the Talmud, but gives no dates and he is not
mentioned in the EB. From the text, the
problem would seem to be sometime in the 1-5 C.]
The 5C Saxon mercenary, Hengist
or Hengest, is said to have requested from Vortigern: "as much land as can
be encircled by a thong". He
"then took the hide of a bull and cut it into a single leather thong. With this thong he marked out a certain
precipitous site, which he had chosen with the greatest possible cunning." This is reported by Geoffrey of Monmouth in
the 12C and this is quoted by the editor in:
The Exeter Book Riddles; 8-10C
(the book was owned by Leofric, first Bishop of Exeter, who mentioned it in his
will of 1072); Translated and edited by
Kevin Crossley-Holland; (As: The Exeter
Riddle Book, Folio Society, 1978, Penguin, 1979); Revised ed., Penguin, 1993; pp. 101-102.
Lucca 1754. c1330.
Ff. 8r‑8v, pp. 31‑33.
Several problems, e.g. a city 1
by 24 has perimeter 50
while a city 8 by 8 has perimeter 32 but is 8/3
as large; stitching two sacks together
gives a sack 4 times as big.
Calandri. Arimethrica. 1491. F. 97v. Joining sacks which hold 9
and 16 yields a sack which holds
49!!
Pacioli. Summa.
1494. Part II, ff. 55r-55v. Several problems, e.g. a cord of length 4
encloses 100 ducats worth, how much does a cord of length 10 enclose? Also stitching bags together.
Buteo. Logistica. 1559. Prob. 86, pp. 298-299. If 9 pieces of wood are bundled up by 5½
feet of cord, how much cord is needed to bundle up 4 pieces? 5 pieces?
Pitiscus. Trigonometria. Revised ed., 1600, p. 223.
??NYS -- described in: Nobuo Miura; The applications of trigonometry in
Pitiscus: a preliminary essay; Historia Scientarum 30 (1986) 63-78. A square of side 4 and triangle of
sides 5, 5, 3 have the same perimeter but different areas. Presumably he was warning people not to be
cheated in this way.
J. Kepler. The Six‑Cornered Snowflake, op. cit.
in 6.AT.3. 1611. Pp. 6‑11 (8‑19). Discusses hexagons and rhombic interfaces,
but only says "the hexagon is the roomiest" (p. 11 (18‑19)).
van Etten. 1624.
Prob. 90 (87). Pp. 136‑138
(214‑218). Compares fields 6 x 6
and 9 x 3. Compares 4 sacks of diameter 1 with 1 sack
of diameter 4. Compares 2 water pipes
of diameter 1 with 1 water pipe of diameter 2.
Ozanam. 1725.
Question
1, 1725: 327. Question 3, 1778:
328; 1803: 325; 1814: 276;
1840: 141. String twice as long
contains four times as much asparagus.
Question
2, 1725: 328. If a cord of length 10
encloses 200, how much does a cord of length 8 enclose?
Question
3, 1725: 328. Sack 5 high by 4 across
versus 4 sacks 5 high by 1 across.
c= Q. 2, 1778: 328;
1803: 324; 1814: 276; 1840: 140-141, which has sack 4 high by 6
around versus two sacks 4 high by 3 around.
Question
4, 1725: 328‑329. How much water
does a pipe of twice the diameter deliver?
Les Amusemens. 1749.
Prob.
211, p. 376. String twice as long
contains four times as much asparagus.
Prob.
212, p. 377. Determine length of string
which contains twice as much asparagus.
Prob.
223-226, pp. 386-389. Various problems
involving changing shape with the same perimeter. Notes the area can be infinitely small.
Ozanam‑Montucla. 1778.
Question
1, 1778: 327; 1803: 323-324; 1814: 275-276; 1840: 140. Square versus
oblong field of the same circumference.
Prob.
35, 1778: 329-333; 1803: 326-330; 1814: 277-280; 1840: 141-143. Les
alvéoles des abeilles (On the form in which bees construct their combs).
Jackson. Rational Amusement. 1821.
Geometrical Puzzles.
No.
30, pp. 30 & 90. Square field
versus oblong (rectangular?) field of the same perimeter.
No.
31, pp. 30 & 90-91. String twice as
long contains four times as much asparagus.
Magician's Own Book (UK
version). 1871. To cut a card for one to jump through, p.
124, says: "The adventurer of old,
who, inducing the aborigines to give him as much land as a bull's hide would
cover, and made it into one strip by which acres were enclosed, had probably
played at this game in his youth."
See 6.BA.
M. Zacharias. Elementargeometrie und elementare
nicht-Euklidische Geometrie in synthetischer Behandlung. Encyklopädie der Mathematischen
Wissenschaften. Band III, Teil 1,
2te Hälfte. Teubner, Leipzig,
1914-1931. Abt. 28: Maxima und
Minima. Die isoperimetrische
Aufgabe. Pp. 1118-1128. General survey, from Zenodorus (-1C) and
Pappus onward.
6.AD.1. LARGEST PARCEL ONE CAN POST
New section. I have just added the problem of packing a fishing rod as the
diagonal of a box. Are there older
examples?
Richard A. Proctor. Greatest content with parcels' post. Knowledge 3 (3 Aug 1883) 76. Height + girth £ 6 ft. States that a
cylinder is well known to be the best solution. Either for a cylinder or a box, the optimum has height = 2,
girth = 4, with optimum
volumes 2 and 8/π = 2.54... ft3.
R. F. Davis. Letter:
Girth and the parcel post.
Knowledge 3 (17 Aug 1883) 109-110, item 897. Independent discussion of the problem, noting that length
£ 3½ ft is specified, though this doesn't affect the
maximum volume problem.
H. F. Letter: Parcel post
problem. Knowledge 3 (24 Aug 1883) 126,
item 905. Suppose 'length' means
"the maximum distance in a straight line between any two points on its
surface". By this he means the diameter
of the solid. Then the optimum shape is
the intersection of a right circular cylinder with a sphere, the axis of the
cylinder passing through the centre of the sphere, and this has the 'length'
being the diameter of the sphere and the maximum volume is then 2⅓ ft3.
Algernon Bray. Letter:
Greatest content of a parcel which can be sent by post. Knowledge 3 (7 Sep 1883) 159, item 923. Says the problem is easily solved without
calculus. However, for the box, he says
"it is plain that the bulk of half the parcel will be greatest when [its]
dimensions are equal".
Pearson. 1907.
Part II, no. 20: Parcel post limitations, pp. 118 & 195. Length
£ 3½ ft; length + girth £ 6 ft. Solution is a
cylinder.
M. Adams. Puzzle Book. 1939. Prob. B.86: Packing
a parcel, pp. 79 & 107. Same as
Pearson, but first asks for the largest box, then the largest parcel.
Philip Kaplan. More Posers. (Harper & Row, 1964);
Macfadden-Bartell Books, 1965.
Prob. 18, pp. 27 & 89.
Ship a rifle about 1½ yards long when the post office does not
permit any dimension to be more than
1 yard.
T. J. Fletcher. Doing without calculus. MG 55 (No. 391) (Feb 1971) 4‑17. Example 5, pp. 8‑9. He says only that length + girth £ 6 ft.
However, the optimal box has length 2, so the maximal length restriction
is not critical.
I have looked at the current
parcel post regulations and they say
length £ 1.5m and
length + girth £ 3m, for which the largest box is
1 x ½ x ½, with volume 1/4 m3. The largest cylinder has length
1 and radius 1/π
with volume 1/π m3.
I have also considered the
simple question of a person posting a fishing rod longer than the maximal
length by putting it diagonally in a box.
The longest rod occurs at a boundary maximum, at 3/2 x 3/4 x 0 or 3/2 x 0 x 3/4, so one can post a rod of length 3Ö5/4 = 1.677...
m, which is about 12% longer than
1.5m. In this problem, the use
of a cylinder actually does worse!
6.AE. 6" HOLE THROUGH SPHERE LEAVES CONSTANT VOLUME
Hamnet Holditch. Geometrical theorem. Quarterly J. of Pure and Applied Math. 2
(1858) ??NYS, described by Broman. If a
chord of a closed curve, of constant length
a+b, be divided into two parts
of lengths a, b respectively, the difference between the
areas of the closed curve, and of the locus of the dividing point as the chord
moves around the curve, will be
πab. When the closed curve
is a circle and a = b, then this is the two dimensional version
given by Jones, below. A letter from
Broman says he has found Holditch's theorem cited in 1888, 1906, 1975 and 1976.
Richard Guy (letter of 27 Feb
1985) recalls this problem from his schooldays, which would be late 1920s-early
1930s, and thought it should occur in calculus texts of that time, but could
not find it in Lamb or Caunt.
Samuel I. Jones. Mathematical Nuts. 1932. P. 86. ??NYS.
Cited by Gardner, (SA, Nov 1957) = 1st Book, chap. 12, prob.
7. Gardner says Jones, p. 93, also
gives the two dimensional version: If
the longest line that can be drawn in an annulus is 6" long, what is the
area of the annulus?
L. Lines. Solid Geometry. Macmillan, London, 1935;
Dover, 1965. P. 101, Example
8W3: "A napkin ring is in the form
of a sphere pierced by a cylindrical hole.
Prove that its volume is the same as that of a sphere with diameter
equal to the length of the hole."
Solution is given, but there is no indication that it is new or recent.
L. A. Graham. Ingenious Mathematical Problems and
Methods. Dover, 1959. Prob. 34: Hole in a sphere, pp. 23 & 145‑147. [The material in this book appeared in
Graham's company magazine from about 1940, but no dates are provided in the
book. (??can date be found out.)]
M. H. Greenblatt. Mathematical Entertainments, op. cit. in
6.U.2, 1965. Volume of a modified
bowling ball, pp. 104‑105.
C. W. Trigg. Op. cit. in 5.Q. 1967. Quickie 217: Hole
in sphere, pp. 59 & 178‑179.
Gives an argument based on surface tension to see that the ring surface
remains spherical as the hole changes radius.
Problem has a 10" hole.
Andrew Jarvis. Note 3235:
A boring problem. MG 53 (No.
385) (Oct 1969) 298‑299. He calls
it "a standard problem" and says it is usually solved with a triple
integral (??!!). He gives the standard
proof using Cavalieri's principle.
Birtwistle. Math. Puzzles & Perplexities. 1971.
Tangential
chord, pp. 71-73. 10" chord in an annulus. What is the area of the annulus? Does traditionally and then by letting inner
radius be zero.
The
hole in the sphere, pp. 87-88 & 177-178.
Bore a hole through a sphere so the remaining piece has half the volume
of the sphere. The radius of the hole
is approx. .61 of the radius of the sphere.
Another
hole, pp. 89, 178 & 192.
6" hole cut out of
sphere. What is the volume of the
remainder? Refers to the tangential
chord problem.
Arne Broman. Holditch's theorem: An introductory
problem. Lecture at ICM, Helsinki,
Aug 1978. Broman then sent out
copies of his lecture notes and a supplementary letter on 30 Aug 1978. He discusses Holditch's proof (see above)
and more careful modern versions of it.
His letter gives some other citations.
6.AF. WHAT COLOUR WAS THE BEAR?
A
hunter goes 100 mi south, 100 mi east and 100 mi north and finds himself where
he started. He then shoots a bear --
what colour was the bear?
Square
versions: Perelman; Klamkin, Breault & Schwarz; Kakinuma, Barwell & Collins; Singmaster.
I
include other polar problems here. See
also 10.K for related geographical problems.
"A Lover of the
Mathematics." A Mathematical
Miscellany in Four Parts. 2nd ed., S.
Fuller, Dublin, 1735. The First Part
is: An Essay towards the Probable
Solution of the Forty five Surprising PARADOXES, in GORDON's Geography,
so the following must have appeared in Gordon.
Part I, no. 10, p. 9.
"There is a particular Place of the Earth where the Winds (tho'
frequently veering round the Compas) do always blow from the North Point."
Philip Breslaw (attrib.). Breslaw's Last Legacy; or the Magical
Companion: containing all that is Curious, Pleasing, Entertaining and Comical;
selected From the most celebrated Masters of Deception: As well with Slight of
Hand, As with Mathematical Inventions.
Wherein is displayed The Mode and Manner of deceiving the Eye; as
practised by those celebrated Masters of Mirthful Deceptions. Including the various Exhibitions of those
wonderful Artists, Breslaw, Sieur, Comus, Jonas, &c. Also the Interpretation of Dreams,
Signification of Moles, Palmestry, &c.
The whole forming A Book of real Knowledge in the Art of Conjuration. (T. Moore, London, 1784, 120pp.) With an accurate Description of the Method
how to make The Air Balloon, and inject the Inflammable Air. (2nd ed., T. Moore, London, 1784,
132pp; 5th ed., W. Lane, London, 1791,
132pp.) A New Edition, with great
Additions and Improvements. (W. Lane,
London, 1795, 144pp.) Facsimile from
the copy in the Byron Walker Collection, with added Introduction, etc., Stevens
Magic Emporium, Wichita, Kansas, 1997.
[This was first published in 1784, after Breslaw's death, so it is
unlikely that he had anything to do with the book. There were versions in
1784, 1791, 1792, 1793, 1794, 1795, 1800, 1806, c1809, c1810, 1811,
1824. Hall, BCB 39-43, 46-51. Toole Stott 120-131, 966‑967. Heyl 35-41.
This book went through many variations of subtitle and contents -- the
above is the largest version.]. I will
cite the date as 1784?.
Geographical
Paradoxes.
Paradox
I, p. 35. Where is it noon every half
hour? Answer: At the North Pole in
Summer, when the sun is due south all day long, so it is noon every moment!
Paradox
II, p. 36. Where can the sun and the
full moon rise at the same time in the same direction? Answer: "Under the North Pole, the sun
and the full moon, both decreasing in south declination, may rise in the equinoxial
points at the same time; and under the North Pole, there is no other point of
compass but south." I think this
means at the North Pole at the equinox.
Carlile. Collection.
1793. Prob. CXVI, p. 69. Where does the wind always blow from the north?
Jackson. Rational Amusement. 1821.
Geographical Paradoxes.
No. 7,
pp. 36 & 103. Where do all winds
blow from the north?
No. 8,
pp. 36 & 110. Two places 100
miles apart, and the travelling directions are to go 50
miles north and 50 miles south.
Mr. X [cf 4.A.1]. His Pages.
The Royal Magazine 10:3 (Jul 1903) 246-247. A safe catch. Airship
starts at the North Pole, goes south for seven days, then west for seven
days. Which way must it go to get back
to its starting point? No solution given.
Pearson. 1907.
Part
II, no. 21: By the compass, pp. 18 & 190.
Start at North Pole and go
20 miles southwest. What direction gets back to the Pole the
quickest? Answer notes that it is hard
to go southwest from the Pole!
Part
II, no. 15: Ask "Where's the north?" -- Pope, pp. 117 & 194. Start
1200 miles from the North Pole
and go 20 mph due north by the compass.
How long will it take to get to the Pole? Answer is that you never get there -- you get to the North Magnetic
Pole.
Ackermann. 1925.
P. 116. Man at North Pole
goes 20 miles south and 30 miles west.
How far, and in what direction, is he from the Pole?
Richard Guy (letter of 27 Feb
1985) recalls this problem (I think he is referring to the 'What colour was the
bear' version) from his schooldays in the 1920s.
H. Phillips. Week‑End. 1932. Prob. 8, pp. 12
& 188. = his Playtime Omnibus,
1933, prob. 10: Popoff, pp. 54 & 237.
House with four sides facing south.
H. Phillips. The Playtime Omnibus. Faber & Faber, London, 1933. Section XVI, prob. 11: Polar conundrum, pp.
51 & 234. Start at the North Pole,
go 40
miles South, then 30 miles West.
How far are you from the Pole.
Answer: "Forty miles. (NOT thirty, as one is tempted to
suggest.)" Thirty appears to be a
slip for fifty??
Perelman. FFF.
1934. 1957: prob. 6, pp. 14-15
& 19-20: A dirigible's flight;
1979: prob. 7, pp. 18-19 & 25-27: A helicopter's flight. MCBF: prob. 7, pp. 18-19 & 25-26: A
helicopter's flight.
Dirigible/helicopter starts at Leningrad and goes 500km
N, 500km E, 500km
S, 500km W.
Where does it land? Cf Klamkin
et seq., below.
Phillips. Brush.
1936. Prob. A.1: A stroll at the
pole, pp. 1 & 73. Eskimo living at
North Pole goes 3 mi south and 4 mi east.
How far is he from home?
Haldeman-Julius. 1937.
No. 51: North Pole problem, pp. 8 & 23. Airplane starts at North Pole, goes 30 miles south, then 40 miles
west. How far is he from the Pole?
J. R. Evans. The Junior Week‑End Book. Gollancz, London, 1939. Prob. 9, pp. 262 & 268. House with four sides facing south.
Leopold. At Ease!
1943. A helluva question!, pp.
10 & 196. Hunter goes 10
mi south, 10 mi west, shoots a bear and drags it 10
mi back to his starting point.
What colour was the bear? Says
the only geographic answer is the North Pole.
E. P. Northrop. Riddles in Mathematics. 1944.
1944: 5-6; 1945: 5-6; 1961: 15‑16. He starts with the house which faces south on all sides. Then he has a hunter that sees a bear 100
yards east. The hunter runs 100 yards
north and shoots south at the bear -- what colour .... He then gives the three‑sided walk
version, but doesn't specify the solution.
E. J. Moulton. A speed test question; a problem in
geography. AMM 51 (1944) 216 &
220. Discusses all solutions of the
three-sided walk problem.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. No. 50: A fine outlook, pp. 54-55. House facing south on all sides used by an
artist painting bears!
Leeming. 1946.
Chap. 3, prob. 32: What color was the bear?, pp. 33 & 160. Man walks
10 miles south, then 10
miles west, where he shoots a bear.
He drags it 10 miles north to his base. What color .... He gives only one solution.
Darwin A. Hindman. Handbook of Indoor Games &
Contests. (Prentice‑Hall,
1955); Nicholas Kaye, London,
1957. Chap. 16, prob. 4: The bear
hunter, pp. 256 & 261. Hunter
surprises bear. Hunter runs 200
yards north, bear runs 200 yards east, hunter fires south at bear. What colour ....
Murray S. Klamkin,
proposer; D. A. Breault & Benjamin
L. Schwarz, solvers. Problem 369. MM 32 (1958/59) 220 &
33 (1959/60) 110 & 226‑228. Explorer goes 100 miles north, then east, then south, then
west, and is back at his starting point.
Breault gives only the obvious solution. Schwartz gives all solutions, but not explicitly. Cf Perelman, 1934.
Benjamin L. Schwartz. What color was the bear?. MM 34 (1960) 1-4. ??NYS -- described by Gardner, SA (May 1966) = Carnival, chap.
17. Considers the problem where the
hunter looks south and sees a bear
100 yards away. The bear goes 100 yards east and the
hunter shoots it by aiming due south.
This gives two extra types of solution.
Ripley's Puzzles and Games. 1966.
Pp. 18, item 5. 50 mi N, 1000 mi W,
10 mi S to return to your
starting point. Answer only gives the
South Pole, ignoring the infinitely many cases near the North Pole. Looking at this made me realise that when
the sideways distance is larger than the circumference of the parallel at that
distance from the pole, then there are other solutions that start near the
pole. Here there are three solutions
where one starts at distances 109.2,
29.6 or 3.05 miles from the South Pole,
circling it 1, 2 or 3 times.
Yasuo Kakinuma, proposer; Brian Barwell and Craig H. Collins,
solvers. Problem 1212 -- Variation of
the polar bear problem. JRM 15:3 (1982‑83)
222 & 16:3 (1983-84) 226‑228.
Square problem going one mile south, east, north, west. Barwell gets the explicit quadratic
equation, but then approximates its solutions.
Collins assumes the earth is flat near the pole.
David Singmaster. Bear hunting problems. Submitted to MM, 1986. Finds explicit solutions for the general
version of Perelman/Klamkin's problem.
[In fact, I was ignorant of (or had long forgotten) the above when I remembered
and solved the problem. My thanks to an
editor (Paul Bateman ??check) for referring me to Klamkin. The Kakinuma et al then turned up also.] Analysis of the solutions leads to some
variations, including the following.
David Singmaster. Home is the hunter. Man heads north, goes ten miles, has lunch,
heads north, goes ten miles and finds himself where he started.
Used
as: Explorer's problem by Keith Devlin
in his Micromaths Column; The Guardian (18 Jun 1987) 16 &
(2 Jul 1987) 16.
Used
by me as one of: Spring term puzzles;
South Bank Polytechnic Computer Services Department Newsletter (Spring 1989)
unpaged [p. 15].
Used
by Will Shortz in his National Public Radio program 6? Jan 1991.
Used
as: A walk on the wild side, Games 15:2
(No. 104) (Aug 1991) 57 & 40.
Used
as: The hunting game, Focus 3 (Feb
1993) 77 & 98.
Used
in my Puzzle Box column, G&P, No. 11 (Feb 1995) 19 &
No. 12 (Mar 1995) 41.
Bob Stanton. The explorers. Games Magazine 17:1 (No. 113) (Feb 1993) 61 & 43. Two explorers set out and go 500
miles in each direction. Madge
goes N, W, S, E, while Ellen goes E, S, W, N.
At the end, they meet at the same point. However, this is not at their starting point. How come?
and how far are they from their starting point, and in what
direction? They are not near either
pole.
Yuri B. Chernyak & Robert S.
Rose. The Chicken from Minsk. BasicBooks, NY, 1995. Chap. 11, prob. 9: What color was that bear?
(A lesson in non-Euclidean geometry), pp. 97 & 185-191. Camper walks south 2 km, then west 5 km,
then north 2 km; how far is he from his starting point? Solution analyses this and related problems,
finding that the distance x satisfies
0 £ x £ 7.183, noting that there are many minimal cases
near the south pole and if one is between them, one gets a local maximum, so
one has to determine one's position very carefully.
David Singmaster. Symmetry saves the solution. IN: Alfred S. Posamentier & Wolfgang
Schulz, eds.; The Art of Problem Solving: A Resource for the Mathematics
Teacher; Corwin Press, NY, 1996, pp. 273-286.
Sketches the explicit solution to Klamkin's problem as an example of the
use of symmetric variables to obtain a solution.
Anonymous. Brainteaser B163 -- Shady matters. Quantum 6:3 (Jan/Feb 1996) 15 & 48. Is there anywhere on earth where one's
shadow has the same length all day long?
There are several versions of
this. The simplest is moving a ladder
or board around a corner -- here the problem is two-dimensional and the ladder
is thin enough to be considered as a line.
There are slight variations -- the corner can be at a T or +
junction; the widths of the
corridors may differ; the angle may not
be a right angle; etc. If the object being moved is thicker -- e.g.
a table -- then the problem gets harder.
If one can use the third dimension, it gets even harder.
H. E. Licks. Op. cit. in 5.A, 1917. Art. 110, p. 89. Stick going into a circular shaft in the ceiling. Gets
[h2/3 + d2/3)]3/2 for maximum length, where h is
the height of the room and d is the diameter of the shaft. "A simple way to solve a problem which
has proved a stumbling block to many."
Abraham. 1933.
Prob. 82 -- Another ladder, pp. 37 & 45 (23 & 117). Ladder to go from one street to another, of
different widths.
E. H. Johnson, proposer; W. B. Carver, solver. Problem E436. AMM 47 (1940) 569 & 48 (1941) 271‑273. Table going through a doorway. Obtains
6th order equation.
J. S. Madachy. Turning corners. RMM 5 (Oct 1961) 37, 6
(Dec 1961) 61 & 8 (Apr 1962) 56. In 5, he asks for the greatest length of board which can be moved
around a corner, assuming both corridors have the same width, that the board is
thick and that vertical movement is allowed.
In 6, he gives a numerical answer for his original values and asserts
the maximal length for planar movement, with corridors of width w
and plank of thickness t, is
2 (wÖ2 ‑ t). In vol. 8, he says no two solutions have
been the same.
L. Moser, proposer; M. Goldberg and J. Sebastian, solvers. Problem 66‑11 -- Moving furniture
through a hallway. SIAM Review 8 (1966)
381‑382 & 11 (1969) 75‑78 &
12 (1970) 582‑586.
"What is the largest area region which can be moved through a
"hallway" of width one (see Fig. 1)?" The figure shows that he wants to move around a rectangular
corner joining two hallways of width one.
Sebastian (1970) studies the problem for moving an arc.
J. M. Hammersley. On the enfeeblement of mathematical skills
.... Bull. Inst. Math. Appl. 4 (1968)
66‑85. Appendix IV -- Problems,
pp. 83‑85, prob. 8, p. 84.
Two corridors of width 1 at a corner. Show the largest object one can move around it has area < 2 Ö2 and that there is an object of
area ³ π/2 + 2/π
= 2.2074.
Partial
solution by T. A. Westwell, ibid. 5 (1969) 80, with editorial comment thereon
on pp. 80‑81.
T. J. Fletcher. Easy ways of going round the bend. MG 57 (No. 399) (Feb 1973) 16‑22. Gives five methods for the ladder problem
with corridors of different widths.
Neal R. Wagner. The sofa problem. AMM 83 (1976) 188‑189.
"What is the region of largest area which can be moved around a
right‑angled corner in a corridor of width one?" Survey.
R. K. Guy. Monthly research problems, 1969‑77. AMM 84 (1977) 807‑815. P. 811 reports improvements on the sofa
problem.
J. S. Madachy &
R. R. Rowe. Problem 242 --
Turning table. JRM 9 (1976‑77)
219‑221.
G. P. Henderson, proposer; M. Goldberg, solver; M. S. Klamkin, commentator. Problem 427. CM 5 (1979) 77 & 6 (1979) 31‑32 & 49‑50. Easily finds maximal area of a rectangle
going around a corner.
Research news: Conway's sofa problem. Mathematics Review 1:4 (Mar 1991) 5-8 &
32. Reports on Joseph Gerver's almost
complete resolution of the problem in 1990.
Says Conway asked the problem in the 1960s and that L. Moser is the
first to publish it. Says a group at a
convexity conference in Copenhagen improved Hammersley's results to 2.2164.
Gerver's analysis gives an object made up of 18 segments with area 2.2195.
The analysis depends on some unproven general assumptions which seem
reasonable and is certainly the unique optimum solution given those
assumptions.
A. A. Huntington. More on ladders. M500 145 (Jul 1995) 2-5.
Does usual problem, getting a quartic.
The finds the shortest ladder.
[This turns out to be the same as the longest ladder one can get around
a corner from corridors of widths
w and h, so 6.AG is related to
6.L.]
A goat is grazing in a circular field
and is tethered to a post on the edge.
He can reach half of the field.
How long is the rope? There are
numerous variations obtained by modifying the shape of the field or having
buildings within it. In recent years,
there has been study of the form where the goat is tethered to a point on a
circular silo in a large field -- how much area can he graze?
Upnorensis, proposer; Mr. Heath,
solver. Ladies Diary, 1748-49 = T.
Leybourn, II: 6-7, quest. 302. [I have
a reference to p. 41 of the Ladies' Diary.]
Circular pond enclosed by a circular railing of circumference 160
yards. Horse is tethered to a post of
the railing by a rope 160 yards long.
How much area can he graze?
Dudeney. Problem 67: Two rural puzzles -- No. 67: One
acre and a cow. Tit‑Bits 33 (5
Feb & 5 Mar 1898) 355
& 432. Circular field opening onto a small
rectangular paddock with cow tethered to the gate post so that she can graze
over one acre. By skilful choice of
sizes, he avoids the usual transcendental equation.
Arc. [R. A. Archibald]. Involutes of a circle and a pasturage
problem. AMM 28 (1921) 328‑329. Cites Ladies Diary and it appears that it
deals with a horse outside a circle.
J. Pedoe. Note 1477:
An old problem. MG 24 (No. 261)
(Oct 1940) 286-287. Finds the relevant
area by integrating in polar coordinates centred on the post.
A. J. Booth. Note 1561:
On Note 1477. MG 25 (No. 267)
(Dec 1941) 309‑310. Goat tethered
to a point on the perimeter of a circle which can graze over ½, ⅓, ¼ of the area.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968.
No. 8:
"Don't fence me in", pp. 87.
Equilateral triangular field of area 120. Three goats tethered to the corners with ropes of length equal to
the altitude. Consider an area
where n goats graze as contributing
1/n to each goat. What area does each goat graze over?
No.
53: Around the silo, pp. 71 & 112-113.
Goat tethered to the outside of a silo of diameter 20 by a rope of
length 10π, i.e. he can just get to the other side of
the silo. How big an area can he
graze? The curve is a semicircle
together with two involutes of a circle, so the solution uses some calculus.
Marshall Fraser. A tale of two goats. MM 55 (1982) 221‑227. Gives examples back to 1894.
Marshall Fraser. Letter:
More, old goats. MM 56 (1983)
123. Cites Arc[hibald].
Bull, 1998, below, says this
problem has been discussed by the Internet newsgroup sci.math some years
previously.
Michael E. Hoffman. The bull and the silo: An application of
curvature. AMM 105:1 (Jan 1998) ??NYS
-- cited by Bull. Bull is tethered by a
rope of length L to a circular silo of radius R.
If L £
πR, then the grazeable area
is L3/3R + πL2/2. This paper considers the problem for general
shapes.
John Bull. The bull and the silo. M500 163 (Aug 1998) 1-3. Improves Hoffman's solution for the circular
silo by avoiding polar coordinates and using a more appropriate variable,
namely the angle between the taut rope and the axis of symmetry.
Keith Drever. Solution 186.5 -- Horse. M550 188 (Oct 2002) 12. A horse is tethered to a point on the
perimeter of a circular field of radius 1.
He can graze over all but
1/π of the area. How long is the rope? This turns out to make the problem almost
trivial -- the rope is Ö2 long and the angle subtended at the tether
is π/2.
S&B,
pp. 146‑147, show several types.
These
are often made in two contrasting woods and appear to be physically
impossible. They will come apart if one
moves them in the right direction. A
few have extra complications. The
simplest version is a square cylinder with dovetail joints on each face --
called common square version below.
There are also cases where one thinks it should come apart, but the wood
has been bent or forced and no longer comes apart -- see also 6.W.5.
See Bogesen in 6.W.2 for a
possible early example.
Johannes Cornelus Wilhelmus
Pauwels. UK Patent 15,307 -- Improved
Means of Joining or Fastening Pieces of Wood or other Material together,
Applicable also as a Toy. Applied: 9
Nov 1887; complete specification: 9 Aug
1888; accepted: 26 Oct 1888. 2pp + 1p diagrams. It says Pauwels is a civil engineer of The Hague. Common square version.
Tom Tit, vol. 2. 1892.
Assemblage paradoxal, pp. 231-232.
= K, no. 155: The paradoxical coupling, pp. 353‑354. Common square version with instructions for
making it by cutting the corners off a larger square.
Emery Leverett Williams. The double dovetail and blind mortise. SA (25 Apr 1896) 267. The first is a trick T‑joint.
T. Moore. A puzzle joint and how to make it. The Woodworker 1:8 (May 1902) 172. S&B, p. 147, say this is the earliest
reference to the common square version -- but see Pauwels, above. "... the foregoing joint will doubtless
be well-known to our professional readers.
There are probably many amateur woodworkers to whom it will be a
novelty."
Hasluck, Paul N. The Handyman's Book. Cassell, 1903; facsimile by Senate (Tiger Books), Twickenham, London, 1998. Pp. 220‑223 shows various joints. Dovetail halved joint with two bevels, p.
222 & figs. 703-705 of pp. 221-222.
"... of but little practical value, but interesting as a puzzle
joint."
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. Shows the common
square version "given to me some ten years ago, but I cannot say who first
invented it." He previously
published it in a newspaper. ??look in
Weekly Dispatch.
Samuel Hicks. Kinks for Handy Men: The dovetail
puzzle. Hobbies 31 (No. 790) (3 Dec
1910) 248-249. Usual square dovetail,
but he suggests to glue it together!
Dudeney. AM. 1917. Prob. 424: The
dovetailed block, pp. 145 & 249.
Shows the common square version -- "... given to me some years ago,
but I cannot say who first invented it."
He previously published it in a newspaper. ??as above
Anon. Woodwork Joints, 1918, op. cit. in 6.W.1. A curious dovetail joint, pp. 193, 195. Common square version. Dovetail puzzle joint, pp. 194‑195. A singly mortised T‑joint, with an
unmortised second piece.
E. M. Wyatt. Woodwork puzzles. Industrial‑Arts Magazine 12 (1923) 326‑327. Doubly dovetailed tongue and mortise T‑joint
called 'The double (?) dovetail'.
Sherman M. Turrill. A double dovetail joint. Industrial‑Arts Magazine 13 (1924) 282‑283. A double dovetail right angle joint, but it
leaves sloping gaps on the inside which are filled with blocks.
Collins. Book of Puzzles. 1927. Pp. 134‑135:
The dovetail puzzle. Common square
version.
E. M. Wyatt. Puzzles in Wood, 1928, op. cit. in 5.H.1.
The
double (?) dovetail, pp. 44‑45.
Doubly dovetailed tongue and mortise T‑joint.
The
"impossible" dovetail joint, p. 46.
Common square version.
Double‑lock
dovetail joint, pp. 47‑49. Less
acceptable tricks for a corner joint.
Two‑way
fanned half‑lap joint, pp. 49‑50.
Corner joint.
A. B. Cutler. Industrial Arts and Vocational Education
(Jan 1930). ??NYS. Wyatt, below, cites this for a triple
dovetail, but I could not not find it in vols. 1‑40.
R. M. Abraham. Prob. 225 -- Dovetail Puzzle. Winter Nights Entertainments. Constable, London, 1932, p. 131. (= Easy‑to‑do Entertainments and
Diversions with coins, cards, string, paper and matches; Dover, 1961, p.
225.) Common square version.
Abraham. 1933.
Prob. 304 -- Hexagon dovetail;
Prob. 306 -- The triangular dovetail, pp. 142‑143 (100 &
102).
Bernard E. Jones, ed. The Practical Woodworker. Waverley Book Co., London, nd [1940s?]. Vol. 1: Lap and secret dovetail joints, pp.
281‑287. This covers various
secret joints -- i.e. ones with concealed laps or dovetails. Pp. 286-287 has a subsection: Puzzle
dovetail joints. Common square version
is shown as fig. 28. A pentagonal
analogue is shown as fig. 29, but it uses splitting and regluing to produce a
result which cannot be taken apart.
E. M. Wyatt. Wonders in Wood. Bruce Publishing Co., Milwaukee, 1946.
Double‑double
dovetail joint, pp. 26‑27.
Requires some bending.
Triple
dovetail puzzle, pp. 28‑29. Uses
curved piece with gravity lock.
S&B, p. 146, reproduces the
above Wyatt and shows a 1948 example.
W. A. Bagley. Puzzle Pie.
Op. cit. in 5.D.5. 1944. Dovetail deceptions, p. 64. Common square version and a tapered T joint.
Allan Boardman. Up and Down Double Dovetail. Shown on p. 147 of S&B. Square version with alternate dovetails in
opposite directions. This is
impossible!
I have a set of examples which
belonged to Tom O'Beirne. There is a
common square version and a similar hexagonal version. There is an equilateral triangle version
which requires a twist. There is a
right triangle version which has to be moved along a space diagonal! [One can adapt the twisting method to n-gons!]
Dick Schnacke (Mountain Craft
Shop, American Ridge Road, Route 1, New Martinsville, West Virginia, 26155,
USA) makes a variant of the common square version which has two dovetails on
each face. I bought one in 1994.
There are a great many illusions. This will only give some general studies and
some specific sources, though the sources of many illusions are unknown.
An exhibition by Al Seckel says
there are impossible geometric patterns in a mosaic floor in the Roman villa at
Fishbourne, c75, but it is not clear if this was intentional.
Anonymous 15C French illustrator
of Giovanni Boccaccio, De Claris Mulieribus, MS Royal 16 Gv in the British
Library. F. 54v: Collecting cocoons and
weaving silk. ??NYS -- reproduced in:
The Medieval Woman An Illuminated Book
of Postcards, HarperCollins, 1991. This
shows a loom(?) frame with uprights at each corner and the crosspieces joining
the tops of the end uprights as though front and rear are reversed compared to
the ground.
Seckel, 2002a, below, p. 25 (=
2002b, p. 175), says Leonardo da Vinci created the first anamorphic picture,
c1500.
Giuseppe Arcimboldo
(1537-1593). One of his paintings shows
a bowl of vegetables, but when turned over, it is a portrait. Seckel, 2000, below, fig. 109, pp. 120 &
122 (= 2002b, fig, 107, pp. 118 & 120), noting that this is the first
known invertible picture, but see next entry.
Topsy turvy coin, mid 16C. Seckel, 2002a, fig. 65, p. 80 (omitted in
2002b), shows an example which shows the Pope, but turns around to show the
Devil. Inscription around edge
reads: CORVI MALUM OVUM MALII.
Robert Smith. A Compleat System of Opticks in Four
Books. Cambridge, 1738. He includes a picture of a distant windmill for
which one cannot tell whether the sails are in front or behind the mill,
apparently the first publication of this visual ambiguity. ??NYS -- cited by: Nicholas J. Wade; Visual
Allusions Pictures of Perception;
Lawrence Erlbaum Associates, Hove, East Sussex, 1990, pp. 17 & 25, with a
similar picture.
L. A. Necker. LXI. Observations on some remarkable optical
phœnomena seen in Switzerland; and on an optical phœnomenon which occurs on
viewing a figure of a crystal or geometrical solid. Phil. Mag. (3) 1:5 (Nov 1832) 329-337. This is a letter from Necker, written on 24 May 1832. On pp. 336-337, Necker describes the visual
reversing figure known as the Necker cube which he discovered in drawing
rhomboid crystals. This is also quoted
in Ernst; The Eye Beguiled, pp. 23-24].
Richard L. Gregory [Mind in Science; Weidenfeld and Nicolson, London,
1981, pp. 385 & 594] and Ernst say that this was the first ambiguous figure
to be described.
See Thompson, 1882, in 6.AJ.2,
for illusions caused by rotations.
F. C. Müller-Lyer. Optische Urtheilstusehungen. Arch. Physiol. Suppl. 2 (1889) 263-270. Cited by Gregory in The Intelligent Eye. Many versions of the illusion. But cf below.
Wehman. New Book of 200 Puzzles. 1908.
The cube puzzle, p. 37. A 'baby
blocks' pattern of cubes, which appears to show six cubes piled in a corner one
way and seven cubes the other way. I
don't recall seeing this kind of puzzle in earlier sources, though this pattern
of rhombuses is common on cathedral floors dating back to the Byzantine era or
earlier.
James Fraser. British Journal of Psychology (Jan
1908). Introduces his 'The Unit of
Direction Illusion' in many forms.
??NYS -- cited in his popular article in Strand Mag., see below. Seckel, 2000, below, has several
versions. On p. 44, note to p. 9 (=
2002b, p. 44, note to p. 9), he says Fraser created a series of these illusions
in 1906.
H. E. Carter. A clever illusion. Curiosities section, Strand Mag. 378 (No. 219) (Mar 1909)
359. An example of Fraser's illusion
with no indication of its source.
James Fraser. A new illusion. What is its scientific explanation? Strand Mag. 38 (No. 224) (Aug 1909) 218-221. Refers to the Mar issue and says he
introduced the illusion in the above article and that the editors have asked
him for a popular article on it. 16
illustrations of various forms of his illusion.
Lietzmann, Walther &
Trier, Viggo. Wo steckt der
Fehler? 3rd ed., Teubner, 1923. [The Vorwort says that Trier was coauthor of
the 1st ed, 1913, and contributed most of the Schülerfehler (students'
mistakes). He died in 1916 and
Lietzmann extended the work in a 2nd ed of 1917 and split it into Trugschlüsse
and this 3rd ed. There was a 4th ed., 1937. See Lietzmann for a later version combining
both parts.] II. Täuschungen der
Anschauung, pp. 7-13.
Lietzmann, Walther. Wo steckt der Fehler? 3rd ed., Teubner, Stuttgart, (1950),
1953. (Strens/Guy has 3rd ed., 1963.) (See: Lietzmann & Trier. There are 2nd ed, 1952??; 5th ed, 1969; 6th ed, 1972. Math.
Gaz. 54 (1970) 182 says the 5th ed appears to be unchanged from the 3rd
ed.) II. Täuschungen der Anschauung,
pp. 15-25. A considerable extension of
the 1923 ed.
Williams. Home Entertainments. 1914.
Colour discs for the gramophone, pp. 207-212. Discusses several effects produced by spirals and eccentric
circles on discs when rotated.
Gerald H. Fisher. The Frameworks for Perceptual Localization.
Report of MOD Research Project70/GEN/9617, Department of Psychology, University
of Newcastle upon Tyne, 1968. Good
collection of examples, with perhaps the best set of impossible figures.
Pp.
42‑47 -- reversible perspectives.
Pp.
56‑65 -- impossible and ambiguous figures.
Appendix
6, p.190 -- 18 reversible figures.
Appendix
7, pp. 191‑192 -- 12 reversible silhouettes.
Appendix
8, p. 193 -- 12 impossible figures.
Appendix
14, pp. 202‑203 -- 72 geometrical illusions.
Harvey Long. "It's All In How You Look At
It". Harvey Long & Associates,
Seattle, 1972. 48pp collection of
examples with a few references.
Bruno Ernst [pseud. of J. A. F.
Rijk]. (Avonturen met Onmogelijke
Figuren; Aramith Uitgevers, Holland, 1985.)
Translated as: Adventures with
Impossible Figures. Tarquin, Norfolk,
1986. Describes tribar and many
variations of it, impossible staircase, two‑pronged trident. Pp. 76‑77 reproduces an Annunciation
of 14C in the Grote Kerk, Breda, with an impossible perspective. P. 78 reproduces Print XIV of Giovanni
Battista Piranesi's "Carceri de Invenzione", 1745, with an impossible
4‑bar.
Diego Uribe. Catalogo de impossibilidades. Cacumen (Madrid) 4 (No. 37) (Feb 1986) 9‑13. Good summary of impossible figures. 15 references to recent work.
Bruno Ernst. Escher's impossible figure prints in a new
context. In: H. S. M. Coxeter, et
al., eds.; M. C. Escher -- Art and Science; North‑Holland (Elsevier),
Amsterdam, 1986, pp. 124‑134.
Pp. 128‑129 discusses the Breda Annunciation, saying it is
15C and quoting a 1912 comment by an art historian on it. There is a colour reproduction on
p. 394. P. 130 shows and discusses
briefly Bruegel's "The Magpie on the Gallows", 1568. Pp. 130‑131 discusses and illustrates
the Piranesi.
Bruno Ernst. (Het Begoochelde Oog, 1986?.) Translated by Karen Williams as: The Eye Beguiled. Benedikt Taschen Verlag, Köln, 1992. Much expanded version of his previous book, with numerous new
pictures and models by new artists in the field. Chapter 6: Origins and history, pp. 68-93, discusses and quotes
almost everything known. P. 68 shows a
miniature of the Madonna and Child from the Pericope of Henry II, compiled by
1025, now in the Bayersche Staatsbibliothek, Munich, which is similar in form
to the Breda Annunciation (stated to be 15C).
(However, Seckel, 1997, below, reproduces it as 2©
and says it is c1250.) P. 69 notes that
Escher invented the impossible cube used in his Belvedere. P. 82 is a colour reproduction of Duchamp's
1916-1917 'Apolinère Enameled' - see 6.AJ.2.
Pp. 83-84 shows and discusses Piranesi.
Pp. 84-85 show and discuss Hogarth's 'False Perspective' of 1754. Reproduction and brief mention of Brueghel
(= Bruegel) on p. 85. Discussion of the
Breda Annunciation on pp. 85-86. Pp.
87-88 show and discuss a 14C Byzantine Annunciation in the National Museum,
Ochrid. Pp. 88-89 show and discuss
Scott Kim's impossible four-dimensional tribar.
J. Richard Block &
Harold E. Yuker. Can You Believe
Your Eyes? Brunner/Mazel, NY, 1992. Excellent survey of the field of illusions,
classified into 17 major types -- e.g. ambiguous figures, unstable figures,
..., two eyes are better than one. They
give as much information as they can about the origins. They give detailed sources for the following
-- originals ??NYS. These are also
available as two decks of playing cards.
W. E. Hill. My wife and my mother-in-law. Puck, (6 Nov 1915) 11. [However, Julian Rothenstein & Mel
Gooding; The Paradox Box; Redstone Press, London, 1993; include a reproduction
of a German visiting card of 1888 with a version of this illusion. The English caption by James Dalgety
is: My Wife and my Mother-in-law. Cf Seckel, 1997, below.] Ernst, just above, cites Hill and says he
was a cartoonist, but gives no source.
Long, above, asserts it was designed by E. G. Boring, an American
psychologist.
G. H. Fisher. Mother, father and daughter. Amer. J. Psychology 81 (1968) 274-277.
G. Kanisza. Subjective contours. SA 234:4 (Apr 1976) 48-52. (Kanisza triangles.)
Al Seckel, 1997. Illusions in Art. Two decks of playing cards in case with notes. Deck 1 -- Classics. Works from Roman times to the middle of the 20th
Century. Deck 2 -- Contemporary. Works from the second half of the 20th
Century. Y&B Associates, Hempstead,
NY, 1997. This gives further details on
some of the classic illusions -- some of this is entered above and in 6.AU and
some is given below.
10¨: Rabbit/Duck. Devised by Joseph (but notes say Robert) Jastrow, c1900. Seckel, 2000, below, p. 159 (= 2002b, p.
156), says Joseph Jastrow, c1900.
10§: My Wife and My Mother-in-Law, anonymous,
1888. However, in an exhibition,
Seckel's text implies the 1888 German card doesn't have a title and the title
first occurs on an 1890 US card.
Seckel, 2000, below, p. 122 (= 2002b, p. 120), says Boring took it from
a popular 19C puzzle trading card.
Al Seckel, 2000. The Art of Optical Illusions. Carlton, 2000. 144 well reproduced illusions with brief notes. All figures except 69-70 are included in
Seckel, 2002b.
J. Richard Block. Seeing Double Over 200 Mind-Bending Illusions.
Routledge, 2002. Update of Block
& Yuker, 1992.
Edgar
Rubin. Rubin's Vase. 1921.
This is the illusion where there appears to be a vase, but the outsides
appear to be two face profiles. [Pp.
8-11.] But Seckel, 2000, above, p. 122
(= 2002b, p. 120), says Rubin's inspiration was a 19C puzzle card.
My
wife and my mother-in-law. P. 17 says
Hill's version may derive from a late 1880s advertising postcard for
Phenyo-Caffein (Worcester, Massachusetts), labelled 'My Girl & Her Mother',
reproduced on p. 17.
P. 18
has G. H. Fisher's 1968 triple image, labelled 'Mother, Father and Daughter-in-Law'.
P. 44
says that Rabbit/Duck was devised by Joseph Jastrow in 1888.
Al Seckel, 2002a. More Optical Illusions. Carlton, 2002. 137 well reproduced illusions with brief notes, different than in
Seckel, 2000, above. All figures except
65-66, 86-87, 95-95, 137 are included in Seckel, 2002b, but with different
figure and page numbers.
Al Seckel, 2002b. The Fantastic World of Optical
Illusions. Carlton, 2002. This is essentially a combination of Seckel,
2000, and Seckel, 2002a, both listed above.
The Introduction is revised.
Figures 69-70 of the first book and 65-66, 86-87, 94-95, 137 of the
second book are omitted. The remaining
figures are then numbered consecutively.
The page of Further Reading in the first book is put at the end of this
combined book.
Here I make some notes about
origins of other illusions, but I have fewer details on these.
The Müller-Lyer Illusion -- <->
vs >---< was proposed by Zollner in 1859 and described
by Johannes Peter Müller (1801-1858) & Lyer in 1889. This seems to be a confusion, as the 1889
article is by F. C. Müller-Lyer, cf above.
Lietzmann & Trier, p. 7, date it as 1887.
The Bisection Illusion -- with a
vertical segment bisecting a horizontal segment, but above it -- was described
by Albert Oppel (1831-1865) and Wilhelm Wundt (1832-1920) in 1865.
Zollner's Illusion -- parallel
lines crossed by short lines at 45o, alternately in opposite directions -- was
noticed by Johann K. F. Zollner (1834-1882) on a piece of fabric with a similar
design.
Hering's Illusion -- with
parallel lines crossed by numerous lines through a point between the lines --
was invented by Ewals Hering (1834-1918) in 1860.
I have invented this name as it is
more descriptive than any I have seen.
The object or a version of it is variously called: Devil's Fork; Three Stick Clevis;
Widgit; Blivit; Impossible Columnade;
Trichometric Indicator Support;
Triple Encabulator for Tuned Manifold;
Hole Location Gage; Poiyut; Triple-Pronged Fork with only Two
Branches; Old Roman Pitchfork.
Oscar Reutersvård. Letters quoted in Ernst, 1992, pp. 69-70,
says he developed an equivalent type of object, which he calls impossible
meanders, in the 1930s.
R. L. Gregory says this is due to
a MIT draftsman (= draughtsman) about 1950??
California Technical
Industries. Advertisement. Aviation Week and Space Technology 80:12 (23
Mar 1964) 5. Standard form. (I wrote them but my letter was returned
'insufficient address'.)
Hole location gage. Analog Science Fact • Science Fiction 73:4
(Jun 1964) 27. Classic Two pronged
trident, with some measurements given.
Editorial note says the item was 'sent anonymously for some reason' and
offers the contributor $10 or a two year subscription if he identifies
himself. (Thanks to Peter McMullen for
the Analog items, but he doesn't recall the contributor ever being named.)
Edward G. Robles, Jr. Letter (Brass Tacks column). Analog Science Fact • Science Fiction 74:4
(Dec 1964) 4. Says the Jun 1964 object
is a "three-hole two slot BLIVIT" and was developed at JPL (Jet
Propulsion Laboratory, Pasadena) and published in their Goddard News. He provides a six-hole five-slot BLIVIT, but
as the Editor comments, it 'lacks the classic simple elegance of the Original.' However, a letter of inquiry to JPL resulted
in an email revealing that Goddard News is not their publication, but comes
from the Goddard Space Flight Center. I
have had a response from Goddard, ??NYR.
D. H. Schuster. A new ambiguous figure: a three‑stick
clevis. Amer. J. Psychol. 77 (1964)
673. Cites Calif. Tech. Ind. ad. [Ernst, 1992, pp. 80-81 reproduces this
article.]
Mad Magazine. No. 93 (Mar 1965). (I don't have a copy of this -- has anyone got one for
sale?) Cover by Norman Poiyut (?) shows
the figure and it is called a poiyut.
Miniature reproduction in: Maria
Reidelbach; Completely Mad -- A History of the Comic Book and Magazine; Little,
Brown & Co., Boston, 1991, p. 82.
Shows a standard version. Al
Seckel says they thought it was an original idea and they apologised in the
next issue -- to all of the following!
I now have the relevant issue, No. 95 (Jun 1965) and p. 2 has 15 letters
citing earlier appearances in Engineering Digest, The Airman (official journal
of the U.S. Airforce), Analog, Astounding Science Fact -- Science Fiction (Jun
1964, see above), The Red Rag (engineering journal at the University of British
Columbia), Society of Automotive Engineers Journal (designed by by Gregory
Flynn Jr. of General Motors as Triple Encabulator Tuned Manifold), Popular Mechanics, Popular Science (Jul
1964), Road & Track (Jun 1964).
Other letters say it was circulating at: the Engineering Graphics Lab of the University of Minnesota at
Duluth; the Nevada Test Site; Eastman Kodak (used to check
resolution); Industrial Camera Co. of
Oakland California (on their letterhead).
Two letters give an impossible crate and an impossible rectangular frame
(sort of a Penrose rectangle).
Sergio Aragones. A Mad look at winter sports. Mad Magazine (?? 1965); reprinted in: Mad Power; Signet, NY,
1970, pp. 120‑129. P. 124 shows a
standard version.
Bob Clark, illustrator. A Mad look at signs of the times. Loc. cit. under Aragones, pp. 167‑188. P. 186 shows standard version.
Reveille (a UK weekly magazine)
(10 Jun 1965). ??NYS -- cited by
Briggs, below -- standard version.
Don Mackey. Optical illusion. Skywriter (magazine of North American Aviation) (18 Feb
1966). ??NYS -- cited by Conrad G.
Mueller et al.; Light and Vision; Time-Life Books Pocket Edition, Time-Life
International, Netherlands, 1969, pp. 171 & 190. Standard version with nuts on the ends.
Heinz Von Foerster. From stimulus to symbol: The economy of
biological computation. IN: Sign Image Symbol; ed. Gyorgy Kepes; Studio
Vista, London, 1966, pp. 42-60. On
p. 55, he shows the "Triple-pronged fork with only two branches"
and on p. 54, he notes that although each portion is correct, it is impossible
overall, but he gives no indication of its history or that it is at all new.
G. A. Briggs. Puzzle and Humour Book. Published by the author, Ilkley, 1966. Pp. 17-18 shows the unnamed trident in a
version from Adcock & Shipley (Sales) Ltd., machine tool makers in
Leicester. Cites Reveille, above. Standard versions.
Harold Baldwin. Building better blivets. The Worm Runner's Digest 9:2 (1967) 104‑106. Discusses relation between numbers of slots
and of prongs. Draws a three slot
version and 2 and 4 way versions.
Charlie Rice. Challenge!
Op. cit. in 5.C. 1968. P. 10 shows a six prong, four slot version,
called the "Old Roman Pitchfork".
Roger Hayward. Blivets; research and development. The Worm Runner's Digest 10 (Dec 1968) 89‑92. Several fine developments, including two
interlaced frames and his monumental version.
Cites Baldwin.
M. Gardner. SA (May 1970) = Circus, pp. 3‑15. Says this became known in 1964 and cites Mad
& Hayward, but not Schuster.
D. Uribe, op. cit. above, gives
several variations.
6.AJ.2. TRIBAR AND IMPOSSIBLE STAIRCASE
Silvanus P. Thompson. Optical illusions of motion. Brain 3 (1882) 289-298. Hexagon of non‑overlapping
circles.
Thomas Foster. Illusions of motion and strobic
circles. Knowledge 1 (17 Mar 1882)
421-423. Says Thompson exhibited these
illusions at the British Association meeting in 1877.
Pearson. 1907.
Part II, no. 3: Whirling wheels, p. 3.
Gives Thompson's form, but the wheels are overlapping, which makes it
look a bit like an ancestor of the tribar.
Marcel Duchamp (1887-1968). Apolinère Enameled. A 'rectified readymade' of 1916-1917 which
turned a bedframe in an advertisement for Sapolin Enamel into an impossible
figure somewhat like a Penrose Triangle and a square version thereof. A version is in the Philadelphia Museum of
Art and is reproduced and discussed in Ernst; The Eye Beguiled, p. 82. (Duchamp's 'readymades' were frequently
reproduced by himself and others, so there may be other versions of this.)
Oscar Reutersvård. Omöjliga Figure [Impossible Figures -- In
Swedish]. Edited by Paul Gabriel. Doxa, Lund, (1982); 2nd ed., 1984. This seems to be the first publication of his work, but he has
been exhibiting since about 1960 and some of the exhibitions seem to have had
catalogues. P. 9 shows and discusses
his Opus 1 from 1934, which is an impossible tribar made from cubes. (Reproduced in Ernst, 1992, p. 69 as a
drawing signed and dated 1934. Ernst
quotes Reutersvård's correspondence which describes his invention of the form
while doodling in Latin class as a schoolboy.
A school friend who knew of his work showed him the Penroses' article in
1958 -- at that time he had drawn about 100 impossible objects -- by 1986, he
had extended this to some 2500!) He has
numerous variations on the tribar and the two‑pronged trident. An exhibition by Al Seckel says Reutersvård
had produced some impossible staircases, e.g. 'Visualized Impossible Bach
Scale', in 1936-1937, but didn't go far with it until returning to the idea in
1953.
Oscar Reutersvård. Swedish postage stamps for 25, 50, 75
kr. 1982, based on his patterns from
the 1930s. The 25 kr. has the tribar
pattern of cubes which he first drew in 1934.
(Also the 60 kr.??)
L. S. & R. Penrose. Impossible objects: A special type of visual
illusion. British Journal of Psychology
49 (1958) 31‑33. Presents tribar
and staircase. Photo of model
staircase, which Lionel Penrose had made in 1955. [Ernst, 1992, pp. 71-73, quotes conversation with Penrose about
his invention of the Tribar and reproduces this article. Penrose, like the rest of us, only learned
about Reutersvård's work in the 1980s.]
Anon.(?) Don't believe it. Daily Telegraph (24 Mar 1958) ?? (clipping found in an old
book). "Three pages of the latest
issue of the British Journal of Psychology are devoted to "Impossible
Objects."" Shows both the
tribar and the staircase.
M. C. Escher. Lithograph:
Belvedere. 1958.
L. S. & R. Penrose. Christmas Puzzles. New Scientist (25 Dec 1958) 1580‑1581 & 1597. Prob. 2: Staircase for lazy people.
M. C. Escher. Lithograph:
Ascending and Descending. 1960.
M. C. Escher. Lithograph:
Waterfall. 1961.
Oscar Reutersvård, in 1961,
produced a triangular version of the impossible staircase, called 'Triangular
Fortress without Highest Level'.
Joseph Kuykendall. Letter.
Mad Magazine 95 (Jun 1965) 2. An
impossible frame, a kind of Penrose rectangle.
S. W. Draper. The Penrose triangle and a family of related
figures. Perception 7 (1978) 283‑296. ??NYS -- cited and reproduced in Block,
2002, p. 48. A Penrose rectangle.
Uribe, op. cit. above, gives several
variations, including a perspective tribar and Draper's rectangle.
Jan van de Craats. Das unmögliche Escher-puzzle. (Taken from: De onmogelijke Escher-puzzle; Pythagoras (Amsterdam) (1988).) Alpha 6 (or: Mathematik Lehren / Heft 55 -- ??) (1992) 12-13. Two Penrose tribars made into an impossible
5-piece burr.
This is the illusion seen in
alternatingly coloured staggered brickwork where the lines of bricks distinctly
seem tilted. I suspect it must be
apparent in brickwork going back to Roman times.
The illusion is apparent in the
polychrome brick work on the side wall inside Keble College Chapel, Oxford, by
William Butterfield, completed in 1876 [thanks to Deborah Singmaster for
observing this].
Lietzmann & Trier, op. cit.
at 6.AJ, 1923. Pp. 12-13 has a striking
version of this, described as a 'Flechtbogen der Kleinen'. I can't quite translate this -- Flecht is
something interwoven but Bogen could be a ribbon or an arch or a bower,
etc. They say it is reproduced from an
original by Elsner. See Lietzmann,
1953.
Ogden's Optical Illusions. Cigarette card of 1927. No. 5.
Original ??NYS -- reproduced in:
Julian Rothenstein & Mel Gooding; The Paradox Box; Redstone Press,
London, 1993 AND in their: The Playful Eye; Redstone Press, London,
1999, p. 56. Vertical version of
this illusion.
B. K. Gentil. Die optische Täuschung von Fraser. Zeitschr. f. math. u. naturw. Unterr. 66
(1935) 170 ff. ??NYS -- cited by
Lietzmann.
Nelson F. Beeler & Franklyn
M. Branley. Experiments in Optical
Illusion. Ill. by Fred H. Lyon. Crowell, 1951, p. 42, fig. 39, is a good
example of the illusion.
Lietzmann, op. cit. at 6.AJ,
1953. P. 23 is the same as above, but
adds a citation to Gentil, listed above.
Leonard de Vries. The Third Book of Experiments. © 1965, probably for a Dutch edition. Translated by Joost van de Woestijne. John Murray, 1965; Carousel, 1974. Illusion 10, pp. 58-59, has a clear picture
and a brief discussion.
Richard L. Gregory &
Priscilla Heard. Border locking and the
café wall illusion. Perception 8 (1979)
365‑380. ??NYS -- described by
Walker, below. [I have photos of the
actual café wall in Bristol.]
Jearl Walker. The Amateur Scientist: The café‑wall
illusion, in which rows of tiles tilt that should not tilt at all. SA 259:5 (Nov 1988) 100‑103. Good summary and illustrations.
New section, due to reading Glass's
assertion as to the inventor, who is different than other names that I have
seen.
Don Glass, ed. How Can You Tell if a Spider is Dead? and
More Moments of Science. Indiana Univ
Press, Bloomington, Indiana, 1996. Now
you see it, now you don't, pp. 131-132.
Asserts that Christopher Tyler, of the Smith-Kettlewell Eye Research
Institute, San Francisco, is the inventor of stereograms.
This is like a Necker Cube where all
the edges are drawn as wooden slats in an impossible configuration.
Escher. Man with Cuboid, which is essentially a
detail from Belvedere, both 1958, are apparently the first examples of this
impossible object.
Chuck Mathias. Letter
Mad Magazine 95 (Jun 1965) 2.
Gives an impossible crate.
Jerry Andrus developed his
actual model in 1981 and it appeared on the cover of Omni in 1981. But Al Seckel's exhibition says the first
physical example was The Feemish Crate, due to C. F. Cochran.
Seckel, 2002a, figs. 27 A&B,
pp. 36-37 (= 2002b, figs. 169 A&B, pp. 186-187), shows and discusses
Andrus' crate from two viewpoints.
6.AK. POLYGONAL PATH COVERING N x N LATTICE OF POINTS, QUEEN'S TOURS, ETC.
For
magic circuits, see 7.N.4.
3x3 problem: Loyd (1907), Pearson, Anon., Bullivant,
Goldston, Loyd (1914), Blyth,
Abraham, Hedges, Evans,
Doubleday - 1, Piggins &
Eley
4x4 problem: King,
Abraham, M. Adams, Evans,
Depew, Meyer, Ripley's,
Queen's tours: Loyd (1867, 1897, 1914), Loyd Jr.
Bishop's tours: Dudeney (1932), Doubleday - 2, Obermair
Rook's tours: Loyd (1878), Proctor, Loyd
(1897), Bullivant, Loyd (1914), Filipiak, Hartswick, Barwell,
Gardner, Peters, Obermair
Other versions: Prout,
Doubleday - 1
Trick solutions: Fixx,
Adams, Piggins, Piggins & Eley
Thanks
to Heinrich Hemme for pointing out Fixx, which led to adding most of the
material on trick solutions.
Loyd. ??Le Sphinx (Mar 1867 -- but the Supplement to Sam Loyd and His
Chess Problems corrects this to 15 Nov 1866).
= Chess Strategy, Elizabeth, NJ, 1878, no. or p. 336(??). = A. C. White; Sam Loyd and His Chess
Problems; 1913, op. cit. in 1; no. 40, pp. 42‑43. Queen's circuit on 8 x 8 in 14
segments. (I.e. closed circuit,
not leaving board, using queen's moves.)
No. 41 & 42 of White give other solutions. White quotes Loyd from Chess Strategy, which indicates that Loyd
invented this problem. Tit‑Bits
No. 31 & SLAHP: Touring the chessboard, pp. 19 & 89, give No. 41.
Loyd. Chess Strategy, 1878, op. cit. above, no. or p. 337 (??) (= White, 1913, op. cit. above, no. 43,
pp. 42‑43.) Rook's circuit
on 8 x 8 in 16 segments.
(I.e. closed circuit, not leaving board, using rook's moves, and without
crossings.)
Richard A. Proctor. Gossip column. Knowledge 10 (Dec 1886)
43 & (Feb 1887) 92. 6 x 6 array of cells. Prisoner in one corner can exit from the opposite corner if he
passes "once, and once only, through all the 36 cells." "... take the prisoner into either of
the cells adjoining his own, and back into his own, .... This puzzle is rather a sell,
...." Letter and response [in
Gossip column, Knowledge 10 (Mar 1887) 115-116] about the impossibility of any
normal solution.
Loyd. Problem 15: The gaoler's problem. Tit‑Bits 31 (23 Jan
& 13 Feb 1897) 307 &
363. Rook's circuit on 8 x 8
in 16 segments, but beginning and ending on a central square. Cf The postman's puzzle in the Cyclopedia,
1914.
Loyd. Problem 16: The captive maiden.
Tit‑Bits 31 (30 Jan
& 20 Feb 1897) 325 &
381. Rook's tour in minimal
number of moves from a corner to the diagonally opposite corner, entering each
cell once. Because of parity, this is
technically impossible, so the first two moves are into an adjacent cell and
then back to the first cell, so that the first cell has now been entered.
Loyd. Problem 20: Hearts and darts.
Tit‑Bits 31 (20 Feb,
13 & 20 Mar 1897) 381, 437, 455. Queen's tour on 8 x 8, starting in a
corner, permitting crossings, but with no segment going through a square where
the path turns. Solution in 14
segments. This is No. 41 in
White -- see the first Loyd entry above.
Ball. MRE, 4th ed., 1905, p. 197.
At the end of his section on knight's tours, he states that there are
many similar problems for other kinds of pieces.
Loyd. In G. G. Bain, op. cit. in 1, 1907. He gives the 3 x 3 lattice in four lines as the Columbus Egg
Puzzle.
Pearson. 1907.
Part I, no. 36: A charming puzzle, pp. 36 & 152‑153. 3 x 3
lattice in 4 lines.
Loyd. Sam Loyd's Puzzle Magazine (Apr 1908) -- ??NYS, reproduced
in: A. C. White; Sam Loyd and His Chess
Problems; 1913, op. cit. in 1; no. 56, p. 52.
= Problem 26: A brace of puzzles -- No. 26: A study in naval warfare;
Tit‑Bits 31 (27 Mar 1897) 475
& 32 (24 Apr 1897) 59. = Cyclopedia, 1914, Going into action,
pp. 189 & 364. = MPSL1, prob. 46,
pp. 44 & 138. = SLAHP: Bombs to
drop, pp. 86 & 119. Circuit on 8 x 8
in 14 segments, but with two lines of slope 2. In White, p. 43, Loyd
says an ordinary queen's tour can be started "from any of the squares
except the twenty which can be represented by
d1, d3 and d4." This
problem starts at d1. However I think White must have mistakenly
put down twenty for twelve??
Anon. Prob. 67. Hobbies 31 (No. 782) (8 Oct 1910) 39 &
(No. 785) (29 Oct 1910) 94.
3 x 3 lattice in 4
lines "brought under my notice some time back".
C. H. Bullivant. Home Fun, 1910, op. cit. in 5.S. Part VI, Chap. IV.
No. 1:
The travelling draught‑man, pp. 515 & 520. Rook's circuit on 8 x
8 in 16 segments, different than
Loyd's.
No. 3:
Joining the rings. 3 x 3 in 4 segments.
Will Goldston. More Tricks and Puzzles without Mechanical
Apparatus. The Magician Ltd., London,
nd [1910?]. (BMC lists Routledge &
Dutton eds. of 1910.) (There is a 2nd
ed., published by Will Goldston, nd [1919].)
The nine‑dot puzzle, pp. 127‑128 (pp. 90‑91 in
2nd ed.).
Loyd. Cyclopedia, 1914, pp. 301 & 380. = MPSL2, prob. 133 -- Solve Christopher's egg tricks, pp. 93
& 163 (with comment by Gardner). c=
SLAHP: Milkman's route, pp. 34 & 96.
3 x 3 case.
Loyd. Cyclopedia, 1914, pp. 293 & 379. Queen's circuit on 7 x
7 in
12 segments.
Loyd. The postman's puzzle.
Cyclopedia, 1914, pp. 298 & 379.
Rook's circuit on 8 x 8 array of points, with one point a bit out of
line, starting and ending at a central square, in 16 segments. P. 379 also shows another 8 x 8
circuit, but with a slope 2 line.
See also pp. 21 & 341 and SLAHP, pp. 85 & 118, for two more
examples.
Loyd. Switchboard problem.
Cyclopedia, 1914, pp. 255 & 373.
(c= MPSL2, prob. 145, pp. 102 & 167.) Rook's tour with minimum turning.
Blyth. Match-Stick Magic.
1921. Four-way game, pp.
77-78. 3 x 3 in 4 segments.
King. Best 100. 1927. No. 16, pp. 12 & 43. 4 x 4
in 6 segments, not closed, but easily can be closed.
Loyd Jr. SLAHP. 1928. Dropping the mail, pp. 67 & 111. 4 x 4 queen's tour in 6
segments.
Collins. Book of Puzzles. 1927. The star group puzzle,
pp. 95-96. 3 x 3 in
4 segments.
Dudeney. PCP.
1932. Prob. 264: The fly's tour,
pp. 82 & 169. = 536, prob. 422, pp.
159 & 368. Bishop's path, with
repeated cells, going from corner to corner in
17 segments.
Abraham. 1933.
Probs. 101, 102, 103, pp. 49 & 66 (30 & 118). 3 x 3,
4 x 4 and 6 x 6
cases.
The Bile Beans Puzzle Book. 1933.
No. 4: The puzzled milkman. 3 x
3 array in four lines.
Sid G. Hedges. More Indoor and Community Games. Methuen, London, 1937. Nine spot, p. 110. 3 x 3.
"Of course it can be done, but it is not easy." No solution given.
M. Adams. Puzzle Book. 1939. Prob. C.64: Six
strokes, pp. 140 & 178. 4 x 4 array in
6 segments which form a closed
path, though the closure was not asked for.
J. R. Evans. The Junior Week‑End Book. Op. cit. in 6.AF. 1939. Probs. 30 & 31,
pp. 264 & 270. 3 x 3 &
4 x 4 cases in 4 & 6
segments, neither closed nor staying within the array.
Depew. Cokesbury Game Book.
1939. Drawing, p. 220. 4 x 4
in 6 segments, not closed, not staying within the array.
Meyer. Big Fun Book. 1940. Right on the dot, pp. 99 & 732. 4 x 4
in 6 segments.
A. S. Filipiak. Mathematical Puzzles, 1942, op. cit. in
5.H.1, pp. 50‑51. Same as
Bullivant, but opens the circuit to make a 15 segment path.
M. S. Klamkin, proposer and
solver; John L. Selfridge, further
solver. Problem E1123 -- Polygonal path
covering a square lattice. AMM 61
(1954) 423 & 62 (1955) 124 & 443. Shows
N x N can be done in 2N‑2
segments. Selfridge shows this is
minimal.
W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. Joining the stars, pp. 41 & 129. 5 x 5
array of points. Using a line of
four segments, pass through 17 points.
This is a bit like the 3 x
3 problem in that one must go outside
the array.
R. E. Miller & J. L.
Selfridge. Maximal paths on rectangular
boards. IBM J. Research and Development
4:5 (Nov 1960) 479-486. They study
rook's paths where a cell is deemed visited if the rook changes direction
there. They find maximal such paths in
all cases.
Ripley's Puzzles and Games. 1966.
Pp. 72-73, item 2. 4 x 4 cases with closed solution symmetric both
horizontally and vertically.
F. Gregory Hartswick. In:
H. A. Ripley & F. Gregory Hartswick, Detectograms and Other Puzzles,
Scholastic Book Services, NY, 1969.
Prob. 4, pp. 42‑43 & 82.
Asks for 8 x 8 rook's circuit with minimal turning and
having a turn at a central cell.
Solution gives two such with
16 segments and asserts there
are no others.
Doubleday - 1. 1969. Prob. 60: Test case, pp. 75 & 167. = Doubleday - 4, pp. 83-84.
Two 3 x 3 arrays joined at a corner, looking like the
Fore and Aft board (cf 5.R.3), to be covered in a minimum number of
segments. He does it in seven segments
by joining two 3 x 3 solutions.
Brian R. Barwell. Arrows and circuits. JRM 2 (1969) 196‑204. Introduces idea of maximal length rook's
tours. Shows the maximal length on
a 4 x 4 board is 38 and finds there are 3
solutions. Considers also
the 1 x n board.
Solomon W. Golomb & John L.
Selfridge. Unicursal polygonal paths
and other graphs on point lattices. Pi
Mu Epsilon J. 5 (1970) 107‑117.
Surveys problem. Generalizes
Selfridge's 1955 proof to M x N for which
MIN(2M, M+N‑2) segments
occur in a minimal circuit.
Doubleday - 2. 1971.
Path finder, pp. 95-96. Bishop's
corner to corner path, same as Dudeney, 1932.
James F. Fixx. More Games for the Superintelligent. (Doubleday, 1972); Muller, (1977), 1981. 6.
Variation on a variation, pp. 31 & 87.
Trick solution in three lines, assuming points of finite size.
M. Gardner. SA (May 1973) c= Knotted, chap. 6. Prob. 1: Find rook's tours of maximum length
on the 4 x 4 board. Cites
Barwell. Knotted also cites Peters,
below.
Edward N. Peters. Rooks roaming round regular rectangles. JRM 6 (1973) 169‑173. Finds maximum length on 1 x N
board is N2/2 for
N even; (N‑1)2/2 + N‑1 for
N odd, and believes he has
counted such tours. He finds tours on
the N x N board whose length is a formula that reduces to 4 BC(N+1, 3) ‑ 2[(N‑1)/2]. I am a bit unsure if he has shown that this
is maximal.
James L. Adams. Conceptual Blockbusting. Freeman, 1974, pp. 16-22. 3rd ed., (A-W, 1986), Penguin, 1987, pp.
24-33. Trick solution of 3 x 3
case in three lines, assuming points of finite size, which he says was
submitted anonymously when he and Bob McKim used the puzzle on an ad for a talk
on problem-solving at Stanford. Also
describes a version using paperfolding to get all nine points into a line. The material is considerably expanded in the
3rd ed. and adds several new versions.
From the references in Piggins and Eley, it seems that these all
appeared in the 2nd ed of 1979 -- ??NYS.
Cut
out the 3 x 1 parts and tape them into a straight line.
Take
the paper and roll it to a cylinder and then draw a slanting line on the
cylinder which goes through all nine, largish, points.
Cut
out bits with each point on and skewer the lot with a pencil.
Place
the paper on the earth and draw a line around the earth to go through all nine
points. One has to assume the points
have some size.
Wodge
the paper, with large dots, into a ball and stick a pencil through it. Open up to see if you have won -- if not,
try again!
Use
a very fat line, i.e. as thick as the spacing between the edges of the array.
David J. Piggins. Pathological solutions to a popular
puzzle. JRM 8:2 (1975-76) 128-129. Gives two trick solutions.
Three
parallel lines, since they meet at infinity.
Put
the figure on the earth and use a slanting line around the earth. This works in the limit, but otherwise
requires points of finite size, a detail that he doesn't mention.
No
references for these versions.
David J. Piggins & Arthur D.
Eley. Minimal path length for covering
polygonal lattices: A review. JRM 14:4 (1981‑82) 279‑283. Mostly devoted to various trick solutions of
the 3 x 3 case. They cite Piggins'
solution with three parallel lines.
They say that Gardner sent them the trick solution in 1973 and then cite
Adams, 1979. They give solutions using
points of different sizes, getting both three and two segment solutions and
mention a two segment version that depends on the direction of view. They then give the solution on a sphere,
citing Adams, 1979, and Piggins. They
give several further versions using paper folding, including putting the
surface onto a twisted triangular prism joined at the ends to make the surfaces
into a Möbius strip -- Zeeman calls this a umbilical bracelet or a Möbius bar.
Obermair. Op. cit. in 5.Z.1. 1984.
Prob.
19, pp. 23 & 50. Bishop's path
on 8 x 8 in 17 segments, as in Dudeney, PCP, 1932.
Prob.
41, p. 72. Rook's path with maximal
number of segments, which is 57. [For the
2 x 2, 3 x 3, 4 x 4
boards, I get the maximum numbers are
3, 6, 13.]
Nob Yoshigahara. Puzzlart.
Tokyo, 1992. Section: The wisdom
of Solomon, pp. 40-47, abridged from an article by Solomon W. Golomb in Johns
Hopkins Magazine (Oct 1984).
Classic 3 x 3 problem.
For the 4 x 4 case: 1) find four closed paths; 2), says there are about 30 solutions and
gives 19 beyond the previous 4. Find
the unique 5‑segment closed path on the
3 x 4. Gives 3 solutions on 5 x 5.
10-segment solution on 6 x
6 which stays on the board. Loyd's 1867? Queen's circuit. Queen's circuit on 7 x 7, attributed to
Dudeney, though my earliest entry is Loyd, 1914 -- ??CHECK.
This
has such an extensive history that I will give only a few items.
C. L. Lehmus first posed the problem
to Jacob Steiner in 1840.
Rougevin published the first
proof in 1842. ??NYS.
Jacob Steiner. Elementare Lösung einer Aufgabe über das
ebene und sphärische Dreieck.
J. reine angew. Math. 28 (1844) 375‑379 & Tafel III. Says Lehmus sent it to him in 1840 asking
for a purely geometric proof. Here he
gives proofs for the plane and the sphere and also considers external
bisectors.
Theodor Lange. Nachtrag zu dem Aufsatze in Thl. XIII, Nr.
XXXIII. Archiv der Math. und Physik 15
(1850) 221‑226. Discusses the
problem and gives a solution by Steiner and two by C. L. Lehmus. Steiner also considers the external
bisectors.
N. J. Chignell. Note 1031: A difficult converse. MG 16 (No. 219) (Jul 1932) 200-202. [The author's name is omitted in the article
but appears on the cover.] 'Three
fairly simple proofs', due to: M. J.
Newell; J. Travers, improving J. H.
Doughty, based on material in Lady's and Gentleman's Diary (1859) 87-88 &
(1860) 84-86; Wm. Mason, found by
Doughty, in Lady's and Gentleman's Diary (1860) 86.
H. S. M. Coxeter. Introduction to Geometry. Wiley, 1961. Section 1.5, ex. 4, p. 16.
An easy proof is posed as a problem with adequate hints in four lines.
M. Gardner. SA (Apr 1961) = New MD, chap. 17. Review of Coxeter's book, saying his brief
proof came as a pleasant shock.
G. Gilbert & D.
MacDonnell. The Steiner‑Lehmus
theorem. AMM 70 (1963) 79‑80. This is the best of the proofs sent to
Gardner in response to his review of Coxeter.
A later source says this turned out to be identical to Lehmus' original
proof!
Léo Sauvé. The Steiner‑Lehmus theorem. CM 2:2 (Feb 1976) 19‑24. Discusses history and gives 22 references,
some of which refer to 60 proofs.
Charles W. Trigg. A bibliography of the Steiner‑Lehmus
theorem. CM 2:9 (Nov 1976) 191‑193. 36 references beyond Sauvé's.
David C. Kay. Nearly the last comment on the Steiner‑Lehmus
theorem. CM 3:6 (1977) 148‑149. Observes that a version of the proof works
in all three classical geometries at once and gives its history.
This
also has an extensive history and I give only a few items.
T. Delahaye and H. Lez. Problem no. 1655 (Morley's triangle). Mathesis (3) 8 (1908) 138‑139. ??NYS.
E. J. Ebden, proposer; M. Satyanarayana, solver. Problem no. 16381 (Morley's theorem). The Educational Times (NS) 61 (1 Feb 1908)
81 & (1 Jul 1908) 307‑308
= Math. Quest. and Solutions from "The Educational Times" (NS)
15 (1909) 23. Asks for various related
triangles formed using interior and exterior trisectors to be shown
equilateral. Solution is essentially
trigonometric. No mention of Morley.
Frank Morley. On the intersections of the trisectors of
the angles of a triangle. (From a
letter directed to Prof. T. Hayashi.)
J. Math. Assoc. of Japan for Secondary Education 6 (Dec 1924) 260‑262. (= CM 3:10 (Dec 1977) 273‑275.
Frank Morley. Letter to Gino Loria. 22 Aug 1934. Reproduced in: Gino
Loria; Triangles équilatéraux dérivés d'un triangle quelconque. MG 23 (No. 256) (Oct 1939) 364‑372. Morley says he discovered the theorem in
c1904 and cites the letter to Hayashi.
Loria mentions other early work and gives several generalizations.
H. F. Baker. Note 1476:
A theorem due to Professor F. Morley.
MG 24 (No. 261) (Oct 1940) 284‑286. Easy proof and reference to other proofs. He cites a related result of Steiner.
Anonymous [R. P.] Morley's trisector theorem. Eureka 16 (Oct 1953) 6-7. Short proof, working backward from the
equilateral triangle.
Dan Pedoe. Notes on Morley's proof of his theorem on
angle trisectors. CM 3:10 (Dec 1977)
276‑279. "... very tentative
... first steps towards the elucidation of his work."
C. O. Oakley & Charles W.
Trigg. A list of references to the
Morley theorem. CM 3:10 (Dec 1977)
281‑290 & 4 (1978) 132. 169 items.
André Viricel (with Jacques
Bouteloup). Le Théorème de Morley. L'Association pour le Développement de la
Culture Scientifique, Amiens, 1993.
[This publisher or this book was apparently taken over by Blanchard as
Blanchard was selling copies with his label pasted over the previous
publisher's name in Dec 1994.] A
substantial book (180pp) on all aspects of the theorem. The bibliography is extremely cryptic, but
says it is abridged from Mathesis (1949) 175
??NYS. The most recent item
cited is 1970.
6.AN. VOLUME OF THE INTERSECTION OF TWO CYLINDERS
Archimedes. The Method:
Preface, 2. In: T. L. Heath; The Works of Archimedes, with a
supplement "The Method of Archimedes"; (originally two works, CUP,
1897 & 1912) = Dover, 1953. Supplement, p. 12, states the result. The proof is lost, but pp. 48‑51
gives a reconstruction of the proof by Zeuthen.
Liu Hui. Jiu Zhang Suan Chu Zhu (Commentary on the
Nine Chapters of the Mathematical Art).
263. ??NYS -- described in Li
& Du, pp. 73‑74 & 85. He
shows that the ratio of the volume of the sphere to the volume of Archimedes'
solid, called mou he fang gai (two square umbrellas), is π/4,
but he cannot determine either volume.
Zu Geng. c500.
Lost, but described in: Li
Chunfeng; annotation to Jiu Zhang (= Chiu Chang Suan Ching) made c656. ??NYS.
Described on pp. 86‑87
of: Wu Wenchun; The out‑in
complementary principle; IN: Ancient China's Technology and Science;
compiled by the Institute of the History of Natural Sciences, Chinese Academy
of Sciences; Foreign Languages Press, Beijing, 1983, pp. 66‑89. [This is a revision and translation of parts
of: Achievements in Science and
Technology in Ancient China [in Chinese]; China Youth Publishing House,
Beijing(?), 1978.]
He
considers the shape, called fanggai, within the natural circumscribed cube and
shows that, in each octant, the part of the cube outside the fanggai has cross
section of area h2 at distance
h from the centre. This is equivalent to a tetrahedron, whose
volume had been determined by Liu, so the excluded volume is ⅓
of the cube.
Li
& Du, pp. 85‑87, and say the result may have been found c480 by Zu
Geng's father, Zu Chongzhi.
Lam Lay-Yong & Shen
Kangsheng. The Chinese concept of
Cavalieri's Principle and its applications.
HM 12 (1985) 219-228. Discusses
the work of Liu and Zu.
Shiraishi Chōchū. Shamei Sampu. 1826. ??NYS -- described
in Smith & Mikami, pp. 233-236.
"Find the volume cut from a cylinder by another cylinder that
intersects is orthogonally and touches a point on the surface". I'm not quite sure what the last phrase
indicates. The book gives a number of
similar problems of finding volumes of intersections.
P. R. Rider, proposer; N. B. Moore, solver. Problem 3587. AMM 40 (1933) 52 (??NX) &
612. Gives the standard proof by
cross sections, then considers the case of unequal cylinders where the solution
involves complete elliptic integrals of the first and second kinds. References to solution and similar problem
in textbooks.
Leo Moser, solver; J. M. Butchart, extender. MM 25 (May 1952) 290 &
26 (Sep 1952) 54. ??NX. Reproduced in Trigg, op. cit. in 5.Q: Quickie
15, pp. 6 & 82‑83. Moser
gives the classic proof that
V = 16r3/3.
Butchart points out that this also shows that the shape has surface
area 16r2.
NOTATION: (a, b, c) denotes the
configuration of a points in
b rows of c
each. The index below covers
articles other than the surveys of Burr et al. and Gardner.
( 5, 2, 3): Sylvester
( 6, 3, 3): Mittenzwey
( 7, 6, 3): Criton
( 9, 8, 3): Sylvester; Carroll;
Criton
( 9, 9, 3): Carroll; Bridges; Criton
( 9, 10, 3): Jackson; Family Friend; Parlour Pastime;
Magician's Own Book;
The Sociable;
Book of 500 Puzzles;
Charades etc.;
Boy's Own Conjuring Book;
Hanky Panky; Carroll; Crompton;
Berkeley & Rowland;
Hoffmann;
Dudeney (1908);
Wehman; Williams; Loyd Jr;
Blyth; Rudin; Young World; Brooke; Putnam; Criton
(10, 5, 4): The Sociable; Book of 500 Puzzles; Carroll;
Hoffmann;
Dudeney (1908);
Wehman; Williams; Dudeney (1917); Blyth; King; Rudin;
Young World; Hutchings
& Blake; Putnam
(10, 10, 3): Sylvester
(11, 11, 3): The Sociable; Book of 500 Puzzles; Wehman
(11, 12, 3): Hoffmann; Williams;
Young World
(11, 13, 3): Prout
(11, 16, 3): Wilkinson -- in Dudeney (1908
& 1917); Macmillan
(12, 4, 5) -- Trick version of a hollow 3 x 3 square with doubled
corners, as in 7.Q: Family Friend (1858); Secret Out;
Illustrated Boy's Own Treasury;
(12, 6, 4): Endless
Amusement II; The Sociable; Book of 500 Puzzles; Boy's Own Book; Cassell's;
Hoffmann; Wehman; Rudin;
Criton
(12, 7, 4) -- Trick version of a
3 x 3 square with doubled
diagonal: Secret Out; Hoffmann (1876); Mittenzwey;
Hoffmann (1893), no. 8
(12, 7, 4): Dudeney
(1917); Putnam
(12, 19, 3): Macmillan
(13, 9, 4): Criton
(13, 12, 3): Criton
(13, 18, 3): Sylvester
(13, 22, 3): Criton
(15, 15, 3): Jackson
(15, 16, 3): The Sociable; Book of 500 Puzzles; H. D. Northrop; Wehman
(15, 23, 3): Jackson
(15, 26, 3): Woolhouse
(16, 10, 4): The Sociable; Book of 500 Puzzles; Hoffmann;
Wehman
(16, 12, 4): Criton
(16, 15, 4): Dudeney (1899, 1902, 1908); Brooke;
Putnam; Criton
(17, 24, 3): Jackson
(17, 28, 3): Endless Amusement II; Pearson
(17, 32, 3): Sylvester
(17, 7, 5): Ripley's
(18, 18, 4): Macmillan
(19, 19, 4): Criton
(19, 9, 5): Endless
Amusement II; The Sociable; Book of 500 Puzzles; Proctor;
Hoffmann; Clark; Wehman;
Ripley; Rudin; Putnam;
Criton
(19, 10, 5): Proctor
(20, 12, 5): trick method: Doubleday - 3
(20, 18, 4): Loyd Jr
(20, 21, 4): Criton
(21, 9, 5): Magician's
Own Book; Book of 500 Puzzles;
Boy's Own Conjuring Book; Blyth; Depew
(21, 10, 5): Mittenzwey
(21, 11, 5): Putnam
(21, 12, 5): Dudeney (1917); Criton
(21, 30, 3): Secret Out; Hoffmann
(21, 50, 3): Sylvester
(22, 15, 5): Macmillan
(22, 20, 4): Dudeney (1899)
(22, 21, 4): Dudeney (1917); Putnam
(24, 28, 3): Jackson; Parlour Pastime
(24, 28, 4): Jackson; Héraud; Benson; Macmillan
(24, 28, 5): Jackson
(25, 12, 5): Endless Amusement II; Young Man's Book; Proctor; Criton
(25, 18, 5): Bridges
(25, 30, 4): Macmillan
(25, 72, 3): Sylvester
(26, 21, 5): Macmillan
(27, 9, 6): The
Sociable; Book of 500 Puzzles; Hoffmann;
Wehman
(27, 10, 6): The Sociable; Book of 500 Puzzles; Wehman
(27, 15, 5): Jackson
(29, 98, 3): Sylvester
(30, 12, 7): Criton
(30, 22, 5): Criton
(30, 26, 5): Macmillan
(31, 6, 6) -- with 7 circles of 6:
The Sociable; Book of 500
Puzzles;
Magician's Own Book (UK version); Wehman
(31, 15, 5): Proctor
(36, 55, 4): Macmillan
(37, 18, 5): Proctor
(37, 20, 5): The Sociable; Book of 500 Puzzles;
Illustrated Boy's Own Treasury; Hanky Panky; Wehman
(49, 16, 7): Criton
Trick versions -- with doubled
counters: Family Friend (1858); Secret Out;
Illustrated Boy's Own Treasury; Hoffmann (1876);
Mittenzwey;
Hoffmann (1893), nos. 8 & 9; Pearson; Home Book ....;
Doubleday - 3. These could also be
considered as in 7.Q.2 or 7.Q.
A different type of
configuration problem is considered by Shepherd, 1947.
Jackson. Rational Amusement. 1821.
Trees Planted in Rows, nos. 1-10, pp. 33-34 & 99-100 and plate IV,
figs. 1-9. [Brooke and others say this
is the earliest statement of such problems.]
1. (9, 10, 3).
Quoted in Burr, below.
"Your aid I want, nine trees to
plant
In rows just half a score;
And let there be in each row three.
Solve this: I ask no
more."
2. (n, n, 3),
He does the case n = 15.
3. (15, 23, 3).
4. (17, 24, 3).
5. (24, 24, 3)
with a pond in the middle.
6. (24, 28, 4).
7. (27, 15, 5)
8. (25, 28, c)
with c = 3, 4, 5.
9. (90, 10, 10) with equal spacing -- decagon with 10 trees on each side.
10.
Leads to drawing square lattice in perspective with two vanishing points, so
the diagonals of the resulting parallelograms are perpendicular.
Endless Amusement II. 1826?
Prob.
13, p. 197. (19, 9, 5). = New Sphinx, c1840, p. 135.
Prob.
14, p. 197. (12, 6, 4). = New Sphinx, c1840, p. 135.
Prob.
26, p. 202. (25, 12, 5). Answer is a
5 x 5 square array.
Ingenious artists, how may I dispose
Of five-and-twenty trees, in just twelve rows;
That every row five lofty trees may grace,
Explain the scheme -- the trees completely
place.
Prob.
35, p. 212. (17, 28, 3). [This is the problem that is replaced in the
1837 ed.]
Young Man's Book. 1839.
P. 239. Identical to Endless
Amusement II.
Crambrook. 1843.
P. 5, no. 15: The Puzzle of the Steward and his Trees. This may be a configuration problem -- ??
Boy's Own Book. 1843 (Paris): 438 & 442, no. 15:
"Is it possible to place twelve pieces of money in six rows, so as to have
four in each row?" I. e. (12, 6, 5).
= Boy's Treasury, 1844, pp. 426 & 429, no. 13. = de Savigny, 1846, pp. 355 & 358, no.
11.
Family Friend 1 (1849) 148 &
177. Family Pastime -- Practical
Puzzles -- 1. The puzzle of the stars.
(9, 10, 3).
Friends
of the Family Friend, pray show
How
you nine stars would so bestow
Ten rows to form -- in each row three
--
Tell
me, ye wits, how this can be?
Robina.
Answer
has
Good-tempered
Friends! here nine stars see:
Ten rows there are, in each row three!
W. S. B. Woolhouse. Problem 39.
The Mathematician 1 (1855) 272.
Solution: ibid. 2 (1856) 278‑280. ??NYS -- cited in Burr, et al., below, who
say he does (15, 26, 3).
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles.
No. 1,
p. 176 (1868: 187). (9, 10, 3).
Ingenious artist pray disclose,
How I nine trees can so dispose,
That these ten rows shall formed be,
And every row consist of three?
No.
12, p. 182 (1868: 192-193). (24, 28,
3), but with a central pond breaking 4
rows of 6 into 8 rows of 3.
Magician's Own Book. 1857.
Prob.
33: The puzzle of the stars, pp. 277 & 300. (9, 10, 3),
Friends one and all, I pray you show
How you nine stars would so bestow,
Ten rows to form -- in each row three --
Tell me, ye wits, how this can be?
Prob.
41: The tree puzzle, pp. 279 & 301.
(21, 9, 5), unequally spaced on
each row. Identical to Book of 500
Puzzles, prob. 41.
The Sociable. 1858.
= Book of 500 Puzzles, 1859, with same problem numbers, but page numbers
decreased by 282.
Prob.
3: The practicable orchard, pp. 286 & 302.
(16, 10, 4).
Prob.
8: The florist's puzzle, pp. 289 & 303-304. (31, 6, 6) with 7 circles
of 6.
Prob.
9: The farmer's puzzle, pp. 289 & 304.
(11, 11, 3).
Prob.
12: The geometrical orchard, p. 291
& 306. (27, 9, 6).
Prob. 17:
The apple-tree puzzle, pp. 292 & 308.
(10, 5, 4).
Prob.
22: The peach orchard puzzle, pp. 294 & 309. (27, 10, 6).
Prob.
26: The gardener's puzzle, pp. 295 & 311.
(12, 6, 4) two ways.
Prob.
27: The circle puzzle, pp. 295 & 311.
(37, 20, 5) equally spaced along
each row.
Prob.
29: The tree puzzle, pp. 296 & 312.
(15, 16, 3) with some bigger
rows. Solution is a 3 x 4
array with three extra trees halfway between the points of the middle
line of four.
Prob.
32: The tulip puzzle, pp. 296 & 314.
(19, 9, 5).
Prob.
36: The plum tree puzzle, pp. 297 & 315.
(9, 10, 3).
Family Friend (Dec 1858)
359. Practical puzzles -- 2. "Make a square with twelve counters,
having five on each side." (12, 4,
5). I haven't got the answer, but
presumably it is the trick version of a hollow square with doubled corners, as
in 7.Q. See Secret Out, 1859 &
Illustrated Boy's Own Treasury, 1860.
Book of 500 Puzzles. 1859.
Prob. 3, 9, 12, 17, 22, 26, 27,
29, 32, 36 are identical to those in
The Sociable, with page numbers decreased by 282.
Prob.
33: The puzzle of the stars, pp. 91 & 114.
(9, 10, 3), identical to
Magician's Own Book, prob. 33.
Prob.
41: The tree puzzle, pp. 93 & 115.
(21, 9, 5), identical to
Magician's Own Book, prob. 41. See
Illustrated Boy's Own Treasury.
The Secret Out. 1859.
To
place twelve Cards in such a manner that you can count Four in every direction,
p. 90. (12, 7, 4) trick of a
3 x 3 array with doubling along
a diagonal. 'Every direction' must
refer to just the rows and columns, but one diagonal also works.
The
magical arrangement, pp. 381-382 = The square of counters, (UK) p. 9. (12, 4, 5) -- trick version. Same as Family Friend & Illustrated
Boy's Own Treasury, prob. 13.
The
Sphynx, pp. 385-386. (21, 30, 3). = Hoffmann, no. 15.
Charades, Enigmas, and
Riddles. 1860: prob. 13, pp. 58 &
61; 1862: prob. 13, pp. 133 &
139; 1865: prob. 557, pp. 105 & 152. (9, 10, 3).
(The 1862 and 1865 have slightly different typography.)
Sir
Isaac Newton's Puzzle (versified).
Ingenious
Artist, pray disclose
How
I, nine Trees may so dispose,
That
just Ten Rows shall planted be,
And
every Row contain just Three.
Boy's Own Conjuring Book. 1860.
Prob.
40: The tree puzzle, pp. 242 & 266.
(21, 9, 5), identical to
Magician's Own Book, prob. 41.
Prob.
42: The puzzle of the stars, pp. 243 & 267. (9, 10, 3), identical to
Magician's Own Book, prob. 33, with commas omitted.
Illustrated Boy's Own
Treasury. 1860.
Prob.
2, pp. 395 & 436. (37, 20, 5), equally spaced on each row, identical to The
Sociable, prob. 27.
Prob.
13, pp. 397 & 438. "Make a
square with twelve counters, having five on each side." (12, 4, 5).
Trick version of a hollow square with doubled corners. Presumably identical to Family Friend,
1858. Same as Secret Out.
J. J. Sylvester. Problem 2473. Math. Quest. from the Educ. Times 8 (1867) 106‑107. ??NYS -- Burr, et al. say he gives (10, 10, 3), (81, 800, 3) and (a, (a‑1)2/8, 3).
Magician's Own Book (UK
version). 1871. The solution to The florist's puzzle (The
Sociable, prob. 8) is given at the bottom of p. 284, apparently to fill
out the page as there is no relevant text anywhere.
Hanky Panky. 1872.
To
place nine cards in ten rows of three each, p. 291. I.e. (9, 10, 3).
Diagram
with no text, p. 128. (37, 20, 5), equally spaced on each line as in The
Sociable, prob. 27.
Hoffmann. Modern Magic. (George Routledge, London, 1876); reprinted by Dover, 1978.
To place twelve cards in rows, in such a manner that they will count
four in every direction, p. 58. Trick
version of a 3 x 3 square with extras on a diagonal, giving a
form of (12, 7, 4). Same as Secret Out.
Lewis Carroll. MS of 1876.
??NYS -- described in: David
Shulman; The Lewis Carroll problem; SM 6 (1939) 238-240.
Given two rows of five dots,
move four to make 5 rows of 4. Shulman
describes this case, following Dudeney, AM, 1917, then observes that since
Dudeney is using coins, there are further solutions by putting a coin on top of
another. He refers to Hoffmann and
Loyd. The same problem is in
Carroll-Wakeling, prob. 1: Cakes in a row, pp. 1-2 & 63, but undated and
the answer mentions the possibility of stacking the counters.
(9, 10, 3). Shulman quotes from Robert T. Philip; Family
Pastime; London, 1852, p. 30, ??NYS, but this must refer to the item in Family
Friend, which was edited by Robert Kemp Philp.
BMC indicates Family Pastime which may be another periodical. Shulman then cites Jackson and Dudeney. Carroll-Wakeling, prob. 2: More cakes in a
row, pp. 3 & 63, gives the problems
(9, 8, 3), (9, 9, 3), (9, 10, 3), undated.
Mittenzwey. 1880.
Prob.
151, pp. 31 & 83; 1895?: 174, pp.
36 & 85; 1917: 174, pp. 33 &
82. (6, 3, 3) in three ways.
Prob.
152, pp. 31 & 83; 1895?: 175, pp.
36 & 85; 1917: 175, pp. 33 &
82. Arrange 16 pennies as a 3 x 3
square so each row and column has four in it. Solution shows a 3 x 3 square with extras on the diagonal -- but
this only uses 12 pennies! So this the
trick version of (12, 7, 4) as in Secret Out & Hoffmann (1876).
Prob.
153, pp. 31 & 83; 1895?: 176, pp.
36 & 85; 1917: 176, pp. 33 &
82. (21, 10, 5).
Cassell's. 1881.
P. 92: The six rows puzzle. =
Manson, 1911, p. 146.
J. J. Sylvester. Problem 2572. Math. Quest. from the Educ. Times 45 (1886) 127‑128. ??NYS -- cited in Burr, below. Obtains good examples of (a, b,
3) for each a. In most cases, this is
still the best known.
Richard A. Proctor. Some puzzles; Knowledge 9 (Aug 1886) 305-306
& Three puzzles; Knowledge 9 (Sep 1886) 336-337. (19, 9, 5).
Generalises to (6n+1, 3n, 5).
Richard A. Proctor. Our puzzles. Knowledge 10 (Nov 1886)
9 & (Dec 1886) 39-40. Gives
several solutions of (19, 9, 5) and asks for (19, 10, 5). Gossip
column, (Feb 1887) 92, gives another solution
William Crompton. The odd half-hour. The Boy's Own Paper 13 (No. 657) (15 Aug 1891) 731-732. Sir Isaac Newton's puzzle (versified). (9, 10, 3).
Ingenious
artist pray disclose
How
I nine trees may so dispose
That
just ten rows shall planted be
And
every row contain just three.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
Card Puzzles No. IV, p. 3. (9,
10, 3).
Hoffmann. 1893.
Chap. VI, pp. 265‑268 & 275‑281 = Hoffmann-Hordern,
pp. 174-182, with photo.
No. 1:
(11, 12, 3).
No. 2:
(9, 10, 3).
No. 3:
(27, 9, 6).
No. 4:
(10, 5, 4).
No. 5:
(12, 6, 4). Photo on p. 177 shows
L'Embarras du Brigadier, by Mauclair-Dacier, 1891‑1900, which has a board
with a 7 x 6 array of holes and 12 pegs.
The horizontal spacing seems closer than the vertical spacing.
No. 6:
(19, 9, 5).
No. 7:
(16, 10, 4).
No. 8:
(12, 7, 4) -- Trick version of a 3 x
3 square with extras on a diagonal as
in Secret Out, Hoffmann (1876) & Mittenzwey.
No. 9:
9 red + 9 white, form 10 + 8
lines of 3 each. Puts a red and
a white point at the same place, so this is a trick version.
No.
11: (10, 8, 4) -- counts in 8 'directions', so he counts each
line twice!
No.
12: (13, 12, 5) -- with double counting as in no. 11.
No.
15: (21, 30, 3) -- but points must lie on a given figure, which
is the same as in The Secret Out.
Clark. Mental Nuts. 1897, no.
19: The apple orchard; 1904, no. 91:
The lovers' grove. (19, 9,
5). 1897 just has "Place an
orchard of nineteen trees so as to have nine rows of five trees
each." 1904 gives a poem.
I
am required to plant a grove
To
please the lady whom I love.
This
simple grove to be composed
Of
nineteen trees in nine straight rows;
Five
trees in each row I must place,
Or
I shall never see her face.
Cf
Ripley, below.
Dudeney. A batch of puzzles. Royal Magazine 1:3 (Jan 1899) &
1:4 (Feb 1899) 368-372. (22, 20,
4) with trees at lattice points of
a 7 x 10 lattice. Compare with AM,
prob. 212.
Anon. & Dudeney. A chat with the Puzzle King. The Captain 2 (Dec? 1899) 314-320 &
2:6 (Mar 1900) 598-599
& 3:1 (Apr 1900) 89. (16, 15, 4). Cf 1902.
Dudeney. "The Captain" puzzle corner. The Captain 3:2 (May 1900) 179. This gives a solution of a problem called
Joubert's guns, but I haven't seen the proposal. (10, 5, 4) but wants the
maximum number of castles to be inside the walls joining the castles. Manages to get two inside. = Dudeney; The puzzle realm; Cassell's
Magazine ?? (May 1908) 713-716; no. 6: The king and the castles. = AM, 1917, prob. 206: The king and the
castles, pp. 56 & 189.
H. D. Northrop. Popular Pastimes. 1901. No. 11: The tree
puzzle, pp. 68 & 73. = The
Sociable, no. 29.
Dudeney. The ploughman's puzzle. In:
The Canterbury Puzzles, London Magazine 9 (No. 49) (Aug 1902) 88‑92 &
(No. 50) (Sep 1902) 219.
= CP; 1907; no. 21, pp. 43‑44 & 175‑176. (16, 15, 4). Cf 1899.
A. Héraud. Jeux et Récréations Scientifiques -- Chimie,
Histoire Naturelle, Mathématiques.
Baillière et Fils, Paris, 1903.
P. 307: Un paradoxe mathématique.
(24, 28, 4). I haven't checked
for this problem in the 1884 ed.
Pearson. 1907.
Part
I, no. 77: Lines on an old sampler, pp. 77 & 167. (17, 28, 3).
Part
II, no. 83: For the children, pp. 83 & 177. Trick version of (12, 4,
5), as in Family Friend (1858).
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. He says (9, 10, 3)
"is attributed to Sir Isaac Newton, but the earliest collection of
such puzzles is, I believe, in a rare little book that I possess -- published
in 1821." [This must refer to
Jackson.] Says Rev. Mr. Wilkinson gave (11, 16, 3)
"some quarter of a century ago" and that he, Dudeney,
published (16, 15, 4) in 1897 (cf under 1902 above). He leaves these as problems but doesn't give
their solutions in the next issue.
Wehman. New Book of 200 Puzzles. 1908.
P. 4:
The practicable orchard. (16, 10,
4). = The Sociable, prob. 3.
P. 7:
The puzzle of the stars. (9, 10,
3). = Magician's Own Book, prob. 33.
P. 8:
The apple-tree puzzle. (10, 5, 4). = The Sociable, prob. 17.
P. 8:
The peach orchard puzzle. (27, 10, 6). = The Sociable, prob. 22.
P. 8:
The plum tree puzzle. (9, 10, 3). = The Sociable, prob. 36.
P. 12:
The farmer's puzzle. (11, 11, 3). = The Sociable, prob. 9.
P. 19:
The gardener's puzzle. (12, 6, 4) two ways.
= The Sociable, prob. 26.
P. 26:
The circle puzzle. (37, 20, 5) equally spaced along each row. = The Sociable, prob. 27.
P. 30:
The tree puzzle. (15, 16, 3) with some bigger rows. = The Sociable, prob. 29.
P. 31:
The geometrical orchard. (27, 9,
6). = The Sociable, prob. 12.
P. 31:
The tulip puzzle. (19, 9, 5). = The Sociable, prob. 32.
P. 41:
The florist's puzzle. (31, 6, 6) with seven circles of six. = The Sociable, prob. 8.
J. K. Benson, ed. The Pearson Puzzle Book. C. Arthur Pearson, London, nd [c1910, not in
BMC or NUC]. [This is almost identical
with the puzzle section of Benson, but has 13 pages of different
material.] A symmetrical plantation, p.
99. (24, 28, 4).
Williams. Home Entertainments. 1914.
Competitions with counters, p. 115.
(11, 12, 3); (9, 10,
3); (10, 5, 4).
Dudeney. AM.
1917. Points and lines problems,
pp. 56-58 & 189-193.
Prob.
206: The king and the castles. See The
Captain, 1900.
Prob.
207: Cherries and plums. Two (10, 5, 4)
patterns among 55 of the points of an
8 x 8 array.
Prob.
208: A plantation puzzle. (10, 5,
4) among 45 of the points of a 7 x 7
array.
Prob.
209: The twenty-one trees. (21, 12, 5).
Prob.
210: The ten coins. Two rows of
five. Move four to make (10, 5, 4).
Cf Carroll, 1876. Shows there
are 2400 ways to do this. He shows that
there are six basic solutions of the
(10, 5, 4) which he calls: star, dart, compasses, funnel, scissors,
nail and he describes the smallest arrays on which they can fit.
Prob.
211: The twelve mince-pies. 12 points
at the vertices and intersections of a Star of David. Move four to make (12, 7,
4).
Prob.
212: The Burmese plantation. (22, x,
4) among the points of a 7 x 7
array. Finds x = 21.
Cf 1899.
Prob.
213: Turks and Russians, pp. 58 & 191‑193. Complicated problem leading to
(11, 16, 3) -- cites his Afridi
problem in Tit-Bits and attributes the pattern to Wilkinson 'some twenty years
ago', cf 1908.
Blyth. Match-Stick Magic. 1921.
Four
in line, p. 48. (10, 5, 4).
Three
in line, p. 77. (9, 10, 3).
Five-line
game, pp. 78-79. (21, 9, 5).
King. Best 100. 1927. No. 62, pp. 26 & 54. = Foulsham's no. 21, pp. 9 & 13. (10, 5, 4).
Loyd Jr. SLAHP.
1928. Points and lines puzzle,
pp. 20 & 90. Says Newton proposed (9, 10, 3). Asks for
(20, 18, 4) on a 7 x 7
array.
R. Ripley. Believe It or Not! Book 2. Op. cit. in 5.E,
1931. The planter's puzzle, p. 197,
asks for (19, 9, 5) but no solution is given. See Clark, above, for a better version of
the verse.
"I
am constrained to plant a grove
For
a lady that I love.
This
ample grove is too composed;
Nineteen trees in nine straight
rows.
Five trees in each row I must place,
Or
I shall never see her face."
Rudin. 1936. Nos. 105-108, pp.
39 & 99-100.
No.
105: (9, 10, 3).
No.
106: (10, 5, 4) -- two solutions.
No.
107: (12, 6, 4) -- two solutions.
No.
108: (19, 9, 5).
Depew. Cokesbury Game Book.
1939. The orange grower, p.
221. (21, 9, 5).
The Home Book of Quizzes, Games
and Jokes. Op. cit. in 4.B.1,
1941. P. 147, prob. 1 & 2. Place six coins in an L or
a cross and make two rows of four, i.e.
(6, 2, 4), which is
done by the simple trick of putting a coin on the intersection.
R. H. Macmillan. Letter:
An old problem. MG 30 (No. 289)
(May 1946) 109. Says he believes Newton
and Sylvester studied this. Says he has
examples of (11, 16, 3), (12, 19,
3), (18, 18, 4), (24, 28, 4), (25, 30, 4), (36, 55,
4), (22, 15, 5), (26, 21, 5), (30, 26, 5).
G. C. Shephard. A problem in orchards. Eureka 9 (Apr 1947) 11-14. Given
k points in n‑dimensions, the general problem is
to draw N(k, n) hyperplanes to produce k
regions, each containing one point.
The most common example is k =
7, n = 2, N = 3. [See Section 5.Q
for determining k as a function of n and N.]
The author investigates the question of determining the possible
locations of the seventh point given six points. He gives a construction of a set
T such that being in T is
necessary and sufficient for three such lines to exist.
J. Bridges. Potter's orchard. Eureka 11 (Jan 1949)
30 & 12 (Oct 1949) 17. Start
with an orchard (9, 9, 3). Add 16 trees to make (25, 18, 5). The nine trees are three points in a triangle, with the three
midpoints of the sides and the three points halfway between these. Six of the new trees are one third of the
way along the sides of the original triangle; another six are one third of the
way along the lines joining the midpoints of the original triangle; one point
is the centre of the original triangle and the last three are easily seen.
W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. Thirteen rows of three, pp. 45 &
132. (11, 13, 3).
Young World. c1960.
Pp. 10-11.
Three
coin lines. (9, 10, 3).
Five
coin lines. (10, 5, 4).
Eleven
coin trick. (11, 12, 3).
Maxey Brooke. Dots and lines. RMM 6 (Dec 1961) 51‑55.
Cites Jackson and Dudeney. Says
Sylvester showed that n points can be arranged in at least (n‑1)(n‑2)/6 rows of three. Shows (9, 10, 3) and
(16, 15, 4).
R. L. Hutchings &
J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 20-21 & 34-35. Prob. F.
(10, 5, 4) with points in the
centres of cells of a chess board.
Actually only needs a 7 x 7 board.
Ripley's Puzzles and Games. 1966.
Pp. 18-19, item 4. (17, 7, 5).
Doubleday - 3. 1972.
Count down, pp. 125-126. Start
with a 4 x 4 array of coins. Add four
coins so that each row, column and diagonal has the same number. Solution doubles the coins in the 1, 3, 4, 2
positions in the rows.
S. A. Burr, B. Grünbaum & N.
J. A. Sloane. The orchard problem. Geometria Dedicata 2 (1974) 397‑424. Establishes good examples of (a, b, 3)
slightly improving on Sylvester, and establishes some special better
examples. Gives upper bounds for b in
(a, b, 3). Sketches history and tabulates best values and upper bounds for b in
(a, b, 3), for a = 1 (1) 32.
The
following have the maximal possible value of
b for given a
and c.
(3,
1, 3); (4, 1, 3); (5, 2, 3);
(6, 4, 3); (7, 6, 3); (8, 7, 3);
(9, 10, 3); (10, 12, 3); (11, 16, 3); (12, 19, 3); (16, 37, 3).
The
following have the largest known value of
b for the given a
and c.
(13,
22, 3); (14, 26, 3); (15, 31, 3); (17, 40, 3); (18, 46, 3); (19, 52, 3); (20, 57, 3); (21, 64,
3); (22, 70, 3); (23, 77, 3); (24, 85, 3); (25, 92,
3); (26, 100, 3); (27, 109, 3); (28, 117, 3); (29, 126,
3); (30, 136, 3); (31, 145, 3); (32, 155, 3).
M. Gardner. SA (Aug 1976). Surveys these problems, based on Burr, Grünbaum &
Sloane. He gives results for c = 4.
The
following have the maximal possible value of
b for the given a
and c.
(4,
1, 4); (5, 1, 4); (6, 1, 4);
(7, 2, 4); (8, 2, 4); (9, 3, 4);
(10, 5, 4); (11, 6, 4); (12, 7, 4).
The
following have the largest known value of
b for the given a
and c.
(13,
9, 4); (14, 10, 4); (15, 12, 4); (16, 15, 4); (17, 15,
4); (18, 18, 4); (19, 19, 4); (20, 20, 4).
Putnam. Puzzle Fun.
1978. Nos. 17-23: Bingo
arrangements, pp. 6 & 29-30. (21,
11, 5), (16, 15, 4), (19, 9, 5),
(9, 10, 3), (12, 7, 4), (22, 21, 4), (10, 5, 4).
S. A. Burr. Planting trees. In: The Mathematical
Gardner; ed. by David Klarner; Prindle, Weber & Schmidt/Wadsworth, 1981. Pp. 90‑99. Pleasant survey of the 1974 paper by Burr, et al.
Michel Criton. Des points et des Lignes. Jouer Jeux Mathématiques 3 (Jul/Sep 1991)
6-9. Survey, with a graph showing c
at (a, b). Observes that some solutions have points
which are not at intersections of lines and proposes a more restrictive kind of
arrangement of b lines whose intersections give a
points with c points on each
line. He denotes these with square
brackets which I write as [a, b,
c]. Pictures of (7, 6, 3),
[9, 8, 3], (9, 9, 3), (12, 6, 4),
[13, 9, 4], (13, 12, 3), (13, 22, 3), (16, 12, 4), (19, 19, 4), (19, 19, 5), (20, 21, 4), [21, 12,
5], (25, 12, 5), (30, 12, 7), (30, 22, 5), (49, 16,
7) and mentions of (9, 10, 3),
(16, 15, 4),
6.AO.1. PLACE FOUR POINTS EQUIDISTANTLY = MAKE FOUR TRIANGLES WITH SIX MATCHSTICKS
I am adding the problem of making
three squares with nine matchsticks here a it uses the same thought process --
see Mittenzwey and see the extended discussion at Anon., 1910.
Pacioli. De Viribus.
c1500. Ff. 191r - 192r. LXXX. Do(cumento). commo non e possibile piu
ch' tre ponti o ver tondi spere tocarse in un piano tutti (how it is not
possible for more than three points or discs or spheres to all touch in a
plane). = Peirani 252-253. Says you can only get three discs touching
in the plane, but you can get a fourth so they are all touching by making a
pyramid.
Endless Amusement II. 1826?
Prob. 21, p. 200. "To place
4 poles in the ground, precisely at an equal distance from each other." Uses a pyramidal mound of earth.
Young Man's Book. 1839.
P. 235. Identical to Endless
Amusement II.
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles, no. 6, p. 178 (1868:
189). Plant four trees at equal
distances from each other.
Frank Bellew. The Art of Amusing. 1866.
Op. cit. in 5.E. 1866: pp. 97-98
& 105-106; 1870: pp. 93‑94
& 101‑102.
Mittenzwey. 1880.
Prob.
161, pp. 32 & 84; 1895?: 184, pp.
37 & 86; 1917: 184, pp. 34 &
83. Use six sticks to make four
congruent triangles. Solution is a
rectangle (should be a square) with its diagonals, but then two of the sticks
have to be longer than the others.
Prob.
163, pp. 32 & 84; 1895?: 186 &
194, pp. 37 & 86-87; 1917: 186
& 194, pp. 34 & 83-84. Use six
equally long sticks to make four congruent triangles -- solution is a
tetrahedron. The two problems in the
1895? are differently phrased, but identical in content, while the first
solution is a picture and the second is a description.
Prob.
171, pp. 33 & 85; 1895?: 195, pp.
38 & 87; 1917: 195, pp. 34 &
84. Use nine equal sticks to make three
squares. Solution is three faces of a
cube.
F. Chasemore. Loc. cit. in 6.W.5. 1891.
Item 3: The triangle puzzle, p. 572.
Hoffmann. 1893.
Chap.
VII, no. 15, pp. 290 & 298 = Hoffmann-Hordern, pp. 195. Four matches.
Chap.
X, no. 19: The four wine glasses, pp. 344 & 381 = Hoffmann-Hordern,
pp. 238‑239, with photo on p. 239 of a version by Jaques & Son,
1870-1900. I usually solve the second
version by setting one glass on top of the other three, but here he wants the
centre of the feet of the glasses to be equally spaced and he turns one glass
over and places it in the centre of the other three, appropriately spaced.
Loyd. Problem 34: War‑ships at anchor. Tit‑Bits 32 (22 May
& 12 Jun 1897) 135 &
193. Place four warships
equidistantly so that if one is attacked, the others can come to assist
it. Solution is a tetrahedron of points
on the earth's oceans.
Parlour Games for
Everybody. John Leng, Dundee &
London, nd [1903 -- BLC], p. 30.
"With 6 matches form 4 triangles of equal size."
Pearson. 1907.
Part III, no. 77: Three squares, p. 77.
Make three squares with nine matches.
Solution is a triangular prism!
Anon. Prob. 66. Hobbies 31 (No. 781) (1 Oct 1910) 2 &
(No. 784) (22 Oct 1910) 68. Use
nine matches to make three squares.
"... the only possible solution" is to make two adjacent
squares with seven matches, then bisect each square to produce a third square
which overlaps the other two.
I
re-invented this problem in Apr 1999 and posted it on NOBNET on 19 Apr
1999. Solution (1) is the idea I had
when I made up the puzzle, but various friends gave more examples and then I
found solution (3).
(1). Arrange the nine matches to form the
following.
_
|_| |_|
|
|
Then 4 is
a square, 9 is a square and 49 is a square.
(2). Use the matches to form a triangular
prism. One may object that this also
makes two triangles.
(3). Make three squares forming three faces of a
cube, all meeting at one corner. Cf
Mittenzwey 171.
(4). Make two adjacent squares with seven of the
matches. Now bisect each of the squares
with a match parallel to the common edge of the squares. This produces a row of four adjacent
half-squares as below. The middle two
form a new square. Here one may object
that the squares are overlapping.
───
───
│
│ │ │ │
───
───
(5). Use the matches to make the figures 0,
1 and 4.
One
can use the matches to make squares whose edge is half the match length, but
one only needs eight matches to make three squares.
There
are other solutions which use the fact that matches have squared off ends and
have square cross-section, but these properties do not hold for paper matches
torn from a matchbook or for other equivalent objects like toothpicks and hence
I don't consider them quite reasonable.
Anon. Prob. 76. Hobbies 31
(No. 791) (10 Dec 1910) 256
& (No. 794) (31 Dec 1910)
318. Make as many triangles as possible
with six matches. From the solution, it
seems that the tetrahedron was expected with four triangles, but many submitted
the figure of a triangle with its altitudes drawn, but only one solver noted
that this figure contains 16 triangles!
However, if the altitudes are displaced to give an interior triangle, I
find 17 triangles!!
Williams. Home
Entertainments. 1914. Tricks with matches: To form four triangles
with six matches, p. 106.
Blyth. Match-Stick
Magic. 1921. Four triangle puzzle, p. 23.
Make four triangles with six matchsticks.
King. Best 100. 1927.
No. 59, pp. 24 & 53. =
Foulsham's no. 20, pp. 8 & 12. Use
six matches to make four triangles.
6.AO.2. PLACE AN EVEN NUMBER ON EACH LINE
See also
section 6.T.
Sometimes the diagonals are considered, but it is not always clear what is intended.
Leske. Illustriertes Spielbuch für Mädchen. 1864?
Prob.
564-31, pp. 254 & 396. From a 6 x 6
array, remove 6 to leave an even number in each row. (The German 'Reihe' can be interpreted as
row or column or both.) If we consider
this in the first quadrant with coordinates going from 1 to 6, the removed
points are: (1,2), (1,3), (2,1), (2,2),
(6,1), (6,3). The use of the sixth
column is peculiar and has the effect of making both diagonals odd, while the
more usual use of the third column would make both diagonals even.
Prob.
583-5, pp. 285 & 403: Von folgenden 36 Punkten sechs zu streichen. As above, but each file ('Zeile') in 'all
four directions' has four or six points.
Deletes: (1,1), (1,2),
(2,2), (2,3), (6,1), (6,3) which makes
one diagonal even and one odd.
Mittenzwey. 1880.
Prob. 154, pp. 31 & 83;
1895?: 177, pp. 36 & 85;
1917: 177, pp. 33 & 82.
Given a 4 x 4 array, remove 6 to leave an even number in
each row and column. Solution removes
a 2 x 3 rectangle from a corner.
[This fails -- it leaves two rows and a diagonal with an odd
number. One can use the idea mentioned
for Leske 564-31 to get a solution with both diagonals also being even.]
Hoffmann. 1893.
Chap. VI, pp. 271-272 & 285 = Hoffmann-Hordern, pp. 186-187.
No.
22: The thirty‑six puzzle. Place
30 counters on a 6 x 6 board so each horizontal and each vertical
line has an even number. Solution
places the six blanks in a
3 x 3 corner in the
obvious way. This also makes the
diagonals have even numbers.
No.
23: The "Five to Four" puzzle.
Place 20 counters on a
5 x 5 board subject to
the above conditions. Solution puts
blanks on the diagonal. This also makes
the diagonals have even number.
Dudeney. The puzzle realm. Cassell's Magazine ?? (May 1908) 713-716. The crack shots. 10 pieces in a 4 x 4 array making the maximal number of even
lines -- counting diagonals and short diagonals -- with an additional
complication that pieces are hanging on vertical strings. The picture is used in AM, prob. 270.
Loyd. Cyclopedia. 1914. The jolly friar's puzzle, pp. 307 &
380. (= MPSL2, no. 155,
pp. 109 & 172. = SLAHP: A
shifty little problem, pp. 64 & 110.)
10 men on a 4 x 4 board -- make a maximal number of even rows,
including diagonals and short diagonals.
This is a simplification of Dudeney, 1908.
King. Best 100. 1927. No. 72, pp. 29 & 56. As in Hoffmann's No. 22, but specifically
asks for even diagonals as well.
The Bile Beans Puzzle Book. 1933.
No. 19: Thirty-six coins. As in
Hoffmann's No. 22, but specifically asks for even diagonals as well.
Rudin. 1936. No. 151, pp. 53-54
& 111. Place 12 counters on a 6 x 6
board with two in each 'row, column and diagonal'. Reading the positions in each row, the
solution is: 16, 34, 25, 25, 34,
16. Some of the short diagonals and
some of the broken diagonals are empty, so he presumably isn't including these,
or he meant to ask for each of these to have an even number of at most two.
M. Adams. Puzzle Book. 1939. Prob. C.179: Even
stars, pp. 169 & 193. Same as Loyd.
Doubleday - 1. 1969.
Prob. 61: Milky Way, pp. 76 & 167.
= Doubleday - 5, pp. 85-86. 6 x
6 array with two opposite corners
already filled. Add ten more counters
so that no row, column or diagonal has more than two counters in it. Reading the positions in each row, the
solution is: 13, 35, 12, 67, 24,
46. Some short diagonals are empty or
have one counter and some broken diagonals have one or four counters, so he
seems to be ignoring them. Hence this
is the same problem as Rudin, but with a less satisfactory solution.
Obermair. Op. cit. in 5.Z.1. 1984. Prob. 37, pp. 38
& 68. 52 men on an 8 x 8
board with all rows, columns and diagonals (both long and short) having
an even number.
6.AP . DISSECTIONS OF A TETRAHEDRON
Richard A. Proctor. Our puzzles; Knowledge 10 (Feb 1887) 83
& Solutions of puzzles; Knowledge 10 (Mar 1887) 108-109. "Puzzle XIX. Show how to cut a regular tetrahedron (equilateral triangular
pyramid) so that the face cut shall be a square: also show how to plug a square
hole with a tetrahedron." Solution
shows the cut clearly.
Edward T. Johnson. US Patent 2,216,915 -- Puzzle. Applied: 26 Apr 1939; patented: 8 Oct 1940. 2pp + 1p diagrams. Described in S&B, p. 46.
E. M. Wyatt. Wonders in Wood. Op. cit. in 6.AI.
1946. Pp. 9 & 11: the
tetrahedron or triangular pyramid. P. 9
is reproduced in S&B, p. 46.
Donovan A. Johnson. Paper Folding for the Mathematics
Class. NCTM, 1957, p. 26, section 62:
Pyramid puzzle. Gives instructions for
making the pieces from paper.
Claude Birtwistle. Editor's footnote. MTg 21 (Winter 1962) 32.
"The following interesting puzzle was given to us recently."
Birtwistle. Math. Puzzles & Perplexities. 1971.
Bisected tetrahedron, pp. 157-158.
Gives the net so one can make a drawing, cut it out and fold it up to
make one piece.
These
dissections usually also work with a tetrahedron of spheres and hence these are
related to ball pyramid puzzles, 6.AZ.
The
first version I had in mind dissects each of the two pieces of 6.AP.1 giving
four congruent rhombic pyramids.
Alternatively, imagine a tetrahedron bisected by two of its midplanes,
where a midplane goes halfway between a pair of opposite edges. This puzzle has been available in various versions
since at least the 1970s, including one from Stokes Publishing Co., 1292
Reamwood Avenue, Sunnyvale, California, 94089, USA., but I have no idea of the
original source. The same pieces are
part of a more complex dissection of a cube, PolyPackPuzzle, which was produced
by Stokes in 1996. (I bought mine from
Key Curriculum Press.)
In
1997, Bill Ritchie, of Binary Arts, sent a quadrisection of the tetrahedron
that they are producing. Each piece is
a hexahedron. The easiest way to
describe it is to consider the tetrahedron as a pile of spheres with four on an
edge and hence 20 altogether.
Consider a planar triangle of six of these spheres with three on an edge
and remove one vertex sphere to produce a trapezium (or trapezoid) shape. Four of these assemble to make the
tetrahedron. Writing this has made me
realise that Ray Bathke has made and sold these 5-sphere pieces as
Pyramid 4 for a few years. However, the
solid pieces used by Binary Arts are distinctly more deceptive.
Len Gordon
produced another quadrisection of the
20 sphere tetrahedron 0 0
using the planar shape at the right. This was c1980?? 0 0
0
David Singmaster. Sums of squares and pyramidal numbers. MG 66 (No. 436) (Jun 1982) 100-104. Consider a tetrahedron of spheres with 2n
on an edge. The quadrisection
described above gives four pyramids whose layers are the squares 1, 4, ..., n2. Hence
four times the sum of the first
n squares is the tetrahedral
number for 2n, i.e. 4 [1 + 4 + ...
+ n2] = BC(n+2, 3).
6.AQ. DISSECTIONS OF A CROSS, T OR H
The
usual dissection of a cross has two diagonal cuts at 45o to the
sides and passing through two of the reflex corners of the cross and yielding
five pieces. The central piece is six-sided,
looking like a rectangle with its ends pushed in and being symmetric. Depending on the relative lengths of the
arms, head and upright of the cross, the other pieces may be isosceles right
triangles or right trapeziums. Removing
the head of the cross gives the usual dissection of the T
into four pieces -- then the central piece is five-sided. Sometimes the central piece is split in
halves. Occasionally the angle of the
cuts is different than 45o. Dissections of an H have the same basic idea
of using cuts at 45o -- the result can be a bit like two Ts
with overlapping stems and the number of pieces depends on the relative
size and positioning of the crossbar of the
H -- see: Rohrbough.
S&B, pp. 20‑21, show
several versions. They say that crosses
date from early 19C. They show a 6‑piece
Druid's Cross, by Edwards & Sons, London, c1855. They show several T‑puzzles
-- they say the first is an 1903 advertisement for White Rose Ceylon Tea, NY --
but see 1898 below. They also show some
H‑puzzles.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. F. 4r is "Analysis of the
Essay of Games". F. 4v has a cross
cut into 5 pieces in the usual way.
Endless Amusement II. 1826?
Prob. 30, p. 207. Usual five
piece cross. The three small pieces are
equal. = New Sphinx, c1840, pp.
139-140.
Crambrook. 1843.
P. 4.
No.
10: Five pieces to form a Cross.
No.
11: The new dissected Cross.
Without pictures, I cannot tell what
dissections are used??
Boy's Own Book. 1843 (Paris): 435 & 440, no. 2. Usual five piece cross, very similar to
Endless Amusement. One has to make
three pieces of fig. 2. = Boy's
Treasury, 1844, pp. 424 & 428. = de
Savigny, 1846, pp. 353 & 357, no. 1.
Family Friend 2 (1850) 58 &
89. Practical Puzzle -- No. II. = Illustrated Boy's Own Treasury, 1860, No.
32, pp. 401 & 440. Usual five piece
cross to "form that which, viewed mentally, comforts the
afflicted." Three pieces of fig.
1.
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles, no. 7, p. 178-179 (1868:
189). Five piece dissection of a cross,
but the statement of the problem doesn't say which piece to make multiple
copies of.
Magician's Own Book. 1857.
Prob. 17: The cross puzzle, pp. 272 & 295. Usual 5 piece cross, essentially identical to Family Friend,
except this says to "form a cross."
= Book of 500 Puzzles, 1859, prob. 17, pp. 86 & 109. = Boy's Own Conjuring Book, 1860, prob. 16,
pp. 234 & 258.
Charades, Enigmas, and
Riddles. 1860: prob. 33, pp. 60 &
66; 1862: prob. 33, pp. 136 &
143; 1865: prob. 577, pp. 108 & 156. Usual five piece cross, showing all five
pieces.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 584-12, pp.
288 & 406: Ein Kreuz. Begins as the
usual five piece cross, but the central piece is then bisected into two mitres
and the base has two bits cut off to give an eight piece puzzle.
Frank Bellew. The Art of Amusing. 1866.
Op. cit. in 5.E. 1866: pp.
239-240; 1870: pp. 236‑238. Usual five piece cross.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 1: The
cross puzzle, pp. 27 & 30. Usual
five piece cross, but instructions say to cut three copies of the wrong piece.
Mittenzwey. 1880.
Prob. 188, pp. 35 & 88;
1895?: 213, pp. 40 & 91;
1917: 213, pp. 37 & 87. This
is supposed to be a 10 piece dissection of a cross obtained by further
dissecting the usual five pieces.
However, pieces 3 & 4 are drawn as trapezoids in the problem and
triangles in the solution and piece 2 in the solution is half the size given in
the problem. Further, pieces 1 & 2
appear equilateral in the problem, but are isosceles right triangles in the
solution. One could modify this to get
a 9 piece version where eight of the pieces are right trapezoids -- four having
edges 1, 1, 2, Ö2 and four having edges Ö2, Ö2, 2Ö2, 2, but the arms would be twice as long as they
are wide. Or one can make the second
four pieces be Ö2, Ö2,
2 isosceles right triangles. In either case, the ninth piece would be a
rectangle.
Lemon. 1890. A card board
puzzle, no. 33, pp. 8 & 98. Usual
five piece cross.
Hoffmann. 1893.
Chap. III, no. 12: The Latin cross puzzle, pp. 93 & 126
= Hoffmann‑Hordern, pp. 82-83, with photo. As in Indoor & Outdoor.
Photo on p. 83 shows two versions: one in metal by Jaques & Sons,
1870-1895; the other in ivory, 1850-1900.
Hordern Collection, p. 59, shows a Druid's Cross Puzzle.
Lash, Inc. -- Clifton, N.J. --
Chicago, Ill. -- Anaheim, Calif. T Puzzle.
Copyright Sept. 1898. 4‑piece T
puzzle to be cut out from a paper card, but the angle of the cuts is
about 35o instead of
45o which makes it
less symmetric and less confusing than the more common version. The resulting T is somewhat wider than
usual, being about 16% wider than it is tall. It advertises: Lash's Bitters The
Original Tonic Laxative. Photocopy sent
by Slocum.
Benson. 1904.
The cross puzzle, pp. 191‑192.
Usual 5 piece version.
Wehman. New Book of 200 Puzzles. 1908.
The cross puzzle, p. 17. Usual 5
piece version.
A. Neely Hall. Op. cit. in 6.F.5. 1918. The T‑puzzle,
pp. 19‑20. "A famous old
puzzle ...." Usual 4‑piece
version, but with long arms.
Western Puzzle Works, 1926
Catalogue. No. 1394: Four pieces to
form Letter T. The notched piece is less symmetric than
usual.
Collins. Book of Puzzles. 1927. The crusader's
cross puzzle, pp. 1-2. The three small
pieces are equal.
Arthur Mee's Children's
Encyclopedia 'Wonder Box'. The
Children's Encyclopedia appeared in 1908, but versions continued until the
1950s. This looks like 1930s?? Usual 5 piece cross.
A. F. Starkey. The
T puzzle. Industrial Arts and Vocational Education 37
(1938) 442. "An interesting
novelty ...."
Rohrbough. Puzzle Craft. 1932. The "H"
Puzzle, p. 23. Very square H --
consider a 3 x 3 board with the top and bottom middle cells
removed. Make a cut along the main
diagonal and two shorter cuts parallel to this to produce four congruent
isosceles right triangles and two odd pentagons.
See Rohrbough in 6.AS.1 for a
very different T puzzle.
6.AR. QUADRISECTED SQUARE PUZZLE
This
is usually done by two perpendicular cuts through the centre. A dissection proof of the Theorem of
Pythagoras described by Henry Perigal (Messenger of Mathematics 2 (1873) 104)
uses the same shapes -- cf 6.AS.2.
The
pieces make a number of other different shapes.
Crambrook. 1843.
P. 4, no. 17: Four pieces to form a Square. This might be the dissection being considered here??
A. Héraud. Jeux et Récréations Scientifiques -- Chimie,
Histoire Naturelle, Mathématiques.
(1884); Baillière, Paris, 1903. Pp. 303‑304: Casse‑tête. Uses two cuts which are perpendicular but
are not through the centre. He claims
there are 120 ways to try to assemble it, but his mathematics is shaky -- he
adds the numbers of ways at each stage rather than multiplying! Also, as Strens notes in the margin of his
copy (now at Calgary), if the crossing is off-centre, then many of the edges have
different lengths and the number of ways to try is really only one. Actually, I'm not at all sure what the
number of ways to try is -- Héraud seems to assume one tries each orientation
of each piece, but some intelligence sees that a piece can only fit one way
beside another.
Handy Book for Boys and
Girls. Op. cit. in 6.F.3. 1892.
P. 14: The divided square puzzle.
Crossing is off-centre.
Tom Tit, vol 3. 1893.
Carré casse-tête, pp. 179-180. =
K, no. 26: Puzzle squares, pp. 68‑69.
= R&A, Puzzling squares, p. 99.
Not illustrated, but described:
cut a square into four parts by two perpendicular cuts, not necessarily
through the centre.
A. B. Nordmann. One Hundred More Parlour Tricks and
Problems. Wells, Gardner, Darton &
Co., London, nd [1927 -- BMC]. No. 77:
Pattern making, pp. 69-70 & 109.
Make five other shapes.
M. Adams. Puzzle Book. 1939. Prob. C.12: The
broken square, pp. 125 & 173. As
above, but notes that the pieces also make a square with a square hole.
6.AS. DISSECTION OF SQUARES INTO A SQUARE
Lorraine Mottershead. Investigations in Mathematics. Blackwell, Oxford, 1985. P. 102 asserts that dissections of squares
to various hexagons and heptagons were known c1800 while square to rectangle
dissections were known to Montucla -- though she illustrates the latter with
examples like 6.Y, she must mean 6.AS.5.
6.AS.1. TWENTY 1, 2, Ö5 TRIANGLES MAKE A SQUARE OR FIVE EQUAL SQUARES TO A SQUARE
The
basic puzzle has been varied in many ways by joining up the 20 triangles into various
shapes, but I haven't attempted to consider all the modern variants. A common form is a square with a skew # in
it, with each line joining a corner to the midpoint of an opposite side, giving
the 9 piece version. This has four of
the squares having a triangle cut off.
For symmetry, it is common to cut off a triangle from the fifth square,
giving 10 pieces, though the assembly into one square doesn't need this. See Les Amusemens for details.
Cf
Mason in 6.S.2 for a similar puzzle with twenty pieces.
If
the dividing lines are moved a bit toward the middle and the central square is
bisected, we get a 10 piece puzzle, having two groups of four equal pieces and
a group of two equal pieces, called the Japan square puzzle. I have recently noted the connection of this
puzzle with this section, so there may be other examples which I have not
previously paid attention to -- see:
Magician's Own Book, Book of 500
Puzzles, Boy's Own Conjuring Book, Illustrated Boy's Own Treasury, Landells,
Hanky Panky, Wehman.
Les Amusemens. 1749.
P. xxxii. Consider five 2 x 2
squares. Make a cut from a
corner to the midpoint of an opposite side on each square. This yields five 1, 2, Ö5 triangles and five pieces comprising three such triangles. The problem says to make a square from five
equal squares. So this is the 10 piece
version.
Minguet. 1755.
Pp. not noted -- ??check (1822: 145-146; 1864: 127-128). Not in 1733 ed. 10 piece version. Also a
15 piece version where triangles
are cut off diagonally opposite corners of each small square leaving
parallelogram pieces as in Guyot.
Vyse. Tutor's Guide. 1771? Prob. 6, 1793: p. 304, 1799: p. 317 &
Key p. 357. 2 x 10 board to be cut into five pieces to make
into a square. Cut into a 2 x 2
square and four 2, 4, 2Ö5 triangles.
Ozanam‑Montucla. 1778.
Avec cinq quarrés égaux, en former un seul. Prob. 18 & fig. 123, plate 15, 1778: 297; 1803: 292-293; 1814: 249-250; 1840:
127. 9 piece version. Remarks that any number of squares can be
made into a square -- see 6.AS.5.
Catel. Kunst-Cabinet. 1790.
Das
mathematische Viereck, pp. 10-11 & fig. 15 on plate I. 10 piece version with solution shown. Notes these make five squares.
Das
grosse mathematische Viereck, p. 11 & fig. 14 on plate I. Cut the larger pieces to give five more 1, 2, Ö5 triangles and five Ö5, Ö5, 2 triangles.
Again notes these make five squares.
Guyot. Op. cit. in 6.P.2.
1799. Vol. 2: première
récréation: Cinq quarrés éqaux étant sonnés, en former un seul quarré, pp. 40‑41
& plate 6, opp. p. 37. 10 piece
version. Suggests cutting another
triangle off each square to give 10 triangles and 5 parallelograms.
Bestelmeier. 1801.
Item 629: Die 5 geometrisch zerschnittenen Quadrate, um aus 5 ein
einziges Quadrat zu machen. As in Les
Amusemens. S&B say this is the
first appearance of the puzzle. Only
shown in a box with one small square visible.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles.
No. 8,
pp. 25 & 84 & plate I, fig. 5, no. 1.
= Vyse.
No.
10, pp. 25 & 84-85 & plate I, fig. 7, no. 1. Five squares to one. Nine
piece version.
Rational Recreations. 1824.
Feat 35, pp. 164-166. Usual 20
piece form.
Manuel des Sorciers. 1825.
Pp. 201-202, art. 18. ??NX Five squares to one -- usual 10 piece form and
15 piece form as in Guyot.
Endless Amusement II. 1826?
[1837
only] Prob. 35, p. 212. 20 triangles to form a square. = New Sphinx, c1840, p. 141, with
problem title: Dissected square.
Prob.
37, p. 215. 10 piece version. = New Sphinx, c1840, p. 141.
Boy's Own Book. The square of triangles. 1828: 426;
1828-2: 430; 1829 (US):
222; 1855: 576; 1868: 676.
Uses 20 triangles cut from a square of wood. Cf 1843 (Paris) edition, below.
c= de Savigny, 1846, p. 272: Division d'un carré en vingt triangles.
Nuts to Crack IV (1835), no.
195. 20 triangles -- part of a long
section: Tricks upon Travellers. The
problem is used as a wager and the smart-alec gets it wrong.
The Riddler. 1835.
The square of triangles, p. 8.
Identical to Boy's Own Book, but without illustration, some consequent
changing of the text, and omitting the last comment.
Crambrook. 1843.
P. 4.
No. 7:
Egyptian Puzzle. Probably the 10 piece
version as in Les Amusemens. See
S&B below, late 19C. Check??
No.
23: Twenty Triangles to form a Square.
Check??
Boy's Own Book. 1843 (Paris): 436 & 441, no. 5:
"Cut twenty triangles out of ten square pieces of wood; mix them together,
and request a person to make an exact square with them." As stated, this is impossible; it should be
as in Boy's Own Book, 1828 etc., qv. =
Boy's Treasury, 1844, pp. 425 & 429.
= de Savigny, 1846, pp. 353 & 357, no. 4. Also copied, with the error, in:
Magician's Own Book, 1857, prob. 29: The triangle puzzle; Book of 500 Puzzles, 1859, prob. 29: The
triangle puzzle; Boy's Own Conjuring
Book, 1860, prob. 28: The triangle puzzle.
c= Hanky Panky, 1872, p. 122.
Magician's Own Book. 1857.
How to
make five squares into a large one without any waste of stuff, p. 258. 9 piece version.
Prob.
29: The triangle puzzle, pp. 276 & 298.
Identical to Boy's Own Book, 1843 (Paris).
Prob.
35: The Japan square puzzle, pp. 277 & 300. Make two parallel cuts and then two perpendicular to the first
two so that a square is formed in the centre.
This gives a 9 piece puzzle, but here the central square is cut by a
vertical through its centre to give a 10 piece puzzle. = Landells, Boy's Own Toy-Maker, 1858, pp.
145-146.
Charles Bailey (manufacturer in
Manchester, Massachusetts). 1858. An Ingenious Puzzle for the Amusement of
Children .... The 10 pieces of Les
Amusemens, with 19 shapes to make, a la tangrams. Sent by Jerry Slocum -- it is not clear if there were actual
pieces with the printed material.
The Sociable. 1858.
Prob.
10: The protean puzzle, pp. 289 & 305-306. Cut a 5 x 1 into 11 pieces to form eight shapes, e.g. a
Greek cross. It is easier to describe the
pieces if we start with a 10 x 2. Then three squares are cut off. One is halved into two 1 x 2
rectangles. Two squares have
two 1, 2, Ö5
triangles cut off leaving triangles of sides 2, Ö5, Ö5.
The remaining double square is almost divided into halves each with
a 1, 2, Ö5 triangle cut off, but these two triangles remain connected along
their sides of size 1, thus giving a 4, Ö5, Ö5
triangle and two trapeziums of sides
2, 2, 1, Ö5. = Book of 500 Puzzles, 1859, prob. 10, pp. 7 & 23-24.
Prob.
42: The mechanic's puzzle, pp. 298 & 317.
Cut a 10 x 2 in five pieces to make a square, as in Vyse. = Book of 500 Puzzles, 1859, prob. 16, pp.
16 & 35.
Book of 500 Puzzles. 1859.
Prob.
10: The protean puzzle, pp. 7 & 23-24.
As in The Sociable.
Prob.
42: The mechanic's puzzle, pp. 16 & 35.
As in The Sociable.
How to
make five squares into a large one without any waste of stuff, p. 72. Identical to Magician's Own Book.
Prob.
29: The triangle puzzle, pp. 90 & 113.
Identical to Boy's Own Book, 1843 (Paris).
Prob.
35: The Japan square puzzle, pp. 91 & 114.
Indoor & Outdoor. c1859.
Part II, prob. 11: The mechanic's puzzle, pp. 130-131. Identical to The Sociable.
Boy's Own Conjuring Book. 1860.
Prob.
28: The triangle puzzle, pp. 238 & 262.
Identical to Boy's Own Book, 1843 (Paris) and Magician's Own Book.
Prob.
34: The Japan square puzzle, pp. 240 & 264. Identical to Magician's Own Book.
Illustrated Boy's Own
Treasury. 1860.
Prob.
9, pp. 396 & 437. [The Japan square
puzzle.] Almost identical to Magician's
Own Book.
Optics:
How to make five squares into a large one without any waste of stuff, p.
445. Identical to Book of 500 Puzzles,
p. 72.
Vinot. 1860. Art. LXXV: Avec
cinq carrés égaux, en faire un seul, p. 90.
Nine piece version.
Leske. Illustriertes Spielbuch für Mädchen. 1864?
Prob.
174, pp. 87-88. Nine piece version.
Prob.
584-6, pp. 287 & 405. Ten piece
version of five squares to one.
Hanky Panky. 1872.
The
puzzle of five pieces, p. 118. 9 piece
version.
Another
[square] of four triangles and a square, p. 120. 10 x 2 into five pieces
to make a square.
[Another
square] of ten pieces, pp. 121-122.
Same as the Japan square puzzles in Magician's Own Book.
[Another
square] of twenty triangles, p. 122.
Similar to Boy's Own Book, 1843 (Paris), but with no diagram and less
text, making it quite cryptic.
Mittenzwey. 1880.
Prob. 175, pp. 33-34 & 85;
1895?: 200, pp. 38 & 87; 1917:
200, pp. 35 & 84. 10 pieces as in
Les Amusemens. See in 6.AS.2 and 6.S.2
for the use of these pieces to make other shapes.
See Mason, 1880, in 6.S.2 for a
similar, but different, 20 piece puzzle.
S&B, pp. 11 & 19, show a
10 piece version called 'Egyptian Puzzle', late 19C?
Lucas. RM2. 1883. Les vingt triangles, pp. 128‑129. Notes that they also make five squares in
the form of a cross.
Tom Tit, vol. 2. 1892.
Diviser un carré en cinq carrés égaux, pp. 147‑148. = K, no. 2: To divide a square into five
equal squares, pp. 12-14. = R&A,
Five easy pieces, p. 105. Uses 9
pieces, but mentions use of 10 pieces.
Hoffmann. 1893.
Chap.
III, no. 21: The five squares, pp. 100 & 132‑133
= Hoffmann-Hordern, p. 94, with photo.
9 piece version, as in Magician's Own Book. Photo on p. 94 shows: an ivory version,
1850-1900;, and a wood version, named Egyptian Puzzle, by C. N. Mackie,
1860-1890; both with boxes.
Chap.
III, no. 24: The twenty triangles, pp. 101 & 134 = Hoffmann-Hordern,
pp. 96-97, with photo. As in Boy's Own
Book. Photo on p. 97 shows The Twenty
Triangle Puzzle, with box, by Jaques & Son, 1870-1895. Hordern Collection, p. 64, shows
Apollonius, with box, by W. X., Paris, 1880-1900, in a solution very different
to the usual one.
Chap.
III, no. 30: The carpenter's puzzle -- no. 1, pp. 103 & 136‑137
= Hoffmann‑Hordern, p. 101.
Cut a 5 x 1 board into five pieces to make a square.
Chap.
X, no. 25: The divided square, pp. 346 & 384 = Hoffmann-Hordern, p.
242. 9 piece puzzle as a
dissection of a square which forms 5 equal squares. He places the five squares together as a 2 x 2 with an adjacent 1 x
1, but he doesn't see the connection
with 6.AS.2.
Montgomery Ward & Co. Catalog No 57, Spring &
Summer, 1895. Facsimile by Dover, 1969,
??NX. P. 237 shows the Mystic Square,
as item 25463, which is the standard 10 piece version.
Benson. 1904.
The
carpenter's puzzle (No. 2), p. 191. =
Hoffmann, p. 103.
The
five‑square puzzle, pp. 196‑197.
= Hoffmann, p. 100.
The
triangle puzzle, p. 198. = Hoffmann, p.
101.
Wehman. New Book of 200 Puzzles. 1908.
P. 3:
The triangle puzzle. 20 pieces. = Boy's Own Book, omitting the adjuration to
use wood and smooth the edges
P. 12:
The protean puzzle. c= The Sociable,
prob. 10, with the instructions somewhat clarified.
P. 14:
The Japan square puzzle. c= Magician's
Own Book.
P. 19:
To make five squares into a large one.
10 piece version.
P. 27:
The mechanic's puzzle. = The Sociable,
prob. 42.
J. K. Benson, ed. The Pearson Puzzle Book. C. Arthur Pearson, London, nd [c1910, not in
BMC or NUC]. [This is almost identical
with the puzzle section of Benson, but has 13 pages of different
material.] Juggling geometry, pp.
97-98. Five triangles, which should be
viewed as 2, 4, 2Ö5. Cut one from the midpoint of the hypotenuse
to the midpoint of the long leg and assemble into a square, so this becomes a
six-piece or five-piece version as in Vyse, etc.
I have seen a 10 piece French
example, called Jeu du Carré, dated 1900‑1920.
I have seen a 9(?) piece English
example, dated early 20C, called The Euclid Puzzle.
Dudeney. Perplexities column, no. 109: A cutting-out
puzzle. Strand Magazine 45 (No. 265) (Jan 1913) 113 &
(No. 266) (Feb 1913) 238. c= AM,
prob. 153 -- A cutting-out puzzle, pp. 37 & 172. Cut a 5 x 1 to make a square. He shows a solution in five pieces and asks for a solution in
four pieces. AM states the generalized
form given in 6.AS.5.
Rohrbough. Puzzle Craft. 1932. Square
"T", p. 23 (= The "T" Puzzle, p. 23 of 1940s?). 1 x 1
square and two 1 x 2 rectangles cut diagonally can be formed into
a square or into a T.
Gibson. Op. cit. in 4.A.1.a. 1963.
Pp. 71 & 76: Square away. A
five piece puzzle, approximately that formed by drawing parallel lines from two
diagonally opposite corners to the midpoints of opposite sides and then cutting
a square from the middle of the central strip.
As drawn, the lines meet the opposite sides a bit further along than the
midpoints.
Ripley's Puzzles and Games. 1966.
Pp.
58-59, item 6. Five right triangles to
a square. Though not specified, the
triangles have sides proportional to 1,
2, Ö5.
Solution is as in Benson.
Pp.
60-61, item 5. Start with the large
square, which is 2Ö5 on
a side. Imagine the 9 piece puzzle
where one line goes from the upper left corner to the midpoint of the right
side. Number the outer pieces clockwise
from the upper left, so that pieces 1,
3, 5, 7 are the small triangles
and 2, 4, 6, 8 are the trapezia. The pieces of this puzzle are as follows: combine pieces 1, 8, 7 into a Ö5, 2Ö5, 5
triangle; combine pieces 2, 3, 4
into an irregular hexagon; take
separate pieces 5 and 6 and the central square. These five pieces form: a
square; a Greek cross; a 5 x 4 rectangle;
a triangle of sides 2Ö5, 4Ö5, 10; etc.
I think there are earlier versions of this, e.g. in Loyd, but I have
just observed the connection with this section.
6.AS.1.a. GREEK CROSS TO A SQUARE
Note
that a proper Greek cross is formed from five equal squares.
Lucas. RM2. 1883. Loc. cit. in 6.AS.1. Uses 20 triangles.
Lemon. 1890. The Maltese cross
squared, no. 369, pp. 51 & 111. Cut
a Maltese cross (really a Greek cross) by two cuts into four pieces that make a
square.
Hoffmann. 1893.
Chap. III, no. 13: The Greek cross puzzle, pp. 94 & 126
= Hoffmann‑Hordern, pp. 82 & 84. Has four pieces made by two cuts.
Loyd. Tit‑Bits 31 (10,
17 & 31 Oct 1896) 25, 39 & 75.
= Cyclopedia, 1914, p. 14. Four
pieces as in Hoffmann.
Loyd. Problem 23: A new "square and cross" puzzle. Tit‑Bits 31 (13 Mar 1897) 437 &
32 (3 Apr 1897) 3. =
Cyclopedia, 1914, pp. 58, 270 & 376.
Four congruent pieces.
Loyd. Problem 27: The swastika problem. Tit‑Bits 32 (3
& 24 Apr 1897) 3 &
59. = Cyclopedia, 1914, p.
58. Quadrisect square to make two equal
Greek crosses.
Loyd. Problem 30: The Easter problem.
Tit‑Bits 32 (24 Apr
& 15 May 1897) 59 &
117. Dissect square into five
pieces to make two unequal Greek crosses.
Dudeney. Problem 56: Two new cross puzzles. Tit‑Bits 33 (23 Oct &
13 Nov 1897) 59 & 119.
Dissect a half square (formed by cutting a square either vertically or
diagonally) to a Greek cross. Solutions
in 3 and 4 pieces. [The first case = Loyd, Cyclopedia, 1914, Easter 1903, pp.
46 & 345.]
Benson. 1904.
The Greek cross puzzle, p. 197.
= Hoffmann, p. 94.
Dudeney. Cutting-out paper puzzles. Cassell's Magazine ?? (Dec 1909) 187-191
& 233-235.
States
that the dissection with four pieces in two cuts is relatively 'recent'. c= AM, 1917, p. 29, which dates this to 'the
middle of the nineteenth century'.
Fold a
Greek cross so that one cut gives four congruent pieces which form a
square. = AM, 1917, prob. 145, pp.
35 & 169.
M. Adams. Indoor Games. 1912. The Greek cross, p.
349 with figs. on p. 347.
6.AS.1.b. OTHER GREEK CROSS DISSECTIONS
See
also 6.F.3 and 6.F.5.
Dudeney. A batch of puzzles. Royal Magazine 1:3 (Jan 1899) &
1:4 (Feb 1899) 368-372. Squares
and cross puzzle. = AM, 1917, p.
34. Dissect a Greek cross into five
pieces which make two squares, one three times the edge of the other. If the squares in the Greek cross have
edge Ö2, then the cross has area
10 and the two squares have
areas 1 and 9. The dissection arise by joining the
midpoints of the edges of the central square of the cross and extending these
lines in one direction symmetrically.
Dudeney. AM.
1917. Greek cross puzzles, pp.
28-35. This discusses a number of
examples and gives a few problems.
Collins. Book of Puzzles. 1927. The Greek cross
puzzle, pp. 98-100. Take a Greek cross
whose squares have side 2, so the cross has area 10. Take another cross of area 5 and place it inside the large cross. If this is done centrally and the small one
turned to meet the edges of the large one, there are four congruent heptagonal
pieces surrounding the small one which make another Greek cross of area 5.
Eric Kenneway. More Magic Toys, Tricks and Illusions. Beaver Books (Arrow (Hutchinson)), London,
1985. On pp. 56-58, he considers a Greek
cross cut by two pairs of parallel lines into nine pieces which would make five
squares. The lines join an outer corner
to the midpoint of an opposite segment.
This produces a tilted square in the centre. By pairing the other pieces, he gets four identical pieces which
make a square and a Greek cross in a square.
6.AS.2. TWO (ADJACENT) SQUARES TO A SQUARE
The
smaller square often has half the edge of the larger, which connects this with
6.AS.1, but this is not essential. The
two squares are usually viewed as one piece, i.e. a P‑pentomino. These
items are dissection proofs of the Theorem of Pythagoras -- see Yates
(op. cit. in 6.B, pp. 38-39) for some other examples of this point.
See
Ripley's for a similar example, but the
2 x 2 square has a Ö2, Ö2,
2 triangle attached to an edge.
Another
version has squares of area 1 and 8.
The area 8 square is cut into
four pieces which combine with the area 1 square to make an area 9 square. I call this the 4 - 5 piece square.
Walther Karl Julius Lietzmann
(1880-1959). Der Pythagoreische
Lehrsatz. Teubner, (1911, 2nd ed.,
1917), 6th ed., 1951. [There was a 7th
ed, 1953.] Pp. 23-24 gives the standard
dissection proof for the Theorem of Pythagoras. The squares are adjacent and if considered as one piece, the dissection
has three pieces. He says it was known
to Indian mathematicians at the end of the 9C as the Bride's Chair (Stuhl der
Braut). (I always thought this name
referred to the figure of the Euclid I, 47 -- ??)
Thabit ibn Qurra (= Thābit
ibn Qurra). c875. Gives the standard dissection proof for the
Theorem of Pythagoras. The squares are
adjacent and if considered as one piece, the dissection has three pieces. [Q. Mushtaq & A. L. Tan; Mathematics:
The Islamic Legacy; Noor Publishing House, Farashkhan, Delhi, 1993, pp. 71-72]
give this and cite Lietzmann. Greg
Frederickson [email of 18 Oct 1996] cites
Aydin Sayili; Thabit ibn Qurra's generalization of the Pythagorean theorem;
Isis 51 (1960) 35-37.
Abu'l 'Abbas al-Fadhl ibn Hatim
al-Narizi (or Annairizi). (d.
c922.) Ed. by Maximilian Curtze, from a
translation by Gherardo of Cremona, as: Anaritii In decem libros priores
Elementorum Euclidis Commentarii, IN:
Euclidis Opera Omnia; Supplementum; Teubner, Leipzig, 1899. ??NYS -- information supplied by Greg
Frederickson.
Johann Christophorus Sturm. Mathesis Enumerata, 1695, ??NYS. Translated by J. Rogers? as: Mathesis Enumerata: or, the Elements of the Mathematicks; Robert Knaplock, London, 1700, ??NYS --
information provided by Greg Frederickson, email of 14 Jul 1995. Fig. 29 shows it clearly and he attributes
it to Frans van Schooten (the Younger, who was the more important one), but
this source hasn't been traced yet.
Les Amusemens. 1749.
Prob. 216, p. 381 & fig. 97 on plate 8: Réduire les deux quarrés en
un seul. Usual dissection of two
adjacent squares, attributed to 'Sturmius', a German mathematician, i.e the
previous entry.
Ozanam‑Montucla. 1778.
Diverses démonstrations de la quarante-septieme du premier livre
d'Euclide, ..., version 2. Fig. 27,
plate 4. 1778: 288; 1803: 284;
1814: 241-243;
1840: 123-124. This is a
version of the proof that (a + b)2
= c2 + 4(ab/2), but the
diagram includes extra lines which produce the standard dissection of two
adjacent triangles.
Crambrook. 1843.
P. 4, no. 19: One Square to form two Squares -- ??
E. S. Loomis. The Pythagorean Proposition. 2nd ed., 1940; reprinted by NCTM, 1968.
On pp. 194‑195, he describes the usual dissection by two cuts as
Geometric Proof 165 and gives examples back to 1849, Schlömilch.
Family Friend 2 (1850) 298 &
353. Practical Puzzle -- No. X. = Illustrated Boy's Own Treasury, 1860,
Prob. 11, pp. 397 & 437. The larger
square has twice the edge of the smaller and is shown divided into four, so
this is clearly related to 6.AS.1, though the shape is considered as one piece,
i.e. a P-pentomino, to be cut into
three parts to make a square.
Magician's Own Book. 1857.
To form a square, p. 261. = Book
of 500 Puzzles, 1859, p. 75. An abbreviated
version of Family Friend. Refers to
dotted lines in the figure which are drawn solid.
Charades, Enigmas, and
Riddles. 1860: prob. 31, pp. 60 &
65; 1862: prob. 32, pp. 136 &
142; 1865: prob. 576, pp. 108 & 155. Dissect a
P-pentomino into three parts which make a square. Usual solution.
Peter
Parley, the Younger. Amusements of
Science. Peter Parley's Annual for
1866, pp. 139‑155.
Pp.
143-144: "To form two squares of unequal size into one square, equal to
both the original squares." Usual
method, with five pieces. On pp. 146-148,
he discusses the Theorem of Pythagoras and shows the dissection gives a proof
of it.
P.
144: "To make two smaller squares out of one larger." Cuts the larger square along both diagonals
and assembles the pieces into two squares.
Hanky Panky. 1872.
To form a square, pp. 116-117.
Very similar to Magician's Own Book.
Henry Perigal. Messenger of Mathematics 2 (1873) 104. ??NYS -- described in Loomis, op. cit.
above, pp. 104-105 & 214, where some earlier possible occurrences are
mentioned. He gives a dissection proof
of the theorem of Pythagoras using the shapes that occur in the quadrisection
of the square -- Section 6.AR. For
sides a < b, perpendicular cuts through the centre are
made in the square of side b so they meet the sides at distance (b-a)/2
from a corner. These pieces then
fit around the square of side a to make a square of side c.
I
invented a hinged version of this, in the 1980s?, which is described in: Greg
N. Frederickson; Hinged Dissections: Swinging & Twisting; CUP, 2002, pp.
33-34. I am shown demonstrating this on
Frederickson's website: www.cs.purdue.edu/homes/gnf/book2/Booknews2/singm.html
.
I
have seen the assembly of these four pieces and the square of edge a
into the square on the hypotenuse in a photo of the Tomb of Ezekiel in
the village of Al-Kifil, near Hillah, Iraq.
Mittenzwey. 1880.
Prob.
176, pp. 34 & 85; 1895?: 201, pp.
38 & 88; 1917: 201, pp. 35 &
84. Use the 10 pieces of 6.AS.1, as in
Les Amusemens, to make squares of edge 1 and edge 2.
Prob.
180, pp. 34 & 86; 1895?: 205, pp.
39 & 89; 1917: 205, pp. 35-36 &
85. Cut a 2 x 2 and a 4 x 4
into five pieces which make a square.
Both the problem and the solutions are inaccurately drawn. The smaller square has a 1, 2, Ö5 cut off, as for 6.AS.1.
The larger square has the same cut off at the lower left and a 2, 4, 2Ö5 cut
off at the lower right -- these two touch at the midpoint of the bottom edge --
leaving a quadrilateral with edges 4, 2Ö5, Ö5, 3 and two right angles. This is a variant of the standard five piece
method.
Alf. A. Langley. Letter:
Three-square puzzle. Knowledge 1
(9 Dec 1881) 116, item 97. Cuts two squares
into five pieces which form a single square.
Alexander J. Ellis. Letter:
The three-square puzzle. Knowledge
1 (23 Dec 1881) 166, item 146.
Usual dissection of two adjacent squares, considered as one piece, into
three parts by two cuts, which gives Langley's five pieces if the two squares
are divided. Suppose the two squares
are on a single piece of paper and are
ABCD and DEFG,
with E on side CD of the larger square ABCD.
He notes that if one folds the paper so that B and F
coincide, then the fold line meets the line ADG at the point H
such that the desired cuts are
BH and HF.
R. A. Proctor. Letter or editorial reply: Three square puzzle. Knowledge 1 (30 Dec 1881) 184, item
152. Says there have been many replies,
cites Todhunter's Euclid, p. 266 and notes the pieces can be obtained by
flipping the large square over and seeing how it cuts the two smaller ones.
R. A. Proctor. Our mathematical column: Notes on Euclid's first book. Knowledge 5 (2 May 1884) 318. "The following problem, forming a
well-known "puzzle" exhibits an interesting proof of the 47th
proposition." Gives the usual
three piece form, as in Ellis.
B. Brodie. Letter:
Superposition. Knowledge 5 (30
May 1884) 399, item 1273. Response to
the above, giving the five piece version, as in Langley.
Hoffmann. 1893.
Chap. III, no. 11: The two squares, pp. 93 & 125‑126
= Hoffmann-Hordern, pp. 82-83, with photo. Smaller square has half the edge. The squares are viewed as a single piece. Photo on p. 83 shows The Five Squares Puzzle
in paper with box, by Jaques & Son, 1870‑1895, and an ivory version,
with box, 1850-1900.
Loyd. Tit‑Bits 31 (3,
10 & 31 Oct 1896) 3, 25 & 75.
General two cut version.
Herr Meyer. Puzzles.
The Boy's Own Paper 19 (No. 937) (26 Dec 1896) 206 &
(No. 948) (13 Mar 1897) 383.
As in Hoffmann.
Benson. 1904.
The two‑square puzzle, pp. 192‑193.
Pearson. 1907.
Part II, no. 108: Still a square, pp. 108 & 182. Smaller square has half the edge.
Loyd. Cyclopedia. 1914. Pythagoras' classical problem, pp. 101 &
352. c= SLAHP, pp. 15‑16
& 88. The adjacent squares are
viewed as one piece of wood to be cut.
Uses two cuts, three pieces.
Williams. Home Entertainments. 1914.
Square puzzle, p. 118.
P-pentomino to be cut into three pieces to make a square. No solution given.
A. W. Siddons. Note 1020: Perigal's dissection for the
Theorem of Pythagoras. MG 16 (No. 217)
(Feb 1932) 44. Here he notes that the
two cutting lines of Perigal's 1873 dissection do not have to go through the
centre, but this gives dissections with more pieces. He shows examples with six and seven pieces. These cannot be hinged.
The Bile Beans Puzzle Book. 1933.
No.
37: No waste. Consider a square of side
2 extended by an isosceles triangle of hypotenuse 2. Convert to a square using two cuts.
No.
39: Square building. P-pentomino to
square in two cuts.
Slocum. Compendium.
Shows 4 - 5 piece square from Johnson Smith catalogue, 1935.
M. Adams. Puzzle Book. 1939. Prob. C.120: One
table from two, pp. 154 & 185. 3 x
3 and
4 x 4 tiled squares to
be made into a 5 x 5 but only cutting along the grid lines. Solves with each table cut into two
pieces. (I think there are earlier
examples of this -- I have just added this variant.)
Ripley's Puzzles and Games. 1966.
Pp. 58-59.
Item
5. Two joined adjacent squares to a
square, using two cuts and three pieces.
Item
6. Consider a 2 x 2 square with a Ö2, Ö2, 2
triangle attached to an edge.
Two cuts and three pieces to make a square.
6.AS.2.a. TWO EQUAL SQUARES TO A SQUARE
Further
subdivision of the pieces gives us 6.AS.4.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles, no. 13, pp. 25 & 85-86. Cut two equal squares each into two pieces
to make a square.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 172, p.
87. Cut one square into pieces to make
two equal squares. Cuts along the diagonals.
Mittenzwey. 1880.
Prob. 241, pp. 44 & 94;
1895?: 270, pp. 48 & 96;
1917: 270, pp. 44 & 92. As
is Leske.
6.AS.3. THREE EQUAL SQUARES TO A SQUARE
Crambrook. 1843.
P. 4, no. 21: One [Square to form] three [Squares] -- ??
"Student". Proposal [A pretty geometrical
problem]. Knowledge 1 (13 Jan 1882)
229, item 184. Dissect an L-tromino into a square. Says there are 25 solutions -- editor
says there are many more.
Editor. A pretty geometrical problem. Knowledge 1 (3 Mar 1882) 380. Says only the proposer has given a correct
solution, which cuts off one square, then cuts the remaining double square into
three parts, so the solution has four pieces.
Says there are several other ways with four pieces and infinitely many
with five pieces.
Hoffmann. 1893.
Chap. III, no. 23: The dissected square, pp. 101 & 134
= Hoffmann‑Hordern, pp. 96-97, with photo. Cuts three squares identically into three pieces to form one
square. Photo on p. 97 shows The
Dissected Square, with box, by Jaques & Son, 1870-1895. Hordern Collection, p. 63, shows Arabian
Puzzle, with box and some problem shapes to make, 1870‑1890.
Loyd. Problem 3: The three squares puzzle. Tit‑Bits 31 (17 Oct,
7 & 14 Nov 1896) 39, 97 & 112.
Quadrisect 3 x 1 rectangle to a square. Sphinx (i.e. Dudeney) notes it also can be
done with three pieces.
M. Adams. Indoor Games. 1912. The divided square,
p. 349 with figs. on pp. 346-347.
3 squares, 4 cuts, 7 pieces.
Loyd. Cyclopedia. 1914. Pp. 14 & 341. = SLAHP: Three in one, pp. 44 & 100. Viewed as a
3 x 1 rectangle, solution uses 2
cuts, 3 pieces. Viewed as 3 squares,
there are 3 cuts, 6 pieces.
Johannes Lehmann. Kurzweil durch Mathe. Urania Verlag, Leipzig, 1980. No. 6, pp. 61 & 160. Claims the problem is posed by Abu'l-Wefa,
late 10C, though other problems in this section are not strictly as posed by
the historic figures cited. Two of the
squares to be divided into 8 parts so all nine parts make a square. The solution has the general form of the
quadrisection of the square of side Ö2 folded around to surround a square of side 1
(as in Perigal's(?) dissection proof of the Theorem of Pythagoras), thus
forming the square of side Ö3. The four quadrisection pieces are cut into
two triangles of sides: 1, Ö3/2,
(1+Ö2)/2 and 1, Ö3/2, (Ö2-1)/2. Two of each shape assemble into a square of
side 1 which can be viewed as having a diagonal cut and then cuts from the
other corners to the diagonal, cutting off
(Ö2-1)/2 on the
diagonal.
6.AS.3.a. THREE EQUAL 'SQUARES' TO A HEXAGON
Catel. Kunst-Cabinet. Vol. 2,
1793. Das Parallelogramm, pp. 14-15
& fig. 249 a,b,c,d on plate XII.
This shows three squares, each dissected the same way into 4 pieces
which will make a hexagon or two equal equilateral triangles. Consider a hexagon and connect three
alternate vertices to the centre. Join
up the same vertices and drop perpendiculars from the centre to three of the
sides of the hexagon. However, close
examination shows that the squares have dimensions 3/2 by Ö3. The figure of the three adjacent squares has the divisions
between them hard to make out.
Bestelmeier. 1801.
Item 292/293 -- Das Parallelogram.
Almost identical to Catel, except the diagrams are reversed, and worse,
several of the lines are missing.
Mathematical part of the text is identical.
6.AS.4. EIGHT EQUAL SQUARES TO A SQUARE
Divide
four of the squares in half diagonally.
Magician's Own Book. 1857.
Prob. 8: The accommodating square, pp. 269 & 293. c= Landells, Boy's Own Toy-Maker, 1858,
p. 144. = Book of 500 Puzzles, 1859,
prob. 8, pp. 83 & 107. = Boy's
Own Conjuring Book, 1860, prob. 7, pp. 230 & 256. = Illustrated Boy's Own Treasury, 1860, no. 24, pp. 399
& 439.
Hanky Panky. 1872.
[Another square] of four squares and eight triangles, p. 120.
Cassell's. 1881.
Pp. 92-93: The accommodating square.
= Manson, 1911, p. 131.
Handy Book for Boys and
Girls. Op. cit. in 6.F.3. 1892.
Pp. 321-322: Square puzzle.
Hoffmann. 1893.
Chap. III, no. 20: Eight squares in one, pp. 100 & 132
= Hoffmann‑Hordern, p. 94.
Wehman. New Book of 200 Puzzles. 1908.
The accommodating square, p. 13.
c= Magician's Own Book.
6.AS.5. RECTANGLE TO A SQUARE OR OTHER RECTANGLE
New
section. See comment at 6.AS. The dissection of a 5 x 1
into five pieces which make a square is explicitly covered in 6.AS.1,
and the other cases above can be viewed as dissections of 2 x 1,
3 x 1 and 8 x 1.
There must be older examples of the general case??
Ozanam‑Montucla. 1778.
Avec
cinq quarrés égaux, en former un seul.
Prob. 18 & fig. 123, plate 15, 1778: 297; 1803: 292-293; 1814:
249-250; 1840: 127. 9 pieces.
Remarks that any number of squares can be made into a square.
Prob.
19 & fig. 124-126, plate 15 & 16, 1778: 297-301; 1803: 293-296; 1814: 250‑253;
1840: 127-129. Dissect a
rectangle to a square.
Prob.
20 & fig. 125-126, plate 15 & 16, 1778: 301-302; 1803: 297;
1814: 253; 1840: 129. Dissect a square into 4, 5, 6, etc. parts which form a rectangle.
"Mogul". Proposal [A pretty geometrical
problem]. Knowledge 1 (13 Jan 1882)
229, item 184. Dissect a rectangle into
a square. Editor's comment in (3 Mar
1882) 380 says only the proposer has given a correct solution but it will be held
over.
"Mogul". Mogul's Problem. Knowledge 1 (31 Mar 1882) 483.
Gives a general construction, noting that if the ratio of length to
width is £ 2, then it takes two cuts;
if the ratio is in the interval
(2, 5], it takes three cuts; if the ratio is in (5, 10], it takes four
cuts; if the ratio is in (10, 17], it takes
five cuts. In general if the ratio is
in (n2+1, (n+1)2+1], it takes
n+2 cuts.
Richard A. Proctor. Our puzzles; Knowledge 10 (Nov 1886)
9 & (Dec 1886) 39-40
& Solution of puzzles; Knowledge 10 (Jan 1887) 60-61. "Puzzle XII. Given a rectangular carpet of any shape and size to divide it
with the fewest possible cuts so as to fit a rectangular floor of equal size
but of any shape." He says this
was previously given and solved by "Mogul". Solution notes that this is not the problem posed by
"Mogul" and that the shape of the second rectangle is assumed as
given. He distinguishes between the
cases where the actual second rectangular area is given and where only its
shape is given. Gives some solutions,
remarking that more cuts may be needed if either rectangle is very long. Poses similar problems for a parallelogram.
Tom Tit, vol. 3. 1893.
Rectangle changé en carré. en deux coups de ciseaux, pp. 175-176. = K, no. 24: By two cuts to change a
rectangle into a square, pp. 64-65.
Consider a square ABCD of side one. If you draw AA' at angle
α to AB
and then drop BE perpendicular to AA', the resulting three pieces make a rectangle of size sin α
by csc α, where
α must be ³ 450, so the rectangle cannot be more than twice
as long as it is wide. If one starts
with such a rectangle ABCD, where
AB is the length, then one
draws AA' so that DA' is the geometric mean of AB
and AB - AD. Dropping
CE perpendicular to AA'
gives the second cut.
Dudeney. Perplexities column, no. 109: A cutting-out
puzzle. Strand Magazine 45 (No. 265) (Jan 1913) 113 &
(No. 266) (Feb 1913) 238. c= AM,
prob. 153 -- A cutting-out puzzle, pp. 37 & 172. Cut a 5 x 1 into four pieces to make a square. AM states the generalized form: if length/breadth is in
[(n+1)2, n2),
then it can be done with
n+2 pieces, of which n-1
are rectangles of the same breadth but having the desired length. The cases
1 x (n+1)2 are exceptional in that one of pieces vanishes, so
only n+1 pieces are needed. He
doesn't describe this fully and I think one can change the interval above
to ((n+1)2, n2].
Anonymous. Two dissection problems, no. 1. Eureka
13 (Oct 1950) 6 & 14 (Oct 1951) 23. An n-step is formed by n lines of unit squares
of lengths 1, 2, ..., n, with all lines aligned at one end. Hence a
1-step is a unit square, a 2-step is an L-tromino and an n‑step is what is left when an (n-1)-step is removed from a corner of
an n x n square. Show any n-step
can be cut into four pieces to make a square, with three pieces in one
case. Cut parallel to a long side at
distance (n+1)/2 from it.
The small piece can be rotated
180o about a corner to make an n x (n+1)/2
rectangle. Dudeney's method cuts
this into three pieces which make a square, and the cuts do not cut the small
part, so we can do this with a total of four pieces. When n = 8, the rectangles is 8 x 9/2, which is similar
to 16 x 9 which can be cut into two pieces by a staircase cut, so the
problem can be done with a total of three pieces. A little calculation shows this is the only case where n x (n+1)/2
is similar to k2 x
(k-1)2.
Lorraine Mottershead. Investigations in Mathematics. Blackwell, Oxford, 1985. P. 105.
Dissect a 2 x 5 rectangle into four pieces that make a
square.
6.AT. POLYHEDRA AND TESSELLATIONS
These
have been extensively studied, so I give only the major works. See 6.AA for nets of polyhedra.
Gwen White. Antique Toys and Their Background. Batsford, 1971. (Reprinted by Chancellor Press, London, nd [c1989].) P. 9 has a sketch of "Ball of stone,
Scotland", which seems to be tetrahedral and she says: "... one of the
earliest toys known is a stone ball.
Perhaps it is not a plaything, no one knows why it was made, but it is a
convenient size to hold in the hand."
Dorothy N. Marshall. Carved stone balls. Proc. Soc. of Antiquaries of Scotland 108
(1976-7) 40-72. Survey of the Scottish
neolithic carved stone balls. Lists 387
examples in 36 museums and private collections, mostly of 70mm diameter and
mostly from eastern Scotland.
Unfortunately Marshall is not interested in the geometry and doesn't
clearly describe the patterns -- she describes balls with 3, 4, 5, 6, 7, 8, 9, 10 - 55 and
70 - 160 knobs, but emphasises the decorative styles. From the figures, there are clearly
tetrahedral, cubical, dodecahedral(?) and cubo-octahedral shapes. Many are in the National Museum of Antiquities
of Scotland (= Royal Museum, see below), but the catalogue uses a number of
unexplained abbreviations of collections.
Royal Museum of Scotland, Queen
Street, Edinburgh, has several dozen balls on display, showing cubical,
tetrahedral, octahedral and dodecahedral symmetry, and one in the form of the
dual of the pentagonal prism. [This
museum has now moved to a new building beside its other site in Chambers Street
and has been renamed the Museum of Scotland.
When I visited in 1999, I was dismayed to find that only three of the
carved stone balls were on display, in a dimly lit case and some distance
behind the glass so that it was difficult to see them. Admittedly, the most famous example, the
tetrahedral example with elaborate celtic decorative spirals, NMA AS10 from
Glasshill, Towie, Aberdeenshire, is on display -- photo in [Jenni Calder; Museum
of Scotland; NMS Publishing, 1998, p. 21].
They are on Level 0 in the section called In Touch with the Gods.]
Ashmolean Museum, Oxford, has
six balls on display in case 13a of the John Evans Room. One is tetrahedral, three are cubical, one
is dodecahedral and one is unclear.
Keith Critchlow. Time Stands Still -- New Light on Megalithic
Science. Gordon Fraser, London,
1979. Chap. 7: Platonic spheres -- a
millennium before Plato, pp. 131‑149.
He discusses and depicts Neolithic Scottish stones carved into rounded
polyhedral shapes. All the regular
polyhedra and the cubo‑octahedron occur.
He is a bit vague on locations -- a map shows about 50 discovery sites
and he indicates that some of these stones are in the Ashmolean Museum, Dundee
City Museum and 'in Edinburgh'.
Likewise, the dating is not clear -- he only says 'Neolithic' -- and
there seem to be no references.
D. V. Clark. Symbols of Power at the Time of
Stonehenge. National Museum of
Antiquities, Edinburgh, 1985. Pp. 56-62
& 171. ??NYS -- cited by the
Christie's Catalogue, below.
Robert Dixon. Mathographics. Blackwell, 1987, fig. 5.1B, p. 130, is a good photo of the Towie
example.
Anna Ritchie. Scotland BC. HMSO, Edinburgh, for Scottish Development Department -- Historic
Buildings and Monuments, 1988.
P. 8
has a colour photo of a neolithic cubical ball from the Dark Age fort of
Dunadd, Argyll.
P. 14
has a colour photo of a cubical and a tetrahedral ball from Skara Brae, Orkney
Islands, c-2800.
Simant Bostock of Glastonbury
has made a facsimile of the Towie example, casts of which are available from
Glastonbury Film Office, 3 Market Place, Glastonbury, Somerset, BA6 9HD;
tel: 01458-830228. You can also
contact him at 24 Northload Street, Glastonbury, Somerset, BA6 9JJ; tel:
01458-833267 and he has a mail order catalogue. Since he worked from photographs, there are some slight
differences from the original, and the facsimile is slightly larger.
Three examples of tetrahedral
stone balls were in Christie's South Kensington antiquities sale of 12 Apr
2000, lots 124 and 125 (2 balls), p. 62, with colour photo of item 124 and the
better example in lot 124 on p. 63.
(Thanks to Christine Insley Green for a copy of the catalogue.) The descriptive text says 'their exact use
is unclear'. Cites Clark, above.
The British Museum has
icosahedral dice from Egypt, dated -700/-500.
Moritz Cantor. Vorlesungen über Geschichte der
Mathematik. Vol. I, 4th ed., 1906,
pp. 174‑176. He feels all
the regular solids were known to Pythagoras, with the tetrahedron, cube and
octahedron having been known long before.
Says to see various notices by Count Leopold Hugo in CR 77 for a bronze
dodecahedron, a work by Conze on a Celtic bronze example and the paper of
Lindemann, below, for a north Italian example.
However, he says the dates of these are not determined and I think these
are now all dated to later Roman times -- see below. He also notes that moderately regular dodecahedra and icosahedra
occur in mineral deposits on Elba and in the Alps and wonders if Pythagoras
could have known of these.
Thomas L. Heath. Note about Scholium 1 of Book XIII of
Euclid. The Thirteen Books of Euclid's
Elements; trans. & ed. by Thomas L. Heath; (1908?); 2nd ed., (1926); Dover,
3 vols., 1956, vol. 3, p. 438.
"And it appears that dodecahedra have been found, of bronze or
other material, which may belong to periods earlier than Pythagoras' time by
some centuries (for references see Cantor's Geschichte der Mathematik I3,
pp. 175-6)."
HGM I 160 cites Hugo and
Lindemann, dating the Monte Loffa example as -1000/-500.
HGM I 162 discusses the
Scholium, giving it as: "the five
so-called Platonic figures, which, however, do not belong to Plato,
three of the five being due to the Pythagoreans, namely the cube, the pyramid,
and the dodecahedron, while the octahedron and icosahedron are due to
Theaetetus". He cites Heiberg's
Euclid, vol. v., p. 654.
Thomas, SIHGM I 223 says "A
number of objects of dodecahedral form have survived from pre-Pythagorean
days." But he gives no details or
references. Cf Heath's note to Euclid,
above.
Plato (‑427/‑347). Timaeus.
c-350. Page references are based
on the 1578 edition of Plato which has been used for all later references: pp.
54-56. I use the version in: Edith
Hamilton and Huntington Cairns, eds; The Collected Dialogues of Plato including
the Letters; Bollingen Series LXXI, Pantheon Books (Random House), (1961),
corrected 2nd ptg, 1963, which is the translation of Benjamin Jowett in his The
Dialogues of Plato, OUP, (1872), 4th revised ed, 1953. Discusses the regular polyhedra, describing
the construction of the tetrahedron, octahedron, icosahedron and cube from
triangles since he views the equilateral triangle as made from six 30-60-90
triangles and the square made from eight 45-45-90 triangles. "There was yet a fifth combination
which God used in the delineation of the universe with figures of
animals." He then relates the
first four to the elements: tetrahedron
-- fire; octahedron -- air; icosahedron -- water; cube -- earth. [These associations are believed to derive from the
Pythagoreans.] However, these
associations contribute very little to the rest of the dialogue. The incidental appearance of the
dodecahedron lends support to the belief that it was discovered or became known
after the initial relation between regular polyhedra and the elements had been
established and had to be added in some ad hoc manner. [It is believed that the later Pythagoreans related
it to the universe as a whole.]
Scholium 1 of Book XIII of
Euclid. Discussed in: The Thirteen
Books of Euclid's Elements; op. cit. above, vol. 3, p. 438. Heath's discussion of the Scholia in vol. 1,
pp. 64-74, indicates this may be c600.
The Scholium asserts that only the tetrahedron, cube and dodecahedron
were known to the Pythagoreans and that the other two were due to
Theaetetus. Heath thinks the
Pythagoreans had all five solids (cf his note to IV.10, vol. 2, pp. 97-100) and
the Scholium is taken from Geminus, who may have been influenced by the fact
that Theaetetus was the first to write about all five solids and hence the
first to write much about the latter two polyhedra.
Stefano de'Stefani. Intorno un dodecaedro quasi regolare di
pietra a facce pentagonali scolpite con cifre scoperto nelle antichissime
capanne di pietra del Monte Loffa. Atti
del Reale Istituto Veneto di Scienze e Lettere, (Ser. 6) 4 (1885) 1437-1459 +
plate 18. Separately reprinted by G.
Antonelli, Venezia, 1886, which has pp. 1-25 and Tavola 18. This describes perhaps the oldest known
reasonably regular dodecahedron, in the Museo Civico di Storia Naturale,
Palazzo Pompei, Largo Porta Vittoria 9, Verona, Italy, in the central case of
Sala XIX. This is discussed by Herz-Fischler
[op. cit. below, p. 61], I have been to see it and the Director, Dr.
Alessandra Aspes, has kindly sent me a slide and a photocopy of this article.
The
dodecahedron was discovered in 1886 at Monte Loffa, NE of Verona, and has been
dated as far back as -10C, but is currently considered to be -3C or -2C [Herz‑Fischler,
p. 61]. Dr. Aspes said the site was
inhabited by tribes who had retreated into the mountains when the Romans came
to the area, c-3C. These tribes were
friendly with the Romans and were assimilated over a few centuries, so it is
not possible to know if this object belongs to the pre-Roman culture or was due
to Roman influence. She dates it as ‑4C/‑1C. The stone apparently was cut with a bronze
saw and these existed before the Roman incursion (stated in Lindemann,
below). It is clearly not perfectly
regular -- some of the face angles appear to be 90o and some edges
are clearly much shorter than others.
But it also seems clear that it is an attempt at a regular dodecahedron
-- the faces are quite flat. Its faces
are marked with holes and lines, but their meaning and the function of the
object are unknown. de'Stefani
conjectures it is a kind of die.
Lindemann notes that the symbols are not Etruscan nor Greek, but
eventually gets to an interpretation of them, which seems not too fanciful,
using the values: 3, 6,
9, 10, 12, 15, 16,
20, 21, 24,
60, 300. (Are there any ancient Greek models of the regular polyhedra?) But see also the carved stone balls
above.
About 90 examples of a Roman
dodecahedron have been found at Roman sites, north of the Alps, from Britain to
the Balkans, dating about 200-400.
These are bronze and hollow, but also each face has a hole in it, almost
always circular, and each corner has a knob at it, making it look like it could
be used for Hamilton's Icosian game!
The shape is quite precise.
The Society of Antiquaries,
London, has the largest extant specimen, dug up on the north side of the Church
of St. Mary, Carmarthen, in 1768 and presented to the Society about 1780. The edge length is 2 1/5 in (= 56mm) and,
unusually, has plain faces -- almost all examples have some incised decoration
on the faces. [Rupert Bruce-Mitford;
The Society of Antiquaries of London
Notes on Its History and Possessions; The Society, 1951, p. 75, with
photograph as pl. XXIV (b) on p. 74.
The Gallo-Romeins Museum
(Kielenstraat 15, B-3700 Tongeren, Belgium; Tel: 12-233914) has an example
which is the subject of an exhibition and they have produced facsimiles for
sale. "The precise significance and
exact use of this object have never been explained and remain a great
mystery." Luc de Smet says the
bronze facsimile is slightly smaller than the original and that the Museum also
sells a tin and a bronzed version of the original size.
Other examples are in the Newcastle
University Museum (only about half present) and the Hunt Museum, Limerick.
C. W. Ceram. Gods, Graves and Scholars. Knopf, New York, 1956, pp. 26-29. 2nd ed., Gollancz, London, 1971, pp. 24-25. In the first edition, he illustrated this as
an example of the mysterious objects which archaeologists turn up and said that
it had been described as a toy, a die, a model for teaching measurement of
cylinders, a candleholder. His picture
shows one opening as being like a key-hole.
In the second edition, he added that he had over a hundred suggestions
as to what it was for and thinks the most probable answer is that it was a
musical instrument.
Jacques Haubrich has recently
sent an example of a hollow cubical stone object with different size holes in
the faces, apparently currently made in India, sold as a candleholder.
See Thomas, SIHGM I 216‑225,
for brief references by Philolaus, Aëtius, Plato, Iamblichus.
Euclid. Elements.
c-300. Book 13, props. 13‑18
and following text. (The Thirteen Books
of Euclid's Elements, edited by Sir Thomas L. Heath. 2nd ed., (CUP, 1925??),
Dover, vol. 3, pp. 467‑511.)
Constructs the 5 regular polyhedra in a sphere, compares them. In Prop. 18, he continues "I say
next that no other figure, besides the said five figures, can be constructed
which is contained by equilateral and equiangular figures equal to one
another."
Leonardo Pisano, called
Fibonacci (c1170->1240). La Practica
Geometriae. 1221. As:
La Practica Geometriae di
Leonardo Pisano secondo la
lezione del Codice Urbinate no. 202 della Biblioteca Vaticana. In: Scritti di Leonardo Pisano; vol. II, ed.
and pub. by B. Boncompagni; Tipografia delle Scienze Matematiche e
Fisiche, Rome, 1862, pp. 1‑224.
On p. 159, he says there are many polyhedra and mentions there are ones
with 8, 12 and 20 faces which Euclid constructs in a sphere in his book
XIIII. On pp. 161-162, he
describes division in mean and extreme ratio and the construction of the
regular pentagon in a circle, then says you can construct, in a sphere, a solid
with 20 equilateral triangular faces or with 12 pentagonal faces. After some discussion, he says you can also
construct solids with 4, 6, 8, 12, 20 faces, in a sphere. Division in mean and extreme ratio and the
construction of the icosahedron are later covered in detail on pp.
196-202. His only drawings of solids
are of cubes and pyramids.
Drawings of all the regular
polyhedra are included in works, cited in 6.AA, 6.AT.2 and 6.AT.3, by della
Francesca (c1480 & c1487), Pacioli (1494), Pacioli & da Vinci (1498),
Dürer (1525), Jamnitzer (1568), and Kepler (1619).
F. Lindemann. Zur Geschichte der Polyeder und der
Zahlzeichen. Sitzungsber. der math.‑phys.
Classe k. b. Akademie der Wissenschaften zu München 26 (1896) 625‑758
& plates I-IX. Discusses and
illustrates many ancient polyhedra.
Unfortunately, most of these are undated and/or without provenance. He generally dates them as -7C/5C.
A
bronze rhombic triacontahedron, which he dates as first half of the first
millennium AD.
Roman
knobbed dodecahedra, which he describes as Celtic, going back to the La Tène
period (Bronze Age) -- these are now dated to late Roman times. He lists 26 examples listed from the works
of Conze and Hugo (cf Cantor, above).
A
dodecahedral die; an irregular
rhombi-cubo-octahedral die; a bronze
dodecahedral die (having two 1s, three 2s, two 3s, one 4 and four 5s).
The
Verona dodecahedron (from de'Stefani), which he dates as -1000/-500.
An
enamelled icosahedron in Turin with Greek letters on the faces.
An
octagonal bipyramid (elongated) from Meclo, South Tyrol, marked with a form of
Roman numerals in a somewhat irregular order.
It is dated to before the Barbarian migrations.
Three
bronze cubo-octahedra.
He
then does a long analysis of north-Italian culture and its relations to other
cultures and of their number symbols, eventually obtaining an interpretation of
the symbols on the Monte Loffa dodecahedron, which he then justifies further
with Pythagorean number relations.
Roger Herz-Fischler. A Mathematical History of Division in
Extreme and Mean Ratio. Wilfrid Laurier
University Press, Waterloo, Ontario, 1987.
Corrected and extended as: A Mathematical History of the Golden Number;
Dover, 1998. P. 61 discusses the history
of the dodecahedron and refers to the best articles on the history of
polyhedra. Discusses the Verona
dodecahedron, see above.
Judith V. Field. Kepler's Geometrical Cosmology. Athlone Press, London, 1988. This gives a good survey of the work of Kepler
and his predecessors. In particular,
Appendix 4: Kepler and the rhombic solids, pp. 201-219, is most
informative. Kepler described most of
his ideas several times and this book describes all of them and the
relationships among the various versions.
The regular polyhedra in four
dimensions were described by Ludwig Schläfli, c1850, but this was not
recognised and in the 1880s, several authors rediscovered them.
H. S. M. Coxeter. Regular skew polyhedra. Proc. London Math. Soc. (2) 43 (1937)
33-62. ??NYS -- cited and discussed by
Gott, qv.
J. R. Gott III. Pseudopolyhedrons. AMM 74:5 (May 1967) 497-504.
Regular polyhedra have their sum of face angles at a vertex being less
than 360o and approximate to surfaces of constant positive
curvature, while tessellations, with angle sum equal to 360o,
correspond to surfaces of zero curvature.
The pseudopolyhedra have angle sum greater than 360o and
approximate to surfaces of negative curvature.
There are seven regular pseudopolyhedra. Each is a periodic structure.
He subsequently discovered that J. F. Petrie and Coxeter had discovered
three of these in 1926 and had shown that they were the only examples
satisfying an additional condition that arrangement of polygons at any vertex
have rotational symmetry, and hence that the dihedral angles between adjacent
faces are all equal. Coxeter later
refers to these structures as regular honeycombs. Some of Gott's examples have some dihedral angles of 180o. Two of these consist of two planes, with a
regular replacement of pieces in the planes by pieces joining the two
planes. The other five examples go to
infinity in all directions and divide space into two congruent parts. He makes some remarks about extending this
to general and Archimedean pseudopolyhedra.
6.AT.2 STAR AND STELLATED POLYHEDRA
I have heard it stated that
Kircher was the first to draw star polygons.
Paolo Uccello (1397‑1475). Mosaic square on the floor at the door of
San Pietro in San Marco, Venice. 1425‑1430. (This doorway is not labelled on the maps
that I have seen -- it is the inner doorway corresponding to the outer doorway
second from the left, i.e. between Porta di Sant'Alipio and Porta di San
Clemente, which are often labelled.)
This seems to show the small stellated dodecahedron {5/2, 5}. This mosaic has only recently (1955 & 1957) been attributed
to Uccello, so it can only be found in more recent books on him. See, e.g., Ennio Flaiano & Lucia
Tongiorgi Tomase; L'Opera Completa di Paolo Uccello; Rizzoli, Milan, 1971 (and
several translations). The mosaic is
item 5.A: Rombo con elementi geometrici
in the Catalogo delle Opere, with description and a small B&W picture on p.
85. [Bokowski & Wills, below, give
the date 1420.]
Coxeter
[Elem. der Math. 44 (1989) 25‑36] says it "is evidently intended to
be a picture of this star polyhedron."
However,
J. V. Field tells me that the shape is not truly the small stellated
dodecahedron, but just a 'spiky' dodecahedron.
She has examined the mosaic and the 'lines' of the pentagrams are not straight. [The above cited photo is too small to
confirm this.] She says it appears to
be a direct copy of a drawing in Daniele Barbaro; La Practica della
Perspettiva; Venice, 1568, 1569, see below, and is most unlikely to be by
Uccello. See Field, Appendix 4, for a
discussion of early stellations.
In
1998, I examined the mosaic and my photos of it and decided that the 'lines'
are pretty straight, to the degree of error that a mason could work, and some
are dead straight, so I agree with Coxeter that it is intended to be the small
stellated dodecahedron. I now have a
postcard of this. However, I have
recently seen a poster of a different mosaic of the same shape which is distinctly
irregular, so the different opinions may be based on seeing different mosaics!
Both
mosaics are viewed directly onto a pentagonal pyramid, but the pyramids are
distinctly too short in the poster version.
The only spiky dodecahedron in Barbaro is on p. 111, fig. 52, and
this is viewed looking at a common edge of two of the pyramids and the pyramids
are distinctly too tall, so this is unlikely to be the source of the
mosaics. The 'elevated dodecahedron' in
Pacioli & da Vinci, plate XXXI, f. CVI-v, has short pyramids and looks quite
like the second mosaic, but it is viewed slightly at an angle so the image does
not have rotational symmetry. If anyone
is in Venice, perhaps they could check whether there are two (or more?) mosaics
and get pictures of them.
Luca Pacioli & Leonardo da
Vinci. Untitled MS of 1498,
beginning: Tavola dela presente opera e
utilissimo compendio detto dela divina proportione dele mathematici discipline
e lecto -- generally called De divina proportione. Ill. by Leonardo da Vinci. See the entry in 6.AT.3 for fuller details
of the facsimiles and details about which plates are in which of the editions.
Discussed
by Mackinnon (see in 6.AT.3 below) and Field, pp. 214-215. Clearly shows the stella octangula in one of
the superb illustrations of Leonardo, described as a raised or elevated
octahedron (plates XVIIII & XX).
Field, p. 214, gives the illustration. None of the other raised shapes is a star, but the raised
icosahedron is close to a star shape.
Barbaro, Daniele
(1514-1570). La Practica della
Perspettiva. Camillo & Rutilio Borgominieri,
Venice, 1568, 2nd ptg, 1569. (Facsimile
from a 1569 copy, Arnaldo Forni, Milan, 1980.
The facsimile's TP doesn't have the publication details, but they are
given in the colophon. Various
catalogues say there are several versions with dates on the TP and colophon
varying independently between 1568 and 1569.
A version has both dates being 1568, so this is presumed to be the first
appearance. Another version has an
undated title in an elaborate border and this facsimile must be from that version.) P. 111 has a dodecahedron with pyramids
on each face, close to, but clearly not the stellated dodecahedron. P. 112 has an icosahedron with pyramids on
each face, again close to, but clearly not the stellated icosahedron. I would have expected a reasonably accurate
drawing, but in both drawings, several of the triples of segments which should
lie on a single straight line clearly do not.
P. 113 shows an icosi-dodecahedron with pyramids on the triangular faces. If the pyramids extended the edges of the
pentagons, this would produce the dodecahedron! But here the pyramids distinctly point much further out and the
overall perspective seems wrong.
[Honeyman, no. 207, observing that some blocks come from the 1566
edition of Serlio which was dedicated to Barbaro.]
Wentzel Jamnitzer (or
Jamitzer). Perspectiva Corporum
Regularum. With 50 copper plates by
Jost Amman. (Nürnberg, 1568.) Facsimile by Akademische Druck- u.
Verlagsanstalt, Graz, 1973. [Facsimiles
or reprints have also been issued by Alain Brieux, Paris, 1964 and Verlag
Biermann und Boukes, Frankfurt, 1972.]
This
includes 164 drawings of polyhedra in various elaborations, ranging from the 5
regular solids through various stellations and truncations, various skeletal
versions, pseudo-spherical shapes and even rings. Some polyhedra are shown in different views on different
pages. Nameable objects, sometimes part
of larger drawings, include:
tetrahedron, cubo-octahedron, truncated tetrahedron, stella octangula,
octahedron, cube, truncated octahedron, rhombi-cuboctahedron, compound of a
cube and an octahedron (not quite correct), great rhombi-cuboctahedron,
icosahedron, great dodecahedron, dodecahedron, icosi-dodecahedron,
rhombi-icosi-dodecahedron, truncated cube, and skeletal versions of: stella octangula, octahedron, cube,
icosahedron, dodecahedron, icosi-dodecahedron.
There are probably some uniform polyhedra, but I haven't tried to
identify them, and some of the truncated and stellated objects might be
nameable with some effort.
J. Kepler. Letter to Herwart von Hohenberg. 6 Aug 1599.
In: Johannes Kepler Gesammelte
Werke, ed. by M. Caspar, Beck, Munich, 1938.
Vol 14, p. 21, letter 130, line 457.
??NYS. Cited by Field, op. cit.
below. Refers to (small??) stellated
dodecahedron.
J. Kepler. Letter to Maestlin (= Mästlin). 29 Aug 1599. Ibid. Vol. 14,
p. 43, letter 132, lines 142-145.
??NYS. Cited by Field, below,
and in [Kepler's Geometrical Cosmology; Athlone Press, London, 1988, p.
202]. Refers to (small??) stellated
dodecahedron.
J. Kepler. Harmonices Mundi. Godfrey Tampach, Linz, Austria, 1619; facsimile: (Editions)
Culture et Civilization, Brussels, 1968 (but my copy is missing three
plates!) [Editions probably should have
É, but my only text which uses the word Editions is a leaflet in English.] = Joannis Kepleri Astronomi Opera Omnia; ed.
Ch. Frisch, Heyder & Zimmer, Frankfurt & Erlangen, 1864, vol. 5. = Johannes Kepler Gesammelte Werke; ed. by
M. Caspar, Beck, Munich, 1938, vol. 6, ??NYS. Book II. Translated by J. V. Field;
Kepler's star polyhedra; Vistas in Astronomy 23 (1979) 109‑141.
Prop.
XXVI, p. 60 & figs. Ss & Tt on p. 53.
Describes both stellated dodecahedra,
{5/2, 5} and {5/2, 3}.
This is often cited as the source of the stella octangula, but the translation
is referring to an 'eared cube' with six octagram faces and the stella
octangula is clearly shown by Pacioli & da Vinci and by Jamnitzer.
Louis Poinsot. Mémoire sur les polygones et les
polyèdres. J. de L'École Polytechnique
4 (1810) 16‑48 & plate opp. p. 48.
Art. 33‑40, pp. 39‑42, describe all the regular star
polyhedra. He doesn't mention Kepler
here, but does a few pages later when discussing Archimedean polyhedra.
A. L. Cauchy. Recherches sur les polyèdres. J. de L'École Polytechnique 16 (1813) 68‑86. ??NYS.
Shows there are no more regular star polyhedra and this also shows there
are no more stellations of the dodecahedron.
H. S. M. Coxeter, P. Du Val, H.
T. Flather & J. F. Petrie. The
Fifty-Nine Icosahedra. Univ. of Toronto
Press, 1938; with new Preface by Du
Val, Springer, 1982. Shows that there
are just 59 stellations of the icosahedron.
They cite earlier workers: M.
Brückner (1900) found 12; A. H. Wheeler
(1924) found 22.
Dorman Luke. Stellations of the rhombic dodecahedron. MG 41 (No. 337) (Oct 1957) 189‑194. With a note by H. M. Cundy which says that
the first stellation is well known (see 6.W.4) and that the second and third
are in Brückner's Vielecke und Vielfläche, but that the new combinations shown
here complete the stellations in the sense of Coxeter et al.
J. D. Ede. Rhombic triacontahedra. MG 42 (No. 340) (May 1958) 98‑100. Discusses Coxeter et al. and says the main
process generates 8 solids for the icosahedron. He finds that the main process gives 13 for the rhombic
triacontahedron, but makes no attempt to find the analogues of Coxeter et al.'s
59.
Archimedes discovered the
Archimedean solids, -3C, but his work is lost.
Heron quotes some of it and Pappus summarises it. See HGM II 98-101.
Hero of Alexandria (c150). Definitiones. IN: Heronis Alexandrini
Opera quae supersunt omnia; Vol.
IV, Heronis Definitiones Cum Variis Collectionibus Heronis Quae Feruntur Geometrica, ed. by J. L. Heiberg, Teubner,
1912, pp. 64-67. Heath, HGM I 294-295
has a translation, but it doesn't give the complete text which seems open to
two interpretations. The German
goes: Archimedes aber sagt, es gebe in
ganzen dreizehn Körper, die in einer Kugel eingeschreiben werden können, indem
er ausser den genannten fünf noch acht hinzufügt; von diesen habe auch Platon
das Tessareskaidekaeder gekannt, dies aber sei ein zweifaches, das eine aus
acht Dreiecken und sechs Quadraten zusammengesetzt, aus Erde und Luft, welches
auch einige von den Alten gekannt hätten, das andere umgekehrt aus acht
Quadraten und sechs Dreiecken, welches schwieriger zu sein scheint. My translation: But Archimedes said, there are in total 13 bodies, which can be
inscribed in a sphere, as he added eight beyond the named five [regular solids,
which he had just defined]; but of these Plato knew the 14-hedron, however this
is a double, one is composed of eight triangles and six squares, from Earth and
Air, which some of the ancient also knew, the other conversely [is composed] of
eight squares and six triangles, which seems to be more difficult.
Note
that Hero has got the numbers wrong - Archimedes found 13 more than the 5
regular solids. Secondly, the 'more
difficult' solid does not exist! Heath
notes this and suggests that either the truncated cube or the truncated
octahedron was intended. The question
of interpretation arises at the first semicolon -- is this continuing the
statement of Archimedes or is Hero commenting on Archimedes' results? Heath seems to say Archimedes is making the
attribution to Plato, but see below.
MacKinnon, below, seems to be say this is being made by Hero. Heath's discussion on HGM II 100 says
"We have seen that, according to Heron, two of the semi-regular [i.e.
Archimedean] solids had already been discovered by Plato" undoubtedly
using the method of truncation.
However, I don't see that Heron is saying that Plato discovered the
cubo-octahedron and the other solid, only that he knew it. Mackinnon says "Plato is said by Heron
to have discovered the cuboctahedron by making a model of it from a
net." But I don't see that Heron
says this.
Pappus. Collection.
c290. Vol. 19. In:
SIHGM II 194‑199.
Describes the 13 Archimedean solids.
"..., but also the solids, thirteen in number, which were
discovered by Archimedes and are contained by equilateral and equiangular, but
not similar, polygons." He then
describes each one. Pappus' work has
survived in a single MS (Vat. gr. 218) of the 10C in the Vatican and was
not copied until 1550, but see Mackinnon, pp. 175-177, on whether it had been
seen by Piero. For the history of this
MS, see also: Noel M. Swerdlow; The recovery of the exact sciences of
antiquity: mathematics, astronomy, geography; IN: Anthony Grafton, ed.; Rome Reborn
The Vatican Library and Renaissance Culture Catalog of an exhibition at the Library of Congress, Washington,
D.C., Jan. 6 ‑ Apr. 30, 1993; Library of Congress, Washington &
Yale University Press, New Haven & London; in association with the
Biblioteca Apostolica Vaticana; 1993; pp. 137-139. [This exhibition is on-line at
www.ibiblio.org/expo/vatican.exhibit/vatican.exhibit.html.]
R. Ripley. Believe It Or Not. 18th series, Pocket Books, NY, 1971. P. 116 asserts the Romans used dice in the shape of cubo‑octahedra.
The British Museum, Room 72,
Case 9, has two Roman cubo-octahedral dice on display.
F. Lindemann, op. cit. in
6.AT.1, 1896, describes and illustrates an antique rhombic triacontahedron,
possibly a die, possibly from the middle of the Byzantine era.
Nick Mackinnon. The portrait of Fra Luca Pacioli. MG 77 (No. 479) (Jul 1993) plates 1-4 &
pp. 129-219. Discusses the various
early authors, but has mistakes.
della Francesca. Trattato.
c1480. Ff. 105r - 117v (224-250)
treats solid bodies, discussing all the regular polyhedra, with figures, though
Arrighi gives only a projection of the octahedron. Discusses and gives good diagrams of the truncated tetrahedron
and cubo‑octahedron, apparently the first drawings of any Archimedean
polyhedra. Jayawardene refers to the
cubo-octahedron as a truncated cube.
Davis
notes that Pacioli's Summa, Part II, ff. 68v - 73v, prob. 1-56, are essentially
identical to della Francesca's Trattato, ff. 105r - 120r.
Piero della Francesca. Libellus de Quinque Corporibus
Regularibus. c1487 [Davis, p. 44, dates
it to 1482-1492]. Piero would have
written this in Italian and it is believed to have been translated into Latin
by Matteo da Borgo [Davis, p. 54], who improved the style. First post-classical discussion of the
Archimedean polyhedra, but it was not published until an Italian translation
(probably by Pacioli) was printed in Pacioli & da Vinci, qv, in 1509, as:
Libellus in tres partiales tractatus divisus quae corpori regularium e
depēdentiū actine perscrutatiōis ..., ff. 1‑27. A Latin version was discovered by J.
Dennistoun, c1850, and rediscovered by Max Jordan, 1880, in the Urbino
manuscripts in the Vatican -- MS Vat.Urb.lat. 632; the Duke of Urbino was a
patron of Piero and in the MS, Piero asks that it be placed by his De Prospectiva
Pingendi in the Duke's library. This
was published by Girolamo Mancini in: L'opera "De corporibus
regularibus" di Pietro Franceschi detto della Francesca usurpata da Fra
Luca Pacioli, Memorie della Reale Accademia dei Lincei, Classe di Scienze
Morali, Storiche e Filologiche (5) 14:7B (1915) 441-580 & 8 plates, also separately published by Tipografia
della Reale Accademia dei Lincei, Rome, 1916.
Davis identifies 139 problems in this, of which 85
(= 61%) are taken from the
Trattato. There is debate as to how
much of this work is due to Piero and how much to Pacioli. The Latin text differs a bit from the
Italian. See the works of Taylor and
Davis in Section 1 under Pacioli and the discussion on della Francesca's
Trattato and Pacioli's Summa in the common references.
He
describes a sphere divided into 6 zones and
12 sectors. Mackinnon says Piero describes seven of the
Archimedean polyhedra, but without pictures, namely: cuboctahedron, truncated tetrahedron, truncated cube, truncated octahedron,
truncated dodecahedron, truncated icosahedron, rhombi-cuboctahedron. Field, op. cit. in 6.AT.1, p. 107, says
Piero gives six of the Archimedean polyhedra.
In recent lectures Field has given a table showing which Archimedean
polyhedra appear in Piero, Pacioli, Dürer and Barbaro and this lists just the
first six of the above as being in Piero.
I find just the five truncated regular polyhedra -- see above for the
cubo-octahedron -- and there is an excellent picture of the truncated
tetrahedron on f. 22v of the printed version.
Mancini gives different diagrams than in the 1509 printed version,
including clear pictures of the truncated icosahedron and the truncated
dodecahedron. della Francesca clearly
has the general idea of truncation. An
internet biographical piece, apparently by, or taken from, J. V. Field,
(http://www‑history.mcs.st‑andrews.ac.uk/history/Mathematicians.Francesca.html),
shows that the counting is confused by the presence of the cubo-octahedron in
the Trattato but not in the Libellus.
So della Francesca rediscovered six Archimedean polyhedra, but only five
appear in the Libellus. The work of
Pappus was not known at this time.
Pacioli. Summa.
1494.
f.
4. Brief descriptions of the
cubo-octahedron, truncated tetrahedron, icosidodecahedron, truncated
icosahedron. No drawings.
Part
II, ff. 68v - 72r, sections 2 (unlabelled) - 35. Discussion and some crude drawings of the regular polyhedra, the
truncated tetrahedron and the cubo-octahedron.
Mackinnon says these are the first printed illustrations of any
Archimedean polyhedra. Davis notes that
Part II, ff. 68v - 73v, prob. 1-56, are essentially identical to della
Francesca's Trattato, ff. 105r - 120r.
Jacopo de'Barbari or Leonardo da
Vinci. Portrait of Fra Luca
Pacioli. 1495. In the Museo Nazionale di Capodimonte,
Naples. The upper left shows a glass
rhombi-cuboctahedron half filled with water.
Discussed by Mackinnon, with colour reproduction on the cover. Colour reproduction in Pacioli, Summa, 1994
reprint supplement.
Luca Pacioli & Leonardo da
Vinci. Untitled MS of 1498,
beginning: Tavola dela presente opera e
utilissimo compendio detto dela divina proportione dele mathematici discipline
e lecto -- generally called De divina proportione. Three copies of this MS were made. One is in the Civic Library of Geneva, one
is the Biblioteca Ambrosiana in Milan and the third is lost. Three modern versions of this exist.
Transcription
published as Fontes Ambrosiani XXXI, Bibliothecae Ambrosianae, Milan,
1956. This was sponsored by Mediobanca
as a private edition. There is a copy
at University College London.
Colour
facsimile of the Milan copy, Silvana Editoriale, Milano, (1982), 2nd ptg,
1986. With a separate booklet giving
bibliographical details and an Introduzione di Augusto Marinoni, 20pp +
covers. The booklet indicates this is
Fontes Ambrosiani LXXII.
Printed
version: [De] Divina proportione Opera
a tutti glingegni perspicaci e curiosi necessaria Ove ciascun studioso di Philosophia: Prospectiva Pictura Sculptura:
Architectura: Musica: e altre Mathematice: suavissima: sottile: e admirabile
doctrina consequira: e delectarassi: cōvarie questione de secretissima
scientia. Ill. by Leonardo da Vinci. Paganino de Paganini, Brescia, 1509. Facsimile in series Fontes Ambrosiana, no.
XXXI, Milan, 1956; also by Editrice
Dominioni, Maslianico (Como), 1967. (On
f. 23r, the date of completion of the original part is printed as 1497,
but both MSS have 1498.)
The
printed version was assembled from three codices dating from 1497‑1498
and contains the above MS with several additional items. However, the diagrams in the text are
simplified and the plates are in a different order. The MS has 60 coloured plates, double sided; the printed version
has 59 B&W plates, single sided.
There are errors of pagination and plate numbering in both
versions. On f. 3 of the printed
version is a list of plates and one sees that plate LXI should be numbered
LVIIII and that plates LX, LXI were omitted and were to have been a hexagonal pyramid
in solid and framework views (the framework view is in the MS, but the solid
pyramid is not).
NOTE. Simon Finch's Catalogue 48, item 4,
describes the copy that was in the Honeyman Collection and says it has 59
printed plates of geometric figures and is unique in having two contemporary
additional MS plates showing the hexagonal pyramid (numbered LX and LXI), which
are given in the list of plates, but which do not appear in any other known
copy. It seems that these figures were
overlooked in printing and that the owner of the Honeyman copy decided to make
his own versions, or, more likely, got someone to make versions in the original
style. There is a framework hexagonal
pyramid in the MS, and this makes it seem likely that these figures had been
prepared and were omitted in printing -- indeed the Honeyman leaves could be
the overlooked drawings. That leaves
the question of whether there was a solid hexagonal pyramid in the MS?
Most
pictures come in pairs -- a solid figure and then a framework figure. There are the five regular polyhedra, the
following six Archimedean polyhedra: truncated tetrahedron, cubo-octahedron,
truncated octahedron, truncated icosahedron, icosi-dodecahedron,
rhombi-cubo-octahedron and also the stella octangula. There are raised or elevated versions of the tetrahedron, cube,
cubo-octahedron, icosahedron, dodecahedron, icosi-dodecahedron,
rhombi-cubo-octahedron. Also
triangular, square, pentagonal and hexagonal prisms and tall triangular, square
and pentagonal pyramids. Also a
triangular pyramid not quite regular and a sphere divided into 12 sectors and 6
zones. There are also a solid sphere, a
solid cylinder, a solid cone and a framework hexagonal pyramid (the last is not
in the printed version). Mackinnon says
they give the same seven Archimedean polyhedra as Piero, but Piero gives five
or six and Pacioli & da Vinci gives six, with only four common
polyhedra. Pacioli & da Vinci
assert that the rhombi-cuboctahedron arises by truncating a cuboctahedron, but
this is not exactly correct.
Part
of the printed version is Libellus in tres partiales tractatus divisus quae
corpori regularium e depēdentiū actine perscrutatiōis ..., which
is an Italian translation (probably by Luca Pacioli) of Piero della Francesca's Libellus de quinque corporibus
regularibus. There is debate as to
whether this was actually written by Pacioli or whether Pacioli plagiarized it
and whether it actually appeared in the 1509 printing or was added to a later reprinting,
etc.
Davis
[p. 65] says the drawings were made from models prepared by Da Vinci. Davis [p. 74] cites Summa, Part II, f. 68v,
and she quotes part of it on pp. 100-101.
This is also referred to by MacKinnon [p. 170] and Taylor [p. 344],
neither giving details and no two of the three agreeing on what the passage
means. I have not been able to make
complete sense of the passage, but it seems clearly to say that in Apr 1489,
Pacioli presented models of at least the regular solids to the Duke of Urbino
at the palace of Pacioli's protector [Cardinal Giuliano della Rovere, later
Pope Julius II] in Rome. He then says
many other dependants [= variations] of the regular solids can be made, and
models were made for Pietro Valletari, Bishop of Carpentras. There is no reference to the number of
models, nor their material, nor to a set being given to the Cardinal, nor
whether the Cardinal was present when the models were given to the Duke. Due to a missing right parenthesis, ), the sense of one statement involving 'his
own hands' could mean either that Pacioli presented the models to the Duke's
own hands or that Pacioli had decorated the models himself. I doubt whether there were many other solids
at this time, otherwise he would have mentioned them in the Summa -- the Summa
only describes four of the Archimedean solids and only two of them are in Part
II. I suggest that he didn't start
developing the other shapes until about 1494, or later, in 1496 when he went to
Milan and met Leonardo.
However,
on f. 28v of De Divina Proportione, Pacioli is clearer and says that he
arranged, coloured and decorated with his own hands 60 models in Milan and two
other sets for Galeazzo Sanseverino in Milan and for Piero Soderini in
Florence. This refers to his time in Milan,
which was 1496-1499, though the Soderini set might have been made after Pacioli
and da Vinci moved to Florence.
Albrecht Dürer. Underweysung der messung .... 1525 & 1538. Op. cit. in 6.AA. Figures
29‑43 (erroneously printed 34) (pp. 316-347 in The Painter's Manual,
Dürer's 1525 ff. M-iii-v - N-v-r) show a net of each of the regular polyhedra,
an approximate sphere (16 sectors and 8 zones), truncated tetrahedron,
truncated cube, cubo-octahedron, truncated octahedron, rhombi-cubo-octahedron,
snub cube, great rhombi-cubo-octahedron, polyhedron of six dodecagons and
thirty-two triangles (having a pattern of four triangles replacing each
triangle of the cubo-octahedron, so a sort of truncated cubo-octahedron -- not
an Archimedean solid and not correctly drawn) and an elongated hexagonal bipyramid
(not even regular faced). This
gives 7 of the 13 non-regular
Archimedean polyhedra. Mackinnon says
Figures 29‑41 show a net of each of the regular polyhedra and the same
seven Archimedean ones as given by Pacioli & da Vinci, but they give
six and there are only four common ones.
In the revised version of 1538, figure 43 is replaced by the truncated
icosahedron and icosi-dodecahedron (figures 43 & 43a, pp. 414‑419
in The Painter's Manual), giving 9 of the
13 non-regular Archimedean
solids. P. 457 shows the remaining four
Archimedean cases from an 1892 edition.
Albrecht Dürer. Elementorum Geometricorum (?). 1534. Op. cit. in 6.AA. Liber quartus, fig. 29-43, pp. 145-158
shows the same material as in the 1525 edition.
See Barbaro, 1568, in 6.AT.2,
pp. 45-104 for drawings and nets of 11 of the 13 Archimedean solids - he omits
the two snub solids.
See Jamnitzer, 1568, in 6.AT.2
for drawings of eight of the 13 Archimedean solids.
J. Kepler. Letter to Maestlin (= Mästlin). 22 Nov 1599. In: Johannes Kepler
Gesammelte Werke, ed. by M. Caspar, Beck, Munich, 1938. Vol 14, p. 87, letter 142, lines 21-22. ??NYS.
Described by Field, p. 202.
Describes both rhombic solids.
J. Kepler. Strena seu De Nive Sexangula [A New Year's
Gift or The Six‑Cornered Snowflake].
Godfrey Tampach, Frankfurt am Main, 1611. (Reprinted in: Johannes Kepler Gesammelte Werke; ed. by M.
Caspar, Beck, Munich, 1938, vol. 4, ??NYS.)
Reprinted, with translation by C. Hardie and discussion by B. J. Mason
& L. L. Whyte, OUP, 1966. I will
cite the pages from Kepler (and then the OUP pages). P. 7 (10‑11).
Mentions 'the fourteen Archimedean solids' [sic!]. Describes the rhombic dodecahedron and
mentions the rhombic triacontahedron. The
translator erroneously adds that the angles of the rhombi of the dodecahedron
are 6Oo and 120o.
Kepler adds that the rhombic dodecahedron fills space. Kepler's discussion is thorough and gives no
references, so he seems to feel it was his own discovery.
J. Kepler. Harmonices Mundi, 1619. Book II, opp. cit. above. Prop. XXVII, p. 61. Proves that there are just two rhombic 'semi‑regular'
solids, the rhombic dodecahedron and the rhombic triacontahedron, though the
cube and the 'baby blocks' tessellation can also be considered as limiting cases. He illustrates both polyhedra. Def. XIII, p. 50 & plate (missing in
facsimile). Mentions prisms and
antiprisms. Prop. XXVIII, pp. 61‑65. Finds the 13 Archimedean solids and
illustrates them -- the first complete set -- but he does not formally show
existence.
J. Kepler. Epitome Astronomiae Copernicanae. Linz, 1618-1621. Book IV, 1620. P. 464. = Johannes Kepler Gesammelte Werke; ed.
by M. Caspar, Beck, Munich, 1938, vol. 7, p. 272. Shows both rhombic solids.
The following table shows which
Archimedean polyhedra appear in the various early books from della Francesca to
Kepler.
dF = della Francesca, c1487, folio
T
indicates the object appears in the Trattato of c1480.
P = Pacioli, Summa, Part II, 1494, folio.
P&dV = Pacioli & da Vinci,
1498, plate number of the solid version; the framework version is
the
next plate
D = Dürer, 1525, plate
+
indicates the object is added in the 1538 edition.
B = Barbaro, 1568, page
J = Jamnitzer, 1568, plate
K = Kepler, 1619, figure
Truncated tetrahedron T,22v 4v,II-69v 3 35 56 A3 2
Truncated cube 22r 36 61 G2 1
Truncated dodecahedron 21r 76 3
Truncated octahedron 21v 17 38 68 B2,B4 5
Great rhombi-cubo-octahedron 41 88 B6 6
Great rhombi-icosidodecahedron 100 7
Truncated icosahedron
(football!) 20v 4v 23 + 81 4
Cubo-octahedron T 4v,II-69v 9 37 58 B6,F1 8
Icosidodecahedron 4v 29 71 D4,F6 9
Rhombi-cubo-octahedron 35 39 64 B6 10
Rhombi-icosidodecahedron 94 D4 11
Snub cube 40 12
Snub dodecahedron 13
NUMBER 5(6) 4 6 7(9) 11 8 13
Richard Buckminster Fuller. Centre spread card version of his Dymaxion
World map on the cubo-octahedron. Life
(15 Mar 1943). Reproduced in colour,
with extended discussion, in: Joachim
Krausse & Claude Lichtenstein, eds;
Your Private Sky R. Buckminster
Fuller The Art of Design Science [book
accompanying a travelling exhibition in 2000];
Lars Müller Publishers, Bade, Switzerland, 1999, pp. 250-275. (This quotes a Life article on 1 Mar
1943 and a Fuller article, Fluid Geography, of 1944 -- ??NYS. It also reproduces a 1952 colour example of
the icosahedral version.)
Richard Buckminster Fuller. US Patent 2,393,676 -- Cartography. Filed: 25 Feb 1944; granted: 29 Jan 1946. 3pp + 5pp diagrams. His world map on the cubo-octahedron. It was later put on the icosahedron. One page is reproduced in: William Blackwell; Geometry in Architecture;
Key Curriculum Press, Berkeley, 1984, p. 157.
J. H. Conway. Four-dimensional Archimedean polytopes. Proc. Colloq. Convexity, Copenhagen, 1965
(1967) 38-39. ??NYS -- cited by Guy,
CMJ 13:5 (1982) 290-299.
H. S. M. Coxeter, M. S. Longuet‑Higgins & J. C. P.
Miller. Uniform polyhedra. Philos. Trans. Roy. Soc. 246A (1954) 401‑450. They sketch earlier work and present 53
uniform polyhedra, beyond the 5 Platonic, 13 Archimedean and 4 Kepler‑Poinsot
polyhedra and the prisms and anti‑prisms. Three of these uniform polyhedra are actually infinite
families. "... it is the authors'
belief that the enumeration is complete, although a rigorous proof has still to
be given."
S. P. Sopov. Proof of completeness of the list of uniform
polyhedra. Ukrain. Geometr. Sb. 8
(1970) 139-156. ??NYS -- cited in
Skilling, 1976.
J. S. Skilling. The complete set of uniform polyhedra. Philos. Trans. Roy. Soc. London Ser. A
278 (1975) 111‑135. Demonstrates
that the 1954 list of Coxeter, et al., is complete. If one permits more than two faces to meet at an edge, there is
one further polyhedron -- the great disnub dirhombidodecahedron.
J. S. Skilling. Uniform compounds of uniform polyhedra. Math. Proc. Camb. Philos. Soc. 79 (1976)
447-468. ??NYS -- I am told it determines
that there are 75 uniform compounds and also cites Sopov.
6.AT.5. REGULAR‑FACED POLYHEDRA
O. Rausenberger. Konvexe pseudoreguläre Polyeder. Zeitschr. für math. und naturwiss.
Unterricht 46 (1915) 135‑142.
Finds the eight convex deltahedra.
H. Freudenthal & B. L. van
der Waerden. Over een bewering van
Euclides [On an assertion of Euclid] [in Dutch]. Simon Stevin 25 (1946/47) 115‑121. ??NYS.
Finds the eight convex deltahedra -- ignorant of Rausenberger's work.
H. Martyn Cundy. "Deltahedra". MG 36 (No. 318) (Dec 1952) 263‑266. Suggests the name
"deltahedra". Exposits the
work of Freudenthal and van der Waerden, but is ignorant of Rausenberger. Considers non‑convex cases with two
types of vertex and finds only 17 of them.
Considers the duals of Brückner's trigonal polyhedra.
Norman W. Johnson. Convex polyhedra with regular faces. Canad. J. Math. 18 (1966) 169‑200. (Possibly identical with an identically
titled set of lecture notes at Carleton College, 1961, ??NYS.) Lists 92 such polyhedra beyond the 5 regular
and 13 Archimedean polyhedra and the prisms and antiprisms.
Viktor A. Zalgaller. Convex polyhedra with regular faces [in
Russian]. Seminars in Mathematics, V.
A. Steklov Mathematical Institute, Leningrad, vol. 2 (1967). ??NYS.
English translation: Consultants
Bureau, NY, 1969, 95pp. Gives details
of computer calculations which show that Johnson's list is complete. Defines a notion of simplicity and shows
that the simple regular‑faced polyhedra are the prisms, the antiprisms
(excepting the octahedron) and 28 others.
Names all the polyhedra and gives drawings of the simple ones.
Albrecht Dürer. Underweysung der messung .... 1525 & 1538. Op. cit. in 6.AA. Figures
22‑27 (pp. 156‑169 in The Painter's Manual, Dürer's 1525
ff. E‑vi‑v ‑ F-iii-v) show: the three regular tessellations; the quasi-regular one, 3636,
and some of its dual; several
irregular ones, including some partial tessellations with pentagons; and the truncated square lattice, 482. In the revised version of 1538, he adds some tilings by rhombuses
(figures 23a & 24, pp. 410‑411 in The Painter's Manual).
Albrecht Dürer. Elementorum Geometricorum (?). Op. cit. in 6.AA, 1532. Book II, fig. 22‑27, pp. 62-67,
is the material from the 1525 version.
J. Kepler. Letter to Herwart von Hohenberg. 6 Aug 1599.
Op. cit. in 6.AT.2. Field, p.
105, says Kepler discusses tessellations here and this is the earliest of his
writings to do so.
J. Kepler. Harmonices Mundi. 1619. Book II. Opp. cit. above. Prop. XVIII, p. 51 & plate (missing in my facsimile). Shows there are only three regular plane
tessellations and mentions the dual of 3636, which is the 'baby blocks'
tessellation. Prop. XIX‑XX,
pp. 51‑56 & four plates (three missing in my facsimile). Finds the 8 further Archimedean
tessellations and 7 of the 10 further ways to fill 360 degrees with corners of
regular polygons. He misses 3,7,42;
3,8,24; 3,9,18 despite computing, e.g., that a triangle and
a heptagon would leave a gap of
40/21 of a right angle. Field, p. 109, notes that Kepler doesn't
clearly have all vertices the same in some pictures -- e.g. he has both 3366
and 3636 patterns in his figure R.
Koloman Moser. Ver Sacrum.
1902. This Viennese art nouveau
drawing is considered to be the first tessellation using life-like figures. It has trout and the pattern has symmetry
(or wallpaper) group pg and isohedral type IH2.
Branko Grünbaum & G. C.
Shephard. Tilings and Patterns. Freeman, 1986. I haven't examined this thoroughly yet, but it clearly is the
definitive work and describes everything known to date.
6.AT.6.a. TESSELLATING WITH CONGRUENT FIGURES
This
is a popular topic which I have just added.
Gardner's article and addendum in Time Travel gives most recent results,
so I will just give just some highlights.
The facts that any triangle and any quadrilateral will tile the plane
must be very old, perhaps Greek, but I have no early references. Generally, I will consider convex polygons
and most items only deal with the plane.
David Hilbert. Mathematische Probleme. Göttinger Nachrichten (= Nachrichten der
K. Gesellschaft der Wissenschaften zu Göttingen, Math.‑phys. Klasse)
3 (1900) 253‑297. This has been
reprinted and translated many times, e.g. in the following.
R.
Bellman, ed. A Collection of Modern
Mathematical Classics -- Analysis.
Dover, 1961. Pp. 248‑292
[in German].
Translated
by M. W. Newson. Bull. Amer. Math. Soc.
8 (1902) 437‑479. Reprinted
in: F. E. Browder, ed. Mathematical Developments Arising from
Hilbert Problems. Proc. Symp. Pure
Math. 28 (1976) 1‑34.
Problem
18: Aufbau des Raumes aus kongruenten Polyedern [Building up of space from
congruent polyhedra]. "The
question arises: Whether polyhedra also exist which do not appear as
fundamental regions of groups of motions, by means of which nevertheless by a
suitable juxtaposition of congruent copies a complete filling up of space is
possible." Hilbert also asks two
other questions in this problem.
The
problem is discussed by John Milnor in his contribution to the Symposium, but
he only shows non‑convex 8‑ &
10‑gons which fill the plane.
K. Reinhardt. Über die Zerlegung der Ebene in
Polygone. Dissertation der Naturwiss.
Fakultät, Univ. Frankfurt/Main, Borna, 1918.
??NYS -- cited by Kershner.
Finds the three types of hexagons and the first five types of pentagons
which fill the plane.
Max Black. Reported in: J. F. O'Donovan; Clear
thinking; Eureka 1 (Jan 1939) 15 & 20.
Problem 2: which quadrilaterals can tile the plane? Answer: all!
R. B. Kershner. On paving the plane. AMM 75:8 (Oct 1968) 839‑844. Says the problem was posed by Hilbert. Gives exhaustive lists of hexagons and a
list of pentagons which he claimed to be exhaustive. Cites previous works which had claimed to be exhaustive, but he
has found three new types of pentagon.
J. A. Dunn. Tessellations with pentagons. MG 55 (No. 394) (Dec 1971) 366‑369. Finds several types and asks if there are
more.
M. M. Risueño, P. Nsanda Eba
& Editorial comment by Douglas A. Quadling. Letters: Tessellations
with pentagons. MG 56 (No. 398) (Dec
1972) 332‑335. Risueño's letter
replies to Dunn by citing Kershner. Eba
constructs a re‑entrant pentagon.
[This is not cited by Gardner.]
Gardner. On tessellating the plane with convex
polygon tiles. SA (Jul 1975). Much extended in Time Travel, chap. 13.
Ivan Niven. Convex polygons that cannot tile the
plane. AMM 85 (1978) 785-792. n‑gons, with n > 6, cannot tile the
plane.
Doris Schattschneider. In praise of amateurs. In:
The Mathematical Gardner; ed. by David A. Klarner; Wadsworth, Belmont,
California, 1981, pp. 140‑166 & colour plates I‑V between 166
& 167. Surveys history after
Kershner, describing contributions of James & Rice.
Gardner. On tessellating the plane with convex
polygon tiles. [Originally: SA (Jul 1975).] Much extended in Time Travel, 1988, chap. 13. The original article generated a number of
responses giving new pentagonal tilings, making 14 types in all. Good survey of the recent literature.
New
section.
John Gorham. A System for the Construction of Crystal Models
on the Type of an Ordinary Plait: Exemplified by the Forms Belonging to the Six
Axial Systems in Crystallography. E.
& F. N. Spon, London, 1888.
Gorham's Preface says he developed the idea and demonstrated it to the
Royal Society some 40 years earlier.
A. R. Pargeter. Plaited polyhedra. MG 43 (No. 344) (May 1959) 88‑101. Cites and quotes Gorham. Extends to plaiting dodecahedron,
icosahedron and some archimedean, dual and stellated examples.
J. Brunton. The plaited dodecahedron. MG 44 (No. 347) (Feb 1960) 12‑14. With comment by Pargeter. Obtains a 3‑plait which almost
completes the dodecahedron.
New
section -- I know of other articles claiming to 'solve' the problem.
Albrecht Dürer. Melencolia I. 1514. Two impressions are
in the British Museum. In the back left
is an octahedron whose exact shape is the subject of this section. It looks like a cube truncated at two
opposite corners, but the angles do not quite look like 90o.
Albrecht Dürer. Dresden Sketchbook. Facsimile as The Human Figure, the complete
Dresden Sketchbook; Dover, NY, 1972.
??NYS -- cited by Sharp. This
has a sketch of the solid with hidden lines indicated, so the combinatorial
shape is definitely known and is a hexahedron of six equal faces, truncated at
two opposite corners.
E. Schröder. Dürer Kunst und Geometrie. Birkhäuser, Basel, 1980. ??NYS -- cited by Sharp and MacGillavry.
Caroline H. MacGillavry. The polyhedron in A. Dürer's Melencolia
I An over 450 years old puzzle solved? Koninklijke Nederlandse Akademie van
Wetenschappen Proc. B 84:3 (28 Sep 1981) 287-294. The rhombohedral angle, i.e. the angle between edges at the
truncated top and bottom vertices of the rhombohedron, was estimated as 72o
by Grodzinski. She determines it is 79o
± 1o. She then built and
photographed such a polyhedron and then computed its projection, both of which
seem identical to Dürer's picture.
Crystallographers believe Dürer was drawing an actual crystal, with a
form of calcite having rhombohedral angle of 76o being the closest
known shape, though it is not known to have been studied in Dürer's time, so
others have suggested fluorite, though fluorite has two standard forms, neither
of this form, but Dürer's 'hybrid' artistic version could have been derived
from them.
Terence Lynch. The geometric body in Dürer's engraving Melencolia
I. J. Warburg and Courtauld
Institutes 45 (1982) 226-232 & plate a on p. 37?. Lots of references to earlier work. Notes that perspective was not sufficiently advanced for Dürer to
construct a general drawing of such an object.
After many trials, he observes that a parallel projection of the solid
fits onto a 4 x 4 grid -- like the magic square in the picture
-- and that symmetry then permits the construction with straight edge and
compass (which are both shown in the picture).
This shows that the original faces are rhombuses whose diameters are in
the ratio 2 : Ö3. And the dihedral angle between the
triangular faces and the cut off rhombuses is 30o Further, the actual drawing can then be made
by one of the simplified perspective techniques known to Dürer. However, Dürer has taken a little bit off
the top and bottom of the figure and this distortion has misled many previous
workers.
John Sharp. Dürer's melancholy octahedron. MiS (Sep 1994) 18-20. Asserts that the shape was first determined
by Schröder in 1980 and verified by Lynch.
New
section.
Stuart Robertson. The twenty-two cuboids. Mathematics Review 1:5 (May 1991)
18-21. This considers polyhedra with
six quadrilateral faces and determines what symmetries are possible -- there
are 22 different symmetry groups.
6.AU. THREE RABBITS, DEAD DOGS AND TRICK MULES
See
S&B, p. 34.
Loyd's
Trick Mules has two mules and two riders which can only be placed correctly by
combining each front with the other rear.
Earlier
forms showed two dead dogs which were brought to life by adding four lines. The
resulting picture is a pattern, generally called 'Two heads, four children' and
can be traced back to medieval Persian, Oriental and European forms.
The
three rabbits problem is: "Draw three rabbits, so that each shall appear
to have two ears, while, in fact, they have only three ears between
them." Until about 1996, I only
knew this from the 1857 Magician's Own Book and the many books which copied
from it. Someone at a conference at
Oxford in 1996 mentioned that the pattern occurs in a stained glass window at
Long Melford, Suffolk. Correspondence
revealed that the glass is possibly 15C and the pattern was apparently brought
from Devon about that time. More
specifically, it comes from the east side of Dartmoor and inquiries there have
turned up numerous examples as roof bosses from 13‑16C. Totally serendipitously, I was reading a
guide book to Germany in 1997 and discovered the pattern occurs in stonework,
possibly 16C, at Paderborn, Germany. A
letter led to receiving a copy of Schneider's article (see below) which
described the pattern occurring at Dunhuang, c600. I am indebted to Miss Y. Yasumara, the Art Librarian at the
School of Oriental and African Studies, for directing me to several works on
Dunhuang. However, I have not examined
all these works in detail (the largest is five large volumes), so I may not
have found all the examples of this pattern.
Miss Yasumara also directed me to Roderick Whitfield, of the School of
Oriental and African Studies, who tells me there is no other example of this
pattern in Chinese art, and to Susan Whitfield (no relation), head of the
International Dunhuang Project at the British Library. However, Greeves (see his articles, below)
has found other examples of the pattern in Europe, Iran and Tibet and found
that modern carpets with the pattern are being made in China. A student of his recently went to Dunhuang
and the locals told her that the pattern came from 'the West', meaning India,
which opens up a whole new culture to examine.
In
1997, I visited the Dartmoor area, seeing several examples and finding a
reference (Hambling, below) to Tom Greeves' article. In 2000, I again visited the area, seeing more churches and
meeting Tom Greeves and his associate Sue Andrew. Later in 2000, I visited Paderborn. Later in 2000, I showed this material to Wei Zhang and Peter
Rasmussen, leading collectors of Chinese puzzles and they have begun to
investigate the Chinese material much more thoroughly than I have done -- see
below. In 2001, I went again to the
area.
I
have read that rabbits were introduced to England in 1176 by the Normans and
became common in the 13C, though I believe they weren't really wild for some
time after that. E.g. [J. A. R.
Pimlott; Recreations; Studio Vista, 1968, p. 18] says: "The rabbit was not
known in Britain until the thirteenth century and did not become plentiful
until the fifteenth", citing Elspeth M. Veale; The English Fur Trade in
the Later Middle Ages; 1966. I have
read that hares were introduced between -500 and +500, but I have just seen a
mention that bones of a hare found in Ireland have been carbon dated to ‑26,000. So it is probable that the animals in many
of the images are hares, though some of the images are distinctly more
rabbit-like than hare-like.
The
material in this section has grown so much that it is now divided into seven
subsections: China; Other Asia;
Paderborn; Medieval Europe; Modern Versions of the Three Rabbits
Puzzle; Dead Dogs; Trick Mules.
I
have about a dozen letters and emails which have not yet been processed.
Rabbits
going clockwise: Dunhuang (14 caves -- all except 407 & 420); Goepper; St.
Petersburg; Iranian tray; Paderborn; Münster; Bestiary; Lyon; Throwleigh;
Valentine; Clyst Honiton; Hasloch am Main; Michelstadt; Collins; Greeves
letterhead; Urumqi;
Rabbits
going anti-clockwise: Dunhuang (2 caves -- 407 & 420); Corbigny; Lombard's
Gloss; North Bovey; Long Crendon; Chester; Widecombe; Long Melford; Chagford;
Best Cellars, Chagford; Tavistock; Broadclyst; Sampford Courtenay;
Spreyton; Paignton; South Tawton; Valentine; Schwäbisch Hall; Baltrušaitis Fig.
97; Child; Magician's Own Book et al; Warren Inn; Best Cellars,
Chagford; Newman; Lydford; Trinity Construction Services;
The
choice of clockwise versus anticlockwise seems to be random! Except the Chinese clearly preferred
clockwise.
FOUR
RABBITS versions. Baltrušaitis; Wilson; Goepper. Bestiary; Andrew/Lombard; Lyon; Hamann-MacLean;
MORE
FIGURES. Chichester Cathedral. Boxgrove Priory.
Jurgis Baltrušaitis. Le Moyen Age Fantastique Antiquités et exotismes dans l'art
gothique. (A. Colin, Paris, 1955, 299
pp.) Revised, Flammarion, Paris, 1981,
281 pp. Thanks to Peter Rasmussen for
telling me about this. Supposedly an
English edition was published in 1998, but I have found an entry in the Warburg
catalogue for The Fantastic in the Middle Ages, published by Boydell &
Brewer, Woodbridge, Suffolk, 2000, marked 'order cancelled' -- so the
publication seems to have never happened.
This is a major source used for When Silk Was Gold, below. Pp. 132‑139 of the 1981 edition have
many examples of three and four rabbits, four boys, etc. He gives a small illustration (Fig. 96B)
from Dunhuang (Touen-houang) (6C-10C), the oldest example he knows, and many
others. See other sections for more
details.
[Huang Zu'an ?? -- Schneider,
below, gives this author, but there is no mention of an author in the entire
issue.] Dunhuang -- Pearl of the Silk
Road. China Pictorial (1980:3) 10-23
with colour photo on p. 22. 9th article
in a series on the Silk Road. Colour
photo of the three rabbits pattern with caption: "A ceiling design. The three rabbits with three ears and the
apsarases seem to be whirling. Cave
407. Sui Dynasty." The Sui dynasty was from 581 to 618, so we
can date this as c600. The image is
rather small, but the three hares can be made out. There is no discussion of the pattern in the article.
Chang Shuhong &
Li Chegxian. The Flying Devis of
Dunhuang. China Travel and Tourism
Press, Beijing, 1980. Unpaginated. In the Preface, we find the following. "What is particularly novel is the
full-grown lotus flower painted in the centre of the canopy design on the
ceiling of Cave 407. In the middle of
the flower there are three rabbits running one after the other in a
circle. For the three rabbits only
three ears are painted, each of them borrowing one ear from another. This is an ingenious conception of the
master painter." From this, it
seems that this pattern is uncommon.
The best picture of the pattern that I have located is in this book, in
the section on the Sui period. I have
now acquired a copy with its dust jacket and find a painting of the pattern is
on the back of the dust jacket -- this is the painting also given in Li Kai et
al, below, p, 27. [Incidentally, a devi
or apsaras is a kind of Buddhist angel.
The art of Dunhuang is quite lovely.]
The Dunhuang Institute for
Cultural Relics. The Mogao Grottos of
Dunhuang. 5 vols. + Supplement. Heibonsha Ltd., Tokyo, 1980-1982. (In Japanese, with all the captions given in
English at the end of the Supplementary volume. Fortunately the plate and cave numbers are in western
numerals. A Chinese edition was
planned.) Vol. 2, plate 94, is a
double-page spread of the ceiling of Cave 407, with the page division running
right through the middle of the rabbits pattern! Vol. 2, plate 95, is a half-page plate of the ceiling of Cave
406, and shows the rabbits pattern, but it seems rather faded. The English captions simply say
"Ornamental ceiling decoration".
R. Whitfield &
A. Farrer. Caves of the Thousand
Buddhas. British Museum, 1990, esp. pp.
12 & 16. Though cited by Greeves,
these pages only have general material on Dunhuang and the book does not
mention any of the relevant caves.
Duan Wenjie. Dunhuang Art Through the Eyes of Duan Wenjie.
Indira Gandhi National Centre for the Arts, Abhinav Publications, New
Delhi, 1994. This gives much more
detail about the caves. Pp. 400-401
describes Caves 406-407. Peter
Rasmussen examined the book in detail and found it mentions 12 other caves with
the pattern. Peter has found that the
book is accessible on-line at
www.ignca.nic.in/ks_19.htm. This
provides the facility to download a font which will display diacritical marks
and this is worth doing before you start to browse.
In
late 2002, Peter Rasmussen and Wei Zhang were able to inspect 10 caves that
they hadn't seen before. Peter sent
notes of their impressions of 12 caves on 24 Nov 2002 and I will add some of
Peter's comments and additions in [ ] and marked PR. This will make the following list the basic list of all sixteen
of the caves.
Cave
No. 127. Late Tang (renovated in Five
Dynasties and Qing). "The ceiling
exhibits lotus and three rabbits (joining as one) in the centre." [PR: tan on turquoise, going clockwise.]
Cave
No. 139. Late Tang. "The ceiling shows the three rabbits
(joining as one) and lotus designs in the centre." [PR: this is a small cave off the entry to
Cave 138. Tan on light green, going
clockwise, in excellent condition.
"Rabbits beautifully drawn in pen-like detail, showing toes, eyes
(with eyeballs!), all four legs, tail, nose, mouth, outline of thigh muscle,
and hair on stomach, breast, legs and top of head." Peter says this is by far the finest of the
images; he is applying to get it photographed.]
Cave
No. 144. Middle and Late Tang
(renovated during the Five Dynasties and Qing). "The centre of the ceiling shows the three rabbits (joining
as one) and floral designs." [PR:
white on aqua green, going clockwise.]
Cave
No. 145. Late Tang (renovated during
the Five Dynasties and Song). "The
ceiling of the niche on the west wall shows lotus and three rabbits (joining as
one), chess‑board and floral patterns." [Zhang & Rasmussen's letter of 1 Jun 2001 says the rabbits
are going clockwise.]
Cave
No. 147. Late Tang. "Main Hall: The ceiling shows three
rabbits (joining as one) and lotus designs in the centre". [PR: tan or turquoise green, going
clockwise. Paint on ears has peeled
off.]
Cave
No. 200. Middle Tang. "The ceiling shows three rabbits
(joining as one) and round petalled lotus designs in the centre". [PR: white on turquoise green, going
clockwise.]
Cave
No. 205. Early and High Tang (renovated
during Middle Tang and the Five Dynasties).
Main Hall: "The ceiling has the three rabbits (joining as one)
designs drawn in Early Tang".
[Zhang & Rasmussen's letter of 1 Jun 2001 says the rabbits are going
clockwise.]
Cave
No. 237. Middle Tang (renovated in
Western Xia and Qing). Main Hall:
"The ceiling shows the three rabbits (joining as one) and round petalled
lotus designs in the centre."
[Zhang & Rasmussen's letter of 1 Jun 2001 says the rabbits are going
clockwise.]
[PR:
Cave No. 305. Sui (renovated during the
Five Dynasties and Qing). "Rabbit
paint is gone (only white silhouette remains on faint rusty red
background)." See: Decorative Patterns
in the Dunhuang Art; Li Kai et al; Zhang & Rasmussen's letter of 1 Jun
2001, which says the rabbits are going clockwise]
Cave
No. 358. Middle Tang (renovated during
the Five Dynasties, Western Xia and Qing).
"Main Hall: The caisson ceiling shows the three rabbits (joining as
one) and round petalled lotus in the centre." [PR: white on turquoise, going clockwise, faded.]
Cave
No. 383. Sui (renovated during Song,
Western Xia and Qing Dynasties).
"Main Hall: The centre of the caisson ceiling shows the three
rabbits (joining as one) and lotus flower designs". [PR: brown with white outlines, going
clockwise, fair condition.]
Cave
No. 397. Sui and Early Tang (renovated
during the Five Dynasties and Qing).
"Main Hall: The caisson ceiling shows the three rabbits (joining as
one) and lotus in the centre".
[PR: white on peeled-off aqua green, going clockwise. Poor to fair
condition.]
Cave
No. 406. Sui (renovated in Song and
Qing). "The centre of the caisson
ceiling shows four designs of a set of three rabbits (joining as one) and
lotus". I don't quite understand
his phrasing -- there is a picture of a pattern of three rabbits in the centre
of a lotus, as in Cave 407, but perhaps there are other patterns which are not
reproduced?? [PR: white on tan, going
clockwise, faded, fair condition. Peter
says nothing to clarify Duan's text.]
Cave
No. 407. Sui (renovated in Song and
Qing). "Main Hall: The caisson
ceiling is covered with the three rabbits, lotus designs and flying figures
drawn in Sui." [PR: Black with
white outline on turquoise green background, going anticlockwise, good to
excellent condition.]
[PR:
Cave No. 420. Sui. Zhang & Rasmussen's letter of 1 Jun 2001
says the rabbits are going anticlockwise.]
Cave
No. 468. Middle Tang (renovated during
the Five Dynasties). "Main Hall:
The centre of caisson ceiling has three rabbits (joining as one) and a lotus
design". [PR: white on turquoise,
going clockwise.]
CHRONOLOGY Tang is 618-907, but there is a Later Tang
(923-936).
Sui
(581-618, so c600): 305, 383, 406, 407,
420.
Sui
and Early Tang (c620): 397.
Early/Mid
Tang (7-8C): 205.
Mid
Tang (8C): 200, 237, 358, 468.
Mid/Late
Tang (8-9C): 144.
Late
Tang (9-10C): 127, 139, 145, 147.
Roderick Whitfield. Dunhuang
Caves of the Singing Sands.
(Revision of a Japanese book by NHK, 1992.) Textile & Art Publications, London, 1995. On pp. 59 & 238, plates 66 &
361-362, are pictures of the roof of cave 420 which may be showing a three
rabbit pattern, but it is too faded and too small to really be sure.
Decorative Patterns in the
Dunhuang Art. 1996. This has no English text except for the book
title. The dust jacket has a colour
picture of a painting of the ceiling of cave 407. Otherwise it gives only black and white drawings of
patterns. P. 31 is an introduction to a
section and has an unidentified three rabbits pattern, probably from cave
407. P. 32 is cave 305. P. 33 is cave 397.
Li Kai, chief designer; Zhao Le Nin
& Luo Ke Hua, eds. The Selections of Copied Art Works of Dun
Huang Sunk Panel. 1997. This presents various artists' paintings of
the ceiling panels.
Plate
21 is: Sui Dynasty "three rabbits and lotus flower" sunk panel (cave
number 305).
Plate
26 is: Sui Dynasty "three rabbit and flying Apsarase" sunk panel
(cave number 407) copied by Duan Wen Jie.
Plate
27 is: Sui Dynasty "three rabbits and flying Apsarase" sunk panel
(cave number 407) copied by Guo Shi Qing and Chang Sha Na. This is reproduced on the cover of the book,
but it has been reversed in printing in both places. The difference in coloration and even in the outlines show the
difficulty of seeing what is present in a rather faded and damaged cave where
the early artists had only daylight.
Neither of these is the same as the cover of the previous item! See also Shuhong & Chegxian above for
another version.
Plate
32 is: Sui Dynasty sunk panel (the cave number is not recorded).
Plate
26 is: Tang Dynasty sunk panel (cave number 205).
Wei Zhang &
Peter Rasmussen. Letter of 1 Jun
2001 reporting on their research and visit to Dunhuang. They have made good contacts there and were
given a lengthy special tour. They have
now found the three rabbits pattern occurs in 16 caves: 127, 139, 144, 145,
147, 200, 205, 237, 305, 358, 383, 397, 406, 407, 420 and 468, all in the Mogao
caves and all as central ceiling panels.
They were able to see 145 (late Tang, 848-906), 205 (Early Tang,
618-704), 237 (mid Tang, 781-847), 305 (Sui, 581-618), 407 (Sui, 581‑618),
420 (Sui, 581-618). In the first four,
the rabbits are going clockwise, in the last two, they are going
anticlockwise. They also visited the
Western Thousand Buddha Caves, but there was no sign of the three rabbits
there. However, none of the researchers
there has investigated the three rabbits pattern. Unfortunately, photography is not allowed, but they sent the
previous two books and two reproductions from an otherwise unidentified book:
Dunhuang of China (2000).
P.
20 is: Hall With Inverted Funnel Shaped Ceiling Shape of Cave 305 (Sui Dynasty)
and gives a good impression of the shape of these caves -- this one is roughly
cubical with several statues, apparently life-size, on a central plinth, with a
decorated ceiling with the Three Rabbits in the centre. The central part of the ceiling usually is a
panel which is sunk into the ceiling, i.e. higher than the rest of the ceiling,
but some have two steps and some are more rounded. [Greeves (2001)] suggests this is to represent a cloth canopy.
Pp.
40-41 is: Pattern (ceiling) Cave 407 (Sui Dynasty), showing the whole ceiling
and the tops of the walls.
Anna Filigenzi, of the Istituto
Universitario Orientale di Napoli and the Italian Archaeological Mission in
Pakistan reports and has sent an image of a three rabbits plaque found at
Bir-kot-ghwandai, Swat, Pakistan and dating from 9C-11C. The publication is: P. Callieri et al.;
Bir-kot-Ghwandai 1990-92 A Preliminary
Report; Supplemento n. 73 of Annali dell'Istituto Universitario Orientale de
Napoli 52:4 (1992) 45 -- ??NYS. [emails
of 30 Oct & 10 Nov 2001.]
In the Museum für Islamische
Kunst, Berlin, is a glass medallion with the Three Hares pattern, attributed to
Afghanistan. It is 52mm in
diameter. The museum purchased it from
a dealer and there is no record of its origin.
Peter Rasmussen & Wei Zhang were shown it in mid 2002 by Dr. Jens
Kröger, the Curator of Islamic Art at the museum. [Email from Rasmussen, 17 Aug 2002.] There is a description of it in Kröger's book and in Stefano
Carboni; Glass from Islamic Lands; Thames & Hudson, 2001, pp. 272-280. This also describes a fragmentary glass
piece with four rabbits in the al-Sabah Collection in Kuwait, supposed to come
from Ghazni, Afghanistan. [Email from
Rasmussen, 20 Aug 2002.]
Eva Wilson. Islamic Patterns (British Museum Pattern
Books). British Museum Publications,
1988. Plate 42, bottom picture, is a
four rabbits pattern. The notes on p.
19 say: Engraved design in the centre
of a brass plate. Diameter 7cm. Iran, 12th century. British Museum, London (1956 7‑26.12).
[Greeves (2000)] reports a light
blue glass seal from Afghanistan, c1200.
In the Hermitage Museum, St.
Petersburg, is an oriental silver flask of 12/13C with the three rabbits on the
base. Discussed and illustrated in
Hamann-MacLean, below. A small
illustration is in Baltrušaitis (Fig. 96D), cf under China, above, who says it
comes from near Perm and has a Kufic inscription on it. Cf next entry.
Vladislav P. Darkevich. Khudozhestvennyi metall Vostoka VII‑XIII. Nauka, Moskva, 1976, 195 pp. ??NYS – described by Peter Rasmussen [email
of 8 Jan 2002] and photocopy of pl. 34 sent by him. See pp. 16, 17, 115 and
Таблуца 34: Серебряные
чаша у флакон
5 - 8: Селянино
Озеро (No. 19) [Tablutsa 34: Serebryanye chasha
u flakon: 5 - 8: Selyanino Ozero (Plate 34: Silver basin and flagon: 5 - 8:
Salt(?) Lake)]. This is the Hermitage
Museum silver flask. Fig. 5 is an
overall view; fig. 6 is the bottom, which is rounded, rather like part of a
sphere and has the three rabbits pattern; figs. 7 & 8 are the Kufic
inscriptions.
[Greeves (2000)] reports a fine
metal tray from Iran, c1200, in the Keir Collection and gives a fine photo of
it. He notes that Sassanian culture is
believed to have spread outward and could have been the source for the
Himalayan and Chinese versions as well as the western versions. Sue Andrew tells me that similar trays are
in the Victoria and Albert Museum, London.
[Greeves (2000)] reports that
the pattern has been found in Nepalese temples and in c1200 wall paintings at
the temple complex of Alchi in Ladakh, Jammu & Kashmir. I can't find Alchi on maps, but a guide book
says it is on the Indus about 1½ hours drive from Leh.
Roger Goepper. Alchi: Buddhas, Göttinnen, Mandalas:
Wandmalerei in einem Himalaya‑Kloster.
DuMont Buchverlag, Köln, 1982, 110 pp.
??NYS -- photocopies sent by Peter Rasmussen. Another version appeared in 1996 -- ??NYS, described by Rasmussen
[email of 21 Dec 2001], see below -- but the earlier book has different
pictures. Rasmussen says this is about
Sumtsek, the Buddhist temple at Alchi, Ladakh, presumably the temple mentioned
by Greeves, above. Goepper dates the
temple as c1200, but another author suggests 11C or 12C [Lionel Fournier; The
Buddhist Paradise: The Murals at Alchi; 1982].
On p. 127 of the 1996 book is a photo of a 4.63 m high statue of
Maitreya (the Buddha) wearing a dhoti on which Rasmussen finds no less than 50
(fifty!) examples of the three hares pattern.
A close-up on p. 128 shows three complete examples. Goepper refers to these examples as 'three
or four deer-like animals, the three long ears being shared ...', but Rasmussen
could not see any examples with four animals and the text only refers to three
ears. However, plate 7 of the 1982 book
shows the same two roundels as on page 128 of the 1996 Alchi, but this
illustration shows FOUR rabbits (sorry, hares) or deer or bulls (take your
pick) in the upper left corner above Maya that were cropped out of the 1996
illustration, so Goepper's reference on page 126 of his 1996 book to
"three or four deer‑like animals" was correct. In a note on p. 278 of the 1996 book,
Goepper says the occurrence of the same pattern at Paderborn is 'hardly
anything more than a coincidence'.
Rasmussen finally notes that Goepper said the photographer, Jaroslav
Poncar, and his group took about 3000 transparencies, a 'virtually complete
documentation of the Alchi murals', but only 300 occur in the 1996 book. I have seen a B&W copy of Plate 7 of the
1982 edition. This has two images, each
of a central character in a roundel surrounded by four frames in the form of a
Greek cross with extra squares at the corners.
(The colour image from the 1993 book, see below, shows these are
adjacent, indeed overlapping, images.)
These frames contain patterns of three and, in one case, four animals,
but the identity of the animals is not clear.
The four-fold pattern seems to be rabbits sharing ears, but others seem
to be horses (or bulls) sharing ears or deer sharing antlers or possibly bulls
sharing horns. All the animals are
going clockwise.
Sue
Andrew has been in contact with Goepper & Poncar. Goepper said "just about a week ago I hinted at the
'intercultural' character of this strange motif during my lectures at Cologne
University".
Pratapaditya Pal. A Buddhist Paradise: The Murals of Alchi
Western Himalayas. Ravi Kumar for
Visual Dharma Publications Ltd., Hong Kong, 1982, 67 pp. ??NYS -- described by Peter Rasmussen [email
of 4 Jan 2002]. Plate S9 shows
Maitreya's full dhoti, S10 and S11 are closeups of the individual legs, and S12
and S13 are closeups of details. Lionel
Fournier's photography is very poor in comparison to Poncar's. On pp. 51‑52, Pal states
"Interspersed with the rondels [sic] are little cruciform blocks adorned
with leaping bulls, whose exact function is not clear, but which remind one of
similar though more naturalistic bulls on the ceilings of Ajanta." So Goepper's deer are Pal's BULLS!!! But the interesting part of the sentence is
the reference to the ceilings of Ajanta.
It's not clear whether Pal means the Ajanta bulls are leaping in threes,
but this statement reminded us that Terese Bartholomew of the San Francisco
Asian Art Museum told us she thought she remembered the three‑animal
motif being somewhere in the Ajanta caves.
In an email of 8 Jan, Rasmussen reports that he has gone through all the
Ajanta books at UC Berkeley and could only find an image of four deer sharing a
single head -- details are in the Dead Dogs section.
Roger Goepper. The 'Great Stupa' at Alchi. Artibus Asiae 53:1/2 (1993) 111-143. ??NYS -- photocopy sent by Peter
Rasmussen. Figure 9: Deities in outer
triangles of the ceiling. Photo by
Poncar. This shows a number of three
and four beast roundels forming a decorative band going out of the
picture. as with the other Alchi
images, it is hard to tell whether the beasts are rabbits, hares, bulls, etc.,
but they are clearly sharing ears here.
all the beasts are going clockwise.
James C. Y. Watt &
Anne E. Wardwell. When Silk Was
Gold Central Asian and Chinese
Textiles. The Metropolitan Museum of
Art in Cooperation with The Cleveland Museum of Art, dist. by Abrams, nd [mid
1990s?, after 1993]. Section 45: Cloth
of gold with rabbit wheels, p. 158, with a colour plate opposite (no number on
my copy). This shows a square array of
circles with four rabbits in each circle.
The direction of the rabbits alternates from one row of circles to the
next. It is from the 'Eastern Iranian
world, second quarter to mid-13th century', i.e. c1240. The cloth has a green-gold colour, while the
patterns are outlined with red silk, giving red-gold lines. A footnote says to consult Roes 1936-37,
pp. 85‑105 for the history of the motif of the animal wheel. Other notes cite Dunhuang, metalwork in
Khurasan (1150-1225), the dhoti of the Maitreya at Sumtsek Temple in Alchi and
the ceiling paintings of the Great Stupa in Alchi. The section goes on to discuss the 'two heads, four boys' motif -
see under Dead Dogs, below.
Sue
Andrew, via Peter Rasmussen [email of 26 Dec 2001], reported finding this, but
only gives Wardwell as author - perhaps she wrote this part, but this is not
indicated on the photocopies sent by Rasmussen. Peter Rasmussen [email of 4 Jan 2002] says Wardwell's main source
was Baltrušaitis, under China, above.
However, Baltrušaitis doesn't mention several of the cited areas.
[Greeves (2001)] reports a
pre-Islamic Mongol coin from north Iran, dated 1281, with the three hares on
one side.
Jurgis Baltrušaitis. Le Moyen Age Fantastique,.... Op. cit. under China, above. Pp. 132‑139 of the 1981 edition have
many examples of three and four rabbits, four boys, etc. After discussing Dunhuang, he says the motif
was taken on by Islam and cites the Petersburg cup. He says there is a Mogul (school of Akbar) miniature with the
pattern.
Roger Goepper. Alchi
Ladakh's Hidden Buddhist Sanctuary.
Serindia Publications, London, nd [1993?]. Colour image sent by Peter Rasmussen showing the two overlapping
images of the dhoti of Maitreya shown in his 1982 book, above. However, this image is more centralised and
hence shows only one of the three animals patterns completely, with parts of
several others.
In Cairo, the Museum of Islamic
art has a fragment from the bottom of a bowl with a three rabbits pattern using
three colours! It is their item 6939/1,
coming from Egypt or Syria in the 12-13C.
In 2001-2002, it was a featured item at the exhibition: L'Orient de Saladin L'Art des Ayyoubides at the Institut du Monde Arabe, Paris. It is reproduced in the catalogue: Éric Delpont et al. L'Orient de Saladin L'Art des Ayyoubides. [Catalogue for] Exposition présentée à
l'institut du monde arabe, Paris du 23
octobre 2001 au 10 mars 2002. Institut
du monde arabe / Éditions Gallimard, Paris, 2001. Item 111, p. 123, is 'Tesson aux trois lièvres'. A black on yellow version was adopted as the
logo of the exhibition and hence appeared on many other items associated with
the exhibition and on the advertisements for it.
In the cloister (Kreuzgang) of
the Cathedral (Dom) of Paderborn, Nordrhein-Westfalen, Germany, is the
"Three Hares Window" (Dreihasenfenster), with hares instead of
rabbits. This faces the outside, i.e.
into the central garden of the cloister.
I learned of this from the Michelin Green Guide - Germany (Michelin et
Cie, Clermont-Ferrand, 1993, p. 229) and wrote a letter of enquiry. A response from Dr. Heribert Schmitz in the
Archbishopric states that the present form of the cloister dates from the early
16C and this is the date given on a postcard he included (and in a local
guidebook). ([Greeves (2000); Greeves
(2001)] says probably 15C.) He also
included a guide to the Cathedral, a poster and a copy of the parish magazine
with the article by Schneider, see below.
On 12 Jul 2000, I was able to visit Paderborn, see the window, meet Dr.
Schmitz and obtain much more material.
The image is actually carved stone tracery in the arch over one of the
triple windows of the cloister, and is about 3 ft (1 m) across. The central stone image is supported only by
the three rear feet of the hares which are on a circular rim -- the
intermediate spaces are filled with leaded glass. The local guidebook refers to the mason as 'crafty', perhaps
implying that he saved having to carve three more ears. Several of the photos show the bodies of the
hares supported by metal rods, but there are presently no rods. Later inquiry revealed that the original
version is now in the Cathedral Museum and the version in the cloister is a
recent copy. The Cathedral guidebook
refers to the 'well known' window and says the symbol is an old land-mark of
the city and the poster describes it as a famous landmark to be studied and
developed in a workshop for children.
One guide book shows three people dressed as hares who are a regular
feature of parties and celebrations.
The Cathedral guidebook says the 'motif is also to be found in other
buildings, but elsewhere is mostly smaller and less conspicuous', but no
references are given and Dr. Schmitz's letter says that he knows of no other
examples than Long Melford and the article by Schneider. Greeves, below, notes that St. Boniface, the
Apostle of the Germans, came from Crediton, Devon, some 10 mi east of Dartmoor! Further, he consecrated a bishop at
Paderborn.
Hans Schneider. Symbolik des Hasenfensters in Nordwestchina
entdeckt. Die Warte (Heimatzeitschrift
für die Kreise Paderborn und Höxter) 32 (Dec 1981) 9. This was kindly provided by Dr. Schmitz of the Archbishopric of
Paderborn. First Schneider gives
various interpretations of the symbolism of the three hares pattern: old German fertility symbols from the myths
of the gods; the Easter rabbit as a
symbol of the eternal power of nature;
a symbol of the Trinity. In
recent years, it has been connected to the patron saint of Paderborn, St.
Liborius, by viewing his name as Leporius, which means 'hare man'. But Schneider has discovered the article of
Huang and gives a B&W reproduction of the picture. Schneider notes that Paderborn had
connections with the Islamic world -- e.g. Achmed el Taruschi and a delegation
came to Paderborn in 970, [and we know Charlemagne's court had contacts with
Constantinople, Córdoba and Baghdad].
Hence it is possible that the Chinese symbol could have been transmitted
to Paderborn [and elsewhere].
A small illustration is in
Baltrušaitis (Fig. 96D), under China, above.
On 12 Jul 2000 I was able to
visit Paderborn and meet Dr. Schmitz.
The way to the Dreihasenfenster is clearly signposted in the Cathedral
and we found images of it elsewhere in the town, and the local guidebook
mentions further locations, e.g. the Drei Hasen restaurant at 55
Königstrasse. I got an English version
of the Cathedral guide and a children's guide to the Cathedral. I obtained two more postcards featuring the
window and several multi-image postcards of Paderborn with the window as one
image. I also bought a stained glass
roundel of the pattern, 225mm (8 3/4") across. The city information office has the pattern on many of their
guide books and I also got stickers, transfers, etc.
I
had met Michael Freude from Münster and he had recalled there was an example in
Münster and that there was a children's rhyme about it, though he could not
remember it, nor could his family. We
stayed with Michael Freude and Hanno Hentrich in Münster and they had located
the example in Münster, which is a roof boss in the southwest corner of the
south transept of the Dom (Cathedral) (St. Paul's), over the organ. It is very high and I was unable to get a
good picture of it. [Greeves (2000)]
notes that it is stunning, but he told me that he also had been unable to get a
good picture. Since the Dom was much
restored after the War, I thought it might be a post-war addition, but Hentrich
checked in a Münster history and photos showed this part of the Dom had
survived. [Greeves (2000)] says it is
early 16C. I told Dr. Schmitz of this
example when I visited him and he did not know of it. Freude & Hentrich thought it likely that the Paderborn
example was a replacement after the War.
[Greeves
(2000)] says there is an example in the cloisters of a former monastery at
Hardehausen, S of Paderborn (but I can't locate this on my maps).
Theodore Fockele & Ewald
Regniet. Domführer für Jungen und
Madchen. Metropolitan-Kapitel,
Paderborn, (1982), 6th ptg, 1999. On p.
21 is a brief description and a drawing with caption being the rhyme: "Der Hasen und der Löffel drei, /
und doch hat jeder Hase zwei."
[The hares and ears are three,
/ and yet each hare has two[,
you see].] This rhyme also occurs in
the local guidebook and on one of the available postcards.
Verkehrsverein Paderborn
[Paderborn Tourist Information].
Paderborn A short guide to the
old city. Paderborn, 1998, p. 13 and
back cover. This says the window is
early 16C and gives an English version of the rhyme: "Count the ears.
There are but three. But still
each hare has two, you see?" and I have now inserted 'you see' into my
translation above. The pattern is
printed on the outside covers of this booklet
One of the postcards available
in Paderborn has a longer poem.
Viribus
Auribusque Unitis (Mit vereinten Kräften und Ohren)
Jedweder
Hase hat zwei Ohren.
Und
hier ging jedem eins verloren.
Das
Soll ist sechs, das Ist nur drei.
Und
Schein und Sein sind zweierlei.
Was
führt der Steinmetz wohl im Schilde?
Welch
ein Gedanke liegt im Bilde?
Die
Ohren sitzen an der Stirne,
Gehörtes
fliess in drei Gehirne.
Drittselbst
wird hier somit bedacht,
was
Sorgen oder Freude macht.
Vereint
geht manches leichtes eben
im
Hasen- wie in Menschenleben.
Und
überdies ist, was ihr seht,
'ne
Spielart von der Trinität.
[With
united powers and ears Every hare
has two ears. And here each has lost
one. There should be six, there are
only three. And appearance and being
are different. How can the stone mason
make an emblem? What thought is in the
picture? The ears sit on the forehead,
which flow into three heads. A third
itself is here thus considered, which makes fear or joy. United, many things go easily even in the
life of hares as in the life of men.
And moreover this is, as you see, a playful image of the Trinity.]
Annemarie Schimmel. The Mystery of Numbers. (As: Das Mysterium der Zahl; Eugen
Diederichs Verlag, Munich, 1984. Based
on: Franz Carl Endres (1878-1954); Das Mysterium der Zahl, last edition of
1951. It's not clear when Schimmel's
work was done -- the Preface is dated Sep 1991 and her © is dated 1993, so
perhaps 1984 refers to the last printing of the original Endres book??) OUP, 1993, p. 63 has a drawing of the
Paderborn three hares, but with no indication of the puzzle aspect. "Hares, symbols of the tri-unity that
is always awake, seeing and hearing everything. Their ears form a triangle.".
There
are 17 churches with 28 roof bosses of the Three Rabbits in Devon. The dating of these is not very exact and is
not always given in the church guides.
I have now discovered Cave's 1948 book which mentions many of
these. See also Jenkins' book of 1999.
Carl Schuster &
Edmund Carpenter. Patterns That
Connect Social Symbolism in Ancient
& Tribal Art. Abrams, NY,
1996. Pp. 158-159, fig. 453, is a
12C European wind chart with a central demon face with four mouths, but it has
four pairs of eyes.
A stone boss supposed to come
from a church demolished c1200 is built into a cellar of a house in Corbigny,
Nièvre, Bourgogne, about 50km NE of Nevers.
[Greeves (2000) with colour photo.]
[Greeves (2001)] changes the name to Corbenay, but this is not in my
French atlas.]
[Greeves (2000); Greeves (2001)]
says the pattern occurs on a bell from the early 13C in the great abbey church
at Kloster Haina. I can't locate this
on my maps, but there is a Haina, Hessen, about 40km SW of Kassel.
Bestiary. MS Bodley 764, 1220-1250. ??NYS.
Translated by Richard Barber as: Bestiary Being an English version of the Bodleian Library, Oxford M.S.
Bodley 764 with all the original
miniatures reproduced in facsimile; Folio Society, 1992; The Boydell Press, Woodbridge, Suffolk, 1993; PB, 1999.
Pp. 66-67, the entry for hare is preceded by a 'four rabbits'
pattern. There were many medieval
versions of the bestiary and the BL MS Harley 4751 is very similar to
this. However, Barber [p. 13] notes
that the entry for the hare is not in earlier texts and rarely reappears in
later texts. Thanks to Sue Andrew for
this reference and a copy of the photo she had done from the original MS.
Sue Andrew also showed me an
illuminated initial Q, also with four rabbits, from a c1285 MS, a French
copy of Peter Lombard's Gloss on the Psalter, Bodleian MS Auct.D.2.8.
f. 115r, commentary on Psalm 51.
A roof boss in the south choir
aisle of Chichester Cathedral, dated to the first half of the 13C, shows six
'Green Men' sharing eyes. The Green Men
have foliage coming out of their mouths.
My thanks to Marianna Clark for noticing this and sending me an example
of the colour postcard of it which is labelled: Six heads with six eyes between
them. Colin Clark, the Chief Guide to
the Cathedral, told me of the example in Boxgrove Priory. See Cave, 1930 & 1948.
A roof boss in Boxgrove Priory
is roughly contemporary with that in Chichester Cathedral, but shows eight
faces sharing eyes. Photo 300 in Cave,
who includes it on a page of Foliate Heads.
On p. 184, he says alternate heads have a stem from the mouth, but this
is very small, leading one to wonder if the foliage has been broken off. [Jenkins, pp. 686-687] says: "The
second boss from the altar is so crafted that each of eight faces comes
complete with two eyes, yet there are only eight eyes in all." See Cave, 1948.
A roof boss in the Chapter House
of a former Benedictine Abbey, now the sacristy of the church of Saints Peter
and Paul, Wissembourg, Bas-Rhin, Alsace, is dated to c1300. Nearby bosses include a Green Man, one of
the common figures in Devon bosses.
[Greeves (2000); Greeves (2001).]
On the right side of the
southern west doorway of St. John's Cathedral (Primatiale St-Jean) in Lyon,
there is a quatrefoil panel with four rabbits, from about 1315. The rabbits are going clockwise. Discussed and illustrated in Hamann-MacLean,
below, and in Baltrušaitis (Fig. 96A), cf under China, above. I photographed this in Lyon and my picture
is much better than that in Hamann-MacLean, possibly because his photograph was
taken before the facade was cleaned in 1982.
However, there is no postcard of the pattern and I could not find a
picture in any of the material available -- but see the next item.
John
Winterbottom & Diana Hall have sent a photo of a similar four rabbits
pattern, but going anti-clockwise, at Bord du Forêt, near Lyon. Here the rabbits are leaping upward, so the
central square of ears is smaller and tilted by about 30o.
N. Reveyron. Primatial Church of Saint-John-the-Baptist,
Cathedral of Lyon. Translated by
Valérie Thollon & Diana Sarran.
Association Lyon Cathédrale, Lyon, nd [c2000]. Pp. 22-29 describe the western doorways, saying the
sculptures were made during 1308‑1332.
The South doorway is described on pp. 28-29, and the third section, on
p. 28, headed Bestiary, includes: "The four hares, arranged like a
swastika." This is the only time I
have seen the four rabbits pattern compared to a swastika and I don't think
there is any similarity!
In the Church of St. John the
Baptist, North Bovey, Devon, there is a carved wood roof boss of the Tinners'
Hares which possibly dates to the 13C.
The leaflet guide to the Church says the pattern was an emblem of the
tin-miners of the 14C and is thought by some to refer to the Trinity. The symbol is used by a number of local
firms as a logo -- though I didn't see any on my visit in 1997. Thanks to Harry James, Churchwarden, for his
letter of 25 Apr 1997, his drawing of the pattern and a copy of the guide.
In the Church of St. Mary the
Virgin, Long Crendon, Buckinghamshire, about halfway between Aylesbury and
Oxford, is a mid-14C tile showing the Three Rabbits pattern. About a third of it is missing. It is just at the altar step, which has
preserved it somewhat from wear. The
church guide says the tile was made in nearby Penn and is unique. This is an encaustic tile with the rabbits
in yellow on a pale orange background.
(This church is generally locked; try telephoning the Vicar on
01844-208363 or the Churchwarden on 01844-208665 if you want to get in. My thanks to Avril Neal, the Churchwarden,
for letting me in.) The pattern is
reproduced in B&W in Carol Belanger
Grafton; Old English Tile Designs; Dover, 1985 [selected from Haberly, cf
below], p. 91, but there is no text or indication of the source of
it. My thanks to Sue Andrew for finding
the tile and the Grafton reproduction.
She has persuaded a potter to make facsimiles of the tile. The potter is Diana Hall, Anne's Cottage,
Wimborne St. Giles, Dorset, BH21 5NG; tel: 01725‑517475. I have now purchased some of these and they
are very fine.
Loyd
Haberly. Mediaeval English
Pavingtiles. Blackwell for Shakespeare
Head Press, Oxford, 1937. Shows the
complete pattern in red and white on the TP.
Shows it in B&W as fig. CXIII on p. 168, saying it occurs in
Long Crendon Church and Notley.
"The design is also found elsewhere in glass and carved on
stone. Some say it symbolizes the
Trinity, others the Trivium, or three Liberal Arts of Grammar, Rhetoric and
Logic. The conies, these say, represent
the scholars, a feeble folk, who have an ear for each of the three Arts. One writer thinks the donor of the this
design was therefore a man of academic distinction." Thanks to John Winterbottom and Diana Hall
for copies of this material.
[Greeves
(2000)] discusses this and says it is the earliest known British example,
presumably because the North Bovey example is not precisely dated. [Greeves (2001)] thinks this is roughly
contemporary with the Chester tile, see below, and another tile in Anglesey.
In the Grosvenor Museum,
Chester, is a floor tile from the Cathedral with an inlaid design of the Three
Rabbits pattern. Sue Andrew has found
that this is shown in Jane A. Wight; Mediaeval Floor Tiles; John Baker, London,
1975 and she has kindly sent a photocopy.
On p. 48 is fig. 15: "Trinity Rabbits: Narrow inlaid design of
linked rabbits, symbolising the Holy Trinity, in Chester Cathedral. (About 5⅜ inches square.)" These are just outline rabbits, looking much
as though Matisse had drawn them. The
outer edges of the ears are curved to produce a circle, so the usual delta
shape is here very curved. On
pp. 12‑13, Wight discusses the pattern.
Some
signs are ambiguous, hovering between religion and magic, like the three
rabbits or hares linked by their shared ears, that may act as a symbol of the
Holy Trinity. In The Leaping Hare
by George Ewart Evans and David Thomson (1972) it is pointed out that these are
correctly hares, 'joined in a kind of animated Catherine-wheel' and 'another
instance of a pre-Christian symbol being adopted by the church'. (On roof bosses in the Dartmoor churches,
financed by money from the stanneries or mines, the creatures appear as the
craft-badge of the tin-miners.) This
'Holy Trinity' is found on tiles in Chester Cathedral and in Buckinghamshire.
It
is clear that Wight and Evans & Thomson have no knowledge of the puzzle
aspect of the pattern. The connection
with the tin miners is now known to be a modern myth -- see Greeves, below. Hares certainly have pre-Christian
associations, but I don't know of any pre-Christian example of the Three
Rabbits pattern except at Dunhuang and possibly St. Petersburg, see above. [Greeves (2000)] mentions this example and
gives more non-Christian examples.
[Greeves (2001)] suggests this is early 14C, roughly contemporary with
the Long Crendon example and gives a photo of the actual tile on p. 62 --
this is slightly damaged in the middle, but does not have the circularity I
noted above which is thus an artist's liberty.
Greeves, Andrew & Chapman recently visited Chester, but the only
tiles shown to them were from excavations in the Cathedral nave in the 1990s,
which makes us wonder if the tile described by Wight is still somewhere in the
Cathedral??
Sue Andrew (Nov 2002) reports that
Diana Hall has learned of a 14C tile from the Stadion's Prebendary Court in
Constance, now in the Zurich Schweizerisches Landesmuseum.
In St. Pancras Church,
Widecombe, Devon, is a roof boss of the three rabbits pattern, probably from
the late 14C. The guide book to the
church says it is "a symbol of the Trinity connected with
tin-mining." This is shown on a
postcard by Judges of Hastings, card number c11933X, where it is called the
Hunt of Venus. A separate guide to the
roof bosses also calls it the Hunt of Venus and suggests the tinners took the
imagery from either the mines being like rabbit burrows, or from Venus, the
goddess of Cyprus, the island which produced the copper that the tin was
combined with. Thanks to the Rector of
the Church, Derek Newport, for the material.
[Jenkins, 1999, p. 142] says: "This rare symbol of the Trinity is
formed of three animal heads sharing just three ears."
At The Great Church of the Holy
Trinity, Long Melford, Suffolk, there is an example of the three rabbits pattern
in the 15C(?) stained glass.
Christopher Sansbury, the Rector, wrote on 3 Jun 1996 that the motif is
common on the east side of Dartmoor and that it may have been brought to
Suffolk by the Martyn family c1500. He
says it is old glass, older than the church, which was built in 1484, but
doesn't specify a date for it, nor does the commercial postcard (Jarrold &
Sons, Norwich, no. CKLMC 6). The
pattern is considered to be an emblem of the Trinity. In a later latter, he cited Chagford, North Bovey and Widecombe
as churches with the pattern near Dartmoor and cites Greeves' article. Greeves, below, says the pattern seems to be
about 5.5cm in diameter, but I wonder if he is measuring a picture as when I
visited the Church, it seemed to be perhaps 5" or 6". Sansbury wrote that it was less than a foot
across. Greeves also says that the
Cornish merchant family of Martin came to Long Melford c1490, so the glass is
more likely to be 16C. [Jenkins, 1999,
pp. 658-659.]
The Church of St. Michael the
Archangel in Chagford, Devon, has a roof boss in the chancel depicting the
Tinners' Rabbits, from the 15C. My
photo is not very clear. The Rector, P.
Louis Baycock, states that there are one or two other bosses in the wooden
ceilings, but they are dark and were obscured by the lighting so I was unable
to locate them. On a later visit, I
managed to see one at the top of the aisle vault a bit to the left of the
entrance door -- one needs to shield one's eyes from the light bulb near
by. There is a kneeler in the church
with the pattern and it figures at the lower right of a large modern embroidery
'Chagford through the ages' hanging at the back of the church (postcard and
explanatory leaflet available). The
leaflet says that a 'rabbit' was a tool used in tin-mining. On a later visit, I found also a pew seat
cover with four examples of the pattern on it.
James Dalgety tells me that there was a gift shop called Tinners'
Rabbits in Chagford some years ago.
[Mary Gray; Devon's Churches; James Pike Ltd, St. Ives, 1974, pp. 7
& 15] only mentions the pattern at Chagford, describing it as 'the old
tinners' design'.
The
local wine shop, called Best Cellars, has the pattern on its window. On my 2001 visit, the owner said the shop
had previously been the Tinners' Rabbits.
The Newsagent's in the main square sells a Chagford plate featuring the
Tinners Rabbits and with the usual mythology on the back. There is a new building, Stannary Place,
with a modern crest of three rabbits over the doorway, on New Street going away
from the Church.
The 15C house of Cardinal
Jouffroy at Luxeuil-les-Bains, Haute-Saône, Franche-Comté, about 40km NW of
Belfort, has the pattern carved under a balcony. [Greeves (2000); Greeves (2001).]
The church of St. Andrew,
Sampford Courtenay, Devon, has two roof bosses in the Three Rabbits
pattern. A letter from D. P. Miles says
these date from c1450. Cf [Jenkins,
1999, pp. 137-138.]
St. Michael's Church, Spreyton,
Devon, has a roof boss of the Three Rabbits pattern. The sign at the entrance to the churchyard is a large and
beautiful version of the pattern painted by Helen Powlesland, the nicest I have
seen. Thanks to Rev. John Withers for
information. Mrs. Powlesland informs me
that the roof boss is 15C. [Greeves
(2000); Greeves (2001)] says the roof is dated 1451. [Greeves (2001)] has a photo.
Spreyton has produced a Millennium cup showing the school and an
inscription surrounded by small squares containing the three rabbits pattern in
white on blue and blue on white -- sent by Helen Powlesland. In 2003, they produced a cup for the Church
with one side having the three rabbits pattern -- kindly sent by Helen
Powlesland. Cave, p. 211, says there
are two bosses, one in the chancel and one in the nave, and that there is an
inscription on the roof of the chancel dated c1450.
The Chapel of Saints Cyr and
Julitta (now called St. Anne and St. Catherine), Cotehele House, Cornwall, was
built in 1485-1489 and has an example of the pattern as a roof boss -- see
[Greeves (2000); Greeves (2001)]. The boss
is on the midline, one in from the E end of the Chapel. A chandelier hangs from it. It is unpainted and so dark that one really
needs binoculars to make out the pattern.
One can get a better view from a squint off the South Room on the first
floor.
In Scarborough, North Yorkshire,
there is a c1350 building in Sandgate (or Sandside) on the seafront of South
Bay called the King Richard III Inn because he is reputed to have stayed
there. On the ceiling of one of the
upper rooms is a 'Three Rabbits' pattern, but this is in the landlord and
landlady's rooms and the landlady was unwilling to let me see it. Inquiry to the Scarborough Museums and
Gallery Officer elicited a photo held by the Planning Department in which the
pattern can just be discerned (though I can't see which way it is going) and
the information that it is in 16C plasterwork apparently done by Italian
workers. [Greeves (2000)] describes
this and notes that Richard III's wife's family (the Nevills) owned the manor
of North Bovey in the early 16C.
Greeves, Andrew & Chapman have been to see and photograph it -- the
pattern has been painted red.
[Greeves (2000)] reports two
occurrences of plaster ceilings with the pattern in Devon. A 16C example at Treasbeare Farm, near Clyst
Honiton, and a mid-17C(?) example at Upcott Barton, Cheriton Fitzpaine. [Greeves (2001)] says the Clyst Honiton
example is 17C and gives a fine photo of it.
Baltrušaitis. Op. cit. below. Fig. 97 is a 1576 Dutch engraving of three rabbits.
In Throwleigh, Devon, a few miles
from Chagford, there is a roof boss of the Tinners' Rabbits in the 16C north
aisle of the Church of St. Mary the Virgin.
(Thanks to the Rector of Chagford and Throwleigh, P. Louis Baycock, for
directing me to this site.) Photo in
Sale, below, p. 63.
[Greeves (2000)] has a picture
of the example in the parish church, Tavistock, Devon, but only dated as
medieval.
[Greeves (2000)] reports there
is a fine stone lintel from Charmois-L'Orgueilleux (I can't locate this) in the
Musée d'Art Ancien et Contemporain in Epinal, Vosges, Lorraine, dated as 16C,
but Greeves thinks the carving may be more primitive than 16C. He also reports two medieval stone roof
bosses at Ingwiller, Bas-Rhin, Alsace, about 35km NW of Strasbourg, and
Xertigny, Vosges, Lorraine, about 15km S of Epinal. Greeves says there are other French
examples: in the chapel of the Hotel de Cluny (= Cluny Museum), Paris
([Greeves (2001)] says this is 15C; I have a good photo); at Vienne, Isère, Rhône-Alpes; at St. Bonnet le Chateau. Loire, Rhône-Alpes; and that potters at Soufflenheim, Bas-Rhin,
Alsace, use the pattern in current production.
In 1997, an old trunk and crates
in the Statens Museum for Kunst (National Gallery), Copenhagen were opened and
found to contain over 20,000 prints which had been stored during
re-organization in the 1830s. One of
these is a three rabbits engraving, very similar to the 1576 Dutch engraving
shown in Baltrušaitis, below. Here the
rabbits are going clockwise. This
material is the basis of an exhibition continuing until 16 Feb 2003. This picture is being used to advertise the
exhibition and appears in Copenhagen This Week for Jan 2003 with a short
English text, on www.ctw.dk/Sider/Articles.html
. The museum's site is www.smk.dk , but I cannot see an English
catalogue available. Information and
photo from Diana Hall and John Winterbottom -- John's son saw it in Copenhagen
on a stopover at the airport.
The Church of St. John the
Baptist, Broadclyst, Devon, has nine roof bosses of the Three Rabbits pattern. (Greeves (1991) erroneously has eight.) They were made in 1833 but are said to be
careful plaster copies of the medieval examples, but my sources give no date
for the originals. There is one three
rabbits boss in the central aisle -- Chris Chapman thinks some of the bosses in
the central aisle may be originals, but this one has been recently painted and
doesn't look old to me.
[Greeves (2000); Greeves (2001)]
mentions two early 17C plaster ceilings with the pattern at Burg Breuberg in
the Odenwald (I can't locate this on my maps, but the Odenwald is where Hessen,
Baden-Württemberg and Bayern meet)
and at Seligenstadt, Hessen,
about 20km SE of Frankfurt.
Basil Valentine. (He may be catalogued as Basilius Valentinus
(or Valentis) and entered under B rather than
V.) De Macrocosmo, oder von der
grossen Heimlichkeit der Welt und ihrer Artzney, dem Menschen zugehorig. c1600.
??NYS -- reproduced and briefly discussed in Greeves. This shows the Hunt of Venus, with three
hares going clockwise, but with each hare pursued by an unconnected dog and
Greeves notes that the dogs are an essential part of a hunt symbol. Inside the triangle of ears is the
astrological/alchemical symbol for Mercury, which is claimed to be similar to a
symbol used for tin (or Jupiter) -- but I have now looked at a book on alchemy
and there is no similarity; further both symbols are included in the
surrounding text and they are clearly distinct. The pattern is drawn inside a circle of text which contains the
astrological symbols of all seven planets and hence is rather cryptic -- see
Roob, below, for the text. The top of
the picture has a flaming heart pierced by an arrow and the legend VENUS. So there is no real connection with tin,
though this is the source cited by the first book on Dartmoor to mention the
symbol, in 1856. Greeves notes that the
three ears produce a delta-shape and this has connotations of fertility, both
as the Nile delta and as the female pubic triangle.
Valentine
is a (semi-?) mythical character. He
was allegedly a Benedictine monk of the early 15C, but no trace of his writings
occurs before the 17C. (However, de
Rola (below) asserts that Antonius Guainerius (d. 1440) praised Valentine and
that Valentine himself says he was a Benedictine monk at the monastery of St.
Peter in Erfurt. But his real name is
unknown and so he cannot be traced in the records of the monastery or at
Erfurt. De Rola quotes a 1675 report
that Valentine was at St. Peter's in 1413.)
Legend says his works were discovered when a pillar in the Cathedral of
Erfurt split open (a variant of a story often used to give works a spurious
age) -- cf the next item. Despite their
uncertain origin, the works were well received and remained popular for about
200 years, with the pictures of his 'Die zwölff Schlüssel' (The Twelve Keys)
being used to the present day. He was
probably an alias of Johann Thölde (fl. 1595-1625). The work cited may occur in his Last Will
and Testament (1st English ed of 1657) or in his Chymische Schriften (Gottfried
Liebezeite, Hamburg, 1700).
Basilius Valentinus. The Last Will and Testament of Basil
Valentine, Monke of the Order of St. Bennet ... To which is added Two
Treatises: .... Never before Published
in English. Edward Brewster, London,
1671. ??NYS -- seen in a bookdealer's
catalogue, 2003.
Basilius Valentinus. Letztes Testament / Darinnen die Geheime
Bücher vom groffen Stein der uralten Weifen / und anderen verborgenen
Geheimnüssen der Natur Auss dem
Original, so zu Erfurt in hohen Altar / unter einem Marmorsteineren Täflein
gefunden worden / nachgeschrieben: Und
nunmehr auf vielfältiges Begehren / denen Filiis Doctrinæ zu gutem / neben
angehengten XII. Schlüsseln / und in Kupffer gebrachten Figuren ie. deffen Innhalt
nach der Vorrede zu sehen / zum vierdtenmahl ans Liecht gegehen / deme
angehänget ein Tractätlein von der ALCHIMIE, Worinnen von derselben Usprung /
Fortgang und besten Scriptoribus gehandelt / auff alle Einwürffe der
Adversariorum geantwortet / und klar bewiesen wird / dass warhafftig durch die
Alchimie der rechte Lapis Philosophorum als eine Universal Medicin Könne
bereitet werden / von Georg Philips Nenter.
Johann Reinhold Dulssecker, Strasburg, 1712. This has several parts with separate pagination. About halfway through is a new book with TP
starting: Von dem Grossen Stein der
Uhralten / Daran so viel tausend Meister anfangs der Welt hero gemacht
haben. Neben angehängten Tractätlein /
derer Inhalt nach der Vorrede zu finden.
Den Filis doctrinæ zu gutem publicirt / und jetzo von neuen mit seinen
zugehörigen Figuren in Kupffer and Leight gebracht. Strassburg / Im Jahr M.
DCC. XI. Part V. of this is: Von der
grossen Heimligkeit der Welt / und ihrer Artzney / dem Mensche zugehörig --
from Greeves, it appears this is also called Macrocosmo. On p. 140 is a smaller and simpler
version of the three hares picture, with 'folio 222' on it. This has the hares going anticlockwise, the
only example of Basilius' picture with this feature that I have seen. Further, the word 'recht' is missing from
the text around the picture and has been written in. Below the picture is a lengthy poem, starting "Ein
Venus-Jagt ist angestallt". I
translate the first part as: "A
Hunt of Venus is prepared. The hound
catches, so the hare will not now grow old.
I say this truly that Mercury will protect well when Venus begins to
rage, so there occur fearfully many hares.
Then Mars guards with your [sic, but must be his] sword, so that Venus does not become a whore."
Part
II. of Von dem Grossen Stein der Uhralten is
Die zwölff Schlüssel ... and Der neundte Schlüssel is on p. 51,
with 'folio 70' on it. In the centre of
the circular part of this is a pattern of three hearts with serpents growing
out of them and biting the next heart.
Above this are a man and a woman, each nude and in a sitting position,
with bottoms almost touching.
Baltrušaitis mentions this in his discussion of 'two heads, four bodies'
pictures, but I don't feel this is really a picture of that type. A different and less clear version of this
picture in reproduced in the following.
Stanislas
Klossowski de Rola. The Golden
Game Alchemical Engravings of the
Seventeenth Century. Thames &
Hudson, 1998. This reproduces the
entire Twelve Keys from: Michael Maier;
Tripus aureus; Lucas Jennis, Frankfurt, 1618, and gives some discussion. P. 119 is the title page of the Valentine;
p. 103 includes the Ninth Key; pp. 125-126 give explications of the Keys.
Roob
(below), p. 678, from: D. Stolcius v.
Stolcenberg; Viridarium chymicum; Frankfurt, 1624.
Adam
McLean. The Silent Language The Symbols of Hermetic Philosophy Exhibition in the Bibliotheca Philosophica
Hermetica. In de Pelikaan, Amsterdam,
1994, p. 47, reproduced from: Johann Grasshoff; Dyas chymica tripartita;
Lucas Jennis, Frankfurt, 1625.
Baltrušaitis,
above, Fig. 103, but he cites: B.
Valentino; Les Sept Clefs de
Sagesse; vers 1413; via an 1891 book.
This
three heart/serpent pattern is known as the Ouroboros or Ouroborus. McLean explains that man has three hearts:
physical, soul and spiritual.
Basilius Valentius. Chymische Schriften. Hamburg, 1717. ??NYS -- reproduced in:
Alexander Roob; Alchemie & Mystik
Das Hermetische Museum; Taschen, Köln, et al., 1996, p. 676. Same picture as in De Macrocosmo, with some
descriptive text. The circle of text is
translated as: "Sol and Luna and Mars mit Jupiter jagen / Saturnus muss
die Garne trage / Steltt Mercurius recht nach dem Wind / so wird Frau Venus
Kind" (Sun and Moon and Mars hunt
with Jupiter / Saturn must carry the net (or decoy) / Place Mercury correctly
according to the Wind / then Frau Venus captures a child) and Roob says this is an alchemical
description of the preparation of the 'Universal Medicine' from copper vitriol,
which Basilius called the highest of all salts. Externally it is green, but inside it is fiery red from its
father Mars, an oily balsam. He then
gives lines 5-8 of the poem and says the hare is a known symbol for the
fleetness of Mercury.
At Schwäbisch Hall,
Baden-Württemberg, about 60km NE of Stuttgart, the pattern appears painted on
the ceiling of a synagogue, from 1738/9 [Greeves (2000); Greeves (2001), with
picture.] Sol Golomb notes that the
commandment against making graven images is interpreted as referring to solid
images and excepts decorative patterns even if they use animals.
The town of Hasloch (am Main),
Bayern, about 70km SE of Frankfurt, uses the pattern as its town crest (colour
photo of an 1842 example in [Greeves (2000)]).
C. J. P. Cave. The roof bosses in Chichester
Cathedral. Sussex Archaeology 71 (1930)
1-9 & plate opp. p. 1. Photo
11 is a B&W picture, but not as sharp or clear as the current postcard. Discussion on p. 6 says the Boxgrove example
is 'more boldly and better carved'. See
Cave, 1948.
Jurgis Baltrušaitis. Le Moyen Age Fantastique,.... Op. cit. under China, above. Pp. 132‑139 of the 1981 edition have
many examples of three and four rabbits, four boys, etc. Fig. 96 gives small rabbit
illustrations from Dunhuang (10th c.), Islamic vase from the Hermitage Museum
(12th‑13th c.), Lyon Cathedral (1310‑20), Paderborn (15th c.), and
Fig 97 is a large Dutch engraving, Lièvres a Oreilles Communes, of 1576 in the
BN, Paris, with rabbits going anti-clockwise.
He cites examples at Saint-Maurice de Vienne (15C); at the Hôtel de
Cluny, at Saint-Benoit-le-Château (Loire) and says it is frequent in the east
of France (Thiélouse, Xertigny), Switzerland (Abbey of Muototal), and in
Germany (Munster, Paderborn). He cites
a 1928 book for the pattern being a symbol of the Trinity. By the beginning of the 16C, he says it was
used as a vignette by the printer Jacques Arnollet and as the sign of the Three
Rabbits Inn (L'Hostellerie aux trois lapins).
He gives some lines from a poem of the time:
Tournez
et retournez et nous tournerons aussi,
Afin
qu'a chacun de vous nous donnions du plaisir.
Et
lorsque nous aurez tournés faites compte de nos oreilles,
C'est
là que, sans rien déguiser, vous trouverez une merveille.
and
says these or similar occur frequently in prints of the 16C and 17C, referring
to the large Dutch engraving which has similar lines in Dutch and French around
the border - but his reproduction (or the original) is truncated on the
left. He cites an 1879 history of inn
signs and seems to say this inn is in Lyon and that the author had also seen a
version with three deer.
See
Dead Dogs for further material.
C. J. P. Cave. Roof Bosses in Medieval Churches An Aspect of Gothic Sculpture. CUP, 1948.
This has several hundred B&W plates. I have a reference to Cave's collection of photographs of
roof-bosses and an index thereof at the Society of Antiquaries, so I looked up
Cave at the Warburg and found this book.
Chapter
VI: Beasts, birds and fish, pp. 69-75.
On p. 71, he says: "There
are a few rabbits or hares, it is difficult to tell which; the most curious are
the three rabbits with only three ears between them, each rabbit sharing an ear
with its neighbours; this device is found at Broadclyst, North Bovey (49),
Chagford, Sampford Courtenay (182), Spreyton, South Tawton, Tavistock, and
Widecombe-in-the-Moor, all villages on or not far from Dartmoor; at Selby there
is a similar arrangement, but there is a fourth rabbit unconnected with the
three.1
1 The three rabbits occur in stained glass at Long Melford. The same motif occurs in Paderborn Cathedral
in Germany, on a window known as the Hare Window; see Walter Hotz, Mittelalterliche
Groteskplastic (Leipsig), p. 47 (text) and p. 68 (plate); this window is
even mentioned in Baedeker's Guide to North Germany."
Plate
1. Tickencote. This has three heads in a circle, but there
is no interconnection.
Plate
30. Bristol Cathedral, north transept:
Fish. This has three fish arranged in a
triangular pattern, each overlapping the next.
Plate
49. North Bovey; rabbits.
Plate
182. Sampford Courtenay. Rabbits.
Plate
200. Selby. Triple face. [This has
three faces looking forward, a bit to the right and a bit to the left, sharing
four eyes. I have a photo of a similar
keystone in Citta di Castello where the side faces are facing directly right
and left. This seems to be a
development from the Roman double-sided heads of Janus.]
Plate
300. Foliate Heads. Boxgrove.
Appendix
I List of churches containing roof
bosses. Pp. 181-222. He begins with a note: This list does not
aim at completeness, .... Names of
places which I have not visited personally are in square brackets.] He lists 208 sites, including: Bovey (North); Boxgrove Priory;
Broadclyst ('evidently modern copies of old designs', but he doesn't say how
many there are); Chagford ('two bosses ..., one much restored'); Cheriton
Bishop; Chichester Cathedral (cites his: The roof bosses in Chichester
Cathedral, Sussex Archaeological Collections, vol. LXXI, 1940, above);
Havant ('Two early thirteenth-century bosses in the chancel with affinities to
those as Chichester and Boxgrove.' ??); Oxford, Christ Church ('four lion's
bodies joined to one head'); Portsmouth Cathedral ('Two bosses in the north
aisle of the quire in the Chichester-Boxgrove style.' ??); Portsmouth, SS John and Nicholas (Garrison Chapel) ('The
first and third bosses are early conventional foliage in the
Chichester-Boxgrove style.' ??);
Sampford Courtenay (cites two examples, in the chancel and the nave); Selby
Abbey (says many bosses are medieval, but says nothing about the Triple face
(plate 200)); Spreyton (cites two examples, in the chancel and the nave);
Tavistock; Tawton (South); Widecombe-in-the-Moor. The nine three rabbits examples were known, but Chichester and
Boxgrove are new and there are a number of possibles that seem worth
investigating, but Plates 1 and 30, now added above, are the only previously
unnoted pictures of marginal interest.
Richard Hamann-MacLean. Künstlerlaunen im Mittelalter. IN: Friedrich Möbius & Ernst Schubert,
eds.; Skulptur des Mittelalters; Hermann Böhlaus Nachfolger, Weimar, 1987, pp.
385-452. The material of interest is on
pp. 400-403. He discusses and illustrates: the Paderborn example, saying it is
15C; the silver flask in the Hermitage; the four rabbit pattern at Lyon. He notes that the last cannot be considered
a symbol of the Trinity! He mentions a
cave painting in Chinese Turkestan, 10C?, which is probably the Dunhuang
example. He gives a number of
references to earlier articles -- ??NYS.
Tom Greeves. The Tinners' Rabbits -- chasing hares?. Dartmoor Magazine 25 (Winter 1991) 4-6. This was certainly the most informative discussion
of the topic until his later articles.
Greeves is an archaeologist and an authority on the Dartmoor tin
industry. He has now been joined in
this research by Sue Andrew and the photographer Chris Chapman, comprising the
Three Hares Project. Greeves says that
it is claimed to be the emblem of the medieval tinners, and various connections
between tinners and rabbits have been adduced, e.g. it is claimed that the
pattern was the medieval alchemical symbol for tin. It is also called the Hunt of Venus and/or an emblem of the
Trinity. However, the earliest
reference to the pattern on Dartmoor is an 1856 description of Widecombe Church
which only says that the roof was connected with the tinners and that the
pattern had an alchemical connection.
Later guides to Dartmoor are still pretty vague, e.g. a 1956 writer connects
the symbol with copper, not tin. It is
not until 1965 that the symbol is specifically called The Tinners'
Rabbits. See his 2000 article for more
early references.
There
is no particular Dartmoor mythology connected with rabbits, but there is much
mythology of hares. See the note at the
beginning of this section about the introduction of rabbits to Britain. Three unconnected rabbits do occur in some
English crests. Greeves reproduces and
discusses the c1600 Valentine picture -- see above.
Since
no list of occurrences of the pattern had ever been compiled, Greeves examined
almost all the churches in the area and discovered roof bosses with the pattern
in 12 churches -- see [Greeves (2000)] for a longer list). These are all on the east side of Dartmoor
or to the north, except Tavistock is on the west side and Broadclyst is some 20
mi further east. Bridford, Iddesleigh,
Sampford Courtenay and Spreyton have no significant tin-mining
connections. No examples are known from
the much more important tin-mining area of Cornwall, but Greeves has since
found an example at Cotehele, just over the border into Cornwall.
Greeves
then discusses the examples at Long Melford and Paderborn, giving comments
which are mentioned above. He then
briefly describes the Dunhuang example, citing Whitfield & Farrar, and then
the St. Petersburg example. He then
notes some modern versions: a wooden teapot stand from Scandinavia and a
Victorian(?) carving in Holy Trinity, Fareham, Hampshire.
Early
Christianity took over many pagan symbols and the three hares or rabbits (like
the Green Men) could have been adapted to represent the Trinity.
In
a letter of 3 Jun 1997, Greeves says he has located further examples of the
three rabbits in Cheriton Bishop and Paignton in Devon, Cotehele in Cornwall
and in Wales, Scarborough (North Yorkshire), France, Germany, Switzerland,
Bohemia and modern China, where the pattern is still woven into carpets. See [Greeves (2000); Greeves (2001)] for
more details of these.
See
below for the continuation articles [Greeves (2000); Greeves (2001)].
Paul Hambling. The Dartmoor Stannaries. Orchard Publications, Newton Abbott, 1995,
pp. 38-39. This gives a short
summary of Greeves' work. He adds that
a story is that the tinners adopted the rabbit as their emblem in allusion to
their common underground mode of life.
Tinners are also said to have been responsible for some rabbit warrens,
but there were lots of other warrens and they would have been too common to be
specifically associated with the tinners.
He notes that the symbol of three intertwined fishes was a common
Christian symbol.
Simon Jenkins. England's Thousand Best Churches. Allen Lane, 1999; slightly revised, Penguin, 2000.
This mentions the Three Rabbits pattern in several churches.
Boxgrove,
pp. 686-687. "The second boss from
the altar is so crafted that each of eight faces comes complete with two eyes,
yet there are only eight eyes in all."
Long
Melford, pp. 658-659. "... three
rabbits sharing three ears, representing the Trinity."
Sampford
Courtenay, pp. 136-137.
Widecombe
in the Moor, pp. 142. "This rare
symbol of the Trinity is formed of three animal heads sharing just three
ears."
Richard Sale. Dartmoor
The Official National Park Guide.
Photos by Chris Chapman.
Pevensey Guides (David & Charles), 2000. Pp. 63-64 says the symbol of 'three rabbits each with only one
ear' was adopted by the tinners and may be an allusion to the Holy
Trinity. He also says the rabbits
became a pest and myxomatosis was introduced in 1954 and rabbit warrening was
abolished in 1956. P. 63 has a photo of
the boss at Throwleigh.
Tom Greeves. Chasing three hares. Dartmoor Magazine No. 61 (Winter 2000)
8-10. P. 11 contains several
advertisements for hotels and tours featuring the three rabbits motif. This reports on information discovered since
his previous article -- see above. Many
bits of information are incorporated above, citing this as Greeves (2000).
The
earliest known connection of the pattern with the tinners is given in the
Torquay Directory (25 Nov 1925), but the first popular usage seems to be Sylvia
Sayers' The Outline of Dartmoor's Story (1925), p. 24.
He
now has found 17 Devon churches with 28 examples of the pattern: Ashreigney, Bridford, Broadclyst (9 examples
from 1833 said to be careful copies of medieval bosses -- Greeves (1991)
erroneously has 8), Chagford (2 examples), Cheriton Bishop, Iddesleigh,
Ilsington, Kelly (2 examples, one a modern copy), Newton St. Cyres,
North Bovey, Paignton, Sampford Courtenay (2 examples), South Tawton,
Spreyton (2 examples), Tavistock, Throwleigh,
Widecombe(-in-the-Moor). Excepting the
recent copies, there are 19 medieval (i.e. pre-1500) examples, all wooden roof
bosses. These stretch well beyond the
Dartmoor area into mid and east Devon.
Many of these sites have no significant tin-mining connections. No examples are known from the much more
important tin-mining area of Cornwall, except an example at Cotehele, Cornwall,
just over the Cornish border. There are
also examples, probably 15C, on the roof of the Lady Chapel in St. David's
Cathedral, St. David's, Pembrokeshire,
and on an arch of St. Aidan's
church, near Llawhaden, Dyfed.
Greeves
writes that his group are organising an exhibition on the pattern at the High
Moorland Visitor Centre, Princetown, Devon, for several weeks from 22 Nov 2001.
Tom Greeves. Three hares -- a Medieval Mongol
Mystery. Devon Today (Apr 2001)
58-63. Many bits of information are
incorporated above, citing this as Greeves (2001). Notes that Easter is believed to derive from the festival of the
pagan goddess Eastre, whose familiar spirit was the hare. Reports a possibly 15C example at Corfe
Mullen, Dorset. Says the pattern was
used in 18C Slovakian pottery. Gives
photos of examples at Paignton and South Tawton and of a modern carpet from Urumqi.
MODERN VERSIONS OF THE THREE RABBITS PUZZLE
Child. Girl's Own Book. Puzzle
10. 1833: 163; 1839: 143;
1842: 264; 1876: 221. "Can you draw three rabbits, so that
they will have but three ears between them; yet each will appear to have the two
that belongs to it?" (1839, 1842
and 1876 have belong instead of
belongs.)
Magician's Own Book. 1857.
Prob. 7: The three rabbits, pp. 269 & 293. "Draw three rabbits, so that each shall appear to have two
ears, while, in fact, they have only three ears between them." The drawing is similar to, but reasonably
different than that in Girl's Own Book.
Book of 500 Puzzles. 1859.
Prob. 7: The three rabbits, pp. 83 & 107. Identical to Magician's Own Book.
Illustrated Boy's Own
Treasury. 1860. Practical Puzzles, No. 3, pp. 395 &
436. Identical to Magician's Own Book,
prob. 7.
Boy's Own Conjuring Book. 1860.
Prob. 6: The three rabbits, pp. 230 & 255. Identical to Magician's Own Book.
The Drei Hasen hotel,
Michelstadt, Hessen, about 40km NE of Heidelberg, uses 19C and later versions
of the pattern. [Greeves (2000);
Greeves (2001) has a photo of a 19C stained glass window.]
Hanky Panky. 1872.
P. 87: The one-eared hares. Very
similar to Magician's Own Book.
Wehman. New Book of 200 Puzzles. 1908.
The three rabbits, p. 21. c=
Magician's Own Book.
Collins. Book of Puzzles. 1927. The Manx rabbit
puzzle, p. 153. Says it was invented by
a Manxman. Shows three rabbits, each
with two ears, and one has to assemble them to have just three ears.
The Warren House Inn is at one
of the highest passes over Dartmoor, Devon, on the B3212 about halfway between
Princetown and Moretonhampstead. The
pub sign shows the Three Rabbits and they sell a polo shirt with the
pattern. [Thanks to Tom Greeves for directing
me to this.]
Marjorie Newman. The Christmas Puzzle Book. Hippo (Scholastic Publications, London,
1990. Kangaroos' ears, pp. 69 &
126. Like Magician's Own Book, prob. 7,
but with kangaroos.
The Castle Inn, Lydford, Devon,
has a fine stained glass window of the Three Rabbits by James Paterson
(1915-1986). ??NYS -- described and
illustrated in colour in Greeves (2000).
Jan Misspent (??sp). Design Sources for Symbolism. Batsford, 1993, p. 18. Shows the three rabbits, going
anticlockwise, among other examples of three-fold rotational symmetry. Sent by Diana Hall.
Tom Greeves (see above) uses a
three rabbits logo as his letterhead.
Holy Trinity Church, Long
Melford, Suffolk, uses a version of their stained glass as a letterhead.
Trinity Construction Services,
London and Essex, uses a three rabbits logo as their letterhead. A former director saw the pattern in Devon
and liked it.
Laurie Brokenshire reports that
the chaplain at HMS Raleigh, the naval training station near Plymouth,
has a vestment with the three rabbits emblem.
He saw it many years ago in the area and thought it was an excellent
symbol of the Trinity and had the vestment made, probably for use on Trinity
Sunday.
Martin and Philip Webb run a
company called Fine Stone Miniatures (www.finestoneminiatures.com) which makes
miniatures of medieval beasties from cathedrals, etc. They have recently introduced two versions of the three rabbits.
G. Yazdani. Ajanta
Monochrome Reproductions of the Ajanta Frescoes Based on Photography Part I.
1930; reprinted by Swati Publications, Delhi, 1983. Photocopies sent by Peter Rasmussen. P. 3 has the following. "A good example of the artistic fancy
of the sculptors of Ajanta is the delineation of four deer on the capital of a
column in this cave (Plate XLb). They
have been so carved that the one head serves for the body of any of the
four. The poses of the bodies are most
graceful and absolutely realistic, showing close study of nature combined with
high artistic skill." He dates the
cave to the end of the 5C. Footnote 1
says the motif of the four deer also occurs in a cave at Ghatotkach, which have
an inscription dating them to the end of the 5C. Plate IV is a general view of Cave I, but I cannot recognise the
image in it. Plate XLb is "Four
deer with a common head" but very unclear.
Narayan Sanyal. Immortal Ajanta. Hrishikes Barik, Calcutta, 1984.
Photocopies sent by Peter Rasmussen.
P. 18, fig. 4.1, is a Plan of Cave I with Exhibits. P. 19 has a List of Exhibits in Cave I --
entry 17 is "Four deer with one head" and notes that it is shown in
Yazdani. On the plan, 17 points to the
two middle columns on the right side.
P. 38 describes this: "On another [pillar-capital] there are four
deer with a common head. The local
guide would often invite the attention of the tourist to this and similar
freaks. But the magnanimity of Ajanta
lies not in such frivolities ...."
Benoy K. Behl, text and
photographs. The Ajanta Caves Artistic Wonder of Ancient Buddhist
India. Abrams, NY, 1998, 255pp. [= Benoy K. Behl, text and photographs. The Ajanta Caves: Ancient Paintings of
Buddhist India. Thames and Hudson,
London, 1998, 256pp.] P. 18 has a
colour photo of a relief of four deer
sharing a single head. Peter Rasmussen
sent a B&W copy, which is not good, and then an enlarged colour picture
which does show the effect, but it is still not very good. Behl says " Inside the main hall on the
right-hand side one of the capitals is subtly carved to create the illusion of
four recumbent and standing deer sharing a single head." This is apparently the only example at
Ajanta as other authors refer to it as 'artistic fancy', 'freaks',
'frivolities'. The Ajanta Caves date
from -2C to 6C.
Carl Schuster &
Edmund Carpenter. Patterns That
Connect Social Symbolism in Ancient
& Tribal Art. Abrams, NY,
1996. P. 34, fig. 68, shows a painted
pottery pattern from Panama with two heads and four legs, but rather more like
Siamese twins than our 'two heads, four boys' pattern.
The Peterborough Psalter,
Brussels. c1310. At the bottom of f. 48v, Psalm 68, is a
somewhat crude example of 'two heads, four horses', where the two vertical
horses have their front and back hooves touching so they are very curved while
the horizontal horses have bellies on the ground. B&W in: Lucy Freeman
Sandler; The Peterborough Psalter in Brussels and Other Fenland Manuscripts;
Harvey Miller, London, 1974, plate 45, p. 28 (see p. 9 for the date of the
MS). Also in Baltrušaitis. below.
Anna Roes. "Tierwirbel" in JPEK: Jahrbuch für
prähistorische und ethnographische Kunst: Jahrgang 1936/37. De Gruyter, Berlin, 1937, 182 pp. ??NYS -- described by Peter Rasmussen [email
of 8 Jan 2002]. Pp. 85‑105 gives
a history of the animal wheel motif, with many illustrations, but none with
three hares.
Jurgis Baltrušaitis. Le Moyen Age Fantastique,.... Op. cit. under China, above. Pp. 132‑139 of the 1981 edition have
many examples of three and four rabbits, four boys, etc.
He
discusses the animal wheel motif where several animals share the same
head. This sort of image is rather more
common than either the Three Rabbits or the Dead Dogs type of image and I have
not tried to chronicle it. But I will include
a few early examples. Baltrušaitis
shows (Fig. 98) several examples of three fish with one head: from Egypt (XVIII-XX dynasty, i.e. c‑1500); by Villard de Honnecourt, c1235; a pavement at Hérivaux, 13C; an Arabised plate from Paterna, 13C-14C. He cites examples in Italian and French
ceramics.
He
then goes on to 'two heads, four animals' patterns, showing (Fig. 99) a version
of 'two heads, four horses' by the Safavid artist Riza Abbasi, signed and dated
20 Oct 1616, though the authenticity of the signature and date have been
disputed. He notes that another version
is in the Boston Museum of Fine Arts (cf below) and there are definite
differences in the two versions.
He
then shows (Fig. 99) the example of 'two heads, four horses' in the
Peterborough Psalter. His note 160
cites J. van den Gheyn, op. cit., pl. xxi.
Considerable searching finds the citation as part of note 68: J. van den
Gheyn; Le Psautier de Peterborough; Haarlem, nd, ??NYS. There are a number of Peterborough Psalters,
including one in the Society of Antiquaries in London, one now MS. 12 in the
Fitzwilliam Museum, Cambridge, and the famous example in Brussels (which seems
to be the one studied by van den Gheyn).
Baltrušaitis
then shows (Fig. 100) an example of 'two heads, four men' from the Library
Portal of the Cathedral of Rouen, 1290-1300.
This is a relief, nicely enclosed in a quatrefoil frame. In his text, the author says the pattern
also occurs on a misericord in the Cathedral and on the Palace of Justice in
Rouen. These are now the earliest known
examples of this idea. Baltrušaitis
cites an expert on Iranian art who says the tradition is very old in Iran and
could well have inspired these examples.
Baltrušaitis says the pattern also occurs: in the Church at Rosny, Aube (15C); on a cathedral stall at Vendôme;
at New College, Oxford.
He
then shows the Two Apes on Horseback (Fig. 101), apparently the form 1985n of
Schreiber. He says several engravings
of the 15C show these musical apes with the movable centre and his note cites
similar examples with 'amours' and men by Cornelius Reen (1560) and Adrien
Herbert (1576) and very frequently in the 17C.
He then has a photo (Fig. 102) of a oriental scrollweight, and says it
undoubtedly comes from the same Islamic source as the other compositions with
interchangeable elements. He then shows
the Ninth Key of Basil Valentine, cf under Medieval Europe, above, but I don't
feel it is a picture of the present type.
He then reproduces (Fig. 104) Reen's picture, Symbol d'un Amour
Inconstant, from the BN, dated as c1561, with two cherubs. And then he gives (Fig. 105) a 'two heads,
four men' picture with a drinking man and a harlequin horn-player. This is another 1576 Dutch engraving in the
BN, in a circular frame with French and Dutch text.
Zwei Affen als Kunstreiter
(Verwandlungsbild). This has two apes
on a horse with a small bit of paper on a pivot which shifts the mid-body
connections so either head is connected to either legs. Two versions described in W. L. Schreiber;
Handbuch der Holz- und Metallschnitte des XV.Jahrhunderts. Band IV: Holzschnitte, Nr. 1783-2047. Verlag Karl W. Hiersemann, Leipzig,
1927. P. 120. Schreiber pasted up examples of all the woodcuts and metalcuts
described in his Handbuch in four mammoth volumes. This unique set was presented to the Warburg Institute, where
they are RR 240 1-4. The following
plates are in plate vol. 3. Schreiber
describes coloured examples of each version, but he only has uncoloured examples
in his plate volume.
1985m. This was supposedly found in Ulm when a
small church was demolished. Schreiber
says the painting points to a Swiss origin and guesses a date of
1460-1480. His example shows the
rotating piece in one of its positions.
He describes a coloured example in the Germanisches Museum, Nürnberg.
[This
picture is used in: Jasia Reichardt, ed.; Play Orbit [catalogue of an
exhibition at the ICA, London, and elsewhere in 1969‑1970]; Studio
International, 1969, p. 45. It is
described as "Paper toy from an Ulm woodcut, 1470." No further details are given in Reichardt
and she tells me that it was found by a research assistant, probably at the
V&A. However, it seems likely it
was found at the Warburg, though it is possible that the V&A also has an
example of the same print.]
[Ray
Bathke [email of 20 Aug 1998] says the Ulm woodcut, 1470, appears in: Karl
Gröber; Children's Toys of Bygone Ages; Batsford, 1928, 1932. ??NYS.]
1985n. Basically the same picture, but elaborated
and with much more decorative detail, clearly added to an earlier version (an
extra column lacks a base), again probably from Switzerland, but Schreiber
makes no estimate of a date. His copy
lacks the rotating piece and shows no indication of its existence. He describes a coloured version at Zürich
Zentralbibliothek. Baltrušaitis, above,
gives a version as Fig. 100.
Wolfgang Brückner. Populäre Druckgraphik Europas Deutschland
Vom 15. bis zum 20.Jahrhundert.
(As: Stampe Popolari Tedesche; Verlag Electra, Milan, 1969); Verlag Georg D. W. Callwey, München,
1969. Pp. 24-25 (Abb. 17 is on p. 25),
203. Coloured example of 1985n from
Zürich Zentralbibliothek, described as Swiss, 1460/80. Cites Schreiber. He says there is a replica at Nürnberg, but this must be
confusing it with 1985m. This example
is slightly different than Schreiber's picture in that the circle where the
rotating piece would rotate does show, but the print is pasted to another sheet
and one back line of an ape has been drawn in on the backing sheet. I have a slide.
Thomas Eser. Schiefe Bilder Die Zimmernsche Anamorphose und andere Augenspiele aus den
Sammlungen des Germanischen Nationalmuseums [catalogue of an exhibition in
1998]. Germanisches Museum, Nürnberg,
1998, pp. 86-87. The picture is a
B&W version of a coloured example of 1985m, from the Museum Graphische Sammlung,
H5690, Kapsel 8. This is probably the
same example described by Schreiber, but is described as a coloured woodcut,
Swiss or Schwabian, 1460/70 and probably the oldest surviving example of a
picture which the viewer can change. He
cites Schreiber and Brückner.
Barbara Maria Stafford
& Frances Terpak. Catalog for the exhibition Devices of
Wonder, at the Getty Museum, early 2002.
??NYS - information sent by William Poundstone. This shows (in colour) and discusses an oil
painting of Hermes and Aphrodite, like the above apes image, with a rotating
piece in the middle which covers one of the pairs of waists. It dates from early 17C Bohemia, probably
the school of Prague some time after Rudolf II's death in 1612.
James C. Y. Watt &
Anne E. Wardwell. When Silk Was
Gold Central Asian and Chinese
Textiles. The Metropolitan Museum of
Art in Cooperation with The Cleveland Museum of Art, dist. by Abrams. Section 45: Cloth of gold with rabbit
wheels, p. 158. After discussing the
four rabbits textile (see above under Other Asia), this goes on to discuss the
'two heads, four boys' motif, with Figure 73 showing a Ming example in the
Shanghai Museum dated to 16‑17 C.
These were often used as toggles.
This says that the pattern is called 'two boys make four images' and is
a rebus for part of a famous line from the Yi-ching: "The primal 'one' [taiji] begets the two opposites [yi],
the two opposites beget the four elements [xiang], and the four elements
beget the eight trigrams [gua]."
The word yi is a homonym for the word for 'boy' in some
pronunciations and xiang means both 'element' and 'image'.
Rza (or Riza) Abbasi (1587‑1628). Drawing:
Four Horses. Drawing given
in: Kh. S. Mamedov;
Crystallographic Patterns. Comp. &
Maths. with Appls. 12B:1/2 (1986) [= I. Hargittai, ed., Symmetry -- Unifying
Human Understanding, as noted in 6.G.] 511‑529, esp. 525‑526. Two heads and four bodies. This seems to be an outline made from the
original, probably by Mamedov? See
below for a possible original version.
Cf Baltrušaitis, above.
Early 17C Persian drawing: Four Horses: Concentric Design. Museum of Fine Arts, Boston. Reproduced in: Gyorgy Kepes; The New Landscape in Art and Science; Paul
Theobold, Chicago, 1956, fig. 44 on p. 53 with caption on p. 52; and in:
S&B, p. 34. This picture and
the drawing above differ in the position of the feet and other small details,
so it is not clear if the above has been copied from this picture. Seckel, 1997 (op. cit. in 6.AJ), reproduces
it as 6© and says it is 18C.
Baltrušaitis (Fig. 99), above, has a similar, but different version.
Itsuo Sakane. A Museum of Fun (in Japanese). Asahi Shimbun, Tokyo, 1977. Chap. 54-55, pp. 201-207. Seven examples, but I haven't had the text
translated.
Itsuo Sakane et al. The Expanding Visual World -- A Museum of
Fun. Catalogue of a travelling
exhibition with some texts in English.
Asahi Shimbun, Tokyo, 1979. Section
IV: Visual Games, no. 12-15, pp. 96-99, with short texts on p. 170. Seven examples, mostly the same as in the previous
book, including the following. I give
the page numbers of the previous book in ( ) when the picture is in both books.
IV-12,
p. 96 (p. 205), both B&W. Sadakage
Gokotei. Five Children, Ten
Children. Edo era (= 1603-1867). Seckel, 1997 (op. cit. in 6.AJ), reproduces
it in colour as 5© with the same data and
Seckel, 2000 (op. cit. in 6.AJ), reproduces it in colour as fig. 44, p. 55 (=
2002a, fig, 44, p. 55), with no data.
However, the same picture is reproduced in colour in Julian Rothenstein &
Mel Gooding; The Paradox Box; Redstone Press, London, 1993; with a caption by
James Dalgety, saying that it is a painting by Yamamoto Hisabei, c1835, based
on an earlier Chinese image, and giving the title as Ten Children with Five
Heads.
Unnumbered,
p. 97 (pp. 202-203, figs. 2, 3, 4), both B&W, but the titles have no
English versions. Though no credit is
given, the top item is is the early 17C Persian drawing in the Museum of Fine
Arts, Boston.
The second item seems to be from
Renaissance Europe and can be called 'three heads, seven children'. There are three heads and upper bodies
alternating with three legs and lower bodies in a flattened hexagonal pattern
so the the top head can also connect to the bottom legs giving a third
arrangement of the pieces, though the other two children occur in the previous
arrangements, so one gets a total of seven children rather than nine. I have only recently discovered this is a
European item and Bill Kalush has an example of a lead medallion with the same
pattern which is dated to c1610 Prague -- I have a photo. Edward Hordern's collection has a wooden box
with this pattern on the cover, dated as 16C, but it looks later to me, though
I have only seen photos.
The third item is a version of the
'five heads, ten chldren' picture described above. However, the references to this chapter mention Lietzmann and I
have found it there, where it is stated to be a Japanese matchbox -- see below.
IV-14,
p. 99 (colour) (p. 207, B&W).
Kuniyoshi Ichiyusai [= Utagawa Kuniyoshi]. Stop Yawning. Late Edo era
[mid 19C].
IV-15,
p. 98 (colour). Anon. Four Heads, Twelve Horses. Probably Persian. This item belongs to Martin Gardner, having been left to him by
M. C. Escher. It seems to be a leather
cushion cover, probably Persian.
Seckel, 1997 (op. cit. in 6.AJ), reproduces it as the 6ª and says it is 18C. Seckel, 2000 (op. cit. in 6.AJ), reproduces
it as fig. 88, pp. 99 & 122 (= 2002b, fig. 86, pp. 97 & 120) and says
sometime in the 17C.
Metal scrollweights of the 'two
heads, four children' pattern have been made in China since at least the
17C. I have three modern examples which
Peter Hajek obtained for me in Hong Kong.
Edward Hordern's collection has a version from c1680 and a porcelain
version, about six or eight inches across, among other examples -- I have the
date somewhere. James Dalgety also has
an example of the porcelain version.
"Three Boys -- Nine
Torsos". Anonymous painting on
silk from 1700-1710 in City Palace Museum, Jaipur, India. Edward Hordern's collection has a modern
replica. This is similar to the Three
Heads, Seven Children version mentioned above.
I have photos from Hordern's replica and a copy of his information sheet
on it and another painting. Reproduced
from Hordern's example in: Rothenstein & Gooding, below, p. 16.
Mohammad
Bagheri has sent me some souvenir material from the Museum, now named the
Maharaja Sawai Man Singh II Museum, City Palace, Jaipur. This shows a different version of the
pattern, with the same basic geometry but very different figures and
colours.
Family Friend 1 (1849) 148 &
178. Practical puzzles -- No. 3 -- Dead
or alive? "These dogs are dead you
well may say:-- Add four lines
more, they'll run away!"
Answer has "See now the
four lines.
"Tally-ho!" We've touch'd
the dogs, and away they go!"
Julian Rothenstein & Mel
Gooding. The Playful Eye. Redstone Press, London, 1999. They include a number of Japanese prints. There are brief notes on p. 100.
P.
16. "Three Boys -- Nine
Torsos". Photo from Hordern's facsimile,
see above.
P.
20. Three heads, six bodies. Woodblock print, 1860s.
P.
20. Five heads, ten bodies, similar to
the version given by Sakane, above, but with different costumes. Woodblock print, 1860s.
P.
21. Two or three bodies, one head. Woodblock print, c1855. This has six groups of bodies where one head
or hat can be associated with two or three of the bodies. There is a set of four upper bodies which
can be associated with one pair of legs.
P.
22. Five heads, ten bodies, almost
identical to the version given by Sakane, above, but in colour with MADE IN JAPAN on the bottom edge. The
notes say it is a Japanese matchbox label, c1900.
P.
22. Six heads, twelve bodies. Two rings of three heads, six bodies, with
elephants. Japanese matchbox label,
c1900.
P.
24. Five heads, ten bodies, similar to
the version given by Sakane, above, but with women. Woodblock print, c1850.
Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles, no. 5, p. 177 (1868:
188): Alive or Dead. "These dogs
are dead, perhaps you'll say; Add four
lines and then they'll run away."
Magician's Own Book. 1857.
Prob. 27: The dog puzzle, pp. 275 & 298. "The dogs are, by placing two lines upon them, to be suddenly
aroused to life and made to run. Query,
How and where should these lines be placed, and what should be the forms of
them?" S&B, p. 34.
Book of 500 Puzzles. 1859.
Prob. 27: The dog puzzle, pp. 89 & 112. Identical to Magician's Own Book.
Illustrated Boy's Own
Treasury. 1860. Practical Puzzles, No. 33: Dead or Alive,
pp. 401 & 441 "These dogs are
dead you well may say:-- Add four
lines more, they'll run away!"
Boy's Own Conjuring Book. 1860.
Prob. 26: The dog puzzle, pp. 237 & 261. Identical to Magician's Own Book.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 583-7, p.
285. Dead dogs.
Magician's Own Book (UK
version). 1871. The solution drawing is given at the bottom
of p. 231, apparently to fill out the page as there is no relevant text
anywhere. The drawing is better than in
the 1857 US book of the same name.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 6: The
dog puzzle, pp. 28 & 31. "By
connecting the dogs with four lines only they will suddenly start into life, and
commence running. Where should the
lines be placed?" However, he
omits to give a picture!
Hoffmann. 1893.
Chap. 10, no. 32: The two dogs, pp. 348 & 387
= Hoffmann-Hordern, pp. 244-245.
No poetry, but the solution notes that you have to view the dogs
sideways.
Mr. X [cf 4.A.1]. His Pages.
The Royal Magazine 9:5 (Mar 1903) 490-491. Trick donkeys. "Here
are two apparently very dead donkeys.
To bring them to life it is only necessary to fill in the dotted lines
and then turn the page half way round."
Benson. 1904.
The dead dogs puzzle, pp. 256‑257. Prose version.
Pearson. 1907.
Part III, no. 83: Rousing dead dogs -- A good old puzzle, p. 83 . "These dogs are dead, we all should
say; Give them four strokes, they run
away."
Wehman. New Book of 200 Puzzles. 1908.
The dog puzzle, p. 22. c=
Magicians Own Book.
W. Lietzmann. Lustiges und Merkwürdiges von Zahlen und
Formen. 1922. I can't find it in the 2nd ed. of 1923. 4th ed, F. Hirt, Breslau, 1930, p. 208, unnumbered figure, shows
the 'five heads, ten children' pattern mentioned as the third item on p. 97 of
the Sakane book above, labelled:
Japanese Matchbox How many
people, how many heads, how many legs, how many arms are in this picture?
The
material is also in the 6th ed., 1943, p. 200, fig. 46; 7th & 10th eds., 1950 & 1969, p.
196, fig. 37.
Collins. Book of Puzzles. 1927. The dead dogs
puzzle, p. 152.
A two bodied woman. In Seckel, 2002a, op. cit. in 6.AJ, fig. 2,
pp. 11 & 44 (= 2002b, fig. 144, pp. 161 & 194). A real photo of Lady Bird Johnson greeting a
woman friend which shows just one head on two embracing bodies.
Loyd. P. T. Barnum's Trick Mules.
1871. Loyd registered this in
1871 and sold it to Barnum shortly thereafter.
Barnum used it in his Advance Courier.
See S&B, p. 34, for an illustration of Barnum's version and two
recent versions. See SLAHP: Out for a
gallop, pp. 65 & 110. See
Gardner, SA (Aug 1957) c= 1st Book,
chap. 9. In a 1907 interview, it was
stated that thousands of millions of copies of the puzzle had been printed,
with Loyd taking orders for a million at a time!
Gaston Tissandier. Jeux et Jouets du jeune age Choix de récréations amusantes & instructives. Ill. by Albert Tissandier. G. Masson, Paris, nd [c1890]. Le mulet rigolo, pp. 36-37, with elegant
coloured plate. No reference to its
history.
Mel Stover. 1980s??
Trick zebras puzzle. This has
two identical cards with two zebras and two riders. The instructions say to cut one card into three parts along the dotted
lines and put the riders on the zebras.
However, one zebra is facing the opposite way to the usual case and it
takes some time to realise how to solve the problem.
Seckel, 2000 (op. cit. in 6.AJ),
gives a nice colour version as fig. 105, pp. 116 & 122 (= 2002b, fig. 103,
pp. 114 & 120), but only says it is due to Loyd.
6.AV. CUTTING UP IN FEWEST CUTS
Mittenzwey. 1880.
Prob. 191, pp. 36 & 88-89;
1895?: 216, pp. 40 & 91;
1917: 216, pp. 37 & 87. Cut
a 2 x 4 into eight unit squares with three cuts. First cuts into two squares, then overlays
them and then cuts both ways.
Perelman. FFF.
1934.
Sectioning
a cube. Not in the 1957 ed. 1979: prob. 122, pp. 170-171 & 182. MCBF: prob. 122, pp. 171-172 & 186. How many cuts to cut a cube into 27 cubes?
More
sectioning. Not in the 1957 ed. 1979: prob. 123, pp. 171 & 182-183. MCBF: prob. 123, pp. 172-173 & 186. How many cuts to cut a chessboard into 64
squares?
Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. Prob. k, pp. 39,
189 & 191. Cut a cube into 27
cubelets.
6.AW. DIVISION INTO CONGRUENT PIECES
Polyomino
versions occur in 6.F.4.
Quadrisecting
a square is 6.AR.
See
also: 6.AS.1, 6.AT.6.a, 6.AY, 6.BG.
For
solid problems, see: 6.G.3, 6.G.4, 6.AP,
6.AZ?, 6.BC.
See: Charades, Enigmas, and
Riddles, 1862, in 6.AW.2 for a quadrisection with pieces not congruent to the
original.
Take
a square and cut from two corners to the centre to leave ¾ of
the square. The problem is to
quadrisect this into four congruent parts.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. F. 4r is "Analysis of the
Essay of Games". F. 4v has an
entry "8½ a Prob of figure" followed by the L‑tromino. 8½
b is the same with a mitre and there
are other dissection problems adjacent -- see 6.F.3, 6.F.4, 6.AQ, 6.AY -- so it
seems clear that he knew this problem.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles.
No.
15, pp. 26 & 86 & plate II, fig. 11.
2 squares, one double the size of the other, to be cut into four pieces
to make a mitre. Just cut each along
the diagonal.
No.
18, pp. 27 & 87 & plate II, fig. 14.
Six equal squares to form a mitre.
Cut each diagonally. [Actually
you only need to cut three of the squares.]
Endless Amusement II. 1826?
Prob. 5, pp. 192-193. Mitre
puzzle -- says the pieces are not precisely equal. = New Sphinx, c1840, pp. 131-132.
Magician's Own Book. 1857.
Prob.
12: The quarto puzzle, pp. 269 & 294.
Solution is a bit crudely drawn, but the parts are numbered to make it
clear how they are combined.
= Illustrated Boy's Own Treasury, 1860, No. 41, pp. 403 & 442.
Prob.
28: Puzzle of the two fathers, pp. 275-276 & 298. One father has L‑tromino
(see 6.F.4), the other has the mitre.
Solution carefully drawn and shaded.
c= Landells, Boy's Own Toy-Maker, 1858, pp. 148-149.
Book of 500 Puzzles. 1859.
Prob.
12: The quarto puzzle, pp. 83 & 108. Identical to Magician's Own Book.
Prob.
28: Puzzle of the two fathers, pp. 89-90 & 112. Identical to Magician's Own Book..
Boy's Own Conjuring Book. 1860.
Prob.
11: The quarto puzzle, pp. 231 & 257.
Identical with Magician's Own Book.
Prob.
27: Puzzle of the two fathers, pp. 237‑238 & 262. Identical to Magician's Own Book.
Hanky Panky. 1872.
The one‑quarterless square, p. 132
Hoffmann. 1893.
Chap. X, no. 29: The mitre puzzle, pp. 347 & 386
= Hoffmann-Hordern, pp. 243 & 247. Photo on p. 247 shows Enoch Morgan's Sons Sapolio
Color-Puzzle. This says to arrange the
blocks 'in four equal parts so that each part will be the same size color and
shape.' It appears that the blocks are
isosceles right triangles with legs equal to a quarter of the original
square. There are 6 blue triangles, 6
yellow triangles and 12 red triangles.
I think there were originally 6 red and 6 orange, but the colors have
faded, and I think one wants each of the four parts having one color.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co.,
NY), nd [1895]. P. 5: The mitre
puzzle. Similar to Hoffmann. No solution.
Loyd. Origin of a famous puzzle -- No. 19: The mitre puzzle. Tit‑Bits 31 (13 Feb &
6 Mar 1897) 363
& 419. Nearly 50 years ago someone told him to
quadrisect ¾ of a square into congruent figures. The L‑tromino was
intended, but young Loyd drew the mitre shape instead. He says it took him nearly a year to solve
it. But see Dudeney's comments below.
Clark. Mental Nuts.
1904, no. 31. Dividing the
land. Quadrisect an L‑tromino and a mitre.
Pearson. 1907.
Part II, no. 87: Loyd's mitre puzzle, pp. 87 & 178.
Dudeney. The world's best puzzles. Op. cit. in 2. 1908. He says he has
traced it back to 1835 (Loyd was born in 1841) and that "strictly
speaking, it is impossible of solution, but I will give the answer that is
always presented, and that seems to satisfy most people." See also the solution to AM, prob. 150,
discussed in 6.AY. Can anyone say what the
1835 source might be -- a version of Endless Amusement??
Wehman. New Book of 200 Puzzles. 1908.
P. 43:
Puzzle of the two fathers. c=
Magician's Own Book, with cruder solution.
P. 47:
The quarto puzzle. c= Magician's Own
Book, without the numbering of parts.
Loyd. Cyclopedia. 1914. A tailor's problem, pp. 311 & 381. Quadrisect half of a mitre. This has a solution with each piece similar
to to the half mitre.
Loyd Jr. SLAHP.
1928. Wrangling heirs, pp. 35
& 96. Divide mitre into 8 congruent
parts -- uses the pattern of Loyd Sr.
Putnam. Puzzle Fun.
1978. No. 106: Divide the shape,
pp. 16 & 39. "Divide the given
shape into four pieces, such that each and every piece is the same
area." This is much easier than
the usual version. Put two half-size
mitres on the bottom edge and two trapeziums are left.
Here one is cutting a shape into
congruent pieces similar to the original shape. Section 5.J is a version with similar but non‑congruent
pieces.
Charades, Enigmas, and Riddles. 1862: prob. 34, pp. 137 & 143; 1865: prob. 578, pp. 109 & 156. Divide isosceles right triangle and a
hatchet-shaped 10-omino into four congruent pieces. In the first case, they are congruent to the original, but in the
second case, the pieces are right trapeziums of height 1
and bases 2 and
2½.
Dudeney. The puzzle realm. Cassell's Magazine ?? (May 1908) 713-716. No. 3: An easy dissection puzzle. Quadrisect a right trapezium in form of
square plus half a square. = AM,
1917, prob. 146, pp. 35 & 170.
C. Dudley Langford. Note 1464:
Uses of a geometric puzzle. MG
24 (No. 260) (Jul 1940) 209‑211.
Quadrisects: L‑tromino, P‑pentomino, right trapezium
(trapezoid), L‑tetromino,
isosceles trapezium, another right trapezium, two squares joined at a
corner. Gives some 9‑sections and
asks several questions, including asking about 3‑D versions.
R. Sibson. Note 1485:
Comments on Note 1464. MG 24
(No. 262) (Dec 1940) 343. Says some of
Langford's 4‑sections also give 9‑sections. Mentions some 3‑D versions and 144‑sections.
Howard D. Grossman. Fun with lattice points. 13.
A geometric puzzle. SM 14 (1948)
157‑159. Cites Langford &
Sibson. Gives two 9-sections obtained
from Langford's 4‑sections as asserted by Sibson. Gives alternative 4- & 9-sections. Gives a method of generating infinitely many
examples on both square and triangular lattices.
Gardner. SA (May 1963) = Unexpected, chap. 19. Says Golomb started considering rep‑tiles
in 1962 and wrote three private reports on them (??NYS). Gardner describes the ideas in them.
Solomon W. Golomb. Replicating figures on the plane. MG 48 (No. 366) (Dec 1964) 403‑412. Cites Langford and adds numerous
examples. Defines rep‑k
and shows all k can occur.
Roy O. Davis. Note 3151:
Replicating boots. MG 50 (No.
372) (May 1966) 175.
Rochell Wilson Meyer. Mutession: a new tiling relationship among
planar polygons. MTg 56 (1971) 24‑27. A
and B mutually tessellate if each tiles an enlargement of the other.
6.AW.3. DIVIDING A SQUARE INTO CONGRUENT PARTS
In the 1960s, a common trick was to
give someone a number of quadrisection problems where the parts happen to be
congruent to the original figure -- e.g. the quadrisection of the square or
the L-tromino. Then ask her to divide a square into five
congruent parts. She usually tries to
use square pieces in some way and takes a long time to find the obvious
answer. c1980, Des MacHale told me that
it was a serious question as to whether there was any non-trivial dissection of
a square into five or even three congruent pieces. Sometime later, I found a solution -- slice the square into 6
equal strips and say part A consists of the 1st and 4th strips,
part B is the 2nd and 5th,
part C is the 3rd and 6th. However
this is not what was intended by the problem though it leads to other
interesting questions. Since then I
have heard that the problem has been 'solved' negatively several times on the
backs of envelopes at conferences, but no proof seems to have ever
appeared. Very little seems to be
published on this, so I give what little I know. Much of this applies to rectangles as well as squares. QARCH is an occasional problem sheet issued
by the Archimedeans, the Cambridge (UK) student mathematics society.
Gardner, in an article: My ten favorite brainteasers in Games
(collected in Games Big Book of Games, 1984, pp. 130-131) says the dissection
of the square into five congruent parts is one of his favorite problems. ??locate
David Singmaster. Problem 12.
QARCH III (Aug 1980) 3. Asks if
the trisection of the square is unique.
David Singmaster. Response to Problem 12 and Problem 21. QARCH V (Jan 1981) 2 & 4. Gives the trisection by using six strips and
unconnected parts. In general, we can
have an n-section by cutting the square
into kn strips and grouping them regularly. For n = 2, k = 4,
there is an irregular dissection by using the parts as strips 1, 4, 6, 7
and 2, 3, 5, 8. If
p is an odd prime, are there any
irregular p-sections?
John Smith, communicated in a
letter from Richard Taylor, editor of QARCH, nd, early 1981? Smith found that if you slice a square into
9 strips, then the following parts are congruent, giving an irregular
trisection. 1, 2, 6; 3, 7, 8;
4, 5, 9.
David Singmaster. Divisive difficulties. Nature 310 (No. 5977) (9 Aug 1984) 521 &
(No. 5979) (23 Aug 1984) 710.
Poses a series of problems, leading to the trisections of the cube. No solutions were received.
Angus Lavery asserts that he can
trisect the cube, considered as a 3 x 3
x 3 array of indivisible unit
cubes. I had sought for this and was
unable to find such a trisection and had thought it impossible and I still
haven't been able to do it, but Angus swears it can be done, with one piece
being the mirror image of the other two.
George E. Martin. Polyominoes -- A Guide to Puzzles and
Problems in Tiling. MAA Spectrum
Series, MAA, 1991. Pp. 29-30. Fig. 3.9 shows a 5 x 9 rectangle divided
into 15 L-trominoes. Shrinking the length 9
to 5 gives a dissection of the square into 15 congruent pieces which
are shrunken L-trominoes. Prob. 3.10 (very hard) asks for a rectangle
to be dissected into an odd number of congruent pieces which are neither
rectangles nor shrunken
L-trominoes. He doesn't give an
explicit answer, but on p. 76 there are several rectangles filled with an odd
number of L, P and Y
pentominoes. One might argue
that the L and P have the shape of some sort of shrunken L-tromino, but the Y-pentomino is certainly not.
Prob. 3.11 (unsolved) asks if a rectangle can be dissected into three
congruent pieces which are not rectangles.
Prob. 3.12 is a technical generalization of this and hence is also
unsolved.
On 19 Jun 1996, I proposed the
trisection of the square and Lavery's problem on NOBNET. Michael Reid demonstrated that Lavery's
problem has no solution and someone said Lavery had only conjectured it. Reid also cited the following two proofs that
the square trisection is impossible.
I. N. Stewart & A.
Wormstein. Polyominoes of order 3 do
not exist. J. Combinatorial Theory A 61
(1992) 130-136. ??NYS -- Reid says they
show that if a rectangle is dissected into three congruent polyominoes, then each
is a rectangle.
S. J. Maltby. Trisecting a rectangle. J. Combinatorial Theory A 66 (1994)
40-52. ??NYS -- Reid says he proves the
result of Stewart & Wormstein without assuming the pieces are polyominoes.
Martin Gardner. Six challenging dissection tasks. Quantum (May/Jun 1994). Reprinted, with postscript, in Workout,
chap. 16. Trisecting the square into
congruent parts is his first problem.
Cites Stewart & Wormstein.
Then asks if one can have three similar parts, with just two, or none,
congruent. Then asks the same questions
about the equilateral triangle. All
except the first question are solved, but the solutions for the fifth and sixth
are believed to be unique. In the
postscript, Gardner says Rodolfo Kurchan and Andy Liu independently suggested
the problems with four parts and cites Maltby.
6.AW.4. DIVIDING AN L-TROMINO INTO CONGRUENT PARTS
See
also 6.F.4.
F. Göbel. Problem 1771: The L‑shape
dissection problem. JRM 22:1 (1990) 64‑65. The
L‑tromino can be dissected into
2, 3, or 4 congruent parts. Can it be divided into 5 congruent parts?
Editorial
comment -- The L-shaped dissection
problem. JRM 23:1 (1991) 69-70. Refers to Gardner.
Comments
and partial solution by Michael Beeler.
JRM 24:1 (1992) 64-69.
Martin Gardner. Tiling the bent tromino with n
congruent shapes. JRM 22:3
(1990) 185‑191.
This requires placing 12
unit discs snugly into a circular dish of radius 1 + 2 Ö3 = 4.464.
Tissandier. Récréations Scientifiques. 5th ed., 1888, Le secret d'un emballeur, pp.
227-229. Not in the 2nd ed. of 1881 nor
the 3rd ed. of 1883. Illustration by
Poyet. He shows the solution and how to
get the pieces into that pattern. No
dimensions given. = Popular Scientific
Recreations; [c1890]; Supplement: The packer's secret, pp. 855‑856.
Hoffmann. 1893.
Chap. X, no. 48: The packer's secret, pp. 356 & 394
= Hoffmann-Hordern, p. 255. Says
the problem is of French origin. Gives
dimensions 3½ and ¾, giving a ratio of 14/3 = 4.667. "The
whole are now securely wedged together ...." [I think this would be a bit loose.]
"Toymaker". The Japanese Tray and Blocks Puzzle. Work, No. 1447 (9 Dec 1916) 168. Says to make the dish of radius 7
and the discs of radius 1½, again giving a ratio of 14/3 = 4.667. Makes "a firm immovable job ...."
6.AY. DISSECT 3A x 2B TO MAKE 2A x 3B, ETC.
This
is done by a 'staircase' cut. See 6.AS.
Pacioli. De Viribus.
c1500. Ff. 189v - 191r. Part 2.
LXXIX. Do(cumento). un tetragono saper lo longare con restregnerlo
elargarlo con scortarlo (a tetragon knows lengthening and contraction,
enlarging with shortening ??) = Peirani
250-252. Convert a 4 x 24
rectangle to a 3 x 32 using one cut into two pieces. Pacioli's
description
is cryptic but seems to have two cuts, making d c
three
pieces. There is a diagram at the
bottom of f. 190v, badly k f
e
redrawn
on Peirani 458. Below this is a
inserted note which Peirani
252
simply mentions as difficult to read, but can make sense. The g
points
are as laid out at the right. abcd is the original 4 x 24 h a
o b
rectangle. g is
one unit up from a and
e is one unit down from c.
Cut
from c
to g and from e parallel to the base, meeting cg
at f. Then move cdg to
fkh and move fec
to hag. Careful rereading of Pacioli now seems to
show he is using a trick! He cuts
from e
to f to g. then turns over the upper piece and slides
it along so that he can continue his cut from
g to h, which is where f to c is
now. This gives three pieces from a
single cut! Pacioli clearly notes that
the area is conserved.
Although
not really in this topic, I have put it here as it seems to be a predecessor of
this topic and of 6.P.2.
Cardan. De Rerum Varietate. 1557, ??NYS. = Opera Omnia, vol. III, p. 248 (misprinted 348 and with running
head Lib. XII in the 1663 ed.). Liber
XIII. Shows 2A x 3B to 3A x 2B and half of
3A x 4B to 4A x 3B
and discusses the general process.
Kanchusen. Wakoku Chiekurabe. 1727. Pp. 11-12 & 26‑27. 4A x 3B
to 3A x 4B, with the latter being square. Solution asserts that any size of paper can
be made into a square: 'fold lengthwise into an even number and fold the width
into an odd number' -- cf Loyd (1914) & Dudeney (1926) below.
Minguet. 1733.
Pp. 117-119 (1755: 81-82; 1822: 136-137; 1864: 114-115). 3 x 4
to 4 x 3. Shows a straight tetromino along one side
moved to a perpendicular side so both shapes are 4 x 4.
Charles Babbage. The Philosophy of Analysis -- unpublished
collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more
details. F. 4r is "Analysis of the
Essay of Games". F. 4v has an
entry "8½ c Prob of figure" followed by a staircase piece. F. 145-146 show two pieces formed into both
rectangles. There are other dissection
problems adjacent on F. 4v -- see 6.F.3, 6.F.4, 6.AQ, 6.AW.1.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles.
No. 7,
pp. 24 & 83-84 & plate I, fig. 4.
9 x 16 to 12 x 12.
No.
12, pp. 25 & 85 & plate I, fig. 9.
4 x 9 to 6 x 6.
No.
14, pp. 26 & 86 & plate I, fig. 10.
10 x 20 to 13 1/3 x 15.
Endless Amusement II. 1826?
Prob.
1, p. 188. 5A x 6B to
6A x 5B and to 4A x 7B
with two A x B projections. The 6A x 5B looks to be square. = New Sphinx, c1840, p. 136.
Prob.
22, pp. 200-201. 16 x 9 to fill a
12 x 12 hole. Does it by cutting in four pieces --
one 12 x 9 and three 4 x 3. = New Sphinx, c1840, pp. 136-137.
Prob.
33, pp. 210-211. Take a rectangle of
proportion 2 : 3 and cut it into two pieces to make a
square. Uses cut from 4A x 5B to 5A x 4B, but if we make the rectangle 4 x 6,
this makes A = 1, B = 6/5 and the 'square' is 5 x 24/5.
= New Sphinx, c1840, p. 140.
Nuts to Crack II (1833), no.
124. 9 x 16 to fill a 12 x 12 hole using four pieces. = Endless Amusement II, prob. 22.
Young Man's Book. 1839.
Pp. 241-242. Identical to
Endless Amusement II, prob. 22.
Boy's Own Book. 1843 (Paris): 436 & 441, no. 6. 5A x 6B to 6A x 5B and to
4A x 7B with two A x B
projections. The 6A x 5B
looks to be square. = Boy's
Treasury, 1844, pp. 425 & 429. = de
Savigny, 1846, pp. 353 & 357, no. 5.
Cf de Savigny, below.
de Savigny. Livre des Écoliers. 1846.
P. 283: Faire d'une carte un carré.
View a playing card as a 5A x 4B rectangle and make a staircase cut and shift
to 4A x 5B, which will be nearly square.
[When applied to a bridge card,
3.5 x 2.25 in, the result is
2.8 x 2.8125 in.]
Magician's Own Book. 1857.
Prob. 2: The parallelogram, pp. 267 & 291. Identical to Boy's Own Book, 1843 (Paris).
The Sociable. 1858.
Prob. 19: The perplexed carpenter, pp. 292 & 308. 2 x 12
to 3 x 8. = Book of 500 Puzzles, 1859, prob. 19,
pp. 10 & 26. = The Secret Out,
1859, p. 392.
Book of 500 Puzzles. 1859.
Prob.
19: The perplexed carpenter, pp. 10 & 26.
As in The Sociable.
Prob.
2: The parallelogram, pp. 81 & 105.
Identical to Boy's Own Book, 1843 (Paris).
Charades, Enigmas, and
Riddles. 1860: prob. 29, pp. 60 &
64; 1862: prob. 30, pp. 136 &
142; 1865: prob. 574, pp. 107 &
155. 16 x 9 to 12 x 12. All the solutions have an extraneous line in
one figure.
Boy's Own Conjuring Book. 1860.
Prob. 2: The parallelogram, pp. 229 & 254. Identical to Boy's Own Book, 1843 (Paris).
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 584-4, pp.
286 & 404. Looks like 3 x 4
to 2 x 6. The rectangles are formed by trimming a
quarter off a playing card. The
diagrams are not very precise, but it seems that the card is supposed to be
twice as long as wide. If we take the
card as 4 x 8, then the problem is 3 x 8
to 4 x 6.
Hanky Panky. 1872.
The parallelogram, p. 107.
"A parallelogram, ..., may be cut into two pieces, by which two
other figures can be formed."
Shows 5A x 4B cut, but no other figures.
Mittenzwey. 1880.
Prob. 253 & 255, pp. 45-46 & 96-97; 1895?: 282 & 284, pp. 49 & 98-99; 1917: 282 & 284, pp. 45 & 93-94. 4 x 9
to 6 x 6. 16 x 9
to 12 x 12.
Cassell's. 1881.
The carpenter's puzzle, p. 89. =
Manson, 1911, p. 133. 3 x 8 board to cover 2 x 12 area.
Richard A. Proctor. Some puzzles; Knowledge 9 (Aug 1886) 305-306
& Three puzzles; Knowledge 9 (Sep 1886) 336-337. Cut
4 x 3 to 3 x 4. Discusses general method for
nA x (n+1)B to (n+1)A x nB
and notes that the shape can be oblique as well as rectangular.
Lemon. 1890. Card board puzzle,
no. 58, pp. 11‑12 & 99. c=
The parallelogram puzzle, no. 620, pp. 77 & 120 (= Sphinx, no. 706, pp. 92 & 121). Same as Boy's Own Book, 1843 (Paris). In the pictures, A seems to be equal
to B.
Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 136, no. 8. 9 x
15 to
12 x 12. No solution.
Tom Tit, vol. 2. 1892.
Les figures superposables, pp. 149-150.
3 x 2 to 2 x 3.
Berkeley & Rowland. Card Tricks and Puzzles. 1892.
Chinese Geometrical Puzzles No. 1, pp. 108 & 111. Same as Boy's Own Book, 1843 (Paris).
Hoffmann. 1893.
Chap.
III, no. 8: The extended square, pp. 91 & 124‑125
= Hoffmann-Hordern, p. 80. As in
Boy's Own Book, 1843 (Paris), but
A is clearly not equal to B.
Chap.
III, no. 31: The carpenter's puzzle -- no. 2, pp. 103 & 137
= Hoffmann-Hordern, p. 101.
12 x 36 to 18 x 24.
Clark. Mental Nuts. 1897, no.
27. The leaking ship. 12 x 12
to 9 x 16.
Benson. 1904.
The
extended square, p. 190. 5A x 6B square, but the other two figures are 8A x 7B
and 8A x 5B with two
A x B projections.
The
carpenter's puzzle (No. 1), pp. 190‑191.
= Hoffmann, p. 103.
Anon [possibly Dudeney??] Breakfast Table Problems No. 331: A
carpenter's dilemma. Daily Mail (31
Jan & 1 Feb 1905) both p. 7. 16
x 9 to
12 x 12.
Pearson. 1907.
Part II, no. 1: The carpenter's puzzle, pp. 1‑2 & 185‑186. 2 x 12
to 3 x 8.
Wehman. New Book of 200 Puzzles. 1908.
P. 6:
The perplexed carpenter. 2 x 12 to 3
x 8. c= The Sociable.
P. 9:
The carpenter's puzzle. "A plank
was to be cut in two: the carpenter cut it half through on each side, and found
he had two feet still to cut. How was
it?" This is very vague and can
only be recognised as a version of our present problem because the solution
looks like cutting a 2 x 6 to make a
3 x 4.
P. 16:
The parallelogram. Identical to Boy's
Own Book, 1843 (Paris).
P. 17:
Another parallelogram. Takes a
parallelogram formed of a square and a half a square and intends to form a
square. He cuts 5A x 4B
and makes 4A x 5B. But for this to be a square, it must be 20 x 20
and then the original was 25 x
16, which is not quite in the given
shape.
M. Adams. Indoor Games. 1912. A zigzag puzzle, p.
349, with figs. on 348. 5A x 6B square, but the other two figures are 5A x 4B
and 7A x 4B with two
A x B projections.
Loyd. Cyclopedia. 1914.
The
smart Alec puzzle, pp. 27 & 342. (=
MPSL1, prob. 93, pp. 90‑91 & 153‑154.) Cut a mitre into pieces which can form a
square. He trims the corners and
inserts them into the notch to produce a rectangle and then uses a staircase
cut which he claims gives a square using only four pieces. Gardner points out the error, as carefully
explained by Dudeney, below. Since
Dudeney gives this correction 1n 1911, he must have seen it in an earlier Loyd
publication, possibly OPM?
The
carpenter's puzzle, pp. 51 & 345.
Claims any rectangle can be staircase cut to make a square. Shows
9 x 4 to 6 x 6
and 25 x 16 to
20 x 20. Cf Kanchusen (1727) and
Dudeney (1926).
Dudeney. Perplexities. Strand Magazine 41 (No. 246) (Jun 1911) 746 &
42 (No. 247) (Jul 1911) 108.
No. 45: Dissecting a mitre.
"I have seen an attempt, published in America, ..." Sketches Loyd's method and says it is wrong. "At present no solution has been found
in four pieces, and one in five has not apparently been published."
Dudeney. AM.
1917. Prob. 150: Dissecting a
mitre, pp. 35‑36 & 170‑171.
He fully describes "an attempt, published in America", i.e. Loyd's
method. If the original square has
side 84, then Loyd's first step gives a
63 x 84 rectangle, but the
staircase cut yields a 72 x 73½ rectangle, not a square. Dudeney gives a 5 piece solution and says
"At present no solution has been found in four pieces, and I do not
believe one possible."
Dudeney. MP.
1926. Prob. 115: The carpenter's
puzzle, pp. 43‑44 & 132‑133.
= 536, prob. 338, pp. 116‑117 & 320‑321. Shows
9 x 16 to 12 x 12.
"But nobody has ever attempted to explain the general law of the
thing. As a consequence, the notion
seems to have got abroad that the method will apply to any rectangle where the
proportion of length to breadth is within reasonable limits. This is not so, and I have had to expose
some bad blunders in the case of published puzzles ...." He discusses the general principle and shows
that an n‑step cut dissects n2 x (n+1)2 to a square of side n(n+1).
Gardner adds a note referring to AM, prob. 150. Cf Kanchusen (1727) & Loyd (1914)
Stephen Leacock. Model Memoirs and Other Sketches from Simple
to Serious. John Lane, The Bodley Head,
1939, p. 299. Mentions 12 x 12
to 9 x 16.
Harry Lindgren. Geometric Dissections. Van Nostrand, 1964. P. 28 discusses Loyd's mitre dissection
problem and variations. He also thinks
a four piece solution is impossible.
This is a cube dissected into 6 pieces which form 6 cuboids, each
of which can be 'staircased' in two ways.
There is a 6-cycle through the cuboids, with relative sizes: 12 x 12 x 12, 8 x 12 x 18, 8 x 9 x 24, 12 x 6 x
24, 16 x 6 x 18, 16 x 9 x 12. I have a fine example using six different woods, that had been
made for Tom O'Beirne, from Mrs. O'Beirne.
Richard K. Guy. Op. cit. in 5.H.2. 1960. Pp. 151-152
describes O'Beirne's invention.
T. H. O'Beirne. Puzzles and paradoxes -- 9: A six-block
cycle for six step-cut pieces. New
Scientist 9 (No. 224) (2 Mar 1961) 560-561.
This
appears to be a 7 x 5 flag with a Greek cross X X X X X O O X X X X X
of 5 cells removed from the
middle as in the first figure X
X X O O O X X X O O
at the right. One has to cut it into two pieces to make a X X O O
X X O O O
perfect square. This is done by cutting along a 'staircase' X X X O O
O X X X O O
as shown. However, this seems to produce a 5 x 6
flag, X X O O
O O O X X O O O
not a square. But there is usually a swindle -- the
diagram O
O O O O
is not drawn with the unit cells
square, but instead the unit
cells
are 6/5 as wide as they are tall.
Normally the reader would not recognize this and the diagrams are often
rather imprecise.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co.,
NY), nd [1895]. P. 5: The flag
puzzle. Starts with a square and asks
to make a Swiss flag. The square is
actually 31.5 mm = 1¼ in on an edge and the flag is 26 x 37 mm
= 1 x 1½ in with the Greek cross formed from squares of
edge 5 mm = ¼ in, so the areas do not add up! No solution.
Loyd. Problem 4: The Swiss flag puzzle. Tit-Bits 31 (31 Oct 1896)
75 & (21 Nov 1896) 131. c=
Cyclopedia, pp. 250 & 373. Swiss
flag puzzle with the flag at an angle and a slight wiggle in the edges, so the
solution requires an extra cut to make it square.
Loyd. Cyclopedia, 1914, pp. 14 & 341: A Swiss puzzle -- part 2.
= MPSL2, no. 144, pp. 101 & 166.
= SLAHP, pp. 48 & 102: How was this flag made? Starts with the Swiss flag which is 47 x 30 mm
and the Greek cross has cells
6.5 x 5.5 mm, so this is
approximately the correct shape to make a square, but the resulting square is
not drawn.
This section is largely based on
Gordon's Notes, cf below. See also
6.AP.2 for dissections of a tetrahedron in general. This section is now expanding to consider all polysphere puzzles.
See
also S&B, p. 42, which mentions Hein and some other versions.
Piet Hein. Pyramystery. Made by Skjøde of Skjern, Denmark, 1970. With leaflet saying it was "recently
invented by Piet Hein.... Responding to
numerous requests, the inventor has therefore obliged the many admirers of the
puzzle by also inventing its history".
He then gives a story about Cheops.
Peter Hajek and Jerry Slocum have different examples!!
Hajek's
example has four planar rectangular pieces: 1 x 4, 2 x 3, 3 x 2,
4 x 1 rectangles. It is the same as Tut's Tomb -- see
below. It has a 4pp English leaflet
marked © Copyright Piet Hein 1970.
Slocum's
example has 6 planar pieces: 4 3-spheres and 2 4-spheres. The leaflet is 34pp (?? -- Slocum only sent
me part of it) with 3pp of instructions in each of 9 languages and then 6pp of
diagrams of planar and 3-D problems. It
is marked © 1970 Aspila, so perhaps this is a later development from
the above?? The story part of the text
is very similar to the above, but slightly longer. The pieces make an order
4 tetrahedron or two order 3
tetrahedra or two order 4 triangles and one can also divide them into
two groups of three pieces such that one group makes an order 3
tetrahedron, but the other does not.
Advertising
leaflet for Pyramystery from Piet Hein International Information Center, ©1976,
describes the puzzle as having six pieces.
Mag‑Nif Inc. Tut's Tomb.
c1972. Same as the first
Pyramystery.
Akira Kuwagaki & Sadao
Takenaka. US Patent 3,837,652 -- Solid
Puzzle. Filed: 1 May 1973; patented: 24 Sep 1974. 2pp + 4pp diagrams. Four planar
3-spheres and a 2-sphere to make
a square pyramid of edge 3. 11
planar 4-spheres to make an
octahedron shape of edge 4. Cites a 1936 Danish patent -- Hein ??NYS
Len Gordon. Perplexing Pyramid. 1974.
Makes a edge 4 tetrahedron with 6 planar right-angled
pieces: domino; straight and L trominoes; I,
L, Y tetrominoes.
Patrick A. Roberts. US Patent 3,945,645 -- Tangential Spheres
Geometric Puzzle. Filed: 28 Jun
1976; patented: 29 Nov 1977. 3pp + 3pp diagrams. 8
4-spheres and a 3-sphere to make
a tetrahedron of side 5. 5 of
the 4-spheres are non-planar.
Robert E. Kobres. US Patent 4,060,247 -- Geometric
Puzzle. Filed: 28 Jun 1976; patented: 29 Nov 1977. 1p + 2pp diagrams. 5 pieces which make a 4 x 5
rhomboid or a tetrahedron. Two
pieces have the form of a 2 x 3 rhombus;
two pieces are 2-spheres and the last piece is the linear 4-sphere.
Len Gordon. Some Notes of Ball‑Pyramid and Related
Puzzles. Revised version, 10 Jul 1986,
14pp. Available from the author, 2737
N. Nordic Lane, Tucson, Arizona, 85716, USA.
Ming S. Cheng. US Patent 4,988,103 -- Geometric Puzzle of
Spheres. Filed: 2 Oct 1989; patented: 29 Jan 1991. Front page, 5pp diagrams, 4pp text. A short version is given in Wiezorke, 1996,
p. 64. 7 planar 5‑spheres to
make a tetrahedron; a hexagon with
sides 3, 4, 4, 3, 4, 4; an equilateral triangle lacking one vertex.
Bernhard Wiezorke. Puzzling with Polyspheres. Published by the author (Lantzallee 18, D‑4000
Düsseldorf 30, Germany), Mar 1990, 10pp.
Bernhard Wiezorke. Compendium of Polysphere Puzzles. (1995);
Second Preliminary Edition, as above, Aug, 1996. 64pp, reproducing the short versions of the
above patents. Despite Wiezorke's
searches, nothing earlier than Hein's 1970 puzzles has come to light.
Torsten Sillke & Bernhard
Wiezorke. Stacking identical
polyspheres. Part 1: Tetrahedra. CFF 35 (Dec 1994) 11-17. Studies packing of tetrahedra with identical
polysphere pieces, with complete results for tetrahedra of edges 4 - 8
and polyspheres of 3, 4, 5 spheres.
Some of the impossibility results have only been done by computer, but
others have been verified by a proof.
6.BA. CUTTING A CARD SO ONE CAN PASS THROUGH IT
Ozanam. 1725.
1725: vol. IV, prob. 34, pp. 436‑437 & fig. 40, plate 12 (14).
Minguet. 1733.
Pp. 115-117. (1755: 83-84; 1864:
112-113; not noted in 1822, but it's likely to be at p. 138.) Similar to Ozanam, 1725.
Alberti. 1747.
Art. 34, p. 208-209 (110) & fig. 42, plate XI, opp. p. 210
(109). Copied from Ozanam, 1725.
Family Friend 3 (1850) 210 &
241. Practical puzzle -- No. XVII. Shows a 3 inch by 5 inch card. Repeated as Puzzle 15 -- The wonder puzzle
in (1855) 339 with solution in (1856) 28.
Magician's Own Book. 1857.
Prob. 10: The cardboard puzzle, pp. 269 & 294. Problem shows 3 inch by 5 inch card. Answer calls it "the cut card
puzzle". c= Landells, Boy's Own
Toy-Maker, 1858, p. 142. = Book of 500
Puzzles, 1859, pp. 83 & 108. =
Boy's Own Conjuring Book, 1860, prob. 9, pp. 230 & 256.
Indoor & Outdoor. c1859.
Part II, prob. 7: The cardboard puzzle, p. 129. No diagram, so the solution is a bit
cryptic.
The Secret Out. 1859.
How to Cut a Visiting Card for a Cat to Jump through it, p. 382.
Illustrated Boy's Own
Treasury. 1860. No. 27, pp. 400 & 440. Identical to Magician's Own Book, but
solution omits the sentence: "A
laurel leaf may be treated in the same manner."
Magician's Own Book (UK
version). 1871. To cut a card for one to jump through, p.
124. He adds: "The adventurer of
old, who, inducing the aborigines to give him as much land as a bull's hide
would cover, and made it into one strip by which acres were enclosed, had
probably played at this game in his youth." See 6.AD.
Elliott. Within-Doors. Op. cit. in 6.V.
1872. Chap. 1, no. 2: The
cardboard puzzle, pp. 27 & 30‑31.
No diagram, so the solution is a bit cryptic.
Lemon. 1890. Cardboard puzzle,
no. 140, pp. 23 & 102. = Sphinx,
no. 467, pp. 65 & 113.
J. B. Bartlett. How to walk through a laurel leaf. The Boy's Own Paper 12 (No. 587)
(12 Apr 1890) 440.
Hoffmann. 1893.
Chap. X, no. 28: The cut playing‑card, pp. 346 & 385‑386
= Hoffmann‑Hordern, p. 243.
Benson. 1904.
The elastic cardboard puzzle, pp. 200‑201.
Dudeney. Cutting-out paper puzzles. Cassell's Magazine ?? (Dec 1909) 187-191
& 233-235. With photo of Dudeney
going through the card.
Collins. Book of Puzzles. 1927. Through a playing
card, pp. 16-17.
6.BB. DOUBLING AN AREA WITHOUT CHANGING ITS HEIGHT OR WIDTH
The area is usually a square, but
other shapes are possible. If one views
it as a reduction, one can reduce the area to any fraction of the original!
The Sociable. 1858.
Prob. 41: The carpenter puzzled, pp. 298 & 316. 3 x 3
square of wood with holes in it forming a 4 x 4 array with the
corner holes at the corners of the board.
Claims one can cut 1/4 of the board out of the centre without
including any holes. But this only
gets 2/9 of the area -- double the central square. = Book of 500 Puzzles, 1859, prob. 41, pp.
16 & 34. = Secret Out, 1859, pp.
386-387.
Indoor & Outdoor. c1859.
Part II: prob. 14: The carpenter puzzled, pp. 133‑134. Almost identical with The Sociable.
Hanky Panky. 1872.
P. 226 shows the same diagram as the solution in The Sociable, but there
is no problem or text.
Lewis Carroll. Letter of 15 Mar 1873 to Helen Feilden. = Carroll-Collingwood, pp. 212-215 (Collins
154-155), without solution. Cf
Carroll-Wakeling, prob. 28: The square window, pp. 36-37 & 72. Halve the area of a square window. Wakeling and Carroll-Gardner, p. 52, give
the surname as Fielden, but it is Feilden in Carroll-Collingwood and in Cohen,
who sketches her life. Wakeling writes
that Feilden is correct.
Mittenzwey. 1880.
Prob. 216-217, pp. 38-39 & 90;
1895?: 241-242, pp. 43 & 92;
1917: 241‑242, pp. 39 & 88.
Divide a rectangle or square into two pieces with the same height and
width as the square. Solution is to draw
a diagonal.
Lemon. 1890. A unique window,
no. 444, pp. 58 & 114. The
philosopher's puzzle, no. 660, pp. 82 & 121.
Don Lemon. Everybody's Scrap Book of Curious
Facts. Saxon, London, 1890. P. 82 quotes an article from The New
York World describing this as 'an excellent, if an old, puzzle'.
Hoffmann. 1893.
Chap. IX, no. 28: A curious window, pp. 319 & 327
= Hoffmann-Hordern, pp. 211-212.
Notes that either a diamond or a triangle in appropriate position can be
so doubled.
Clark. Mental Nuts. 1897, no.
40. The building lot. "Have a lot 50 x 100. Want to build a
house 50 x 100 and have the yard same size. How?"
Solution shows 50 x 100 with a diagonal drawn.
Pearson. 1907.
Part II, no. 79: At a duck pond, pp. 79 & 176. A square pond is to be doubled without
disturbing the duckhouses at its corners.
Wehman. New Book of 200 Puzzles. 1908.
The carpenter puzzled, p. 39. =
The Sociable.
Will Blyth. Handkerchief Magic. C. Arthur Pearson, London, 1922. Doubling the allotment, pp. 23-24.
Hummerston. Fun, Mirth & Mystery. 1924.
Some queer puzzles, Puzzle no. 76, part 6, pp. 164 & 183. Solution notes that a window in the shape of
a diamond or a right triangle or an isosceles triangle can be doubled in area
without changing its width or height.
King. Best 100. 1927.
No. 2,
pp. 7‑8 & 38. Same as Indoor
& Outdoor, with the same error.
No. 4,
pp. 8 & 39. Halve a square
window. See Foulsham's.
Foulsham's Games and Puzzles
Book. W. Foulsham, London, nd
[c1930]. No. 2, pp. 5 & 10. Double a window without changing its height
or width. (This is one of the few cases
where the problem is not quite identical to King.)
M. Adams. Puzzle Book. 1939. Prob. B.117:
Enlarging the allotment, pp. 86 & 110.
Double a square allotment without disturbing the trees at the corners.
This
consists of 27 blocks,
a x b x c, to make into a
cube a+b+c on a side. It was first
proposed by Dean Hoffman at a conference at Miami Univ. in 1978. See S&B, p. 43. The planar version, to use 4
rectangles a x b to make a square of side a + b
is easy. These constructions are
proofs of the inequality of the arithmetic and geometric means. Sometime in the early 1980s, I visited David
Klarner in Binghamton and Dean Hoffman was present. David kindly made me a set of the blocks and a three-sided corner
to hold them.
D. G. Hoffman. Packing problems and inequalities. In:
The Mathematical Gardner, op. cit. in 6.AO, 1981. Pp. 212‑225. Includes photos of Carl Klarner assembling the first set of the
blocks. Asks if there are analogous
packings in n dimensions.
Berlekamp, Conway &
Guy. Winning Ways. 1982.
Vol. 2, pp. 739‑740 & 804‑806. Shows all 21 inequivalent solutions.
6.BD. BRIDGE A MOAT WITH PLANKS
In
the simplest case, one has a 3 x 3 moat with a
1 x 1 island in the centre. One wants to get to the island using two
planks of length 1 or a bit less than 1. One plank is laid
diagonally across the corner of the moat and the second plank is laid from the
centre of the first plank to the corner of the island. If the width of the moat is D
and the planks have length
L, then the method works if 3L/2 > DÖ2, i.e. L > 2Ö2
D/3 = .94281.. D. One really
should account for the width of the planks, but it is not clear just how much
overlap is required for stability.
Depew is the only example I have seen to use boards of different
lengths. With more planks, one can
reach across an arbitrarily large moat, but the number of planks needed gets
very large. In this situation, the case
of a circular moat and island is a bit easier to solve.
The
Magician's Own Book (UK version) version is quite different and quite
erroneous.
Magician's Own Book (UK
version). 1871. The puzzle bridge, p. 123. Stream 15 or 16 feet across, but none of the
available planks is more than 6 feet long.
He claims that one can use a four plank version of the three knives make
a support problem (section 11.N) to make a bridge. However the diagram of the solution clearly has the planks nearly
as long as the width of the stream. In
theory, one could build such a bridge with planks slightly longer than half the
width of the stream, but to get good angles (e.g. everything crossing at right
angles or nearly so), one needs planks somewhat longer than Ö2/2 of the width. E.g. for a
width of 16 ft, 12 ft planks would be
adequate.
Mittenzwey. 1880.
Prob. 298, pp. 54 & 105;
1895?: 330, pp. 58 & 106;
1917: 330, pp. 52‑53 & 101. 4 m gap bridged with two boards of length 3¾
m. He only gives a diagram. In fact this doesn't work because the ratio
of lengths is 15/16 = .9375
Lucas. RM2. 1883. Le fossé du champ carré. Bridge the gap with two planks whose length
is exactly 1. Notes this works because
3/2 > Ö2.
Hoffmann. 1893.
Chap. VII, no. 9, pp. 289 & 295.
Matchstick version = Hoffmann-Hordern, p. 193.
Benson. 1904.
The moat puzzle, p. 246. Same as
Hoffmann, but the second plank is shown under the first!!
Dudeney. CP.
1907. No. 54: Bridging the
ditch, pp. 83-85 & 204. Eight 9'
planks to cross a 10' ditch where it makes a right angle.
Pearson. 1907.
Part I, no. 34: Across the moat, pp. 122 & 186.
Blyth. Match-Stick Magic.
1921. Boy Scouts' bridge, p.
21. Ordinary version done with matchsticks.
J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Boy Scouts' bridge, pp. 68-69. As in Blyth.
Depew. Cokesbury Game Book.
1939. Crossing the moat, pp.
225-226. Square moat 20
feet wide to be crossed with boards of width 18 and 15
ft. In fact this doesn't work --
one needs L1 + L2/2
> DÖ2.
"Zodiastar". Fun with Matches and Matchboxes. Op. cit. in 4.B.3. Late 1940s? The bridge,
pp. 66-67 & 83. Matchstick version
of the square moat & square island problem.
F. D. Burgoyne. Note 3106:
An n plank problem. MG 48 (No.
366) (Dec 1964) 434‑435. The
island is a point in the centre of a 2
x 2 lake. Given n planks of length s, can you get to the
island? He denotes the minimal length
as s(n) and computes s(1) = 1,
s(2) = 2 Ö2/3, s(3) = .882858... and
says s(¥) = Ö2/2, [but I believe it is 0,
i.e. one can get across an arbitrarily large moat with a fixed length of
plank].
Jonathan Always. Puzzling You Again. Tandem, London, 1969.
Prob. 10:
A damsel in distress, pp. 15 & 70‑71. Use two planks of length
L to reach a point in the centre
of a circular moat of radius R. He finds one needs L2 ³ 4R2/5.
Prob.
11: Perseus to the rescue again, pp. 15‑16 & 71‑72. Same with five planks. The solution uses only four and needs L2 ³ 2R2/3.
C. V. G.[?] Howe? Mathematical Pie 75 (Summer 1975) 590 &
76 (Autumn 1975) 603. How big a square
hole can be covered with planks of unit length? Answer says there is no limit, but the height of the pile increases
with the side of the square.
Highlights for Children
(Columbus, Ohio). Hidden Picture
Favorites and Other Fun. 1981. Brain Buster 4, pp. 12 & 32: Plink,
plank, kerplunk? Two children arrive at
a straight(!) stream 4m wide with two planks 3m
long. Solution: extend one plank
about 1¼ m over the stream and one
child stands on the land end. The
second child carries the other plank over the stream and extends it to the
other side and crosses. He then pulls
the plank so it extends about 1¼ m over the stream. The first child now extends her plank out to rest on the second
plank and crosses, pulling up her plank and taking it with her. I theory this technique will work if L > ⅔D, but one needs some overlap space, and the children may not have
the same weight.
Richard I. Hess. Puzzles from Around the World. The author, 1997. (This is a collection of 117 puzzles which he published in Logigram,
the newsletter of Logicon, in 1984-1994, drawn from many sources. With solutions.) Prob. 64. Usual problem
with D = 10, but he says the board have width
W = 1 and so one use the
diagonal of the board in place of
L. In my introduction, we saw
that the standard version leads to L2
³ 8D2/9,
so Hess's version leads to L2 ³ 8D2/9 - W2 and we can get across with slightly shorter
planks, but we have to tread very carefully!
6.BE. REVERSE A TRIANGULAR ARRAY OF TEN CIRCLES
One
has a triangle of ten coins with four on an edge. Reverse its direction by moving only three coins. New section -- I'm surprised not to have
seen older examples.
Sid G. Hedges. More Indoor and Community Games. Methuen, London, 1937. The triangle trick, p. 54. Uses peas, buttons or nuts.
M. Adams. Puzzle Book. 1939. Prob. B.34:
Pitching camp, pp. 66 & 103. Array
of tents.
Evelyn August. The Black-Out Book. Op. cit. in 5.X.1. 1939. The General
inspects the balloons, pp. 106 & 214.
Array of 10 barrage balloons.
Leopold. At Ease!
1943. 1, 2, 3 -- shift!, pp. 20
& 198. Thanks to Heinrich Hemme for
the lead to this.
Joseph Leeming. Games with Playing Cards Plus Tricks and
Stunts. Franklin Watts, 1949. ??NYS -- but two abridged versions have
appeared.
Games
and Fun with Playing Cards. Dover, NY,
1980. This contains everything except
the section on bridge.
Tricks
and Stunts with Playing Cards Plus Games of Solitaire. Gramercy Publishing, NY, nd [1960s?]. This includes all the tricks, stunts,
puzzles and solitaire games.
25
Puzzles with Cards, 8th puzzle. Tricky
triangle. Dover: pp. 154-155 &
172. Gramercy: pp. 45-46 & 65. Both have fig. 25 & 42.
Young World. c1960.
P. 7: fifteen coin problem.
Reverse a triangle with five on a side by moving five coins.
Robert Harbin. Party Lines. Op. cit. in 5.B.1.
1963. Birds in flight, p. 34. Says this problem is described by Gardner,
but gives no specific source.
Maxey Brooke. (Fun for the Money, Scribner's, 1963); reprinted as: Coin Games and Puzzles, Dover, 1973. Prob. 4: Bottoms up, pp. 15 & 75. On p. 6, he acknowledges Leopold as his source. Thanks to Heinrich Hemme for this reference.
D. B. Eperson. Triangular (old) pennies. MG 54 (No. 387) (Feb 1970) 48‑49. The number of pennies which must be moved to
reverse a triangle with n on a side is [T(n)/3], where T(n) is the
n‑th triangular number, which is the number in the array.
James Bidwell. The ten‑coin triangle. MTg 54 (1971) 21‑22. How many coins must be moved to reverse the
triangle with n on an edge?
His students find the same value as Eperson, but they weren't sure they
had proved it.
Putnam. Puzzle Fun.
1978. No. 25: Triangular
reverse, pp. 6 & 31. Usual 10 coin
triangle.
6.L might be considered as part of this
section. There are some examples of
problems with ladders which look like crossed ladders, but are simple
Pythagorean problems.
See
also 6.AS.2 for dissection proofs of the theorem of Pythagoras. I will include here only some interesting
ancient examples. See Elisha Scott
Loomis; The Pythagorean Proposition; 2nd ed., NCTM, 1940, for many proofs.
Aryabhata
I, v. 17, states the Theorem of Pythagoras and the related theorem that if ABC
is a diameter of a circle and
LBM is a chord perpendicular to
it, then LB2 = AB x BC;
Bhaskara I's commentary applies the latter in several forms where modern
algebra would make it more natural to use the former. Brahmagupta, v. 41, states
LM2 = AB x BC.
I
had overlooked the examples in Mahavira -- thanks to Yvonne Dold for pointing
them out.
Fibonacci. 1202.
Pp. 397‑398 (S: 543-544) looks like a crossed ladders problem but
is a simple right triangle problem.
Vyse. Tutor's Guide. 1771?
Prob.
9, 1793: p. 178, 1799: p. 189 & Key p. 224. A ladder 40 long in a roadway can reach 33
up one side and, from the same point, can reach 21
up the other side. This is
actually a simple right triangle problem.
There is a misprint of 9 for
6 in the answer.
Prob.
17 (in verse), 1793: 179, 1799: p. 190 & Key p. 228. A variation of the Broken Bamboo problem, cf
below, with D = 30, H - X = 63,
which is a simple right triangle problem.
Hutton. A Course of Mathematics. 1798?
Prob.
VIII, 1833: 430; 1857: 508.
= Vyse, prob. 19.
Prob.
IX, 1833: 430; 1857: 508.
= Vyse, prob. 17 with D =
15, H - X = 39.
A
bamboo (or tree) of height H breaks at height X from the ground so that
the broken part reaches from the break to the ground at distance D
from the foot of the bamboo. In
fact the quadratic terms drop out of the solution, leaving a linear
problem. This may be of Babylonian
origin?? The hawk and rat problems of
6.BF.3 are geometrically the same problem viewed sideways.
In
all cases below, H and
D are given and X is
sought, so I will denote the problem by
(H, D).
See
Tropfke, p. 620.
Chiu Chang Suan Ching (Jiu Zhang
Suan Shu). c‑150? Chap. IX, prob. 13, p. 96. [English in Mikami, p. 23 and in Swetz &
Kao, pp. 44‑45, and in HM 5 (1978) 260.]
(10, 3).
Bhaskara I. 629.
Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97‑103; English
version of the examples is on pp. 296-300.
The material of interest is examples 4 and 5. In the set-up described under 6.BF.3, the bamboo is BOC
which breaks at O and the point C reaches the ground
at L.
Ex.
4: (18, 6). Shukla notes this is used by Chaturveda.
Ex.
5: (16, 8).
Chaturveda. 860.
Commentary to the Brahma‑sphuta‑siddhanta, chap. XII,
section IV, v. 41, example 2. In
Colebrooke, p. 309. Bamboo: (18, 6).
Mahavira. 850.
Chap. VII, v. 190-197, pp. 246-248.
v.
191. (25, 5), but the answer has H -
X rather than X.
v.
192. (49, 21), but the answer has H - X rather than X.
v.
193. (50, 20), but with the problem reflected so the known
leg is vertical rather than horizontal.
v.
196. This modifies the problem by
imagining two trees of heights H and
h, separated by D.
The first, taller, tree breaks at height X from the ground and
leans over so its top reaches the top of the other tree. If we subtract h from X and H,
then │X - h│ is the solution of the problem (H - h, D).
Because the terms are squared, it doesn't matter whether X is
bigger or smaller than h. He does the case H, h, D = 23, 5, 12.
Bhaskara II. Lilavati.
1150. Chap. VI, v. 147‑148. In Colebrooke, pp. 64‑65. (32, 16).
Bhaskara II. Bijaganita.
1150. Chap. IV, v. 124. In Colebrooke, pp. 203‑204. Same as Lilavati.
Needham, p. 28, is a nice
Chinese illustration from 1261.
Gherardi. Libro di ragioni. 1328. Pp. 75‑76:
Regolla di mesura. (40, 14).
Pseudo-dell'Abbaco. c1440.
No. 166, p. 138 with B&W reproduction on p. 139. Tree by stream. (60, 30). I have a colour
slide of this.
Muscarello. 1478.
F. 96v, pp. 224-225. Tree by a
stream. (40, 30).
Calandri. Aritmetica.
c1485. Ff. 87v-88r, pp.
175-176. Tree by a stream. (60, 30).
= Pseudo-dell'Abbaco.
Calandri. Arimethrica. 1491. F. 98r. Tree by a river. (50, 30). Nice woodcut
picture. Reproduced in Rara, 48.
Pacioli. Summa.
1494. Part II, f. 55r, prob.
31. (30, 10). Seems to say this very beautiful and subtle invention is due to
Maestro Gratia.
Clark. Mental Nuts. 1897, no.
78. The tree and the storm. (100, 30).
[I have included this as this problem is not so common in the 19C and
20C as in earlier times.]
N. L. Maiti. Notes on the broken bamboo problem. Gaņita-Bhāratī [NOTE: ņ
denotes an n with an underdot] (Bull. Ind. Soc. Hist.
Math.) 16 (1994) 25-36 -- ??NYS -- abstracted in BSHM Newsletter 29 (Summer
1995) 41, o/o. Says the problem is not
in Brahmagupta, though this has been regularly asserted since Biot made an
error in 1839 (probably a confusion with Chaturveda -- see above). He finds eight appearances in Indian works,
from Bhaskara I (629) to Raghunath-raja (1597).
6.BF.2. SLIDING SPEAR = LEANING REED
A
spear (or ladder) of length H stands against a wall. Its base moves out B from the wall, causing
the top to slide down D. Hence
B2 + (H - D)2 = H2.
The
leaning reed has height H. It reaches
D out of the water when it is
straight up. When it leans over, it is
just submerged when it is B away from its upright position. This is identical to the sliding spear
turned upside down.
In
these problems, two of H, B,
D are given and one wants the
remaining value. I will denote them, by
e.g. H, D = 30, 6.
See
Tropfke, p. 619 & 621.
BM 85196. Late Old Babylonian tablet in the British
Museum, c‑1800. Transcribed,
translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte
II; Springer, Berlin, 1935, pp. 43++. Prob.
9 -- translation on pp. 47-48, commentary on p. 53. Quoted in B. L. van der Waerden; Science
Awakening; OUP, 1961, p. 76. See
also: J. Friberg; HM 8 (1981)
307-308. Sliding beam(?) with H, D = 30, 6 and with H, B = 30,
18.
BM 34568. Seleucid period tablet in the British
Museum, c‑300. Transcribed,
translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte
III, Springer, Berlin, 1937, pp. 14-22 & plate 1. Prob. 12 -- translation on p. 18, commentary on p. 22. Quoted in van der Waerden, pp. 76‑77. See also:
J. Friberg, HM 8 (1981) 307-308.
Sliding reed or cane, B, D = 9,
3.
Papyri Cairo J. E. 89127‑30 &
89137‑43. c‑260. Shown and translated in: Richard A. Parker; Demotic Mathematical
Papyri; Brown Univ. Press, Providence, 1972; pp. 1, 3‑4, 35‑40
& Plates 9‑10.
Prob.
24‑26: H, B =
10, 6; 14½, 10; 10, 8.
Prob.
27‑29: H, D =
10, 2; 14½, 4;
10, 4.
Prob.
30‑31: B, D =
6, 2; 10, 4.
Chiu Chang Suan Ching (Jiu Zhang
Suan Shu). c‑150? Chap. IX.
Prob. 6,
p. 92. [English in Mikami, p. 22.] Leaning reed, B, D = 5, 1.
Prob.
7, p. 93. Version with a rope hanging
and then stretched giving B, D = 8, 3.
Prob.
8, p. 93. [English in Mikami, p. 22 and
in Swetz & Kao, pp. 30‑32 and in HM 4 (1977) 274.] Ladder, but with vertical and horizontal
reversed. B, D = 10, 1. Mikami misprints the answer as 55
rather than 50.5.
Bhaskara I. 629.
Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97‑103, with
reproductions of original diagrams on pp. 101-102; English version of the
examples is on pp. 296-300. The
material of interest is examples 6 and 7.
In the setup of 6.BF.3, the lotus is
OBA and LBM
is the water level.
Ex.
6: B, D = 24, 8. Shukla notes this is used by Chaturveda.
Ex.
7: B, D = 48, 6.
Chaturveda. 860.
Commentary to the Brahma‑sphuta‑siddhanta, chap. XII,
section IV, v. 41, example 3. In
Colebrooke, pp. 309‑310. Leaning
lotus: B, D = 24, 8.
Tabari. Miftāh al-mu‘āmalāt. c1075.
P. 124, no. 42. ??NYS - cited by
Tropfke 621.
Bhaskara II. Lilavati.
1150. Chap. VI, v. 152‑153. In Colebrooke, pp. 66. Leaning lotus: B, D = 2, ½.
Bhaskara II. Bijaganita.
1150. Chap. IV, v. 125. In Colebrooke, p. 204. Same as Lilavati.
Fibonacci. 1202.
P. 397 (S: 543). Sliding
ladder: H, B = 20, 12.
Leonardo Fibonacci. La Practica di Geometria. Volgarizzata da Cristofano di Gherardo di
Dino, cittadino pisano. Dal Codice 2186
della Biblioteca Riccardiana di Firenze, 1448.
Ed. by Gino Arrighi, Domus Galilaeana, Pisa, 1966. P. 37 and fig. 18. Same as Fibonacci 1202.
Zhu Shijie. Siyuan Yujian (Precious Mirror of the Four
Elements). 1303. ??NYS -- English given in Li & Du, p.
179. Questions in verse, no. 1. Two reeds,
14 apart, which reach 2½
and 1 out of the water. When
they lean together, they just touch at the water surface. The water is assumed to have the same depth
at both reeds, which reduces the problem from two variables to one variable.
Gherardi?. Liber habaci. c1310. Pp. 139‑140. H, B = 20, 12
Gherardi. Libro di ragioni. 1328. Pp. 77‑78. Ship's mast with H, D = 131, 4.
B&W picture on p. 77, from f. 46v.
Columbia Algorism. c1350.
No.
135, pp. 138‑139. Sliding
ladder: H, B = 10, 6.
No.
140, pp. 149‑150. Leaning
tree: H, B = 20, 10.
(I have colour slides of the illustrations to
these problems.)
Pseudo-dell'Abbaco. c1440.
No. 167, pp. 138-140, with B&W picture on p. 139. Sliding spear: H, D = 30, 4. I have
a colour slide of this.
della Francesca. Trattato.
c1480. F. 44r (108). B, D = 6, 2. English in Jayawardene.
Pacioli. Summa.
1494. Part II, ff. 54v-55v.
Prob.
25 (misprinted 52). B, D = 6, 2. = della Francesca.
Prob.
26. H,
H - D = 10, 8.
Prob.
27. H, D = 10, 4.
Prob.
28. H,
D = 10, B/3.
Prob.
29. D + H = 12, DH = 12.
Prob.
30. H = D + 4, DH = 12.
Prob.
41. Tree of height 40
with a rope of length 50 tied to the top which reaches to the ground
at a point 30 away. Length 10
of the rope is pulled, causing the tree to lean. How high is the top of the tree now? We now have a triangle of sides 40, 40, 30
and want the altitude to the side of length 30.
PART II.
F.
68r, prob. 98. Tree of height 30
has a rope of length 50 tied to the top, so it reaches the
ground 40 away. How much rope has
to be pulled to move the top of the tree to being 8 away from the
vertical? He neglects to give the
value 50 in the problem statement and the ground distances of 8
and 32 are interchanged in the diagram.
van Etten. 1624. Prob. 89 (86), part V (4), p. 135 (214). Sliding ladder: H, B =
10, 6.
Ozanam. 1694.
Prob. 42 & fig. 48, plate 10, 1696: 123-124; 1708: 129;
1725: 320-321 & plate 10 (11).
Ladder 25 long with foot 7 from wall. Foot is pulled out 8 more -- how much does
the top come down?
Vyse. Tutor's Guide. 1771? Prob. 16 (in verse), 1793: p. 179, 1799: p.
190 & Key p. 228. H, B = 100,
10.
Mittenzwey. 1880.
Prob. 294, pp. 53-54 & 104;
1895?: 324, pp. 57 & 106;
1917: 324, pp. 52 & 100.
Leaning reed, B, D = 5,
1. [I have included this and the next
entry as this problem is not so common in the 19C and 20C as in earlier times.]
Clark. Mental Nuts. 1904, no.
69; 1916, no. 95. The boatman's puzzle. Leaning pole, B, D = 12,
6. Find H - D.
6.BF.3. WELL BETWEEN TWO TOWERS
The
towers have heights A, B
and are D apart.
A well or fountain is between them and equidistant from the tops of the
towers. I denote this by (A, B, D).
Vogel, in his DSB article on Fibonacci, says the problem is Indian, and
Dold pointed me to Mahavira. Pseudo‑dell'Abbaco
introduces the question of a sliding weight or pulley -- see Pseudo‑dell'Abbaco,
Muscarello, Ozanam-Montucla, Tate, Palmaccio, Singmaster.
I
have just found that Bhaskara I gives several unusual variations on this.
See
Tropfke, p. 622. See also 10.U.
INDEX
of A, B, D problems, with A £ B.
0 4 8 Chaturveda
0 9 27 Bhaskara II
0 12 24 Bhaskara I
0 18 81 Bhaskara I
5 6 12 Bhaskara I
10 10 12 Bhaskara I
13 15 14 Mahavira
18 22 20 Mahavira
20 24 22 Mahavira
20 30 30 Gherardi?
20 30 50 Perelman
30 40 50 Fibonacci, Muscarello, Cardan
30 50 100 Bartoli
30 70 100 Tate
40 50 30 Muscarello
40 50 70 Pseudo-dell'Abbaco
40 50 100 della Francesca
40 60 50 Lucca 1754
60 80 100 Calandri c1485
70 100 150 Columbia Algorism, Pacioli
80 90 100 Calandri 1491
Bhaskara I. 629.
Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97‑103; English
version of the examples is on pp. 296-300.
The material of interest is examples 2 and 3.
These
are 'hawk and rat problems'. A hawk is
sitting on a wall of height A and a rat is distance D
from the base of the wall. The
rat tries to get to its hole, in the wall directly under the hawk. The hawk swoops, at the same speed as the
rat runs, and catches the rat when it hits the ground. Hence this is the same as our two tower
problem, but with B = 0, so I will denote this version by (A, 0, D).
Bhaskara I attributes this type of problem to unspecified previous
writers. Shukla adds that later writers
have it, including Chaturveda and Bhaskara II, qqv.
Ex.
2: (12, 0, 24).
Ex.
3: (18, 0, 81). Bhaskara I explains the solution in detail
and Shukla gives an English precis of it.
Let ABOC be the horizontal diameter of a circle and
let LBM be a vertical chord.
LB is our pole, with the hawk
at L,
and the rat is at C and wants to get to B.
The point of capture is O, because
LO = OC. From LB2 = AB x BC, we can determine AB and hence the other
values.
Looking
at Chaturveda (below), I now see that turning this sideways gives the same
diagram as the broken bamboo problem -- the tree was BC and breaks at O to
touch the ground at L. So the broken bamboo problem (H, D)
is the same as the two towers or hawk and rat problem (D, 0, H).
Bhaskara I. 629.
Ibid. Examples 8 and 9 are
'crane and fish problems'. A fish is at
the NE corner of a rectangular pool and a crane is at the NW corner and they
move at the same speeds. The fish swims
obliquely to the south side, but the crane has to walk along the edge of the
pool. The fish unfortunately gets to
the side just as the crane reaches the same point and gets eaten. This again like our two tower problem, but
with one pigeon unable to fly, so it has to walk down the tower and
across. Because the pool is
rectangular, the two values A and
B are equal.
Ex.
8: (6, 6, 12).
Ex.
9: (10, 10, 12). The meeting point is 3 3/11
from the SW corner.
Chaturveda. 860.
Commentary to the Brahma‑sphuta‑siddhanta, chap. XII,
section IV, v. 41, example 4. In
Colebrooke, p. 310. Cat and rat, where
the cat behaves like the hawk of Bhaskara I:
(4, 0, 8).
Mahavira. 850.
Chap. VII, v. 201-208, pp. 249-251.
He
gives several problems, but he usually also asks for the equal distance from
the top
of
each tower to the fountain.
v.
204. Two pillars, with a rope between
them which touches the ground but with equal lengths to the tops. (13, 15, 14).
v.
206. Two hills with mendicants who are
able to fly along the hypotenuses.
(22, 18, 20)
v.
208. Same context. (20, 24, 22).
Bhaskara II. Lilavati.
1150. Chap. VI, v. 149‑150. In Colebrooke, pp. 65‑66. Peacock and snake version of the hawk and
rat problem: (9, 0, 27).
Fibonacci. 1202.
De duabus avibis [On two birds], pp. 331‑332 (S: 462-463). (40, 30, 50). He does the same problem differently on pp. 398‑399 (S:
544-545).
Gherardi?. Liber habaci. c1310. P. 139. (20, 30, 30).
Lucca 1754. c1330.
F. 54v, pp. 120-121. (60, 40,
50).
Columbia Algorism. c1350.
No. 136, pp. 139‑140. (70,
100, 150).
Bartoli. Memoriale.
c1420. Prob. 10, f. 76r (=
Sesiano 138-139 & 148-149, with reproduction of the relevant part of f. 76r
on p. 139). (50, 30, 100).
Pseudo-dell'Abbaco. c1440.
Prob.
80, p. 72, with picture on p. 71. (50,
40, 70).
Prob.
158‑159, pp. 129‑133, with illustrations on pp. 130 & 132, deal
with the related problem where a rope with a sliding weight hangs between two
towers, and the diagram clearly shows the weight in the air, not reaching the
ground, so that the resulting triangles are similar. [I found it an interesting question to determine when the rope
was long enough to reach the ground, and if not, how much above the ground the
weight was -- see Muscarello, Ozanam-Montucla, Singmaster below.]
Prob. 158 has A, B, D = 40, 60, 40 and a rope of length L = 110,
so the rope is more than long enough for the weight to reach the ground,
but all he does is show that the two parts of the rope are 66
and 44, which is a bit dubious as there is slack in
the rope. The diagram clearly shows the
weight in the air. I have a colour
slide of this.
Prob. 159 has A, B = 40, 60 with L = 120 such that the weight just touches the ground
-- find the distances of the weight to the towers.
Prob.
160, p. 133. (40, 30, 50).
Muscarello. 1478.
Ff.
95r-95v, pp. 222-223. A, B, D = 50, 40,
30. Place a rope between the towers
just long enough to touch the ground.
Ff.
95v-96r, pp. 223-224. A, B, D = 30, 20,
40. A rope of length L = 60
with a sliding weight is stretched between them -- where does the weight
settle?
F.
99r, pp. 227-228. Fountain between
towers for doves: (40, 30, 50).
della Francesca. Trattato.
c1480. F. 22r (72). (40, 50, 100). English in Jayawardene.
Calandri. Aritmetica.
c1485. Ff. 89r-89v, pp. 178‑179. (60, 80, 100). (Tropfke, p. 599, shows the illustration in B&W.)
Calandri. Arimethrica. 1491. F. 100v. Well between two towers. (80, 90, 100). Nice double size woodcut picture.
Pacioli. Summa.
1494. Part II.
F.
59v, prob. 62. (70, 100, 150) = Columbia Algorism.
Ff.
59v-60r, prob. 63. A, B, D = 30, 40,
20 with rope of length 25
between the towers with a sliding lead weight on it. How high is the weight from the ground.
Ff.
61r-61v, prob. 66. Three towers of
heights
A, B, C = 125, 135, 125, with distances AB, AC, BC = 150, 130,
140. Find the point on the ground
equidistant from the tops of the towers.
Cardan. Practica Arithmetice. 1539.
Chap. 67.
Section
9, f. NN.vi.r (p. 197). (40, 30, 50).
Section
10, ff. NN.vi.r - NN.vii.v (pp. 197-198).
Three towers of heights
A, B, C = 40, 30, 70, with distances AB, AC, BC = 50, 60, 20. Find the point on the ground equidistant
from the tops of the towers. Same idea
as Pacioli, prob. 66.
Ozanam‑Montucla. 1778.
Vol. II, prob. 7 & fig. 5, plate 1.
1778: 11; 1803: 11-12; 1814: 9‑10; 1840: 199.
Rope between two towers with a pulley on it. Locate the equilibrium position.
Uses reflection.
Carlile. Collection.
1793. Prob. XLV, p. 25. Find the position for a ladder on the ground
between two towers so that leaning it each way reaches the top of each
tower. (80, 91, 100). He simply states how to do the
calculation, x = (D2 + A2
- B2)/2D for the distance
from the base of tower B.
T. Tate. Algebra Made Easy. Chiefly Intended for the Use of Schools. New edition. Longman, Brown, Green, and Longman, London, 1848. P. 111.
No.
36. A, B, D = 30, 70, 100. Locate
P such that the towers subtend
the same angle, i.e. the two triangles are similar. Clearly P divides
D as 30 to 70.
No.
37. Same data. Locate
P so the distance to the tops is
the same. This gives 70
to 30 easily because A + B = D.
No.
38. Same data. Locate
P so the difference of the
squares of the distances is 400. Answer is
68 from the base of the shorter
tower.
Perelman. MCBF.
1937. Two birds by the
riverside. Prob. 136, pp. 224-225. (30, 20, 50). "A problem of an Arabic mathematician of the 11th
century."
Richard J. Palmaccio. Problems in Calculus and Analytic
Geometry. J. Weston Walch, Portland,
Maine, 1977. Maximum-Minimum Problems,
No. 3, pp. 9 & 70-71. Cable supported
pulley device over a factory. A, B,
D =
24, 27, 108 with cable of length L = 117. Find the lowest point. He
sets up the algebraic equations corresponding to the the two right triangles,
assumes the distance from one post and the height above ground are implicit
functions of the length of the cable from the pulley to the same post and
differentiates both equations and sets equal to zero, but it takes half a page
to get to the answer and he doesn't notice that the two triangles are similar
at the lowest point.
David Singmaster, proposer; Dag Jonsson & Hayo Ahlburg,
solvers. Problem 1748: [The two
towers]. CM 18:5 (1992) 140 &
19:4 (1993) 125-127. Based on
Pseudo-dell'Abbaco 158-159, but the solution by reflection was later discovered
to be essentially Ozanam‑Montucla.
David Singmaster. Symmetry saves the solution. IN:
Alfred S. Posamentier & Wolfgang Schulz, eds.; The Art of Problem
Solving: A Resource for the Mathematics Teacher; Corwin Press, NY, 1996, pp.
273-286. Gives the reflection solution.
Yvonne Dold-Samplonius. Problem of the two towers. IN: Itinera mathematica; ed. by R. Franci,
P. Pagli & L. Toti Rigatelli.
Siena, 1996. Pp. 45-69. ??NYR.
The earliest example she has found is Mahavira and it was an email from
her in about 1995 that directed me to Mahavira.
A
railway rail of length L and ends fixed expands to length L + ΔL. Assuming the rail makes two hypotenuses, the
middle rises by a height, H, satisfying
H2 = {(L+ΔL)/2}2 ‑ (L/2)2, hence
H @ Ö(LΔL/2).
However,
one might assume the rail buckled into an arc of a circle of radius r.
If we let the angle of the arc be
2θ, then we have to solve rθ
= (L + ΔL)/2; r sin θ
= L/2. Taking sin θ @ θ - θ3/6, we get
r2 @ (L + ΔL)3/ 24
ΔL. We have H = r (1 - cos θ) @
rθ2/2 and combining
this with earlier equations leads to H @ Ö{3(L+ΔL)ΔL/8} which is about Ö3 / 2 = .866... as big as the estimate in the linear case.
The Home Book of Quizzes, Games
and Jokes. Op. cit. in 4.B.1,
1941. P. 149, prob. 12. L = 1 mile, ΔL = 1 ft or 2 ft -- text is not
clear. "Answer: More than 54
ft." However, in the linear
case, ΔL = 1 ft gives
H = 51.38 ft and ΔL = 2 ft gives H = 72.67
ft, while the exact answers in the
circular case are 44.50 ft and
62.95 ft.
Sullivan. Unusual.
1943. Prob. 15: Workin' on the
railroad. L = 1 mile, ΔL = 2 ft. Answer: about 73
ft.
Robert Ripley. Mammoth Believe It or Not. Stanley Paul, London, 1956. If a railroad rail a mile long is raised 200
feet in the centre, how much closer would it bring the two ends? I.e.
L = 1 mile,
H = 200 ft.
Answer is: "less than 6
inches". I am unable to
figure out what Ripley intended.
Jonathan Always. More Puzzles to Puzzle You. Tandem, London, 1967. Gives the same question as Ripley with
answer "approximately 15 feet".
The exact answer is 15.1733..
feet or 15 feet 2.08 inches.
David Singmaster,
submitter. Gleaning: Diverging lines. MG 69 (No. 448) (Jun 1985) 126.
Quotes from Ripley and Always.
David Singmaster. Off the rails. The Weekend Telegraph (18 Feb 1989) xxiii &
(25 Feb 1989) xxiii.
Gives the Ripley and Always results and asks which is correct and
whether the wrong one can be corrected -- cf Ripley above.
Phiip Cheung. Bowed rail problem. M500 161 (?? 1998) 9. ??NYS.
Paul Terry, Martin S. Evans, Peter Fletcher, solvers and
commentators. M500 163 (Aug 1998) 10-11.
L = 1 mile, ΔL = 1 ft. Terry treats the bowed rail as circular and
gets H = 44.49845 ft. Evans takes
L = 1 nautical mile of 6000 ft and gets almost exactly H = 50 ft. Fletcher says it took 15 people to lift a 60ft length of rail, so
if someone lifted the 1 mile rail to insert the extra foot, it would need about
1320 people to do the lifting.
6.BF.5. TRAVELLING ON SIDES OF A RIGHT TRIANGLE.
New
section. See also the Mittenzwey
example in 10.A.6.
Brahmagupta. Brahma‑sphuta‑siddhanta. 628.
Chap. XII, sect. IV, v. 39. In
Colebrooke, p. 308. Rule for the
problem illustrated by Chaturveda.
Mahavira. 850.
Chap. VII, v. 210-211, pp. 251-252.
A slower traveller goes due east at rate v. A faster traveller
goes at rate V and starts going north. After time
t, he decides to meet the other
traveller and turns so as to go directly to their meeting point. How long,
T, do they travel? This gives us a right triangle with
sides vT, Vt, V(T-t) leading to a quadratic in T
whose constant term drops out, yielding
T = 2t V2/(V2-v2). If we set
r = v/V, then T = 2t/(1-r2), so we can determine T
from t and r without actually knowing V
or v. Indeed, if we let ρ
= d/D, we get 2ρ = 1 - r2. v, V, t = 2, 3, 5.
Chaturveda. 860.
Commentary to the Brahma‑sphuta‑siddhanta, chap. XII,
section IV, v. 39. In Colebrooke, p.
308. Two ascetics are at the top of a
(vertical!) mountain of height A. One, being a wizard, ascends a distance X
and then flies directly to a town which is distance D
from the foot of the mountain.
The other walks straight down the mountain and to the town. They travel at the same speeds and reach the
town at the same time. Example
with A, D = 12, 48.
Bhaskara II. Lilavati.
1150. Chap. VI, v. 154-155. In Colebrooke, pp. 66-67. Similar to Chaturveda. Two apes on top of a tower of height A
and they move to a point D away.
A, D = 100, 200.
Bhaskara II. Bijaganita.
1150. Chap. IV, v. 126. In Colebrooke, pp. 204-205. Same as Lilavati.
6.BG. QUADRISECT A PAPER SQUARE WITH ONE CUT
This
involves careful folding. One can also
make mn rectangles with a single cut.
Mittenzwey. 1880.
Prob. 193 & 194, pp. 36 & 89;
1895?: 218 & 219, pp. 41 & 91;
1917: 218 & 219, pp. 37 & 87.
Make four squares. Make four
isosceles right triangles.
Walter Gibson. Big Book of Magic for All Ages. Kaye & Ward, Kingswood, Surrey,
1982. Tick-tack-toe, pp. 68-69. Take a
4 x 4 array and mark alternate
squares with Xs and
Os. By careful folding and
cutting, one produces eight free squares and a connected lattice of the other
eight squares, with the free squares being either all the Os
or all the Xs, depending on how the final part of the cut
is made.
David Singmaster. Square cutting. Used in my puzzle columns.
Weekend
Telegraph (18 & 25 Mar 1989) both p. xxiii.
G&P,
No. 16 (Jul 1995) 26. (Publication
ceased with No. 16.)
I
have not yet found any real history of this topic. One popular book says moirés were first made in 15C China. The OED has several entries for Moire and
Moiré. It originally refers to a type
of cloth and may be a French adaptation of the English word mohair -- Pepys
refers to 'greene-waterd moyre' and this is the earliest citation. In the early 19C, the term began to be used
for the 'watered' effect on cloth and metal.
At some point, the term was transferred to the optical phenomena, but
the OED does not have this meaning.
Journal of Science and Arts 5
(1818) 368. On the Moiré Metallique, or
Fer blanc moiré. ??NYS -- cited in OED
as their first citation for the noun use of the term.
John Badcock. Domestic Amusements, or Philosophical
Recreations, ... Being a Sequel Volume to Philosophical Recreations, or Winter
Amusements. T. Hughes, London, nd
[Preface dated Feb 1823]. [BCB 16-17;
OCB, pp. 180 & 196. Heyl 21. Toole Stott 78‑80. Wallis 34 BAD. HPL [Badcock]. These give
dates of 1823, 1825, 1828.]
Pp. 139-141, no. 169: Moiré Metal, or Crystallised Tin &
no. 170: Moiré Watering, by other Methods. "Quite new and splendid as this art is, .... M. Baget, a Frenchman, however, claims
the honour of a discovery of this process, attributing the same to accident,
...." Cited in the OED as the
first adjectival use of the term, though the previous entry seems to also have
an adjectival usage.
Rational Recreations. 1824.
Experiment 16, p. 15: Metallic watering, or, fer blanc moire. Says it is of Parisian invention and gives the
method of applying sulphuric acid to tin.
Endless Amusement II. 1826?
Pp. 24-25: Application of the moiré métallique to tin-foil. This deals with obtaining a moiré effect in
tin-foil and is quite different than Badcock.
Young Man's Book. 1839.
Pp. 312-314. Identical to Endless
Amusement II.
Tom Tit, vol. 2. 1892.
Le papier-canevas et les figures changeantes, pp. 137-138. Uses perforated card.
Hans Giger. Moirés.
Comp. & Maths. with Appls. 12B:1/2 (1986) [= I. Hargittai, ed., Symmetry
-- Unifying Human Understanding, as noted in 6.G.] 329‑361. Giger says the technique of moiré fabrics
derives from China and was first introduced into France in 1754 by the English
manufacturer Badger (or Badjer). He
also says Lord Rayleigh was the first to study the phenomenon, but gives no
references.
6.BI. VENN DIAGRAMS FOR N SETS
New topic. I think I have seen more papers on this and Anthony Edwards has
recently sent several more papers.
Martin Gardner. Logic diagrams. IN: Logic Machines and
Diagrams; McGraw‑Hill, NY, 1958, pp. 28-59. Slightly amended in the 2nd ed., Univ. of Chicago Press, 1982,
and Harvester Press, Brighton, 1983, pp. 28-59. This surveys the history of all types of diagrams. John Venn [Symbolic Logic, 2nd ed., ??NYS]
already gave Venn diagrams with 4 ovals and with 4 ovals and a disconnected
set. Gardner describes various binary
diagrams from 1881 onward, but generalised Venn diagrams seem to first occur in
1909 and then in 1938-1939, before a surge of interest from 1959. His references are much expanded in the 2nd
ed. and he cites most of the following items.
John Venn. On the diagrammatic and mechanical
representation of propositions and reasonings.
London, Edinburgh and Dublin Philos. Mag. 10 (1880) 1-18. ??NYS -- cited by Henderson.
John Venn. Symbolic Logic. 2nd ed., Macmillan, 1894.
??NYS. Gardner, p. 105,
reproduces a four ellipse diagram.
Lewis Carroll. Symbolic Logic, Part I. 4th ed., Macmillan, 1897; reprinted by Dover, 1958. Appendix -- Addressed to Teachers, sections
5 - 7: Euler's method of diagrams; Venn's method of diagrams; My method of
diagrams, pp. 173-179. Describes
Euler's simple approach and Venn's thorough approach. Reproduces Venn's four-ellipse diagram and his diagram for five
sets using four ellipses and a disconnected region. He notes that Venn suggests using two five-set diagrams to deal
with six sets and does not go further.
He then describes his own method, which easily does up to eight
sets. The diagram for four sets is the
same as the common Karnaugh diagram used by electrical engineers. For more than four sets, the regions become
disconnected with the cells of the four-set case being subdivided, using a
simple diagonal, then his 2-set, 3-set and 4-set diagrams within each cell of
the 4-set case. ?? -- is this in the
1st ed. -- ??NYS date??
Carroll-Gardner,
p. 61, says this is in the first ed. of 1896.
William E. Hocking. Two extensions of the use of graphs in
elementary logic. University of
California Publications in Philosophy 2:2 (1909) 31(-??). ??NYS -- cited by Gardner who says Hocking
uses nonconvex regions to get a solution for any n.
Edmund C. Berkeley. Boolean algebra and applications to
insurance. Record of the Amer. Inst. of
Actuaries 26:2 (Oct 1937) & 27:1 (Jun 1938). Reprinted as a booklet by Berkeley and Associates, 1952. ??NYS -- cited by Gardner. Uses nonconvex sets.
Trenchard More Jr. On the construction of Venn diagrams. J. Symbolic Logic 24 (Dec 1959)
303-304. ??NYS -- cited by Gardner. Uses nonconvex sets.
David W. Henderson. Venn diagrams for more than four
classes. AMM 70:4 (1963) 424-426. Gives diagrams with 5
congruent irregular pentagons and with
5 congruent quadrilaterals. Considers problem of finding diagrams that
have n-fold rotational symmetry and
shows that then n must be a prime. Says he has found an example for
n = 7, but doesn't
know if examples can be found for all prime
n.
Margaret E. Baron. A note on the historical development of
logic diagrams: Leibniz, Euler and Venn.
MG 53 (No. 384) (May 1969) 113‑125. She notes Venn's solutions for
n = 4, 5. She gives
toothed rectangles for n = 5, 6.
K. M. Caldwell. Multiple‑set Venn diagrams. MTg 53 (1970) 29. Does n = 4 with rectangles and then uses indents.
A. K. Austin, proposer; Heiko Harborth, solver. Problem E2314 -- Venn again. AMM 78:8 (Oct 1971) 904 &
79:8 (Oct 1972) 907-908. Shows
that a diagram for 4 or more sets cannot be formed with
translates of a convex set, using simple counting and Euler's formula. (The case of circles is in Yaglom &
Yaglom I, pp. 103-104.) Editor gives a
solution of G. A. Heuer with 4 congruent rectangles and more complex
examples yielding disconnected subsets.
Lynette J. Bowles. Logic diagrams for up to n classes. MG 55 (No. 394) (Dec 1971) 370‑373. Following Baron's note, she gives a binary
tooth‑like structure with examples for
n = 7, 8.
Vern S. Poythress & Hugo S.
Sun. A method to construct convex
connected Venn diagrams for any finite number of sets. Pentagon (Spring 1972) 80-83. ??NYS -- cited by Gardner.
S. N. Collings. Further logic diagrams in various
dimensions. MG 56 (No. 398) (Dec 1972)
309‑310. Extends Bowles.
Branko Grünbaum. Venn diagrams and independent families of
sets. MM 48 (1975) 12‑22. Considers general case. Substantial survey of different ways to
consider the problem. References to
earlier literature. Shows one can use 5
identical ellipses, but one cannot use ellipses for n > 5.
B. Grünbaum. The construction of Venn diagrams. CMJ 15 (1984) 238‑247. ??NYS.
Allen J. Schwenk. Venn diagram for five sets. MM 57 (1984) 297. Five ovals in a pentagram shape.
A. V. Boyd. Letter:
Venn diagram of rectangles. MM
58 (1985) 251. Does n = 5
with rectangles.
W. O. J. Moser & J. Pach. Research Problems in Discrete Geometry. Op. cit. in 6.T. 1986.
Prob. 27: On the extension of Venn diagrams. Considers whether a diagram for
n classes can be extended to one
for n+1 classes.
Mike Humphries. Note 71.11:
Venn diagrams using convex sets.
MG 71 (No. 455) (Mar 1987) 59.
His fourth set is a square;
fifth is an octagon.
J. Chris Fisher, E. L. Koh &
Branko Grünbaum. Diagrams Venn and
how. MM 61 (1988) 36‑40. General case done with zig‑zag
lines. References.
Anthony W. F. Edwards. Venn diagrams for many sets. New Scientist 121 (No. 1646) (7 Jan 1989)
51-56. Discusses history, particularly
Venn and Carroll, the four set version with ovals and Carroll's four set
version where the third and fourth sets are rectangles. Edwards' diagram starts with a square
divided into quadrants, then a circle.
Fourth set is a two-tooth 'cogwheel' which he relates to a Hamiltonian
circuit on the 3-cube. The fifth set is
a four-tooth cogwheel, etc. The result
is rather pretty. Edwards notes that
the circle in the n set diagram meets the 2n-1 subsets of the n-1 sets other than that given by the circle,
hence travelling around the circle gives a sequence of the subsets of n-1
objects and this is the Gray code (though he attributes this to Elisha
Gray, the 19C American telephone engineer -- cf 7.M.3). The relationship with the n-cube leads to a partial connection between
Edwards' diagram and the lattice of subsets of a set of n
things.
New
Scientist (11 Feb 1989) 77 has: Drawing
the lines -- letters from Michael Lockwood -- describing a version with
indented rectangles -- and from Anthony Edwards -- noting some errors in the
article.
Ian Stewart. Visions mathématiques: Les dentelures de
l'esprit. Pour la Science No. 138
(Apr 1989) 104-109. c= Cogwheels
of the mind, IN: Ian Stewart; Another
Fine Math You've Got Me Into; Freeman, NY, 1992, chap. 4, pp. 51-64. Exposits Edwards' work with a little more
detail about the connection with the Gray code.
A. W. F. Edwards &
C. A. B. Smith. New 3-set Venn
diagram. Nature 339 (25 May 1989)
263. Notes connection with the family
of cosine curves, y = 2-n
cos 2nx on [0, π]
and Gray codes and with the family of sine curves, y = 2-n sin 2nx on [0, 2π] and ordinary binary codes.
Applying a similar phase shift to Edwards' diagram leads to diagrams
where more than two set boundaries are allowed to meet at a point.
A. W. F. Edwards. Venn diagrams for many sets. Bull. Intern. Statistical Inst., 47th
Session, Paris, 1989. Contrib. Papers,
Book 1, pp. 311-312.
A. W. F. Edwards. To make a rotatable Edwards map of a Venn
diagram. 4pp of instructions and
cut-out figures. The author, Gonville
and Caius College, Cambridge, CB2 1TA, 21 Feb 1991.
A. W. F. Edwards. Note 75.39:
How to iron a hypercube. MG 75
(No. 474) (1991) 433-436. Discusses his
diagram and its connection with the
n-cube.
Anthony Edwards. Rotatable Venn diagrams. Mathematics Review 2:3 (Feb 1992)
19-21. + Letter: Venn revisited. Ibid. 3:2 (Nov 1992) 29.
This
will cover a number of cases which are not very mathematical. I will record just some early examples. See also 6.G (esp. 6.G.1), 6.N, 6.W (esp.
6.W.7), 6.AP for special cases. The
predecessors of these puzzles seem to be the binomial and trinomial cubes
showing (a+b)3 and
(a+b+c)3, which I
have placed in 6.G.1. Cube dissections
with cuts at angles to the faces were common in the 19C Chinese puzzle chests,
often in ivory. I have only just
started to notice these. It is hard to
distinguish items in this section from other burr puzzles, 6.W.7, and I have
tried to avoid repetition, so one must also look at that section when looking
at this section.
Catel. Kunst-Cabinet. 1790. Der Vexierwürfel, p. 11 & fig. 32 on
plate II. Figure shows that there are
some cuts at angles to the faces, so this is not an ordinary cube dissection,
but is more like the 19C Chinese dissected cubes.
C. Baudenbecher. Sample book or catalogue from c1850s. Op. cit. in 6.W.7. One whole folio page shows about 20 types of wooden interlocking
puzzles, including most of the types mentioned in this section and in 6.W.5 and
6.W.7. Until I get a picture, I can't
be more specific.
Slocum. Compendium.
Shows: Wonderful "Coffee
Pot"; Magic "Apple"; Magic "Pear"; Extraordinary "Cube"; Magic "Tub" from Mr. Bland's Illustrated Catalogue of
Extraordinary and Superior Conjuring Tricks, etc.; Joseph Bland, London,
c1890. He shows further examples from
1915 onward.
Hoffmann. 1893.
Chap. II, pp. 107-108 & 141-142 = Hoffmann-Hordern, pp.
106-107, with photo.
No.
37: The Fairy Tea-Table. Photo on p.
107 shows a German example, 1870-1895.
No.
38: The Mystery. Photo on p. 107 shows
a German example, 1870-1895, with instructions.
Western Puzzle Works, 1926 Catalogue.
No.
5075. Unnamed -- Fairy Tea-Table.
Last
page shows 20 Chinese Wood Block Puzzles, High Grade. These are unnamed, but the shapes include various burr-like
objects, cube, spheres, egg, barrel, tankard, pear and apple.
P. M. Grundy. A three-dimensional jig-saw. Eureka 7 (Mar 1942) 8-10. Consider a
2 x 2 x 2 array of unit cubes. He suggests removing and/or adding lumps on
the interior faces to make a jig-saw.
He then considers lumps in the form of a isosceles right triangular prism
with largest face being a unit square.
He finds there are 25 such pieces, subject to the conditions that there
are most two removals, and that when there are two, they must be parallel. He gives a graphical view of a 2 x 2 x 2
formed from such pieces which gives some necessary, but not sufficient,
conditions for a set of eight such pieces to be able to make a cube. One solution is shown. [If one assumes there is just one removal
and one addition, I find just four pieces, which form a 1 x 2 x 2
block. One could view this as a
3-dimensional matching puzzle, where the internal faces have to match both like
a head to tail matching, but also with correct orientation. MacMahon thought of notching pieces as an
alternative to colouring edges, but his pieces were two-dimensional.]
New
section.
Gardner. SA (Sep 1965) = Carnival, chap. 18. Describes how the problem arose in the
design of Sergel's Square, Stockholm, in 1959.
Addendum in Carnival gives the results given by Gridgeman, below. Also says road engineers used such curves
with n = 2.2, called '2.2 ellipses', from the 1930s for bridge arches.
N. T. Gridgeman. Lamé ovals.
MG 54 (No. 387) (Feb 1970) 31‑37.
Lamé (c1818) seems to be the first to consider (x/a)n + (y/b)n = 1. Hein's design in Stockholm uses a/b = 6/5
and n = 5/2. Gerald Robinson used a/b = 9/7
and n = 2.71828..., which he determined by a survey asking
people which shape they liked most.
Gridgeman studies curvature, area, perimeter, evolutes, etc.
6.BL. TAN-1 ⅓ + TAN-1 ½ = TAN-1 1, ETC.
This
problem is usually presented with three squares in a row with lines drawn from
one corner to the opposite corners of the squares. New section. Similar
formulae occur in finding series for
π. See 6.A and my
Chronology of π.
L. Euler. Introductio in Analysin Infinitorum. Bousquet, Lausanne, 1748. Vol. 1, chap. VIII, esp. § 142 (??NYS). = Introduction to the Analysis of the
Infinite; trans. by John D. Blanton; Springer, NY, 1988-1990; Book I, chap.
VIII: On transcendental quantities which arise from the circle, pp. 101-115,
esp. § 142, pp. 114-115. Developing
series to calculate π, he considers angles a, b
such that a + b = π/4, then examines the formula for tan (a + b)
and says: "If we let tan a = ½,
then tan b = ⅓ .... In this way we calculate ... π,
with much more ease than ... before." Conway & Guy give some more details.
Carroll. ?? -- see Lowry (1972) and Conway & Guy
(1996).
Størmer. 1896.
See Conway & Guy.
Gardner. SA (Feb 1970) = Circus, chap. 11, prob.
3. Says he received the geometric
problem from Lyber Katz who had been given it when he was in 4th grade in
Moscow.
C. W. Trigg. A three‑square geometry problem. JRM 4:2 (Apr 1971) 90‑99. Quotes a letter from Katz, dating his 4th
year as 1931-32. Trigg sketches 54
proofs of the result, some of which generalize.
H. V. Lowry. Note 3331:
Formula for π/4. MG 56 (No.
397) (Oct 1972) 224-225. tan‑1 1/a = tan-1 1/b + tan-1
1/c implies a(b+c) = bc - 1,
hence (b‑a)(c‑a) = a2 + 1, whence all integral solutions can be
determined. Conway & Guy say this
was known to Lewis Carroll.
J. R. Goggins & G. B.
Gordon. Note 3346: Formula for π/4 (see Note 3331, Oct
1972). MG 57 (No. 400) (Jun 1973) 134. Goggins gets π/4 = Σn=1 tan-1 1/F2n+1, where
Fn is the n‑th Fibonacci number. [I think this formula was found by Lehmer
some years before??] Gordon also
mentions Eureka No. 35, p. 22, ??NYS, and finds recurrences giving tan-1 1/pn + tan-1
1/qn = tan-1 1/rn.
Douglas A. Quadling. Classroom note 304: The story of the three squares
(continued). MG 58 (No. 405) (Oct 1974)
212‑215. The problem was given in
Classroom note 295 and many answers were received, including four proofs
published by Roger North. Quadling
cites Trigg and determines which proofs are new. Trigg writes that tan‑1 1/F2n+2
+ tan-1 1/F2n+1
= tan-1 1/F2n, which is the basis of Goggins' formula.
Alan Fearnehough. On formulas for π involving inverse
tangent functions and Prob. 23.7.
MS 23:3 (1990/91) 65-67 & 95.
Gives four basic theorems about inverse tangents leading to many
different formulae for π/4. The problem gives a series using inverse
cotangents.
John H. Conway & Richard K.
Guy. The Book of Numbers. Copernicus (Springer-Verlag), NY, 1996. Pp. 241-248 discusses relationships among
values of tan-1 1/n which they denote as tn and call Gregory's numbers.
Euler knew t = t2 + t3, t1 = 2 t3 + t7 and
t1 = 5 t7 + 2 t18 - 2 t57 and used them to compute π
to 20 places in an hour. They
say Lewis Carroll noted that tn
= tn+c + tn+d if
and only if cd = n2 +
1. In 1896, Størmer related Gaussian
integers to Gregory numbers and showed how to obtain a Gregory number as a sum
of other Gregory numbers. From this it
follows that the only two-term expressions for
π/4 are t2 + t3, 2 t2 - 4 t7, 2 t3 + t7 and
4 t5 - t239.
This is described in Conway & Guy, but they have a misprint of 8
for 18 at the bottom of p. 246.
6.BM. DISSECT CIRCLE INTO TWO HOLLOW OVALS
Consider a circle of radius 2. Cut it by two
perpendicular diameters and by the circle of radius 1 about the centre. Two of the outer pieces (quarters of the
annulus) and two of the inner pieces (quadrants) make an oval shape, with a
hollow in the middle. The problem often
refers to making two oval stools and the hollows are handholds! After the references below, the problem
appears in many later books.
Jackson. Rational Amusement. 1821.
Geometrical Puzzles, no. 9, pp. 25 & 84 & plate I, fig. 6. Mentions handholes. Solution is well drawn.
Endless Amusement II. 1826?
Mentions handholes. Solution is
well drawn.
Crambrook. 1843.
P. 4, no. 6: A Circle to form two Ovals. Check??
Magician's Own Book. 1857.
Prob. 36: The cabinet maker's puzzle, pp. 277 & 300. Solution is a bit crudely drawn. = Book of 500 Puzzles, 1859, pp. 91 &
114. = Boy's Own Conjuring Book, 1860,
prob. 35, pp. 240 & 265.
Family Friend (Dec 1858)
359. Practical puzzle -- 4. I don't have the answer.
The Secret Out. 1859.
The Oval Puzzle, pp. 380-381.
Asks to 'produce two perfect ovals.'
Solution is a bit crudely drawn, as in Magician's Own Book, but the text
and numbering of pieces is different.
Illustrated Boy's Own
Treasury. 1860. Prob. 34, pp. 401 & 441. Same crude solution as Magician's Own Book,
but with different text, neglecting to state that the stools have handholes in
their centres.
Magician's Own Book (UK
version). 1871. On p. 282, in the middle of an unrelated
problem, is the solution diagram, very poorly drawn -- the pieces of the oval
stools are shown as having curved edges almost as though they were circular
arcs. There is no associated text.
Hanky Panky. 1872.
The oval puzzle, p. 123. Same
crude solution as Magician's Own Book, but different text, mentioning
handholes.
Mittenzwey. 1880.
Prob. 256, pp. 46 & 97;
1895?: 285, pp. 50 & 99;
1917: 285, pp. 45 & 94. The
stools are very poorly drawn, with distinct wiggles in what should be straight
lines.
Hoffmann. 1893.
Chap. II, no. 32: The cabinet maker's puzzle, pp. 104 & 137‑138
= Hoffmann-Hordern, p. 102.
Mentions hand holes. Well drawn
solution.
Benson. 1904.
The cabinet‑maker's puzzle, p. 201. Mentions hand holes. Poor
drawing.
6.BN. ROUND PEG IN SQUARE HOLE OR VICE VERSA
Wang Tao‑K'un. How to get on. Late 16C. Excerpted and
translated in: Herbert A. Giles; Gems
of Chinese Literature; 2nd ed. (in two vols., Kelly & Walsh, 1923), in one
vol., Dover, 1965, p. 226. "...
like square handles which you would thrust into the round sockets ..."
Sydney Smith. Sketches of Moral Philosophy. Lecture IX.
1824. "If you choose to
represent the various parts in life by holes upon a table, of different shapes,
-- some circular, some triangular, some square, some oblong, -- and the persons
acting these parts by bits of wood of similar shapes, we shall generally find
that the triangular person has got into the square hole, the oblong into the
triangular, and a square person has squeezed himself into the round hole. The officer and the office, the doer and the
thing done, seldom fit so exactly that we can say they were almost made for
each other." Quoted in: John Bartlett; Familiar Quotations; 9th ed.,
Macmillan, London, 1902, p. 461 (without specifying the Lecture or date). Irving Wallace; The Square Pegs;
(Hutchinson, 1958); New English Library, 1968; p. 11, gives the above quote and
says it was given in a lecture by Smith at the Royal Institution in 1824. Bartlett gives a footnote reference: The right man to fill the right place --
Layard: Speech, Jan. 15, 1855. It is
not clear to me whether Layard quoted Smith or simply expressed the same idea
in prosaic terms . Partially quoted,
from 'we shall ...' in The Oxford Dictionary of Quotations; 2nd ed. revised,
1970, p. 505, item 24. Similarly quoted
in some other dictionaries of quotations.
I have located other quotations
from 1837, 1867 and 1901.
William A. Bagley. Paradox Pie. Vawser & Wiles, London, nd [BMC gives 1944]. No. 17: Misfits, p. 18. "Which is the worst misfit, a square peg
in a round hole or a round peg in a square hole?" Shows the round peg fits better. He notes that square holes are hard to make.
David Singmaster. On round pegs in square holes and square
pegs in round holes. MM 37 (1964) 335‑337. Reinvents the problem and considers it
in n
dimensions. The round peg fits
better for n < 9. John L. Kelley pointed out that there must
be a dimension between 8 and
9 where the two fit equally
well. Herman P. Robinson kindly
calculated this dimension for me in 1979, getting 8.13795....
David Singmaster. Letter:
The problem of square pegs and round holes. ILEA Contact [London] (12 Sep 1980) 12. The two dimensional problem appears as a SMILE card which was
attacked as 'daft' in an earlier letter.
Here I defend the problem and indicate some extensions -- e.g. a circle
fits better in a regular n‑gon
than vice‑versa for all n.
I
have generally avoided classical geometry of this sort, but Bankoff's paper
deserves inclusion.
Leon Bankoff. The metamorphosis of the butterfly
problem. MM 60 (1987) 195‑210. Includes historical survey of different
proofs. The name first appears in the
title of the solution of Problem E571, AMM 51 (1944) 91 (??NYS). The problem first occurs in The Gentlemen's
Diary (1815) 39‑40 (??NYS).
6.BP. EARLY MATCHSTICK PUZZLES
There are too many matchstick puzzles
to try to catalogue. Some of them occur
in other sections, e.g. 6.AO.1. Here I
only include a few very early other examples.
At first I thought these would date from mid to late 19C when matches
first started to become available, but the earliest examples refer to slips of
paper or wood. The earliest mention of
matches is in 1858.
Rational Recreations. 1824.
Exer. 23, p. 132. Double a sheep
pen by adding just two hurdles. I have
just realised this is a kind of matchstick puzzle and I suspect there are other
early examples of this. It is a little
different than most matchstick puzzles in that one is usually not given an
initial pattern, but must figure it out.
(A hurdle is a kind of panel woven from sticks or reeds used by
shepherds to make temporary pens.)
Family Friend 2 (1850) 148 &
179. Practical Puzzle -- No. V. = Illustrated Boy's Own Treasury, 1860,
Prob. 46, pp. 404 & 443. "Cut
seventeen slips of paper or wood of equal lengths, and place them on a table,
to form six squares, as in the diagram.
...."
Magician's Own Book. 1857.
Prob. 20: Three square puzzle, pp. 273 & 296. (I had 87 & 110 ??) Almost identical to Family Friend, with a
few changes in wording and a different drawing, e.g. "Cut seventeen slips
of cardboard of equal lengths, and place them on a table to form six squares,
as in the diagram. ...."
The Sociable. 1858.
Prob. 2: The magic square, pp. 286 & 301. "With seventeen pieces of wood (lucifer matches will answer
the purpose, but be careful to remove the combustible ends, and see that they
are all of the same length) make the following figure: [a 2
x 3 array of squares]", then
remove 5 matches to leave three squares.
Book of 500 Puzzles. 1859.
Prob.
2: The magic square, pp. 4 & 19. As
in The Sociable.
Prob.
20: Three square puzzle, pp. 87 & 110.
Identical to Magician's Own Book.
Boy's Own Conjuring Book. 1860.
Three-square puzzle, pp. 235 & 259.
Identical to Magician's Own Book.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 5: The
three squares, pp. 28 & 31. Almost
identical to Magician's Own Book, prob. 20, with a slightly different diagram.
Mittenzwey. 1880.
Prob. 156-171, 202-212, 240, 242, pp. 32-33, 37-38, 44 & 83-85, 90,
94; 1895?: 179-196, 227-237, 269, 271,
pp. 37-38, 41-42, 48 & 85-87, 92, 96;
1917: 179‑196, 227-237, 269, 271, pp. 33-34, 38-39, 44 &
82-84, 88, 92. This is the first puzzle
book to have lots of matchstick problems, though he doesn't yet use the name,
calling them 'Hölzchen' (= sticks).
Several problems occur elsewhere, e.g. in 6.AO.1. The last problem is the same as in Jackson.
Sophus Tromholt. Streichholzspiele. Otto Spamer, Leipzig, 1889;
5th ed., 1892; 14th ed.,
Leipzig, 1909; slightly revised and
with a Preface by Rüdiger Thiele, Zentralantiquariat der DDR, Leipzig,
1986; Hugendubel, 1986. [Christopher 1017 is Spamer, 1890. C&B give 1890. There is an edition by Ullstein, Frankfurt, 1990.] This is the earliest book I know which is
entirely devoted to matchstick puzzles.
Gaston Tissandier. Jeux et Jouets du jeune age Choix de récréations amusantes & instructives. Ill. by Albert Tissandier. G. Masson, Paris, nd [c1890]. P. 40, no. 4-5: Le problème des
allumettes. Make five squares with nine
matches. Solution has four small
squares and one large one.
Clark. Mental Nuts. 1897, no.
56. The toothpicks. Use 12 toothpicks to make a 2 x 2
array of squares. Move three
picks to form three squares.
H. D. Northrop. Popular Pastimes. 1901. No. 1: The magic
square, pp. 65 & 71. = The Sociable.
6.BQ. COVERING A DISC WITH DISCS
The general problem is too complex to
be considered recreational. Here I will
mainly deal with the carnival version where one tries to cover a circular spot
with five discs. In practice, this is
usually rigged by stretching the cloth.
Eric H. Neville. On the solution of numerical functional
equations, illustrated by an account of a popular puzzle and of its solution. Proc. London Math. Soc. (2) 14 (1915)
308-326. Obtains several possible
configurations, but says "actual trial is sufficient to convince"
that one is clearly the best, namely the elongated pentagon with 2-fold
symmetry. This leads to four trigonometric
equations in four unknown angles which theoretically could be solved, but are
difficult to solve even numerically. He
develops a modification of Newton's method and applies it to the problem,
obtaining the maximal ratio of spot radius to disc radius as 1.64091.
Described by Gardner and Singleton.
Ball. MRE. 10th ed., 1922. Pp. 253-255: The five disc problem. Sketches Neville's results.
Will Blyth. More Paper Magic. C. Arthur Pearson, London, 1923.
Cover the spot, pp. 66-67. "This
old "fun of the fair" game has been the means of drawing many pennies
from the pockets of frequenters of fairs." Says the best approach is an elongated pentagon which has only
2-fold symmetry.
William Fitch Cheney Jr,
proposer; editorial comment. Problem E14. AMM 39 (1932) 606
& 42 (1935) 622. Poses the problem. Editor says no solution of this, or its equivalent, prob. 3574,
was received, but cites Neville.
J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Cover the spot, pp. 132-134. Uses discs of diameter 1 5/8 in
to cover a circle of diameter 2
1/2 in. This is a ratio of 20/13 = 1.538..., which should be fairly easy to cover?? His first disc has its edge passing through the centre of the
circle. His covering pattern has
bilateral symmetry, though the order of placing the last two discs seems
backward.
Walter B. Gibson. The Bunco Book. (1946); reprinted by
Citadel Press (Lyle Stuart Inc.), Secaucus, New Jersey, 1986. Spotting the spot, pp. 24-25. Also repeated in summary form, with some
extra observations in: Open season on
chumps, pp. 97-106, esp. pp. 102‑103.
The circles have diameter
5" and the discs "are
slightly more than three inches in diameter." He assumes the covering works exactly when the discs have
five-fold symmetry -- which implies the discs are 3.090... inches in diameter
-- but that the operator stretches the cloth so the spot is unsymmetric and the
player can hardly ever cover it -- though it still can be done if one plays a
disc to cover the bulge. "There is
scarcely one chance in a hundred that the spectator will start correctly
...." On p. 103, he adds that
"in these progressive times" the bulge can be made in any direction
and that shills are often employed to show that it can be done, though it is
still difficult and the operator generally ignores small uncovered bits in the
shills' play in order to make the game seem easy.
Martin Gardner. SA (Apr 1959)?? c= 2nd Book, chap. 13.
Describes the five disc version as Spot-the-Spot. Cites Neville.
Colin R. J. Singleton. Letter:
A carnival game -- covering disks with smaller disks. JRM 24:3 (1992) 185-186. Responding to a comment in JRM 24:1, he
points out that the optimum placing of five discs does not have pentagonal
symmetry but only bilateral. Five discs
of radius 1 can then cover a disc of radius
1.642.., rather than 1.618...,
which occurs when there is pentagonal symmetry. He cites Gardner and E. H. Neville. His
1.642 arises because Gardner had
truncated the reciprocal ratio to three places.
6.BR. WHAT IS A GENERAL TRIANGLE?
David & Geralda Singmaster,
proposers; Norman Miller, solver. Problem E1705 -- Skewness of a
triangle. AMM 71:6 (1964) 680 &
72:6 (1965) 669. Assume a £ b £
c. Define the skewness of the triangle
as S = max {a/b, b/c, c/a} x min {a/b,
b/c, c/a}. What triangles have maximum
and minimum skewness? Minimum is S = 1
for any isosceles triangle.
Maximum occurs for the degenerate triangle with sides 1, φ, 1+φ, where
1 + φ = φ2, so φ = (1 + Ö5)/2 is the golden mean.
Baruch Schwarz & Maxim
Bruckheimer. Let ABC
be any triangle. MTr 81 (Nov
1988) 640-642. Assume AB < AC < BC and ÐA
< 90o. Drawing BC
and putting A above it leads to a small curvilinear
triangular region where A can be.
Making A equidistant from the three boundaries leads
to a triangle with sides proportional to
Ö33, 7, 8 and
with angles 44.5o, 58.5o, 77o. The sides
are roughly in the proportion 6 :
7 : 8.
Gontran Ervynck. Drawing a 'general' triangle. Mathematics Review (Nov 1991). ??NYS -- cited by Anon., below. Notes that if we take an acute triangle with
angles as different as possible, then we get the triangle with angles 45o, 60o, 75o.
Anon. [possibly the editor, Tom
Butts]. What is a 'general'
triangle? Mathematical Log 37:3 (Oct
1993) 1 & 6. Describes above two
results and mentions Guy's article in 8.C.
Gives an argument which would show the probability of an acute triangle
is 0.
Anon. [probably the editor,
Arthur Dodd]. A very scalene
triangle. Plus 30 (Summer 1995) 18-19
& 23. (Content says this is
repeated from a 1987 issue -- ??NYS.)
Uses the same region as Schwarz & Bruckheimer, below. Looks for a point as far away from the
boundaries as possible and takes the point which gives the angles 45o, 60o, 75o.
In 1995?, I experimented with
variations on the definition of skewness given in the first item above, but
have not gotten much. However,
taking a = 1, we have 1 £ b £ c £ b
+ 1. Plotting this in the b, c
plane gives us a narrow strip extending to infinity. For generality, it would seem that we
want
c = b + ½,
but there is no other obvious condition to select a central point in
this region. As fairly random points, I
have looked at the case where c = b2, which gives
b = (1 + Ö3)/2 = 1.366.., c = 1.866..
-- this triangle has angles
about 31.47o, 45.50o, 103.03o
-- and at the case where b = 3/2,
which gives a triangle with sides proportional to 2, 3, 4
with angles about 28.96o, 46.57o, 104.46o. In
Mar 1996, I realised that the portion of the strip corresponding to an acute
triangle tends to 0 !!
I have now (Mar 1996) realised
that the situation is not very symmetric.
Taking c = 1, we have
0 £ a £ b £ 1 £ a
+ b and plotting this in the a, b
plane gives us a bounded triangle with vertices at (0, 1),
(½, ½), (1, 1). There are various possible central points of
this triangle. The centroid is at (1/2, 5/6),
giving a triangle with sides
1/2, 5/6, 1 which is similar
to 3, 5, 6, with angles 29.93o, 56.25o, 93.82o. An
alternative point in this region is the incentre, which is at (½, ½{-1 + 2Ö2}), giving a triangle similar to
1, -1 + 2Ö2, 2 with angles 29.85o, 65.53o, 84.62o. The probability
of an acute triangle in this context is
2 - π/2 = .429.
6.BS. FORM SIX COINS INTO A HEXAGON
O
O O O O
Transform
O O O into O O in three moves.
O O
New section -- there must be
older versions.
Young World. c1960.
P. 13: Ringing the change. He
starts with the mirror image of the first array.
Robert Harbin. Party Lines. Op. cit. in 5.B.1.
1963. The ring of coins, p. 31. Says it is described by Gardner, but gives
no details. Notes that if you show the
trick to someone
O O O and
then give the coins in the mirror image pattern shown at the left, he will
O
O O not be able to do it.
Putnam. Puzzle Fun.
1978.
No. 12: Create a space, pp. 4
& 27. With four coins, create the
pattern OO
on
the right. [Takes two moves from a
rhombic starting pattern.] O O
No. 13: Create a space again,
pp. 4 & 38. Standard hexagon
problem.
6.BT. PLACING OBJECTS IN CONTACT
New section. The objects involved are usually common objects such as coins or
cigarettes, etc. The standard
recreation is to have them all touching one another. However, the more basic question of how many spheres can touch a
sphere goes back to Kepler and perhaps the Greeks. Similar questions have been asked about cubes, etc.
Endless Amusement II. 1826?
Problem II, p. 189. "Five
shillings or sixpences may be so placed over each other, as to be all visible
and all be in contact." Two
solutions. The first has two coins on
the table, then two coins on top moved far enough onto one of the lower coins
that a vertical coin can touch both of them and the two lower coins at
once. The second solution has one coin
with two coins on top and two slanted coins sitting on the bottom coin and touching
both coins in the second layer and then touching each other up in the air. [I have recently read an article analysing
this last solution and showing that it doesn't work if the coin is too thick
and that the US nickel is too thick.] =
New Sphinx, c1840, p. 131.
Will Baffel. Easy Conjuring without apparatus.
Routledge & Dutton,
nd, 4th ptg [c1910], pp. 103-104.
Six matches, each touching all others.
Make a V with two matches and place a third match in
the notch to make a short arrow. Lie
one of these on top of another.
Will Blyth. Money Magic. C. Arthur Pearson, London, 1926.
Five in contact, pp. 98-101.
Same as Endless Amusement II.
Rohrbough. Puzzle Craft. 1932. Six Nails, p. 22 (=
p. 22 of 1940??). As in Baffel.
Meyer. Big Fun Book. 1940. Five coins, p. 543. Same as the first version in Endless
Amusement II.
Philip Kaplan. More Posers. (Harper & Row, 1964);
Macfadden-Bartell Books, 1965.
Prob. 29, pp. 35 & 92.
Second of the forms given in Endless Amusement II.
Ripley's Puzzles and Games. 1966.
P. 36. Six cigarettes, as in
Baffel.
I recall this is in Gardner and
that a solution with 6 cigarettes was improved to 7.
New
Section. This is really a proper
geometric topic, but there is some recreational interest in two aspects.
A. Attempts to construct regular n-gons for impossible values of n,
e.g. n = 7, either by ruler and compass or by origami or
by introducing new instruments -- see 6.BV.
B. Attempts to construct possible cases,
e.g. n = 5, by approximate methods.
Aspect A is closely related to
the classic impossible problems of trisecting an angle and duplicating a cube
and hence some of the material occurs in books on mathematical cranks -- see
Dudley. Further, there were serious
attempts on both aspects from classic times onward.
Abu’l-Jūd. 11C.
He "devised a geometrical method to divide the circle into nine
equal parts." [Seyyed Hossein
Nasr; Islamic Science -- an Illustrated Study; World of Islam Festival
Publishing Co., London?, 1976, p. 82.
Q. Mushtaq & A. L. Tan; Mathematics: The Islamic Legacy; Noor
Publishing House, Farashkhan, Delhi, 1993, p. 70.]
Pacioli. De Viribus.
c1500. These problems are
discussed by Mackinnon, op. cit. in 6.AT.3, pp. 167, 169, citing Agostini, p.
5. Let
ln be the side of a
regular n-gon inscribed in a unit
circle.
Ff.
146r-147r, XXIII afare la 7a fia dicta nonangolo. cioe de
.9. lati difficile (XXIII to make the 7th figure called nonagon, that is of 9
sides, difficult) = Peirani 198‑199. Asserts
l9 = (l3 + l6)/4. Mackinnon computes this gives .6830
instead of the correct .6840.
Ff.
148r-148v, XXV. Documento della 9 fia recti detta
undecagono (XXV. on the 9th rectilinear figure called undecagon) = Peirani 200. Asserts l11 =
φ (l3 + l6)/3,
where φ is the golden mean: (1 - Ö5)/2. Mackinnon computes this gives .5628
instead of the correct .5635.
F.
148v, XXVI. Do. de' .13. (XXVI. on the 13th)
= Peirani 200. Asserts l13 = (1 φ)·5/4. Mackinnon computes this gives .4775
instead of the correct .4786.
Ff.
149r-149v, XXVIII. Documento del .17. angolo cioe fia de .17. lati
(XXVIII on the 17-angle, that is the figure of 17 sides) = (Peirani 201-202). Peirani says some words are missing in the
second sentence of the problem and Agostini says the text is too corrupt to be
reconstructed. MacKinnon suggests l17 = (l3 - l6)/2 which gives .3660 instead of the correct .3675.
Barbaro, Daniele. La Practica della Perspectiva. Camillo & Rutilio Borgominieri, Venice,
(1569); facsimile by Arnaldo Forni,
1980, HB. [The facsimile's TP doesn't
have the publication details, but they are given in the colophon. Various catalogues say there are several
versions with dates on the TP and colophon varying independently between 1568
and 1569. A version has both dates being
1568, so this is presumed to be the first appearance. Another version has an undated title in an elaborate border and
this facsimile must be from that version.]
Pp. 26-27 includes discussion of constructing a regular heptagon, but it
just seems to say to divide the circumference of a circle into seven equal
parts -- ??
Christian Huygens. Oeuvres Complètes. Vol. 14, 1920, pp. 498-500: problem dated 1662, "To inscribe
a regular heptagon in a circle."
??NYS -- discussed by Archibald.
R. C. Archibald. Notes (to Problems and Solutions section)
24: Problems discussed by Huygens. AMM 28 (1921) 468‑479 (+??). The third of the problems discussed is the
construction of the heptagon quoted above.
Archibald gives an extensive survey of the topic on pp. 470-479. A relevant cubic equation was already found
by an unknown Arab writer, c980, and occurs in Vieta and in Kepler's Harmonices
Mundi, book I, Prop. 45, where Kepler doubts that the heptagon can be
constructed with ruler and compass.
An
approximate construction was already given by Heron of Alexandria and may be
due to Archimedes -- this says the side of the regular heptagon is
approximately half the side of the equilateral triangle inscribed in the same
circle. Jordanus Nemorarius (c1230)
called this the Indian method. Leonardo
da Vinci claimed it was exact. For the
central angle, this approximation gives a result that is about 6.5'
too small.
Archibald
then goes on to consider constructions which claim to work or be good
approximations for all n-gons. The earliest seems to be due to Antoine de
Ville (1628), revised by A. Bosse (1665).
In 1891, A. A. Robb noted that a linkage could be made to construct the
heptagon and J. D. Everett (1894) gave a linkage for n-gons.
Italo Ghersi; Matematica Dilettevoli
e Curiosa; 2nd ed., Hoepli, 1921; pp. 425-430: Costruzioni approssimate.
He
says the following construction is given by Housel; Nouvelles Annales de
Mathématiques 12 (1853) 77-?? with no indication of its source. Ghersi says it also occurs in Catalan's
Trattato di Geometria, p. 277, where it is attributed to Bion. However, Ghersi says it is due to Rinaldini
(probably Carlo Renaldini (1615-1698)).
Let AOB be a horizontal diameter of a circle of
radius 1 and form the equilateral triangle ABC with C
below the diameter. Divide AB
into n equal parts and draw the line through C and the point 4/n
in from B. Where this line hits the circle, say P,
is claimed to be 1/n of the way around the circumference from B.
Ghersi obtains the coordinates of
P and the angle BOP
and computes a table of these values compared to the real values. The method works for n = 2, 3, 4, 6. For n = 17, the error is 36'37".
On
pp. 428-430, he discusses a method due to Bardin. Take AOB as above and draw the perpendicular
diameter COD. Divide the diameter into
n equal parts and extend both
diameters at one end by this amount to points
M, N. Draw the line MN
and let it meet the circle near
B at a point P.
Now the line joining P to the third division point in from B is
claimed to be an edge of the regular
n-gon inscribed in the circle.
Ghersi computes this length, finding the method only works for n ³ 5, and gives a table of values
compared to the real values. This is
exact for n = 6 and is substantially more accurate than
Renaldini's method. For n = 17,
the error is 1'10.32".
The "New" School of
Art Geometry, Thoroughly Remodelled so as to Satisfy all the Requirements of
the Science and Art Department for
Science Subject I. Sections I. and II,
Practical Plane and Solid Geometry,
(Cover says: Gill's New School
of Art Geometry Science Subject I.) George Gill and Sons, London, 1890.
Pp.
26-27, prob. 66 -- To describe any regular Polygon on a given straight line,
AB. He constructs the centre of a
regular n-gon with AB as one edge. Taking the side AB as 1,
the height hn of the centre is given by hn = (n‑4) Ö3/4 ‑ (n-6)/4, while the correct answer is ½ cot π/n. For large n, the relative error approaches 14.99%.
He gives no indication that the method is only approximate and doesn't
even work for n = 5.
Pp.
74-75, prob. 188 -- To inscribe any Regular Polygon in a given circle. He gives three methods. The first is to do it by trial! The second requires being able to construct
the regular 2n-gon! The third construction is Renaldini's, which
he does indicate is approximate.
R. C. Archibald, proposer; H. S. Uhler, solver. Problem 2932. AMM 28 (1921) 467 (??NX)
& 30 (1923) 146-147. Archibald gives De Ville's construction and
asks for the error. Uhler gives values
of the error for n = 5, 6, ...,
20, and the central angles are
about 1o too large, even for n = 6,
though the error seems to be slowly decreasing.
T. R. Running. An approximate construction of the side of a
regular inscribed heptagon. AMM 30
(1923) 195-197. His central angle
is .000061" too small.
W. R. Ransom, proposer; E. P. Starke, solver. Problem E6.
AMM 39 (1932) 547 (??NX) & 40 (1933) 175-176. Gives Dürer's method for the pentagon and asks if it is
correct. Starke shows the central angle
is about 22' too large.
C. A. Murray, proposer; J. H. Cross, E. D. Schell, Elmer Latshaw,
solvers. Problem E697 -- Approximate
construction of regular pentagon. AMM
52 (1945) 578 (??NX) & 53 (1946) 336-337. Describes a method similar to that of de Ville - Bosse and asks
if it works for a pentagon. Latshaw
considers the general case. The formula
is exact for n = 3, 4,
6. For
n = 5, the central angle is 2.82'
too small. For n > 6,
the central angle is too large and the error is increasing with n.
J. C. Oldroyd. Approximate constructions for 7, 9, 11,
13-sided polygons. Eureka 18 (Oct 1955)
20. Gives fairly simple constructions which
are accurate to a few seconds.
Marius Cleyet-Michaud. Le Nombre d'Or. Presses Universitaires de France, Paris, 1973. Méthode dite d'Albert Dürer, pp. 45-47. Describes Dürer's approximate method for the
pentagon and says it fails by 22'.
Underwood Dudley. Mathematical Cranks. MAA Spectrum, 1992. This book discusses many related problems,
e.g. duplication of the cube, trisection of the angle. The chapter: Nonagons, Regular, pp.
231‑234 notes that there seem to be few crank constructors of the
heptagon but that a nonagoner exists -- Dudley does not identify him. Actually he constructs 10o with an error of about
.0001', so he is an excellent
approximater, but he claims his construction is exact.
Robert Geretschläger. Euclidean constructions and the geometry of
origami. MM 68:5 (Dec 1995)
357-371. ??NYS -- cited in next
article, where he states that this shows that all cubic equations can be solved
by origami methods.
Robert Geretschläger. Folding the regular heptagon. CM 23:2 (Mar 1997) 81-88. Shows how to do it exactly, using the result
of his previous paper.
Dirk Bouwens, proposer; Alan Slomson & Mick Bromilow,
independent solvers. An early protractor. M500
171 (Dec 1999) 18 & 173 (Apr 2000) 16-17. Draw a semicircle on diameter AOB.
Draw the perpendicular through
O and extend it to C so
that BC = BA (the diameter)
[this makes c = OC = Ö3]
. If
P divides AO in the ratio λ : 1-λ, then
draw CP to meet the semicircle at
D and OD divides the arc ADE
in approximately the same ratio.
He finds the exact value of the angle
AOD and finds that the maximum
error in the process is only .637o (when
λ @ .18). Second author provides a graph of the error and says the maximum
error is at about 18o. [I
get .637375o at
.181625o.]
Ken Greatrix. A better protractor. M500 175 (Aug 2000) 14-15. Taking
c = 1.67721 in the previous
construction gives a more accurate construction, with maximum error about .32o at about 13o. [I get
.324020o at .130164o.]
New Section. This is really a proper geometric topic, but there is some
recreational interest in it, so I will cite some general references.
Robert C. Yates. Geometrical Tools. (As: Tools; Baton Rouge,
1941); revised ed., Educational Publishers,
St. Louis, 1949. Pp. 82-101 &
168-191. Excellent survey. After considering use of straightedge and
compasses, he considers: compasses
only; folds and creases; straightedge only; straight line linkages;
straightedge with fixed figure (circle, square or parallelogram); straightedge with restricted compasses
(collapsible compass, rigid (or rusty) compass or rigid dividers); parallel and angle rulers; higher order devices (marked ruler,
carpenter's square, tomahawk, compasses of Hermes, two right angle rulers,
straightedge with compasses and fixed conic);
plane linkages in general. Each
section has numerous references.
6.BW. DISTANCES TO CORNERS OF A SQUARE
New
section. If ABCD is a rectangle, P is a point and a, b, c, d are the
distances of P from the corners of the rectangle, then the
basic relation a2 + c2
= b2 + d2 is
easily shown. This leads to a number of
problems. A little research has found
references back to 1896, but the idea might be considerably older.
AMM 3 (1896) 155. ??NYS -- this is the earliest reference
given by Trigg, cf below.
SSM 15 (1915) 632. ??NYS -- cited by Trigg, below.
AMM 35 (1928) 94. ??NYS -- cited by Trigg, below.
SSM 32 (1932) 788. ??NYS -- cited by Trigg, below.
NMM 12 (1937) 141. ??NYS -- cited by Trigg, below.
AMM 47 (1940) 396. ??NYS -- cited by Trigg, below.
NMM 16 (1941) 106. ??NYS -- cited by Trigg, below.
NMM 17 (1942) 39. ??NYS -- cited by Trigg, below.
AMM 50 (1943) 392. ??NYS -- cited by Trigg, below.
SSM 46 (1946) 89, 783. ??NYS -- cited by Trigg, below.
SSM 50 (1950) 324. ??NYS -- cited by Trigg, below.
SSM 59 (1959) 500. ??NYS -- cited by Trigg, below.
"A. Polter Geist",
proposer; Joseph V. Michalowicz, Mannis
Charosh, solvers, with historical note by Charles W. Trigg. Problem 865 -- Locating the barn. MM
46:2 (Mar 1973) 104 & 47:1 (Jan 1974) 56-59. For a square with an interior point, a, b, c
= 13, 8, 5. How far is
P from the nearest side? First
solver determines the side, s, by applying the law of cosines to
triangles BPA and BPC and using that angles ABP
and BPC are complementary. This gives a fourth order equation which is
a quadratic in s2. Second solver uses a more geometric approach
to determine s. The distances to the sides are then easily
determined. Trigg gives 13 references
to earlier versions of the problem -- see above.
Anonymous. Puzzle number 35 -- Eccentric lighting. Bull. Inst. Math. Appl. 14:4 (Apr 1978) 110 &
13:5/6 (May/Jun 1978) 155. Light
bulb in a room, with distances measured from the corners. a, b, c
= 9, 6, 2. Find
d. Solution uses the theorem of
Apollonius to obtain the basic equation.
David Singmaster, proposer and
solver. Puzzle number 40 -- In the
beginning was the light. Bull. Inst.
Math. Appl. 14:11/12 (Nov/Dec 1978)
281 & 15:1 (Jan 1979) 28.
Assuming P is inside the rectangle, what are the
conditions on a, b, c for there to be a rectangle with these
distances? When is the rectangle
unique? When can P be
on a diagonal? Solution first obtains
the basic relation, which does not depend on
P being in the rectangle. Reordering the vertices if necessary, assume b is
the greatest of the distances.
Then a2 + c2
³ b2 is
necessary and sufficient for a rectangle to exist with these distances. This is unique if and only if equality
holds, when P = D. If the distances are all equal, then P is
at the centre of the rectangle, which can have a range of sizes. If the distances are not all equal, there is
a unique rectangle having P on a diagonal and it is on the diagonal
containing the largest and smallest distances.
Marion Walter. Exploring a rectangle problem. MM 54:3 (1981) 131-134. Takes
P inside the rectangle with a, b, c
= 3, 4, 5. Finds the basic
relation, noting P can be anywhere, and determines d.
Then observes that the basic relation holds even if P is
not in the plane of ABCD. Ivan Niven pointed out that the problem
extends to a rectangular box. Mentions
the possibility of using other metrics.
James S. Robertson. Problem 1147 -- Re-exploring a rectangle
problem. MM 55:3 (May 1982) 177
& 56:3 (May 1983) 180-181. With
P inside the rectangle and a, b, c
given, what is the largest rectangle that can occur? Observes that a2 + c2 > b2 is necessary and sufficient for a rectangle
to exist with P interior to it. He then gives a geometric argument which seems to have a gap in
it and finds the maximal area is
ac + bd.
I. D. Berg, R. L. Bishop & H. G. Diamond,
proposers. Problem E 3208. AMM 94:5 (May 1987) 456-457. Given
a, b, c, d satisfying the basic
relation, show that a rectangle containing
P can have any area from zero up
to some maximum value and determine this maximum.
Problem 168.2. M500 168 ??? Seven solvers, M500 170 (Oct 1999) 15-20. Given
P inside a square and a, b, c = 5, 3, 4, find the side of the square.
7. ARITHMETIC & NUMBER‑THEORETIC RECREATIONS
We
use the standard form: F0 =
0, F1 = 1, Fn+1 = Fn + Fn-1, with the auxiliary Lucas numbers being given
by: L0 = 2, L1 = 1, Ln+1 = Ln + Ln-1.
Parmanand Singh. The
so-called Fibonacci numbers in ancient and medieval India. HM 12 (1985) 229-244. In early Indian poetry, letters had weights
of 1
or 2 and meters were classified both by the number of letters and by
the weight. Classifying by weight gives
the number of sequences of 1s and
2s which add to the weight n
and this is Fn+1.
Pińgala
[NOTE: ń denotes n with an overdot.] (c-450) studied prosody
and gives cryptic rules which have been interpreted as methods for generating
the next set of sequences, either classified by number of letters or by weight
and several later writers have given similar rules. The generation implies Fn+1
= Fn + Fn-1.
Virahāńka [NOTE:
ń denotes n
with an overdot.] (c7C) is slightly more explicit. Gopāla (c1134) gives a commentary on
Virahāńka [NOTE: ń denotes
n with an overdot.] which
explicitly gives the numbers as 3, 5,
8, 13, 21. Hemacandra (c1150) states
"Sum of the last and last but one numbers ... is ... next." This is repeated by later authors.
The
Prākŗta [NOTE: ŗ denotes
r with an underdot]
Paińgala [NOTE: ń denotes
n with an overdot.] (c1315)
gives rules for finding the k-th
sequence of weight n and for finding the position of a particular
sequence in the list of sequences of weight
n and the positions of those
sequences having a given number of 2s (and hence a given number of letters). It also gives the relation Fn+1 = Σi
BC(n-i,i).
Narayana
Pandita (= Nārāyaņa Paņdita’s [NOTE: ņ
denotes n with an overdot and the d
should have an underdot.]) Gaņita[NOTE: ņ denotes n
with an underdot.] Kaumudī
(1356) studies additive sequences in chap. 13, where each term is the sum
of the last q terms. He gives rules
which are equivalent to finding the coefficients of (1 + x + ... + xq-1)p and relates to ordered partitions using 1, 2, ..., q.
Narayana Pandita (= Nārāyaņa Paņdita
[NOTE: ņ denotes n with an overdot and the d
should have an underdot.]).
Gaņita[NOTE: ņ denotes
n with an underdot.] Kaumudī (1356). Part I, (p. 126 of the Sanskrit ed. by P.
Dvivedi, Indian Press, Benares, 1942), ??NYS -- quoted by Kripa Shankar Shukla
in the Introduction to his edition of: Narayana Pandita (=
Nārāyaņa Paņdita); Bījagaņitāvatamsa [NOTE:
the ņ denotes an n with an under dot and there should be a dot
over the m.]; Part I; Akhila Bharatiya
Sanskrit Parishad, Lucknow, 1970, p. iv.
"A cow gives birth to a calf every year. The calves become young and themselves begin giving birth to
calves when they are three years old.
Tell me, O learned man, the number of progeny produced during twenty
years by one cow."
WESTERN
HISTORIES
H. S. M. Coxeter.
The golden section, phyllotaxis, and Wythoff's game. SM 19 (1953) 135‑143. Sketches history and interconnections.
H. S. M. Coxeter.
Introduction to Geometry. Wiley,
1961. Chap. 11: The golden section and
phyllotaxis, pp. 160-172. Extends his
1953 material.
Maxey Brooke.
Fibonacci numbers: Their history
through 1900. Fibonacci Quarterly 2:2
(1964) 149‑153. Brief sketch,
with lots of typographical errors.
Doesn't know of Bernoulli's work.
Leonard Curchin & Roger Herz-Fischler. De quand date le premier rapprochement entre
la suite de Fibonacci et la division en extrême et moyenne raison? Centaurus 28 (1985) 129-138. Discusses the history of the result that the
ratio Fn+1/Fn approaches
φ. Pacioli and Kepler,
described below, seem to be the first to find this.
Roger Herz‑Fischler.
Letter to the Editor. Fibonacci
Quarterly 24:4 (1986) 382.
Roger Herz-Fischler.
A Mathematical History of Division in Extreme and Mean Ratio. Wilfrid Laurier University Press, Waterloo,
Ontario, 1987. Retitled: A Mathematical
History of the Golden Number, with new preface and corrections and additions,
Dover, 1998. Pp. 157-162 discuss early
work relating the Fibonacci sequence to division in extreme and mean
ratio. 15 pages of references.
Georg Markovsky.
Misconceptions about the Golden Ratio.
CMJ 23 (1992) 2-19. This surveys
many of the common misconceptions -- e.g. that
-- appears in the Great Pyramid,
the Parthenon, Renaissance paintings and/or the human body and that the Golden
Rectangle is the most pleasing -- with 59 references. He also discusses the origin of the term 'golden section',
sketching the results given in Herz-Fischler's book.
Thomas Koshy.
Fibonacci and Lucas Numbers with Applications. Wiley-Interscience, Wiley, 2001.
Claims to be 'the first attempt to compile a definitive history and
authoritative analysis' of the Fibonacci numbers, but the history is generally
second-hand and marred with a substantial number of errors, The mathematical work is extensive, covering
many topics not organised before, and is better done, but there are more errors
than one would like.
Ron Knott has a huge website on Fibonacci numbers and their
applications, with material on many related topics, e.g. continued fractions,
π, etc. with some history.
www.ee.surrey.ac.uk/personal/r.knott/fibonacci/fibnat.html .
Fibonacci.
1202. Pp. 283‑284 (S:
404-405): Quot paria coniculorum in uno anno ex uno pario germinentur [How many
pairs of rabbits are created by one pair in one year]. Rabbit problem -- the pair propagate in the
first month so there are Fn+2 pairs at the end of the n-th month. (English translation in:
Struik, Source Book, pp. 2‑3.) I have colour slides of this from L.IV.20 & 21 and Conventi
Soppresi, C. I. 2616. This is on
ff. 130r-130v of L.IV.20, f. 225v
of L.IV.21, f. 124r of CS.C.I.2616.
Unknown early 16C annotator. Marginal note to II.11 in Luca Pacioli's copy of his 1509 edition
of Euclid. Reproduced and discussed in
Curchin & Herz-Fischler and discussed in Herz-Fischler's book, pp. 157-158. II.11 involves division in mean and extreme
ratio. Uses 89, 144, 233 and
that 1442 = 89 * 233 +
1. Also refers to 5, 8, 13.
Gori. Libro di
arimetricha. 1571. F. 73r (p.81). Rabbit problem as in Fibonacci.
J. Kepler. Letter of
Oct 1597 to Mästlin. ??NYS -- described
in Herz-Fischler's book, p. 158. This
gives a construction for division in extreme and mean ration. On the original, Mästlin has added his
numerical calculations, getting
1/φ = .6180340, which
Herz-Fischler believes to be the first time anyone actually calculated this
number.
J. Kepler. Letter of
12 May 1608 to Joachim Tanckius. ??NYS
-- described in Herz-Fischler (1986), Curchin & Herz-Fischler and
Herz-Fishler's book, pp. 160-161. Shows
that he knows that the ratio Fn+1/Fn approaches
φ and that Fn2 + (-1)n = Fn-1Fn+1.
J. Kepler. The Six‑Cornered
Snowflake. Op. cit. in 6.AT.3. 1611.
P. 12 (20‑21).
Mentions golden section in polyhedra and that the ratio Fn+1/Fn approaches
φ. See Herz-Fischler's
book, p. 161.
Albert Girard, ed.
Les Œuvres Mathematiques de Simon Stevin de Bruges. Elsevier, Leiden, 1634. Pp. 169‑170, at the end of Stevin's
edition of Diophantos (but I have seen other page references). Notes the recurrence property of the
Fibonacci numbers, starting with 0, and asserts that the ratio Fn+1/Fn approaches the ratio of segments of a line
cut in mean and extreme ratio, i.e.
φ, though he doesn't even
give its value -- but he says
13, 13, 21 'rather
precisely constitutes an isosceles triangle having the angle of a
pentagon'. Herz-Fischler's book, p.
162, notes that Girard describes it as a new result and includes 0 as
the starting point of the sequence.
Abraham de Moivre.
The Doctrine of Chances: or, A Method of Calculating the Probability of
Events in Play. W. Pearson for the
Author, London, 1718. Lemmas II &
III, pp. 128‑134. Describes
how to find the generating function of a recurrence. One of his illustrations is the Lucas numbers for which he
gets:
x +
3x2 + 4x3 + 7x4 + 11x5 + ... =
(2x + x2)/(1 - x - x2). However, he does not have the Fibonacci numbers and he does not
use the generating function to determine the individual coefficients of the
sequence. In Lemma III, he describes
how to find the recurrence of p(n) an where
p(n) is a polynomial. Koshy [p. 215] says De Moivre invented
generating functions to solve the Fibonacci recurrence, which seems to be
reading much more into De Moivre than De Moivre wrote. The second edition is considerably revised,
cf below.
Daniel Bernoulli.
Observationes de seriebus quae formantur ex additione vel substractione
quacunque terminorum se mutuo consequentium, ubi praesertim earundem insignis
usus pro inveniendis radicum omnium aequationum algebraicarum ostenditur. Comm. Acad. Sci. Petropolitanae 3
(1728(1732)) 85‑100, ??NYS. = Die
Werke von Daniel Bernoulli, ed. by L. P. Bouckaert & B. L. van der Waerden,
Birkhäuser, 1982, vol. 2, pp. 49+??.
Section 6, p. 52, gives the general solution of a linear recurrence when
the roots of the auxiliary equation are distinct. Section 7, pp. 52‑53, gives the 'Binet' formula for Fn. [Binet's presentation is so much less clear that I suggest the
formula should be called the Bernoulli formula.]
Abraham de Moivre.
The Doctrine of Chances: or, A Method of Calculating the Probability of
Events in Play. 2nd ed, H. Woodfall for
the Author, London, 1738. Of the
Summation of recurring Series, pp. 193-206.
This is a much revised and extended version of the material, but he says
it is just a summary, without demonstrations, as he has given the
demonstrations in his Miscellanea Analytica of 1730 (??NYS). Gives the generating functions for various
recurrences and even for a finite number of terms. Prop. VI is: In a recurring series, any term may be obtained
whose place is assigned. He assumes the
roots of the auxiliary equation are real and distinct. E.g., for a second order recurrence with
distinct roots m, p, he says the general term is Amn + Bpn where he has given A and B in
terms of the first two values of the recurrence. He even gives the general solution for a fourth order recurrence
and expresses A, B, C, D in terms of the first four values of the
recurrence. Describes how to take the
even terms and the odd terms of a recurrence separately and how to deal with
sum and product of recurrences.
R. Simson. An explication
of an obscure passage in Albert Girard's commentary upon Simon Stevin's
works. Phil. Trans. Roy. Soc. 48 (1753)
368‑377. Proves that Fn2 + (‑1)n = Fn-1Fn+1. This says that the triple Fn-1, Fn, Fn+1 "nearly express the segments of a line
cut in extreme and mean proportion, and the whole line;" from which he
concludes that the ratio Fn+1/Fn does converge to φ. Herz-Fischler's
book, p. 162, notes that his proof is essentially an induction. (He also spells the author Simpson, but it
is definitely Simson on the paper.)
[Koshy, p. 74, says Fn2 + (‑1)n = Fn-1Fn+1 was discovered in 1680 by Giovanni Domenico
Cassini, but he gives no reference and neither Poggendorff nor BDM help to
determine what paper this might be.]
Ch. Bonnet. Recherches
sur l'usage des feuilles dans les plantes.
1754, pp. 164‑188.
Supposed to be about phyllotaxis but only shows some spirals without any
numbers. Refers to Calandrini. Nice plates.
Master J. Paty (at the Mathematical Academy, Bristol),
proposer; W. Spicer, solver. Ladies'
Diary, 1768-69 = T. Leybourn, II: 293, quest. 584. Cow calves at age two and every year
thereafter. How many offspring in 40
years? Answer is: 0 + 1 + 1 + 2 + 3 + ... + F39. We would give this as: F41 - 1, but he gets it as: 2F39 + F38 - 1.
Eadon.
Repository. 1794. P. 389, no. 47. Same as the previous problem.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions, no. 32, pp.
22 & 82. Similar to Fibonacci, with
a cow and going for 20 generations.
Martin Ohm. Die
reine Elementar-Mathematik. 2nd ed.,
Jonas Verlags-Buchhandlung, Berlin, 1835.
P. 194, footnote to Prop. 5.
??NYS -- extensively discussed in Herz-Fischler's book, p. 168. This is the oldest known usage of 'goldene
Schnitt'. It does not appear in the 1st
ed. of 1826 and here occurs as:
"... nennt man wohl auch den goldenen Schnitt" (... one also
appropriately calls [this] the golden section). The word 'wohl' has many, rather vague, meanings, giving
different senses to Ohm's phrase.
Herz-Fischler interprets it as 'habitually', which would tend to imply
that Ohm and/or his colleagues had been using the term for some time. I don't really see this meaning and
interpreting 'wohl' as 'appropriately' would give no necessity for anyone else
to know of the phrase before Ohm.
However the term is used in several other German books by 1847. [Incidentally, this is not the Ohm of Ohm's
Law, but his brother.]
A. F. W. Schimper & A. Braun. Flora. 1835. Pp. 145 & 737. ??NYS
J. Binet. Mémoire
sur l'integration des équations linéaires aux différences finies, d'un ordre
quelconque, à coefficients variables.
(Extrait par l'auteur). CR Acad.
Sci. Paris 17 (1843) 559‑567.
States the Binet formula as an example of a general technique for
solving recurrences of the form: v(n+2)
= v(n+1) + r(n)v(n), but the general
technique is not clearly described, nor is the linear case.
B. Peirce.
Mathematical investigation of the fractions which occur in
phyllotaxis. Proc. Amer. Assoc. Adv.
Sci. 2 (1849) 444‑447. Not very
interesting.
Gustav Theodor Fechner.
Vorschule der Ästhetik.
Breitkopf & Härtel, Leipzig, 1876.
??NYS. Origin of the aesthetic
experiments on golden rectangles.
Koshy [p. 5] says Lucas originally called Fn the 'série de Lamé', but introduced the name Fibonacci numbers in
May 1876. However, he doesn't give a
reference. There are several papers by
Lucas which might be the desired paper.
Note
sur le triangle arithmétique de Pascal et sur la série de Lamé. Nouvelle Correspondence Mathématique 2
(1876) 70-75; which might be the desired paper.
L'arithmétique,
la série de Lamé, le problème de Beha-Eddin, etc. Nouvelles Annales de Mathématiques 15 (1876) 20pp.
Édouard Lucas.
Théorie des fonctions numériques simplement périodiques. AJM 1 (1878) 184-240 (Sections 1-23) &
289-321 (Sections 24-30). [There
is a translation by Sidney Kravitz of the first part as: The Theory of Simply Periodic Numerical
Functions, edited by Douglas Lind, The Fibonacci Association, 1969. Dickson I 400, says this consists of 7
previous papers in Nouv. Corresp. Math. in 1877-1878 with some corrections and
additions. Robert D. Carmichael; Annals
of Math. (2) 15 (1913) 30-70, ??NX, gives corrections.] The classic work which begins the modern
study of recurrences.
Koshy, p. 273, says Adolf Zeising's Der goldene Schnitt of
1884 put forth the theory that "the golden ratio is the most artistically
pleasing of all proportions ...."
But cf Fechner, 1876.
Pearson. 1907. Part II, no. 63: A prolific cow, pp. 126
& 203. Same as Fibonacci's rabbits,
but wants the total after 16 generations.
Koshy, p. 242, asserts that Mark Barr, an American
mathematician, introduced the symbol
φ (from Phidias) for the
Golden Ratio, (1 + Ö5)/2, about 1900, but he gives no reference.
Coxeter, 1953, takes
τ from the initial letter
of τoμη, the Greek word for section, but I have no
idea if this was used before him.
There is a magic trick where you ask someone to pick two
numbers and extend them to a sequence of ten by adding the last two numbers
each time. You then ask him to add up
the ten numbers and you tell him the answer, which is 11 times the seventh
number. In general, if the two starting
numbers are A and B, the
n-th term is Fn-2A + Fn-1B and the sum of the first 2n
terms is F2nA + (F2n+1-1)B = Ln
(FnA + Fn+1B),
but only the case n = 5 is interesting! I saw Johnny Ball do this in 1989 and I have found it in: Shari
Lewis; Abracadabra! Magic and Other
Tricks; (World Almanac Publications, NY, 1984); Puffin, 1985; Sum trick!,
p. 14, but it seems likely to be much older.
7.B. JOSEPHUS OR SURVIVOR PROBLEM
See
Tropfke 652.
This
is the problem of counting out every k‑th
from a circle of n. Early versions counted out half the
group; later authors and the Japanese
are interested in the last man -- the survivor. Euler (1775) seems to be the first to ask for the last man in
general which we denote as L(n, k). Cardan, 1539, is the first to associate this
process with Josephus. Some later
authors derive this from the Roman practice of decimation.
For
last man versions, see the general entries and: Michinori?,
Kenkō, Cardan, Coburg,
Bachet, van Etten, Yoshida,
Muramatsu, Schnippel, Ozanam (1696 & 1725), Les Amusemens, Fujita,
Euler, Miyake, Matuoka,
Boy's Own Book, Nuts to
Crack, The Sociable, Indoor & Outdoor, Secret Out (UK), Leske, Le Vallois, Hanky Panky, Kemp, Mittenzwey, Gaidoz,
Ducret, Lemoine, Akar et al,
Lucas, Schubert, Busche,
Tait, Ahrens, Rudin,
MacFhraing, Mendelsohn, Barnard,
Zabell, Richards, Dean,
Richards,
2 to last, counted
by 9s: Boy's Own Book,
3 to last, counted
by 9s: Boy's Own Book,
4 to last, counted
by 9s: Boy's Own Book,
5 to last, counted
by 9s: Boy's Own Book,
6 to last, counted
by 9s: Boy's Own Book,
7 to last, counted
by 9s: Boy's Own Book,
9 to last, counted
by 9s: Boy's Own Book,
10 to last, counted by 9s:
Boy's Own Book,
11 to last, counted by 9s:
Boy's Own Book,
12 to last, counted by 9s:
Boy's Own Book, Secret Out
(UK),
12 to last, counting number unspecified: Coburg,
13 to last, counted by 2s:
Ducret, Leeming,
13 to last, counted by 9s:
Boy's Own Book, Secret Out
(UK), Leske, Rudin,
14 to all!, counted by 6s:
Secret Out,
14 to last, counted by 10s:
Mittenzwey,
17 to last, counted by 3s:
Barnard,
21 to last, counted by 5s:
Hyde,
21 to last, counted by 7s:
Nuts to Crack, The
Sociable, Indoor & Outdoor, Hanky Panky, H. D. Northrop,
21 to last, counted by 8s:
Mittenzwey,
21 to last, counted by 10s:
Hyde,
24 to last, counted by 9s:
Kemp,
28 to last, counted by 9s:
Kemp,
30 to last, counted by 9s:
Schnippel,
30 to last, counted by 10s:
see entries in next table for 15
& 15 counted by 10s
40 to last man, counted by 3s: van Etten (erroneous),
41 to last man, counted by 3s: van Etten, Ozanam
(1725), Vinot, Ducret,
Lucas (1895),
General case: Euler, Lemoine,
Akar et al., Schubert, Busche,
Tait, Ahrens, MacFhraing,
Mendelsohn, Robinson, Jakóbczyk,
Herstein & Kaplansky,
Zabell, Richards,
There
are a few examples where one counts down to the last two persons -- see
references to Josephus and:
Pacioli, Muramatsu, Mittenzwey,
Ducret, Les Bourgeois Punis.
Almost
all the authors cited consider 15 &
15 counted by 9s, so I will only index
other versions.
2 & 2 counted by 3s: Ball (1911),
2 & 2 counted by 4s: Ball (1911),
3 & 3 counted by 7s: Ball (1911),
3 & 3 counted by 8s: Ball (1911),
4 & 4 counted by 2s: Leeming,
4 & 4 counted by 5s: Ball (1911),
4 & 4 counted by 9s: Ball (1911),
5 & 5 counted by ??: Dudeney (1905), Pearson,
Ball (1911), Ball (1920), Shaw,
6 & 6 counted by ??: Dudeney (1900),
8 & 2 counted by ??: Les Bourgeois
Punis,
8 & 8 counted by 8s: Kanchusen,
8 & 8 counted by ??: Dudeney (1899),
12 & 12 counted by
6s: Harrison,
15 & 15 counted by
3s: Tartaglia, Alberti,
15 & 15 counted by
4s: Tartaglia,
15 & 15 counted by
5s: Tartaglia,
15 & 15 counted by
6s: AR, Codex lat. Monacensis 14908, Tartaglia,
15 & 15 counted by
7s: Tartaglia, Schnippel,
15 & 15 counted by
8s: Codex lat. Monacensis
14836, AR, Codex lat. Monacensis 14908,
Tartaglia, Alberti,
15 & 15 counted by 10s:
Michinori?, Reimar von
Zweiter, AR, Codex lat. Monacensis 14908, Chuquet,
Tartaglia, Buteo, Hunt,
Yoshida, Muramatsu, Wingate/Kersey, Schnippel, Alberti, Shinpen Kinko-ki, Fujita, Miyake, Matuoka,
Sanpo Chie Bukuro,
Hoffmann, Brandreth, Benson,
Williams, Collins, Dean.
(Almost all of these actually continue to the last person.)
15 & 15 counted by 11s:
Tartaglia, Schnippel,
15 & 15 counted by 12s:
AR, Codex lat. Monacensis
14908, Tartaglia,
15 & 15 counted by other values, not specified --
??check: Codex lat. Monacensis
14836, Meermanische Codex, at‑Tilimsâni, Bartoli,
Murray 643, Chuquet, Keasby,
17 & 15 counted by 10s:
Schnippel,
17 & 15 counted by 12s:
Mittenzwey,
18 & 2 counted
by 12s: Pacioli, Rudin,
18 & 6 counted
by 8s:
Manuel des Sorciers,
18 & 18 counted by
9s: Chuquet,
24 & 24 counted by
9s: Chuquet,
30 & 2 counted
by 7s:
Pacioli,
30 & 2 counted
by 9s:
Pacioli,
30 & 6 counted by
10s: Ducret, 30 & 10 counted by 12s:
Endless Amusement II, Magician's
Own Book, The Sociable, Boy's Own Conjuring Book, Lucas (1895),
30 & 30 counted by 12s:
Sarma,
36 & 4 counted
by 10s: Jackson,
n2-n+1
& n-1 counted by
n: Lucas (1894), Cesarò,
Franel, Akar,
Many authors provide a mnemonic for the case of 15 and 15 counted by 9s. In this case, the longest group of the same
type is five, so a common device is to encode the numbers 1, 2, 3, 4, 5 by the vowels a, e,
i, o, u and then produce a phrase with
the vowels in the correct order. I will
call this a vowel mnemonic. The most
popular form is: Populeam virgam Mater
Regina ferebat, giving the numerical
sequence: 4, 5, 2, 1, 3, 1, 1, 2,
2, 3, 1, 2, 2, 1. The first group of 4
are good guys, followed by 5 bad guys, etc.
Below I list the mnemonics and where they occur, but I did not always
record them in my notes below, so I must check a number of the sources again ??
-- the classification was inspired by seeing that Franci (op. cit. in 3.A)
describes a vowel mnemonic in Benedetto da Firenze which I had overlooked. Ahrens gives many more verse and vowel
mnemonics -- to be added below. Hyde gives
an Arabic mnemonic due to al-Safadi using the first letters of the Arabic alphabet:
a, b, gj, d, h.
Dahbagja Ababgja Baba
[= Da h b a gj a A ba b gj a
Ba b a]: Hyde from al-Safadi.
Unspecified(?) verse mnemonic: ibn Ezra
Populea irgam mater
regina reserra: Pacioli;
Populea virga pacem regina ferebat: Minguét
Populeam jirgam mater Regina ferebat: Badcock
Populeam virgam mater Regina ferebat: van Etten;
Hunt; Schnippel; Ozanam 1725; Les Amusemens;
Hooper; Jackson; Manuel des Sorciers; Boy's Own Book; The Sociable; Le
Vallois; Gaidoz; Lucas
Populeam virgam mater
regina reserrat: Agostini's version of Pacioli;
Populeam virgam Mater Regina tenebat: Hyde from Wit's Interpreter; Murphy;
Schnippel/Bolte
Mort tu ne failliras pas en me liurant le trespas: van Etten
Mort, tu ne falliras pas
En me livrant au trépas: Manuel
des Sorciers;
Mort, tu ne falliras pas.
En me livrant le trépas:
Schnippel/Bolte; Ozanam
1725; Les Amusemens; The Sociable; Le Vallois (without the first comma); Ducret; Lucas
On tu ne dai la pace ei la rendea: Schnippel/Bolte
Gott schuf den Mann in Amalek, der (or den) Israel
bezwang: Schnippel
Gott schlug den Mann in Amalek, den Israel bezwang: Schnippel/Bolte
So du etwan bist gfalln hart, Stehe widr, Gnade erwart: Schnippel/Bolte
Non dum pena minas a te declina degeas: Schnippel/Bolte
Nove la pinta dà e certi mantena: Benedetto da Firenze
From member's aid and art, Never will fame depart: Schnippel/Bolte
From numbers, aid and art / Never will fame depart: Wingate/Kersey
From numbers, aid, and art, Never will fame depart: Ingleby;
Jackson; Rational Recreations
From number's aid and art, Never will fame depart: Gaidoz
From numbers aid and art / Never will fame depart: The Sociable
I have only
one example of a mnemonic for 15 & 15 counted by 10s.
Rex Paphicum Gente Bonadat Signa Serena: Hunt
See 5.AD for
the general problem of stacking a deck to produce a desired effect.
Josephus. De Bello
Judaico. c80. Book III, chap. 8, sect. 7.
(Translated by Whiston or by Thackeray (Loeb Classical Library,
Heinemann, London, 1927, vol. 2, pp. 685‑687.)) (Many later authors cite Hegesippus which is a later version of
Josephus.) This says that Josephus
happened to survive "by chance or
God's providence".
H. St. J. Thackeray.
Josephus, the Man and the Historian.
Jewish Institute Press, NY, 1929, p. 14. Comments on the Slavonic text, which says that Josephus "counted the numbers with cunning and
thereby misled them all" but gives
no indication how.
Ahrens. MUS II. 1918.
Chap. XV: Das Josephsspiel, pp. 118-169. This is the most extended and thorough discussion of this problem
and its history. I have used it as the
basis of this section. He gives a
rather complex method, based on work of Busche, Schubert and Tait, for determining
the last man, or any other man in the sequence of counting out, which I never
worked through, but which is clearly explained under Richards (1999/9).
Gerard Murphy. The
puzzle of the thirty counters.
Béaloideas -- The Journal of the Folklore of Ireland Society XII (1942)
3‑28. In this work and the
material cited (mostly ??NYS), the problem of
15 and 15 counted by 9s is shown
to have the medieval name Ludus Sancti
Petri = St. Peters‑Spiel = St. Peter's Lake (lake being an Old
English word for game [or Anglo-Saxon for 'to play']) = Sankt Peter Lek or Sankt Päders Lek (in Swedish). Murphy cites Schnippel & Bolte to assert
that it was known to the Arabs in the 14C -- cf below. He says the usual European version has
Christians and Jews on a ship, with St. Peter present and suggesting the
counting out process. [I had forgotten
that such versions occur in Ahrens, MUS II 130.] However, Murphy was unable to consult MUS, so his background is
not as complete as it might be.
Murphy
demonstrates that the problem was recently well‑known in both Scots and
Irish Gaelic in a form where a woman has to choose between two groups of
warriors seated in a circle, with emotional reasons for her preference. The solution is given in a vernacular
mnemonic, using actual numbers as in early Latin forms, while later Latin and
vernacular forms used vowel mnemonics.
He gives an Irish reconstruction, with English translation, based on
several 18C MSS whose texts he estimates as 13C to 17C, probably 16C. This is titled: Goid Fhinn Agus Dubháin Anso (Here is the Thieving of Fionn and
Dubhán). One MS has the Latin
subtitle: Populeam virgam Mater Regina tenebat, which is a common Latin vowel mnemonic. One of Murphy's sources says this refers to
the Queenly Mary appearing to the ship's captain and holding a poplar rod.
Murphy
also gives an extended Irish story (3pp) built around the problem: Ceann Dubhrann na Ndumhchann Bán (Ceann
Dubhrann of the White Sandhills). The
Gaelic names Fionn and
Dubhán are derived from 'fionn'
and 'dub' meaning
'white' and 'black'.
Murphy gives a contemporary Irish version on board a ship with a white
captain and a black wife and a crew of 15 and 15, with half having to go
overboard due to lack of food. He
sketches numerous other Irish and Scots version with varying combinations of
details, but using essentially the same verse mnemonic.
Murphy
cites a study by Manitius of a 9C MS (Bib. Nat. Paris, No. 13029) where the
problem begins "Quadam nocte niger
dub nomine, candiduus alter" (One night a black man named Dub and another
[named] White). They have to choose
between the blacks and the whites to keep watch. Cf. Codex Einsidelensis No. 326 below. The MS ends with the prose line
"These two Irish soldiers, one named 'Find' the other 'Dub', were
engaged in hunting. 'Find' means
"white", 'dub' "black"." The 12C Rouen MS No. 1409 attributes the problem to a
Clemens Scottus, which Murphy interprets as Clement the Irishman. The 12C MS Bib. Nat. Paris No. 8091
attributes it to a Thomas Scottus.
Murphy
concludes that the problem has an Irish origin, c800. He gives what he believes to be the earliest Latin form,
basically Bib. Nat. Paris No. 13029, and opines there must have been an Irish
predecessor.
Codex Einsidelensis No. 326. 10C. F. 88'. Latin verse. Published by Th. Mommsen, Handschriftliches. Zur lateinischen Anthologie. Rheinischen Museum für Philologie (NS) 9
(1854) 296‑301, with material of interest on pp. 298-299. Latin given in: M. Curtze, Bibliotheca Math. (2) 9 (1895) 34‑35. Latin & German in MUS II 123‑125. Begins:
"Quadam nocte niger dux nomine, candidus alter". 15 white
& 15 black soldiers, half to
keep watch, counted off by 9. The
colours refer to clothing, not skin!
Codex lat. Monacensis 14836. 11C. F. 80' gives rules
for 15 and 15 counted by 9 (though this
value is not specified) and mentions
counting by 8 and other values. No
mention of what is being counted.
Quoted and discussed by: M.
Curtze; Zur Geschichte der Josephspiels; Bibliotheca Math. (2) 8 (1894)
116 and in: Die Handschrift No. 14836 der Königl. Hof‑ und Staats‑bibliothek
zu München; AGM 7 (1895) 105 & 111‑112 (Supplement to Zeitsch. für Math. und Physik 40 (1895)).
Codex Bernensis 704.
12C. Published by: Hermann Hage; Carmina medii aevi maximam
partem inedita; Ex Bibliothecis Helveticus collecta; Bern, 1877; no. 85, pp.
145‑146. ??NYS. Latin in:
Curtze, op. cit. at Codex Einsidelensis, pp. 35‑36; and in:
MUS II 127. Jews &
Christians.
Ahrens, MUS II 118‑147, gives many further references
from 10‑13C. Originals ??NYS.
Meermanische
Codex, 10C. Mentions counting by other
values.
Leiden
Miscellancodex, 12C
Basel
Miscellancodex, 13C
Michinori Fujiwara (1106-1159). This work is lost, but has been conjectured to contain a form of
the problem -- see under Kenkō, c1331, and Yoshida, 1634.
Rabbi Abraham ben Ezra.
Ta'hbula (or Tachbûla), c1150.
??NYS -- described in: Moritz
Steinschneider; Abraham ibn Ezra (Abraham Judaeus, Avenare); Zur Geschichte der
mathematischen Wissenschaft im XII Jahrhundert; Zeitschr. für Math. und Physik 25 (1880): Supp: AGM 3, Part II
(1880). The material is Art. 20, pp.
123‑124. 15 students and 15
good-for-nothings on a ship, counted by 9s.
This seems to be the first extant example on board a ship. Verse mnemonic, which Steinschneider says is
not original. Steinschneider cites
further sources. Smith & Mikami, p.
84, say ben Ezra died in 1067 -- ??
Reinmar von Zweter.
Meisterlied: "Ander driu,
wie man juden und cristen ûz zelt".
13C. (In MUS II 128.) Jews and Christians on a ship, counts by 10.
Kenkō, also known as
Kenkō Yoshida or Urabe no Kaneyoshi or Yoshida no
Kaneyoshi (1283-1350). Tsurezuregusa. c1331. Translated by
Donald Keene as: Essays in Idleness The
Tsurezuregusa of Kenkō; Columbia Univ. Press, NY, 1967. (Kenkō was the author's monastic
name. His lay name was Urabe no Kaneyoshi. He lived for a long time at Yoshida in
Kyoto.)
This
book is one of the classics of Japanese literature, consisting of 243 essays,
ranging from single sentences to several pages. The most common themes of these relate to the impermanence of
life and the vanity of man.
In
Japanese, the Josephus problem is called
Mamakodate or Mamako-date San (or Mama-ko tate no koto
- cf Matŭ-oka, 1808) or Mamagodate
(Scheme to benefit the step-children
or Stepchild disposition). It is said to have been in the lost work of
Michinori Fujiwara (1106‑1159), qv.
The word Mamagodate first occurs in essay 137 of Kenkō, pp.
115-121 in Keene's version (including a double-page illustration which doesn't
depict the problem), whose beginning is characteristic of Kenkō's
style: "Are we to look at cherry
blossoms only in full bloom, the moon only when it is cloudless? To long for the moon while looking on the
rain, to lower the blinds and be unaware of the passing of the spring -- these
are even more deeply moving." The
passage of interest is toward the end, on p. 120 of Keene: "When you make
a mamagodate1 with backgammon counters, at first you cannot
tell which of the stones arranged before you will be taken away. Your count then falls on a certain stone and
you remove it. The others seem to have
escaped, but as you renew the count you will thin out the pieces one by one,
until none is left. Death is like
that." The footnote refers to
counting 15 and 15 by 10s, so that 14 white stones are eliminated, then the
counting is reversed and all the black stones are eliminated. "The Japanese name mamagodate
(stepchild disposition) derives from the story of a man with fifteen children
by one wife and fifteen by another; his estate was disposed of by means of the
game, one stepchild in the end inheriting all." Kenkō's text clearly shows he was familiar with the process
of counting to the last man and the use of the name indicates that he was
familiar with the version mentioned in the footnote, though its earliest
explicit appearance in Japan is in Yoshida, 1634, qv. My thanks to Takao Hayashi for the reference to Keene.
Thomas Hyde.
Mandragorias seu Historia Shahiludii, .... (= Vol. 1 of De Ludis Orientalibus, see 4.B.5 for vol. 2.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Prolegomena curiosa. The
initial material and the Prolegomena are unpaged but the folios of the
Prolegomena are marked (a), (a 1), ....
The material is on (e 1).v - (e
2).v, which are pages 34-36 if one starts counting from the beginning of the Prolegomena. Cited by Bland (loc. cit. in 5.F.1 under
Persian MS 211, p. 31); Ahrens (MUS II
136) & Murray 280. Several
citations are to ii.23, which may be to the 1767 reprint of Hyde's
works.
Hyde
asserts that the problem of the ship
with 15 Moslems and 15 Christians on a ship, counted by 9s, was given by
al-Safadi (Şalâhaddîn aş‑Şafadî [NOTE: Ş,
ş denote S,
s, with an underdot and the h
should have an underdot.] =
al-Sâphadi = AlSáphadi) (d. 1363)
in his Lâmiyato ’l Agjam (variously printed in the text). This must be his Sharh [the h should have an underdot] Lâmîyat al‑‘Ajam of c1350.
Hyde gives an Arabic mnemonic using the first five letters of the Arabic
alphabet, which he transliterates as:
Dahbagja Ababgja Baba [= Da h b a gj a A ba b gj a Ba b a]. He says the problem occurs in an English
book called Wit's Interpreter (??NYS) (8oS.87.Art) where the
mnemonic Populeam virgam mater Regina
tenebat is given. He then says that the problem is also
described in 'Megjdium & Abulphedam' -- p. 43 of his main text identifies
Abulpheda as a prince born in 672 AH -- ??
Shihâbaddîn Abû’l‑‘Abbâs Ahmad [the h
should have an underdot] ibn Yahya [the
h should have an underdot] ibn
Abî Hajala [the H should have an underdot] at‑Tilimsâni
alH‑anbalî [the H should have an underdot]. Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shaţranj
[NOTE: ţ denotes a t with an underdot] (Book of the examples of
warfare in the game of chess). Copied
by Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî in 1446.
This
is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray. Murray 280 says "Man. 36‑45
relate to as‑Safadî's problem of the ship (see Hyde, ii.23)",
described by Murray as 15 Christians and 15 Muslims counted by n.
Bland has "the well-known problem of the Ship, first as described
by Safadi, and then in other varieties.
(Hyde, p. 23.)"
Murray 620 says the problem is of Muslim origin and says it
appears in the c1530 Italian version of the Bonus Socius collection which
Murray denotes It. (See 5.F.1 under
Bonus Socius.) Murray refers to 15 & 15 counted by 9s, but it is not clear if this refers to this
particular MS.
Murray 622 cites MS
Sloan 3281 in the BM, 14C, as giving the Latin mnemonic solution.
Bartoli.
Memoriale. c1420. F. 100r (= Sesiano p. 135). Il giuocho de' Cristiani contra Saracin. 15 Christians and 15 Saracens -- the text ends in the middle of the
statement of the problem.
AR. c1450. Prob. 80, pp. 52, 181‑182 &
229. 15 Christians and 15 Jews. Gives only mnemonics for counting by 10, 9, 8, 6 or 12.
Codex lat. Monacensis 14908. c1460. F. 76 gives
mnemonics for 15 Jews and 15 Christians counted by 6, 8, 9, 10, 12. Quoted
and discussed by Curtze, opp. cit. under Cod. lat. Mon. 14836, above. [In the first paper, the codex number is
misprinted as 14809.]
Benedetto da Firenze.
c1465. Pp. 142‑143. 15 Christians and 15 Jews on a boat counted
by 9s. Vowel mnemonic: Nove la pinta dà e certi mantena. Diagrammatic picture on p. 143.
Murray 643 says the MS Lasa version, c1475, of the Civis
Bononiae collection (described in 5.F.1 under Civis Bononiae) has "16 diagrams of the 'ship' puzzle under
different conditions".
Chuquet. 1484. Prob. 146.
English in FHM 228-230, with reproduction of the original on p.
229. 15 Jews and 15 Christians on a
ship, counting by 9s. Says one can have
18 or 24 of each and can count by 10s, etc.
The reproduction on FHM 229 shows a circle marked out, with Populeam virgam matre regina tenebat written in the middle. The commentary says this "problem is
comparatively rare in fifteenth century texts", which doesn't seem like a
fair assessment to me.
Calandri.
Aritmetica. c1485. Ff. 102v-103v, pp. 205‑207. Coloured plate opp. p. 192 of the text
volume. (Tropfke 654 gives this in
B&W.) Franciscans and Camoldensians
on a boat: 15 & 15 counted by 9s.
Pacioli. De
Viribus. c1500. Probs. 56‑60.
Ff.
99r - 102r. LVI. (Capitolo) de giudei
Chri'ani in diversi modi et regole. a farne quanti se vole etc (Of Jews and
Christians in diverse methods and rules, to make as many as one wants, etc.). = Peirani 140-143. Does 2 & 30 by 9s --
there is a diagram for this in the margin of f. 100r, but it is not in the
transcription and Peirani says another diagram is lacking. Pacioli suggests counting the passengers on
shore and doing the counting out with coins or pebbles in case one will need to
know the arrangement in a hurry. He
also says one might count by 8s, 7s,
6s, 13s, etc., with any number of Christians and Jews.
Ff.
102r - 102v. [Unnumbered.] de .18. Giudei et .2. Chri'ani. = Peirani 144. 2 & 18 by 7s.
F.
102v. LVII. C(apitolo). de .30. Giudei
et .2. contando per .7. ch' toca va in aqua (Of 30 Jews and 2 counting by 7
with the touched going in the water). =
Peirani 144.
Ff.
102v - 103r. LVIII. C(apitolo). de .15.
Giudei et .15. Chri'ani per .9. in aqua (Of 15 Jews and 15 Christians by 9 in
the water.) = Peirani 144-145. 15 & 15 by 9s. In order to remember the arrangement, he says to see the next
section.
Ff.
103r. LIX. C(apitolo). Quater quinque.
duo. unus. tres unus. et unus. bis duo. ter unus. duo duobus un' (4, 5, 2, 1,
3, 1, 1, 2, 2, 3, 1, 2, 2, 1). =
Peirani 145. Pattern for 15 & 15
counted by 9s.
Ff.
103r - 103v. LX. C(apitolo). si da
unaltro verso viz. Populea. irga. mater regina. reserra. ['unaltro' is in the margin with a mark
showing where it is to go.] Vowel
mnemonic for 15 & 15 counted by 9s, which he explains in detail. Agostini says this is intended to be: Populeam
virgam mater regina
reserrat but both Pacioli's
heading and his discussion have
Populea irgam mater
regina reserra. The Index has LVIII-LX under one heading
which only refers to 'versi memorevili'.
Elias Levita der Deutsche.
Ha‑Harkabah. Rome,
1518. ??NYS. Attributes to ben Ezra, c1150??. Smith & Mikami, p. 84, say this seems to be the first printed
version of the problem.
Cardan. Practica
Arithmetice. 1539. Chap. LXI, section 18, ff. T.iiii.r ‑
T.v.r, but the material of interest is just a few sentences on f. T.iv.v (p.
113). Very brief description of 15
white and 15 black as 'ludus Josephus', saying one can work out any numbers
with some pebbles. MUS says this is
first to relate the problem to Josephus as the last man, but he doesn't give
any numerical details.
Hans Sachs (1494-1576).
Meisterleid: 'Historia Die XV Christen und XV Türcken, so auff dem
meer furen'. (MUS II 132‑133
gives text.)
Tartaglia. General
Trattato, 1556, art. 203, pp. 264v‑265r.
15 whites and 15 blacks (or Turks and Christians) counted out by 3, 4, ..., 12. No reference to Josephus.
Buteo.
Logistica. 1559. Prob. 89, pp. 303-304. 15 Christians and 15 Jews on a ship counted
by 10s. [Mentioned in H&S 52.]
Simon Jacob von Coburg.
Ein new und Wolgegründt Rechenbuch ....
1565 or 1612 (in quarto, not to be confused with octavo versions of 1565
and 1613 which do not contain the problem), f. 250v. ??NYS -- described in MUS II 133-134. 12 drinkers deciding who shall pay the bill. Ahrens doesn't specify the counting
number. Ahrens & Bolte (below) say
this is the earliest example, after Cardan, of finding the last man. Ahrens describes numerous later examples of
this type from 1693 on.
Prévost. Clever and
Pleasant Inventions. (1584), 1998. Pp. 183-185. This seems like a version of the Josephus problem but isn't. Place ten counters in a circle and then ten
on top of them. Start anywhere and
count off five and remove the top counter.
He says to count five again -- starting on the place where the counter
was removed, so we now would say he is counting four -- and remove the top counter. Continue in this way, counting the places
where a top counter has been removed and you manage to remove all the top
counters. In fact this is impossible,
but after removing five counters, you subtly start counting from the next
position rather than where the top counter was removed! Hence you remove the top counters in the
order 5, 9, 3, 7, 1, 6, 10, 4, 8,
2. Your audience will not observe this
and hence cannot reproduce the effect.
The
mathematical description is simpler if one counts by fours, removing 4, 8, 2, 6, 10, 5, 9, 3, 7,
1. The first five values are the values
of 4a (mod 10) for
a = 1, 2, 3, 4, 5.
Because GCD (4, 10) = 2, this sequence repeats with period 5.
Your trick shifts from the even values to the odd values and then you
can count out the five odd values.
Bachet.
Problemes. 1612. Préface, 1624: A.5.v - A.7.r; 1884: 8-9
&
prob. XX, 1612: 103‑106; prob. XXIII, 1624: 174-177; 1884: 118‑121. Turks & Christians -- discusses Josephus
as last man.
van Etten.
1624. Prob. 7 (7), pp. 7‑9
(16‑19). 15 Turks and 15
Christians counted by 9s.
Mnemonics: Populeam virgam mater
Regina ferebat; Mort tu ne failliras
pas en me liurant le trespas. Discusses
other cases, Roman decimation and Josephus as
40 counted by 3s. In the 1630 edition, 40
is changed to 41. Henrion's Notte, pp. 9‑10, refers to
Bachet's prob. 23 and mentions the correction of 40 to 41.
Hunt. 1631
(1651). Pp. 266-269 (258-261). 15 Christians & 15 Turks counted by 9s;
mnemonic: Populeam virga mater regina
ferebat. Then does the same counting by
10s and gives the mnemonic: Rex
Paphicum Gente Bonadat Signa Serena.
Yoshida (Shichibei) Kōyū (= Mitsuyoshi Yoshida)
(1598-1672). Jinkō‑ki. Additional problems in the 2nd ed.,
1634. Op. cit. in 5.D.1. ??NYS.
Shimodaira (see the entry in 5.D.1) discusses the Josephus problem on
pp. 12-14. He gives some of the
information on the Japanese names and on Michinori (1106-1159) and Kenkō,
c1331, which is presented under them. I
have a transcription of (some of?) Yoshida into modern Japanese which includes
this material as prob. 3 on pp. 66-67.
15 children
(in black) and 15 stepchildren (in white) counted by 10s. When
14 stepchildren are eliminated,
the last stepchild says the arrangement was unfair and requests the counting to
go the other way from him (so that he is number 1 in the counting). His stepmother agrees and thereby eliminates
all her own children.
This
is discussed in Smith & Mikami, pp. 80-84.
They quote a slightly later version by
Seki Kōwa (1642-1708) where
the stepmother simply reverses the order due to overconfidence. (On p. 121, they identify the source as Sandatsu Kempu, a MS of Kōwa.) This
is also discussed in MUS II 139‑140, where it says that the change in
counting was an error on the stepmother's part. Needham, p. 62, gives a picture from the 1634 ed. of Yoshida, but
this is different than the picture in my modern transcription. I have a photocopy from an 1801 ed. Dean, 1997, gives the picture, discusses
this and provides some additional details, citing the Heibonsha encyclopaedia
for the version with the intelligent stepchild. Dean, 1997, also gives an illustration from a 1767 version called
Shinpen Jinko-ki, cf at 1767.
Ahmed el-Qalyubi (d. 1659).
Naouadir (or Nauadir), c1650?, published at Boulaq (a suburb of Cairo),
1892, hist. 176, p. 82. ??NYS --
described in Basset (1886-1887 below) and MUS II 136. 15 Moslems and 15 infidels on a ship counted by 9s.
Muramatsu Kudayū Mosei. Mantoku Jinkō‑ri.
1665. ??NYS -- described in
MUS II 139 and Smith & Mikami, pp. 80-84. Smith & Mikami, p. 81, and Dean, 1997,
give Muramatsu's schematic diagrams.
The top diagram is for the classic 15 and 15 counted by 10s. The second has 32 people counted by 10s to
the last two, though the first 15 are coloured black and the second 15 are
coloured white, with the last two drawn as squares marked by dice patterns for
5 and 6.
Wingate/Kersey.
1678?. Prob. 3, pp.
531-532. 15 & 15 counted by 9s or
10s or any other. Christians and Turks.
From numbers, aid and art / Never will
fame depart. Discusses Josephus.
Thomas Hyde.
Historia Nerdiludii, hoc est dicere, Trunculorum; .... (= Vol. 2 of De Ludis Orientalibus, see
above for vol. 1.) From the Sheldonian
Theatre (i.e. OUP), Oxford, 1694. De
Ludo Charîgj seu Charîtch, pp. 225-226.
Says it is an Arabic game, i.e. Ludus Exeundi & Eliminadi. The description is very vague, but it seems
to involve counting out in a circle.
The diagram shows a circle of 21 and the text mentions counting by ten
or by five. No reference to any other
version of the process. ??need to read
the Latin more carefully.
Emil Schnippel (& Johannes Bolte). Das St. Peters-Spiel (with a Nachtrag by
Bolte). Zeitschrift für Volkskunde 39
(1929) 190-192 (& 192-194).
Schnippel describes the appearance of solutions of the St. Peters-Spiel = Sankt Päders Lek =
Saint Peter's Lake on 17-18C rune
calendars from Sweden, which mystified academics until identified by G.
Stephens in 1866. He gives the
vowel-mnemonic: Populeam virgam mater
regina ferebat. The rune marks are X
for
Χριστιαvoι (Xristianoi) and
I for ’Ioυδαîoι
(Ioudaioi). He gives the German
vowel-mnemonic: Gott schuf den Mann in
Amalek, der (or den) Israel bezwang. He
cites other writers (??NYS) who describe a 1497 MS with 15 & 15 Christians
and Jews counted by 10s, and versions
counted by 7s and 11s, and a version
with 17 & 15 Christians and Jews counted by 12s. He cites: a 1604
reference to Josephus but without specific numbers; a 1703 version with 15 & 15 French and Germans; and a 1782 version with 30 deserters, 15 to
be pardoned.
[Nigel
Pennick; Mazes and Labyrinths; Robert Hale, London, 1990, p. 37, says that in
Finland, stone labyrinths are sometimes called
"Pietarinleikki (St Peter's Game).
The latter name refers to a traditional numerical sequence which appears
to be related to the lunar cycle. It is
known from rock carvings and ancient Scandinavian calendars and as an
anti-semitic folk-tale." Can
anyone provide details of a connection to the lunar cycle or its appearance in
rock carvings??]
Bolte's
Nachtrag cites Gaidoz et al. (below at 1886-1887) and MUS and an article by
himself in Euphorion 3 (1896) 351-362, ??NYS -- cited MUS II 132. He sketches the history as given by Ahrens. Mentions the Japanese versions and
reproduces Matuoka's picture. He adds
three citations including a 1908 Indian version with 15 honest men and 15
thieves counted by 9s to the last man (??).
He gives vowel-mnemonics in Latin, French, German, English and Italian
as follows.
Non
dum pena minas a te declina degeas.
Populeam
virgam mater regina ferebat.
Mort,
tu ne falliras pas. En me livrant le
trépas.
So
du etwan bist gfalln hart, Stehe widr, Gnade erwart.
Gott
schlug den Mann in Amalek, den Israel bezwang.
From
member's (sic) aid and art, Never will fame depart.
On
tu ne dai la pace ei la rendea.
Ozanam. Murphy, note
4, says the problem is not in the 1694 ed. -- but see below which could explain
why Murphy didn't find it here.
Ozanam. 1696. Preface to vol. 2 -- first and second of
unnumbered pages, which are pp. 269‑270. 1708: Author's Preface -- second and third of unnumbered pp. Discusses Josephus, citing Bachet.
Ozanam. 1725. Prob. 45, 1725: 246‑250. Prob. 17, 1778: 168-171; 1803: 168-171; 1814: 148-150. Prob.
16, 1840: 76-77. 15 Turks and 15
Christians counted by 9s. Gives two
verse mnemonics: Mort, tu ne failliras
pas, En me livrant le trépas; Populeam
virgam mater Regina ferebat. Discusses
decimation. Quotes Bachet on Josephus
and asserts Hegesippus says Josephus used the method and suggests 41
counted by 3s (however, Hegesippus doesn't say this!).
Kanchusen. Wakoku
Chiekurabe. 1727. Pp. 8 & 35. 8 and 8 counted by 8s.
This is pointing out the remarkable fact that one can count out either
set first by starting at different points, in different directions. See:
Dudeney, 1899 & 1905; Shaw,
1944?
Minguet. 1733. Pp. 152-154 (1755: 110-111; 1822: 169-171;
1864: 146-148). 15 & 15 by 9s,
whites and blacks. Populea virga
pacem regina ferebat.
Alberti. 1747. 'Modo di disporre 30 cose ...', pp. 132‑134
(77‑78). 15 Christians and 15
Turks or Jews, counted by 3, 8, 9, 10.
Les Amusemens.
1749. Prob. 16, p. 138: Tiré de
Josephe l'Historien. 15 and 15 counted
by 9s. French and Latin mnemonics: Mort tu ne failliras pas En me livrant le trépas; Populeam Virgam Mater Regina ferebat.
Shinpen Jinko-ki (New Edition of the Jonko-ki), more
correctly entitled Sanpo Shinan Guruma (A Mathematical Compass). 1767.
BL ORB 30/3411. ??NYS --
illustration reproduced in Dean, 1997.
Fujita Sadasuke.
Sandatsu Kaigi. 1774. ??NYS -- cited in a draft version of Dean,
1997, as a Japanese commentary on the problem.
Hooper. Rational
Recreations. Op. cit. in 4.A.1. 1774.
Vol. 1, recreation XIII, pp. 42-43.
30 deserters of whom 15 are to be punished, counted by 9s. Populeam virgam mater regina ferebat.
L. Euler.
Observationes circa novum et singulare progressionum genus. (Novi Comment. Acad. Sci. Petropol. 20 (1775
(1776)) 123‑139.) = Opera Omnia
(1) 7, (1923) 246‑261. Gets the
recurrence for the last man: L(n) º
L(n-1) + k (mod n).
Miyake Kenryū.
Shojutsu Sangaku Zuye.
1795. ??NYS. (Described in MUS II 142‑143.) First(?) to modify Yoshida's problem (1634?)
so that the last stepchild sees his imminent fate and asks for the count to
restart with him. Smith, History II 543
and Smith & Mikami, p. 82, give a poorish picture from this. Dean, 1997, is a better picture.
Matuoka (= Matŭ-oka ??= Matsuoka Nōichi). 1808.
??NYS -- translated by Le Vallois, with reproductions of the pictures,
cf below. Ahrens, MUS II 140‑142,
discusses this, based on Le Vallois and reproduces the main picture from Le
Vallois. Gives Miyake's version. Le Vallois gives the title as: Mama-ko tate
no koto (Problème des beaux-fils (i.e. step-sons)). There is a diagram showing the counting-out processes.
Ingleby. Ingleby's
Whole Art of Legerdemain, containing all the Tricks and Deceptions, (Never
before published) As performed by the Emperor of Conjurors, at the Minor
Theatre, with copious explanations; Also, several new and astonishing
Philosophical and Mathematical Experiments, with Preliminary Observations,
Including directions for practicing the Slight of Hand. T. Hughes
& C. Chaple, London, nd [1815]. Trick L. The Turks and
Christians, pp. 104-106. 15 & 15
counted by 9s. "This ingenious
trick, which is scarcely known, ...."
"From numbers, aid, and art, /
Never will fame depart."
Sanpo Chie Bukuro (A Bag of Mathematical Wisdom). 1818.
BL ORB 30/3411. ??NYS -
illustration reproduced and discussed in Dean, 1997. Here a man and a woman are studying a set of 29 black and white
go stones and the text describes the problem and how to arrange the children.
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
Pp. 83-85, no. 130: Thirty soldiers having deserted, so to place them in
a ring, that you may save any fifteen you please, and it shall seem the effect
of chance. 15 & 15 by 9s. Populeam jirgam mater Regina ferebat. (jirgam
must be a misprint of
virgam.) Says Josephus and 'thirty
or forty of his soldiers' hid in a cave and Josephus arranged to be one of the
last.
Jackson. Rational
Amusement. 1821. Arithmetical Puzzles.
No.
16, pp. 4-5 & 54-56. 15 Turks and
15 Christians counted by 9s. Solution
gives: From numbers, aid, and art,
Never will fame depart and Populeam virgam mater regina ferebat.
No.
39, pp. 9-10 & 61-62. Decimation of
a troop of 40 to be counted by 10s -- where to place the four ringleaders so
they will be the four to be shot.
Rational Recreations.
1824. Feat 27, p. 103. 15 Turks and 15 Christians counted by
9s. Gives: From numbers, aid, and art,
/ Never will fame depart.
Manuel des Sorciers.
1825. ??NX Pp. 79-80, art. 40. 15 & 15 by 9s.
Populeam virgam mater regina ferebat.
Mort, tu ne falliras pas En me
livrant au trépas. Says one can also
do 18 & 6 by 8s, etc. Cf Gaidoz, below, col. 429.
Endless Amusement II.
1826? P. 117 (misprinted 711 in 1826?):
Predestination illustrated. 30 and 10
counted by 12s.
Boy's Own Book.
The
slighted lady. 1828: 411‑412; 1828-2: 417-418; 1829 (US): 210-211;
1855: 565‑566; 1868:
670. 13 counted down to last person by
9s. Before 1868, it gives the survivor for 2, 3, ..., 13, counted out by 9s.
The
partial reprieve. 1828: 417‑418; 1828-2: 422; 1855: 571;
1868: 672-673;
1881: 214. 30 criminals
counted by 9s to eliminate 15. Populeam
virgam mater regina ferebat.
Nuts to Crack XIV (1845), no. 72. 21 counted by 7s
to the last man.
Magician's Own Book.
1857.
The
fortunate ninth, pp. 221-222. 15
oranges and 15 apples, counted by 9s.
English mnemonics based on vowel coding.
Another
decimation of fruit, p. 224-225. 30
apples and 10 oranges, counted by 12s in order to get the oranges first.
The Sociable. 1858.
Prob.
31: The puzzle of the Christians and the Turks, pp. 296 & 312-314. From numbers aid and art /
Never will fame depart. Mort, tu
ne faillras pas / en me livrant le trepas. Populeam Virgam Mater regina ferebat. Then considers counting out 10 from 40,
counting by 12s. Discusses Josephus,
citing Hegesippus, and suggests counting by 3s. = Book of 500 Puzzles, 1859, prob. 31, pp. 14 & 30-32.
Prob.
39: The landlord tricked, pp. 298 & 316.
21 counted by 7s to the last man.
= Book of 500 Puzzles, 1859, prob. 39, pp. 16 & 34. = Wehman; New Book of 200 Puzzles; 1908, p.
51.
The Secret Out.
1859. The Circle of Fourteen
Cards, p. 87. This appears to be
counting out all 14 cards by 6s (it says by 7s, but it takes the counted out
card as one for the next stage), but it's not clear what the object is. This seems to be a corruption of an earlier
version??
Indoor & Outdoor.
c1859. Part II, prob. 19: The
landlord tricked, p. 136. Identical to
The Sociable.
Boy's Own Conjuring Book.
1860.
The
fortunate ninth, pp. 190‑191.
Identical to Magician's Own Book.
Another
decimation of fruit, p. 194. Identical
to Magician's Own Book.
Vinot. 1860. Art. XXVI: De l'historien Josèphe, pp.
55-56. Gives the Josephus story and
does it as counting from 41 by
3s to the last man.
The Secret Out (UK).
c1860. A delicate distribution,
p. 12. Count 13 by 9s to the last
person (different context than Leske).
Mentions counting 12 by 9s.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-30, pp. 254 & 396: Aus 12
Dreizehn machen. Count 13 by 9s to the
last person.
M. le Capitaine Le Vallois.
Les Sciences exactes chez les Japonais.
With comments by Louis de Zélinski & M. Sédillot. Congrès International des Orientalists (=
International Congress of Orientalists).
Compte-Rendu de la première session, Paris, 1873. Maisonneuve et Cie., Paris, 1874. T. 1, pp. 289‑299, with comments on
pp. 299-303. The material of interest
is on pp. 294-298. Gives a translation
of Matŭ-oka, 1808, and reproduces the pictures, cf above. Discusses Bachet, Ozanam, Mort tu ne failliras pas En me livrant le trépas, Populeam virgam mater Regina tenebat, Josephus (saying Josephus arranged to be
last).
Hanky Panky. 1872. The landlord tricked, pp. 129-130. Identical to The Sociable, prob. 39.
Kamp. Op. cit. in
5.B. 1877. No. 7, pp. 323‑324.
28 counted by 9s until one is left.
Footnote seems to refer to 24 counted by 9s.
Mittenzwey.
1880. Prob. 282-285, pp. 50-52
& 100-101; 1895?: 311-314, pp.
54-56 & 102-103; 1917: 311‑314,
pp. 49‑50 & 97‑98.
282
(311): 15 Negroes and 17 Europeans on a
ship, counted by 12s.
283
(312): 7 students and a crafty Jew who
wishes to make two of the students,
A & B,
pay the bill since they had been rude to him. Initially he is not included.
Starting with A and counting clockwise by 3s, A &
B are left. Starting with B and counting anti‑clockwise
by 3s, A & B are left.
Then the Jew is included and starts with himself, counting anticlockwise
by 3s and again A &
B are left.
284
(313): 14 counted by 10s to the last
man.
285
(314): 21 counted by 8s to the last
man.
Cassell's.
1881. P. 103: To reward the
favourites, and show no favouritism. =
Manson, 1911, p. 256. 15 & 15
counted by 9s.
Henri Gaidoz, Israël Lévi & René Basset. Le jeu de Saint‑Pierre -- Amusement
arithmétique. This is a series of five
notes in Mélusine 3 (1886‑87).
Gaidoz. Part I. Col. 273‑274. Gives classical version with St. Peter, 15 Christians
& 15 Jews counted by 9s. He
then gives two versions from Ceylon.
One version is called Yonmaruma
-- The massacre of the Moors -- and has
15 Portuguese & 15 Moors with a Singhalese verse mnemonic. The second version involves the Portuguese siege
of Kandy in 1821, again 15 &
15 by 9s, but different versions have
the Portuguese winning or losing. These
versions come from: The Orientalist 2
(1885) 177, ??NYS. The editor of The
Orientalist added a version learned from an Irish soldier with the vowel‑mnemonic: From number's aid and art, Never will fame
depart. Gaidoz says he cannot venture a
source for the puzzle.
Gaidoz. Part
II. Col. 307‑308. Comments on correspondence generated by Part
I which provided: 'Populeam virgam
mater regina ferebat'; the version with
the Virgin instead of St. Peter; a
version with negroes and whites and a negro captain; a version with French and English; references to Josephus, Bachet and Ozanam.
Lévi. Part III. Col. 332.
Says ibn Ezra's "Tahboula" (Stratagem), c1150, is devoted to
this game. Cites Schwenter (1623); Steinschneider's 1880 article discussed above at Ezra; Steinschneider's Catalog librorum hebr.
Biblioth. Bodleianae, col. 687 -- all ??NYS.
Previously Steinschneider opined the game derived from Jahia ibn al‑Batrik's
Secret of Secrets (8C), but Lévi says that that is a different amusement
involving 9.
Gaidoz. Part
IV. Col. 429. Cites: Le Manuel des
Sorciers, Paris, 2nd ed., 1802, p. 70 for a version with French and
English. ??NYS, but see the 1825 ed
above.
Basset. Part V. Col. 528.
Describes el-Qalyubi, c1650? -- cf above.
Robert Harrison. UK
Patent 15,105 -- An Improved Puzzle or Game.
Applied: 25 Sep 1889; accepted:
2 Nov 1889. 2pp + 1p diagrams. 12 whites and 12 blacks on a boat with a
lifeboat that will hold 12, counted by 6s, called The Captain's Dilemma.
É. Ducret.
Récréations Mathématiques. Op.
cit. in 4.A.1. 1892?
Pp.
105-106: Une dame pas contente. 13
counted by 2s to last person.
Pp.
118-119: Stratagéme de Joséphe. 41
counted by 3s to last two, claimed to be the method used by Josephus.
Pp.
120-121: Les marauders punis. 15 &
15 counted by 9s. Officers and soldiers
to be executed.
Pp.
121-122: Les naufrages. Same numbers,
with Turks and Christians on a boat.
Mort, tu ne failliras pas, En me livrant au trépas.
P.
122: Les Élections perfectionnées. 36
counted by 10s -- want the first six chosen.
Hoffmann. 1893. Chap. 4, pp. 156-157 & 210-211 =
Hoffmann-Hordern, pp. 134-135, with photo.
No.
54: Tenth man out. 15 whites and
15 blacks on a ship, counted
by 10s, but first 15 get to go into the lifeboats. Photo on p. 135 shows L'Equipage Decime,
with box and instructions, by Watilliaux, 1874-1895.
No.
55: Ninth man out. Same, counted
by 9s.
Hoffmann cites Bachet and gives a Latin mnemonic. Photo on p. 135 shows La Question des
Boches, with box having instructions on base, 1914-1918.
É. Lucas. Problem
32. Intermed. des Math. 1 (1894)
9. n2 persons, counted by n
until n‑1 are left.
"Problème dit de Caligula".
E. Cesarò. Solution
to 32. Ibid., pp. 30‑31.
J. Franel. Deuxième
réponse [to Problem 32]. Ibid., p.
31. Cites: Busche, CR 103, pp. 118, ??NYS.
Adrien Akar.
Troisième réponse [to Problem 32].
Ibid., pp. 189‑190.
E. Lemoine. Problem 330. Ibid, pp. 184‑185. Asks for last man of n
counted by p.
Adrien Akar; H.
Delannoy; J. Franel; C. Moreau.
Independent solvers of Lemoine's problem. Ibid., 2 (1895) 120‑122
& 229‑230. Akar refers to Josephus, Bachet, etc. Moreau has the clearest form of the
recurrence.
Brandreth Puzzle Book.
Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 7: The tenth man out. Almost identical to Hoffmann, no. 54. No solution.
Lucas.
L'Arithmétique Amusante.
1895. Pp. 12-18.
Le
stratagème de Josèphe, pp. 12-17. Prob.
VI. 15 Christians and 15 Turks, counted
by 9s. Vowel mnemonics: Mort, tu ne falliras pas, En me livrant le
trépas!; Populeam virgam mater Regina
ferebat. Discusses and quotes Bachet's
1624 Préface which gives the Josephus story and the idea of counting 41
by 3s.
Prob.
VII: Le procédé de Caligula, pp. 17-18.
6 and 30 counted by 10s
so as to count the 6 first.
H. Schubert. Zwölf
Geduldspiele. 1895. P. 125.
??NYS -- cited by Ahrens; Mathematische Spiele; Encyklopadie article,
op. cit. in 3.B; 1904.
E. Busche. Ueber die
Schubert'sche Lösung eines Bachet'schen Problems. Math. Annalen 47 (1896) 105‑112.
Clark. Mental
Nuts. 1897, no. 13; 1904, no. 22. The ship's crew. 1897 has
the usual 15 and 15 counted by
9s, starting with the captain,
involving whites and blacks on a ship and half being thrown overboard. 1904 has
14 whites and 15
blacks and the captain must discharge
15 at a port. He joins the crew and starts counting from
himself and wants to discharge the 15
blacks.
P. G. Tait. On the
generalization of Josephus' problem.
Proc. Roy. Soc. Edin. 22 (1898) 165‑168. = Collected Scientific Papers, vol. II, pp. 432‑435. Says the Josephus passage is "very
obscure, ..., but it obviously suggests deliberate fraud of some kind on
Josephus' part." Develops a way of
computing the last man.
Les Bourgeois Punis.
Puzzle from c1900, shown in S&B, p. 133. 8 and 2 counted by ?? to leave the 2 at the end.
Dudeney. A batch of
puzzles. Royal Magazine 1:3 (Jan
1899) & 1:4 (Feb 1899) 368-372.
The prisoners of Omdurman. 8
Europeans followed by 8 Abyssinians in a ring.
Start counting with the first European.
Determine the counting-out number to eliminate the Abyssinians in
sequence. Doing it in reverse sequence
works for any multiple of 16, 15, 14,
..., 9. The LCM is 720720.
But doing it in forward sequence can be done with 360361.
Since the pattern is symmetric in the two types of people, a change of
initial position, but keeping the same direction, will count out the others
first.
Dudeney. "The
Captain" puzzle corner. The
Captain 3:2 (May 1900) 97 & 179
& 3:4 (Jul 1900)
303. The "blacks" and
"whites" puzzle or The twelve schoolboys. 6 consecutive "blacks" and 6
consecutive "whites" in a circle.
What is the smallest number to count out by which will count out the
"whites" first? You can start
anywhere and in any direction. Answer
is 322
and one starts counting on the fourth "white" in the direction
of counting.
Sreeramula Rajeswara Sarma.
Mathematical literature in Telugu: An overview. Sree Venkateswara University Oriental
Journal 28 (1985) 77-90. Telugu is one
of the Dravidian languages of south India, spoken in the area north of Madras,
and is the state language of Andhra Pradesh.
On pp. 83-87 & 90, he reports finding examples in Telugu in the
notebooks of the schoolmaster Panakalu Rayudu (1883-1928) who was a collector
of material from many sources.
Unfortunately there is no indication of where Rayudu obtained these
examples and Sarma knows of no Indian versions. He has 15 thieves and
15 brahmins counted by 9s,
then 30 thieves and
30 brahmins counted by 12s.
Solutions are given in some literary form. The second problem is new to me.
In his notes, Sarma cites the German mnemonic Gott schuf den Mann in Amalek, der (or den) Israel bezwang given by Schnippel, and that he has learned
that the problem occurs in the Peddabālaśikşa [NOTE: ş
denotes an s with an underdot.], a work which is unknown
to me.
H. D. Northrop.
Popular Pastimes. 1901. No. 8: the landlord tricked, pp. 67-68 &
72. = The Sociable, no. 39.
Ahrens.
Mathematische Spiele.
Encyklopadie article, op. cit. in 3.B.
1904. Pp. 1088‑1089
discusses Schubert's work and its later developments.
Benson. 1904. The black and white puzzle, pp. 225‑226. As in Hoffmann, no. 54, but first 15 get
thrown overboard. Solution is linear
rather than circular.
Dudeney. Tit‑Bits
(14 Oct & 28 Oct 1905). ??NYS --
described by Ball; MRE, 5th ed., 1911, pp. 25‑27. 5 and 5 arranged so that one method
eliminates one group while another method eliminates the other group. Determine the two starting points and
counts. Ball doesn't give these values,
but seems to imply that both counts go in the same direction, and this is the
case in the examples given below. Ball
asks if the starting points can ever be the same for two groups of C?
He gives solutions for C =
2 (counted by 4 & 3), 3 (counted
by 7 & 8), 4 (counted by 9 & 5). He believes
this question is new. Note on p. 27
gives the solution for C = 5, but with different starting points. See MRE, 10th ed., 1920, for a general
solution with the same starting points.
See: Kanchusen, 1727; Dudeney, 1899; Shaw, 1944?
Pearson. 1907. Part II, no. 62, pp. 126 & 203. 15 Christians, including St. Peter, who does
the counting, and 15 Jews, counted by 9s.
Ball. MRE, 5th
ed. 1911. See under Dudeney, 1905.
Loyd.
Cyclopedia. 1914. Christians and Turks, pp. 198 &
365. = MPSL2, prob. 42, pp. 30‑31
& 134. Like Dudeney's 1905 version
with a different arrangement of 5 and 5.
Williams. Home
Entertainments. 1914. A decimation problem, pp. 122-124. 15 whites & 15 blacks counted by
10s. Half have to go over because of
shortage of provisions. Simple circular
picture with man counting in middle.
Ball. MRE, 10th ed.,
1920, pp. 26-27. See under Dudeney,
1905, for the previous version.
Incorporates the solution for the case
C = 5 into the text and adds a
general solution due to Swinden.
See: Will Blyth; Money
Magic; 1926 for a related problem.
Collins. Book of
Puzzles. 1927. Sailors don't care puzzle, pp. 70-71. 15 whites & 15 blacks counted by 10s.
Captain has to throw half over because of shortage of provisions. Diagram of 15 circles in a row above a
picture with 15 circles in a row below, but normally numbered -- it would seem
natural in this problem to have the lower row numbered backward to simulate a
circle.
William P. Keasby.
The Big Trick and Puzzle Book.
Whitman Publishing, Racine, Wisconsin, 1929. Counting out problem, pp. 71 & 95. 15 & 15 shown in a
circle with the starting point and direction given. Determine the counting number.
Rudin. 1936. Nos. 102-103, pp. 37-38 & 98.
No.
102. 13 counted by 9s until last man.
No.
103. 17 and 15 counted by 12s to
eliminate the 15 first.
Ern Shaw. The Pocket
Brains Trust -- No. 2. Op. cit. in
5.E. c1944. Prob. 50: Poser with pennies.
Pattern of 5 Hs
and 5 Ts given -- determine the
count to count the Hs first, which turns out to be 11.
Though not mentioned, the pattern of
Hs is equivalent to that
for Ts, so one can count out the
Ts first by starting at a
different point in the opposite direction.
The pattern is the same as Dudeney (1905).
Robert Gibbings.
Lovely is the Lee. Dent, London,
1945. Pp. 111-114. He says he was shown the puzzle by an old
man on the Aran Islands. Cites Murphy,
but his version is different than anything in Murphy. ??NYS -- information sent by Michael Behrend in an email of 12
Jun 2000.
Le Rob Alasdair MacFhraing (= Robert A. Rankin, who tells me
that the Gaelic particle 'Le' means
'by' and is not part of the
name). Àireamh muinntir Fhinn is
Dhubhain, agus sgeul Iosephuis is an dà fhichead Iudhaich (The numbering of
Fionn's men and Dubhan's men, and the story of Josephus and the forty Jews) (in
Scots Gaelic with English summary).
Proc. Roy. Irish Acad. Sect. A. 52:7 (1948) 87-93. A more detailed description than in the
Summary appears in Rankin's review in:
Math. Reviews 10, #509b (= A99-5 in:
Reviews in Number Theory). I had
assumed that this was a development from Murphy's article, but Rankin writes
that he had not heard of Murphy's article until I wrote for a reprint of
Rankin's article in 1991. He gives a
Scottish Gaelic version which is clearly a variant of those studied by
Murphy. He then studies the problem of
determining the last man, citing Tait.
For counting out by 2s, the rule is simple. [There is a story that the the number of mathematicians fluent in
Scots Gaelic is so small and the author's name is so obscured that the journal
sent the paper to Rankin to referee, not knowing he was actually the
author. The story continues that the
referee made a number of suggestions for improvement which the author
gratefully accepted. However, Rankin
told me that he was not the referee.
But he did review it for Math. Reviews!]
Joseph Leeming.
Games with Playing Cards Plus Tricks and Stunts. Op. cit. in 6.BE, 1949. ??NYS -- but two abridged versions have
appeared which contain the material -- see 6.BE.
24
Stunts with Cards, 7th & 19th stunts.
A surprising card deal
& All in order. Dover: pp. 124 & 131. Gramercy: pp. 11 & 18. Both stunts involve dealing cards by putting
one out face up, then the next is put at the bottom of the deck, then the next
is dealt face up, .... This process is
the same as counting out by 2s. The object is to produce the cards in a particular
order. The first has 8 cards and wants
an alternation of face cards and non-face cards. The second has the 13 cards of a suit and wants them produced in
order. This is the only example that I
can recall of the use of the Josephus idea as a card trick, though other forms
of counting out are common, e.g. by counting
1, 2, 3, ..., or by spelling the
words one, two, three, ....
N. S. Mendelsohn, proposer;
Roger Lessard, solver. Problem E
898 -- Discarding cards. AMM 57
(1950) 34-35 & 488-489. Basically counting out by 2s.
Determine the position of the last card discarded. No mention of Josephus, though the editor
asks what happens if every r-th card is
discarded and gets the recurrence
f(N) º r + f(N‑1) (mod N).
W. J. Robinson. Note
2876: The Josephus problem. MG 44 (No. 347) (Feb 1960) 47‑52. Analyses what sequences of persons can be
removed by varying the count. Applies
to Dudeney's problem.
Barnard. 50 Observer
Brain-Twisters. 1962. Prob. 36: Circle of fate, pp. 41‑42,
64‑65 & 95. Princess counts
out from 17 suitors by 3s. She sees
that her favourite will be the next one out, so she reverses direction and then
the favourite is the survivor.
F. Jakóbczyk. On the
generalized Josephus problem. Glasgow
Math. J. 14 (1973) 168-173. Gives a
method of determining when the i-th man
is removed and which is the k-th to be
removed. Somewhat similar to Rankin's
method.
D. Woodhouse The
extended Josephus problem. Revista
Matematica Hispano-Americana 33 (1973) 207-218. Gets recurrences for the last person, but unnecessarily
complicates the process by considering the starting point. By combining the recurrences, he gets
an n‑fold iteration for the
result, but this doesn't really clarify anything. Only cites Josephus.
Israel N. Herstein & Irving Kaplansky. Matters Mathematical. 1974;
slightly revised 2nd ed., Chelsea, NY, 1978. Chap. 3, section 5: The Josephus permutation, pp. 121-128. They study the permutation where f(i) = number of i-th man eliminated, but restrict to the case where one counts
by 2s,
which has considerable structure.
Gives a substantial bibliography, mostly included here.
Sandy L. Zabell.
Letter [on the history of the Josephus problem]. Fibonacci Quarterly 14 (1976) 48 &
51. Sketches the history.
I. M. Richards. The
Josephus problem. MS 24 (1991/92)
97-104. Studies the case of counting
out by 3s. Shows the 'Tait numbers',
i.e. n
such that L(n) = 1 or 2, are given by [η(3/2)i + 1/3],
where η =
1.216703..., and obtains a formula
for L(n). Presumably this could be extended to the general case??
Michael Dean.
Josephus and the Mamako-date san (Scheme to benefit the
step-children). International Netsuke
Society Journal 17:2 (Summer 1997) 41-53.
There are inro boxes from late 17C Japan which have pictures of the 15
children and 15 stepchildren
problem. These initially
mystified the art historians, but eventually they discovered the Josephus
problem and its Japanese forms, but only as far back as Bachet. Dean gives a brief history for the benefit
of art collectors, with references to a number of Japanese sources (some of
which I have not seen) -- see above at 1767, 1795, 1818 -- and some photos of
the inro boxes (including a fine late 17C example from the collection of
Michael and Hiroko Dean) and other material.
Ian M. Richards. The
Josephus Problem and Ahrens arrays. MS
31:2 (1998/9) 30-33. He has finally
obtained a copy of Ahrens' work, but from the first edition, and states the
result clearly. For n
persons, labelled 1, 2, ...,
n, counted out by k,
if we want to locate the e-th
man counted out, form a sequence starting with
1 + k(n-e) and then form each
next term by multiplying a term by
k/(k-1) and rounding the result
up to an integer. (I.e. xn+1 = éxn
* k/(k-1)ù.) Then the position
number of the e-th person eliminated is
the difference between kn + 1 and the largest term in the sequence less
than kn + 1. The sequence is giving the points where L(n, k) is zero in some
sense. Note that when e = n, so we are looking for the last person, then the sequence starts
at 1,
which is because we start counting with the first person as one. kn + 1
is the total amount counted in counting
n people by k,
For other values of e, the change of the starting point of the
sequence compensates for the fact that one only counts k(n-e) + 1
to eliminate the e-th
person. Ahrens then examined the
sequences obtained, with rational multipliers, and found some nice properties
which Richards states. Richards
generalises to arbitrary multipliers and finds connections with Beatty
sequences, ëanû.
Ian M. Richards. Towards
an analytic solution of the Josephus problem.
Unpublished preprint sent to me on 21 Mar 1999, 12pp. (Available from the author, 3 Empress
Avenue, Penzance, Cornwall, TR18 2UQ.)
Gets formulae for the case k =
4 which give the result with a maximum
error of ±1.
David Singmaster.
Adjacent survivors in the Josephus Problem. Nov 2003, 5pp, but may be extended. This was inspired by the first example in Pacioli's De Viribus,
which has 2 'good guys' and 30 'bad guys' arranged in a circle and every 9th
person is thrown overboard. I was
struck by the fact that the two survivors were adjacent in the original circle
as clearly marked in the marginal diagram.
Offhand it seems an unlikely result, but one soon observes that this
remains true as the counting out takes place.
That is, if the two survivors in counting off N by Ks
are adjacent, then this is also true for counting off n
by Ks for 3 £ n
< N. This paper investigates the
maximal N for which counting out by
Ks leaves two last survivors who
were originally adjacent.
The
basic problem is to represent a given fraction as a sum of fractions with unit
numerators and distinct denominators, as done by the Egyptians.
NOTE:
Dating of early Egyptian documents is rather uncertain and sources can vary by
several hundred years. I will tend to
use dates of Neugebauer and Parker, as given in Gillings. This dates the composition of the Rhind Papyrus
and the Moscow Papyrus as 13th Dynasty, c‑1785, but other sources say the
Moscow Papyrus is several hundred years older and other sources date the
composition of the Rhind Papyrus to the 12th Dynasty, c-1825.
Papyrus Rhind, composed c-1785 (or c-1825), copied c‑1650
(or c-1700). A. B. Chace, ed. (1927‑29); c= NCTM, 1978. Pp.
21‑22, 50‑51.
Moscow Mathematical Papyrus. c-1785. W. W. Struve, ed;
Mathematischer Papyrus des Staatlichen Museums der Schönen Künste in Moskau;
Quellen und Studien zur Geschichte der Mathematik, Abt. A: Quellen, Band 1;
Springer, 1930.
Fibonacci. 1202. Pp. 77‑83 (S: 119-126): ... de
disgregatione partium in singulis partibus [... on the separation of fractions
into unit fractions]. He clearly has
the idea of taking the smallest n such that
1/n £ a/b, but he doesn't
prove that this gives a finite sequence.
J. J. Sylvester. On
a point in the theory of vulgar fractions.
Amer. J. Math. 3 (1880) 332‑335 & 388‑389.
M. N. Bleicher. A
new algorithm for the expansion of Egyptian fractions. J. Number Theory 4 (1972) 342‑382. The Introduction, pp. 342‑344,
outlines the history. Pp. 381‑382
give 41 references.
E. J. Barbeau.
Expressing one as a sum of distinct reciprocals. CM 3:7 (1977) 178‑181. Bibliography of 20 items.
Paul J. Campbell. A
"practical" approach to Egyptian fractions. JRM 10 (1977-78) 81-86.
Discusses Fibonacci & Sylvester's methods, etc. 22 references.
Charles S. Rees.
Egyptian fractions. Math.
Chronicle 10 (1981) 13‑30. Survey
with 47 references.
R. J. Gillings.
Mathematics in the Time of the Pharaohs. Dover, 1982. He has a
long discussion on the Egyptian approach to this topic, discussing and
comparing the work in the various sources:
Reisner Papyri (c-2134); Rhind
Papyrus (c-1785); Moscow Papyrus
(c-1785); Kahun Papyri (c-1785, but
later than the previous two items);
Egyptian Mathematical Leather Roll (c-1647), but he certainly devotes most space to the Rhind Papyrus and the
Leather Roll.
S. Newcomb. Note on
the frequency of use of the different digits in natural numbers. Amer. J. Math. 4 (1881) 39‑40. Obtains the law by simply considering
logarithms.
F. Benford. The law
of anomalous numbers. Proc. Amer. Phil
Soc. 78 (1938) 551‑572.
E. H. Neville. Note
2540: On even distribution of
numbers. MG 39 (No. 329) (Sep 1955)
224‑225. Says the problem is not
precisely defined. (Not cited in
Raimi.)
R. A. Fairthorne.
Note 2541: On digital
distribution. Ibid., p. 225. Cites earlier results (see Raimi) and says
the law is "a consequence of the way we talk about [numbers]." (Not cited in Raimi.)
R. A. Raimi. The
first digit problem. AMM 83 (1976) 521‑538. Extensive survey and references.
G. T. Q. Hoare & E. E. Wright. Note 70.5: The
distribution of first significant digits.
MG 70 (No. 451) (Mar 1986) 34‑37.
Generates numbers as ratios of reals uniformly distributed on (0, 1).
Finds explicit and surprisingly simple probabilities for initial digits
of these numbers, which are reasonably close to Benford's probabilities.
Peter R. Turner. The
distribution of l.s.d. and its implications for computer design. MG 71 (No. 455) (Mar 1987) 26‑31. l.s.d. = leading significant digit. Cites some recent articles.
7.E. MONKEY AND COCONUTS PROBLEMS
Most
of these problems are determinate.
Mahavira gives two indeterminate problems, but the next are in Ozanam,
with the classic version of the problem first reappearing in
Carroll, 1888; Ball, 1890; Clark, 1904; and Pearson, 1907, qv.
NOTATION. The
classic coconuts problem has the following recurrence for the number of
coconuts remaining:
Ai+1 =
(n-1)/n [Ai - 1],
i.e. each sailor removes 1
(given to the monkey) and
1/n of the rest. There are two common endings of the problem.
Ending
0 --
the n‑th man leaves a multiple of n,
so the monkey doesn't get a final coconut. See: Mahavira: 131,
132; Williams; Moritz;
Meynell; Leeming.
Ending
1 --
the n‑th man leaves one more than a multiple of n,
so the monkey gets another coconut.
See: Carroll-Wakeling; Ball;
Clark; Pearson; Roray; Collins; Kraitchik; Phillips; Home Book; Leeming; Devi;
Allen.
One can extend this to
Ending E -- the
n-th man leaves a number º E (mod n).
Other indeterminate versions: Ozanam; Dudeney; Weber (Dirac); Rudin.
For the solution with
-(n-1) coconuts, see: Roray; Weber (Dirac); Birkhoff & Mac
Lane; Anonymous in Eureka; Gardner; Pedoe, Shima & Salvatore; Singmaster.
See Morris
(1988); Singmaster (1993) for the alternative division form where the
pile is divided equally and the monkey takes one from the remainder, i.e. each
sailor takes 1/n of the pile and then the monkey then
takes 1 from the remainder, so the recurrence is
Ai+1 = (n‑1)Ai/n
- 1. This is similar to the form of
recurrence occurring in the determinate versions of the problem, where division
takes place first and then some more is included. Comparing this with the standard form, we see that the forms can
be described by the number of coconuts
(mod n) at each stage. In the classic form, each Ai º 1 (mod n), and in Morris's form, each Ai º 0 (mod n), so we can
conveniently name these Form 1 and
Form 0. Unless specified, all
examples have Form 1.
It is easy to generalize to giving c coconuts to the monkey
at each stage, in either Form, which I call Forms 1c and 0c, but only Anonymous in Eureka; Kircher; Pedoe, Shima
& Salvatore; Singmaster consider this.
Only Kircher considers giving variable amounts to the monkey
and he even permits negative values,
e.g. if the monkey is adding coconuts to the pile!
Birkhoff & Mac Lane; Herwitz; Pedoe, Shima &
Salvatore consider a variation where no ending is specified except that there
is an integral number left after the
n-th division. A discussion of
this version has now been added to Singmaster.
Jackson gives a simple form with no monkey. Edwards gives a form where the monkey only
gets a coconut at the end.
See Tropfke 582. See
also 7.S.1.
Hermelink, op. cit. in 3.A, says there are Egyptian
versions, presumably meaning some of the simpler determinate types of heap or
'aha' problems in the Rhind Papyrus.
Old Babylonian tablet YBC 4652. c-1700?. Transcribed,
translated and commented on in:
O. Neugebauer & A. Sachs; Mathematical Cuneiform Texts;
American Oriental Society and American Schools of Oriental Research, New Haven,
1945, pp. 100-103, plate 13 & photo on plate 39. This has fragments of 22 simple problems, of which six can be
restored. The authors say the dating of
the tablets discussed in the book is quite uncertain, only stating "they
are to be dated to the centuries around 1700 B.C."
No. 7
is reconstructed as: I found a stone,
but did not weigh it; after I added one-seventh and added one-eleventh, I
weighed it: 1 ma-na. What was the
original weight of the stone? In modern
notation, this is: x + x/7 + (x
+ x/7) / 11 = 1, or simply: x (8/7) (12/11) = 1 which is a simple 'aha' problem.
No. 8
leads to x - x/7 + (x
- x/7) / 11 = 1.
No. 9
leads to x - x/7 + (x
- x/7) / 11 - [x - x/7 + (x - x/7)/11] / 13
= 1.
No. 19
leads to 6x + 2 +
(6x + 2)·24/21 = 1.
No. 20
leads to 8x + 3 +
(8x + 3)·21/39 = 1.
No. 21
leads to x - x/6 + (x
- x/6) / 24 = 1.
Old Babylonian tablet YBC 4669. c-1700?. Neugebauer and
Sachs continue on p. 103 with a new analysis of this table which Neugebauer had
previously treated in Mathematische Keilschrift-texte III, op. cit. in 6.BF.2,
p. 27. It leads to (2/3) (2/3) x + 10 = x/2.
Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c‑150.
Chap.
VI, prob. 27, p. 69. Man carrying rice
through customs pays 1/3, then
1/5, then 1/7
and has 5 left.
Chap.
VI, prob. 28, pp. 69‑70. Man
pays 1/2, 1/3, 1/4, 1/5, 1/6, making
1 paid out.
Chap.
VII, prob. 20, pp. 79‑80.
Man gains 30%
and sends home 14000; then
gains 30% and sends
13000; then 30%
and 12000; then
30% and 11000;
then 30% and
10000; leaving 0.
Capital was 30468 84876/371293. (English in Lam & Shen, HM 16 (1989) 113.)
Zhang Qiujian (= Chang Chhiu‑Chien = Chang Ch'iu Chien = Zhang Yo Chien). Zhang Qiujian Suan Jing (= Chang Chhiu‑Chien Suan Ching)
(Zhang Qiujian's Mathematical Manual).
468. ??NYS. Chap. II, no. 17. Man gains 40%
and withdraws 16000; then gains
40% and withdraws 17000;
then gains 40% and withdraws 18000; then gains 40%
and withdraws 19000; then gains
40% and withdraws 2000;
leaving 0. Capital was
35326 5918/16807. (English in Lam & Shen, HM 16 (1989)
117.)
Anania Schirakatzi (= Ananias of Shirak). Arithmetical problems. c640.
Translated by: P. Sahak Kokian
as: Des Anania von Schirak arithmetische
Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919) 112-117. Kokian cites several versions and editions
of this Armenian MS as well as some studies on Ananias, but I haven't been able
to determine just where Shirak was. The
title varies on the different MSS and Kokian heads the text with one version
translated into German: Des Anania
Vardapet Schirakuni Frage und Auflösung [Questions and Solutions of the Priest
Ananias of Shirak.] There are 24
problems, mostly of the 'aha' or 'heap' type.
Only the numerical solutions are given -- no methods are given. There are several confusing errors which may
be misprints or may be errors in the MS, but Kokian says nothing about
them. One problem seems to have omitted
an essential datum of the number of grains of barley in a 'kaith'. I cannot reconcile one solution with its
problem (see 7.H).
Prob.
11. Spend 5/6 thrice, leaving 11.
Answer: 2376.
Prob.
13. Spend 3/4 thrice, leaving 5.
Answer: 320.
Prob.
19. (Double and give away 25)
thrice to leave zero.
Answer: 21⅞. Kokian notes that this and prob. 22 are the
earliest occurrences of fraction signs in Armenian. Hermelink, op. cit. in 3.A, points out that here the doubling is
done by God in response to prayer in churches -- then the Arabic world converts
the churches to mosques, and then the West reverts to churches, while in the
Renaissance, the doubling is by winning at gambling. In fact, during the Renaissance, it often was by profit from
trade.
Prob.
21. Give away 1/2, then 1/7,
then 1/8, then
1/14, then 1/13,
then 1/9, then
1/16, then 1/20,
leaving 570. Answer:
2240.
Papyrus of Akhmim.
c7C. Jules Baillet, ed. Le Papyrus Mathématique d'Akhmîm. Mémoires publiés par les membres de la
Mission Archéologique Français au Caire, vol. IX, part 1, (1892) 1‑89. Brief discussion of the following problems
on pp. 58-59.
Prob.
13, p. 70. Take 1/13th,
then 1/17th of the rest, leaving 150.
Answer: 172 + 1/2 + 1/8 + 1/48 + 1/96. (Also given in HGM II 544. Kaye I 48, op. cit. under Bakhshali MS,
discusses the Akhmim problem and says both Heath and Cantor give misleading
references, but I don't see what he means.)
Prob.
17, p. 72. Take 1/17th,
then 1/19th of the rest, leaving 200.
Answer: 224 + 1/4 + 1/18.
Bakhshali MS.
c7C.
In: G. R. Kaye, The Bakhshāli
manuscript. J. Asiatic Soc. Bengal (2)
8:9 (Sep 1912) 349‑361. P. 358:
Sutra 25: example -- merchant pays customs of
1/3, 1/4, 1/5 and finds he has paid 24.
= Kaye III 205, f. 14r.
Hoernle,
1888, op. cit. under Bakhshali MS, p. 277 gives the above and the
following. Merchant gains 1/3,
1/4, 1/5, 1/6
and finds he has gained 40. (Kaye III 205, f. 14r gives this in less
detail and it is not clear if Hoernle's statement is what is intended.) Merchant loses 1/3, 1/4, 1/5
for a total loss of 27. (Kaye III 205, f. 14v says the remainder
is 27
but gives the original amount as
45, so he seems to have loss and
remainder interchanged.). Merchant
loses 1/3, 1/4, 1/5 leaving
20 (can't find in Kaye III ??).
Kaye
I 48, section 89, says there are 17 examples of this general form, some with
the initial value given and the final result wanted, others with the final
result given and the initial value wanted.
Gives the first example above and two others with the same rates and a
payment of 280 (Kaye III 165, ff. 52r-52v) or a result
of 2x - 32, where x is the initial value (Kaye III 207, f.
15r). Kaye III 204, f. 13v: start with
60, lose 1/2, gain
1/3, lose 1/4,
gain 1/5. Kaye III 208, f. 16r: give
2/3, then 2/5,
then 2/7, then
2/9, leaving 3.
How much was given?
See
also Datta, op. cit. under Bakhshali MS, pp. 44 & 52‑53. He says the Akhmim problems give the
remainder, while the Bakhshali MS and Mahavira problems give the amount paid --
but above we have seen both kinds.
Datta, p. 46, says (Kaye III 184,) f. 70v has a badly damaged
problem about a king who gives away 1/2, 1/3
and 1/4 of his money, making 65
given away. Datta says that the
king had only 60 to start!!
But if this is a problem of the type being treated here, then the
fractions are applied to the amount left after the previous stage and the king
would have 1/4 of his original amount left and he must have
had 86 1/3 to start.
Ripley's Puzzles and Games.
1966. P. 78 asserts that Premysl
of Staditze won the kingdom of Bohemia by solving the following. Give (half and one more) twice, then half
and three more to leave zero. Typically
Ripley's gives no details. The
Encyclopædia Britannica says the origin of the Premysl dynasty is obscure,
deriving from a plowman who married the Princess Libuse, but giving no date,
though apparently by the 9C. [Rob
Humphreys; Prague The Rough Guide; The
Rough Guides, London, (1992), 3rd ed, 1998, p. 249] gives the legends of the
founding of Prague. The maiden queen
Libuše, in the 7C or 8C, fell into a trance and told her followers to seek a
ploughman with two oxen. Such a man,
named Přemysl (meaning ploughman) was found and produced the dynasty. He makes no mention of the problem, nor does
the Blue Guide for Prague.
Mahavira. 850. Chap. III, v. 129-140, pp. 67-69 are simple
problems of this general type, involving sums of numbers diminished by
fractions -- I give just some examples.
Chap. VI, v. 112, 114, 130,
131, 132, pp. 116 & 123‑125.
Chap. III.
133. x(3/4)(4/5)(5/6) + y(1/2)(5/6)(4/5) +
z(3/5)(3/4)(5/6) = 1/2.
This reduces to: x/2 + y/3
+ 3z/8 = 1/2 and he arbitrarily
picks two of the values, getting:
1/3, 1/4, 2/3.
134. x(1/2)(5/6)(4/5)(7/8)(6/7) =
1/6.
Chap. VI.
112. Double and subtract 5,
triple and subtract 5, ...,
quintuple and subtract 5, leaving
0 (Datta & Singh I 234, note
this gives 43/12 flowers and hence replace 5 by 60
to give 43).
114. Less regular problem, leaving 0.
130. Gives a general technique.
131. Two sons and mangoes -- (subtract
1 and halve) twice,
leaving some even number -- i.e.
Ending 0 with 2
men. Cf Pearson, 1907.
132. Man placing flowers in a temple -- (subtract 1 and delete
⅓) thrice, leaving some multiple of 3
-- i.e. Ending 0
with 3 men. Cf Pearson, 1907.
There
are some simpler problems in Chap. IV, v. 29‑32, pp. 74‑75.
Chaturveda.
860. There are some simple
examples on pp. 282‑283 of Colebrooke.
Sridhara. c900. V. 74(i), ex. 97, pp. 59‑60 &
96. Give away 1/2, then 2/3,
then 3/4, then
4/5, leaving 3.
Tabari. Miftāh
al-mu‘āmalāt. c1075. ??NYS.
Pp.
177f. & 128, no. 28 & 45.
Tropfke 585 says these are business trips.
P. 127,
no. 44. Hermelink, op. cit. in 3.A,
says this is Fibonacci's seven gate problem of p. 278, with oranges instead of
apples. Tropfke 585 says it is a
problem with porters at an orange orchard.
Abraham. Liber
augmenti et diminutionis. Translated from
Arabic in 12C (Tropfke 662 says early 14C).
Given in: G. Libri; Histoire des
Sciences Mathématiques en Italie; vol. 1, pp. 304-376, Paris, 1838. ??NYR -- cited by Hermelink, op. cit. in
3.A.
Bhaskara II.
Bijaganita. 1150. Chap. 4, v. 114. In Colebrooke, pp. 196‑197. (Lose 10, double,
lose 20) thrice to triple original.
Fibonacci.
1202. Pp. 258‑267, 278,
313‑318 & 329 (S: 372-383, 397-398, 439-445, 460‑461) gives
many versions! Chap. 12, part 6, pp.
258‑267 (S: 372-383): De viagiorum propositionibus, atque eorum similium
[On problems of travellers and also similar problems] is devoted to such
problems.
P. 258
(S: 372-373). (Double & spend 12)
thrice to leave 0 or
9.
Answers: 10½, 11⅝. H&S 60 gives English.
P. 259
(S: 374). Start with 10½,
(double and spend x) thrice to leave 0. H&S 60 gives
English.
P. 259
(S: 374). Same, starting with 11⅝
and leaving 9.
P. 259
(S: 374-375). (Triple and spend 18)
four times to leave 0. Answer:
8 8/9.
P. 260
(S: 375). Start with 8 8/9,
(triple and spend x) four times to leave 0.
Pp.
260‑261 (S: 375-376). (Triple and
spend 18) four times to leave 12, or the original amount, or original amount + 20.
P. 261
(S: 376-377). Three voyages, making
profits of 1/2, 1/4,
1/6 and spending 15
each time to leave final profit of
1/2. Answer: 24 6/7.
Same, with initial amount 24
6/7, find the common expenditure. Same, with 'leave final profit 1/2'
replaced by 'leave 21'.
Pp.
261‑266 (S: 377-383). Many
variations.
Pp. 266‑267
(S: 383). Start with 13,
(double and spend 14) X
times to leave 0. H&S 60 gives English. He gets
X = 3¾ voyages, by linear
interpolation between 3 and
4. Exact answer is log2 14 = 3.80735.
P. 278
(S: 397-398). De illo qui intravit in
viridario pro pomis collegendis [On him who went into the pleasure garden to
collect apples]. Man collects apples in
a garden with 7 gates.
(Subtract half and one more)
seven times to leave 1. H&S 60 and Sanford 221 give
English. Answer: 382.
Pp.
313‑316 (S: 439-443). Man starts
with 100 and spends 1/10 twelve times. This is not strictly of the type we are looking at, but it is
notable that he computes 100 (.9)12 using a form of decimal fraction,
getting 28.2429536481. See 7.L for related problems.
Pp.
316‑318 (S: 443-445). Exit from a
city with 10 gates. He pays 2/3
of his money and 2/3 more, then
1/i of his money and 1/i
more for i = 2, ...,
10, leaving 1.
P. 329
(S: 460). Same as on p. 258, done by
false position.
P. 329
(S: 460-461). Start with 12,
(double and spend x) thrice to leave 0.
Abbot Albert.
c1240. Prob. 8, p. 334. (Double and subtract 1)
thrice, leaving 0.
Chu Shih‑Chieh (= Zhu Shijie). Ssu Yuan Yü Chien (= Siyuan Yujian)
(Precious Mirror of the four Elements =
Jade Mirror of the Four Unknowns).
1303. Questions in Verse, prob.
4. ??NYS. English in Li & Du,
p. 179. (Double and drink 19)
four times to leave 0.
BR. c1305.
No.
89, pp. 108‑109. (Double and
spend 35) thrice leaving 0.
No.
119, pp. 134‑135. (Double and
spend 40) thrice leaving 0.
Folkerts.
Aufgabensammlungen. 13-15C.
(Double
and give a) n times to leave
nothing. 17 sources. Folkerts notes the solution is a - a/2n and the MSS give a general rule.
(Give
half and one more ) n times to leave c. 12 sources, with the
MSS giving a general rule. Two sources
where half is replaced by a
quarter. One irregular example: Lose half, gain 2; lose half, gain 4; lose
half, gain 6; to leave 4.
Munich
14684, XXXIV. 6 sources.
Cites
a number of other sources, almost all cited in this section (two items are
NYR).
Gherardi?. Liber
habaci. c1310. Pp. 144‑145. Three porters -- i‑th
takes half plus i, leaving none.
Gherardi. Libro di
Ragioni. 1328.
Pp. 47‑48. Man gathering apples. Four porters -- i‑th takes half plus
5 ‑ i, leaving 1.
P.
100. Man makes 12d
on his first trip. He earns at
the same rate on his second trip and then has
100d. This leads to a quadratic
and he finds the positive solution. See
Van Egmond, op. cit. in Common References, pp. 168, 177 & 185 for Italian,
English and algebraic versions and some corrections.
Lucca 1754.
c1330.
Ff.
26r‑27r, pp. 64‑65.
Multiply by 6/5 and spend
12, multiply by 5/3
and spend 17, double and spend 20, leaving 0.
F.
59v, p. 135. (Double and spend 12)
thrice to leave 3.
Paolo dell'Abbaco.
Trattato di Tutta l'Arte dell'Abacho.
1339. The first work in the
codex Plimpton 167 in the Plimpton collection, Columbia University, New York,
is a c1445 copy. ??NYS -- described in
Rara, 435‑440 and Van Egmond's Catalog 254-255. Van Egmond 365 lists 9 MSS of this work. MS B 2433, Biblioteca
Universitaria, Bologna, is a c1513 copy of just the problems of this work --
Dario Uri has kindly sent a copy of this, but it is somewhat blurry and often
illegible; he has now sent a version on a CD which is clearer. It is dated 1339. See: Van Egmond's Catalog 67-68.
Rara
438 calls it the Dagomari Manuscript and reproduces a figure of a garden with
three gates and guards. Only the first
line of the text of the problem is included, but the text is on ff. 25r-25v of
B 2433. (Halve and subtract 1)
thrice to leave 3.
Munich 14684. 14C.
Prob.
V, p. 78. (Double and subtract 2)
some times to leave 0 -- determines initial values for various
numbers of times as 2(1 - 1/2n). The text seems to also consider (Double and subtract 5).
Prob.
VI, p. 78. (Halve and subtract 1)
thrice to leave 3.
Prob.
XXXIV, p. 84. (Double and subtract 100)
thrice, then (double and
subtract 50) thrice, leaving 0. Answer:
92 31/32,
Bartoli.
Memoriale. c1420. Prob. 9, f. 75v (= Sesiano pp. 138 &
148). Man going into a garden to get
apples. Gives 3/4 plus 3
more; 2/3 plus
2 more; 1/2
plus 1 more; to leave 1.
Provençale Arithmétique.
Written (or more likely copied) at Pamiers, c1430. MS in Bibliothèque Nationale, Paris, fonds
français, nouvelle acquisition 4140.
Previously in the collections of Colbert (no. 5194) and the King (no.
7937). Partially transcribed/translated
and annotated by Jacques Sesiano; Une Arithmétique médiévale en langue
provençale; Centaurus 27 (1984) 26-75.
The problems are not numbered, so I will give the folios and the pages
in Sesiano. However the indications of
the original folios have not come through on a few pages of my copy and I then
only give Sesiano's page.
P.
58. Man doubles his money and
spends 1, triples and spends 2, quadruples and spends 2,
leaving him with 3.
F.
113v-114r, p. 60. (Sell 1/2
and one (or 1/2 ??) more)
three times to leave 3. The author gives a general solution as
starting with the final result, (adding the extra number and double) three times to get the original number.
Pseudo-dell'Abbaco.
c1440.
Prob.
47, p. 44 with plate on p.45. (Halve
and subtract 2) thrice to leave 7.
Prob.
71, pp. 65‑67 with plate on p. 66.
(Lose ⅓ and
6 more) thrice to leave 24. (The illustrations
are very different from that in Rara (see previous entry). Rara does not show enough text to see if the
numbers used are the same as here, though the wording is clearly
different.) I have a colour slide of
this.
AR. c1450. No. 185 & 187, pp. 87‑88, 173‑174
& 220.
185 = Fibonacci, p. 258.
187. Double and spend 6, double and spend 12,
double and spend 15, leaving the initial amount.
Muscarello.
1478.
Ff.
78v-79r, pp. 194-196. Lose 1/2
and 6 - i more,
for i = 1, 2, 3, 4, 5, leaving
1.
Ff.
84r-85r, pp. 201-204. Merchant starts
with 79 and makes profits of 17%, 19%,
21%, 23% at four fairs.
della Francesca.
Trattato. c1480.
F. 23r
(73-74). Gain 1/3 + 1/4 and 20
more. Then spend 1/4 + 1/5
and 20 more to leave
24.
F. 37v
(97-98). Double and spend 11, triple and
spend 47, double and spend 34, double and spend 16 to leave 0. English in Jayawardene.
F. 41v
(104). Identical to f. 23r.
Chuquet. 1484.
Prob.
30. (Double and subtract 12)
thrice, leaving 0. English in FHM 206.
Prob.
31-33 are generalized versions. E.g.
Prob. 31 is double and spend 5,
triple and spend 9, quadruple and spend 12
to leave 8.
Prob.
95, English in FHM 219. Merchant makes
a profit of 1/3 and
i more on his i-th journey. He makes as many journeys as he has money to start with. When does he have 15? This gives a messy
equation: (4/3)x = 3
- 9/(x+12). Chuquet uses some
interpolation to estimate X = Ö(50 16297/16384) - 4 13/128 [FHM misprints this] =
3.03949414, but I get
3.045827298. Chuquet says
ordinary interpolation is not valid.
Calandri.
Arimethrica. 1491.
F.
66v. (Double and spend 2) thrice
leaving 0.
F.
74r. Double and then gain 50%
giving 1000. Woodcut of merchant on horse.
Pacioli. Summa. 1494.
F.
105v, prob. 20. (Give half and one
more) thrice leaving 1.
(See also H&S 58.)
F.
105v, prob. 22. (Double and spend 12)
thrice leaving 0.
F.
187r, prob. 8. Start with x
and double x times to get 30. This gives us x 2x = 30, whose answer is 3.21988.... He
interpolates both factors linearly on the third day, getting (3+y)(8+8y) = 30, so 3+y = 1 + Ö(19/4) = 3.17945....
He
approaches the following problems similarly.
Ff.
187r-187v, prob. 9. Start with x
and make 25% on each of
x trips to make 40%
overall. This gives x (1.25)x = 1.4 x.
F.
187v, prob. 10. Leads to x (1.2)x = x2.
F.
187v, prob. 11. Leads to x (1.4)x = 6x.
F.
188r, prob. 13. Start with 13,
(double and spend 14) x
times to leave 0. He observes that each iteration doubles the
distance from 14, so the problem leads to 2x = 14, but again he has to interpolate on the third
day.
Calandri, Raccolta.
c1495. Prob. 16, pp. 17‑18. Merchant gains ⅓ of his money
plus i
on the i-th trip. After three trips he has 15.
Pacioli. De
Viribus. c1500. Ff. 120r - 120v, 111r - 111v (some pages are
misbound here). C(apitolo). LXVII. un signore ch' manda un servo a coglier pome
o ver rose in un giardino (A master who sends a servant to gather apples or
roses in a garden). = Peirani
156-158. (Lose half and one more) three times to leave 1.
Discusses the problem in general and also does (Lose half and one more)
five times to leave 1; (Lose half and one more) three times to leave 3.
Blasius. 1513. Ff. F.iii.r - F.iii.v: Decimaquinta
regula. Sack of money. First man takes half and returns 100;
second takes half and returns
50; third takes half and
returns 25; leaving 100 in the sack.
Johannes Köbel.
Rechenbiechlein auf den linien mit Rechenpfeningen. Augsburg, 1514. With several variant titles, Oppenheim, 1518; Frankfort, 1531, 1537, 1564. 1564 ed., f. 89r, ??NYS. Lose half, gain 100, lose half, gain 50,
lose half, gain 25 to yield
100. (H&S 58‑59 gives
German and English.)
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 29, f. 66v. Start with 100 and have to bribe ten guards with 1/10
each time. Computes the exact
residue, i.e. 100 x .910. (H&S 59-60).
Tonstall. De Arte
Supputandi. 1522.
Quest.
43, p. 173. Give half plus i+1,
for i = 1, 2, 3, leaving
1. (H&S 61 cites this to the
1529 ed., f. 103)
Quest.
44, pp. 173-174. Give half and get
back 2i for
i = 1, 2, 3,
leaving 12.
P.
246. (Double and spend 12)
thrice to leave 0.
Riese. Die
Coss. 1524. Several examples -- no. 35, 53, 55, 56, 58, 59, 60, 61, 62. I describe a few.
No.
35, p. 45. Man stealing apples: (give half and one more) four times, leaving 1.
No.
53, pp. 47‑48. (Double and
spend 12) thrice, leaving 0.
No.
55, p. 48. (Double and spend i)
for i = 1, 2, 3, leaving
10.
No.
58, p. 48. x + (4x+1) +
(3(4x+1)+3) = 56.
No.
61, p. 48. (Give half plus 2+2i
more) for i = 1, 2, 3, leaving 0.
No.
62, p. 49. (Give half less 2i)
for i = 1, 2, 3, leaving
12.
Tartaglia. General
Trattato. 1556. Book 12, art. 34, p. 199v. Book 16, art. 47 & 113-116,
pp. 246r & 253v‑254r.
Book 17, art. 9 & 20, p. 268v & 271r. Final remainder specified in each case.
12-34. (Take half plus i more) for
i = 1, 2, 3, 4, leaving 1. Cf
16-115.
16-47. Take
1/2 and 1
more, 1/3 and
2 more, 1/4
and 4 more, leaving 26.
16-113. (Double and subtract 20)
thrice, leaving 0 (H&S 61 gives Latin and English of this
one and says it appears in van der Hoecke (1537), Stifel (quoting Cardan)
(1544), Trenchant (1566) and Baker (1568) (but see below).
16-114. (Halve and subtract 1)
thrice, leaving 1, 2, ....
16-115. Halve and subtract 1, then 2,
3, 4, leaving 1. Cf 12-34.
16-116. Lose
1/2 and 3
more, lose 2/3
and get back 10, lose
3/4 and 6
more, lose 4/5
and get back 16, leaving
24.
17-9. Double & spend 4, double &
spend 8, leaving 24.
17-20. Double and spend 18, double and spend 24,
double and spend 36, leaving
280.
Buteo.
Logistica. 1559.
Prob.
6, pp. 334-335. Lose 1/2
and 3 more, lose 1/3
and 4 more, lose 1/4
and gain 1,
to leave 100.
Prob.
13, pp. 342-343. Start with X,
gain 40. Make the same rate of profit twice again and
then the second of these gains is 90.
Prob.
19, p. 347. Double and spend 12,
triple and spend 15, quadruple and spend 14,
leaving 12.
Prob.
20, pp. 347-348. Gain 1/4
and spend 7, gain
1/3 and spend 10,
lose 3/7 and spend
8, leaving 0.
Prob.
21, pp. 348-350. (Double and spend 10)
X times to leave 0.
He makes an error at X = 8 and deduces
X = 7.
Baker. Well Spring
of Sciences. 1562?
Prob.
7, 1580?: ff. 192r-193r; 1646: pp. 302‑304; 1670: pp. 344-345. Lose half and gain
12, lose half and gain 7,
lose half and gain 4, leaving
20.
Prob.
8, 1580?: ff. 193r-193v; 1646: p. 304; 1670: p. 345. (Double and
spend 10) thrice leaving 12.
Gori. Libro di
arimetricha. 1571. F. 72r (pp. 77‑78). (Lose half and one more) four times to leave 3.
Book of Merry Riddles.
1629? (Take half and half more)
thrice, leaving one.
Wells. 1698. No. 118, p. 209. Soldiers take half of a flock of sheep and half a sheep more,
thrice, leaving 20.
Ozanam. 1725. Prob. 28, question 1, 1725: 211‑212. (Give half the eggs and half an egg) thrice.
He doesn't specify the remainder and says that 8n‑1 eggs will
leave n‑1 and that one can replace 8
by 2k if one does the process k
times. Montucla replaces this by
some determinate problems -- see below.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XV, pp. 86-87 (1790: prob. XXVII, pp. 88-89.
Shepherd loses (half and 1/2
more) thrice to leave 5
(or 1) sheep.
Les Amusemens.
1749.
Prob.
108, p. 249. (Double and give away 6)
thrice to leave 0.
Prob.
111, pp. 252-253. Double and spend 20,
triple and spend 27, double and spend 19, leaving
250.
Prob.
118, p. 260. (Halve and give 1/2
more) thrice to leave 0.
Walkingame. Tutor's
Assistant. 1751.
1777:
p. 82, prob. 7; 1860: p. 111, prob.
7. Stealing apples. Give half and get back 10,
give half and get back 4, give half, get back 1,
leaving 24.
1777:
pp. 174-175, prob. 86; 1860: p. 183, prob. 85.
Sheep fold robbed of half its sheep and half a sheep more, thrice,
leaving 20.
Euler. Algebra. 1770.
I.IV.III: Questions for practice, no. 9, p. 204.
(Spend 100, then gain ⅓) thrice to double
original value.
Vyse. Tutor's
Guide. 1771?
Prob.
18, 1793: p. 32; 1799: p. 36 & Key p. 29.
Sheep fold. (Lose half and ½
more) thrice, leaving 20.
Prob.
31, 1793: p. 57; 1799: p. 62 & Key p. 69.
(Gain ⅓, less
100) for 3¼
years to yield £3179 11s
8d. Solution assumes the final quarter
year is the same as
(gain 1/12, less 25),
but it is not obvious how to determine an appropriate expression for the
quarterly effect. In general,
repeating ax - b four times gives a4x - b(a4-1)/(a-1) and setting this equal to 4x/3 - 100
gives a = 1.075.., b = 22.371. However, the 100 is the expenses of
the merchant's family and he may not be able to reduce it in one quarter.
Prob.
33, 1793: pp. 57-58; 1799: pp. 62-63 & Key p. 70. Lose ½, get back
10; lose ⅓, get
back 2; lose ½, get back 1;
leaving 12.
Prob.
2, 1793: p. 129; 1799: p. 137 & Key p. 179. (Double and spend 6) thrice, leaving 0. Solution by double
position.
Prob.
4, 1793: p. 129; 1799: p. 137 & Key p. 180. Lose ½, gain
10; lose ½,
gain 4; lose
½, gain 1;
yielding 18. Solution by double position.
Dodson. Math. Repository. 1775.
P. 10,
quest. XXIV. Double and spend 6; triple
and spend 12; quadruple and spend 18;
leaving 30.
P. 47,
quest. C. Shepherd loses ¼ of his flock
and ¼ of a sheep; then ⅓ of his flock and ⅓ of a sheep; then ½ of
his flock and ½ of a sheep; and has 25 sheep left.
P. 48,
quest CI. Man (spends 50 and gains
⅓ on the remainder) thrice, yielding double his original amount.
P. 49,
quest. CII. Lose ¼, win 3, lose
⅓, win 2, lose 1/7, yielding 12.
Ozanam‑Montucla.
1778.
Prob.
15, part a, 1778: 207-208; 1803:
203. Prob. 14, 1814: 175‑176; 1840: 91.
(Sell half the eggs and half an egg more) thrice to leave 36. This was one of the more popular forms of
the puzzle after this time -- see: Jackson, Endless Amusement II, Nuts to
Crack, Young Man's Book, Boy's Own Book, Magician's Own Book, Boy's Own
Conjuring Book, Wehman, Collins, Sullivan.
Prob.
15, part b, 1778: 208-209; 1803:
203-204. Prob. 15, 1814: 176-177; 1840: 91.
(Spend half and 1/2 more)
thrice leaving 0. Gives the rule for the problem with more
iterations.
Bonnycastle.
Algebra. 1782. P. 86, no. 20 (1815: p. 107, prob. 30). Lose
1/4 of what he has, win 3,
lose 1/3, win
2, lose 1/7,
leaving 12.
Eadon.
Repository. 1794.
P.
296, no. 9. Man loses 1/4,
then gains 3, then loses
1/3, then gains 2,
then loses 1/7, then has
12.
P.
296, no. 10. Man (spends 50
and then gains 1/3) thrice to double his money.
P.
297, no. 14. Man sends out 1/3
and 25 more of his men, leaving
1/2 and 100
more.
Bonnycastle.
Algebra. 10th ed., 1815. P. 106, no. 19. (Double and pay 1) four times, leaving 0.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions.
No.
10, pp. 16 & 73. (Give one
half) seven times, leaving 1.
No.
13, pp. 17 & 74. (Give half
plus i) for i = 1, 2, 3, leaving
1.
No.
18, pp. 18 & 75. (Spend half plus
half a guinea) four times, leaving 0.
No.
19, pp. 18-19 & 75-76. (Sell half,
receive 10); (sell third, receive 2); (sell half, receive 1);
leaving 12.
No.
23, pp. 19-20 & 77. Same as
Ozanam-Montucla, prob. 15a.
Endless Amusement II.
1826? Prob. 27, pp.
202-203. Identical to the 1803 English
of Ozanam-Montucla, prob. 15a. = New
Sphinx, c1840, p. 138.
Nuts to Crack II (1833), no. 126. Same as Ozanam-Montucla, prob. 15a.
Young Man's Book.
1839. Pp. 234-235. Identical to the 1803 English of
Ozanam-Montucla, prob. 15a.
Boy's Own Book. 1843
(Paris): 344. Same as Ozanam-Montucla,
prob. 15a. = Boy's Treasury, 1844,
p. 301. = de Savigny, 1846, p. 289: La
paysanne et les œufs.
Magician's Own Book.
1857.
Countrywoman
and eggs, pp. 238-239. Almost identical
to Boy's Own Book.
The
old woman and her eggs, p. 240. (Give
half and half an egg more) thrice,
leaving 1.
The
apple woman, p. 252. Sell half,
gain 10, sell third, gain 2, sell half, gain 1, have 12 remaining.
= Book of 500 Puzzles, 1859, p. 66.
= Illustrated Boy's Own Treasury, 1860, prob. 15, pp. 428 & 431-432.
Boy's Own Conjuring Book.
1860.
Countrywoman
and eggs, p. 205. Almost identical to Boy's
Own Book.
The
old woman and her eggs, p. 212.
Identical to Magician's Own Book.
The
apple woman, p. 223. Identical to
Magician's Own Book.
Vinot. 1860. Art. LV: Les œufs, pp. 72-73. (Sell half the eggs and half an egg
more) thrice to leave 0.
Lewis Carroll.
Letter of 22 Jan 1878 to Jessie Sinclair. = Carroll-Collingwood, pp. 205-207. Cf Carroll-Wakeling, prob. 26, pp. 34 & 72. "Tell Sally it's all very well to say
she can do the two thieves and the five apples, ...." Wakeling omits the number of apples since it
is the answer to the problem he poses.
Cohen and Wakeling give a possible version of the problem, provided by
Peter Heath, as: steal half and half an
apple more, then the second thief steals half of what the first thief stole and
half an apple more, leaving none, for which the answer is 5.
Cohen says John Fisher [The Magic of Lewis Carroll; op. cit. in 1, p.
79] cites a similar problem from the notebooks of Samuel Taylor Coleridge, but
this is: (sell half and half an egg
more) thrice, leaving three. Another
possibility, which seems much more likely to me, would be (steal half and half an apple more) twice,
leaving 5, for which the answer is
23. This is the more common form
of the problem, whereas Heath's version takes
5 as the answer rather than the
data. In Carroll-Gardner, pp. 77‑78,
Gardner gives a totally different explanation, saying this is an old magic
trick and explaining it.
Mittenzwey.
1880.
Prob.
103, pp. 21 & 73; 1895?: 120, pp.
25-26 & 75; 1917: 120, pp. 24 &
72-73. Egg woman sells (half of her
eggs and half an egg more) four times, leaving
1. In 1917, the solution is
expanded and he notes that starting with
2k - 1, four stages
bring you to 2k-4 - 1, but he doesn't seem to understand how to
solve the problem in general.
Prob.
120, pp. 24-25 & 76; 1895?: 138,
pp. 28-29 & 79; 1917: 138, pp. 26
& 76‑77. Three men
successively taking 1/3 of a pile of potatoes. Remainder is 24. Observes that the
second person is entitled to 3/8 of the remainder and the third person gets
the rest [but this is unnecessary information]. How many potatoes were there?
William J. Milne.
The Inductive Algebra Embracing a Complete Course for Schools and
Academies. American Book Company, NY,
1881. Pp. 138 & 332, no. 81. (Double and lose one) thrice to get triple original amount.
Hoffmann. 1893. Chap. IV gives several deterministic
examples: nos. 27, 39, 46, 67,
76 (? see 7.E.1), 111, 112.
Lucas.
L'Arithmétique Amusante.
1895. P. 184: Prob. XLII: La
marchande d'Œufs.
(Sell 1/2
plus half an egg more) n times to leave 0.
Carroll-Wakeling.
1888 Prob. 17: Four brothers and
a monkey, pp. 21 & 68. This has a
pile of nuts on a table and is Form 1, Ending 0. Wakeling gives the solution
765 and says there are other
solutions, citing 2813 and
5885, i.e. 765 + 2048 k, but the general solution is actually 765 + 1024 k.
This
is one of the problems on undated sheets of paper that Carroll sent to
Bartholomew Price. Wakeling said he
will look for a watermark on it, but the date is now pretty definite. Wakeling says there is no other mention of
the problem in Carroll's work, MSS or correspondence. Carroll-Gardner, p. 53, mentions Carroll-Wakeling and cites his
1958 article.
On
28 May 2003, Wakeling kindly sent me copies of three items of the Carroll/Price
material. First is Carroll's solution
of the problem, which is typewritten, 'probably using Dodgson's Hammond
typewriter, purchased in 1888.' This
solution is grossly erroneous -- he only takes three stages and obtains the
answer 61 + 64k. Most importantly, Wakeling sent a note from
John (later Sir John) Evans to Price, dated 15 Oct 1888, thanking Price for his
solution of the problem and saying that his attempt had gotten to a value of 1789
(which is a correct solution!).
Evans then adds that he cannot make Price's solution work. Price must have given 253,
but after the fourth brother, there remain 78 which is not divisible
by four (nor is it one more than a multiple of four). Evans than says that 509
(= 253 + 256) and 765 (= 253 + 512) also fail, 'I think'. Wakeling also sent the statement, only, of
the problem, in Evans' handwriting, headed
Four Brothers & the Family Monkey -- this differs from the version
in Carroll-Wakeling.
Though
this is a moderately messy problem, it is depressing to discover that three
competent mathematicians were unable to get the correct solution and failed to
check the solutions that they had obtained!!
However, we now know that the problem was in circulation in 1888, and
the fact that wrong answers were being obtained shows that the problem was new
at that time.
W. W. Rouse Ball.
Elementary Algebra. CUP, 1890
[the 2nd ed. of 1897 is apparently identical except for minor changes at the
end of the Preface]. Prob. 12, p. 260
& 475. Three Arab jugglers and
their monkey, on their way to Mecca, buy a basket of dates. Ending 1.
He gives no background to the problem nor any indication that it is
novel. The solution is just the
numerical answer.
The presence of two versions by c1890 would indicate the
problem must have been known to at least a few other people. The fact that both Carroll and Ball knew of
the problem leads one to conjecture a possible mode of transmission. After the appearance of Alice in
1865, Carroll's reputation was immense.
Ball's A Short Account of the History of Mathematics appeared in
1888 and was well-known in the English-speaking world and Ball was becoming
known as an authority on the history of mathematics and on mathematical
recreations. Either or both might have
been sent the problem from India (or anywhere in the world), perhaps by the
translator of Mahavira when he first came on the problem. However, I have examined the Ball material
at Trinity College Cambridge and it is clear that he (or his heirs) disposed of
his correspondence, so this conjecture cannot be verified. We hope 'something will turn up' to
elucidate this. Something has turned up
-- see the further material under Carroll-Wakeling -- but this does not
determine the source of the problem.
Clark. Mental
Nuts. See also at 1904.
1897,
no. 11; 1904, no. 19; 1916, no. 17. The man and his money.
(Spend 1/2 and
1/2 more) four times to leave 0.
1897,
no. 20; 1904, no. 92; 1916, no. 8. The man and the stores.
(Double and pay 10) thrice to leave 0.
1897,
no. 29; 1904, no. 40. The farmer and his horses. (Pay
1, then sell 1/2,
then pay 1 more)
four times to leave 1.
Dudeney. Problem
522. Weekly Dispatch (8 &
22 Nov 1903) both p. 10.
Multiple of 25 eggs.
Sell half and half an egg more until all gone.
Clark. Mental
Nuts. 1904, no. 99. Three boys and basket of apples. Coconuts -- Ending 1 with
3 people. (This is not in the 1897 or 1916 eds. This complicates the possible connection
with Mahavira -- cf the discussion under Pearson, below.)
Pearson. 1907. Part II.
Several determinate versions, and the following.
No.
29: The men, the monkey, and the mangoes, pp. 119 & 197. Coconuts -- Ending 1 with
3 people. Gives only one solution.
No.
94: One for the parrot, pp. 133‑134 & 210. Coconuts -- Ending 1
with 4 boys, a bag of nuts and a parrot. Gives only one solution.
The connection of these with Mahavira, 850, bemused me and I
conjectured that Pearson might have heard of Mahavira's work, though the
translation didn't appear until 1912.
Kaye's note (see under Mahavira in the Abbreviations) shows that an
advance version of the translation was produced in 1908, which makes my
conjecture much more likely. The Frontispiece
of Pearson's book shows him as a clergyman of about 40-50 years old, and
another of his books describes him as MA of Balliol College, Oxford and Rector
of Drayton Parslow, Buckinghamshire, with a stamp underneath giving Springfield
Rectory, Chelmsford [Essex], so he might have been a missionary or had Indian
contacts. (Incidentally, the publisher
Cyril Arthur Pearson was his son.)
HOWEVER, I have now seen Carroll, Ball and Clark and this makes the
connections less clear.
Wehman. New Book of
200 Puzzles. 1908.
P. 50:
The sheepfold robbery. (Lose 1/2
and 1/2 a sheep more) thrice leaving 2.
P. 51:
The maid and her apples. c= Magician's
Own Book, p. 252.
P. 57:
The countrywoman and her eggs. Same as
Ozanam-Montucla, prob. 15a.
Nelson L. Roray, proposer;
A. M. Harding, Norman Anning and the proposer, solvers. Problem 288. SSM 12 (1912) 235
& 520‑521. Coconuts -- Ending 1 with
3 men. Anning shows that the solution is ‑2 (mod 34),
but none of the solvers generalise to
n men.
Loyd. Newsboys
puzzle. Cyclopedia, 1914, pp. 116 &
354. (= MPSL2, prob. 9, pp. 8 &
123. = SLAHP: Family rivalry, pp. 51
& 103.) Complex specification of
one amount.
Loyd. A study in
hams. Cyclopedia, 1914, pp. 268 &
375. (= SLAHP: The Ham peddler,
pp. 81 & 117.) (Half plus half
a ham more) four times, (half a ham plus half), (half plus half a ham), leaving
0.
R. L. Weber. A
Random Walk in Science. Institute of
Physics, London & Bristol, 1973.
P. 97 excerpts a Russian book on humour in physics which states
that P. A. M. Dirac heard a version of the problem with three fishermen and a
pile of fish, but only three divisions, at a mathematical congress while he was
a student (at Cambridge?) and gave the solution, -2. In fact, he only came
to Cambridge as a graduate student in 1923 and became a fellow in 1927, so that
the story, if true and if it refers to his time at Cambridge, relates to the
mid 1920s.
Ben Ames Williams.
Coconuts. Saturday Evening Post
(9 Oct 1926) 10,11,186,188. Reprinted
in: Clifton Fadiman, ed.; The
Mathematical Magpie; Simon & Schuster, NY, 1962, pp. 196-214. Ending 0
with 5 men.
R. S. Underwood, proposer;
R. E. Moritz, solver. Problem
3242. AMM 34 (1927) 98 (??NX) &
35 (1928) 47‑48. General
version of coconuts problem, Ending
0 with
n men.
Wood. Oddities. 1927.
Prob.
52: A goose problem -- not for geese to solve, pp. 43 & 44. Sell a half and half a goose more; sell a third and a third of a goose more; sell a quarter and 3/4 of a goose more; sell a fifth and a fifth of a goose
more; leaving 19.
Prob.
57: Eggs this time, p. 46. Sell half
and half an egg more; sell a third and
a third of an egg more; sell a quarter
and a quarter of an egg more; sell a
fifth and a fifth of an egg more;
leaving a multiple of 13.
Determine the least number of possible eggs. Gives answer 719. Complete answer is 719 (mod 780).
Collins. Book of
Puzzles. 1927. The basket of eggs puzzle, p. 77. Same as Ozanam-Montucla, prob. 15a.
Collins. Fun with
Figures. 1928. The parrot talks, pp. 183-185. Four boys and a parrot and a bag of
nuts. Ending 1 with
n = 4. = Pearson 94.
Kraitchik. La
Mathématique des Jeux, 1930, op. cit. in 4.A.2. P. 13.
Prob.
39. Monkey and mangoes problem, Ending 1
with 3 men. = Pearson 29. = MR, 1942, prob. 35, pp. 32-33.
Prob.
40. (Take ⅓) thrice,
leaving 8. = MR, 1942, prob. 36, p. 32.
He asserts these are Hindu problems but gives
no source.
Rudin. 1936. No. 5, pp. 3 & 76. Three men.
(Take away ⅓) thrice, then
divide in thirds. He gives only the
answer 81, though any multiple of
81 works.
Hubert Phillips.
Question Time. Op. cit. in
5.U. 1937. Prob. 203: Adventure island, pp. 137 & 246. Ending 1
with 5 men and Friday instead of a monkey.
Francis & Vera Meynell.
The Week‑End Book.
Nonesuch Press, 1924 and numerous printings and editions. I have 8th printing, 2nd ed., Mar 1925, and
a 5th(?) ed., in 2 vols., Penguin, 1938.
The earlier edition has some extra text surrounding the problems, but
has only 8 of the 12 problems in the Penguin ed. This problem is not in the 2nd ed. 5th?? ed., prob. seven, p. 408: Three men and a monkey. Ending 0,
with 3 men. No solution.
Joseph Bowden.
Special Topics in Theoretical Arithmetic. Published by the author, Lancaster, Pennsylvania, 1936. The problem of the dishonest men, the
monkeys and the coconuts, pp. 203-212.
??NYS - cited by Pedoe, Shima & Salvatore.
McKay. At Home
Tonight. 1940.
Prob.
33: The niggers and the orchard, pp. 69 & 82. Three men and apples.
Ordinary division with the extra thrown away and Ending 1.
Prob.
35: Dividing nuts, pp. 70 & 83.
Divide nuts among 5 girls with one left over. One girl divides hers among the rest, with
one left over. Then another girl
divides hers among the rest, with one left over.
The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. Pp. 149‑150: The bag of
peanuts. Five Italians, a bag of
peanuts and a monkey. Coconuts --
Ending 1 with 5 people. The given
answer is 3121, but it should be 15621. 3121 would be the answer for Ending 0 -- cf
Leeming, 1946.
Garrett Birkhoff & Saunders Mac Lane. A Survey of Modern Algebra. Macmillan, (1941, ??NYS), revised,
1953. Prob. 11, p. 26 (p. 28 in the 4th
ed. of 1977). Usual five men and a
monkey form, but no ending is specified.
That is, nothing is stated about the number left over in the morning
except that it is an integer. No answer
given, but there is a hint to try
-4 coconuts. Somewhat surprisingly, the answer is the
same as for the Ending 0 problem -- see Pedoe, Shima & Salvatore.
Leeming. 1946. Chap. 5, prob. 16, pp. 57‑58 &
177. Five Italian organ grinders, their
monkey and a pile of peanuts. Coconuts
-- Ending 1 with 5 people.
The given answer is 3121, but it should be 15621. 3121 would be the answer for Ending 0 -- cf Home
Book, 1941.
Sullivan.
Unusual. 1947. Prob. 40: Another old problem. Sell (half plus half an egg more) thrice, leaving 36. = Ozanam-Montucla,
prob. 15a.
Paul S. Herwitz. The
theory of numbers. SA 185:1 (Jul 1951)
52-55. With letter and response in SA
185:3 (Sep 1951) 2 & 4. Gives the
Birkhoff & Mac Lane version with
n = 3 and solves it by
explicitly computing the number remaining in terms of the initial number,
obtaining one linear diophantine equation in two unknowns. He states the equation for the case m = 4,
but doesn't give the solution, and for the general case, though he doesn't
sum the geometric progression that appears.
The letter requests the solution for the case m = 5 and Herwitz outlines
how to find the solution by the Euclidean algorithm, obtaining 3121.
Anon.?? Monkeys and
coconuts. Mathematics Teacher 54 (Dec
1951) 560-562. ??NYS - cited by Pedoe,
Shima & Salvatore.
Anonymous. The
problems drive. Eureka 17 (Oct 1954)
8-9 & 16-17. No. 2. Three men and cigarettes guarded by a Boy
Scout. Two are given to the scout at
each division and at the end, so this is Form 1c, Ending c, with c = 2.
Solution observes that
-(n-1)c is a fixed point and the
solution is -(n-1)c (mod nn+1), as seen in Singmaster, but the solution here
doesn't give any proof.
Ron Edwards. The
cocoanut poker deal. The Cardiste (Mar
1958) 5-6. Uses the three person
problem as the basis of a card trick.
He states the original problem in a novel form -- each hunter finds the
pile evenly divisible by three, so the monkey doesn't get any coconuts until
the morning division when he gets one.
But in the trick, Edwards uses the classic form, with a variation. The
52 cards are dealt into 3
piles, with one extra put in a
discard pile. The spectator places one
of the end piles on the middle giving a new deck of 34. The process is repeated
twice more leaving 22, then
14 cards. These
14 are dealt into three piles,
but now there are two extras which are discarded and then the five discards are
found to be a royal flush! (The
Cardiste was a mimeographed magic magazine which ran from 1957 to 1959. My thanks to Max Maven for remembering and
finding this and sending a copy.)
M. Gardner. SA (Apr
1958) = 2nd Book, chap. 9. Describes the Williams story of 1926 and
says the Post received 2000 letters the first week after publication and the
editor telegrammed: "For the love
of Mike, how many coconuts? Hell
popping around here." Gardner says
Williams modified the older problem of Ending 1 with 5 men.
He gives the solution of
-4, but says he could not trace
its origin. His addendum cites Anning
(1912) for this.
Roger B. Kircher.
The generalized coconut problem.
AMM 67:6 (Jun/Jul 1960) 516-519. Generalizes by taking any number of sailors, any number of
divisions and allowing the i‑th
division to discard a variable amount Vi before taking away 1/n of the rest, even
allowing negative Vi, e.g. if the monkey is adding coconuts to the
pile! Sadly, his basic recurrence
equations (1) and (2) are misprinted.
He solves this by use of difference calculus.
Nathan Altshiller Court.
Mathematics in Fun and in Earnest.
Op. cit. in 5.B. 1961. Prob. c, pp. 188-190. Take
⅓ thrice, leaving 8.
How to divide the remainder to make things equal?
Philip Haber. Peter
Pauper's Puzzles & Posers. Peter
Pauper Press, Mount Vernon, NY, 1963.
Prob. 125, pp. 34 & 57.
Basket of pears divided among four people. First gets ¼ of the total plus ¼ of a pear. Second gets
⅓ and ⅓.
Third gets ½ and
½. Fourth gets remainder, which
is half of what the first got.
Harold H. Hart. Grab
a Pencil No. 3. Hart Publishing, NY,
1971. The horse trader, pp. 41 &
118. (Pay 1, then half, then 1
more) thrice, leaving 1.
Birtwistle. Math.
Puzzles & Perplexities. 1971.
The
Arabs, the monkey and the dates, pp. 50-52.
Four people, Ending 1. Gets 1021.
Baling
out, pp. 52, 169 & 189. (Lose
⅓ of a load of bales of hay and ⅓ of a bale more) four times,
leaving an integral number and no broken bales. Solution is to start with
80 bales.
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 144: The mango thieves, pp. 89 &
135. Ending 1 with three boys, but no monkey -- the boys each eat a mango when
they steal ⅓ and there is an extra mango when they divide
in the morning.
Michael Holt. Figure
It Out -- Book One. Dragon (Granada),
London, 1978. Problem 14 (no page
number) gives a Russian version involving a man who sells his soul to the
Devil. (Double and spend 8)
thrice to leave 0.
Birtwistle.
Calculator Puzzle Book.
1978. Prob. 27: Shipwreck, pp.
21 & 80. Four sailors, the ship's
cat and a box of biscuits, but only three of the sailors take a fourth, with
one for the cat, before the final division into four, with none for the
cat. Answer is 61,
but this should be taken (mod
256).
Dan Pedoe, Timothy Shima & Gali Salvatore. Of coconuts and integrity. CM 4:7 (Aug/Sep 1978) 182‑185. General discussion of history and examples,
based on Gardner (1958). They
generalise to consider giving c to the monkey and to having m
divisions. They first do
Birkhoff & Mac Lane's version and
Ending 1, effectively noting that
the latter is the same as the former but with an extra division step. I.e., the Birkhoff & Mac Lane problem
has m = n, while Ending 1 has
m = n+1. Examination of the
results shows that the Birkhoff & Mac Lane problem with odd n
has the same answer as the
Ending 0 problem, but for
even n, it corresponds to an
Ending 2 problem, which turns
out to be one greater than the Form,
Ending = 0, 1 problem. They separately
solve the Ending 0 problem, again with general c.
Ben Hamilton.
Brainteasers and Mindbenders.
(1979); Prentice‑Hall,
Englewood Cliffs, NJ, 1981. Problem for
March 29, pp. 36 & 156. i‑th
customer buys i+1 plus
1/(i+1) of the rest. How many customers can be served?
Scot Morris. The
Next Book of Omni Games. Plume (New
American Library), NY, 1988. The monkey
and the coconuts, pp. 30-31 & 182-183.
Sketches usual history. He then
notes that the usual process has the pile
º 1 (mod n), so the
monkey essentially gets one, then the pile is divided into n
parts. But one could
alternatively have the pile º 0
(mod n), so the pile is divided
into n
parts, the sailor takes his part and then the monkey takes 1
from the remainder. I.e. rather
than removing one and 1/n of the rest, each step removes 1/n
plus one more. I have now termed
these Form 1 and Form 0. In this situation he doesn't allow the
monkey to get one in the final division, i.e. he considers Ending 0,
but Ending 1 could be permitted, as studied by me below.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
It really takes the biscuit, pp. 50-52 & 119. Four boys and their cat dividing
biscuits. Form 1, Ending 1 with
4 boys.
David Singmaster.
Coconuts. The history and
solutions of a classic diophantine problem.
Technical Report SBU-CISM-93-02, School of Computing, Information
Systems and Mathematics, South Bank Univ., Dec 1993, 21pp, Feb 1994. Submitted to Mathematics Review (Univ. of
Warwick), 1993, but the journal closed before using it. Revised in 1996 and again as the 3rd ed. of
11 Sep 1997, 21pp. Extensive history,
based on the material in this section.
Following Morris's comment, I consider both division Forms and both
Endings, giving four basic problems rather than the three that he
mentions. In the 3rd ed., I changed the
terminology to Form and Ending and have converted the material in this section
to conform with this. A little
reflection shows that the solution for
Form 0, Ending 0 is one less
than for Form 1, Ending 1. Actual calculation shows that one of the
four cases has the same sequence of pile sizes as another, but shifted by one
stage. When n is odd, Form 0, Ending 0 is the same as Form 0,
Ending 1, but starting one stage
earlier. When n is even, Form 1, Ending 1 is the same as Form 1,
Ending 0, but starting one stage
earlier. Although these results are
easily seen from the algebraic expressions, I cannot see any intuitive reason
for these last equalities.
Having
now seen Pedoe, Shima & Salvatore, I have added two supplementary pages
discussing the Birkhoff & Mac Lane problem and relating it to the standard versions.
7.E.1. VERSIONS WITH ALL GETTING THE SAME
The i‑th child gets some linear function
of i
applied to the remainder, but all wind up with the same amount.
See Tropfke 586.
Fibonacci. 1202.
P. 279
(S: 399). i-th gets i
+ 1/7 of rest. (Sanford 219
gives the English; H&S 61‑62
gives Latin & English.)
P. 279
(S: 399). i-th gets i
+ 2/11 of the rest, but he doesn't ask for the number of children.
Pp.
279‑280 (S: 399-401). i-th
gets (3i‑1) +
6/31 of the rest.
Pp.
280‑281 (S: 401). i-th gets (2i+1)
+ 5/19 of the rest.
Maximus Planudes.
Ψηφηφoρια κατ'
Ivδoυσ η Λεγoμεvη
Μεγαλη [Psephephoria kat' Indous e Legomene
Megale (Arithmetic after the Indian method)].
c1300. (Greek ed. by Gerhardt,
Das Rechenbuch des Maximus Planudes, Halle, 1865, ??NYS [Allard, below, pp.
20-22, says this is not very good].
German trans. by H. Waeschke, Halle, 1878, ??NYS [See HGM II 549; not mentioned by Allard].) Greek ed., with French translation by A.
Allard; Maxime Planude -- Le Grand Calcul selon des Indiens; Travaux de la
Faculté de Philosophie et Lettres de l'Univ. Cath. de Louvain -- XXVII, Louvain‑la‑Neuve,
1981.
i-th
gets i
+ 1/7 of the rest, pp. 191‑194 & 233‑234. On pp. 233‑234, Allard discusses the
history of the problem, citing:
Fibonacci; BR; Parisinus supp. gr. 387 &
Scoriolensis Φ I 16.
BR. c1305. No. 84, pp. 102‑105. i-th gets
i + 1/7 of the rest.
Gherardi. Libro di
ragioni. 1328. Pp. 37‑38. i-th daughter gets i
+ 1/10 of the rest.
Lucca 1754.
c1330. F. 82v, p. 199. i‑th gets i + 1/10
of rest. (This problem is not
clearly expressed.)
Bartoli.
Memoriale. c1420. Prob. 9, f. 75v (= Sesiano pp. 138 &
147-148). i-th gets i
+ 1/7 of the rest.
Pseudo-dell'Abbaco.
c1440.
Prob.
168, p. 140. i-th gets 1000 i
+ 1/10 of the rest.
Prob.
169, pp. 140‑141. i-th gets 1/6
plus 10 i.
AR. c1450. No. 114, 115, 352, pp. 64‑65, 154, 173‑174
& 220.
114: i-th gets
i + 1/10 of rest.
115: i-th gets
i + 1/6 of rest.
352: i-th gets
1/5 of remainder + i ‑
1.
Muscarello.
1478. F. 85v, pp. 204-205. i-th gets
i + 1/9 of the rest.
Chuquet. 1484. Probs. 129‑141. English of prob. 129 in FHM 224-225, with
some description of the others. i-th
child gets (ai + b) plus
r of the rest; i-th child gets r of amount and a + bi
more. Many problems have non‑integral
number of children and amounts received -- e.g. prob. 133 has 2 5/6
children receiving 6 2/3, with the
5/6 getting 5 5/9.
HB.XI.22. 1488. Pp. 44‑45 (= Rath 247). i-th gets
i + 1/9 of rest.
Calandri.
Arimethrica. 1491. F. 65r.
i-th gets 1/10 +
1000 i
Calandri, Raccolta.
c1495. Prob. 26, pp. 25‑26. i-th gets
1000 i + 1/10 of the
rest.
Ghaligai. Practica
D'Arithmetica. 1521.
Prob. 24,
ff. 65v-66r. i-th gets 1000 i
+ 1/7 of rest.
Prob.
25, f. 66r. i-th gets 1/7
+ 1000 i.
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 65, f. FF.ii.r
(p. 155). i-th child gets 1/7
of remainder + 100 i
Tartaglia. Quesiti,
et Inventioni Diverse. Venice,
1546. Book 9, quest. 2, pp. 98r‑98v. i-th child gets i + 1/8
of the rest.
Tartaglia. General
Trattato, 1556, art. 46, pp 245v‑246r.
i-th child gets i +
1/6 of the rest; discusses same with 1/7
and 1/13.
Buteo. Logistica. 1559.
Prob. 78, pp. 286-288. i-th
child gets 1/6 + 100 i.
Bachet.
Problemes. 1612. Addl. prob. VII: Un homme venant à mourir
..., 1612: 149-154; 1624: 221-226; 1884: 158‑161. i-th child gets i + 1/7
of the rest; also ai + 1/n and
1/n + ai. Asserts
some cases are impossible, contrary to Chuquet's approach. Labosne has much revised the entire problem.
Ozanam. 1725. Prob. 10, question 9, 1725: 67‑68. Prob. 1, 1778: 185; 1803: 182-183; 1814: 159; 1840: 82. i-th gets
10000 i + 1/7
of the remainder.
Les Amusemens.
1749.
Prob.
55, pp. 187-188. i-th gets 1000 i
+ 1/7 of the rest.
Prob.
177, p. 328. i-th gets 1000 i
+ 1/5 of the rest.
Euler. Vollständige
Anleitung zur Algebra. (St. Petersburg,
1770) Part 2, sect. 1, chap. 3, art. 42.
(= Opera Omnia (1) 1 (1911), pp. 226‑228. = Algebra; 1770; I.IV.III.604: Question 21,
pp. 202‑203.) i-th gets 100 i
+ 1/10 of rest.
Hutton. A Course of
Mathematics. 1798? Prob. 10,
1833: 214-217; 1857:
218-221. Father with three sons
leaves ai + 1/n of the remainder to the i-th and this exhausts the fortune (but they
do not get equal amounts except when n
= 4). Finds algebraic expressions for
the total and each portion, e.g. the total fortune is
(6n2
- 4n + 1)a/(n - 1)2.
Augustus De Morgan.
Arithmetic and Algebra. (1831
?). Reprinted as the second, separately
paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin &
Craddock, London, 1836. Art. 114, pp.
29-30. First gets 2 +
1/6 of rest; second gets 3 + 1/6
of rest; they find they got the
same amount.
Bourdon.
Algèbre. 7th ed., 1834. Art. 48, pp. 66-71 is the same as
Hutton's 1798? problem. Art. 49, pp.
71-73 then treats the usual problem of the same form, finding n-1
children and a fortune of a(n-1)2.
Fred Barlow. Mental
Prodigies. Hutchinson, London, (1951),
corrected reprint, 1952. On
pp. 38-41, he describes Henri Mondeaux (1826-1862), an illiterate who
developed powers of mental calculation, and was then taught by a schoolmaster,
M. Jacoby. At some time, he was asked
to solve the problem of people taking
100i + 1/13 of the rest
and he found the answer in a few seconds by taking 12 x 100
as the amount of each person and
12 x 12 x 100 as the total
amount. I wonder if he knew this type
of problem beforehand??
Vinot. 1860. Art. XXXIX: Du testament, pp. 38-39. i-th gets
1000 i + 1/5
of the rest.
Hoffmann. 1893. Chap. IV, no. 76: Another eccentric
testator, pp. 166 & 221‑222 = Hoffmann-Hordern, p. 148. First son gets 1/6 plus 240,
second son gets 1/5 of the remainder plus 288,
..., fifth gets 1/2
of the remainder plus 720 and all wind up with the same amount.
Lucas.
L'Arithmétique Amusante.
1895. Prob. XL: Le Testament du
nabab, pp. 144-147. i‑th
gets i
+ 1/7 of the rest. Gives
general solution of: i-th gets i
+ 1/n of the rest, using a rectangular layout of markers similar to
ancient Indian multiplication tables and Lucas thinks the ancient Indians must
have known this problem and solution.
7.F. ILLEGAL OPERATIONS GIVING CORRECT RESULT
'Two digit'
refers to illegal cancellation with two digit numbers, e.g. 16/64 = 1/4, etc.
A. Witting. Ernst
und Scherz im Gebiete der Zahlen. Zeitschr.
math. u. naturw. Unterricht 41 (1910) 45‑50. P. 49 gives the rule of Ahrens, below, pp. 75‑76, for the
case k = 2. He also gives three of the two digit examples: 26/65,
16/64, 19/95 --
omitting 49/98.
Ahrens. A&N,
1918.
Pp. 73‑74
finds the two digit solutions and some with more digits.
Pp. 75‑76
studies (a + m/n)1/k = a
* (m/n)1/k and finds
that m = a, n = ak ‑ 1 works.
W. Lietzmann.
Lustiges und Merkwürdiges von Zahlen und Formen. 1922.
2nd
ed., F. Hirt, Breslau, 1923. Pp. 103‑104. Gives
26/65 and 16/64
and the general rule of Ahrens, pp. 75‑76, citing Witting and
Ahrens.
4th
ed, same publisher, 1930, p. 153, says there are more two digit examples and
gives also 266/665, etc.
The
material is also in the 6th ed. (1943), but not in the 7th ed. (1950).
R. K. Morley, proposer;
Pincus Schub, solver. Problem
E24. AMM 40 (1933) 111 &
425‑426. 2 digit versions.
G. [presumably the editor, Jekuthiel Ginsburg]. Curiosa 31 -- Another illegal
operation. SM 5 (1939) 176. Cites Morley. Refers to E. Nannei presenting several larger examples in 1935,
some involving cancellation of several digits.
William R. Ransom.
Op. cit. in 6.M. 1955. Freak cancellations, pp. 100‑102. Finds the 2‑digit versions and give
examples of several 3‑digit forms:
138/345 = 18/45, 163/326 =
1/2, 201/603 = 21/63.
B. L. Schwartz, proposer;
C. W. Trigg, solver. Problem 434
-- Illegal cancellation. MM 34 (1961)
??NYS & 367‑368. 3 digit
versions.
Anon. Curiosa 122 --
A common illegal operation. SM 12
(1946) 111. (a2 ‑ b2)/(a ‑ b) = a + b.
Alan Wayne, proposer;
solution ??NYS. Problem
3568. SSM 75:2 (No. 660) (Feb 1975)
204. (3/2)2 - (1/2)2 gives the right answer when the exponents
are interpreted as multipliers.
Ben Hamilton. Op.
cit. in 7.E, 1979. Problem for April 9,
pp. 40 & 157‑158.
249/498 gives 24/48
correctly, which gives 2/8 wrongly.
R. P. Boas.
Anomalous cancellation. In: R. Honsberger, ed.; Mathematical Plums; MAA,
1979. Chap. 6, pp. 113‑129. Surveys the problem and studies the two
digit case for other bases, e.g. 32/13
= 2/1 in base 4. Cites the SM report, 1939.
R. P. Boas.
Generalizations of the
64/16 problem. JRM 12 (1979‑80) 116‑118. Summarises the above paper and poses
problems.
7.G.1. HALF + THIRD + NINTH, ETC.
This
is usually called 'The 17 camels', etc.
Early
versions of this problem simply divided the amount proportionally to the given
numbers, regardless of whether the numbers added to one or not. I mention a few early examples of this
below. By about the 15C, people began
objecting to such proportions, though Tartaglia (qv) and others see no
difficulty with the older idea. Sanford
218‑219 says Tartaglia was among the first to suggest the 18th camel --
but I have not found this in Tartaglia so far.
H&S 87 says that the use of the 18th camel is really a modern
problem. I haven't found it occurring
until late 19C, when several authors claim it is centuries old and comes from
the Arabic world or India or China! See
1872 below. Not everyone is happy with
the problem, even to this day -- see Ashley, 1997 -- and many strange
explanations have been given.
1/2,
1/3 Pseudo-dell'Abbaco; Tagliente;
Buteo
1/3,
1/4 Jackson
9,
8, 7 Papyrus of Akhmim
5/6,
7/12, 9/20 Recorde
4/5,
3/4, 2/3 Tartaglia
44
2/3,
1/6, 1/8 Eperson
2/3,
1/2, 1/4 Chuquet; Apianus;
1/2,
1/3, 1/4 Bakhshali
MS; Lucca 1754; AR 204;
Calandri; Tagliente; Riese;
Recorde; Tartaglia 42; Buteo;
Ozanam; Les Amusemens; Decremps;
Bullen; Collins; Always;
1/2,
1/3, 1/6 AR,
286
1/2,
1/3, 1/9 AR,
170, 286; Hanky Panky; Cassell's;
Proctor; Cole; Lemon;
Hoffmann; Brandreth Puzzle
Book; Loyd; H. D. Northrop;
Benson; White; Ball‑FitzPatrick; Dudeney;
Kraitchik; McKay; Sullivan;
Doubleday - 1; Ashley;
1/2,
1/4, 1/5 Lemon; Clark;
Ernst; King; Foulsham
1/2,
1/4, 1/6 Clark;
1/2,
1/4, 1/8 Bath
2/5,
1/3, 1/4 AR,
202; Wagner
1/3,
1/4, 1/5 BR; Riese;
Tartaglia 43; Jackson
1/4,
1/5, 1/6 Apianus; W. Leybourn
2/3,
1/2, 1/3, 1/4 Papyrus
Rhind; Pike; D. Adams, 1835
1/2,
1/3, 1/4, 1/5 Chaturveda; Blasius
1/2,
1/3, 1/4, 1/6 Mahavira
1/2,
1/3, 1/6, 1/19 Parlour
Pastime
1/3,
1/4, 1/5, 1/6 Walton; Simpson;
Dodson; J. King
1/3,
1/4, 1/6, 1/8 D.
Adams, 1801;
6, 5,
4, 3, 2 Mahavira
7/2,
5/2, 15/4, 25/4, 4 Papyrus
of Akhmim
1/2,
1/3, 1/4, 1/5, 1/6 Tonstall
1/3,
1/4, 1/5, 1/6, 1/7 Walkingame; Vyse
1/3,
1/4, 1/6, 1/8, 1/9 Meyer; Haldeman-Julius; Leeming
Papyrus Rhind, c‑1650, loc. cit. in 7.C. Problem 63, p. 101 of vol. 1 (1927) (= p. 53
(1978)). Divide 700
loaves in proportion
⅔ : ½ : ⅓ : ¼.
Papyrus of Akhmim.
c7C. Jules Baillet, ed. Le Papyrus Mathématique d'Akhmîm. Mémoires publiés par les membres de la
Mission Archéologique Français au Caire, vol. IX, part 1, (1892) 1‑89. Brief discussion of this type of problem on
p. 56. Probs. 3, 4, 10, 11, 28?, 47, 48, 49
are of this type. I describe two
examples.
Prob.
3, pp. 64-65. Divide 1000
in proportion
3 + 1/2 : 2 + 1/2 : 3 + 1/2 + 1/4 : 6 + 1/4
: 4.
Prob.
11, pp. 68-69. Divide 3 + 1/2 + 1/4 in proportion 7 : 8 : 9.
Bakhshali MS.
c7C. See in 7.E, where a king
gives away ½ + ⅓ + ¼ of his money!
Mahavira. 850. Chap. VI, v. 80, 86, pp. 110-111. Divide
120 in proportion 1/2 : 1/3 : 1/4 : 1/6. Divide
480 in proportion 2 : 3 : 4 : 5 : 6.
BR. c1305. No. 71, pp. 94‑95. 1/3 + 1/4 + 1/5.
Lucca 1754.
c1330. F. 61r, p. 140. Divide into
½ + ⅓ + ¼. He divides in
proportion
6 : 4 : 3.
Pseudo-dell'Abbaco.
c1440. Prob. 122, pp. 97‑98. Divide into
½ + ⅓. He divides in the
ratio 3 : 2.
AR. c1450. Probs. 170, 202-204, 207, 229-230, 286. Pp. 81‑82, 94‑96, 106-107, 130,
160‑161, 166‑167, 211‑213.
170: 1/2 + 1/3 + 1/9.
202: 1/3 + 1/4 + 2/5.
203: Divide 384 into 2/3 and 6 more, 3/5 and 8
more, 5/6 and 10 more, 7/8 and 6 more. He takes a common denominator of
360 and finds 2/3
of it is 240 and then adds 6 to get 246.
Similarly, he gets 224, 310,
321 and then divides in the
proportion 246 : 224 : 310 : 321. This is actually indeterminate as it depends
on the choice of common denominator.
Vogel says the problem is unclear and the solution is false and notes
that dividing 387 instead of
384 would give an integral
solution. He cites a number of other
occurrences of this problem -- cf. Widman below.
204: ½ + ⅓ + ¼.
207: Divide 100 into (1/3 ‑ 1/4) + (1/4 ‑ 1/5) + (1/5 ‑ 1/6).
229: Divide 20 into 1½, 2½, 1, 1, 1, 1. Does
as
3 : 5 : 2 : 2 : 2 : 2.
230: Divide 20 into 1½ + ⅓, 2½ + ¼, 1, 1, 1, 1. Does as
22 : 33 : 12 : 12 : 12 : 12.
286
discusses problems where one removes fractions and deals with the
remainder. Notes that 1/2 + 1/3 + 1/6 leaves nothing, but 1/2 +
1/3 + 1/9 leaves something.
Chuquet. 1484. Triparty, part 1. English in FHM 75.
"I wish to divide 100 into three parts of such proportion as
are 1/2, 2/3, 1/4 ..."
Ulrich Wagner.
Untitled text known as "Das Bamberger Rechenbuch". Heinrich Petzensteiner, Babenberg (=
Bamburg), 1483. Reproduced, with transcription
and notes by Eberhard Schröder as: Das
Bamberger Rechenbuch von 1483. Akademie‑Verlag,
Berlin, DDR, 1988.
Pp. 69‑70
& 200. = AR, no. 202.
Pp.
71-72 & 201-202. = AR, no. 203.
Calandri.
Aritmetica. c1485. Ff. 93v-94r, pp. 187‑188. Same as Lucca 1754.
Johann Widman.
Behēde und hubsche Rechnung auff allen kauffmanschafft. Conrad Kacheloffen, Leipzig, 1489. ??NYS.
(Rara 36‑40. This is
extensively described by: J. W. L.
Glaisher in Messenger of Mathematics 51 (1921‑22) 1‑148, but he gives
the title as: Behēde und hubsche
Rechenung ....) Smith and Glaisher
give Widman, but Knobloch (7.L.2.c) uses
Widmann and Behende und hubsche Rechenung ....
F.
195v (Glaisher 14-15 & 122). = AR,
no. 230.
Ff.
195v-196 (Glaisher 15 & 122). = AR,
no. 207.
F.
196v (Glaisher 18-19, 38, 45, 122, 130).
= AR, no. 203.
Glaisher notes that Pacioli's Summa,
(see below), gives a more natural determinate interpretation for similar
problems. In this example, this would
first subtract 6 + 8 + 10 + 6 from
384, leaving 354
which would be divided in the proportion 2/3 : 3/5 : 5/6 : 7/8.
He also notes the appearance of Widman's problem and solution in Huswirt
(1501) ??NYS and of problems similar to
Widman and done in the same way, in Arithmetice Lilium (a book of c1510, ??NYS)
(divide 100 into 1/2 + 3, 1/3 + 2,
1/5 + 4) and Tonstall. Rudolff's Kunstliche Rechnung of 1526,
??NYS, does (divide into 1/2 and 6, 1/3
and 4, 1/4 less 2) in Pacioli's
manner. Cf Apianus for a similar
version. Riese's Rechenung nach der
Lenge (1550?, ??NYR) does (divide
124½ into 2/3 less 12, 1/4 and 10, 5/6 less
24, 3/8 and 6, 2/5 less 7) in Pacioli's way.
Pacioli. Summa. 1494.
These give the more natural interpretation of this type of problem.
F.
150r, prob. 3. Divide 100 as 1/2 plus 5;
1/3 less 4. Subtracts 1 from 100
and divides the resulting 99 in the proportion
3 : 2.
Ff.
150r-150v, prob. 4. Divide 100 as 1/2 plus 3;
1/3 less 5. Divides 102 as 3 : 2.
F.
150v, prob. 5. Divide 100 as 1/2 less 4;
1/3 less 2. Divides 106 as 3 : 2.
F.
150v, prob. 6. Divide 30 as 1/2 plus 2;
1/3 plus 3. Divides 25 as 3 : 2.
F.
150v, prob. 7. Divide 10 as 1/2 less 3;
1/3 plus 4. Divides 9 as 3 : 2.
F.
150v, prob. 8. Divide 1046 as 1/2 less 2;
1/3 less 1; 1/4 plus 5. No working or answer.
Blasius. 1513. F. F.ii.r: Prime regula. Man leaves
6000 to be divided 1/2 + 1/3 + 1/4 + 1/5. There is an error in the calculation.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob.
94, part 2, ff. 48v-49r. Divide 120
into ½ + ⅓.
Prob.
95, ff. 48v-49v. Divide 12
into ½ + ⅓ + ¼.
Riese.
Rechnung. 1522.
1544
ed. -- pp. 81‑82; 1574 ed. -- pp.
55r‑55v. 1/3 + 1/4 + 1/5.
1544
ed. -- pp. 98‑99; 1574 ed. -- p.
66r. Three men take 1/2 + 1/3 + 1/4 of the profits, making 50
all told. What was the
profit? Answer: 50 x 12/13.
Tonstall. De Arte
Supputandi. 1522.
Quest.
17, pp. 147-149. Divide into 1/2 + 1/3 + 1/4 + 1/5 + 1/6. Uses
4350 as common denominator.
Quest.
18, pp. 149-150. Same as Quest. 17,
done with denominator 60.
Quest.
21, pp. 151-152. Divide into
parts: 1/3 + 1/4; 1/4 + 1/5;
1/5 + 1/6.
Quest.
22, pp. 153-154. Divide 600
into: 2/3 plus 9; 3/5 plus 8; 5/6 plus 7;
7/8 plus 6. See: AR;
Widman; Pacioli for discussion
of this type of problem. Takes common
denominator of 120 and then divides as 89 : 80 : 107 : 111 which is not the way I read the
problem. I would divide 570
as
80 : 72 : 100 : 105, as done by Pacioli.
Apianus. Kauffmanss
Rechnung. 1527.
F.
H.v.r. Divide 1300 as 1/2 plus 8,
1/3 less 5, 1/4 less 12. Does in Pacioli's manner, dividing 1309
in the proportion 12 : 8 :
6 and then amending by +8, -5, -12.
F.
H.v.r. Divide 58 as 1/2 + 2/3 + 1/4. Cf Chuquet.
F.
H.vii.r. Divide 40
as 1/4 + 1/5 + 1/6.
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 91, second part, f. HH.i.v
(p. 166). First has ⅓
plus 7; second has
¼ plus 13; third has ½ minus 28. How much was there?
Recorde. Second
Part. 1552. Pages from 1668 ed.
Pp.
284-289: A question of building; An impossible question; The former question of building now
possible. Divide 3000 into: 1/2 plus 6;
1/3 plus 12;
2/3 less 8; 1/4 plus
20. He first says it is impossible,
then rephrases it and solves by Pacioli's method.
P.
293: Another question of a testament.
Divide 7851 into ½ + ⅓ + ¼. Usual solution.
P.
294: Another like question. Divide 450
into 1/2 + 1/3, 1/3 + 1/4,
1/4 + 1/5. Usual solution.
Tartaglia. General
Trattato. 1556. Book 12, art. 42-44, pp. 200r-201r.
Art.
42. ½ + ⅓ + ¼. Long discussion of an error of Luca Pacioli
and others who assert that such problems are impossible or illegal. Tartaglia says simply to divide in the
proportion 6 : 4 : 3 and can't understand why others are making
such a fuss.
Art.
43. 1/3 + 1/4 + 1/5.
Art.
44. 4/5 + 3/4 + 2/3.
(Sanford
218‑219 says Tartaglia was among the first to suggest the 18th camel --
but I see nothing of this here. H&S
87 says that this is really a modern problem in that previously property was
divided in proportion to fractions, regardless of whether they summed to
unity. Tartaglia's discussion of
Pacioli and others makes it clear that people were starting to object to this
at this time, but examples continue and I don't see the modern version
occurring until late 19C.)
Buteo.
Logistica. 1559.
Prob.
5, pp. 203-204. Divide 77 into ½ + ⅓ + ¼.
Prob.
74, pp. 283-284. Divide 30 into ½ + ⅓. Discusses the solution.
Prob.
75, pp. 284-285. Divide 15 as 1/2 plus 2;
1/4 + 3. He divides 10 into ½ + ⅓, as in Pacioli. See: AR;
Widman; Pacioli for discussion of this type of problem.
Prob.
76, pp. 285-286. Divide 24 as 1/3 less 7;
1/4 less 4.
Prob.
77, p. 285. Divide 12 as 2/3 less 3;
1/6 plus 4.
Prob.
22, pp. 350-351. Divide 60 as 1/4;
1/3 plus 4; 3/4 less 8.
Prob.
26, pp. 353-354. Divide 30 as 1/2 plus 2;
1/3 less 3.
Prob.
28, p. 355. Divide 224 as 1;
6/5 plus 4.
Izaak Walton. The
Compleat Angler. (R. Marriott, London,
1653); Everyman edition, Dent, London,
1906, et seq. Chap. V -- The Fourth
Day, pp. 101‑102. The World's
Classics, OUP, 1935, Chap. V, pp. 114-116.
Divide 20 into
1/3 + 1/4 + 1/5 + 1/6.
Leaves one left over.
W. Leybourn.
Pleasure with Profit. 1694. Prob. 11, pp. 37-38. £6000 divided 1/4 + 1/5 + 1/6. He divides in the proportion
15 : 12 : 10. Cf Apianus.
Ozanam. 1725. Prob. 24, question 5, 1725: 179. Divide
26000 into ½ + ⅓ + ¼. Takes in proportion 12 : 8 : 6.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XIV, pp. 85-86 (1790: prob. XXIV, pp. 86-87). Divide 20 into 1/3 +
1/4 + 1/5 + 1/6, done by proportion.
Les Amusemens.
1749. Prob. 52, p. 184. Divide 78 into ½ + ⅓ + ¼,
by using proportions.
Walkingame. Tutor's
Assistant. 1751. 1777: p. 177, prob. 119; 1835: p. 180, prob. 58; not in 1860. Divide £500 into
1/3 + 1/4 + 1/5 + 1/6 + 1/7. Gives exact answer as integer plus a
fraction. 1835 reduces the fractions to
lowest terms.
Vyse. Tutor's
Guide. 1771? 1793: p. 105; 1799: Prob. 6, p. 113 & key p. 151. Same as Walkingame. Solution gives answers rounded to farthings
and never gives the exact fractions.
Dodson. Math.
Repository. 1775. P. 32, quest. LXXVIII. Divide 20s in proportion: 1/3, 1/4, 1/5,
1/6.
Pike.
Arithmetic. 1788.
P.
335, no. 4. "Being a little
dipped, they agreed that A should pay
2/3, B 1/2, C 1/3, and
D 1/4." = D. Adams, 1835. Cf D. Adams, 1801.
P.
355, no. 39. A, B, C do a job. A and B do
3/11 of it, A and C do
5/13, B and C do 4/14.
(Also entered at 7.H.)
Henri Decremps.
Codicile de Jérôme Sharp, ....
Op. cit. in 4.A.1. 1788. Avant-propos, pp. 18-19 mentions ½ + ⅓ + ¼, but there is no solution.
Samuel Bullen. A New
Compendium of Arithmetic ... Printed
for the author, London, 1789. Chap. 38,
prob. 3, p. 239. Divide into ½ + ⅓ + ¼, phrased as
2A = 3B = 4C.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795
writer, Twickenham, 1995. The 1795 has
dividing 100 into
1/3 + 1/4 + 1/5 + 1/6. Gives usual solution.
D. Adams. Scholar's
Arithmetic. 1801. P. 206, no. 35. Four men divide a purse of
$12 as 1/3 + 1/4 + 1/6 + 1/8.
Divides in the proportion: 8 : 6
: 4 : 3. Cf his 1835 book.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions, no. 16, pp.
18 & 75. Division in the proportion
of 1/3, 1/4, and 1/5, but last person dies. Solution indicates this as standard
practice.
D. Adams. New
Arithmetic. 1835. P. 249, no. 132. = Pike, no. 4. Divides in
the proportion: 8 : 6 : 4 : 3. which is the same proportion as in his 1801
version.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Arithmetical puzzles, no. 4, p. 173 (1868: 184). Pay
20s with only 19s
by dividing into 1/2 + 1/3 + 1/6
+ 1/19. "This, however, is only a
payment upon paper."
Hanky Panky.
1872. A Chinese puzzle, pp.
73-74. 17 elephants to be divided
1/2 + 1/3 + 1/9.
Cassell's.
1881. P. 102: The clever lawyer. = Manson, 1911, p. 255. 17
horses divided
1/2 + 1/3 + 1/9. =
Rohrbough; Puzzle Craft; 1932, p. 7.
Richard A. Proctor.
Some puzzles. Knowledge 9 (Aug
1886) 305-306. "... the familiar
puzzle [of] the farmer, ignorant of numbers, who left 17 horses to his three
sons (or, equally well it may be, an Arab sheik who left 17
camels)". Points out that
if there were 35 camels, then the Cadi could also be left a
camel.
E. W. Cole. Cole's
Fun Doctor. The Funniest Book in the
World. Routledge, London &
E. W. Cole, Melbourne, nd
[HPL gives 1886 and lists the author as
Arthur C. Cole]. P. 224: A
Chinese puzzle. 17 elephants left by a
Chinaman to be divided
1/2 + 1/3 + 1/9.
Says it is in the Galaxy for August, which might have been an
Australian publication by Proctor, who had connections there.
Lemon. 1890. The legacy, no. 652, pp. 81 & 121. 19
camels divided 1/2 + 1/4 +
1/5.
Don Lemon.
Everybody's Pocket Cyclopedia.
Revised 8th ed., 1890. Op. cit.
in 5.A. P. 135, no. 2. 17
horses: 1/2 + 1/3 + 1/9. Dervish loans them his horse.
Hoffmann. 1893. Chap. IV, no. 11: An unmanageable legacy,
pp. 147 & 191‑192 = Hoffmann-Hordern, p. 119. 1/2 + 1/3 + 1/9. Answer says "this expedient is
frequently employed" in "the Mahomedan Law of Inheritance".
Mittenzwey.
1895?. Prob. 164, pp. 34 &
82; 1917: 164, pp. 31-32 & 80. 17 camels,
1/2 + 1/3 + 1/9,
dervish loans them his camel.
Brandreth Puzzle Book.
Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 3: An unmanageable legacy, with nice
colour picture. Identical to
Hoffmann. No solution.
Loyd. Problem 37: A
queer legacy. Tit‑Bits 32 (5
Jun & 3 Jul 1897) 173
& 258. = Cyclopedia, 1914, The herd of camels,
pp. 57 & 346. 17 horses in
proportion 1/2 : 1/3 :
1/9. Says the use of proportion makes
the solution actually correct.
Clark. Mental
Nuts. 1897, no. 25; 1904, no. 5; 1916, no. 18. The heirs
and the sheep. 1897 has 1/2 + 1/4 + 1/6 of 15 sheep. This seems
to have been a miscopying of the question with 11 sheep. He says to borrow a sheep and
distribute 8, 4, 3, returning one. 1904 & 1916 amend this to
1/2 + 1/4 + 1/5 of 19 sheep.
H. D. Northrop.
Popular Pastimes. 1901. No. 18: The clever lawyer, pp. 69 &
74. = Cassell's.
Benson. 1904. The lawyer's puzzle, p. 225. 1/2 + 1/3 + 1/9. There originally were 18 horses, but one died.
William F. White.
Op. cit. in 5.E. 1908. Puzzle of the camels, p. 193. 17 camels divided 1/2 + 1/3 + 1/9.
Ball-FitzPatrick.
2nd ed., 1908‑1909. Part
1, p. 111, footnote says the problem is Arabic. The material is not in the 1st ed., nor in Ball, 5th ed. A&N, pp. 84‑85, cites this but
says it has been in German oral tradition for a long time. He gives it with 17 horses.
E. Ernst.
Mathematische Unterhaltungen und Spielereien. Vol. 2, Otto Maier, Ravensburg, 1912. P. 15: Das geschente Weinfass.
Divide 19 in
1/2 + 1/4 + 1/5.
Dudeney. MP. 1926.
Prob. 89: The seventeen horses, pp. 33-34 & 123-124. = 536, prob. 172, pp. 54‑55 &
266‑267. Discusses interpretation
of proportion, as in Loyd, in detail.
King. Best 100. 1927.
No. 21, pp. 14 & 43. 19
horses into 1/2 + 1/4 +
1/5.
Collins. Book of
Puzzles. 1927.
The
lady bookmaker's problem, pp. 72-73.
Because 1/2 + 1/3 +
1/4 = 13/12, one can offer odds in a
three horse race of: even money, 2 to 1
and 3 to 1.
The
sheik and his camels, pp. 77-78. Usual
form. Cadi loans them his camel.
M. Kraitchik. La
Mathématique des Jeux, 1930, op. cit. in 4.A.2, chap. 1, prob. 47, p.
15. 17 sheep into 1/2 + 1/3 + 1/9. Says it is of Hindu origin.
Anon. Foulsham's New
Fun Book. W. Foulsham, London, nd
[1930s?]. Pp. 85‑86: The farmer's
horses. Identical to King, 1927.
The Bile Beans Puzzle Book.
1933. No. 26: A farmer's
will. 19 horses divided
1/2 + 1/4 +
1/5.
McKay. Party
Night. 1940. No. 35, p. 184. "It
is said that an Arab had 17 cattle."
Sullivan.
Unusual. 1943. Prob. 12: Will trouble. 17 horses into 1/2 + 1/3 + 1/9.
Jerome S. Meyer. Fun
for the Family. (Greenberg Publishers,
1937); Permabooks, NY, 1959. No. 30: Think cow it is done, pp. 42-43
& 241. Herd to be divided 1/3 + 1/4 + 1/6 + 1/8
+ 1/9. Neighbour loans two cows and
everything divides up properly with two cows left over for the neighbour. How many cows were there?
Haldeman-Julius.
1937. No. 131: Cow problem, pp.
15 & 27. Same as Meyer, asking what
is wrong with the problem and answering that "The problem is coo-coo
because all the fractions do not add up to unity."
Leeming. 1946. Chap. 5, prob. 24: The herd of cattle, pp.
61 & 179‑180. Same as Meyer.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
No. 66: The vicar's garden, pp. 25 & 52. 7s divided 1/2 + 1/4 +
1/8 by adding an extra shilling. Solution doesn't seem to understand this and
claims there really should be 7/8
s left over.
Doubleday - 1.
1969. Prob. 25: Milk shake, pp.
36 & 159. = Doubleday - 5, pp.
35-36. 17 cows divided 1/2 + 1/3 + 1/9. He states the usual solution and then asks what is wrong with
it. His solution notes that the fractions
add to 17/18 and then says 'So, in making his will, the farmer hadn't
distributed his entire herd.' This
seems confused to me as the entire herd has been distributed to the sons.
Jonathan Always.
Puzzles for Puzzlers. Tandem,
London, 1971. Prob. 35: An odd bequest,
pp. 22 & 71. Divide 13 camels
as 1/2 + 1/3 + 1/4. Executor borrows one camel, so there are now 12
camels, then he gives the first son
6 = 12/2 and the second son 4 = 12/3, leaving 2.
But the third son ought to have
3 = 12/4, so the executor
returns the borrowed camel and gives the third son his three! The answer says this is an amusing way of
arriving at the intended division in the proportion 6 : 4 : 3. See
Singmaster, 2000, for an extension.
D. B. Eperson.
Puzzles, pastimes and problems.
MiS 3:6 (Nov 1974) 12‑13 & 26‑27. Prob. 6: The Shah's Rolls‑Royces. Divide 23 Rolls‑Royces into 2/3 + 1/6 + 1/8. The answer erroneously asserts this works for n º ‑1 (mod 24).
David Singmaster. A
Middle Eastern muddle. 41 oil wells to be divided into 1/2 + 1/3 + 1/7. But then I ask if there are values other than 2, 3, 7, 41
which produce such a puzzle problem.
There are 12 such quadruples. I recall seeing this when I was a student but I haven't relocated
it. Appeared in my puzzle columns as
follows.
"Well,
well, well." Brain Twister. Weekend Telegraph (27 Feb 1988) xv
(misprinted), (5 Mar 1988) (corrected)
xv & (12 Mar 1988) xv.
Reprinted,
with no title, in: The Daily Telegraph
Braintwisters No. 1; Pan Books, London,
1993; with Barry R. Clarke, Rex Gooch
and Angela Newing. Prob. 25, pp. 27, 76
& 117.
"A
Middle Eastern muddle." The Puzzle
Box. Games & Puzzles 12 (Mar 1995)
18-19 & 13 (Apr 1995) 40.
John P. Ashley.
Arithmetickle. Arithmetic Curiosities,
Challenges, Games and Groaners for all Ages.
Keystone Agencies, Radnor, Ohio, 1997.
P. 8: Omar divides 17 horses among 3 sons. The Answer says: "It was a great solution, but it was not
correct mathematics. The sum of the
fractional parts: 1/2, 1/3 and 1/9 do not add up to 1 but to 17/18. Therefore, each of the heirs got a bit more
than the will intended."
David Singmaster.
The seventeen camels and the thirteen camels. Draft paper written in Dec 2000.
This discusses the classic 17 camels problem and the 13 camels problem
given by Always (1971) finding all solutions for two, three or four sons.
A
man dies, leaving a pregnant wife and a will explaining how his estate is to be
divided between the wife and a son or a daughter. The wife produces twins, one boy and one girl. How is the estate to be divided?
The
most common version has son : wife =
wife : daughter = 2 : 1
given in the will and derives
son : wife : daughter = 4 : 2 : 1.
I will denote this as the "usual form". Other proportions cited by Smith are: 4 : 3 : 2;
2 : 2 : 1; 9 : 6
: 4.
Alcuin &
BR proceed by dividing the
estate in half first.
See Brooks for some odd versions.
See
Tropfke 655.
Moritz Cantor.
Vorlesungen über Geschichte der Mathematik. Vol. 1, 3rd ed., Teubner, Leipzig, 1907. Pp. 561-563 sketches the history of this
problem. This is the basis of Smith's
discussion below. This asserts that the
problem is based on Roman Lex Falcidia of
‑40 which required that at
least ¼ of an estate should go to the legal heir. He says it first appears in the works of
Celsus and quotes Julianus. He also
cites Caecilius Africanus (c100) and Julius Paulus (3C). Describes the common case and says that if
the will was invalidated, then only the children would inherit.
D. E. Smith. Op.
cit. in 3. Based on Cantor, cites Lex
Falcidia, Celsus, Julianus, Africanus.
He cites 22 medieval references, including Vander Schuere and Recorde of
those below.
See also: Sanford
218‑219; H&S 87‑88; A&N 24‑26.
Juventius Celsus. De
istituzione uxoris et postumi et postumae.
c75. ??NYS -- cited by
Smith. Julianus cites him.
Salvianus Julianus.
c140. Lex 13 principio. Digestorum lib. 28, title 2. ??NYS -- quoted by Cantor and cited by Buteo
& Smith. Cantor says this reports a
case, though the quoted text isn't very specific. Usual form. Julianus
cites Celsus.
Caecilius Africanus.
c150. Lex 47 §1. Digestorum lib 28, title 5. ??NYS -- cited by Cantor & Smith. Cantor says this refers to a case.
Julius Paulus.
3C. Lex 81 principio. Digestorum lib. 28, title 5. ??NYS -- cited by Cantor & Smith. Cantor says this refers to a case.
Alcuin. 9C. Prob. 35: Propositio de obitu cujusdam
patrisfamilias. Problem of posthumous
twins. Ratios are 3 : 1
for son : mother and
5 : 7 for mother : daughter. He takes half the estate and shares it 3 : 1 and then the other
half is shared 5 : 7. This gives
9 : 8 : 7. Ahrens,
A&N, p. 26, suggests 15 : 5 :
7, which is the result of the usual
Roman process.
BR. c1305. No. 91, pp. 110‑111. Son : wife
= wife : daughter = 3
: 2. Divides in halves, as in Alcuin,
and divides each half as 3 : 2, giving
son : wife : daughter = 3 : 5 : 2.
Gherardi?. Liber
habaci. c1310. P. 145.
Usual form.
Gherardi. Libro di
ragioni. 1328. P. 37.
Son : wife = 3 : 1;
wife : daughter = 2 : 1.
Divides as 6 : 2 : 1.
Lucca 1754.
c1330.
F.
60r, pp. 136‑137. Posthumous
triplets, 2 boys and a girl with usual ratios.
He divides in proportion 4 : 4 :
2 : 1 for boy : boy : mother: girl.
F.
83r, pp. 200‑201. Posthumous
twins. Usual form.
Pseudo-dell'Abbaco.
c1440. Prob. 100, p. 85 with
plate on p. 86. Posthumous twins. Usual form.
I have a colour slide of this.
AR. c1450. Prob. 209, pp. 97, 176, 223. Man has son, wife and two daughters and
gives the usual ratios, hence divides in the proportion 4 : 2 : 1 : 1.
Muscarello.
1478. Ff. 75r-76r, pp.
189-191. Posthumous twins. Usual form.
Wagner. Op. cit. in
7.G.1. 1483. Pp. 73‑75 & 202‑203. Usual will, but wife produces a son and two daughters. Divides as in AR.
Chuquet. 1484. Prob. 205.
English & discussion in FHM 205.
Usual form. FHM say it
"goes back to the Roman emperor and legislator Justinian" and quotes
Recorde.
HB.XI.22. 1488. P. 44 (Rath 247). Posthumous twins.
Pacioli. Summa. 1494.
F.
158r, prob. 80. Usual posthumous
twins. Then says that Nofrio Dini of
Florence, a respectable merchant in Pisa, at the shop of Giuliano Salviati,
told him about such a will on 16 Dec 1486.
After a bequest to the church, there was an estate of 800 to be
divided. If a son was born, the mother
was to get 400; if a daughter, the
mother was to get 300. Twins were born
and he says to divide as 3 : 3 :
5. Says one can deal similarly with
similar problems. Of the Biographical
Sources listed in Section 1, Taylor, p. 149 & Fennell, p. 11, mention this
problem.
F.
158v, prob. 82. Selling a pregnant cow
which bears twins. Gives some rules
which determine the relative values.
Blasius. 1513. F. F.ii.v: Quarta regula. Dying man with pregnant wife. If she has a son, he gets 3/5
and the mother and the church get
1/5 each. If she has a daughter, the daughter and the
mother get 2/5 each and the church gets 1/5.
She has a son and a daughter. He
divides in proportion 3 : 2 : 2 :
1, but gives no reason. Offhand, I would think that 1/5
should go to the church -- since this is specified in either case -- and
then the remaining 4/5 should be divided in the proportion 3 : 1 : 1, giving overall proportions of 12 : 4 : 4 : 5. He says one can similarly deal with two sons or two daughters or
two sons and one daughter.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 100, ff. 50v-51r. Usual
form.
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 23, f. 65v. Usual posthumous twins.
Riese.
Rechnung. 1522. 1544 ed. -- p. 80; 1574 ed. -- p. 54v.
Father leaves a widow, a son and two daughters. Divides as in AR.
Tonstall. De Arte
Supputandi. 1522. Quest. 16, pp. 146-147. Usual form.
Then considers 3 sons and 2 daughters!
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 87, ff. GG.vi.v - GG.vii.r
(p. 164). Posthumous twins. Son : wife =
4 : 1;
wife : daughter = 2 : 1; divides as
8 : 2 : 1
Giovanni Sfortunati.
Nuovo lume. Venice, 1545. F. 58v.
??NYS -- described by Franci, op. cit. in 3.A, p. 38. Franci's discussion is about several
extended versions, but it seems to indicate that Sfortunati deals with a
hermaphrodite.
Recorde. Second
Part. 1552. Smith, op. cit. in 3.A, p. 69, quotes 1558, fol. X 8
(??NYS). 1668, pp. 289-293: A
question of a Testament.
Man
with fortune of 3600 and a pregnant wife makes a will and dies. If she has a son, the son get ½ and she gets
⅓; if she has a daughter, she gets ½ and the daughter gets ⅓. "If some cunning lawyers had this
matter in scanning, they would determine this testament to be quite voyde, and
so the man to die intestate, because the testament was made unsufficient." (The 1668 has identical wording, except it
uses 'void' and 'insufficient'.) He
divides in the proportion 9 : 6 : 4.
Tartaglia. General
Trattato. 1556. Book 12, art. 35-41, pp. 199v-200v.
Art.
35-40 give various ratios S : M =
son : mother and M : D = mother :
daughter, then divides in the usual way to get the proportions s : m : d
such that s/m = S/M and
m/d = M/D. E.g. when S : M = 2 : 1 = M : D, then
s : m : d = 4 : 2 : 1.
Art.
35. S : M = 2 : 1; M : D = 2 : 1.
Art.
36. S : M = 5 : 3; M : D = 1 : 1.
Art.
37. S : M = 2 : 1; M : D = 1 : 1.
Art.
38. S : M = 2 : 1; M : D = 2 : 1, in a different context than Art. 35.
Art.
39. S : M = 2 : 1; M : D = 3 : 1.
Art.
40. S : M = 3 : 1; M : D = 2 : 1.
Art.
41. S : M = 2 : 1; M : D = 2 : 1, but quadruplets are produced -- two sons and two daughters. He divides in proportion 4 : 4 : 2 : 1 : 1.
Buteo.
Logistica. 1559.
Prob.
60, pp. 264-266. Usual form. Cites Julianus.
Prob.
12, pp. 341-342. Selling a pregnant
cow, where the value depends on the sex of the calf. Cow + daughter is worth
40, while cow + son is worth 45. This is insufficient to determine the
relative values, but he then adds excessive information: C = 3D = 2S. The cow produces twins -- one son and one daughter.
Gori. Libro di
arimetricha. 1571.
Ff.
75r‑75v (pp. 83-84). Usual form.
F. 75v
(p. 84). Posthumous quintuplets --
divides in same proportions, though there is some confusion in the text of the
solution.
Jacob Vander Schuere.
Arithmetica, oft Reken‑const.
G. Kooman, Haarlem, 1600.
??NYS. [Smith, Rara, 421‑423.] F. 98 is quoted in Smith, op. cit. in 3, p.
69, note 7. Posthumous triplets: boy, girl and hermaphrodite. Divides in proportion 12 : 4 : 2 : 7 = son : wife : daughter :
hermaphrodite. Smith doesn't give the
original ratios, but they were probably
son : wife = 3 : 1, wife :
daughter = 2 : 1.
Schott. 1674.
Ænigma
VII, pp. 559-560. Usual form.
Ænigma
X, p. 560. Son : wife = wife : daughter
= 2 : 1, but he interprets this as the
son getting 2/3 of the estate, the wife getting 2/3
of the rest with residue going to the daughter, leading to son : wife : daughter = 6 : 2 : 1.
W. Leybourn.
Pleasure with Profit. 1694. Prob. 15, pp. 39-40. Posthumous triplets: boy, boy, girl. Usual ratios. Divides 4 : 4 : 2 : 1.
Ozanam. 1725. Prob. 24, 1725: 179. Prob. 4, 1778: 187-188; 1803: 185-185; 1814: 160‑161;
1840: 83. 1725 gives just
posthumous triplets -- two girls and a boy.
He divides 4 : 2 : 1 : 1. Montucla does usual form, then remarks that
one could have posthumous triplets, e.g. two sons and a daughter, and that he
thinks that the will would be declared legally void.
Les Amusemens.
1749. Prob. 54, p. 186. Usual case.
Vyse. Tutor's
Guide. 1771? Prob. 19, 1793: pp. 156-157; 1799: p. 167 & Key p. 209. Usual case.
Dodson. Math.
Repository. 1775. P. 13, Quest. XXXIV. Usual case.
Charles Hutton. A
Complete Treatise on Practical Arithmetic and Book-keeping, .... New edition, corrected and enlarged by
Alexander Ingram. [c1780?] G. & J. Ross, Edinburgh, 1804. [BMC earliest entry is 7th ed., 1785, then
14th ed., 1815.] Prob. 53, p. 137. Usual form, but states that the mother
thereby loses 2400£ compared to the case of just having a
girl. What would she have got if she
had only had a son? Answer is 2100£
which assumes the usual 4 : 2 :
1 division for the case of twins.
Vinot. 1860. Art. XLI: Testament à interpréter, pp.
61-62. First gives usual solution. The says the problem is not serious because
French legislation gives a solution.
Since the wife receives at least a third in either case mentioned by the
husband, she must receive a third in any case.
The author then suggests the rest be divided equally among the children
if more than one is born.
Edward Brooks. The Normal
Mental Arithmetic A Thorough and
Complete Course by Analysis and Induction.
Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough
and Complete Course by Analysis and Induction; Sower, Potts & Co., Philadelphia, 1873. Several examples in an unusual context.
1863
-- p. 128, no. 15; 1873 -- pp. 167-168,
no. 5. Man left $26,000
to wife, son and daughter. If
the daughter dies before coming of age, the widow gets ¼;
if the son dies before coming of age, the widow gets ¾;
what happens if all live?
Unusually for this book, this problem has a remark which says the
division should be in the proportion
son : wife : daughter = 9 : 3 :
1.
1863
-- p. 128, no. 18; 1873 -- p. 168, no.
9. Man with children abroad and wife at
home. If the son does not return, the
widow gets ⅔; if the daughter does not return, the widow
gets ⅓; both return and it is found that the son gets $3000
more than the daughter. What was
the estate?
1863
-- p. 128, no. 19; 1873 -- p. 168, no.
10. A, B, C are thinking of buying a farm.
They agree that if A and B buy it, then A pays 2/5
and if B and C buy it, then B pays 2/5.
All three buy it together and
C is found to pay $500
more than A. What was the cost?
Susan Cunnington.
The Story of Arithmetic. Swan
Sonnenschein,, London, 1904. Prob. 11,
p. 212. Usual form. Asserts it is a Roman problem of +300, but
gives no references.
Collins. Fun with
Figures. 1928. Then he put in his other foot, pp.
236-237. Usual form. He adds:
A further complication -- triplets, two boys and a girl. "The easiest way to find out is to let
the lawyers decide it, and it is the one best bet that they will get it
all."
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. Pp. 53-54: The judge's dilemma. Ratios are
2 : 1 for son : mother and
3 : 1 for mother :
daughter. Divides as 6 : 3 : 1.
7.H. DIVISION AND SHARING PROBLEMS -- CISTERN PROBLEMS
See
Tropfke 578.
The
earliest sources in this group include what I call 'assembly problems'. In these, there are several processes which
constitute a unit of work. The rates
for the processes are given and one has to determine the number of units which
can be done in a day (or how long some number of units will take). See the Babylonian examples below and: Chiu Chang Suan Ching; Heron;
Metrodorus 134, 136; Bakhshali
MS; BR 97, 98; AR 57, 75;
Muscarello; Borghi; Riese;
Cardan; Tartaglia; Pike;
Treatise, 1850; Chambers; Bullen;
Pearson. I am indebted to
Eleanor Robson for the Old Babylonian examples. She has also provided most of the references to the source
material which I not yet seen. The
dating of these examples is generally pretty vague.
Note. Cistern problems with two pipes have the
same form as meeting problems, cf. 10.A.
NOTATION: (a, b, c, ...) means that different pipes, etc. can do the job in a, b, c, .... How long for all together?
Negative
values indicate outlets.
a b ...
30 20 Lucca
1754; Wingate/Kersey; Wells;
20 16 Tartaglia
20 15 Benedetto
da Firenze
18 12 Calandri,
c1485
16 10 De
Morgan, 1831
15 12 Calandri,
1491; Tagliente; De Morgan, 1836;
14 12 Hutton-Rutherford
13 10 Vyse; Bonnycastle; Colenso
12 7 Pike
10 8 Pseudo-dell'Abbaco
10 7 Muscarello
10 6 BR
99
10 -15 D.
Adams, 1835
9 -12 Les
Amusemens
8 6 Calandri,
c1485; Hutton; Mittenzwey
7 5 De
Morgan, 1836; Sonnenschein & Nesbit
6 4 Benedetto da
Firenze; Calandri, 1491; Tagliente;
Gori
5 4 Pseudo-dell'Abbaco
4 3 Lucca
1754; D. Adams, 1835; Burnaby
4 1 Heron; BR 98
4 -11 Calandri,
1491; Tonstall
14/4 5/2 Lacroix
3 2 Gherardi?
3 -9 BR
70
2 -3 Buteo
60 30 20 Meichsner
55 45 -30 Pike
50 40 -25 Hutton, c1780?; Eadon;
Colenso
27 15 12 D. Adams, 1801
24 12 8 Buteo
20 15 12 Unger 520
20 15 10 Calandri, c1485
18 12 6 Calandri, c1485
16 12 10 Simpson
16 12 8 Calandri, c1485
15 12 10 Benedetto da
Firenze; Calandri, 1491 (twice);
Dictator
Roffensis;
12 10 9 Sonnenschein &
Nesbit
12 10 8 Mittenzwey
12 10 6 Calandri, c1485
12 9 6 Borghi; Ozanam
12 8 4 Metrodorus 131
12 8 -10 Unger 521
10 9 8 Milne
10 8 4 Pacioli
10 5 4 Pacioli; Tartaglia
9 7 -2 Pike
9 6 -4 Mittenzwey
8 6 4 Calandri, 1491; Tonstall
8 6 3 Pacioli
7 5 6 King
7 5 4 AR 70
7 5 3 Gherardi
6 5 4 Fibonacci
6 4 3 Chuquet
6 4 2 Metrodorus 135; Ozanam-Montucla
6 4 -4 Sonnenschein &
Nesbit
6 3 1 Faulhaber
5 4 3 Lucca 1754; Gori;
Les Amusemens
5 3 2 Calandri, c1485
4 3 2 Gherardi?; AR 98;
Wagner; Faulhaber
4 2 1 Gori
3 8/3 12/5 Newton; Dodson;
Eadon; Colenso
3 2 1 Metrodorus 133; Anania(?);
al-Karkhi; BR 64; AR 51, 97;
Calandri,
1491; Blasius; Tonstall;
Riese; Vyse; King
3 1 2/5 Metrodorus 132
5/3 1/2 1/3 Chaturveda
1 3/4 1/2 Wingate/Kersey
1 1/2 1/4 AR 281; Tonstall
1/2 1/3 1/4 Columbia Alg.; Pike
1/2 -10/7 -7/3 Wingate/Kersey
80 40 20 10 D.
Adams, 1801
72 60 20 12 Levi
ben Gershon
27 24 9 6 Fibonacci
6 8 9 12 Recorde
6 5 4 3 Bartoli
6 5 3 2 Muscarello
6 4 3 2 W.
Leybourn
4 3 2 1 Metrodorus
130; Fibonacci; Tonstall
4 3 2 1/2 Metrodorus
7; BR 65; van Etten;
Wingate/Kersey;
4 3 2 1/4 Schott; Ozanam
1 1/2 1/3 1/6 Bhaskara
II
1 1/2 1/4 1/5 Chaturveda
1/2 1/3 1/4 1/5 Mahavira
1/2 1/4 1/5 1/6 Sridhara
5 3 5/2 1 1/3 Chiu Chang Suan Ching
4 3 2 -4 -6 BR 116
3 2 -3 -4 -5 della Francesca
1/2 1/3 1/4 1/5 1/6 Columbia Alg.
1/2 1/3 1/5 1/7 1/9 BR 25
6 5 4 3 2 1 Bartoli
4 3 2 -3 -4 -5 della Francesca
3 2 1 -3/4 -4 -5 Cardan
3 2 1 -2 -3 -4 Pacioli
12 10 8 6 -3 -4 -5 -6
Bullen; Treatise, 1850
General
solution -- see: Levi ben Gershon; Wells;
Newton; Simpson; Dodson;
Bonnycastle; Hutton; Lacroix;
De Morgan; Bourdon; Young;
Mittenzwey; Milne.
The
earliest forms derive joint rates from individual rates. Deriving individual rates from joint rates
seems to begin in the 14C.
NOTATION: (A, x) in B
means the first can do it in
A and the first and second
together can do it in B. How long would it take the second? For such problems, see: BR;
Gherardi; Pseudo-dell'Abbaco; AR;
Treviso Arith.; Chuquet; Calandri, 1491; Tonstall;
Gemma Frisius;
Tartaglia; Buteo; Wingate/Kersey; Wells; Simpson; Euler;
Vyse; Dodson; Ozanam‑Montucla; Bonnycastle; Pike; Bullen; Eadon;
Hutton, 1798?;
Bonnycastle, 1815;
Jackson;
Nuts to Crack; D.
Adams, 1835; Family Friend; Treatise, 1850; Colenso;
Docharty; Thomson; Brooks.
(50,
x) in
36 Gherardi
(48,
x) in
24 Docharty
(36,
x) in
30 Gherardi
(36,
x) in
24 Docharty
(30,
x) in
12 Dodson; Bonnycastle; Hutton, 1798?; Nuts to
Crack
(20,
x) in
60 Silvester
(20,
x) in
14 Gemma Frisius
(20,
x) in
12 Wingate/Kersey; Wells;
Euler; Dodson; Pike;
Bonnycastle, 1815;
Mittenzwey
(20,
x) in
8 Treviso Arith.
(18,
x) in
11 Vyse
(35/2,
x) in 40 Docharty
(gives a negative x!)
(16,
x) in
10 Treatise, 1850
(15,
x) in
18 Thomson (gives a negative
x!)
(15,
x) in
10 Treatise, 1850
(13,
x) in
9 AR 76
(13,
x) in
8 Pike
(12,
x) in
3 Family Friend, 1849
(10,
x) in
7 Colenso
(
9, x) in 5 Pseudo-dell'Abbaco
(
8, x) in 5 Buteo; Eadon
(
7, x) in 5 D. Adams, 1835
(
5, x) in
15/8 BR 67
(
3, x) in
4/3 BR 66
(
3, -x) in 9/2 BR
69 (negative value!)
(-9,
x) in
9/2 BR 68
(80,
60, x) in 30 Tartaglia
(44,
32, x) in 16 Eadon
(40,
30, x) in 15 Calandri,
1491; Tonstall; Wingate/Kersey
(37,
23, x) in 15 Pike
(34,
24, x) in 12 Vyse
(17/2,
21/4, x) in 6/5 Treatise,
1850
(8, 6, x) in 3 Brooks
(5/2,
9/4, x) in 1 Treatise,
1850
For the general solution of: (x, y) in A, (y, z) in
B, (x, z) in C, see:
della Francesca; Simpson; Euler;
Ozanam-Montucla; Bonnycastle; Hutton;
De Morgan, 1836; Colenso; Singmaster.
For examples of this form, see also:
Muscarello; Dodson; D. Adams, 1835; Docharty; Todhunter; Sonnenschein & Nesbit. This is a form of the type III problem in
Section 7.R.1, where the inverses of the variables are used. Singmaster asks how to choose A, B, C
so that x, y, z and the time for all three together are all
integers -- the case with data 20, 15,
12 is by far the simplest example and
none of the other examples have this property.
A B C
60 4 -40 Colenso
30 20 15 AR 182
20 15 12 Docharty; Todhunter;
Singmaster
15 12 10 della Francesca
14 12 21/2 Colenso
10 9 8 Simpson; Euler;
Dodson; Ozanam-Montucla; Bonnycastle;
Hutton;
Docharty; Vinot;
9 8 6 Sonnenschein &
Nesbit
5 4 3 Muscarello
4 6 5 D. Adams, 1835
Vyse,
Docharty and Thomson are the only examples I have seen with four people and you
know how long it takes each set of three.
Fish has five workers and you know how long each four take. If you use the reciprocals of the times,
then these are like type III problems in 7.R.1. That is, if A, B, C,
D take
A, B, C, D days, their rate of
work is a = 1/A per day, etc. Then saying that A, B, C can do it in d4 days becomes a + b + c =
1/d4, etc.
For
problems where the combinations involve one tap or worker working only part of
the time that the other does, see:
Fibonacci; Gherardi; Chuquet;
Cardan; Buteo; Pike;
Jackson; Treatise, 1850; Colenso;
Young; Chambers; Brooks;
André; Sonnenschein & Nesbit.
For
problems like (x, x/2, x/3)
in 2, see: di Bartolo; Buteo; Todhunter.
For
problems like (x, x‑5) in
12, which lead to quadratic equations,
see: Di Bartolo; Buteo;
Tate; Todhunter; Briggs & Bryan.
Sonnenschein
& Nesbit has a version where pumps can work at half or full power.
I
have included a few direct rate problems as comparisons -- these usually
involve money -- see: Bakhshali; Chaturveda;
Pike; Chambers.
See
Clairaut for the use of this context to discuss negative solutions.
See
Smith, op. cit. in 3. See also 7.E &
H&S 69‑71.
5.W.1
can be viewed as parodies of this problem.
COMPARISON of assembly and cistern
problems. Consider the cistern-type
problem (a1, a2 ,
...). In the unit of time, the pipes
do 1/a1, 1/a2,
... of the work, so all together they
do S = Σ 1/ai per unit time and so the whole job takes
time 1/S.
In
an assembly-type problem, we can do ai units of process i per unit of time. Hence it takes 1/ai time to
do one unit of process i. If each process has to be done the same
number of times, then it takes S =
Σ 1/ai time to do a
unit of work and so 1/S units can be done in a unit of time. In the Babylonian problems, the unit of work
may require varying amounts of the different units. If the unit of work requires
bi units of process i,
then we take S = Σ bi/ai.
Hence
the problems are mathematically the same, though the formulations are
different.
YBC 7164. Old
Babylonian problem tablet at Yale, problems 6 & 7, c‑1700? Transcribed, translated and commented on in
Neugebauer & Sachs, op. cit. in 7.E, 1945, pp. 81-88 & plate 10 &
photo plate 35. On pp. 148-149, a
linguistic analysis says it probably comes from Larsa, in southern Mesopotamia.
Problem
7. A canal has to be cleaned to 3 kùš
deep. A man can clear 20 gín of silt from
the top kùš in a day or he can clear 10 gín from the lower level in a day. How much can he clear in a day? Here
a1 = 20, a2
= 10, and we can take b1 = 1, b2 = 2,
because the lower level is twice as thick as the upper level.
Problem
6. This is the same, but with
depth 4½ kùš divided into three levels with the rate of doing the bottom
level from 3 to 4½ deep being only 7½ gín per day. So we just add a3 = 7½
and b3 = 1½ to the previous problem.
BM 85196. Late Old
Babylonian tablet in the British Museum, prob. 16, c‑1700?. Transcribed, translated and commented on by
O. Neugebauer; Mathematische Keilschrift-texte II; Springer, Berlin, 1935, pp.
45+, 49, 56+ -- ??NX. [See 6.BF.2 for
another problem from this tablet.] But
Neugebauer was not able to make sense of it until he saw the above problems, so
it is reconsidered in Neugebauer & Sachs, pp. 88‑90. Robson says it is definitely from Sippar
(middle Mesopotamia) and cites Thureau-Dangin; Revue d'Assyriologie 32 (1935)
1+ for another publication of the text, ??NYS.
This problem and those of YBC 7164 are more recently discussed by Marvin
A. Powell; Evidence for agriculture and waterworks in Babylonian mathematical
texts; Bulletin on Sumerian Agriculture 4 (1988) 161-172, ??NYS
Like
problem 6 above, with each level of depth 1 kùš and rates of 20, 10,
6⅔ gín per day.
In Spring 1994, I mentioned the assembly problems from the
Chiu Chang Suan Ching (see below) in a lecture at Oxford. Eleanor Robson told me that such problems
occur in Old Babylonian times and she sent me details, including the above
references, and later provided more details and references. She described four further examples, without
specific dates, and the next four examples are simplified from her letter. The simplifications are basically to avoid
use of coefficients giving the number of bricks per unit of weight, etc.
Haddad 104, c-1770.
Tablet from Tell Haddad, near Baghdad, found in the destruction layer
from when Hammurabi conquered the site -- usually dated at -1762. The tablet is in Baghdad. See:
Farouk al-Rawi & Michael Roaf; Ten Old Babylonian mathematical
problems from Tell Haddad; Sumer 43 (1984) 175-218.
Prob.
ix -- Making bricks. One man can
dig 1/3 sar of earth in a day, or he can mix 1/6 sar or he can mould
1/3 sar into bricks. If 1 sar makes 1620 bricks, how many
bricks can a team of three make in a day?
For one man, we get
S = 3 + 6 + 3, so he can process
1/12 sar per day, or 135
bricks, so three men can make
405 bricks.
Prob.
x -- Carrying earth to make bricks.
Same problem as the previous, but the earth must be carried 5 nindan
from the digging site to the works. The
amount one man can carry in a day is given somewhat cryptically. The simplest interpretation is that one man
can carry 1/3 sar of earth over the 5 nindan in a day, but there still are
three workers in the group. Here we get S = 3 + 3 + 6 + 3, so one man can process
1/15 sar per day or 108
bricks and three men make
324 bricks.
YBC 4669. This and
the following tablet are in the same hand, but have no provenance. See Neugebauer, vol. III, pp. 28-29 &
plate 3, ??NYS. Reverse, col. 3, lines
7-17, c-1800 -- Demolishing walls.
A man can knock down 1/15 sar of wall in 1/5 of a day and he can
carry away 1/12 sar in a day. How much wall can he demolish and carry away in a day; and what part of the day is devoted to each
task? Here a1 = (1/15)/(1/5) = 1/3, so S = 3 + 12 and he can do 1/15 sar per day.
YBC 4673. See
Neugebauer, vol. III, pp. 30 & 32 & plate 3, ??NYS. Obverse, col. 2, lines 10-18, c-1800 --
Constructing a pile of bricks. A man
can carry 1/18 sar of earth (bricks??) in a day. He can pile up 1 sar of bricks in 14 2/5
days. If a sar makes 5184
bricks of this size, how many bricks can he carry and pile up in a day?
Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c‑150? Chap. VI.
Prob.
22, p. 67. Man can do two processes at
rates of 38 in 3 days and
76 in 2 days. How many of both together can he do in one
day? Answer is given as 25½
but Vogel's note on the calculation shows 25⅓ was meant, and
this is erroneous -- the correct answer is
9½. The error arises from
taking 38 and 76 as rates per day.
Prob.
23, p. 67. Three processes at
rates 50, 30, 15 per day, how many together in a day? Correct answer, 8⅓, is
obtained. (Arrow shafts, arrow feathering,
arrow heading.)
Prob.
25, p. 68. Three processes at
rates 7, 3, 5 per day, how many together in a day? Correct answer,
105/71, is obtained.
Prob.
26, pp. 68‑69. Cistern: (1/3, 1, 5/2, 3, 5). Correct answer. Vogel says this is the first appearance of the problem.
Heron (attrib.).
c150. Περι
Μετρov (Peri Metron).
In: J. L. Heiberg, ed.; Heronis
Alexandrini Opera Quae Supersunt Omnia, vol. V; Teubner, Leipzig, 1914;
reprinted 1976, pp. 176‑177.
Greek and German texts.
Problem
20: Μετρησισ
χιστερvασ (Metresis xisternas)
[Vermessung einer Zisterne]. (1,4) in hours, but he computes 1 + 4 = 5
and then sets the cistern = 12
ft and computes 12/5 as the number of hours. See BR, c1305, prob. 98, for the explanation
of this.
Problem
21: Αλλωσ η
μετρεσισ (Allos e metresis) [Die
Vermessung in anderer Weise]. + 1/7 ‑
1/11, how long to make 100?
He treats the ‑ as
+ and gets the correct answer
for that case, though Heiberg says the calculation is senseless.
Smith, History II 538, quotes from Bachet's Diophantos,
implying a date of c275, citing the 1570 edition with Fermat's notes, but
Smith's citation is to the part of Bachet taken from Metrodorus! It is Art. 130 of Metrodorus.
Sanford
216 also cites Diophantos, but her discussion is based on Smith's AMM article
(op. cit. in 3), which is the basis of the section in Smith's History
containing Smith's quote. The problem
is nowhere in Heath's edition of Diophantos.
However,
Tropfke 578 gives a reference to the Tannery edition of Diophantos, vol. 2, p.
46 -- ??NYS.
Metrodorus.
c510. 8 cistern‑type
problems.
Art.
7, pp. 30‑31. "I am a brazen
lion." (2, 3, 4, 1/2), where 6 hours is counted as 1/2
day, i.e. a day has 12 hours.
Art.
130, pp. 96‑97. (1, 2, 3, 4).
Art.
131, pp. 96‑97. (4, 8, 12).
Art.
132, pp. 96‑97. "This is
Polyphemus, the brazen cyclops."
(3, 1, 2/5).
Art.
133, pp. 96‑99. (1, 3, 2).
Art.
134, pp. 98‑99. Three spinners can
do 1, 4/3, 1/2 unit per day, how long for all three to do
one unit?
Art.
135, pp. 98‑99. "We three
Loves" (or Cupids). (2, 4, 6).
Art.
136, pp. 98‑101.
'Brickmakers.' Three brickmakers
can make 300, 200, 250 per day.
How long for all three to make 300?
Bakhshali MS.
c7C. Kaye I 49‑52
discusses several types, e.g. first gives 5/2 dinars in 3/2 days; next gives 7/2 in 4/3; third gives 9/2 in 5/4; how long for all three to give 500 dinars?
(= Kaye III 192, ff. 21v-22r). Kaye III
191 has three rates of
1/(1/3), 1/(1/2), 3/5
-- how long to give 100? I 51 (= III 233-234, ff. 44v-44r) is an
example with an income, some capital and three rates of expenditure. On I 50 (= III 234-235, ff. 44r-43v) is
an example with an income, some capital and seven rates of expenditure!
Anania Schirakatzi (= Ananias of Shirak). Arithmetical problems. c640.
Translated by: P. Sahak
Kokian as: Des Anania von Schirak
arithmetische Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919)
112-117. See 7.E for description. Prob. 24 is a cistern with pipes: (1, 2, 3),
but he gives the answer: 1/4 +
1/6 + 1/16 + 1/18, which = 77/144, which is close to the correct answer of 6/11. No working is shown
and I am unable to see how 77/144 can arise, even allowing for a possible
misprint.
H&S 70 says the cistern problem appears in Alcuin, 9C,
but the only possible problem is a trivial problem (8: Propositio de cupa)
which mentions a barrel.
Mahavira. 850. Chap. VIII, v. 32, p. 266. Cistern:
(1/2, 1/3, 1/4, 1/5).
Chaturveda.
860. Commentary on Brahma‑sphuta‑siddhanta,
chap. XII, sect. 1, art. 9.
In Colebrooke, p. 282.
Cistern: (1, 1/2, 1/4, 1/5). (Datta & Singh, I, 234 and others cite
this as Brahmagupta.)
Bestowing
alms: (1/3, 1/2, 5/3).
Sridhara. c900. V. 69, ex. 91, pp. 55‑56 &
95. Cistern: (1/2, 1/4, 1/5, 1/6).
al‑Karkhi.
c1010. Sect II, no. 15‑16,
p. 83. Cistern: (1, 2, 3).
No. 16 asks how often the cistern will be filled in 5 days.
Bhaskara II.
Lilavati. 1150. Chap. IV, sect. II, v. 95. In Colebrooke, p. 42. Cistern: (1, 1/2, 1/3,
1/6). (Datta & Singh, I, 234,
erroneously say this is the same problem as Brahmagupta, i.e. Chaturveda.)
Fibonacci. 1202.
P. 182
(S: 279-280): De Leone et leopardo et
urso [On the lion and leopard and the bear].
Lion, leopard and bear eating a sheep:
(4, 5, 6).
P. 182
(S: 280): De duabus navibus ... [On two
ships ...] is in 10.A.
P. 183
(S: 281): cistern problems (1, 2, 3, 4) & (6, 9, 24, 27).
Pp.
183‑186 (S: 282-285) -- several problems with water butts having
different size openings at different heights.
E.g., pp. 183‑184 has four openings at 1/4, 2/4, 3/4, 4/4
of the way down, which could drain the whole butt in 4, 8, 12, 16 days.
How long to drain a full butt with all holes open? Answer: 7 267/2275 days.
BR. c1305.
No.
25, pp. 44‑45. Ship with 5
sails: (1/2, 1/3, 1/5, 1/7, 1/9). Vogel says this is the first example of the
formulation of a ship with sails.
No.
64, pp. 88‑89. Cistern, (1, 2, 3).
No.
65, pp. 88‑91. Cistern -- 'I am a
noble lion', (2, 3, 4, 1/2). = Metrodorus 7.
No.
66, pp. 90‑91. Cistern, (3, x)
in 4/3.
No.
67, pp. 90‑93. Cistern, (5, x)
in 15/8.
No.
68, pp. 92‑93. Cistern, (x, ‑9) in
9/2.
No.
69, pp. 92‑93. Cistern, (3, ‑x) in
9/2.
No. 70,
pp. 94‑95. Cistern, (3, ‑9).
No.
96, pp. 114‑117. 3 cisterns of
volumes 30, 60, 120 with pipes that fill them in 6, 4, 3. Using all three pipes, how long to fill all three cisterns?
No.
97, pp. 116‑117. Cistern, + 1/7 ‑ 1/11 to yield 100. This is Heron's prob. 21 and is done in the
same way -- as though it were + 1/7 +
1/11, -- but one MS is worded so this
is the correct method, as noted by Vogel.
No.
98, pp. 118‑119. Cistern, (1, 4).
This is Heron's prob. 20, again noted by Vogel. The text says to set the cistern equal
to 12
and then divides by 5 = 1 +
4. Vogel notes that this does not give
the time, but 12/5 is the volume delivered by the smaller pipe.
No.
99, pp. 118‑119. Cistern, (6, 10).
No.
116, pp. 132‑133. Cistern, (2, 3, 4, ‑6, ‑4).
Gherardi?. Liber
habaci. c1310 Pp. 143‑144: Compangnia et viaggio. Baratti xvii. Three workers, (2, 3,
4). Ship with two sails, (2, 3).
Levi ben Gershon.
Maaseh Hoshev (= Ma‘aseh hosheb (the
h should have an underdot))
(Work of the Computer), also known as Sefer ha mispar (Book of Number). 1321 or 1322. ??NYS -- translation of the following by Shai Simonson, sent by
David E. Kullman.
"Question:
A certain container has various holes in it, and one of the holes lets all the contents
drain out in a given times. And so on
for each of the holes. How much time
will it take to empty the container when all the holes are opened?
First,
calculate how much drains from each hole in an hour, add them all up, and note
the ratio to the full container. This
ratio is equal to the ratio of one hour to the time it will take t empty the
container when all holes are open."
He
then does (72, 60, 20, 12).
Gherardi. Libro di
ragioni. 1328.
Pp. 44‑45: Volare una bocte. (3,5), but the 5 is
half‑way down the barrel while the
3 is at the bottom.
P.
45: Ship with three sails, (3, 5, 7).
Pp. 56‑57: Uno chavaleri che vuole far fare uno
pallagio. Three workers, (50, B) in 36, (50, B, C) in 30.
Lucca 1754.
c1330. F. 59r, p. 134.
Ship
with 3 or 2 sails: (3, 4, 5), (3, 4).
Two
couriers meeting: (20, 30).
Columbia Algorism.
c1350.
Prob.
66, pp. 87‑88. Cask can be
emptied in 1/2, 1/3, 1/4, 1/5, 1/6 of a day.
(See also Cowley 399.)
Prob.
141, p. 150. Same with 1/2, 1/3, 1/4.
Giovanni di Bartolo.
Certi Chasi. c1400. Copied by Maestro Benedetto (da Firenze), in
Cod. L.IV.21, Biblioteca degli Intronati di Siena, 1463. Edited by M. Pancanti, Quaderni del Centro
Studi della Matematica Medioevale, No. 3, Univ. di Siena, 1982. Cf Van Egmond's Catalog 189-190 which
doesn't mention this material.
Prob.
1, pp. 4‑5. (x, x/2) in
5.
Prob.
2‑8, pp. 5‑17 are more complex examples, often leading to quadratic
equations, e.g. (x, x‑5) in
12.
Bartoli.
Memoriale. c1420.
Prob.
14, f. 76v (= Sesiano, pp. 140-142 & 149, with facsimile of the relevant
part of f. 76v on p. 141. Cask
with four taps: (3, 4, 5, 6).
Prob.
33, f. 79r (= Sesiano, pp. 146 & 150).
Six workers building a wall:
(1, 2, 3, 4, 5, 6). He correctly finds the total rate is 137 [/ 60], but then
uses 36 instead of 60 -- "Now make 6 times 6,
which is 36, because there are 6 workers." Sesiano describes this as fantasy.
Pseudo-dell'Abbaco.
c1440.
Prob.
23, p. 32. (4, 5).
Prob.
50, p. 49 with plate on p. 50. Ship
with two sails, (8, 10). I have a colour slide of this.
Prob.
62, p. 59. (9, x) in
5.
AR. c1450. Prob. 51, 57, 70, 75, 76, 97, 98, 182,
281. Pp. 42, 44‑45, 48‑50,
58‑59, 85‑86, 128‑129, 157, 160‑161, 165‑166,
175, 211‑213, 221.
51: cistern with three drains, (1, 2, 3),
erroneously done -- see 97.
57: three mills, but gives amounts each can do
per day.
70: three builders, (7, 5, 4).
75: three tailors, but gives amounts each can do
per day.
76: (13, B)
in 9.
97: cistern with three drains, (1, 2, 3),
= Metrodorus 133.
98: ship with three sails, (2, 3, 4).
182: three scribes; (A, B) in 20, (A, C) in
30, (B, C) in 15, how long for each?
281: barrel with three taps, (1, ½, ¼).
Benedetto da Firenze.
c1465.
Pp. 90‑91: ship with three sails (10, 12, 15).
P.
91: cistern (4, 6).
P.
91: two workers (20, 5).
"The Treviso Arithmetic" = Larte de labbacho
(there is no actual title). Treviso,
1478. Translated by David Eugene Smith,
with historical commentary by Frank J. Swetz, as: Capitalism and Arithmetic; Open Court, La Salle, Illinois, (1987),
improved ed., 1989. This is discussed
in: D. E. Smith; The first printed
arithmetic (Treviso, 1478); Isis 6 (1924) 310‑331. Facsimile edition, from the copy at the
Diocese of Treviso, with commentary booklet by Giuliano Romano, (Editore
Zoppelli, sponsored by Cassa di Risparmio della Marca Trevigian, Treviso,
1969); updated ed., Libreria Canova,
CalMaggiore 31, Treviso (tel: 0422-546253), 1995 [Swetz, p. 324, cites the
1969 ed.] See: www.calion.com/cultu/abbacho/abbacen.htm for a description of the book and how to
order it. The facsimile has taken its
title from the end of the opening sentence.
Romano's commentary calls it:
L'Arte dell'Abbacho. The text
nowhere gives a publisher's name.
Smith, Rara, pp. 3-7, says it was probably published by Manzolo or
Manzolino, while Swetz, p. 26, specifies Michael Manzolo or Manazolus. Romano says it was published by Gerardus de
Lisa. There was a copy in the Honeyman
Collection, with title Arte
dell'Abbaco and publisher [Gerardus de
Lisa], who is described as the prototypographer at Treviso from 1471. The entry says only ten copies are known --
the web page says nine.
F.
57r (pp. 162‑163 in Swetz).
Carpenters, (20, x) in
8.
Muscarello.
1478.
F.
58r, p. 161-162. (B, C) in
3, (A, C) in
4, (A, B) in
5. There are two copying errors
in the MS answers.
F.
62v, pp. 169-170. Four workers, (2, 3, 5, 6).
Ff.
77r-77v, pp. 192-193. Three mills can
do 9, 8, 5 per day. How long will it
take them to do 6 and how much does each do?
F.
77v, p. 193. Ship with two sails, (7, 10).
F.
81r, p. 196. Cask with three spouts
which can let out 6, 7, 8 per hour.
How long will it take to empty a cask of 23?
della Francesca.
Trattato. c1480.
Ff.
15r-15v (61-62). Basin with three
inlets and three outlets, (2, 3, 4, -3,
-4, -5). Plug the +4
pipe, which gives (2, 3, -3, -4,
-5). English in Jayawardene.
F.
127r (269). Three workers: A & B in 15; A & C in 12; B &
C in 10. English in Jayawardene.
Wagner. Op. cit. in
7.G.1. 1483. Regel von einem Fass, pp. 114 & 224. Cask with three taps (2, 3, 4).
Chuquet. 1484.
Prob.
21. English in FHM 204. Cistern emptying, (3, 4, 6).
Prob.
53. English in FHM 209-210. The first says: "If you help me
8 days, I will build it in 20".
The second responds: "If
you help me 10 days, I can do it in 15".
How long for each alone?
Prob.
54. Same as prob. 53 with
parameters 5, 17; 6, 24.
Borghi.
Arithmetica. 1484.
F.
106v (1509: ff. 91r-91v). Three mills
can grind 6, 9, 11 per day.
How long to do 100?
F.
109r (1509: ff. 91v-92r). Ship with
three sails, (6, 9, 12). (H&S 70 gives Latin and English.)
Calandri.
Aritmetica. c1485.
Johann Widman. Op.
cit. in 7.G.1. 1489. (On pp. 131-132, Glaisher mentions the
following.) Ff. 136r‑138v: Eyn fasz mit dreyen Czapfen; Von der Mulen; Leb, wolff, hunt; Schiff. (Cistern problem; 3 mills; lion, wolf, dog
eating a sheep; ship with 3 sails.)
Calandri.
Arimethrica. 1491.
Pacioli. Summa. 1494.
See also Buteo.
Blasius. 1513. F. F.iii.r: Decimatertia regula. Three rivers can water a field in (1, 2, 3)
days. Gets 13 1/11
hours for all three -- so he is using
24 hour days.
Tagliente. Libro de
Abaco. (1515). 1541.
Tonstall. De Arte
Supputandi. 1522.
Riese. Die
Coss. 1524.
Cardan. Practica
Arithmetice. 1539.
Gemma Frisius.
Arithmetica. 1540. (20, x)
in 14 -- man & wife drinking a cask of wine. ??NYS -- Latin given in H&S, p. 71.
Recorde. Second
Part. 1552. 1668, pp. 329-330: A question of water, the eighth example. (6, 8, 9, 12).
Tartaglia. General
Trattato, 1556, art. 74, p. 248v; art.
176‑177, p. 261v; art. 187‑188,
pp. 262r‑262v.
Buteo.
Logistica. 1559.
Gori. Libro di
arimetricha. 1571.
Johann Faulhaber.
Arithmetischer Wegweiser ....
Ulm, 1614. ??NYS. A 1708 ed. is quoted in Hugo Grosse;
Historische Rechenbücher des 16. und 17. Jahrhunderts; (1901); reprinted by Sändig, Wiesbaden, 1965, p.
120.
van Etten.
1624. Prob. 83 (76): Du Lyon de
Bronze posé sur une fontaine avec cette epigraphe, pp. 94‑95 (140). (2, 3, 4, 1/2) = Metrodorus, art. 7.
Georg Meichsner.
Arithmetica Historica.
Hieronymus Körnlein, Rotenburg/Tauber, 1625. No. 68, p. 209.
??NYS. Quoted in Hugo Grosse,
op. cit. under Faulhaber, above, p. 77.
Three men with devices to pump out flooded lands in Holland, (60, 30, 20).
Schott. 1674. Ex. 1, pp. 570-571. Cistern:
(2, 3, 4, 1/4) done several
ways. Cites Clavius for the lion
fountain (Metrodorus 7).
Wingate/Kersey.
1678?.
W. Leybourn.
Pleasure with Profit. 1694. Prob. 14, p. 39. Cistern emptying: (6, 4, 3, 2).
Wells. 1698.
Isaac Newton.
Arithmetica Universalis, 1707.
??NYS. English version: Universal Arithmetic, translated by Mr.
Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James
Maguire and Theaker Wilder; W. Johnston, London, 1769. (De Morgan, in Rara, 652‑653, says
there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of
1720 and 1728, ??NYS.) Resolution of
Arithmetical Questions, Problem VII, pp. 184‑185. "The Forces of several Agents being
given, to determine x the Time, wherein they will jointly perform
a given Effect d." Gives general approach for three
workers. Example is (3, 8/3, 12/5), where the Force of the second is expressed as saying he can do
the work "thrice in 8 weeks".
Ozanam. 1725.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790).
Alexis-Claude Clairaut.
Elémens d'Algèbre. 1746. Vol. I, art. LVI (my source quotes from the
6th ed. of 1801). ??NX. He uses the context of this type of problem
to study the meaning of negative solutions.
A cistern of size a is filled by a source running for time b
together with another source running for time c. Another reservoir of
size d
is filled by the sources in times
e and f. Determine the rate of
each source.
Les Amusemens.
1749.
"By his Holiness the Pope". The Gentleman and Lady's Palladium (1750)
22. Qn. 11. (??NYS, cited by E. H. Neville; Gleaning 1259: On Gleaning 1146;
MG 23 (No. 254) (May 1939) 149. "If a Cardinal can pray a soul out of
purgatory ...." See Welch, 1833,
below.
Dictator Roffensis, proposer; Steph. Hodges & Will. Smith, solvers. Ladies' Diary, 1750-51 = T.
Leybourn, II: 45-46, quest. 334. [??NX
of p. 46.] Three drinkers: (10, 12, 15) for 12 hour days, how
long together for 10 hour days.
Arthur Young. Rural
Oeconomy: or, Essays on the Practical Parts of Husbandry. Dublin, 1770, p. 32. ??NYS - described in: Keith Thomas; Children
in early modern England; IN: Gillian Avery & Julia Briggs; Children and Their Books A Celebration of the Work of Iona and Peter Opie; OUP, (1989), PB
ed, 1990, pp. 45-77, esp. pp. 66 & 76.
Proverb: one boy, one day's work; two boys, half a day's work; three
boys, no work at all.
Euler. Algebra. 1770.
I.IV.III: Questions for practice.
Vyse. Tutor's
Guide. 1771?
Dodson. Math.
Repository. 1775.
Ozanam-Montucla.
1778.
Charles Hutton. A
Complete Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2. [c1780?] 1804: prob. 70,
pp. 139-140. Cistern: (40, 50, -25).
Bonnycastle.
Algebra. 1782.
Pike.
Arithmetic. 1788.
Bullen. Op. cit. in
7.G.1. 1789. Chap. 38.
Eadon.
Repository. 1794.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795
writer, Twickenham, 1995. P. 100: (1, 2, 3);
(7, 5, 6).
Hutton. A Course of
Mathematics. 1798?
D. Adams. Scholar's
Arithmetic. 1801. P. 125, nos. 26 & 27. (80, 40, 20, 10) &
(27, 15, 12).
Bonnycastle.
Algebra. 10th ed., 1815. P. 226, no. 6. Cistern: (20, x) in
12 hours.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions.
Silvestre François Lacroix.
Élémens d'Algèbre, a l'Usage de l'École Centrale des
Quatre-Nations. 14th ed., Bachelier,
Paris, 1825. Section 14, ex. 1, pp.
25-27. (5/2, 15/4) and
(a, b) in general.
Augustus De Morgan.
Arithmetic and Algebra.
(1831?). Reprinted as the
second, separately paged, part of: Library of Useful Knowledge -- Mathematics
I, Baldwin & Craddock, London, 1836.
Arts. 3 & 112, pp. 1-2 & 28-29.
In Art. 3, he mentions the problem
(10, 16) as an example of algebraic formulation. In Art. 112, he solves it and (a, b).
Welch. Improved American
Arithmetic. 1833 ed. This must be Oliver Welch's American
Arithmetic, first published in 1812 and which went through at least eight
editions up to 1847. [Halwas 459-465,
of which 1833 is 462.] "If a
Cardinal can pray a soul out of purgatory by himself in 1 hour, a bishop in 3
hours, a Priest in 5 hours, a Friar in 7 hours, in what time can they pray out
3 souls, all praying together?" In
the 1842 ed., this was changed to steam, water, wind and horse power. ??NYS -- quoted in Gleaning 1146, MG 21 (No.
245) (Oct 1937), 258. See above at 1750
for an earlier version.
Nuts to Crack II (1833), no. 129. (30, x) in 12.
Identical to Bonnycastle, 1782, no. 25.
Bourdon.
Algèbre. 7th ed., 1834. Art. 57, p. 85. General solution for (a/b, c/d, e/f).
D. Adams. New
Arithmetic. 1835.
Augustus De Morgan.
On The Study and Difficulties of Mathematics. First, separately paged, part of: Library of Useful Knowledge --
Mathematics I, Baldwin & Craddock, London, 1836. P. 31. (12, 15), then does
(a, b).
Augustus De Morgan.
Examples of the Processes of Arithmetic and Algebra. Third, separately paged, part of: Library of
Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836.
Unger. Arithmetische
Unterhaltungen. 1838.
Hutton-Rutherford. A
Course of Mathematics. 1841? Prob. 8,
1857: 81. Two workers: (12, 14).
T. Tate. Algebra
Made Easy. Op. cit. in 6.BF.3. 1848.
Family Friend 1 (1849) Answers to correspondents, pp. 4
& 6. Questions requiring answers.
No. 1. (12, x) in
3.
Anonymous. A
Treatise on Arithmetic in Theory and Practice: for the use of The Irish
National Schools. Third edition,
revised, improved and enlarged.
Published by direction of the Commissioners of National Education in
Ireland, Dublin, 1850. Many examples,
of which the following are the more interesting.
John William Colenso (1814-1883). Arithmetic Designed for the Use of Schools .... New edition. Longman, Brown, Green, and Longmans, 1853. ??NX - Wallis 246 COL. I have 1857, which seems identical for pages
1 - 164, then adds a chapter on decimal coinage on pp. 165-171. I also have 1871, which rearranges the material
at the end and adds Notes and Examination-Papers -- the advertisement on p. v
says the additional material was added by J. Hunter in 1864. This gives a large number of variations of
the problem which I have included here as representative of mid 19C texts.
Miscellaneous
Examples, pp. 122‑136, with answers on pp. 161-163 (1871: 211‑213).
Examination--Paper VIII, (1864), 1871: pp. 170‑172, with answers
on p. 214.
John Radford Young (1799-1885). Simple Arithmetic, Algebra, and the Elements of Euclid. IN: The Mathematical Sciences, Orr's Circle
of the Sciences series, Orr & Co., London, 1854. [An apparently earlier version appeared as Circle of the
Sciences, vol. (2 &) 3, with no TP or other details except two editorial
remarks referring to Professor Young as author of the Arithmetic and Algebra
sections. I have vol. 3, which begins
with the last sheet of Arithmetic and then covers the other material. (I saw a vol. 8? on zoology? in the same
bookshop.) This version includes
detailed solutions and some plates not connected with the text. The 1854 text is identical except that the
order of topics has been changed and there are some consequent changes to the
text.] This is a typical mid 19C text
with a number of cistern problems, of which the more interesting are the
following.
Gerardus Beekman Docharty.
A Practical and Commercial Arithmetic: .... Harper & Brothers, NY, 1854.
Many examples on pp. 166-167, 242-243, 247, including Simpson's
XXXI; the same problem with values 12, 20, 15;
(35/2, x) in 40
and the following.
Vinot. 1860. Art. LVII: Les trois Ouvriers, pp.
74-75. Same as the example in Simpson
XXXI.
James B. Thomson.
Higher Arithmetic; or the Science and Application of Numbers; .... Designed for Advanced Classes in Schools and
Academies. 120th ed., Ivison, Phinney
& Co, New York, (and nine copublishers), 1862. Lots of straightforward examples and the following.
Edward Brooks. The
Normal Mental Arithmetic A Thorough and
Complete Course by Analysis and Induction.
Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough
and Complete Course by Analysis and Induction;
Sower, Potts & Co., Philadelphia, 1873. Lots of examples. I
mention only those of some novelty to illustrate mid/late 19C texts.
Boy's Own Magazine 2:2 (No. 8) (Aug 1863) 183 &
2:4 (No. 11) (Nov 1863) 367.
(Reprinted as (Beeton's) Boy's Own Magazine 3:8 (Aug 1889) 351 &
3:10 (Oct 1889) 431.)
Mathematical question 87.
Complicated version, typical of its time. Bacchus drinks from a cask for
⅔ of the time it would
take Silenus to drink the whole cask.
Silenus then finishes it off and the total time is two hours longer than
if they had drunk together. But if they
had drunk together, Bacchus would only have drunk half as much as he left for
Silenus.
[Robert Chambers].
Arithmetic. Theoretical and
Practical. New Edition. Part of: Chambers's Educational Course --
edited by W. & R. Chambers. William
and Robert Chambers, London and Edinburgh, nd, [1870 written on fep]. [Though there is no author given, Wallis 242
CHA is the same item, attributed to Robert Chambers, with 1866 on the fly-leaf,
so I will date this as 1866? --??check in BMC.]
Stoddard, John F.
The American Intellectual Arithmetic: Containing An Extensive Collection
of Practical Questions on the General Principles of Arithmetic. With Concise and Original Methods of
Solution, Which Simplify Many of the Most Important Rules of Arithmetic. Sheldon & Company, New York &
Chicago, 1866. P. 120, no. 30. "If a wolf can eat a sheep in ⅞
of an hour, and a bear can eat it in
¾ of an hour, how long would it
take them together to eat what remained of a sheep after the wolf had been
eating ½ of an hour?" Thanks
to David E. Kullman for sending this.
Todhunter. Algebra,
5th ed. 1870. Many examples -- the less straightforward are the following.
Miscellaneous Examples, no. 48, pp. 548 & 604. Simpson's XXXI with values 12, 15, 20. Cf Docharty.
Bachet-Labosne.
Problemes. 3rd ed., 1874. Supp. prob. V, 1884: 187-188. Poetic and complicated form. In
3/10 of the time that Silenus
would take to drink the whole amphora, Bacchus drinks 1/4 of what he leaves for
Silenus to finish. But if they drank it
all together, they would finish it in two hours less than the previous time.
Daniel W. Fish, ed.
The Progressive Higher Arithmetic, for Schools, Academies, and
Mercantile Colleges. Forming a Complete
Treatise of Arithmetical Science, and its Commercial and Business
Applications. Ivison, Blakeman, Taylor
& Co., NY, nd [but prefaces give: 1860; Improved Edition, 1875]. P. 419, no. 80. Five persons building a house.
All but A can do in 14 days; all
but B can do in 19; all but C can do in
12; all but D can do in 15; all but E can do in 13. How long for all five and who is the fastest
worker? Answer: 11 4813/12137.
M. Ph. André.
Éléments d'Arithmétique (No 3) a l'usage de toutes les
institutions .... 3rd ed., Librairie
Classique de F.-E. André-Guédon, Paris, 1876.
Several straightforward problems and the following. Prob. 99, p. 62. Four companies of workers can do a job in 45, 9, 27, 36 days. How long will it take 2/5
of the first company, 3/4 of the second company, 1/2 of the third company and 1/3
of the fourth company?
Fred Burnaby. On
Horseback Through Asia Minor. Sampson
Low, et al., London, 1877. Vol. 1, pp.
208‑210. One man can mow in 3
days, the other in 4.
How long together?
Mittenzwey.
1880.
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E.
William Briggs & George Hartley Bryan. The Tutorial Algebra, based on the Algebra
of Radhakrishnan -- Part II -- Advanced Course. W. B. Clive, London, (1898), 1900. Exercises X, prob. 6, pp. 123 & 579. Two reapers, (x, x‑5) in 6.
A. Sonnenschein & H. A. Nesbit. The New Science and Art of Arithmetic For
the Use of Schools. A. & C. Black,
London, 1903. This is a typical text of
its time and has a number of variations on the basic problem.
Pearson. 1907. Part II, no. 162, pp. 145 & 223. A
brings a pint every 3 minutes,
B a quart every 5
minutes and C a gallon every 7 minutes. How long to fill a 55 gallon drum and who
finishes the job?
Stephen Leacock. A,
B, and C. IN: Literary Lapses,
(1910). The book has been frequently
reprinted and the piece has been widely anthologised. It is pp. 237-245 in my 9th English ed.
Collins. Fun with
Figures. 1928. According to Hoyle, not arithmetic, pp.
32-33. "[I]f your father can build
a chicken coop in 7 days and your Uncle George can build it
in 9
days, how long ...."
"They'd never get it done;
they'd sit down and swap stories of rum runners, and bootleggers and
hijackers."
C. Dudley Langford.
Note 1558: A graphical method of
solving problems on "Rate of Work" and similar problems. MG 25 (No. 267) (Dec 1941) 304-307. +
Note 2110: Addition to Note
1558: "Rate of Work"
problems. MG 34 (No. 307) (Feb 1950)
44. Uses a graph to show (a, b)
problems as meeting problems.
Also solves problems
(A, x) in B
and (a, -b), the latter appearing as an overtaking
problem. The Addition gives a clearer
way of viewing (a, ‑b) problems as overtaking problems.
David Singmaster.
How long is a brick wall? (my title is: Three bricklayers). Weekend Telegraph (30 May 1992) xxxii &
(7 Jun 1992) xxx. Al and Bill
can build a wall in 12 days; Al &
Charlie in 15; Bill & Charlie in
20. How long does it take each individually
and how long does it take all three together?
This is well known, but then I ask how can you determine integer data to
make all the results come out integers?
Let A, B, C denote the amounts each can build in a
day. To make all the data and results
come out as integers, we have to have all of
A, B, C, A+B, A+C,
B+C, A+B+C be fractions with unit numerators. To combine them easily, imagine that all
these fractions have been given a common denominator d, so we can
consider A = a/d, B = b/d,
etc., and we want a, b,
c, a+b, a+c,
b+c, a+b+c to all divide d. We can achieve this
easily by taking any three integers
a, b, c, and letting d be
the least common multiple of a, b,
c, a+b, a+c,
b+c, a+b+c. Taking
a, b, c = 3, 2, 1, we find d = 60 and the given problem is by far the simplest example with
distinct rates A, B,
C.
I
feel this is based on my remembering the problem from somewhere, but the only
previous use of this data is in Docharty, but he doesn't consider the
diophantine problem, and none of the other data has this property. I suspect this was an AMM or similar problem
some years ago.
John Silvester recently asked me if I knew the following
version, which he heard from John Reeve.
A man can pack his bag to go to a meeting in 20 minutes. But if his wife helps him, it takes an
hour. How long would it take his wife
on her own? I.e. (20, x) in 60. The answer is -30 minutes!
7.H.1. WITH GROWTH -- NEWTON'S CATTLE PROBLEM
The
example of Ray has led me to re-examine the relation between this topic and the
cistern problems. Ray's problem can be
recast as follows.
There
is a cistern with an input pipe and a number of equal outlet taps. When
a taps are turned on, the
cistern empties in time c; when
d taps are turned on, the
cistern empties in time f; how long
[h] will it take to empty
when x
taps are turned on?
Despite the similarity, we
do not have the times for the individual inlets and outlets to fill or empty
the cistern and so it is much easier to use rates rather than times.
Isaac Newton.
Arithmetica Universalis, 1707.
??NYS. English version: Universal Arithmetic, translated by Mr.
Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James
Maguire and Theaker Wilder; W. Johnston, London, 1769. (De Morgan, in Rara, 652‑653, says
there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of
1720 and 1728, ??NYS.) Resolution of
Arithmetical Questions, Problem 11, pp. 189‑191. (Sanford 165 quotes from the 1728 English
edition and it is the same as the 1769.)
"If the Number of Oxen a eat up the Meadow b in the time c;
and the Number of Oxen d eat up as good a Piece of Pasture e in
the Time f, and the Grass grows uniformly;
to find how many Oxen [x] will eat up the like Pasture g in
the Time h." Gives a general solution: (gbdfh ‑ ecagh ‑ bdcgf + ecfga)/(befh
- bceh) and an example with
a, b, c; d, e, f; g, h = 12, 3⅓, 4;
21, 10, 9; 24, 18. One easily finds
that the rate of grass growth per unit area per unit time is G
= (ace - bdf)/cf(bd - ae) and the rate of grass consumption per ox
per unit time is E =
be(c-f)/cf(bd - ae). There are
actually three unknowns since we also don't know the initial amount of grass
per unit area, G0. We can either take proportions of these or
we can adopt a unit of grass such that the initial amount of grass per unit
area is 1. In the second case, the basic equation b (G0 + Gc) = aEc becomes
b (1 + cG) = acE.
Walkingame. Tutor's
Assistant. 1751. 1777: p. 177, prob. 121; 1860: p. 185, prob. 114. Identical to Newton.
Eadon.
Repository. 1794.
Stoddard, John F.
Stoddard's Practical Arithmetic.
The Practical Arithmetic, Designed for the use of Schools and Academies;
Embracing Every Variety of Practical Questions Appropriate to Written
Arithmetic, with Original, Concise and Analytic Methods of Solution. Sheldon & Company, NY, 1852. Pp. 281-282, no. 23.
a, b, c; d, e, f; g, h = 14,
2, 3; 16, 6, 9; 24, 6.
Thanks to David E. Kullman for sending this.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Arithmetical puzzles, no. 6, p. 174 (1868: 185): Sir Isaac Newton's
problem. Gives the numerical values
from Newton, but 3⅓ is often mis-set as 3½,
and when so done, is identical in all three editions.
Vinot. 1860. Art. LI: Les boeufs de Newton, pp.
67-68.
a, b, c; d, e, f; g, h = 3, 2, 2; 2, 2, 4; 6, 6.
Answer: 5.
A. Schuyler. A
Complete Algebra for Schools and Colleges.
Van Antwerp, Bragg & Co., Cincinnatti & NY. Pp. 99-100, nos. 31-32, gives Newton's
specific problem, then the general version.
Thanks to David E. Kullman for sending this.
M. Ph. André.
Éléments d'Arithmétique (No 3) a l'usage de toutes les
institutions .... 3rd ed., Librairie
Classique de F.-E. André-Guédon, Paris, 1876.
Same as Vinot, but no answer.
Joseph Ray, revised by J. M. Greenwood. Ray's New Higher Arithmetic. American Book Co., Cincinnati, 1880. ??NYS -- quoted and discussed in: David E. Kullman; Story problems with a
flavor of the old Northwest; preprint of 8pp sent by the author in Apr 1999, p.
4. "There is coal now on the dock,
and coal is running on also, from a shoot [sic], at a uniform rate. Six men can clear the dock in one hour, but
11 men can clear it in 20 minutes: how long would it take 4 men? Ans: 5 hr." This is like Newton's problem, but assuming b = e = g,
and solving for h rather than
x. In a letter Kullman notes that this problem is not in Ray's original
edition, called Higher Arithmetic, date not given, but Ray died in 1855.
G. H. Mapleton, proposer;
Charles Hammond, solver.
Arithmetical problem. Knowledge
1 (30 Dec 1881) 191 & (20 Jan 1882) 258, item 9.
a, b, c; d, e, f; g, h = 12, 10, 16; 18,
10, 8; 40, 6. Cites Newton, 1722 ed., p. 90.
Charles Pendlebury.
Arithmetic. Bell, London,
(1886), 30th corrected and expanded printing, 1924. Section XLIV: Pasture with growing grass, pp. 336q-336s &
answers, part II, p. xv. (These
pages were added after the 6th ed. of 1893, by the 10th ed. of 1897, possibly
for the 10th ed?) This section carefully
works an example to answer different questions, then gives 12 problems. In all cases, all the fields have the same
size, which much simplifies things.
I have read an article (in HM??) which described this as a
common problem in 19C textbooks.
Walter Percy Workman.
The Tutorial Arithmetic.
University Tutorial Press, (1902);
2nd ed., 1902. [There is a 3rd
ed., (c1908), c1928, which contains a few more pages.] Section IX [= Chap. XXXI in the 1928
ed.], examples CXLV, prob. 59‑60, pp. 430 & 544 [= 436 & 577 in
the 1928 ed.].
Wehman. New Book of
200 Puzzles. 1908. An ox problem, p. 55. Gives Newton's problem with just the
numerical values.
Loyd. Cyclopedia,
1914, pp. 47 & 345. = MPSL1, prob.
48, pp. 46 & 138‑139. Cow,
goat and goose.
Wood. Oddities. 1927.
Prob. 71: Ox and grass, p. 55.
a, b, c; d, e, f; g, h = 6, 10, 16; 18, 10,
8; 40, 6. Gets 88, but is confused about the growing of the grass
-- Newton's formula gives 104. Says Newton divides the oxen into those that
eat the accumulated grass and those that eat the increase, but he doesn't apply
this correctly. Indeed, for this data,
the grass grows at a negative rate!
Undoubtedly intended to be the data of Mapleton, 1881, for which the
answer is 88.
Perelman. MCBF. 1937.
A. I. Ostrovsky.
Oxen grazing in a field. MG 50
(No. 371) (Feb 1966) 46‑48.
Quotes Newton and gives a graphical solution which converts this into an
overtaking problem, where the grass starts growing 12 weeks before the cows
are put in.
John Bull. Grazing
Oxen. M500 165 (Dec 1998) 1-4. He is unhappy with some of the limiting
situations and proposes a different basic equation. However, his unhappiness is really due to not understanding the
basic equation properly.
NOTATION: (a, b, c)
among n means to divide a full, b
half‑full and c empty casks among n people so that each has
an equal amount of contents and of casks.
Dividing kn
casks containing 1, 2, ...,
kn among k people so each gets the
same amount of contents and of casks
-- see: Albert;
Munich 14684; AR; Günther (1887); Singmaster (1998).
See Tropfke 659.
Alcuin. 9C.
Abbot Albert. c1240. Prob. 3, p. 333. Divide 9 casks containing 1, 2, ..., 9 among
three. He gives only one of the two
solutions: 1, 5, 9; 2, 6, 7;
3, 4, 8. See Singmaster, 1998,
for a generalization.
BR. c1305. No. 40, pp. 58‑61. 300
ewes, 100 each with
1, 2, 3 lambs, to be divided
among three sons so each son has the same number of ewes and lambs and no lamb
is separated from its mother. This is
the same as (100,100,100) among
3. He gives one solution: 0, 100, 0;
50, 0, 50; 50, 0, 50. [There are
234 solutions!]
Munich 14684.
14C. Prob. XXIV, f. 32r. Same as Abbot Albert and with the same
solution.
Folkerts.
Aufgabensammlungen. 13-15C. 10 sources of Abbot Albert's problem. Also cites Albert, AR, Günther.
AR. c1450. Prob. 351, pp. 154, 182. Same as Abbot Albert, with the same
solution, but arranged in columns.
Vogel comments that this makes a 'half‑magic square' and cites
Günther, 1887, as having already noted this.
Tartaglia. General
Trattato, 1556, art. 130‑131, p. 255v.
Bachet.
Problemes. 1612. Addl. prob. IX: Trois hommes ont à partager
21 tonneaux ...., 1612: 161‑164;
1624: 233-236; 1884: 168‑171.
Labosne adds: (5,
11, 8) among 3 -- giving all three
solutions.
(Ahrens,
A&N, 29, says that the 1st ed. also does
(5, 5, 5) among 3.)
van Etten.
1624. Prob. 89 (86), part IV,
pp. 134‑135 (213). (7, 7, 7) among
3. One solution: 3, 1, 3;
3, 1, 3; 1, 5, 1.
Hunt. 1651. Pp. 284-285. Of three men that bought wine.
(7, 7, 7) among 3.
Two answers.
Ozanam. 1725. Prob. 44, 1725: 242‑246. Prob. 24, 1778: 182-184; 1803: 180-182; 1803: 158-159; 1814:
158-159; 1840: 81-82.
Les Amusemens.
1749. Prob. 15, p. 137: Les
Tonneaux.
Bestelmeier.
1801. Item 717: Die sonderbare
Weintheilung unter 3 Erben. Says there
are 21
casks, so presumably (7, 7,
7) among 3.
Jackson. Rational
Amusement. 1821. Arithmetical Puzzles.
Endless Amusement II.
1826? Prob. 20, pp.
199-200. (7, 7, 7) -- two solutions.
Young Man's Book.
1839. Pp. 239-240. Identical to Endless Amusement II.
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. P. 211.
(7, 7, 7) with both solutions.
Magician's Own Book.
1857. The wine and the tables,
p. 225. (7, 7, 7), (8, 8, 8),
(9, 9, 9) among 3 --
gives two solutions for each. = Boy's
Own Conjuring Book, 1860, p. 195.
Vinot. 1860. Art. XL: Un partage curieux, pp. 59-60. (7, 7, 7)
-- two solutions.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 593, pp. 299 & 410: Sechs
Knacknüsse -- part 1. (7, 7, 7) -- one solution shown diagrammatically:
3, 1, 3; 3, 1, 3; 1, 5, 1.
Mittenzwey.
1880. Prob. 104, pp. 21 &
73; 1895?: 121, pp. 26 & 75; 1917: 121, pp. 24 & 73. (7, 7, 7)
among 3. Gives three solutions,
but one is a rearrangement of another.
Solution asks, if pouring is allowed, can all three get the same
inheritance? It says to pour two half
barrels into two other half barrels, obtaining
(9, 3, 9).
Siegmund Günther.
Geschichte des mathematischen Unterrichts im deutschen Mittelalter bis
zum Jahre 1525. Monumenta Germaniae
Paedagogica III. 1887. Facsimile reprint by Sändig Reprint Verlag,
Vaduz, Liechtenstein, 1969. He
discusses Abbot Albert and his problem on pp. 35-36, noting that the solution
can be viewed as a set of lines in a magic square so that the perpendicular
lines give a second solution, but that magic squares were then unknown in
Europe. He gives no other examples.
Lucas.
L'Arithmétique Amusante.
1895. Prob. XV: Jeux de tonneau,
pp. 50-51. (7, 7, 7) among
3. Gives both solutions. Notes that
half empty = half full, so
doubling gives us empty = full!
Ahrens.
A&N. 1918. Pp. 29‑33. Gives all solutions for
(n, n, n) among 3
for
n = 5
(1) 10; (8, 8, 8) among 4
& 6; (11, 5, 8)
among 3;
(5, 11,
8) among 3; (4, 12, 8) among 3.
McKay. Party
Night. 1940. No. 14, p. 178. (7, 7,
7) among three. One solution.
M. Kraitchik.
Mathematical Recreations, 1943, op. cit. in 4.A.2, chap. 2, prob.
34, pp. 31‑32. 9 full,
9 three-quarter full, 9 half, 9 quarter full, 9
empty among 5.
David Singmaster.
Triangles with integer sides and sharing barrels. CMJ 21 (1990) 278‑285. Shows the number of ways of sharing (N, N, N)
among 3 is the same as the number of integer sided
triangles of perimeter N. Shows this is the number of partitions
of N/2
or (N‑3)/2 into
3 parts. Finds necessary and sufficient conditions
for sharing (a, b, c) among
k people.
David Singmaster.
Fair division of the first
kn integers into k
parts. Written in 1998. This generalizes the problem of Abbot Albert
and determines that there is a partition of the first kn integers into k
sets of n values with each set having the same sum if
and only if n is even or n > 1
and k is odd. I call this a
fair division of the first kn integers into k parts. The number of such divisions is large, but
might be worth examining.
7.H.3. SHARING UNEQUAL RESOURCES -- PROBLEM OF THE PANDECTS
NOTATION: (a, b, ...; c) means
persons contribute a, b, ... which is shared equally among themselves
and an extra person who pays c. See Clark for a variant formulation with the
same result.
See McKay; Party Night; 1940 in
7.H.5 for a related form.
INDEX of (a, b, ...; c) problems with a £ b £
....
2 3 3 Kraitchik
2 3 5 Fibonacci; al-Qazwînî; Bartoli; Calandri
2 3 35 McKay
3 4 4 Kraitchik
3 4 7 Kraitchik
3 4 14 Gherardi
3 5 4 Vinot
3 5 5 Benedetto da Firenze
3 5 8 D. Adams, 1801; Jackson;
Badcock; New Sphinx; Magician's Own Book; Bachet‑Labosne; Mr. X;
Pearson; Clark; Kraitchik;
Rohrbough; Sullivan
3 5 10 AR
3 5 80 Kraitchik
4 5 4 Pseudo-dell'Abbaco
4 6 10 Mittenzwey
5 7 12 Kraitchik
7 8 30 Kraitchik
10 14 6 Hummerston
31 50 40 Tagliente
90 120 70 Kraitchik
11 14 17 42 Kraitchik
2 3 6 9 4 Kraitchik
Fibonacci.
1202. P. 283 (S: 403-404). (3, 2; 5).
Qazwini = Zakariyâ
ibn Muhammad ibn Mahmûd [the h should have an underdot] abû Yahya [the h
should have an underdot] al-Qazwînî.
(= al-Kazwînî [the K should have an underdot] = Zakariyyā' b. Muhammad b. Mahmūd
[the h
should have an underdot] Abū Yahya [the h should have an
underdot] al-Kazwīnī [the
K should have an underdot] =
Zakarīyā ibn Muhammad [the
h should have an underdot]
al-Qazwīnī). (Kitâb) ‘Ajâ’ib
al‑Makhlûqât wa Gharâ’ib al-Mawjûdât (= Adjāyib
al-Makhlūkāţ [NOTE:
ţ denotes a t
with an underdot and the second
k should hve an underdot.] wa
Ghārā'ib al-Mawdjūdāţ [NOTE: ţ denotes a t
with an underdot.] = ‘Ajā’ib al‑makhlūqāt
wa-gharā’ib al-mawjūdāt) ((The Book of the) Wonders of the
Creation and Unique [Phenomena] of the Existence = Prodigies of Things Created and Miraculous Aspects of Things
Existing = The Wonders of Creation and
the Peculiarities of Existing Things =
The Cosmography). c1260. ??NYS -- the earliest dated copy, of 1458,
and several others are in the Wellcome Institute; BL has a page from a 14C copy
on display. Part 8: On the arts; chap. 9: On reckoning. In:
J. Ruska; Kazwīnīstudien[the
K should have an underdot]; Der Islam 4 (1913) 14‑66 & 236‑262. German translation (Arabic omitted) of this
problem on pp. 252‑253.
(3, 2; 5). Story says
one proposes a 3 : 2 split, but
4 : 1 is found to be
correct. [Qazwini also wrote a
Geography, in two editions, and its titles are slightly similar to the
above. I previously had reference to
the Arabic titles of the other book, but rereading of Ruska and reference to
the DSB article shows the above is correct.]
Gherardi. Libro di
ragioni. 1328. Pp. 40‑41: Chopagnia. (3, 4; 14).
Bartoli.
Memoriale. c1420. Prob. 26, ff. 77v-78r (= Sesiano, pp. 144
& 149). (2, 3; 5), correctly solved.
Pseudo-dell'Abbaco.
c1440. Prob. 94, p. 81 with
plate on p. 82. (5, 4; 5). I have a colour slide of this.
AR. c1450. Prob. 212, p. 98. (5, 3; 10) correctly
solved. (Unusually, Vogel's notes,
pp. 160‑161 & 211‑213, say nothing about this problem.)
Benedetto da Firenze.
c1465. c1480. P. 106.
Part of the text is lacking, but it must be (3, 5; 5).
Calandri.
Arimethrica. 1491. F. 63v. (2, 3; 5).
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 130, ff. 60v-61r. Philippo
and Jacomo share lunch with Constanzo --
(50, 31; 40).
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 27, ff. 66r-66v. Three men have 3, 2, 1 loaves of bread and
other foods worth 8, 6, 4. A fourth comes and shares with them,
paying 9. How much should each of the three get? The total value of the food is
4 x 9 = 36,
so the bread is worth 36 - 8 - 6
- 4 = 18, or 3 per loaf. So the first
should get 9, the second
3 and the third owes them 2 !!
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 16, ff. CC.iv.v - CC.v.v
(pp. 140-141). Three men with bread,
wine and fish in the amounts:
3, 4, 0; 0, 5, 6; 2, 0, 7; which are considered of equal value. A fourth man with one bread comes and they
share the meal and the fourth man pays
5.
Tartaglia. General
Trattato. 1556. Book 12, art. 33, pp. 199r-199v. Three men have quails and bread. The first has 6 quails and 2s
worth of bread; second has 4
quails and 3s worth of bread; third has 2 quails and
5s worth of bread. They share with a fourth person who
pays 8s.
Buteo.
Logistica. 1559.
D. Adams. Scholar's
Arithmetic. 1801. P. 210, no. 6. (3, 5; 8). Doesn't give
any attempts at division, nor a solution.
Jackson. Rational
Amusement. 1821. Arithmetical Puzzles, no. 12, pp. 3-4 &
53-54. (3, 5; 8). Says it appears in an Arabian
manuscript. The man with five loaves
divides 5, 3; the other protests and divides
4, 4; judge divides 7, 1.
Why? = Magician's Own Book (UK
version), 1871, Arithmetical paradox, pp. 28-29.
John Badcock.
Domestic Amusements, or Philosophical Recreations. Op. cit. in 6.BH. [1823]. Pp. 186-187, no.
255: Arithmetical paradox. Same as
Jackson, saying it appears in an Arabic manuscript.
The New Sphinx.
c1840. (3, 5; 8). Three Greeks. Extra person divides his money
3, 5, but the second was
dissatisfied and had the matter referred to Solon, who gave the right division.
Magician's Own Book.
1857. The three travellers, pp.
225-226. (3, 5; 8). = Boy's Own Conjuring Book, 1860, pp. 195‑196.
Vinot. 1860. Art. LX: Chacun son écot, p. 77. (3, 5; 4).
Edward Brooks. The
Normal Mental Arithmetic A Thorough and
Complete Course by Analysis and Induction.
Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough
and Complete Course by Analysis and Induction;
Sower, Potts & Co., Philadelphia, 1873. Several ordinary examples and some unusual examples.
Bachet-Labosne.
Problemes. 3rd ed., 1874. Supp. prob. I: Deux Arabes allaient dîner:
..., 1884: 181. (3, 5; 8).
Mittenzwey.
1880. Prob. 61, pp. 12-13 &
64; 1895?: 67, pp. 17 & 66; 1917: 67, pp. 16 & 63. (6, 4; 10).
One suggest dividing 6, 4, the other suggests 5, 5. Gives correct
solution.
Hoffmann. 1893. Chap IV, no. 74: The three Arabs, pp. 165
& 220 = Hoffmann-Hordern, p. 147.
(3, 5; 8). First says to
divide 3, 5; second says 4, 4; third says both are wrong.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
10:6 (Oct 1903) 530-531. The stranger's
dinner. (3, 5; 8). Refers to Arabians. Similar to Jackson.
Pearson. 1907. Part II, no. 138, pp. 141 & 218. (3, 5; 8).
Ball‑FitzPatrick.
Footnote cited in 7.G.1.
1908. Says the problem is
Arabic, but gives no reason.
Clark. Mental
Nuts. 1916, no. 22. Real estate. A invests $5000, B invests
$3000. They buy three houses of
equal value. They each take one and then
sell the third for $8000. How do they divide the money? Answer is
$7000 and $1000.
His 1904, no. 25; 1916, no. 39 is a standard version of
(3, 5; 8) with sandwiches.
Hummerston. Fun,
Mirth & Mystery. 1924. A partnership problem, Puzzle no. 3, pp. 19
& 172. (10, 14; 6), then the second uses his receipts to buy
more which are shared equally and his colleagues pay him -- how many more could
he now buy?
Kraitchik. La
Mathématiques des Jeux. Op. cit. in
4.A.2. 1930. Chap. 1, pp. 7‑8: Problème des Pandects. He gives several examples and says they come
from Unterrichtsblätter für Mathematik und Naturwissenschaften 11, pp. 81‑85,
??NYS.
Rohrbough. Brain
Resters and Testers. c1935. The Travelers' Dinner, pp. 23-24. Arabs,
(3, 5,; 8).
McKay. Party
Night. 1940. No. 25, p. 182. A &
B give a party and invite 2
and 3 guests. The party
costs 35s -- how do they divide the expense? Initial reaction is in the ratio
2 : 3, but it should be 3 : 4.
(Also entered in 7.H.5)
Kraitchik.
Mathematical Recreations. Op.
cit. in 4.A.2. 1943. The problem of the Pandects, pp. 28‑29. c= No. 26 of Math. des Jeux.
Sullivan.
Unusual. 1943. Prob. 19: An Arab picnic. (3, 5; 8).
7.H.4. EACH DOUBLES OTHERS' MONEY TO MAKE ALL EQUAL, ETC.
See 7.R for
related problems.
See Tropfke
647‑648.
Diophantos.
Arithmetica. c250. Book I.
Mahavira. 850. Chap. VI, v. 259‑267, pp. 160‑162.
Fibonacci.
1202. Pp. 287‑293 (S:
409-415) gives several versions.
Fibonacci.
Flos. c1225. In Picutti, pp. 326-332, numbers VIII-X.
Lucca 1754.
c1330. Ff. 61v‑62r, p.
142. 4 & 3 men. Each doubles the others to make all
equal. In the 4 man case, he specifies
the total money is 400.
Giovanni di Bartolo.
Op. cit. in 7.H. c1400. Prob. 9, pp. 17‑18. Four men, each doubles the others' money and
the product of the results is 1000. He
assumes, for no clear reason, that the original amounts are proportional
to 8 : 4 : 2 : 1.
AR. c1450. Prob. 231, pp. 107‑108 & 169‑171. Two men, each doubles the others' money and then
both have 13½.
Muscarello.
1478. Ff. 78r-78v, pp.
193-194. Four men, each doubles the
others' money, then all are equal.
Answer: 33, 17, 9, 5.
Chuquet. 1484. Prob. 148.
3 people. Mentioned in passing
on FHM 230.
Calandri.
Aritmetica. c1485. Ff. 101r-102r, pp. 202‑204. 6 men.
Each doubles the others to make all equal.
Pacioli. Summa. 1494.
Tonstall. De Arte
Supputandi. 1522. P. 245.
First wins 1/2 of the second's; second wins 1/3 of the third's; third wins 1/5 of the first's to make all equal 100.
This is a correct version of Pacioli.
He gives (500, 1000,
1200)/9 and shows the calculation which
implies all three calculations are done at once -- that is, the 1/5
of the first's money is based on what he had to start, not what he has
after winning from the second.
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 91, ff. GG.viii.v - HH.i.v
(pp. 165‑166). A situation
somewhat similar to 7.P.7, where the first two take money leaving the third
with 5. Friend says the first is to give
10 and ⅓ of what he has
left to the second and the second is then to give 7 and ¼ of
what he has left to the third to make their amounts proportional to (3, 2, 1).
Answer is x = 172, y = 39
and the total sum is 216.
Tartaglia. General
Trattato, 1556, art. 11, 18, 37, 38, pp. 240v, 241v, 244v‑245r. 2 and 3 people versions.
Buteo.
Logistica. 1559. Prob. 63, pp. 270-273. Three players -- first wins 1/2
of second's, second wins 1/3
of third's, third wins 1/4
of what the first had originally,
and all wind up with 100. That is, we have x + y/2 ‑ x/4 = y ‑ y/2 + z/3 = z ‑ z/3 + x/4 = 100.
Bachet.
Problemes. 1612. Addl. prob. VIII, 1612: 154-160; 1624: 226‑233; 1884: 162‑167. 3 people; also with tripling. Labosne adds the general case.
van Etten.
1624. Prob. 57 (52), pp. 52‑53
(78). 3 people version used as a kind
of divination.
Ozanam. 1694. Prob. 26, 1696: 81; 1708: 72.
Prob. 47, 1725: 253. Prob. 17,
1778: 209; 1803: 204. Prob. 16, 1814: 177; 1840: 91.
Three person version, resulting in each having 8, used as a kind of
divination.
Euler. Algebra. 1770.
I.IV,IV.616: Question 4, pp. 211‑212. Three players, all winding up with 24.
Hutton. A Course of
Mathematics. 1798? Prob. 42,
1833: 223; 1857: 227. Five players, all ending up with £32.
Silvestre François Lacroix.
Élémens d'Algèbre, a l'Usage de l'École Centrale des
Quatre-Nations. 14th ed., Bachelier,
Paris, 1825. Section 82, ex. 9, p.
123. Three players, each doubles
others, all ending with 120.
Thomas Grainger Hall.
The Elements of Algebra: Chiefly Intended for Schools, and the Junior
Classes in Colleges. Second Edition:
Altered and Enlarged. John W. Parker,
London, 1846. Pp. 130-131, ex. 16. Each
3/2's others, with total given
as 162. Cf Mahavira 266.
G. Ainsworth & J. Yeats. A Treatise on the Elements of Algebra. H. Ingram, London, 1854.
Exercise XXXVIII, pp. 81‑83 & 178.
Vinot. 1860. Art. LXXI: Les trois joueurs, pp.
86-87. Three person version, all ending
up with 24.
Boy's [Own] Magazine 2:5 (No. 11) (Nov 1863) 459 [answer
would be in Jan 1864, ??NYS].
(Reprinted as (Beeton's) Boy's Own Magazine 3:11 (Nov 1889) 479. Mathematical question 137. Three boys playing. Each pays the winner half of what he
has. Each one wins once and then they
have 30d, 60d, 120d. How much did they have to start? [In fact, they had the same amounts. In general, if they start with 1, 2, 4,
then the amounts after the first, second, third wins are: 4, 1, 2;
2, 4, 1; 1, 2, 4.]
Todhunter. Algebra,
5th ed. 1870. Examples XIII, no. 26, pp. 104 & 578. 3
men, each doubles the others to make all equal to 16.
Mittenzwey.
1880.
Lucas.
L'Arithmétique Amusante.
1895. Prob. XLIII: Les jouers,
pp. 184-185. Three players winding up
with 12 each. Does general solution
for n
players winding up with a each.
Workman. Op. cit. in
7.H.1. 1902. Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 45,
pp. 428 & 544 (434 & 577 in c1928 ed.). Version with five people.
Hermann Schubert.
Beispiel‑Sammlung zur Arithmetik und Algebra. 3rd ed., Göschen, Berlin, 1913. Section 17, no. 107, pp. 66 & 140. Three people. Each gives away half his money to be equally shared by the
others, and then they all have 8.
Solution: 4, 7, 13.
Loyd.
Cyclopedia. 1914. Sam Loyd's Mystery Puzzle, pp. 226 &
369. (= MPSL2, prob. 85, pp. 61
& 148. = SLAHP: An initiation fee,
pp. 63 & 109.) 3 people
version, resulting in the first person having lost 100.
Collins. Book of
Puzzles. 1927. The five gamblers puzzle, p. 75. They all end up with $32.
7.H.5. SHARING COST OF STAIRS, ETC.
See Tropfke 529. The simplest form is given by Sridhara, BR,
Pseudo-dell'Abbaco, Gori, von Schinnern.
Mahavira. 850. Chap. VI, v. 226‑232, pp. 151‑153.
Sridhara. c900.
BR. c1305. No. 36, pp. 54‑57. Divide oil among 12 lamps which are to
burn 1, 2, ..., 12 hours.
Pseudo-dell'Abbaco.
c1440. Prob. 92, pp. 78‑81. House rented to 1 person the first
month, who shares with a 2nd the
2nd month, who share with a 3rd the
3rd month, ..., who share with a 12th the 12th month. How much does each pay? He says many obtain 12/78, 11/78, ...,
1/78, but that it is more
correct if the first pays
(1 + 1/2 + 1/3 + ... + 1/12) *
1/12, the second pays (1/2 + 1/3 + ... + 1/12)
* 1/12, ..., the 12th pays 1/12 * 1/12.
Gori. Libro di
arimetricha. 1571. F. 73r (p. 80). Same as Prob. 92 of Pseudo-dell'Abbaco, but with only the first solution.
Bullen. Op. cit. in
7.G.1. 1789. Chap. 38, prob. 32, p. 243.
Four men hire a coach to go
130 miles (misprinted as 100).
After 40 miles, two more men join. How much does each pay?
Clemens Rudolph Ritter von Schinnern. Ein Dutzend mathematischer
Betrachtungen. Geistinger, Vienna,
1826, pp. 14‑16. Discusses
general problem of sharing a cost of
n for lighting a x
floor staircase. Does the
case n = 48, x = 4, getting 3, 7, 13, 25, which are the same proportions as Sridhara, ex. 86‑87.
Dana P. Colburn.
Arithmetic and Its Applications.
H, Cowperthwait & Co., Philadelphia, 1856. Miscellaneous Examples No. 66, p. 365. A and B hire a horse and carriage for $7 to go 42 miles and
return. After 12 miles, C joins them,
and after 24 miles, D joins them. How
much should each pay? No solution is
given, but there is a Note after the question saying there are two common ways
to do the allocation. First is in
proportion to the miles travelled, so A
: B : C : D = 42 : 42 : 30 : 18. Second
is to divide the cost of each section among the number of riders, which here
gives
A : B : C : D = 29 : 29 : 17
: 9
Clark. Mental
Nuts. 1897, no. 95; 1904, no. 71; 1916, no. 14. The livery
team. I hire a livery team for $4 to go
to the next city, 12 miles away. At the
crossroads 6 miles away, I pick up a rider to the city who then rides back to
the crossroads. How much should he
pay? Answer is $1.
M. Adams. Puzzle
Book. 1939. Prob. C.135: Answer quickly!, pp. 158 & 187. Man hires car to go to theatre. He picks up and drops off a friend who lives
half way to the theatre. How do they
divide the fare? Answer is 3 : 1.
Depew. Cokesbury
Game Book. 1939. Passenger, p. 219. Similar to M. Adams.
McKay. At Home
Tonight. 1940. Prob. 13: Sharing the cost, pp. 65 &
79. Similar to M. Adams.
McKay. Party
Night. 1940. No. 25, p. 182. A &
B give a party and invite 2
and 3 guests. The party
costs 35s -- how do they divided the expense? Initial reaction is in the ratio
2 : 3, but it should be 3 : 4.
(Also entered in 7.H.3.)
Doubleday - 3.
1972. Fair's fair, pp.
61-62. Man hires a taxi to go to the
city and pays in advance. Halfway
there, he picks up a friend. Later they
go back, with the friend dropped at the halfway point. How should they share the fare? Answer says the friend should pay one third,
because the man 'had only hired the taxi to take him into town ... not for a
round-journey.' This may be introducing
the following extra feature. Normally
the friend would pay 1/4 of the total cost. But when the taxi has got halfway back, the fare shown will only
be 3/4
of the total cost and the friend should pay 1/3 of that amount. The problem does say that the same driver
has been used and there was no charge for the waiting time. But I wonder whether taxi-meters can be
stopped and restarted in this way. It
would be more natural if they caught another taxi back. Then at the halfway point, the fare shown
would be 1/4 of the total cost and the friend show pay all of the amount
shown!
New section. I have seen other examples. A grindstone of radius R is
to be shared between two (or k) buyers -- one grinding until his share is
used. An inner circle of radius r is
unusable. The first man should grind to
radius x where x2 - r2 = (R2
- r2)/k or x2 = [(k-1)R2
+ r2]/k. It is
straightforward to adapt this to the case when the buyers contribute unequal
amounts to the purchase price.
Clark gives a problem of sawing
through a tree which uses the fact that the area of a segment of a circle of
radius R and segment height H is R2cos‑1(R‑H)/R ‑ (R‑H)Ö(2RH‑H2). Though well-known, this seems about on the
border of what I consider to be recreational.
Anonymous proposer; solution lacking. Ladies' Diary, 1709-10 = T.
Leybourn, I: 5-6, quest. 9. [??NX of p.
6.] Share a grindstone among seven
people. 2R = 60".
Carlile.
Collection. 1793. Prob. XXIV, p. 16. Three men buy a grindstone of radius 20 for 20s.
They pay 9s, 6s,
5s respectively. How much should each man get to grind? He makes no allowance for wastage.
Hutton-Rutherford. A
Course of Mathematics. 1841? Prob. 22,
1857: 558. Share a grindstone
among seven people. 2R = 60". He takes
r = 0.
Dana P. Colburn.
Arithmetic and Its Applications.
H, Cowperthwait & Co., Philadelphia, 1856. Miscellaneous Examples No. 43, p. 362. Four men sharing a grindstone 4 ft in diameter. No indication of inner wastage. No solution given.
Clark. Mental
Nuts. 1904, no. 93. Cutting trees. Three men cutting through a tree of diameter
three. One cuts in one from one side,
the next cuts in one from the opposite side.
How much is left? Answer is 41.64%,
which is correct. [The obvious
question is how far should the first two men get so that all three cut an equal
area? I find it should be .7350679153972 of the radius.]
Collins. Fun with
Figures. 1928. A grindstone dispute, p. 22. 2R = 5 ft 6 in; 2r = 18 in.
7.H.7. DIGGING PART OF A WELL.
Recently separated from 7.H.5.
See Tropfke 529.
The 'digging a well' problem has a
contract to dig a well a deep for payment b, but the digging stops
at c.
How much should be paid? The
value of b is not always given and then only the ratio of the part payment
to the total payment is sought.
NOTATION -- this problem is
denoted (a, b; c).
If the
difficulty is proportional to depth, then integration yields that the payment
should be proportional to (a/c)2. A common medieval approach is use the proportion
1 + ... + c : 1 + ... +
a. We let Ta = 1 + ... + a, so this proportion is Tc : Ta.
Benedetto da Firenze; della Francesca; Calandri, 1491, are the
only cases where c > a.
Della Francesca begins the inverse
problem -- if the contract for depth
a is worth b
and work stops at x such that the value of the dug hole is d,
what is x? Denote this situation as (a, b; x) worth
d. See: della Francesca; Pacioli; Calandri, Raccolta; Buteo.
Ozanam, Vyse and Jackson are the
only ones to consider use of other arithmetic progressions.
Berloquin gives a simple argument
that work is proportional to a2/2.
In Neugebauer & Sachs, op. cit. in 7.E, the problems
discussed on pp. 81-91 involve digging out ditches and the cost or difficulty
of digging increases with the depth, but none of these are like the problem
considered here, though Tropfke 529 notes a resemblance.
Tabari. Miftāh
al-mu‘āmalāt. c1075. ??NYS - quoted and discussed by Tropfke
529.
Qazwini = al‑Qazwînî. Loc. cit. in 7.H.3. c1260.
P. 253. German translation only
-- man contracts to dig a well 10 ells deep for 10 dirhems. He stops at
9 ells, so we have (10, 10; 9). Man asks for 9 dirhems, but an expert says only 8
and somewhat more.
BR. c1305. No. 22, pp. 40‑43. Man contracts to dig 10 x 10 x 10 cistern but only does
5 x 5 x 5. Text gives
him 1/8 of the value.
Lucca 1754. c1330. F. 64v, p.
152. Man digging a well, (10, b; 8).
He divides in ratio T8 : T10
= 36 : 55.
Pseudo-dell'Abbaco.
c1440. Prob. 102, p. 87 with
plate on p. 88. Man contracts to dig a
well 20 deep and stops at 14, i.e.
(20, b; 14). Author divides in
ratio T14 : T20 =
1 : 2. But he says he doesn't think
this is a correct method, though he doesn't know a better one. I have a colour slide of this.
Benedetto da Firenze.
c1465. Pp. 115‑116. If a well
12 deep is worth 12,
how much is a well 14 deep
worth? This is (12, 12; 14). He takes values proportional to Td.
Muscarello.
1478. Ff. 66v-67r, pp.
176-177. Man to dig a hole but hits
water and has to stop, (10, b; 7). Divides in the ratio T7 : T10 =
28 : 55.
della Francesca.
Trattato. c1480. F. 52r (121-122). Men agree to dig a well of depth
4 for 10, but no water is found
and they continue until the cost is 11 more.
I.e. (4, 10; x) worth 21.
Since T4 = 10 and
T6 = 21, they dig
to 6.
English in Jayawardene.
Calandri.
Arimethrica. 1491. F. 65v.
(12, 12; 16).
Pacioli. Summa. 1494.
Calandri.
Raccolta. c1495. Prob. 38, pp. 33‑34. If a well
24 deep is worth 24,
how deep a well is worth
40? I.e. (24, 24; x)
worth 40. He takes values proportional to Td.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 110, ff. 55r-55v. Dig a
well, (9, 24; 5). Divides in ratio T5 : T9 = 15 : 45.
Apianus. Kauffmanss
Rechnung. 1527. F. D.vi.v.
Mason to build a tower 100 high
for 100. He falls ill after 84. Divides in ratio T84 : T100.
Cardan. Practica
Arithmetice. 1539. Chap. 66.
Buteo.
Logistica. 1559.
Ozanam. 1694, 1725.
Les Amusemens.
1749. Prob. 44, p. 176. Mason to dig a well, (10, b; 4).
Divides as T4 : T10
= 10 : 55.
Vyse. Tutor's
Guide. 1771? Prob. 17, 1793: p. 136; 1799: pp. 144-145 & Key p. 188. Same as Ozanam, prob. 7, but with depth 30.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions, no. 25, pp.
20 & 79. Dig 20
yards for £20. Man falls sick after 8
yards. "How much was then
due to him, on a supposition that the labour increases in arithmetical
proportion as the depth?"
I.e. (20, 20; 8). Solution notes that the data does not
determine what the arithmetic progression is and chooses 5s
as cost of the first yard -- see Ozanam.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 182 & 263, no. 693. Man digging a well 49 feet deep. First foot costs 15, but each successive
foot costs 6 more than the previous.
Find cost of last foot and total cost.
So this is really an arithmetic progression problem, but I haven't seen
others of those using this context.
Vinot. 1860. Art. CVII: Problème du puits et du Maçon,
pp. 126-127, (20, 400; 10). He assumes the cost of digging up a unit
depth is 5 and that lifting the i-th
unit raises it from its centre of gravity, so is given by A, 3A, 5A, ..., 39A, where
A has to be determined from the
total cost. He finds A = 3/4
and d = 125.
Pierre Berloquin.
The Garden of the Sphinx, op. cit. in 5.N. 1981. Prob. 117: A
bailout fee, pp. 66 & 166. Man
contracts to bail out a 20 yard deep well for $400 and gives up at 10
yards. Answer says value is
proportional to the depth d times the average distance lifted, i.e. d/2,
hence value is proportional to d2/2 and this is the result that integration
produces.
Express an integer using four 4s,
etc. Cupidus Scientiae, 1881, seems to
be the first to ask for solutions to a lot of the integers, rather than a few
specific examples. The next examples of
the general form are Cunningham & Wiggins (1905), Pearson (1907), Ball
(1911), Ball (1912). Dawson (1916) is
the first to ask for four R's, where
R is indeterminate, e.g. 3 = (R+R+R)/R. I have included examples where a set of numbers
and operations is given and one has to obtain a given value. This overlaps a bit with 7.I.1, where the
object is to find the maximum possible value, and with 7.AC.3-6, where one uses
all nine or ten of the digits and I have included problems of inserting signs
into 12...9 to make 100 in 7.AC.3.
Dilworth.
Schoolmaster's Assistant.
1743. Part IV: Questions: A
short Collection of pleasant and diverting Questions, p. 168.
Les Amusemens.
1749. Pp. 52-54. Several problems leading to: 7 7/7,
33 3/3, 55 5/5, 99 9/9, 77 77/77, 2222
2222/2222, 11 1/1, etc. See the entry in 7.AN.
Vyse. Tutor's
Guide. 1771? Prob. 1, 1793: p. 155; 1799: p. 165 & Key p. 206. "Four Figures of nine may be so placed
and disposed of as to denote and read for
100, neither more nor less. Pray how is that to be done?"
Pike.
Arithmetic. 1788. P. 350, no. 16. "Said Harry to Edmund, I can place four 1's
so that, when added, they shall make precisely 12; Can you do so
too?"
Jackson. Rational
Amusement. 1821. Arithmetical Puzzles.
Endless Amusement II.
1826? Prob. 23, p. 201. "Put down four nines, so that they will
make one hundred."
Child. Girl's Own
Book. Arithmetical puzzles, no. 5. 1832: 170 & 179; 1833: 184 & 193; 1839: 164 & 173; 1842: 282 & 291; 1876: 231 & 244. "Place four nines together, so as to
make exactly one hundred. In the same
way, four may be made from three threes, three may be made from three twos,
&c." The 1833 solution is
printed rather oddly as 199 9‑9,
while the 1839 and 1842 solution is 99
9-9 and the 1876 solution is 99 9¸9.
Nuts to Crack III (1834), no. 211. "Write down four nines so as to make a hundred."
Family Friend 1 (1849) 150 & 178. Problems, arithmetical puzzles, &c. --
4. "Put four fives in such a
manner, that they shall make 6½. --
D. F." Answer is 5 5/5 + .5. = The Sociable, 1858, Prob. 46: A dozen quibbles, part 12: pp.
300 & 318. = Book of 500 Puzzles,
1859, prob. 46: part 12, pp. 18 & 36.
Boy's Own Book. To
place four figures of 9 in such a manner as to represent 100. 1855: 601.
Magician's Own Book.
1857. Quaint questions, p.
253. [No. 4] -- "Place three sixes
together, so as to make seven."
[No. 6] -- "Place four fives so as to make six and a
half." [Boy's Own Conjuring Book,
1860, pp. 224-225, has the Quaint Questions, but omits these two questions!]
Book of 500 Puzzles.
1859.
Charades, Enigmas, and Riddles. 1860: prob. 30, pp. 60 & 64;
1862: prob. 31, pp. 136 & 142;
1865: prob. 575, pp. 108 & 155.
"Write a Hundred with 4 nines." (1862 & 1865 have slightly different typography.)
Illustrated Boy's Own Treasury. 1860.
Leske. Illustriertes
Spielbuch für Mädchen. 1864?
Magician's Own Book (UK version). 1871. Paradoxes [no. 3],
p. 37. "With four fives make 6½? -- 5 5/5 .5." where the
5/5 is written as a 5 over a 5
with no fraction bar. Cf Jackson and
Magician's Own Book.
Hugh Rowley. More
Puniana; or, Thoughts Wise and Other-Why's.
Chatto & Windus, London, 1875. P. 300. "Write down
one hundred with four nines."
Mittenzwey.
1880. Prob. 1, pp. 1 &
58; 1895?: 1, pp. 7 & 62; 1917: 1, pp. 7 & 56. Write
100 with six equal digits.
"Cupidus Scientiae" (possibly the editor, Richard
A. Proctor). Four fours, singular
numerical relation. Knowledge 1 (30 Dec
1881) 184, item 151. A bit vague as to
what operations are permitted, but wants four
4s to make various values. Says he has not been able to make 19.
H. Snell. Singular
property of number 4. Knowledge 1 (6 Jan 1882) 209, item 178. 19 = 4! ‑ 4 -
4/4. Editor says 4!
is not reasonable for the problem as posed.
Solutions from various contributors. Four fours.
Knowledge 1 (13 Jan 1882) 229, item 184. Numerous solutions for
1 through 20,
except 19. Solutions for 19 are: 4/.4 + 4/.4; 4! - 4 - 4/4; 4/Ö.4 ‑ 4/4
("manifestly erroneous"); (4
+ 4 - .4)/.4; (x + x ‑ .x)/.x in general.
Four 3s give same results as three 5s,
except for 17.
Albert Ellery Berg, ed.
Op. cit. in 4.B.1. 1883. P. 373.
"Place three sixes together so as to make seven."
Lemon. 1890.
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Card Puzzles No. II: A
curious addition sum, p. 2. Mentions
"writing down 100 with four nines" as 99 9/9.
(Sam Loyd.) One
hundred pounds for correct answer to a puzzle.
Tit-Bits (14 Oct 1893) 25.
"Find How to Arrange the Figures
· 4 · 5 · 6 · 7 · 8 · 9 · 0 · in
an Arithmetical Sum which Adds up the Nearest to 82." "Mr. Loyd
is confident that no one will find it out." Indeed, Loyd will be paid £100 only if no correct answer is
received.
(Sam Loyd.) Solution
of Mr. Sam Loyd's one hundred pound puzzle.
Tit-Bits (18 Nov 1893) 111. 80·5
+ ·97 + ·46 = 82. (There are points
over the 5, 9, 7, 4, 6, but my printer may not print these clearly.) Here the mid-line dot (·) is used for a
decimal point. Because of the number of
correct solutions, ten extra names were drawn from them for additional £5
prizes. "It seems that not a
single person in the whole of America has sent the correct answer when a prize
was offered there, but here we have received a very large number actually
correct." [See MRE for another
solution.]
Report on the 82 puzzle appeared in 25 Nov and letter from Loyd appeared about two weeks
later - photocopies on order.
Hoffmann. 1893. Chap. IV, no. 18: Another way to make a
hundred, pp. 148 & 193 = Hoffmann-Hordern, p. 120. Use six
9s to make 100.
Ball. MRE, 3rd ed.,
1896. P. 13. "... a question which attracted some attention in London in
October, 1893, ...." [See Loyd
above.] He says that the problem is to
make 82 with the seven digits 9,
8, 7, 6, 5, 4, 0 and gives one solution
as 80·69 + ·74 + ·5 (with points over the 9, 4, 5
-- there should also be points over the
6 and 7).
H. D. Northrop.
Popular Pastimes. 1901. No. 10: A dozen quibbles, no. 12, pp. 68
& 73. Express 6½
by four 5s. Answer is:
5 5/5 . 5, which seems pretty
poor to me. c= Jackson, no. 17.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
10:5 (Sep 1903) 426-427. "How to
arrange four 9's to make 100."
Ball. MRE, 4th ed.,
1905. P. 14. Repeats the material in the 3rd ed of 1896, again omitting two
points, and adds further questions. Use
the 10
digits to total 1 -- a solution is 35/70 + 148/296 -- or to
total 100 -- a solution is 50 + 49
+ 1/2 + 38/76. Use the 9
digits to make four numbers which total
100 -- a solution is 78 + 15 + 2Ö9 +
3Ö64.
A. Cunningham & T. Wiggins. ?? Math. Quest. Educ.
Times 7 (1905) 43‑46. ??NYS --
cited in Dickson I 460, item 45d.
Expressions using four 9s and four
4s.
Pearson. 1907.
Wehman. New Book of
200 Puzzles. 1908. P. 26.
= Magician's Own Book, no. 4.
Ball. MRE, 5th ed.,
1911. Pp. 13-14. Briefly restates the material in the 4th ed.
as "questions which have been propounded in recent years. ...
To the making of such questions of this kind there is no limit, but
their solution involves little or no mathematical skill."
"Another
traditional and easy recreation .... I
have never seen this recreation in print, but it seems to be an old and
well-known question." Deals just
with four 4s and says one can get up through 170. G. N. Watson has pointed out that one can get further by using
factorials and subfactorials. (The
subfactorial of n is
n¡ = n![1/0! ‑ 1/1! + 1/2! - 1/3! +
... ± 1/n!].) The topic is not in
earlier editions.
W. W. Rouse Ball.
Four fours. Some arithmetical
puzzles. MG 6 (No. 98) (May 1912) 289‑290. "An arithmetical amusement, said to
have been first propounded in 1881, ...."
[This would seem to refer to Knowledge, above.] Studies various forms of the problem. Says it occurs in his MRE -- see above. MRE 6th ed., 1914, p. 14, cites this
article.
Ball. MRE, 6th ed.,
1914. Pp. 13-14. He now splits the material into three
sections.
Empirical
Problems. Restates the material in the
5th ed. as "... numerous empirical problems, ..." and omits Loyd's
problem. "To the making of such
questions there is no limit, but their solution involves little or no
mathematical skill."
He
then introduces the "Four Digits Problem". "I suggest the following problem as being more
interesting." Using the digits 1, 2, ..., n, express the integers from
1 up using four different digits
and the operations of sum, product, positive integral power and base-10 notation (or also allowing iterated square
roots and factorials). With n = 4, he can get to 88 or to
264. With n = 5,
he can get to 231 or
790. Using 0, 1, 2, 3, he can get to 36 (or
40).
Under
Four Fours Problem, he discusses what operations are permitted and says he can
get to 112, or to 877 if subfactorials are permitted (citing his
MG article for this). Mentions four 9s
and four 3s problems.
Williams. Home
Entertainments. 1914. The six 9's, p. 119. "Express the number 100
by means of six 9's."
Thomas Rayner Dawson.
1916. ??NYS. Cited in:
G&PJ 3 (Jan 1988) 45
& 4 (Mar 1988) 61. Asks for four R's, where R is
indeterminate, e.g. 3 = (R+R+R)/R.
Ball. MRE, 7th ed.,
1917. Pp. 13-14. The material of the first two sections is
repeated, but under "Four Fours Problem", he discusses the operations
in more detail. With +,
-, x, ¸, brackets and base-10
notation, he can get to 22. Allowing also finitely iterated square
roots, he can get to 30. Allowing also factorials, he can get to 112.
Allowing also integral indices expressible by 4s and infinitely
iterated square roots, he can get to
156. Allowing also
subfactorials, he can get to 877. (In the 11th ed., 1939, pp. 15-16, two
footnotes are added giving expressions for
22 in the first case and 99
in the third case.) Gives some
results for four 2s, four
3s, four 5s,
four 9s. Mentions the general problem of n
ds.
Smith. Number
Stories. 1919. Pp. 112‑113 & 140‑141. Use four
9s to make 19,
2 and 20.
Ball. MRE, 9th ed.,
1920. Pp. 13-14. In the "Four Digits Problem", he
considers n = 4, i.e. using
1, 2, 3, 4, and discusses the
operations in more detail. Using sum,
product, positive integral power and base-10
notation, he can get to 88. Allowing also finitely iterated square roots
and factorials, he can get to 264. Allowing also negative integral indices, he
can get to 276. Allowing also fractional indices, he can get
to 312. He then mentions using 0,
1, 2, 3 or four of the five digits 1, ..., 5.
Under
"Four Fours Problem", he repeats the material of the 7th ed., but
adds some extra results so he has results for four ds, d = 1, 2, 3, 5, 6, 7,
8, 9.
Ball. MRE, 10th ed.,
1922. Pp. 13-14. In the "Four Digits Problem", he
repeats the material of the 9th ed., but at the end he adds that using all of
the five digits, 1, ..., 5, he has gotten to 3832 or 4282,
depending on whether negative and fractional indices are excluded or allowed.
Hummerston. Fun,
Mirth & Mystery. 1924. Some queer puzzles, Puzzle no. 76, part 2,
pp. 164 & 183. "Express
100 by using the same figure six times."
Dudeney. MP. 1926.
Prob. 58: The two fours, pp. 23‑24 & 114. = 536, prob. 109, pp. 34 & 248‑249,
with extensive comments by Gardner.
King. Best 100. 1927.
Heinrich Voggenreiter.
Deutsches Spielbuch Sechster
Teil: Heimspiele. Ludwig Voggenreiter,
Potsdam, 1930. P. 112. Write 1000 with seven or five equal digits.
Perelman. FFF. 1934.
1957: probs. 98, 100 & 102, pp. 137 & 143-144; 1979: probs. 101, 103 & 105, pp. 166-167
& 174-175. = MCBF, probs. 101,
103 & 105, pp. 167 & 176‑178.
Perelman. MCBF. 1937.
Any number via three twos. Prob.
202, pp. 398-399. "A witty
algebraic brain-teaser that amused the participants of a congress of physicists
in Odessa."
n = ‑ log2 log2
Ön2, where Ön means n‑fold
iterated square root.
Haldeman-Julius.
1937. No. 16: Adding fives, pp.
5 & 21. Use four 5s to make 6½.
Answer is: 5 + 5/5 + .5.
M. Adams. Puzzle
Book. 1939. Prob. B.83: Figure juggling, part 3, pp. 78 & 107. Use a digit
8 times to make 1000.
Answer uses 8s.
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
Number, please!, pp. 20 & 210.
Depew. Cokesbury
Game Book. 1939.
McKay. Party
Night. 1940. No. 9, p. 177.
Meyer. Big Fun
Book. 1940. A half dozen equals 12, pp. 119 & 738. Use six
1s to make 12.
Answer: 11 + 11/11.
George S. Terry. The
Dozen System. Longmans, Green &
Co., NY, 1941. ??NYS -- quoted in: Underwood Dudley; Mathematical Cranks;
MAA, 1992, p. 25. What numbers can be
expressed with four 4s duodecimally? About 5 dozen. How many
numbers can be expressed using each of the digits 1, 2, 3,
4 once only, again duodecimally? About 9 dozen and nine. Terry (or Dudley) also gives the results for
decimal working as 22 and 88.
Sullivan.
Unusual. 1943. Prob. 17: Five of a kind. Write 100 with the same figure five times "and
the usual mathematical symbols".
Says it can be done with
1s, 2s, 5s (two ways), 9s and perhaps others.
J. A. Tierney, proposer;
Manhattan High School of Aviation Trades, D. H. Browne,
H. W. Eves, solvers. Problem
E631 -- Two fours. AMM 51 (1944)
403 & 52 (1945) 219. Express 64
using two 4s.
Vern Hoggatt & Leo Moser, proposers and solvers. Problem E861 -- A curious representation of
integers. AMM 56 (1945) 262 &
57 (1946) 35. Represent any
integer with p a's,
for any p ³
3 and any a ¹ 1. Solution for ±n uses
log to base ÖÖ...Öa, with
n radicals.
S. Krutman. Curiosa
138: The problem of the four n's. SM 13 (1947) 47.
Sullivan.
Unusual. 1947. Prob. 28: A problem in arithmetic. What is the smallest number of eights which
make 1000?
G. C. S[hephard, ed.]
The problems drive. Eureka 11
(Jan 1949) 10-11 & 30.
Anonymous. The
problems drive. Eureka 13 (Oct 1950) 11
& 20-21.
Anonymous. The
problems drive. Eureka 17 (Oct 1954)
8-9 & 16-17. No. 5. Use four 4s to express 37; 57; 77; 97; 123.
D. G. King‑Hele.
Note 2509: The four 4's
problem. MG 39 (No. 328) (May 1955)
135. n = log[{log
4}/{log Ön 4}]/log Ö4 expresses any positive integer n in
terms of three 4s. A slight variation expresses n in
terms of four x's, for any real x ¹ 0, 1. 1 can be used by taking x = .1. He also
expresses n in terms of m x's
for real x ¹ 0,
1 with
m > 5 and also with m = 5.
Anonymous. Problems
drive. Eureka 18 (Oct 1955) 15-17 &
21. No. 6. Use four 4s to make 7;
17; 37; 3,628,800.
Anonymous. Problems
drive, 1958. Eureka 21 (Oct 1958) 14-16
& 30. No. 10. Use
1, 2, 3, 4, 5, in order to
form 100; 3 1/7; 32769.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
M. R. Boothroyd
& J. H. Conway. Problems drive, 1959. Eureka 22 (Oct 1959) 15-17 & 22-23. No. 3.
Use three 1s to make the integers from one to twelve, using only
arithmetic symbols. (No trigonometric
functions or integer parts allowed.)
Young World.
c1960. P. 54. Use five
9s to make 1000.
B. D. Josephson
& J. M. Boardman. Problems drive 1961. Eureka 24 (Oct 1961) 20-22 & 24. Prob. K.
Use three 7s to express 1, ...,
11, using only arithmetic symbols.
J. H. Conway
& M. J. T. Guy. π
in four 4's. Eureka 25 (Oct
1962) 18‑19. Cite Eureka 13
(1950). Note that π = Ö[(‑Ö4/4)!]4, if non‑integral factorials are
allowed. Show that any real number can
be arbitrarily well approximated using four
4s.
R. L. Hutchings
& J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 20-21 & 34-35. Prob. H.
Use four identical digits to represent 100 in as many ways as possible,
but not using representations which are independent of the digit used,
like (5x5)/(.5x.5). The give eight examples, using 9, 5, 5, 4,
4, 3, 3, 9, 1, and say there are more.
D. E. Knuth.
Representing numbers using only one
4. MM 37 (1964) 308‑310.
Gardner. SA (Jan
1964) adapted as Magic Numbers, chap. 5.
Cites Knowledge as the origin.
Magic Numbers gives numerous other citations.
Marjorie Bicknell & Verner E. Hoggatt. 64 ways to write 64 using four 4's.
RMM 14 (Jan‑Feb 1964) 13‑15.
Jerome S. Meyer.
Arithmetricks. Scholastic Book
Services, NY, 1965. Juggling numbers,
no. 3, pp. 83 & 88.
Ripley's Puzzles and Games.
1966. Pp. 16-17, item 2. Use
13 3s to make 100.
Steven Everett.
Meanwhile back in the labyrinth.
Manifold 10 (Autumn 1971) 14‑16.
(= Seven Years of Manifold 1968‑1980; ed. by I. Stewart &
J. Jaworski; Shiva Publishing, Cheshire, 1981, pp. 64‑65.) n
= ‑Ö4 *
log4 log4 Ön4, where log4 means log to the base 4 and Ön means n‑fold
iterated square root. This is a variant
of King‑Hele's form. This article
is written in a casual style and seems to indicate that this formula was
devised by Niels Bohr. He gets a form
for e,
but it uses infinitely many factorial signs!!!!...
Editor's
note on p. 2 (not in the collection) gives an improvement due to Michael
Gerzon, but it is unclear what is intended.
The note gives another method due to Professor Burgess using sec tan-1 Öm = Ö(m+1) and
1 = Önn which expresses n by one
1.
[Henry] Joseph and Lenore Scott. Master Mind Pencil Puzzles.
1973. Op. cit. in 5.R.4. Numbers-numbers, part 3, pp. 109-110. Use
13 3s to make 100. The give
33 + 33 + 33 + (3/3)3 + 3x3
+ 3x3. I found 33 + 33 + 33 + 33/3 - 3x3 - 3/3, which seems simpler.
Ball. MRE, 12th ed.,
1974. Pp. 15-17. Under "Four fours problem", the
material of the 9th ed. and the footnotes mentioned at 7th ed are repeated, but
the bound for four 9s is increased.
Bronnie Cunningham.
Funny Business. An Amazing
Collection of Odd and Curious Facts with Some Jokes and Puzzles Too. Puffin, 1978. Pp. 38 & 142. Arrange
three 9s to make 20. Answer:
(9 + 9)/.9.
Putnam. Puzzle
Fun. 1978.
P. Grammer, I. McFiggans, N. Blacknell, T. Joyce, J. Anstey
& A. Devonald. Counting in
fours. MiS 9:4 (Sep 1980) 21-22. Uses four 4s to express 1, ..., 50.
Says 51 - 100 will appear in
next issue, but they didn't.
J. Bellhouse. Four
fours. MiS 14:1 (Jan 1985) 15. Says the promised table for 51 - 100
(see Sep 1980 above) had not appeared, so his students found their own.
Anne Williamson.
1985. MiS 14:4 (Sep 1985)
7. Use the four digits 1, 9, 8, 5
to express integers 1 - 100. Unhappy with expressions for 24, 31, 65
which use !.
Ken Lister.
Letter. MiS 15:2 (Mar 1986)
47. Responding to Bellhouse (Jan
1985). Corrects and improves some
values, but says 71 and 73 have not been done. Expresses
a/b, for single digits a, b,
by use of four 4s.
Angie Aurora.
Letter. MiS 15:3 (May 1986)
48. Improvements for Williamson's
problem -- Sep 1985 above.
Joyce Harris.
Letter: Four fours. MiS 15:3 (May 1986) 48. Responding to Lister (Mar 1986), gives
expressions for 71 and 73.
Bob Wasyliw.
Letter: Four 4's -- the ultimate
solution. MiS 15:5 (1986) 39. Adapts Everett's 1971 method to include non‑positive
integers.
Simon Gray & Colin Abell. Letters: Four fours
again. MiS 16:2 (1987) 47. Gray notes that 4 = Ö4 * Ö4, so that 'four 4s' is the same as 'at most
four 4s'. He gives π = Ö4 * sin‑1(4/4) and more complex forms. Abell gives
π = ‑ Ö(‑4/4) * log (‑4/4) [The first minus sign is ambiguous??]
and π = ‑ Ö(4*4)
* Tan-1(4/4) [The minus sign
is wrong.]
Tim Sole. The Ticket
to Heaven and Other Superior Puzzles.
Penguin, 1988.
Number play (ii)
‑ (iv), pp. 15, 29 & 178‑180.
Three of the
best -- (iii), pp. 17 & 32. Some
solutions with one 4, using
!, Ö, [n] = INT(n) and Σ(n).
Tony Forbes.
Fours. M500 116 (Nov 1989) 4‑5. Says someone (possibly Marion Stubbs?) gave
a simple variation of King‑Hele's and Everett's formulae to use exactly
four 4s to yield n. Forbes suggests using one 4
and the three operations:
!, Ö and
INT. He already gets stuck at
12.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
Take two, pp. 96 & 139. Use
five 2s to express 1, 2, ...,
26, particularly 17 and 26.
P. H. R. Fawlty
keys. Mathematical Pie, 129 (Summer
1993) 1023 & Notes, p. 1. Calculator whose only keys are
5, 7, +, -, x, ¸ and
=. Make numbers from 0 to
20. Solution notes you can make any
number by adding enough terms of the form
5 ¸ 5 and then gives
short solutions for 1 through 20.
Clifford A. Pickover.
Phi in four 4's. Theta (Crewe) 7:2 (Autumn 1993) 5-8. In Sep 1991 he asked for good approximations
to φ using four 4s, either with as many symbols as you want or
with each symbol used at most four times.
Says he was inspired by Conway & Guy's paper of 1962. Brian Boutel produced φ
= (Ö4 +
Ö{4!-4})/4. Pickover
then extended the question and various solvers got φ in five 5s,
seven 6s, eight
8s, nine 9s and
2k-5 ks.
John Seldon.
Fours. M500 136 (Jan 1994) 15‑16. Answers Forbes' 1989 problem of
expressing 1 ‑ 100 with one
4 and any number of
factorials, x!, and integer square roots, [Öx].
David Crawford and students. 1999 the end of an
era. MiS 28:4 (Sep 1999) 25. Uses
1, 9, 9, 9 to make all integers
from 1
to 100. Notes that
2000, 2001, ... are not going to
be very useful for such puzzles!
Derek Ball. Four
4s. MTg 173 (Dec 2000) 18. Says his fifth year teacher discovered the
following for n in terms of four 4s: n = logÖ4/4 log4 Ön 4, where Ön denotes n-fold repeated square root. Cf Perelman, 1937; King-Hele, 1955; Everett,
1971 -- Everett is very close to this and the others are not quite that close.
7.I.1. LARGEST NUMBER USING FOUR ONES, ETC.
Mittenzwey. 1880.
Prob. 142, pp. 30 & 79-80;
1895?: 162, pp. 34 & 82;
1917: 162, pp. 31 & 79-80.
Find largest number using four digits.
Gets 9^9^9^9 and tries to contemplate its size. 9^9
is given as 387,420,488 (last
digit should be 9), so 9^9^9
has 369,693,100 digits.
James Joyce.
Ulysses. (Dijon, 1922); Modern Library (Random House), NY, 1934,
apparently printed 1946. P. 684
(Gardner says the 1961 ed. has p. 699).
Bloom estimates that
9^(9^9) would occupy "33
closely printed volumes of 1000 pages each", but he erroneously phrases
the number as "the 9th power of the 9th power of 9", which is
only 981.
King. Best 100. 1927.
No. 35, pp. 19 & 46. Largest
number using two 4s. Gives
44 = 256.
Perelman. FFF. 1934.
Four 1's. 1957: prob. 103, pp.
137 & 144-145; 1979: prob. 106,
pp. 167 & 175. = MCBF, prob.
106, pp. 167 & 178 = MCBF, prob.
131, p. 217. "What is the biggest
number that can be written with four
1's?
Perelman. MCBF. 1937.
Probs. 128-132, pp. 214-219.
Largest numbers with: Three
twos; Three threes; Three fours; Four ones; Four twos.
Sullivan.
Unusual. 1947. Prob. 30: Not 999. Largest number that
can be written with three integers [sic!].
Answer: 9^(9^9).
G. C. S[hephard, ed.]
The problems drive. Eureka 11
(Jan 1949) 10-11 & 30. No. 5. Find the largest numbers expressible using
four 2s or four 4s, no symbols allowed.
Answers: 2^2^22; 4^4^4^4.
Leroy F. Meyers. An
integer construction problem. AMM 66:7
(Aug/Sep 1959) 556-561. This deals with
Ball's "Four Digits Problem" (see MRE, 6th ed., 1914 in 7.I) and
generalizations. In particular, he
shows that if one uses 1, 2, 3, 4, with operations +, ‑, x
and brackets, then one can obtain precisely the following: 1, 2, ..., 28, 30, 32, 36. In general he obtains the largest integer
expressible using a given multiset of integers (i.e. one is allowed a fixed
number of repeats of a value) using the operations +, x and brackets. He also shows that allowing also
-, for both negation and
subtraction, does not increase the maximum obtainable value. He conjectures that allowing also ¸, for both reciprocation and division, does not increase the
maximum obtainable, but Meyers has written that a student once showed him a
counterexample, but he cannot remember it.
He applies his general results to show that no other values are
obtainable when using 1, 2, 3, 4.
Problematical Recreations 4. Problem 1 and its answer, pp. 3 & 36. (This is one of a series of booklets issued
by Litton Industries, Beverly Hills, California, nd [c1963], based on the
series of the same name in Aviation Week and Electronic News during
1959-1971. Unfortunately, neither the
date nor location nor author is given and the booklet is unpaginated. The answer simply states the maximum value
with no argument.) Reproduced with a
proper solution in: Angela Dunn;
Mathematical Bafflers; (McGraw-Hill, 1964, ??NYS); revised and corrected 2nd ed., Dover, 1980, pp. 119 & 132 and with just the answer in: James F. Hurley; Litton's Problematical
Recreations; Van Nostrand Reinhold, NY, 1971, chap. 7, prob. 8, pp. 238 &
329. "What is the largest number
which can be obtained as the product of positive integers which add up to
100?" (This type of problem must
be much older than this?? Meyers writes
that he first encountered such problems as an undergraduate in 1947. If one looks at maximizing the LCM instead
of the product of the terms, this is the problem of finding a permutation
of 100
letters with maximum order.)
Sol Golomb. Section
13.8 The minimization of the cost of a
digital device (the juke-box problem).
IN: Ben Nobel; Applications of undergraduate mathematics in
engineering. MAA & Macmillan, 1967,
pp. 284-286. This considers the problem
of Problematical Recreations in the inverse form. We have r k-state devices which allow kr choices and the cost is proportional to rk. Minimize the ratio of
cost to capacity or maximize the ratio of capacity to cost. [Another way to express this is to ask which
base is best for a computer to use? I recall
this formulation from when I was a student in the early 1960s. The answer is e, but here only integer
values are used. One can generalise to:
given a value, find the smallest sum of numbers whose product is the given
value.] The connection with juke-boxes
is that they typically have two rows of
12 buttons and one has to press
two buttons to make a selection of one from
144 records. One can do much better with five rows of
three buttons, but asking a customer to punch five buttons may be unreasonable,
so perhaps three rows of five or six buttons might be best.
The Fortieth William Lowell Putnam Mathematical Competition,
1 Dec 1979. Problem A-1. Reproduced in: Gerald L. Alexanderson, Leonard F. Klosinski & Loren C.
Larson; The William Lowell Putnam
Mathematical Competition Problems and
Solutions: 1965‑1984; pp. 33
& 109. "Find positive
integers n and a1, a2, ..., an such that a1 +
a2 + ... + an = 1979
and the product a1a2...an is as large as possible."
Cliff Pickover & Ken Shirriff. The terrible twos problem.
Theta (Crewe) 6:2 (Autumn 1992) 3-7.
They study the problem of making numbers using just +,
-, x, ^, and 1s
(or 1s and 2s). For a given
n, what is the least number of
digits required? They later permit concatenation,
e.g. 11 or 12 is permitted. They report results from various programs and mention some
related problems.
Bryan Dye. 1, 2, 3,
4 -- four digits that dwarfed the universe.
Micromath 10:3 (Aut 1994) 12‑13.
Says a version appeared in SA a few years ago and is discussed in: Clifford A. Pickover; Computers and the
Imagination; Alan Sutton, 1991. Dye's
version is to make the largest number using
1, 2, 3, 4 once and the signs +, -, x, ¸, ( ),
. (i.e. decimal point).
Exponentiation was not considered a sign and was permitted. Pickover's version allowed only the
signs -, ( ), . (i.e. decimal
point). The largest value found
actually fits Pickover's conditions:
.3^-(.2^-{.1^-4}) has 106990 digits. The largest
number using just exponentiation was
2^(3^41)) with 1019 digits.
It is better to get a rise of 5
every six months than a rise of
20 every year. The interpretation of the first phrase is
somewhat ambiguous -- see Mills (1993).
If the salary is S every six months, the usual interpretation
of the first phrase is that the half-yearly payments are: S, S
+ 5, S + 10, S + 15, S + 20, S + 25,
..., while the second phrase
gives payments of: S, S, S
+ 10, S + 10, S + 20, S + 20, ...,
and the former gets 5 extra every year.
Ball. MRE, 3rd ed.,
1896. pp. 26‑27. £20 per year
versus £5 every half year. He says this
is a question "which I have often propounded in past years." It is not in the 1st ed.
Workman. Op. cit. in
7.H.1. 1902. Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 16,
pp. 425 & 544 (431 & 577 in c1928 ed.). Compares rise of £15 per year every 3 years with £5 every
year. This represents a precursor of
the puzzle version.
A. Sonnenschein & H. A. Nesbit. The New Science and Art of Arithmetic For
the Use of Schools. A. & C. Black,
London, 1903. Prob. 12 & 14, pp.
341 & 489. "A youth entered an
office at the age of 15 at a salary of £40 a year, with an annual rise of
£12. ..." "... What total sum would he have
received in 30 years? and what would he have received if the increase had been
at the rate of £1 per month?"
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
10:4 (Aug 1903) 336-337. Not too
obvious. Half-yearly rise of £5 versus
£20 a year. No solution given.
Susan Cunnington.
The Story of Arithmetic. Swan
Sonnenschein,, London, 1904. Prob. 35,
p. 217. £5 raise each six months versus
£10 raise each year.
Dudeney?? Breakfast
Table Problems No. 330: Smith's salary.
Daily Mail (30 & 31 Jan 1905) both p. 7. Raise of
10 per year versus 2½
every six months.
Pearson. 1907. Part II, no. 87, pp. 132 & 208. As in Ball.
Loyd. Salary
puzzle. Cyclopedia, 1914, pp. 312 &
381. = MPSL1, prob. 84, pp. 81 &
150‑151. = SLAHP: The
stenographer's raise, pp. 60 & 108.
Raise of 100 per year versus 25 every half year. Interpreting the raise of 25
as worth only 12.50 in a half‑year, this option loses,
contrary to all other approaches.
Clark. Mental
Nuts. 1916, no. 6. The two clerks. $25 rise each six months versus $100 rise each year.
Dudeney. AM. 1917.
Prob. 26: The junior clerk's puzzle, pp. 4 & 150. Two clerks getting £50 per year with one
getting a raise of £10 per year versus the other getting a raise of £2 10s
every six months with a complicated further process of savings at
different rates for five years.
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 276, pp. 103 & 171: A problem in
salaries. £20 rise every six months
versus £80 rise each year.
T. O'Conor Sloane.
Rapid Arithmetic. Quick and
Special Methods in Arithmetical Calculation Together with a Collection of
Puzzles and Curiosities of Numbers. Van
Nostrand, 1922. [Combined into: T.
O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures;
Van Nostrand, 1939.] The two clerks, p.
168. Raise of $50 every six months
versus $200 each year.
F. & V. Meynell.
The Week‑End Book, op. cit. in 7.E. 1924. Prob. one, p. 274
(2nd ed.), pp. 406‑407 (5th? ed.).
Raise of 20 per year versus 5 every half year.
Peano. Giochi. 1924.
Prob. 16, p. 5. 1000 per year
with rise of 20 each year versus 500 each half-year with rise of 5 each
half-year.
Wood. Oddities. 1927.
Prob. 31: A matter of incomes, p. 31.
$1000 per year with $20 per year increase versus "$5 each half year
increase".
Collins. Fun with
Figures. 1928. Do figures really lie?, pp. 35-36. $50 every six months versus $200 per year.
R. Ripley. Believe
It Or Not! Book 2. (Simon & Schuster, 1931); Pocket Books, NY, 1948. P. 123: Figure your raise in pay. A raise of one every day is better than a
raise of 35 every week. (Assumes a six
day week.)
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 20, pp. 16 & 189‑190. Raise of 2000 per year versus 500 half‑yearly.
Phillips.
Brush. 1936. Prob. G.1: The two clerks, pp. 20 &
87. Raise of 200 annually versus 50
half‑yearly.
McKay. At Home
Tonight. 1940. Prob. 4: A choice of rises, pp. 63 &
76. £5 per half-year versus £15 per
year. Solution is unclear and seems to
be wrong. "£5 each six months is
£5 in the first half-year and £10 in the second -- that is, £15 per year. But the man who gets £5 per six months gets
£5 in the first year, and of course he keeps this advantage year by
year." I get that the first case
is ahead by £5n in the
n-th year.
Sullivan.
Unusual. 1943. Prob. 5: Raising the raise question. Raise of $20 per year versus $5 every half
year.
Birtwistle.
Calculator Puzzle Book.
1978. Prob. 56: The interview,
pp. 39 & 99-100. £3050 yearly plus
£100 each year versus £1500
half-yearly plus £50 each half-year.
D. J. Hancox, D. J.
Number Puzzles For all The
Family. Stanley Thornes, London,
1978. Puzzle 11, pp. 4 & 48.. £1 rise per month every month versus £144 rise
per year every year. Says the first
gives £66 extra in the first year and £210 extra in the second year, while the
second gives no extra in the first years and £144 extra in the second year. He then says: "Hence he would never get
the £66 he would have received in the first year." In fact, he loses £66 every year.
Stuart E. Mills.
Dollars and sense. CMJ 24 (1993)
446-448. Raise of $1000 per year versus
$300 every half year. Discusses various
interpretations of the second phrase and gives some recent references.
Comments
by myself and various others seem to have appeared (??NYS) as they are included
in the collection of these columns:
Edward J. Barbeau; Mathematical Fallacies, Flaws, and Flimflam; Spectrum
Series, MAA, 2000, pp. 11-14. This gives
references to various recent appearances of the problem in 1943, 1983, 1992.
John P. Ashley.
Arithmetickle. Arithmetic
Curiosities, Challenges, Games and Groaners for all Ages. Keystone Agencies, Radnor, Ohio, 1997. P. 80: The best job offer. $20,000 per year plus a $500 raise every six
months versus $20,000 per year plus a raise of $1000 each year. He says the six-monthly payments of the
first are 10,000, 10,500,
11,000, 11,500, ... and the payments for the second are 10,000,
10,000, 11,000, 11,000,
.... But the second is a raise
of $2,000 per year!
The Friday Night Book
(A Jewish Miscellany). Soncino
Press, London, 1933. Mathematical
Problems in the Talmud: The Divisibility Test, p. 137. Hebrew law requires fields to lie fallow
every seventh year, and this is to hold for all fields at once! Rabbi Huna gave the following rule. Write the year as y = 100a + b. Form 2a + b.
Then the year is a Sabbatical year if
7 divides 2a + b.
No explanation is given in the Talmud, but we clearly have y º 2a + b (mod 7). [The Talmud was compiled in the period -300
to 500. This source says Rabbi Huna is
one of the few mathematicians mentioned in the Talmud, but gives no dates and
he is not mentioned in the EB. From the
text of another problem attributed to him (cf in Section 6.AD), the problem
would seem to be sometime in the 1-5 C.]
See Smith, History II 151‑154
for a detailed discussion. He says it
appears in al‑Khowarizmi and al‑Karkhi and that it is generally
assumed to come from India, but his earliest Indian source is Lilavati,
1150. G. R. Kaye; References to Indian mathematics in certain
Mediæval works; J. Asiatic Society of
Bengal (NS) 7:11 (Dec 1911) 801-816 notes the appearances in al-Khwârizmî,
Avicenna and Maximus Planudes [Arithmetic after the Indian method; c1300; op.
cit. in 7.E.1] but asserts it does not occur in early Indian sources -- but cf
Aryabhata II, 950.
Dickson I, chap. XII, pp. 337‑346,
especially p. 337, gives a concise history.
He says al‑Karkhi was the first to use a (mod 11) check.
See Tropfke, pp. 165-167.
I have recently realised that
certain puzzle problems should be listed here, but so far I have only noted
Boy's Own Book, Boy's Own Book (Paris), Carroll, Peano, Parlour Games for
Everyone -- there must be many more 19C and even 18C examples. Basically these involve getting someone to
produce a number divisible by nine and asking him to delete one digit and tell you
the others -- you tell him the missing digit.
There are many of these and I probably won't try to record all of them,
but subtracting a number from its reversal may be the forerunner of the 1089
puzzle of 7.AR.
St. Hippolytus.
Κατα Πασωv
Αιρεσεωv
Ελεγχoσ (??= Philosphumena) (= Refutatio Omnium
Haeresium = Refutation of all
Heresies). c200. Part iv, c. 14. ??NYS. Discusses adding
up digits corresponding to letters (mod
9) and mentions considering it (mod 7).
(HGM I 115‑117) See
also: Smith II 152, Dickson I 337 and Saidan (below), p. 472.
Hippolytus doesn't use the method to check any arithmetic. (St. Hippolytus may be the only antipope to
be counted a saint! A reference says he
was Bishop of Portus and the MS was discovered at Mt. Athos in 1842.)
Iamblichus. On
Nicomachus's Introduction to Arithmetic.
c325. ??NYS. In:
SIHGM I 108‑109.
Special case. (See also HGM I
114‑117.)
Muhammad [the h should have an underdot] ibn Mûsâ al‑Khwârizmî. c820.
Untitled Latin MS of 13C known as Algorismus or Arithmetic, Cambridge Univ.
Lib. MS Ii.6.5. Facsimile ed., with
transcription and commentary by Kurt Vogel as:
Mohammed ibn Musa Alchwarizmi's Algorismus, Das früheste Lehrbuch zum
Rechnen mit indischen Ziffern; Otto Zeller Verlagsbuchhandlungen, Aalen,
1963. English translation by John N.
Crossley & Alan S. Henry as:
Thus spake al‑Khwārizmī: A translation of the
text of Cambridge University Library Ms.Ii.vi.5; HM 17 (1990) 103‑131. [Crossley & Henry name this author Abū Ja‘far Muhammad [the h
should have an underdot] ibn Mūsā al‑Khwārizmī, but I have seen no other authority
giving Abū Ja‘far -- several give Abū ‘Abdallāh
and Rosen's translation The Algebra of Mohammed ben Musa specifically
says our author "must therefore be distinguished from Abu Jafar Mohammed ben Musa, likewise a mathematician and astronomer, who
flourished under the Caliph Al Motaded" (c900). F. 108r = Vogel
p. 25 = Crossley & Henry p.
117. Describes casting out 9s
in doubling and in multiplication.
Aryabhata II. Mahâ‑siddhânta. 950.
Edited by M. S. Dvivedi, Braj Bhushan Das & Co., Benares, 1910. English Introduction, pp. 21‑23; Sanskrit text, p. 245. Casting out
9s for multiplication, division,
squaring, cubing, and taking square and cube roots. (Datta & Singh I 181 give the text in English.)
Abû al-Hassan [the
H should have an underdot]
Ahmad[the h should have an underdot] Ibn Ibrâhîm al-Uqlîdisî. Kitâb al Fuşûl [NOTE: ş
denotes an s with an underdot.] fî al‑Hisâb[the H
should have an underdot] al‑Hindî. 952/953.
MS 802, Yeni Cami, Istanbul.
Translated and annotated by A. S. Saidan as: The Arithmetic of Al‑Uqlīdisī; Reidel, 1978. Book II, chap. 13, pp. 153‑155 and
Book III, chap. 7‑8, pp. 195‑201 deal with checking by casting
out 9s, which is given only briefly, apparently being well‑known. He applies it to division and square
roots. The method is also mentioned in
Book II, chap. 2. On pp. 468‑472,
Saidan discusses the appearance of various rules in early texts. His earliest Indian example is Lilavati,
1150, but he gives no reference.
Kūshyār ibn Labbān = Abū ăl-Hasan [the
H should have an underdot]
Kūšyār ibn Labbān ibn Bāšahri al-Ğīlī. Kitāb fī usūl Hisāb
al-Hind [Principles of Hindu Reckoning].
c1000. Facsimile with
translation by Martin Levey & Marvin Petruck. Univ. of Wisconsin Press, Madison, 1965.
Ibn Sina =
Avicenna. Treatise on Arithmetic. c1020.
??NYS. Complete rules for
checking operations by casting out
9s, attributed to the Hindus,
(Smith, Isis 6 (1924) 319). (See also Cammann
-- 3 (cited in 7.N); Datta & Singh,
I, 184; and Kaye, above, who cite
F. Woepcke; Mémoire sur la
propagation des chiffres indiens; J.
Asiatique (6) 1 (1863) 27-529; p.
502, ??NYS.) The DSB entry indicates that the material is in Ibn Sina's Al‑Shifâ'
(The Healing) and there doesn't appear to be a translation. Suter 89 mentions some Latin translations
but I'm not clear whether they are this book or a related book.
Saidan's
discussion says Woepcke (p. 550 [sic]) construes ibn Sina as saying that the
method is Indian, but this is a contentious interpretation. Kaye, above, says Woepcke is wrong. Smith, History II 151, says the expression
"has been variously interpreted".
Bhaskara II.
Lilivati. 1150. Smith, History II 152, cites this in Taylor's
edition, p. 7, but the method is not in Colebrooke and neither Dickson nor
Datta & Singh cite it, so perhaps it is an addition in the text Taylor
used??
Fibonacci.
1202. Pp. 8-9, 20, 39, 45 (S:
24-26, 41, 67, 74) uses checks (mod 7,
9 and 11). On p. 8 (S: 24), he implies
that if the 'proof' is right, then the calculation is correct -- see comments
at Kūshyār above.
Maximus Planudes. Ψηφηφoρια
κατ' Ivδoυσ η
Λεγoμεvη Μεγαλη
(Psephephoria kat' Indous e Legomene Megale) (Arithmetic after the Indian method). c1300.
(Greek ed. by Gerhardt, Das Rechenbuch des Maximus Planudes, Halle,
1865, ??NYS [Allard, below, pp. 20-22, says this is not very good]. German trans. by H. Waeschke, Halle, 1878,
??NYS [See HGM II 549; not mentioned by
Allard].) Greek ed., with French
translation by A. Allard; Maxime Planude -- Le Grand Calcul selon des Indiens;
Travaux de la Faculté de Philosophie et Lettres de l'Univ. Cath. de Louvain --
XXVII, Louvain‑la‑Neuve, 1981.
Proofs by casting out 9s are given in the material on the operations of
arithmetic.
Narayana Pandita (= Nārāyaņa Paņdita
[NOTE: ņ denotes n with an overdot and the d
should have an underdot.]).
Gaņita[NOTE: ņ denotes
n with an underdot.] Kaumudī (1356). Edited by P. Dvivedi, Indian Press, Benares,
1942. Introduction in English,
p. xv, discusses the material.
Allows any modulus. (English in
Datta & Singh I 183.)
The Treviso Arithmetic = Larte de labbacho. Op. cit. in 7.H. 1478. F. 4v onward (p. 46
in Swetz) uses casting out 9s as a check on many examples. Swetz (p. 189) refers to Avicenna and
the Hindus. On f. 10v (Swetz
p. 59), the anonymous author says that proving a subtraction by addition "is
more rapid and also more certain than the proofs by 9s" and he makes
similar statements regarding multiplication and division.
On
p. 323 of his Isis article, Smith says the author "gives a proof by
casting out sevens". This would be
on or near f. 17v (Swetz p. 73). I
can find nothing of the sort -- the author has an example of multiplication
by 7,
but he checks it by casting out
9s.
Borghi.
Arithmetica. 1484. Ff. 8r-9r (1509: ff. 9r-9v). Casting out
7s and 9s. This is applied over
the next few sections, but I don't see any indication that casting out 9s
is not a certain test. However
he uses casting out 7s more often than 9s which may indicate
that he was aware that 7s is a more secure test than 9s.
Chuquet. 1484. Triparty, part 1. English in FHM 41-42.
"There are several kinds of proofs such as the proof by 9,
by 8, by 7, and so on by other individual figures down
to 2,
.... Of these only the proof
by 9,
because it is easy to do, and the proof by 7, because it is even
more certain than that by 9 are treated here." He then notes that these proofs are not
always certain.
Pacioli. Summa. 1494.
Ff. 20v-23v. Discusses casting
out 9s
and 7s and notes that these tests are not sufficient.
Apianus. Kauffmanss
Rechnung. 1527. Gives numerous examples of testing by 9s,
and also by 8s, 7s
and 6s, in his sections on the four arithmetic
operations and also under arithmetic progressions.
Recorde. First
Part. 1543. Discusses the proof by nines in his chapters on: addition, ff. D.i.r - D.iii.v (1668: 29-32:
The proof of Addition); subtraction,
ff. F.iii.r - F.iiii.r (omitted in 1668);
multiplication, ff. G.vi.r - G.vi.v (1668: 70-72: Proof of
Multiplication); and division, ff.
H.iii.v - H.iiii.v (1668: 82-84: Proof of Division).
Hutton. A Course of
Mathematics. 1798? 1833 & 1857: 6-12. In his discussion of the basic arithmetic
operations, we find on p. 7 under To
Prove Addition, "Then, if the
excess of 9's in this sum, ...., be equal to the excess of 9's in the total sum
..., the work is right." A
footnote explains the idea and is less clear as to the direction of implication
being asserted: "it is plain that this last excess must be equal to the
excess of 9's contained in the total sum". The note concludes: "This rule was first given by Dr Wallis
in his Arithmetic, published in the year 1657." However, Hutton does not mention the rule under subtraction and
under multiplication on pp. 10-11, he says the "remainders must be equal
when the work is right." All in
all, it seems that he is surprisingly unclear for his time.
Boy's Own Book.
Boy's Own Book. 1843
(Paris): 342. "To make any number
divisible by nine, by adding a figure to it." Only appends or inserts the necessary digit. = Boy's Treasury, 1844, p. 299.
Lewis Carroll. Diary
entry for 8 Feb 1856, In Carroll-Gardner,
pp. 43-44. Observes that a number minus
its reverse is divisible by nine, so you can ask someone to delete a digit and
show you the rest and you announce the deleted digit. Gardner points out that one can subtract any permutation of the
original digits.
Mittenzwey.
1880.
Parlour Games for Everybody. John Leng, Dundee & London, nd [1903 -- BLC], p. 42: The
expunged figure. Have someone write a
number, form the sum of its digits and subtract that from the given
number. Get him to strike out any
figure and tell you the sum of the remaining figures. Says that if the result is a multiple of nine, then a nine was
struck out.
Peano. Giochi. 1924.
Prob. 50, p. 13. Take x,
form 10x and subtract x. Cancel a non-zero
figure from the result and tell me the other figures. I will tell what number you cancelled.
See Tropfke 628.
These also occur in 7.M, 7.S.1 and
10.A.
I am starting to include early
problems which involve interpolation in a geometric series here -- these were
normally solved by linear interpolation.
From about 1400, such problems arise in compound interest but I will
omit most such problems. See Chuquet
here and Chiu Chang Suan Ching &
Cardan in 10.A.
H. V. Hilprecht.
Mathematical, Metrological and Chronological Tablets from the Temple
Library of Nippur. Univ. of
Pennsylvania, Philadelphia, 1906. Pp.
13, 28‑34, 62, 69, pl. 15, PL. IX, are about a tablet which has a
geometric progression from c‑2300.
The progression is double: an
= 125 * 2n and 604/an for
n = 0, 1, ..., 7.
There is no summation.
Tablet K 90 of the British Museum. A moon tablet deciphered by Hincks containing 5, 10, 20, 40, 80 followed by
80, 96, 112, 128, ..., 240.
Described in The Literary Gazette (5 Aug 1854) -- ??NYS. Cited and described in: Nicomachus of Gerasa: Introduction to
Arithmetic; Translated by Martin Luther D'Ooge, with notes by Frank Egleston
Robbins and Louis Charles Karpinski; Macmillan, London, 1926; p. 12.
Euclid. IX: 35,
36. This gives the general rule for the
sum of a geometric progression.
The Friday Night Book
(A Jewish Miscellany). Soncino
Press, London, 1933. Mathematical
Problems in the Talmud, pp. 132-133.
The Talmud says that any one visiting a sick person takes away a
sixtieth of his illness. This led to
the question of what happened if sixty people visited the person. This was answered by saying that the visitor
took away a sixtieth of the illness that the person had, i.e. the patient was
left with 59/60 of his illness, so that 60
visits left him still with
(59/60)60 of his
illness. The text quoted in the source
says this 'is still approximately one-quarter of the original illness', but it
is .36479. The modern compiler adds that 'The Talmud does not indicate the
method of working out the remainder after each visitor, and it is to be noted
that although the summation of series was known to the Greeks, there is no
mention of it anywhere in the Talmud.'
To me, this shows some confusion as I don't see that summation of series
is needed! [The Talmud was compiled in
the period -300 to 500, but nothing in the source gives any more precise dating
for this problem.]
Zhang Qiujian . Zhang
Qiujian Suan Jing. Op. cit. in
7.E. 468, ??NYS. Mikami 42 gives: "A horse, halving its speed every day,
runs 700 miles in 7 days. What are his
daily journeys?" -- i.e.
x*(1 + 1/2 + ... + 1/64) = 700. Solved by adding up.
Mahavira. 850.
Fibonacci.
1202. Pp. 313‑316 (S:
439-443). Man has 100
and gives away 1/10 of his wealth 12 times. This has been described under 7.E.
Lucca 1754.
c1330. F. 10v, pp. 36‑37. Computes
240 & 2100 by repeated squaring.
Columbia Algorism.
c1350. Prob. 63, pp. 84‑85. Same as the Fibonacci, but he converts to
pounds, shillings and pence!
Folkerts.
Aufgabensammlungen. 13-15C. Four sources with progressions with ratio 7
and seven sources with ratio 12.
Chuquet. 1484. He gives a number of such problems -- see
also 7.E.
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 29, f. 66v. Same as Fibonacci.
(H&S 59-60.)
Buteo.
Logistica. 1559.
Jacques Chauvet Champenois.
Les Institutions de L'Arithmetique.
Hierosme de Marnef, Paris, 1578, p. 70.
??NYS. Problem of tailor and
robe involving 4888 divided by
2 twenty times. (French quoted in H&S 14‑15.)
van Etten.
1624. Prob. 87 (84): Des
Progressions & de la prodigieuse multiplication des animaux, des Plantes,
des fruicts de l'or & de l'argent quand on va tousjours augmentant par
certaine proportion, pp. 111‑118
(177‑183). Numerous examples
including horseshoe problem and chessboard problem, with ratios 1000, 4,
50. Henrion's Notte, p. 38,
observes that there are many arithmetical errors which the reader can easily
correct. In part X: Multiplication des
Hommes, he considers one of the children of Noah, says a generation takes 30
years and that, when augmented to the seventh, one family can easily produce
800,000 souls. The 1674 English ed.
has: "... if we take but one of the Children of Noah, and suppose
that a new Generation of People begin at every 30 years, and that it be
continued to the Seventh Generation, which is 200 years; ... then of one only
Family there would be produced 111000 Souls, 305 to begin the World: ... which
number springing onely from a simple production of one yearly ...."
W. Leybourn.
Pleasure with Profit. 1694. Chap. VI, pp. 24-28: Of the Increase of
Swine, Corn, Sheep, &c. Examples
with ratios 4, 40, 2, 1000, 2, mostly taken from van Etten. Then art. VI: Of Men, discusses the
repopulation of the world from Noah's children: "... if we take but one of
the Children of Noah, and suppose that a New Generation of People begin
at every 30 years, and that it be continued to the seventh Generation, which is
210 years; ... then, of one only family there would be produced 111305, that
is, One hundred and eleven thousand, three hundred and five Souls to
begin the World .... ... such a number
arising only from a simple production of only One yearly ...." I cannot work out how 111305 arises -- the
fact that he spells it out makes it unlikely to be a misprint.
Ozanam. 1694. Prob. 8, 1696: 33-35; 1708: 29-32. Prob. 11, 1725: 68-75.
Section II, 1778: 68-74; 1803:
70-76; 1814: ??NYS; 1840: 34-36. A discussion of geometric progression and a mention of 1, 2, 4, .... 1778 et seq. also mention 1, 3, 9, ....
Ozanam. 1725. Prob. 11, questions 6 & 7, 1725: 79‑82. Prob. 3, parts 1-3, 1778: 80-82; 1803: 82-84; 1814: 72-75; 1840:
38-39. Examples of population growth in
Biblical and biological contexts. In
1725, he has ratios of 2, 50, 3, 4, 1000, The examples vary a bit between 1725
and 1778.
Walkingame. Tutor's
Assistant. 1751. The section Geometrical Progression gives
several problems with powers of 2 and the following less common types.
Vyse. Tutor's
Guide. 1771? The section Geometrical Progression, 1793: 35, pp. 138-143;
1799: XXXV, pp. 146-151 & Key pp. 190-192, gives several examples with
doublings and triplings as well as examples with ratios of 3/2
and 10. There is a major error in the solution of
prob. 7, to find 2 + 6 +
18 + ... + 2 x 319.
Pike.
Arithmetic. 1788. Pp. 237-239. Numerous fairly standard examples, mostly doubling, but with
examples of powers of 3 and of
10 and the following. D. Adams, 1835, copies two examples, but not
the following.
Pp.
239-240, no. 8. One farthing placed
at 6%
compound interest in year 0 is worth what after 1784
years? And supposing a cubic
inch of gold is worth £53 2s 8d, how much gold does this make? This is very close to 2150 farthings and makes about
4 x 1014 solid gold
spheres the size of the earth!
Eadon.
Repository. 1794. P. 241, ex. 3. Doubling 20 times from a farthing.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795
writer, Twickenham, 1995. P. 100. 10 + 102 + ... + 1011 grains of wheat, converted to bushels and
value at 4s per bushel.
(Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255 &
3:8 (Aug 1889) 351. (This is
undoubtedly reprinted from Boy's Own Magazine 1 (1863).) Mathematical question 59. Seller of
12 acres asks 1
farthing for the first acre,
4 for the second acre, 16 for the third acre, .... Buyer offers
£100 for the first acre, £150 for the second acre, £200
for the third acre, .... What is
the difference in the prices asked and offered? Also entered in 7.AF.
Lewis Carroll.
Sylvie and Bruno Concluded.
Macmillan, London, 1893. Chap.
10, pp. 131‑132. Discusses
repeated doubling of a debt each year as a way of avoiding paying the debt --
"You see it's always worth while waiting another year, to get twice
as much money!"
= Carroll-Wakeling II, prob. 5: A new way to pay old debts, pp. 9
& 66, where Wakeling adds some problems based on repeated doubling and
gives the chessboard problem.
7.L.1. 1 + 7 + 49 + ... & ST. IVES
See Tropfke
629.
Papyrus Rhind, c‑1650, loc. cit. in 7.C. Problem 79, p. 112 of vol. 1 (1927) (= p. 59
(1978)). 7 + 49 + 343 + 2401 +
16807. (Sanford 210 and H&S 55 give
Peet's English.) Houses, cats, mice,
ears of spelt, hekats.
L. Rodet. Les
prétendus problèmes d'algèbre du manuel du calculateur Égyptien (Papyrus
Rhind). J. Asiatique 18 (1881) 390‑459. Appendice, pp. 450‑454. Discusses this problem and its appearance in
Fibonacci (below).
F. Cajori. History
of Mathematics. 2nd ed., Macmillan,
1919; Chelsea, 1980. P. 90 gives the legend that Buddha was once
asked to compute 717.
Shakuntala Devi. The
Book of Numbers. Orient Paperbacks
(Vision Books), Delhi, 1984. This gives
more details of the Buddha story, saying it occurs in the Lalitavistara and
Buddha finds the number of atoms (of which there are seven to a grain of dust)
in a mile, obtaining a number of 'about 50 digits'. Note: 758 =
1.04 x 1049.
Alcuin. 9C. Prob. 41: Propositio de sode et scrofa. This has sows which produce 7
piglets, but this results in a GP of ratio 8.
Fibonacci.
1202.
Munich 14684.
14C. Prob. XXXI, pp. 83‑84. Refers to
7 + 49 + ... + 117649.
AR. On p. 227, Vogel
refers to an example in CLM 4390 which has not been published.
Peter van Halle. MS.
3552 in Royal Library Brussels, beginning "Dit woort Arithmetica coomt
uuter griexer spraeken ...."
1568. F. 23v. "There were 5 women and each woman had
5 bags but in each bag were 5 cats and each cat had 5 kittens question how many
feet are there to jump with?" Copy
of original Dutch text and English translation provided by Marjolein Kool, who
notes that van Halle only counts the feet on the kittens.
Josse Verniers. MS.
684 in University Library of Ghent, beginning "Numeration heet tellen ende
leert hoemen die ghetalen uutghespreken ende schrijven ...." 1584.
F. 7v. "Item there is a
house with 14 rooms and in each room are 14 beds and each bed lay 14 soldiers
and each soldier has 14 pistols and in each pistol are 3 bullets Question when
they fire how many men, pistols and bullets are there" Copy of original Dutch text and English
translation provided by Marjolein Kool.
Harley MS 7316, in the BM.
c1730. ??NYS -- quoted in: Iona & Peter Opie; The Oxford Dictionary
of Nursery Rhymes; OUP, (1951); 2nd
ed., 1952, No. 462, p. 377. The Opies
give the usual version with 7s, but their notes quote Harley MS 7316
as: "As I went to St. Ives I met
Nine Wives And every Wife had nine Sacs And every Sac had nine Cats And every
Cat had Nine Kittens." The Opies'
notes also cite Mother Goose's Quarto (Boston, USA, c1825), a German version
with 9s and a Pennsylvania Dutch version with 7s.
Halliwell,
James Orchard. Popular Rhymes &
Nursery Tales of England. John Russell
Smith, London, 1849. Variously
reprinted -- my copy is Bodley Head, London, 1970. P. 19 refers to "As I was going to St. Ives" in MS.
Harl. 7316 of early 18C, but doesn't give any more details.
D. Adams. Scholar's
Arithmetic. 1801.
As
I was going to St. Ives,
I
met seven wives,
Every
wife had seven sacks,
Every
sack had seven cats,
Every
cat had seven kits,
Kits,
cats, sacks and wives,
How
many were going to St. Ives?
No solution.
Child. Girl's Own
Book. 1842: Enigma 35, pp.
233-234; 1876: Enigma 27, pp.
196-197. "As I was going to St.
Ives, I chanced to meet with nine old
wives: Each wife had nine sacks, Each sack had nine cats, Each cat had nine kits; Kits, cats, sacks, and wives, Tell me how many were going to St.
Ives?" Answer is "Only myself. As I met all the others, they of course were coming from St.
Ives." The 1876 has a few
punctuation changes.
=
Fireside Amusements. 1850: No. 48, pp.
114 & 181; 1890: No. 34, p.
102. The 1850 solution is a little
different: "Only myself. As I
was going to St. Ives, of course all the others were coming from
it." The 1890 solution differs a
little more: "Only myself. As I
was going to St. Ives, all the others I met were coming from it."
Kamp. Op. cit. in
5.B. 1877. No. 20, p. 327. 12 women, each with 12 sticks, each with 12
strings, each with 12 bags, each with 12 boxes, each with 12
shillings. How many shillings?
Mittenzwey.
1880. Prob. 20, pp. 3 &
59; 1895?: 24, pp. 9 & 63; 1917: 24, pp. 9 & 57. Man going to Stötteritz meets 9
old women, each with 9 sacks, each with 9 cats, each with 9
kittens. How many were going to
Stötteritz? Answer is one.
Heinrich Voggenreiter.
Deutsches Spielbuch Sechster
Teil: Heimspiele. Ludwig Voggenreiter,
Potsdam, 1930. Pp. 105-106: Wieviel
Füsse sind es? Hunter going into the
woods meets an old woman with a sack which has six cats, each of which has six
young. How many feet all together were going
into the woods? His answer is 'Only
one', which has confused 'feet' with 'walker' -- this may be an obscure German
usage, but I can't find it in my dictionaries.
Joseph Leeming.
Riddles, Riddles, Riddles.
Franklin Watts, 1953; Fawcett
Gold Medal, 1967. P. 150, no. 11:
"As I was going to St. Ives, I chanced to meet with nine old wives;
...." [I don't recall any other
contemporary examples using 9 as multiplier.]
Mary & Herbert Knapp.
One Potato, Two Potato ... The
Secret Education of American Children.
Norton, NY, 1976. Pp. 107-108
gives a modern New York City version:
"There once was a man going to St. Ives Place. He had seven wives; each wife had seven sacks; each sack had seven cats; each cat had seven kits. How many altogether were going to St. Ives
Street? One." St. Ives has become attached to a location
in New York!
Colin Gumbrell.
Puzzler's A to Z. Puffin,
1989. Pp. 9 & 119: As I was going
...
"As I
was going to St Ives, I met a man with
seven wives; And every wife had seven
sons; But they were not the only
ones, For every son had seven
sisters! Bewildered by so many misters And by so many misses too, I quickly cried: 'Bonjour! Adieu!'
And hurried to another street,
Away from all their trampling feet.
Now, here's the point that puzzles me yet: Just how many people had I met?" Answer is 106, or 64 if there are just seven girls who are
half-sisters to all the 49 sons.
About 2000, someone told me that the answer to the classic
St. Ives riddle ought to me 'None' as it asks how many of the kits, cats,
sacks, and wives were going to St. Ives.
Ed Pegg Jr, Marek Penszko and Michael Kleber circulated a
new version in early 2001 on the Internet.
Tim Rowett has adapted one of the St Ives postcards with this new text.
As
I was going to St Ives
I
met a man with seven wives
Every
wife had seven sacks
Every
sack had seven cats
Every
cat had seven kits
We
traded bits
Each
cat, sack, wife, and he
Took
a kit. The rest for me.
So
now I have a kit supply.
How
many kits did I just buy?
The St. Ives of the riddle is usually thought to be the one
in Cornwall, but there are also St. Ives in Cambridgeshire (near Huntingdon)
and in Dorset (near Ringwood) and a St. Ive in eastern Cornwall (near
Liskeard).
Darrell Bates. The
Companion Guide to Devon and Cornwall.
Collins, 1976. P. 301 says the
Cornish St. Ives is named for a 6C Irish lady missionary named St. Ia who
crossed the Irish Sea on a leaf. John
Dodgson [Home Town What's behind the
name; Drive Publications for the Automobile Association, Basingstoke, 1984; p.
40] agrees.
Gilbert H. Doble.
St. Ives Its Patron Saint and
its Church. Cornish Parish Histories
No. 4, James Lanham Ltd, St. Ives, 1939.
This says the lady was named Ya, Hya or Ia. The 'v' was probably inserted due to the influence of the Breton
St. Yves, with the first appearance of the form 'Ives' being in 1571. The earliest reference to St. Hya is c1300
and says she was an Irish virgin of noble birth, who found her friends had
departed for Cornwall. As she prayed
she saw a leaf in the water and touched, whereupon it grew big enough to
support her and she was wafted to Cornwall, arriving before her friends. Doble suggests that Ireland refers to Wales
here. The next mention is in 1478 and
says she was the sister of St. Erth and St. Uny and was buried at St. Hy. There is only one other old source, a
mention in 1538. There seems to be very
little, if anything, known about this saint!
The Michelin Green Guide to Brittany (3rd ed., Michelin et
Cie, Clermont‑Ferrand, 1995, pp. 178 & 237-238) describes the
Breton St. Yves, which I had assumed to be the eponym of the Cornish St.
Ives. St. Yves (Yves Helori (1253‑1303))
was once parish priest at Louannec, Côtes-d'Armor, where a chasuble of his is
preserved in the church. His tomb is in
the Cathedral of St. Tugdual in Tréguier, Côtes-d'Armor. His head is in a reliquary in the
Treasury. He worked as a lawyer and is
the patron saint of lawyers. He was
born in the nearby village of Minihy-Tréguier and his will is preserved in the
Chapel there. The Chapel cemetery
contains a monument known as the tomb of St. Yves, but this is unlikely to
contain him. Attending his festival,
known as a 'pardon', is locally known as 'going to St. Yves' -- !! The Cornish and Breton stories may have
influenced each other.
See Høyrup in 7.L.2.a for other early examples of doubling
30 times.
Chiu Chang Suan Ching.
c‑150? Chap. III, prob. 4,
pp. 28‑29. Weaver weaves
a (1 + 2 + 4 + 8 + 16), making
5 in all. (English in Needham, pp. 137‑138. Needham says this problem also occurs in Sun
Tzu (presumably the work cited in 7.P.2, 4C), ??NYS.)
Alcuin. 9C. Prob. 13: Propostio de rege et de ejus
exercitu. 1 + 1 + 2
+ 4 + ... + 229 = 230. Calculations are suppressed in the Alcuin text, but given in the
Bede. Murray 167 wonders if there is
any connection between this and the Chessboard Problem (7.L.2.a).
Bhaskara II. Lilavati. 1150.
Chap. V, sect. II, v. 128. In
Colebrooke, pp. 55‑56.
2 + 4 + ... + 230.
W. Leybourn.
Pleasure with Profit. 1694. See in 7.L.
Wells. 1698. No. 103, p. 205. Weekly salary doubles each week for a year: 1 + 2 + 4 + ... + 251.
Ozanam. 1725. Prob. 11, question 5, 1725: 78‑79. 1 + 2 + ... + 231.
Walkingame. Tutor's
Assistant. 1751. The section Geometrical Progression gives
several problems with straightforward doublings -- see 7.L and 7.L.2.b for some
more interesting examples.
Vyse. Tutor's
Guide. 1771? Same note as for Walkingame.
Eadon.
Repository. 1794. P. 241, ex. 3. Doubling 20 times from a farthing.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795 writer,
Twickenham, 1995. P. 115. For
20 horses, is starting with a
farthing and doubling up through the
19th horse, with the 20th
free, more or less expensive than
£20 per horse?
Boy's Own Book.
Curious calculation.
1868: 433. 1 + 2 + ... + 251 pins would weigh 628,292,358 tons and
require 27,924 ships as large as the Great Eastern to carry
them.
Arithmetical [sic] progression. 1868: 433. 1 + 2 + ... +
299 farthings. Answer is wrong.
Ripley's Believe It or Not, 4th Series, 1957. P. 15 asserts that the Count de Bouteville
directed that his widow, age 20, should receive one gold piece during the first
year of widowhood, the amount to be doubled each successive year she remained
unmarried. She survived 69 years
without marrying! Ripley says the Count
'never suspected the cumulative powers of arithmetical [sic!] progression'.
See Tropfke
630. See 5.F.1 for more details of
books on the history of chess.
Jens Høyrup.
Sub-scientific mathematics: Undercurrents and missing links in the
mathematical technology of the Hellenistic and Roman world. Preprint from Roskilde University, Institute
of Communication Research, Educational Research and Theory of Science, 1990,
Nr. 3. (Written for: Aufsteig und
Niedergang der römischen Welt, vol. II 37,3 [??].) He discusses this type of problem, citing al-Uqlīdisī
[Abû al-Hassan [the H should have an underdot] Ahmad[the h
should have an underdot] Ibn Ibrâhîm al-Uqlîdisî. Kitâb al Fuşûl [NOTE: ş
denotes an s with an underdot.] fî al‑Hisâb[the H
should have an underdot] al‑Hindî.. 952/953.
MS 802, Yeni Cami, Istanbul.
Translated and annotated by A. S. Saidan as: The Arithmetic of Al‑Uqlīdisī; Reidel, 1978. ??NYS.
P. 337] as saying: "this is a question many people ask. Some ask about doubling one 30 times, and
others ask about doubling it 64 times".
Høyrup says that doubling 30 times is found in Babylonia, Roman Egypt,
Carolingian France, medieval Damascus and medieval India.
On
pp. 23-24, he describes the first two examples mentioned above and then
mentions Alcuin and al-Uqlīdisī.
The last example is probably Bhaskara II.
A
cuneiform tablet from Old Babylonian Mari [Denis Soubeyran; Textes
mathématiques de Mari; Revue d'Assyriologie 78 (1984) 19-48. ??NYS.
P. 30] has, in Høyrup's translation: "To one grain, one grain has
been added: Two grains on the first
day; Four grains on the second
day; ...." this goes on to 30 days. The larger amounts are not computed as
numbers, but converted to larger units.
Old Babylonian is c-1700.
Papyrus
Ifao 88 [B. Boyaval; Le P. Ifao 88: Problèmes de conversion monétaire;
Zeitschrift für Papyrologie und Epigraphik 7 (1971) 165-168, Tafel VI, ??NYS]
starts with 5 and doubles 30 times, again using larger units for the later
stages. Høyrup says this is a
Greco-Egyptian papyrus 'probably to be dated to the Principate but perhaps as
late as the fourth century' -- I am unable to determine what the Principate
was.
Perelman. FFF. 1934.
1957: prob. 52, pp. 74-80; 1979:
prob. 55, pp. 92-98. = MCBF,
prob. 55, pp. 90-98. This
describes a Roman version where the general Terentius can take 1
coin the first day, 2 the second day, 4 the third day, ..., until he can't carry any more, which
occurs on the 18th day.
A footnote says this is a translation "from a Latin manuscript in
the keeping of a private library in England." ??
Murray mentions the problem on pp. 51‑52, 155, 167,
182 and discusses it in detail in his Chapter XII: The Invention of Chess in
Muslim Legend, pp. 207‑219. He
discusses various versions of the invention of chess, some of which include the
doubling reward. He describes the
doubling legends in the following.
al‑Ya‘qûbî (c875).
al-Maş‘udi [NOTE:
ş denotes an s
with an underdot.] (943).
Firdawsî's Shâhnâma (1011).
Kitâb ash‑shaţranj [NOTE: ţ
denotes a t with an underdot.] [= AH] (1141) as AH f. 3b
(= Abû Zakarîyâ [= H] f. 6a).
BM MS Arab. Add. 7515 (Rich) [= BM] (c1200?).
von Eschenbach (c1220).
BM Cotton Lib. MS Cleopatra, B.ix [= Cott.] (13C).
ibn Khallikan (1256).
Dante (1321).
Shihâbaddîn at‑Tilimsâni [= Man.] (c1370), which gives
five versions.
Kajînâ [= Y] (16C?).
Murray 218
mentions two treatises on the problem:
Al‑Missisî.
Tad‘îf buyût ash‑shaţranj [Note: ţ denotes a t with
an underdot and the d should have a dot under it.]. 9C or 10C.
Al‑Akfânî.
Tad‘îf ‘adad ruq‘a ash‑shaţranj [Note: ţ
denotes a t with an underdot and the first d
should have a dot under it.] .
c1340.
On
p. 217, Murray gives 10 variant spellings of Sissa and feels that Bland's
connection of the name with Xerxes is right.
On
p. 218, he says the reward of corn would cover England to a depth of 38.4
feet.
Murray 218.
"This calculation is undoubtedly of Indian origin, the early Indian
mathematicians being notoriously given to long‑winded calculations of the
character." He suggests the
problem may be older than chess itself.
Al‑Ya‘qûbî.
Ta’rîkh. c875. Ed. by Houtsma, Leyden, 1883, i, 99‑105. ??NYS.
Cited by Murray 208 & 212.
"Give me a gift in grains of corn upon the squares of the
chessboard. On the first square one
grain (on the second two), on the third square double of that on the second,
and continue in the same way until the last square." [Quoted from Murray
213.]
al-Maş‘udi [NOTE:
ş denotes an s
with an underdot.] (= Mas'udi = Maçoudi). Murûj adh‑dhahab [Meadows of
gold]. 943. Translated by: C. Barbier
de Meynard & P. de Courteille as:
Les Prairies d'Or; Imprimerie Impériale, Paris, 1861. Vol. 1, Chap. VII, pp. 159‑161. "The Indians ascribe a mysterious
interpretation to the doubling of the squares of the chessboard; they establish a connexion between the First
Cause which soars above the spheres and on which everything depends, and the
sum of the square of its squares. This
number equals
18,446,744,073,709,551,615 ...."
[Quoted from Murray 210. The
French ed. has two typographical errors in the number.] No mention of the Sessa legend.
Muhammad ibn Ahmed Abû’l-Rîhân
(the h
should have an underdot) el-Bîrûnî
(= al‑Bîrûnî =
al-Biruni). Kitâb al‑âtâr
al‑bâqîya ‘an al‑qurûn al‑halîya (= al‑Âthâr al‑bâqiya min al‑qurûn al
khâliya = Athâr‑ul‑bákiya) (The Chronology of Ancient Nations). 1000.
Arabic (and/or German??) ed. by E. Sachau, Leipzig, 1876 (or
1878??), pp. 138‑139. ??NYS. English translation by E. Sachau,
William H. Allen & Co., London, 1879, pp. 134‑136. An earlier version is: E. Sachau; Algebraisches über das Schach bei
Bîrûnî; Zeitschr. Deut. Morgenländischen Ges. 29 (1876) 148‑156, esp. 151‑155.
Wieber, pp. 113‑115, gives
another version of the same text.
Computes 1 + 2 + 4 + ... + 263 as 264
‑ 1 by repeated squaring. Doesn't mention Sessa. He shows the total is 2,305
mountains. "But
these are (numerical) notions that the earth does not contain." Murray 218 gives: "which is more
than the world contains." but I'm not sure if al-Biruni means the
mountains or the numbers are more than earth can contain.
BM MS Arab. Add. 7515 (Rich). Arabic MS with the spurious title "Kitâb ash‑Shaţranj [NOTE: ţ
denotes a t with an underdot.] al Başrî [NOTE: ş
denotes an s with an underdot.]", perhaps c1200. Copied in 1257. Described
by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray on
p. 173. Murray denotes it BM.
Bland,
p. 26, says p. 6 of the MS gives the story of
Súsah ben Dáhir and the
reward. Bland, p. 62, says the various
forms of the name Sissah are corruptions of Xerxes. Forbes, pp. 74‑76, does not mention the story or the
reward.
Murray
217 says all the Arabic MSS include the reward problem as part of one of their
stories of the invention of chess, but on pp. 173 & 211‑219, he
doesn't mention the story in this MS specifically. However, on p. 173, he notes that the spurious first page gives
the calculation in 15C Arabic and again in Turkish.
Fibonacci.
1202. Pp. 309‑310 (S:
435-437). He induces the repeated
squaring process and gets 264
‑ 1. He computes the equivalent
number of shiploads of grain -- there is a typographic error in his result.
Wolfram von Eschenbach.
Willehalm. c1220. Ed. by Lachmann, p. 151, ??NYS -- quoted by
Murray 755. "Ir hers mich bevilte,
der Zende ûz zwispilte ame schâchzabel ieslîch velt mit cardamôm."
Murray 755 gives several other medieval European references.
(Al-Kâdi Shemseddîn Ahmed) Ibn Khallikan. Entry for:
Abû Bakr as‑Sûli. In: Kitab wafayât al‑a‘yân. 1256.
Translated by MacGuckin de Slane as:
Biographical Dictionary; (London, 1843‑1871;) corrected reprint,
Paris, 1868. Vol. III, p. 69‑73. Sissah ibn Dâhir, King Shihrâm and the
chessboard on pp. 69‑71. An
interpolation(?) mentions King Balhait.
BM Cotton Lib. MS Cleopatra, B.ix. c1275. Anglo‑French
MS of c1275, described by Murray 579‑580, where it is denoted Cott. No. 18, f. 10a, gives doubling.
Dante. Divina
Commedia: Paradiso XXVIII.92.
1321. "Ed eran tante che'l
numero loro Piu che'l doppiar degli scacchi s'imila." [Quoted from Murray 755.]
Paolo dell'Abbaco.
Trattato di Tutta l'Arta dell'Abacho.
1339. Op. cit. in 7.E. B 2433 f. 21v has an 8 x 8
board with two columns filled in with powers of two, starting with 2.
Below he gives 264 and treats it as farthings and converts to
danari, soldi, libri??, soldi d'oro, libri?? d'oro and then a further step that
I cannot understand. No mention of
chess or a reward.
Thomas Hyde.
Mandragorias seu Historia Shahiludii, .... (= Vol. 1 of De Ludis Orientalibus, see 4.B.5 for vol. 2.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Prolegomena curiosa. The
initial material and the Prolegomena are unpaged but the folios of the
Prolegomena are marked (a), (a 1), ....
The material is on (d 1).r - (d
4).v, which are pages 25-32 if one starts counting from the beginning of the
Prolegomena. He mentions Wallis (see below) and arithmetic (sic!)
progressions and says the story is given in
al-Safadi (Şalâhaddîn aş‑Şafadî [NOTE: Ş,
ş denote S,
s, with an underdot and the h
should have an underdot.] =
al-Sâphadi = AlSáphadi) (d. 1363)
in his Lâmiyato ’l Agjam (variously printed in the text). This must be his Sharh [the h should have an underdot] Lâmîyat al‑‘Ajam of c1350.
Hyde gives some Arabic text and a Latin translation. Wallis gives the full Arabic text and
translation. This refers to Ibn
Khallikan. In his calculation, he uses
various measures until he takes 239 grains as a granary, then 1024
granaries (= 249 grains) as a city, so the amount on the 64th square is 16384 (= 214) cities, “but you know there are not so many cities in the whole
world". He then gives 264 - 1 correctly and converts it into cubic miles, but seems off by a
factor of ten -- see Wallis, below, who gives details of the units and
calculations involved, noting that al‑Safadi is finding the edge (=
height) of a square pyramid of the volume of the pile of wheat. Hyde then adds a fragment from a Persian
MS, Mu’gjizât, which gives the story with drachmas instead
of grains of wheat, but the calculations are partly illegible. In his main text, pp. 31-52 are on the
invention of the game and he gives various stories, but doesn't mention the
reward.
Folkerts.
Aufgabensammlungen. 13-15C. 7 sources.
Shihâbaddîn Abû’l‑‘Abbâs Ahmad [the h
should have an underdot] ibn Yahya [the
h should have an underdot] ibn
Abî Hajala [the H should have an underdot] at‑Tilimsâni
alH‑anbalî [the H should have an underdot]. Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shaţranj
[NOTE: ţ denotes a t with an underdot] (Book of the examples of
warfare in the game of chess). Copied
by Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî in 1446.
This is the
second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray. Described by Bland and Forbes, loc. cit. in
5.F.1 under Persian MS 211, and by Murray 175-177 (as Man) & 207‑219. Gives five versions of the chessboard
story. The first is that of ibn
Khallikan; others come from ar-Râghib's
Muhâdarât (the h and
d should have underdots)
al-Udabâ’; from Quţbaddîn [NOTE: ţ
denotes t with an underdot.] Muhammad (the h
should have an underdot) ibn ‘Abdalqâdir's Durrat al-Mudî’a (the
d should have an underdot) and from
al-Akfânî. One calculates in
lunar years and another version calculates in miles!!
Columbia Algorism.
c1350. Prob. 88, pp. 106‑107. Chessboard.
Uses repeated squaring. Copying
error in the final value.
Persian MS 211. Op.
cit. in 5.F.1. c1400. Bland, loc. cit., p. 14, mentions "the
well known story of the reward asked in grain". Forbes' pp. 64‑66 is a translation of the episode of the
Indian King Kaid's reward to Sassa. On
p. 65, Forbes mentions various interpretations of the total.
AR. c1450. Prob. 319, pp. 141, 180, 227. Chessboard, with very vague story.
Benedetto da Firenze.
Trattato di Praticha d'Arismetrica.
Italian MS, c1464, Plimpton 189, Columbia University, New York. ??NYS.
Chessboard. (Rara, 464‑465;
Van Egmond's Catalog 257-258.)
Pacioli. Summa. 1494.
Ff. 43r-43v, prob. 28. First
mentions 1, 2, 6, 18, 54, ..., where
each cell has double the previous total.
Then does usual chessboard problem, but with no story. Computes by repeated squaring. Converts to castles of grain. Shows how to do 1 + 2 + 6 + 18 + ...
for 64 cells and computes the result.
Muhammad ibn ‘Omar Kajînâ.
Kitab al‑munjih fî‘ilm ash‑shaţranj [NOTE: ţ
denotes a t with an underdot and the second h
should have an underdot.] (A book to lead to success in the knowledge of
chess). 16C? Translated into Persian by
Muhammad ibn Husâm ad‑Daula,
copied in 1612. Described by
Bland and Forbes and more correctly by Murray on p. 179, where it is identified
as MS BM Add. 16856 and denoted Y, since it was a present from Col. Wm. Yule.
Bland,
p. 20, mentions Sísah ben Dáhir al
Hindi and the reward claimed in
grain. "The geometrical
progression of the sixty four squares ... is computed here at full length,
commencing with a Dirhem on the first square, and amounting to two thousand
four hundred times the size of the whole globe in gold."
Forbes
describes this on pp. 76‑77 and in the note on p. 65, where he computed
the reward to make a cube of gold about 6 miles along an edge. He says the above Persian value is wrong
somewhere, but he hasn't been able to see the original. [I can't tell if he means the Persian or the
Arabic MS. If a dirhem was the size of
an English 2p coin or an American quarter, the reward is about 2 x 104 km3, compared to earth's volume of about 1012
km3. The reward would make a
cube about 27 km on an edge or about 17
miles on an edge.] Murray doesn't refer
to this MS specifically.
Ian Trenchant. L'Arithmetique. Lyons, 1566, 1571, 1578, ... ??NYS.
1578 ed., p. 297.
1, 3, 9, 27.
(H&S 91 gives French and English and says similar appear in Vander
Hoecke (1537), Gemma Frisius (1540) and Buteo (1556).)
Clavius. ??NYS. Computes number of shiploads of wheat
required. (H&S 56.)
van Etten.
1624. Prob. 87, pp. 111‑118
(not in English editions). Includes
chessboard as part XI, on p. 117.
Henrion's Notte, p. 38, observes that there are many arithmetical errors
in prob. 87 which the reader can easily correct.
John Wallis.
(Mathesis Universalis. T.
Robinson, Oxford, 1657. Chap. 31.) = Operum Mathematicoroum. T. Robinson, Oxford, 1657. Part 1, chap. 31: De progressione
geometrica, pp. 266-285. This includes
the story of Sessa and the Chessboard in Arabic & Latin, taken from al-Safadi,
c1350, giving much more text than Hyde (see above) does and explaining
the units and the calculation, showing that
al-Safadi's 60 miles should be
about 6 miles and this is the edge and height of a square pyramid of the same
volume as the wheat. He then computes
all the powers of two up to the 63rd
and adds them! John Ayrton Paris
[Philosophy in Sport made Science in Earnest; (Longman, Rees, Orme, Brown, and
Green; London, 1827);, 8th ed., Murray, 1857, p. 515] says Wallis got 9
English miles for the height and edge.
Anonymous proposer; a Lady, solver, with Additional
Solution. Ladies' Diary, 1709-10 = T.
Leybourn, I: 3-4, quest. 6. 64 diamonds
sold for 1 + 2 + 4 + ... + 263 grains of wheat. Suppose a pint of wheat contains 10,000 grains, a bushel of wheat
weighs half a hundredweight [a hundredweight is 112 lb], the value is 5s per
bushel, a horse can carry 1000 lb and a ship can carry 100 tons, then how much
is the payment worth and how many horses or ships would be needed to carry it?
Euler. Algebra. 1770.
I.III.XI: Questions for practice, no. 3, p. 170. Payment to Sessa, converted to value.
Ozanam‑Montucla.
1778. Prob. 3, 1778: 76-78; 1803: 78-81; 1814: 70-72; 1840: 37‑38. Problem wants the results of doublings, with
no story. Discussion gives the story
of Sessa, taken from Al‑Sephadi. Gives various descriptions of the pile of
grain, citing Wallis for one of these and says it would cover three times the
area of France to a depth of one foot.
Eadon.
Repository. 1794. Pp. 369-370, no. 11. Indian merchant selling 64
diamonds to a Persian king for grains of wheat, in verse. Supposing a pint holds 10000
grains and a bushel of 64 pints weighs 50 pounds, how many horse
loads (of a thousand pounds each) does this make? How many ships of
100 tons capacity?
Manuel des Sorciers.
1825. P. 84. ??NX
Chessboard.
The Boy's Own Book.
The sovereign and the sage.
1828: 182; 1828‑2: 238; 1829 (US): 106; 1855: 393; 1868:
431. Uses 63 doublings for no reason.
Hutton-Rutherford. A
Course of Mathematics. 1841? Prob. 26,
1857: 82. Reward to Sessa for
inventing chess. Takes a pint as 7680
grains and 512 pints as worth 27/6 to value the reward at 6.45 x 1012 £.
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. P. 200.
King conferring reward on a general.
Computes number of seers, which contain 15,360 grains, and the value if
30 seers are worth one rupee.
Nuts to Crack XIV (1845), no. 73. The sovereign and the sage.
Almost identical to Boy's Own Book.
Magician's Own Book.
1857. The sovereign and the
sage, pp. 242-243. A simplified version
of Ozanam-Montucla. = Book of 500
Puzzles, 1859, pp. 56-57. = Boy's Own
Conjuring Book, 1860, p. 213.
Vinot. 1860. Art. XVIII: Problème des échecs, pp.
36-37. Uses grains of wheat and says
there are 20,000 grains in a litre. Says the reward would cover France to a depth of 1.6 m.
He gives the area of France as
9,223,372 km2.
[Chambers].
Arithmetic. Op. cit. in
7.H. 1866? P. 267, quest. 54. Story
of Sessa with grains of wheat. Suppose 7680
grains make a pint and a quarter is worth £1 7s 6d, how much was
the wheat worth?
James Cornwell & Joshua G. Fitch. The Science of Arithmetic: .... 11th ed., Simpkin, Marshall, & Co.,
London, et al., 1867. (The 1888 ed. is
almost identical to this, so I suspect they are close to identical to the 2nd
ed. of 1856.) Exercises CXLIII, no. 6,
pp. 299 & 371. Chessboard problem
with no story, assumes 7680 grains to a pint.
Mittenzwey.
1880. Prob. 94, pp. 19 &
68; 1895?: 109, pp. 24 & 71; 1917: 109, pp. 22 & 68. King Shehran rewarding Sessa Eba Daher,
according to Asephad.
Cassell's.
1881. P. 101: Sovereign and the
sage. Uses sage's and king's common age
of 64, with no reference to chessboard.
Lucas.
L'Arithmétique Amusante.
1895. Le grains du blé de Sessa,
pp. 150-151. Says it would take 8 times
the surface of the earth to grow enough grain.
Berkeley & Rowland.
Card Tricks and Puzzles. 1892. The chess inventor's reward, pp. 112‑114. Assumes
7489 29/35 grains to the pound,
with 112 pounds to the cwt.,
20 cwt. to the ton and 1024
tons to the cargo, getting
1,073,741,824 cargoes, less one
grain. The number of grains is chosen
so that a ton contains exactly 224 grains of rice and the answer is 230 cargoes less one grain.
7.L.2.b. HORSESHOE NAILS PROBLEM
See Tropfke
632.
AR. c1450. Prob. 274, 317, 318, 353. Pp. 125, 140, 154, 180, 227.
Riese. Rechenung
nach der lenge .... 1525. (Loc. cit. under Riese, Die Coss.) Prob. 32, p. 20. 32 horseshoe nails.
Christoff Rudolff.
Künstliche rechnung mit der ziffern und mit den zal pfenninge. Vienna, 1526; Nürnberg, 1532, 1534, et seq.
F. N.viii.v. ??NYS. 32
horseshoe nails.
(H&S 56 gives German.)
Apianus. Kauffmanss
Rechnung. 1527. Ff. D.vi.r - D.vi.v. 32 horseshoe nails.
Anon. Trattato
d'Aritmetica, e del Misure. MS, c1535,
in Plimpton Collection, Columbia Univ.
??NYS. Horseshoe problem: 1 + 2 + 4 + ... + 223. (Rara, 482‑484, with reproduction on
p. 484.)
Recorde. First
Part. 1543. Ff. L.ii.r - L.ii.v (1668: 141-142: A question of an Horse). 24 horseshoe nails.
Buteo.
Logistica. 1559. Prob. 34, pp. 237-238. 24
horseshoe nails. (H&S 56.)
van Etten.
1624. Prob. 87, pp. 111‑118
(not in English editions).
Includes 24 horseshoe nails problem as part VII on p.
115. Henrion's Notte, p. 38, observes
that there are many arithmetical errors in prob. 87 which the reader can easily
correct.
Wells. 1698. No. 102, p. 205. 24 nails.
Ozanam. 1725. Prob. 11, question 4, 1725: 77‑78
& 80. Part of prob. 3, 1778:
79-80; 1803: 81; 1814: 72;
1840: 38. 24 nails -- first asks for the price of
the 24th, then the total.
Dilworth.
Schoolmaster's Assistant.
1743. P. 96, no. 1. 32
nails, starting with a farthing.
Walkingame. Tutor's
Assistant. 1751. Geometrical Progression, prob. 6, 1777: p.
95; 1835: p. 103; 1860: p. 123. 32 nails, one farthing for the first, wants total, which he gives in £ s/d.
Mair. 1765? P. 493, ex. III. "What will a horse cost by tripling the 32 nails in his
shoes with a farthing?" I.e., 32 horseshoe
nails, but with tripling!
Euler. Algebra. 1770.
I.III.XI.511, p. 166. Horse to
be sold for the value of 32 nails,
1 penny for the first, ....
Vyse. Tutor's
Guide. 1771? Prob. 2, 1793: p. 140; 1799: p. 148 & Key p. 190. 36
horseshoe nails. Want value of
last one, starting ¼, ½, 1, ....
Bullen. Op. cit. in
7.G.1. 1789. Chap. 31, prob. 1, p. 215.
Same as Walkingame.
Bonnycastle.
Algebra. 10th ed., 1815. P. 80, no. 6. Same as Walkingame.
Manuel des Sorciers.
1825. P. 84. ??NX
24 horseshoe nails.
The Boy's Own Book.
The horsedealer's bargain. 1828:
182; 1828-2: 238; 1829 (US): 106; 1843 (Paris): 346; 1855:
393‑394; 1868: 431-432. Wants value of 24th nail, starting with a farthing. = Boy's Treasury, 1844, p. 304. = de Savigny, 1846, p. 292: Le marché aux chevaux.
Nuts to Crack XIV (1845), no. 74. The horsedealer's bargain.
Almost identical to Boy's Own Book.
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. P. 199.
32 nails.
[Chambers].
Arithmetic. Op. cit. in
7.H. 1866? P. 268, quest. 64.
24 nails, starting with a
farthing. Finds total.
Mittenzwey.
1880. Prob. 93, pp. 19 &
68; 1895?: 108, pp. 23-24 &
71; 1917: 109, pp. 21-22 & 68. 32 horseshoe nails, starting at 1 pf.
Cassell's.
1881. P. 101: The horse‑dealer's
bargain. 24 nails, unclear, but uses
223 farthings as the
answer.
7.L.2.c. USE OF 1, 2, 4, ... AS WEIGHTS, ETC.
See Tropfke 633.
A special case of this is the use of
such amounts to make regular unit payments, e.g. rent of one per day. See:
Knobloch; Fibonacci; BR;
Widman; Tartaglia; Gori;
Les Amusemens.
Eberhard Knobloch.
Zur Überlieferungsgeschichte des Bachetschen Gewichtsproblems. Sudhoffs Archiv 57 (1973) 142-151. This describes the history of this topic and
7.L.3 from Fibonacci to Ozanam (1694).
He gives a table showing occurrences of: powers of two, powers of
three, weight problem, payment problem. I am not entirely clear what he means in the first three cases --
I would have two kinds of weight problem corresponding to the first two cases
and perhaps some of his references in the first case are listed under
7.L.2. However, the last case clearly
corresponds to the problem of making a payment of one unit per day as in
Fibonacci. He lists this as occurring
in Fibonacci, BR, Widmann
and Tartaglia and notes that Sanford, H&S 91, only
noticed Fibonacci. Knobloch notes that
Ball's citations are not very good and that Ahrens' note about them does not go
much deeper. I have a number of
references listed below which were not available to Knobloch.
Fibonacci.
1202. P. 298 (S: 421). Uses
5 ciphi of value 1, 2, 4, 8, 15 to pay a man at rate of
1 per day for 30
days.
BR. c1305. No. 93, pp. 112‑113. Use of
1, 2, 4 as payments at rate of
one per year for 7 years.
Widman. Op. cit. in
7.G.1. 1489. Ff. 138v-139r. ??NYS --
Knobloch says he uses values of 1, 2,
4, 8, 16 to pay for 31
days.
Pacioli. Summa. 1494.
Ff. 97v-98r, no. 35. Use five
cups to pay daily rent for 30 days.
Uses cups of weight 1, 2, 4, 8,
15. In De Viribus, c1500, F. XIIIv,
item 86 in the Indice for the third part is: De 5 tazze, diversi pesi ogni di
paga l'oste (Of 5 cups of diverse weights to pay the landlord every day) = Peirani 20, but at the end Pacioli says this problem is in 'libro nostro',
i.e. the Summa. Cf Agostini, p. 6.
Tartaglia. General
Trattato, 1556, part 2, book 1, chap. 16, art. 32: Di una particolar proprieta
della progression doppia geometrica, p. 17v.
Weights: 1, 2, 4, 8, ... (See MUS I 89.). Also does payments with
1, 2, 4, 8, 16, 29. Knobloch
also refers to art. 33-35 -- ??NYS -- and notes that the folios are
misnumbered, but miscites 'doppia' as 'treppia' here. This covers the powers of
3 also.
Buteo. Logistica. 1559.
Prob. 91, pp. 309-312. Use
of 1, 2, 4, 8, 16, ... as weights.
(Cited by Knobloch.)
Knobloch also cites
Ian Trenchant (1566), Daniel
Schwenter (1636), Franz van Schooten
(1657).
Gori. Libro di
arimetricha. 1571. Ff. 71r‑71v (p. 76). Use of cups weighing 1, 2, 4
to make all weights through
7, to pay for days at one per
day.
Bachet.
Problemes. 1612. Addl. prob. V & V(bis), 1612:
143-146; as one prob. V, 1624: 215-219; 1884: 154-156. Mentions
1, 2, 4, 8, 16
and cites Tartaglia, art. 32 only.
This was omitted in the 1874 ed.
Knobloch cites 1612, pp. 127 & 143-146, but but p. 127 is Addl.
prob. I, which is a Chinese Remainder problem?
van Etten/Henrion.
1630. Notte to prob. 53, pp. 20‑21. Refers to Bachet and compares with ternary
weights.
Ozanam. 1694.
Les Amusemens.
1749. Prob. 8, p. 128. Coins of value 1, 2, 4, 8, 15 to pay for
a room at a rate of 1 per day for
30 days.
The Bile Beans Puzzle Book.
1933. No. 42: Money
juggling. Place £1000 in 10 bags so any
amount can be paid without opening a bag.
Solution has bags of:
1, 2, 4, 8,
16, 32, 63, 127, 254, 493. I cannot see
why the solution isn't:
1, 2, 4, 8,
16, 32, 64, 128, 256, 489.
7.L.3. 1 + 3 + 9 + ... AND OTHER SYSTEMS OF WEIGHTS
See MUS I
88-98; Tropfke 633.
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 125ff., no. 43. ??NYS -- Hermelink, op. cit. in 3.A, and Tropfke 634-635 say this
gives 1, 3, 9, ..., 19683 = 39 to weigh up to 10,000.
Fibonacci.
1202.
Gherardi. Libro di
ragioni. 1328. P. 53.
Weights 1, 3, 9, 27, 80 to weigh up through 120.
Columbia Algorism.
c1350. Prob. 71, pp. 92‑93. Weights
1, 3, 9, 27.
AR. c1450. Prob. 127, pp. 67 & 182. 1, 3, 9, 27.
Chuquet. 1484. Prob. 142.
Pacioli. Summa. 1494.
Ff. 97r-97v, no. 34. General
discussion of 1, 3, 9, 27, 81, 243,
.... In De Viribus, c1500, F. XIIIv,
item 85 in the Indice for the third part is: De far 4 pesi che pesi fin 40 (To
make four weights which weigh to 40) =
Peirani 20, but at the end Pacioli says
this problem is in 'libro nostro', i.e. the Summa. Cf Agostini, p. 6.
Cardan. Practica
Arithmetice. 1539. Chap. 65, section 12, ff. BB.vii.r -
BB.vii.v (p. 136). Weights 1, 3, 9, 27, ....
Knobloch also cites:
Giel vanden Hoecke (1537); Gemma
Frisius (1540);
Michael Stifel (1553);
Simon Jacob (1565); Ian
Trenchant (1566);
Daniel Schwenter (1636);
Kaspar Ens (1628); Claude Mydorge
(1639); Frans van Schooten
(1657).
Tartaglia, 1556 -- see in 7.L.2.c.
Buteo.
Logistica. 1559. Prob. 91, pp. 309-312. Use of
1, 3, 9, 27, ... as
weights. (Cited by Knobloch.)
John [Johann (or Hanss) Jacob] Wecker. Eighteen Books of the Secrets of Art &
Nature Being the Summe and Substance of Naturall Philosophy, Methodically
Digested .... (As: De Secretis Libri XVII; P. Perna, Basel,
1582 -- ??NYS) Now much Augmented and
Inlarged by Dr. R. Read. Simon Miller,
London, 1660, 1661 [Toole Stott 1195, 1196];
reproduced by Robert Stockwell, London, nd [c1988]. Book XVI -- Of the Secrets of Sciences:
chap. 19 -- Of Geometricall Secrets: To poyse all things by four Weights,
p. 289. 1, 3, 9, 27; 1, 3, 9, 27, 81; 1, 3, 9, 27, 81, 243.
Cites Gemma Frisius.
Bachet.
Problemes. 1612. Addl. prob. V & V(bis), 1612: 143-146; as one prob. V, 1624: 215-219; 1884: 154‑156.
Weights: 1, 3, 9, 27, ..., and the general case via the sum of a
GP. In the 1612 ed., Bachet only does
the cases 40 and 121, then does the general case. Knobloch cites 1612, pp. 127 & 143-146,
but p. 127 is Addl. prob. I, which is a Chinese Remainder problem. He also says this is the first proof of the
problem, excepting Chuquet, though I don't see such in Chuquet.
van Etten.
1624. Prob. 53 (48), pp. 48‑49
(72). 1, 3, 9, 27; 1, 3, 9, 27, 81; 1, 3, 9, 27, 81, 243. Henrion's Notte, pp. 20‑21, refers to
Bachet and compares this with binary weights.
Ozanam. 1694.
Les Amusemens.
1749. Prob. 18, p. 140: Les Poids. Weights
1, 3, 9, 27, 81, 243.
Vyse. Tutor's
Guide. 1771? Prob. 2, 1793: p. 303; 1799: p. 316 & Key pp. 356-357. Weights
1, 3, 9, 27.
Bonnycastle.
Algebra. 1782. P. 202, no. 13. 1, 3, 9, 27, 81, 243, 729, 2187
to weigh to 29 hundred weight --
an English hundred weight is 112 pounds.
c= 1815: p. 230, no. 33. 1, 3,
9, 27, 81 to weigh to a hundred weight.
Eadon.
Repository. 1794. Pp. 297-298, no. 1. 1, 3, 9, ..., 313.
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
P. 50, no. 77: To find the least Number of Weights that will weigh from
One Pound to Forty. 1, 3, 9, 27.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions. No. 24, pp. 20 & 78‑79. 1, 2, 4, 8, 16, ... and
1, 3, 9, 27, 81, ....
Rational Recreations.
1824. Exer. 22, p. 131. 1, 3, 9, 27.
Endless Amusement II.
1826?
Young Man's Book.
1839. P. 242. To name five weights, .... Identical to Endless Amusement II, p. 201.
Boy's Own Book. 1843
(Paris): 346-347. "To find the
least number of weights which will weigh any intermediate weight, from one
pound to forty, exclusive of fractions.
Indicates that one can continue the progression." = Boy's Treasury, 1844, p. 304.
Family Friend 1 (1849) 150 & 178. Problems, arithmetical puzzles, &c.
-- 5. How can one divide 40
lb into four weights to weigh every value from 1 to 40?
Proposer says he can't do it.
Usual answer, but one solver gives weights 6, 10, 11, 13. However,
the latter weights will not weigh 22,
25, 26, 31, 32, 33, 35, 36, 37, 38, 39.
Boy's Own Book. To
weigh from one to forty pounds with four weights. 1855: 392;
1868: 430. 1, 3, 9,
27. No generalizations.
Magician's Own Book.
1857. The mathematical blacksmith,
p. 230. 1, 3, 9, 27 and this can be continued. = Boy's Own Conjuring Book, 1860, p. 200.
Todhunter. Algebra,
5th ed. 1870. Miscellaneous Examples, no. 175, p. 559. Show that
1, 1, 5, 5, 25, 25, 125, 125
can weigh any integral amount up to 312. No solution given.
F. J. P. Riecke. Op.
cit. in 4.A.1, vol. 3, 1873. Art. 3:
Die Zauberkarten, p. 13. Uses balanced
ternary for divination. See under 7.M.4
Mittenzwey.
1880.
P. A. MacMahon.
Certain special partitions of numbers.
Quart. J. Math. 21 (1886) 367‑373. Very technical.
P. A. MacMahon.
Weighing by a series of weights.
Nature 43 (No.1101) (4 Dec 1890) 113‑114. Less technical description of the above
work.
Lucas.
L'Arithmétique Amusante.
1895. Pp. 166-168. Notes that pharmacists, etc. use weights: 1, 1, 2, 5, 10, 10, 20, 50, 100, 100, 200,
500, 1000, 1000, 2000, 5000, ....
Discusses ternary.
Wehman. New Book of
200 Puzzles. 1908. P. 49.
1, 3, 9, 27, 81.
Ahrens. MUS I. 1910.
Pp. 88-98 discusses this and some generalizations like MacMahon's.
In 2002, Miodrag Novaković told me that a student had
told him how to determine twice as many integral weights with the same number
of weights. E.g., he used weights 2, 6, 18, 54 to determine integral weights
1, 2, ..., 81. One can get exact
balancing for the even values: 2, 4,
..., 80. The odd values fall between
two consecutive even amounts, so if a package weighs more than 6 but less than
8, we deduce it weighs 7!
7.M. BINARY SYSTEM AND BINARY RECREATIONS
The binary system has several
origins.
a) Egyptian & Russian peasant
multiplication.
b) Weighing -- see 7.L.2.c.
c) Binary divination -- see 7.M.4.
d) The works below.
See also: 5.E.2 for Memory Wheels;
5.F.4 for circuits on the n‑cube; 5.AA for an application to
card-shuffling; 7.AA.1 for Negabinary.
Anton Glaser.
History of Binary and Other Nondecimal Numeration. Published by the author, 1971; (2nd ed., Tomash, Los Angeles, 1981). General survey, but has numerous omissions
-- see the review by Knuth at Harriot, below, and MR 84f:01126. He has no references to early Chinese
material.
Shao Yung.
c1060. Sung Yuan Hsüeh An, chap.
10. Fu‑Hsi diagram of the 64
hexagrams of the I‑Ching, in binary order. A version appears in Leibniz‑Briefe 105 (Bouvet) Bl.
27r/28r in the Niedersächsische Landesbibliothek, Hannover. Needham, vol. 2, p. 341, notes that this had
only been published in Japanese and Chinese up to 1956. See Zacher & Kinzô below for
reproductions. Also reproduced in: E. J. Aiton; Essay Review [of Zacher,
below]; Annals of Science 31 (1974) 575‑578.
Chu Hsi. Chou I Pen
I Thu Shuo. 12C. Fu‑Hsi Liu‑shih‑ssu Kua
Tzhu Hsü (Segregation Table of the symbols of the Book of Changes) --
reproduced in Hu Wei's I Thu Ming Pien.
An illustration is given in Needham, vol. 2, fig. 41 = plate XVI, opp.
p. 276 -- he says it is based on the original chart of Shao Yung and that
Tshai Chhen (c1210) gave a simplified version.
Also in Kinzô and in Aiton & Shimao, below. Shows the alternation of 0 and 1 in each
binary place.
Thomas Harriot.
Unpublished MS. c1604. Described by J. W. Shirley; Binary numeration before Leibniz; Amer. J. Physics
19 (1951) 452‑454; and by D. E. Knuth; Review of 'History of Binary
and Other Nondecimal Numeration'; HM 10 (1983)) 236‑243. This shows some binary calculation. Shirley reproduces BM: Add MSS 6786, ff.
346v‑347r. Knuth cites 6782, 1r,
247r; 6786, 243v, 305r, 346v, 347r,
516v; 6788, 244v.
Francis Bacon. Of
the Advancement of Learning. 1605. ??NYS.
Describes his binary 5‑bit coding.
Francis Bacon. De
augmentis scientarum. 1623. ??NYS.
Full description of his coding.
He does not have any arithmetic content, so he is not really part of the
development of binary.
John Napier.
Rabdologiae. Edinburgh,
1617. ??NYS. Describes binary as far as extracting square roots. William F. Hawkins; The Mathematical Work of
John Napier (1550‑1617); Ph.D. thesis, Univ. of Auckland, 1982, ??NYS,
asserts this is THE invention of the binary system.
G. W. Leibniz. De
Progressione Dyadica. 3pp. Latin MS of
Mar 1679. Facsimile and translation
into German included in: Herrn von
Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, (1966), 2nd
corrected printing, 1969; facsimile between pp. 20 & 21 and
translation on pp. 42-47, with an essay by Hermann J. Greve: Entdeckung der
binären Welt on pp. 21-31. His first,
unpublished, MS on the binary system, showing all the arithmetic processes.
H. J. Zacher. Die
Hauptschriften zur Dyadik von G. W. Leibniz.
Klosterman, Frankfurt, 1973.
Gathers almost all the Leibniz material, notably omitting the above 1679
paper. He does reproduce the Fu‑Hsi
diagram sent by Bouvet (cf. Shao Yung above).
However Leibniz's letter of 2 Jan 1697 to Herzog Rudolf Augustus, in
which he gives his drawing of his plan for a medallion commemorating the binary
system, is now lost, but it was published in 1734.
G. W. Leibniz. Two
Latin letters on the binary system, 29 Mar 1698 & 17 May 1698,
recipient not identified, apparently the author of a book in 1694 which
occasioned Leibniz's correspondence with him.
Opera Omnia, vol. 3, 1768, pp. 183-190.
Facsimile and translation into German included in: Herrn von Leibniz' Rechnung mit Null und
Eins; Siemens Aktiengesellschaft, 2nd corrected printing, 1966; facsimile
between pp. 40 & 41 and translation on pp. 53-60. On p. 26, it seems to say the second letter
was sent to Johann Christian Schulenberg.
G. W. Leibniz.
Explication de l'arithmètique binaire.
Histoire de l'Academie Royale des Sciences 1703 (1705) 85-89. Facsimile and translation into German
included in: Herrn von Leibniz'
Rechnung mit Null und Eins; Siemens Aktiengesellschaft, 2nd corrected printing,
1966; facsimile between pp. 32 & 33 and translation on pp. 48-52. Illustrates all the arithmetic operations
and discusses the Chinese trigrams of 'Fohy' and his correspondence with Father
Bouvet in China.
G. W. Leibniz.
Letter of 1716 to Bouvet. ??NYS
-- cited in Needham, vol. 2, p. 342.
Fourth section is: Des Caractères dont Fohi, Fondateur de l'Empire
Chinois, s'est servi dans ses Ecrits, et de l'Arithmétique Binaire.
Gorai Kinzô. Jukyô
no Doitsu seiji shisô ni oyoboseru eikyô (Influence of Confucianism on German
Political Thought [in Japanese]).
Waseda Univ. Press, Tokyo, 1929.
??NYS. First publication of
Leibniz's correspondence with Bouvet which led to the identification of the Fu
Hsi diagram with the binary numbers.
Gives a redrawn Fu‑Hsi diagram and a segregation table.
E. J. Aiton & Eikon Shimao. Gorai Kinzô's study of Leibniz and the I Ching hexagrams. Annals of Science 38 (1981) 71‑92. Describes the above work. Reproduces Kinzo's Fu‑Hsi diagram and
segregation table.
Ahrens. MUS I. 1910.
24-104 discusses numeration systems in general and numerous properties
of binary and powers of 2.
Gardner. SA (Aug
1972) c= Knotted, chap. 2. General survey of binary recreations. The material in the book is much expanded
from the SA column.
See MUS I
61-72; S&B 104‑107, 111 &
135.
See also
4.A.4, 11.K.1.
In various
places and languages, the following names are used:
Chinese Rings
Chainese Rings [from
www.tama.or.jp/~tane, via Dic Sonneveld, 13 Nov 2002]
Cardan's Rings, but
Cardan called it Instrumentum ludicrum
Ryou-Kaik-Tjyo
or Lau Kák Ch'a =
Delay-guest-instrument
Kau Tsz' Lin Wain
= Nine connected rings
Chienowa = Wisdom rings
Kyūrenkan
= Nine connected rings
Lien nuan chhuan
[from www.roma.unisa.edu.au, via Dic Sonneveld, 13 Nov 2002]
Tarriers or Tarriours
Tiring Irons or Tyring Irons or Tarrying Irons
The Puzzling Rings
The Devil's Needle
Complicatus Annulis
[Wallis]
Baguenaudier spelled
various ways, e.g. Baguenodier
Juego del ñudo Gordiano
Меледа [Meleda]
Наран-шина [Naran-shina] (stirrup ring toy)
Zauberkette
Magische Ringspiel
Nürnberger Tand
Grillenspiel
Armesünderspiel
Zankeisen
Nodi d'anelli
S. N. Afriat. The
Ring of Linked Rings. Duckworth,
London, 1982. This is devoted to the
Chinese Rings and the Tower of Hanoi and gives much of the history.
Sun Tzu. The Art of
War. c-4C. With commentary by Tao Hanzhang.
Translated by Yuan Shibing.
(Sterling, 1990); Wordsworth,
London, 1993. In chap. 5: Posture of
Army, p. 109, the translator gives:
"It is like moving in a endless circle" In the commentary, p. 84, it says: "their interaction as endless as that of
interlocked rings." Though
unlikely to refer to the puzzle,, this and the following indicate that
interlocked rings was a common image of the time.
Needham, vol. 2, pp. 189-197, describes the paradoxes of Hui
Shih (‑4C). P. 191 gives
HS/8: Linked rings can be
sundered. On p. 193, Needham gives
several explanations of this statement and a reference to the Chinese Rings in
vol. III, but he does not claim this statement refers to the puzzle.
Stewart Culin.
Korean Games. Op. cit. in
4.B.5. Section XX: Ryou‑Kaik‑Tjyo
-- Delay Guest Instrument (Ring Puzzle), pp. 31‑32. Story of Hung Ming (181‑234) inventing
it. (Wei-Hwa Huang says this is
probably Kong Ming (= Zhuge Liang), a famous war strategist, to whom many
inventions were attributed.) States the
Chinese name is Lau Kák Ch'a (Delay Guest Instrument) or Kau Tsz' Lin Wain (Nine Connected Rings).
Says there a great variety of ring puzzles in Japan, known as Chie No Wa
(Rings of Ingenuity) and illustrates one, though it appears to be just
10 rings joined in a chain -- possibly a puzzle ring?? He says he has not found out whether the
Chinese rings are known in Japan -- but see Gardner below.
Ch'ung‑En Yü.
Ingenious Ring Puzzle Book. In
Chinese: Shanghai Culture Publishing
Co., Shanghai, 1958. English
translation by Yenna Wu, published by Puzzles -- Jerry Slocum,
Beverly Hills, Calif., 1981. P. 6. States it was well known in the Sung (960‑1279). [There is a recent version, edited into
simplified Chinese (with some English captions, etc.) by Lian Huan Jiu, with some commentary by
Wei Zhang, giving the author's
name as Yu Chong En, published by China Children's Publishing
House, Beijing, 1999.]
The Stratagem of Interlocking Rings. A Chinese musical drama, first performed
c1300. Cited in: Marguerite Fawdry; Chinese Childhood;
Pollock's Toy Theatres, London, 1977, pp. 70-72. Otherwise, Fawdry repeats information from Culin and the story
that it was used as a lock.
Needham. P. 111
describes the puzzle as known in China at the beginning of the 20C, but says
the origins are quite obscure and gives no early Chinese sources. He also cites his vol. 2, p. 191, for
an early possible reference -- see above.
Pacioli. De
Viribus. c1500. Ff. 211v-212v, Part 2, Capitulo CVII.
Do(cumento), cavare et mettere una strenghetta salda in al quanti anelli saldi.
dificil caso (Remove and replace a joined string a number of joined rings - a
difficult thing). = Peirani
290-292. Dario Uri says this describes
the Chinese Rings. It is hard to make
out, but it appears to have six rings.
Uri gives several of the legends about its invention and says Cardan
called it Meleda, but that word is not in Cardan's text. He lists 27 patents on the idea in five
countries.
Cardan. De
subtilitate. 1550. Liber XV.
Instrumentum ludicrum, pp. 294‑295. = Basel, 1553, pp. 408‑409. = French ed., 1556, et les raisons d'icelles; Book XV, para. 2,
p. 291, ??NYS. = Opera Omnia,
vol. 3, p. 587. Very cryptic
description, with one diagram of a ring.
In England, the Chinese Rings were known as Tarriers or
Tarriours or Tiring or Tyring or Tarrying Irons. The OED entry at Tiring-irons gives 5 quotations from the
17C: 1601, 1627, 1661, 1675, 1690.
John II Tradescant (1608-1662). Musæum Tradescantianum.
1656. Op. cit. in 6.V. P. 44: "Tarriers of Wood made like our
Tyring-Irons." (The following
entry is: "Tarriers of Wood like Rolles to set Table-dishes on." -- I
cannot figure out what this is.)
Gardner. Knotted,
chap. 2, says there are 17C Japanese haiku about it and it is used in Japanese
heraldic emblems.
Kozaburo Fujimura
& Shigeo Takagi. Pazuru no Genryū (The Origins of
Puzzles, in Japanese). Daiyamondo Sha,
Tokyo, 1975. ??NYS -- information
kindly sent by Takao Hayashi. Chap. 9
is on the Chinese Rings.
This
says the oldest datable record in Japan is in
Osaka Dokugin Shū, a book
of haiku compiled in 1675. A haiku of Saikaku Ihara is "Chienowa ya
shijōdōri ni nukenuran",
where the first word means 'wisdom rings' and denotes the Chinese Rings
puzzle.
Another
poetry book, Tobiume Senku, of 1679 has
"Tenjiku shintan kuguru chienowa".
The
Chinese Rings was used as a family crest.
The first known reference is in the description of a kimono worn by a
character in Monzaemon Chikamatsu's joruri,
Onna Goroshi Abura Jigoku, first
staged in 1721. See next item for more
details.
Another
word for the puzzle is
kyūrenkan, which is
borrowed from Chinese and means 'nine connected rings'. It is explained in a Chinese lexicon, Meibutsu rokujō, by
Choin Ito, c1725.
The
mathematician Yasuaki Aida
(1747-1817), in his unpublished
autobiography Jizai Butsudan of 1807, says he solved the puzzle when he
was nine, i.e. c1756.
Yoriyuki
Arima (1714-1783), another mathematician, treats the Chinese Rings in his
mathematical work Shūki
Sampō, of 1769.
Gennai
Hiraga (1728-1779) unlocked a bag locked with a Chinese Rings belonging to
Captain Jan Crans of the Dutch factory (i.e. trading post) in c1769. This is related in Genpaku Sugita's Rangaku
Kotohajime of 1815.
Chienowa is recorded in the 1777 Japanese
dictionary Wakun no Shiori.
Dictionary of Representative
Crests. Nihon Seishi Monshō
Sōran (A Comprehensive Survey of Names and Crests in Japan), Special issue of Rekishi Dokuhon (Readings in History), Shin Jinbutsu Oraisha, Tokyo, 1989, pp. 271-484. Photocopies of relevant pages kindly sent by
Takao Hayashi. Crest 3447 looks like a
Chinese Rings with five rings and 3448 looks like one with four rings, but both
are simplified and leave out one of the bars.
John Wallis. De
Algebra Tractatus. 1685, ??NYS. = Opera Math., Oxford, 1693, vol. II,
chap. CXI, De Complicatus Annulis, 472‑478. Detailed description with many diagrams.
Ozanam. 1725: vol.
4. No text, but the puzzle with 7
rings is shown as an unnumbered figure on plate 14 (16). Ball, MRE, 1st ed., 1892, p. 80, says the
1723 ed., vol. 4, p. 439 alludes to it.
The text there is actually dealing with Solomon's Seal (see 11.D) which
is the adjacent figure on plate 14 (16).
Minguet. 1733. Pp. 55-57 (1755: 27-28; 1822: 72-74; 1864:
63-65): Juego del ñudo Gordiano, ò lazo de las sortijas enredadas. 7
ring version clearly drawn.
Alberti. 1747. No text, but the puzzle is shown as an
unnumbered figure on plate XIII, opp. p. 214 (111), copied from Ozanam, 1725.
Catel.
Kunst-Cabinet. 1790. Der Nürnberger Tand, p. 15 & fig. 41 on
plate II. Figure shows 7
rings, text says you can have
7, 9, 11 or 13.
Bestelmeier.
1801. Item 298: Der Nürnberger Tand. Diagram shows 6 rings, but text refers
to 13
rings. Text is partly copied
from Catel.
Endless Amusement II.
1826? Prob. 29, pp.
204-207. Cites Cardan as being very
obscure. Shows example with 5
rings and seems to imply it takes 63 moves.
The Boy's Own Book. The
puzzling rings. 1828: 419‑422; 1828‑2: 424‑427; 1829 (US): 216-218; 1855: 571‑573; 1868: 673-675. Shows 10 ring version and says it takes 681 moves. Cites Cardan.
Crambrook.
1843. P. 5, no. 9: Puzzling
Rings, or Tiring Irons.
Magician's Own Book.
1857. Prob. 45: The puzzling
rings, pp. 279-283. Identical to Boy's
Own Book, except 1st is spelled out first, etc. = Book of 500 Puzzles, 1859, pp. 93-97. = Boy's Own Conjuring Book, 1860, prob. 44, pp. 243‑246.
Magician's Own Book (UK version). 1871. The tiring-irons,
baguenaudier, or Cardan's rings, pp. 233-235.
Quite similar to Boy's Own Book, but somewhat simplified and gives a
tabular solution.
L. A. Gros. Théorie
du Baguenodier. Aimé Vingtrinier, Lyon,
1872. (Copy in Radcliffe Science
Library, Oxford -- cannot be located by them.)
??NYS
Lucas. Récréations
scientifiques sur l'arithmétique et sur la géométrie de situation. Troisième récréation, sur le jeu du
Baguenaudier, ... Revue Scientifique
de la France et de l'étranger (2) 26 (1880) 36‑42. c= La Jeu du Baguenaudier, RM1, 1882, pp.
164‑186 (and 146‑149). c=
Lucas; L'Arithmétique Amusante; 1895; pp. 170-179. Exposition of history back to Cardan, Gros's work, use as a lock
in Norway. He says that Dr. O.-J.
Broch, former Minister and President of the Royal Norwegian Commission at the
Universal Exposition of 1878, recently told him that country people still used
the rings to close their chests and sacks.
RM1 adds a letter from Gros.
The French term 'baguenaudier' has long mystified me. A 'bague' is a ring. My large Harrap's French‑English
dictionary defines 'baguenaudier' as "trifler, loafer, retailer of idle
talk; ring‑puzzle, tiring
irons; bladder‑senna", but
none of the related words indicates how 'baguenaudier' came to denote the
puzzle. However, Farmer & Henley's
Dictionary of Slang gives 'baguenaude' as a French synonym for 'poke', so
perhaps 'baguenaudier' means a 'poker' which has enough connection to the
object to account for the name?? MUS I
62-63 discusses Gros's use of 'baguenodier' as unreasonable and quotes two
French dictionaries of 1863 and 1884 for 'baguenaudier' which he identifies as
an ornamental garden shrub, Colutea arborescens L.
Cassell's.
1881. Pp. 91-92: The puzzling
rings. = Manson, 1911, pp. 144-145:
Puzzling rings. Shows 7
ring version and discusses
10 ring version, saying it takes
681 moves. Discusses the Balls and
Rings puzzle.
Peck & Snyder.
1886. P. 299: The Chinese
puzzling rings. 9 rings.
Mentions Cardan & Wallis.
Shown in Slocum's Compendium.
Ball. MRE, 1st ed.,
1892, pp. 80-85. Cites Cardan, Wallis,
Ozanam and Gros (via Lucas). P. 85
says: "It is said -- though a
priori the fact would have seemed very improbable -- that Chinese rings are
used in Norway to fasten the lids of boxes, .... I have never seen them employed for such purposes in any part of
the country in which I have travelled."
This whole comment is dropped in the 3rd ed.
Hoffmann. 1893. Chap. X, no. 5: Cardan's rings, pp. 334‑335
& 364‑367 = Hoffmann‑Hordern, pp. 222-225, with
photo. Cites Encyclopédie Méthodique
des Jeux, p. 424+. Photo on p. 223
shows The Puzzling Rings, by Jaques & Son, 1855‑1895, with
instructions, and Baguenaudier, with box, 1880-1895. Hordern Collection, p. 92, shows the Jaques example, an ivory
example with an elaborate handle and another of ivory or bone, all dated
1850-1900. I now have an example of the
Jaques version which has rings coloured red, white and blue.
H. F. Hobden. Wire
puzzles and how to make them. The Boy's
Own Paper 19 (No. 945) (13 Feb 1896) 332-333.
Magic rings (= Chinese rings) with
10 rings, requiring 681 moves. (I think it should be 682.)
Dudeney. Great
puzzle crazes. Op. cit. in 2. 1904.
"... it is said to be used by the Norwegians as a form of lock for
boxes and bags ..."
Ahrens.
Mathematische Spiele.
Encyklopadie article, op. cit. in 3.B.
1904. Note 60, p. 1091, reports
that a Norwegian professor of Ethnography says the story of its use as a lock
in Norway is erroneous. He repeats this
in MUS I 63.
M. Adams. Indoor
Games. 1912. Pp. 337‑341 includes The magic rings.
Bartl. c1920. P. 309, no. 80: Armesünderspiel oder
Zankeisen. Seven ring version for sale.
Collins. Book of
Puzzles. 1927. The great seven-ring puzzle, pp. 49-52. Cites Cardan and Wallis. Says it is known as Chinese rings, puzzling
rings, Cardan's rings, tiring irons, etc.
Says 3 rings takes 5 moves, 5 rings takes 21 and 7 rings takes 85.
Rohrbough. Puzzle
Craft. 1932. The Devil's Needle, p. 7 (= p. 9 of 1940s?). Cites Boy's Own Book of 1863.
R. S. Scorer, P. M.
Grundy & C. A. B. Smith. Some
binary games. MG 28 (No. 280) (Jul
1944) 96‑103. Studies the binary
representations of the Chinese Rings and the Tower of Hanoi. Gives a triangular coordinate system representation
for the Tower of Hanoi. Studies Tower
of Hanoi when pegs are in a line and you cannot move between end pegs. Defines an
n‑th order Chinese Rings and gives its solution.
E. H. Lockwood. An
old puzzle. With Editorial Note by H.
M. Cundy. MG 53 (No. 386) (Dec 1969)
362‑364. Derives number of moves
by use of a second order non‑homogeneous recurrence. Cundy mentions the connection with the Gray
code and indicates how the Gray value at step
k, G(k), is derived from the binary representation of k,
B(k). [But he doesn't give the
simplest expression, given by Lagasse, qv in 7.M.3.] This easily gives the number of steps.
Marvin H. Allison Jr.
The Brain. This is a version of
the Chinese Rings made by Mag-Nif since the 1970s. [Gardner, Knotted.]
William Keister. US
Patent 3,637,215 -- Locking Disc Puzzle.
Filed: 22 Dec 1970; patented: 25
Jan 1972. Abstract + 3pp + 1p
diagrams. This is a version of the
Chinese Rings, with discs on a sliding rod producing the interaction of one
ring with the next. Described on the
package. Keister worked on puzzles of
this sort since the 1930s. It was first
produced by Binary Arts in 1986 under the name Spin Out.
Ed Barbeau. After
Math. Wall & Emerson, Toronto,
1995. Protocol, pp. 163-166. Gives seating and standing problems which
lead to the same sequence of moves as for the Chinese rings, but one is in
reverse order.
Anatoli Kalinin says that the Chinese Rings are a old folk
puzzle called Меледа [Meleda],
especially popular among the Kalmyks near the Caspian Sea, where it is
called
Наран-шина [Naran-shina] (stirrup ring
toy). The name Меледа is derived from a verb
which is no longer in Russian.
See MUS I 52-61, S&B 135.
See also 5.F.4 for connection with Hamiltonian
circuits on the n‑cube and 5.A.4 for the Panex Puzzle.
All the following have three pegs
unless specified otherwise.
The Conservatoire National des Arts et Métiers -- Musée
National des Techniques, 292 Rue St. Martin, Paris, has two examples -- No.
11271 & 11272 -- presented by Edouard Lucas, professeur de mathématiques au
Lycée Saint-Louis à Paris, in 1888. The
second is a 'grand modèle pour les cours publics' 1.05 m high! Elisabeth Lefevre has kindly sent details
and photocopies of the box, instruction sheet (one sheet printed on both sides)
and an article. Both versions have 8
discs. I am extremely grateful
to Jean Brette, at the Palais de la Decouverte, who told me of these examples
in 1992.
The
box is 157 x 180 mm and has an elaborate picture with the
following text:
La
Tour d'Hanoï / Veritable casse‑Téte Annamite / Jeu / rapporté du Tonkin /
par le Professeur N. Claus (de Siam) / Mandarin / du College / Li‑Sou‑Stian
/ Brevete / S. G. D. G.
This cover is
shown in Claus [Lucas] (1884) and Héraud (1903). I will refer to this as the original cover. (The Museum does not know of any patent --
they have looked in 1880-1890. S. G. D.
G. stands for Sans garantie du gouvernement.)
The bottom of the box has an ink inscription: Hommage del'auteur Ed
Lucas Paris 1888 -- but the date is
not clearly legible on the photocopy.
Inside the cover, apparently in the same hand (that is, in Lucas's
writing), is an ink inscription:
La
tour d'Hanoï, --
Jeu
de combinaison pour
expliquer
le systeme de la numération
binaire,
inventé par M. Edouard Lucas,
(novembre
1883). -- donné par l'auteur.
The
Museum describes the puzzle as 15 cm
long by 14.5 cm wide by 10 cm
high. There is no photo available, but
the examples are shown in catalogues of 1906 and 1943.
The
instruction sheet reads as follows.
La
Tour d'Hanoï
Véritable
casse‑tête annamite
Jeu
rapporté du Tonkin
par
le Professeur N. Claus (de Siam)
Mandarin
du Collège Li‑Sou‑Stian!
....
Paris,
Pékin, Yédo et Saïgon
....
1883
The
sheet mentions the Temple of Bénarès where there are 64 discs. A prize of a million (= a thousand thousand)
francs is offered for a demonstration of the solution with 64
discs! The second sheet of the
instructions gives the rules and the number of moves for 2, 3, ..., 8 discs and the general rule.
It also refers to RM.
Edward Hordern's collection has an example with the original
instruction sheet, but in a simple box with just 'La Tour d'Hanoï' in Chinese‑style
lettering, not like the box described above.
Also, my recollection is that it is much smaller than the example
above. It has 8 discs.
G. de Longchamps.
Variétés. Journal de
Mathématiques Spéciales (2) 2 (1883) 286-287.
(The article is only signed G. L., but the author is further identified
in the index on p. 290. Copy
provided by Hinz.) Solves the
recurrence relation un = 2un-1
+ 1, u0 = 0. Says he was 'inspired by a letter which we
have recently received from professor N. Claus.' Describes the Tower of Hanoi briefly and says the above solution
gives the number of moves when there are
n discs.
Henri de Parville.
Column: Revue des sciences. Journal des Débats Politiques et Littéraires
(27 Dec 1883) 1-2. On p. 2, he reports
receiving an example in the post with a cover like that of the original. Gives the Benares story. Wonders who the mandarin could be and notes
the anagrams on Lucas d'Amiens and
Saint-Louis.
N. Claus (de Siam) [= Lucas (d'Amiens)]. La tour d'Hanoï. Jeu de calcul. Science et
Nature 1:8 (19 Jan 1884) 127-128. Says
it takes 2n - 1 moves
"que M. de Longchamps l'a démontré (1)." "(1) Journal de mathématiques
spéciales.". Observes that
each of the discs always moves in the same cycle of pegs and hence gives the
standard rule for doing the solution, which is attributed to the nephew of the
inventor, M. Raoul Olive, student at the Lycée Charlemagne. Asks for the minimum number of moves to
restore an arbitrary distribution of discs to a Start position. Says this is a complex problem in general,
depending on binary and refers to RM 1 for this idea.
(This
paper is not in Harkin's bibliography (op. cit. in 1). Hinz, 1989, cites it.)
Henri de Parville.
Récréations Mathématiques: La
Tour d'Hanoï et la question du Tonkin.
La Nature (Paris) 12 (No. 565, part 1) (29 Mar 1884) 285‑286. Illustration by Poyet. Asserts Lucas is the inventor.
R. E. Allardice & A. Y. Fraser. La Tour d'Hanoï. Proc. Edin. Math. Soc. 2 (1883‑1884) 50‑53. Includes de Parville from J. des
Débats. Then derives number of moves.
Anton Ohlert. US
Patent 303,946 -- Toy. Applied: 24 Jul
1884; patented: 19 Aug 1884. 1p + 1p diagrams. 8
discs. Ohlert is a resident of
Berlin.
Edward A. Filene, No. 4 Winter St., Boston Mass. Eight Puzzle. Copyrighted in 1887.
??NYS -- described and illustrated in S&B, p. 135. 8 disc
advertising version.
Tissandier.
Récréations Scientifiques. 5th
ed., 1888, La tour d'Hanoï et la question du Tonkin, pp. 223-228. Not in the 2nd ed. of 1881 nor the 3rd ed.
of 1883. Essentially de Parville's
article with same illustration, but introduces it with comments saying it has
had a great success and comes in a box labelled: "la Tour d'Hanoï, véritable casse-tête annamite,
rapporté du Tonkin par le professeur N. Claus (de Siam), mandarin du collège
Li-Sou-Stian". This would seem to
be the original box.
=
Popular Scientific Recreations; [c1890]; Supplement: The tower of Hanoï and the
question of Tonquin, pp. 852‑856.
Anon. Jeux, Calculs
et Divertissements. Récréations
Mathématiques. La Tour d'Hanoï. Liberté (9 Dec 1888) no page number on
clipping. Gives the story of N. Claus
and says it was invented by Lucas and comes in a box decorated with annamite
illustrations. This would seem to be
the original box.
Lucas. Nouveaux jeux
scientifiques de M. Édouard Lucas. La
Nature 17 (1889) 301‑303.
Describes a series of games under the title of the next item, so the
next may refer to the game or its instruction booklet. A later version of the Tower of Hanoi is
described on pp. 302‑303, having
5 pegs, and is shown on p.
301. He says you can have 3, 4 or 5
pegs.
Lucas. Jeux
scientifiques pour servir à l'histoire, à l'enseignement et à la pratique du
calcul et du dessin. Première série:
No. 3: La Tour d'Hanoï. Brochure,
Paris, 1889. ??NYS. (Not listed in BNC, but listed in Harkin,
op. cit. in Section 1. I wonder if
these were booklets that accompanied the actual games?? Hinz says he has found some of these, but
not the one on the Tower of Hanoi. The
booklet for La Pipopipette (= Dots and Boxes) is reproduced in his
L'Arithmétique Amusante of 1895 -- see 4.B.3.)
Jeux Scientifiques de Ed. Lucas. Advertisement by Chambon & Baye (14 rue Etienne-Marcel,
Paris) for the 1re Serie of six games. Cosmos. Revue des
Sciences et Leurs Applications 39 (NS No. 254) (7 Dec 1889) no page number on
my photocopy.
B. Bailly [name not given, but supplied by Hinz]. Article on Lucas's puzzles. Cosmos.
Revue des Sciences et Leurs Applications. NS, 39 (No. 259) (11 Jan 1890) 156-159. Shows 'La nouvelle Tour d'Hanoï', which has five pegs and 15?
discs. I need pp. 156‑157.
Alfred Gartner & George Talcott. UK Patent 20,672 -- Improvements in Games
and Puzzles. Applied: 18 Dec 1890; patented: 21 Feb 1891. 2pp + 1p diagrams. Shunting puzzle equivalent to Tower of Hanoi with 7 discs.
Ball. MRE, 1st ed.,
1892, pp. 78-79. "... described by
M. Tissandier as being common in France but which I have never seen on sale in
England." Gives the Benares story
from De Parville in La Nature.
Hoffmann. 1893. Chap. X, no. 4: The Brahmin's Rings, pp.
333-334 & 361-364
= Hoffmann/Hordern, pp. 220-222, with photo. Gives Benares story and then gives the
problem with 8 discs, noting that it is made by Messrs.
Perry & Co. Photo on p. 221 shows
an example, with box with instructions on the top, by Perry & Co.,
1880-1900. Hordern Collection, p. 90,
shows a version with box, by R. Journet, 1905-1920.
Lucas.
L'Arithmétique Amusante.
1895. La tour d'Hanoï, pp.
179-183. Description, including Benares
story. Says the nephew of the inventor,
Raoul Olive, has noted that the smallest disc always moves in the same
direction. Says the whole idea of the
mandarin and his story was invented a dozen years ago at 56 rue Monge, which was built on the site where Pascal died. [I visited this site recently -- it is a
hotel and they knew nothing about Pascal.
Another source says Pascal died at
67 Rue Cardinal Lemoine, which
is several blocks away, but also is not an old building.]
Ball. MRE, 3rd ed.,
1896, pp. 99-101. Omits the Tissandier
reference and says: "It was
brought out in 1883 by M. Claus (Lucas)."
A. Héraud. Jeux et
Récréations Scientifiques -- Chimie, Histoire Naturelle, Mathématiques. Not in the 1884 ed. Baillière et Fils, Paris, 1903. Pp. 300‑301 shows the original cover.
Burren Loughlin
& L. L. Flood. Bright-Wits
Prince of Mogador. H. M.
Caldwell Co., NY, 1909. The five
shields, pp. 20-24 & 59. Five
discs.
Tom Tit??. In Knott,
1918, but I can't find it in Tom Tit.
No. 165: The tower of Hanoi, pp. 382‑383. Describes it with cards A ‑ 10 on piles labelled with J,
Q, K.
Robert Ripley.
Believe It Or Not! Book 2. (Simon & Schuster, 1931); Pocket Books, NY, 1948, pp. 52‑53. = Believe It or Not! Two volumes in one; (Simon & Schuster,
1934); Garden City Books, 1946, pp. 222-223.
= Omnibus Believe It Or Not!; Stanley Paul, London, nd [c1935?],
pp. 256‑257. The Brahma
Pyramid. Outlines the Benares story,
says he didn't locate the temple when he was in Benares, but it 'really
exists', and that it will take 264 moves, but he then writes out 264 ‑ 1. He says the Brahmins have been at it
for 3000 years!!
B. D. Price. Pyramid
patience. Eureka 8 (Feb 1944) 5-7. Straightforward development of basic properties.
Scorer, Grundy & Smith.
1944. Op. cit. in 7.M.1. They develop the graph of all positions of
the Tower of Hanoi.
Donald W. Crowe.
The n-dimensional cube and the
tower of Hanoi. AMM 63 (1956)
29-30. Describes the connection with
Hamiltonian circuits and binary ruler markings.
M. Gardner. The
Icosian Game and the Tower of Hanoi.
(SA (May 1957)) = 1st Book, chap. 6. Describes Crowe's work.
A. J. McIntosh.
Binary and the Tower of Hanoi.
MTg 59 (1972) 15. He sees the
connection between binary and which disc is to be moved, but he wonders how to
know which pegs are involved. [This is
a valid query -- though we know each disc moves cyclically, alternate ones in
alternate directions, I don't know any easy way to translate a particular step
number into the positions of all the discs -- Hinz (1989) gives a method which
may be as simple as possible, but I have a feeling it ought to be easier. Actually, I have now seen a fairly easy way
to do this.]
Andy Liu & Steve Newman, proposers and solvers. Problem 1169 (ii) -- The two towers. CM 12 (1986) 179 &
13 (1987) 328‑332. Three
pegs, two identical piles of size
n on two of them. The object is to interchange the bottom
discs and reform the piles (though the smaller discs may or may not be interchanged). They find it takes 7*2n+1 ‑ 3n ‑ (10 or
11)/3 steps, depending on whether n is
odd or even.
Andreas M. Hinz. The
Tower of Hanoi. L'Enseignement Math. 35
(1989) 289-321. Surveys history and
current work. 50 references. Finds many properties, particularly the
average distance from having all discs on a given peg and the average distance
between legal positions. He also
studies illegal positions. Uses the Grundy,
Scorer & Smith graph. Gives general
results, such as Schwenk's below. He
asserts that the minimality of the classic solution was not proven until 1981,
but I think the classic method clearly implies the proof of its minimality.
Hugh Noland, proposer;
Norman F. Lindquist, David G. Poole & Allen J. Schwenk,
solvers. Prob. 1350 -- Variation on the
Tower of Hanoi. MM 63:3 (1990) 189 &
64:3 (1991) 199-203. Three pegs, 2n
discs, initially with the evens on one peg and the odds on another. How many moves to get all onto the empty
peg? Answer is ë(5/7) 4nû. Schwenk gives a solution for any starting
position of N discs and shows the average number of moves to get to a single
pile is (2/3)(2N-1).
Andreas M. Hinz.
Pascal's triangle and the Tower of Hanoi. AMM 99 (1992) 538-544.
Shows the Grundy, Scorer & Smith graph is equivalent to the pattern
of odd binomial coefficients in the first
2n rows and hence to
Sierpiński's fractal triangle.
Gives some life of Lucas.
David Poole. The
bottleneck Towers of Hanoi problem. JRM
24:3 (1992) 203-207. Studies the
problem when big discs can go on smaller discs, but not too much smaller
ones. 11 references to recent work
on variations of the classical problem.
Ian Stewart. Four
encounters with Sierpiński's gasket.
Math. Intell. 17:1 (1995) 52-64.
This discusses the connections between the graph of the Tower of Hanoi,
the pattern of odd binomial coefficients, Sierpiński's gasket and
Barnsley's iterated fractal systems.
Lots of references, including 11 on the Tower of Hanoi.
Vladimir Dubrovsky.
Nesting Puzzles -- Part I: Moving oriental towers. Quantum 6:3 (Jan/Feb 1996) 53-59 &
49-51. Outlines the history and theory
of the Tower of Hanoi. Misha Fyodorov, a
Russian high school student, observed that the peg not used in a move always
moves in the same direction. Discusses
Kotani's modification which prevents placing some discs on a particular
peg. Also discusses the Panex Puzzle --
cf Section 5.A.4.
Jagannath V. Badami.
Musings on Arithmetical Numbers
Plus Delightful Magic Squares.
Published by the author, Bangalore, India, nd [Preface dated 9 Sep
1999]. Section 4.15: The Tower of
Brahma, pp. 123-124. "The author
has lived in Banaras for a number of years and does not find any basis for this
legend."
David Singmaster.
The history of some combinatorial recreational problems. Draft of a chapter for History of
Combinatorics, ed. Robin J. Wilson. Jan
2001. This gives a detailed development
of the distances di of a position from the three perfect
positions and the complementary distances
d'i = (2n-1) - di, leading to
THEOREM 4. The Scorer, Grundy
& Smith graph of positions in the Tower of Hanoi with n
discs and with adjacency between positions one move apart, is isomorphic
to the graph of triples of binary n‑tuples (d'0, d'1, d'2) satisfying
(*') Σi Dk(d'i) = 1 for each k,
considered as triangular coordinates in a triangle of edge length 2n-1 and with adjacency being adjacency in the lattice.
7.M.2.a. TOWER OF HANOI WITH MORE PEGS
Lucas. Nouveaux jeux
scientifiques ..., op. cit. in 4.B.3, 1889.
(See discussion in 7.M.2.)
Dudeney. The Reve's
Puzzle. The Canterbury Puzzles. London Mag. 8 (No. 46) (May 1902) 367‑368 &
8 (No. 47) (Jun 1902) 480. = CP,
prob. 1, pp. 24‑25 & 163‑164.
4 pegs, 8, 10 or 21
discs.
Dudeney.
Problem 447. Weekly Dispatch (25 May, 15 Jun, 1902) both
p. 13. 4 pegs,
36 discs.
Dudeney. Problem
494. Weekly Dispatch (15 Mar, 29 Mar, 5
Apr, 1903) all p. 13. 5 pegs,
35 discs.
B. M. Stewart, proposer;
J. S. Frame & B. M. Stewart, solvers. Problem 3918. AMM 48 (1941)
216‑219. k pegs,
n discs. General solution, but editorial note implies
there is a gap in each solver's work.
Scorer, Grundy & Smith.
Op. cit. in 7.M.1. 1944. They give some variations on the Tower of
Hanoi with four pegs.
Doubleday - 2. 1971. Keep count, pp. 91-92. 15
discs, 6 pegs
-- solved in 49 moves.
Ted Roth. The tower
of Brahma revisited. JRM 7 (1974) 116‑119. Considers
4 pegs.
Brother Alfred Brousseau.
Tower of Hanoi with more pegs.
JRM 8 (1975/76) 169‑176.
Extension of Roth, with results for
4 and 5 pegs.
The Diagram Group.
Baffle Puzzles -- 3: Practical Puzzles.
Sphere, 1983. No. 10. 6 pegs, 15 discs. Gives a solution in 49 moves.
Joe Celko. Puzzle
Column: Mutants of Hanoi. Abacus 1:3 (1984) 54‑57. Discusses variants: where a disc can only move to an adjacent
peg in a linear arrangement; with two
or three colours of discs; with several
piles of discs; where a disc can only
move forward in a circular arrangement.
Grame Williams.
In: Joe Celko; Puzzle column
replies; Abacus 5:2 (1988) 70‑72.
Table of minimum numbers of moves for
k pegs, k = 3, ..., 8 and
n discs, n = 1, ..., 10.
Andreas Hinz. An
iterative algorithm for the Tower of Hanoi with four pegs. Computing 42 (1989) 133-140. Studies the problem carefully. 17 references.
A. D. Forbes.
Problem 163.2 -- The Tower of Saigon.
M500 163 (Aug 1998) 18-19. This
is the Tower of Hanoi with four pegs.
Quotes an Internet posting by Bill Taylor giving an algorithm and its
number of moves up to 12 discs. Asks if
this is optimal.
See Gardner
under 7.M.
L. A. Gros. Op. cit.
in 7.M.1, 1872. ??NYS. (Afriat.)
J. Émile Baudot.
c1878. ??NYS. Used Gray code in his printing telegraph.
(Described by F. G. Heath; Origins of the binary code; SA (Aug 1972) 76‑83.)
Anon. Télégraphe
multiple imprimeur de M. Baudot.
Annales Télégraphiques (3) 6 (1879) 354‑389. Says the device was presented at the 1878
Exposition and has been in use on the Paris‑Bordeaux line for several
months. See pp. 361‑362 for
diagrams and p. 383 for discussion.
George R. Stibitz.
US Patent 2,307,868 -- Binary Counter.
Applied: 26 Nov 1941; granted:
12 Jan 1943. 3pp + 1p
diagrams. Has an electromechanical
binary counter using the Gray code with no comment or claims on it.
Frank Gray. US
Patent 2,632,058 -- Pulse Code Communication.
Applied: 13 Nov 1947, patented: 17 Mar 1953. 9pp + 4pp diagrams.
Systematic development of the idea and its uses.
A. J. Cole. Cyclic
progressive number systems. MG 50 (No.
372) (May 1966) 122‑131. These
systems are Gray codes to arbitrary bases -- e.g. in base 4,
the sequence begins: 0, 1, 2, 3,
13, 12, 11, 10, 20, 21, 22, 23, 33, 32, 31, 30, 130, 131, .... For odd bases, the sequence is harder. He gives conversion rules and rules for
arithmetic.
J. Lagasse. Logique
Combinatoire et Séquentielle. Maîtrise
d'E.E.A. C3 -- Automatique. Dunod, Paris, 1969. Pp. 14-18 discuss the Gray code (code
réfléchi) stating that the Gray value at step
k, G(k), is given by
G(k) = B(k)
EOR B(ëk/2û). [I noted this a few years ago and am
surprised that it does not appear to be old.
Gardner's 1972 article describes it but not so simply.] My thanks to Jean Brette for this reference.
William Keister. US Patent
3,637,216 -- Pattern-Matching Puzzle.
Filed: 11 Dec 1970; patented: 25
Jan 1972. Abstract + 4pp + 2pp
diagrams. This has a bar to remove from
a frame -- one has to move various bits in the pattern of the Gray code (or
similar codes) to extract the bar. Made
by Binary Arts since about 1986.
The classic cards for this process
have the numbers 1 - 2n on them, the i-th card containing those numbers whose binary expression has
a 1
in its (i-1)-st place -- e.g.
the first card contains all the odd numbers.
Then one adds up the smallest numbers, i.e. 2i-1 on
the i‑th card, on the chosen
cards to get the number thought of. If
one replaces the numbers by holes in the corresponding positions, one can
overlay the cards to read off the answer.
This takes a little more work though -- one has to have each card
containing its set of holes to be used if it is chosen and also containing the
complementary holes to be used if it is not chosen. This can be achieved if the holes are centrally located -- then
turning the card around produces the complementary set of holes.
A very similar principle is used as
a kind of logical device. See: Martin Gardner; Logic Machines and Diagrams;
McGraw‑Hill, NY, 1958, pp. 117-124;
slightly extended in the 2nd ed., Univ. of Chicago Press, 1982, and
Harvester Press, Brighton, 1983, pp. 117-124;
for discussion of this idea and references to other articles.
A simple form of binary division is
used to divine a card among sixteen cards arranged in two columns, but it is
surprisingly poorly described. This is
related to the 21 card trick which is listed in 7.M.4.b. I have only recently added this topic and
may not have noticed many versions.
Pacioli. De
Viribus. c1500. Ff. 114r - 116r. C(apitolo). LXIX. a trovare una moneta fra 16 pensata (To find a
coin thought of among 16). = Peirani
161-162. Divides 16 coins in half 4
times, corresponding to the value of the binary digits. Pacioli doesn't describe the second stage
clearly, but Agostini makes it clear.
Bachet.
Problemes. 1612. Prob. XVI, 1612: 87-92. Prob. 18; 1624: 143-151; 1884: 72-83. 15 card trick. His Avertissement mentions that other versions are possible and
describes divining from sixteen cards in two columns and in four columns, but
with no diagrams!
Yoshida (Shichibei) Kōyū (= Mitsuyoshi Yoshida)
(1598-1672). Jinkō‑ki. 2nd ed., 1634 or 1641??. Op. cit. in 5.D.1. ??NYS
Shimodaira (see entry in 5.D.1) discusses this on pp. 2-12 since there
are several Japanese versions of the idea.
The Japanese call these Metsukeji (Magic Cards). The binary version is discussed on pp. 4-7
where it is said that they are known since the 14C or even earlier. The Japanese magic card shown on p. 6
has 1, 2, 4, 8, 16 associated with branchings on a picture of a
tree used to divine one of the 21 characters written on the flowers and
leaves. The other kinds of magic cards
are more complex, not involving binary, but just memorisation. The recent transcription of part of Yoshida
into modern Japanese does not include this problem.
Ozanam. 1725. Prob. 39, 1725: 231-233. Prob. 15, 1778: 164-165; 1803: 165-166; 1814: ??NYS. Prob.
14, 1840: 74-75. Sixteen counters being
disposed in two rows, to find that which a person has thought of. Similar to Bachet, but with some diagrams.
Ozanam-Hutton. 1814:
124-126; 1840: 64. (This is an addition which was not in the
1803 ed.) Six cards to divine up
through 63.
Endless Amusement II.
1826? Pp. 180-181. Sixteen Cards being disposed in Two Rows, to
tell the Card which a Person has thought of.
c= Ozanam, with 'counter' replaced by 'card'.
Young Man's Book.
1839. Pp. 202-203. Identical to Endless Amusement II.
Crambrook.
1843. P. 7, no. 5: A pack of
cards by which you may ascertain any person's age. Not illustrated, but seems likely to be binary divination -- ??
Magician's Own Book.
1857. The mathematical fortune
teller, pp. 241‑242. Six cards
each having 30 numbers used to divine a number up
through 60. Some cards have duplicate numbers in the 30th position. = Boy's Own Conjuring Book, 1860, pp. 211-212. = Illustrated Boy's Own Treasury, 1860,
prob. 38, pp. 402 & 442.
Book of 500 Puzzles.
1859. The mathematical fortune
teller, pp. 55-56. Identical to
Magician's Own Book. However, my
example of the book has 61 in the last cell of the last card, but a
photocopy sent by Sol Bobroff has this as a
41, as do all other versions of
this problem.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 563-VI, pp. 249-250: Überraschungen
mittels sieben Zauberkarten. Seven
cards used to divine up to 100.
Adams & Co., Boston.
Advertisement in The Holiday Journal of Parlor Plays and Pastimes, Fall
1868. Details?? -- photocopy sent by
Slocum. P. 5: Magic Divination
Cards. For telling any number thought
of, or a person's age. Amusing,
curious, and sometimes "provoking."
Not illustrated, but seems likely to be binary divination -- ??
Magician's Own Book (UK version). 1871. The numerical
fortune teller, pp. 89-90. Very similar
to Magician's Own Book, pp. 241-242, with the same cards, but different text.
Bachet-Labosne.
Problemes. 3rd ed., 1874. Supp. prob. X, 1884: 196-197. Divine a number up to 100
with 7 cards.
F. J. P. Riecke. Op.
cit. in 4.A.1, vol. 3, 1873. Art. 3:
Die Zauberkarten, pp. 11‑13.
Describes 5 cards giving values up to 31,
with explanation. Describes how
to use balanced ternary to construct
7 cards giving values up to 22.
The limitation to 22 is due to the size of the cards -- the
method works up to 40.
Mittenzwey.
1880. Prob. 37, pp. 6-8; 1895?: 43, pp. 12-13; 1917: 43, pp. 11‑13. Seven cards. Calls them "Boscos Zauberkarten" -- Bosco (1793- ) was a noted conjurer of the early 19C.
Hoffmann. 1893. Chap. IV, no. 68: The magic cards, pp. 160‑161
& 216‑217 = Hoffmann‑Hordern, pp. 141-142, with
photo. Seven cards. Photo on p. 141 shows an ivory set of six
cards, 1850-1900, and two German sets of seven cards from a box of puzzles
called Hokus Pokus: Zauber‑Karten and
Ich weiss wie alt du bist, both
1870‑1890. These have their own
boxes or wrappers. Hordern Collection,
p. 73, shows the same(?) Zauber-Karten, with its box or wrapper, dated 1860‑1890,
and fully spread out so the instruction card is legible.
Lucas.
L'Arithmétique Amusante.
1895. L'éventail mystérieux, pp.
168‑170. Shows five cards for
divining 1 through 31 and notes it is based on binary.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
10:2 (Jun 1903) 140-141. To tell a
lady's age. Six cards to divine up
through 63.
Ahrens. MUS I. 1910.
Pp. 39-40 does 5 cards.
Pp. 43-48 develops a set of four see-through templates.
M. Adams. Indoor
Games. 1912. How to divine ages, pp. 349-350.
The Quaker Oats Age Prediction Cards. (I have a facsimile of a set, made by Dave
Rosetti for G4G5. He says it was
invented in 1921.) 8 cards with various
holes and numbers for divining from 1 to 99.
You turn the card round if the number is not on it but put it on the
pile. At the end, you turn the pile
over and the age and the slogan 'Never Known to Fail' can be read through the
holes.
Card
1 is just a viewing window with four windows that are five digits wide. Cards 6, 7, 8 have numbers ingeniously
arranged so that groups of five digits are in the window positions. Reading two digits at a time, these groups
give four possible values, hence 16 for each configuration of cards 6, 7, 8 --
except that when both 6 and 7 are up, then card 8 is not visible, and when
cards 6 and 7 are turned, there are only four different values -- this results
in just the 100 values: 00, 01, ..., 99.
Cards 2 and 3 select which window is open. Cards 4 and 5 select which pair of digits in the window can be
seen.
William P. Keasby.
The Big Trick and Puzzle Book.
Whitman Publishing, Racine, Wisconsin, 1929. Magic columns, pp. 176-177.
Applies the idea with letters, so you add the column heads to get the
number of the letter.
Rohrbough. Brain
Resters and Testers. c1935. How to Mystify People, pp. 10-11. = Keasby, whom he cites elsewhere.
John Fisher. John
Fisher's Magic Book. Muller, London,
1968. Think‑a‑Drink, pp.
27-29. 5 cards with cut out
holes to divine 32 types of drink. The last card seems to need one side reversed.
This is much less common than binary
divination and I have just added it.
Gardner, op. cit. at beginning of 7.M.4, describes one set of
triangular cards.
Riecke. 1873. See in 7.M.4.
Martin Hansen. Mind
probe. MiS 21:1 (Jan 1992) cover &
2-6. Adapts binary divination to locate
a number among 1 - 80 with four cards -- but each card has its
numbers half in black and half in white.
Describes how to make 4 x 4 and
8 x 8 binary cards with holes so
the chosen number will appear in the hole.
Adapts to ternary to produce triangular cards with holes so that the
chosen number appears in the hole.
Hansen has kindly given me a set of these: 'The Kingswood Mathemagic Club's Window Cards'. He also describes 'logic cards' which
display the truth values of three basic quantities which are consistent with
various statements -- these were previously described by Gardner and Cundy
& Rollett -- see the Gardner item at the beginning of 7.M.4 for details and
references.
7.M.4.b. OTHER DIVINATIONS USING BINARY OR TERNARY
New section. Discussion with Bill Kalush has revealed
that the classic 21-card trick is based on the ternary system. Another trick involving twice taking 3/2
of a number, which goes back to at least Pseudo-Bede, makes a little use
of binary.
The
21-card trick can also be done with
15 or 27 cards and it is
easiest to explain for 27 cards.
One deals 27 cards out, face up, into three columns and
asks the spectator to mentally choose one card and tell you which column it is
in. You pick up the three columns,
carefully placing the chosen column in the middle of the other two, then deal
out the deck into three columns again.
Ask the spectator to tell you what column the card is now in and pick up
the cards again with the chosen column in the middle and deal them out
again. Repeat the whole process a third
time, then deal out the cards face down and ask the spectator to turn over the
middle card, which will be his chosen card.
The first process puts the chosen card among the positions whose initial
ternary digit is a 1. The second process puts it among the
position whose ternary representation begins
11. The third process puts it at
the position whose ternary representation is
111. For any number of cards
which is a multiple of three, each process puts the chosen card in the 'middle
third' of the previously determined portion of the deck and the trick works,
though the patterns are less systematic than with 27 cards. When the trick is done with, e.g. 24
cards, there is no middle card and you have to expose the chosen card
yourself or develop some other way to do it.
One can also adapt the idea to other numbers of columns -- the two
column case will usually be connected to binary and may be listed in 7.M.
In the 3/2 method, you ask a
person to think of a number, x, then take
3/2 of it (or itself plus half
of itself). If there is a half present,
he is to round it up and let you know.
Do this again on the result. Now
ask how many nines are contained in his result. You then tell the number thought of. The process actually converts
x = a + 2b + 4c to 3a + 5b + 9c, where
a, b = 0
or 1. We have that a = 1
iff there is a rounding at the
first stage and b = 1 iff there is a rounding
at the second stage and the person
then tells you c. We will denote this as PB2.
Sometimes the second stage is
omitted or the division by two in it is omitted, which makes things
simpler. The latter case takes a + 2b
to 6a + 9b and the person gives you b
and there is a remainder if and only if
a = 1. We will denote this by
PB1.
I'm including some early examples of
simple algebraic divination here.
Pseudo-Bede. De
Arithmeticis propositionibus. c8C,
though the earliest MS is c9C.
IN:
Venerabilis Bedae, Anglo‑Saxonis Presbyteri. Opera Omnia: Pars Prima, Sectio II -- Dubia et Spuria: De
Arithmeticis propositionibus. Tomus 1,
Joannes Herwagen (Hervagius), Basel, 1563, Columns 133-135, ??NYS. Folkerts says Hervagius introduced the title
De Arithmeticis propositionibus.
Revised
and republished by J.‑P. Migne as:
Patrologiae Cursus Completus: Patrologiae Latinae, Tomus 90, Paris,
1904, columns 665‑668.
Critical
edition by Menso Folkerts. Pseudo-Beda:
De arithmeticis propositionibus. Eine
mathematische Schrift aus der Karolingerzeit.
Sudhoffs Archiv 56 (1972) 22-43.
A friend of Bill Kalush has made an English translation of the German
text, 1998?, 11pp.
See
also: Charles W. Jones; Bedae Pseudepigrapha: Scientific writings falsely
attributed to Bede. Cornell Univ. Press
& Humphrey Milford, OUP, 1939, esp. pp. 50‑53.
This
has three divination problems and Folkerts says these are the first known
western examples.
1. Triple and halve, then triple and tell
quotient when divided by nine and whether there is a remainder in this. I.e. PB1.
2. 3/2
twice and tell if there is a rounding up at each stage as well as the
quotient when divided by 9. I.e. PB2.
3. Divine a digit a from a * 2 + 5 * 5 * 10, where the operations are performed
sequentially from left to right. The
result is 100a + 250. This is not of the type considered in this
section, but is the prototype of most later divination methods.
Folkerts
mentions several later occurrences of these methods.
(The
fourth and last part of the text is probably slightly later in the 9C and
describes adding positive and negative numbers in a way not repeated in the
west until the 15C.)
Fibonacci.
1202. Pp. 303-304 (S:
427-428). Take 3/2
twice, i.e. PB2.
Folkerts.
Aufgabensammlungen. 13-15C.
Chuquet. 1484. Prob. 155.
English in FHM 230-231.
PB1. Text only indicates what
happens when x is even.
Marre notes that this appears in de la Roche,
1520: ff. 218v ‑ 219r; 1538: ff. 150v - 151r. FHM say Chuquet gives an example, but it is
not in Marre.
Pacioli. De
Viribus. c1500.
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 34-35, f. 67r-67v. Take
3/2 twice, but he rounds down
each time and he is not at all clear how the roundings relate to finding the
number, and it is not nearly as elegant as when one rounds up.
Apianus. Kauffmanss
Rechnung. 1527. Ff. M.vii.r - M.viii.r. PB1.
Tartaglia. General
Trattato. 1556. Book 16, art. 197-198, f. 263v-264r. Divination by 1 + 1/2 twice
and by 3/2 twice, i.e. PB2.
Recorde-Mellis.
Third Part. 1582. Ff. Yy.ii.r - Yy.ii.v (1668: 473-474). PB2, done with one example of 7.
Prévost. Clever and
Pleasant Inventions. (1584), 1998.
Io. Baptiste Benedicti (= Giambattista Benedetti). Diversarum Speculationum Mathematicarum,
& Physicarum Liber. Turin, (1580),
Nicolai Bevilaquæ, Turin, 1585; (Venice, 1599). [Rara 364. Graves
141.f.16.] Theorema CXVI, pp.
78-79. PB2.
Bachet. Problemes. 1612.
Hunt. 1631
(1651).
Pp. 217-218
(209-210): A sixth way to find out a number thought. PB2.
Pp. 221
(misprinted 212) - 222 (213-214): An eighth way to find out a number thought.
PB1.
Schott. 1674.
Ozanam. 1694.
Henry Dean. The
Whole Art of Legerdemain, or Hocus Pocus in Perfection. 11th ed., 1790? ??NX -- seen at UCL Graves 124.b.36. Pp. 89-90 (89 is misprinted as 87). 21 card trick. "This
trick may be done by an odd number of cards that may may be divided by
three."
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
Pp. 81-83, no. 129: To tell the Number a Person has fixed upon, without
asking him any question. A variant of
the 3/2 method. Pick a number
from 1
to 15, add 1 to it.
(Triple the number; if it is odd, add
1; then halve it) thrice. Finally, if the result is odd, add one, and
again halve. From observing when the
person has to add one, you determine the number. If
n + 1 = 8a + 4b + 2c + d, where the coefficients are zero or one, the
process is straightforward up to the third tripling which yields 54a + 27b + 15c + 9d which is odd if and only if b + c + d
is odd. The next steps are no
longer simply expressible in terms of the coefficients. One has to add one at the third stage if and
only if n + 1 º 1, 2, 4, 7 (mod
8) and one has to add one at the fourth
stage if and only if
n + 1 = 1, 6, 8, 10, 11, 12, 13, 15. He gives tables to determine n
from the observations.
Manuel des Sorciers.
1825. ??NX
Boy's Own Book. To
tell a card thought of blindfold.
1828-2: 390-391; 1829 (US):
197; 1855 & 1859:
542-543; 1868: 640-641; 1880: 670-671; 1881 (NY): 196.
21 card trick.
Boy's Own Book. 1843
(Paris).
Magician's Own Book.
1857. Which counter has been
thought of out of sixteen?, pp. 226‑227. Like Prevost but clearer.
Uses 16 cards and repeated halving to bring the
thought of card to the top in four steps.
Says it can be extended to
32 cards. = Boy's Own Conjuring Book, 1860, pp.
196-197.
Vinot. 1860.
Indoor Tricks and Games.
Success Publishing, London, nd [1930s??]. The wonderful twenty-seven
& Variants of the 27-card trick, pp. 47-51. Uses ternary to locate and position a card
among 27 -- but I find the description of how to perform the trick a bit
cryptic.
7.M.5. LOONY LOOP = GORDIAN KNOT
This is a configuration of wire
loops attached to a rod with a loop of string to be removed. The solution method uses the binary pattern
of the Chinese Rings. I have a note of
an early 19C version, but no details -- ??
George E. Everett, via George Barker. UK Patent 15,971 -- An Improved Puzzle. Applied: 21 Sep 1891; accepted: 24 Oct 1891. 1p + 1p diagrams. Barker states that the invention was communicated by Everett, of
Grand Island, Nebraska. There is no
indication of the binary pattern in the puzzle.
Though this puzzle does not appear in Hoffmann, 1893,
Hordern has included it in a photo of Victorian puzzles omitted by Hoffmann on
p. 256 of Hoffmann‑Hordern.
Unfortunately, this example has no date associated with it.
It
appears in Jaques puzzle boxes of c1900, named The Canoe Puzzle. I have an example and Dalgety has several
examples of these boxes with the puzzle and the solution which says that if the
cord gets entangled, it can be cut and replaced!
M. Adams. Indoor
Games. 1912. Pp. 337‑341. The
double link (= Loony Loop or Satan's Rings).
Ch'ung‑En Yü.
Ingenious Ring Puzzle Book.
1958. Op. cit. in 7.M.1. P. 21 shows a simple version.
New section.
The problems here are usually
electronic puzzles with an array of lightable buttons which can take on two
states -- lit and unlit. Pressing a
button toggles the lights on a certain pattern of buttons. The earliest commercial example I know is
the XL-25. The box says Patent No.
122-8201 061. However, there is an
older similar electrical switching problem by Berlekamp -- see below. There are also mechanical versions, such as
Game Jugo from Japan (mid 1980s?), which has 15 petals such that turning one
over turns over some others, and Orbik -- see below. A number of further versions have appeared and I have seen some
incorporated in game packages on computers.
Rubik's Clock is essentially the same kind of problem except that the
states of the clocks and the possible turns are (mod 12) instead of (mod 2),
though the main interest centres on whether a clock is correct or not. Likewise Orbik is essentially the same, but
with values (mod 4).
Elwyn Berlekamp.
Switching Game. Devised and
actually built at Bell Labs, c1970.
This is an n x n array of lights with 2n
switches which will invert the lights in a row or column. Berlekamp's version is n = 10.
See Fishburn & Sloane below.
Lazlo Meero. The
XL-25. Exhibited at the London Toy
Fair in 1983 and marketed by Vulcan Electronics of London. The box says Patent No. 122-8201 061.
David Singmaster.
The XL-25. Cubic Circular 7/8
(Summer 1985) 39-42. The XL-25 has
a 5 x 5 array of buttons and two choices of toggle patterns -- a simple
cross of 5 cells and the pattern of a cell and the cells which are a
knight's move away. In both cases,
cells off the edge are simply lost. I
show that such problems with n buttons can be specified by an n x n
input/output or transition binary matrix A = (aij),
with aij = 1 if pressing the j-th button toggles the
i-th button; otherwise aij = 0. If
x is a binary column vector
showing which buttons are pushed, then
Ax is a binary column vector showing
which buttons have been toggled. From a
random start, the object is to light all buttons. So the general solution of the problem is obtained by inverting A (mod 2). For the
5 x 5 array, the
knight version is invertible and I sketch its inverse. But the cross version gives a matrix of
rank 23. I note that when the matrix is singular, the Gauss-Jordan
elimination method for the inverse yields the null space and range and a kind
of pseudo‑inverse, allowing for complete analysis -- in theory. (More recently, I have seen that this is a
proper pseudo-inverse and all the pseudo-inverses of A are computable by this
process.) Again I sketch the solution
process. (I now incorporate this idea
in my teaching of linear algebra as a handout
"On trying to invert a singular matrix".) I corresponded with Meero who had obtained
some similar results and showed that each feasible pattern for the cross
version could be obtained in at most
15 moves. He also showed that if the input/output
matrix A is symmetric and reflexive (i.e.
aii = 1),
then one can turn on all the lights, starting with them all off. He studied the cross version on n x n
boards up to
n = 100. A friend of
Meero studied fixed point patterns such that
Ax = x and anti-fixed points
where Ax is the vector complementary to
x. Meero asserts the knight
version has A invertible if n ¹ 6,
7, 8 (mod 9). I wondered what happens
for cylindrical or toroidal boards, but made no attempt to study them.
In Spring 1986, I applied this method to a Japanese puzzle
called Game Jugo which Edward Hordern had shown me. This has
15 two-sided 'petals' around a
centre which has four pointers. When
one pointer points to petal 1, the others point to 4, 8, 11
and turning one of these over turns over the others in this set. Since the sum of all rows of the transition
matrix is the zero vector, it follows that the matrix is singular. I found that it had rank 12,
so there are just 212
= 4096 achievable patterns, each of
which has 23 = 8 solutions.
It turns out that there are three groups of 5 petals, e.g. {1, 4, 7, 10, 13}, such that the sum of the turns in a group must be 0
for an achievable pattern. From
this, I showed that any achievable pattern could be reached in at most 6 moves, determined how many patterns required each number of moves
and showed that the average number of moves was 4.6875. I am very keen to
get an example of this and/or its instructions (preferably with an English
translation).
Donald H. Pelletier.
Merlin's magic square. AMM 94:2
(Feb 1987) 143‑150. Merlin is a
product of Parker Brothers and provides several games, including a 3 x 3
binary button game. If we number
the cells 1, 2, 3; 4, 5, 6;
7, 8, 9; then: pressing
1 toggles 1, 2, 4, 5; pressing
2 toggles 1, 2, 3;
pressing 5 toggles
2, 4, 5, 6, 8 (same as the cross
on the XL-25). The object is to light
all but button 5. He develops the binary vectors as above,
finds that the transition matrix is invertible, computes the inverse and
answers a few simple questions.
T. E. Gantner. The
game of quatrainment. MM 61:1 (Feb
1988) 29-34. Considers a game on a 4 x 4
field where a corner move reverses the six cells closest to the
corner; an edge move reverse the the
neighbouring cells and a centre move reverse the cell and its four neighbours
(i.e. the + pattern). Sets up the
matrix approach and shows the transition matrix is invertible, finding inputs
which reverse just one cell. Modifies
the moves and finds versions of the game with ranks 12 and 14.
Wiggs, Christopher C.
& Taylor, Christopher J.
C. US Patent 4,869,506 -- Logical
Puzzle. Filed: 29 Jul 1988; patented: 26 Sep 1989. Cover page + 2pp + 6pp diagrams. This is the patent for what was marketed by
Matchbox as Rubik's Clock in 1988. The
address of Wiggs and Taylor is just across the street from Tom Kremer's firm
which is Rubik's agent. This is
essentially the same kind of problem except that the states of the clocks and
the possible turns are (mod 12) instead of
(mod 2), though the main
interest centres on whether a clock is correct or not. 18
clock dials -- 9 on front and 9 on back, both in a 3 by 3
pattern. Four drive wheels on
the edges, connected to the corner clocks, but their effects on other clocks
are determined by the positioning of
4 buttons in the middle of the
puzzle, giving 30 types of move. The four corner front clocks are connected to the four corner
rear clocks, so there are 14 independent motions to make and the
input/output matrix is 30 by 14.
Daniel L. Stock.
Merlin's magic square revisited.
AMM 96:7 (Aug/Sep 1989) 608-610.
He gives an easy algorithm for solving the problem by doing edges, then
corners, then middle.
P. C. Fishburn & N. J. A. Sloane. The solution to Berlekamp's switching
game. Discrete Mathematics 74 (1989)
263-290. They describe Berlekamp's
game, with photo, as a coding theory problem.
The transition matrix A here is
2n x n2. For any
given initial state x0, consider all the states that can be achieved
from it, say S(x0) = { x0 + Ax | x Î Z22n
}. We might expect S(x0) to have 22n states, but reversing all rows is the same
as reversing all columns -- and there is no other dependence -- so there
are 22n‑1 states.
Among all these states, there is one with a minimal number, f(x0), of lights turned on. The
covering radius R of the code formed by the 2n
rows and columns, considered as words in Z2n^2, is the maximum of these minimal numbers, i.e. min { f(x0) | x0
Î Z2n^2 }. These codes are called 'light‑bulb'
codes and have been investigated since c1970 since they have the smallest known
covering radius. From our game point of
view, the problem corresponds to finding the most-unsolvable position and R is
a measure of unsolvability. The values
for R
were known for n £
5. The authors use extensive hand
computing to extend this up through n =
9 and then a lot of computer time to
get to n = 10. The values of R for n = 1, .., 10 are:
0, 1, 2, 4, 7, 11, 16, 22, 27, 34. That is, for the 10 x
10 game, there is an array of 34
turned-on lights which cannot be reduced to a smaller number of
turned-on lights by any inversion of rows and columns.
Orbik. Orbik is a
ring of 12 wheels, each having
4 colours but just one colour
can be seen through the top cover of windows.
There are three marks. When one
mark is at 1, the others are at 4 and
8. When the top cover is turned
ahead, the marked wheels move forward one colour. A backward turn leaves everything fixed, but moves the position
of the marks. Made by James
Dalgety. I believe it was Edward
Hordern's exchange gift for a puzzle party, c1993.
Edward Hordern.
Orbik. CFF 29 (Sep 1992)
26-27. ??NYR. Orbik is a ring of
12 wheels, each having 4
colours but just one colour can be seen through the top cover of
windows. There are three marks. When one mark is at 1,
the others are at 4 and
8. When the top cover is turned
ahead, the marked wheels move forward one colour. A backward turn leaves everything fixed, but moves the position
of the marks.
Ralph Gasser.
Orbik. CFF 32 (Aug 1993)
26-27. He counts both forward and
backward turns and finds there are
60 antipodal positions
requiring 54 moves to solve. The
shortest processes for moving a single wheel by 1, 2, 3 colours take 29, 28, 25
moves. If a sequence of turns in
the same direction is counted as a single move, there are 4
antipodal positions requiring 23 moves to solve and the single wheel
processes take 9, 7, 9 moves.
Revital Blumberg, Michael Ganor & Avish J. Weiner. US Patent 5,417,425 -- Puzzle Device. Filed: 8 Apr 1994; patented: 23 May 1995.
Cover page + 3pp + 1p correction + 4pp diagrams. This is the patent for Lights Out, which is
essentially identical to XL-25, but has some additional patterns. Patents 5,573,245 & 5,603,500, granted
to different groups of people, continue this.
No reference to Meero or any Hungarian patent, but cites Parker
Brothers' Merlin as undated.
Dieter Gebhardt.
Cross pattern piling. CFF 33
(Feb 1994) 14-17. Notes that Dario Uri
independently invented the XL-25 idea with the cross pattern -- he called it
Matrix of Lights. Gebhardt modifies the
problem by making two ons remain on.
Thus his computation of Ax is an ordinary matrix product and he wants
results with each entry the same. If
one thinks of the cross shapes as five cubes piled onto the board, the sought
result is a uniformly stacked board of some height h. This also allows for
some cell to be turned on several times.
Thus we are trying to solve Ax =
hJ, where J is the vector of
all 1s, h is a positive integer
and x
is a vector with non-negative integer entries. Obviously the minimal value of
h is wanted. He determines solvability and all minimal
solutions up to 8 x 8, with
9 x 9 given as a contest.
Tiger Electronics, 980 Woodlands Parkway, Vernon Hills,
Illinois, 60061, USA & Belvedere House, Victoria Avenue, Harrogate,
UK. Lights Out. Model 7-574, 1995. Essentially the same concept as the XL-25 with its 'cross'
pattern. With lots of preprogrammed
puzzles, random puzzles and option to input your own puzzles. The longest solution is 15
moves, as found by Meero for the
XL-25.
Uwe Mèffert produced Orbix (or Light Ball) in 1995 for
Milton Bradley. I advised a bit on the
design of the games. This is a sphere
with 12 light buttons in the pattern of a dodecahedron. There are four different games. The object is to turn all lights on, but in
some games, one can also get all lights off.
However, only the first game is a linear transformation in the sense
discussed above. The later games have
rules where the effect of a button depends on whether it is lit or not and even
on whether the opposite button is lit or not.
Nonetheless all examples are solvable in 12 moves or less.
Edward Hordern.
What's up? CFF 38 (1995)
38. ??NYR. Discusses Tiger Electronics'
Lights Out.
Dieter Gebhardt & Edward Hordern. How to get the lights of "Lights
Out" out. CFF 39 (1996)
20-22. ??NYR. Sketches a solution.
Edward Hordern.
What's up? CFF 41 (Oct 1996)
42. Discusses Tiger Electronics' Deluxe Lights Out which has a 6 x 6 array with several options -- one can affect
five lights in the form of a + or of a
x; a button can have effect only
if it is lit, or alternately lit/unlit.
4 9 2 The 3 x 3
magic square is usually given in the form on the left.
3 5 7 We denote each of the 8 possible forms
by its top row. I.e. this is the
8 1 6 492
form. All Chinese material seems
to give only this form, called the
Lo
Shu [Lo River Writing].
7
2 An unrelated
diagram, shown on the left, is called the Ho Thu
8 3 5 4 9 diagram [River Plan]. See 7.N.5 for magic versions of this shape.
1
6
Semi‑magic denotes a square whose rows and columns add
to the magic constant, but not the diagonals.
Pandiagonal means that the 'broken diagonals' also add
to the magic constant. Lucas called
these diabolic and they are also called Nasik,
as they were studied by Frost, who was then living in Nasik, India.
Associated or complementary means that two cells symmetric with respect
to the centre add to n2 + 1.
See 7.AC.3 for related pan-digital
sums.
The early history of magic squares remains
rather obscure. In particular, the
first four Chinese sources below are extremely vague! Cammann‑4 argues that magic squares had powerful mystic
meanings to the Chinese, Indians and Arabs, hence were not explicitly described
in writing. However, some modern
scholars doubt if the Chinese had any magic square before 10C! -- cf Gardner,
1996.
There are several surveys of some or
all of the history of magic squares which I list first for later
reference. These provide many more
references.
17‑20C material has generally
been omitted, but see Bouteloup. Smith
& Mikami discuss several workers in Japan, but I've omitted some of them.
Wilhelm Ahrens - 1.
Studien über die "magischen Quadraten" der Araber. Der Islam 7 (1917) 186‑250.
Wilhelm Ahrens - 2.
Die "magischen Quadrate" al-Būnī's. Der Islam 12 (1922) 157‑177.
Schuyler Cammann ‑ 1.
The evolution of magic squares in China. J. Amer. Oriental Soc. 80 (1960) 116‑124.
Schuyler Cammann ‑ 2.
The magic square of three in old Chinese philosophy and religion. History of Religions 1 (1961) 37‑80. ??NYR
Schuyler Cammann ‑ 3.
Old Chinese magic squares.
Sinologia 7 (1962) 14‑53.
Schuyler Cammann ‑ 4.
Islamic and Indian magic squares I
& II. History of Religions 8 (1968‑69) 181‑209 & 271‑299.
Bibhutibhusan Datta & Avadhesh Narayan Singh. Magic squares in India. Indian J. History of Science 27:1 (1992)
51-120. All references to Datta &
Singh in this section are to this paper, not their book.
Menso Folkerts. Zur
Frühgeschichte der magischen Quadrate in Westeuropa. Sudhoffs Archiv 65:4 (1981) 313-338.
Heinrich Hermelink.
Die ältesten magischen Quadrate höher Ordnung und ihre
Bildungsweise. Sudhoffs Arch. 42 (1953)
199‑217.
Lam Lay Yong.
1977. See under Yang Hui
below. Her commentary surveys the
history.
Needham. 1958. Pp. 55‑61. See also: vol. 2, 1956,
pp. 393 & 442; Vol. 5, Part IV,
1980, pp. 462-472.
Jacques Sesiano ‑ I
& II. Herstellungsverfahren magischer Quadrate aus
islamischer Zeit (I) & (II).
Sudhoffs Arch. 64 (1980) 187‑196
& 65 (1981) 251‑265.
A number of the entries in 5.A --
The Fifteen Puzzle -- refer also to magic squares. See: Loyd (1896);
Cremer (1880); Tissandier (1880
& 1880?); Cassell's (1881); Hutchison (1891).
Some entries in 5.A and here give
problems of sliding the Fifteen Puzzle into a magic square. See:
Dudeney (1898); Anon &
Dudeney (1899); Loyd (1914); Dudeney (1917); Gordon (1988) in 5.A
and Ollerenshaw & Bondi
below.
Anon. Shu
Ching. c‑650. Part V, book IV, The Great Plan --
commentary and book XXII, The Testamentary Charge. IN: J. Legge, trans. The Texts of Confucianism, Translated: Part
I. The Shû King, the religious portions
of the Shih King, the Hsiâo King. OUP,
1879, pp. 138‑139 & 239. P.
138 discusses the Lo Shu and says it does not occur. On p. 139, we see:
"To [Yu], Heaven gave the Great Plan with its nine divisions
..." Various commentators,
from Gan‑Kwo on, have asserted that this was the Lo‑shu
which appeared on the back of a tortoise in the river Lo. Legge says there is no evidence to connect
the Lo‑shu with the Great Plan and that the commentators are indulging in
leger‑de‑plume. P. 239
mentions the River Plan.
(See
also: J. Legge, trans.; The Chinese
Classics, etc.; Vol. III -- Part II; Trübner, London, 1865; pp. 321‑325
& 554. This gives the Chinese and
the English, with extensive notes.)
At
this time, the number 'nine' was used to describe the largest number and hence
does not necessarily imply 32.
Anon. Lun Yu
(Confucian Analects). c‑5C. Book IX, Tsze Han; chap. VIII. IN:
J. Legge, trans. The Chinese
Classics, etc. vol. 1, Confucian Analects, The Great Learning, and the Doctrine
of the Mean. Trübner, London, 1861, p.
83. = The Life and Teachings of
Confucius; Trübner, London, 1869, pp. 169‑170, ??NX. Also in:
A. Waley; The Analects of Confucius; Allen & Unwin, London,
1949, p. 140. "The river sends
forth no map."
Chuang Tzu (= Kwang‑Sze). The Writings of Kwang‑Sze.
c‑300. Part II, sect.
VII = Book XIV, Thien Yu (The
Revolution of Heaven). IN: J. Legge, trans. The Texts of Tâoism. OUP, 1891.
Vol. 1, p. 346. Refers to
"the nine divisions of the writing of Lo."
Anon. Ta Chuan (=
Hsi Tzhu Chuan) (The Great Commentary on the I Ching [= Yi Jing]). c‑300?? (Needham, vol. 2, p. 307, says
c‑100 and vol. 5:IV, pp. 462-463, says -2C) IN: J. Legge, trans. The Texts of Confucianism, Part II: The Yî
King. OUP, 1882. Appendix III, sect. 1, chap. 12,
art. 73, p. 374 & note on p. 376.
[There is a 1963 Dover ed. of Legge's 1899 edition.] Also as:
Part I, chapter IX -- On the Oracle.
IN: The I Ching, translated by
R. Wilhelm and rendered into English by C. F. Baynes, 3rd ed., 1968, Routledge
and Kegan Paul, London, pp. 308‑310.
The text is: "The Ho gave forth the map, and the Lo the writing, of
(both of) which the sages took advantage." This occurs just after paragraphs on the origin of the hexagrams
and legend says the Ho Thu inspired the creation of the 8 trigrams. Legge says the original Ho Thu map was
considered to be lost in the -11C and the earliest reconstruction of it was
presented during the reign of Hai Zung in the Sung Dynasty (1101-1125). The I Ching is often cited but only this
later commentary mentions an association of numbers with concepts. Later commentators interpret this
association as referring to the Ho Thu and Lo Shu diagrams, though this is not
obvious from the association -- the names were not associated with the diagrams
until about the 10C -- see Xu Yiu below.
See Needham, vol. 2, pp. 393 & 442 for discussion of the
interpolation of the diagrams into the I Ching.)
E. John Holmyard.
Alchemy. Penguin, 1957, pp.
36-38, discusses magic squares in relation to Chinese alchemy and Taoism. He says the Taoist emphasis on the number 5
is related to its being the central value of the order 3 magic square. He says this relation has been studied by H.
E. Stapleton, but gives no reference.
Stapleton says the square of order 3 was the ground plan of the
Ming-Tang or Hall of Distinction. This
Hall was used for the proclamation of monthly ordinances and the Calendar
(which was partly lunar and hence of variable length). When in the Ming-Tang, the Emperor became
the incarnation of the god and hence the ground plan became of major importance
in Chinese alchemy. Stapleton
conjectures that the original numbering of the
3 x 3 array of rooms in the Hall
may not have been magic, but would have had 5 in the middle and that the magic
numbering may have occurred at some time and been recognised as having special
properties. Holmyard indicates the
Ming-Tang arose about -1000. All in
all, these statements do not agree with most of the other material in this section
and it would be good to locate the work of Stapleton (??NYS), which is presumably
well-known to students of ancient chemistry/alchemy.
Needham, Vol. 5:IV, 1980, pp. 462-472. Cites Stapleton on p. 462 and indicates his
work is a bit cranky, but I haven't got the details yet. He goes on to discuss why 9 was so important
to the Chinese. He describes the tour
of the pole-star sky-god Thai I which went through the nine cells of the Lo Shu
in the order: 5, 1, 2, 3, 4, 5, pause, 5, 6, 7, 8, 9, 5. His fig. 1535 shows this from a Tang
encyclopedia, though this has the 2 7
6 orientation of the square. The Chinese could see Yin and Yang (= even
and odd), the Four Seasons and the Five Elements, and the Nine Directions of
space, all in the Lo Shu. Consequently
it was not revealed to the general public until the end of the Tang (618‑907). Needham then discusses the influence of the
Lo Shu on Arabic alchemical thought.
Nâgârjuna. c1C. Order 4 squares, including one later called
Nâgârjunîya after him, described in a MS on magic called Kakşapuţa
[NOTE: ş, ţ
denote s, t
with underdot.], nd. ??NYS. [A. N. Singh; History of magic squares in
India; Proc. ICM, 1936, 275‑276.
Datta & Singh.] Datta &
Singh say Nâgârjuna gives several rules for forming magic squares of order 4,
but all the examples given do not use consecutive values, much less the first
16 positive integers, e.g.
n-3, 1, n-6,
8; n-7, 9, n-4,
2; 6, n-8, 3, n-1;
4, n-2, 7, n-9,
which has
magic constant 2n. The Nâgârjuna square is the following, with
an unrelated structure and with constant 100.
30,
16, 18, 36; 10, 44, 22, 24; 32, 14, 20, 34; 28, 26, 40, 6.
There is also
no reference to Nâgârjuna or his book.
Can anyone provide this?
Tai the Elder. Ta
Tai Li Chi (Record of Rites). c80.
Chap. 67, Ming Thang. ??NYS
(See Needham, p. 58.) Chap. 8,
p. 43 of Szu‑pu ts'ung‑k'an edition, Shanghai, 1919‑1922. Describes the 492 form. (See Cammann‑2.) (Cammann‑1, Lam and Hayashi
say this is the first clear reference.)
Anon. I Wei Chhien
Tso Tu. c1C. Chap. 2, p. 3a.
??NYS (Translated in Needham, p.
58.)
Anon. Lî Kî. c2C.
Book VII -- Lî Yun, sect. IV.
IN: J. Legge, trans. The Texts of Confucianism, Part III:
The Li Ki, I‑X. OUP,
1885. Pp. 392‑393. "The Ho sent forth the horse with the
map (on his back) 1. 1
The famous 'River Map' from which, it has been fabled, Fû-hsî
fashioned his eight trigrams.
See vol. xvi, pp. 14-16."
This last reference is ??NYS.
Theon of Smyrna.
c130. Part B: Βιβλιov
τα τησ εv Αριθμoσ
Μoυσικησ
θεωρηματα
Περιεχov
(Biblion ta tes en Arithmos Mousikes Theoremata Periechon). Art. 44.
IN: J. Dupuis, trans.; Théon de Smyrne; Hachette, Paris,
1892; pp. 166‑169. (Greek &
French.) Natural square -- often
erroneously cited as magic and used to 'prove' the Greeks had the idea of magic
squares.
Xu Yiu (= Hsu Yo =
Xu Yue). Shu Shu Ji Yi (= Shu Shu Chi I) (Memoir on Some Traditions of Mathematical Art). 190(?).
??NYS. Ho Peng Yoke [Ancient Chinese Mathematics; IN:
History of Mathematics, Proc. First Australian Conf., Monash Univ., 1980;
Dept. of Math., Monash Univ., 1981, pp. 91‑102], p. 94, says that this is
the earliest Chinese text to give the order 3 square.
The date and authorship of this is
contentious. Current belief is that
this was written by Zhen Luan (= Shuzun)
in c570, using the name of Xu
Yue. Li & Du, pp. 96‑97, say
that this work first introduces the diagram.
The diagram was called the "nine houses computation". The diagram was connected with the Yi Jing
commentary in the 10C and then renamed
Lo Shu. After the 13C, magic
squares were called zong heng tu (row and column diagrams).
Needham,
vol. 5:IV, p. 464, considers this as being c190, referring the Chen Luan as a
commentator on it. He calls the diagram
"Nine Hall computing method".
Varahamihira (= Varāhamihira (II)). Bŗhatsamhitā [NOTE: ŗ
denotes r with an underdot it and the m should have an underdot.]. c550.
Hayashi, below, cites a Sanskrit edition (NYS) and the following.
M.
Ramakrishna Bhat.
Varāhamihira's Bŗhat
Samhitā [NOTE: ŗ denotes
r with an underdot it and the
m should have an overdot.] with
English Translation, Exhaustive Notes and Literary Comments. 2 vols, Motilal Banarsidass, Delhi, 1981‑1982. Vol. II: Chapter LXXVII -- Preparation of
perfumes, pp. 704-718. On pp. 714-715
is the description of a 4 x 4 array:
2,
3, 5, 8; 5, 8, 2, 3; 4, 1, 7, 6; 7, 3, 6, 1, with some
cryptic observations that any mixture totalling 18 is permitted, e.g.
"by combining the four corners, or four things in each corner, or the
central four columns, or the four central ones on the four sides." As given, many of the groups indicated do
not add up to 18, nor do the columns. However, the following article notes that
the bottom row should read
7, 6, 4, 1!!
(Datta & Singh have this correct.)
This
material is described and analysed in:
Takao Hayashi; Varāhamihira's pandiagonal magic square of the order
four; HM 14 (1987) 159-166. He gives
the book's name as
Bŗhatsamhitā [NOTE:
ŗ denotes r
with an underdot it and the m
should have an underdot.] and
says the material is in Chapter 76 (Combinations of perfumes). He gives the correct form of the array. He notes that the array is a pandiagonal
magic square with constant 18, except the entries are 1, ..., 8
repeated twice. Hayashi believes
that Varāhamihira must have known one of the actual magic squares which
yield this square when the numbers are taken
(mod 8). He shows there are
only 4
such magic squares, two of which are pandiagonal. One of the pandiagonal squares is a rotation
of:
8, 11, 14,
1; 13, 2,
7, 12; 3, 16, 9,
6; 10, 5,
4, 15,
which he
describes as the most famous Islamic square of order 4 described in Ahrens‑2. Hayashi feels that order 4
squares must originate in India, contrary to Cammann's thesis. (See also Singh, op. cit. at Nâgârjuna,
above, and Ikhwān al-
Şafā’ [NOTE: Ş denotes
S with an underdot.], below.) Datta & Singh just say that
Varāhamihira gives a magic square.)
Datta
& Singh give the square beginning
2, 3, 5, 8, and says that it is
a special case of
n-7, 3, n-4,
8; 5, n-1, 2, n-6;
4, n-8, 7, n-3; n-2,
6, n-5, 1.
Taking n = 9
gives Varāhamihira's square and taking n = 17 gives the square
with 8, 11, 14, 1 as
right hand column. One can also take
the other set of the first eight integers as given, getting a square
starting 2, n-6, 5, n-1.
Varāhamihira's square has many magic properties and he called it
'Sarvatobhadra' (Magic in all respects).
They say these properties are fully described by Varāhamihira's
commentator Bhaţţotpala [NOTE:
ţ denotes t
with an underdot.] in 966.
Jabir ibn Hayyan (=
Jâbir ibn Hayyân = Geber)
(attrib.). Kitâb al‑Mawâzin (Book of the Balances). c800.
??NYS -- discussed in Ahrens-1.
The Arabic and a French translation are in: M. Berthelot; La Chimie au Moyen Age: Vol. 3 -- "L'Alchimie
Arabe"; Imprimerie Nationale, Paris, 1893. The text is discussed on pp. 19-20, where he refers to the magic
square of Apollonius, with a footnote saying 'De Tyane'. The text is given on p. 118 (Arabic section)
& 150 (French section). Gives 3 x 3
square in form 492.
"Here
is a figure divided into three compartments, along the length and along the
width. Each line of cells gives the
number 15 in all directions.
Apollonius affirms this is a magic tableau formed of nine cells. If you draw this figure on two pieces of
linen [or rags], which have never been touched by water, and which you place
under the feet of a woman, who is experiencing difficulty in childbirth, the
delivery will occur immediately."
The
French is also in Ahrens-1, who notes that the square does not appear in the
few extant writings of Apollonius of Tyana (c100).
Hermelink
mentions this as the earliest Arabic square, but gives no details. Needham, vol. 5:IV, p. 463, says Cammann
gave this, and dates this as c900.
Folkerts gives this as the first Arabic example.
Suter,
pp. 3‑4, doesn't mention magic squares for ibn Hayyan, but this
appearance is simply in a list of questions on properties of animals,
vegetables and minerals, so hardly counts as mathematics.
Holmyard
[op. cit. above, pp. 74-75] discusses the work of Kraus and Stapleton on
Jabir. Jabir considers the numbers 1, 3, 5, 8
as of great importance -- these are the entries in the lower left 2 x 2
part of his magic square. These
add to 17 and everything in the world is governed by this number! He also attaches importance to 28
which is the sum of the other entries.
Holmyard asserts this magic square was known to the Neo-Platonists of 3C
-- an assertion which I have not seen elsewhere. Jabir uses ratios
1/3 and 5/8
extensively in his alchemical theories.
‘Ali ibn Sahl Rabbān al-Tabarī (d. 860). Paradise of Wisdom. This is a gynaecological text discovered by
Siggel. ??NYS -- described in Needham,
vol. 5:IV, p. 463. Example of a magic
square used as a charm in cases of difficult labour. Needham thinks this is the earliest Arabic magic square.
Tâbit
ibn Qorra (= Thâbit ibn Qurra). c875.
This is the first reference to magic squares in Suter, on pp. 34‑38,
but he seems to say that the work has not survived and Ahrens-1 confirms
this. Needham, vol. 5:IV, p. 463, says
Cammann wonders if this ever existed.
Ikhwān al‑Şafā’ [NOTE: Ş
denotes S with an underdot.]. Rasā’il (Encyclopedia) (??*).
c983. Cairo edition, 1928, p.
69. ??NYS. Paris MS Arabe 2304 (formerly 1005) of this is the work
translated by F. Dieterici as: Die
Propaedeutik der Araber im zehnten Jahrhundert; Mittler & Sohn, Berlin,
1865; reprinted as vol. 3 of F.
Dieterici; Die Philosophie bei den Araben im X.Jahrhundert n. Chr.; Olms,
Hildesheim, 1969. Pp. 42‑44 (of
the 1969 ed.) shows squares of orders
3, 4, 5 and 6. The order 3
square is in the form 276. The text refers to orders 7, 8 and 9
and gives their constants. On p.
44, the translator notes that the Arabic text has some further incomplete
diagrams which are not understandable.
Hermelink and Cammann‑4 say that the Cairo ed. is the only version
to give these diagrams. Ahrens-1 says
it continues with a cryptic description of the use of a 9 x 9
square on two sherds, which have not been sprinkled with water, for
easing childbirth. The prescription has
several more details than ibn Hayyan's.
The Arabic text and rough translation are
given in: van der Linde; Geschichte und
Literatur des Schachspiels; op. cit. in 5.F.1, vol. 1, p. 203. This is a description of the 3 x 3
magic square, form 492 or
294, in terms of chess
moves. Ahrens-1 says that Ruska tells
him that much, if not all, of the magic square and adjacent material in
Dieterici was added later to the Encyclopedia.
Ruska says there are many errors in the translation and Ahrens cites
several further errors in nearby material.
Hermelink
describes the methods and reconstructs the squares of orders 7, 8, 9
from the 1928 Cairo ed. and van der Linde. Cammann‑4 says he obtained the same squares independently,
but he doesn't agree on all the interpretations. He feels there are Chinese influences, possibly via India, and
gives his interpretations.
The
square of order 4 is given by Hayashi, op. cit. above at
Varāhamihira, as:
4, 14, 15,
1; 9, 7,
6, 12; 5, 11, 10, 8;
16, 2, 3, 13. This is not
pandiagonal. The square of order 7 is
doubly bordered -- the first such.
(Abû ‘Alî el‑Hasan ibn el Hasan) (the Hs
should have dots under them) ibn el‑Haitam. c1000.
??NYS. Suter, p. 93, says he
wrote: Über die Zahlen des magischen
Quadrates. He cites Woepcke, ??NYS, for
MS details. Ahrens-1 indicates that the
work does not exist.
J. H. Rivett‑Carnac.
Magic squares in India. Notes
and Queries (Aug 1917) 383. Quoted
in: Bull. Amer. Math. Soc. 24 (1917)
106, which is cited by: F. Cajori; History of Mathematics; op. cit.
in 7.L.1, pp. 92‑93. The square
is in the ruins of a Hindu temple at Dudhai, Jhansi, attributed to the
11C. It is 4 x 4, and each 2 x 2
subsquare also adds to 34, but
the full square is not given.
Cammann‑4,
p. 273, says this is the same as the Jaina square at Khajuraho described below
and cites the archaeological report, ??NYS.
He is dubious about the date.
(Muhammed ibn Muhammed ibn
Muhammed, Abû Hâmid,) el‑Ġazzâlî
(= al‑Ghazzali).
Mundiqh. c1100. ??NYS -- described by Ahrens-1. Ahrens cites two differing French editions
which give 3 x 3 forms
492 and 294.
He says the latter is a transcription error. Al‑Ghazzali's text
is very similar to ibn Hayyan's, though one translator says the cloths are
moistened. Ahrens discusses this point. He says that amulets with this magic square,
called 'seal of Ghazzali' are still available in the Middle East.
Lam,
p. 318, cites this as an early Arabic magic square, but doesn't give
details. Suter, p. 112, doesn't mention
magic squares.
Abraham ibn Ezra.
Sepher Ha‑Schem (Book of
Names), 12C, and Jesod Mora,
1158. ??NYS -- both are
described in: M. Steinschneider; op.
cit. in 7.B and excerpted in the next
item. The material is art. 13, pp.
95++. The 672 form is shown on p.
98. Steinschneider, p. 98, also gives
the 492 form and says it appears in
Jesod Mora, described on pp. 99‑101. Ahrens-1 only mentions that Sepher Ha‑Schem gives an order 3 square.
Abraham ibn Ezra.
Sêfer ha‑Echad.
c1150. Translated and annotated
by Ernest Müller as: Buch der Einheit;
Welt-Verlag, Berlin, 1921, with excerpts from:
Jessod Mora, Sefer ha‑Schem, Sefer ha-Mispar and his Bible commentary.
Abraham ibn Ezra.
Ta'hbula. c1150. ??NYS.
Some source says this has magic squares, but Lévi's comments in 7.B
indicate that this book is only concerned with the Josephus problem. Steinschneider's description of Tachbula,
pp. 123‑124 of the above cited article, makes no mention of a
magic square.
Anon. Arabic MS,
Fatih 3439. c1150. ??NYS.
Described in Sesiano‑I.
Construction of squares of almost all orders. Describes: a method
of ibn al‑Haytham (c1000) for odd orders; a method of
al‑Isfarâ’inî (c1100) for
evenly even orders; a method of ibn al‑Haytham for oddly even squares which only works for
order º 2 (mod 8). Suter, p. 93, mentions ibn al‑Haitam -- see above, c1000.
Tshai Yuan‑Ting.
Lo‑Shu diagram, c1160.
??NYS -- Biggs cites this as being in Needham, but the only references
to Tshai in Needham refer to indeterminate analysis (p. 40) and geology (p.
599). Paul Carus [Reflections on magic
squares, IN: W. S. Andrews, op. cit. in 4.B.1.a, pp. 113-128, esp. p.
123] says that Ts'ai Yüan-Ting (1135-1198) gives the Lo-Shu diagram 'but
similar arithmetical diagrams are traceable as reconstructions of primitive
documents among scholars that lived' during 1101-1125. Datta & Singh cite this and say this is
the earliest Chinese interpretation of the Lo-Shu as a magic square. This ignores Tai the Elder, I Wei
Chhien Tso Tu, and Xu Yiu,
though the first two are a bit vague.
(Ahmed (the h should have an underdot) ibn ‘Alî ibn Jûsuf)
el‑Bûnî, (Abû'l‑‘Abbâs, el‑Qoresî) = Abu‑l‘Abbas al‑Buni (??= Muhyi'l‑Dîn Abû’l-‘Abbâs al‑Bûnî -- can't relocate my source of this
form.) Kitâb et‑chawâşs
[NOTE: ş denotes an s with an underdot.] (= Kitab al Khawass
or Sharkh ismellah el‑a‘zam??) (The Book of Magic Properties). c1200.
Suter, p. 136, mentions magic squares.
??NYS -- described in: Carra de
Vaux; Une solution arabe du problème des carrés magiques; Revue Hist. Sci. 1
(1948) 206‑212. Construction of
squares of all orders by bordering.
Hermelink refers to two other books of
al‑Buni, ??NYS.
al‑Buni. Sams al‑ma‘ârif = Shams al‑ma‘ârif al‑kubrâ = Šams al-ma‘ārif. c1200.
??NYS. Ahrens-1 describes this
briefly and incorrectly. He expands and
corrects this work in Ahrens-2, which mainly deals with 3 x 3
and 4 x 4, the various sources and the accumulated
errors in most of the squares. He notes
that a 4 x 4 can be based on the pattern of two orthogonal Latin squares of
order 4, and Al-Buni's work indicates knowledge of such a pattern,
exemplified by the square (discussed by Hayashi under Varāhamihira, c550)
8, 11, 14,
1; 13, 2,
7, 12; 3, 16, 9,
6; 10, 5,
4, 15 considered (mod 4).
Al‑Buni gives several 4 x 4's, including that of
Ikhwān al‑Şafā’ (the Ş should be an S
with a dot under it), c983, which
does not have the above pattern. He
also has Latin squares of order 4 using letters from a name of God. He goes on to show 7 Latin squares of
order 7, using the same 7 letters each time -- though four are
corrupted. (Throughout, the Latin
squares also have 'Latin' diagonals.)
These are arranged so each has a different letter in the first place. It is conjectured that these are associated
with the days of the week or the planets.
In Ahrens-1, Ahrens reported that he had recently been told that Al-Buni
had an association of magic squares of orders 3 through 9
with the planets, but he had not been able to investigate this. In Ahrens-2, he is clear that al-Buni
has no such association -- indeed, there is no square of order 9
anywhere in the standard edition of the works of al-Buni.
But Folkerts says such an association was made by the Arabs, perhaps
referring to the Nadrûnî, below. See
14C & 15C entries below.
Cammann‑4,
p. 184, says this text is "deliberately esoteric ... to confuse
people" and the larger squares are so garbled as to be
incomprehensible. On pp. 200‑201,
he says this has the knight's move method for odd orders. Later it was noted that any number could be
in the centre and 1 was popular, giving the 'unit centred'
square of symbolic importance. These
squares are also pandiagonal. Al‑Buni gives many variant 4 x 4 squares with the
top row spelling some magical word -- e.g. one of the 99 names of God. He mentions a "method of the
Indians", possibly the lozenge method described in Narayana, 1356.
BM Persian MS Add. 7713.
1211? Described in Cammann‑4,
pp. 196ff. On p. 201, Cammann says p.
23 gives unit centred squares of orders
5 & 9,
pp. 112‑114 gives a rule for singly even order and p. 164 has an
order 20 square. This also has odd
order lozenge squares -- see Narayana, 1356.
It also has some examples of a form of the system of broken reversions.
Persian MS.
1212. Garrett Collection, No.
1057, Princeton Univ. See Cammann‑1 & Cammann‑4, p.
196. ??NYS
Cammann‑4,
pp. 196ff, says the above two MSS show new developments and describes
them. Diagonal rules for odd orders
first appear here and give an associated square with centre (n2 + 1)/2 which acquired mystic significance as a
symbol of Allah.
(Jahjâ (the h should have a dot under it) ibn Muhammed ibn
‘Abdân ibn ‘Abdelwâhid, Abû Zakarîjâ Neġm ed‑dîn,) known as
Ibn el‑Lubûdî (= Najm al‑Din (or
Abu Zakariya) al‑Lubudi. c1250.
Essay on magic squares dedicated to al‑Mansur. ??NYS.
Mentioned in Suter, p. 146.
Yang Hui. Hsü Ku
Chai Ch'i Suan Fa (= Xugu Zhaiqi
Suanfa) (Continuation of Ancient
Mathematical Methods for Elucidating the Strange [Properties of Numbers])
(Needham, vol. 5:IV, p. 464, gives: Choice Mathematical Remains collected to
preserve the Achievements of Old).
1275. IN: Lam Lay
Yong; A Critical Study of the Yang Hui Suan Fa; Singapore Univ. Press, 1977.
Book III, chap. 1, Magic Squares, pp. 145‑151 and commentary,
pp. 293‑322. This is the
only source for older higher order squares in China. (See Cammann‑1 and Cammann‑3 for details of
constructions.) Bordered squares of
order 5 and 7. Magic squares of orders 3
through 10, the last being only
semimagic. Methods are given for
orders 3 and 4 only.
Gives some magic circles and other forms. (Lam's commentary, p. 313, corrects the first figure on p.
150. Lam also discusses the
constructions.)
Li
& Du, pp. 166‑167, say that 6
x 6 and 7 x 7 'central' (=
bordered) squares arrived from central Asia at about this time. The
6 x 6 example on their p. 172
and the 7 x 7 example on their p. 167 are bordered.
Cammann‑4 says one of the order 8
squares is based on a Hindu construction.
Jaina square. Inscription at Khajuraho, India.
12‑13C. Ahrens-1
(p. 218) says it first appears as:
Fr. Schilling; Communication to the Math. Gesellschaft in Göttingen
[Mitteilung zur Math. Ges. in Göttingen], 31 May 1904; reported in: Jahresber. Deutschen Math.‑Verein. 13 (1904) 383‑384. (Schilling is reporting a communication from
F. Kielhorn.)
7, 12,
1, 14; 2, 13, 8, 11;
16, 3, 10, 5;
9, 6, 15, 4.
It is
pandiagonal and associated. Cammann‑4,
p. 273, says this is the same as the square reported by Rivett-Carnac (qv
above) and there claimed to be 11C?
Cammann feels it may derive from an Islamic source. See also:
Smith History II 594 and Singh, op. cit. at Nâgârjuna above. Datta & Singh give this and date it as
11C.
Five cast iron plates with
6 x 6 magic squares, late
13C(??), were found at Xian in 1956.
The numerals are similar to East Arabic numerals so these reflect the
Arabic influence on the Mongol dynasty.
Li & Du, p. 172, reproduces one.
This is on display in the new Provincial History Museum in Xian. Jerry Slocum has given me a facsimile, in
reduced size, of the same one. Can any
one supply more information about the others??
Datta & Singh give another Jaina square, from 'not later
than the fourteenth century', 'probably a very old one'. This is the following, but with all entries
multiplied by five to give a magic constant of 170, which 'is closely connected
with an ancient Jaina mythology'.
5, 16,
3, 10; 4, 9,
6, 15; 14, 7, 12,
1; 11, 2, 13,
8.
Folkerts discusses an anonymous and untitled
astrological-magic treatise which appears to derive from the court of Alfonso
the Wise in Madrid, where a similar work of al-Magriti
(al-Mağrīţī [NOTE:
ţ denotes t
with an underdot.]) (10C) had been translated in the 13C and developed
under the name Picatrix. Picatrix
refers to astrological amulets but gives no instance of a magic square on
one. Though there is no known direct
connection, Folkerts considers this treatise to be in the tradition of the
Picatrix and names it 'Picatrix-Tradition'.
He finds seven MSS of it, from early 14C onward. This specifically associated magic squares
and planets according to Folkerts' System I.
Folkerts
discusses these associations and calls them Systems I and II. For
n = 3, 4, ..., 9, they are as follows.
I: Saturn,
Jupiter, Mars, Sun,
Venus, Mercury, Moon.
II: Moon,
Mercury, Venus, Sun,
Mars, Jupiter, Saturn.
System I is
almost universal, only a 1446 Arabic MS and Cardano use system II.
Not
all writers use the same squares, but there are generally only two examples for
each order. Folkerts gives a table of
these and which authors used which squares.
Basically, there are two sets of squares, one used by Picatrix-Tradition
and Pacioli, the other by Agrippa and Cardano (in reverse).
Folkerts
says this association was done by the Arabs, but Nadrûnî (cf below) is the
earliest Arabic source I have.
In
Codex Vat. Reg. lat. 1283 is a 13C(?) fragment with a 5 x 5 magic square
associated to Mars. Paolo dell'Abbaco's
Trattato di Tutta l'Arte dell'Abacho, 1339, op. cit. in 7.E, has order 6 and 9
squares with their associations (sun and moon) given. A 15C Frankfurt MS (UB, Ms. lat. oct. 231), has some examples and
Paracelsus (1572) copied squares from different sources.
Folkerts then
discusses various constructions due to al-Buni, Moschopoulos, al‑Haitham,
etc. The 15C Frankfurt MS is the first
attempt at a theory, followed by Ries and Stifel.
Μαvoυηλ
Μoσχoπoυλoυ (Manuel Moschopoulos
-- variously spelled in Greek and variously transliterated). c1315.
MS 2428, Bibliothèque Nationale, Paris.
Paolo dell'Abbaco. Trattato
di Tutta l'Arta dell'Abacho. 1339. Op. cit. in 7.E. B 2433, ff. 20v - 21r, gives order 6 and order 9 magic
squares. The latter may be associated
with the moon, but my copy is not quite legible. Dario Uri (email of 31 Oct 2001) said he had found this MS (B
2433), which has 6x6 and
8x8 magic squares, but the
latter must be a misreading.
‘Abdelwahhâb ibn Ibrâhîm, ‘Izz eddîn el‑Haramî el
(the H
should have a dot under it) Zenġânî = ‘Abd al‑Wahhâb ibn Ibrâhîm al‑Zinjânî. Arabic MS,
Feyzullah Ef. 1362. c1340. ??NYS -- Described in Sesiano‑II. Suter, p. 144, doesn't mention magic
squares. Construction of bordered
squares of all orders.
Muhammad [the h should have an underdot) ibn
Yūnis. Compendium on construction
of bordered magic squares in MS Hüsrev Pasa 257 in the Süleymaniye Library,
Istanbul, ff. 32v-37v. Translated and
discussed in: Jacques Sesiano; An Arabic treatise on the construction of
bordered magic squares; Historia Scientiarum 42 (1991) 13-31. The actual MS was compiled in the 12th
century of the Hegira, i.e. c18C), but the treatise is undated. Sesiano compares the methods with other
medieval Arabic material, e.g. al-Buni, al-Karagi, al-Buzjani, al‑Zanjani,
so he seems to think it dates from a similar period.
Narayana Pandita (= Nārāyaņa Paņdita
[NOTE: ņ denotes n with an overdot and the d
should have an underdot.]).
Gaņita[NOTE: ņ denotes
n with an underdot.] Kaumudī (1356). Edited by P. Dvivedi, Indian Press, Benares,
1942. Part II: Introduction -- magic squares,
pp. xv‑xvi (in English); Chap. 14: Bhadra gaņita [NOTE: ņ
denotes an n with an underdot.], esp. pp. 384‑392
(in Sanskrit). Shows orders 6, 10, 14.
Shows the 8 forms of order 3. Obviously an extensive
section -- is there an English translation of this material??. (Editor refers to earlier sources: Bhairava
and Ŝiva Tāndava
[the d
should have a dot under it] Tantras, ??NYS. Cammann‑4
cites other MS sources. Singh, op. cit.
under Nâgârjuna, c1C, above, says this is the first mathematical
treatment. He says it classifies into odd, evenly even and oddly
even; gives the superposition method of
de la Hire; gives knight's move method
for 4n
and filling parallel to diagonal for odd, attributing both to previous
authors.
Cammann‑4,
pp. 274‑290 discusses this in more detail. He gives another diagonal rule, sometimes beginning and ending at
the middle of a side. He then gives a
quite different rule based on use of x
+ y with x = 0, n, 2n, ..., (n‑1)n, y
= 1, 2, ..., n with both sets of values cycling in the row,
then reversing the xs. E.g., for
n = 5, his first row of y
values is: 4 5 1 2
3 and the second is: 5 1 2
3 4. His first two rows of x values are: 15 20 0 5
10 and 20
0 5 10 15. Reversing the xs and adding gives rows: 14
10 1 22 18 and
20 11 7 3 24.
This process gives a central lozenge (or diamond) pattern of the odds
and has an extended knight's move pattern.
He extends this to doubly even squares.
He also gives the 'method of broken reversions' for singly‑even
squares in three forms -- cf. C. Planck; The Theory of Reversions, IN: W. S. Andrews, op. cit. in
4.B.1.a, pp. 295‑320.
Datta
& Singh give a lengthy (51pp) description of Narayana's work, including about
19 other magic figures. Many of
Narayana's methods are novel.
Ahrens-1 gives references to further Arabic mentions of
magic squares, usually as amulets, notably to
ibn Khaldun (c1370). He also gives many 14C and later examples
of 3 x 3 and 4 x 4 squares, with rearrangement and/or constants
added, used for magical purposes.
Nadrûnî. Qabs al‑Anwâr. pre-1384.
??NYS -- described in Ahrens-1, but not mentioned in Ahrens-2. Ahrens only knows of this from a modern
article in Arabic. This gives the
association of planets by System I. See
Folkerts, above.
Arabic MS, 1446, ??NYS.
Discussed in Ahrens-1 and Ahrens-2, citing: W. Ahlwardt; Verzeichniss der arab. Handschr. der Königl.
Bibliothek zu Berlin; Berlin, 1891; Vol. III, pp. 505-506 (No. 4115). This gives the System II association of
planets with magic squares, later given by Cardan in 1539, with the unique
addition of a 10 x 10 square for the zodiac coming after Saturn.
Dharmananda. 15C
Jaina scholar. Datta & Singh
present his 8 x 8 square and say his method works for the
evenly even case in general, extending Narayana.
Sundarasūri.
c15C Jaina scholar. Datta &
Singh say he gives some novel methods, extending Narayana.
Jagiellonian MS 753.
15C Latin MS in Cracow.
Described in Cammann‑4, pp. 291‑297. Earliest European set of magic squares of
orders 3 through 9 associated with the planets in System I --
but see Folkerts above, c13C. The
order 4 square is Dürer's. These
squares later appear in Paracelsus.
Sûfî Kemal al‑Tustarî. Ghayat al‑Murâd.
1448. MS at Columbia. ??NYS -- cited by Cammann‑4, p.
192. On p. 196 Cammann says this
represents a Persian Sufi tradition which was lost in sectarian warfare and the
Mongol invasion. On p. 201 he says this
has a unit‑centred square of order
7. On pp. 205‑206 are
squares of orders 20, 29, 30. He describes two bordering methods
beyond al‑Buni's.
Hindu square in a temple at Gwalior Fort, 1483. Cited by Cammann‑4, p. 275, where the
original source is cited -- ??NYS.
Pacioli. De Viribus. c1500.
Ff. 118r - 118v, 121r - 122v (some folios are wrongly inserted in the
middle). C.A. [i.e. Capitolo] LXXII.
D(e). Numeri in quadrato disposti secondo astronomi ch' p(er) ogni verso fa'no
tanto cioe per lati et per Diametro figure de pianeti et amolti giuochi
acomodabili et pero gli metto (Of numbers arranged in a square by astronomers,
which total the same in all ways, along sides and along diagonals, as symbols
of the planets and suitable for many puzzles and how to put them ??). Gives magic squares of orders 3
through 9 associated with planets in System I, usually
attributed to Agrippa (1533), but see Folkerts, above. Ff. 121v and 122r have spaces for diagrams,
but they are lacking. He gives the
first two lines of the order 4 square as
16, 3, 2, 13; 5, 10, 11,
8; so it must be the same square
as shown by Dürer, below.
Albrecht Dürer.
Melencolia. 1514. Two impressions are in the British
Museum. 4 x 4 square with 15, 14 in the bottom centre cells. Surprisingly, this is the same as the 4 x 4 appearing in Ikhwān
al‑Şafā’ [NOTE:
Ş denotes S
with an underdot.] (c983), with
the two central columns interchanged and the whole square reflected around a
horizontal midline, i.e.
16, 3,
2, 13; 5, 10, 11, 8;
9, 6, 7, 12; 4, 15, 14, 1.
This is the same as that described by Pacioli. There is some belief that the association with Jupiter relates to
the theme of the picture. This is the
first printed 4 x 4 magic square.
Riese.
Rechnung. 1522. 1544 ed. -- pp. 106‑107; 1574 ed. -- pp. 71v‑72v. Gives
3 x 3 square in 672
form and how to construct other
3 x 3 forms. Also gives a 4 x 4 square, like Dürer's
but with inner columns interchanged.
Riese. Rechenung
nach der lenge .... Op. cit. under
Riese, Die Coss. 1525. ??NYR.
Cammann‑4, p. 294, says pp. 103r‑105v gives a diagonal rule
for odd orders. A quick look shows the
material starts on p. 102v.
Cornelius Agrippa von Nettesheim. De Occulta Philosophia.
Cologne, 1531, ??NYS. Included
in his Opera, vol. 1, and available in many translations. 2nd Book, Chap. 22. Gives association of planets with magic
squares in System I -- as previously done by Pacioli, c1500, but with different
squares. See Folkerts, above, and
Cammann‑4, p. 293-294. The squares
do not appear in the 1510 draft version of this book. Bill Kalush has kindly sent Chap. 22 from a 1913 English version,
but it doesn't have any squares -- perhaps it was from the wrong Book??. He gives each square twice, with Arabic and
Hebrew numerals. His 3 x 3
is the 492 version.
His 4 x 4 is the same as that of Ikhwān al‑Şafā’
[NOTE: Ş denotes S with an underdot.], c983.
Cardan. Practica
Arithmetice. 1539.
Michael Stifel.
Arithmetica Integra. Nuremberg,
1544. ??NYS -- discussed in Cammann‑4,
p. 194. Pp. 25‑26a shows
some some bordered squares.
Consequently he is sometimes credited with inventing the concept, but
see Ikhwān al‑Şafā’
[NOTE: Ş denotes S with an underdot.] (c983), al‑Buni (c1200),
Yang Hui (1275), ‘Abdelwahhâb (c1340), Sûfî Kemal al‑Tustarî (1448)
and ibn Yūnis, above.
M. Mersenne. Novarum
observationum physico‑mathematicarum.
Paris, 1647. Vol. 3,
chap. 24, p. 211. ??NYS. States Frenicle's result. (MUS II #29.)
Isomura Kittoku.
Ketsugi-shō. 1660, revised
in 1684. ??NYS -- described in Smith
& Mikami, pp. 65-77. He gives magic
squares of orders up to order 10. The
order 9 square contains the order 3 square, in the 618 form, in the top
middle section. He gives magic circles
with n
rings of 2n about a central value of 1,
for n = 2 - 6. The values are
symmetrically arranged, so corresponding pairs add to 2n2 + 3 and
each ring adds up to n (2n2
+ 3), while each diameter adds to one
more than this. In the 1684 edition, he
gives some magic wheels, but these are simply a way of depicting magic squares,
though it is not clear where the diagonals are.
Muramatsu Kudayū Mosei. Mantoku Jinkō‑ri.
1665. ??NYS -- described in
Smith & Mikami, pp. 79-80. Gives a
magic square of order 19. Gives a magic
circle of Isomura's type for n =
8. Smith & Mikami, p. 79, gives
Muramatsu's diagram with a transcription on p. 80. The central 1 is omitted and the corresponding pairs no
longer add to 131, but the pairs adding to 131
lie on the same radius.
Bernard Frénicle de Bessy.
Des Quarrez ou Tables Magiques,
including: Table generale des
quarrez de quatre. Mem. de l'Acad. Roy.
des Sc. 5 (1666‑1699) (1729) 209‑354. (Frénicle died in 1675.
Ollerenshaw & Bondi cite a 1731 edition from The Hague??) (= Divers Ouvrages de Mathématique et de
Physique par Messieurs de l'Académie des Sciences; ed. P. de la Hire; Paris,
1693, pp. 423‑507, ??NYS. (Rara,
632). = Recueil de divers Ouvrages de
Mathematique de Mr. Frenicle; Arkstèe & Merkus, Amsterdam & Leipzig,
1756, pp. 207-374, ??NX.)
Shows
there are 880 magic squares of order
4 and lists them all. Cammann‑4, p. 202, asserts that they
can all be derived from one square!!
The
list of squares has been reprinted in the following.
M. Gerardin. Sphinx‑Oedipe
-- supplement 4 (Sep‑Oct 1909) 129‑154. ??NX
K. H. de Haas.
Frenicle's 880 basic Magic Squares of 4 x 4
cells, normalized, indexed, and inventoried (and recounted as 1232).
D. van Sijn & Zonen, Rotterdam, 1935, 23pp.
Seki Kōwa.
Hōjin Yensan. MS revised in
1683. Known also as his Seven Books. ??NYS -- described in Smith & Mikami, pp. 116-122. Describes how to border squares of all
sizes. Gives an easy method for writing
down a magic circle of Isomura's type.
Thomas Hyde.
Mandragorias seu Historia Shahiludii, .... (= Vol. 1 of De Ludis Orientalibus, see 4.B.5 for vol. 2.) From
the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Prolegomena curiosa. The
initial material and the Prolegomena are unpaged but the folios of the
Prolegomena are marked (a), (a 1), ....
The material is on (d 4).v - (e
1).v, which are pages 32-34 if one starts counting from the beginning of the
Prolegomena.
Seems
to believe magic squares come from Egypt and gives association of orders of
squares with planets as in Pacioli and Agrippa, but he only gives one example
of a magic square -- an 8 x 8 which is associated.
W. Leybourn.
Pleasure with Profit. 1694. Prob. 10, pp. 4-5. Gives the 294 form and then says that each line can be
rearranged four ways, e.g. 294, 492,
924, 942. He writes these out for all
eight lines, but I can't see any pattern in the way he chooses his four of the
six possible permutations.
Ozanam. 1694. 1696: Prob. 9: Des quarrez magiques,
36-41. Prob. 9: Of magical squares,
1708: 33-36. Prob. 12: Des quarrez
magiques, 1725: 82-102. Chap. 12: Des
quarrés magiques, 1778: 217-244. Chap.
12: Of magic squares: 1803:
211-240; 1814: 183-207 &
366-367; 1840: 94-105. Extended discussion, but contains little new
-- except some comments on Franklin's squares -- see Ozanam-Hutton (1803). Associates squares with planets, as done by
Pacioli.
Wells. 1698. No. 119, pp. 209-210. Studies the
3 x 3 square carefully, showing
that the centre cell must be 5 and the sum of each pair of adjacent side cells
is double the value in the opposite corner -- e.g. 9 + 7 is twice 8. I
don't recall ever seeing this result before.
Philippe de la Hire.
Sur les quarrés magiques.
Mémoires de l'Académie Royale des Sciences (1705 (1706)) 377‑378. Gives a method for singly‑even
squares, but it uses so many transpositions that it is hard to see if it works
in general. ??NYS -- described in
Cammann‑4, p. 286.
Muhammed ibn Muhammmed.
A Treatise on the Magical Use of the Letters of the Alphabet. Arabic MS of 1732, described and partly reproduced in: Claudia Zaslavsky; Africa Counts; Prindle, Weber & Schmidt,
Boston, 1973; chap. 12, pp. 137‑151.
Several of his magic squares are deliberately defective, presumably
because of the Islamic belief that only God can create something perfect. I do not recall any other mention of this
feature.
Minguet. 1733. Pp. 169-172 (1755: 122-123; 1864: 158-160;
not noticed in 1822, but probably about p. 180.) Magic squares of order three with various sums, made by laying
out cards.
Benjamin Franklin.
1736-1737. Discovery of some
large magic squares and circles. He described
these in letters to Peter Collinson whose originals do not survive. I. Bernard Cohen [Benjamin Franklin Scientist and Statesman; DSB Editions,
Scribner's, 1975, pp. 18-19] dates them as above, citing Franklin's
Autobiography, but my copy is an abridged edition without this material --
??. He also reproduces them. They were first published in the following.
James
Ferguson. Tables and Tracts, Relative
to Several Arts and Sciences.
A. Millar & T. Cadell, London, 1767. Pp. 309-317. ??NYS. Ferguson may be indicating that he is the
first person to whom Franklin showed them.
B.
Franklin. Experiments and Observations
on Electricity. 4th ed., London,
1769. Two letters to Peter Collinson,
pp. 350-355(??). ??NYS, but reprinted
in: Albert Henry Smyth; The Writings of
Benjamin Franklin; Vol. II, Macmillan, 1907, pp. 456‑461 and Plates
VII (opp. p. 458) and VIII (opp. p. 460).
The
squares are of order 8 and
16, but are only semi-magic (see
Ozanam-Hutton (1803) and Patel (1991)), and the circle has 8
rings and 8 radii.
Franklin said he could make these squares as fast as he could write down
the numbers!
Dilworth.
Schoolmaster's Assistant.
1743. Part IV: Questions: A
short Collection of pleasant and diverting Questions, p. 168. Problem 4.
Asks for a 3 x 3 magic square.
Caietanus Gilardonus.
9 x 9 square on a marble plaque
on the Villa Albani, near Rome, dated 1766.
The square and the accompanying inscription are given in: E. V. R.; Arranged squares; Knowledge 1 (27
Jan 1882) 273, item 231. These are also
given by Catalan; Mathesis 1, p. 151
(??NYS) and Lucas; L'Arithmétique Amusante; 1895; pp. 224-225. Lucas says they were discovered in 1881, and
that the villa is now owned by Prince Torlonia and is outside the Porta Salaria.
Catel.
Kunst-Cabinet. 1790.
Bestelmeier. 1801.
Ozanam-Hutton.
1803. Chap. 12: Remarks, 1803:
237-240 & fig. 1, plate 4. 1814:
203-207 & corrections: 366-367 & fig. 1, plate 4 & additional plate
5. 1840: 104-105, with no figure. The 1814 corrections note that Franklin's
square is only semi-magic and gives another large example -- 1840 omits this.
Rational Recreations.
1824.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 238-243, no. 907-926. This gives lots of straightforward exercises
-- e.g. find a 5 x 5 magic square with sum 96,
which he does by adding 6
1/5 to each entry of a normal example.
Young Man's Book.
1839. P. 232. The Magical Square. "The Chinese have discovered mystical
letters on the back of the tortoise, which is the common magical square, making
each way 15, viz." Gives the 294
form, while all Chinese forms have
492. Pp. 236-238 is a
straightforward section on Magic Squares.
Boy's Own Book. 1843
(Paris).
Indoor & Outdoor.
c1859. Part II, prob. 13:
Franklin's magic square, pp. 132-133.
Gives Franklin's 16 x 16 square and states some of its properties,
very similar to Ozanam‑Hutton.
Vinot. 1860. Art. CLIV: Des carrés magiques, pp.
188-201. On p. 190, he gives an
association of squares with planets, as given by Pacioli, but with the addition
of: 1 -- God; 2 -- matter.
Magician's Own Book (UK version). 1871. The magic square
oraculum, pp. 94-98. This shows a
square of order 11 and says it is "a magic square of eleven, with one in
the centre", but it is not at all magic.
Initially it appears to be bordered, but it is a kind of arithmetical
square. 1 is in the middle.
Then 2 - 9 are wrapped around the central square, going
clockwise with the 2 above the
1. Then 10 - 25
are wrapped around the central 3
x 3 area, with the 10
above the 2, etc.
The 'oracle' consists of thinking of a number and consulting a list of
fortunes, so the 'magic square' is never used!
Carroll-Wakeling.
c1890? Prob. 8: Magic postal
square, pp. 10-11 & 65. The first
nine values of postage stamps in Carroll's time had values 1, 2, 3, 4, 5, 6, 7, 8, 10 in units of half-pence. But the total of the values in a magic
square is three times the magic constant, and these values add up to 46. So Carroll allows one of the values to be
repeated and the ten values now have to be placed to make a 3 x 3
magic square. Surprisingly, the
value to be repeated is uniquely determined and there is just one such square.
T. Squire Barrett.
The magic square of four.
Knowledge 14 (Mar 1891) 45-47
& Letter (Apr 1891)
71 & Letter (Aug 1891) 156.
Says he hasn't seen Frenicle's list.
Classifies the 4 x 4 squares into 12 types, and
obtains 880 squares, but doesn't guarantee to have found all of them. The first letter notes that he erred in
counting one type, getting 48 too many, but a friend has found 16
more. Second letter notes that
the missing squares have been found by another correspondent who has compared
them with Frenicle's list and then Barrett corrects some mistakes. The other correspondent notes that Frenicle
had proceeded by a trial and error method and probably had found all examples.
Ball. MRE, 1st ed.,
1892, pp. 108-121: Chap. V: Magic squares.
Later editions amplify this material, but the material is too detailed
and too repetitive to appeal to me at the moment.
Hoffmann. 1893. Chap. IV, pp. 146 & 183‑191
= Hoffmann-Hordern, pp. 114-118, with photos on pp. 133 & 143.
Stewart Culin. Chinese
Games with Dice and Dominoes. From the
Report of the U. S. National Museum for 1893, pp. 489‑537. On pp. 536-537, he discusses the Lok Shü,
citing Legge. He reports that the 618
version is popular as a charm with both Hindus and Moslems in India,
while the Chinese 492 form is used in Tibet.
Dudeney. A batch of
puzzles. Royal Magazine 1:3 (Jan
1899) & 1:4 (Feb 1899) 368-372.
The eight clowns puzzle. = CP;
1907; prob. 81, pp. 128 & 126. 3 x
3 array with pieces x 2 3;
4 5 6; 7 8 9 to be rearranged into a magic square, the
blank being counted as a 0. Answer is that this is impossible, but the
clown marked 9 is juggling balls which make his number .9,
i.e. .9 recurring,
which is 1!
Dudeney. The magic
square of sixteen. The Queen (15 Jan
1910) 125-126. Good derivation of
the 880 squares of order 4, classified into 12 types. A condensed version with some extra
information is in AM, pp. 119-121.
Loyd.
Cyclopedia. 1914. The 14‑15 puzzle in puzzleland, pp.
235 & 371. = MPSL1, prob. 21,
pp. 19‑20 & 128. c= SLAHP:
The "14‑15" magic square, pp. 17‑18 & 89. Given the
15 Puzzle with the 14
and 15 interchanged, move to a magic square. The blank counts zero, so the magic constant is 30.
Collins. Book of
Puzzles. 1927. Magic squares and other figures, pp.
79-94. Brief survey. Gives a number of variant forms.
D. N. Lehmer. A
complete census of 4 x 4 magic squares. Bull. Amer. Math. Soc. 39 (1933) 764-767. Here he is dealing with semi-magic squares
and then any permutation of the rows or columns or transposition of the whole
array preserves the row and column sums.
Hence there are 2(n!)2 arrays in each equivalence class and he
describes a normalized form for each class.
For the 3 x 3 case, there are 72 semimagic squares and
one normalized form. For the 4 x 4
case, there are 468 normalized forms and hence 468 x 2 x 242 =
539,136 semimagic squares.
D. N. Lehmer. A
census of squares of order 4, magic in rows, columns, and diagonals. Bull. Amer. Math. Soc. 39 (1933)
981-982. Here he discusses Frenicle's
enumeration of 880 4 x 4
magic squares and points out that there are additional equivalences
beyond the symmetries of the square so that Frenicle only needed to find and
list 220 squares. Using ideas like
those in his previous article, he finds
220 such squares, confirming
Frenicle's result once again.
J. Travers. Rules
for bordered magic squares. MG 23 (No.
256) (Oct 1939) 349‑351. Cites
Rouse Ball, MRE (no details), as saying no such rules are known. He believes these are the first published
rules.
Anonymous. A Book of
Fun with Games and Puzzles. One of a
set of three 12pp booklets, no details, [1940s?]. P. 7: Here is the magic square.
Gives a 4 x 4 square using the numbers 3, ..., 17
and asks for it to be dissected along the lines into four pieces which
can be rearranged into a magic square.
D. H. Hallowes.
On 4 x 4 pan‑magic squares. MG 30 (No. 290) (Jul 1946) 153‑154. Shows there are only 3
essentially different 4 x 4 pan‑magic squares.
Ripley's Puzzles and Games.
1966. P. 52. The magic cube. This actually shows only the front of a cube and really comprises
three magic squares as the faces have no relation to each other. What is really being attempted is to arrange
the numbers 1, 2, ..., 27 into three
3 x 3 magic squares. Then the magic sum must be 42.
The given arrangement has
22 of the 24
lines adding to 42 -- two of the diagonals fail. ?? -- is such an arrangement possible? Ripley's says the pattern adds to 42
in 44 directions -- apparently they count each of the 22
lines in each direction.
Gardner. SA (Jan
1976) c= Time Travel, chap. 17. Richard Schroeppel, of Information
International, used a PDP‑10 to find
275,305,224 magic squares of
order 5, inequivalent under the
8 symmetries of the square. If one also considers the 'eversion'
symmetries, there are 32 symmetries and 68,826,306 inequivalent
squares. (Gardner says there is an Oct
1975 report on this work by Michael Beeler, ??NYS, and gives Schroeppel's
address: 835 Ashland Ave., Santa
Monica, Calif., 90405 -- I believe I wrote, but had no reply??)
K. Ollerenshaw & H. Bondi. Magic squares of order four.
Phil. Trans. Roy. Soc. Lond. A306 (1982) 443‑532. (Also available separately.) Gives a new approach to Frénicle's
results. Relates to Magic Card Squares
and the Fifteen Puzzle.
Lee C. F. Sallows.
Alphamagic squares: I &
II. Abacus 4:1 (Fall 1986) 28‑45 &
4:2 (Winter 1987) 20-29 & 43.
Reprinted in: The Lighter Side of Mathematics; ed. by R. K. Guy & R.
E. Woodrow; MAA, 1994, pp. 305-339.
Introduces notion of alphamagic square -- a magic square such that the
numbers of letters in the words for the numbers also form a magic square. Simplest example is: 5, 22, 18; 28, 15, 2; 12,
8, 25. He asserts that this
appears in runes in an 1888 book describing a 5C charm revealed to King Mi of
North Britain. (This seems a bit far‑fetched
to me or mi?) Asks if there can be an
alphamagic square using the first n2 numbers and shows that n ³ 14. Notes some interesting results on formulae for 4 x 4
squares, including one with minimum number of symbols. There was a letter and response in Abacus
4:3 (Spring 1987) 67-69.
Martin Gardner.
Prime magic squares. IN: The
Mathematical Sciences Calendar for 1988; ed. by Nicholas J. Rose, Rome Press,
Raleigh, North Carolina, 1987.
Reprinted with postscript in Workout, chap. 25. Says Akio Suzuki found a 35 x 35
magic square using the first odd primes in 1957. I have a poster of this which Gardner gave
me. Offers $100 for the first 3 x 3
magic square using consecutive primes.
The postscript says Harry L. Nelson won, using a Cray at Lawrence
Livermore Laboratories. He found 22
examples. The one with the lowest
constant has smallest value
1,480,028,129 and the values all
have the same first seven digits and their last three digits are: 129, 141, 153, 159, 171,
183, 189, 201,
213.
Lalbhai D. Patel.
The secret of Franklin's 8 x
8 'magic' square. JRM 23:3 (1991) 175-182. Develops a method to make Franklin's squares
as fast as one can write down the numbers!
Jacques Bouteloup.
Carrés Magiques, Carrés Latins et Eulériens. Éditions du Choix, Bréançon, 1991. Nice systematic survey of this field, analysing many classic
methods.
Lee Sallows.
Alphamagic squares. CFF 35 (Dec
1994) 6-10. "... a brief synopsis
of [his above article] which handles the topic in very much greater
detail."
Martin Gardner. The
magic of 3 x 3. Quantum 6:3 (Jan-Feb 1996) 24-26; with
addendum in (Mar-Apr 1996). Reprinted
with a postscript in Workout, chap. 22.
Says that modern scholars doubt if the pattern in China is older than
10C! Mentions his 1987 prize and
Nelson's least solution. Says Martin
LaBar [CMJ (Jan 1984) 69] asked for a 3
x 3 magic square whose entries are all
positive squares. Gardner reiterates
this and offers $100 for the first example -- in the postscript, he extends the
prize to include a proof of impossibility.
He gives examples of 3 x 3 squares with various properties and the
lowest magic sum, e.g. using primes in arithmetic progression.
Lee Sallows. The
lost theorem. Math. Intell. 19:4 (1997)
51-54. Gives an almost solution to
Gardener's problem, but one diagonal fails to add up correctly. Gives an example of Michael Schweitzer which
is magic but contains only six squares.
Using Lucas' pattern, where the central number, c,
is one third of the magic sum and two adjacent corners are c + a
and c + b, he observes that these can be vectors in the
plane or complex numbers, which allows one to correspond classes of eight magic
squares with parallelograms in the plane.
This leads to perhaps the most elegant magic square by taking c = 0,
a = 1, b = i.
Kathleen Ollerenshaw, Kathleen & David S. Brée. Most-perfect Pandiagonal Magic Squares Their construction and enumeration. Institute of Mathematics and its
Applications, Southend-on-Sea, 1998. A
most-perfect square is one which is magic and pandiagoanal and all 2 x 2
subsquares have the same sum, even when the square is considered on a
torus. They find a formula for the
number of these in general. Summary
available on www.magic-squares.com or
www.most-perfect.com .
Lee Sallows. Email
of 11 Jun 1998. He asks if any set of
16 distinct numbers can produce more than
880 magic squares of order 4.
He finds that -8, -7, ..., -2,
-1, 1, 2, ..., 7, 8 gives 1040
magic squares. He has not found
any other examples, nor indeed any new examples with as many as 880,
though he has looked at other types of values, even Gaussian integers.
K. Pinn & C. Wieczerkowski. Number of magic squares from parallel tempering Monte Carlo. Intern. J. Modern Physics C 9:4 (1998)
541-546. ??NYS - cited by Chan &
Loly, below. They estimate there
are 1.77 x 1019 magic squares of order 6.
Harvey D. Heinz
& John R. Hendricks. Magic Square Lexicon: Illustrated. Harvey D. Heinz, 15450 92A Avenue, Surrey,
British Columbia, 2000.
Frank J. Swetz.
Legacy of the Luoshu - The 4000 Year Search for the Meaning of the Magic
Square of Order Three. Open Court,
2002. ??NYS - cited by Chan & Loly,
below.
Wayne Chan
& Peter Loly. Iterative compounding of square matrices to
generate large-order magic squares.
Mathematics Today 38:4 (Aug 2002) 113-118. Primarily they develop programs for doing compounding to produce
very large squares.
Lee Sallows.
Christmas card for 2003: Geometric magic square. Consider the magic square:
11 9
9 7; 6 10 8 12; 6 8
10 12; 13 9 9 5
with magic constant 36. Each quadrant also adds up to 36.
Sallows uses sixteen polyominoes having these numbers of unit squares
and arranged in this pattern so that each quadrant forms a 6 x 6
square. These polyominoes can be
assembled into many other squares.
Note. Historically, a k3 has been called magic when all the 3k2 lines parallel to the axes and the 4 space diagonals have
the same sum. But there are also 6k 2‑dimensional
diagonals -- if these also have the same sum, we will say that the cube is
perfectly magic. Pandiagonal
(= pan‑n‑agonal) refers just to the space diagonals. Perfectly pandiagonal refers to all the
diagonals. In higher space, the simpler
words refer to the 2n-1 'space' diagonals and perfect will include
all the diagonals in intermediate dimensions.
A
k-agonal is a line which varies in
k coordinates, so a 1-agonal is a row or column, etc., the
2-agonals of a cube include the diagonals of the faces, while the
3-agonals of a 3‑cube are
the space diagonals.
Associated or complementary means that two cells symmetric with respect
to the centre add to kn + 1.
Pierre de Fermat.
Letter to Mersenne (1 Apr 1640).
Oeuvres de Fermat. Ed. by P.
Tannery & C. Henry. Vol. 2,
Gauthier‑Villars, Paris, 1894, pp. 186‑194. Gives a magic 43.
A
shorter, undated, version, occurs in Varia Opera Mathematica D. Petri de
Fermat, Toulouse, 1679; reprinted by
Culture et Civilization, Brussels, 1969; pp. 173‑176. The version in the Oeuvres has had its
orthography modernized.
On p. 174 of the Varia (= p. 190 of the
Oeuvres), he says: "j'ay trouvé
une regle generale pour ranger tous les coubes à l'infiny, en telle façon que
toutes les lignes de leurs quarrez tant diagonales, de largeur, de longeur, que
de hauteur, fassent un méme nombre, & determiner outre cela en combien de
façons differentes chaque cube doit étre rangé, ce qui est, ce me semble, une
des plus belles choses de l'Arihmetique [sic]..." He describes a assembly of four squares
making a magic cube. [The squares are
missing in the Varia.] He says that the
magic sum occurs on 72 lines, but it fails to have the magic sum on 8 of
the 2‑agonals and all 4 of
the 3‑agonals.
Lucas. Letter. Mathesis 2 (1882) 243‑245. First publication of the magic 43 described by Fermat above.
Says it will appear in the Oeuvres.
E. Fourrey. Op. cit.
in 4.A.1. 1899. Section 317, p. 257. Notes that Fermat's magic cube has only 64
magic lines.
Lucas.
L'Arithmétique Amusante.
1895. Note IV: Section IV: Cube magique de Fermat,
pp. 225‑229. Reproduces
the 43 from his Mathesis letter and gives a
generalization by V. Coccoz, for which the same diagonals fail to have the
magic sum, though he implies they do have the sum on p. 229.
Pierre de Fermat.
Letter to Mersenne, nd [Jun? 1641].
Opp. cit. above: Oeuvres, vol.
2, pp. 195‑199; Varia Opera,
pp. 176‑178. On p. 177 of the
Varia (= p. 197 of the Oeuvres), he says:
"Pource qui est des cubes, je n'en sçay pas plus que Monsieur
Frenicle, mais pourtant je puis les ranger tous à la charge que les Diagonales
seules de quarrez que nous pouvons supposer paralleles à l'Horizon, seront
égales aux côtez des quarrez, ce qui n'est pas peu de chose. En attendant qu'une plus longue meditation
découvre le reste, je dresseray celuy de
8. 10. ou 12. à ces conditions si Monsieur de Frenicle me
l'ordonne."
Joseph Sauveur.
Construction générale des quarrés magiques. Mémoires de l'Académie Royale des Sciences (1710 (1711)) 92‑138. ??NYS -- mentioned by Brooke (below), who
says Sauvier [sic] presented the first magic cube but gives no reference. Discussed by Cammann‑4, p. 297, who
says Sauveur invented magic cubes and Latin squares. This paper contains at least the latter and an improvement on de
la Hire's method for magic squares, but Cammann doesn't indicate if this
contains the magic cube.
Charles Babbage.
Notebooks -- unpublished collection of MSS in the BM as Add. MS
37205. ??NX. See 4.B.1 for more details.
F. 308: Essay towards forming a Magick Cube, c1840?? Very brief notes.
Gustavus Frankenstein.
[No title]. Commercial (a daily
paper in Cincinnati, Ohio) (11 Mar 1875).
??NYS. Perfect 83. Described by Barnard, pp. 244‑248.
Hermann Scheffler.
Die magischen Figuren. Teubner,
Leipzig, 1882; reprinted by Sändig,
Wiesbaden, 1981. Part III: Die magische
Würfel, pp. 88‑101 & plates I & II, pp. 113 & 115. He wants all 2‑ and 3‑agonals
to add up to the magic constant, though he doesn't manage to construct any
examples. He gives a magic 53 which has the magic sum on
14 of the 30 2‑agonals
and many of the broken 2‑agonals. He also gives a 43 and a 53, but I haven't checked how successful they are.
F. A. P. Barnard.
Theory of magic squares and of magic cubes. Memoirs of the National Academy of Science 4 (1888) 209‑270. ??NYS. Excerpted, including the long
footnote description of Frankenstein's
83, in Benson &
Jacoby (below), pp. 32‑37, with diagrams of the result on pp. 37‑42.
C. Planck, on pp. 298 & 304 of Andrews, op. cit. in
4.B.1.a, says the first magic 63 was found by W. Firth of Emmanuel College,
Cambridge in 1889.
Pao Chi-shou. Pi Nai
Sahn Fang Chi (Pi Nai Mountain Hut
Records). Late 19C. ??NYS -- described by Lam (in op. cit. in
7.N under Yang Hui), pp. 321-322, who says it has magic cubes,
spheres and tetrahedrons. See also
Needham, p. 60.
V. Schlegel. ?? Bull. Soc. Math. France (1892) 97. ??NYS.
First magic 34. Described by Brooke (below).
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Magic cubes, pp.
99-100. 33 magic cube with also the 6
2-dimensional diagonals through the centre having the same sum, so there
are 37
lines with the magic sum. 43 magic cube -- this has the 3 x 16 + 4 = 52 expected magic lines, but he asserts it has 68
magic lines, though I can only find
52. The perfect case would
have 76 magic lines.
C. Planck. Theory of
Path Nasiks. Privately printed, Rugby,
1905. ??NYS. (Planck cites this on p. 363 of Andrews, op. cit. in 4.B.1.a, and
says there are copies at BM, Bodleian and Cambridge.) The smallest Nasik (= perfectly pandiagonal) kn has k = 2n. If the cube is also associated, then k = 2n + 1. He quotes these results on p. 366 of Andrews
and cites earlier erroneous results. On
p. 370 of Andrews, he says that a perfect
k4 has k ³ 8.
Collins. Book of
Puzzles. 1927. A magic cube, pp. 89-90. 33, different than that in Berkeley & Rowland, but with the same
properties.
J. Barkley Rosser & Robert J. Walker. MS deposited at Cornell Univ., late
1930s. ??NYS. (Cited by Gardner, loc. cit. below, and Ball, MRE, 11th ed., p.
220; 12th ed., p. 219.) Finds a Nasik 83 and shows
that Nasik k3 exist precisely for the multiples of 8
and for odd k > 8.
G. L. Watson. Note
2100: To construct a symmetrical, pandiagonal
magic cube of oddly even order 2n ³
10. MG 33 (No. 306) (Dec 1949) 299‑300.
Maxey Brooke. How to
make a magic tessarack. RMM 5 (Oct
1961) 40‑44. Cites Sauvier and
Schlegel. Believes this is the first
English exposition of Schlegel.
The 33 he develops is magic, but only the 6 2‑agonals
through the centre have the magic sum.
The resulting 34 is magic but not perfect.
Harry Langman. Play
Mathematics. Hafner, 1962. ??NYS -- cited by Gardner below. Pp. 75‑76 gives the earliest
known perfect 73.
Birtwistle. Math.
Puzzles & Perplexities. 1971. The magic cube, pp. 157 & 199. 33 which is associated, but just the 6 2-agonals through the
centre have the magic sum; the other
12 2-agonals do not. The space diagonals also have the magic sum.
John Robert Hendricks.
The third‑order magic cube complete. JRM 5:1 (1972) 43‑50.
Shows there are 4 magic
33, inequivalent
under the 48 symmetries of the cube.
None of these is perfect. (The author
has published many articles on magic cubes in JRM but few seem appropriate to
note here.)
Gardner. 1976. Op. cit. in 7.N. Gives Richard Lewis Myers Jr.'s proof that a perfect 33 does not exist, and Richard Schroeppel's 1972 proof that a perfect 43 does not exist. (Gardner
says Schroeppel published a memorandum on this, ??NYS.) Says that perfect cubes of edge 5, 6, 7
are unknown and gives a perfect, associated 83 found by
Myers in 1970. The Addendum in Time
Travel cites Planck and Rosser & Walker for earlier 83 and says that many readers found a perfect 73 and refers to Langman.
Also 93, 113 and higher orders were found.
Johannes Lehmann.
Kurzweil durch Mathe. Urania
Verlag, Leipzig, 1980. No. 8, pp. 61
& 160. Arrange 0 - 15
on the vertices of a 24 hypercube so that each 2-dimensional face has the sum 30.
John R. Hendricks.
The perfect magic cube of order 4.
JRM 13 (1980‑81) 204‑206.
Shows it does not exist.
William H. Benson & Oswald Jacoby. Magic Cubes -- New Recreations. Dover, 1981. Summarises all past results on p. 5. There are perfect n3 for
n ¹ 6 (mod 12), n ³
7, except n = 10. It is not clear
if they have proofs for n º 4
(mod 8) or n º 2 (mod 4). They are unable to show the non‑existence
for n º 5 (cf. p. 29). There are pandiagonal n3 for
n ¹ 6 (mod 12), n ³ 4, though it is not clear if they have a proof for n º 2 (mod 4) (cf. p. 102).
Rudolf Ondrejka.
Letter: The most perfect (8 x 8
x 8) magic cube? JRM 20:3 (1988) 207‑209. Says Benson & Jacoby sketch a perfectly
pandiagonal 83. He gives it in detail and discusses it.
Joseph Arkin, David C. Arney & Bruce J. Porter. A perfect
4‑dimensional hypercube of order 7 -- "The Cameron
Cube". JRM 21:2 (1989) 81‑88. This is the smallest known order in four
dimensions.
Allan William Johnson Jr.
Letter: Normal magic cubes of
order 4M+2. JRM 21:2 (1989) 101‑103.
Refers to Firth, 1889, who is mentioned by Planck in Andrews. Gives a program to compute (4M+2)3 cubes and gives a 63 in base 10
& base 6.
John Robert Hendricks.
The magic tessaracts of order
3 complete. JRM 22:1 (1990) 15‑26. Says there are 58 of them and gives some
history.
Jacques Bouteloup.
Carrés Magiques, Carrés Latins et Eulériens. Éditions du Choix, Bréançon, 1991. Nice systematic survey of this field, analysing many classic
methods. Includes some material on
magic cubes.
There are quite a number of possible
types here and I have not been very systematic in recording them.
Frenicle de Bessy.
Letter to Mersenne, Mar 1640.
In: Oeuvres de Fermat, op. cit.
in 7.N.1, vol. 2, pp. 182‑185.
Discusses a magic triangle.
Scheffler. Op. cit.
in 7.N.1. 1882. Part II: Das magische Polygon, pp. 47‑88
& Plate I, p. 113. He considers
nested n‑gons with the number of
numbers on each edge increasing
1, 3, 5, ... or 2, 4, 6, ..., such that each edge of length k > 2
has the same sum and the diameters all have the same sum, though these
sums are not all the same. He develops
various techniques and gives examples up to
26‑gons and 5‑level
pentagons.
H. F. L. Meyer.
Magic Triangles. In: M. Adams; Indoor Games; 1912; pp. 357‑362. He divides a triangle by lines so the
triangle of order three has rows of 1,
3, 5 cells. He gets some lines of
2 and of 3 to
add to the same value, and then considers hexagons of six cells, but doesn't
really get anywhere.
Peano. Giochi. 1924.
Collins. Book of
Puzzles. 1927.
Collins. Fun with
Figures. 1928.
Perelman. FFF. 1934.
1957: probs. 46-48, pp. 56-57 & 61-62; 1979: probs. 49-51, pp. 70‑71 & 77-78. = MCBF, probs. 49-51, pp. 69-70 & 74-75.
Ripley's Puzzles and Games.
1966. Pp. 34-35, item 2. P;ace the fifteen pool balls so each edge
and the central three balls total to the same sum. Ripley's gives examples with magic sum of 34, 35, 36. One can construct examples with magic sum of
32, 33, ..., 39. A sum of 40 initially
seems possible but further analysis shows it is not possible. If one rules out trivial rearrangements
leaving the rows having the same elements, there are 2716 distinct
solutions. Each of these has 1296
trivial rearrangements.
Jaime Poniachik, proposer;
Henry Ibstedt, solver. Prob.
1776 -- Connected differences. JRM 22:1
(1990) 67 & 23:1 (1991) 74-75. Triangular lattice with edge
2. Place numbers 1, ..., 15
on the 6 points and
9 edges so that each edge is the
difference of its end points. 19 solutions found by computer. [Not sure where to put this item??]
7.N.3. ANTI‑MAGIC SQUARES AND TRIANGLES
An antimagic n x n
has its 2n+2 sums all distinct. A consecutively antimagic
n x n has its 2n+2
sums forming a set of 2n+2 consecutive integers. Berloquin calls these heterogeneous and
antimagic, respectively. A heterosquare
has all the 4n sums along rows, columns and all broken
diagonals being different.
Loyd Jr. SLAHP. 1928.
Magic square reversed, pp. 44 & 100. 3 2 7; 8 5 9;
4 6 1 has all eight sums
different.
Dewey Duncan.
?? MM (Jan 1951) ??NYS -- cited
by Gardner. Defines a heterosquare as
an arrangement of 1, 2, ..., n2 such that the 4n sums along the rows,
columns and all broken diagonals are all different. (However, in the next item it appears that only the main
diagonals are being considered??) Asks
for a proof that the 2 x 2 case is impossible and for a 3 x 3
example -- which turns out to be impossible.
John Lee Fults.
Magic Squares. Open Court, La
Salle, Illinois, 1974. On p. 78, he
asserts that Charles W. Trigg posed the problem of non‑existence of anti‑magic 2 x 2's
in 1951, that it was solved by Royal Heath and that Trigg gave the
spiral construction for anti‑magics of odd order. Unfortunately Fults gives no source, only
noting that Trigg was editor at the time.
There is nothing in Heath's MatheMagic.
I now see that this is a corruption of the preceding item. Madachy (see below at Lindon) refers to the
problem in MM (1951) without further details, but restricted to just the main
diagonals, and then says "An exchange of correspondence between Charles W.
Trigg, then "Problems and Questions" editor for the magazine, and the
late Royal V. Heath, ..., soon established some basic properties of potential
heterosquares."
C. W. Trigg, proposer;
D. C. B. Marsh, solver. Prob.
E1116 -- Concerning pandiagonal heterosquares.
AMM 61 (1954) 343 & 62 (Jan 1955) 42. The solution is also in:
Trigg; op. cit. in 5.Q; Quickie 160: Pandiagonal heterosquare, pp.
45 & 151. There is no arrangement
of 1, 2, ..., n2 such that the 4n sums along the rows,
columns and all broken diagonals are consecutive numbers.
Charles F. Pinska.
?? MM (Sep/Oct 1965)
250-252. ??NYS -- cited by
Gardner. Shows there are no 3 x 3
heterosquares, but gives two 4 x
4 examples.
Gardner. SA (Jan
1961) c= Magic Numbers, chap. 2. Notes
that 1 2 3;
8 9 4;
7 6 5 is anti‑magic,
i.e. all 8 sums are different, and it is also a rook's tour. In Magic Numbers, Gardner says he had not
known of anti-magic squares before seeing this one, but later discovered the
Loyd example. He summarises the
knowledge up to 1971.
J. A. Lindon. Anti‑magic
squares. RMM 7 (Feb 1962) 16‑19. Summarised and extended in Madachy;
Mathematics on Vacation, op. cit. in 5.O, (1966), 1979, pp. 101-110. Author and editor believe this is the first
such article. Wants the 2n+2
sums for an n x n square to be all different and also to be a
set of consecutive integers. There are
no such for n = 1, 2, 3, but they do exist for n > 3.
M. Gardner.
Letter. RMM 8 (Apr 1962)
45. Points out his SA article and notes
that
9 8 7, 2 1
6, 3 4 5 is even more anti‑magic in that
the 8
lines and the 4 2 x 2
subsquares all have different sums.
Pierre Berloquin.
The Garden of the Sphinx, op. cit. in 5.N. 1981.
C. W. Trigg. Special
antimagic triangular arrays. JRM 14
(1981‑82) 274‑278. Says
Gardner gives the following definition in a letter to Trigg on 22 Dec
1980. Consider a triangular array of
the numbers 1, 2, ..., n(n+1)/2. We say this is anti‑magic if the sum
of the three vertices and the 3(n‑1) sums of rows of two or more, parallel to the
sides, are all distinct. Gardner then
asks if these 3n‑2 sums can be consecutive. Trigg asks when these sums can be in
arithmetic progression and finds that this requires n = 2, 3, 4, 10, 24 or n
> 99. He resolves the existence
problems for n = 2, 3, 4.
M. Gardner. Puzzles
from Other Worlds. Vintage (Random
House), NY, 1984. Problem 8: Antimagic
at the number wall, pp. 19‑20, 96‑97 & 142‑143. Notes his examples (which are complementary)
are the only rook‑wise connected anti‑magic squares of order 3
and discusses anti‑magic triangles.
Note: for
the 8 x 8, the magic constant is
260.
G. P. Jelliss.
Special Issue -- Magic Tours; Chessics 26 (Summer 1986) 113‑128 &
Notes on Chessics 26 (Magic Tours); Chessics 29 & 30 (1987)
163. Says Kraitchik (L'Echiquier,
1926), ??NYS, showed there is no magic
tour unless both sides are even.
(Mentioned in his Math. des Jeux, op. cit. in 4.A.2, p. 388.) Jelliss considers tours by other pieces
including various generalized chess pieces.
He gives 8 x 8 magic king's and queen's tours. Gives
97 inequivalent semi‑magic
knight's paths of which 29 are tours.
These are derived from MSS of H. J. R. Murray at the Bodleian Library,
Oxford. He gives Beverley's square --
see below. He says magic tours (paths?)
exist on 8k x 8k boards for
k ³ 2 and gives a 48 x 48
magic tour. He attributes the
method to Murray (Fairy Chess Review, Aug 1942).
The
Notes report that one of the magic queen's tours was miscopied and that Tom
Marlow has found that one of the 8 x
8 semi‑magic knight's tours is
wrong.
William Beverley. On
the magic square of the knight's march.
(Letter of 5 Jun 1847.) (The
London, Dublin and Edinburgh) Philosophical Magazine & Journal (of Science)
?? (Aug 1848) 1-5. Semimagic
knight's path with diagonals of 210 and
282. See: G&PJ 1 (Sep 1987) 11 &
2 (Nov 1987) 17, which
describe what is known about the problem.
Tom Marlow reports that he has found that there are 101
examples, but he doesn't seem to consider the diagonals, so these are
semimagic. See also: Murray, 1936, below.
C. F. de Jaenisch.
Op. cit. in 5.F.1. 1862. Vol. 2, pp. 151-189. ??NYS.
Semi‑magic knight's tour, with diagonal sums of 256
and 264. [Given in Dickins, p. 27 -- which Dickins??.]
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Magic squares in chess --
The knight's tour, pp. 101-102.
Semi-magic knight's tour with diagonals of 264 & 256,
taken from a BM MS: Bibl. Reg. 13, A. xviii., Plut. xx. c. -- ??NYS, but
same as de Jaenisch's example. Notes
that symmetric cells differ by 32. Gives a cryptic argument that this solution
can be used to produce examples starting at
48 of the cells of the board.
Ball. MRE, 5th ed.,
1911, p. 133 gives a magic king's tour, which is hence a magic queen's
tour. The 11th ed., 1939, p. 185 refers
to Ghersi, below.
Italo Ghersi.
Matematica Dilettevoli e Curiosa.
2nd ed., Hoepli, 1921. Pp. 320‑321,
fig. 261 & 265. Magic king's tour
-- different than Ball's.
H. J. R. Murray.
Beverley's magic S-tour and its plan -- probs. 2106-2108. Problemist Fairy Supplement (later known as
Fairy Chess Review) 2:16 (Feb 1936) 166.
Discusses Beverley's 1848 semi-magic path and says the method leads to 28
solutions and many, including Beverley's example, have the property that
each quarter of the board is also [semi-]magic.
H. J. R. Murray. A
new magic knight's tour -- Art. 68, prob. 5226. Fairy Chess Review 5:1 (Aug 1942) 2-3. A 16 x 16 semi-magic tour. Cites Beverley and Roget.
Implies that Kraitchik asserted that no such 16 x 16 tours were
possible.
Joseph S. Madachy.
Mathematics on Vacation. Op.
cit. in 5.O, (1966), 1979. Pp.
87-89. Order 16 magic knight's tour.
Stanley Rabinowitz.
A magic rook's tour. JRM 18:3
(1985‑86) 203‑204. Gives
one. Also gives Ball's magic king's
tour. Says the magic knight's tour is
still unsolved.
David Marks.
Knight's Tours. M500 137 (Apr
1994) 1. Brief discussion of magic
knight's tours, giving a semi-magic example due to Euler? and a magic example
made up of 2 tours of 32 squares due to Roget.
See also 7.Q
and 7.Q.1.
Yang Hui. Hsü Ku
Chai Ch'i Suan Fa. Op. cit. in
7.N. 1275. Book 3, chap. 1, Magic squares, pp. 149-151, with Lam's
commentary on pp. 311-322. This
includes 6 magic 'circles' which are diagrams of overlapping circles of
values such that each circle adds to a constant value. In two cases, the centres of the circles are
used and in one case some lines also give the same total. Needham, pp. 60-61, and Lam's commentary
describe later work by: Ch'êng
Ta-Wei (1593), Fang Chung-Thung (1661), Chang Ch'ao (c1680),
Ting I-tung (Sung dynasty), Wang Wên-su
(Ming Dynasty) and Pao Ch'i‑shou (late 19C), who has magic cubes, spheres and
tetrahedrons -- see 7.N.1.
Kanchusen. Wakoku
Chiekurabe. 1727. Pp. 5 & 18-21 show two kinds of magic
circles. The first has two rings
of 4
and one in the centre so that each ring adds to 22
and each diameter adds to
23. This is achieved by
putting 1 in the centre and then symmetrically placing the numbers in the
pairs 2-9, 3-8, 4-7, 5-6. The second example uses the same pairing
principle to give three rings of six, with
1 in the centre, so each ring
adds to 63 and each diameter to 64.
See Franklin, c1750, in 7.N for a magic circle.
Curiosities for the Ingenious selected from The most
authentic Treasures of Nature, Science and Art, Biography, History, and General
Literature. 2nd ed., Thomas Boys,
London, 1822. Magic circle of circles,
pp. 56-57 & plate IV, opp. p. 56.
12 - 75 arranged in 8
rings of 8 sectors, with another 12
in the centre, so that each ring and each radius, with the 12
in the centre, makes 360.
Rational Recreations.
1824. Exer. 4, pp. 24-25. = Curiosities for the Ingenious.
Manuel des Sorciers.
1825. Pp. 82-83. ??NX
4 x 8 semi-magic rectangle,
associated. The row sums are 132 while
the column sums are 66. I don't ever
recall seeing a magic rectangle like this before. For an A x B rectangle, we want
1 + ... + AB = AB(AB+1)/2 to be divisible by both A
and B which holds if and only if
A º B (mod 2). Since it doesn't make sense to talk about
diagonal sums, this can only give semi-magic shapes, hence they should be
easier to produce. I find essentially
one solution for the 2 x 4 case.
The Secret Out.
1859. The Twelve-cornered
Arithmetical Star, pp. 374-375. This is
the case n = 6 of the general problem of arranging 1, 2, ..., 2n around a circle so that ai + ai+1 = an+i + an+i+1 for each
i. This immediately leads
to ai ‑ an+i = ‑ (ai+1 ‑ an+i+1) and this requires that n be
odd. The given solution fails to
work at several points. See Devi, 1976,
and Singmaster, 1992, below.
Mittenzwey. 1880.
Prob. 101, pp. 20-21 & 72;
1895?: 117, pp. 25 & 74; A
1917: 117, pp. 23 & 71. Gives the triangular form at the right and
asks for B C
S
= A + B + D = A
+ C + F = D + E + F = B + C
= B + E. This easily D
E F
forces D = A,
C = E = 2A, B = A + F, S = 3A + F.
The original pattern had A = 4,
F = 5, S = 17, and asks
for a solution with S = 13. The solution gives A = 2, F = 7, but there
are five solutions corresponding to A =
0, 1, 2, 3, 4.
Hoffmann. 1893. Chap IV, no. 10: The 'twenty‑six'
puzzle, pp. 146‑147 & 191 = Hoffmann‑Hordern, pp. 118-119,
with photo on p. 133. 4 x 4 square with corners deleted. Arrange
1 ‑ 12 so the 4
horizontal and vertical lines of
4 give the magic sum 26,
and so do the central 4 cells and hence the two sets of opposite
outer cells. Gives two solutions and
says there may be more. Photo shows a
German version labelled with a large
26, comprising a board with
square holes and 12 numbered cubes, 1880-1895.
I. G. Ouseley.
Letters: Pentacle puzzle.
Knowledge 19 (Mar 1896) 63
& (Apr 1896) 84. Consider a pentagram, its five points and
the five interior intersections. Place
the numbers 1 to 10 on these so that each line of five has the
same sum and the five internal values shall add to the same sum, while the five
outer values shall add to twice as much.
Second letter says it seems to be unsolvable and Editorial Note points
out that the sum of all the numbers is
55, which is not divisible
by 3,
so the problem as stated is unsolvable.
[But what if we take the numbers
0 to 9 ??]
Pearson. 1907. Part I, no. 35: A magic cross, p. 35. Same pattern as Hoffmann, with numbers
differently arranged. Says there are 33
combinations that add to 26.
Williams. Home
Entertainments. 1914. The cross puzzle, pp. 117-118. Shape of the Ho Thu (the River Plan, see
beginning of 7.N) to make have the same sum of
23 across and down. [In fact one can have magic sums of 23, 24, ..., 27.]
Wood. Oddities. 1927.
Prob. 8: A magic star, pp. 9-10.
Make an 8-pointed star by
superimposing two concentric squares, one twisted by 45o. There
are 8
vertices of the squares and
8 points of intersection, so
there are 4 points along each square edge.
Arrange the numbers 1 - 16 on these points so each edge adds up to the
same value. This forces the sums of the
4 vertices to be the same (which he states as a given) and the magic constant
to be 34. He gives one solution and says there are 18
solutions, which I have confirmed by computer -- doing it by hand must
have been tedious or I have overlooked some simplifications.
Collins. Book of
Puzzles. 1927.
Meyer. Big Fun Book. 1940. Mathematical fun,
pp. 98 & 731. Make the ABC
figure at the right magical. I've called this a Magic Hourglass -- see
below. D
standard arguments show that the magic
constant must be 12 and
D = 4. EFG
The solution
is essentially unique, with one horizontal line containing
7, 3, 2.
Anonymous. The
problems drive. Eureka 12 (Oct 1949)
7-8 & 15. No. 3. Place the numbers 1, ..., 20 at the
vertices of a dodecahedron so that the sum of the numbers at the corners of
each face is the same. Answer: it cannot
be done! [Similar argument shows that
the only regular polyhedron that can be so labelled is the cube. Then one sees that the sum on an edge is the
same as on the edge symmetric with respect to the centre. Then one finds that an edge must have the
numbers 1, 8 on it and hence all the edges parallel to it have a sum of 9,
hence must have the pairs 2, 7; 3, 6;
4, 5. Putting these vertically
and putting 1 in the top face, we find the top face must contain {1, 4, 6, 7} and there are three distinct ways to place these. The cubo-octahedron cannot be done, but I'm
not sure about the rhombic dodecahedron.]
Ripley's Puzzles and Games.
1966. P. 48. Magic cross-cube. Consider a 2 x 2 x 2 cube.
This has the numbers 1 -
12, 14 - 25 on the facelets so that each face totals 52
and the three facelets at each corner total 39. Such an arrangement
requires the total of all numbers be a multiple of 24, but 1 + ... + 24 = 300 is an odd multiple of 12.
312 is the next multiple of 24
and leads to the numbers used.
Doubleday
- 2. 1971. Ups and downs, pp. 121-122.
Gives the figure at 3 3
3
right, with intermediate lines making
three rows horizontally, 5
5 5
one row vertically and four rows
diagonally. Rearrange the 7 7
7
numbers
present so all these rows total
15. I wondered if one
can put the
numbers 1, ..., 9 on this figure to get the same
sum on all
these lines. However, the magic sum
must then be 15 and the middle number of the top and bottom
rows must occur in four different sums
15 and only the digit 5
occurs in four such sums.
Indeed, letting B be the middle digit in the top row and
adding these four sums gives 4 x
15 =
45 + 3B, so B = 5.
Similarly the middle digit of the bottom row must also be 5.
So the magic figure is impossible.
Further, this argument works for Doubleday's problem, forcing the vertical
line to be all 5s. There are then just two solutions, depending
on whether the top row is 3 5 7 or 7
5 3. Doubleday gives one solution,
saying you may be able to find others.
Birtwistle. Math. Puzzles & Perplexities. 1971. A
Pp. 11 & 13. Consider the pattern at the right. Place the digits B
1, 2, ..., 8 so that the vertical and horizontal quadruples and the E F
G H
inner and outer circles (i.e. A, H, D, E
and B, G, C, F) all have the C
same sum. By exchanging C, D with
G, H, we have a cube with four D
magic faces,
but the pattern has more automorphisms than the cube. One
sees
that A + D = F + G, B + C = E + H, so these pairs can be interchanged. Also, one can interchange
B and C, etc. We can then assume A = 1, B <
C, B < E < H, F < G
and then there are 6 solutions.
Each of these gives 16 solutions with A = 1 and hence 128
solutions allowing any value of
A. Hence there are 768
solutions in total. He gives
one.
Shakuntala Devi.
Puzzles to Puzzle You. Orient
Paperbacks (Vision Press), Delhi, India, 1976.
Prob. 102: The circular numbers, pp. 65 & 125. The case
n = 5 of the arithmetical star
of The Secret Out, 1859. Asks for and
gives just one solution. See Singmaster,
1992.
Gareth Harries.
Going round in triangles. M500
128 (Jul 1992) 11-12. Consider the
lattice triangle of side two. This
has 4
triangles, 6 vertices and 9 edges. Place the numbers 1, .., 15 on the vertices
and edges so that each edge number is the difference of the numbers at its
ends. He says his computer found 19
solutions. For the side one
problem, there are just two solutions.
David Singmaster.
Braintwister: Correct sum, rounded up.
The Weekend Telegraph (27 Jun 1992) xxx &
(4 Jul 1992) xxviii. Based on
the version in Devi, I asked for all the solutions for n = 5
and n = 4. Using the argument I gave under The Secret
Out, there are no solutions for n =
4 or any even n. For n = 5,
the common difference
d = │ai - an+i│ must be either 1 or 5 and in either
case, the antipodal pairs are determined.
Fixing a1 = 1 gives
4!/2 distinct solutions for each
value of d, where the divisor factors out mirror images. In a note to my solution, which was not
published, I showed that for general
n, d must be a divisor of n, and each such divisor gives (n-1)!/2
distinct solutions. Now that I
have found the version in The Secret Out, I am somewhat surprised not to have
found more examples of this problem.
David Singmaster.
Braintwister: Give the hour-glass some time. The Weekend Telegraph (6 Feb 1993) xxxii &
(13 Feb 1993) xxxvi. The Magic
Hourglass problem, as in Meyer, though I don't recall where I saw the problem
-- possibly in one of Meyer's other books.
I recall that my source gave the value of D or S.
Mirko Dobnik (Rošnja 5, 62205 Starše, Slovenia). Letter and example of 31 Aug 1994. Take a solid cube and make each face into
a 2 x 2 array of squares. He puts
the numbers 1, 2, ..., 24 into these cells as shown below. Each face adds up to 50
and numerous bands of 8 around the cube (at least 18)
add up to 100. As he notes, the roundness of these values
is notable.
10 13
11 16
1
22 8 17
3 24 20 5
4
23 19 6
2 21 7 18
14 9
15 12
Mirko Dobnik (Rošnja 5, 62205
Starše, Slovenia). Letter and example
of 24 Jan 1995. Amends two faces of the
above array to the following
10 13
11 16
1
22 8 19
3 24 6 17
4
23 5 18
2 21 7 20
14 9
15 12
The unique solution is given at the
right. It is often presented 18 17
3
with
a point up. I will describe versions by
saying which point or 11 1
7 19
edge
value is up and whether it is a reflection or not. 9 6
5 2 16
E.g.
This has 17 up. 14 8
4 12
15 13
10
[Ernst] von Haselburg (of Stralsund). MS of problem and solution in the City
Archives of Stralsund, dated 5 May 1887, with note of 11 May 1887 saying it was
sent to Illustrierten Zeitung, Leipzig.
Xerox kindly provided by Heinrich Hemme.
[Ernst] von Haselburg (of Stralsund), proposer and
solver. Prob. 795. Zeitschrift für mathematischen und
naturwissenschaftlichen Unterricht 19 (1888) 429 & 20 (1889) 263‑264. Poses the problem for side 3
hexagon. Solution deals with the
three sums of six symmetric points to show the central number is at most 8
and then finds only 5 is feasible and it gives a unique
solution. He has 18
up. Reported by Martin Gardner,
1988, without a diagram; see also
Hemme, Bauch.
William Radcliffe.
38 Puzzle. UK & US patents,
1896. ??NYS -- cited by Gardner,
1984. I couldn't find the US patent in
Marcel Gillen's compilation of US puzzle patents, but a version is reproduced
in Tapson and in Hemme, which has
3 up.
The Pathfinder. (A
weekly magazine from Washington, DC.)
c1910. ??NYS Trigg says Clifford Adams saw the problem
here, but with a row sum of 35!
Tom Vickers. Magic
Hexagon. MG 42 (No. 342) (Dec 1958)
291. Simply gives the solution,
with 13 up, reflected.
M. Gardner. SA (Aug
1963) = 6th Book, chap. 3. Describes Clifford Adams' discovery of
it. Shows 15 up, reflected.
C. W. Trigg. A
unique magic hexagon. RMM 14 (Jan‑Feb
1964) 40‑43. Shows the uniqueness
in much the same way as von Haselburg, but in a bit more detail. Shows
15 up, reflected.
Ross Honsberger.
Mathematical Gems. MAA,
1973. Chap. 6, section 2, pp.
69-76. Describes Adams' work, as given
in Gardner, and outlines Trigg's proof of the uniqueness. A postscript adds that a Martin Kühl, of
Hannover, found a solution c1940, but it was never published and cites Vickers
note.
M. Gardner. Puzzles
from Other Worlds. Vintage (Random
House), NY, 1984, p. 141. Describes
Radcliffe's work -- see above. Gardner
says he was a school teacher at the Andres School on the Isle of Man and
discovered the hexagon in 1895 and patented it in the US and UK. He shows the pattern with 15 up, reflected.
Frank Tapson. Note
71.25: The magic hexagon: an historical
note. MG 71 (No. 457) (Oct 1987)
217‑220. Says he has a book of
correspondence of Dudeney's containing:
two letters from Radcliffe in 1902;
Dudeney's copy of Frénicle's letter to Fermat (or Mersenne?? -- see
7.O.1 & 7.N.2) and a reproduction
of Radcliffe's published solution, labelled
'Discovered 1895 Entered at
Stationers Hall 1896'. He shows 13
up, reflected, and a reproduction of Radcliffe's reflected form, which
is 3
up. Radcliffe's letters refer to
the similar problem discussed by Dudeney in Harmsworth's Magazine (Jan 1902) and
London Magazine (Feb 1902) -- see under Dudeney in 7.O.1. [These are the same magazine -- it changed
its name.]
Martin Gardner.
Letter: The history of the magic
hexagon. MG 72 (No. 460) (Jun 1988)
133. Describes von Haselburg, with no
diagram, as communicated to him by Hemme.
Heinrich Hemme. Das
Kabinett: Das magische Sechseck. Bild der Wissenschaft (Oct 1988) 164-166. Shows the solution is unique. Describes Adams's discovery and Gardner's
article. Says R. A. Cooper then
discovered Vickers' note, then Tapson discovered Radcliffe's version in
1973. Hemme says Tapson said Radcliffe
was a teacher at the Andreas-Schule on the Isle of Man and got a UK
patent. Then in the mid 1980s, Ivan
Paasche of Stockdorf saw the German translation of Gardner and recalled
Haselburg's work which he was able to locate.
Paasche found that there was a 'Stadtbaurat' with an interest in
mathematics named von Haselberg in Stralsund at the time. Stadtbaurat has several meanings -- it could
be a member of the local planning board or a city architect, but it also was an
honorific for a distinguished architect.
Hans F. Bauch. Zum
magischen Sechseck von Ernst v. Haselberg.
Wissenschaft und Fortschritt 40:9 (1990) 240-242 & a cover
side. This is the first source to give
von Haselberg's given name and to give a picture of him. He was born in 1827 and died in 1905. The original MS of the problem and solution
have been located in the City Archives of Stralsund (see above) and Bauch
sketches the original method -- there is a fair amount of trial and error --
and reproduces some of the MS. There
are a number of sub‑configurations which must add up to the constant 38
and Bauch shows these, including a Star of David configuration -- see
7.O.1. Von Haselberg submitted his MS to
the Illustrierten Zeitung of Leipzig, but they didn't use it. He was 'Stadtbaumeister' (= City
Architect) in Stralsund and restored the facade of the City Hall. He also published a five volume work on
local architectural monuments.
Ed Barbeau. After
Math. Wall & Emerson, Toronto,
1995. A magic hexagon, pp.
151-152. When he gave this problem in
his puzzle column, a solution came in only two weeks later and a second
solution arrived six weeks after that.
Neither solver seems to have used a computer. He cites only Honsberger.
The first solver obtained a number of identities which would simplify
the solution.
Frenicle de Bessy.
Loc. cit. in 7.N.2. 1640. Discusses a magic hexagon of White's form II
(below).
I. G. Ouseley.
Letter: The puzzle of
"26". Knowledge 18 (1895)
255-256. Arrange 1 ‑ 12 on the points and crossings of a Star of David so that the sum on each line =
sum of vertices of each large triangle
= 26. Says this has come from Mr. T. Ordish, who has arranged the
numbers to sum to 26 in
30 different ways. A number of these are consequences of the
above conditions, e.g. the sum on the
inner hexagon = the sum of the vertices of a rhombus which is 26, as is the sum along certain angles, but I can't understand what is meant by '6 obtuse angles' or '4
rhomboids'. "I believe
there are at least six ways ...."
In
the next issue, p. 278, are three letters with a comment by Ouseley. Un Vieil Étudiant sends four solutions with
the sum of the vertices of the large triangles being 13, and hence the sum on
the inner hexagon being 52. These are complementary to the solutions
requested, though neither he nor Ouseley notes this. These complementary solutions are rather easier to find though
and I have indeed found six solutions, as does Ahrens below. J. Willis sends one solution of the original
form, possibly implying that it is unique.
T. sends one different solution.
Ouseley notes that in a solution, the value at a point is the sum of the
values at the two opposite crossings.
?? -- possibly more letters in the next issue.
In
the next volume, pp. 35 & 84, are two notes saying that the publishers (T.
Ordish & Co., London) and the proprietors (Joseph Wood Horsfield & Co.,
Dewsbury) of the puzzle have complained that the above notes are an
infringement of their copyright in the "26" Puzzle and Knowledge
apologizes for this.
Dudeney. Puzzling
times at Solvamhall Castle. London
Magazine 7 (No. 42) (Jan 1902) 580‑584
& 8 (No. 43) (Feb 1902)
53-56. The archery butt. = CP, prob. 35, pp. 60-61 & 187-188. Hexagon of
19 numbers so that the 6
radii from the centre to the corners and the 6 sides each add to 22.
Problem is to rearrange them so each adds to 23. Solution says one can
get any number from 22 to
38, except 30.
William F. White.
Op. cit. in 5.E. 1908. Magic squares -- magic hexagons, pp. 187‑188.
Ahrens.
A&N. 1918. Chap. XII: "Die wunderbare 26", pp. 133‑140. Consider a hexagram (or six pointed star)
formed from two triangles. This
has 12
vertices. He finds 6
ways to place 1, 2, ..., 12 on these vertices so that each set of four
along a triangle edge adds up to the same value (which must then be 26),
and the six corners of the inner hexagon also add to 26.
I.e. this is Ordish/Ouseley's problem, but with a slightly different
statement of conditions.
On
p. 134, a note says the puzzle "Wunderbare 26" is made and sold by Züllchower Anstalten, Züllchow bei
Stettin, and they have registered designs 42,768 and 45,600 for it, though the
'Rustic 26' has already been on sale
for many years. S&B, p. 39,
shows an English example with no identification.
Collins. Fun with
Figures. 1928.
Perelman. 1934. See in 7.N.2 for the Star of David pattern
with just one solution.
Birtwistle. Math.
Puzzles & Perplexities. 1971.
Mathe
mit Energie -- Energie mit Mathe.
Verlag Leipziger Volkszeitung, 1981.
A
Magische Rosette, p. 34 &
solutions p. 17. Place numbers 1, ..., 13
on F L G B
the centred Star of David pattern at
the right so that each rhombus from K
M H
the centre to an outer vertex has the
same sum 35. They give one E J I
C
solution. In 2000, I found three, after factoring out the symmetries of the D
figure. A little work shows that the magic sum
satisfies 20 £ S £
36 and
a simple
program shows that 21 £ S £
35 and for S = 21, 22, ..., there
are
3, 5, 9, 20, 23, 18, 8, 26, 8, 18, 23, 20, 9, 5,
3 solutions, giving a total of 198
solutions.
Bell & Cornelius.
Board Games Round the World. Op.
cit. in 4.B.1. 1988. Marvellous
'26', p. 79. Says this was sold
for 6d
by T. Ordish "probably in about 1920". [Above we see that it was known in
1895.] They quote the instructions from
the box: "each of the six sides as
well as the six spaces around the centre total up to 26 with perhaps the
finding of several additional 'twenty sixes'." This is Ordish/Ouseley's problem.
Bauch. Op. cit. in
7.O, 1990. In this, he gives the Star
of David problem, which is a subproblem of the Magic Hexagon. He asserts it has 96 'classical' solutions,
but he gives no discussion or reference and it is not clear if this is for all
possible magic sums -- he shows one solution with magic sum 26
and the central hexagon having total
26.
David Singmaster.
Magic Stars of David and the
26 Puzzle. Draft written in May 1998. There are
960 solutions for a magic Star
of David. There is clearly an
equivalence given by the symmetries of the regular hexagon, so these solutions
fall into 80 equivalence classes. Six
of these are solutions of the 26 Puzzle (i.e. the sum of the central hexagon
is also 26) and six are solutions of
the complemented problem (i.e. the sum of the outer points is 26).
One can use complementation to reduce the number of classes to 40.
However, there are more symmetries of the magic Star of David when one
ignores the additional constraint of the
26 Puzzle -- indeed the pattern
is isomorphic to labelling the edges of a cube so the sum of the edges around
each face is 26. Hence one can use the symmetries of the cube
to produce 20 equivalence classes of solutions, but these symmetries do not
interact simply with the additional sums used in the 26 Puzzle.
See also 7.E,
7.R.1, 7.R.2, 7.U.
7.P.1. HUNDRED FOWLS AND OTHER LINEAR PROBLEMS
See Tropfke 565, 569, 572 & 613.
NOTATION: n for p
at a, b, c means
n items of three types,
costing a, b, c were bought for a total of p.
I.e. we want
x + y + z = n;
ax + by + cz = p, with the
conditions that x, y, z are positive (or non-negative) integers.
(a, b) solutions means a non‑negative solutions, including b
positive solutions -- so a £
b. I have checked these with a computer
program. I also have a separate
numerical index to these problems which enables me to tell whether problems are
the same.
When there are just two types of
fowl, one gets two equations in two unknowns, but I have generally omitted such
problems except when they seem to be part of an author's development or they
represent a different context.
See: MS Ambros. P114; Tartaglia 17‑25 &
26; Hutton; Ozanam-Hutton 9;
Williams; Collins.
Della Francesca gives two problems
where the value of two different combinations of two fruits or three animals
are given. These are not Hundred Fowls
Problems, but in the second, elimination of one unknown leads to one equation
in two unknowns, just as the Hundred Fowls Problem does. Pacioli gives a similar problem with no
integral solutions. I don't recall
seeing other examples of this type of problem.
The medieval problems of alligation
are related, but the solution need not be integral. See: Fibonacci; Lucca 1754;
Bartoli; della Francesca; Borghi;
Apianus; Tropfke 569 for discussion and examples. See:
Devi for a modern version with
integral solutions. See Williams
for a history of this aspect.
Some monetary problems naturally
occur here. Paying a sum with
particular values and a specified number of pieces is just our ordinary
problem. Paying a sum with particular
values, without specifying the number of pieces, leads to one equation in
several unknowns. This is the same as
asking for the number of ways to change the total value. Perhaps more interesting are problems where
one person has to pay a debt to another and they only have certain values,
which leads to problems like ax - by =
c.
Ordinary problem in monetary terms
-- see: Riese (1524); Dodson (1747?); Ozanam‑Montucla (1778);
Ozanam-Hutton (1803); Hall
(1846); Colenso (1849); Perelman (1934); M. Adams (1939); Depew (1939); Hedges
(1950?);
Little Puzzle Book (1955); Ripley's (1966); Scott
(1973); Holt (1977);
Paying a sum or making change: ax + by + ... = n -- see:
Simpson (1745 & 1790);
Dodson (1747? & 1753); Euler (1770);
Moss (1773);
Ozanam-Montucla (1778);
Bonnycastle (1782 & 1815); Hutton, 1798?;
Ozanam-Hutton (1803); De Morgan
(1831?); Bourdon (1834); Unger (1838); Hall (1846); Clark
(1916);
Paying a debt with limited values:
usually ax - by = c -- see:
Euler (1770); Ozanam‑Montucla
(1778); Bonnycastle (1782); Stewart (1802); Ozanam-Hutton (1803);
De Morgan (1831?); Unger
(1838); Todhunter (1870); McKay (1940);
McKay (1940); Little Puzzle Book (1955) use the context of buying postage stamps.
G.F. (1993) uses the context of wheels of vehicles.
Simpson (1745); Dodson (1747?); Euler (1770);
Ozanam-Montucla (1778); Bonnycastle
(1782 & 1815); Ozanam-Hutton
(1803); Bourdon (1834); Todhunter (1870); Clark (1904); McKay
(At Home Tonight, 1940) are the only
examples here which consider problems with one equation in two unknowns.
Problems involving heads and feet of
a mixture of birds and beasts:
Clark; Williams; Collins;
Ripley's.
Impossible problems -- sometimes a
problem is impossible only if positive solutions are required. Abu Kamil;
Fibonacci 1202 & 1225;
Tartaglia; Buteo; Simpson;
Euler; Ozanam‑Montucla; Bonnycastle; Perelman; Depew; Little Puzzle Book; Scott;
Holt.
I have recently realised that the
relatively modern problem of asking how to hit target values to make a
particular value is a problem of this general nature, especially if the number
of shots is given. E.g. a target has
areas of value 16, 17, 23, 24, 39; how does one achieve a total of 100?
These occur in Loyd, Dudeney, etc., but I haven't recorded them. I may add some of them.
Zhang Qiujian. Zhang
Qiujian Suan Jing. Op. cit. in
7.E. 468. Chap. 3, last problem, pp. 37a ff (or 54 f ??). ??NYS.
Hundred Fowls: 100 for 100 at 5,
3, ⅓. (Cocks, hens, chicks.) This has (4, 3) solutions -- he gives (3, 3)
of them, but only states the relation between the solutions -- no indication of
how he found a solution. (Translation
in Needham 121-122 (problem only),
Libbrecht 277, Mikami 43, Li & Du 99.)
Ho Peng Yoke. The
lost problems of the Chang Ch'iu‑chien
Suan Ching, a fifth‑century
Chinese mathematical manual. Oriens
Extremus 12 (1965) 37‑53. On pp.
46‑48, he identifies prob. 31 of
Yang Hui (below at 1275) as
being from Chang (= Zhang). The
solution involves choosing one value arbitrarily.
Zhen Luan (= Chen Luan).
Op. cit. in 7.N, also called
Commentary on Hsü Yo's Shu Shu Chi I. c570.
??NYS. [See Li & Du, p.
100.] 100 for 100 at 5, 4, ¼.
(Cocks, hens, chicks.) This has
(2, 1) solutions -- he gives (1, 1).
Mikami says Chen's method would give one solution to Chang's
problem. Libbrecht, pp. 278‑279,
gives the method, which is indeed nonsense, and states that other scholars
noted that Chen's method is fortuitous.
He says Chen also gives 100 for 100 at
4, 3, ⅓, which has (3, 2)
solutions, of which Chen gives (1, 1).
Liu Hsiao‑sun.
Chang ch'iu‑chien suan‑ching hsi‑ts'ao (Detailed solutions of [the problems] in
the Chang ch'iu‑chien suan‑ching). c600.
??NYS. Described in Libbrecht
pp. 279‑280 as nonsense!!
Bakhshali MS. c7C. Sūtra C7 (VII 11-12). Hayashi 648-650 studies this. General discussion and first example are too
mutilated to restore.
Li Shun‑fêng.
Commentary on Chang chiu‑chien. 7C.
??NYS. Libbrecht (p. 280)
describes his comments as "unmitigated nonsense".
L. Vanhée. Les cent
volailles ou l'analyse indéterminée en Chine.
T'oung Pao 14 (1913) 203‑210
& 435‑450. On pp. 204‑210, he gives 24 problems
in Chinese and French, but doesn't identify the sources!!
Alcuin. 9C.
Mahavira. 850. Chap. III, v. 133, pp. 67-68 is related to
this general type. Chap VI, v. 143‑153,
pp. 130‑135.
Chap. III.
Chap. VI.
Pseudo-Alcuin.
9C. ??NYS -- cited by Hermelink,
op. cit. in 3.A.
Sridhara. c900. V. 63‑64, ex. 78‑80, pp. 50‑52
& 95.
Abu Kamil [Abū Kāmil Shujā‘ ibn Aslam ibn
Muhammad (the h should have an underdot) ibn Shujā‘,
al-Hāsib al-Mişrī].
Kitāb al‑ţara’if [NOTE: ş, ţ denote
s, t with an underdot.] fi’l‑hisāb (the h
should have an underdot) (Book
of Rare Things in the Art of Calculation).
c900. Trans. by H. Suter
as: Das Buch der Seltenheiten der
Rechenkunst von Abū Kāmil el‑Mişrī; Bibl. Math. (3) 11 (1910‑1911) 100‑120. (I have a reference to an Italian
translation by G. Sacerdote; IN:
Festschrift zum 80 Geburtstag M. Steinschneiders; Leipzig, 1896, pp. 169‑194,
??NYS.) (Part is in English in Ore;
Number Theory and Its History; 139‑140.)
Six problems of 100 for 100 at the following.
al‑Karkhi.
c1010. Sect II, no. 10, p.
82. Mix goods worth 5, 7, 9 to make one worth 8.
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 110f., part IV, no. 20. ??NYS - cited by Tropfke 614, who says it
has (cows, sheep, hens).
Bhaskara II.
Bijaganita. 1150. Chap VI, v. 158‑159. In Colebrooke, pp. 233‑235. Same as Mahavira's 152. (Doves, cranes, geese, peacocks.) (This also has
5x + 8y + 7z + 92 = 7x
+ 9y + 6z + 62. ??)
Fibonacci.
1202. Chap. 11: de consolamine
monetarum [on the alloying of monies], pp. 143‑166 (S: 227-257)
deals with this problem in the more general form of combining metals. (Other versions involve mixing spices,
wines, etc.) These lead to a1x1 + ... + anxn =
b(x1 + ... + xn),
often with x1 + ... +
xn specified. xi is the weight and ai is the purity or gold content, etc., of
the i-th metal. We are mixing the metals to produce a total
of weight x1 + ... + xn with purity
b. Hence there is no need to
consider only integral values but he usually gives one (or a few) integral
solutions. I describe a few
examples. A denotes the vector of ai's.
Fibonacci.
Epistola. c1225. In Picutti, pp. 332-336, numbers XI -
XIII. One of the problems is briefly
mentioned in: M. Cantor; Mathematische
Beiträge zum Kulturleben des Völker; Halle, 1863; reprinted by Olms,
Hildesheim, 1964; p. 345. Surprisingly,
none of these problems have appeared elsewhere!
Abbot Albert. c1240.
Yang Hui. Xu Gu Zhai
Qi Suan Fa (= Hsü Ku Chai Ch'i Suan
Fa). 1275. Loc. cit. in 7.N, pp. 165‑166, prob. 29‑31.
Gherardi. Libro di ragioni. 1328.
Lucca 1754. c1330.
Munich 14684.
14C. Prob. XI, p. 79.
Narayana Pandita (= Nārāyaņa Paņdita
[NOTE: ņ denotes n with an overdot and the d
should have an underdot.]).
1356. Op. cit. in 7.N, p. 1,
lines 2‑5, p. 93. (Same as Mahavira's
152.) ??NYS -- see Bag, op. cit. under
Bakhshali MS, p. 92.
Folkerts.
Aufgabensammlungen. 13-15C.
Bartoli.
Memoriale. c1420. Ff. 89v - 95r (= Sesiano, pp. 134-135). Eleven problems of alligation, many
identical to Lucca 1754.
Jamshid al-Kāshī
= Ğamšīd ibn Mas‘ūd ibn Mahmūd (the h
should have an underdot), Ġijāt ed-dīn
al-Kāšī = Ghiyāth al-Din
al-Kāshī. Miftāh
al-hisāb (the h should have an underdot) (The Calculator's
Key). 1426. Ed. by A. Demerdash & M. H. Hifna (the H
should have an underdot) ; Cairo, nd.
Facsimile and Russian translation by:
B. A. Rozenfeld', V. S. Segal' & A. P. Juškevič
as: Ključ Arifmetiki -- Traktat ob
Okružnosti; Moscow, 1956. ??NYS --
Hermelink, op. cit. in 3.A, says he gives a version with three fowls.
Provençale Arithmétique.
c1430. Op. cit. in 7.E.
Pseudo-dell'Abbaco.
c1440. Prob. 190, pp. 150‑151. 48 for 48 at 4, 2, ¼. (1, 1) solution,
given.
AR. c1450. Prob. 45, 81, 102, 119‑126, 309. Pp. 40, 52, 60‑61, 66‑67, 137‑138,
175‑176, 221‑222.
Correspondence of Johannes Regiomontanus, 1463?-1465. Edited by Maximilian Curtze as: Der Briefwechsel Regiomontan's mit Giovanni
Bianchini, Jacob von Speier und Christian Roder. Part II of: Urkunden zur Geschichte der Mathematik im Mittelalter
und der Renaissance. AGM 12 (1902).
Benedetto da Firenze.
c1465. Pp. 151‑152. Part of the text is lacking, but the problem
must be 60 for 60 at 6, ½,
⅓. (Cows, calves, pigs.) This has (2, 2) solutions, (1, 1) is given.
Gottfried Wolack.
1468. Dresden MS C80, ff.
301'-303. This is a MS of lectures
given at Erfurt, 1467 & 1468,
Transcribed and discussed in: E.
Wappler; Zur Geschichte der Mathematik im 15. Jahrhundert; Zeitschrift für
Math. und Physik -- Hist.-Litt. Abteilung 45 (1900) 47-56. P. 51 has:
100 for 100 at 2, 1, ½. (Men, women, children.) There is an obscure calculation leading to
dividing the people in the proportion 4
: 2 : 1, so there are 57 1/7
men! This may be getting
confused with 7.G.2 -- Posthumous twins.
della Francesca.
Trattato. c1480.
Luca Pacioli.
Aritmetica. c1480. ??NYS -- described in Sesiano. F. 238r.
Chuquet. 1484.
Triparty, part
1. Of apposition and remotion. English in FHM 88-90. Several examples treated as problems in
numbers -- no mention of buying anything.
Gives a method of finding one positive solution.
Appendice.
Borghi.
Arithmetica. 1484. Ff. 93v-101v. Several problems of alligation, getting up to mixing five grains
of values 44, 48, 52, 60, 66 per measure to mix to produce a product of
value 50.
Johann Widman. Op.
cit. in 7.G.1. 1489. ??NYS.
Glaisher, pp. 14 & 121, gives:
F. 109v: mix wines worth 20, 15,
10, 8 to make one worth 12. Gives one solution: 6, 6, 11, 11.
Pacioli. Summa. 1494.
Riese.
Rechnung. 1522. 1544 ed. -- pp. 104‑106; 1574 ed. -- pp. 70r‑71v. The 1544 ed. calls this section 'Regula
cecis oder Virginum'; the 1574 ed.
calls it 'Zech rechnen'. There is first
a simple problem with only two types, hence determinate.
Tonstall. De Arte
Supputandi. 1522. P. 240.
Repeats Pacioli's prob. 14, except there is a misprint -- in the second
case, he has 3, 5, 7 costing
94⅔.
Riese. Die
Coss. 1524. No. 67, p. 49. 100 for
460 at 3, 5. (Coins.)
H&S 93 says a tavern version is in Rudolff (1526?).
Apianus. Kauffmanss
Rechnung. 1527.
Sesiano cites a 16C MS Ambros. P114 sup which gives 40 for 100 at 1/5, 1/10. Answer: 960, -920.
Cardan. Practica
Arithmetice. 1539.
Tartaglia. General
Trattato. 1556. Book 16, art. 117‑129, pp. 254r‑255v &
Book 17, art. 25, 26, 43, 44,
pp. 272v & 277r-277v. 18
versions. The objects being bought are
mostly not in the Italian dictionaries that I have consulted. They are apparently 16C Italian, probably
Venetian dialect. Several Italian or
Italian-speaking friends have helped to determine these -- my thanks to
Jennifer Manco, Ann Maury, Ann Sassoon and especially Maria Grazia Enardu and a
student of hers.
Buteo.
Logistica. 1559.
Baker. Well Spring
of Sciences. 1562? Prob. 11,
1580?: ff. 194r-195r; 1646:
pp. 305‑306; 1670: pp.
346-347. 20 for 240 at 20, 15, 8.
(Payments to men, women, children.)
(1, 1) solutions, which he gives.
Bachet.
Problemes. 1612. Addl. prob. X, 1612: 164-172; 1624: 237-247; 1884: 172‑179.
Book of Merry Riddles.
1629? 12 for 12 at 4, 2, ½, ¼.
(Capons, hens, woodcocks, larks).
(5, 1) solutions, with the positive one given.
John Wallis. A
Treatise of Algebra, both Historical and Practical. John Playford for Richard Davis, Oxford, 1685. (Not = De Algebra Tractatus.) Chap. LVIII, pp. 216-218. ??NX
W. Leybourn.
Pleasure with Profit. 1694. Chap. XIII: Of Ceres and Virginum, pp.
51-55.
Anonymous proposer and solver. Ladies' Diary, 1709-10
= T. Leybourn, I: 5, quest.
8. Mix wines worth 32, 20, 16
to make 56 worth
22. I.e. 56 for 1232 at 32, 20, 16. (12, 11)
solutions of which the positive ones are given.
Adrastea, proposer; anonymous solver. Ladies' Diary, 1721-22 = T.
Leybourn, I: 112-113, quest. 89. 24
passengers of four ranks pay £24, with their fares in the proportion 16 : 8 : 2 : 1. The solutions depend on whether one takes
fares as whole numbers of pounds, shillings or pence. The cheapest fare,
R, is readily seen to
satisfy 15d £ R £ 1£. The solution says a Mr. Evans collected 100
true answers. I found (1, 0)
solutions in pounds and (89,
43) solutions in shillings, then wrote
a special program to solve the 24 cases which occur in pence, finding (201, 101)
solutions. Presumably Mr. Evans
missed one of the positive solutions.
Using farthings gives just one more case with (8, 3) solutions.
Ozanam. 1725. [This is one of the few topics where the
1725, 1778 and 1803 editions vary widely -- see each of them.]
Dilworth.
Schoolmaster's Assistant.
1743. Part IV: Questions: A
short Collection of pleasant and diverting Questions, p. 168. Problem 3.
20 for 20 at 4, ½, ¼. (Pigeons, larks, sparrows.) This has (2, 1) solutions, none given. = Pacioli 18.
Simpson.
Algebra. 1745. Section XIII, quest. 2-8, 10-12, pp.
170-181 (1790: prob. II‑VII, IX,
XI‑XV, pp. 183‑200).
Dodson. Math.
Repository. (1747?); 1775. He also has several mixture problems and
some simple problems which are mentioned at the entry for vol. II, below.
Les Amusemens.
1749. Prob. 164, p. 310. 20 for 60 at 6, 4, 1. (Men, women,
valets.) (3, 2) solutions -- he gives
(1, 1).
James Dodson. The
Mathematical Repository. Vol. II. Containing Algebraical Solutions of A great
Number of Problems, In several Branches of the Mathematics. I. Indetermined Questions, solved
generally, by an elegant Method communicated by Mr. De Moivre. II.
Many curious Questions relating to Chances and Lotteries. III.
A great Number of Questions concerning annuities for lives, and their
Reversions; wherein that Doctrine is illustrated in a Multitude of interesting
Cases, with numerical Examples, and Rules in Words at length, for those who are
unacquainted with the Elements of these Sciences, &c. J. Nourse, London, 1753. On pp. 1 - 63, he deals with 13 examples of
determining the number of solutions of a linear equation in two (5 cases),
three (6 cases) or four (2 cases) unknowns.
Because of the unusual extent of this, I will describe them all. (I had to completely revise my programs for
counting the number of solutions in order to deal with these. The revision introduced double precision
values for the counts and a way of computing the number of solutions for the
two variable problem which speeded up the programs up by a factor of about 100,
but one problem still took 2½ days!) P.
1 has a subheading: The Solution of
indetermined Questions in Affirmative Integers, communicated by Mr. Abraham De
Moivre, Fellow of the Royal Societies of London and Berlin.
Euler. Algebra. 1770.
II.I, pp. 302‑310.
II.II, pp. 311‑317.
II.III: Questions
for practice, p. 321.
Mr. Moss, proposer and solver. Ladies' Diary, 1773-74
= T. Leybourn, II: 374-376,
quest. 658. Pay 50£ (= 1000s) with pistoles (17s), guineas (21s), moidores (27s) and
six-and-thirties (36s). (529, 412) solutions, of which the positive ones are
given.
Ozanam-Montucla.
1778. [This is one of the few
topics where the 1725, 1778 and 1803 editions vary widely -- see each of them.]
The following are contained in the
Supplement and have no solutions given.
Bonnycastle.
Algebra. 1782. Pp. 135-137 give a number of problems of
finding some or all the integral solutions of one linear equation in two or
three variables.
Hutton. A Course of
Mathematics. 1798? Prob. 19,
1833: 221; 1857: 225. Pay
£120 (= 2400s) with
100 coins using guineas (= 21s) and moidores (= 27s), i.e. 100 for 2400 at 27, 21.
John Stewart. School
exercise book of 1801‑1802.
Described by: W. More; Early
nineteenth century mathematics; MG 46 (No. 355) (Feb 1962) 27‑29. "Having nothing on me but guineas and
having nothing on him but pistoles, I wish to pay him a shilling." = Euler II.III.8. Least solution given.
Ozanam-Hutton.
1803. [This is one of the few
topics where the 1725, 1778 and 1803 editions vary widely -- see each of them.]
The following are
contained in the Supplement and have no solutions given.
Bonnycastle.
Algebra. 10th ed., 1815.
Augustus De Morgan. Arithmetic
and Algebra. (1831 ?). Reprinted as the second, separately paged,
part of: Library of Useful Knowledge -- Mathematics I, Baldwin & Craddock,
London, 1836. Arts. 310-317, pp.
103-110, cover indeterminate problems in general.
Bourdon.
Algèbre. 7th ed., 1834.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 145-172 & 259-262, nos.
541-645. He treats the topic at great
and exhausting length, starting with solving
ax = by, then x + y = c, then
ax + y = c, then ax = by + 1
and ax = by + c, also phrased as ax º c (mod b). He then does Chinese Remainder problems, but
with just two moduli. He continues
with ax + by = c, initially giving just one solution, but then
asking how many solutions there are. He
varies this in an uncommon way -- solve
x + y = e with x º a (mod b) and
y º c (mod d). He also
varies the problem by letting a, b be rationals. After 20 pages and 90 problems, he finally gets to one problem,
no. 631, of the present type. He then
goes on to ax + by = c (x + y), but returns with 12 problems of our type,
no. 634-645. Of these, 640 and 645 have
an answer with a zero, but he only gives the positive answers. None of these problems are the same as any
others that I have seen.
Thomas Grainger Hall.
The Elements of Algebra: Chiefly Intended for Schools, and the Junior
Classes in Colleges. Second Edition:
Altered and Enlarged. John W. Parker,
London, 1846.
John William Colenso (1814-1883). The Elements of Algebra Designed for the Use of Schools. Part I.
Longman, Brown, Green, Longmans & Roberts, London, (1849), 13th ed.,
1858 [Advertisement dated 1849].
Exercises 63, p. 114 & Answers, p. 14.
Family Friend (Dec 1858) 357. Arithmetical puzzles -- 3.
Same as Alcuin's 39 with (oxen, sheep, geese). I haven't got the answer.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-44, pp. 256 & 397:
Arithmetisches Rätsel. 100 for 100
at 10, 3, ½. (Geese, hares, partridges.)
(2, 1) answers, only (1, 1) given.
Todhunter. Algebra,
5th ed. 1870. Many straightforward examples, of which a I give a few.
Mittenzwey.
1880.
Editorial answer to F. Chapman, a correspondent. Knowledge 2 (17 Nov 1882) 409. 100 for 100 at 10, 3, ½.
(Bullocks, sheep, pigs.) (2, 1)
answers, only (1, 1) given.
c= Leske.
Hoffmann. 1893. Chap. IV, no. 35: Well laid out, pp. 152
& 200‑201 = Hoffmann-Hordern, pp. 125-126. 21 for 24 at 2, 3, ½, 4.
(6, 2) solutions -- he gives (2, 2).
Clark. Mental
Nuts. 1904, no. 49; 1916, no. 63. Turkeys and sheep.
"A drove of turkeys and sheep have 100 heads and feet. How many are there of each?" This leads to 3T + 5S = 100,
which has (7, 6) answers. He
says there are 6 answers and gives one example. His
1897, no. 12; 1904, no. 20; 1916, no. 93 and 1904, no. 67 are ordinary versions, with only one
positive answer in each case.
Pearson. 1907. Part II, no. 155, pp. 144 & 222. = Alcuin 39. Only the positive answer is given.
Loyd.
Cyclopedia. 1914. Sam Loyd's candy puzzle, pp. 121 &
354. = SLAHP: Assorted postcards,
pp. 45 & 101. = Pacioli 18. Only the positive solution is given.
Williams. Home
Entertainments. 1914. The menagerie, p. 127. Menagerie of birds and beasts has 36
heads and 100 feet.
I.e. 36 for 100 at 2, 4.
Clark. Mental
Nuts.
Hummerston. Fun,
Mirth & Mystery. 1924. The cinema puzzle, Puzzle no. 36, pp. 92
& 177. How many ways to change 6d
using 6d, 3d, 1d, ½d, ¼d? (67,0) solutions -- he says 67.
Collins. Fun with
Figures. 1928. Four out of five have it, p. 183. Hunter has shot birds and rabbits and
has 36
heads and 100 feet.
I.e. 36 for 100 at 2, 4.
Loyd Jr. SLAHP. 1928.
He gives a number of examples, usually with some extra feature.
Perelman. FFF. 1934.
Hundred rubles for five. 1957:
prob. 37, pp. 54 & 57-58;
1979: prob. 40, pp. 69‑70 & 72‑73. = MCBF: prob. 40, pp. 67 & 70-71. Magician asks for 20 for 500 or 300 or 200 at
50, 20, 5. These are all
impossible!
M. Adams. Puzzle
Book. 1939. He has several straightforward problems, which I omit, and the
following. Prob. C.10, pp. 125 &
173. Use 26d in florins (= 24d), shillings (= 12d),
sixpence (= 6d), pennies (= 1d) and
half‑pences (= ½ d) to
measure
5⅝ inches. The
widths of the coins, in 16ths of an inch, are:
18, 15, 12, 19, 16, respectively. This leads to: 24a + 12b
+ 6c + d + e/2 = 26,
18a + 15b + 12c + 19d + 16e =
90. (1, 0) solution,
which is given.
Depew. Cokesbury
Game Book. 1939.
McKay. At Home
Tonight. 1940.
McKay. Party
Night. 1940. No. 24, pp. 181-182. 6
for 6 at 2, 1, ½. (Men, women, children eating loaves of
bread.) Gives (1, 1) of the (3, 1)
solutions.
Sid G. Hedges. The
Book of Stunts & Tricks. Allenson
& Co., London, nd [c1950?].
Shilling change, p. 45. Change a
shilling into 12 coins without using pennies, i.e. 12 for 12 at 6, 3, ½, ¼. (2, 1)
solutions -- he gives (1, 1).
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. This has a number of examples which I
include as illustrating mid 20C usage.
Ripley's Puzzles and Games.
1966.
[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers.
1973. Op. cit. in 5.E.
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 143: The farmer and the animals, pp.
88 & 134. Buy animals at 50, 40, 25, 10 (mules, sheep, goats, pigs)
to produce an average value of
30. This is essentially an
alligation problem as discussed under Fibonacci. She gives one answer: 1,
1, 2, 1 and says "Other answers
are possible" -- somewhat of an understatement since it has a three
parameter set of solutions and two of the parameters can range to infinity!
Michael Holt. Math
Puzzles and Games. Walker Publishing
Co., NY, (1977), PB ed., 1983. Hundred
dollars for five, pp. 37-38 & 101-102.
10 for 500 at 10, 25, 50 -- making change, but requiring each type of
coin to be used. (1, 0) answers, hence
impossible if each type of coin must be used.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
Takeaway pay, pp. 83 & 133.
Nine employees of three types, earning
£5.00, £3.75 and £1.35 per hour
earn £333.60 in a shift of a whole number of hours. How long was the shift?
This gives us
x + y + z = 9, 500x + 375y + 135z =
33360/n, where n is
the number of hours in a shift. In the
problem, x is given as 1, but this information is not needed -- I have
used this generalization for one of my problems.
Tom Bullimore.
Sherlock Holmes Puzzles.
(Originally: Baker Street Puzzles; Ravette; ©1992 with Knight
Features.) Orient Paperbacks (Vision
Books), New Delhi. India, 1998. No. 34,
PP. 38 & 125. 2000 for 7500 at 3, 4, 5.
(Goblet, candlesticks, figurines.)
This would have (751, 749) answers, but he states that the sum of two of
the numbers sold must be 506, which leads to just one solution. [Question: can you find values for the sum
of two of the numbers which give two or three solutions?]
G. F. The
crash. Mathematical Pie, 129 (Summer
1993) 1022 & Notes, p. 1. Determinate version involving vehicles having 2, 4, 8
wheels (motorcycles, cars,
lorries), giving m + c + l
= 11, 2m + 4c + 8l = 44,
m = 4. Omitting the last
equation would give us 11 for 44 at (2,
4, 8), which has (4, 3) solutions.
David Singmaster.
The hundred fowls, or how to count your chickens. Accepted by Mathematics Review (Univ. of
Warwick) for 1994, but it closed before the article was published. General survey of the history.
J. Williams.
Mathematics and the alloying of coinage
1202-1700. Annals of Science 52
(1995) 213-234 & 235-263. ??NYS --
abstracted in BSHM Newsletter 29 (Summer 1995) 42 -- o/o. Surveys the problem from Fibonacci through
Kersey, etc. in the 17C.
Vladimir Dubrovsky.
Brainteaser B161 - Two-legged, three legged, and four legged. Quantum 6:3 (Jan/Feb 1996) 15 & 48. Room contains a number of people sitting on
three-legged stools and four-legged chairs.
There are no spare seats and the total number of legs is 39.
This gives x = y + z, 2x + 3y + 4z = 39.
7.P.2. CHINESE REMAINDER THEOREM
See Tropfke 636.
There are two forms of this. The ancient Indians describe them as the
residual pulveriser, where one is given the residues to several moduli, and the
non-residual pulveriser, where one has
ax - c º 0 (mod b) which is
ax - by = c. The residual form
is the classic Chinese Remainder Theorem, whose solution is generally found by
reducing to the non-residual form, solved by the Euclidean algorithm.
Notation. Multiple congruences are written in an abbreviated notation. E.g.
n º 1 (mod 2) and
n º 2 (mod 3) is written n º 1 (2),
2 (3) or even more abbreviatedly
as n
º -1 (2, 3).
Standard problem types.
A‑k.
n º 1 (2, 3, 4, 5, ..., k‑1),
0 (k).
A-5. See:
Tartaglia; Baker; Dilworth;
Jackson.
A-7. See:
Bhaskara I; Ibn al-Haitam; Fibonacci;
AR; Benedetto da Firenze;
della
Francesca; Chuquet; HB.XI.22;
Pacioli; Tagliente; Cardan;
Tartaglia;
Buteo; Baker;
van Etten; Ozanam, 1725; Vyse;
Dodson; D. Adams, 1801;
Badcock; New Sphinx;
Magician's Own Book;
Wehman.
Solution
is 301 + 420k, but early examples give just 721.
A-10. See: Pacioli.
A-23. See: Pacioli.
B-k. n º 0 (2, 3, 4, ..., k).
B-9. See:
Euler; Bonnycastle 1782; Jackson.
B-10.
See: Ripley's.
C-k. n º -1 (2, 3, ..., k).
C-5. See:
Mahavira.
C-6. See:
Ladies' Diary, 1748; Vyse; Bonnycastle 1782.
C-9. See:
Tartaglia; W. Leybourn; Carlile.
C-10.
See: Pseudo-dell'Abbaco; Muscarello;
Ripley's.
C-20.
See: Gentlemen's Diary, 1747; Vyse.
C. n º ‑1 (3, 4, 5, 6).
See: Brahmagupta; Bhaskara I
??; Bhaskara II.
D-k. n º ‑1 (2, 3, 4, 5, ..., k‑1), 0 (k).
D-5. See: Baker;
Illustrated Boy's Own Treasury.
D-7. See: Fibonacci; Marliani; Benedetto da
Firenze; Pacioli; Ghaligai;
Cardan; Tartaglia;
Buteo; Baker;
van Etten; Ozanam-Montucla; Dodson.
D-9. See: Pacioli.
D-11. See: Pacioli.
D-23. See: Pacioli.
E. General result
for 3, 5, 7. See: Sun Zi; Fibonacci;
Yang Hui; AR; Pacioli;
Tartaglia.
F. General result
for 5, 7, 9. See: Fibonacci; Pacioli.
Cases
with for moduli 28, 19, 15 arise in computing the Julian year -- see:
Simpson; Dodson; Bonnycastle, 1782; Todhunter.
I have
a separate index of problems.
Sun Zi (= Sun
Tzu). Sun Zi Suan Ching (Master Sun's Arithmetical Manual). 4C. [This
is not the famous general and writer of The Art of War, c-4C.] Chap. 3, prob. 26, p. 10b: There is an
unknown number of things. ??NYS. n º 2 (3),
3 (5), 2 (7) & problem E. Only the least answer
is given. (See Needham, pp. 34 &
119. English in Needham 119, Mikami 32
and Li & Du 93; Chinese and English in Libbrecht 269.)
The problem has been transmitted as a folk rhyme in China
and Japan. It is known as Sun Zi Ge
(The Song of Master Sun) or Han
Xin Dian Bin (General Han Xin's Method
of Counting Soldiers). (Han Xin was a general of c‑200.) The rhyme is cryptic but gives the three
multipliers 70, 21, 15 for problem
E. An English version of a 1592
version of the rhyme and of Sun
Zi's text is in: Li Wenlin and Yuan Xiangdong; The Chinese
Remainder Theorem, pp. 79‑110 of
Ancient China's Technology and Science, op. cit. in 6.AN. Li & Yuan also note that such problems
arose and were undoubtedly solved in calendrical calculations in the 3C.
Li & Du, pp. 93‑94, discuss several Chinese
versions over the next centuries, including the rhyme of Li & Yuan. Li & Du's p. 94 mentions the calculation
of the Da Ming Calendar by Zu Chongzhi in 462 which probably used 11
simultaneous congruences, but the method has not survived. (Li & Yuan say it was 10
congruences.)
Aryabhata. 499. Chap. II, v. 32‑33, pp. 74-84. (Clark edition: pp. 42‑50.) Rule for the residual form, i.e. the Chinese
Remainder Theorem. The text is rather
brief, but Shukla makes it clear that it is giving the Euclidean algorithm for
the problem with two residues. Shukla
does an example with three residues and gives the general solution, though the
text stops with one solution. Shukla
gives an alternative translation which would apply to the non-residual form of
the problem and notes that later writers recognised the relation between the
two forms. (See Libbrecht 229, Datta & Singh II 87‑99 & 131‑133 and
Bag, op. cit. under Bakhshali MS in 7.P.1, pp. 193‑204.)
Brahmagupta. Brahma‑sphuta‑siddhanta. 628.
Chap. XVIII, sect. 1, art. 7. In
Colebrooke, pp. 326‑327. After
some astronomical data, he gives Problem
C as an example.
The
earlier part of Section 1 (pp. 325-326) deals with the general rule and the
later part (pp. 327-338) gives astronomical examples.
Bhaskara I.
Mahā‑Bhāskarīya.
c629. Edited and translated by
Kripa Shankar Shukla. Lucknow Univ.,
1960. Chap 1, v. 41‑52, Sanskrit
pages 7-9; English pages 29‑46.
These deal with the non-residual 'pulveriser' in its astronomical
applications and it seems clearly illustrated.
Illustrative examples are provided by Shukla, from other works of
Bhaskara, or from Chap 8 of stated, but not worked, examples to this work,
e.g. 44789760000 x ‑ 101
= 210389 y (chap. 8, no. 13).
Bhaskara I.
629. Commentary to Aryabhata,
chap. II, v. 32-33. Sanskrit is on pp.
132-171; English version of the examples is on pp. 309-332. Some further applications are discussed on
pp. 191-196, 199-201, 332-334. Bhaskara
I gives 26 examples (and four further examples). I will just give the simpler ones. Shukla, pp. lxxxi-lxxxii, says Bhaskara I was the first to
distinguish the residual and non-residual forms of the problem. He also gives tables of solutions of ax - 1 = by
for values of a, b of astronomical significance -- Shukla gives
these in Appendix II, pp. 335-339.
Shukla asserts that the Indians were the first to find the general
solution for the Chinese Remainder Problem and it was transmitted to China,
c700. However, see Li & Du above.
Ex.
1: n
º 1 (5), 2 (7).
Ex.
2: n
º 5 (12), 7 (31).
Ex.
3: n º 1 (7), 5 (8),
4 (9).
Ex.
4: A-7. First appearance; see beginning of section for list of later
appearances. Answer: 721.
Ex.
5: 8x + 6 º 0
(13).
Ex.
6: 11x - 3 º 0
(23).
Ex.
7: 576x - 86688 º 0
(210389).
All the later
examples are similar, arising from astronomical calculations.
(Bag, loc.
cit., cites Examples 2 and 4 above and says
C occurs, but it is not
here. Datta & Singh may imply that C
occurs here. Cf
Brahmagupta.) (See Libbrecht 231‑232 and
Datta & Singh II 133‑135.)
Bhaskara I. Laghu‑Bhāskarīya. c629.
Edited and translated by Kripa Shankar Shukla. Lucknow Univ., 1963. Chap
8, v. 17-18, Sanskrit pages 26-27; English pages 99‑102. This is a simplified version of his earlier
Mahā‑Bhāskarīya, so it does not deal with the pulveriser,
but these verses give some astronomical problems with number theoretic
conditions which lead to uses of the pulveriser, e.g. 36641x - 24 º 0 (394479375).
Mahavira. 850. Chap. VI, v. 121‑129, pp. 122‑123. 6 simple and 3 more complex examples, e.g.
the following.
Ibn al‑Haitam.
c1000. ??NYS. Problem
A‑7 (= Bhaskara I). (English in Libbrecht, p. 234.) (See also:
E. Wiedemann; Notiz über ein vom
Ibn al Haitam gelöstes
arithmetisches Problem; Sitzungsber. der phys. Soz. in Erlangen 24 (1892)
83. = Aufsätze zur Arabischen
Wissenschaftsgeschichte; Olms, Hildesheim, 1970, vol. 2, p. 756.) Answer: 721.
Bhaskara II.
Bijaganita. 1150. Chap. VI, v. 160 & 162. In Colebrooke, pp. 235‑237 & 238‑239.
Fibonacci. 1202.
Chhin Chiu‑shao
(= Ch'in Chiu Shao = Qin
Jiushao). Shu Shu Chiu Chang (Mathematical Treatise in Nine Sections). 1247.
Complete analysis. (See
Libbrecht, passim. See also Mikami 65‑69 and
Li & Yuan (op. cit under Sun Zhi).
(n º 32 (83), 70 (110), 30 (135) is given by Mikami 69.)
Yang Hui. Hsü Ku
Chai Ch'i Suan Fa. 1275. Loc. cit. in 7.N, pp. 151‑153,
problems 1‑5.
Giovanni Marliani.
Arte giamata arismeticha. In
codex A. II. 39, Biblioteca Universitaria de Genova. Van Egmond's Catalogue 139 dates it c1417. Described and partly transcribed by Gino
Arrighi; Giuochi aritmetici in un "Abaco" del Quattrocento Il matematico milanese Giovanni
Marliani. Rendiconti dell'Istituto
Lombardo. Classe di Scienze (A) 99
(1965) 252‑258. Prob. IV: D-7.
Pseudo-dell'Abbaco.
c1440.
AR. c1450. Prob. 268, 311, 349. Pp. 120‑121, 138‑139, 153, 181,
228‑229.
Correspondence of Johannes Regiomontanus, 1463?-1465. Op. cit. in 7.P.1.
Benedetto da Firenze.
c1465. Pp. 68‑69. Problems
A‑7 (= Bhaskara I), D-7 (=
Fibonacci). He indicates the general
answers.
Muscarello.
1478. Ff. 69r-69v, p. 180. n º -1 (2, 3, 4, ..., 10) (= Pseudo-dell'Abbaco).
della Francesca.
Trattato. c1480. F. 122v (261). A-7. Answer: 721.
English in Jayawardene.
Chuquet. 1484.
HB.XI.22. 1488. Pp. 52‑53 (Rath 247). Prob. A‑7 (= Bhaskara I).
Answer: 721. Editor notes that 301 is the smallest
solution.
Pacioli. De
Viribus. c1500. Problems XXII - XXV.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 116, f. 57v. Woman with
basket of eggs -- problem A‑7 (=
Bhaskara I).
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 26, f. 66r. Basket of eggs -- problem
D‑7 (= Fibonacci).
Cardan. Practica
Arithmetice. 1539. Chap. 66, sections 63 & 64, ff. FF.i.v -
FF.ii.r (pp. 154‑155).
Problems A-7 (= Bhaskara I) &
D-7 (= Fibonacci).
Tartaglia. General
Trattato, 1556, art. 146‑150, pp. 257v‑258v; art. 199, p.
264r.
Buteo.
Logistica. 1559. Prob. 70, pp. 279-280. Problem
D-7 (= Fibonacci). Discusses
problem A‑7 (= Bhaskara
I) and Cardan.
Baker. Well Spring
of Sciences. 1562?
Bachet.
Problemes. 1612. Prob. V: Faire encore le même d'une autre façon,
1612: 37-45. Prob. VI, 1624:
84-93; 1884: 34‑37. General solution for 3, 4, 5
used for divination. Labosne
adds case 2, 3, 5, 7 and a general approach. 1612 cites Forcadel, Gemma Frisius,
Tartaglia, etc.
van Etten.
1624. Prob. 51‑52 (46‑47),
pp. 46‑48 (69‑71).
Problems A‑7 (= Bhaskara
I) and
D-7 (= Fibonacci). French ed.
refers to Bachet for more detailed treatment.
Henrion's 1630 Notte to prob. 52, p. 18, says that Bachet has
treated this problem.
Seki Kōwa.
Shūi Shoyaku no Hō. 1683. ??NYS -- described in Smith & Mikami,
pp. 123‑124. Studies ax - by = 1. n º 1 (5),
2 (7) (= Bhaskara I). Then extends to any number of congruences. Then does the system 35 n
º 35 (42), 44 n
º 28 (32) and
45 n º 35 (50).
W. Leybourn.
Pleasure with Profit. 1694. Prob. 17, pp. 40-41. Prob.
C-9. Constructs a table of 2·n! - 1,
n = 2, ..., 9, and says 2·9! - 1
is the least solution. But he
then gives 2519 + 2520k, k = 0, ..., 7 and says these are some of the infinitely many other solutions
Ozanam. 1694. Prob. 23, 1696: 74-77; 1708: 65‑67. Prob. 27, 1725: 188‑198.
Prob. 10, 1778: 195-198;
1803: 192-195; 1814:
167-169; 1840: 86-87. 1696 gives many examples, too numerous to
detail, and some general discussion. The
following is common to all editions:
n º 1 (2, 3, 5), 0 (7). 1725 has problem A-7 (= Bhaskara I). 1778
et seq. has problem D-7 (=
Fibonacci) and then notes that Ozanam
would solve this as 119 (mod 5040) rather than
119 (mod 420) -- but the
1696 or 1725 ed. only have relatively prime moduli.
Dilworth.
Schoolmaster's Assistant.
1743. Part IV: Questions: A
short Collection of pleasant and diverting Questions, p. 168. Problem 2.
Problem A-5 (= Tartaglia).
Simpson.
Algebra. 1745. Section XIII.
Gentlemen's Diary, 1747 -- see Vyse, below. C-20.
Ladies' Diary, 1748 -- see Vyse, below. C-6.
Les Amusemens.
1749. Prob. 171, p. 318. n º 1 (2, 3, 4, 6), 4 (5), 0 (7).
Euler. Algebra. 1770.
II.I.
II.III:
Questions for practice.
Vyse. Tutor's
Guide. 1773? The following are in a supplement in the Key only, but some are
referred to earlier sources.
Dodson. Math.
Repository. 1775.
Bonnycastle.
Algebra. 1782. Pp. 137-140 (c= 1815: pp. 159-162) discuss
the general method and give the following examples and problems.
Carlile.
Collection. 1793. Prob. LXVIII, pp. 39-40. n º ‑1 (2, 3, 4, 5, ..., 9) (= Tartaglia, C-9). He gives the answer:
2519.
D. Adams. Scholar's
Arithmetic. 1801. P. 208, no. 65. A-7. Answer: 721.
Bonnycastle.
Algebra. 10th ed., 1815. Pp. 159-162 is similar to the 1782 ed., but
has the following different problems.
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
Pp. 49-50, no. 76: A quantity of eggs being broken, to find how many
there were, without remembering the number.
Problem A-7 (= Bhaskara I).
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions.
Rational Recreations.
1824. Exer. 11, p. 56. A-7 (= Bhaskara I). Answer: 301.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 145-172 & 259-262, nos.
541-645. He treats the topic of 7.P.1
and 7.P.2 at great and exhausting length, starting with solving ax = by,
then x + y =
c, then ax + y = c, then ax = by + 1
and ax = by + c, also phrased as ax º c (mod b). On pp. 156-158 & 260, nos. 588-597, are
Chinese Remainder problems, but with just two moduli, so I will not describe
them.
The New Sphinx.
c1840. No. 45, pp. 24 &
121. Problem A-7 in verse, with bunch
of walnuts.
Magician's Own Book.
1857. The basket of nuts, pp.
245-246. Problem A-7 (= Bhaskara I), general solution. = Book of 500 Puzzles, 1859, pp. 59-60. = Boy's Own Conjuring Book, 1860, p. 217.
Illustrated Boy's Own Treasury. 1860. Prob. 14, pp. 428
& 431. Problem D-5, one solution (= Baker).
Todhunter. Algebra,
5th ed. 1870. Examples XLVI, nos. 22-24, pp. 394 & 596.
Wehman. New Book of
200 Puzzles. 1908. The basket of nuts, p. 55. A-7 (= Bhaskara I) = Magician's Own Book, except that a line of the statement
has been dropped by the typesetter, making the problem unintelligible.
Ripley's Puzzles and Games.
1966.
7.P.3. ARCHIMEDES' CATTLE PROBLEM
Archimedes? Letter to Eratosthenes, c‑250.
Greek
text, with commentaries, first published by Gotthold Ephraim Lessing in
Beiträge zur Geschichte und Literatur 1 (1773) 421‑???.
Archimedes. Opera Omnia. Ed. by. J. L. Heiberg. 2nd ed., vol. II, Teubner, 1913. Problema Bovinum, pp. 527‑534. Heiberg gives the same classical references
as Dijksterhuis (below), cites Lessing as the first editor of the problem (from
Gud. Graec. 77, f. 415v), gives the later commentators and editors and says the
problem also appears in Cod. Paris Gr. 2448, f. 57. He then gives the Greek and a Latin translation.
The first edition of Heiberg's edition is
the basis of:
T. L. Heath. The Works of Archimedes (CUP, 1897) + The Method of
Archimedes (CUP, 1912); reprinted in
one volume by Dover, 1953. The Cattle‑Problem,
pp. 319‑326, discusses the problem and the attempts at solving it.
Dijksterhuis, p. 43, says Lessing's
article occurs in the Zweiter Beitrag (1773), 2nd ed., Braunschweig, 1773
(??). = Sämtliche Schriften; ed.
by K. Lachmann, vol. IX, p. 285+; 3rd ed., much corrected by F. Muncker,
Leipzig, 1897, vol. XIII (or 12??), pp. 99-115.
English verse version in H. Dorrie; 100
Great Problems of Elementary Mathematics; Dover, 1965, pp. 5‑6. Dorrie also cites the 19C historians on the
question of authenticity.
Greek
and English in SIHGM II 202-207. Thomas
notes that the epigram is unlikely to have been actually written by
Archimedes. SIHGM I 16-17, is a
Scholium to Plato's Charmides 165 E which states "logistic ... treats on
the one hand the problem called by Archimedes the cattle-problem" and
Thomas gives some of the standard references in a note.
English
translation in D. H. Fowler; Archimedes Cattle Problem and the Pocket
Calculating Machine; Preprint, 1980, plus addenda. (Based on SIHGM.)
Dijksterhuis.
Archimedes. Op. cit. in
6.S.1. P. 398 gives some classical
references to the problem: a scholium
to Plato's Charmides; Heron; two references in Cicero -- all ??NYS.
T. L. Heath.
Diophantos of Alexandria. Op.
cit. as Diophantos. 1910. Pp. 121‑124 discusses the problem.
J. F. Wurm. Review
of J. G. Hermann's pamphlet: De
archimedis problemate bovino; Leipzig, 1828.
In: Jahn's Jahrbücher für
Philologie und Pädagogik 14 (1830) 195-??.
??NYS -- cited by Archibald.
Solves the easier interpretation, getting 5,916,837,175,686 cattle
in all.
B. Krumbiegel. Das
problema bovinum des Archimedes.
Zeitschrift für Mathematik und Physik -- hist.-litterar. Abt. 25 (1880)
121-136. ??NYS -- cited by
Archibald. Survey of earlier historical
work.
A. Amthor. ???. Ibid., pp. 153-171. ??NYS -- cited by Archibald. Survey of the mathematics.
H. E. Licks. Op.
cit. in 5.A. 1917. Art. 54: The cattle problem of Archimedes,
pp. 33‑39. discusses the work of Amthor & A. H. Bell (AMM
(May 1895), ??NYS), who started the calculation of the answers. This is an abridgement of an article by
Mansfield Merriman in Popular Science Monthly (Nov 1905) ??NYS, o/o, but
omitting the author's name, which leads us to believe that H. E. Licks is a
pseudonym of Mansfield Merriman.
[R. C. Archibald.]
Topics for club programs -- 14: The cattle problem of Archimedes. AMM 25 (1918) 411-414. He gives a detailed history, but some
details vary from Dijksterhuis. He
cites Krumbiegel and Amthor as the basic works, Wurm as the first solver with
the simpler interpretation and numerous other works.
H. C. Williams, R. A. German & C. R. Zarnke. Solution of the cattle problem of
Archimedes. Math. Comp. 19 (1965)
671. Plus comments by D. Shanks, ibid.,
686‑687. Describes full solution,
but doesn't print it.
Harry L. Nelson. A
solution to Archimedes' cattle problem
& Note. JRM 13 (1980‑81) 162‑176 &
14 (1981‑82) 126. First
printed version of the solution --
206,545 digits. The note clarifies the formulation of the
problem.
Bakhshali MS. c7C.
Mahavira. 850. Chap. VI, v. 162‑166, pp. 137‑138. Gupta (op. cit. under Bakhshali MS) says the
rule given is similar to the Bakhshali rule.
Sridhara. c900. Gupta (op. cit. under Bakhshali MS) says
that Sridhara gives the same rule as the Bakhshali MS, but allowing n
people. This rule is quoted in
the Kriyākramakari, a 1534 commentary on the Lilavati and Gupta quotes and
translates it. The Kriyākramakari
also quotes Mahavira, without attribution.
Gupta cannot locate this rule in Sridhara's extant works.
Bhaskara II. Lilavati. 1150.
Chap. IV, sect. IV, v. 100. In
Colebrooke, p. 45. Also in his
Bijaganita, chap. IV, v. 111, pp. 195‑196. 8, 10, 100, 5 giving one
each to others and all are equal.
7.P.5 . SELLING DIFFERENT AMOUNTS 'AT SAME PRICES' YIELDING THE SAME
NOTATION: (a, b, c, ...) means the
sellers initially have a, b, c,
.... They all sell certain amounts at
one price, then sell their remnants at a second price so that each receives the
same amount. Western versions give a, b, c, ... and sometimes the amount each receives. The Indian versions give the proportion a : b : c : ... (by
stating each person's capital, but not the cost price of the items; they invest
their capital in the items and then sell them) and the larger price for selling
the remnant of the items. Further, the
price for selling the first part of the items is the reciprocal of an
integer. (However the remnant price is
sometimes a fraction.) In both
versions, the problem is indeterminate, with a
3 parameter solution set, but
scaling or similarity or fixing the yield reduces this to 2.
There are also non‑negativity and integrality conditions. The Indian version has infinitely many
solutions, while the Western version gives a finite number of solutions. I have recently found a relatively simple
way to generate and count the solutions in the Western version, which is
basically a generalization of Ozanam's example -- see my paper below. The article by Glaisher discusses many of
these problems. As in 7.P.1, (a, b)
solutions means a non-negative solutions of which b
are positive solutions.
Versions where the earnings are
different: Ghaligai.
See Tropfke 651.
Index of western
versions.
(10,
20) Abraham
(10,
30) Fibonacci
(12,
32) Fibonacci
(12,
33) Fibonacci
(18,
40) Labosne
( 7, 18, 29) McKay
( 8, 17, 26) Blasius
(10,
12, 15) Labosne
(10,
16, 22) Amusement
(10,
20, 30) Pacioli
(10,
25, 30) Ozanam
(10,
30, 50) Munich
14684, Folkerts, Marliani, Provençale Arithmétique, Pseudo‑dell'Abbaco,
Chuquet, HB.XI.22?, Widman, Demaundes Joyous, Tagliente, Ghaligai, Tartaglia,
Jackson, Badcock, Rational Recreations, Boy's Own Book, Rowley, Hoffmann
(11,
33, 55) Tartaglia
(16,
48, 80) Tartaglia
(18,
40, 50) Labosne
(19,
25, 27) Williams
& Savage
(20,
25, 32) Bachet
(20,
30, 40) Bachet,
van Etten, Hunt
(20,
40, 60) Tagliente
(27,
29, 33) Leske,
Mittenzwey, Hoffmann, Pearson
(30,
56, 82) Widman
(31,
32, 37) Labosne
(60,
63, 66) Bath
(17,
68, 119, 170) Widman
(20,
30, 40, 50, 60) Dudeney
(305,
454, 603, 752, 901) Widman
(20,
40, ..., 140) Glaisher,
Gould
(10,
20, ..., 90) Tartaglia
Mahavira. 850.
Chap. VI, v. 102‑110, pp. 113‑116. He gives a rule which gives one special solution of Sridhara's
set of solutions.
Sridhara. c900. V. 60‑62, ex. 76‑77, pp. 44‑49
& 94. The verses are brief rules,
which are expanded by editorial algebra, giving a one parameter family of solutions.
Bhaskara II.
Bijaganita. 1150. Chap. 6, v. 170. In Colebrooke, pp. 242‑244. Capitals
6, 8, 100; remnant
price 5. Solution given is 3294,
4392, 54900, which is one solution from
Sridhara's set of solutions, but not by the same method as Mahavira. The method is not clearly described. Bhaskara says: "Example instanced by ancient authors .... This, which is instanced by ancient writers
as an example of a solution resting on unconfirmed ground, has been by some
means reduced to equation; and such a supposition introduced, as has brought
out a result in an unrestricted case as in a restricted one. In the like suppositions, when the
operation, owing to restriction, disappoints; the answer must by the
intelligent be elicited by the exercise of ingenuity."
Fibonacci.
1202. Pp. 298‑302 (S:
421-423): De duobus hominibus, qui habuerunt poma [On two men who had
apples]. He clearly states that there
are two forums where the same prices are different.
Abraham. Liber
Augmenti et Diminutionis. Early
14C. ??NYR -- cited by Tropfke
651. (10, 20). This has
(55, 36) solutions.
Munich 14684.
14C. Prob. XIII, pp. 79‑80. (10, 30, 50). Gives the solution with prices
1/7 and 3. There are
(25, 16) solutions.
Folkerts.
Aufgabensammlungen. 13-15C. 21 sources for (10, 30, 50). Says the
problem goes back to Fibonacci, but Fibonacci only has examples with two
vendors.
Giovanni Marliani.
Arte giamata arismeticha. In
codex A. II. 39, Biblioteca Universitaria de Genova. Van Egmond's Catalogue 139 dates it c1417. Described and partly transcribed by Gino
Arrighi; Giuochi aritmetici in un "Abaco" del Quattrocento Il matematico milanese Giovanni Marliani. Rendiconti dell'Istituto Lombardo. Classe di Scienze (A) 99 (1965) 252‑258. Prob. V: (10, 30, 50).
Provençale Arithmétique.
c1430. Op. cit. in 7.E. F. 116v, pp. 62-63. (10, 30, 50). Gives the solution with prices
3 and 1/7 and then gives a
solution with three prices!
Pseudo-dell'Abbaco.
c1440. Prob. 101, pp. 85‑87. (10, 30, 50). Gives the solution with prices
3 and 1/7.
Chuquet. 1484. English in FHM 227-228. Prob. 145.
(10, 30, 50). Same two solutions
as in Provençale Arithmétique. FHM says
it appears in one of Dudeney's books, where he expresses "grave
dissatisfaction with the answer".
HB.XI.22. 1488. P. 54 (Rath 247). Rath doesn't give the numbers and says it is similar to Munich
14684 and p. 73v of Cod. Vindob. 3029.
Glaisher dates Cod. Vindob. 3029 as c1480.
Johann Widman. Op.
cit. in 7.G.1. 1489. F. 134v+. ??NYS -- discussed by Glaisher, pp. 1‑18. (10, 30, 50) -- one solution with prices
3 and 1/7. He then generalises
this example to construct single solutions for other examples: (30, 56, 82), (17, 68, 119, 170), (305, 454, 603, 752, 901),
which have (225, 196), (45, 35),
(11552, 11400) solutions
respectively. Glaisher then describes
several examples that Widman might have constructed.
Pacioli. De
Viribus. c1500. Ff. 119r - 119v.
Anon. Demandes
joyeuses en manière de quodlibets. End
of 15C. ??NYS. Selected and translated as: The Demaundes
Joyous. Wynken de Worde, London,
1511. [The French had 87 demandes, but
the English has 54. This is the oldest
riddle collection printed in England, surviving in a single example in
Cambridge Univ. Library. Often
attributed to de Worde. Santi 9 uses
Yoyous and Wynkyn and list de Worde as author.] Facsimile with transcription and commentary by John Wardroper,
Gordon Fraser Gallery, London, 1971, reprinted 1976. Prob. 50, pp. 6 of the facsimile, 26-27 of the
transcription. (10, 30, 50) apples.
One solution with prices 3 and
1/7.
Blasius. 1513. F. F.iii.r: Decimaquarta regula. Selling eggs -- (8, 17, 26). There
are (16, 9) solutions. He give one
with prices 2 and 1/5 and each sold as many batches of 5 as
possible. Discussed by Glaisher.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 115, ff. 57r-57v. Women selling
eggs -- (10, 30, 50). One solution with prices 1/7
and 3. Glaisher, below and in op. cit. in 7.G.1, cites Hieronymus Tagliente and says the 1515 &
1527 editions give
(10, 30, 50) with
solution at prices 1/7 &
3, and the 1525 ed. has (20, 40, 60) with solution at prices
3 and 1/7. The latter has (100, 81)
solutions.
Ghaligai. Practica
D'Arithmetica. 1521.
Tartaglia. General
Trattato, 1556, art. 136‑139, pp. 256r‑256v.
Bachet.
Problemes. 1612. Prob. XXI, 1612: 106-115. Prob. XXIV, 1624: 178-186; 1884: 122‑126. He gives a new "general and infallible
rule", which is fairly general -- Glaisher says it produces a selection of
the solutions. He applies the idea to (20, 30, 40), exhibiting
4 solutions. It has
(100, 81) solutions. Cf. van Etten. 1612 also does (20, 25,
32).
In
the 5th ed., the general material is dropped and replaced by some vague
algebra. Labosne gives two solutions
for (18, 40), but one of them uses fractions.
It has (171, 36) solutions.
He then considers (18, 40,
50) and gives one fractional solution
-- there are (3, 1) solutions.
He then makes some discussion of
(10, 12, 15) (which has (7, 4)
solutions) and (31, 32, 37) (which has
(70, 60) solutions).
van Etten.
1624. Prob. 69 (62), pp. 64‑65
(90‑91). (20, 30, 40). Gives one solution with prices 3
and 1. There are (100, 81) solutions.
Cf. Bachet. Henrion's 1630
Notte, p. 22, states that Bachet found many other solutions and gives a
solution with prices
2 & 7.
Hunt. 1651. Pp 282-283: Of three women that sold
apples. (20, 30, 40). Gives one solution with prices 1
and 3.
Ozanam. 1694. Prob. 24, 1696: 77-80; 1708: 68‑70. Prob. 28, 1725: 201‑210.
Prob. 12, 1778: 199-204;
1803: 196-201; 1814:
170-174; 1840: 88-90. (10, 25, 30). Glaisher describes the material in the 1696 ed. and says that
Ozanam first considers the same general form that Bachet considered and then
applies it to the example. Glaisher
indicates that Ozanam's and Bachet's methods are essentially the same, but
Ozanam certainly gets all solutions, while I am not sure that Bachet can do
so. 1696 gives two solutions at
prices 7 and 2 and at prices 6 and 1.
1725 et seq. gives a general method and finds all 10
solutions of the specific problem.
(Glaisher notes that the Remarques on pp. 203‑210 are new to the
1723 ed. They give an algebraic form of
the solution.) 1725 refers to the
second part of Arithmétique Universelle, p. 456 (more specifically identified
as by M. de Lagny [1660-1734] in 1778 et seq.), where 6 solutions are
found. 1725 says there are 10
solutions and 1778 says de Lagny is mistaken -- but in fact, there
are (10, 6) solutions and de Lagny probably meant just the positive
ones. 1778 drops the preliminary
general general form.
Amusement for Winter Evenings. A New and Improved Hocus Pocus; or Art of Legerdemain: Explaining
in a Clear and Comprehensive Manner Those Apparently Wonderful and Surprising
Tricks That are performed by Slight of Hand and Manual Dexterity: Including
Several Curious Philosophical Experiments.
M. C. Springsguth, London, nd [c1800 -- HPL], 36pp. Pp. 22-23: Of three sisters. (10, 16, 22) sold at 7 a penny and then a penny apiece, i.e. prices 1/7
and 1, each earning 4.
Bestelmeier.
1801. Item 718: Das Eyerverkauf. Three women sell different numbers of eggs
and make the same. Further details not
given.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions. No. 14, pp. 17 & 74. (10, 30, 50). Gives the solution with prices
3 and 1/7.
John Badcock.
Domestic Amusements, or Philosophical Recreations. Op. cit. in 6.BH. [1823]. P. 155, no. 192:
Trick in reasoning. (10, 30, 50) -- gives the solution with prices 3
and 1/7.
Rational Recreations.
1824. Exer. 21, p. 98. (10, 30, 50) eggs.
Boy's Own Book. 1843
(Paris): 341. "Three country-women
and eggs." (10, 30, 50) -- gives the solution with prices 3
and 1/7. = Boy's Treasury, 1844, p. 299. = de Savigny, 1846, p. 289: Les trois
paysannes et les œufs.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-21, pp. 254 & 395-396: Die
Blumenmädchen. (27, 29, 33). This has
(117, 100) solutions -- she
gives one with prices ⅓ and
1, each earning 13.
Hugh Rowley. More
Puniana; or, Thoughts Wise and Other-Why's.
Chatto & Windus, London, 1875.
P. 179. Man with three daughters
and lots of apples -- (10, 30,
50). Solution with prices 3
and 1/7.
Mittenzwey.
1880. Prob. 122, pp. 25 &
76; 1895?: 140, pp. 29 & 79; 1917: 140, pp. 27 & 77. Three sisters selling bunches of
violets. (27, 29, 33). This has
(117, 100) solutions -- he gives
one with prices ⅓ and
1, each earning 13.
= Leske.
Hoffmann. 1893. Chap. IV, pp. 160 & 215-216
= Hoffmann-Hordern, p. 140.
Dudeney?? Breakfast
Table Problems No. 328: "How were
the oranges sold". Daily Mail
(27 & 28 Jan 1905) both
p. 7. (20, 30, 40, 50, 60). Gives the solution with prices 1 and 1/11. There are (45, 36)
solutions.
Pearson. 1907. Part II, no. 37, pp. 121 & 199. (27, 29, 33) c= Leske.
J. W. L. Glaisher.
On certain puzzle-questions occurring in early arithmetical writings and
the general partition problems with which they are connected. Messenger of Mathematics 53 (1923-24)
1-131. Discusses the versions in
Blasius, Widman, Tagliente and attempts
to explain the methods used.
Unfortunately he is rather prolix and I often get lost in the many
examples and special cases, but he seems to have general solutions. On p. 12, he mentions that Tagliente's
problem could be extended to (20, 40,
..., 140) -- cf. Gould below. On p. 77, he says more results will appear
in a later paper -- check index of Messenger??
A. A. Krishnaswami Ayyangar. A classical Indian puzzle-problem. J. Indian Math. Soc. 15 (1923-24) 214-223. Responding to Glaisher, he analyses the
Indian version, obtaining a simple complete solution with two parameters having
an infinite range and a third parameter being bounded. I found this a bit confusing since he
sometimes uses price to mean the number of items per unit cost. He says many of the solutions are not in
Glaisher's system given on p. 19, but I can't tell if Glaisher intends this to
be a complete solution.
Rupert T. Gould. The
Stargazer Talks. Geoffrey Bles, London,
1944. A Few Puzzles -- write up of a
BBC talk on 10 Jan 1939, pp. 106-113.
Seven applewomen with
20, 40, 60, 80, 100, 120, 140. One solution with prices 3
and 1/7 -- cf. Glaisher. There are (27, 21) solutions.
McKay. At Home
Tonight. 1940. Prob. 16: Extraordinary sales, pp. 65 &
81. (7, 18, 29) eggs.
This has (12, 6) solutions.
He asks for a solution where all make
10d. He gives one solution with
prices 1/4 and 3 and selling as many eggs in batches of 4 as
possible.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940. No. 75: A falling
market, pp. 43 & 128.
Cauliflowers (19, 25, 27) with each making 85d, both prices being
integral and each sells some at the lower price, not to be less than 2d.
Actually there is only one solution with integral prices and each
making 85d.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
No. 45: Potatoes for sale, pp. 19 & 47. (60, 63, 66). This
has (900, 841) solutions.
He wants solutions where all make
9/6 = 114d. There are two
solutions with integral prices, but he gives the solution with prices 2d
and 12/7 d, i.e.
7 for 1s, but he doesn't require
maximum numbers of batches of 7 to be sold.
David Singmaster.
Some diophantine recreations.
In: Papers Presented to Martin
Gardner on the occasion of the opening of the exhibition: Puzzles: Beyond the
Borders of the Mind at the Atlanta International Museum of Art and Design; ed.
by Scott Kim, 16 Jan 1993, pp. 343-356
AND The Mathemagician and Pied
Piper A Collection in Tribute to Martin
Gardner; ed. by Elwyn R. Berlekamp
& Tom Rodgers;. A. K. Peters,
Natick, Massachusetts, 1999, HB, pp. 219-235.
Sketches some history, gives complete solutions for the Western and
Indian cases (filling a gap in Ayyangar) and finds a new simple formula for the
number of solutions in the Western case.
7.P.6. CONJUNCTION OF PLANETS, ETC.
See Tropfke 642.
Some overtaking problems in 10.A
take place on a circular track and are related to or even identical to these
problems. In particular, if two persons
start around an island of circumference
D, from the same point and in the same direction at rates a, b,
this is the same as O-(a, b;
D) of Section 10.A. This is easily adapted to dealing with
different starting points and going in opposite directions (which gives a
meeting problem). Clock problems, 10.R,
are also related to these.
Sun Zi. Sun Zi Suan
Ching. Op. cit. in 7.P.2. 4C.
??NYS. Sisters come home every 5, 4, 3 days, when do they all come together?. (Mikami 33 gives English.)
Zhang Qiujian. Zhang
Qiujian Suan Jing. Op. cit. in
7.E. 468, ??NYS -- translated on p. 139 of: Shen Kangsheng; Mutual‑subtraction algorithm and its
applications in ancient China; HM 15 (1988) 135‑147. Circular route of length 325. Three persons start with speeds 150, 120, 90. When do they all meet at the start?
Brahmagupta. Brahma‑sphuta‑siddhanta. 628.
Several of the sections of Chap. XVIII discuss astronomical versions of
this problem, but with complicated values and unclear exposition.
Gherardi. Libro di
ragioni. 1328. P. 47.
Two men start going in a circuit.
One can do it in 4 days, the other in 5½ days. When do they meet again? Same as
O-(1/4, 1/5½), D = 1, in the notation of Section 10.A.
AR. c1450. Prob. 148, pp. 72, 164‑165, 214. Though titled 'De planetis' and described as
conjunction by Vogel, this is really just an overtaking problem -- see 10.A.
H&S 74‑75 says sun and moon problems are in van
der Hoecke and Trenchant (1556).
Cardan. Practica
Arithmetice. 1539. Chap. 66, sections 20-24, ff. CC.vii.v
- DD.i.r (pp. 142‑143).
Several versions concerning conjunctions of planets, including
irrational ratios and three planets.
Examples with periods: Ö7, Ö5; Ö18, Ö30; Ö8, Ö20; 1000, 999; and Ö5, Ö4, Ö3. In section 23, he gives periods of Saturn
and Jupiter as 30 &
12 years and periods of Jupiter
and Mars as 144 &
23 months. (H&S 75 gives English and some of the
Latin.)
Cardan. Opus Novum
de Proportionibus Numerorum.
Henricpetrina, Basil, 1570, ??NYS.
= Opera Omnia, vol. IV, pp. 482-486. General discussion and examples, e.g. with periods 2, 3, 7.
Wells. 1698.
Vyse. Tutor's
Guide. 1771? Prob. 20, 1793: p. 79; 1799: p. 85 & Key p. 110-111. Island
73 in circumference; three persons set out in the same direction
at speeds 5, 8, 10. When do they all meet again?
Bonnycastle.
Algebra. 1782. P. 86, no. 23. Identical to Vyse.
Pike.
Arithmetic. 1788. P. 353, no. 31. Island 50 in circumference. Three walkers start in the same direction at speeds 7, 8, 9.
When and where do they meet again?
= D. Adams; Scholar's Arithmetic; 1801, p. 208, no. 66.
Hutton. A Course of
Mathematics. 1798? Prob. 37,
1833: 223; 1857: 227. Identical to Vyse.
Kaida Anmuyo.
c1800. Problem given on pp. 139‑140
of Shen Kangsheng, loc. cit. under Zhang Qiujian above. Assume
365¼ degrees in a circle. Five stars are in a line and travel at
speeds of 28 13/16, 19 1/4,
13 5/12, 11 1/7, 2 7/9
degrees per day. When do they
meet at the starting point again?
D. Adams. New
Arithmetic. 1835. P. 244.
Anonymous. A
Treatise on Arithmetic in Theory and Practice: for the use of The Irish National
Schools. 3rd ed., 1850. Op. cit. in 7.H. P. 360, no. 48. Three
walkers start to circle an island of circumference 73, at rates 6, 10, 16.
When do they meet again?
James B. Thomson.
Higher Arithmetic; or the Science and Application of Numbers; .... Designed for Advanced Classes in Schools and
Academies. 120th ed., Ivison, Phinney
& Co, New York, (and nine copublishers), 1862. Prob. 94, p. 308 & 422.
Same as Anon: Treatise.
Daniel W. Fish, ed.
The Progressive Higher Arithmetic, for Schools, Academies, and
Mercantile Colleges. Forming a Complete
Treatise of Arithmetical Science, and its Commercial and Business
Applications. Ivison, Blakeman, Taylor
& Co., NY, nd [but prefaces give: 1860; Improved Edition, 1875]. P. 418, no. 64. Island 120 in circumference. Seven men start walking around it from the same point at
speeds 5, 25/4, 22/3, 33/4, 19/2,
41/4, 45/5 per day. When are they all together again? Answer: after 1440 days.
Mittenzwey.
1880. Prob. 117, pp. 23-24 &
76; 1895?: 135, pp. 27-28 &
79; 1917: 135, pp. 25-26 & 76. Seven colleagues return to a forest inn
every 1, 2, 3, 4, 5, 6, 7 days.
when will they all return again?
Lemon. 1890. The Maltese cross, no. 483, pp. 63 &
115. Walkers complete 6, 9, 12, 15 circuits per hour -- when are they all again at start?
Hoffmann. 1893. Chap. IV.
Clark. Mental
Nuts. 1897, no. 51. When will we three meet again. Three bicycle riders can ride around a one
mile track in 2 1/2, 2 3/5,
3 1/4 minutes. If they all start together, when will they
all meet again at the starting point?
Dudeney. "The
Captain" puzzle corner. The
Captain 3:2 (May 1900) 97 & 179
& 3:4 (Jul 1900)
303. No. 2: The seven money boxes. Boy puts a penny in i-th box on
i-th day, where day 1 is 1
Jan 1900. When he has to put in seven
pennies, he will then open them all up.
When is this and how much will he have?
Answer: 420 days =
24 Feb 1901 and £4 10s 9d.
Depew. Cokesbury
Game Book. 1939. Bicycle racers, p. 221. One can travel around the track in 6
minutes, the other in 9 minutes.
When are they together again?
Men keep money together and divide
it into amounts x1, x2, ... -- usually by
robbing the common fund. They put
fractions aixi into a pool and divide the pool in
proportion b1 : b2
: .... They then have money in the
proportion c1 : c2
: ..., or actual amounts d1, d2, ....
I use A for (a1, a2, a3), etc.
Kurt Vogel. Ein
unbestimmtes Problem al-Karağī in Rechenbüchern des Abendlands. Sudhoffs Archiv 61 (1977) 66-74. Gives the history of this problem,
particularly the transmission to Fibonacci via John of Palermo and the different
methods of solving it -- Fibonacci gives three methods. He mentions all the entries below except
Gherardi and Calandri, which had not been published when he wrote, and Pacioli,
which is sad because Pacioli is not very clear!
Jacques Sesiano, op. cit. under 7.R, 1985, discusses this
problem along with the problems in 7.R.
He calls it "The disloyal partners". He cites Abū Kāmil's Algebra, ff. 100r-101r, ??NYS, and
says it is solved two ways there.
Martin Levey's 1966 edition of the Algebra does not give the problem and
states that the unique Arabic MS ends on f. 67r, and uses a 'better' Hebrew
text. The Arabic MS actually continues
and the third part of the book, ff. 79r-111r, was treated by Schub & Levey
in 1968 & 1970. Sesiano; Les
méthodes d'analyse indéterminée chez Abū Kāmil; Centaurus 21 (1977)
89-105 is scathing about the work of Levey and of Schub & Levey, saying the
Hebrew MS is third-rate, and the translators have made serious mathematical and
philological errors. Sesiano studies
some of Abu Kamil's problems in this article, but unfortunately the problem of
this section is not among them.
al‑Karkhi.
c1010. Sect I, no. 45 &
47; sect III, no. 6; pp. 80‑81 & 90.
Fibonacci.
Flos. c1225. Pp. 234-236: De tribus hominibus pecuniam
comunem habentibus. = Fibonacci, below,
pp. 293-294. = al‑Karkhi's III‑6. Problem II in Picutti, pp. 310‑312.
Cantor;
op. cit. under Fibonacci in 7.P.1; 1863; p. 345, discusses the contest between
Fibonacci and John of Palermo before Frederick II at Pisa in 1225 (or 1226) and
says this problem is the third and last of the contest problems and this is how
I read the text. However Picutti has
this as the second problem. Licks, op.
cit. in 5.A, says it was problem 5 in the contest. Vogel doesn't mention which problem it was.
Leonardo
says he later found three further methods of solution, which are "in libro
uestro, quem de numero composui, patenter inserui". Leonardo is here addressing the Emperor, so
Vogel interprets 'libro uestro' as a book dedicated to the Emperor. Vogel interprets this as referring to the
material in Liber Abbaci, so I have now dated the next entry as 1228 rather
than 1202, although there is no mention of the contest or the Emperor in Liber
Abbaci.
The
solution here is different than below and Vogel calls this Fibonacci's third
method, the shortest and cleverest, and which Fibonacci described as
"exceedingly beautiful"'.
Vogel notes the remarkable hybrid notations: XXX3 for 33;
XXIII 1/1 for 23½;
X 1/1 for 10½ in this 15C MS.
Fibonacci. 1228 --
see above entry. Pp. 293‑297 (S:
415-420). Several versions. He often notes that the values xi can be multiplied through by any value.
Gherardi. Libro di ragioni.
1328. Pp. 54‑56. A = (½, ⅓, ¼), B = (1,1,1), C = (6, 4, 3). Answer is 240 : 93 :
44. He gives 4/29 of these values by
starting with x1 + x2 + x3 =
52.
Lucca 1754.
c1330. Ff. 60v‑61r, pp.
138‑139. (= al‑Karkhi III‑6.) Gives some explanation, but Vogel says only
the beginning makes sense.
Columbia Algorism.
c1350. Prob. 5, pp. 34‑35. (= al‑Karkhi III‑6.) Gives only the answer with no
explanation. Vogel's Introduction, p.
22, sketches the history.
Pacioli. Summa. 1494.
Gives a number of versions.
Calandri, Raccolta.
c1495. Same as Fibonacci, pp.
296‑297. Calandri simply says it
is "insolubile".
Tonstall. De Arte
Supputandi. 1522. Pp. 244-245. Same as Pacioli, prob. 11.
Cardan. Practica
Arithmetice. 1539. Chap. 66.
Recorde. Second
Part. 1552. Pp. 330-335: A question of partners, the ninth example. A = (¾, ⅓) or
(⅓, ¾), the person with
the larger xi to give back ¾. B = (1, 1), D = (180, 120). Solution: (1680, 1620)/11. The second person took the more money.
Buteo.
Logistica. 1559.
Vogel says that Clavius; Epitome Arithmeticae; Rome, 1595,
pp. 249-252, gives a simple example with two persons and that then the problem
vanishes from the literature.
7.Q. BLIND ABBESS AND HER NUNS -- REARRANGEMENT ALONG SIDES OF A 3 x 3 SQUARE TO CONSERVE SIDE TOTALS
This is a kind of magic figure,
except that here we generally have repeated values.
There are three trick versions of
6.AO which might be classified here or in 7.Q.2.
(12, 4, 5) -- Trick version of a hollow 3 x 3 square with doubled
corners, as in 7.Q:
Family Friend (1858),
Secret Out,
Illustrated Boy's Own Treasury.
Van Etten and Mittenzwey are the
only inverse examples, where the total number remains fixed but the number on
each side changes.
Shihâbaddîn Abû’l‑‘Abbâs Ahmad [the h
should have an underdot] ibn Yahya [the
h should have an underdot] ibn
Abî Hajala [the H should have an underdot] at‑Tilimsâni
alH‑anbalî [the H should have an underdot]. Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shaţranj
[NOTE: ţ denotes a t with an underdot] (Book of the examples of
warfare in the game of chess). Copied
by Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî in 1446. This is the second of Dr. Lee's MSS,
described in 5.F.1, denoted Man. by Murray.
Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211,
and by Murray 207‑219.
Murray
280 says No. 46‑49 give the problems of arranging 32, 36, 40, 44 men along the walls and corners so the total along each edge
is 12.
Pacioli. De
Viribus. c1500.
Hunt. 1631
(1651). Pp. 264-266 (256-258). General and guards. 24
guards become 20 then
28.
van Etten. English
ed., 1653, prob. 72: Of the game of square formes, pp. 124‑125. 24
men on sides of a fort, becoming
28 and 20. Discusses case
of 12
men making 3, 4
or 5 on a side.
Anon. Schau‑Platz
der Betrieger: Entworfen in vielen List‑ und Lustigen Welt‑Händeln. Hamburg & Frankfurt, 1687, pp. 543‑545. ??NYS
(A&N, p. 5.)
Ozanam. 1694. Prob. 1, 1696: 1-2; 1708: 1‑2; 1725: 1-3. Prob. 20,
1778: 172-174; 1803: 172-174; 1814: 151-153. Prob. 19, 1840: 77‑78.
Blind abbess and 24 nuns with
9 on a side. 1696 gives three arrangements with 24, 28, 20
on a side. 1725 adds another
arrangement with 32. 1778 says Ozanam has presented this in a
rather indecent manner to excite the curiosity of his readers and adds arrangements
with 36 and 18 on a side.
The last has 5 and
4 in the corners and none in the
side cells, but can be done in other ways.
1803 drops the 'indecent' reference.
Dilworth.
Schoolmaster's Assistant.
1743. Part IV: Questions: A
short Collection of pleasant and diverting Questions, p. 168. Problem 1.
General stationing guards around a castle, wanting 18
on a side, starting with 48 men and changing to 56,
then 40.
Les Amusemens.
1749. Prob. 11, p. 131: Les
rangs de Neuf. Wine merchant with 32
bottles, 9 on a side, reduced to 28, 24, 20.
Catel.
Kunst-Cabinet. Vol. 2,
1793. Die Nonnenlist (The nuns'
strategem), pp. 15-16 & fig. 251 on plate XII. The diagram shows the eight outside cells
with 5
spots in the form of a 5 on a die and one spot in the centre. However, the text says there are 25
cones or pieces and one must read the instructions to learn the game. The number of pieces seems peculiar and I'm
not entirely sure this is our problem, despite its name.
Bestelmeier.
1801. Item 191: Die
Nonnenlist. Picture is an obscure copy
of Catel. Text is copied from part of
Catel, but says there are 15 pieces!
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
Pp. 10-11, no. 20: The blind abbess and her nuns. 9 on
a side, starts with 24 and changes to 28, then 20.
Jackson. Rational
Amusement. 1821. Arithmetical Puzzles.
John Badcock.
Domestic Amusements, or Philosophical Recreations. Op. cit. in 6.BH. [1823]. Pp. 156-157, no.
194: Dishonest contrivance. 32 sheep, with
12 on a side, reduced to 28.
Rational Recreations.
1824. Exer. 18, pp. 94-95: The
convent. 24 increased to 28 then
reduced to 20.
Manuel des Sorciers.
1825. Pp. 75-78, art. 38. ??NX
Blind abbess. Gets totals
of 32,
28, 24, 20.
Endless Amusement II.
1826?
The Boy's Own Book.
The wine merchant and his clerk.
1828: 412; 1828‑2: 418; 1829 (US): 211; 1855: 565; 1868: 669. 32 bottles, reducing to 20.
The Riddler.
1835. The wine merchant and his
clerk, pp. 4-5. Identical to Boy's Own
Book.
Crambrook.
1843. P. 10, no. 23: The Blind
Abbess and her Nuns, a laughable trick.
Magician's Own Book.
1857.
Landells. Boy's Own
Toy-Maker. 1858. Pp. 149-150. c= Magician's Own Book, prob. 24.
The Sociable.
1858. Prob. 21: The blind abbot
and the monks, pp. 292-294 & 309.
24 monks, 9 on
a side, changed to 20, 28, 32, 36,
18. = Book of 500 Puzzles,
prob. 21.
Book of 500 Puzzles.
1859.
Boy's Own Conjuring book.
1860.
Illustrated Boy's Own Treasury. 1860. Prob. 29, pp. 429
& 434. Very similar to Magician's
Own Book, prob. 24.
Vinot. 1860. Art. LVIII: Les étrennes du Commissaire, pp.
75-76. 140 bottles of wine
arranged 1, 34, 1 on each side. Clerk steals four bottles sixteen times, reducing to 17, 2, 17.
The Secret Out (UK).
c1860. Both of the following are
presented with cards.
Leske. Illustriertes
Spielbuch für Mädchen. 1864?
Bachet-Labosne.
Problemes. 3rd ed., 1874. Supp. prob. VI, 1884: 189-191. 60
bottles diminished to 44.
Kamp. Op. cit. in
5.B. 1877. No. 16, pp. 325‑326.
Nine wine bottles on each side on the square. Servant steals some.
Mittenzwey.
1880.
Cassell's.
1881. Pp. 98‑99: The
twenty‑four monks. = Manson,
1911, pp. 249-250.
Handy Book for Boys and Girls. Op. cit. in 6.F.3.
1892. Pp. 38-39: The counter
puzzle. "In an old book published
over half a century ago, I came across this puzzle ...." Rearranges
24 as 20, 28, 32.
Somerville Gibney.
So simple! V. -- A batch of
match tricks. The Boy's Own Paper 20
(No. 988) (18 Dec 1897) 188-189.
24 changed to 25, 20, 28, 32, 30.
Dudeney. Problem 70:
The well and the eight villas -- No. 70: The eight villas. Tit‑Bits 33 (5 Mar 1898) 432 &
34 (2 Apr 1898) 8. How many ways
can numbers be placed in the 8 cells to make 9 along each side? Answer is
2035. He gives a general
formula.
H. D. Northrop.
Popular Pastimes. 1901. No. 3: The blind abbot and the monks, pp.
66-67 & 72. = The Sociable.
Dudeney. The monk's
puzzle. London Mag. 9 (No. 49) (Aug
1902) 89‑91 & 9 (No. 50) (Sep 1902) 219. (= CP, prob. 17, pp. 39‑40 & 172‑173.) How many ways can numbers be placed in
the 8
cells to make 10 along each side?
Benson. 1904. The dishonest servant puzzle, p. 228. 28
bottles, 9 on each edge, reduced to 24,
then 20.
Wehman. New Book of
200 Puzzles. 1908.
M. Adams. Indoor
Games. 1912. The cook and the jam, pp. 353-354. 36 jars of jam.
Blyth. Match-Stick
Magic. 1921. Escaping from Germany, pp. 75-76. 32 with 9 on
a side -- arranged 1, 7, 1 -- reduced to 28, 25 and
24.
Will Blyth. Money
Magic. C. Arthur Pearson, London,
1926. The stolen tarts, pp. 91-95. 9 on a side, start with 32,
reduce to 28, 24, 20.
Rohrbough. Brain
Resters and Testers. c1935. Ba Gwa, p. 18 (= pp. 18‑19 of
1940s?). 3 x 3 frame.
Two player game. Start with 7 in
each corner and 1 in each edge. A player places a extra counter in the frame and the other tries
to rearrange to preserve 15 in each outside row. It says you can get 56
men on the board. [I can get 60
if the corners can be empty.]
Jeffrey J. F. Robinson.
Musings on a problem. MTg 37
(1966) 23‑24. A farmer has 41
cows and wants to see 15 along each side of his house which is in the
centre of a 3 x 3 array.
How many solutions are there if only
1, 2, ..., 8 different values
can be used? He finds all the solutions
in some cases.
Ripley's Puzzles and Games.
1966. Pp. 64-65, item 1. 16
pigs in 8 pens with
6 along each side. Add four pigs.
J. A. Dixon & Class 3T.
Number squares. MTg 57 (1971) 38‑40. Use the digits 1 ‑ 8 so the four
side totals are the same. They find
that the sum can only be: 12, 13, 14,
15, with 1, 2, 2, 1
solutions. They then use 8
digits from 1 ‑ 9 and find
35 solutions.
7.Q.1. REARRANGEMENT ON A CROSS
The counts from the base to the top
and to the end of each arm remain constant though some (usually 2)
of the pearls or diamonds have been removed. Trick versions with doubling up are in 7.Q.2.
Versions with a T
or Y: Secret Out; M. Adams;
Pacioli. De
Viribus. c1500. Ff. 116r - 117v. Cap. LXX. D(e). un prete ch' in pegno la borscia del corporale
con la croci de p(er)le al Giudeo (Of a priest who pledges to a Jew the burse
of the corporale with a cross of pearls).
15 with three on each arm, one
counts to nine from the base to each arm end.
This is reduced to 13. Asks how one can add one pearl and produce a
count of ten -- answer is to put it at the base.
Thomas Hyde.
Historia Nerdiludii, hoc est dicere, Trunculorum; .... (= Vol. 2 of De Ludis Orientalibus, see 7.B
for vol. 1.) From the Sheldonian
Theatre (i.e. OUP), Oxford, 1694. De
Ludo dicto Magister & Servus, pp. 234-236.
Cross with 28 thalers with 16 from the base to the
end of each arm. He points out that the
picture has an error causing the count to the ends of the side arms to be 17.
Discusses general solution. Says
it is known to the Arabs of the Holy Land.
Les Amusemens.
1749. P. xxvi. 13
markers reduced to 11.
Manuel des Sorciers.
1825. Pp. 136-138, art. 16. ??NX
Cross of coins reduced from
13 to 11.
Endless Amusement II.
1826? Prob. 11, pp.
195-196. 15 diamonds reduced to 13.
The Boy's Own Book.
The curious cross. 1828:
414; 1828-2: 420; 1829 (US): 213; 1855: 568; 1868:
628. 13 markers, reducing to 11.
Nuts to Crack III (1834), no. 80. The curious cross. Almost
identical to Boy's Own Book.
The Riddler.
1835. The curious cross, p.
6. Identical to Boy's Own Book.
Young Man's Book.
1839. P. 60. Easy Method of Purloining without
Discovery. Identical to Endless
Amusement II, except with a title.
The New Sphinx.
c1840. The curious cross, p.
143. Same as Boy's Own Book, with a few
words changed.
The Sociable.
1858. Prob. 25: The dishonest
jeweller, pp. 295 & 310. 15 diamonds, reducing to 13.
= Book of 500 Puzzles, 1859, prob. 25, pp. 13 & 28. = Wehman, New Book of 200 Puzzles, 1908, p.
7.
The Secret Out.
1859. The Dishonest Servant, pp.
78-80. T shape. 16 coins reduced to 14. Presentation of the problem is done with
cards.
Boy's Own Conjuring Book.
1860. Easy method of purloining
without discovery, p. 295. 15 diamonds, reducing to 13.
Very similar to The Sociable.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 585-5, pp. 296 & 409:
Juwelenkreuz. 18 jewels reduced to 16.
Mittenzwey.
1880. Prob. 226, pp. 41-42 &
92; 1895?: 252-253, pp. 45-46 &
94; 1917: 252‑253, pp. 42 &
90. 15
jewels reduced to 13. The 1895? addition has 10
reduced to 8.
Lucas. La croix de
perles. RM2, 1883, pp. 134‑135. c= Lucas; L'Arithmétique Amusante; 1895; pp.
10-11. 15 reduced to 13 and discussion.
Lemon. 1890. The puzzling pearls, no. 535, pp. 69 &
117. 15 reduced to 13.
Don Lemon.
Everybody's Pocket Cyclopedia.
Revised 8th ed., 1890. Op. cit.
in 5.A. P. 136, no. 7. Diamond cross reduced from 15
to 13. No solution.
É. Ducret.
Récréations Mathématiques. Op.
cit. in 4.A.1. 1892? P. 104: Les croix de jetons. Rearrange a cross of 13
to 11.
H. D. Northrop.
Popular Pastimes. 1901. No. 4: The dishonest jeweller, pp. 67 &
72. = The Sociable.
M. Adams. Indoor
Games. 1912 The dishonest steward, pp. 22 & 24. Rearrangement of a Y.
Dudeney. AM. 1917.
Prob. 423: The ruby brooch, pp. 144-145 & 249. Complicated version. Brooch is a circle with 8
radii and the owner used to count from the centre to the circle,
along ⅛ of the circumference and then back in, getting 8
each time. Dishonest jeweller
reduces stones from 45 to
41 in a symmetric pattern of one
in the middle, two on each arm and three on each arc. What was it before?
Blyth. Match-Stick
Magic. 1921. Counting the cross, p. 33.
Rearrange a cross of 15 to
13.
7.Q.2. REARRANGE A CROSS OF SIX TO MAKE TWO LINES OF FOUR, ETC.
For the standard version, one has to
put one marker on top of the one at the crossing. See 6.AO and 7.Q for some similar trick versions. The one which is closest to this section is:
(12, 7, 4) -- Trick version of a 3 x 3
square with doubled diagonal:
Hoffmann (1876),
Mittenzwey, Hoffmann (1893), no.
8.
See also in 6.AO, Hoffmann (1893),
no. 9.
Les Amusemens.
1749. P. xxv. A cross of six -- four crossing three --
rearranged to count four both ways.
Blyth. Match-Stick
Magic. 1921. Five by count, pp. 33-34.
A cross of seven -- five crossing three -- rearranged to count five both
ways.
J. F. Orrin. Easy
Magic for Evening Parties. Jarrolds,
London, nd [1930s??]. The five puzzle,
pp. 66-67. As in Blyth.
Sid G. Hedges. More
Indoor and Community Games. Methuen,
London, 1937. Penny puzzle, pp.
51-52. Four pennies in the shape of
a T
or Y tetromino. Make two rows
of three. Put one from an end of the
row of three on the crossing.
Depew. Cokesbury Game
Book. 1939. Seven coins, p. 223. As
in Blyth.
Ripley's Puzzles and Games.
1966. Pp. 18-19, item 2. An
L with four in one leg and three
in the other rearranged to have four on both lines.
7.R. "IF I HAD ONE FROM YOU, I'D HAVE TWICE YOU"
Jacques Sesiano. The appearance of negative solutions in
mediaeval mathematics. Archive for the
History of the Exact Sciences 32 (1985) 105-150. In this, he discusses problems of the types given here and in
7.P.1, 7.R.1, 7.R.2 and 7.P.7. He pays
particular attention to whether the author discusses the problems in general or
recognizes conditions for positivity or consistency, covering this in more
detail than I do here.
See Tropfke 609.
In the medieval period, all these
problems are extended in various ways to lead to quadratic and higher
equations, but I think these become non-recreational although they certainly
were not practical at the time, except for a few involving compound interest.
I include a few examples of unusual
cases of two equations in two unknowns here.
NOTATION.
(a,
b; c, d) denotes the general form for
two people.
"If I had a
from you, I'd have b times you."
"And if I had c
from you, I'd have d times you."
I‑(a, b; c, d; ...) denotes the
case for more people where 'you' means all the others.
II‑(a,
b; c, d; ...) denotes the same where 'you' means just the next person,
taken cyclically.
With two people, there is no need to
distinguish these cases.
See Alcuin for an example where the second
statement is interpreted as occurring after the first is carried
out.
Sometimes a = 0 -- see: Kelland
(1839); Hummerston (1924). If a = c = ... = 0, this can interpreted as a form of 7.R or 7.R.1 -- see: Dodson
(1775).
See Ghaligai, Cardan for versions
with "If I had a
times yours from you, I'd have b times you", i.e. x + ay = b(y - ay), ....
Versions giving (x-a)/(y-a) = c; (x+b)/(y+b) = d,
etc. are forms of Age Problems
and are generally placed in 7.X. But
see: Dodson.
See Hall for a version where the
first equation is x + a = by.
There are versions where one asks
for a fraction x/y such that
(x+a)/y = c/d, x/(y+b) = e/f. These are forms of 7.R.1. See: Dodson; Hall.
Let
T be the total of the
amounts. Then I‑(a1,b1; a2,b2;
...) with n people has n
equations xi + ai = bi(T ‑ xi
- ai), which can be
rewritten as xi + ai(1+bi) = bi(T
- xi), so we see that this
is the same problem as discussed in 7.R.1 below where men find a purse, but
with variable known purses, pi
= ai(1+bi). We
get xi = biT/(1+bi)
- ai. Adding these for
all i
gives one equation in the one unknown
T, T [Σ {bi/(1+bi}
- 1] =
Σ ai.
For
II‑(a1,b1; a2,b2;
...), systematic elimination in
the n
equations xi + ai = bi (xi+1 ‑ ai) leads to
x1 [b1b2...bn ‑ 1] = a1(b1+1) + a2b1(b2+1) + a3b1b2(b3+1) +
.... ,
and any other value can be found by shifting the starting point of the
cycle.
Verse versions: Euclid;
Wingate/Kersey; Ozanam; Vinot;
Euclid. c-325. Opera.
Ed. by J. L. Heiberg & H. Menge, Teubner, Leipzig, 1916. Vol. VIII, pp. 286‑287. Ass and mule in Greek and Latin verse. (1, 2;
1, 1). (Sanford 207 gives
English of Clavius's 1605 version. Cf
Wingate/Kersey.)
Diophantos.
Arithmetica. c250. Book I.
Metrodorus.
c510. Art. 145‑146, p.
105. (10, 3; 10, 5); (2, 2; 2, 4).
Earliest example with non-integral answers.
Alcuin. 9C. Problem 16: Propositio de duobus homines
boves ducentibus. (2, 1; 2, 2),
but the second statement in the problem is interpreted as
happening after the first is actually carried out. If a problem with parameters
(a, b; c, d) is interpreted this way, it is the same as
our usual problem with parameters (a,
b; c-a, d).
Mahavira. 850. Chap. VI, v. 251‑258, pp. 158‑159.
al‑Karkhi.
c1010. Sect. III, no. 5, p.
90. II‑(1, 2; 2, 3;
3, 4; 4, 5).
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 145f., no. 7. ??NYS -- cited by Tropfke 611.
Bhaskara II.
Bijaganita. 1150. Chap. IV, v. 106. In Colebrooke, p. 191.
(100, 2; 10, 6).
Fibonacci.
1202. Pp. 190‑203 (S:
289-305). Numerous versions, getting up
to five people, some inconsistent examples and types where the second clause
is "I'd have b1 times you plus b2 more."
Lucca 1754.
c1330. Ff. 29r‑30v, pp. 68‑70. (12, 2;
17, 3). II‑(15, 2; 18, 3;
21, 5).
Munich 14684.
14C. Prob. XV, p. 80. (1, 1;
2, 2) and (n, 1;
n, 2).
Folkerts.
Aufgabensammlungen. 13-15C. Many sources. Almost all have (a,
1; a, 2), with the objects being exchanged being, gold, animals, coins,
nuts, often noting that the answer is
a times the answer for a = 1.
Examples of (a, 1; a, d)
with d = 3, 5, 9. 17. An example of (a, ½; a, d). Folkerts
cites Metrodorus, Alcuin, Fibonacci, AR, etc.
Three
sources of the following. Ask a person
to put the same amount of money into each of her hands. Tell her to transfer n
coins from the right hand to the left.
Now transfer enough from the left to double what is in the right
hand. This leaves 2n
in the left hand.
Giovanni di Bartolo.
Op. cit. in 7.H. c1400. He gives complex examples in probs. 10‑14,
54, 56, 57 on pp. 18‑27, 101‑107.
E.g. prob. 10, pp. 18‑21.
"If I had the square root of your money, I'd have 3
times you." "And if I
had the square root of your money, I'd have
4 times you."
Provençale Arithmétique.
c1430. Op. cit. in 7.E.
Pseudo-dell'Abbaco.
c1440.
AR. c1450. Prob. 138‑147, 220, 334‑336. Pp. 70‑71, 102, 146‑147, 169‑171,
218.
Benedetto da Firenze.
c1465. Pp. 153. (60, 6;
50, 13); II‑(10, 2; 19, 4;
15, 9); I‑(16, 2; 30, 8; 26,
7/2).
Muscarello. 1478. Ff. 81v-82r, p. 198. (2, 2;
2, 1).
della Francesca.
Trattato. c1480.
Chuquet. 1484. Prob. 57, 58, 59, 60. Prob. 61‑79 extend in various
ways. English of 69, 70 78 in FHM
210-212. 78 is indeterminate.
Calandri.
Arimethrica. 1491. F. 66r.
(20, 2; 30, 3). This has the
unusual feature that x = y and I do not recall any other such
example. The condition for x = y
is more complex than one might expect:
a (b+1)/(b-1) = c (d+1)/(d-1).
Pacioli. Summa. 1494.
He has numerous problems, sometimes mixing amounts and parts and
sometimes mixing this topic with 7.R.1 and 7.R.2, often saying "that part
of yours that 12 is to mine", i.e.
y(12/x) {cf. della Francesca and
7.R.1}, and he often continues into problems where one gives the square root of
what one has or says something about the square of an amount.
Calandri, Raccolta.
c1495.
Hans Sachs (attrib.).
Useful Table-talk, or Something for all; that is the Happy Thoughts, good
and bad, expelling Melancholy and cheering Spirits, of Hilarius Wish-wash,
Master-tiler at Kielenhausen. No
publisher, place or cover, 1517, ??NYS -- discussed and quoted in: Sabine
Baring-Gould; Strange Survivals Some
Chapters in the History of Man; (1892), 3rd ed., Methuen, 1905, pp.
220-223. [Not in Santi.] Baring-Gould, p. 221 has (1, 2;
1, 1).
Ghaligai. Practica
D'Arithmetica. 1521. He has a series of problems of this type, of
increasing complexity, all involving men and money. I omit the more complex cases.
He also uses parts as in Pacioli.
Riese.
Rechnung. 1522. 1544 ed. -- p. 93; 1574 ed. -- pp. 62v‑63r.
(1, 1; 1, 3).
Riese. Die
Coss. 1524. No. 25‑30, p. 44 & No. 63‑64, p. 49.
Cardan. Practica
Arithmetice. 1539. Chap. 61.
Recorde. Second
Part. 1552. Pp. 322-324: The fourth example.
(2, 4; 3, 1).
Tartaglia. General
Trattato. 1556.
Buteo.
Logistica. 1559. Prob. 59, p. 264. i-th says "I have ai times as
much as the rest of you."
with (ai) =
(1, 1/2, 1/5). This could be
considered as I-(0, 1; 0, 1/2;
0, 1/5). This is indeterminate
with general solution proportional to
(3, 2, 1). He assumes x = 24 and gets y = 16, z = 8.
van Etten.
1624. Prob. 83 (76), parts a
& c, pp. 90‑92 (134‑136).
Ass & mule -- (1, 2; 1, 1)
= Euclid. (10, 3; 10, 5),
(2, 2; 2, 4) = Metrodorus.
Hunt. 1651. Pp. 280-281: Of the mule and the ass. (1, 2;
1, 1).
Schott. 1674. Ænigma V, pp. 553. (1, 1; 1, 2) = Euclid.
Cites Euclid, Clavius and Lantz.
Wingate/Kersey.
1678?. Quest. 39, pp.
502-503. Ass and mule in Latin verse -
cf Euclid. (1, 2; 1, 1)
Edward Cocker.
Arithmetic. Ed. by John
Hawkins. T. Passinger & T. Lacy,
London, 1678. [De Morgan states "I
am perfectly satisfied that Cocker's Arithmetic is a forgery of Hawkins"
and then spends several pages detailing this charge and showing that the book
is a rather poor compilation from several better books. However Ruth Wallis [Ruth Wallis; Edward
Cocker (1632?-1676) and his Arithmetick: De Morgan demolished; Annals of
Science 54 (1997) 507-522] has argued that De Morgan is wrong. Inspection of a 1st ed. at the Graves
collection and a 3rd ed., 1680, at Keele shows no noticeable difference in the
texts other than resetting which makes the book smaller with time -- all the
editions seen have the same 32 chapters.
The 1st and 3rd eds. seem to have identical pagination so I will not
cite the 1680.] 1st ed., 1678 & 3rd
ed., 1680, both T. Passinger & T. Lacy, London. = 33rd ed., Eben. Tracy, London, 1715. = Revised by John Mair; James & Matthew Robertson, Glasgow,
1787. Chap. 32, quest. 4. 1678: p. 333; 1715: p. 215; 1787: p.
186. (1, 5; 1, 1).
Wells. 1698. No. 104, p. 206. Ass & mule: (1,
1; 1, 2); (a, 1; a, 2).
Ozanam. 1725. De l'asne et du mulet, prob. 24, question 3,
1725: 176‑178.
Prob. 6, 1778: 189-190;
1803: 186-188; 1814:
162-163; 1840: 84-85. (1, 2;
1, 1). 1725 gives two versions
and solutions in Latin verse. 1778 et
seq. gives just one version and solution, but with slight differences, and
refers to the Metrodorus problems in Bachet's Diophantos, though these are not
the numbers in Metrodorus.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XXIII, p. 93 (not in 1790 ed.). A says to
B, if I had a of
your money, I'd have as much as you together with half of C,
etc., giving: x + a =
y-a + ½z, y + b =
z-b + ⅓x, z + c =
x-c + ¼y. Finds general solution
and does case a = b = c = 5.
Walkingame. Tutor's
Assistant. 1751. 1777: p. 175, prob. 98; 1860: p. 184, prob. 97. (4, 1; 4, 2), flocks of sheep.
Mair. 1765? P. 458, ex. 7. Two men with money, like ass & mule: (1, 5;
1, 1).
Euler. Algebra. 1770.
I.IV.IV.612: Question 3, pp. 208‑209. Ass and mule: (1, 2; 1, 3).
Vyse. Tutor's
Guide. 1771? Prob. 11, 1793: p. 130; 1799: p. 138 & Key p. 183. (1, 1;
1, 2) (= Euclid).
Dodson. Math.
Repository. 1775.
Eadon.
Repository. 1794. P. 296, no. 8. (5, 1; 5, 3).
D. Adams. Scholar's
Arithmetic. 1801. P. 209, no. 2. (1, 1; 1, 2).
Augustus De Morgan.
Arithmetic and Algebra. (1831
?). Reprinted as the second, separately
paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin &
Craddock, London, 1836. Art. 121, p.
32. (10, 2; 10, 3).
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 117-119 & 256-257, nos. 443‑450. (5½, 2; 6½, 5), (13½, 7;
16½, 3), (10, 4; 7½, 2),
(10, 4; 12, 3), (20⅓, 3; 12⅔, 4),
(9½, 4; 15, 2),
(33½, 2; 16½, 3), (6, 5;
4 2/5, 8).
Philip Kelland. The
Elements of Algebra. A. & C. Black,
Edinburgh, et al., 1839. ??NX. P. 134: "A's money or debt
is a
times B's; if
A lose £10 to B,
it will be b times
B's." (Also entered in
7.X.)
The New Sphinx.
c1840. No. 46, pp. 24 &
122. Women with baskets of eggs: (1, 2; 1, 1).
Fireside Amusements.
1850: No. 2, pp. 101 & 180;
1890: No. 2, p. 96. = New
Sphinx. c1840.
The Family Friend (1856) 376, Enigmas, Charades, &c.
176 Arithmetical Puzzle. Standard
(1, 2; 1, 1) given in a four stanza poem involving two
costermongers with barrows of apples.
Signed G. M. F. G.
Thomas Grainger Hall.
The Elements of Algebra: Chiefly Intended for Schools, and the Junior
Classes in Colleges. Second Edition:
Altered and Enlarged. John W. Parker,
London, 1846. P. 132, ex. 7. Appears to be (50, 2; 50, 1), but it reads: "A says to B, if you give
me £50, I shall have twice as much as you had; but if I give you £50, each will
have the same sum." The use of
'had' means the first equation is x +
50 = 2y, while the second equation is the usual x - 50 = y + 50.
Answer: 250, 150.
Magician's Own Book.
1857. The two drovers, p.
246. (1, 1; 1, 2). = Book of 500
Puzzles, 1859, p. 60. = Boy's Own
Conjuring Book, 1860, p. 218.
Vinot. 1860. Art. XLVI: L'Anesse et le Mulet, pp.
64-65. Gives a French translation of
the Latin verse. (1, 1; 1, 2).
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-33, pp. 255 & 396. (1, 1;
1, 2). Notes that the solution
to (a, 1; a, 2) is just a
times the solution of the original.
(Beeton's) Boy's Own Magazine 3:4 (Apr 1889) 175 &
3:6 (Jun 1889) 255. (This is
undoubtedly reprinted from Boy's Own magazine 1 (1863).) Mathematical question 34. (5, 1; 10, 2) with postage stamps.
Mittenzwey.
1880.
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E.
Hoffmann. 1893. Chap. IV, no. 26: A rejected proposal, pp.
150 & 195 = Hoffmann-Hordern, p. 122.
(1, 3; 2, 1), but the first person also says he already
has twice the second, so this is an overdetermined problem. (Hoffmann's other example, no. 25, is just
the classic (1, 2; 1, 1), which is not overdetermined, so I
have omitted it here.)
Hummerston. Fun,
Mirth & Mystery. 1924. Pocket money, Puzzle no. 4, pp. 21 &
172. (0, 2; ½, 3).
Sullivan.
Unusual. 1947. Prob. 36: A problem old enough to be
considered new. (1, 1; 1, 2).
David Singmaster.
Some diophantine recreations.
Op. cit. in 7.P.5. 1993. Sketches some history; finds condition for
integer data in (a, b; c, d) to produce an integer solution, namely (bd - 1)/(b+1, d+1) divides
a + c where (b+1, d+1)
is the Greatest Common Divisor of
b+1 and d+1.
A letter from S. Parameswaran interpreted the problem given in Alcuin as
though the second statement was also made by the first animal. This gives
x + a = b (y - a);
x + c = d (y - c)
and similar reasoning finds the integrality conditions for this variant.
David Singmaster. A
variation of the ass and mule problem.
CM 28: 4 (May 2002) 236-238. The
2001 Maritime Mathematics Contest had a specific case of the following. The first person says: "If I had a from you, I'd have b
times you, but if I gave c to you, I'd have d times you." This leads to the equations: x + a
= b (y - a); d (y + c)
= x - c. The integral d of the classic problem
has been changed to 1/d with
d integral. Note that
b > d for reasonable
solutions. Reasoning similar to the
previous article finds the condition for integer data to produce an integer
solution and the condition is simpler than for that problem.
Tomislav Došlić.
Fibonacci in Hogwarts? MG 87
(No. 510) (Nov 2003) 432-436. Observes
that Fibonacci does the case (7,
5; 5, 7), whose solution is non-integral:
(121, 167)/17. He considers the
case (a, b; b, a) and finds there
are 16
positive pairs a, b which give integral solutions. If we set
a £ b, these are a, b
= 1, 2; 1, 3;
1, 5; 2, 2; 2, 3;
2, 8; 3, 3; 3, 7;
5, 8.
David Singmaster.
Integral solutions of ass and mule problems. Gives a simpler solution for Došlić's result and finds all
integral solutions when positivity is not required.
7.R.1. MEN FIND A PURSE AND 'BLOOM OF THYMARIDAS'
See Tropfke 604 & 606.
Algebraically, 7.R.1 and 7.R.2
differ only in signs.
NOTATION. Finding a purse has two forms.
I‑(a1,
a2, ..., an)
-- i-th says "If I had the
purse, I'd have ai times the rest of you".
II‑(a1,
a2, ..., an)
-- i-th says "If I had the
purse, I'd have ai times the
i+1-st person".
There are two related problems which
I call forms III and
IV.
III‑(a1,
a2, ..., an) --
the sum of all amounts except the i‑th
is ai. See the discussion under
Iamblichus. A number of problems in 7.H lead to this
type of problem when one uses
reciprocal times as work rates.
IV-(a1,
a2, ..., an)
-- xi + xi+1
= ai. (This is determinate
only if n is odd.)
For
n = 3, types III and IV
are the same, though the constants or the variables are taken in a
different order, so that III‑(a, b, c)
= IV-(c, a, b) if we keep the variables
in the same order.
I give answers as a list of the
amounts, in order; then the purse.
Let
p be the value of the purse and
let T
be the total of the amounts.
Then I‑(a1,a2,...) with
n people has n
equations xi + p = ai(T ‑ xi), so we see that this is the same problem as
discussed in 7.R.2 below where men buy a horse, but with the value of the horse
and the multipliers all negative, which makes this version have fewer sign
complications in its solution. Thus we
get xi = (aiT-p)/(1+ai). Adding these for all i
gives one equation in the two unknowns
T and p. However, letting C = T + p
leads to the simplest equation:
(n‑1)T = [Σ 1/(1+ai)]
C.
For
II‑(a1,a2,...), systematic elimination in the
n equations xi + p = aixi+1 leads to
x1 [a1a2...an - 1] = p
[1 + a1 + a1a2 + ... + a1a2...an], and any other value can be found by shifting
the starting point of the cycle.
In either case, the solution can be
adapted to variable purses -- see 7.R.
In some problems, gaining the purse is replaced by paying out, so the
purse can be treated as having a negative value -- see Unger, 1838.
Other versions of the problem has
several horses and a saddle (or other equipage) or several cups and a cover. The
i-th horse with the saddle is worth
ai times the others
or the next. I don't seem to have
entered any of these until recently.
There are versions where one asks
for a fraction x/y such that
(x+a)/y = b, x/(y+c) = d, where
a, c are integral (possibly
non-positive) and b, d are rationals. This is a mixture between 7.R and 7.R.1, but fits better here as
we can think of a, c as purses.
I will denote this as:
V-(a,
b; c, d). This has solution x
= β(a + αb)/(α -
β); y = (a + βb)/(α -
β).
See: Dodson; Todhunter.
Some
of the cistern problems in 7.H are of type III
Diophantos.
Arithmetica. c250. Book I.
Iamblichus. On
Nicomachus's 'Introductio Arithmetica'.
c325. Pp. 62‑63,
??NYS. Partly given in SIHGM I 138‑141. Describes the 'Bloom of Thymaridas' which
has n+1 unknowns x, x1, ..., xn and we
know x + xi = ai and
x + x1 + ... + xn
= s. Then
x = (a1 + ... + an
‑ s)/(n‑1). (Iamblichus
uses n
unknowns.) Heath
(HGM I 94‑96) says Iamblichus continues and applies the Bloom
to I‑(a1, ..., an), with integral ai, by letting x be
the value of the purse and s = (a1+1)...(an+1), which yields x + xi = sai/(ai+1). (We can take n-1 times the value
of s
to insure integer solutions.)
For rational ai, we let
s be the LCM
of the denominators of ai/(ai+1). Iamblichus gives the problems I‑(2, 3, 4) and I‑(3/2, 4/3,
5/4). See Chuquet for an indeterminate
version of the Bloom.
This
is closely related to problems like the following: x + y = a,
y + z = b,
z + x = c,
i.e. III‑(b, c, a) =
IV-(a, b, c). We set T = x + y + z and so
T ‑ z = a, T ‑ y =
b, T ‑ x = c. This is a case of the 'bloom' for n = 3,
with x = T, x1 = ‑z, a1 = a, etc., and
s = 0. In general, this
gives us x = T = (a1 + ...
+ an)/(n‑1).
Aryabhata. 499. Chap. II, v. 29, pp. 71-72. (Clark edition: pp. 40‑41.) III‑(a1, ..., an). Gives
T = (a1+...+an)/(n‑1).
Bakhshali MS.
c7C. Kaye I 39-42, sections
78-79 and Datta, pp. 45‑46
discuss two types of related systems for n amounts x1, x2, ..., xn. See the discussion under Iamblichus.
Bhaskara I.
629. Commentary to Aryabhata,
chap. II, v. 29. Sanskrit is on pp.
125-127; English version of the examples is on pp. 307-308.
Ex.
1: III-(30, 36, 49, 50).
Ex.
2: III-(28, 27, 26, 25, 24, 23, 21).
Mahavira. 850. Chap. VI, v. 159, 233‑250, pp. 136‑137,
153‑158.
al‑Karkhi.
c1010. Sect. III, no. 24‑25
& 29‑30, pp. 95 & 98.
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 129, no. 46. ??NYS -- Hermelink, op. cit. in 3.A, says this is a problem with
two persons. Tropfke 607 cites this
with no details.
Fibonacci.
1202. Pp. 212‑228 (S:
317-337), Chap. XII, part 4: De inventione bursarum [On the finding of a
purse]. Many problems, going up to five
men, 4 purses and an example with a negative solution.
Fibonacci.
Flos. c1225. In Picutti, pp. 316-319, numbers IV-V.
Jordanus de Nemore.
De Numeris Datis. c1225. Critical edition and translation by Barnabas
Hughes. Univ. of Calif. Press,
Berkeley, 1981. Prob. II‑24, pp.
150‑151. General version of
type I
with purse given. Example: I‑(1/9, 1/3, 3/5, 1) with purse
6. Answer: 2, 14, 24, 34.
Ibn Badr =
Abenbéder = Abu ‘Abdallah Muhammad
(the h
should have an underdot) ibn ‘Umar ibn Muhammad (the h
should have an underdot).
c1225. Arabic text with Spanish
translation by José A. Sánchez Pérez as:
Compendio de Álgebra de Abenbéder; Centro de Estudios Históricos,
Madrid, 1916. Tercer problema análogo,
pp. 109-111 (pp. 70-71 of the Arabic).
(4, 7). Answer: (8, 5; 27)/9.
BR. c1305. No. 61, pp. 84‑87. (7, 11).
Answer given is 12, 8; 76, but should be 8, 12; 76.
Gherardi. Libro di
ragioni. 1328. P. 53.
3 men find a purse. I‑(2,
3, 4). Answer: 7, 17, 23; 73.
Lucca 1754.
c1330. F61r, pp.139‑140. II‑(2, 3, 4, 5). Answer:
33, 76, 65, 46; 119.
(= Fibonacci, pp. 218‑220.)
Bartoli.
Memoriale. c1420. Prob. 4, f. 75r (= Sesiano, pp. 136-137
& 147. I‑(2, 3, 4). Answer:
7, 17, 23; 73. (= Iamblichus's
1st.)
Pseudo-dell'Abbaco.
c1440. Prob. 125, p. 100. I‑(2, 3, 4). Answer: 7, 17, 23;
73. (= Iamblichus' 1st problem.)
AR. c1450. Prob. 113, 159, 227. Pp. 63‑64, 76, 168‑169, 217.
Correspondence of Johannes Regiomontanus, 1463?-1465. Op. cit. in 7.P.1.
Benedetto da Firenze.
c1465.
Chap. 22: "... huomini che ànno danari et trovano
borse di danari", pp. 181‑183.
Muscarello.
1478. Ff. 65r-65v, pp.
172-173. Men find a purse, I‑(3, 4, 5). Answer: 7, 13, 17; 83.
della Francesca.
Trattato. c1480.
Chuquet. 1484.
Triparty, part
I. FHM
80-81 & 83-85. Sesiano cites
pp. 642 & 646 of Chuquet, ??NYS.
Appendice.
Borghi.
Arithmetica. 1491? -- this
material is additional to the 1484 ed. and Rara indicates that the first ed. to
have 100ff is the 4th of 1491. The
folio numbers are from the 1509 ed.
Calandri. Arimethrica. 1491.
Pacioli. Summa. 1494.
Some of his problems mix this with 7.R.2.
Ghaligai. Practica
D'Arithmetica. 1521. He gives several versions, of increasing
complexity -- the later ones involve various numbers in geometric progressions
or using roots and I omit these.
Tonstall. De Arte
Supputandi. 1522.
Riese. Die
Coss. 1524.
Apianus. Kauffmanss
Rechnung. 1527.
Cardan. Practica
Arithmetice. 1539. Chap. 66.
Recorde. Second
Part. 1552. Pp. 320-322: A question of debt, the third example. IV‑(47, 88, 71). Answer: 15, 32, 56.
Tartaglia. General
Trattato. 1556. Book 16, art. 28‑35 & 40, pp. 243r‑245r.
Buteo.
Logistica. 1559. Prob. 9, p. 209-210. III-(4900, 3760, 4660). Remarks on the case with four people.
Schott. 1674.
Wells. 1698. No. 117, p. 208. Two horses and equipage:
(1, 2) with equipage (equivalent
to a purse) worth 5.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob. V,
pp. 79-80 (1790: prob. VII, p. 80).
III-(17, 16, 15). 1745 gives
two methods of solution;
1790 gives one.
Mair. 1765? Pp. 458-459, ex. 8. Two horses:
(2, 3) with saddle worth 50.
Vyse. Tutor's
Guide. 1771? Prob. 27, 1793: pp. 56-57; 1799: pp. 61-62 & Key p. 67. III‑(33000, 30000, 32000, 28000,
25000).
Dodson. Math.
Repository. 1775.
Hutton. A Course of
Mathematics. 1798? Prob. 30,
1833: 222; 1857: 226. Two horses with saddle. I‑(2, 3) with saddle worth
50. Answer: 30, 40; 50.
(= 10 times Mahavira v. 244.)
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 115-117 & 256, nos. 431-442. All of these have different purses except
no. 435, which is I-(½, ⅓) with a purse of -10, i.e. 10
is taken away rather than gained.
No. 432 uses the context of ages.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Arithmetical puzzles, no. 2, pp. 172-173 (1868: 184). I-(1, 2, 3)
with purse of 11. Answer:
1, 5, 7; 11. See AR 113 &
227.
Todhunter. Algebra,
5th ed. 1870. Section XIII, art. 185; Example XIII, nos. 1, 3, 5, pp. 99-100 & 578.
Mittenzwey.
1880.
Clark. Mental
Nuts. 1897, no. 75. The horses and saddle. I-(1, 2, 3)
with the saddle acting as the purse and the total value of the horses
and saddle being 220. Answer: (1, 5, 7; 11)
* 55/6. See AR 113 & 227; Parlour
Pastime.
Haldeman-Julius.
1937. No. 23: The coin problem,
pp. 5 & 21. Like two men finding a
purse of 25. First says: If I had the
purse, I'd have 3 times what you have. Second says: And if I had the purse, I'd
have five less than half of what you have.
Answer: 230, 85.
Ken Russell & Philip Carter. Intelligent Puzzles.
Foulsham, Slough, 1992. Prob.
57, pp. 45 & Answer 75, p. 164: Money.
Jack has 75p and ¾ of what Jill has;
Jill has 50p and ½ of what Jack has.
This is a straightforward problem, but it is also a concealed version
of I‑(4/3, 2) with a purse of 100.
7.R.2. "IF I HAD 1/3 OF YOUR MONEY, I COULD BUY THE HORSE"
See Tropfke 608.
For related problems, see 7.H.4.
NOTATION. Again there are two forms.
I‑(a1,
a2, ..., an) --
i-th says "If I had ai of what the rest of you have, then I could
buy the
horse".
II‑(a1,
a2, ..., an) --
i-th says "If I had ai of what the i+1-st has, then I could buy the
horse".
The problems are the same when there
are only two people and I will label two person cases as form I.
The problem often replaces horse by
house, ship, etc. In some cases, the
value of the horse, house, etc. is given.
In some cases the value is given with no reference to anything bought
and I say "with h = ..." to indicate the value. Some problems have different values of
horses. If values are simply given, I
say "with h1, h2 =
..." or "with hi
= ...". See: Fibonacci,
Gherardi, Lucca 1754, AR,
Benedetto da Firenze, della
Francesca, Pacioli, Riese: Coss, Tartaglia, Buteo, Schott,
Simpson for examples where there
are different values of horses.
Solution notation as in 7.R.1 with
the horse last.
Let
h be the value of the horse and let T be
the total of the amounts. Then I‑(a1,a2,...) with
n people has n
equations xi + ai(T ‑ xi) =
h, so xi = (h - aiT)/(1-ai). Adding these for all i
gives one equation in the two unknowns
T and h. However, letting C = T - h
leads to the simplest equation:
(n‑1)T = [Σ 1/(1‑ai)]
C.
For
II‑(a1,a2,...), systematic elimination in the
n equations xi + aixi+1
= h leads to x1 [1 + (-1)n+1a1a2...an] = h
[1 - a1 + a1a2 - ... + (-1)na1a2...an], and any other value can be found by shifting
the starting point of the cycle.
In either case, the solution can be
adapted to variable purses -- see 7.R.
Taking negative values of h and all
ai converts this into
7.R.1 -- men find a purse, which is slightly easier to deal with.
INDEX OF SOME COMMON
TYPES.
I have recently compiled this and
was surprised to see how much repetition is present.
I‑(⅔, ¾). Riese: Rechnung; A Lover of the Mathematics;
Euler IV.618; Vyse;
I‑(½, ⅔). Chiu Chang Suan Ching; Sun Zi;
AR 176;
I-(½, ⅓). della Francesca 16r; Peurbach;
Lacroix;
I-(⅓, ¼). al-Karkhi I-42; Fibonacci 228; BR 7; AR 171-175, 177, 221;
della
Francesca 36v; Pacioli 190v; Riese: Rechnung; Schott 562-563;
I-(1/3, 1/5). Calandri;
Calandri: Raccolta;
I‑(1/6, 1/7). Benedetto da Firenze;
I-(½, ⅔, ¾). della Francesca 21v;
I‑(½, ⅓, ¼). Gherardi 46-47; Lucca 1754 58r; Munich
14684; Folkerts;
Provençale
Arithmétique; AR 178, 224, 340; Benedetto da Firenze;
della
Francesca 17v-18r, 39r; Chuquet; Borghi;
Pacioli 105v-106r, 192r, 192v;
Tonstall; Riese: Coss 122, 123; Buteo 81;
Les Amusemens; Euler III.19,
IV.622;
I‑(1/3, 1/4, 1/5). Diophantos 24; al-Karkhi III-26;
Fibonacci 245; AR 341; Buteo 192-193;
Schott
563;
I-(2/3, 3/4, 4/5, 5/6). Chuquet;
I‑(1/2, 1/3, 1/4,
1/5). Fibonacci 245-248. de Nemore II-27; Provençale Arithmétique;
Riese: Coss
124-126; Pearson;
I‑(1/3, 1/4, 1/5,
1/6). Diophantos 25; al-Karkhi III-27;
I-(1/2, 2/3, 3/4, 4/5,
5/6). Chuquet;
I‑(1/2, 1/3, 1/4, 1/5,
1/6). Provençale Arithmétique;
II‑(3/4, 4/5, 5/6). Riese: Coss 121;
II‑(½, ⅔, ¾). AR 180;
Chuquet; Tartaglia 22;
II‑(½, ⅓, ¼). Lucca 1754 58r; AR 179, 223, 339; della
Francesca 36v-37r; Chuquet;
Riese: Rechnung; Riese: Coss 31, 47, 120; Peurbach;
Tartaglia 40, 41; Buteo 81;
Euler
IV.619-629;
II‑(1/3, 1/4, 1/5). Fibonacci 229; BR 114; AR 156, 157,
338; della Francesca 17v; Chuquet;
II‑(1/3, 1/5, 1/4). al-Karkhi III-33;
II‑(1/2, 1/3, 1/4,
1/5). de Nemore II-25; Gherardi 45-46
II‑(1/3, 1/4, 1/5,
1/6). Fibonacci 231-232.
II‑(1/2, 1/3, 1/4, 1/5,
1/6). Fibonacci 327-329.
II‑(1/3, 1/4, 1/5, 1/6,
1/7). Fibonacci 234.
II‑(2/3, 3/4, 4/5, 5/6,
6/7, 7/8, 8/9). Riese: Coss 140;
Buying a horse with a friend or
a found purse. Lucca 1754 61r-61v; della Francesca 40v;
Pacioli
193v-194r; Cardan;
Jens Høyrup.
Sub-scientific mathematics: Undercurrents and missing links in the
mathematical technology of the Hellenistic and Roman world. Preprint from Roskilde University, Institute
of Communication Research, Educational Research and Theory of Science, 1990,
Nr. 3. (Written for: Aufsteig und
Niedergang der römischen Welt, vol. II 37,3 [??].) P. 18 quotes Plato's Republic, 333b-c, "to buy in common or
sell a horse" and feels this may be a reference to this type of
problem. (This seems a bit far-fetched
as there are many simpler types of commercial problem which could be described
by this phrase.)
Chiu Chang Suan Ching.
c‑150. Chap. VIII.
Diophantos.
Arithmetica. c250. Book I.
Sun Zi. Sun Zi Suan
Ching. Op. cit. in 7.P.2. 4C.
??NYS. Chap. III, no. 28. I-(½, ⅔) with h = 48 -- like Chiu Chang Suan Ching, prob. 10. (English in Lam & Shen, HM 16 (1989)
117.)
See 7.P.4 -- Bakhshali MS for a problem which is related,
leading to x1/2 + x2 + x3 + x4 + x5 =
h, etc., where h is the price of a jewel. Solution: 120, 90, 80, 75,
72; 377. Also an example with three
values and diagonal coefficients
-7/12, -3/4, -5/6
and solution: 924, 836, 798;
1095.
Sesiano cites Abū Kāmil's Algebra and
al-Karajī's Kāfī, ??NYS.
Hermelink, op. cit. in 3.A, cites Kāfī &
Beha-Eddin, ??NYS
al‑Karkhi.
c1010.
Sect I, no. 42‑43,
p. 80.
Sect. III, no.
26‑27, 32‑35, pp. 95‑100.
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 133, no. 52. ??NYS -- Hermelink, op. cit. in 3.A, mentions this without
details. Tropfke 609 cites this problem
and also p. 150f., no. 13, as buying a horse.
Fibonacci.
1202. Pp. 228‑258 (S:
337-372): chap. 12, part 5: De emptione equorum inter consocios secundum datam
proportionem [On the purchase of horses among partners according to some given
proportion]. Many examples, getting up
to seven men, up to five horses, an inconsistent example and negative
solutions. I have omitted some of the
variations and some of the more complex problems. Some of the problems cited are followed by some discussion.
See: K. Vogel; Zur Geschichte der linearen
Gleichungen mit mehreren Unbekannten; Deutsche Mathematik 5 (1940) 217‑240; for a thorough study of the problems on pp.
228‑258. Some further versions
occur on pp. 327‑349 (S: 458-484).
Vogel cites several sources, ??NYS:
Codex lat. Monacensis 14908 (1456)
and, for a simpler type of problem,
Mich. Pap. 620.
Fibonacci. Flos. c1225.
In: Picutti, pp. 312-316 &
320-326, numbers III, VI & VII.
Fibonacci.
Epistola. c1225. In Picutti, pp. 338-340, numbers XV &
XVI.
Jordanus de Nemore.
c1225. Op. cit. in 7.R.1.
BR. c1305.
Gherardi. Libro di
ragioni. 1328.
Lucca 1754. c1330.
Munich 14684.
14C. Prob. XVII, p. 80. I‑(½, ⅓, ¼). Answer: 10, 22, 26; 34. Cf Gherardi, pp. 46-47.
Folkerts.
Aufgabensammlungen. 13-15C. He calls it Sperberkauf (sparrow hawk
purchase). 11 sources for I‑(½, ⅓, ¼) = Munich 14684 = Gherardi, pp. 46-47, none
of which give a derivation. It's not
clear if h = 34 is given in some cases. Numerous other citations.
Provençale Arithmétique.
c1430. Op. cit. in 7.E. F. 100r-101v, pp. 49-53.
AR. c1450. Prob. 156, 157, 171‑181, 186, 221,
223, 224, 338‑341. Pp. 74‑75,
82‑85, 87, 102‑103, 149‑150, 171‑173, 218‑219.
Dresden MS C80, 15C, has some of these problems. ??NYR -- mentioned in BR, p. 157.
Benedetto da Firenze.
c1465.
Pp. 155‑157.
Chap. 20: "... uomini che ànno d. et vogliono
chompare chavagli," pp. 168‑171.
Muscarello.
1478. Ff 56v-58r, pp.
158-161. Four men buying a house
worth 100, II‑(2/3, 5/8, 4/5, 7/10). Answer should be: (1250,
1575, 1160, 1425)/23, but there are two
copying errors in the MS. The second
answer is given as 66 11/23 instead of
68 11/23 and the fourth
answer is given as 61 12/23 instead of
61 22/23.
della Francesca.
Trattato. c1480.
Chuquet. 1484. Triparty, part 1. Sesiano cites p. 641, ??NYS
Borghi.
Arithmetica. 1484. Ff. 113v-116v (1509: ff. 95v-98r). I‑(½, ⅓, ¼) with
h = 20. Answer: (100, 220, 260)/17.
Calandri.
Arimethrica. 1491. F. 667v.
Two men buy a lamprey. I-(1/3,
1/5). Takes h = 60 rather arbitrarily
and gets answer: (300, 360)/7.
Francesco Pellos.
Compendion de lo Abaco. Turin,
1492. ??NYS -- see Rara 50-52 &
both Sesiano papers mentioned at Provençale Arithmétique, c1430,
above. Sesiano says ff. 64v-65r
translates the three problems in the Provençale Arithmétique and is the first
printed problem with a negative solution.
He says it may have been composed c1460. Smith doesn't mention either of these points.
Pacioli. Summa. 1494.
Calandri, Raccolta.
c1495. Prob. 25, pp. 24‑25. I‑(1/3, 1/5) with horse worth 60. Answer: (300, 160)/7. Cf Calandri.
Riese.
Rechnung. 1522. 1544 ed. -- pp. 91‑92 & 94‑95; 1574 ed. -- pp. 61v‑62v & 63v‑64r.
Tonstall. De Arte
Supputandi. 1522. Pp. 246-248. I-(½, ⅓, ¼) with
common value 20. Cf Gherardi 36-37.
Riese. Die Coss. 1524.
Many examples, including the following.
Apianus. Kauffmanss
Rechnung. 1527. Ff. M.iii.r - M.iii.v. x + y/3 = y + z/2 = z + 2y/3 = 30.
Answer: (45, 45, 60)/2. Because
x = y, this is actually the same
as II-(1/3, 1/2, 2/3).
Georg von Peurbach.
Elementa arithmetices, algorithmus ....
Joseph Klug, Wittenberg, 1534, ??NYS.
(There were several previous editions back to 1492, with variant
titles. Rara 53‑54. Glaisher, op. cit. in 7.G.1 under Widman,
describes this extensively and gives the following on p. 97. This edition is substantially better than
previous ones, but Peurbach died in 1461!)
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 98, ff. HH.vi.r - HH.vi.v
(p. 169). Men find a purse and
buy a horse, giving:
x + y + p/2 = y + z + p/5 = x + z + p/3 = h. Answer: 6, 10, 15; 30, 31. Cf Lucca 1754 61r-61v; della Francesca 40v;
Pacioli 193v-194r.
Tartaglia. General
Trattato. 1556. Book 17, art. 8, 13, 14, 22, 32, 40, 41, pp.
268v, 269v, 271v, 273v, 275v-277r.
Buteo.
Logistica. 1559.
Schott. 1674.
"A Lover of the Mathematics." A Mathematical Miscellany in Four
Parts. 2nd ed., S. Fuller, Dublin,
1735. Part III, no. 45, p. 94. Men buying a house worth 1200. I-(⅔, ¾). Answer: 800, 600. Cf Riese: Rechnung.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XXV, pp. 95-96 (1790: prob. 34, pp. 94‑95). II‑(1/2, 1/3, 1/4, 1/5) with
hi = a, b, c, d.
Does example with values 357,
476, 595, 714 and answer: 190,
334, 426, 676. 1745 gives a second
method.
Les Amusemens.
1749. Prob. 172, pp.
319-320. I‑(½, ⅓, ¼) with
h = 40. Solves with a
general h. Cf Gherardi 36-37.
Euler. Algebra. 1770.
I.IV.
Vyse. Tutor's
Guide. 1771? Prob. 13, 1793: p. 131; 1799: p. 138 & Key pp. 183-184. I‑(⅔, ¾) buying a horse worth 1200.
Cf Riese: Rechnung.
Silvestre François Lacroix.
Élémens d'Algèbre, a l'Usage de l'École Centrale des Quatre-Nations. 14th ed., Bachelier, Paris, 1825. Section 82, ex. 8, pp. 122-123. Variant version -- he has I-(1/2, 1/3) with h = 50,000 and then says the third could buy if he
had 1/4 of the first's money. I
think this is a corruption of I-(1/2,
1/3, 1/4). Cf della Francesca 16r.
Pearson. 1907. Part II, no. 139: The money‑boxes, pp.
141 & 218. I‑(1/2, 1/3, 1/4,
1/5) with h = 740. Cf Fibonacci
245-248.
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 101, pp. 43 & 144: A garaging
problem. A variation on type IV. a + b + c + d + e = 100; a + b = 52;
b + c = 43;
c + d = 34; d + e
= 30.
New section.
NOTATION: (a, b) means each boy
has a
times as many sisters as brothers, while each girl has b
times as many brothers as sisters.
This only has integer solutions for the integer pairs:
(a, b) = (1, 2),
(1, 3), (2, 1), (2, 2),
(3, 1), with solutions:
(Boys,
Girls) = (4, 3), (3, 2), (3, 4),
(2, 2), (2, 3).
See Wolff for a related problem.
See
Fireside Amusements; Cutler for a related trick problem.
Fireside Amusements.
1850: No. 47, pp. 114 & 181;
1890: No. 33, p. 102. "Mr.
Jones told another gentleman that he had six daughters, and each daughter had a
brother; how many children had Mr. Jones?" Cf Cutler, below.
Mittenzwey.
1880. Prob. 15, pp. 2-3 &
59; 1895?: 17, pp. 8-9 & 63; 1917: 15, pp. 8 & 57. Sisters and brothers. (2, 1).
Peano. Giochi. 1924.
Prob. 55, p. 15. (1, 2).
King. Best 100. 1927.
No. 39, pp. 19 & 47. =
Foulsham's, no. 11, pp. 7 & 11. (2,
1).
Heinrich Voggenreiter.
Deutsches Spielbuch Sechster
Teil: Heimspiele. Ludwig Voggenreiter,
Potsdam, 1930. P. 108. (2, 1).
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia Verlagsbuchhandlung,
Prague, 1937. Prob. 28, pp. 193
& 203. A man in an office says he
has four times as many male colleagues as female colleagues. One of the women answers that she has five
times as many male colleagues as female colleagues. This is (1/4, 5) in the above notation. In general, this would give m - 1 = af,
m = b (f - 1), which has solution f = (b + 1)/(b ‑ a). See Dodson in 7.R for a different phrasing
of the same problem.
Depew. Cokesbury
Game Book. 1939. Sisters and brothers, p. 218. (2, 1).
Owen Grant. Popular
Party Games. Universal, London, nd
[1940s?]. Prob. 8, pp. 36 &
50. (2, 1).
John Henry Cutler.
Dr. Quizzler's Mind Teasers.
Greenberg, NY, 1944. ??NYS --
excerpted in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun
1992) 47 & 43, prob. 27. "Mr.
and Mrs. Twichell have six daughters.
Each of the daughters has a brother.
How many persons are there in the entire family?" Only
9 (counting the parents). Cf Fireside Amusements, above.
Joseph Leeming.
Riddles, Riddles, Riddles.
Franklin Watts, 1953; Fawcett
Gold Medal, 1967. P. 113, no. 40. (1, 2).
Ripley's Puzzles and Games.
1966. P. 12. Man has two boys and a girl. He wants to have 12 boys and for each boy to
have a sister. How many girls are needed?
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
The sum of the siblings, pp. 31 & 110. (2, 1).
7.R.4. "IF I SOLD YOUR EGGS AT MY PRICE, I'D GET ...."
New section.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XL, pp. 106-107 (1790: prob. LIII, p. 108).
Market women bring x and
y eggs to market, with x + y = 100 [c] and sell at prices A and
B such that they get the same,
i.e. Ax = By. First says to the second:
"Had I brought as many eggs as you I should have received 18 [a]
Pence for them". The other
responds: "Had I brought no more
than you, I should have received only 8
[b] Pence for mine". I.e.
Ay = a, Bx = b. This gives
(x/y)2 = b/a,
x = cÖb/[Öa+Öb] =
bc/[b+Öab], y = cÖa/[Öa+Öb].
Mittenzwey.
1880. Prob. 123, pp. 25 &
76; 1895?: 141, pp. 29 & 79; 1917: 141, pp. 27 & 77. Same as Simpson, with it being clearly
phrased "If I had had your eggs and sold them at my price, ...",
with c = 110, a = 250, b = 360. Answer is given as 50, 60, but ought to be
as 60, 50. Here the prices also come out integral.
McKay. At Home
Tonight. 1940. Prob. 11: Buying cows, pp. 64 & 78. Farmers
A and B each buy £350
worth of cows. If A
had bought at B's price, he would have paid £250.
What would B have paid if he bought at A's
price? Letting x, y
be the numbers of cows and A, B be the prices, we have Ax = By = 350, Bx = 250, and we want Ay.
Then y/x =
7/5 and Ay
= Ax(7/5) =
490.
7.S. DILUTION AND MIXING PROBLEMS
See Tropfke 569.
There are a number of problems of
this sort. One type is the same as the
Hundred Fowls problem (7.P.1) where the solutions need not be integers. Lucca 1754, c1330, has a number of
these. Here I consider only some of special
interest and the following.
Recorde. Second
Part. 1552. H&S quotes from the 1579 ed., f. Y.3. "It hath great use in composition of
medicines, and also in myxtures of metalles, and some use it hath in myxtures
of wines. but I wshe [sic] it were lesse used therein than it is now a
daies." The 1668 ed., p. 295: The
Rule of Mixture, has: "And it hath
great use in composition of Medicines, and also in mixtures of Metalls, and
some use it hath in mixtures of Wines : but I wish it were less used therein
then it is now-a-days."
7.S.1. DISHONEST BUTLER DRINKING SOME AND REPLACING WITH WATER
Dodson, Todhunter and Clark are
problems to determine the amount taken off each time.
Papyrus Rhind, op. cit. in 7.C. Prob. 71, p. 108 of vol. 1 (1927) (= p. 57 of 1978
ed.). ¼ is poured off & replaced, what is the strength? (H&S 85 quotes Peet's version.)
Bakhshali MS.
c7C. Kaye I 48; III 201-202, ff.
12r-12v. Man has bottle holding 4
prasthas of wine. (Drinks ¼
and refills with water) four
times. How much wine is left?
Cardan. Practica
Arithmetice. 1539. Chap. 66, sections 36 & 37, ff. DD.v.r -
DD.v.v (p. 146). Drink three
pitchers and replace with water four times leaving wine of half strength. Then the same for three times.
Tartaglia. Quesiti,
et Inventioni Diverse, 1546, op. cit. in 7.E.1, Book 9, quest. 18,
pp. 102v‑103r. (Remove 2 and
replace) thrice to halve strength.
Buteo.
Logistica. 1559. Prob. 85, pp. 296-298. Butler drinking some and replacing with
water. (7/8)5. (H&S 85)
Trenchant. Op. cit.
in 7.L, 1566. 1578 ed., p. 297. ??NYS.
(Remove 1/12th and
replace) six times. (H&S 85 gives French and English. Sanford 209 gives English.)
Les Amusemens.
1749. Prob. 176, pp.
326-327. Sommelier drinks 6
pints from a cask of 360 and replaces with water three times -- how
much wine has he drunk?
Dodson. Math.
Repository. 1775. P. 76, Quest. CXLI. Cask of 81 gallons. (x
is drawn off and replaced by water) four times, leaving 16 gallons of
wine in the mixture. Gives a general
solution.
Ozanam‑Montucla.
1778. Prob. 21, 1778:
212-214; 1803: 207-209. Prob. 20, 1814: 179‑181; 1840: 93.
Dishonest butler (removes 1/100th
and replaces) 30 times.
Notes that it is easier to use logarithms.
Bullen. Op. cit. in
7.G.1. 1789. Chap. 38, prob. 33, p. 243.
Cask of 500 gallons;
(remove 1/10th and replace with water) five times.
Bourdon.
Algèbre. 7th ed., 1834. Art. 222, last problem, p. 364. Barrel of
100 pints of wine. One pint is drunk and replaced by water each
day. How much wine is left after 50
days? When is the wine diluted
to half its strength? One third? One quarter?
Anonymous. A
Treatise on Arithmetic in Theory and Practice: for the use of The Irish
National Schools. 3rd ed., 1850. Op. cit. in 7.H. P. 358, no. 35. Same as
Bullen.
Vinot. 1860. Art. LVI: Le Sommelier infidèle, pp.
73-74. Barrel of 100.
Sommelier drinks 1 and replaces with water 30
times. Computes the amount of
water in the barrel each day.
Todhunter. Algebra,
5th ed. 1870. Examples XXXIII, no. 8, pp. 285 & 590. From
256 gallons of wine, draw
off x
and replace with water, four times to leave only 81
gallons in the container.
M. Ph. André.
Éléments d'Arithmétique (No 3) a l'usage de toutes les
institutions .... 3rd ed., Librairie
Classique de F.-E. André-Guédon, Paris, 1876.
Prob. 98, p. 62. Barrel of wine
holding 210. Remove 45 and replace with water three times. Determine the amount of wine and water.
Mittenzwey.
1880.
Lucas.
L'Arithmétique Amusante.
1895. Prob. XLI: Le tonneau
inépuisable, p. 183. (Remove 1/100
and add water) 20 times -- how much is left? Gives general solution.
Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg,
1960 and says it was reset for the 49th ptg of 1944. Examination Papers XIX, prob. 15, p. 192. A vessel is 17% spirit. When
10 has been drawn off and
replaced with water, it is now 15 1/9 %
spirit. How big is the vessel?
Dudeney. Weekly
Dispatch (8 Feb 1903) 13. (Remove 1/100th
and replace) 30 times.
Clark. Mental
Nuts. 1904, no. 39. Find capacity of the keg. (Fill a keg from a 20 gallon cask and then replace with water) three times to dilute
the cask to half-strength. How big is
the keg?
7.S.2. WATER IN WINE VERSUS WINE IN WATER
Two containers, one of wine and one
of water. One puts an amount of water
in the wine, stirs and then transfers the same amount of the mixture back to
the water. Is there now more water in
the wine or wine in the water?
Todhunter. Algebra,
5th ed. 1870. Miscellaneous Examples, no. 187, pp. 560. Vessels of size A and B
containing wine and water respectively.
C is taken from each and then
put into the other. This is
repeated r times. Determines the
quantity of wine in the second vessel as
AB/(A+B) [1 - βr]
where β = 1
- C/A - C/B.
M. Ph. André.
Éléments d'Arithmétique (No 3) a l'usage de toutes les
institutions .... 3rd ed., Librairie
Classique de F.-E. André-Guédon, Paris, 1876.
A predecessor of this problem -- prob. 92, p. 61. Vase
A has 12 litres of wine and 4
litres of water. Vase B
has 8 litres of wine and 3 litres of water. Take off 4 litres from each and then put the 4
litres from A into
B and the four litres from B
into A. Determine contents in each vase.
Viscount John Allsebrook Simon. [Memory of Lewis Carroll.]
IN: Appendix A of Derek Hudson;
Lewis Carroll, An Illustrated Biography; Constable, (1954); new ed., 1976, pp. 248‑249. = Carroll-Wakeling II, prob. 34:
Two tumblers, pp. 52 & 73.
50 spoonfuls of brandy and
50 spoonfuls of water -- transfer
a spoonful back and forth. He says
Carroll posed this. Mentioned in
Carroll-Gardner, p. 80, who gives the full name. The DNB says he entered Wadham College, Oxford, in 1892,
and his Memory says he met Carroll then.
So this dates from ³1892, but Carroll could have
been propounding it years before.
Ball. MRE, 3rd ed.,
1896, pp. 26‑27. Water in wine
versus wine in water. He says this is a
question "which I have often propounded in past years". Not in the 1st ed of 1892.
W. P. Workman. The
Tutorial Arithmetic, op. cit. in 7.H.1.
1902. Section IX,
examples CXLV, prob. 35, pp. 427 & 544 (= 433 & 577 in c1928 ed.).
A contains 11 pt water and 7 pt
wine, B contains 5 pt water and 13 pt wine. Transfer
2 pt from A
to B and back. Find changes of
water and wine in both A &
B. This is a precursor of the
puzzle idea.
Pearson. 1907. Part II, no. 18, pp. 117 & 194‑195. Butter in lard versus lard in butter.
Loyd. Cyclopedia,
1914, pp. 287 & 378. Forty quarts
milk and forty quarts water with a quart poured back and forth. He says the ratios of milk to water are
then 1 : 40 and 40 : 1, which is correct, but isn't the usual
question.
H. E. Licks. Op.
cit. in 5.A. 1917. Art. 20, pp. 16‑17. Water and wine.
Ahrens. A&N,
1918, p. 89.
T. O'Conor Sloane.
Rapid Arithmetic. Quick and
Special Methods in Arithmetical Calculation Together with a Collection of
Puzzles and Curiosities of Numbers. Van
Nostrand, 1922. [Combined into: T.
O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures;
Van Nostrand, 1939.] Wine and water
paradox, pp. 168‑169.
F. & V. Meynell.
The Week‑End Book. Op.
cit. in 7.E. 1924. Prob. two: 2nd ed., p. 274; 5th?? ed., p. 407. Find "proportion of the amount of water in A to
the amount of milk in B."
Peano. Giochi. 1924.
Prob. 24, p. 7. Water and wine.
Loyd Jr. SLAHP. 1928.
Cheating the babies, pp. 40 & 98.
Two large cans with 10 gallons of milk and 10
of water. Pour 3
gallons back and forth.
"Have I more milk in the water can than I have water in the milk
can?" He works out that each can
has the proportion 7 9/13 : 2
4/13 [= 100 : 30 = 10 : 3].
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 25, pp. 17 & 190. Whisky and water in equal amounts. He asks about proportions rather than
amounts.
Phillips. The
Playtime Omnibus. Op. cit. in
6.AF. 1933. Section XVII, prob. 17: The two vessels, pp. 56 & 238. Same as in his Week‑End.
Abraham. 1933. Prob. 26 -- Whiskey and water, pp. 10 &
24 (7 & 112).
Perelman. FMP. c1935?
Water and wine, pp. 215 & 218.
Phillips.
Brush. 1936. Prob. D.5: Whisky and water, pp. 12 &
82. Same as in his Week‑End.
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia
Verlagsbuchhandlung, Prague, 1937.
Prob. 36, pp. 194 & 204. Man
takes a cup of black coffee. He
drinks 1/6 of it and fills it up with milk.
He then drinks 1/3 of it and fills it up with milk. He then drinks half of it and fills it up
with milk. Then he drinks the whole
cup. Has he drunk more milk than coffee
or vice-versa?
McKay. Party
Night. 1940. No. 6, pp. 176-177. Water
and wine. "I have seen, after
dinner, parties discuss the problem for a long time, ...."
Sullivan.
Unusual. 1943. Prob. 21: Mixtures. Gasoline and alcohol.
E. P. Northrop. Riddles
in Mathematics. 1944. 1944: 14-16; 1945: 13‑15; 1961:
23‑25. Water and milk.
Gamow & Stern.
1958. Gin and tonic. Pp. 103‑104.
The game takes (a, b, c, d) to (½a ‑
b½, ½b - c½, ½c ‑
d½, ½d ‑ a½). See 7.BB for other iterated functions.
C. Ciamberlini & A. Marengoni. Su una interessante curiosità numerica. Period. Mat. (4) 17 (1937) 25‑30. They attribute the problem to E. Ducci.
G [= J. Ginsburg].
Curiosa 30. An interesting
observation. SM 5 (1938) 135. Brief report on above article.
Benedict Freedman.
The four number game. SM 14
(1948) 35‑47. No references. Obtains basic results for n‑number version.
S. P. Mohanty. On a
problem of S. J. Bezuszka & M. J. Kenney on cyclic difference of pairs of
integers. Fibonacci Quarterly 19 (1981)
314‑318. (This and three of its
references are not in Meyers.)
Leroy F. Meyers.
Ducci's four‑number problem:
a short bibliography. CM 8
(1982) 262‑266. See
Ludington-Young, below, for nine additional references.
M. Gardner. Riddles
of the Sphinx and Other Mathematical Puzzle Tales. New Math. Library, MAA, 1987.
Prob. 29: Hustle off to Buffalo, parts 2‑5, pp. 134‑136, 151‑152,
160‑163. Gives a proof that most
quadruples converge to all zeroes and finds the quadruples that cycle.
Joseph W. Creely.
The length of a three-number game.
Fibonacci Quarterly 26:2 (May 1988) 141-143. Solves three and two number versions.
Stanley Bezuszka with Lou D'Angelo & Margaret J.
Kenny. The Wonder Square. Booklet 2, Boston College Math. Inst. Motivated Math Project Activity. Boston College Press, Chestnut Hill, Mass.,
1976. 32pp. Studies the process with various special numbers, e.g.
progressions, Fibonacci, Tribonacci and figurate numbers. The Tribonacci case produces starting sequences
with length n for any n.
Anne Ludington-Young.
The length of the n-number
game. Fibonacci Quarterly 28:3 (Aug
1990) 259-265. Obtains a bound and
solves some cases. Cites Meyers and 9
additional references.
What integers are non‑negative
integral combinations of a, b,
...? In particular, what is the largest
integer which is not such a combination?
This is well known for the case of two values, but remains unknown for
more values. From about 1960 onward, the
case with two values frequently occurs in number theory texts and as a puzzle
problem, but I haven't entered such appearances.
Dickson, vol. II, chap. II is uncharacteristically obscure
about this. It is generally attributed
to Frobenius (1849-1917).
J. J. Sylvester.
Math. Quest. Educ. Times 41 (1884) 21.
??NYS. Solves the problem for
two values.
Alfred Brauer. On a
problem of partitions. Amer. J. Math.
64 (1942) 299‑312. On p. 301, he
says that some of the results are due to I. Schur in a lecture in 1935,
others are due to himself and others are joint work. He cites Sylvester, but doesn't mention Frobenius.
Alfred Brauer & B. M. Seelbinder. On a problem of partitions II. Amer. J. Math. 76 (1954) 343‑346. "A number of years ago, the first of the
authors studied together with I. Schur the following problem of Frobenius:
..."
Alfred Brauer & James E. Shockley. On a problem of Frobenius. J. reine angew. Math. 210 (1962) 215‑220. "G. Frobenius, in his lectures, raised
the following problem repeatedly ....
No result was obtained for many years.
In 1935, I. Schur proved in his last lecture in Berlin the following
result ...."
Ernest S. Selmer. On
the linear diophantine problem of Frobenius.
J. reine angew. Math. 293/294 (1977) 1‑17. Gives 25 references which he believes to be
a complete list. Cites Sylvester, but
the next oldest are the 1942 and 1954 papers above. The 1962 paper above is the first to mention Frobenius in the
title.
7.V. XY = YX AND ITERATED EXPONENTIALS
D. Bernoulli. Letter
to C. Goldbach, 29 Jun 1728. In: P. H. Fuss, ed. op. cit. in 5.F.1, vol. 2,
p. 262. xy = yx in integers.
Goldbach. Letter to
D. Bernoulli, 31 Jan 1729. Ibid., pp.
280‑283. Reply to the above. Setting
y = ax, he gives an easy proof
for the only integer solutions. He says
the fractional solutions are (f/g)g/(f‑g).
L. Euler.
Introductio in Analysin Infinitorum.
Bousquet, Lausanne, 1748. Vol.
2, § 519, pp. 295-296 & Tab. XXV, fig. 103. = Introduction to the Analysis of the Infinite; trans. by
John D. Blanton; Springer, NY, 1988-1990; Book II, pp. 339-340 & 489. Gets
x = (1 + 1/n)n;
y = (1 + 1/n)n+1.
L. Euler. De
formulis exponentialibus replicatus.
Acta Acad. Sci. Petropol. 1 (1777(1778)) 38‑60. = Opera Omnia (1) 15 (1927) 268‑297. Iterated exponentials.
Ahrens. A&N,
1918, pp. 76‑78, discusses the problem and notes that Goldbach's
fractional solution is rational if and only if
g/(f‑g) is integral,
say n,
which gives
x = (1 + 1/n)n.
C. A. B. S[mith].
5-minute problem. Eureka 3 (Jan
1940) 4 & 24. Letting p, m, c
be the number of physicists, chemists and mathematicians at a lecture,
he observes that pm = mc, cm = mp, cp = pc and
m > c. One finds p = c,
hence cm = mc.
R. L. Goodstein.
Note 1725: The equation ab = ba. MG 28 (No. 279) (May 1944) 76. Quick derivation of general solution and the
rational solutions.
Anonymous. The
problems drive. Eureka 14 (Oct 1951)
12-13 & 22. No. 5. Find rational solutions of pq = qp.
R. A. Knoebel.
Exponentials reiterated. AMM 88
(1981) 235‑252. Extensive history
and bibliography.
F. Gerrish. Note
76.25: ab = ba: the positive integer solution. MG 76 (No. 477) (Nov 1992) 403. Short note on the integer case with two
recent references.
R. F. Churchhouse. Solutions
of the equation xy = yx. Bull. Inst. Math. Appl. 31:7/8 (Jul/Aug
1995) 106. Easily finds the real
solutions, then the rational and integer solutions. Notes that x = i, y = -i
is a solution!
Identical cards (or dominoes) of
length 1 can be stacked to reach out from the edge of a cliff. The simplest analysis shows that n
cards can reach out
1/2 + 1/4 + 1/6 + ... + 1/2n @ (log
n)/2. Some authors consider real
dominoes which can be piled in several orientations.
J. G. Coffin, proposer.
Problem 3009. AMM 30 (1923)
76. Asks for maximum overhang for n
cards. (Never solved!)
Max Black. Reported
in: J. F. O'Donovan; Clear thinking; Eureka 1 (Jan 1939) 15 & 20. Problem 1.
Asks for maximum extension from the bottom card with 52
cards, then for n cards.
Solution is 1/2 + 1/4 + 1/6 +
... + 1/(2n-2). For n = 52,
this is about 9/4.
A. S. Ramsey.
Statics. 2nd ed., CUP, 1941,
example 4.68, pp. 47‑48.
Discusses equal spacing with a support at the outer end, e.g. a
staircase. (Is this in the 1st ed. of
1934?? My source indicated that Ramsey
cited a Tripos exam.)
Heinrich Dörrie.
Mathematische Miniaturen.
Ferdinand Hirt, Breslau, 1943;
facsimile reprint by Martin Sändig, Wiesbaden, 1979. Prob. 240, pp. 279-282. Using cards of length 2,
he gets the extension 1/1 + 1/2
+ 1/3 + ... and shows the curve formed
is x = log (y/t), where t is the thickness of the cards
and x
and y are measured from the outer end of the top card -- in the usual
picture, they are both going negatively.
R. T. Sharp, proposer;
C. W. Trigg, solver. Problem
52. Pi Mu Epsilon J. ?? &
(April 1954) 411‑412.
Shows overhang approaches infinity, but the proposal asks for the
largest overhang for n dominoes, which is not answered. Notes that the dominoes can be angled so the
diagonal is perpendicular to the cliff edge.
This is also in Trigg; op. cit. in 5.Q; Quickie 52: Piled dominoes, pp.
17 & 99, but it still doesn't answer the proposal.
P. B. Johnson, proposer;
Michael Goldberg, Albert Wilansky, solvers. Problem E1122 -- Stacking cards.
AMM 61 (1954) 423 & 62 (1955) 123‑124. Both show overhang can go to infinity.
Paul B. Johnson.
Leaning tower of lire. Amer. J.
Physics 23 (Apr 1955) 240. Claims
harmonic series gives greatest overhang!!
P. J. Clarke. Note
2622: Statical absurdity. MG 40 (No. 333) (Oct 1956) 213‑215. Considers homogeneous bricks with weight
bounded above and below and length bounded below. Then one can take such bricks in any order to achieve an
arbitrarily large overhang.
Gamow & Stern.
1958. Building blocks. Pp. 90‑93.
D. St.P. Barnard.
Problem in The Observer, 1962.
??NYS.
D. St.P. Barnard.
Adventures in Mathematics. Chap.
8 -- The Domino Story. (1965); Funk & Wagnalls, NY, 1968, pp. 109‑122. Gets
3.969 for 13
dominoes.
Birtwistle. Math.
Puzzles & Perplexities. 1971. Overbalance, pp. 122-124. Standard piling. Shows the harmonic series diverges.
See A. K. Austin, 1972, in 5.N for a connection with this
problem.
Stephen Ainley.
Letter: Finely balanced. MG 63 (No. 426) (1979) 272. Gets
1.1679 for 4
cards.
[S. N. Collings].
Puzzle no. 47. Bull. Inst. Math.
Appl. 15 (1979) 268 & 312. Shows
that a simple counterbalancing scheme gets
m/2 for 2m ‑ 1 dominoes, so the overhang for n
dominoes is at least ½ log2(n+1).
R. E. Scraton.
Letter: A giant leap. MG 64 (No. 429) (1980) 202‑203. Discusses some history, especially Barnard's
problem.
Nick Lord, proposer;
uncertain solver. Problem 71.E. MG 71 (No. 457) (Oct 1987) 236 &
72 (No. 459) (Mar 1988) 54‑55.
Overhang can diverge even if the lengths converge to zero.
Jeremy Humphries, proposer;
various solvers. Prob. 129.5 --
Planks. M500 129 (Oct 1992) 18 &
131 (Feb 1993) 18. Uniform
planks of lengths 2, 3, 4. Find maximum overhang. Best answer is 3 5/18.
Unnamed solver, probably Jeremy Humphries. Prob. 145.2 -- Overhanging dominoes. M500 145 (?? 1995) ?? & 146 (Sep 1995) 17. He uses approximately real dominoes of
size .7 x 2.2 x 4.4 cm. Given three dominoes, how far out can you
reach? And how far away from the edge
can one domino be? Answer is basically independent
of the shape. Place one domino on its
large face with its centre of gravity at the cliff edge and its diagonal going
straightout. For maximum reach,
similarly place another domino with its centre of gravity at the far corner of
the first domino. To get a domino
maximally far from the edge, place one on edge so its largest face is parallel
to the cliff edge and with its centre of gravity at the far corner of the first
domino. In both cases, counterbalance
by putting the other domino with its centre of gravity at the inner corner of
the first domino. So the answer to the
first problem is the longest face diagonal and the answer to the second problem
is half the longest face diagonal minus half the thickness.
7.X. HOW OLD IS ANN? AND OTHER AGE PROBLEMS
The simplest age problems are 'aha'
problems, like 'Diophantos's age', going back at least to Metrodorus and will
not be considered here -- see Tropfke 575 for these. More complicated, but still relatively straightforward age
problems appear in various 16-19C arithmetic and algebra works, e.g. Schooten;
Recorde; Baker; Cocker;
Pike; American Tutor's Assistant; Ainsworth & Yeats, 1854, op.
cit. in 7.H.4; Colenso. I have included only a sample of these to
show the background and a few earlier versions. Simple problems of Form III then appear from the mid 18C and
later in standard arithmetic books, and later in puzzle books like
Hoffmann; Pearson; Loyd;
Dudeney and Loyd Jr.
These usually lead to two equations in two unknowns, a bit like 7.R, or
one equation in one unknown, depending on how one sets up the algebra. In the 19C, this problem was popular in
discussions of algebra as the problem can have a negative solution, which means
that the second time is before the first rather than after, and so the problem
was used in discussions of the existence and meaning of negative numbers --
see: Hutton, 1798?; Kelland;
De Morgan (1831, 1836, 1840).
About 1900, we get the "How
old is Ann (or Mary)" versions,
forms I and II below.
Form I: "The combined ages of Mary and Ann are 44
years, and Mary is twice as old as Ann was when Mary was half as old as
Ann will be when Ann is three times as old as Mary was when Mary was three
times as old as Ann. How old is
Ann?" Answer is 16½;
Mary is 27½. See:
Pearson; Kinnaird; Loyd;
Bain; White; Dudeney;
Loyd Jr; Grant; Ransom;
Doubleday - 2.
Form II: "Mary is 24. She is twice as old as Ann was when Mary was
as old as Ann is now. How old is
Ann?" Answer is 18.
See: Haldeman-Julius; Menninger;
Ransom; Doubleday - 2.
Other examples of this genre: Baker;
Vinot; Brooks; Gibney;
Lester; Clark; M. Adams; Meyer; Little Puzzle
Book; Dunn.
Form III‑(a, b, c). X is
now a
times as old as Y; after
b years, X is
c times as old as Y.
I.e. X = aY, X + b
= c (Y + b). This can be rephrased depending on the time
of narration -- e.g. X is now
c times as old as Y,
but b years ago, X was only
a times as old as Y.
See Clark for an equivalent problem with candles burning.
If the problem refers to both some
time ago and some time ahead, it is the same as a form usually stated about
increasing or decreasing both the numerator and denominator of a fraction X/Y,
i.e. (X-a)/(Y-a) = b; (X+c)/(Y+c) = d. See: Simpson; Dodson; Murray; Vinot. I have seen other examples of this but didn't note them.
Form IV-(a, b, c). X is
now a;
Y is now b;
when will (or was) X be c times as old as Y? I.e. a + x
= c (b + x).
General solution of Form III occurs
in: Milne; Singmaster.
General solution of Form IV occurs
in: De Morgan (1831).
Examples of type III.
a b c
10/7 2 4/3 Boy's Own Mag.
2 12 8/5 Milne
2 30 7/5 Dilworth
2 60 5/4 Dilworth
7/3 6 13/9 Milne
3 10 2 Vinot
3 10½ 2 Walkingame, 1751
3 14 2 Murray; Hummerston
3 15 2 Ladies' Diary; Mair;
Vyse; Amer. Tutor's Asst.; Eadon;
Bonnycastle, 1815; Child; Walkingame, 1835; New Sphinx;
Charades, etc.;
Pearson;
3 18 2 Dudeney
4 -2 6 Unger
4 4 3 Tate
4 5 3 Young
4 10 5/2 Tate
4 14 2 Simpson
5 5 3 Dudeney
5 5 4 Hutton, c1780?; Hutton, 1798?; Treatise;
6 3 4 Unger; Milne
6 24 2 Clark, 1897
7 3 4 Berkeley & Rowland
7 15 2 Unger
9 3 3 M. Adams
14 10 5⅓ Unger
Examples of type IV.
a b c
a b n De Morgan, 1931?
a b 3 De Morgan, 1831?
a b 4 Bourdon
32 5 10 Perelman
40 8 3 De Morgan, 1831?
40 9 2 Young
40 20 3 Kelland; Carroll-Gardner
42 12 4 Hutton, 1798?
45 12 3 Hoffmann; Dudeney
45 15 4 Bourdon
48 12 3 Vinot
48 12 7 Vinot
50 40 2 De Morgan, 1840
54 18 2 Manuel des Sorciers
56 29 2 De Morgan, 1836
62 30 5 Colenso
71 34 2 Hoffmann
71 34 3 Hoffmann
See Young World for a variant
form equivalent to
2 2 -20
Robert Recorde. The
Whetstone of Witte. John Kyngstone,
London, 1557. Facsimile by Da Capo
Press, NY & Theatrum Orbis Terrarum, Amsterdam,
1969. A question of ages, ff. Gg.i.v -
Gg.ii.r. (The gathering number at the
bottom of folio ii is misprinted G.)
Father and two sons. B = A +
2. C = A + B + 4. A + B + C = 96.
Baker. Well Spring
of Sciences. 1562? Prob. 1,
1580?: ff. 189r‑190r;
1646: pp. 297-299; 1670: pp.
340-341. A is 120. B
says if he were twice his present age, he would be as much older
than A
as A is older than B is now.
C says the same with three times
his age, D with four and E with five.
Frans van Schooten Sr.
MS algebra text, Groningen Univ. Library, Hs. 443, c1624, f. 54r. ??NYS -- reproduced, translated and
described by Jan van Maanen; The 'double--meaning' method for dating
mathematical texts; IN: Vestigia
Mathematica; ed. by M. Folkerts & J. P. Hogendijk; Rodopi, Amsterdam,
1993, pp. 253--263. "A man, wife
and child are together 96 years, that is to say the man and child
together 2 years more than the wife and the wife with the child
together 14 years more than the man.
I ask for the years of each."
Van Maanen feels that this problem probably describes the van Schooten
family and it is consistent with the family situation for a period in
1623--1624, when the man was 41, the wife
47 and the child 8,
thereby giving a reasonable date for the undated MS. Van Schooten's two sons also used the
problem in their works, but reversed the role of man and wife because it was
not common for the wife to be so much older.
Edward Cocker.
Arithmetic. Op. cit. in
7.R. 1678. Several problems, of which the following are the most
interesting.
Anonymous proposer and solver. Ladies' Diary, 1708-09
= T. Leybourn, I: 3, quest.
5. [I have a reference to this as
Question V and Leybourn also gives this number.] III‑(3, 15, 2).
"A person remarked that upon his wedding day the proportion of his
own age to that of his wife was as three to 1; but 15 years afterwards the
proportion of their ages was as 2 to 1.
What were their ages upon the day of marriage."
Also
reproduced in The Diarian Repository, 1774, Wallis 341.5 DIA, ??NX, but I
copied it as follows. [This seems to
indicate that Leybourn has condensed the material -- ??]
When
first the marriage knot was tied
Betwixt
my wife and me,
My
age to her's we found agreed
As
nine doth unto three;
But
after ten and half ten years,
We
man and wife had been,
Her
age came up as near to mine,
As
eight is to sixteen.
Now
tell me if you can, I pray,
What
was our age o' th' wedding day?
Dilworth.
Schoolmaster's Assistant. 1743.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XVI, p. 87 (1790: prob. XXII, pp. 85-86).
Essentially III‑(4, 14,
2), though the present time is halfway
between the times involved in the problem -- i.e. 7 years ago, the father
was 4
times the son's age; in 7
years, he will be 2 times.
Les Amusemens.
1749. Prob. 97, p. 239. Father is three times the age of his son,
when will he be only twice as old? -- i.e.
III-(3, b, 2) -- with
answer b = Y, the son's present age.
Walkingame. Tutor's
Assistant. 1751.
Mair. 1765?
Vyse. Tutor's
Guide. 1771?
Dodson. Math.
Repository. 1775.
Charles Hutton. A
Complete Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2. [c1780?] 1804: prob. 2,
p. 132. III-(5, 5, 4).
Bonnycastle.
Algebra. 1782. Prob. 11, p. 84 (1815: prob. 10, p.
105). A = 2B; B = 3C; A + B +
C = 140.
Pike.
Arithmetic. 1788. P. 335, no. 7. B = 3A/2;
C = 21 (A + B)/10; A + B + C =
93.
Anon. The American
Tutor's Assistant. (1st ed. is
unknown; 2nd ed., Philadelphia,
1791); 1810 ed., Joseph Crukshank,
Philadelphia. ??NYS -- quoted in: Lucas N. H. Bunt et al.; The Historical
Roots of Elementary Mathematics; Prentice-Hall, 1976; p. 33. III‑(3, 15, 2).
When
first the marriage knot was ty'd
Between
my wife and me,
My
age was to that of my bride
As
three times three to three
But
now when ten and half ten years,
We
man and wife have been,
Her
age to mine exactly bears,
As
eight is to sixteen;
Now
tell, I pray, from what I've said,
What
were our ages when we wed?
Answer:
Thy
age when marry'd must have been
Just
forty-five; thy wife's fifteen.
Eadon.
Repository. 1794. P. 297, no. 16. III-(3, 15, 2) in verse.
When
first the marriage knot was ty'd
Betwixt
my wife and me,
My
age did her's as far exceed,
As
three times three doth three;
But
after ten, and half ten years,
We
man and wife had been,
Her
age came up as near to mine,
As
eight is to sixteen.
Now
tell me (you who can) I pray,
What
were our ages on the wedding day?
Hutton. A Course of
Mathematics. 1798? Prob. 1,
1833 & 1857: 80. III-(5, 5,
4). On
1833: 224-231; 1857:
228-235, he has an extensive
discussion Remarks upon Equations of
the First Degree concerning possible
negative roots and considers IV‑(42, 12,
4) whose solution is -2.
Bonnycastle.
Algebra. 10th ed., 1815. P. 104, no. 7. III-(3, 15, 2).
L. Murray. The Young Man's
Best Companion, and Book of
General Knowledge; .... Thomas Kelly, London, (Preface dated 1814;
BL has 1821), 1824. P. 177, example
2. III-(3, 14, 2) stated at the middle of the 14 years.
Manuel des Sorciers.
1825. P. 81, art. 41. ??NX
IV-(54, 18, 2).
Augustus De Morgan.
Arithmetic and Algebra. (1831 ?). Reprinted as the second, separately paged,
part of: Library of Useful Knowledge -- Mathematics I, Baldwin & Craddock,
London, 1836. Art. 113, p. 29. IV-(40, 8, 3) -- "in how many years hence will the father's age be just
three times the son's?" Then gives
general solution of IV-(a, b, 3). Observes that a = 40, b = 18 leads to
-7 and that a = 3b
leads to 0 and discusses the meaning of both
situations. Then solves the general
case IV-(a, b, n).
Child. Girl's Own
Book. Arithmetical Puzzles. 1832: No. 8, pp. 171 & 179; 1833: No. 8, pp. 185 &
193 (answer numbered 6); 1839: No. 8,
pp. 165 & 173; 1842: No. 8, pp. 283
& 291; 1876: No. 6, pp. 231
& 244. III‑(3, 15, 2) given in verse which is the same in the
1832, 1839 and 1876 eds. (Except 1839
& 1842 have her's in line three.)
When
first the marriage knot was tied
Between
my wife and me,
My
age exceeded hers as much,
As
three times three does three.
But
when ten years and half ten years
We
man and wife had been,
Her
age approached as near to mine
As
eight is to sixteen.
The 1833 ed
has the first lines of the second verse garbled as: But when the man and wife had been, / For ten and half ten years.
=
Fireside Amusements; 1850: Prob. 7, pp. 132 &184.
Bourdon.
Algèbre. 7th ed., 1834. Art. 58, pp. 87-89. Discusses the problem IV-(a, b, 4) and the significance of negative solutions, using IV-(45, 15, 4) as an example.
Walkingame. Tutor's
Assistant. 1835. P. 180, prob. 59. III-(3, 15, 2).
When
first the marriage knot was ty'd
Between
my wife and me,
My
age did hers as far exceed,
As
three times three does three;
But
when ten years, and half ten years,
We
man and wife had been,
Her
age came then as near to mine,
As
eight is to sixteen.
= Depew;
Cokesbury Game Book; 1939; Marriage problem, pp. 207-208.
Augustus De Morgan.
On The Study and Difficulties of Mathematics. First, separately paged, part of: Library of Useful Knowledge --
Mathematics I, Baldwin & Craddock, London, 1836. P. 53. IV-(56, 29,
2) -- "when will the father be
twice as old as the son?" Answer
is -2
and he discusses the meaning of this.
Unger. Arithmetische
Unterhaltungen. 1838.
Philip Kelland. The
Elements of Algebra. A. & C. Black,
Edinburgh, et al., 1839. ??NX.
Augustus De Morgan.
Negative and impossible numbers.
IN: Penny Cyclopædia, vol. XVI,
1840, pp. 130-137. ??NX IV-(50, 40, 2) -- "at what date is (was, or will be, as the case may be)
the first twice as old as the second?"
The New Sphinx.
c1840. P. 142. III‑(3, 15, 2).
When
first the marriage knot was tied,
Betwixt
my love and me,
My
age did then her age exceed
As
three times three doth three.
But
when we ten and half ten years
We
man and wife had been,
Her
age came up as near to mine,
As
eight is to sixteen.
Solution.
-- The man was 45, the woman was 15.
T. Tate. Algebra
Made Easy. Op. cit. in 6.BF.3. 1848.
Phineas Taylor Barnum, c1848, is reputed to have given a
friend a problem involving ages: a man
aged 30 with a child aged 1 is
30 times as old as his child,
but in 30 years he is only twice as old, and in another 30
years he is only one-third older, ..., when does the child catch up with
the father? This is given in a cartoon
biography of P. T. Barnum by Walt Kelly in the early 1930s -- reproduced in: Outrageously Pogo; ed. by Mrs. Walt Kelly
& Bill Crouch Jr; Simon & Schuster, NY, 1985, pp. 14-21. Cf Abbot & Costello, 1941.
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. P. 210, No. 4. III‑(3, 15, 2). No
solution.
When first the marriage knot was tied,
betwixt my wife and me;
My
age did hers as far exceed, as three times three does three.
But
when ten years and half ten years, we man and wife had been,
Her
age came then as near to mine, as eight does to sixteen.
John H. Boardman.
Arithmetic: Rules and Reasons.
Macmillan, Cambridge, 1850. P.
98. "A. is now twice as old as
B.; eight years ago he was three
times as old, and one year more; find
the age of each."
Anonymous. A
Treatise on Arithmetic in Theory and Practice: for the use of The Irish
National Schools. 3rd ed., 1850. Op. cit. in 7.H.
John Radford Young (1799-1885). Simple Arithmetic, Algebra, and the Elements of Euclid. IN: The Mathematical Sciences, Orr's Circle
of the Sciences series, Orr & Co., London, 1854. [An apparently earlier version is described in 7.H.]
The Family Friend (1856) 298 & 329. A is
now one-fifth the age of B, but five years ago, A
was one-seventh of B's present age.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Arithmetical puzzles, no. 3, p. 173 (1868: 184). "The square of my age 60
years ago is double my present age."
Magician's Own Book.
1857. December and May, p.
246.
X + Y = 100,
Y = 4X/9. = Boy's Own
Conjuring Book, 1860, p. 216. =
Illustrated Boy's Own Treasury, 1860, prob. 19, pp. 428 & 432, but this has
no title.
Charades, Enigmas, and Riddles. 1860: prob. 17, pp. 58 & 62;
1862: prob. 16, pp. 33 & 139;
1865: prob. 560, pp. 105 & 152.
III‑(3, 15, 2). (1862
and 1865 differ very slightly in typography.)
This is essentially identical to Child.
When
first the marriage knot was tied
Between
my wife and me,
My
age exceeded her's as much,
As
three times three does Three:
But
when Ten years and half ten years
We
man and wife had been,
Her
age approached as near to mine
As
Eight is to sixteen.
Vinot. 1860.
Edward Brooks. The
Normal Mental Arithmetic A Thorough and
Complete Course by Analysis and Induction.
Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough
and Complete Course by Analysis and Induction;
Sower, Potts & Co., Philadelphia, 1873. Lots of examples. I
mention only those of some novelty to illustrate mid/late 19C texts.
(Beeton's) Boy's Own Magazine 3:4 (Apr 1889) 175 &
3:6 (Jun 1889) 255. (This is
undoubtedly reprinted from Boy's Own Magazine 1 (1863).) Mathematical question 33. III‑(10/7, 2, 4/3)
Colenso. Op. cit. in
7.H. These are from the new material of
(1864), 1871.
Lewis Carroll.
Letter of 6 Feb 1873 to Mary MacDonald.
c= Carroll-Gardner, p. 50.
Congratulates her sister Lilia "Lily" on becoming 21 and adds
"Why, last year I was double her age!
And once I was three times her age, but when that was, I
leave you to find out." I.e. IV-(40, 20, 3), cf Kelland, 1839.
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E.
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Pp. 117-118. III‑(7, 3, 4).
Hoffmann. 1893. Chap. IV, nos. 20 and 29, pp. 149 & 194
and 151 & 196 = Hoffmann‑Hordern, pp. 121 & 123. Simple age problems: IV-(71, 34, 3), IV‑(71, 34, 2),
IV-(45, 12, 3)
Clark. Mental
Nuts. 1897, no. 39. Father and child. III-(6, 24, 2).
Somerville Gibney.
So simple! The Boy's Own Paper
20 (No. 992) (15 Jan 1898) 252
& (No. 993) (22 Jan 1898)
267. "When you are as old as I am,
I shall be twice as old as you were when I was as old as you are." This only gives the ages being in the ratio
of 5
to 4.
P. Holland Lester.
Some "B.O.P." puzzles.
The Boy's Own Paper 20 (No. 1017) (9 Jul 1898) 655. "I am twice as old as you were when I
was what you are; when you are what I
am our united ages will be
63." = Vinot. "You are twice as old as I was when you
were what I am; when you are twice what
I shall be when you are twice what I am, our united ages will be 133."
Parlour Games for Everybody. John Leng, Dundee & London, nd [1903 -- BLC], p. 32: Nuts to
crack, no. 8.
A
bachelor tired of single life
Took
to himself a charming wife;
The
damsel's name was Mary Page,
The
bachelor was three times her age.
But
after fifteen years had flown
Her
husband's age was twice her own;
Now
tell me, gentle reader, pray,
Their
ages on their wedding day?
Clark. Mental Nuts.
Pearson. 1907.
Part I, no. 39,
pp. 123‑124 & 186‑187.
III-(3, 15, 2) in verse with
verse solution, very similar to previous examples.
Part II.
Wehman. New Book of
200 Puzzles. 1908. The marriage knot, p. 51. Gives only half of the problem! Seems to be copied from a four-line version
of Walkingame.
Loyd. How Old was
Mary? Cyclopedia, 1914, pp. 53 &
346. (= MPSL2, prob. 10, pp. 8 &
123.) Form I -- he says this is a
companion to his 'famous problem "How old was Ann"'. He gives other age problems.
Tell
mother's age, pp. 84 & 349‑350.
(= MPSL1, prob. 85, pp. 82 & 151.)
Pp.
216 & 367. (= MPSL2, prob. 106: How
old is Jimmy, pp. 75 & 155.)
G. G. Bain. An
Interview with Sam Loyd, 1907, op. cit. in 1, p. 777. Refers to "How old was Mary?", and gives form I as in
the Cyclopedia with slightly different wording but the same illustration.
A. C. White. Sam
Loyd and His Chess Problems. 1913. Op. cit. in 1. P. 101 calls it "How old was Mary?" and gives form
I as in the Cyclopedia.
Dudeney. AM. 1917.
Numerous problems, including the following.
Hummerston. Fun,
Mirth & Mystery. 1924. A problem in ages, Puzzle no. 85, pp. 170
& 185. III-(3,14,2).
M. Adams. Puzzles
That Everyone Can Do. 1931.
James Joyce.
Ulysses. (Dijon, 1922); Modern Library (Random House), NY, 1934,
apparently printed 1946. P. 663
(Gardner says the 1961 ed. has p. 679).
Humorous calculation assuming the ratio of ages could remain fixed
at 17
to 1. The numbers become wrong after a bit.
Ackermann.
1925.
Collins. Book of
Puzzles. 1927. How old is Jane?, pp. 73-74. Form I with Jane and Ann.
Loyd Jr. SLAHP. 1928.
How old is Ann?, pp. 4 & 87.
Gives the "original wording" as form I. He gives numerous other age problems.
Perelman. 1937. MCBF.
The equation does the thinking, p. 244.
IV-(32, 5, 10).
Haldeman-Julius.
1937.
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940. No. 88: The
monkey's mother, pp. 50 & 132-133.
Monkey on a rope over a pulley with a weight on the other side, but the
weight of the rope is equal to the age of the monkey's mother, who was twice as
old as the monkey was .... What was the
length of the rope?
R. S. S[corer].
Problem for poultry farmers.
Eureka 4 (May 1940) 4 &
5 (Jan 1941) 15. "The
chicken was twice as old when when the day before yesterday was to-morrow to‑day
was as far from Sunday as to-day will be when the day after to-morrow is
yesterday as it was when when to-morrow will be to-day when the day before
yesterday is to-morrow yesterday will be as far from Thursday as yesterday was
when to-morrow was to-day when the day after to-morrow was yesterday. On what day was the chicken hatched
out?" Solution is: Friday.
Bud Abbott and Lou Costello. Buck Privates. Universal
Pictures, 1941. Text and stills in
Richard J. Anobile, ed.; Who's On First?
Verbal and Visual Gems from the Films of Abbott & Costello; Darien
House, NY, 1972; Studio Vista, London, 1973.
Pp. 54-59. "You're forty
years old and you're in love with a little girl who's ten years old. ...
You're four times as old ....
... so you wait five years.
... You're only three times as
old .... So wait fifteen years more
.... You're only twice as old .... How long do you have to wait until you and
that little girl are the same age?"
Cf Barnum, c1848.
Clark Kinnaird. Loc.
cit. in 1. 1946. Pp. 265‑266, says Loyd Jr. is best
known for this problem (form I). On p.
267, he says it "made Sam Loyd
[Jr.] famous, although he did not originate it .... Loyd based his version upon
a similar poser which went around like a chain‑letter fad in the early
years of this century ...." He
also says 'References to "How Old
is Ann" have been found as far back as 1789 ....', but he doesn't give any
such references!
Owen Grant. Popular
Party Games. Universal, London, nd
[1940s?]. Prob. 17, pp. 39-40 &
52. Form I with Mary and Ann replaced
by Smith and Robinson.
Meyer. Big Fun
Book. 1940. No. 10, pp. 167 * 753.
"The sum of our ages is
22. I shall be 7
times as old as you are now when I become twice your age."
H. Phillips. News
Chronicle "Quiz" No. 5: Dickens.
News Chronicle, London, 1946.
Pp. 14 & 37. III-(2, 20,
6/5) phrased as seven years ago and
thirteen years hence.
Karl Menninger. (Mathematik in deiner Welt.
Vandenhoek & Ruprecht, Göttingen, 1954; revised, 1958.) Translated
as: Mathematics in Your World; G. Bell,
London, 1961. The complicated problem
of Anne and Mary, pp. 65‑66.
Gives form II with no history.
William R. Ransom.
Op. cit. in 6.M. 1955. How old is Ann?, p. 91; Mrs. M. & Miss A., p. 92. He first gives form II and says "This problem raged throughout the
United States in the early 1900's. It
was concocted by Robert D. Towne, who died at the age of 86 in 1952." He then gives form I and says "This is a much older problem, of the
same type as our "How old is Ann?" which has circulated mostly in
England."
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. P. 62: A problem of age. "When I am as old as my father is now I
shall be five times the age my son is now.
By then my son will be eight years older than I am now. The combined ages of my father and myself
total 100 years. How old is my
son?"
Young World.
c1960. P. 53: A matter of
time. In two years, X will be twice as old as she was
20 years ago.
B. D. Josephson
& J. M. Boardman. Problems drive 1961. Eureka 24 (Oct 1961) 20-22 & 24. Prob. A.
"When A was three times as old as B was the year before A was a
half of B's present age, B was 3 years younger than A was when B was two thirds
of A's present age. A's and B's ages
now total 73. How old are A and
B?"
R. L. Hutchings
& J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 20-21 & 34-35. Prob. K.
"Tom is twice as old as Dick was when Tom was half as old as Dick
will be when Tom is twice as old as Dick was when Tom was a year younger than
Dick is now. Dick is twice as old as
Tom was when Dick was half as old as Tom was when Dick was half as old as Tom
was two years ago. How old are Dick and
Tom?"
L. S. Harris
& J. M. Northover. Problems drive 1963. Eureka 26 (Oct 1963) 10-12 & 32. Prob. A.
"When A was half B's present age, B's age was the square of A's age
when B was born; the sum of their ages is a perfect cube. How old are they both now? (Take (1 year)2 = 1
year.)" Given solution is: 65, 60.
However the problem is quadratic and there is a second solution: 59 3/8,
65 5/8 for which A's age when B
was born is ‑6 1/4.
Doubleday - 2. 1971.
Angela Fox Dunn.
Mathematical Bafflers. Dover,
1980. Selected from Litton's
Problematical Recreations, which appeared from 1971, ©1964. An age problem, pp. 18 & 42. "Lottie and Lucy Hill are both 90 years
old. Mary Jones, on the other hand, is
half again as old as she was when she was half again as old as she was when she
lacked 5 years of being half as old as she is now." The solution is 90, i.e. Mary is as old as
the Hills!
David Singmaster.
Some diophantine recreations.
Op. cit. in 7.P.5. 1993. Determines when integer data in III-(a, b, c) gives integer solutions.
7.Y. COMBINING AMOUNTS AND PRICES INCOHERENTLY
Often called the applesellers'
problem or the market women's problem.
Two persons have the same number of
items which they combine. They average
the number of items per unit cost rather than the cost per item.
See Tropfke 652.
Alcuin. 9C. Prob. 6: Propositio de duobus
negotiatoribus C solidos communis habentibus. Buy 125 & 125 at 5 for 2,
costing 100, sell at
2 for 1 and 3 for 1, leaving 5 & 5 left over, so
they received 100 and the leftovers are worth 4 1/6.
Ibn Ezra, Abraham (= ibn Ezra, Abraham ben Meir = ibn Ezra, Abu Ishaq Ibrahim al‑Majid). Sefer ha‑Mispar. c1163.
Translated by Moritz Silberberg as:
Das Buch der Zahl ein
hebräisch-arithmetisches Werk; J. Kauffmann, Frankfurt a. M., 1895. P. 42.
Buys 100 for 100, sells
50 & 50 at 5 for 4
and 3 for 4, makes
6⅔. Silberberg's note 95,
p. 107, says that the the same problem occurs in Elia Misrachi, c1500.
Fibonacci.
1202. P. 281 (S: 402). Buys
100 for 100 and sells 50 & 50
at 4/5 and 4/3. What profit?
Folkerts.
Aufgabensammlungen. 13-15C. Buy
1000 birds at 4 for 1.
Sell 500 & 500 at 5
for 1 and 3 for 1, i.e. at the same
average price, but he has a profit of
16 and 2 birds. 5 sources for similar problems. Cites Alcuin and AR.
Provençale Arithmétique.
c1430. Op. cit. in 7.E. F. 113r, pp. 59-60. Buy
60 for 24, sell as x & y
at a & b to make profit of 1. How? This is actually an indeterminate question,
complicated by prices which are non-integral.
The author seems to give just one solution: x = y = 30, a = 1/2, b = 1/3.
AR. c1450. Prob. 354, pp. 154, 183, 229‑230. Buy
100 at 5 for 2, sell 50 & 50
at 3 for 1 and
2 for 1, making 1⅔.
Pacioli. De
Viribus. c1500.
Tartaglia. General
Trattato, 1556, art. 160‑162, p. 259v.
Buteo.
Logistica. 1559. Prob. 17, pp. 215-217. Buy
60 for 24, i.e. at 5 for 2.
How were they sold 'at the same price' to make 1? Sell at 2 for 1
and 3 for 1 and gain
1. Says this was proposed by
Stephanus and that the apples aren't sold at the same price as they were
bought. (H&S 53 gives Latin.)
Ozanam. 1725. Prob. 51, art. 1, 1725: 258. Buy
20 at 5 for 2; sell 10 & 9
at 2 for 1 and
3 for 1 to recover cost and
have 1
left over.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XVII, pp. 87-88 (1790: prob. XXV, p. 87).
Buy x & x at
2 for 1 and 3 for 1,
sell at 5 for 2, lose 4 -- what was x?
Vyse. Tutor's
Guide. 1771? Prob. 28, 1793: p. 57; 1799: p. 62 & Key p. 68. Buy
120 & 120 at 2 for 1 and 3 for 1, sell at
5 for 2, lose 4 -- "Pray how comes that about?"
Dodson. Math.
Repository. 1775. P. 24, Quest LXII. Same as Simpson.
Bonnycastle.
Algebra. 1782. P. 82, no. 9 (1815: pp. 101-102, no.
8). Same as Simpson.
Pike.
Arithmetic. 1788. P. 492, no. 9. Same as Simpson.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795
writer, Twickenham, 1995. P. 55. Buy
120 & 120 at 3 for 1d
and 2 for 1d. Do I gain or lose by selling at 5 for 2d?
He finds a loss of 4d. On pp. 152-153, the Editor discusses Alcuin,
where pigs are sold at three different prices, but he knows no example with
four or more different prices. On p.
163, he mentions almost the same problem, with prices multiplied by 12.
[I think he has misinterpreted Alcuin -- the left over pigs are sold at
the same prices as the others.]
Robert Goodacre.
Arithmetic & A Key to R. Goodacre's Arithmetic. 2nd ed., T. Ostell & C. Law, London,
1804. Miscellaneous Questions, no. 128,
p. 205 & Key p. 270. Buy 30 & 30 at 3 for 1 and
2 for 1, sell at 5 for 2.
Does he gain or lose?
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions. No. 21, pp. 19 & 76‑77. Buy
120 & 120 at 2 for 1
and 3 for 1, sell at
5 for 2 and discover a loss
of 4.
c= Magician's Own Book (UK version), 1871, The costermonger's puzzle,
pp. 38-39.
Boy's Own Book.
Profit and loss. 1828: 414; 1828-2: 418-419; 1829 (US): 212;
1843 (Paris): 348;
1855: 566; 1868: 669. Buy
96 & 96 at 3 for 1
and 2 for 1, sell at
5 for 2 giving 2
left over and a loss of 4 and says the loss is a fraction over 3½.
= Boy's Treasury, 1844, p. 306.
= de Savigny, 1846, p. 294: Profit et perte.
Augustus De Morgan.
Examples of the Processes of Arithmetic and Algebra. Third, separately paged, part of: Library of
Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836. Prob. 11, p. 28. Buy 150 at 3
for 1 and 100 at 2 for 1,
sell at 5 for 2 leads to no loss. But buying 150 &
150 at
3 for 1 and 2 for 1,
selling at 5 for 2 leads to a loss. "What was the reason of this?"
Hutton-Rutherford. A
Course of Mathematics. 1841? Prob. 13,
1857: 81. Buy 180 & 180 oranges at 3 for 1d and
2 for 1d. Does he gain or lose
by selling at 5 for 2d? He finds a loss of 6d.
Magician's Own Book.
1857. The astonished farmer, p.
244. Compare selling 30 & 30
at 3 for 1 and
2 for 1 versus selling 60
at 5 for 2, which makes
1 less. = Boy's Own Conjuring Book, 1860, p.
214. = Illustrated Boy's Own Treasury,
1860, prob. 16, pp. 428 & 432.
The Sociable.
1858. Prob. 35: The market
woman's puzzle, pp. 297 & 315. Same
as Jackson. = Book of 500 Puzzles,
1859, prob. 35, pp. 15 & 33. =
Wehman, New Book of 200 Puzzles, 1908, p. 29.
Book of 500 Puzzles.
1859.
[Chambers].
Arithmetic. Op. cit. in
7.H. 1866? P. 250, quest. 6.
Buy x & x at 2 for 1
and 3 for 1. Sell at
5 for 2 and lose 8d.
How many were bought?
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E. Buy
x apples at 5 for 2;
sell half at 2 for 1, the other half at 3 for 1; make 1.
Cassell's. 1881. Pp. 100‑101: The costermonger's
puzzle. Buy 120 & 120
at 2 for 1 and
3 for 1, sell at 5 for 2
and lose 4.
Hoffmann. 1893. Chap. IV, no. 6: A little miscalculation,
pp. 146 & 182 = Hoffmann‑Hordern, p. 114. Buy
120 & 120 at 4 for 1
and 6 for 1, then sell at 10 for 2.
Loyd. The lost
cent. Cyclopedia, 1914, pp. 39 &
344 (plus mention on p. 153). = MPSL1,
prob. 60 -- The missing pennies, pp. 58 & 142 = SLAHP: The apple mystery, pp. 44 & 100. Calls it also the Covent Garden Problem. Two applesellers with x & x
being sold at 2 for 1 and
3 for 1. They combine and sell
at 5 for 2, losing 7. Further the proceeds are divided equally --
how much did the 2 for 1 lady lose?
Mittenzwey. 1917:
137, pp. 26 & 76. Statement is rather
vague, but the solution is the situation of Alcuin, who is cited.
Ahrens. A&N,
1918, pp. 85‑87, discusses this problem, gives Alcuin's problem and the
following. Sell 30 & 30
at 1 for 5 (= 2 for 10) and
3 for 10, combine and
sell 60 at 5 for 20, losing
10 thereby.
Hummerston. Fun,
Mirth & Mystery. 1924. The lost pound, Puzzle no. 35, pp. 91 &
177. 30 & 30 at 3
for 1 and 2 for 1, combine and sell 60
at 5 for 2, losing
1.
Dudeney. PCP. 1932.
Prob. 17: The missing penny, pp. 18 & 129. = 536, prob. 23, pp. 8 & 229.
McKay. Party
Night. 1940. No. 26, p. 182. "A
man bought equal quantities of apples at
2 a penny and at 3 a
penny. He sold them all at 5
for twopence. Did he gain or
lose?" Does with 30 & 30.
Sullivan.
Unusual. 1947. Prob. 33: Another missing dollar. Compare selling 30 & 30 at 3 for 1 and 2 for 1 versus selling 60 at 5 for 2,
which makes 1 less.
7.Y.1. REVERSAL OF AVERAGES PARADOX
Example. Player A gets
1 for 1 and 1 for 2 while player B gets
8 for 9 and 1 for 3. A has averaged better
than B
in each part, but overall A has
2 for 3 while B
has 9 for 12 and has averaged better overall. I have now computed small examples and the
examples using the smallest integer values are: 2 for 1 and 2 for 4
versus 4 for 3 and
1 for 3 and 3 for 1
and 3 for 4 versus
4 for 2 and 1 for 2.
Allowing numerators of 0 gives a simpler? example: 2 for 1
and 1 for 3 versus
3 for 2 and 0 for 1.
By going up to values of
15, one can have A
being twice as good as B in each part but B is twice as good
overall!. E.g. 2 for 1
and 2 for 15 versus
13 for 13 and 1 for 15.
This is sometimes called Simpson's
Paradox.
New section -- having now found Hummerston,
it seems likely there are other early examples.
Hummerston. Fun,
Mirth & Mystery. 1924. Bowling averages, Puzzle no. 77, pp. 165
& 184. A & B both have
42 for 90 (wickets for runs) and
then A
does much better, getting 6 for
54 while B gets 1 for 39,
but they have the same overall average.
Morris R. Cohen & Ernest Nagel. An Introduction to Logic and Scientific
Method. Harcourt, 1934. ??NYS -- cited by Newson below. I have an abridged student's edition which
doesn't seem to have the example described by Newson.
Rupert T. Gould. The
Stargazer Talks. Geoffrey Bles, London,
1944. A Few Puzzles -- write up of a
BBC talk on 10 Jan 1939, pp. 106-113.
Cricket version. A takes
5 wickets for 30
runs, B takes
5 for 31. Then A
takes 3 for 12 and
B takes 7 for 29.
But overall B is better as the totals are 8 for 42
and 12 for 60. On p. 113, his Postscript gives another
version due to a correspondent: 28 for
60 and
28 for 60 combined with 4 for 36
and 1 for 27 give the same total averages of 32 for 96
and 29 for 87.
E. H. Simpson.
?? J. Royal Statistical Society
B, 13:2 (1951) ??NYS -- cited by Newson below.
R. L. Bolt. Class
Room Note 19: Cricket averages. MG 42 (No. 340) (May 1958) 119‑120. In cricket one says, e.g. that a bowler
takes 28 wickets for 60 runs, but one considers the runs per wicket. A
and B both take 28 wickets for
60 runs, then continue with 4 for 36 and 1 for 27. A
seems to be better than B, but both have averaged 3
runs per wicket. He gives a
clear graphic explanation of such perplexities and constructs 10 for 26
and 5 for 44 versus
10 for 22 and 10 for 78
as an example where the second is better at each stage but worse
overall.
Colin R. Blyth. On
Simpson's paradox and the sure-thing principle. J. American Statistical Association 67 (Jun 1972) 364-381. ??NYS -- cited by Gardner below.
Gardner. SA (Apr
1976) c= Time Travel, chap. 19. Time Travel gives a number of more recent
references up to 1985.
Birtwistle.
Calculator Puzzle Book.
1978. Prob. 18: Cricket
commentary, pp. 16 & 75-76. Two
bowlers have each taken 10 wickets for
70 runs, then get 1 for 15
and 2 for 26. The latter has done better in this last
match, but is worse overall.
Clifford Wagner.
Simpson's paradox in real life.
American Statistician 36:1 (Feb 1982) 46‑48. ??NYS -- cited by Gardner in Time Travel and
by Newson below.
Donald Watson. Note
72.23: Combination of ratios. MG 72 (No. 460) (Jun 1988) 126‑127. Graphical and other analyses.
Nick Lord. Note
74.11: From vectors to reversal
paradoxes. MG 74 (No. 467) (Mar
1990) 55‑58. He says the paradox
is called "Simpson's reversal paradox", but gives no reference. [Another source cites E. H. Simpson without
reference.] He discusses various
interpretations of the phenomenon.
Graham Newson.
Simpson's paradox revisited. MG
75 (No. 473) (Oct 1991) 290-293. Cites
the Simpson paper and some other recent papers, but with a lamentable lack of
details. He gives an example from
"old SMP Puzzle Corners" which deals with cricket averages in
1906-07. He quotes three examples from
Wagner, one of which is taken from Cohen & Nagel referring to incidence of
tuberculosis in 1910.
New section. The first
problem might be considered as having some resemblance to 7.E and 7.J, but the
second is novel.
Magician's Own Book.
1857. The unfair division, p.
255. Farmer is to give 2/5
of his yield to the landlord, but the farmer uses 45
bushels of the harvest before they can divide it. He then proposes to give 18
bushels to the landlord and then divide up the rest. Is this correct? = Book of 500 Puzzles, 1859, p. 64. = Boy's Own Conjuring Book, 1860, p. 225, which has some
mathematical misprints.
Clark. Mental
Nuts. 1897, no. 30. The gentleman and his tenant. Landlord and tenant are sharing a harvest
equally. They are carrying away their
shares each day. One day they each have
20 bushels on their wagons when the tenant's wagon breaks down. They shift the 20 bushels onto the
landlord's wagon and take it all to the landlord's. The landlord then says that the tenant should get an extra 20
bushels on the next day to compensate for the 20 taken to the landlord's. Is that correct?
7.Z. MISSING DOLLAR AND OTHER ERRONEOUS ACCOUNTING
The withdrawal version is the
confusion caused by adding the amounts remaining, which has no meaning.
The missing dollar version mixes
payments and refunds. E.g. three people
pay $10 each for a triple room.
The landlord decides they were overcharged and sends $5
back with the bellhop. Perplexed
by dividing $5 by 3, he appropriates $2
and refunds each person $1. Now they have paid $9 each, making $27,
and the bellhop has $2, making
$29 in all. But there was $30
originally. What happened to the
other dollar?
I have just added the swindles.
Walkingame. Tutor's
Assistant. 1751. 1777: p. 177, prob. 118; 1835, p. 180, prob. 57; 1860: p. 185, prob. 116. "If
48 taken from 120
leaves 72, and
72 taken from 91
leaves 19, and
7 taken from thence leaves 12,
what number is that, out of which, when you have taken 48, 72, 19, and 7, leaves 12?" Though this is not the same as the
withdrawal problems below, the mixing of amounts subtracted and remainders
makes me think that this kind of problem may have been the basis of the later
kind.
Mittenzwey.
1880.
Lost Dollar Puzzle.
Puzzle trade card, c1889, ??NX -- seen at Shortz's.
Sigmund Freud. Jokes
and their Relation to the Unconscious.
(As: Der Witz und seine Beziehung zum Unbewussten, 1905); translated and
edited by James Strachey; The Standard Edition of the Complete Psychological
Works of Sigmund Freud, Vol. VIII (The Hogarth Press and R&KP, 1960). The Penguin Freud Library, vol. 6, edited by
Angela Richards, Pelican, 1976, p. 94.
A gentleman entered a pastry-cook's shop and ordered a cake; but he soon
brought it back and asked for a glass of liqueur instead. He drank it and began to leave without
having paid. The proprietor detained
him. "What do you want?"
asked the customer. "You've not
paid for the liqueur." "But I
gave you the cake in exchange for it."
"You didn't pay for that either." "But I hadn't eaten it."
[The
idea of the above is not always humorous.
When I was a student in Berkeley, I went into my bank in the middle of
the day when it was quite busy and went to the first counter, near the
door. An elderly and rather shabby man
was in front of me and handed the cashier some jumbled up one dollar bills,
asking for a ten dollar bill. She
flattened them out and counted them and said there were only nine bills,
setting them between her and the man.
The man apologised, fumbled in his pockets and came out with another one
dollar bill, whereupon she put down a ten dollar bill between them. The man then pushed the pile of ten ones and
the ten toward her and asked for a twenty, which she gave him. He took it and shuffled out the door, at
which point the cashier screamed "I've been done!" But by the time a manager got to the door,
the man was nowhere to be seen. Apparently
this is a famous swindle.]
Cecil B. Read.
Mathematical fallacies. SSM 33
(1933) 575‑589. Gives a version
with $50 in the bank being withdrawn.
Withdraw $20 leaving
$30; withdraw $15
leaving $15; withdraw
$9 leaving $6;
withdraw $6 leaving
$0. But $30 + $15 + $6 = $51.
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
What happened to the shilling?, pp. 82 & 213. Three girls paying 5s each to share a room,
landlord refunds 5s, lift boy appropriates 2s.
Meyer. Big Fun
Book. 1940. Where did the dollar go?, pp. 111 & 735. 3
men paying $30. = Jerome S. Meyer; Fun-to-do; op. cit. in
5.C; 1948; prob. 71: Where did the dollar go?, pp. 55-56 & 195.
Harriet Ventress Heald.
Mathematical Puzzles. Booklet 171, Educational Research Bureau, Washington,
1941. Prob. 9, pp. 6‑7. $50
being withdrawn from a bank.
Bud Abbott and Lou Costello. Buck Privates. Universal
Pictures, 1941. Text and stills in
Richard J. Anobile, ed.; Who's On First?
Verbal and Visual Gems from the Films of Abbott & Costello; Darien
House, NY, 1972; Studio Vista, London, 1973.
Pp. 47-51. "Loan me
$50." "I've only got
$40." "Thanks, now you owe me
$10."
E. P. Northrop.
Riddles in Mathematics.
1944. 1944: 8-9; 1945: 8;
1961: 18. 3 men paying a bill of $30
or 30s.
W. A. Bagley.
Paradox Pie. Op. cit. in
6.BN. 1944. No. 12: The financier, p. 15.
Withdrawing £100.
Leeming. 1946. Chap. 3, prob. 5: What happened to the
dollar?, pp. 19‑20 & 152.
3 men paying $30.
Sullivan.
Unusual.
"Willane".
Willane's Wizardry. Academy of
Recorded Crafts, Arts and Sciences, Croydon, 1947. Easy money, p. 47. Withdrawing £100.
G. A. Briggs. Puzzle
and Humour Book. Published by the
author, Ilkley, 1966. Prob.
4/26 (b): The dishonest waiter, pp. 44 & 85.
This is the use of ‑5, ‑4, ..., 0, 1, ..., 5 instead of
0, 1, ..., 9. See Knuth in
7.AA.1 for a general discussion of positional number systems, including
negative digits, e.g. balanced ternary.
J. Colson. A short
account of negativo‑affirmative arithmetick. Philos. Trans. Roy. Soy. 34 (1726) 161‑173. He describes the use of negative digits
quite clearly. All work is done in the
decimal system. In concluding, he
mentions "the several other species, as duodecimal, sexagesimal,
centesimal, etc."
John Leslie. The
Philosophy of Arithmetic. Constable,
Edinburgh & Longman, London, 1817.
Pp. 33-34, 54, 64-65, 117, 150.
I have the 2nd ed, 1820, ??NYR.
Thomas Fowler (1777-??), of Great Torrington, Devon, built a
calculating machine using base 3 with negative digits, i.e. using the
digits 0, 1, -1 (written T). It was made of wood, 6' x 3' x 1'. He exhibited it in May 1840 at King's College London, where
Babbage, De Morgan, Airy and others came to see it. He proposed using decimal in a later machine because of the
labour of converting to and from ternary, but he suggested using balanced
decimal, which would still require conversions. Memorial window in St. Michael's Church, [Great] Torrington,
showing the machine. Sadly, neither the
machine nor any drawings for it survive, but a working partial model based on
extant descriptions was built in Aug 2000 and is on display in the Torrington
Museum and Archive. [See: John McKay & Pamela Vass; Thomas Fowler;
www.thomasfowler.org.uk . Mark Clusker;
Thomas Fowler's ternary calculating machine: how a nineteenth century
inventor's departure from decimal presaged the modern binary computer, BSHM
Newsletter 46 (Summer 2002) 2-5.]
A. L. Cauchy. CR 11
(1840) 789-798. ??NYS.
Léon Lalanne. CR 11
(1840) 903-905. ??NYS -- described in
Knuth, op. cit. in 7.AA.1. Introduces
balanced ternary. Knuth cites some mid
20C discussions of the system as a possible system for computers.
J. Halcro Johnston.
The Reverse Notation, Introducing Negative Digits with 12
as the Base. Blackie, London,
1937. ??NYS.
C. A. B. Smith. A
new way of writing numbers. Eureka 5 (Jan 1941) 7-9 & 6 (May 1941)
11. General exposition, citing Cauchy,
Johnson. Discusses the method for base
6 and gives a 7 x 6 crossnumber puzzle in this system.
Cedric A. B. Smith.
Biomathematics. (Originally by
W. M. Feldman, 1923. 3rd ed by Smith,
1954.) 4th ed, in two volumes, Hafner,
1966 & 1969. Chap. 23: Colson
notation: Arithmetic made easy, pp. 611-624.
Gives an exposition of the idea.
In the Appendix: Tables, pp. 631-632 & 651-661, he gives Colson
versions of tables of squares, common logarithms, sines and cosines, Woolf's
function (2x ln x, used in testing contingency tables) and common
logarithms of factorials
J. Halcro Johnston.
Two‑way arithmetic. MiS
1:6 (Sep 1972) 10‑12.
C. A. B. Smith.
Looking glass numbers. JRM 7
(1974) 299‑305.
E. Hillman, A. Paul & C. A. B. Smith. History of two‑way numbers &
Bibliography of two‑way numbers.
Colson News 1:4 (Dec 1984) 45‑46
& 47. The bibliography lists 14 items which are
all that are known to the authors. Additional
references in 2:1 (Mar 1985) 1.
[Smith produced about 12 issues of this newsletter devoted to unusual
number systems.]
For binary,
see 7.M.
The
Duodecimal Bulletin has regular articles discussing various bases.
Georg Cantor.
Zeitschrift für Math. und Physik 14 (1869) 121-128. ??NYS -- Knuth, below, says this is the
first general treatment of mixed base systems.
Vittorio Grünwald.
Intorno all'arithmetica dei sisteme numerici a base negativa. Giornale di Matematiche di Battaglini 23
(1885) 203‑221 & 367. ??NYS
-- cited by Glaser, op. cit. in 7.M, pp. 94 & 109..
N. G. de Bruijn. On
bases for the set of integers. Publ.
Math. Debrecen 1 (1950) 232‑242.
??NYS -- cited by Knuth & Gardner, below. Knuth says it has representations with negative bases, but
doesn't do arithmetic.
Donald E. Knuth.
Paper submitted to a Science Talent Search for high-school seniors in
1955. ??NYS -- described in Knuth,
below. Discussed negative bases and
complex bases.
G. F. Songster.
Master's thesis, Univ. of Pennsylvania, 1956. ??NYS -- Knuth, below, says it studies base -2.
Z. Pawlak & A. Wakulicz. Bull. de l'Acad. Polonaise des Sciences, Classe III, 5 (1957) 233‑236;
Série des sciences techniques 7 (1959) 713-721. ??NYS -- Knuth, below, cites this and the next item as the first
mentions of negative base arithmetic in print.
Louis R. Wadel.
Letter. IRE Transactions on
Electronic Computers EC‑6 (1957) 123.
??NYS -- cited by Knuth & Gardner, below.
W. Parry. Acta Math.
(Hungar.) 11 (1960) 401-416. ??NYS --
Knuth says he treats irrational bases.
Donald E. Knuth. The
Art of Computer Programming: Vol. 2:
Seminumerical Algorithms.
Addison-Wesley, Reading, Massachusetts, 1969. Section 4.1: Positional arithmetic, pp. 161-180 is an
exposition of various bases. The fact
that any positive number can be used as base seems to first appear in Pascal's
De numeris multiplicibus of c1658. He
suggested base 12. Erhard Wiegel proposed base 4
from 1673. Joshua Jordaine's
Duodecimal Arithmetick, London, 1687, obviously expounded base 12.
Juan Caramuel Lobkowitz's Mathesis biceps 1 (Campaniae, 1670) 45-48
discussed bases 2, 3, 4, 5, 6, 7, 8, 9,
10, 12 and 60. Charles XII of Sweden
seems to have invented base 8 c1717 and also considered base 64.
John W. Nystrom developed base
16 in: J. Franklin Inst. 46 (1863) 263-275, 337-348 & 402-407. Knuth then discusses negative and complex
bases -- see above items -- and describes bases 2i and i-1
and the use of balanced ternary and negative digits in general.
Gardner. SA (Apr
1973) c= Knotted, chap. 8. Exposits negative bases, which seem to have
been invented c1955 by Donald Knuth.
But the Addendum in Knotted cites Grünwald via Glaser. Gardner cites several other articles.
Daniel Goffinet.
Number systems with a complex base: a fractal tool for teaching
topology. AMM 98 (1991) 249-255. Explores the set of all sums of distinct
powers of the base, using base
1/2, i, etc.
Fred Newhall.
History of the duo-decimal, base
12, dozenal idea,
chronologically. The Duodecimal
Bulletin 37;:2; (No. 73; (11*2) 4‑6 [i.e. 43:2 (No. 87) (1994) 4-6]. Outline chronology with 42 entries.
Paul Gee. Mad
Hatters maths! MTg 174 (Mar 2001)
15. Students asked about using
base -2 and then base Ö2 and then base π.
This is too lengthy a subject to
cover in detail here. Below are a few
landmarks. See Dickson I, ch. I for an
extended history. Heath's notes
summarise the history. I have a separate
file on the history of Mersenne and perfect numbers.
Let
Mn = 2n - 1 and Pn = 2n‑1(2n-1). Mn is known to be prime, and hence
Pn is perfect
for n = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279,
2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497,
86243, 110503, 132049, 216091, 756839, 859433.
Paul Poulet (1918) found two
amicable chains, one starting at 12496, and Henri Cohen found seven more, one
of which starts at 14316 and has 28 links.
Euclid. (The
Thirteen Books of Euclid's Elements, edited by Sir Thomas L. Heath; 2nd ed., (CUP, 1925??); Dover, vol. 2, pp. 278, 293-294 &
421-426.)
Marcus Vitruvius [Pollo].
De Architectura. c-20. Translated by Morris Hicky Morgan as:
Vitruvius The Ten Books on
Architecture; Harvard Univ. Press, 1914 = Dover, 1960. Pp. 73-75.
Rather general discussion of special numbers and mentions six as being
perfect according to the mathematicians and gives some related properties of
ten and sixteen.
Nicomachus of Gerasa (c100). Introduction to Arithmetic.
Translated by Martin Luther D'Ooge, with studies in Greek Arithmetic by
Frank Egleston Robbins and Louis Charles Karpinski. University of Michigan Studies
Humanistic Series Vol. XVI. Macmillan, London, 1926. (The translation was also separately printed
with the same pagination, in: The
Classics of the St. John's Program; St. John's College Press, Annapolis, 1960;
special edition of 250 copies for the College.) Chap. XIV - XVI, pp. 207-212.
Defines abundant (which he calls superabundant), deficient and perfect
numbers. Gives 6, 28, 496, 8128. He implies there is one in each range (i.e. with a given number
of digits) and states that they end alternately in 6 and 8 and that Euclid's
rule gives all of them, correctly noting that
Mn must be a
prime. He seems to be the first to
claim the perfect numbers alternately end in
6 and 8.
Theon of Smyrna (c125) defines abundant and deficient
numbers.
Iamblichus. On
Nicomachus's Introduction to Arithmetic.
c325. ??NYS -- cited in Dickson
I 38. Gives first four perfect
numbers. Earliest known reference to amicable
numbers, attributed to Pythagoras, giving the first pair: 220, 284.
There was Arabic interest in these numbers. Thabit ibn Qurra (Thābit ibn Qurra)
(836-901) gave a complex rule to produce amicable numbers, but apparently could
not find any. Ibn al-Banna (Ibn al-Bannā’) (1256-1321)
discovered the second known pair of amicable numbers: 17296 and 18416,
apparently using Thabit ibn Qurra's rule. [Q. Mushtaq & A. L. Tan; Mathematics: The Islamic
Legacy; Noor Publishing House, Farashkhan, Delhi, 1993, pp. 93-94.]
Jordanus de Nemore.
De Arithmetica. c1225. An edition by Lefèvre d'Etaples was printed
in Paris in 1496. Book VII, prop.
53-60. ??NYS -- described in: Nobuo
Miura; Charles de Bovelles and perfect numbers; Historia Scientarum 34 (1988)
1-10. He knows of the existence of odd
abundant numbers, despite Dickson's assertion to the contrary.
Paolo dell'Abbaco.
Trattato di Tutta l'Arta dell'Abacho.
1339. Op. cit. in 7.E. B 2433, ff. 11r-11v discuss perfect
numbers, mentioning only 6, 28 and 496.
Chuquet. 1484. Triparty, part 1. FHM 57-59 & 360 gives English and discussion. Believes the pattern is regular, giving Pn for n = 2, 3, 5, 7, 9,
11, 13 -- but the 5th
& 6th of these are not perfect.
Pacioli. Summa. 1494.
Ff. 6v-8v. Discussion of perfect
numbers. Cites Euclid. Says
3, 7, 31, 127 are
prime and so 6, 28, 496, 8128 are perfect. Says endings alternate.
Pacioli. De
Viribus. c1500. Ff. 44v - 47r. XXVI effecto a trovare un nů pensato quando sia perfecto
(26th effect to find a number thought of if it is perfect). Gives the first five perfect numbers as 6, 28, 496, 8128, 38836. The last is actually 4 · 7 · 19 · 73 and is so far wrong that I assumed that Peirani had miscopied it,
but it is clear in the MS. We do
have 38836 = 76 M11, so it seems Pacioli erroneously thought M11 = 511 was prime, but the multiplication by 256
was corrupted into multiplication by
76, probably by shifting the
partial product by 2 into alignment with the partial product by 5.
Charles de Bovelles (Carolus Bovillus). Liber de perfectis numeris. This is a small treatise included in an
untitled collection of his works, Paris, 1510 (reprinted Stuttgart/Bad
Cannstatt, 1970), ff. 172r-180r. ??NYS
-- described in: Nobuo Miura; Charles de Bovelles and perfect numbers; Historia
Scientarum 34 (1988) 1-10. This is
probably the first printed work on number theory, but it claims that Pn is prime when n is odd!!
Gives values of the 'perfect' number for n = 1, 2, 3, 5, 7, 9, ..., 39.
Tonstall. De Arte
Supputandi. 1522. Pp. 222-223. Brief discussion of abundant, deficient and perfect numbers, only
mentioning 6 and 28.
Cardan. De Numerorum
Proprietatibus. ??, ??NYS. = Opera Omnia, vol. IV, pp 2-4, sections 4
& 5. (It is possible that this is
the first publication of this item??)
Brief discussion, mentioning 6,
28, 496, 8128 and Euclid.
Cardan. Practica
Arithmetice. 1539. Chap. 42, sections 2 & 3, ff. H.i.r -
H.i.v (p. 52). Mentions 220 & 284, then similar to above.
Robert Recorde. The
Whetstone of Witte. John Kyngstone,
London, 1557. Facsimile by Da Capo
Press, NY & Theatrum Orbis Terrarum, Amsterdam,
1969. Nombers perfecte, ff. A.iv.r
- A.iv.v. Discusses perfect numbers
briefly and asserts Pn is perfect for n = 2, 3, 5, 7, 9, 11,
13, 15.
van Etten.
1624. Prob. 70 (63), parts IIII
& VII, pp. 66-67 (92-93). Mentions
perfect numbers and says they occur for
n = 2, 3, 5, 7, 9, 11, 13,
with 486 for
496. He asserts the endings
alternate between 6 and
28. Henrion's 1630 Notte, pp. 22‑23,
refers to Euclid, corrects 486 to
496, and says someone has
recently claimed 120 is a perfect number. Deblaye, op. cit. in 1, copies 486
as 286! The 1653 English ed copies 486,
gives 120816 for
130816, extends the list
with n = 15, 17, 19 and asserts
n = 39 gives the 20th
perfect number. Describes 220
and 284.
M. Mersenne. Letter
to Descartes, 1631. (??NYS -- cited
in: Ore, Number Theory and Its History,
95 and
Dickson I 33.) Raises question
of multiply perfect numbers and states(?)
120 is 3‑perfect.
Thomas Stanley.
Pythagoras. The Ninth Part of
The History of Philosophy, (1655-1662), collected ed., 1687, pp. 491-576. Reprinted by The Philosophical Research
Society, Los Angeles, 1970. P. 552
discusses the Pythagorean attitude to perfect numbers. "... nor without reason is the
number 6 the foundation of generation, for the Greeks call it τελείov,
we perfect; because its three parts,
1/6 and 1/3
and 1/2 (that is
1, 2, and 3.) perfect it".
W. Leybourn.
Pleasure with Profit. 1694. Observation 2, p. 3. Essentially taken from the English ed. of
van Etten, with the same numerical mistakes and a further mistake in copying
the 20th case. Mentions 220, 284.
Ozanam. 1694. Prob. 5, quest. 17-19, 1696: 14-18; 1708: 13-15. Prob. 8, quest. 17-19, 1725: 29-41. Chap. 3, art. 11-12 & 15, 1778: 32-38; 1803: 35-40; 1814: 32-36;
1840: 19-21. 1696 says 2p‑1(2p-1) is perfect if 2p - 1 is
prime, but asserts this is true for p =
11. 1696 also notes that 120
is 3-perfect and gives several
amicable numbers. 1725 extends the
remarks on amicable numbers. 1778 notes
that p = 11 fails, stating that Ozanam forgot that 2p - 1 had to be a prime, and gives the first 8 perfect numbers. 1778 also gives some amicable pairs.
Manuel des Sorciers.
1825. P. 86. ??NX
Gives the "perfect numbers" corresponding to p = 2, 3, 5, 7, 9, 11, 13, giving
486 for 496
and saying the endings alternate between 6 and 8. Probably copied from van Etten. I've included this because it is
surprisingly late to be so erroneous!
B. N. I. Paganini.
Atti della Reale Accademica delle Scienze di Torino 2 (1866‑1887)
362. ??NYS. Discovery of second smallest amicable pair: 1184, 1210.
Pearson. 1907. Part II: Amicable numbers, pp. 35‑36. Asserts that 220, 284; 17296,
18416; 93 63584, 94 37056 are the only amicable pairs below 10 millions.
Alan Turing and/or colleagues was the first to use a
computer to search for new Mersenne primes on the Manchester Baby in 1949, but
it could not easily deal with numbers greater than M353.
R. M. Robinson wrote a program to search for Mersenne primes
using the Lucas-Lehmer test on the SWAC in late 1951/early 1952. It was his first program. On 30 Jan 1952. it was loaded and ran! It discovered the 13th and 14th Mersenne
primes: M521 (at about 10:00 pm, taking about a minute)
and M607 (just before midnight). M1279, M2203 and M2281 were found in the next months.
The program comprised 184 machine instructions on 24
feet of paper tape and would handle cases up through 2297.
It ran successfully on its first trial!
Lehmer was present when the program was tested on M257, which Lehmer spent some
700 hours in testing c1932,
and the program confirmed this in a fraction of a second. c1982,
Robinson ran his program on an early PC which only ran about twice as
fast as the SWAC.
Alan L. Brown.
Multiperfect numbers -- cousins of the perfect numbers. RMM 14 (Jan‑Feb 1964) 31‑39. Lists all known 3‑, 4‑, 5‑perfects and the first 100
6‑perfects.
Elvin J. Lee
& Joseph S. Madachy. The history and discovery of amicable
numbers -- Parts 1, 2, 3. JRM 5
(1972) 77‑93, 153‑173, 231‑249. Part 1 is the main history.
Parts 2 and 3 give all 1107 amicable pairs known at the time, with notes
explaining the listings.
B. L. van der Waerden.
A History of Algebra. Springer,
Berlin, 1985. Pp. 21‑23 describes
the work of Tabit ibn Qurra (824?‑901) on amicable numbers and its
development by Fermat, Descartes, Euler and Legendre.
Jan P. Hogendijk. Thābit
ibn Qurra and the pair of amicable numbers
17296, 18416. HM 12 (1985)
269-273. This pair is often named for
Fermat, who first mentions it in Europe.
Thābit gives a general rule which would yield this pair as the
second example, though he doesn't give the values. Hogendijk analyses Thābit's work and concludes that he must
have known these values. In the same
issue, a review by Hogendijk (pp. 295-296) mentions that the pair in question
was known in 14C Persia and that the pair
9363584, 9437056, usually
ascribed to Descartes, was known c1600 in Persia.
Ettore Picutti. Pour
l'histoire des septs premiers nombres parfaits. HM 16 (1989) 123-126.
7.AC. CRYPTARITHMS, ALPHAMETICS AND SKELETON ARITHMETIC
A
skeleton problem shows all the working with most digits indicated by the same
symbol, e.g. *, and only a few digits are left in place.
A cryptarithm or alphametic usually
shows just the data and the result with digits replaced by letters as in a
substitution cipher.
The opening section includes some
miscellaneous numerical-alphabetical recreations which I haven't yet classified
in subsections.
C. Dudley Langford.
Some missing figure problems and coded sums. MG 24 (No. 261) (Oct 1940) 247‑253. Lots of examples of various forms.
[J. S. Madachy?]
Alphametics. RMM 6 (Dec 1961)
27, 7 (Feb 1962) 13 &
10 (Aug 1962) 11. Historical
comments. Cites Berwick, Schuh,
Dudeney, Minos, Hunter. Says Strand
Mag. (1921) is first division with letters instead of uniform *.
"Fomalhaut".
Cryptophile cryptofile:
Cryptarithms. World Game Review
8 (Jul 1988) 5‑12. Survey of
various forms of these puzzles and related books and magazines.
Graham Hawes.
Wordplay. M500 116 (Nov 1989) 6‑7. Using the numerological mapping A = 1, B = 2, ..., Z = 26, he finds two numbers, in British usage,
whose numerological value is itself.
One is "two hundred and fifty one". (The use of 'and' is British, but not American.) He has found none in American usage. Again in British usage, the sum 73 + 89 = 162 gives a correct sum for its numerological values: 166 + 116 = 282.
7.AC.1. CRYPTARITHMS: SEND + MORE = MONEY, ETC.
American Agriculturist (Dec 1864). ??NYR -- copy sent by Shortz.
Multiplication problem where the letters for 1 ‑ 0 spell
Palmerston.
Anon. Prob. 82. Hobbies 31
(No. 797) (21 Jan 1911) 395
& (No. 800) (11 Feb 1911)
464. Multiplication laid out: PHSF * XV
= HBAKF + OFPHF_ =
OHSBKF. Solution: 7690 * 48
= 369120.
Loyd.
Cyclopedia. 1914.
Smith. Number
Stories. 1919. See 7.AC.2 for examples with full layout.
Dudeney.
Perplexities: Verbal arithmetic.
Strand Mag. (Jul 1924).
??NYS. SEND + MORE =
MONEY; EIGHT ‑ FIVE
= FOUR; TWO * TWO =
THREE; SEVEN/TWO =
TWO (with full division layout).
Dudeney. Problem ?:
The Arab's puzzle. Strand Mag. (Early
1926?). ??NX. ABCD * EFGHI
= ACGEFHIBD.
Loyd Jr. SLAHP. 1928.
Kindergarten algebra, pp. 48 & 102.
AB * AB = CDDD.
MINOS [Simon Vatriquant].
Sphinx 1 (May 1931) 50.
Introduces word "cryptarithmie". "A charming cryptarithm should (1) make sense in the
given letters as well as the solved digits,
(2) involve all the digits, (3)
have a unique solution, and
(4) be such that it can be
broken by logic, without recourse to trial and error." (Translation by C. W. Trigg in CM 4 (1978)
68.)
C. O. Oakley, proposer;
W. E. Buker, solver. Problem
E7. AMM 39 (1932) 548 ??NYS &
40 (1933) 176. SEND + MORE =
MONEY. Editorial comment in
solution cites L'Echiquier (June 1928) and Sphinx.
H. Phillips. The
Playtime Omnibus. Op. cit. in
6.AF. 1933. Section XVI, prob. 7: The money code, pp. 49‑50 &
234. SEND + MORE =
MONEY.
Rudin. 1936. No. 84, pp. 28-29 & 92. SEND + MORE
= MONEY.
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia
Verlagsbuchhandlung, Prague, 1937.
Prob. 7, pp. 188 & 197-198.
SEND + MORE = MONEY.
M. Adams. Puzzle
Book. 1939. Several straightforward examples and the following. Prob. C.38: Economy, pp. 133 &
176. SAVE + MORE =
MONEY. Four solutions given.
Alan Wayne. The
Cryptogram (c1945) (a US puzzle magazine), ??NYS. Introduces 'doubly true additions', e.g. SEVEN + SEVEN + SIX =
TWENTY. (See Trigg cited above
at MINOS.)
Alan Wayne, proposer;
A. Chulick, solver; editorial
note by Howard Eves. Problem E751 -- A
cryptarithm. AMM 54 (1947) 38 &
412‑414.
FORTY + TEN + TEN
= SIXTY. Editor cites Wayne in The Cryptogram for
several others:
SEVEN + SEVEN + SIX = TWENTY; SEVEN + THREE + TWO = TWELVE; TWENTY + FIFTY + NINE + ONE =
EIGHTY.
Morley Adams. Puzzle
Parade. Faber, London, 1948.
Jerome S. Meyer.
Fun-to-do. Op. cit. in 5.C.
1948. Prob. 27: The Cleveland butcher,
pp. 34 & 186. PORK/CHOP =
C > 2. Unique answer is 9867/3289 = 3. [Can there be answers with
C = 2??]
Anonymous. The
problems drive. Eureka 17 (Oct 1954)
8-9 & 16-17. No. 4. ONE + TWO + FOUR =
SEVEN. One solution (not unique)
given.
J. A. H. Hunter. Fun
with figures. Globe & Mail
(Toronto) (27 Oct 1955) 27 & (28 Oct 1955) 29. "It's just an easy alphametic
today." ABLE/RE =
SIR given in full diagram,
determine the value of MAIL (= 8940). Brooke (below), p. 45, reproduces Hunter's column. Brooke says: 'Hunter received a letter from a reader referring to a
"alphametical problem in which letters take the the place of
figures".'
Anonymous. Problems
drive, 1957. Eureka 20 (Oct 1957) 14-17
& 29-30. No. 2. THIS + IS = EASY in base 7.
G. J. S. Ross
& M. Westwood. Problems drive, 1960. Eureka 23 (Oct 1960) 23-25 & 26. Prob. F.
TWO + TWO = FOUR. Find the minimum
base in which this holds, and a solution.
R. L. Hutchings
& J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 20-21 & 34-35. Prob. E.
THIS + ISSO = HARD in base
eight. Four solutions.
Maxey Brooke. 150
Puzzles in Crypt‑Arithmetic. Dover,
(1963), 2nd ed. 1969. On p. 4, he
asserts that "Arithmetical Restorations" ... "were probably
invented in India during the Middle Ages", but he gives no evidence on
this point.
J. A. H. Hunter.
Note 3104: CROSS + ROADS =
DANGER. MG 48 (No. 366) (Dec
1964) 433‑434. This was posed by
E. A. Maxwell in MG (Feb 1964) 114.
Hunter finds the unique solution.
Jonathan Always.
Puzzles for Puzzlers. Tandem,
London, 1971.
Roy Childs. Letter
of 4 Aug 1999. He has used a computer
to study 'doubly-true additions' with sums up to TWENTY. In 14 cases,
there are unique solutions, of which all but the first have sum TWENTY.
THREE + THREE
+ TWO + TWO + ONE;
FIVE + THREE
+ THREE + THREE + THREE + THREE;
SEVEN + THREE
+ THREE + ONE + ONE + ONE + ONE + ONE + ONE + ONE;
SEVEN + THREE
+ THREE + TWO + TWO + ONE + ONE + ONE;
SEVEN + FIVE
+ TWO + TWO + ONE + ONE + ONE + ONE;
SEVEN + FIVE
+ TWO + TWO + TWO + ONE + ONE;
SEVEN + FIVE
+ FIVE + TWO + ONE;
SEVEN + SEVEN
+ TWO + ONE + ONE + ONE + ONE;
SEVEN + SEVEN
+ TWO + TWO + TWO;
SEVEN + SEVEN
+ SIX;
EIGHT + EIGHT
+ TWO + ONE + ONE;
ELEVEN +
THREE + THREE + ONE + ONE + ONE;
ELEVEN +
THREE + THREE + TWO + ONE;
ELEVEN +
THREE + THREE + THREE
David Singmaster.
Two wrongs can make a right.
WRONG + WRONG = RIGHT. I
find 21 solutions, two of which have
O = 0 and two others have I = 1.
Micromath 14:2 (Summer 1998) 47.
Brain jammer column, The Daily Telegraph, Weekend section, (20 Mar
1999) 27 & (27 Mar 1999) 19.
David Singmaster.
Three lefts make a right. LEFT +
LEFT + LEFT = RIGHT. I find 27
solutions, four of which use only positive digits. In no case does I = 1. Brain jammer
column, The Daily Telegraph, Weekend section, (22 May 1999) 21 &
(29 May 1999) 19.
Victor Bryant. On an
episode of Puzzle Panel in 1999?, he asked: "How is ONE + TWELVE = TWO +
ELEVEN?" Though it initially seems
like an alphametic, it is actually an ingenious anagram.
David Singmaster.
Letter to Victor Bryant, 27 Dec 2002.
I wondered if
ONE + TWELVE = TWO + ELEVEN could be made into an alphametic. This requires some repeated values as we
have to have TWE = ELE, letter by letter, so T = E,
W = L. There are 133
solutions of the resulting alphametic, e.g.
047 + 797917 = 790 + 797174. This seems to be the closest thing to a
triply-true alphametic. I thought
Victor said that
ONE + TWELVE = TWO + ELEVEN was unique, but there are six other such anagrams, such as
FOUR + SIXTEEN = SIX + FOURTEEN, though
one might regard these as fairly trivial anagrammatically. I've tried these examples to see if they
give alphametics as above. In all but
one case, the lengths differ and this rapidly leads to a contradiction. E.g., for the first case cited, we have to
have SIX = 999, FOUR = 1000
and then the units digits lead to
X = R, which is a
contradiction. (This is making the
assumption that the numbers do not have leading zeroes.) But for
FOUR + NINETEEN = NINE + FOURTEEN, the units digits give us R = E
and this forces FOUR = NINE and the problem reduces to
NINE + NINETEEN = NINE + NINETEEN which is trivial, with (10)7 = 10·9·8·7 solutions (this includes the cases with
leading zeroes, but replacing the 10 by a 9 gives the number without leading
zeroes). So this isn't really
satisfactory, but again it seems to be the best one can do.
We
also have 21 + 32 = 22 + 31, etc., as well as 20 + 31 = 21 + 30. Consider the problem as being of the form AC + BD = AD
+ BC, where the first example above would
be 21 + 32 = 22 + 31 or
A = TWENTY,
C = ONE, B =
THIRTY, D = TWO. Let │A│ be the number of letters in the English word
for A0, │C│ be the
number in C, etc., so │B│
= 6, │D│ = 3. Let │A│ be the number of letters in the English word
for A,
etc. We can assume │D│ ³ │C│. It is easily seen that any assignment of
values to letters gives an alphametic solution when │C│ = │D│. But if │D│
> │C│, then we can get
an alphametic if and only if
│A│ = │B│.
These alphametics will generally have some different letters having the
same value.
There
are also possibilities of the form 20 +
31 = 21 + 30, i.e. │C│ =
0. Similar analysis shows this gives an
alphametic solution if and only if
│A│ = │B│.
More
elaborately, we have
67 + 79 + 96 = 76 + 69 + 97 and
679 + 796 + 967 = 697 + 976 +
769. We write this out as
SIXTYSEVEN + SEVENTYNINE + NINETYSIX =
SEVENTYSIX + SIXTYNINE + NINETYSEVEN. The 0-th, 1st, 2nd, 3rd and 4th columns from
the right give no information, but the 5th column (appropriately!) gives us Y + T + E = N + T +
Y, whence we must have E = N.
The 6-th column gives
T + N + N = E + X + T, but
E = N forces E = X.
Carrying on, we get
E = I = N = S = V = X and both sides reduce to EEETYEEEEE + EEEEETYEEEE + EEEETYEEE, which has
(10)3 solutions. Considering the hyphen, -,
as a character, only shifts the argument a bit and one gets both sides
reducing to
EEETY-EEEEE + EEEEETY-EEEE + EEEETY-EEE with
(10)4 solutions. In the second case, we get the same letter
identifications and both sides of the problem reduce to EEEHUEDREDEEEETYEEEEE +
EEEEHUEDREDEEEEETYEEE + EEEEEHUEDREDEEETYEEEE,
with (10)7 solutions.
7.AC.2. SKELETON ARITHMETIC: SOLITARY SEVEN, ETC.
Anonymous, Problems drive, 1957 is the only example I have seen of the
inverse problem of starting with a known situation and finding a skeleton
problem that gives it.
W. P. Workman. The
Tutorial Arithmetic, op. cit. in 7.H.1, 1902.
See also comments in Ackermann, under Berwick, below. Chap. VI -- Examples XIX, probs. 30‑41,
pp. 48‑49 & 503 (= 50‑51 & 529 in c1928 ed.). Simple problems, e.g. prob. 30: 2982**
divided by 456 leaves remainder 1. Probs. 31‑34 are
skeleton multiplications; 35‑37
are skeleton divisions.
W. E. H. Berwick.
School World 8 (Jul & Aug 1906) 280 & 320. ??NYS.
Division with seven sevens given:
7375428413 / 125473 = 58781.
Actually, there are 13 7s
in the layout, so not all the
7s are shown. Ackermann, below, says Berwick composed
this, at age 18, after seeing some examples in Workman.
Pearson. 1907.
Anon. Sol. 38. Hobbies 30 (No. 756) (9 Apr 1910) 37 &
46. (I don't have the problem proposal
-- ??NYS.) ...1 * 7. =
6.... + .6.2._ = ....5.
with solution
8061 * 78 = 628758.
??? J. Indian Math
Club, 1910. ??NYS -- cited by Archibald
who says it has skeleton divisions with
4 digits given.
Smith. Number
Stories. 1919. Pp. 111‑112 & 139‑140. Two cryptarithmic multiplications and two
cryptarithmic divisions, but with full layouts.
W. E. H. Berwick. MG
(Mar 1920) 43. ??NYS -- cited by
Archibald. Four fours.
F. Schuh. Een
tweetal rekenkundige ardigheden. Nieuw
Tijdschrift voor Wiskunde 8 (1920‑1921) 64. Skeleton division with no digits given, but the quotient has a
repeating decimal and the divisor and dividend are relatively prime. The problem is reproduced as Note 16, AMM 28
(1921) 278, signed ARC [= R. C. Archibald], with solution by D. R. Curtiss and
comment by A. A. Bennett in AMM 29 (1922) 210‑213. Bennett shows that relatively primality is
not essential. The problem and solution
are given as Section 258: Repeating division puzzle, pp. 320‑322, in
Schuh's The Master Book of Mathematical Recreations, Dover, 1968. (Originally Wonderlijke Problemen; Leerzaam
Tijdverdrijf Door Puzzle en Spel, Thieme, Zutphen, 1943.)
7752341 / 667334 = 11.6168830001168830001....
R. C. Archibald. AMM
28 (1921) 37 -- sketches the history:
J. Indian Math Club; Berwick's
'seven sevens' and 'four fours' cited to MG (Mar 1920) 43. See The President, 1941.
W. E. H. Berwick.
Problem 555: The four fours (in Dudeney's column). Strand Mag. (1921 or 1922?) ??NYS.
Four fours given in the skeleton of
1200474 / 846 = 1419,
but there are other fours present and there are three other solutions.
Egbert F. Odling.
Problem 627: Solitary seven (in Dudeney's column). Strand Mag. (Nov & Dec 1922),
??NYS. Skeleton of 12128316 / 124 = 97809, with only the 7 given. Unique solution and the 7
only occurs once. Repeated as
Problem 1105: Skeleton sum; Strand Mag. (Jul 1932) 104 & (Aug 1932) 216.
Anon. Note 671. MG 11 (1922‑23) 338. Gives Odling's problem and solution and some
comments.
Ackermann.
1925. Pp. 109‑115. Discusses Berwick's problems, referring to
MG of Mar 1920, Dec 1921 and Jan 1922 (??NYS) for four fours, five fives,
three threes and six sixes. Gives one
of Smith's problems, Berwick's seven sevens and five fives, Schuh's problem
(attributed to Ball) and Odling's problem.
Dudeney. MP. 1926.
Prob. 70: The solitary seven, pp. 26‑27 & 118. (= 536, prob. 144, pp. 43 &
259.) Cites EFO [= Odling]. "It is the first example I have seen
... in which only one figure is given."
A. A. Bennett, proposer;
H. Langman, solver. Problem
3212. AMM 33 (1926) 429 &
34 (1927) 538-540. Skeleton
division xxxxxxxxxxxxxccfx /
xxxxabxxxx = xcxxxxx, with numerous
further positions given by definite letters.
Solution notes that some of the information is not needed, e.g. f
can be replaced by x. Answer is
70900515872010075 / 68253968253
= 1038775.
Collins. Fun with
Figures. 1928. Through a knot-hole, p. 189. *7*9* / 215
= 1** with some further figures given.
H. E. Slaught, proposer;
C. A. Rupp, solver. Problem
E1. AMM 39 (1932) 489 &
40 (1933) 111-112. Odling's
'Solitary Seven' problem. Editor's note
says it has appeared in Le Sphinx.
Perelman. FFF. 1934.
Mysterious division & Another division. 1957: probs. 104 & 105, pp. 138 & 145; 1979: probs. 107 & 108, pp. 167 &
176. = MCBF, probs. 107 & 108, pp.
168 & 178-179. Berwick's 'four
fours' and 'seven sevens', with all four solutions of the latter, rather poorly
attributed to "the American publications School World (1906) and Mathematical
Magazine (1920)".
Perelman. FMP. c1935?
Mysterious division, pp. 256 & 268‑269. Skeleton of
11268996 / 124 = 90879, with only the 7 given.
The quotient is unique, but there are
11 possible divisors: 114, 115, ..., 124, of which
115, 116 and 120 give no
other 7s in the layout.
A. G. Sillito. Note
1424: Division without figures. MG 23 (No. 257) (Dec 1939) 467‑468. Two divisions like Schuh's: 16 / 41
and 81 / 91.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940.
The President. Some
missing-figure divisions. Eureka 6 (May
1941) 21-24. Studies skeleton
divisions. Cites 'the solitary seven'
and 'the numberless decimal division' 'in one of Dudeney's Puzzle Books' --
presumably MP, 1926. Then gives six
problems, four given by Berwick in MG -- ??NYS. I is Four-threes, with complete solution. II is Three-threes, proposed by Berwick, the
same as Four-threes with one three not given, which is a harder version for
which he does not have a satisfactory solution, though the answer is the same
as for case I. III is Berwick's
Four-fours, for which Berwick gives the complete solution so here the four
answers are only stated. IV is
Berwick's Five-fives, again completely solved by Berwick with just the answer
here. V is Four-sixes, apparently
novel, with complete solution. VI is
Seven-sevens, proposed by Berwick, with complete solution here.
Sullivan.
Unusual. 1947. Prob. 37: Lost and found. Skeleton division of
1089708 / 12 = 90809 with only the 8 of the quotient given.
P. L. Chessin, proposer;
???, solver. Problem E1111. AMM 61 (Apr 1954) 712 &
???. ??NYS -- given in the Otto
Dunkel Memorial Problem Book, ed. by Howard Eves and E. P. Starke, AMM
64:7 part II (Aug-Sep 1957) 6, where it is described as the most popular
problem ever published in the AMM, with 70 solvers. Also given by Gardner, SA (May 1959) = 2nd Book, chap. 14,
prob. 5: The lonesome 8, pp. 154-155 & 160‑161. Skeleton division of 10020316 / 124 = 80809 with only the middle 8 of
the quotient given. Answer is unique.
William R. Ransom.
Op. cit. in 6.M. 1955. Only one digit known, p. 134. Skeleton of 11260316 / 124 = 90809 with only the 8 given. Answer is unique and 8
only appears once.
Anonymous. Problems
drive, 1957. Eureka 20 (Oct 1957) 14-17
& 29-30. No. 9. Gives complete layout of 1345 x 32
and asks to find five digits which would determine the rest.
G. A. Guillotte.
Note 2865: Missing digits. MG 43 (No. 345) (Oct 1959) 200. Long division with 17 0s specified.
B. D. Josephson
& J. M. Boardman. Problems drive 1961. Eureka 24 (Oct 1961) 20-22 & 24. Prob. E.
Full skeleton of 5980482 /
498 with a single 8
given in the last line.
Anonymous postcard to The Science Correspondent, "The
Glasgow Herald", 8 May 1963, found in Prof. Lenihan's copy of Gardner's
More Mathematical Puzzles and Diversions and given in Jay Books sale catalogue
129 (Feb? 1992) and 130 (Jun 1992).
Full skeleton of 1062 / 16 =
66.375 with no digits specified. The solution is unique.
L. S. Harris
& J. M. Northover. Problems drive 1963. Eureka 26 (Oct 1963) 10-12 & 32. Prob. J.
Complete skeleton of 32943 /
139 with all six 3s given.
Philip Kaplan. More
Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 69, pp. 70 & 104. Same as Sullivan, 1947.
Birtwistle. Math.
Puzzles & Perplexities. 1971.
These are generally of the form ABC + DEF
= GHI. The digits may be positive or
9 of the 10
digits. One can also have a 10‑digital form, e.g. ABC + DEF
= GHIJ.
Dudeney. Problem 64:
The lockers puzzle. Tit‑Bits 33
(18 Dec 1897 & 5 Feb 1898) 220 &
355. = AM, prob. 79, pp. 14
& 156. Find ABC + DEF
= GHI using 9 of the
10 digits which have the least
result, the greatest result and a result whose digits are distinct from the
first two. Answers: 107 + 249
= 356; 235 + 746 or 324 + 657
= 981; 134 + 586 = 720
or 134 + 568 =
702 or 138 + 269
= 407.
Dudeney. AM. 1917.
Prob. 77: Digits and squares, pp. 14 & 155. For
ABC + DEF = GHI,
he wants
DEF = 2 * ABC,
so GHI =
3 * ABC, using the 9
positive digits. Says there are
four solutions, the tops being
192, 219, 273,
327.
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 89, pp. 38 & 142: Doubles and
trebles. Same as Dudeney, AM, prob. 77.
Morley Adams. Puzzle
Parade. Op. cit. in 7.AC.1. 1948.
No. 12: Figure square, pp. 146 & 150. As in Dudeney, AM, prob. 77.
Says there are four solutions, but wants the one with minimal E.
Solution: 219 + 438 =
657.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
No. 15: Skeleton addition, pp. 10 & 41. Complete ABC + DE7 =
GH8 using all nine positive
digits. Gets four forms by
reversing A and D and/or
B and E.
Jonathan Always. Puzzles
to Puzzle You. Op. cit. in 5.K.2. 1965.
No. 135: All the numbers again, pp. 42 & 90. As in Dudeney, AM, prob. 77. Gives one solution:
192 + 384 = 576.
Ripley's Puzzles and Games.
1966. Arrange the nine positive
digits in two columns with the same sum.
He forms an X shape with
5, 9, 1, 6, 2 down one line
and 4, 8, 1, 7, 3 down the other. Both 'columns' add to 23. [Another solution is to invert the 6
or the 9.]
Wickelgren. How to
Solve Problems. Op. cit. in 5.O. 1974.
Integer‑path‑addition
problem, pp. 130‑132. Wants the
9 positive digits in a
pan-digital sum, so the 1
2 9
resulting 3 x 3 array has each
digit i horizontally or vertically adjacent +
4 3 8
to the digit i+1. Says there is one
solution, shown at the right. = 5 6 7
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 41: The
number and
the square, pp. 31 and 107. As in
Dudeney, AM, prob. 77. Gives all
solutions.
Birtwistle.
Calculator Puzzle Book.
1978. Prob. 87: Three by three
-- part two, pp. 61 & 120‑121.
Which distributions of the nine positive digits as ABC, DEF, GHI have the lowest sum and product?
147 + 258 + 369
= 774 gives the lowest sum and the digits in each position can be
permuted -- e.g. 348 + 257 + 169 gives the same sum. The lowest product is uniquely given by 147 x 258 x 369 = 13994694.
Johannes Lehmann.
Kurzweil durch Mathe. Urania
Verlag, Leipzig, 1980. No. 14, pp. 39
& 139-140. A + B + C = D
+ E + F = G + H + I has just two solutions
using the positive digits.
[Interestingly, one gets no more solutions using the ten digits.]
David Singmaster.
Determination of all pan-digital sums with two summands. JRM 27:3 (1995) 183-190. AB + CDE = FGHI has no solutions with the nine positive digits and ten basic
solutions using nine of the ten digits.
AB + CDEF = GHIJ has nine basic
solutions. (Each basic solution gives
four or eight equivalent solutions.)
ABC + DEF = 1GHI
has 12 basic solutions, which can be paired. ABC + DEF = GHI has 216
basic solutions, but 80 have A =
0. 42 cases use the positive nine
digits. The 216 can be grouped into 72
triples and a canonical example is given for each triple. The cases where the three terms form a
simple proportion are listed.
7.AC.3.a INSERTION OF SIGNS TO MAKE 100, ETC.
I include here problems like
inserting + and - (and perhaps ´ and ¸) signs into
12...9 to yield 100,
which I call 'insertion to
make 100'. This has two quite different sets of answers depending on whether
the operations are carried out sequentially (as on an old calculator) or in
algebraic order of precedence (as on a computer or modern calculator). See also 7.AC.6 for similar problems.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-16, pp. 253 & 395. Make
100 from 1, 2, ..., 9, 0. Answer:
9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1.
(0 isn't used, but could be added.)
Mittenzwey.
1880. Prob. 139, pp. 30 &
79; 1895?: 159, pp. 33 & 82; 1917: 158, pp. 31 & 79. Make
100 from 1, 2, ..., 9 using only multiplication and addition. Same answer as Leske.
Anon. & Dudeney.
A chat with the Puzzle King. The
Captain 2 (Dec? 1899) 314-320; 2:6 (Mar
1900) 598-599 & 3:1 (Apr 1900) 89. Insert as few signs as possible in 12...9 to make 100.
Usual answer is 1 + 2 + 3 +
4 + 5 + 6 + 7 + (8x9) (cf Leske), but he gives 123 ‑ 45 ‑ 67 + 89. Cf. AM, 1917.
Dudeney. AM. 1917.
Prob. 94: The digital century, pp. 16-17 & 159-160. Insert signs into 12...9 to make 100,
using: (1) as few signs as
possible; (2) as few strokes as
possible, with - counting as
1 stroke; +, ()
& ´ counting as
2; ¸ counting as
3. He finds his 1899 result is
best under both criteria. Cf. Anon
& Dudeney, 1899.
Hummerston. Fun,
Mirth & Mystery. 1924. Century making, p. 66.
1 + 2 + 3 + 4 + 5 + 6 + 7 +
(8x9) (cf Leske). 12 + 3 - 4 + 5 + 67 + 8 + 9. 123 + 45 ‑ 67 + 8
- 9. 9x8 + 7 + 6 + 5 + 4 + 3 + 2 +
1 (cf Leske; Hummerston notes this is
the reverse of his first example). 98
- 76 + 54 + 3 + 21. Also the trick
version:
15 + 36 + 47 = 98 + 2 = 100 (cf 7.A.6).
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 262, pp. 98 & 169: A number
puzzle. Insert 'mathematical signs'
into 4 3 2 1 to make 100.
Answer: Ö4 * [(3 + 2) / .1]
.
Perelman. 1934. See in 7.AC.6 for pandigital sum
yielding 1 and 100.
McKay. At Home
Tonight. 1940. Prob. 19: Centuries, pp. 66 & 80. Insertion to make 100. -1x2 - 3 - 4 - 5 +
6x7 + 8x9. -1x2 -3 - 4 + 5x6 +7 +
8x9. 1x2x3x4 + 5 + 6 + 7x8 + 9.
Anonymous. The
problems drive. Eureka 12 (Oct 1949)
7-8 & 15. No. 6. Insert symbols into 1 2 ... 9
to make 1000; 1001; 100.
Answers: 1234 -
(5+6+7+8)x9; (12 x 34) ‑ (5 x
6) + (7 x 89); 123 - 45 - 67 + 89. "These solutions are not unique."
Anonymous. Problems
drive, 1958. Eureka 21 (Oct 1958) 14-16
& 30. No. 10. Use
1, 2, 3, 4, 5, in order to
form 100; 3 1/7; 32769.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
No. 48: A tricky problem, pp. 20 & 48. "Can you replace the asterisks by the digits of the
number 216345879 in this order so the resulting total is 100?
* + * + * + * + * + *"
Answer: 2/1 + 6/3 + 4 + 5 + 8 + 79. His fraction bars are horizontal, but this
problem seems a bit unreasonable to me.
Richard E. Bellman.
On some mathematical recreations.
AMM 69:7 (Aug/Sep 1962) 640‑643. Develops a general theory for the number of ways n
can be obtained by inserting + or
x into a1a2...aN and for determining the minimum number
of +
signs occurring. He computes the
example of inserting into 12...9 to make
100 by use of recursion, finding
that
1x2x3x4 + 5 + 6 + 7x8 + 9 = 100 has the minimal number of +
signs, cf McKay.
Gardner. SA (Oct
1962) c= Unexpected, chap. 15.
Insertion to make 100. Cites Dudeney. Asks for minimum number of insertions into 98...1
to make 100.
Answer with four signs.
Gardner. SA (Jan
1965) c= Magic Numbers, chap. 6.
Considers inserting + and
- signs in 12...9
or 98..1 to yield
100. Says he posed this in SA
(Oct 1962) c= Unexpected, chap. 15 and many solutions for both the ascending
and descending series were printed in Letters in SA (Jan 1963). He gives a table of all the answers for both
cases: 11 solutions for the
ascending series and 15 solutions for the descending series. He extends the problem slightly by allowing
a -
in front of the first term and finds
1 and 3 new solutions.
Jonathan Always.
Puzzles for Puzzlers. Tandem,
London, 1971. Prob. 10: Very simple
arithmetic, pp. 14 & 62. Insert
signs into 1 2 3 4 5 6 to form an equation. He gives 12 ¸ 3 ¸ 4
+ 5 =
6. I find 1 + 2 x 3 + 4 = 5 + 6. which seems more satisfactory.
Birtwistle.
Calculator Puzzle Book.
1978. Prob. 65: Key to the
problem, pp. 47-48 & 104. Using a
calculator, insert operations in
012...9 and 98...10
to produce 100. Gives one example of each:
0 + 1/2 + 3x4x5 + 6 + 7 + 8 + 9; 9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 (cf
Leske).
Steven Kahane. Sign
in, please! JRM 23:1 (1991) 19-25. Considers inserting +
and - signs in 12...n and in
n...21 to produce various
results, e.g. 0, n+1,
n, among others.
Ken Russell & Philip Carter. Intelligent Puzzles.
Foulsham, Slough, 1992.
Loyd. Problem 36: Juggling with figures. Tit‑Bits 32 (5 Jun
& 3 Jul 1897) 173 &
258. 10‑digital product with
smallest result is
3907 * 4 = 15628.
Solution also gives the largest case:
9403 * 7 = 65821. [I have verified that these are correct.]
Dudeney. The
miller's puzzle. The Canterbury
puzzles. London Mag. 8 (No. 46) (May
1902) 367-371 & 8 (No. 47) (Jun 1902) 480-482. = CP, prob. 3, pp. 26 & 164-165. Find a solution of A*BC = D*EF
= GHI using the 9 positive digits and which is closest to a
given pattern. Answer says there are
four solutions, the closest is
2 * 78 = 4 * 39 = 156.
Dudeney. CP. 1907.
Anon. Prob. 50. Hobbies 30
(No. 765) (11 Jun 1910) 228
& (No. 768) (2 Jul 1910)
289. Same pattern as in Dudeney, 1902,
with different set-up.
M. Thié. ?? Nouv. Ann. Math. (4) 11 (1911) 46. ??NYS -- cited by Dickson I 463, item
62a. Found examples with 9
positive digits like 12 * 483 =
5796.
T. C. Lewis. ?? L'Intermédiaire des Math. 19 (1912)
26-27 & 187. ??NYS -- cited by
Dickson I 463, item 66. Examples with
the 10
digits, like 7 * 9403 =
65821 and 3 * 1458 = 6 * 0729.
Dudeney. AM. 1917.
Peano. Giochi. 1924.
Prob. 34, p. 9. Notes 2 * 78 = 39 * 4 =
156. (Cf Dudeney, 1902.)
Hummerston. Fun,
Mirth & Mystery. 1924. Grand-dad's age, Puzzle no. 68, pp. 155
& 182. A product of two 2-digit
numbers contains the same digits and is the birth date of the grandfather,
namely 21 * 87 =
1827. [There are three other
examples, but none has a product in the recent past: 15*93 = 1395; 27*81 =
2187; 35*41 = 1435. After making the first number less than or
equal to the second, eliminating leading and trailing zeroes in the factors, I
find the following numbers of solutions for factors of M, N
digits. 1, 1 : 0; 1, 2 : 3;
1, 3 : 7; 1, 4 : 36; 2, 2 : 4;
2, 3 : 41; 2, 4 : 170; 3, 3 : 119; 3, 4 : 972. Note that
0 * 0 = 00, etc. has been
eliminated. The solutions in the 1, 3
case are: 3*51 = 153; 6*21 = 126;
8*86 = 688.]
Wood. Oddities. 1927.
W. F. Cheney Jr, proposer;
Victor Thébault, solver. Problem
E13. AMM 39 (1932) 606 &
41 (1934) 265‑266. Two
factor products using all the digits just once. Gives all solutions without
0: 2 of form A*BCED
= FGHI (confirming results of Buker in AMM 40 (1933) 559 ??NYS); 7 of
form AB*CDE = FGHI. Gives some solutions with 0: 4
of form A*BCDE = FGHIJ;
3 of form AB*CDE
= FGHIJ.
Perelman. FFF. 1934.
Tricky multiplication. 1957:
prob. 45, pp. 56 & 61; 1979: prob.
48, pp. 71 & 77. = MCBF: prob. 48,
pp. 69 & 74. Gives all 9
solutions without 0 to Dudeney's AM prob. 80.
Victor Thébault, proposer;
G. H. Biucliu & L. Tits, solvers.
Mathesis 44 (1935) 205‑207.
??NYS -- described in CM 9 (1983) 89.
All 94 solutions of
n*ABCDE = FGHIJ.
Jerome S. Meyer.
Arithmetricks. Scholastic Book
Services, NY, 1965. No repeating
digits, pp. 16-17. Says A*BCDEFGHI
= abcdefghi with
A = 9 has four solutions such
that we also have 2*abcdefghi =
abcdefghij.
81274365; 72645831;
58132764; 76125483 (where the
last 3
is misprinted as 4). I find seven other examples: 58463721;
57624831; 85372461; 72534861;
83257614; 82164735; 71465328.
However this doesn't include Dudeney's example in AM 86, because it
doesn't satisfy the extra condition.
Charles L. Baker.
(Presumably in RMM. ??NYS) Reported in Madachy's Mathematical
Recreations, op. cit. in 5.O, (1966), 1979, pp. 183‑185. Confirms Thébault's (1934) and Perelman's
results without 0 and presents all two‑factor products
with 0: 13 of form A*BCED = FGHIJ; 9 of
form AB*CDE = FGHIJ.
Charles W. Trigg, proposer;
Edward Moylan, solver; David
Daykin, commenter. Problem 691 -- A
product of integers. MM 41:3 (1968)
158; 42:1 (Jan 1969) 44-45 &
42:2 (Mar 1969) 102-103.
Solve a = 8b, where
a and b together use the 9
positive digits once each. Must
have the form ABCDE =
8*FGHI and there are 46
solutions, all listed. D. Sumner
assumed that both a and
b contained all nine positive
digits and found a unique solution with
b = 123456789. Daykin gives the
number of solutions of the first problem in base β, with
multiplier m, for 2 £ m
< β £ 15, and also
considering the use of 0 as a digit and the use of just odd or just
even digits. The tables show surprising
irregularity. [Is this really surprising??]
Stewart Metchette. A
note on digital products. JRM 10 (1977‑78)
270‑271. Extends Thébault and
Baker to three factor products and gives all of the following forms: 12
of form A*BC*DE =
FGHI; 10 of form A*BC*DE
= FGHIJ; 2 of
form A*BC*DEF =
GHIJ.
Birtwistle.
Calculator Puzzle Book.
1978.
David Singmaster.
Work of 30 Jul 1998 in response to a letter of J. I. Collings. There are 99 solutions of AB*CDE = FG*HIJ, with A < F. Of these,
35 have a leading zero. Trailing zeroes lead to 13
pairs of related solutions, e.g.
23 * 760 = 76 * 230 = 95 * 184. The largest value of the common product
is
58560 = 64 * 915 = 80
* 732, as given by Dudeney, AM, prob.
82. The smallest common product is 3588
= 04 * 897 = 23
* 156, while the smallest without a
leading zero is 8262 = 18
* 459 = 27 * 306. There are two
cases with the same common product and further one of the two products is the
same:
18 * 465 = 30 * 279 = 45 * 186. Collings notes that there is only one
solution with E or
J being three and no leading
zeroes
Mittenzwey.
1880. Prob. 141 pp. 30 & 79; 1895?: 161, pp. 33 & 82;
1917: 161, pp. 31 & 79.
Use 1, 2, ..., 9 to form three fractions which add to
one. Solutions: 9/12 + 5/34 + 7/68 (cf Yoshigahara, below); 21/84 + 9/63 + 5/7; 21/48 + 7/36 + 5/9; 19/76 + 4/32 + 5/8, etc.
[Are there really others?]
Pearson. 1907. Part II: Juggling with the digits, pp. 40‑41. Examples of ABCD/EFGHI = 1/n for n = 3, 4, ..., 9.
Dudeney. AM. 1917.
Peano. Giochi. 1924.
Prob. 35, p. 10. Gives 6
solutions of 9 =
ABCDE/FGHIJ, three of which have F = 0.
Haldeman-Julius.
1937. No. 112: Half problem, pp.
18 & 26. Use the nine positive
digits to make 1/2. Answer is
7293/14586, which is similar to
Dudeney's prob. 88.
M. Adams. Puzzle
Book. 1939. Prob. B.83: Figure juggling, part 2, pp. 78 & 107. Asks for an example of using the nine
positive digits to make 1/2 and remarks that solutions exist for 1/n
with n = 3, ..., 9. Gives same solution as Haldeman-Julius.
George S. Terry. The
Dozen System. Longmans, Green &
Co., NY, 1941. ??NYS -- quoted in
Underwood Dudley; Mathematical Cranks; MAA, 1992, p. 25. Express unity as a sum of two fractions
which contain all the digits once only, duodecimally. E.g. 136/270 +
48χ/95ε = 1
(χ = 10, ε = 11). Says
about five dozen. Terry (or Dudley)
says the decimal answer is about one dozen.
Ripley's Believe It or Not, 24th series. Pocket Books, NY, 1975. P. 76.
3/6 = 7/14 = 29/58
uses all nine positive digits.
[Are there other examples or other forms??]
Michael Holt. Math
Puzzles and Games. Walker Publishing
Co., NY, (1977), PB ed., 1983. 9 in ten digits, pp. 26 & 98. ABCDE/FGHIJ = 9 has six solutions, all given.
James W. Carroll.
Letter: Computerizing Sam
Loyd. Games 7:5 (May 1983) 6. ABCD/EFGHI
= 1/n for n = 2, 3, ...,
9 has
12, 2, 4, 12, 3, 7, 46, 3 solutions.
Nob Yoshigahara.
Puzzle problem used on his TV(?) program in Japan and communicated to me
at 13th International Puzzle Party, August, 1993. Use the nine positive digits to make A/BC + D/EF + G/HI
= 1. There is a unique solution:
5/34 + 7/68 + 9/12.
7.AC.6. OTHER PAN‑DIGITAL AND SIMILAR PROBLEMS
See also 7.I
and 7.I.1 for related problems.
The Family Friend (1856) 149 & 180. Enigmas, Charades, &c. 87
Mathematical Puzzle. "Take
all the figures, (i.e., 1 2 3 4 5 6 7 8 9 0,) and place them in such a
mode, that, when they are added up, they may be equal to 100." Signed
S. W. S. Answer is 76 + 3 10/5 + 8 + 9 4/2.
Magician's Own Book.
1857. The united digits, p.
246. "Arrange the figures 1
to 9 in such order that, by adding them together, they amount to 100."
15 + 36 + 47 = 98 + 2 = 100. = Book of 500 Puzzles, 1859, p. 60. = Boy's Own Conjuring Book, 1860, p.
216. c= Parlour Games for
Everybody. John Leng, Dundee &
London, nd [1903 -- BLC], p. 41.
Leske. Illustriertes
Spielbuch für Mädchen. 1864?
Boy's Own Book. The
united digits. 1868: 429. "The figures 1 to 9
may be placed in such order that the whole added together make
exactly 100. Thus --
15 + 36 + 47 = 98 + 2 = 100."
Hanky Panky.
1872. The century of cards, p.
294. 15 + 36 + 47 = 98
+ 2 =
100 given with cards.
Mittenzwey.
1880.
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Card Puzzles No. 4:
Century addition, p. 4. Use 1, ..., 9
to add up to 100. Solutions are: 74 + 12 + 3 = 89 + 6 + 5 = 100; same as Book of 500 Puzzles; 19 + 28 + 6 = 53 + 47 =
100.
Anonymous problem proposal with solutions by K. K.;
R. Ichikawa; S. Tamano and two papers by T. Hayashi. J. of the Physics School in Tokyo 5 (1896)
82, 99-103, 153-156 & 266-267, ??NYS
Abstracted in: Yoshio Mikami,
ed.; Mathematical Papers from the Far East; AGM 28 (1910) 16-20; as: A queer
number. In base b,
we have [12...b * (b‑2)]
+ b-1 = b...21. [I wonder about
other solutions of a * x + b =
y, where a, b
are digits and x, y are pandigital expressions (either with
the 9
positive digits or all 10 digits, either separately or together)
or y
is the reversal of x, etc.]
H. D. Northrop. Popular
Pastimes. 1901. No. 17: Magical addition, pp. 69 &
74. "Arrange the figures 1 to 9,
so that by adding them together they will make 100. How can this be done?"
Solution is:
15 + 36 + 47 = 98 + 2 = 100. Cf Magician's Own Book.
T. Hayashi. On the
examination of perfect squares among numbers formed by the arrangements of the
nine effective figures. J. of the
Physics School in Tokyo 5 (1896) 203-206, ??NYS. Abstracted in: Yoshio
Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 23-25. Says Artemas Martin asked which squares
contain all nine positive digits once each and that Biddle found 29
of these. (Cites same J. 5
(1896) 171, ??NYS, for the solutions.)
[What about with all 10 digits?].
Clark. Mental
Nuts. 1897, no. 98; 1904, no. 42; 1916, no. 44. One in
addition. "Place the figures 1 2 3
4 5 6 7 8 9 0 to add 100."
Answer: 50 1/2 + 49
38/76.
Ball. MRE, 4th ed.,
1905. P. 14. Use the 10 digits to total 1 -- a solution is 35/70 + 148/296 -- or to total 100 -- a solution is 50 + 49 + 1/2 + 38/76.
Use the 9 digits to make four numbers which total 100
-- a solution is 78 + 15 + 2Ö9 +
3Ö64.
Ball. MRE, 5th ed.,
1911. Pp. 13-14. Briefly restates the material in the 4th ed.
as "questions which have been propounded in recent years. ...
To the making of such questions of this kind there is no limit, but
their solution involves little or no mathematical skill."
M. Adams. Indoor
Games. 1912. A clever arrangement, p. 353.
Same as Boy's Own Book.
Ball. MRE, 6th ed.,
1914. Pp. 13-14. Restates the material in the 5th ed. as
"... numerous empirical problems, ....
To the making of such questions there is no limit, but their solution
involves little or no mathematical skill."
He
then introduces the "Four Digits Problem". "I suggest the following problem as being more
interesting." Using the
digits 1, 2, ..., n, express the integers from 1 up
using four different digits and the operations of sum, product, positive
integral power and base-10 (or also
allowing iterated square roots and factorials). With n = 4, he can get to 88 or to 264.
With n = 5, he can get to 231 or 790.
Using 0, 1, 2, 3, he can get to 36 (or 40).
Dudeney. AM. 1917.
Prob. 13: A new money puzzle, pp. 2-3 & 148-149. States largest amount of old English money
expressible with all nine positive digits is
£98765 4s 3½d.
Asks for the smallest amount.
Answer: £2567 18s
9¾d.
Ball. MRE, 9th ed.,
1920. Pp. 13-14. In the "Four Digits Problem", he
considers n = 4, i.e. using
1, 2, 3, 4, and discusses the
operations in more detail. Using sum,
product, positive integral power and base-10
notation, he can get to 88. Allowing also finitely iterated square roots
and factorials, he can get to 264. Allowing also negative integral indices, he
can get to 276. Allowing also fractional indices, he can get
to 312. He then mentions using 0,
1, 2, 3 or four of the five digits 1, ..., 5.
Ball. MRE, 10th ed.,
1922. Pp. 13-14. In the "Four Digits Problem", he
repeats the material of the 9th ed., but at the end he adds that using all of
the five digits, 1, ..., 5, he has gotten to 3832 or 4282,
depending on whether negative and fractional indices are excluded or
allowed.
Hummerston. Fun,
Mirth & Mystery. 1924. Century making, p. 66.
15 + 36 + 47 = 98
+ 2 = 100.
Wood. Oddities. 1927.
A. B. Nordmann. One
Hundred More Parlour Tricks and Problems.
Wells, Gardner, Darton & Co., London, nd [1927 -- BMC]. No. 94: The "100" problem, pp. 88
& 114. Use 1, 2, ..., 9 to make
100. Answer: 15 + 37 + 46 =
98 + 2 = 100.
Perelman. FFF. 1934.
1957: prob. 99 & 101, pp. 137 & 144; 1979: probs. 102 & 104, pp. 166-167 & 174-175. = MCBF, probs. 102 & 104, pp. 167 &
177-178.
See: Meyers in 7.I.1 for the largest integer constructible
with various sets of numbers.
Haldeman-Julius.
1937. No. 29: The 700 problem,
pp. 6 & 22. Arrange the ten digits
so "they'll add up to 700."
Answer is: 102 4/8 +
597 3/6.
J. R. Evans. The
Junior Week‑End Book. Op. cit. in
6.AF. 1939. Prob. 32, pp. 264 & 270. Find largest and smallest amounts in pounds, shillings, pence and
farthings using the nine positive digits.
Solutions as in Dudeney.
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. P. 56: One hundred per cent. Form
100 using all 10
digits, but not in order. Gives
just one solution.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
No. 27: Another sum of money, pp. 14 & 48. Least amount of English money using all 10 digits. Answer:
£20567 18s 9¾d.
Robert Harbin. Party
Lines. Op. cit. in 5.B.1. 1963.
1, 2, 3 ... 100, p. 77. Use each of the 10 figures to make a
total of 100. Answer: 57 + 23
= 80 + 1 + 4 + 6 + 9 = 100.
Jonathan Always.
Puzzles to Puzzle You. Op. cit.
in 5.K.2. 1965. No. 91: Monetary matter, pp. 30 &
82. Find smallest amount, as in
Dudeney.
Ripley's Puzzles and Games.
1966. P. 14. Says there are five examples of
integers A, B such that A + B and
A x B have the same digits: 9, 9;
3, 24; 2, 47; 2, 263;
2, 497.
There
are infinitely more examples. One can
show that one of A, B must be a single digit, say A,
and that except for the special case
9, 9, the number of digits in A + B
must be the same as in B, say
n. There are two special
solutions with n = 1, namely
0, 0 and 2, 2.
The number of solutions for n =
2, 3, 4, 5, 6, 7 is 2, 2, 8, 29, 184,
1142. I have found solutions for each
digit A greater than one, except for
A = 4, 7 and I searched up to
nine digit numbers.
Ripley's Believe it or Not, 14th series. Pocket Books, NY, 1968. Unpaged -- about 85% of the way through. AB * CDE
gives a product of five digits which is a permutation of ABCDE. E.g.
14 * 926 = 12964. Asserts there
are just 12 "sets of figures in which the multiplicand and the
multiplier reappear in the product."
Gives six examples. Cf Wood, 1927.
Ripley's Believe it or Not, 15th series. Pocket Books, NY, 1968. Unpaged -- about 30% of the way
through. Gives 7
examples of the above situation, one of which was given above,
saying "The original figures
reappear in the results ...." Cf
Wood, 1927. They do not seem to have
considered other forms, e.g. I find 3
* 51 =
153; 6 * 21 =
126;
8 * 86 =
688.
Doubleday - 3.
1972. Sum total, pp.
129-130. Put 1, 2, 3, 5, 6, 7, 8, 9
into two groups of four with the same sum. Answer: 173 + 5 = 86
+ 92.
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 20: Make a century, pp. 20 &
99-100. Express 100
as a mixed fraction using the nine positive digits, e.g. 81
5643/297. There are 10
of this form and one other:
3 69258/714.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
Fodder for number crunchers, pp. 69 & 128. Find the 10-digit
numbers, using all 10 digits, which are divisible by 1, 2, ..., 18 [i.e. are multiples of
12252240]. There are four such
and none is divisible by 19. [I checked and found no smaller multiples
of 12252240 with distinct digits. I
found that there are 94 multiples of 9! (= 362880) with distinct digits. Of these, there are 2, 3, 16, 73 with
7, 8, 9, 10 digits. (Any multiple of 10! (= 3628800)
has its last two digits equal.)]
Scot Morris. The
Next Book of Omni Games. Op. cit. in
7.E. 1991. Pp. 55 & 192. Form
the digits into two sums: 2 + 6 + 7 +
9 and
1 + 3 + 4 + 5 + 8. Make
the sums equal by moving one number! Cf
Wood, 1927.
Nob Yoshigahara.
Puzzlart. Tokyo, 1992. Pandigital times, pp. 16 & 93. Using two digits for month, day, hour,
minute, second, one can use the nine positive digits as in 8:19:23:46:57. This happens 768 times a year -- what are the earliest and latest
such times. He gets: 3:26:17:48:59 and 9:28:17:56:43 and I have checked this. [?? -- what if we use all ten digits?]
Ed Barbeau. After
Math. Wall & Emerson, Toronto,
1995. Four problems with whole numbers,
pp. 127-130.
1. Find all integers such that the number and
its square contain all nine positive digits just once. Answers:
567 and 854.
3. Find all integers such that its cube and its
fourth power contain all ten digits just once.
Answer: 18.
Jamie & Lea Poniachik.
Cómo Jugar y Divertirse con su Inteligencia; Juegos & Co. &
Zugarto Ediciones, Argentina & Spain, 1978 & 1996. Translated by Natalia M. Tizón as:
Hard-to-Solve Brainteasers. Ed. by
Peter Gordon. Sterling, NY, 1998. Pp. 15 & 71, prob. 21: John Cash. Find a three digit number abc
with abc = 5*bc and
bc = 5*c.
7.AC.7. SELF-DESCRIPTIVE NUMBERS, PANGRAMS, ETC.
New
section. Are there older examples?
Solomon W. Golomb.
Shift Register Sequences.
Holden-Day, 1967. ??NYS -- cited
by a 1996 article, but I cannot locate the material in the revised ed., Aegean
Park Press, Laguna Hills, California, 1982; perhaps it is in some other work of
Golomb ??check. Find a non-decreasing
sequence of positive integers, (ai), i = 1, 2, ..., such that i appears
ai times. Unique answer is: 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6,
7, ....
Douglas Hofstadter.
SA (Jan 1982) c= Metamagical
Themas, Basic Books, NY, 1985, chap. 2, pp. 25-48. In Jan 1981, he had discussed
self-referential sentences, and readers sent in a number of numerical
ones.
Jonathan
Post:
This
sentence contains ten words, eighteen syllables and sixty four letters.
John Atkins:
'Has
eighteen letters' does.
Howard
Bergerson:
In
this sentence, the word and occurs twice, the word eight occurs
twice, the word four occurs twice, the word fourteen occurs four
times, the word in occurs twice, the word seven occurs twice, the
word the occurs fourteen times, the word this occurs twice, the
word times occurs seven times, the word twice occurs eight times
and the word word occurs fourteen times.
Lee Sallows:
Only
the fool would take trouble to verify that his sentence was composed of ten
a's, three b's, four c's, four d's, forty-six e's, sixteen f's, four g's,
thirteen h's, fifteen i's, two k's, nine l's, four m's, twenty-five n's,
twenty-four o's, five p's, sixteen r's, forty-one s's, thirty seven t's, ten
u's, eight v's, eight w's, four x's, eleven y's, twenty-seven commas,
twenty-three apostrophes, seven hyphens, and last, but not least, a single !
Raphael
Robinson asks to fill in the blanks in the following and says there are two
solutions:
In
this sentence, the number of occurrences of
0 is __, of 1
is __, of 2 is
__, of 3 is __,
of 4 is __, of
5 is __, of 6
is __, of 7 is
__, of 8 is __,
and of 9 is
__.
The
supplemental material in the book includes
J. K.
Aronson:
'T'
is the first, fourth, eleventh, sixteenth, twenty-fourth, twenty-ninth,
thirty-third, ....
See below for
further material.
Douglas Hofstadter.
Metamagical Themas, Basic Books, NY, 1985, chap. 3, pp. 49-69. In the Post Scriptum, pp. 68-69, he reports
on Sallows' search for a 'pangram'.
Apparently the first of the type he wants is in Dutch, by Rudy Kousbroek
and Sarah Hart: Dit pangram bevat vijf
a's, twee b's, .... After some months
search, Sallows' computer found:
This
pangram tallies five a's, one b, one c, two d's, twenty-eight e's, eight f's,
six g's, eight h's, thirteen i's, one j, one k, three l's, two m's, eighteen
n's, fifteen o's, two p's, one q, seven r's, twenty-five s's, twenty-two t's,
four u's, four v's, nine w's, two x's, four y's, and one z.
He challenges
people to compute a version starting:
This
computer-generated pangram contains ....
Douglas Hofstadter.
Metamagical Themas, Basic Books, NY, 1985, chap. 16 pp. 364-395. In the Post Scriptum, pp. 387-395, he
continues his discussion of the above material. He notes that Robinson's problem is convergent in the sense that
if one inserts a random sequence of numbers, then counts the occurrences of the
numbers and uses the counts as a new number, etc., then this iterative process usually
converges to a solution. There are two
solutions, but there is also a two term cycle and Hofstadter conjectures all
initial values converge to one of these three situations. Sallows' challenge was given in A. K.
Dewdeney's Computer Recreations column (SA, Oct 1984) and Larry Tesler used an
iterative program on it. Tesler soon
found a loop and modified the program a bit to obtain a solution:
This
computer-generated pangram contains six a's, one b, three c, three d's,
thirty-seven e's, six f's, three g's, nine h's, twelve i's, one j, one k, two
l's, three m's, twenty-two n's, thirteen o's, three p's, one q, fourteen r's,
twenty-nine s's, twenty-four t's, five u's, six v's, seven w's, four x's, five
y's, and one z.
Lee Sallows. In
Quest of a Pangram. Published by the
author, Holland, nd [1985?]. Describes
his search for a pangram.
Lee Sallows.
Reflexicons. Word Ways 25 (1992)
131-141. A 'reflexicon' is a list of
numbers and letters which specifies the number of times the letter occurs in the
list. There are two in English.
fifteen
e's, seven f's, four g's,
six h's, eight i's, four n's,
five o's, six r's, eighteen s's, eight t's, four u's, three v's,
two w's, three x's.
sixteen
e's, five f's, three g's,
six h's, nine i's, five n's,
four o's, six r's, eighteen s's, eight t's, three
u's, three v's, two w's,
four x's.
He
also discusses 'pangrams', which are sentences containing the above kind of
information -- e.g. This sentence
contains one hundred and ninety-seven letters:
four a's, .... The search for
these is described in his booklet cited above.
He then discusses crosswords using the number names.
Tony Gardiner.
Challenge! What is the title of this article? Mathematics Review 4:4 (Apr 1994) 28-29. Following on a previous article in 4:1, he
discusses self-describing sequences, where the description arises by reading
the sequence. E.g. 22
is read as 'two twos'; 31 12 33 15
is 'three ones, one two, three
threes, one five'. He also mentions
self-describing lists, e.g. 1210 contains
'one 0, two 1s, one 2, zero 3s'.
Ed Barbeau. After
Math. Wall & Emerson, Toronto,
1995. Pp. 117-122. Considers self-describing lists of
length n and shows there are only the following: 1210, 2020, 21200
and, for n > 6, (n-4)2100...001000
Lee Sallows. Problem
proposal to Puzzle Panel, 18 Jun 1998.
"How many letters would this question contain, if the answer wasn't
already seventy three?"
7.AD. SELLING, BUYING AND SELLING SAME ITEM
The problem is to determine the
profit in a series of transactions involving the same item, but there is
usually insufficient information. But
see Clark and Sullivan for an unusual answer.
Clark. Mental
Nuts. 1897, no. 1; 1904, no. 2; 1916, no. 24. The horse
question. Sell a horse for $90, buy back at $80, resell at $100.
"What did he make on the transaction?" Answer is $20, but this assumes the horse had no initial cost. If the item has no initial cost and the
prices are a, b, c, then the gain is a - b + c. But if the
question is asking for the profit, then the data are insufficient as the base
cost is not given.
Loyd. The trader's
profit. Cyclopedia, 1914, pp. 291 (no
solution). (= MPSL1, prob. 13 -- What
was the profit?, pp. 12 & 125.)
Sells a bicycle at 50, buys back at 40, sells again at 45.
Lengthy discussion of various 'solutions' of 15, 5 and 10. He says
"the President of the New York Stock Exchange was bold enough to maintain
over his own signature that the profit should be $10." Gardner points
out that there isn't enough information.
H. E. Licks. Op.
cit. in 5.A. 1917. Art. 25, p. 18. Sells at 50, buys back at 45, sells at 60
-- what is the profit? Author
labels it impossible.
Smith. Number
Stories. 1919. Pp. 126‑127 & 146. Buys at
5000, sells at 5000,
buys back at 4500, sells again at 5500.
Loyd Jr. SLAHP. 1928.
The used‑car puzzle, pp. 9‑10 & 88. Sell a used car at 100, buy back at 80,
resell at 90. "This popped into my head one morning
...." Gives arguments for profits
of 30, 10 and 20. Solution says the information is
insufficient.
Collins. Fun with
Figures. 1928. This sticks 'em up, p. 69. Buys at
$55, sells at $55,
buys back at $50, sells again at $60.
Rudin. 1936. No. 160, pp. 57 & 113. Buy for
$70, sell for $80,
buy back for $90, sell for
$100.
Sullivan.
Unusual. 1943. Prob. 2: A business transaction. Sell for
$4000, buy back for $3500,
sell again for $4500. Says the gain is $5000, composed of the
initial $4000 plus the gain on the buying and selling. This is like Clark and markedly different
than most approaches, which refer to profit.
Hubert Phillips.
Something to Think About.
Ptarmigan (Penguin), 1945.
Problem 12: Alf's bike, pp. 15 & 89. More complex version with four persons and each person's
percentage profit or loss given.
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. P. 40: Multiple choice (B). Sell a cow for $100, buy back at $90,
resell at $120. Says the profit is $30.
Viscount John Allsebrook Simon. [Memory of Lewis Carroll.]
Loc. cit. in 7.S.2.
= Carroll-Wakeling II, prob. 33: Going to the theatre, pp. 51
& 73. He says Carroll gave the
following. Man pawns 12d
for 9d and sells the ticket to a friend for 9d. Who loses and how
much? Simon relates that when he said
that the friend lost 6d, Carroll pointed out that pawnbrokers charge
interest. Mentioned in Carroll-Gardner,
p. 80, who gives the full name. The DNB
says he entered Wadham College, Oxford, in 1892, and his Memory says he met
Carroll then. So this dates from ³1892.
Don Lemon.
Everybody's Pocket Cyclopedia.
Revised 8th ed., 1890. Op. cit.
in 5.A. He doesn't have the problem,
but on p. 155 he gives current pawnbrokers' charges. These would be ½d for the ticket and ½d per 2s
or part thereof per month or part thereof. So the above will cost at least
¾d.
Smith. Number
Stories. 1919. Pp. 131 & 148. Pawn $1 for
$.75 and sell ticket for $.50.
Doesn't consider interest.
Lilian & Ashmore Russan. Old London City A
Handbook, Partly Alphabetical. Simpkin,
Marshall, Hamilton, Kent & Co., London, 1924, p. 222. Says the Bank of England once contemplated
setting up a pawn business, to charge
1d per £1 per month.
R. Ripley. Believe
It Or Not! Book 2. Op. cit. in 7.J. 1931. P. 143. "A man owed $3.00. He had a $2.00
bill, which he pawned for
$1.50, and then sold the pawn
ticket to another man for $1.50, who redeemed the $2.00 bill. Who lost?" No answer given.
H. Phillips. The
Playtime Omnibus. Op. cit. in
6.AF. 1933. Section XVII, prob. 7, pp. 54 & 236. Jones pawns
6d for 5d, then sells the ticket
to Brown for 4d. Who lost?
Doesn't consider interest.
John Paul Adams. We
Dare You to Solve This!. Op. cit. in
5.C. 1957? Prob. 52: Losers weepers, pp. 32 & 49. Pawn
$5 for $3, sell ticket to a
friend for $3 and he redeems it. Who
lost and how much? Doesn't consider
interest.
7.AE. USE OF COUNTERFEIT BILL OR FORGED CHEQUE
These problems give one or several
transactions involving a bill or cheque which is then found to be counterfeit
or forged. Who loses and how much? In the classic version of the hatter, the
straightforward answer is that he loses the value of the counterfeit bill. However, there are two values for the hat --
the sale price and the cost price, whose difference is the profit that would be
made in a normal sale. One can argue
that one has only lost the cost price of the hat, so the loss is the value of
the counterfeit bill less the expected profit.
Magician's Own Book.
1857. The unlucky hatter, p.
245. Man buys $8 hat with
counterfeit $50 bill.
"... and in almost every case the first impression is, that the
hatter lost $50 besides the hat, though it is evident he was
paid for the hat...." = Book of
500 Puzzles, 1859, p. 59. = Boy's Own
Conjuring Book, 1860, pp. 215‑216, but this spells out $ as
dollars..
Lemon. 1890. The unlucky hatter, no. 225, pp. 34 &
106. Man pays for $8
hat with counterfeit $50 bill.
"In almost every case the first impression ... is that the hatter
lost $50 beside [sic] the hat..."
Hoffmann. 1893. Chap. IV, no. 43: What did he lose?, pp. 153‑154
& 205‑206 = Hoffmann‑Hordern, p. 129. Man pays for hat with a counterfeit
bill. How much does the seller
lose? Answer says that "The reply
of most people is, almost invariably, that the hatter lost [the change] and the
value of the hat, but a little consideration will show that this is incorrect." He then says that the seller loses the amount
given in change less his profit on the goods sold; "the nett value of the hat, plus such trade profit, being
balanced by the difference ... which he retained out of the proceeds of the
note." This is wrong, as the sale
price is part of the the refund that he has to make to the person who changed
the note. Hordern notes that Hoffmann
is wrong and 'the reply of most people' is indeed correct.
Clark. Mental
Nuts. 1897, no. 2; 1904, no. 21; 1916, no. 7. The shoe
question. Boy pays for $4 pair
of shoes with counterfeit $10 bill.
Answer says he lost $6 and the pair of shoes.
Dudeney. Some much‑discussed
puzzles. Op. cit. in 2. 1908.
Dud cheque used to buy goods and get cash. "Perpetually cropping up in various guises."
H. E. Licks. Op.
cit. in 5.A. 1917. Art. 21, p. 17. Use of a counterfeit bill.
Lynn Rohrbough, ed.
Mental Games. Handy Series, Kit
E, Cooperative Recreation Service, Delaware, Ohio, 1927. Counterfeit Bill, p. 10. Man buys $6 pair of shoes with a phoney $20
bill. How much did seller lose? No solution given.
Ahrens. A&N,
1918, pp. 95‑96 gives such a problem.
Dudeney. PCP. 1932.
Prob. 34: The banker and the note, pp. 21 & 131. = 536; prob. 31: The banker and the
counterfeit bill, pp. 10‑11 & 230.
Counterfeit bill goes in a circle so no one loses.
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 15, pp. 14‑15 & 188. Forged cheque used to settle account and
receive cash.
Robert A. Streeter & Robert G. Hoehn. Are You a Genius? Vol. 1, 1932; vol. 2, 1933, Frederick A. Stokes Co., NY. Combined ed., Blue Ribbon Books, NY,
1936. Vol. 1, p. 46, no. 10:
"Brain twister". Man owes
me 40¢, he gives me a knife worth at least 60¢ and I give him 20¢.
I then find the knife was stolen and I pay the owner 75¢,
which is its value. How much
have I lost? Answer is 60¢.
This is based on my payment of
75¢ being a fair purchase, so my
loses are the 40¢ debt and the 20¢ change, which are now
irrecoverable. However, if the first
transaction is considered fair, then I've lost
75¢. Further, I might consider
the original debt as a past loss, which would reduce the present loss to 20¢
or 35¢.
Phillips. The
Playtime Omnibus. Op. cit. in
6.AF. 1933. Section XVII, prob. 5: Pinchem, pp. 53 & 236. Identical to Week‑End.
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia
Verlagsbuchhandlung, Prague, 1937.
Prob. 3, pp. 187-188 & 196-197. Sell bracelet worth
60 for phoney 100
bill, which is changed by the neighbouring shopkeeper. Author says he gets many answers,
including 140, 200
and even 340, but the right answer is 100.
Depew. Cokesbury
Game Book. 1939. Shoe dealer, p. 211. Sell shoes worth $8 for phoney $20
bill. Loss is $12
plus value of shoes.
McKay. Party
Night. 1940. No. 7, p. 177. Man buys
boots worth 15s with a bad
£1 note. Says he gets answers up to "35s and a
15s pair of boots". Notes that the neighbouring grocer who
changed the bill is irrelevant and the bootseller is simply £1
out of pocket. This ignores his
profit on the boots.
Meyer. Big Fun
Book. 1940. No. 10, pp. 171 & 754.
Same as Streeter & Hoehn.
Doubleday - 1.
1969. Prob. 8: Cash on delivery,
pp. 15 & 157. = Doubleday - 4, pp.
19-20. Forged $5 bill used to settle a
circle of debts. Solution claims
everything returns to the previous state, except for one stage -- this seems
very confused and incorrect.
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 13: The counterfeit note, pp. 17 &
98. Counterfeit note goes in a
circle. She claims all the transactions
are invalid -- but the bill has simply allowed a circuit of debts to be
cancelled and hence no one has lost and there is no reason to cancel anything.
See Tropfke 625.
There are many problems which are
based on this. Some occur in 7.H.7 and
10.A. Here I only include the most
interesting.
Bakhshali MS.
c7C. In: G. R. Kaye, The Bakhshāli
manuscript; J. Asiatic Soc. Bengal (2)
8:9 (Sep 1912) 349‑361; p. 358
and in Kaye I 43 & III
176-177, ff. 4r-5r, sutra 18 and
in Gupta. Consider two APs a, a +
b, ..., and c, c + d, ..., and
suppose the sums after n terms are equal to S. In the notation of
10.A, this is O-(a, b; c, d). Then n = 2(a ‑ c)/(d -
b) + 1. Does examples with (a, b; c, d) =
(4, 3; 6, 1); (2, 3; 3, 2) and
(5, 6; 10, 3).
Kaye
III 174, f. 4v & Gupta.
This is a problem of the same type, but most of it is lost and the
scribe seems confused. Gupta attempts
to explain the confusion as due to using the data a, b; c, d = 3, 4; 1, 2, with the rule
n = 2(c‑a)/(b-d) + 1, where the scribe takes the absolute values
of the differences rather than their signed values. In this way he gets n =
3 rather than n = -1.
Pacioli. Summa. 1494.
F. 44v, prob. 32. 1 + 2 + ... +
10½. He gets 10½ x 11½ / 2.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 182 & 263, no. 693. Man digging a well 49 feet deep. First foot costs 15, but each successive
foot costs 6 more than the previous.
Find cost of last foot and total cost.
So this is really an arithmetic progression problem, but I haven't seen
others of those using this context.
M. Ph. André.
Éléments d'Arithmétique (No 3) a l'usage de toutes les
institutions .... 3rd ed., Librairie
Classique de F.-E. André-Guédon, Paris, 1876.
Prob. 550, p. 239. How many
edges and diagonals does a convex octagon have?
(Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255 &
3:8 (Aug 1889) 351. (This is
undoubtedly reprinted from Boy's Own magazine 1 (1863).) Mathematical question 59. Seller of
12 acres asks 1
farthing for the first acre,
4 for the second acre, 16 for the third acre, .... Buyer offers £100 for the first
acre, £150 for the second acre, £200
for the third acre, .... What is
the difference in the prices asked and offered? Also entered in 7.L.
Perelman. 1937. MCBF, A team of diggers, prob. 195, pp.
372-373. A team can dig a ditch in 24
hours, but just one digger begins and then the others join in at equal
intervals, with the work finished in one interval after the last man
joined. The first man works 11
times as long as the last man.
How long did the last man work?
Perelman finds this noteworthy (and I agree) because the number of men
in the team cannot be determined!
Alcuin. 9C. Prob. 42: Propositio de scala habente gradus
centum. Computes 1 + 2 + ... + 100 as 100 + (1+99) + (2+98)
+ ... + 50.
Pacioli. Summa. 1494.
F. 44v, prob. 31. Collect 100
oranges.
Pacioli. De Viribus.
c1500.
Ff. 122v - 124r. C(apitolo)
LXXIII. D(e). levare .100. saxa a filo (To pick up 100 stones in a line). Wager on the number of steps to pick up 100
stones (or apples or nuts), one pace apart. Gives the number for
50 and 1000 stones.
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 55, f. EE.iiii.r (pp.
150-151). (The 55 is not printed in the
Opera Omnia.) Picking up 100
stones in a line. (H&S 56-57
gives Latin with English summary.)
Buteo.
Logistica. 1559. Prob. 87, pp. 299-300. Ant collecting 100 grains. (H&S 56.)
H&S 56 says this occurs in Trenchant (1566), ??NYS.
Baker. Well Spring
of Sciences. 1562? ??check if this is in the Graves copy of the
1562/1568 ed. Prob. 2, 1580?: f. 36r; 1646: p. 56; 1670: pp.
72-73. 100 stones.
van Etten.
1624. Prob. 87 (84), part IV
(8), p. 114 (184). 100 apples, eggs or stones. Henrion's Notte, p. 38, observes that there
are many arithmetical errors in prob. 87 which the reader can easily correct.
Ozanam. 1694. Prob. 7, question 6, 1696: 53; 1708: 29.
Prob. 10, question 6, 1725: 65‑66. Prob. 1, 1778: 64-65;
1803: 66-67; 1814: 59-60; 1840: 32.
100 apples, becoming stones in
1778 et seq. 1778 describes a bet based
on this process versus a straight run of the same distance.
Wells. 1698. No. 101, p. 205. 20 stones.
Dilworth.
Schoolmaster's Assistant.
1743. P. 94, no. 6. 100
stones 2 yards apart. Answer in miles, furlongs and yards.
Walkingame. Tutor's
Assistant. 1751. Arithmetical Progression, prob. 3, 1777: p.
90; 1835: p. 98; 1860: p. 118. 100 eggs a yard
apart. Answer in miles and yards.
Edmund Wingate (1596-1656).
A Plain and Familiar Method for Attaining the Knowledge and Practice of
Common Arithmetic. .... 19th ed., previous ed. by John Kersey (1616-1677)
and George Shell(e)y, now by James Dodson.
C. Hitch and L. Hawes, et al., 1760.
Quest. 44, p. 366. 100 stones a yard apart.
Mair. 1765? P. 483, ex. II. 100 eggs a yard
apart. Answer: 5 miles, 1300 yards.
Euler. Algebra. 1770.
I.III.IV: Questions for practice, no. 4, p. 139. 100
stones a yard apart.
Answer: 5 miles, 1300 yards.
Vyse. Tutor's
Guide. 1771?
Charles Hutton. A Complete
Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2.
[c1780?]
Bonnycastle.
Algebra. 1782. P. 60, no. 7 (1815: p. 76, no. 8). 100
stones, a yard apart.
Pike.
Arithmetic. 1788. P. 221, no. 3. Stones laid a yard apart over a mile, starting a yard from the
basket. Finds the travel is 1761
miles.
Bullen. Op. cit. in
7.G.1. 1789. Chap. 30, prob. 6, pp. 213‑214. 100 stones, at
2 yard intervals. Converts to miles, furlongs and yards.
Eadon.
Repository. 1794.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795
writer, Twickenham, 1995. P. 100. 100
eggs a yard apart. He also gives
a variation: 100 sheep are priced in arithmetic progression,
with the first costing 1s and the last costing £9 19s (= 199s); what do the sheep cost all together?
Hutton. A Course of
Mathematics. 1798? Prob. 7,
1833: 277; 1857: 313. 100
stones 2 yards apart. Gives answer in miles and yards.
Manuel des Sorciers.
1825. Pp. 83-84. ??NX
120 stones 6 feet apart.
Endless Amusement II.
1826? Pp. 115-116: "If a hundred Stones ...."
Boy's Own Book. The
basket and stones. 1828: 176; 1828-2: 239; 1829 (US): 107;
1843 (Paris): 342;
1855: 394; 1868: 432. 100
stones a yard apart. = Boy's
Treasury, 1844, pp. 299-300. = de
Savigny, 1846, pp. 290-291: Le panier et les petites pierrés, using mètres
instead of yards, except that it ends with '10,100 mètres, ou 21 kilomètres' --
??
Bourdon.
Algèbre. 7th ed., 1834. Art. 190, question 5, p. 319. Loads of sand (or grit) have to be
delivered, one load at a time, to
100 vehicles in a line, 6 meters apart, from a pile 40 meters from the end of the line.
Hutton-Rutherford. A
Course of Mathematics. 1841? Prob. 24,
1857: 82. 100 eggs a yard apart.
Nuts to Crack XIV (1845), no. 76. The basket and stones.
Almost identical to Boy's Own Book.
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. P. 197: "The following is a favorite
old problem." 100 stones, a yard apart.
John Radford Young (1799-1885). Simple Arithmetic, Algebra, and the Elements of Euclid. IN: The Mathematical Sciences, Orr's Circle
of the Sciences series, Orr & Co., London, 1854. [An apparently earlier version is described in 7.H.] No. 2, p. 228. 200 stones.
Magician's Own Book.
1857. The basket and stones, pp.
246-247. 100 stones, one yard apart. =
Book of 500 Puzzles, 1859, pp. 60-61. =
Boy's Own Conjuring Book, 1860, p. 218.
= Indoor & Outdoor, c1859, part II, prob. 20, pp. 136-137.
Illustrated Boy's Own Treasury. 1860. Prob. 30, pp.
429-430 & 434. 100 trees to be watered, five steps apart.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-32, pp. 255 & 396: Am Feste
der bemalten Eier. 100 eggs, a fathom apart. Says that the 'egg gathering' is a
traditional race.
Todhunter. Algebra,
5th ed. 1870. Examples XXX, no. 23, pp. 265 & 588. Basket at origin and n-1
stones, with first stone at
1, then second is 3
further, then 5
further, the 7 further, .... That is, the stones are at positions 1, 4, 9, 17, ..., so the
total distance is 2 [12
+ 22 + ... + (n-1)2]
= n(n-1)(2n-1)/3.
Daniel W. Fish, ed.
The Progressive Higher Arithmetic, for Schools, Academies, and
Mercantile Colleges. Forming a Complete
Treatise of Arithmetical Science, and its Commercial and Business
Applications. Ivison, Blakeman, Taylor
& Co., NY, nd [but prefaces give: 1860; Improved Edition, 1875]. Pp. 412-413. Equivalent to a man picking up stones with the distance to the first
stone being 5, the distance to the last stone being 25 and his total travel
being 180.
M. Ph. André.
Éléments d'Arithmétique (No 3) a l'usage de toutes les
institutions .... 3rd ed., Librairie
Classique de F.-E. André-Guédon, Paris, 1876.
Probs. 551 & 552, p. 239, are similar to Bourdon.
W. W. Rouse Ball.
Elementary Algebra. CUP, 1890
[the 2nd ed. of 1897 is apparently identical except for minor changes at the
end of the Preface]. Prob. 38, pp. 347
& 480. 10 balls equally spaced in a row, starting 12 ft from the
basket. A boy picks them up in the
usual way and find he has travelled ¼
mile. What was the spacing?
Lucas.
L'Arithmétique Amusante.
1895. Prob. XXX: La course des
œufs, pp. 110-111. 100 apples. Says it is played on the beaches with eggs
in good weather.
A. Sonnenschein & H. A. Nesbit. The New Science and Art of Arithmetic For
the Use of Schools. A. & C. Black,
London, 1903. Pp. 342 & 489. Potato race with 50 potatoes a yard apart
to be retrieved.
Pearson. 1907. Part II, no. 143: The stone carrier, pp. 142
& 219. 52 stones, with
spacing 1, 3, 5, ..., 103, to be brought to the first stone,
yielding
2 (1 + 3 + 5 + ... + 103) = 2 * 522.
Wehman. New Book of
200 Puzzles. 1908. An egg problem, p. 53. 100 eggs, a yard apart.
[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers.
1973. Op. cit. in 5.E. More mileage possible, pp. 187-188. Consider
10 points equally spaced on a
line. Salesman starts at one of them
and visits each of them once. What is
the maximum mileage possible? They get
49 miles by going 5 7 4 8 3 9 2 10 1
6. Adapting my computer search
mentioned below, I find this is indeed maximal and there are 1152
maximal solutions. Though not the
same as the above problems, this uses the same set-up, except one does not
return to base after each visit.
David Singmaster.
Problem proposal sent to AMM on 22 Oct 2001. General version of the Scotts' problem with a base office at
position 0 and the salesman has to start and return to it. Conjectures the maximum distance for N
customers is ë(N+1)2/2û, which is verified by exhaustive search up
to N = 10, but there are many such trips, e.g. 28,800 of them for N = 10.
Adding 0 at each end of the Scotts' trip is an example
of a maximal journey. The referee
produced an elegant proof which also gives the number of maximal journeys. The editor suggested it would be more
appropriate for MM and a revised version has been submitted there.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 113, f. 56v.
Rudolff. Künstliche
rechnung, 1526, op. cit. in 7.L.2.b.
1540 ed., f. N.vii.v.
??NYS. (H&S 57 gives the
German.)
Apianus. Kauffmanss
Rechnung. 1527. How many times does a clock strike
during 1 to 12?
H&S 57 says this is in Buteo (1559), ??NYS.
Wingate.
Arithmetic. 1629? Cf Wingate/Kersey. This item is in a section which Kersey did not revise. P. 296 in the 1678? ed. "How many strokes the Clock strikes
betwixt midnight and noon."
Baker. Well Spring
of Sciences. Prob. 1, 1670: p. 71,
??NX. 1 + ... + 12 strokes.
Wells. 1698. No. 97, p. 204. 1 + ... + 12 strokes.
A Manual of Curious and Useful Questions. MS of 30 Jan 1743(OS) owned by Susan
Cunnington (??NYS) and described in her:
The Story of Arithmetic; Swan Sonnenschein, London, 1904, pp.
155-157. "How many strokes do ye
clocks of Venice (which go on to
24 o' th' clock) strike in the
compass of a natural day?" [Where
is this MS??]
Dilworth.
Schoolmaster's Assistant.
1743. P. 93, no. 1. "How many strokes does the hammer of a
clock strike in 12 hours?"
Walkingame. Tutor's
Assistant. 1751. Arithmetical Progression, prob. 1, 1777: p.
90; 1835: p. 98; 1860: p. 118. How many strokes in
12 hours?
Mair. 1765? P. 483, ex. V. How many strokes in
12 hours?
Euler. Algebra. 1770.
I.III.IV: Questions for practice, no. 3, p. 139. "The clocks of Italy go on to 24
hours: how many strokes do they
strike in a complete revolution of the index?"
Vyse. Tutor's
Guide. 1771?
Pike.
Arithmetic. 1788. P. 221, no. 2. "It is required to find out how many strokes the hammer of a
clock would strike in a week, or
168 hours, provided it increased
at each hour?"
Bonnycastle.
Algebra. 1782. P. 60, no. 5 (1815: p. 75, no. 5). "How many strokes do the clocks in
Venice, which go on to 24 o'clock, strike in the compass of a day?" (1815 omits "the compass of".)
Hutton. A Course of
Mathematics. 1798?
D. Adams. New
Arithmetic. 1835. P. 224, no. 12. "How many times does a common clock strike in 12
hours?"
Hutton-Rutherford. A
Course of Mathematics. 1841? Prob. 25,
1857: 82. "The clocks of
Italy go on to 24 hours; then how many strokes do they strike in one complete
revolution of the index?"
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. P. 196.
"How many strokes does a common clock strike in the compas of 12
hours?"
[Chambers].
Arithmetic. Op. cit. in
7.H. 1866? P. 224, quest. 5.
"How many times does a common clock strike in a day?" Answer:
156.
James Cornwell & Joshua G. Fitch. The Science of Arithmetic: .... 11th ed., Simpkin, Marshall, & Co.,
London, et al., 1867. (The 1888 ed. is
almost identical to this, so I suspect they are close to identical to the 2nd
ed. of 1856.) Exercises CXXXVIII, no.
9, pp. 291 & 370. "How many
times does the hammer of a clock strike in a week?"
Lucas.
L'Arithmétique Amusante.
1895. Prob. XXXI: Les quatre
cents coups, p. 111. A clock that goes
to 12,
strikes 78 + 78 = 156 hours in a day. If it also strikes quarters, it makes 240 of those,
totalling 396, nearly
400 blows per day.
Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg,
1960 and says it was reset for the 49th ptg of 1944. Examination Papers III, prob. 181. How many hour strokes does a clock make in a day? How many in the year 1896?
Dudeney. AM. 1917.
Prob. 115: A printer's error, pp. 20 & 162. ABCA = ABCA has the unique solution 2592.
Hubert Phillips.
Question Time. Op. cit. in
5.U. 1937. Prob. 133: Phoney 'phone, pp. 87 & 219. ABCD
= ABCD. Answer asserts that 2592
is the unique solution.
Donald L. Vanderpool.
Printer's "errors".
RMM 10 (Aug 1962) 38. Extends
Dudeney's examples, e.g. 25 * 25/31 =
25 25/31; 34 * 425 =
34425 (which can be multiplied
by any power of 10).
M. H. Greenblatt.
Mathematical Entertainments, op. cit. in 6.U.2, 1968. Expression for which 2,592
is a solution, pp. 15‑16. He
asserts that Dudeney's expression was discovered by his officemate, Larry, in
response to a colleague, B. Rothlein, complaining that he could not remember
his telephone number, EVergreen 2592. Rothlein had this telephone on 41st St.,
Philadelphia, during 1943‑47. [A
likely story??]
7.AH. MULTIPLYING BY REVERSING
For two-digit numbers, reversing and
shifting are the same, but I will consider them here as the transformation ab
to ba seems more like a reversal than a shift.
I have just noted the items by
Langford in 7.AR and the item by Meyer below which make me realise that this
section is connected to 7.AR. Meyer
doubles the result of 1089 which comes up in 7.AR and gets 2178
which I remembered occurs in this section. Upon investigating, we see Langford notes that 1089
= 1100 - 11 so that
k * 1089 = kk00 ‑ kk = k,k‑1,9‑k,10-k in base
10. From this we see that k*1089
is the reverse of
(10-k)*1089. Now 10-k
is a multiple of k for
k = 1, 2, but we get some new
types of solution for
k = 3, 4,
namely: 7 * 3267 = 3
* 7623; 6 * 4356 = 4
* 6534.
This pattern leads to all solutions
in 7.AR and also leads to further solutions here, using 11000 ‑ 11 = 10989, etc., but this does not seem to give a complete
solution here.
Charles Babbage. The
Philosophy of Analysis -- unpublished collection of MSS in the BM as Add. MS
37202, c1820. ??NX. See 4.B.1 for more details. F. 4r is "Analysis of the Essay of
Games". F. 4r has entry 12: "Given product of a number consisting
of n
figures mult. by same [some??] figures in an inverted order query num" Though a bit cryptic, this seems to refer to the problem of this
section.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 136 & 258, no. 525. ab
= 2⅔ * ba. He adds a further condition, but this is not
needed.
Sphinx. 1895. Arithmetical, no. 217, pp. 33 &
104. x2 =
4*(x reversed).
T. C. Lewis.
L'Intermédiaire des Mathematiciens 18 (1911) 128. ??NYS. Dickson, Vol. 1, p. 463, item 70
says he discusses "number divisible by the same number reversed".
Ball. MRE, 6th ed.,
1914, p. 12. Mentions the general
problem and gives 8712 =
4*2178 and 9801
= 9*1089 as examples. Gives four citations to L'Intermédiaire des Maths., none of them
the same as the above!
R. Burg.
Sitzungsber. Berlin Math. Gesell. 15 (1915) 8‑18. ??NYS. Dickson, vol. 1, p. 464, item 83,
says he found those N, base 10, whose reversal is kN, in particular for
k = 9, 4.
Wood. Oddities. 1927.
Prob. 56: Wizard stunts, pp. 45-46.
Notes that
83 * 41096 = 3410968 and asks for more examples of
AB * n = BnA. Finds 86 * 8 = 688 and then says the next example is
'infinitely harder to find':
71 * 16 39344 26229 50819 67213 11475
40983 60655 73770 49180 32787.
Haldeman-Julius.
1937. No. 102: Four-digit
problem, pp. 17 & 26. Asks for a
four digit number whose reversal is four times it. Answer: 2178.
Ball. MRE, 11th ed.,
1939, p. 13. Adds some different types
of examples.
312*221 = 68952;
213*122 = 25986.
See also 7.AJ.
G. H. Hardy. A
Mathematician's Apology. CUP,
1940. Pp. 44-45. "8712
and 9801 are the only four-figure numbers which are
integral multiples of their 'reversals': ....
These are odd facts, very suitable for puzzle columns and likely to
amuse amateurs, but there is nothing in them which appeals to a
mathematician." He cites Ball for
this information.
Morley Adams.
1948. See item in 7.AC.1 for an
example of 6 * ABC =
5 * CBA.
K. Subba Rao. An
interesting property of numbers. The Mathematics
Student 27 (1959) 57‑58. Easily
shows the multiplier must be 1, 4 or
9 and describes all solutions.
Jonathan Always.
Puzzling You Again. Tandem,
London, 1969. Prob. 32: Four different
answers, pp. 23 & 80. AB *
7/4 =
BA has four solutions.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
A tram called Alec, pp. 89 & 137-138. 4 * TRAMS = SMART.
Jerome S. Meyer.
Arithmetricks. Scholastic Book
Services, NY, 1965.
7.AH.1. OTHER REVERSAL PROBLEMS
New
section. There are many forms of this
that I have not recorded before.
William Leighton, proposer;
Rich. Gibbons, solver. Ladies'
Diary, 1751-52 = T. Leybourn, II: 49, quest. 338. ABC
has digits in arithmetic progression, its value divided by the sum of
its digits is 48 and CBA - ABC = 192. This is rather overdetermined as there are only two three digit
numbers whose quotient by the sum of the digits is 48, namely 432
and 864.
Augustus De Morgan.
Arithmetic and Algebra. (1831
?). Reprinted as the second, separately
paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin &
Craddock, London, 1836. Art. 114, pp.
29-30. ab + 18 = ba with
a + b = 6.
Todhunter. Algebra,
5th ed. 1870. Examples XIII, nos. 30-31, pp. 104 & 578.
Haldeman-Julius.
1937. No. 108: A problem in 2's,
pp. 13 & 26. x + 2 is reversal of 2x. Answer: 47.
Birtwistle.
Calculator Puzzle Book. 1978. Prob. 37: In reverse, pp. 28 & 84. Which two digit numbers squared are the
reversals of the squares of their reversals?
These are given by 12 and
13.
7.AI. IMPOSSIBLE EXCHANGE RATES
F. & V. Meynell.
The Week‑End Book. Op.
cit. in 7.E. 1924. 2nd ed., prob. three, p. 274; 5th? ed., prob. six, pp. 407‑408. US & Mexico value each other's dollar
at 58/60 (i.e. at 4s 10d :
5s).
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 36, pp. 20‑21 & 192. Two countries each value the other's money
at 90%
of its own. The solution says
this appeared in the New Statesman and Nation in late 1931, ??NYS.
E. P. Northrop.
Riddles in Mathematics.
1944. 1944: 9; 1945: 8‑9; 1961: 18‑19. As in
Phillips.
W. A. Bagley.
Paradox Pie. Op. cit. in
6.BN. 1944. No. 3: South of the border, pp. 8-9. US & Mexico each value other's dollar at $1.05.
John Fisher. John
Fisher's Magic Book. Muller, London,
1968. Magical shopping, pp. 120‑121. North and South Fantasi value each
other's £ as 19s (i.e.
95%) of its own.
That is, we want solutions of k * a1 ... am b1
... bn = b1 ... bn a1
... am. E.g., for m = 1, 3 * 142857 = 428571;
for n = 1, 4 * 025641
= 102564, 4 * 102564
= 410256, 5 * 142857
= 714285. If
1/n has a repeating decimal of
period n-1, then all multiples of it by
1, 2, ..., n-1 are shifts of it
-- e.g. 1/7 = .142857142857... gives
2 * 142857 = 285714,
etc. Such an n
must have 10 as a primitive root.
Early versions of the idea are
simply observations of the properties of
1/7 etc.
Note that two-digit problems,
i.e. m = n = 1, look more like reversals and are considered
in 7.AH.
Charles Babbage. The
Philosophy of Analysis -- unpublished collection of MSS in the BM as Add. MS
37202, c1820. ??NX. See 4.B.1 for more details. F. 4r is "Analysis of the Essay of
Games". F. 4r has entry 13: "What number is that whose 6
first multiples are the same digits differently arranged".
Ainsworth & Yeats.
Op. cit. in 7.H.4. 1854. Exercise XXXVI, pp. 73 & 176.
Birger Hausted.
?? Tidsskrift for Math. 2 (1878)
28. ??NYS -- cited by Dickson, vol. 1,
pp. 170‑171, item 81. Studies
problem with all shifts of the same number, but starting with the case n = 1.
He finds that k * a1...amB = Ba1...am gives
B/(10k‑1) = (a1...amB)
/ (10m+1 ‑ 1). He
allows k to be rational.
Dickson, vol. 1, pp. 174‑179, items 101, 102, 106,
114, 120, 137, 150.
Hoffmann. 1893. Chap. IV.
L. E. Dickson.
?? Quarterly J. Math. 27 (1895) 366‑377. [Item 106 above.] Shows that all ks are integers and a1 ¹ 0 only for 142857 (in base
10).
Anonymous note. J.
of the Physics School in Tokyo 6? (1897?) ??NYS Abstracted in: Yoshio
Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 21; as:
Another queer number. 3 * 526 31578
94736 84210 = 1578 94736 84210 52630,
which is not quite a pure shift of the first number, but would be if the
final zeroes were dropped. Also
multiplication by 4 or 8 or division by 5 give similar results.
T. Hayashi. On a
number that changes its figures only cyclically when multiplied or divided by
any number. J. of the Physics School in
Tokyo 6 (1897) 148-149. Abstracted
ibid., p. 21. Notes that the above
number is based on the form 10 Σ
(10r)i, which explains and
generalizes its properties.
U. Fujimaki. Another
queer number. J. of the Physics School
in Tokyo 6 (1897) 148-149. Abstracted
ibid., pp. 22-23. Notes that the above
number has the form (10m -
1) / n, which also explains and
generalizes its properties.
Pearson. 1907. Part II, no. 31: A large order, pp. 120
& 197‑198.
n = 1, m = 21, k = b1 = 7.
Schubert. Op. cit.
in 7.H.4. 1913. Section 16, no. 186, pp. 51 & 137. k = 3,
m = 1, a1 = 1, n = 5.
Peano. Giochi. 1924.
Ackermann.
1925. Pp. 107‑108. k = 7,
n = 1, b1 = 7.
W. B. Chadwick. On
placing the last digit first. AMM 48
(1941) 251-255. n = 1. Shows
b1 ³ k and N = a1
... am b1 = b1 (10m+1 - 1) /
(10k-1). Finds all solutions for k = 2, 3, ..., 9. Cites Guttman, AMM 41 (1934) 159 (??NYS) for
properties of these 'cyclic numbers'.
An editorial note adds that the expression for N works for any base.
Jerome S. Meyer.
Fun-to-do. Op. cit. in 5.C.
1948. No. 76: Mathematical whiz, pp.
59-60. Discusses cyclic properties of
the period of 1/17, though he doesn't identify it as arising
from 1/17.
J. Bronowski.
Christmas Teasers. New Statesman
and Nation (24 Dec 1949). ??NYS. Case
m = 1, k = 3/2.
Anonymous. The
problems drive. Eureka 14 (Oct 1951)
12-13 & 22. No. 11. Same as Bronowski. Solution: 428571.
J. H. Clarke. Note
2298: A digital puzzle. MG 36 (No. 318) (Dec 1952) 276. Solves Bronowski's problem, leading to 3 * 10n º 2 (mod 17) for an n+1
digit answer.
D. E. Littlewood.
Note 2494: On Note 2298: A digital puzzle. MG 39 (No. 327) (Feb 1955) 58.
Easy solution of Bronowski's problem, via 20/17.
E. J. F. Primrose.
Note 2495: A digital
puzzle. Ibid., 58‑59. Case
m = 1, k = p/q.
R. Sibson. Note
2496: On Note 2298. Ibid., 59.
Simple but fortuitous solution via
20/17.
R. L. Goodstein.
Note 2600: Digit transfers. MG 40 (No. 332) (May 1956) 131‑132. Shows Littlewood's method gives an easier
solution to Primrose's version.
Gardner. SA (Jan
1961) c= Magic Numbers, chap. 2.
Mentions 4*102564 =
410256 and the general problem
with n = 1 and k = b1. Gives some references to the most general
problem in various standard works and some journal references up to 1968.
Barnard. 50 Observer
Brain-Twisters. 1962. Prob. 23: A safe number, pp. 30, 62 & 85‑86. Wants the smallest number with n = 1
and k = b1, i.e. such that b * a1a2...amb = ba1a2...am. Finds it is 102564.
Charles W. Trigg.
Division of integers by transposition.
JRM 1:3 (1968) 180-182.
Considers the case n = 1, k = b1, i.e. the same as Barnard.
He gives the smallest answers for bases
3, 4, ..., 12 and k = 2, ..., b-1. Cites some earlier versions not given above.
Steven Kahan.
k-transposable integers. MM 49:1
(1976) 27-28. Studies the case m = 1
and shows that then only k =
3 can work. The basic solutions are
142857 and 285714
and all other solutions are obtained by repeating these, e.g. 142857142857.
Steven Kahan.
k-reverse-transposable integers.
JRM 9:1 (1976-77) 15-20. Studies
the case n = 1 and finds all solutions.
Birtwistle. Calculator
Puzzle Book. 1978.
Warren Page. A
general approach to p.q r‑cycles.
In: Warren Page, ed.; Two‑Year
College Mathematics Readings; MAA, 1981, pp. 263‑274. He considers rational k
(= his q/p) and given n
(= his r). He also studies the
case where multiples of a1...amb1...bn
have the same properties. He gives tables of all solutions for n = 1
and n = 2. 11 references to similar work.
Anne L. Ludington.
Transposable integers in arbitrary bases. Fibonacci Quarterly 25:3 (1987) 263-267. Considers case m = 1. Cites Kahan's
results. Considers base g
and shows that there is some
k if and only if g = 5
or g ³
7. Shows that there are only a finite
number of basic solutions for any
k and hence for any g
and describes how to find them.
Anne L. Ludington.
Generalized transposable integers.
Fibonacci Quarterly 26:1 (1988) 58‑63. Considers general case with shift of j (= m in
my notation) and arbitrary base g. Shows there is such a k
for all j ³
0 if and only if g = 5
or g ³
7. For
g = 3, 4, 6, there is
such a k for all j ³ 2. For fixed
j, there are only a finite
number of solutions.
Keith Devlin. Better
by degrees -- Micromaths column in The Guardian (17 Nov 1988) 31. Let
F, C be the same temperature in
Fahrenheit and Centigrade (= Celsius).
When can F = a1b1...bn, C =
b1...bna1?
First answer is 527.
7.AJ.1. MULTIPLYING BY APPENDING DIGITS
New section.
Tony Gardiner.
Challenge! What is the title of this article? Mathematics Review 4:4 (Apr 1994) 28-29. Find an integer A such that adding digits
of 1
at each end multiplies it by
99. If A has n
digits, then this gives us 10n+1
+ 10A + 1 = 99A, so A = (10n+1
+ 1) / 89. He leaves the rest to the
reader.
A worker earns a
for each day he works and forfeits
b for each day he doesn't
work. After c days, he has
gained d. This gives x + y =
c, ax ‑ by = d or
(a+b)x - bc = d. This is a
straightforward problem, giving
x = (bc + d)/(a + b),
y = (ac - d)/(a + b), so I only
give some early examples. There are also
many different forms of problem which give the same equations.
NOTATION: we denote this by (a, b,
c, d).
PROBLEM -- for which integral (a, b, c, d), is it true that x, y are integral? One can generate all quadruples
(a, b, c, d) with integral
solutions as follows. Choose any a, b, c, x
and set d = (a+b)x - bc. This is not the kind of solution of the
Problem that I'd like to have, but it may be best possible since we have only
one relationship.
I have now computed x in
each case and include it in the table below.
One can scale a, b, d by any factor, so we eliminate fractions in a, b, d
and even make GCD(a, b) =
1, but I haven't done any scaling in
the table.
For general solutions, see: Hutton, 1798?; Lacroix; De Morgan.
See Tropfke 603.
a b c d x Sources
2½ 1 30 54 24 Hutton, c1780?; Hutton, 1798?;
4½ 1½ 24 4½ 27/4 Hutton-Rutherford
4½ 1½ 24 78 19 Hutton-Rutherford
5 3 12 28 8 Lacroix;
5 3 28 0 21/2 Riese, 1524;
5 6 30 12 192/11 Chuquet;
5 9 30 10 20 Columbia Alg.;
5 9 30 15 285/14 Gherardi;
5 12 30 99 27 Robinson
6 5 30 0 150/11 Muscarello;
6½ 5¾ 50 18 1222/49 Unger;
7 3 365 0 219/2 Dodson
7 4 30 1 11 Fibonacci; Columbia Alg.;
7 4 30 30 150/11 Fibonacci;
7 5 30 0 25/2 Riese, 1522;
9 11 30 0 33/2 Bartoli
9¼ 6⅓ 70 180 40 Unger;
10 4 30 132 18 BR;
10/30 6/30 30 -2 15/2 al-Karkhi;
10/30 6/30 30 0 45/4 al-Karkhi;
10/30 6/30 30 4 74/4 al-Karkhi;
10 12 30 0 180/11 Benedetto da Firenze; Calandri, 1491;
10 12 40 0 240/11 AR;
Wagner;
10 14 20 15 295/24 Tartaglia;
11 8 36 12 300/19 Riese, 1524
12 8 365 0 146 Schott
12 8 390 0 156 Wells; Vyse
15 5 60 240 27 Todhunter
16 8 12 126 37/4 Eadon;
16 15 30 0 450/31 della Francesca 17r
16 20 30 0 50/3 Eadon;
16 24 36 0 108/5 Tartaglia;
16 24 36 60 231/10 Tartaglia;
18 16 30 0 240/17 Pacioli;
20 8 40 372 173/7 Hutton, 1798?
20 8 40 380 25 Simpson; Bonnycastle
20 10 40 500 30 Vyse
20 16 30 0 40/3 della Francesca 38r
20 28 40 0 70/3 Borghi;
20 28 40 30 575/24 Borghi;
24 12 48 504 30 Bourdon
25 30 40 65 23 Recorde
30 15 40 660 28 Ozanam-Montucla
40 5 60 980 256/9 Les Amusemens;
A vaguely related, but fairly
trivial problem, is the following. A
man offers to work for a master for a time T with payment of a horse
(or cloak) and M money.
After time t, the man quits and is paid the horse and m
money. Letting r be
the rate per unit time and letting
H be the value of the horse,
this gives rT = H + M, rt = H + m,
which gives r(T-t) = M-m. della Francesca f. 43v (107) (English in
Jayawardene) is an early example.
Muhammad (the h should have an underdot) ibn Muhammad
(the h
should have an underdot) ibn Yahyā(the h should have an
underdot) al-Bŭzağānī Abū al-Wafā’ = Abū al-Wafā’ al‑Būzajānī. Arithmetic.
c980. Arabic text edited by Ahmad Salim Saidan; Arabic Arithmetic; Amman, 1971. P. 353.
??NYS -- mentioned by Hermelink, op. cit. in 3.A.
al‑Karkhi.
c1010. Sect. I, no. 12‑14,
p. 83. (10/30, 6/30, 30, d) with
d = 0, 4, ‑2.
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 113, no. 23. ??NYS -- Hermelink, op. cit. in 3.A, gives al‑Karkhi's
problem and then says "This
problem occurs also in Ţabarī [NOTE:
Ţ denotes T
with an underdot.]
...." Tropfke 603 gives the
same reference.
Fibonacci. 1202.
BR. c1305. No. 51, pp. 68‑69. (10, 4, 30, 132).
Gherardi. Libro di
ragioni. 1328. Pp. 48‑49. (5, 9, 30, 15)
Columbia Algorism.
c1350.
Dresden C80. ??NYS
-- asserted in BR, p. 158.
Bartoli.
Memoriale. c1420. Prob. 7, ff. 75r - 75v (= Sesiano, pp. 137
& 147). (9, 11, 30, 0).
AR. c1450. Prob. 183, pp. 86, 176, 222‑223. (10, 12, 40, 0).
Benedetto da Firenze.
c1465. P. 88. (10, 12, 30, 0).
Muscarello. 1478. Ff. 73v-74r, p. 188. Worker building a house, (6, 5, 30, 0).
della Francesca.
Trattato. c1480.
Wagner. Das
Bamberger Rechenbuch, op. cit. in 7.G.1.
1483. Von Tagelohn oder Arbeit,
pp. 98 & 216. (10, 12, 40,
0). (= AR.)
Chuquet. 1484.
Borghi.
Arithmetica. 1484.
Calandri.
Arimethrica. 1491. F. 69v.
(10, 12, 30, 0).
Pacioli. Summa. 1494.
F. 99r, prob. 11. (18, 16, 30,
0).
Riese.
Rechnung. 1522. 1544 ed. -- pp. 89‑90; 1574 ed. -- pp. 60v‑61r. (7, 5, 30, 0).
Riese. Die
Coss. 1524.
Recorde. Second
Part. 1552. Pp. 312-318: A question of Masonry, the first example. (25, 30, 40, 65).
Tartaglia. General
Trattato. 1556. Book 17, art. 38, 39, 42, pp. 275r-275v
& 277r.
Schott. 1674. Ænigma 1, p. 558.
Wells. 1698. No. 113, p. 208. (12, 8, 390, 0).
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
VII, pp. 80-81 (1790: prob. XIX, p. 84).
(20, 8, 40, 380).
Les Amusemens.
1749. Prob. 112, p. 254. (40, 5, 60, 960).
Vyse. Tutor's
Guide. 1771?
Dodson. Math.
Repository. 1775. P. 9, Quest. XXI. (7, 3, 365, 0).
Ozanam‑Montucla.
1778. Prob. 8, 1778:
194-195; 1803: 192; 1814: 166‑167; 1840: 86.
(30, 15, 40, 660). 1790 has 620
for 660, apparently a misprint (check if this is in
1778 -- ??).
Charles Hutton. A
Complete Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2. [c1780?] 1804: prob. 3,
p. 135. (2½, 1, 30, 54).
Bonnycastle.
Algebra. 1782. Prob. 6, p. 80. Same as Simpson.
Eadon.
Repository. 1794.
Hutton. A Course of
Mathematics. 1798?
Silvestre François Lacroix.
Élémens d'Algèbre, a l'Usage de l'École Centrale des Quatre-Nations. 14th ed., Bachelier, Paris, 1825. Section 15, pp. 28-31. (5, 3, 12, 28) and general solution.
Bourdon.
Algèbre. 7th ed., 1834. Art. 47, prob. 2, pp. 64-65. (24, 12, 48, 504) and the general problem.
On pp. 102-104, he discusses the problem algebraically and the meaning
of the signs involved.
D. Adams. New
Arithmetic. 1835. P. 246, no. 105. (.75, .25, 50, 27.50).
Notes that if he worked every day, he would earn 37.50
and that he loses 1.00 from this for every day he didn't work.
Augustus De Morgan.
On The Study and Difficulties of Mathematics. First, separately paged, part of: Library of Useful Knowledge --
Mathematics I, Baldwin & Craddock, London, 1836. Workman earns a each day he works, but has expenses of b
every day. After m
days, he has earned c. This is equivalent to (a-b, b, m, c), giving ax - bm = c. Gives general solution and discusses
problems with negative values.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 41-45 & 248, nos. 176-190. These cover a variety of problems leading to
the same equations.
Hutton-Rutherford. A
Course of Mathematics. 1841?. Prob. 43,
1857: 43. (4½, 1½, 24, 78) and
(4½, 1½, 24, 4½).
Joseph Ray (1807-1855).
Ray's Arithmetic, Third Book.
Practical Arithmetic, by Induction and Analysis. One Thousandth Edition -- Improved. Wilson, Hinkle & Co., Cincinnati &
New York, ©1857, HB. P. 268, no.
57. (2, 1, c, 25) where he works three times as many days as
he idles.
Todhunter. Algebra,
5th ed. 1870. Section X, art. 172, p. 82.
(15, 5, 60, 240).
Horatio N. Robinson.
New Elementary Algebra: Containing the Rudiments of the Science for
Schools and Academies. Ivison,
Bargeman, Taylor & Co., New York, 1875.
This has several versions of the problem, but on p. 77, probs. 77-78 use
the following novel formulation: a boy is to deliver some glass vessels and
will be paid a for each success and will forfeit b
for each breakage; after c vessels, he has earned d.
Does (5, 12, 30, 99) and the general case.
Generally, this is done by
proportion of some sort. In the
simplest case -- if a is
b, what is c?
-- the answer is cb/a. In the more complex case -- if ab = c,
what is de = f? -- the answer is usually f = cde/ab.
Many problems have e and
f given and ask for d,
which is then d = abf/ce. However, some authors say that all answers
should be multiplied by c/ab and hence give d = cf/abe, which I will
call inverted reasoning or the inverted answer -- see: Benedetto da Firenze, p. 65; The Sociable (& Book of 500
Puzzles); Lemon; Hoffmann;
Pearson; Loyd.
Fibonacci.
1202. P. 170 (S: 264). He discusses this and says that it is just a
proportion which is commonly stated in this way. E.g. if 5 is
9, what is 11?
He says 99/5. Also, if
7 is half 12, what is half of 10? He says 35/6.
Gherardi. Libro di
ragioni. 1328. P. 17: Questi sono numeri. "If
9 is the half of 16,
what part is 12 of
25?" Answer is 8/9
of 12/25.
Bartoli.
Memoriale. c1420.
AR. c1450. Prob. 303, 305, pp. 135‑136, 178, 225.
Vogel says these problems also occur in Widman, 1489, ??NYS,
and versions are in al‑Khowarizmi.
However al‑Khowarizmi's examples are straightforward rules of
three, e.g. "If you are told 'ten
for six, how much for four?'...."
Benedetto da Firenze.
c1465.
The Treviso Arithmetic = Larte de labbacho. Op. cit. in 7.H. 1478. Ff. 31r‑33v
(= Swetz, pp. 103‑109).
Three examples in abstract rule of three, e.g. "If 8 should become 11, what would 12
become?" Answer: 12 * 11/8.
Chuquet. 1484. Triparty, part 1. FHM 73-74 gives a number of problems which are treated as
proportions.
Pacioli. Summa. 1494.
Tonstall. De Arte
Supputandi. 1522. Pp. 223-224.
Dilworth.
Schoolmaster's Assistant.
1743. P. 157, no. 3. "If the ⅓ of 6
be 3, what will ¼ of
20 be?" Answer:
7½, which is the direct answer.
Les Amusemens.
1749. P. xxv. See 7.AN for a problem that looks like it
belongs here.
Walkingame. Tutor's
Assistant. 1751. 1777: p. 172, prob. 48; 1835: p. 178, prob. 27; 1860: p. 180, prob. 47. Identical to Dilworth.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions. No. 2, pp. 15 & 71. "If the half of five be seven, What
part of nine will be eleven?"
Answer:
55/126 = (5/2)(11/7)/9.
The Sociable.
1858. Prob. 46: A dozen
quibbles: part. 7, pp. 300 & 319.
"If 5 times
4 are thirty-three, what will
the fourth of twenty be?" Answer
is 8¼
with no explanation, which is the inverted answer. = Book of 500 Puzzles, 1859, prob. 46:
part 7, pp. 18 & 37. = Magician's
Own Book (UK version), 1871, Paradoxes [no. 2], p. 37.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Arithmetical puzzles, no. 7, p. 174 (1868: 185). "If six the third of twenty be, what is
the fourth of thirty‑three?"
Gets (6 * 33/4) / (20/3) = 7
17/40.
Lewis Carroll. Alice
in Wonderland. 1865. Chap. II.
In: M. Gardner; The Annotated
Alice; revised ed., Penguin, 1970, p. 38.
"Let me see: four times
five is twelve, and four times six is thirteen, and four times seven is -- oh
dear! I shall never get to twenty at
that rate!" Gardner's note 3 gives
various explanations, the simplest of which is that
4 * 12 =
19 and Alice only knows tables
up to 12 times. Cf John Fisher; The
Magic of Lewis Carroll; op. cit. in 1; pp. 34-35 & Carroll-Wakeling, prob.
7: Alice's multiplication tables, pp. 8-9 & 64-65.
Lemon. 1890. Quibbles, no. 254(b), pp. 37 & 107 (=
Sphinx, no. 453(b), pp. 63 & 113.)
"If five times four are thirty‑three, what will the fourth of
twenty be?" Inverted answer
of 8¼
with no explanation -- see The Sociable.
Hoffmann. 1893. Chap. IX, no. 34: A new valuation, pp. 320
& 327 = Hoffmann-Hordern, p. 212. Identical to Lemon.
Answer is 8¼, but he gives no reason.
Clark. Mental
Nuts. 1897, no. 76; 1904, no. 77; 1916, no. 73.
Suppose. "Suppose the
one-fourth of twenty was three, what would the one-third of ten be?" Normal answer of 2.
H. D. Northrop.
Popular Pastimes. 1901. No. 10: A dozen quibbles, no. 7, pp. 68
& 73. = The Sociable.
Pearson. 1907. Part II, no. 95, pp. 134 & 210. Same as Lemon.
Loyd.
Cyclopedia. 1914. Sam Loyd's perplexed professor, pp. 332
& 383. = SLAHP: If things were
different, pp. 56 & 106. "If
five times six were 33, what would the half of 20
be?" Answer: "If five times six is 33
-- ten would naturally be 1‑3 of what
30 would be, viz: 11."
This is the inverted answer.
Loyd Jr. gives a bit more explanation.
Loyd.
Cyclopedia. 1914. p. 317 (no solution). As in Lemon.
Perelman. MCBF. 1937.
Imaginary nonsense, prob. 143, pp. 243-244. "What is the number
84 if 8 x 8 is 54?"
Finds that the base is 12, so the answer is 8412 = 100.
Haldeman-Julius.
1937. No. 63:
The-what-is-it-problem, pp. 9 & 24.
"If one third of six be three, what will one third of 29 be?" Answer is
10, with no explanation. This does not fit into either of the
standard versions here, but would be the usual form if 29
were a misprint for 20.
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. P. 8: Mathematical "if". "If a fourth of forty is six; what is a third of twenty?" Says four, basically by saying the result
is 6/10 of the real result.
G. A. Briggs. Puzzle
and Humour Book. Published by the
author, Ilkley, 1966. Prob. 1/9,
pp. 12 & 71. If a third of six
were three, what would the half of twenty be?" Answer is 15, which is the normal answer, but he gives no
reason.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
If, pp. 28 & 109. "If a
third of six was three, what would a quarter of twenty be?" Takes
3/2 of the correct answer, as in
the usual method.
New section
-- there may be older examples, but see Barnard, 1963.
Dudeney. Strand
Magazine (1926). ??NYS -- cited by
Angela Newing.
Richard Hoadley Tingley.
Mathematical Cross Number Puzzle[s].
In: S. Loyd Jr., ed.; Tricks and
Puzzles; op. cit. in 5.D.1, 1927.
Pp. 103‑105 & Answers pp. 11-12. Three examples with 13 x 13
frames. Tingley's name is only
on the first example. "This puzzle
is radically different from the usual type .... We have named this new brain teaser "Cross Number
Puzzle" ...." I found several
errors in the second example.
Dudeney. PCP. 1932.
Prob. 175: Cross‑figure puzzle, pp. 48 & 148. 11 x 11 frame. Erroneous set of
clues and solution. Corrected as
Cross-number puzzle in the revised ed. of 1935?, pp. 48‑49 & 148.
Michael H. Dorey.
"Little Pigley" [or "Little Pigsby"]. 1936.
This is also called "Dog's Mead" -- original ??NYS. A 1939 version with this attribution and
date are given in: Tim Sole; The Ticket
to Heaven and Other Superior Puzzles; Penguin, 1988, pp. 92 & 108. A 1935 version is given in: Williams & Savage, 1940, below. A 1936 version is given in: Philip Carter & Ken Russell; Classic
Puzzles; Sphere, London, 1990, pp. 62-63 & 128. A 1939 version is given as The Little Pigsby Farm Puzzle in: David Ahl & Burchenal Green, eds.; The
Best of Creative Computing, vol. 3; Creative Computing Press, Morristown, NJ,
1980, p. 177; no solution.
Phillips.
Brush. 1936. 4 x 4
numerical crosswords.
Jerome S. Meyer. Fun
for the Family. (Greenberg Publishers,
1937); Permabooks, NY, 1959. No. 28: Family skeleton, pp. 41 &
240. 4 x 4 based on ages in a family.
Haldeman-Julius.
1937. No. 132: Family skeleton
problem, pp. 15 & 27. Same as
Meyer. He says he got it from the Feb
1937 issue of College Humor.
E. P. H[icks].
& C. H. B. A mathematical crossword. Eureka 1 (Jan 1939) 17 &
2 (May 1939) 28. 5 x 5 diagram with central square black and heavy
division lines forming 18 lights. Two
verticals are four digit numbers and the excess single digits do not have down
clues. Some of the clues are pretty
obscure -- e.g. 'Magazine without the printers' refers to the magazine Printers'
Pie, so the solution is 3142.
S. E. W[ood]. A
numerical square. Eureka 3 (Jan 1940)
18. 6 x 6 array with some heavy division lines giving 18 lights. I have not found a solution in this or
following issues.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940.
Anonymous.
Crosswords Five hour or five
minute puzzles. Eureka 4 (May 1940) 16 & 5 (Jan 1941)
15. 8 x 8 array with 26 lights.
C. A. B. Smith. A
new way of writing numbers. Eureka 5 (Jan 1941) 7-9 & 6 (May 1941) 11. General exposition of the negative digit
system -- see section 7.AA. Discusses
the method for base 6 and gives a 7 x 6 crossnumber puzzle in this system.
E. M. White. A crossword
in decimal. Eureka 5 (Jan 1941) 20 & 6 (May 1941) 11. 9 x 9
array with 26 clues.
M. A. Porter. Note
1982: The missing clue. MG 31 (No. 296) (Oct 1947) 237. 4 x 4
puzzle.
Anonymous examples in Eureka.
A. H. Barrass.
Numerical square. Eureka 22 (Oct
1959) 13 & 22. 5 x 5 array with division lines giving 17
lights. Each answer has the form x2 ± y where x is a positive integer and y is
a positive prime. Complex clues for
the x
and y values.
Philip E. Bath. Fun
with Figures. Op. cit. in 5.C. 1959.
No. 100: A number crossword, pp. 38 & 61. 5 x 5
array with 17 clues.
M. R. Boothroyd
& J. H. Conway. Problems drive, 1959. Eureka 22 (Oct 1959) 15-17 & 22-23. No. 5.
2 x 2 x 2 to be filled with
eight distinct digits. Eight clues
saying the value is a square or half a square (where this can be the integer
part of half an odd square).
G. J. S. Ross
& M. Westwood. Problems drive, 1960. Eureka 23 (Oct 1960) 23-25 & 26. Prob. C.
3 x 3 array with two
opposite corners deleted, with six clues in base 7.
B. D. Josephson
& J. M. Boardman. Problems drive 1961. Eureka 24 (Oct 1961) 20-22 & 24. Prob. J.
3 x 3 array with opposite corners
omitted, with six clues being algebraic expressions in four positive integer
variables.
During 1960-1980, R. E. Holmes, "Rhombus",
contributed 45 puzzles to The Listener.
Some, perhaps all, of these were formidable crossnumber puzzles. I have been sent three examples of these,
but there are no dates on them. He also
contributed at least one example to G&P, but my copy has no date on
it. Can anyone provide information
about these puzzles or the setter?
P. E. Knight. Oh,
How I love thee, Dr. Pell. In: H. Phillips; Problems Omnibus II; Arco,
London, 1962; pp. 163‑164 & 228‑229. 3 x 4 array, but complex,
based on Pell's equation.
Barnard. 50 Observer
Brain-Twisters. 1962. Prob. 40: Crossnumber, pp. 46‑47, 65
& 98‑99.
D. St. P. Barnard.
Anatomy of the Crossword. Bell,
London, 1963. On p. 30, he says "The crossword ... has given rise
recently to the Crossnumber Puzzle"
and he refers to his book above.
L. S. Harris
& J. M. Northover. Problems drive 1963. Eureka 26 (Oct 1963) 10-12 & 32. Prob. H.
3 x 3 array with centre
omitted. Clues are 9x2, 4x2; 4x2y, (3x+y)2, where
x, y are integers.
Birtwistle. Math.
Puzzles & Perplexities. 1971.
Birtwistle.
Calculator Puzzle Book.
1978.
K. Heinrich.
Zahlenkreuzrätsel. In: Johannes Lehmann; Kurzweil durch Mathe;
Urania Verlag, Leipzig, 1980; pp. 38 & 138. 6 x 6 grid with 24
clues.
7.AN. THREE ODDS MAKE AN EVEN, ETC.
Alcuin. 9C. Prop. 43: Propositio de porcis. Kill
300 (or 30)
pigs in three days, an odd number each day. "[Haec ratio indissolubilis ad increpandum composita
est.] ... Ecce fabula .... Haec
fabula est tantum ad pueros increpandos."
([This unsolvable problem is set to cause confusion.] This is a fable .... This fable is posed to confuse children.)
Pacioli. De
Viribus. c1500.
W. Leybourn.
Pleasure with Profit. 1694. ??NYS -- described in Cunnington, op. cit.
in 7.G.2, 1904, p. 151 and in De Morgan, Rara, p. 633. "How can you put five odd numbers to
make twenty?" "Write three
nines upside down and two ones." De
Morgan says he does not recall ever seeing this problem, that Leybourn
considers the answer a fallacy, but that he thinks "the question more than
answered, viz. in very odd numbers."
Les Amusemens.
1749.
Philip Breslaw (attrib.).
Breslaw's Last Legacy. 1784? Op. cit. in 6.AF. 1795: 78-81. 'How to rub
out Twenty Chalks at five Times rubbing out, every time an odd one.' Set out marks numbered 1
to 20. Then erase the last four, which are those starting at 17,
which is odd, etc.
Henri Decremps.
Codicile de Jérôme Sharp, ....
Op. cit. in 4.A.1. 1788. Avant Propos, pp. 20-21. Statement is the same as in Breslaw, but
solution is not given.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Parlour magic, no. xx, p. 202 (1868: 203): How to rub out twenty chalks
in five rubs, each time erasing an odd number.
"Begin at the bottom and rub out upwards, four at a
time." See Breslaw for
clarification.
Magician's Own Book.
1857. How to rub out twenty
chalks at five times, rubbing out every time an odd one, p. 239. = Boy's Own Conjuring Book, 1860, How to rub
twenty chalks at five times rubbing out, every time an odd one, pp. 205‑206. Cf Breslaw.
Magician's Own Book (UK version). 1871. The Arabian trick,
p. 313. "To take up twenty cards,
at five times, and each time an odd-numbered one", he lays out twenty
cards, 1 ‑ 10, 1 - 10
(considered as 1 - 20) and proceeds as in Breslaw.
Don Lemon.
Everybody's Pocket Cyclopedia.
Revised 8th ed., 1890. Op. cit.
in 5.A. P. 136, no. 3. "Place
15 sheep in 4
pens, so there will be the same number of sheep in each pen." Though not of the same type as others in
this section, and no solution is given, I think the solution is similar to
other solutions here, namely to put the pens concentrically around the inner
pen and put all the sheep inside the inner pen.
Pearson. 1907. Part II.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
27:1 (Nov 1911) 89: The Ass-tute farmer.
Farmer has 17 asses and a friend bets he can't put an odd number in each
of four stalls. Farmer puts 7, 5, 3,
2. After inspecting the first three
stalls, the friend says the farmer has lost, but the farmer says to go into the
last stall to check, whereupon he shuts and locks the door, announcing there
are three asses in the stall!
Loyd.
Cyclopedia. 1914. The pig sty problem, pp. 37 & 343. = MPSL2, prob. 7, pp. 6‑7 &
123. = SLAHP: Pigs in pairs, pp. 51
& 104. = Pearson's shepherd's
puzzle.
Loyd.
Cyclopedia. 1914. A tricky problem, p. 38. = SLAHP: Torturing Dad, pp. 75 & 115. Five odd figures to make 14.
Gives Pearson's first solution.
Smith. Number
Stories. 1919. Pp. 126 & 146. Put 10 pieces of sugar in three cups so each cup
has an odd number. Put 7
& 3 and put one cup inside another cup.
Blyth. Match-Stick
Magic. 1921. The twenty game, p. 79. As in Breslaw, etc., but more clearly expressed: "The matchsticks have now to be removed
in five lots, .... Each time ... the
last of the group must be an odd number."
Hummerston. Fun,
Mirth & Mystery. 1924. Puzzle no. 64, pp. 149 & 182. "How would you arrange twenty horses in
three stalls so as to have an odd number of horses in each stall?" Arranges as
1, 3, 16 -- "sixteen is a very odd number of horses to put into any
stall."
Wood. Oddities. 1927.
Prob. 50: The lumps of sugar, pp. 42-43. Ten lumps of sugar into three cups so each cup contains an odd
number of lumps. Confusing solution,
but lets one cup be inside another and lists all 15 possible solutions.
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
Number, Please!, pp. 20 & 210.
Use the same odd figure five times to make 14. 1 + 1 + 1 + 11.
McKay. Party
Night. 1940. Pill-taking extraordinary, p. 152. "A man had a box holding
100 pills. He took an odd number of pills on each of
the seven days of the week, and at the
end of the week all the pills were gone.
How could he manage that?"
1 on each of the first 6
days and 94 on the last -- "you must admit
that 94 is a very odd number of pills to take on any day."
Jerome S. Meyer.
Fun-to-do. Op. cit. in 5.C.
1948. Prob. 17: Pigs and pens, pp. 27
& 184. Put nine pigs in four pens,
an odd number in each pen. Three pens
of three with a big pen around them all.
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. P. 19: Pigs in pens. Same as Meyer.
John Paul Adams. We
Dare You to Solve This!. Op. cit. in
5.C. 1957?
Gibson. Op. cit. in
4.A.1.a. 1963.
Doubleday - 1.
1969. Prob. 71: Ups and downs,
pp. 87 & 170. = Doubleday - 5, pp.
97-98. Labourer has to carry 85 bricks
up to a bricklayer in a hod that can hold 16, but he is instructed to always
carry an odd number. How does he do it
in six journeys? Solution is to carry
up 15 on five trips and then bring one down and carry up the remaining 11. To me, the return stages should all be alike
and one could do this by carrying up and carrying down one five times, then
carrying up the remaining 15. Of
course, the last stage cannot be expected to be the same as the others --
indeed he may be sent off on some other task.
[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers. 1973.
Op. cit. in 5.E. Three short
problems -- no. 1, pp. 41-42. Have five
odd numbers add to 14.
11 + 1 + 1 + 1 = 14.
7.AO. DIVINATION OF A PERMUTATION
There are simpler versions of this
used to divine three numbers, e.g. to locate a ring on a person, finger and
digit -- a common one uses:
x *2 +5 *5 +10 +y *10 +z = 350 + 100x + 10y
+ z -- cf Fibonacci, Tagliente
(1515). The operations are performed
from left to right, corresponding to the instructions given. I will make no attempt to trace this very
common but very dull type of problem, but see 7.M.4.b, especially the Folkerts
entry which cites six early sources for this type of divination.
If the ai are a
permutation of 1, 2, ..., n, the method of interest forms P = a1m1 + a2m2 + ... anmn for multipliers mi. By
appropriate choice of the mi, the value of P determines the
permutation. Generally P is
subtracted from some convenient constant.
Sometimes the solution uses a division to yield a1 and a2. Some formulae work even if ai are not a permutation, but are digits or dice values. If we have a permutation, one can
ignore an since it is determined by the others, i.e.
one can let mn = 0.
NOTE. The standard form of this problem has n objects permuted
among n people. The
permutation (ai) can be viewed in two ways.
1. The
more natural view is that ai
= j means the i-th person has object
j. Then the multipliers are
associated with the people.
2. The
inverse view is easier to implement and hence much more common in this
problem. We view ai = j as meaning that the i-th
object is with person j. Then the multipliers are associated with the
objects.
The most common form of the problem
is with three people and 24 counters.
You give the people 1, 2, 3 counters to start and leave the other 18
counters on the table. Let them
secretly permute three items, say A, B,
C among themselves. You tell the person with object A to
take as many counters as he already has; the person with object B to
take twice as many; the person with object
C to take four times as
many. In our inverse view, we are
taking P = a1 + 2a2
+ 4a3 from 18.
NOTATION: M = (m1, m2, ..., mn) denotes this process.
A sequence of multipliers, M,
is suitable if the products
P are all distinct. In the case that the (ai) are a permutation, it is easy to see that the following processes
preserve the suitability of M. (1) Permuting the mi. (2) Shifting all the mi by a
constant. (3) Multiplying all the mi by a non-zero constant.
Thus we can arrange the mi in ascending order and make m1 = 0 and m2 =
1. So for the case n = 3,
any sequence can be brought to the form
(0, 1, m3). By
subtracting from m3 and scaling, we see that this sequence is
equivalent to (0, 1, m3/(m3-1)). A little work shows that either m3 ³ 2 or m3/(m3-1)
³ 2, so we can
assume m3 ³
2. m3 = 2 gives an unsuitable M,
but m3 > 2 always gives a suitable M.
So the simplest possible case is
(0, 1, 3), which is equivalent
to the most common case (1, 2, 4) and to
(3, 4, 6). Cases equivalent
to (0, 1, 4), (0, 1, 6) and (0, 1, 8)
also occur.
For the case of permutations,
subtracting P values from some constant
S is equivalent to changing mi to S/Σai - mi.
As in the Josephus Problem,
mnemonics were constructed. For the
case n = 3, objects were labelled by the vowels a, e, i and mnemonics
were constructed consisting of words (or phrases) with these three vowels in
all six permutations (or having just the first two vowels of each
permutation). See Bachet for an
example. Gardner also gives mnemonics
using consonants.
See Meyer for a slightly more
complex multiplication process, which can be reduced to the usual form.
See Tropfke 648‑651.
COMMON VOWEL MNEMONICS
-- note that spelling and layout vary.
Angeli, Beati, Taliter, Messias, Israel, Pietas. Baker;
Angeli Beati Pariter
Elias Israel Pietas.
Hunt.
Anger, fear, pain, may
be hid with a smile.
Magician's Own Book; Boy's Own Conjuring Book;
Aperi, Premati, Magister,
Nihil, Femina, Vispane,
Vispena. Minguét.
Aperì Prelati Magister
Camille Perina Quid habes
Ribera. Alberti 76-77;
Avec éclat L'Aï
brillant devint libre.
Labosne, under Bachet; Lucas;
Brave dashing sea, like a
giant revives itself.
Magician's Own Book; Boy's Own Conjuring Book;
Graceful Emma, charming she reigns in all circles.
Magician's Own Book; Boy's Own Conjuring Book;
Il a jadis brillé
dans ce petit État.
Lucas;
James Easy admires now
reigning with a bride.
Magician's Own Book; Boy's Own Conjuring Book;
Pallētis
Evandri Sanguine Feritas
Imane (the m should have an overbar) Vigebat. Schott.
Par fer César jadis
devint si grand prince.
Bachet; Ozanam 1725, prob. 46; Alberti 76-77; Les Amusemens; Hooper;
Magician's Own Book; Boy's Own Conjuring Book; Boy's Own Book; Magician's Own
Book (UK version); Lucas;
Pare ella ai segni; Vita, Piè. Alberti 76-77.
Salve certa anima
semita vita quies.
Van Etten; Schott (with one respelling); Ozanam 1725, prob. 46;
Alberti 76-77; Manuel des Sorciers; Endless Amusement; Parlour Pastimes;
Magician's Own Book; Boy's Own Conjuring Book; Magician's Own Book
(UK version);
Take her
certain anise seedlings
Ida quince. Magician's Own Book (UK version);
Tabari. Miftāh
al-mu‘āmalāt. c1075. P. 109, part IV, no. 17. ??NYS - described by Tropfke 648. As in Fibonacci, below, with a = 18
and M = (2, 17, 18).
Fibonacci. 1202.
Abbott Albert.
c1240. Prob. 2, pp. 332‑333. M = (2, 9, 10). He then subtracts from
60 which produces 8a1 + a2. This is identical to Fibonacci's first
example, but Albert gives a complete table of all the partitions of 6
into 3 non‑negative summands and computes 2x + 9y + 10z for each, showing that P
determines the ai if
a1 + a2 + a3 = 6,
whether it is a permutation of 1, 2, 3 or not. No mnemonics. This can also be viewed as a form of 7.P.1.
BR. c1305.
Munich 14684.
14C. No. IX. This is obscure, but is repeated more
clearly as No. XIX. M = (2, 9, 10). IX is followed by a two line verse. Curtze could make no sense of it, but I
wonder if it might be a mnemonic??
Folkerts.
Aufgabensammlungen. 13-15C.
AR. c1450. Prob. 269, p. 122, 180‑181, 227‑228. M = (2, 9, 10). Then 60 ‑ P = 8a1
+ a2. Vogel cites several
earlier appearances, but mentions no mnemonics. ??NYS.
Chuquet. 1484. Prob. 159.
M = (1, 2, 4). Starts with 24,
but 6 are used to label the people, so this is really subtracting
from 18. Table, but no mnemonic.
FHM 232.
Pacioli. De Viribus. c1500.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 142, f. 63v.
x*2+5*5+10+y*10+z = 350 + 100x + 10y + z. Cf Fibonacci.
Cardan. Practica
Arithmetice. 1539. Chap. 61, section 18, f. T.iv.v
(p. 113). Mentions divination of a
permutation of three things by use of
18 counters, so this is
probably M = (1, 2, 4), subtracting from 18, as in Chuquet, Baker,
etc.
Baker. Well Spring
of Sciences. 1562? Prob. 4,
1580?: ff. 197r‑198r;
1646: pp. 310-312; 1670: pp.
354-355. M = (1, 2, 4), subtracting from 18. Vowel mnemonic: Angeli, Beati, Taliter, Messias, Israel,
Pietas with explanatory table.
Recorde-Mellis.
Third Part. 1582. Ff. Yy.v.r - Yy.v.v (1668:. 477-478: Another [divination] of things
hidden. M = (1, 2, 4). No mnemonics.
John Wecker. Op.
cit. in 7.L.3. (1582), 1660. Book XVI -- Of the Secrets of Sciences:
Chap. 20 -- Of Secrets in Arithmetick: To discover to one a thing that is hid,
pp. 289‑290. M = (1, 2,
4). No mnemonics. Cites Gemma Frisius, ??NYS.
Prévost. Clever and
Pleasant Inventions. (1584), 1998.
Bachet.
Problemes. 1612. Prob. XXII: De trois choses et de trois
personnes proposées deviner quelle chose aura été prise par chaque personne,
1612: 115-126. Prob. XXV, 1624: 187-198; 1884: 127‑134. M = (1, 2, 4). Vowel mnemonic: Par fer César jadis
devint si grand prince.
Gives a four person version, S =
78, M = (1, 4, 16, 0), referring to
Forcadel as giving an erroneous method.
Labosne says Diego Palomino (1599) has studied the four person
case. [This must be Jacobo Palomino;
Liber de mutatione aeris in quo assidua et mirabilis mutationis temporum
historia cum suis caussis enarratur. -- Fragmentum quodam ex libro de
inventionibus scientiarum; Madrid, 1597 or 1599, ??NYR.] Labosne adds some explanation, another
mnemonic: Avec éclat
L'Aï brillant devint
libre, and expands on Bachet's
work on the case of four objects, but eliminates reference to Forcadel.
van Etten.
1624. Prob. 8 (8), pp. 9‑11
(19‑22). M = (1, 2, 4). Vowel mnemonics: Salve certa anima
semita vita quies;
Par fer Cesar Iadis
Devint si grand Prince.
Henrion's 1630 Notte, pp. 10‑11, says that Bachet has extended it
to 4
objects.
Hunt. 1631
(1651). Pp. 255-261 (247-253). Usual form with mnemonic: Angeli
Beati Pariter Elias
Israel Pietas. Then does another version.
Schott. 1674. Art. V, p. 58. M = (1, 2, 4). Vowel
mnemonics: Salve Certa
Animæ Semita Vita
Quies; Pallētis Evandri
Sanguine Feritas Imane (the
m should have an overbar)
Vigebat.
Ozanam. 1694. Prob. 28, 1696: 83; 1708: 74.
Prob. 32: 1725: 217-218. Prob.
10, 1778: 154-155; 1803:
154-155; 1814: 136-137. Prob. 9, 1840: 70. M = (6, 4, 3). No
mnemonics.
Ozanam. 1725. Prob. 46, 1725: 250-253. Prob. 12, 1778: 158-161; 1803: 159-161; 1814: 140-142. Prob.
11, 1840: 72. M = (1, 2, 4). Par fer
Cesar jadis devint
si grand Prince and
Salve certa anima semita vita quies.
1778 et seq. has César and
animæ.
Minguet. 1733. Pp. 176-180 (1755: 127-129; not noticed in
1822; 1864: 164-166).
M = (1, 2, 4) as
in Chuquet. Vowel mnemonic: Aperi,
Premati, Magister, Nihil,
Femina, Vispane, Vispena.
Alberti. 1747. Part 2, p. ?? (69). M = (6, 4, 3) translated from Ozanam, 1725, prob. 32.
Alberti. 1747. Part 2, pp. ?? (76-77). M = (1, 2, 4) as in Ozanam, 1725, prob. 46, but he first gives an Italian
mnemonic: Aperì Prelati
Magister Camille Perina
Quid habes Ribera. He explains the usage using 4 = Camille
as example, but later notes that
4 never occurs! Then gives
Salva certa anima semita vita quies;
Perfer Cesar Jadis devint sigrand Prince; Pare ella ai segni; Vita, Piè.
Les Amusemens.
1749. Prob. 13, pp. 134-135: Les
trois Bijoux. M = (1, 2,
4). Par fer, César jadis devint
si grand Prince.
Hooper. Rational
Recreations. Op. cit. in 4.A.1. 1774.
Recreation IX: The confederate counters, pp. 34-36. M = (1, 2, 4). Par fer Cesar jadis
devint si grand prince.
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
Pp. 85-86, no. 131: Three persons having each chosen privately one out
of three things, to tell them which they have chosen. Salve certa anima
semita vita quies.
Manuel des Sorciers.
1825. ??NX
Endless Amusement II.
1826?
Young Man's Book.
1839. Pp. 198-199. Identical to Endless Amusement II, p. 179.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Parlour magic, no. xii, pp. 196-198 (1868: 209-210): How to discover the
possessors of any articles taken from the table during your absence. Salve
cesta animæ semita
vita quies.
Magician's Own Book.
1857.
Boy's Own Conjuring Book.
1860.
Vinot. 1860.
Boy's Own Book.
Divination by cards. 1868:
637-638. Basically M = (1, 2, 4) but set up as (2, 3, 5). Vowel
mnemonic: Par fer César jadis
devint si grand prince.
Magician's Own Book (UK version). 1871. To tell which
article each of three persons took, pp. 35-36.
Mnemonics: Take her certain
anise seedlings Ida
quince; Salve certa
animæe servita vita
quies; Par-fer Cæsar
jadis devint si-grand
prince.
Hanky Panky.
1872. A new three-card trick,
pp. 256-257. M = (6, 4, 3), which is equivalent to (4, 2, 1).
Bachet-Labosne.
Problemes. 3rd ed., 1874. Supp. prob. VII, 1884: 192-193. M = (6, 4, 3), which is the same as (4, 2, 1).
No mnemonics.
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Simple Tricks by
Calculation No. I: To tell three persons which card each one has chosen, pp.
32-33. M = (4, 2, 1).
É. Ducret.
Récréations Mathématiques. Op.
cit. in 4.A.1. 1892? Pp. 107-?? (didn't photocopy the following pages):
Les trois Bijoux. Seems to be usual
form.
Lucas.
L'Arithmétique Amusante.
1895. Prob. VIII-X, pp.
21-25. M = (1, 2, 4) with result subtracted from 18:
Par fer César jadis
devint si grand prince;
Avec éclat l'Aï
brillant devint libre.
M = (0, 1, 3) with result
subtracted from 12: Il a
jadis brillé dans ce
petit État.
Meyer. Big Fun
Book. 1940. The picture card trick, pp. 509-510. Uses K, Q, J. Forms
P = (2a1-1)m1 + (2a2-1)m2 + (2a3-1)m3 for multipliers (mi) = (2, 3, 6).
Then examines the remainder of the deck, which has 49 - P = R
cards. 2 - [R/11] is the person holding the J. A
counting process, which is erroneously described, gives the person with the Q as
congruent to R + [R/11] (mod 3). In fact
3 - R(mod 3) is the position of
the K.
Gardner.
MM&M. 1956. The purloined objects, pp. 57-59. Gives several mnemonics using consonants. E.g. when the objects are Toothpick, Lipstick and Ring, use:
tailor altar trail
alert rattle relate.
If the objects are denoted by A,
B, C, use: Abie's bank account
becomes cash club;
or if they are denoted Small,
Medium, Large, use: Sam
moves slowly (since)
mule lost limb.
7.AP. KNOWING SUM VS KNOWING PRODUCT
Two persons are told the sum and
product of two integers. They then have
some conversation such as: "I
don't know what the numbers are."
"I knew that." "I
now know what they are." "So
do I." What were the numbers? This seems to be a recent problem and I have
only a few references. There are older
versions, often called census-taker problems, where one knows the sum and
product of three numbers (usually ages), but needs more information (often
whether there is an oldest, which eliminates twins).
Thanks to Leroy F. Meyers for many
of these references.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940. No. 96: The
church afloat, pp. 53 & 135. Three
ages: product = 840, sum is twice the curate's age. This is insufficient, but whether the eldest
is older or younger than the vicar is sufficient to decide.
Lester R. Ford, proposer.
Problem E776. AMM 54 (1947) 339
& 55:3 (Mar 1948) 159-160. ??NYS/NX
-- proposal is quoted and the location of the solution given in the Otto Dunkel
Memorial Problem Book (= AMM 64:7, part II, (Aug-Sep 1957) 61 & 89) and the
location is more specifically given by Meyers & See. Four families of different sizes, not enough
to form two baseball teams (i.e. the total is
< 18), and the product of the numbers is the host's house
number. The guest says he needs more
information -- does the smallest family have just one child? The host's answer allows him to determine
the numbers -- what were they?
Meyers writes that he first heard of the census-taker
problem in 1951.
AMM problem E1126, 1954?.
??NYS.
W. A. Hockings, proposer; A. R. Hyde, solver. Problem E1156 -- The dimensions of Jones's
ranch. AMM 62 (1955) 181 (??NX) &
63:1 (Jan 1956) 39-42. This is a
continuation of E1126 and involves four men and their ranches.
Hubert Phillips. My
Best Puzzles in Mathematics. Dover,
1961. Prob. 87: The professor's
daughter, pp. 48 & 101. Youngest
daughter is at least three. The product
of their ages is 1200 and the sum is ten less than the wife's
age. Visitor computes and then makes
two wrong guesses as to the age of the youngest daughter. How old is the wife? The fact that the visitor made two wrong
guesses means there must be at least three sets of ages, all ³ 3, with product 1200 and the same sum, and indeed there is just
one such situation, and this has three daughters. Allowing younger children permits some more complicated
possibilities since we have
1
+ 3 + 20 + 20 = 2 + 2 + 10 + 30 =
3 + 16 + 25
= 4 + 10 + 30
and 2 + 2 + 15 + 20 = 4
+ 15 + 20 = 5 + 10 + 24 = 6 + 8 + 25.
(Phillips had
most of these problems in his newspaper and magazine columns so it is likely
that this will turn up in the period 1930-1950.)
M. H. Greenblatt.
Mathematical Entertainments, op. cit. in 6.U.2, 1968. Chap. 1: "Census-taker" problems,
pp. 1-7. Says he believes these
problems came from some wartime project at MIT. Discusses three similar types.
1:
The neighborhood census. Product of
three ages is 1296 and the sum is the house number. Census taker asks if any of them are older
than than the informant, who says 'no' and then the census taker knows the
ages.
2:
The priest and the banker. Product of
three ages of ladies with the banker is
2450 and the sum is the same as
the priest's age. The priest says this
is insufficient and asks if any of the ladies is as old as the banker. When he says 'no', the priest knows all the
ages. This even determines the banker's
age!
3:
The three Martians. Product of three
ages is 1252 and the sum is the age of the informant Martian's father. The interrogator says this is insufficient
and asks if any of the Martians is as old as the informant. When he answers 'no', the interrogator knows
the ages. When you start on the
problem, you find that
1252 = 22·313
is not a suitable number.
However, the problem shows a drawing of the Martian and one see he only
has three fingers on each hand!
Interpreting 1252 as a base
6 number gives the decimal 320
and the problem proceeds as before.
Gardner. SA (Nov
& Dec 1970) c= Wheels, Chap. 3,
prob. 10: The child with the wart.
Supplied by Mel Stover. The
product of the ages of three children is
36 and the sum is the
questioner's house number. When the
questioner says the information is insufficient, the father says the oldest
child has a wart, which is sufficient to determine the ages. See Meyers & See for a generalization.
A. K. Austin. A
calculus for know/don't know problems.
MM 49:1 (Jan 1976) 12-14. He
develops a set-theoretic calculus for systematically solving problems involving
spots on foreheads, etc. (see 9.D), including problems similar to the present
section. He gives the following problem
considered by Conway and Patterson, but apparently unpublished. Two persons each have a card on their back
bearing a positive integer, visible to the other person, but not to the
person. They are told that the sum of
the two integers is 6 or
7. They are then asked, in turn,
to state whether they know what their own integer is. If they both have 3, what is the sequence of responses? Austin finds there are five 'no' answers,
then the answers alternate yes, no.
David J. Sprows, proposer;
Problem Solving Group, Bern, solver.
Prob. 977 -- Mr. P. and Ms. S.
MM 49:2 (Mar 1976) 96 & 50:5 (Nov 1977) 268. P & S are given the product and sum of
two integers a, b, greater than one, but the sum is £ 100. P says he doesn't know the numbers. S says she knew that. P
replies that he now knows the values and S responds that so does she. The unique answer is a, b = 4, 13. Editor cites the above AMM problems, though they are not quite
the same type. See the discussion
below.
Gardner. SA 241:6
(Dec 1979) 20-24. Problem 1: The
impossible problem. Gives a version of
Sprows' problem and says the problem was sent by Mel Stover and had been
circulating for a year or two. This
version assumes the numbers are greater than
1 but at most 20,
which gives the unique solution
4, 13. However Gardner asserts
that the solution remains the same if the bound is increased to 100 -- but I think there are further
answers, e.g. 4, 61, see the discussion below. Stover says a computer program has checked
and found no further solutions up to
2,000,000 and it may be that
there is no further solution when the upper bound is removed -- this is a
mistake of some sort, see the discussion below. Further, Kiltinen & Young say they had a letter from Gardner
conjecturing that there are infinitely many solutions.
John O. Kiltinen
& Peter B. Young. Goldbach, Lemoine, and a know/don't know
problem. MM 58:4 (Sep 1985)
195-203. This discusses the Sprows
problem, without the bound on S and various generalizations involving more
stages of conversation. This leads to
use of Goldbach's Conjecture and a conjecture of Lemoine that every odd number ³ 7 can be expressed as 2p +
q, where p, q are odd primes.
Friedrich Wille.
Humor in der Mathematik.
Vandenhoek & Ruprecht, Göttingen, (1984), 3rd ed., 1987. Paul und Simon, pp. 62 & 121‑122. This has solution 4, 13. Note to the
solution says there was much correspondence after the 1st ed., leading to a new
solution and the computation of further examples with Sum = 17, 65,
89, 127, 137,
163, 179, 185,
191, 233, 247,
269, 305.
I
recently was sent a version of the Sprows problem, without a bound on S,
by Adrian Seville and solved it before checking this section. Dr. S (= Simon) and Dr. P (= Paul) are given
the sum and product of two integers greater than 1. Dr. S says he doesn't
know the numbers, but he can tell that Dr. P cannot determine the numbers. After a bit, Dr. P says he now knows the
numbers. After a bit more, Dr. S says
he now knows the numbers also. The
version I received implies there is a unique solution, but there are more. Wille's discussion notes that the possible
values of the sum, after Dr. S's statement are the {odd composites + 2},
which much simplifies the analysis.
I found some further values of
Sum: 343, 427,
457 and then extended considerably, finding: 547, 569, 583,
613, 637, 667, 673, 697, 733,
757, 779, 787,
817, 821, 853,
929, 967, 977,
989, 997, giving
36 values less than 1000
and there are another 42 solutions between 1000 and 2000.
In looking at the earlier solutions, it appeared that one of the two numbers
was always a power of 2, with the other number being odd, but for S = 757,
P = 111756, one has
the numbers being 202 and
556. Five more counterexamples
to the initial appearance occur up to
2000.
While
doing this investigation, I wondered what would happen if the condition
'greater than 1' was reduced to 'positive'. After some
calculation, it became apparent and is provable that all solutions have P = S - 1
with the numbers being 1 and
S - 1. Solutions occur for S =
5, 9, 10, 16, 28,
33, 34, 36,
46, 50, 52,
66, 78, 82,
88, 92, 96,
giving 17 values less than 100, and there are
another 87 values between 100 and
1000, making a total of 104
values less than 1000, which is where I stopped. If one assumes Goldbach's Conjecture, or
part of it, one can show there are infinitely many solutions of the form S = pq + 1,
where p, q are primes such that p + q = r + 1 for a prime
r. However, I have not been able
to see if there are infinitely many solutions of the original form of the
problem.
I
later was referred to the MM problem 977 and was surprised to see that the
solution is unique when S £
100 is imposed -- offhand, one might
expect S = 65 and 89 to also be solutions in this case, but the
additional knowledge about S affects the intermediate stages of the
deduction. Rerunning an early version
of my program with additional printout, I find that one needs S < 107
to prevent S = 65 being a solution. I then found Gardner's 1979 version and examination of the same
data indicates that
S = 65 is a solution
when the numbers are bounded by 100. All in all, I find the presence of a bound
has subtle effects and I find it unsatisfying.
Tim Sole. The Ticket
to Heaven and Other Superior Puzzles.
Penguin, 1988. Standard and
Poor, pp. 116 & 126‑128.
Solution is 2, 15.
Leroy F. Meyers & Richard See. The census-taker problem.
MM 63:2 (Apr 1990) 86-88. The
product of the three ages in the next house is
1296 and the sum is the house
number. The census-taker then asks if
any of those people is older than the respondent. When he says 'yes', the census-taker says he knows the ages. This is a slight variant on Greenblatt's
type 1. Authors investigate what other
values, N, can be used instead of
1296. That is, we want N = ABC = DEF with
A + B + C = D + E + F with just
two such triples A, B, C; D, E, F. They determine the simple forms of such N and list the 45
examples £
1296. The first few are: 36,
40, 72, 96,
126, 176, 200,
234, 252, 280,
297, 320, 408,
520, 550, 576,
588, 600. They give a number of references to similar
questions.
Nicolas Guerrero.
Problem 184.7 -- Deux nombres.
M500 184 (Feb 2002) 25. The
problem is given in French, but I will translate and paraphrase. Dr. P is given the product and Dr. S is
given the sum of two integers between 2 and 100. (This is ambiguous -- the range might be [2, 100]
or [3, 99].) Dr. P says he doesn't know the numbers. Dr. S then says he doesn't know the numbers. Dr. P then says he knows them, and Dr. S
then says he knows them.
ADF [Tony Forbes].
Editorial remark on Problem 184.7.
M500 188 (Oct 2002) 26. His
citation is erroneously to M500 185, p. 25.
He says he has not been able to obtain a satisfactory answer, in
particular his answer is not unique.
My
analysis gives four answers for S,
P in the range [2, 100]:
a, b, S,
P =
2, 6, 8, 12; 4, 19,
23, 76; 4, 23, 27, 92; 7, 14, 21, 98. However, it is clear that the intention is for just a, b to be in this range and this leads to so many possible values
of P
that a computer must be used. My
program finds two solutions:
a, b, S,
P =
2, 6, 8, 12; 84, 88, 172,
7392. Adapting the program to the range [2, 200],
the larger solution disappears, but is replaced by six solutions
with P
larger than 30,000. I suspect that the proposer of the problem
may have intended the earlier problem of Sprows, but either misheard or
misremembered the exact details.
If
one takes range [3, 99] for all the variables, the calculation is a little simpler, and
there are three solutions: a, b, S,
P =
3, 8, 11, 24; 9, 9, 18, 81; 9, 11, 20, 99. But there is an extra feature -- the values a, b, S, P
= 6, 13, 19, 78; 8, 11, 19, 88 lead to the above sequence of statements except that Dr. S's last
statement is that he still doesn't know the numbers, but we know his sum!
7.AQ. NUMBERS IN ALPHABETIC ORDER
I recall the problem of figuring out
the reason for the sequence: 8, 5, 4,
9, 1, 7, 6, 3, 2 from 1956 or 1957.
?? Alphabetic Number
Tables, 0 ‑ 1000. MIT, Cambridge,
Massachusetts, 1972. ??NYS.
"Raja" [= Richard & Josephine Andree]. Puzzle Potpourri #3. Raja Books, Norman, Oklahoma, 1976. No. 17: volumes 1, 2, ..., 12 shelved alphabetically.
Harvey & Robert Dubner.
A tabulation of the prime numbers in the range of one to one‑thousand,
in English and in Roman numerals, in alphabetic order. JRM 24 (1992) 89-93.
The basic process is to take a three
digit number, find the difference between it and its reversal, then add that
result to its reversal and one gets
1089. The items by Meyer and
Langford note that 9 * 1089 =
9801, so this is connected to 7.AH.
I now think that this ought to have
originated from the fact that a number minus its reversal is divisible by nine,
cf Berkeley & Rowland below and Section 7.K.1, but the monetary version
seems to have arisen first. I have a
note that I have seen a c1881 reference to the monetary version of this
problem, but I cannot find it.
Prof. Orchard, proposer;
Prof. Anderson, Rev. H. Sewell, et al, solvers. Prob. 10441. Mathematical Questions with Their Solutions from the
"Educational Times" [generally known as Educational Times Reprints]
53 (1890) 78-79. Prove and generalise
the fact that the process done with old English money gives £12 18s 11d. Solution says that if the multipliers of the units are m
and n so that a, b, c denotes
amn + bm + c, then the result of
the calculation is m, n-2, m-1. The proposal makes no mention of the
problems when the number of pounds is
12 or more or when the numbers
of pounds and pence are equal and assumes the reversal is less than the original,
but the solver assumes that a <
m to make the reversal be an ordinary
sum of money and then that
c < a "to make
the subtraction possible". Mr.
Davis notes that this trick has been "current in well-informed
City-circles for some months." If
you perform it more than once, instruct the victim to add some convenient
amounts to the result.
Don Lemon.
Everybody's Scrap Book of Curious Facts. Saxon, London, 1890.
Curious arithmetical puzzle, pp. 302-303. Quotes a letter to the London Globe from a 'constant reader'
asking for an explanation for the monetary version, assuming you start with
less than £11 19s 11d and the number of pence is less than the
number of pounds.
Ball. MRE, 1st ed.,
1892.
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Card Puzzles No. X: A
subtraction sum, p. 10. A 3-digit number minus its reversal has the
form a9c, where a + c = 9.
Somerville Gibney.
An arithmetical flourish for drawing-room shows. The Boy's Own Paper 15 (No. 734) (4 Feb
1893) 299. Monetary version,
getting £12 18s 11d. "An arithmetical curiosity which is
quite new ...."
Somerville Gibney. Re
The arithmetical flourish for drawing-room shows. The Boy's Own Paper 15 (No. 750) (27 May 1893) 556. Explains why it works. Correspondents from various places do
it: in francs and centimes, getting 99 99;
in thalers and grosschen
(30 grosschen = a thaler), getting 29 29; and in avoirdupois, getting 28cwt 2qr 27lb.
Ball‑FitzPatrick.
1st ed., 1898, pp. 14‑15.
Shows the property of 1089. A remark of the translator shows the answer
in base b is (b‑1) * (b+1)2,
which is
b b2
+ (b-2) b + (b-1).
Carroll-Collingwood.
1899. P. 269 (Collins 194). = John Fisher; The Magic of Lewis
Carroll; op. cit. in 1; pp. 244-245. =
Carroll-Wakeling II, prob. 25: Pounds, shillings and pence, pp. 41 &
71, Monetary version, with number of
pounds 'not more than twelve', giving
£12 18s 11d. Carroll assumes the
reversal is less than the original number and says it works "whatever
numbers may have been selected."
Mentioned in Carroll-Gardner, p. 76.
Neither Collingwood nor Wakeling nor Gardner note the difficulty when
numbers of pounds and pence are equal nor the question of whether the reversal
is actually smaller than the original, but Fisher does, and also mentions 1089.
Collingwood says he believes this was invented by Carroll and this seems
reasonable, but see Orchard et al, 1890, above.
Clark. Mental
Nuts. 1904, no. 81; 1916, no. 54. Mental telegraphy. As in Berkeley & Rowland.
Ball. MRE, 4th ed.,
1905, p. 9. Adds a section on 1089
just before the monetary version.
Laisant. Op. cit. in
6.P.1. 1906. Chap. 18: Opérations curieuses: No. 1, p. 43. 1089.
Notes that the two end digits must be different.
E. N. Barisien.
?? Suppl. al Periodico di Mat.
13 (1909) 20-21. ??NYS -- cited by Dickson
I 462, item 53, but the interesting material is cited in the following.
E. Nannei. ?? Suppl. al Periodico di Mat. 14 (1910/11)
17-20. ??NYS -- cited by Dickson
I 462, item 55, as treating a problem of Barisen. Takes a
6-digit number, subtracts its reversal and then adds the reversal of
that to itself. Result is one of 13
values: 0, 9900,
..., 1099989.
M. Adams. Indoor
Games. 1912. Magic money, 354-355.
Monetary version getting
£12 18s 11d.
Clark. Mental
Nuts. 1916, no. 53. Foretelling the answer. Gives the rule and an example, saying the
answer will always be 1089, though he doesn't make any restriction on
the original number.
Dudeney. AM. 1917.
Prob. 33: A puzzle in reversals, pp. 5 & 151. Says most people know that the money process
leads to £12 18s 11d -- provided you start with less than 12 in
the £
place and the number of pounds exceeds the number of pence. If the number of pounds can be 12
or more, then what is the smallest sum for which the process fails and the
largest sum for which it works?
T. O'Conor Sloane.
Rapid Arithmetic. Quick and
Special Methods in Arithmetical Calculation Together with a Collection of
Puzzles and Curiosities of Numbers. Van
Nostrand, 1922. [Combined into: T.
O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures;
Van Nostrand, 1939.] A mystery in
money, pp. 178-180. Uses US money to
get $10.89 and also gives the English result £12 18s 11d.
Peano. Giochi. 1924.
Hummerston. Fun,
Mirth & Mystery. 1924.
A. B. Nordmann. One
Hundred More Parlour Tricks and Problems.
Wells, Gardner, Darton & Co., London, nd [1927 -- BMC]. No. 60: Mental arithmetic trick, pp.
51-52. Monetary version.
Kraitchik. Math. des
Jeux. Op. cit. in 4.A.2. 1930.
Chap. IV, no. 7, p. 55. Notes
that the first result is a multiple of
99, but doesn't find 1089.
Not in his Math. Recreations.
Perelman.
1934?? Guessing a number without
asking anything. FFF (1934). Not in 1957 ed. 1979: prob. 14, pp. 33-35.
= MCBF (1937), prob. 14, pp. 32-33.
Basic property of 1089.
Rohrbough. Brain
Resters and Testers. c1935. Brain Teaser, p. 15. Assumes digits are in descending order.
Freda Holmdahl, ed.
The Rainy Day Book. Nelson,
Edinburgh, 1936. Figure tricks,
pp. 196‑197. Monetary
version.
Depew. Cokesbury
Game Book. 1939. Descending numbers, p. 205. 1089.
"Willane".
Willane's Wizardry. Academy of
Recorded Crafts, Arts and Sciences, Croydon, 1947. Curious figures, p. 45.
Version with English money, with
£ less than 12.
Anonymous. Problems
drive. Eureka 18 (Oct 1955) 15-17 &
21. No. 9. Usual 1089 problem.
P. M. Seeviour & M. Keates. Note 3077: Fibonacci
again. MG 48 (No. 363) (Feb 1964) 78‑79. Cites Kraitchik, Math. des Jeux. Determines all the possible results of the
process for n‑digit numbers in
base b, as being (b2 ‑
1) times certain n-digit numbers formed of 0s
and 1s, corresponding to the borrows in the
subtraction. They find the number t(n)
of such values is given by
t(2n) = t(2n+1) = F2n+1 where
Fm are the Fibonacci
numbers. They do this via the case when
the first and last digits are distinct for which it is stated that a certain
recurrence holds, but this is not immediate -- I spent about half a day before
I could see it and it would still require some work to make a careful
proof. For convenience, I list the first
solutions for the case of different end digits.
n =
2. 99.
n =
3. 1089.
n =
4. 10989, 9999, 10890.
n =
5. 109989, 99099, 109890.
n =
6. 1098900, 1099989, 1090089, 1099890,
1089990, 999999, 990099,
991089.
n =
7. 10998900, 10999989, 10891089, 10999890,
10890990, 10008999, 9900099,
9901089.
C. Dudley Langford.
Note 3102: A party puzzle
extended (Cf Note 3077). MG 48 (No.
366) (Dec 1964) 432. If a + b = 10,
then 1089*a and
1089*b are reversals. 1089
can be replaced by 99 * 11...1 and he asserts that these, up to
trailing 0s, are the numbers obtained in Note 3077. However, his numbers are:
99, 1089, 10989,
109989, 1099989, ..., 109...989,
..., and these do not immediately yield
the above solutions.
Jerome S. Meyer.
Arithmetricks. Scholastic Book
Services, NY, 1965. The 2178
trick, pp. 3-4. Gets 1089,
but he doubles the result at the end to produce 2178.
Michael Holt. What
is the New Maths? Anthony Blond,
London, 1967. Pp. 88-89. Gives a four digit version, saying to choose
the least significant two digits as
< 5 and the most significant
as ³ 5, then the result is always
10890.
Walter Gibson. Big
Book of Magic for All Ages. Kaye &
Ward, Kingswood, Surrey, 1982. Magic
Math & Super Math, pp. 108-109.
After doing 1089, he considers doing the same process with
five digit numbers. For convenience, he
assumes the digits are all distinct.
Then only two possible answers can arise: 99099 or 109890.
No proof given.
Paul Swinford. The
Wondrous World of Numberplay & Wordplay.
A Lecture by Paul Swinford.
Published by the author, 1999.
Pp. 1-4. Covers 1089 with a
variety of ways to use the result. On
pp. 3-4, he considers four digit versions and is unhappy to often get 10890.
He modifies the process so that one only exchanges the end digits and
finds that this always leads to
10989. (For a k-digit number, the process will lead to 1099..989,
with k-3 9s
in the middle.)
New section. There must be older examples. The usual form uses b
butts to make a cigarette and the person has found B = b2 butts which give him
b+1 smokes. The Scotts are the only ones to realise that B
can more generally be such that
b-1 divides B-1,
giving (B‑1)/(b-1) smokes.
I have just found the general result that the number of smokes is ë(B‑1)/(b-1)û. Taking
b = 2 gives the result that a
knockout tennis tournament with B players has
B-1 matches.
On 15 Nov 2001, Willy Moser
mentioned that if the tramp borrows a butt, he gets ëB/(b-1)û smokes with a butt left over to return to
his friend! He thought this was well
known, but I'd never heard or seen it before
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 192, pp. 74 & 158: Scrooge the
miser. Starts with 125
cigarettes, but saves all his butts, using 5 to make a cigarette, so
he gets 31 further cigarettes.
Rudin. 1936. No. 49, pp. 16 & 85. 36
butts, using 6 to make a cigarette.
J. R. Evans. The
Junior Week‑End Book. Op. cit. in
6.AF. 1939. Prob. 47, pp. 266 & 272.
Tramp has 49 butts and uses seven to make a
cigarette. How many cigarettes can he
make?
Harold Hart. The
World's Best Puzzles. Home Recreation
Library, NY, 1943. Smoke gets in your
eyes, pp. 22 & 54. 125 butts and five make a cigarette.
Leopold. At
Ease! 1943. Iffs and Butts, pp. 10-11 & 196. 36 butts, six butts to a
cigarette.
Sullivan.
Unusual. 1943. Prob. 9: A bum cigarette. As in Rudin.
J. S. R. Cameron. A
tennis problem. Eureka 10 (Mar 1948) 18 & 11 (Jan 1949)
31. Complicated system of byes in a
tennis tournament with 100 players.
Solution notes that the system of byes is irrelevant! This must have been known much earlier.
Jonathan Always.
Puzzles to Puzzle You. Op. cit.
in 5.K.2. 1965. No. 131: Cigarettes this way, pp. 41 &
89. As in Evans.
[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers.
1973. Op. cit. in 5.E. Hard times made easier, pp. 35-36. 41
butts, six butts to a cigarette.
David Singmaster.
Cigarette butts. MS 31:2
(1998/9) 40. The number of smokes
is ë(B‑1)/(b-1)û. Taking
b = 2 gives the result that a
knockout tennis tournament with B players has
B-1 matches.
New
section. This must appear in the 19C??
Loyd.
Cyclopedia. 1914. Pp. 327 & 383. = SLAHP, Timing a bookworm, pp. 69 & 112. 2 volumes.
Dudeney. AM. 1917.
Prob. 420: The industrious bookworm, pp. 143-144 & 248-249. 3 volumes.
Peano. Giochi. 1924.
Prob. 10, p. 4. 3 volumes.
King. Best 100. 1927.
No. 3, pp. 8 & 39. = Foulsham's,
no. 1, pp. 5 & 10. 3 volumes.
William P. Keasby.
The Big Trick and Puzzle Book.
Whitman Publishing, Racine, Wisconsin, 1929. Pp. 37 & 63.
Streeter & Hoehn.
Op. cit. in 7.AE. Vol. 1, 1932,
p. 300, no. 10: "Brain
twister".
Haldeman-Julius. 1937. No. 33: The bookworm problem, pp. 6 &
22. Three volumes.
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
Bookworm's pilgrimage, pp. 163 & 215. 4 volumes.
McKay. At Home
Tonight. 1940. Prob. 18: The bookworm, pp. 66 &
79-80. 2 volumes.
The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. P. 148, prob. 4. 2 volumes.
Leopold. At
Ease! 1943. Way of a worm, pp. 9 & 195.
Owen Grant. Popular
Party Games. Universal, London, nd
[1940s?]. Prob. 10, pp. 37 &
50. 3 volumes.
Birtwistle. Math.
Puzzles & Perplexities. 1971. Second, pp. 145 & 196. 4
volumes.
7.AU. NUMBER OF CUTS TO MAKE N PIECES
New section. A typical problem is that it takes twice as
long to make three pieces as it does to make two pieces. This will include other situations where one
has to change from n to
n-1.
If one can overlay the material,
then one can reduce the number of cuts.
If one can fold the material, then one can reduce to one cut as in 6.BG.
I've just added the problem of
cutting a cube into 27 pieces here -- this must be much older??
See also 7.AV.
Alcuin. 9C. Prob. 15: De homine. "How many furrows has a man made ...,
when he has made three turnings at each end of the field?" The Alcuin text gives six, but Bede, Folkerts
and Folkerts & Gericke give seven.
Child. Girl's Own
Book. Arithmetical puzzles. 1832: No. 10, pp. 171 & 179; 1833: No. 10, pp. 185 & 193 (answer
numbered 8); 1839: No. 10, pp. 156
& 173; 1842: No. 10, pp. 284
& 291; 1876: No. 8, pp. 232
& 244. "If you cut thirty
yards of cloth into one-yard pieces, and cut one yard every day, how long will
it take you?" The 1839, 1842 &
1876 texts omit the hyphen in one-yard.
I didn't copy the exact text from the 1832.
=
Fireside Amusements, 1850, Prob. 8, pp. 132 & 184.
The Sociable.
1858. Prob. 46: A dozen
quibbles: part 9, pp. 300 & 318.
"If you cut thirty yards of cloth into one yard pieces, and cut one
yard every day, how long will it take?"
= Book of 500 Puzzles, 1859, prob. 46: part 9, pp. 18 & 36. c= Depew; Cokesbury Game Book; 1939; Cutting
up, p. 216. c= Magician's Own Book (UK
version), 1871, Paradoxes [no. 5], p. 38.
= Wehman, New Book of 200 Puzzles, 1908, p. 50.
Hoffmann. 1893. Chap. IX, no. 24: The draper's puzzle, pp.
318 & 326 = Hoffmann-Hordern, p. 211.
Lucas.
L'Arithmétique Amusante.
1895. Prob. XII: La coupe du
tailleur, p. 26. How long to cut a
piece of length 16 into lengths of length 2?
[This makes the problem more deceptive.]
H. D. Northrop.
Popular Pastimes. 1901. No. 10: A dozen quibbles, no. 9, pp. 68
& 73. = The Sociable.
Benson. 1904. The tailor's puzzle, p. 227.
Hummerston. Fun,
Mirth & Mystery. 1924. Some queer puzzles, Puzzle no. 76, part 4,
pp. 164 & 183. Cut twenty yards
into yard lengths.
King. Best 100. 1927.
No. 24, pp. 15 & 44.
Cut 90 yards of cloth into
90 lengths.
Meyer. Big Fun
Book. 1940.
Birtwistle. Math.
Puzzles & Perplexities. 1971. Fifth, pp. 58 & 190. How many cuts to make 27
cubes from a cube?
Putnam. Puzzle
Fun. 1978. Nos. 93-96: Cube and
27 cubes, pp. 13 & 37. How many cuts to make 27
cubes from a cube? Other questions
deal with colourings.
7.AV. HOW LONG TO STRIKE TWELVE?
New
section. I have now started adding rate
problems of this type.
See also
7.AU.
King. Best 100. 1927.
No. 36, pp. 19 & 47. If a
clock takes 8 sec to strike 8, how long does it take to strike 12?
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 186, pp. 72 & 157: Big Ben. "If it takes Big Ben ten seconds to
strike five o'clock, how long will it take to strike twelve o'clock?"
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
Is your brain working? -- no. 2, pp. 148 & 215. If Big Ben takes 4 seconds to strike 4,
how long does it take to strike midnight?
Depew. Cokesbury
Game Book. 1939. The clock, p. 217. If it takes 7 seconds to strike 7, how long does it take
to strike 11.
Sullivan.
Unusual. 1943. Prob. 25: Caution. If it takes 8 minutes to pass 8 stop lights, how long
does it take to pass 12?
W. A. Bagley. Puzzle
Pie. Op. cit. in 5.D.5. 1944.
No. 25: The patriot's paradox, pp. 30-31. 60 shots in an hour is
not the same as one shot per minute, since the latter requires 61
shots to be fired in a 60 minute interval from the first shot.
Ripley's Puzzles and Games.
1966. P. 70. "If 10 sheep jump over a fence in 10
minutes how many will jump over in an
hour?" Answer: 55.
Barbara Lee. Six
short problems. M500 162 (?? 1998) ??NYS -- cited by solver. A, J. Welton. Solutions to 'Six short problems'. M500 163 (Aug 1998) 12-13.
No. 3: "A clock strikes six in 5 seconds. How long does it take to strike twelve?" "Six hours and five seconds."
There are three (or more) ways of
demonstrating this result -- e.g.
7 doesn't go into 2,
but it goes into 8 once, leaving us with 21
and 7 goes into 21 3
times. The other ways are based
on multiplying 13 by
7 and on adding seven 13s. This is the basis of
a vaudeville routine which was used in the Abbott and Costello movie 'In the
Navy', 1949. I can only recall seeing
it in the following.
Rohrbough. Brain
Resters and Testers. c1935. Tricks for the Toastmaster (no. 2),
pp. 12-13.
W. A. Bagley.
Paradox Pie. Op. cit. in
6.BN. 1944. No. 2: Financial skullduggery, pp. 7‑8.
Bud Abbott and Lou Costello. In the Navy. Universal
Pictures, 1941. Text and stills in
Richard J. Anobile, ed.; Who's On First?
Verbal and Visual Gems from the Films of Abbott & Costello; Darien
House, NY, 1972; Studio Vista, London, 1973.
Pp. 128‑139.
William R. Ransom.
Op. cit. in 6.M. 1955. Fake arithmetic, pp. 9‑11. Shows
there are 22 examples of this phoney arithmetic.
Robert Harbin. Party
Lines. Op. cit. in 5.B.1. 1963.
The Christmas club, p. 62.
The archetypal example is to
determine A, B such that
A + B = A * B, usually with some additional condition. I have not noticed early examples of such problems until recently
reading Riese's Die Coss and later Muscarello, but I recall other later
examples such as Briggs & Bryan.
Muscarello.
1478. Ff. 69r-69v, p. 180. A + B
= A * B. He gives examples: 6 & 6/5; 7 & 7/6; 8 & 8/7; and says it continues.
Calandri.
Arimethrica. 1491. F. 97v.
A + B = A * B.
He finds 5/2, 5/3.
Riese. Die
Coss. 1524.
Ozanam. 1694. Prob. 5, question 14, 1696: 13; 12.
Prob. 8, quest. 14, 1725: 27-28.
A + B = A * B. Gives (a+b)/a and
(a+b)/b, e.g. 5/2
& 5/3.
Eadon.
Repository. 1794.
Manuel des Sorciers.
1825. ??NX P. 85.
A + B = A * B.
Walter Taylor. The
Indian Juvenile Arithmetic .... Op.
cit. in 5.B. 1849. Pp. 193-194. Solves A + B = A
* B as
A = (a+b)/a, B = (a+b)/b for any
a, b.
Mittenzwey.
1880. Prob. 80, pp. 15 &
66; 1895?: 88, pp. 20 & 69; 1917: 88, pp. 18 & 66. A + B
= 3 (A - B); AB = 3 (A + B).
1917 gives an algebraic solution.
Briggs & Bryan.
The Tutorial Arithmetic, Part II.
Op. cit. in 7.H. 1898. Exercises X, prob. 16, pp. 124 &
579.
A + B = A * B = A2 ‑ B2.
C. T. C. Wall. The
(σ, x, p) problem. Eureka 20 (Oct
1957) 20-25. This starts with
finding x, y, p, q such that
x + y = pq, xy = p + q. There are several special solutions: x, ‑1, 1-x, -1; x, -x, 0, -x2 and two sporadic solutions: 3, 2, 5, 1;
2, 2, 2, 2. He then considers
more variables such that the symmetric functions of one set are the same as for
the other set, but in reverse order.
Nathan Altshiller Court.
Mathematics in Fun and in Earnest.
Op. cit. in 5.B. 1961. Prob. e, pp. 189-190. A - B
= A/B.
Jerome S. Meyer.
Arithmetricks. Scholastic Book
Services, NY, 1965. Pp. 35-36. Considers all four cases.
A * B = A
+ B. (n+1)/n and
n+1.
A * B = A
- B. n and n/(n+1).
A / B = A
+ B. n + 1/(n+2) and
(n+1)/(n+2).
A / B = A
- B. n + 1/(n-2) and
n-1.
Bronnie Cunningham.
Funny Business. An Amazing
Collection of Odd and Curious Facts with Some Jokes and Puzzles Too. Puffin, 1978. Pp. 37 & 142. A
- B =
A/B = 4.
A
PDI (Perfect Digital Invariant)
is an n digit number which is equal to the sum of the k-th powers of its digits. If k
= n, it is called a PPDI
(PluPerfect Digital Invariant).
More generally, one can consider the function fk(N) = sum of the
k-th powers of the digits of
N (in base 10
usually) and study its
iterates. It is not hard to see
that fk(N) < N from some point on, so the iterates must
lead to cycles. E.g., for k = 3,
we have:
160 ®
217 ® 352 ® 160.
See Reviews in Number Theory,
sections A62 and A64 for related material.
See 7.BB for other iterated functions of integers.
F. Hoppenot.
Courrier du "Sphinx".
Sphinx 7:4 (Apr 1937) 72.
Gives 153, 371, 407, 8208, 9474 as PPDIs.
M. Fistié. Courrier
du "Sphinx". Sphinx 7:5 (May
1937) 87. Adds 370
to Hoppenot's list and says these are all the PPDIs for k = 3.
Ball. MRE, 11th ed.,
1939, p. 13. Gives the four PPDI's
for k = 3: 153, 370, 371, 407,
citing Sphinx.
G. H. Hardy. A
Mathematician's Apology. CUP,
1940. "There are just four numbers
(after 1) which are the sums
of the cubes of their digits .... These
are odd facts, very suitable for puzzle columns and likely to amuse amateurs,
but there is nothing in them which appeals to a mathematician." He cites Ball for them.
H. Steinhaus. One
Hundred Problems in Elementary Mathematics.
(As: Sto Zadań, PWN -- Polish
Scientific Publishers, Warsaw, 1958.)
Pergamon Press, 1963. With a
Foreword by M. Gardner, Basic Books, NY, 1964.
Prob. 2: An interesting property of numbers, pp. 11-12 & 55-58. Considers iterates of f2(n). He shows this always leads to a cycle and find there are two
cycles, 1 of length 1 starting from 1 and 1 of
length 8 starting from 4.
Kiyoshi Iseki and coworkers extended Steinhaus's idea
to k = 3, 4, 5 in 5 papers during 1960‑1963. See Leveques's Reviews in Number Theory,
A64-12 -- A64-16 for details. For k = 3,
there are 9 cycles
(5 of which have period 1:
1 and the four PPDI's for k = 3). For k = 4, there are
6 cycles (4
of which have length 1: 1,
1634, 8208, 9474).
For k = 5, there are
15 cycles, the longest having
length 28. There are 5 cycles of length 1: 1, 54748,
93084, 92727, 194979.
Max Rumney. Digital
invariants. RMM 12 (Dec 1962) 6-8. Quotes Hardy. General introduction to the ideas. Interested in repeating the process of summing the n-th powers of an n-digit number, so his PDI is what is now called a PPDI. He also considers cases where the process
cycles through a few values, e.g. for the sum of the cubes of the digits, we
get the cycle: 133, 055,
250, 133. There are such cycles starting with 133,
136, 217, 919.
Gardner. SA (Jan
1963) c= Magic Numbers, chap. 3.
Surveys what is known with numerous references up through 1973. In particular, D. St. P. Barnard showed
that the number of PPDIs is finite by noting that 61 * 961 < 1060, so that a PPDI must have < 61
digits.
Harry L. Nelson.
More on PDI's. Publication
UCRL-7614, Univ. of California. 1 Dec
1963. ??NYS -- cited by Schwarz and
others below, who say it proves and improves Barnard's result and that it makes
the distinction between PDI and PPDI as now generally used.
Azriel Rosenfeld, proposer;
Nathan J. Fine, solver. Problem
E1651 -- Sums of squares of digits. AMM
71 (1964) 90 & 1042-1043.
"Prove that no multidigit integer is equal to the sum of the
squares of its digits."
Benjamin L. Schwarz.
Finiteness of a set of self-generating integers. JRM 2:2 (Apr 1969) 79‑83. Proves Barnard's result and improves it
to < 60 digits.
Benjamin L. Schwarz.
Finite bounds on digital invariants -- some conjectures. JRM 3:2 (Apr 1970) 88-92. General discussion of whether the number of
PDIs is infinite. He gives a heuristic
argument that there are infinitely many.
Joseph S. Madachy.
Some new narcissistic numbers.
Fibonacci Quarterly 10:3 (Apr 1972) 295-298. "[R]eports on various narcissistic numbers other than
digital invariants." Gives various
forms and some references.
Benjamin L. Schwarz.
Self-generating integers. MM
46:3 (May 1973) 158‑160. Gives a
simple condition for functions of the digits so that the number of PPDIs (=
SGIs in his notation) must be finite.
This applies to the sum of the factorials of the digits.
Lionel E. Deimel Jr. & Michael T. Jones. Finding pluperfect digital invariants: techniques, results and observations. JRM 14:2 (1981‑82) 87-108. Cites work from 1962. All PPDIs for bases 2
through 10 were found, using months of Data General
MV/8000 time, checking primality for bases
< 10. There are 89
PPDIs in base 10 (but this includes 0 as an example for
exponent 1, though it is not clearly a
1-digit number, so perhaps
88 is a better count).
Mary T. Whalen & Gordon L. Miller. Prime PPDI's. JRM 25:2 (1993) 118‑123.
Says PPDIs are also known as Armstrong numbers. The present authors have checked the above
results and looked for examples of primes in base 10. The base 10
results took a DEC 6310 just over
14 days and three primes (beyond
the exponent 1 cases) were found. Results were checked with Mathematica on a NeXT. They give
88 PPDIs for base 10.
Tony Forbes.
Recurring digital invariants.
M500 165 (Dec 1998) 4-7. After
some comments and results on PDIs in base
10 and loops in the iteration
process, he decides to examine base
3 and finds there are many examples
and gives some very large examples.
7.AZ. DIVINATION OF A PAIR OF CARDS FROM ITS ROWS
A spectator mentally chooses a pair
from ten pairs of cards. You then lay
them out in a 5 by 4 array and ask him to tell you in which rows
his pair appears, whereupon you point out his pair. This is done by laying out the cards in a particular pattern
traditionally given by the Latin words:
MUTUS DEDIT NOMEN
COCIS. In fact, such patterns
are easily found for r(r+1)/2 pairs laid out in r rows -- but finding a
memorable pattern takes some doing.
Since the person's response can be either one row or two rows, the
number of responses is
r + BC(r, 2) = r(r+1)/2
= BC(r+1, 2).
Unger, Secret Out, Ball, Ahrens
and Indoor Tricks and Games deal with triples. With r rows, there are r + BC(r, 2) + BC(r, 3)
= r (r2 + 5)/6 possible responses. However, this does not always give a
possible situation. E.g., for r = 2,
there are 3 possible responses, but one cannot
layout 3 triples in two rows, and the same problem occurs when r is
even. For r = 3, there is an easy
pattern: AAADDEG BBBDFFG
CCCEEFG and I imagine it can be
easily done whenever r is odd.
However, we are getting to more responses than we can handle and it
might be easier to use fewer of them.
See Unger for an approach.
This is actually a combinatorial
problem and I will probably shift this section into Chapter 5.
Bachet.
Problemes. 1612. Prob. XVI: De plusieurs cartes disposées en
divers rangs deviner laquelle on aura pensée: 1612: 87-92. Prob. 18, 1624: 143‑151; 1884: 72-83. This problem covers several card divinations. The present topic has been added to
"cette seconde [impression de ce livre]". He discusses the problem for
r rows and shows the simplest
pattern, but gives no mnemonics.
For 10 pairs, the simplest pattern has rows: AABCD, BEEFG, CFHHI,
DGIJJ.
Wingate/Kersey.
1678?. Prob. 5, pp.
539-542. Does 10 and 15
pairs as in Bachet.
Hooper. Rational
Recreations. Op. cit. in 4.A.1. 1774.
Recreation XXVII: The ten duplicates, pp. 70-71. MUTUS
DEDIT NOMEN COCIS.
Endless Amusement I.
c1818. P. 106: The ten
duplicates. As in Hooper.
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
Pp. 122-123, no. 184: The ten duplicates. c= Endless Amusement I.
Rational Recreations.
1824. Feat 38, p. 168: Cards in
couples.
MUTUS DEDIT NONEM [sic] COCIS.
Manuel des Sorciers.
1825. Pp. 48-51, art. 21. ??NX
MUTUS NOMEN DEDIT
COCIS. Extends to 15 pairs, in a
different way than Bachet, by bordering the usual array of letters with a left
hand column 5, 4, 3, 2, 1 and adding a bottom row which continues from
the 1
just added with 1, 2, 3, 4, 5.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 233-235, nos. 897-901. He first gives the 10 pair case, using ABCBD
EFAGE HGHIC KFKID,
which is isomorphic to
MUTUS NOMEN DEDIT
COCIS, but he never gives any
mnemonics. He then asks which other
arrays can be done and sees that r rows can be used with r(r+1)/2
pairs. He gives a scheme
for r = 3: ABAC BDDE EFCF,
and a scheme for r = 5: ABCDEA
BFGHFI KGLLCM ONNDHK
EIOPMP. He then considers
triples and gives a scheme for 8 triples:
AAA123 BBB124 CCC143
DDD423. This has the feature
that the responses are either one row or three rows, giving
r + BC(r, 3) = r (r2 ‑ 3r
+ 8)/6 responses. He shows an easy method for extending a
scheme to the next number of rows getting
AAA123567 BBB124589 CCC143860
DDD423907 EEE567890 with 15 triples. He notes that one may want to rearrange the columns to make it
less obvious when the cards are laid out.
He really should have started with
AAA1 BBB1 CCC1
and then constructed AAA123 BBB124
CCC134 DDD234 in order to make the construction clearer.
Young Man's Book.
1839. Pp. 205-206. The Ten Duplicates. Identical to Endless Amusement I.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Parlour magic, no. xxii, pp. 202-203 (1868: 213-214): The ten
duplicates. MUTUS DEDIT
NOMEN COCIS.
Magician's Own Book.
1857. No. 21: The pairs
re-paired, pp. 65-66. MUTUS DEDIT
NOMEN COCIS ["Mutus gave a name to the Coci,"
a people who have yet to be discovered.]
= Boy's Own Conjuring Book, 1860, p. 73.
The Secret Out.
1859.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 3, no. 9: The
pairs repaired, pp. 83-84. MUTUS DEDIT
NOMEN COCIS.
Boy's Own Book. The
ten pairs of cards distinguished. 1868:
638. NOMEN MUTUS DEDIT RORIS.
Hanky Panky.
1872. To tell two cards out of
twenty, p. 254. MISAI TATLO
NEMON VESUL.
Berkeley & Rowland.
Card Tricks and Puzzles.
1892. Simple Tricks by
Calculation No. XII: To name several different pairs of cards chosen by one or
more persons, pp. 44-46. SHEER CHAFF
USUAL COLOR; MUTUS
DEDIT NOMEN COCIS;
MISAI TATLO NEMON
VESUL (where V = U).
Ball. MRE. 1st ed., 1892. Determination of a selected pair of cards out of ½n(n+1)
given pairs, pp. 98-101. Gives
general rule for pairs. MATAS DEDIT
NOMEN COCIS. Mentions
8 triples: LANATA
LEVETE LIVINI NOVOTO,
as in Secret Out.
Ahrens. MUS I. 1910.
Pp. 148-152. MUTUS DEDIT
NOMEN COCUS. Discusses generalizations, including
examples with triples, etc. Cites
Unger, 1838, for the first idea of finding triples and quotes the mnemonic from
Ball.
Indoor Tricks and Games.
Success Publishing, London, nd [1930s??].
Phyllis Fraser & Edith Young. Puzzles Quizzes and Games.
Bantam Books, NY, 1947.
No. 2: Card trick, pp. 97-99.
THIGH ATLAS GOOSE
BIBLE.
Stewart Judah. The
divertisements of Stewart Judah. The
New Phoenix, No. 319 (Nov 1954) 83. UNDUE GOANO
TETRA RIGID.
Will Dexter. The
Illustrated Book of Magic Tricks. Abbey
Library, London, nd, but Introduction dated 1957. The Romans had a word for it!, pp. 73-74. DAVID
LOVEL IN YON ABBEY;
MUTUS NOMEN COCIS
DEDIT ("Mutus gave a name to Cocis").
Martin Gardner.
Mutus Nomen. Pallbearers Review
5:7 (May 1970) 338. Reprinted in
his: Martin Gardner Presents; Richard
Kaufman & Alan Greenberg, 1993, pp. 237-238. For 30 cards: LIVELY RHYTHM
MUFFIN SUPPER SAVANT.
For 20 cards: BIBLE ATLAS
GOOSE THIGH.
Walter Gibson. Big
Book of Magic for All Ages. Kaye &
Ward, Kingswood, Surrey, 1982. Twenty
Cards & New Deal Twenty Card Trick, pp. 162-165. Mentions
MUTUS ... and says there are many
English possibilities and gives RUFUS STEEL
TIARA FOLIO. The New Deal version uses a more complicated
laying-out process starting with ten piles of two using MAGIC
STORE, then laying out according
to AGORA METER SCOTS GIMIC.
Karl Fulves. Mutus
Nomen. The author, 1998. A 90pp booklet on variations of this classic
trick. ??NYR.
At a lunch in 2002, Rob Eastaway did the trick and said he
used BIBLE GOOSE ATLAS THIGH
and Jeremy Wyndham said he used
DAVID LOVEL IN YON
ABBEY. I had never met anyone who
used anything other than MUTUS ...
before!
7.BA. CYCLE OF NUMBERS WITH EACH CLOSER TO TEN THAN THE PREVIOUS
New section, due to discovering
Lewis Carroll's version. I have no idea
if this is older than Carroll.
Lewis Carroll. A
Tangled Tale. (1885) = Dover,
1965. The pigs. Knot VIII: De omnibus rebus, pp. 52-57 &
132‑134. (In the answers, this
part of the Knot is denoted §1. The pigs.)
"... place twenty‑four pigs in those four sties, so that, as
she goes round the court, she may always find the number in each sty nearer to
ten than the number in the last."
Answer:
8, 10, 0, 6 --
the key is that "nothing is nearer ten than 10".
7.BB. ITERATED FUNCTIONS OF INTEGERS
See 7.T and 7.AY for some special
forms.
Let
f(n) be a function of an
integer n. The problems in this section concern the behaviour of the sequence of iterations
of f:
n, f(n), f2(n) = f(f(n)), f3(n), .... There are a great number of behaviours that
can occur.
1. The
sequence can diverge to infinity.
2. The
sequence can converge to a fixed point.
3. The
sequence can become cyclic.
4. The
sequence can behave chaotically.
Further
the behaviour may be different for different starting points.
If
f(n) > n in all cases, then
the iteration will diverge to infinity and is not very interesting. Hence we are normally only interested in
functions which have f(n) < n for some non-zero proportion of the
integers.
If
f(n) £ n in all cases,
then the sequence must converge to a fixed point, though different starting
points may lead to different fixed points.
Then the problem is to identify the fixed points and the sets they
attract.
So the most interesting cases
have f(n) sometimes greater and sometimes smaller than n.
Two classic examples are the following.
The proper sum of divisors, Σ(n),
is σ(n) - n, where
σ(n) is the classic number
theoretic function of the sum of all divisors.
The iterates of this have been studied since the 19C as the fixed points
are the perfect numbers and 2-cycles are amicable numbers -- see 7.AB. Some larger cycles have been found -- I
recall a 5 and a 17 cycle. No provably
infinite sequence has been found.
The Syracuse function is the
following. If n is even, halve it; if
n is odd, form
3n + 1. Some of the
sequences have been computed for hundreds of thousands of terms without
termination.
Problems of this sort are genuine
mathematical recreations but are a bit too elaborate for me to include
here. What I will include are functions
whose sequences are always finite.
These generally are of two types.
Functions such that f(n) <
n for all n above some limit --
see, e.g., 7.AY, and functions such that
f(n) has the same number of digits
as n.
If f(n) is the sum of some function g(ai) of the digits ai of n, set
M = max {g(0), g(1), ..., g(9)}. Then for
10d-1 £ n < 10d, we have
f(n) £ (d+1)M and this
will be less than n from some point on.
Joseph S. Madachy.
Mathematics on Vacation. Op.
cit. in 5.O, (1966), 1979. Narcissistic
numbers: Digital invariants, pp. 163‑165. Based on Rumney and Nelson.
Carries on to consider other types of 'narcissistic numbers' which are
numbers which are some function of their digits -- various forms are discussed
on pp. 165-177.
George D. Poole. Integers
and the sum of the factorials of their digits.
MM 44:5 (Nov 1971) 278-279. The
only integers equal to the sum of the factorials of their digits are: 1, 2, 145, 40585.
He has to check up to 2,000,000 before a general method can be used.
In
Dec 2001, Al Posamentier mentioned the following cycle of this function:
169 ®
363601 ® 1454 ® 169. I essentially repeated Poole's work, but found five numbers which
are equal to the sum of the factorials of their digits, excluding leading
zeroes: 0, 1, 2, 145, 40585. Zero is exceptional -- I view it as having
no digits, so the sum involved is empty!
The largest number for which this function is larger than the given number
is 1,999,999 and there are
208,907 such numbers. Using this, one has a limited search to find
all the cycles - there are two other cycles, both 2-cycles: 871 « 45361 and
872 « 45362. Cf Schwartz
and Kiss, below.
Benjamin L. Schwarz.
Self-generating integers. MM
46:3 (May 1973) 158‑160. Gives a
simple condition for functions of the digits so that the number of PPDIs (=
SGIs in his notation) must be finite.
This applies to the sum of the factorials of the digits.
Peter Kiss. A
generalization of a number-theoretic problem [in Hungarian, with English
summary]. Mat. Lapok 25:1-2 (1974
(1977)) 145-149. ??NYS -- described in
Reviews in Number Theory A62-201F (= MR 55, #12612). An apparent English translation is: A generalization of a problem
in number theory; Math. Sem. Notes Kobe Univ. 5:3 (1977) 313-317; ??NYS -
described ibid. A62-201G (= MR 57, #12362).
He demonstrates that if
f(n) is the sum of any function
of the digits of n, then the iteration of f
must lead to a cycle -- as sketched above. He finds all the cycles for three cases, including for the
factorial function. The review mentions
Steinhaus (cf section 7.AY) and Poole.
Roger Cook.
Reversing digits. MS 31:2
(1998/9) 35-37. The initial problem was
to take a four digit number, arrange the digits in ascending order and in
descending order and subtract the first from the second. In all cases except aaaa,
the iterations converge to 6174 (7641 - 1467 = 6174). He discusses other number of digits and
other bases.
7.BC. UNUSUAL DIFFICULTY IN GIVING CHANGE
The typical problem here is that a
customer offers a note to pay a bill and the shopkeeper says he can't give
change, but when the customer offers a larger note, then the shopkeeper can
give change.
Some of the problems in 7.P.1 could
be considered as falling into this topic.
New section. I have seen a 1935 version somewhere.
Carroll-Collingwood.
c1890 Prob. 2, pp. 317-318. = John Fisher; The Magic of Lewis Carroll;
op. cit. in 1; p. 79. =
Carroll-Wakeling II, prob. 31: Coins, pp. 49 & 72-73. A man has a half-sovereign (10s), a florin
(2s) and a 6d and wants to buy goods worth 7s 3d. However the shopkeeper has a crown (5s), a shilling (1s) and a
penny (1d), so cannot make change. A
friend comes into the shop and he has a double florin (4s), a half-crown
(2s 6d), a fourpenny piece (4d) and a threepenny bit (3d). Can they sort out the payment? I consulted a dictionary and found that the
double-florin was first minted in 1887, which puts a lower bound on the date of
this problem, so I have given the date as c1890 rather than c1890?
Clark. Mental
Nuts. 1916, no. 4. The conductor's money. Passenger tenders a $1 note to pay 5¢ (= $.05)
fare, but conductor cannot give change, but he can give change for a $5
note. The answer is that he can give a
$2.50 gold piece, a $2 note and 45¢ in coins.
David Singmaster. No
change! Written up as a problem for my
Telegraph column in summer 1999.
Shopkeeper cannot give change for a £5 note, but can for two £5
notes. How can this happen? How about in the US?
Most of the recreations in
probability are connected with some paradoxical feature. A good exposition of most of these appears
in the following.
Gábor J. Székely.
Paradoxes in Probability Theory and Mathematical Statistics. Akadémiai Kiadó, Budapest and Reidel,
Dordrecht, 1986. [Revised translation
of: Paradoxonok a Véletlen
atematikában; Műszaki Könvkiadó, Budapest, nd.] Translated by Márta Alpár and Éva Unger. ??NYR.
R. E. Miles & J. Serra.
En Matiere d'introduction.
In: Geometrical Probability and
Biological Structures: Buffon's 200th Anniversary. Lecture Notes in Biomathematics, No. 23, Springer, 1978, pp. 3‑28. Historical survey, reproduces main texts.
Buffon. (Brief
commentary). Histoire de l'Acad. des Sci.
Paris (1733 (1735)) 43‑45.
Discusses problem of a disc meeting a square lattice and then the stick
(baguette) problem, but doesn't give the answer.
Buffon. Essai
d'arithmétique morale, section 23.
1777. (Contained in the fourth
volume of the supplement to his Histoire Naturelle, pp. 101‑104??) = Oeuvres Complètes de Buffon; annotated by
M. Flourens; Garnier Frères, Paris, nd [c1820?], pp. 180‑185. (Also in Miles & Serra, pp. 10‑11.)
Laplace. Théorie
Analytique des Probabilitiés.
1812. Pp. 359‑360. ??NYS.
(In Miles & Serra, p. 12.)
3rd ed, Courcier, Paris, 1820, pp. 359-362. Finds the answer to Buffon's problem with needle length 2r
and line spacing a. Then solves the case of two perpendicular
sets of lines, with possibly different spacings.
M. E. Barbier. Note
sur le problème de l'aiguille et le jeu du joint couvert. J. Math. pures appl. (2) 5 (1860) 273‑286. Gives result for arbitrary curves and
considers several grids. Also gives his
theorem on curves of constant width.
M. W. Crofton. On
the theory of local probability.
Philos. Trans. Roy. Soc. 158 (1869) 181‑199. (Excerpted in Miles & Serra, pp. 13‑15.)
A. Hall. On an
experimental determination of
π. Messenger of Mathematics
2 (1873) 113‑114. ??NYS.
Tissandier.
Récréations Scientifiques.
1880? 2nd ed., 1881, 139-145
describes the result and says 10,000
tries with a 50 mm needle on a floor with spacing 63.6 mm,
produced 5009 successes giving π = 3.1421.
= 5th ed., 1888, pp. 204-208. c= Popular Scientific Recreations, 1890?
pp. 729‑731, but the needle is
2 in on a floor of spacing 2½ in
and 5009 is misprinted as £5000 (sic) but 5009
is used in the calculation.
J. J. Sylvester. On
a funicular solution of Buffon's "Problem of the needle" in its most
general form. Acta Math. 14 (1890‑1891)
185‑205.
N. T. Gridgeman.
Geometric probability and the number
π. SM 25 (1959) 183‑195. Debunks experimental results which are often
too good to be true, although they are frequently cited.
J. G. L. Pinhey. The
Comte de Buffon's paper clip. MG 54
(No. 389) (Oct 1970) 288. Being caught
without needles, he used paperclips. He
derives the probability of intersection assuming a paper clip is a rectangle
with semi‑circular ends.
Jack M. Robertson & Andrew F. Siegel. Designing Buffon's needle for a given
crossing distribution. AMM 93 (1986)
116‑119. Discusses various
extensions of the problem.
How many people are required before
there is an even chance that some two have the same birthday?
George Tyson was a retired mathematics teacher when he
enrolled in the MSc course in mathematical education at South Bank in about
1980 and I taught him. He once remarked
that he had known Davenport and Mordell, so I asked him about these people and
mentioned the attribution of the Birthday Problem to Davenport. He told me that he had been shown it by
Davenport. I later asked him to write
this down.
George Tyson. Letter
of 27 Sep 1983 to me. "This was
communicated to me personally by Davenport about 1927, when he was an
undergraduate at Manchester. He did not
claim originality, but I assumed it.
Knowing the man, I should think otherwise he would have mentioned his
source, .... Almost certainly he
communicated it to Coxeter, with whom he became friendly a few years later, in
the same way." He then says the
result is in Davenport's The Higher Arithmetic of 1952. When I talked with Tyson about this, he said
Davenport seemed pleased with the result, in such a way that Tyson felt sure it
was Davenport's own idea. However, I
could not find it in The Higher Arithmetic and asked Tyson about this, but I
have no record of his response.
Anne Davenport.
Letter of 23 Feb 1984 to me in response to my writing her about Tyson's
report. "I once asked my husband
about this. The impression that both my
son and I had was that my husband did not claim to have been the 'discoverer'
of it because he could not believe that it had not been stated earlier. But that he had never seen it
formulated."
I have discussed this with Coxeter (who edited the 1939
edition of Ball in which the problem was first published) and C. A. Rogers (who
was a student of Davenport's and wrote his obituary for the Royal Society), and
neither of them believe that Davenport invented the problem. I don't seem to have any correspondence with
Coxeter or Rogers with their opinions and I think I had them verbally.
Richard von Mises.
Ueber Aufteilungs‑ und Besetzungs‑
Wahrscheinlichkeiten. Rev. Fac. Sci.
Univ. Istanbul (NS) 4 (1938‑39) 145‑163. = Selected Papers of Richard von Mises; Amer. Math. Soc., 1964,
vol. 2, pp. 313‑334. Says the
question arose when a group of 60 persons found three had the same
birthday. He obtains expected number of
repetitions as a function of the number of people. He finds the expected number of pairs with the same birthday is
about 1 when the group has
29 people, while the expected
number of triples with the same birthday is about 1 when there are 103
people. He doesn't solve the
usual problem, contrary to Feller's 1957 citation of this paper.
Ball. MRE, 11th ed.,
1939, p. 45. Says problem is due to H.
Davenport. Says "more than 23"
and this is repeated in the 12th and 13th editions.
P. R. Halmos, proposer;
Z. I. Mosesson, solver. Problem
4177 -- Probability of two coincident birthdays. AMM 52 (1945) 522
& 54 (1947) 170. Several solvers cite MRE, 11th ed. Solution is
23 or more.
George Gamow. One,
Two, Three ... Infinity. Viking, NY,
1947. = Mentor, NY, 1953,
pp. 204‑206. Says 24
or more.
Oswald Jacoby. How
to Figure the Odds. Doubleday, NY,
1947. The birthday proposition, pp.
108-109. Gives answer of 23
or more.
William Feller. An
Introduction to Probability Theory and Its Applications: Vol 1. Wiley, 1950, pp. 29-30. Uses an approximation to obtain 23
or more. The 2nd ed., 1957,
pp. 31‑32 erroneously cites von Mises, above.
J. E. Littlewood. A
Mathematician's Miscellany. Op. cit. in
5.C. 1953. P. 18 (38) mentions the problem and says 23
gives about even odds.
William R. Ransom.
Op. cit. in 6.M. 1955. Birthday probabilities, pp. 38‑42. Studies usual problem and graphs probability
of coincidence as a function of the number of people, but he doesn't compute
the break‑even point. He then
considers the probability of two consecutive birthdays and gets upper and lower
estimates for this.
Gamow & Stern.
1958. Birthdays. Pp. 48‑49. Says the break‑even point is "about twenty‑four".
C. F. Pinzka.
Remarks on some problems in the American Mathematical Monthly. AMM 67 (1960) 830. Considers number of people required to give
greater than 50% chance of having 3, 4 or 5 with the same
birthday. He gets 88, 187, 314 respectively, using a Poisson approximation. He gives the explicit formula for
having 3 with the same birthday.
Charlie Rice.
Challenge! Op. cit. in 5.C. 1968.
Probable probabilities, pp. 32-36, gives a variety of other forms of the
problem.
Howard P. Dinesman.
Superior Mathematical Puzzles.
Op. cit. in 5.B.1. 1968. No. 58: The birthday puzzle, pp. 76, 116-117
& 122. Asks for probability of a
shared birthday among 30 people.
Says the problem was introduced by Gamow. Answer is .70. He adds that 24 or more will give
better than even odds and then asks how many people are necessary for one of
them to have his birthday on a given day -- e.g. today. Here the answer is 253.
E. J. Faulkner. A
new look at the probability of a coincidence of birthdays in a group. MG 53 (No. 386) (Dec 1969) 407‑409. Suggests the probability should be obtained
by the ratio of the unordered selections, i.e.
Prob. (r distinct birthdays) = BC(365, r)/BC(364+r,
r), rather than P(365, r)/365r. But the unordered selections with
repetitions are not equally likely events -- see Clarke & Langford,
below. For his approach, the breakeven
number is r = 17.
Morton Abramson & W. O. J. Moser. More birthday surprises. AMM 77 (1970) 856‑858. If a year has n days, k ³ 1 and p people are chosen at random, what is the
probability that every two people have birthdays at least k
days apart? For n = 365,
k = 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, the breakeven
numbers are
p = 23, 14, 11, 9, 8, 8, 7, 7,
6, 6.
L. E. Clarke & Eric S. Langford. Note 3298 -- I & II: On Note 3244. MG 55 (No. 391) (Feb 1971) 70‑72. Note the non‑equally likely events in
Faulkner.
W. O. J. Moser. It's
not a coincidence, but it is a surprise.
CM 10 (1984) 210‑213.
Determines probability P that in a group of k people, at least two
have birthdays at most w days apart.
This turns out to have a fairly simple expression. To get
P > .5 with w = 0
requires k ³
23, the classical case. With
w = 1, it requires k ³ 14, ....
Tony Crilly & Shekhar Nandy. The birthday problem for boys and girls. MG 71 (No. 455) (Mar 1987) 19‑22. In a group of 16 boys and 16
girls, there is a probability greater than ½ of a boy and a girl
having the same birthday and 16 is the minimal number.
Roger S. Pinkham.
Note 72.25: A convenient
solution to the birthday problem for girls and boys. MG 72 (No. 460) (Jun 1988) 129‑130. Uses an estimate to obtain the value 16
of Crilly and Nandy.
M. Lawrence Clevenson & William Watkins. Majorization and the birthday
inequality. MM 64:3 (1991)
183-188. Do the numbers necessary for P > .5
get bigger if birthdays are not random?
Answer is "no" and it is a result in majorization theory, but
they give an elementary treatment.
S. Ejaz Ahmed
& Richard J. McIntosh. An asymptotic approximation for the birthday
problem. CM 26:2 (Apr 2000)
151-155. Without using Stirling's
approximation, they show that for a calendar of n days and a desired probability p, 0
< p < 1, then the minimum class
size to produce a probability ³
p of two people having the same
birthday is asymptotically Ö[2n
log {1/(1-p)}]. For p = .5,
n = 365, this gives 22.494.
See also 6.BR, esp. the Mathematical
Log article and my comments. I have a large
number of similar results, mostly by myself, in a file. I estimate there are about 20
possible answers, ranging from
0 to 1.
J. J. Sylvester. On
a special class of questions in the theory of probabilities. Birmingham British Association Report (1865)
8. = The Collected Mathematical papers
of James Joseph Sylvester, (CUP, 1908); Reprinted by Chelsea, 1973, item 75,
pp. 480-481. ??NYS, but Guy (below)
reports that it is a discursive article with no results. Attributes problem for three points within a
circle or sphere to Woolhouse but feels the problem is not determinate.
C. Jordan. 1872‑1873. See entry in 8.G.
E. Lemoine.
1882-1883. See entry in 8.G.
L. Carroll. Pillow
Problems. 1893. ??NYS.
4th ed., (1895). = Dover,
1958. Problem 58, pp. 14, 25, 83‑84. Prob (acute) = .639.
C. O. Tuckey. Note
1408: Why do teachers always draw acute‑angled
triangles? MG 23 (No. 256) (1939)
391‑392. He gets Prob(obtuse) varying between .57 and
.75.
E. H. Neville.
Letter: Obtuse angling -- a
catch. MG 23 (No. 257) (1939) 462. In response to Tuckey, he shows Prob(obtuse) = 0 and deduces that
Prob(acute) = 0 (!!!).
Nikolay Vasilyev.
The symmetry of chance. Quantum
3:5 (May/Jun 1993) 22-27 & 60-61.
Survey on geometric probability.
Asks for the probability of an acute triangle when one takes three
points at random on a circle and gets
¼.
Richard K. Guy.
There are three times as many obtuse-angled triangles as there are
acute-angled ones. MM 66:3 (Jun 1993)
175-179. Gives 12 different approaches,
five of which yield Prob(acute) =
¼, with other values ranging from 0 to .361.
He tracked down the Sylvester reference -- see above.
In Mar 1996, I realised that the two approaches sketched in
6.BR give probabilities of acuteness as
0 and 2 - π/2 = .429.
8.D. ATTEMPTS TO MODIFY BOY‑GIRL RATIO
This is
attempted, e.g. by requiring families to stop having children after a girl is
born.
Pierre Simon, Marquis de Laplace. Essai Philosophique sur les Probablitiés (A Philosophical Essay
on Probabilities). c1819, ??NYS. Translated from the 6th French ed. by F. W.
Truscott & F. L. Emory, Dover, 1951.
Chap. XVI, pp. 160‑175, especially pp. 167‑169. Discusses whether the excess of boys over
girls at birth is due to parents stopping having children once a son is born.
Gamow & Stern.
1958. A family problem. Pp. 17‑19.
Nicholas Bernoulli.
Extrait d'une Lettre de M. N. Bernoulli à M. de M... [Montmort] du 9
Septembre 1713. IN: Pierre Rémond de Montmort; Essai d'analyse
sur les jeux de hazards. (1708); Seconde edition revue & augmentee de
plusieurs lettres, (Quillau, Paris,
1713 (reprinted by Chelsea, NY, 1980));
2nd issue, Jombert & Quillau, 1714.
Pp. 401-402. Quatrième
Problème & Cinquième Problème, p. 402.
[See note to Euler; Vera aestimatio sortis in ludis; op. cit. below,
pp. 459‑461.] In the 4th
problem, he proposes pay-offs of 1, 2,
3, 4, ... if the player first throws a
six with a die on throw 1, 2, 3, 4,
... and asks for the expectation. He does not compute it, but I get 6.
In the 5th problem, he asks what happens in the same situation if the
pay-offs are 1, 2, 4, 8, ... or
1, 3, 9, 27, ...
or 1, 4, 9, 16, ... or
1, 8, 27, 64, ... etc. Again, he doesn't give and results, but the
first two give divergent series, while the later two are convergent.
Daniel Bernoulli.
Specimen theoriae novae de mensura sortis. Comm. Acad. Sci. Imp. Petropol. 5 (1730-31(1738)) 175-192, ??NYS. IN: Die Werke von Daniel Bernoulli; ed. by
L. P. Bouckaert & B. L. van der Waerden; Birkhäuser, 1982; pp. 223-234 and
notes by van der Waerden, pp. 195 & 197-200. English translation in Econometrica 22 (1954) 23-36, ??NYS.
Lewis Carroll.
Lionel Tollemache, Reminiscences of Lewis Carroll, Literature (5 Feb
1898), ??NYS, quoted in Carroll-Wakeling II, prob. 28: A good prospect,
pp. 44 & 72. (Tollemache was at
Balliol College, Oxford, in 1856-1860.
He then entered Lincoln's Inn, London, so this must refer to c1858.
He
says that Carroll gave the problem with a coin and pay-offs of 0, 1, 3, 7, 15, ... if the player first throws a tail on
throw 1, 2, 3, 4, 5, .... Neither Tollemache nor Wakeling give any
reference to any other version of the problem.
I would compute the expectation as
Σn=0
(2n-1)(1/2)n+1
= 0/2 + 1/4 + 3/8 + 7/16 + 15/32
+ ....
Wakeling does
it by rewriting the sum as
1/4
+ (1+2)/8 + (1+2+4)/16 + (1+2+4+8)/32 + ...,
and regrouping as:
1/4
+ 1/8 + 1/16 + 1/32 + ... + 2/8 + 2/16 + 2/32 + ... + 4/16 + 4/32 + ... + 8/32
+ ... = 1/2 + 1/2 + 1/2 + 1/2 + ...
and says "Hence, his prospects are ½d. for every successive throw
of a head." Basically, making the
payoff be a sum means that the expectation is a sum over half a quadrant. The normal process would be to first sum the
finite rows and see that each is at least
1/4. Wakeling is first summing
the columns and finding each column sum is
1/2. I wonder if Carroll
carefully chose the peculiar pay-offs in order to make the latter process work.
Leonhard Euler. Vera
aestimatio sortis in ludis. [Op.
postuma 1 (1862) 315-318.] = Op. Omnia
(I) 7 (1923) 458-465.
Tissandier.
Récréations Scientifiques.
1880? 2nd ed., 1881, p. 140
gives a brief unlabelled description, saying this "problème de
Pétersbourg" was discussed by Daniel Bernoulli in "Mémoires de
l'Académie de Russie". Not in the
5th ed. of 1888.
Tissandier. Popular
Scientific Recreations. 1890? Pp. 727-729 discusses the idea and says D.
Bernoulli presented his material on this in "Memoires [sic] de l'Académie
de Russie". This is somewhat
longer than the material in the 2nd French ed of 1881.
Anon. [presumably the editor, Richard A. Proctor]. Strange chances. Knowledge 10 (Oct 1887) 276-278. Brief discussion of "the famous Petersburg problem".
C. S. Jackson. Note
438: The St. Petersburg problem. MG 8 (No. 116) (Mar 1915) 48. Notes that the value of the game is not more
than log2 of the bank's
funds.
Dan Pedoe, The Gentle Art of Mathematics, op. cit. in 5.C,
1958, p. 55, says D. Bernoulli published it in the Transactions (??) of of the
St. Petersburg Academy. ??NYS.
Pedoe, ibid., p. 57, also says that Buffon tested this
with 2048 games and he won
10,057 in them.
Jacques Dutka. On
the St. Petersburg paradox. Archive for
the History of the Exact Sciences 39 (1988) 13-39. ??NYR.
Nick Mackinnon
& 5Ma. Note 74.9:
A lesson on the St. Petersburg paradox.
MG 74 (No. 467) (1990) 51‑53. Suppose a maximum is put on the payment -- i.e. the game stops if
n
heads appear in a row. How does
this affect the expected value? They
find n = 10 gives expected value of
6.
A game consists of n
points. How do you divide the
stake if you must quit when the score is
a to b to ....? This problem was resolved by Fermat and
Pascal in response to a question of the Chevalier de Méré and is generally
considered the beginning of probability theory. The history of this topic is thoroughly covered in the first
works below, so I will only record early or unusual occurrences.
NOTATION: Denote this by (n; a, b,
...).
GENERAL
HISTORIES
Florence Nightingale David.
Games, Gods and Gambling. The
origins and history of probability and statistical ideas from the earliest
times to the Newtonian era. Griffin,
London, 1962.
Anthony W. F. Edwards.
Pascal's Arithmetical Triangle.
Griffin & OUP, London, 1987.
This corrects a number of details in David.
David E. Kullman.
The "Problem of points" and the evolution of probability. Handout from talk given at MAA meeting, San
Francisco, Jan 1991. Uses (6; 5, 3)
as a common example and outlines approaches and solutions of: Pacioli (1494) -- 5 : 3; Cardan (1539) -- 6 : 1; Tartaglia (1539) -- 2 : 1; Pascal (1654) -- 7 : 1; Fermat (1654) -- 7 :
1; Huygens (1657) -- like Pascal. He also describes the approaches of Pascal,
Fermat and Huygens to the three person case and generalizations due to Montmort
(1713) and de Moivre (1718).
Pacioli. Summa. 1494.
Oystein Ore. Pascal
and the invention of probability theory.
AMM 67 (1960) 409‑419. Ore
says Pacioli is the first printed version of the problem. He translates parts of the texts. Ore says he has seen the problem in Italian
MSS as early as 1380, but he doesn't give details (???). He opines that the problem is of Arabic
origin. He discusses Cardan and
Tartaglia and gives some examples from Forestiani, 1603 -- (8; 5, 3), (14; 10, 8, 5) (??NYS). But there is no
proper mathematics until Pascal & Fermat.
Calandri, Raccolta.
c1495.
Cardan. Practica
Arithmetice. 1539.
Tartaglia. General
Trattato, 1556, art. 206, pp. 265r‑265v.
(6; 5, 3). Criticises Luca del
Borgo (= Pacioli) and gives another method.
Giovanni Francesco Peverone. Due brevi e facili trattati, il primo d'Arithmetica, l'altro di
Geometria ..., Gio. Tournes, Lyons, 1558.
Reissued as: Arithmetica e geometria del Sig. Gio. Francesco Peverone di
Cuneo, ibid., 1581. ??NYS -- described
in: Livia Giacardi & Clara Silvia Roero;
Bibliotheca Matematica Documenti
per la Storia della Matematica nelle Biblioteche Torinese; Umberto Allemandi & C., Torino, 1987,
pp. 117-118. They say he includes
correct solutions of some problems of games of chance, in particular the
'divisione della posta', i.e. the problem of points.
Ozanam. 1694. Prob. 10, 1696: 41-52, esp. 45-50; 1708: 37-48, esp. 42‑45. Prob. 13, 1725: 123‑130. Prob. 3, 1778: 117-121; 1803: 116-120; 1814: 102-106; 1840:
54-55. Discusses the problem in general
and specifically (3; 1, 0), but 1778 et seq. changes to (3; 2, 1)
and adds reference to Pascal and Fermat.
Pierre Rémond de Montmort.
Essai d'analyse sur les jeux de hazards. (1708); Seconde edition
revue & augmentee de plusieurs lettres,
(Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); 2nd issue, Jombert & Quillau, 1714. Avertissement (to the 1st ed.), pp. xxi-xxiv
and (to the 2nd ed.) xxv-xxxvii discusses the history of the problem, the work
of Fermat and Pascal and de Moivre's assertion that Huygens had solved it
first.
Chr. Mason, proposer; Rob. Fearnside, solver. Ladies' Diary, 1732-33 = T.
Leybourn, I: 223, quest. 168. (15; 10, 8,
5). I haven't checked the solution, but
the procedure is correct and another solver got the same results. Editor cites De Moivre.
Pearson. 1907. Part II, no. 98, pp. 134 & 210-211. (3; 2, 1).
He divides correctly as 3 : 1.
Hummerston. Fun,
Mirth & Mystery. 1924. Marbles, Puzzle no. 26, pp. 71 &
176. (4; 3, 2). He divides correctly as 3 : 1.
8.G. PROBABILITY THAT THREE LENGTHS FORM A TRIANGLE
See also 8.C.
E. Lemoine. Sur une
question de probabilités. Bull. Soc.
Math. France 1 (1872‑1873) 39‑40.
Obtains ¼ by considering that the stick can be broken
at m
equidistant points and then letting
m increase.
? Halphen. Sur un
problème de probabilités. Ibid., pp.
221‑224. Extends Lemoine to n
pieces, getting 1 ‑ n/2n‑1, by an argument similar to homogeneous
coordinates and by integration.
Camille Jordan.
Questions de probabilités.
Ibid., pp. 256‑258 & 281‑282. Generalizes to find the probability that n of
the m
parts, into which a line is broken, have length > a.
He finds the probability that four points on a sphere form a convex
spherical quadrilateral. Pp. 281‑282
corrects this last result. [Laquière;
Note sur un problème de probabilité; ibid. 8 (1879-80) 79-80 gives a simple
argument.]
M. Laquière.
Rectification d'une formule de probabilité. Bull. Soc. Math. France 8 (1879-80) 74-79. Treats the first problem of Jordan. Observes that Jordan's formula can give a
probability greater than one! Says
Jordan has a confusion between 'some
n' and
'a given n' and he gives a
corrected version.
E. Lemoine. Quelques
questions de probabilités résolues géométriquement. Bull. Soc. Math. France 11 (1882-83) 13-25. Refers to his article in vol. 1 and the many
resulting works. Takes a point in a
triangle and asks for the probability that the three lengths to the sides form
a triangle (an acute triangle). Then
says that breaking a stick corresponds to using an equilateral triangle, giving
probabilities ¼ and
log 8 - 2 = .0794415.... Makes various generalizations.
Takes a point, M, in an equilateral triangle ABC
and asks the probability that
MA, MB, MC form an acute
triangle, getting 4 - 2π/Ö3 =
.3718....
E. Fourrey.
Curiositiés Géometriques, op. cit. in 6.S.1. 1907. Part 3, chap.
1, section 5: Application au calcul des probabilitiés, pp. 360‑362. Break a stick into three pieces. Gets
P = ¼. Cites Lemoine.
G. A. Bull. Note
2016: A broken stick. MG 32 (No. 299) (May 1948) 87‑88. Gets Halphen's result by using homogeneous
coordinates.
S. Rushton. Note 2083: A broken stick. MG 33 (No. 306) (Dec 1949) 286‑288. Repeats Bull's arguments and then considers
making one break, then breaking the larger piece, etc. (It's not clear if he takes the longer of
the two new pieces when trying for a 4‑gon.) Says that the only example of this that he
has seen is Whitworth's DCC Exercises in Choice and Chance, 1897, exer. 677,
??NYR, which has Prob(triangle) =
⅓. Author says this is wrong and
should be 2 log 2 ‑ 1 =
.386... He gets a solution for an n‑gon.
D. N. Smith.
Letter: Random triangles. MiS 19:2 (Mar 1990) 51. Suggests, as a school project, generating
random triangles by rolling three dice and using the values as sides.
Joe Whittaker.
Random triangles. AMM 97:3 (Mar
1990) 228-230. Take a stick and break
it at two random points -- or -- break once at random and then break the longer
part at random. Prob(triangle) = ¼ in the first case and appears to be ⅓
in the second case, but the second analysis assumes an incorrect
distribution. Correcting this leads
to Prob(triangle) = 2
log 2 - 1 = .38..., as in Rushton.
J. Bertrand. Calcul des Probabilités. Gauthier‑Villars, Paris, 1889. Chap. I, art. 2, pp. 2‑3.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 26: Mexican
jumping beans, pp. 40-41 & 96.
Deranged matchboxes of red and black beans -- see 5.K.1. The problem continues by unlabelling the
boxes -- if you draw a red bean, what is the probability that the other bean in
the box is red?
Nicholas Falletta.
The Paradoxicon. Doubleday, NY,
1983; Turnstone Press, Wellingborough,
1985. Probability paradoxes, pp. 116‑125,
esp. pp. 118‑121, which describes:
a three‑card version due to Warren Weaver (1950), the surprise ace paradox of J. H. C.
Whitehead (1938) and a three prisoner paradox.
Ed Barbeau. The
problem of the car and goats. CMJ 24:2
(Mar 1993) 149-154. A version of the
problem involving three doors with a car behind one of them appeared in Marilyn
vos Savant's column in Parade magazine and generated an immense amount of
correspondence and articles. This
article describes 4 (or more?) equivalents and gives 63
references, including Bertrand, but not Dinesman, Falletta, Weaver or Whitehead.
8.H.2. BERTRAND'S CHORD PARADOX
J. Bertrand. Op.
cit. in 8.H.1. 1889. Chap. I, art 5, pp. 4‑5. Gets answers ¼, ⅓ and
½.
F. Garwood & E. M. Holroyd. The distance of a "random chord" of a circle from the
centre. MG 50 (No. 373) (Oct 1966) 283‑286. Take two random points and the chord through
them. This gives an expected distance
from the centre of .2532.
This is the problem where a man can
board equally frequent trains going either way and takes the next one to
appear, but finds himself going one way more often than the other. Why?
New section.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940. No. 42: Bus
times, pp. 28 & 116. Two bus lines running
the same route equally often, but it is twice as likely that the next bus is
Red rather than Green.
Jerome S. Meyer.
Fun-to-do. Op. cit. in 5.C.
1948. Prob. 14: The absent-minded
professor, pp. 25 & 184.
Birtwistle. Math.
Puzzles & Perplexities. 1971. Fifth, pp. 146 & 197.
8.J. CLOCK PATIENCE OR SOLITAIRE
The patience or solitaire game of
Clock has 13 piles of four face-down cards arranged with 12 in
a circle and the 13th pile in the
centre. You turn up a card from the top
of the 13th pile -- if it has
value n, you place it under the
n-th pile and turn up the card from the top of that pile and repeat the
process. You win if you turn over all
the cards. The probability of winning
is precisely 1/13 since the process generates an arbitrary
permutation of the 52 cards and is a win if and only if the last
card is a K (i.e. a 13). In response
to a recent question, I looked at my notes and found there are several papers
on this.
John Reade, proposer;
editorial solution. Problem 46.5
-- Clock patience. M500 46 (1977)
17 & 48 (1977) 16. What is the
probability of winning? Solution says
several people got 1/13 and the problem is actually easy.
Anon. proposal
& solution, with note by David Singmaster. Problem 11.3. MS 11 (1978/79) 28
& 101. (a)
What is the probability of winning?
Solution as in my comment above.
(b) If the bottom cards are a
permutation of A, 2, ..., K, then the game comes out if and only if this
is a cyclic permutation. Singmaster
notes that the probability of a permutation of
13 cards being cyclic is 1/13,
so the probability of winning in this situation is again 1/13.
Eric Mendelsohn & Stephen Tanny, proposers; David Kleiner, solver. Problem 1066 -- The last 1.
MM 52 (1979) 113 & 53 (1980) 184-185. Generalizes to k copies of
L ranks. Asks for probability of winning and for a
characterization of winning distributions.
Solution is not very specific about the distributions, just saying there
is a correspondence between the cards turned up and the original piles.
T. A. Jenkyns & E. R. Muller. A probabilistic analysis of clock solitaire. MM 54 (1981) 202‑208. Says the expected number of cards played
is 42.4. They consider continuing the game after the last K by
restarting with the first available unturned card. They call this the 'second play' and allow you to continue to
third play, etc. They determine the
expected numbers of cards turned in each play and also generalize to m
cards of n ranks.
They show that the number of plays is determined by the relative
positions of the last cards of each rank and show that the probability that the
game takes p plays, but fail to note that this is │s(n,p)│/n!,
where s(n,p) is the Stirling number of the first kind, so
they rederive a number of properties of these numbers.
Michael W. Ecker.
How to win (or cheat) in the solitaire game of "Clock". MM 55 (1982) 42-43. Shows that whether you can win is determined
by the bottom cards of each pile.
Though these values are not a permutation, one can define 'f-cycles' and a distribution comes out if
and only if every f-cycle contains the
initial value.
This covers situations where the
punter can not easily tell if the bet is reasonable or not. These are often used to lure suckers, but
they have also been historically important as incentives to develop
probabilistic methods. As an example,
the Chevalier de Méré knew that the probability of throwing one six in four
throws of a die was better than even, but he thought this should make throwing
a double six in 24 throws also better than even and it is
not. See the histories cited in 8.F for
more on the early examples of these problems.
Of course lotteries come into this category. See 8.L for some other forms.
The classic carnival game of Chuck-a-Luck is an excellent
example of this category. This is the
game with three die. You bet on a
number -- if it comes up once, you win double your bet (i.e. your bet and the
same again); if it comes up twice, you
win triple; if it comes up thrice, you
win quadruple. The relative frequency
of 0, 1, 2, 3 of your number is 125,
75, 15, 1, so the return in 216
throws will be ‑125 + 75 + 30 + 3 = ‑17, giving a
7.9% profit to the operator.
Collins. Fun with Figures. 1928.
How figures can cheat -- The tin-horn gambler, pp. 34-35. Six dice, each marked on just one face. Gambler bets $100 to $1
that the punter will not get the six marked faces all up in 20
throws. Collins asserts that the
probability of winning is 20/66 =
1/2332.3, so the fair odds
should be $2332 to
$1. This is not quite
right. The true probability is 1 - (1 - 1/66)20 =
.00042858 = 1/2333.27.
The fair odds depend on whether the winnings include the punter's $1 or
not -- in this case, they do not, so the odds according to Collins ought to
be $2331.3 to $1, or more exactly $2332.27 to $1.
In the 1980s, I saw stands at school and village fairs,
where one gets one throw with six dice for
£.50. If one gets six 6s,
one wins a new Rover. Since 66 = 46656, the promoter gets an average of £23,328
for each car, which was a tidy profit.
Gardner.
Nontransitive dice and other probability paradoxes. SA (Dec 1970). Extended in Wheels, chap. 5.
Consider two red and two black cards.
Choosing two, what are the odds of getting two of the same colour? Typical naive arguments get ⅔
or ½, but the true answer is
⅓. In the Addendum in
Wheels, S. D. Turner describes the version with R red cards and B
black cards. The probability of
choosing two of the same colour is then
[R(R-1) + B(B-1)]/[(R+B)(R+B-1)]
which is always less than
½.
In the simplest form, we know
that A
can defeat B and B can defeat
C, but this does not imply
that A
can defeat C, so non-transitivity is quite common in real
game playing. Indeed, in the classical
game of Rock, Scissors, Paper, the game situation has C defeating A.
Similar phenomena occur in preferences, particularly voting and
loving. This section will generally
deal with mathematical versions, particularly where the game seems fair, but
making a later choice than your opponent gives you an edge. Hence these versions can be used as the
basis of sucker bets.
In about 1932, the following golf scores were sent to The
Scotsman.
A 4 5 5 6 9 4 5 4 4 ‑ 46
B 5 6 6 7 3 5 6 3 5 ‑ 46
C 6 7 7 8 6 3 3 3 3 ‑ 46
A beats
B by 4 and 3, B
beats C by 5 and 4,
C beats A by
one hole. A correspondent, J. C. Smith,
suggested the following series for three holes.
A 1 2 3 ‑ 6 A one up on B.
B 2 3 1 ‑ 6 B one up on C.
C 3 1 2 ‑ 6 C one up on A.
Similar
results can be obtained for four men playing four holes, and so on.
Reported
by J. W. Stewart as Gleaning 854: Golf Scores; MG 16 (No. 218) (May 1932) 115.
Walter Penney, proposer and solver. Problem 95 -- Penney-ante. JRM 2:4 (Oct 1969) 241 &
7:4 (Fall 1974) 321. Opponent
picks a triple of heads and tails, then you pick a triple. A coin is thrown until the triple
occurs. If he chooses HHH
and you choose HTH, show your probability of winning is 3/5.
Walter Penney and David L. Silverman, proposers and
solver. Problem 96 -- Penney-ante. JRM 2:4 (Oct 1969) 241 &
8:1 (1975) 62-65. As above, but
the opponent and you both pick a triple without the other's knowledge. Elaborate analysis is required to obtain the
optimal mixed strategy which guarantees you a probability of ½.
Gardner.
Nontransitive dice and other probability paradoxes. SA (Dec 1970). Extended in Wheels, chap. 5.
Describes Bradley Efron's sets of 4 nontransitive dice which give the
second chooser a ⅔ chance of winning. Efron says it had been proven that this is the maximum obtainable
with four dice. For three dice, the
maximum is .618, but this requires dice with more than six
faces. As the number of dice increases,
the maximum value approaches ¾. The
Addendum in Wheels describes numerous variants developed by magicians and
mathematicians.
Gardner. Nontransitive
paradoxes. SA (Oct 1974) c= Time Travel, chap. 5. Discusses voting paradoxes and describes
examples of nontransitive behaviour back to mid-20C. Describes Penney's game, giving it with pairs first and showing
that for each choice by the opponent, you can pick a better choice, with
probability of winning being at least
⅔. The bibliography in
Time Travel is extensive and Gardner notes that some of the items give many
further references.
Richard L. Tenney & Caxton C. Foster. Non-transitive dominance. MM 49:3 (May 1976) 115-120. Gives three dice with odds for the second
chooser being 5/9. Gives proofs for the results stated by Efron
in Gardner, above, observing that they arise from results known about voting
paradoxes.
Many
combinatorial recreations can be considered as logical.
9.A. ALL CRETANS ARE LIARS, ETC.
Diogenes Laërtius.
3C. De Clarorum Philosophorum
Vitis, Dogmatibus et Apophtegmatibus, II, Life of Euclides. Ed. by C. G. Cobet; Paris, 1888, p. 108, ??NYS. Translated by C. D. Yonge; Bell,
London, 1894, pp. 97‑98.
Translated by R. D. Hicks; Loeb Classical Library; vol. 1, pp. 236‑237. Refers to Eubulides (c‑330) as the
source of "The Lying One" or "The Liar" -- Ο
Ψεθδoμεvoσ (O Pseudomenos). According to: I. M. Bochenski; Ancient Formal Logic; North Holland,
Amsterdam, 1951, p. 100; Eubulides also
invented: "the swindler", "the concealed", "the heap" (how many grains make a
heap?), "the Electra" and
"the horned" (equivalent to "Have you stopped beating
your wife?").
Bochenski, Ancient Formal Logic, pp. 101‑102, says the
liar paradox was unknown to Plato, but is quoted by Aristotle in his Libro de Sophisticis Elenchis 25, 180 b 2‑7,
??NYS, (See: M. Wallies, ed.; Topica
cum Libro De Sophisticis Elenchis; Leipzig, 1923; ??NYS; and:
Ethica Nicomachea, H3, 1146 a 21‑27; ed. by Fr. Susemihl; Leipzig,
1887; ??NYS.) It is also given in Cicero; Ac. Pr. II, 95, 96 (=? Topica, 57);
??NYS (In: G. Friedrich, ed.; Opera Rhetorica; Leipzig, 1893; ??NYS) and in many later writers.
Athenaeus Naucratica.
c200. The Deipnosophists, Book 9
(end of c.64). Translated by
C. D. Yonge, Bohn, London, 1854, vol. 2, p. 633. Epitaph of Philetas of Cos (c‑340/c‑285). "Traveller, I am Philetas; the argument
called the Liar and deep cogitations by night, brought me to death." Sadly, there is no indication where he died
or was buried. (Bochenski; Ancient
Formal Logic,; p. 102 gives the Greek of Athenaeus. I. M. Bochenski; History of Formal Logic; corrected
ed., Chelsea, 1970, p. 131; gives the English.)
The Stoics. c‑280. Bochenski; Ancient Formal Logic; pp. 100‑102
says they invented several paradoxes, including "the crocodile" who
takes a baby and says he will return it if the mother answers his question
correctly. He then asks "Will I return the baby?" She answers
"No".
Anon. History of the
Warring States. [The Warring States
period is ‑475/‑221 and this history may be ‑2C.] The Elixir of Death. Translated in: Herbert A. Giles; Gems of Chinese Literature; op. cit. in
6.BN, p. 43. Chief Warden swallows an
elixir of immortality which he was supposed to convey to the Prince. The Prince orders the Warden's execution,
but the Warden argues that if the execution succeeds, then the elixir was false
and he is innocent of crime. The Prince
pardons him.
Tung‑Fang So (‑2C, see Giles, ibid., p. 77) is
said to have been in the same situation as the Warden and argued: "If the elixir was genuine, your
Majesty can do me no harm; if it was not,
what harm have I done?"
St. Paul. Epistle to
Titus, I, 12. c50? "One of themselves, even a prophet of
their own, said, The Cretans are always liars, .... This witness is true."
M. Cervantes. Don
Quixote. 1605. Book II, chap. 51. Translated by Thomas Shelton, 1612‑1620, reprinted by the
Navarre Society, London, 1923, vol. 2, pp. 360‑362. Sentinel paradox: truthtellers pass; liars
will be hanged. "I will be
hanged."
Henri Decremps.
Codicile de Jérôme Sharp, ....
Op. cit. in 4.A.1. 1788. Avant Propos, pp. 19-20: Sentinel
paradox: truthtellers pass; liars will be thrown in the river. "You will throw me in the
river." Author says he will give
the answer in another volume.
Henri Decremps. Les
Petits Aventures de Jerome Sharp.
Professor de Physique Amusante; Ouvrage contenant autant de tours
ingénieux que de leçons utiles, avec quelques petits portraits à la maniére
noire. Brussels, 1789; also 1790, 1793. Toole Scott records an English edition, Brussels, 1793. Sentinel paradox. ??NYS. Cited by Dudeney;
Some much‑discussed puzzles; op. cit. in 2; 1908; as the first appearance
of this paradox.
The Sociable.
1858. Prob. 43: The Grecian
paradox, pp. 299 & 317. Protagoras
suing his pupil who had promised to pay for his tuition when he won his first
case. = Book of 500 Puzzles, 1859,
prob. 43, pp. 17 & 35. c=
Magician's Own Book (UK version), 1871, pp. 26-27. c= Mittenzwey, 1895?: prob. 157, pp. 33
& 81; 1917: 157, pp. 30 &
79, which claims the teacher wins.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-21, pp. 253 & 395. Form of the sentinel paradox. To enter a garden, one must make a
statement; if true, one pays 3 marks; if false, one pays 6 marks. "I will pay 6 marks."
Carroll-Wakeling II.
c1890? Prob. 32: Bag containing
tickets, pp. 50 & 73. This is one
of the problems on undated sheets of paper that Carroll sent to Bartholomew
Price. Wakeling reproduces the MS.
A
bag contains 12 tickets, 3 marked 'A', 4 'B', 5 'C'. One is drawn in the presence of 12 witnesses of equal credibility:
three say it was 'A', four 'B', five 'C'.
What is the chance that it was 'A'?
There is no
answer on the Carroll MS. Wakeling
gives an answer.
Let
the credibility of a witness be "a" when telling the truth. Hence, the credibility of a witness when telling
a lie is "1 - a".
If
it was A, then 3 tell the truth, and 9 lie; hence the credibility is 3 in
12, or 1 in 4.
Therefore
the chance that it is A, and no other, is:
3/12
x 1/4 + 4/12 x 3/4 + 5/12 x 3/4
= 5/8
I cannot see
how this last formula arises. Wakeling
writes that he was assisted in this problem by a friend who has since died, so
he does not know how the formula was obtained.
Assuming a is
the probability of a person telling the truth, this probability depends on what
the chosen ticket is and is not really determined by the information given --
e.g., if the ticket was A, then any value of a between 0
and 1 is possible. The
value 3/12 is an estimate of a, indeed the maximum likelihood estimate. If
k of the 12
people are telling the truth, I would take the situation as a binomial
distribution. There are BC (12, k)
ways to select them and the probability of having k
liars is then BC (12, k) ak
(1-a)12-k. Now it seems that
Bayes' Theorem is the most appropriate way to proceed. Our basic events can be denoted A, B, C
and their a priori probabilities are
3/12, 4/12, 5/12. Taking a = k/12,
the a posteriori probabilities
are proportional to
k/12 x BC
(12, k) (k/12)k ({12-k}/12)12-k, for
k = 3, 4, 5.
Dropping the common denominator of
1213, these
expressions are 6.904, 8.504,
10.1913 times 1012. Dividing by the total gives the a posteriori probabilities
of A, B, C as
27.0%,
33.2%, 39.8%.
This
is really a probability problem rather than a logical problem, but it
illustrates that the logical problem seems to have grown out of this sort of
probability problem.
Lewis Carroll. Diary
entry for 27 May 1894. "I have
worked out in the last few days some curious problems on the plan of 'lying'
dilemma. E.g. 'A says B lies; B says C lies; C says A and B lie.'
Answer: 'A and C lie; B speaks
truly'. The problem is quoted in
Carroll-Gardner with his discussion of the result, pp. 22-23. Gardner says this was printed as an
anonymous leaflet in 1894.
Carroll-Wakeling. Prob. 9: Who's telling the truth?, pp. 11
& 65. Wakeling says "This is a
puzzle based on a piece of logic that appears in his diary.
The
Dodo says the Hatter tells lies.
The
Hatter says that the March Hare tells Lies.
The
March Hare says that both the Dodo and the Hatter tell lies.
Who
is telling the truth?"
In
a letter of 28 May 2003, Wakeling quotes the text from the Diary as above, but
continues with it: "And today 'A
says B says C says D lies; D says two
lie and one speaks true.' Answer: 'D
lies; the rest speak truly.' Wakeling
adds that the Diary entry for 2 Jun 1896 (not given in the Lancelyn Green
edition) says: "Finished the
solution of the hardest 'Truth-Problem' I have yet done", but Carroll gives no indication what it was.
Lewis Carroll. The
problem of the five liars. In his
unpublished Symbolic Logic, Part II. He
was working on this after Part I appeared in 1896 and he had some galley proofs
when he died in 1898. Published in
Lewis Carroll's Symbolic Logic, ed. by William Warren Bartley III; Clarkson N.
Potter, NY, 1977,
Cesare Burali Forti.
Una questione sui numeri transfiniti.
Rendiconti del Circolo Matematico di Palermo 9 (1897) 154-164. ??NYS.
This was the first published antinomy of modern set theory. The set of all ordinal numbers is itself an
ordinal! However, Cantor had observed
the paradox in 1895 and communicated it to Hilbert in a letter in 1896.
Irving Anellis. The
first Russell paradox. Paper given at
AMS meeting, Chicago, Mar 1985. ??NYS
-- abstract given in HM 12 (1985)
380. Says it is usually believed that
Russell discovered his paradox in Jun 1901, but he sent a version of it to
Couturat on 8 Dec 1900 (unpublished MS in the Russell Archives, McMaster
Univ.).
Gregory H. Moore. A
house divided against itself: The emergence of first-order logic as the basis
for mathematics. IN: Esther R. Phillips, ed.; Studies in the
History of Mathematics; MAA, 1987, pp. 98-136.
On pp. 114-115, he dates Russell's paradox to May 1901 and says Russell
wrote about it to Frege on 16 Jun 1902.
The first publications are in Russell's Principles of Mathematics and
Frege's Fundamental Laws, vol. 2, both in 1903.
B. Russell. The
Principles of Mathematics. CUP,
1903. ??NYS -- cited in
Garciadiego. He discusses Russell's
paradox and also Cantor's paradox of the greatest cardinal and Burali Forti's
paradox of the greatest ordinal. I
won't consider these much further, but this may have inspired the development
of the more verbal paradoxes described in this section.
G. G. Berry. Letter
to Russell on 21 Dec 1904. In the
Russell Archives, McMaster University.
Quoted in Garciadiego. "...
the least ordinal which is not definable in a finite number of words. But this is absurd, for I have just defined
it in thirteen words." The paradox
of Jules Richard (late Jun 1905) is very similar and similar versions were
found by J. König and A. C. Dixon about the same time, though these all use
Zermelo's well-ordering axiom. Sometime
earlier, Berry had introduced himself to Russell with a note saying "The statement on the other side of
this paper is true" with the other
side reading "The statement on the
other side of this paper is false",
and consequently is also considered the inventor of the "visiting
card paradox".
B. Russell. Les
paradoxes de la logique. Revue de
Métaphysique et de morale 14 (1906) 627‑650. ??NYS -- cited by Garciadiego.
First publication of a modified version of Berry's paradox.
B. Russell.
Mathematical logic as based on the theory of types. Amer. J. Math. 30 (1908) 222‑262. On p. 223, he first gives Berry's
paradox: "the least integer not
nameable in fewer than nineteen syllables". He also reformulates König & Dixon as "the least
indefinable ordinal".
Kurt Grelling & Leonard Nelson. Bemerkungen zu den Paradoxien von Russell
und Burali‑Forti. Abhandlungen
der Fries'schen Schule (NS) 2 (1908) 301‑344. ??NYS. Grelling's
paradox: "Is heterological
heterological?"
A. N. Whitehead & B. Russell. Principia Mathematica.
CUP, 1910. Vol. 1, pp. 63‑64. Discusses several paradoxes and repeats
Berry's paradox.
F. & V. Meynell.
The Week‑End Book. Op.
cit. in 7.E. 1924. 2nd ed., prob. five, p. 275; 5th? ed., prob. nine, p. 408, gives
Russell's paradox as a problem -- and gives no solution!
Hummerston. Fun,
Mirth & Mystery. 1924. The bridge, pp. 68-69. Sentinel paradox -- "I am going to be
hanged on that gallows!" Author
says "It is impossible to answer ... satisfactorily. Perhaps the best plan is to throw the varlet
in the river."
John van Heijenoort.
Logical paradoxes. Encyclopedia
of Philosophy 5 (1967) 45‑51.
Excellent survey of the paradoxes of logic and set theory, but only a
mention of pre‑19C paradoxes.
Alejandro R. Garciadiego.
The emergence of some of the nonlogical paradoxes of the theory of sets,
1903-1908. HM 12 (1985) 337-351. Good survey. He has since extended this to a book: Bertrand Russell and the Origins of the Set-theoretic
"Paradoxes"; Birkhäuser, 1992, ??NYS
At the 19th International Puzzle Party in London, 1999,
Lennart Green told the story of a friend of his who was such a failure in life
that he decided to wrote a book on "How to be a Failure". But if this failed, he would be successful
and if it succeeded, he would have failed.
9.B. SMITH -- JONES -- ROBINSON PROBLEM
See also
5.K.2 for a special form of these problems.
Dudeney. PCP. 1932.
Prob. 49: "The Engine‑Driver's Name", pp. 24 &
132. = 536; prob. 521: "The
Engineer's Name", pp. 214 & 411.
The driver is Smith, but the other two names are not determined.
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 39, pp. 22 & 39. Same as Dudeney.
Phillips.
Brush. 1936. Prob. K.2: The Engine‑driver, pp. 36
& 96. Same as Dudeney.
Rudin. 1936. No. 183, pp. 65 & 119. Similar to Dudeney, but Americanized,
somewhat simplified(?) and asking for the brakeman's (= guard's) name.
Haldeman-Julius. 1937. No. 4:
Robinson-Smith-Jones problem, pp. 3 & 20.
Similar to Dudeney, but Americanized differently than in Rudin, asking
for the engineer's name. Says it was
sent by J. C. Furnas.
James Joyce.
Finnegans Wake. Viking Press,
NY, 1939. P. 302, lines 23-24: "Smith-Jones-Orbison".
J. G. Oldroyd.
Mathematicians in the army.
Eureka 5 (Jan 1941) 6 &
6 (May 1941) 10. Six men of
different ranks from three different schools, colleges and faculties (i.e.
subjects).
Irving Adler.
Thinking Machines. Dobson,
London, 1961. Pp. 111-116: Who is the
engineer? Essentially identical to
Rudin, but asks for the engineer's name.
Gives a systematic solution via boolean algebra.
Doubleday - 2.
1971. Flight plan, pp.
153-154. Same as Dudeney, slightly
reordered.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
Who's who?, pp. 86 & 134-135.
Spaceship version, similar to Dudeney, but more precise.
I now realise
that this is an extension of 9.D.
See
Littlewood, 1953, in 9.D.
Gamow & Stern.
1958. Forty unfaithful
wives. Pp. 20‑23. (Communicated by V. Ambarzuminian.)
Michael Spivak.
Calculus. 2nd ed., Publish or
Perish. ??date, place. P. 35, probs. 27 & 28.
Uri Leron & Mike Eisenberg. On a knowledge-related paradox and its resolution. Int. J. Math. Educ. Sci. Technol. 18 (1987)
761-765.
Ed Barbeau.
Fallacies, flaws and flimflam.
CMJ 22:4 (Sep 1991) 307. Gives a
brief discussion and the reference to Spivak and to Leron & Eisenberg. The paradox has to do with what information
has been provided by the stranger.
See also
9.C. 7.AP is somewhat related.
William Wells Newell.
Games and Songs of American Children.
Harper and Brothers, (1883); 2nd
ed. 1903; reprinted with Editor's Note
of 1883 and new Introduction and Index, Dover, 1963. Chap. IX, No. 77: Laughter games, pp. 136‑137. "In a Swiss game, .... Each child pinches his neighbor's ear; but by agreement, the players blacken their
fingers, keeping two of the party in ignorance. Each of the two victims imagines it to be the other who is the
object of the uproarious mirth of the company." The Notes on p. 274 indicate that this probably comes from: E. L. Rochholz; Alemannisches Kinderleid und
Kinderspiel, Leipzig, 1857, ??NYS
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 13, pp. 14 & 188. Two boys fall down, one gets a dirty face,
the other washes his own face.
Phillips. The
Playtime Omnibus. Op. cit. in
6.AF. 1933. Section XVII, prob. 11: Odd, pp. 55 & 237. Identical to Week‑End.
W. E. Buker, proposer;
Robert Wood and O. B. Rose, solvers.
Science Question 686. SSM 35
(1935) 212 & 429.
3 persons.
A. A. Bennett, proposer;
E. P. Starke and G. M. Clemence, solvers. Problem 3734. AMM 42
(1935) 256 & 44 (1937) 333‑334. n
persons with smudges on foreheads.
Says the 3 person case was suggested by Dr. Church of
Princeton and cites the Buker problem.
Phillips. Brush.
1936.
Rudin. 1936. No. 145, pp. 51-52 & 110. 3
persons with black crosses on foreheads, who whistle when they see a
black cross.
Haldeman-Julius.
1937. No. 76: The circle
problem, pp. 10 & 24. Three persons
with red crosses, who put up their hands if they see a red cross. "This problem promises to become
famous. It has been going the rounds
during the past few weeks -- .... We
printed it several years ago, but believe it deserves reprinting."
McKay. At Home
Tonight. 1940. Prob. 36: The five disks, pp. 70 &
83. Three mandarins and five disks, two
black and three white. Emperor puts
white on each forehead. Answer argues
using the number of blacks.
M. Kraitchik.
Mathematical Recreations. Op.
cit. in 4.A.2. 1943. Chap. 1.
Harold Hart. The
World's Best Puzzles. Op. cit. in
7.AS. 1943. The problem of the marked foreheads, pp. 23 & 55. Three students with blue and green crosses.
Leopold. At
Ease! 1943. Short cut to chevrons, pp. 23-24 & 199. Three men with smudged foreheads.
A. K. Austin. A
calculus for know/don't know problems.
MM 49:1 (Jan 1976) 12-14. He
develops a set-theoretic calculus for systematically solving problems involving
spots on foreheads, etc., including problems with knowing sums, see 7.AP. His typical problem has a man with four red
and three blue stamps and he sticks three on the foreheads of two boys, telling
them they each have at least one red.
The first says he doesn't know what he has; then the second says he
doesn't know; then the first says he does know. What stamps did the first have?
Leeming. 1946. Chap. 3, prob. 9: The three small boys, pp.
21‑22 & 153‑154. Three
boys with smudged foreheads.
Henry Cattan. The
Garden of Joys. (An anthology of
oriental anecdotes, fables and proverbs.)
Namara Publications, London, 1979.
How he won the office of Grand Vizier, pp. 107‑109 &
note 89 on p. 114. Same as
Kraitchik's problem 4, but with the simpler solution based on symmetry. The note says: "This story is anonymous and was heard by the author in
Palestine." A letter from the
author says he heard the story in this form in Palestine before 1948.
Jerome S. Meyer.
Fun-to-do. Op. cit. in 5.C.
1948. Prob. 51: Red hats and green
hats, pp. 46 & 190-191. 2 red hats & 3 green hats. Answer gives the symmetry solution and the
logical solution without clearly recognizing the distinction.
Max Black. Critical
Thinking. 1952. Op. cit. in 6.F.2. Prob. 11, pp. 12 & 432.
Three sons with white marks.
J. E. Littlewood. A Mathematician's
Miscellany. Op. cit. in 5.C. 1953.
Pp. 3‑4 (25‑26).
Three dirty faces. Mentions that
this can be extended to n dirty faces, which "has not got into
the books so far as I know". This
may be the origin of 9.C?
The Little Puzzle Book. Op. cit. in 5.D.5.
1955. Pp. 34-35: Clean and
dirty. Two men fall through a roof. Man with clean face goes to wash.
T. J. Fletcher.
The n prisoners. MG 40 (No.
332) (May 1956) 98‑102. Considers
Kraitchik's problem with n persons,
n white discs and n‑1
black discs. Also studies
various colour distributions and assignments, including < n
white discs and use of three colours.
Gamow & Stern.
1958. Three soot‑smeared
faces. Pp. 77‑79.
Birtwistle. Math.
Puzzles & Perplexities. 1971. Hats Off!, pp. 108 & 193. A, B, C
are seated in a row, so B can see
A while C
can see both B and
A. They all know that there are
three white and two red hats in a bag.
A hat is taken out and put on A's
head, but he can't see it.
Similarly, hat are taken and put onto
B and C. C is now asked it he knows what colour his hat
is and replies that he does not.
B, having heard C's
response, is asked if he knows what colour his hat is and he replies
that he does not. Is A, having heard these, able to know what
colour his hat is? Extends to various
other combinations and to n people.
The answer is "The answer to all the questions is, yes, it is
possible."
Complicated questions of kinship
have arisen due to religious taboos on incest.
Most religions have a list of kinship relations which are not permitted
to marry. These get a bit more involved
than I want to go into. See the items
in the first section below for some typical material.
The second section deals with marrying
a deceased wife's sister.
The third section deals with general
strange families riddles and puzzles, but 'That man's father ...' are collected
in 9.E.1.
Ripley's Believe It or Not! books
give a number of examples of strange families.
I will enter these under the date of the persons involved. BION-xx
denotes the xx-th series of
Believe It or Not!
Problems of this type are generally
put in the form of a riddle, and many of these are collected in the following.
Mark Bryant.
Dictionary of Riddles.
Routledge, 1990. (Based on his
Riddles Ancient and Modern; Hutchinson, 1983.)
GENERAL STUDIES OF KINSHIP RELATIONS
J. Cashdan & Martin D. Stern. Forbidden marriages from a woman's angle. MG 71 (No. 456) (1987). ??NYS -- cited by Stern, 1990.
Martin D. Stern.
Consanguinity of witnesses -- a mathematical analysis. Teaching Mathematics and Its Applications
6:2 (1987). ??NYS -- cited by Stern,
1990.
Martin D. Stern.
Mathematical motivation through matrimony. MM 63:4 (Oct 1990) 231‑233. ??NYS -- reproduced in Robert L. Weber; Science with a Smile;
Institute of Physics, Bristol, 1992, pp. 314-318. Presents a notation for kinship relations and uses it to see that
the table of prohibited degrees of marriage given in the Book of Common Prayer
is symmetric with respect to sex and hence there are no unexpected
prohibitions. However, the Jewish
restrictions on marriage and on testimony by consanguineous relatives are not
symmetric -- cf the above items.
Marcia Ascher.
Ethnomathematics. Op. cit. in 4.B.10. 1991.
Chapter Three: The logic of kinship relations, pp. 66-83. Gives a number of folk puzzles and then
analyses several complicated kinship systems.
Some references.
Martin Stern.
Discrete avoidance of marital indiscretion. Mathematics Review (Univ. of Warwick) 2:3 (Feb 1992) 8-11. He presents a notation for kinship relations
and uses it to describe the prohibited relations in Christian, Jewish and
Islamic traditions. The Jewish
prohibitions are not symmetric between male and female.
Helen Cooper. A
little more than kin [Review of Elizabeth Archibald; Incest and the Medieval
Imagination; OUP, 2001]; The Times Literary Supplement (26 Oct 2001) 27. This notes that the taboos on incest are
very variable. Egyptian Pharaohs
indulged in brother-sister marriages and the Macedonians did not understand why
Oedipus was upset when he discovered he was married to his mother. Medieval Christianity extended the family to
include godparents, who were spiritual siblings, and even in-laws of in-laws,
as well as all the relatives of lovers, since sex made the lovers 'one
flesh'. On the other hand, Henry VIII
married his deceased brother's wife and later divorced her in order to marry
the sister of his mistress. Archibald
notes that since Christ is God, Mary is "Maid and mother, daughter of thy
son!" [Chaucer's translation of Dante].
The medieval legend of Judas makes him a version of Oedipus -- he
unknowingly killed his father and married his mother. The medieval period introduced the double incest version where
the son of a brother-sister relation is sent away and returns as an adult and
unknowingly marries his mother. In 13C
French versions of the Arthurian legend, Mordred, the nephew of Arthur, is made
into his son, the result of a liaison between Arthur and his half-sister, who
did not know of their relationship.
Mordred attempted to marry Arthur's wife. Luther urged that if a wedded couple were later discovered to be
brother and sister or half-sister, or even mother, that the knowledge should be
suppressed lest it drive them to the ultimate sin of despair.
E. S. Turner. Roads
to Ruin -- The shocking history of social reform. Michael Joseph, London, 1950.
Chap. 5: Two wives, one mother-in-law, pp. 98-121. This surveys the British preoccupation with
the legality of marrying a deceased spouse's sibling. Since a couple were considered to become 'one flesh' (Ephesians
5:31), such a marriage was considered incestuous by the Church. Leviticus 18:6 & 16 were interpreted as
prohibiting such marriage, but Leviticus 18:18 was interpreted as saying that
the previous verses stated that a man should not have sisters as wives at the
same time [which is the Islamic interpretation], while Deuteronomy 25:5-10 not
only permits, but even commands, that a man should marry his brother's
widow.
The
English preoccupation with the problem dates from Henry VIII's marriage to his
brother's widow, Catherine of Aragon.
[In fact, he then divorced her to marry his mistress's sister.] This particular question is mentioned in
Shakespeare's Henry VIII and the general question is the basis of Hamlet, whose
mother marries her dead husband's brother.
There was at least one execution, in early 18C Scotland, of a woman who
had sex with her sister's widower.
Up
to 1835, marriage to a deceased wife's sister was permitted, but it could be
voided and the children declared bastards, if an action was brought. But if an action was brought and dropped,
further actions were prevented. In
1835, the Duke of Beaufort, who had married his deceased wife's half-sister,
persuaded the Lord Chancellor to introduce a bill to legitimize such marriages
up to date. The Bishops managed to
amend this to prohibit such marriages in the future. However, such a couple could go to Europe to be married and such
marriages remained legal in places like Jersey, though they were not legitimate
in England. The Catholic Church
generally gave dispensation for such marriages. From 1841 onwards, bills to remove the prohibition were
introduced in almost every Parliament.
Marriage to a deceased husband's brother or to a deceased spouse's
nephew/niece was not sufficiently common to be considered by the reformers. The question was mentioned by Gilbert &
Sullivan (near the end of the first act of Iolanthe (1882), the Queen,
referring to Strephon, says "He
shall prick that annual blister, / Marriage with deceased wife's
sister;"). The journal Moonshine
commented: "To be able to marry
two wives at the cost of but one mother-in-law is something to fight
for." In 1906, the Colonial
Marriages Bill legitimized such marriages made in the colonies. In 1907, the Deceased Wife's Sister Bill was
passed. Canon Law was later changed to
accept this. One man who had married
his deceased wife's sister sued a Canon who refused him Communion and won, with
his win being confirmed by the Court of Appeals and the House of Lords in
1912. However, marriage to a divorced
wife's sister was not permitted while the ex-wife lived. Marriage to a deceased husband's brother was
permitted in 1921. A number of other
marriages were permitted in 1921 and all these acts are consolidated in the
Marriage Act of 1949. Turner is not
clear whether marrying a divorced spouse's sibling was permitted, and I don't
know the further history. A 2000
article says marrying a deceased wife's aunt or niece was permitted by the
Marriage Act of 1931.
A discussion of Strawberry Hill House says marriage to a
deceased husband's brother was prohibited by the 1835 act, so that Frances, the
widow of John Waldegrave, had to go to Scotland to marry his half-brother
George Waldegrave, 7th Earl Waldegrave.
Susan Kelz Sperling.
Tenderfeet and Ladyfingers.
Viking, NY, 1981, p. 98-99. She
gives some details of the Hebrew view.
The Hebrew law of yibbum declares that if a man dies without heir, his
brother or nearest relative is obliged to marry the widow (i.e. she is marrying
her dead husband's brother). However,
he could decline the duty by a ceremony called halitzah, as specified in Deuteronomy,
by putting on a special shoe which the widow removed and then she spat in front
of him to break the contract. This
takes place in Ruth, allowing her to marry Boaz.
BION-11 cites an American example where a woman successively
married the widowers of two of her sisters.
William Holman Hunt, the Victorian painter, married his
deceased wife's sister in 1866, in Switzerland [Judi Culbertson & Tom
Randall; Permanent Londoners; Robson Books, London, 1991, p. 140].
See Dudeney, AM, prob. 52, below, for a complication of this
situation.
Haldeman-Julius.
1937. No. 88: Marriage problem,
pp. 11 & 25. How can a man have
married his widow's sister? (Also
entered under General Family Riddles.)
Mindgames. Frontiers
(a UK popular science magazine) No. 1 (Spring 1998) 107. "My wife's sister is also her
cousin. How can this be?" Solution is that her father married his dead
wife's sister and had another daughter.
The riddles which the Queen of Sheba proposed to Solomon are
not recorded in the biblical account of their meeting (I Kings 10 & II
Chronicles 9), which would be c-960.
Josephus' History of the Jews only mentions that Hiram and Solomon
traded riddles, without giving any of them.
Bryant, p. 19, says the Queen's riddles are given are given in the 2nd
Targum to the Book of Esther and elsewhere in the rabbinical literature. The Targums are commentaries on biblical
books, created after the Babylonian Captivity of ‑587/-538 and written
down from 100 onwards. One of these is
a strange family riddle which occurs in the first few entries below. If this is really due to the Queen of Sheba,
or even actually in the Targums, it would be by far the earliest strange
families riddle known. A variant of the
riddle is given by Yachya Ben Sulieman, c1430, qv below. Ms Zimmels at the library of the London
School of Jewish Studies told me that there is an 1893 German translation:
Targum Shennai(?) zum Buch Esther and that the riddles occur in Ginzberg. However, Rappoport gives more precise
information.
Louis Ginzberg. The
Legends of the Jews. Translated from
the German Manuscript. Vol. IV: Bible
Times and Characters From Joshua to
Esther. Jewish Publication Society of
America, Philadelphia, (1913), 5th ptg, 1947, pp. 142-149. He gives 22 riddles. P. 146, no. 2: 'Then she [the Queen of
Sheba] questioned him [Solomon] further: "A woman said to her son, thy
father is my father, and thy grandfather my husband; thou art my son, and I am
thy sister."
"Assuredly," said he, "it was the daughter of Lot who
spake thus to her son."' However,
Ginzberg gives no source or date for this.
Angelo S. Rappoport.
Myth and Legend of Ancient Israel.
Vol. III. Gresham Publishing
Co., London, 1928. The riddles of the
Queen of Sheba, pp. 125-130. P.
127:
'Said
she: "I will ask thee another question.
A woman once said unto her son: Thy father is my father, thy grandfather
my husband; thou art my son but I am thy sister."
To
which Solomon made answer: "It must surely have been one of Lot's daughters
who thus spoke to her son."
The
similarity of the text with Ginzberg's makes it clear that they are both taken
from the same source. Fortunately
Rappoport is specific as to his sources.
He says the second Targum to Esther (citing Targum Sheni to the Book of
Esther; ed. P. Cassel, Leipzig, 1885; ed. E. David, Berlin, 1898) contains
three riddles (the last three in Ginzberg) and then says that the Midrash
Mishle, or Midrash to the Proverbs (citing Midrash Mishle, ed. S. Buber, Vilna,
1893 and A. Wünsche, Midrash Mishle, Leipzig, 1885), gives four riddles,
which are the first four in Ginzberg, hence include our riddle. For our riddle, he also gives another
reference: J. Lightfoot, Horæ Hebraica, Rotterdam, 1686, II, 527; see also
Yalkut, II, §1085. After these four
riddles, he says the Midrash Hachefez (ed. and translated by S. Schechter,
Folklore, No. 1, pp. 349-358) gives 19 riddles, which are the first 19 of
Ginzberg, so again include our riddle.
However, Rappoport gives no indication of the dates of these Midrashs.
The Exeter Book Riddles.
8-10C (the book was owned by Leofric, first Bishop of Exeter, who
mentioned it in his will of 1072).
Translated and edited by Kevin Crossley-Holland. (As: The Exeter Riddle Book, Folio Society,
1978, Penguin, 1979.) Revised ed.,
Penguin, 1993.
Alcuin. 9C.
Abbot Albert.
c1240. P. 335.
Cooper (above under General Studies of Kinship Relations)
says the medieval period introduced the double incest situation, where the son
of a brother-sister relation is sent away and returns as an adult and
unknowingly marries his mother. Cf
Pacioli, c1500, and the Martham tombstone of 1730 for a triple incest.
Dialogue of Salomon and Saturnus. 14C. ??NYS. Given in Bryant, p. 12. "Tell me, who was he that was never
born, was then buried in his mother's womb, and after death was
baptised?" Answer: Adam. Cf:
Adevineaux Amoureux, 1478; Vyse,
1771?, prob. 2.
In about 1380, the Duke of Gloucester, uncle of Richard II,
opposed the marriage of his brother's son to his wife's younger sister. [John Kinross; Discovering Castles 1. Eastern England; Shire Discovering
Series No. 23, 1969, p. 12.]
Yachya Ben Sulieman.
Hebrew text, c1430. ??NYS. Quoted in Folk‑Lore (1890) ??NYS. Quoted in Tony Augarde, op. cit. in 5.B, p.
3. A riddle attributed to the Queen of
Sheba. "A woman said to her son,
thy father is my father, and thy grandfather my husband; thou art my son, and I am thy
sister." "Assuredly," said he [Solomon], "it was the daughter of Lot who spake thus to her
son." Bryant, no. 1116, pp. 259
& 346 gives the same wording, with an extra level of quotation marks, and
attributes it to the Queen of Sheba with no further details. Cf Queen of Sheba and Exeter Book above.
Adevineaux Amoureux.
Bruges, 1478. ??NYS -- quoted by
Bryant, no. 6, pp. 67-68 & 333.
"Je fus nez avant mon pere / Et engendré avant ma mere, / Et ay
occis le quart du monde, / Ainsi qu'il gist a la reonde, / Et si despucelay ma
taye. / Or pensez se c'est chose vraie."
(Bryant's translation: "I was born before my father, begotten
before my mother and have slain a quarter of the world's population. How can this be?" Answer: Cain. Cf: Dialogue of Salomon
and Saturnus, 14C; Vyse, 1771?, prob.
2.
Chuquet. 1484. Prob. 166.
Same as Alcuin/Bede 11a. FHM 233
mentions it briefly without giving the relationships.
In 1491, the 14 year-old Duchess Anne of Brittany married
Charles VIII, King of France in 1491.
This was slightly complicated because both of them were married already,
indeed Charles was married to the daughter of Anne's husband, the future
Emperor Maximilian of Austria, so he was marrying his own
step-mother-in-law. Fortunately, as was
often the case in those days, both marriages were unconsummated -- indeed the
couples had probably not yet seen each other and such proxy marriages were more
like engagements -- so a little influence at Rome got both marriages
dissolved. Somewhat surprisingly, as
the marriage was more or less forced by Charles' siege of Rennes, the couple
got on very well and developed a definite affection.
Pacioli. De
Viribus. c1500. Part 3.
Ivan Morris. Foul
Play and Other Puzzles of all Kinds.
(Bodley Head, London, 1972);
Vintage (Random House), NY, 1974.
Prob. 21: No incest, pp. 39 & 93.
Quotes Dudeney (??NYS) who gave an authentic 1538 epitaph describing the
situation of Alcuin/Bede 11a with each couple having a child.
Tartaglia. General
Trattato, 1556, art. 135, p. 256r.
2 fathers and 2
sons make only 3 people.
16C(?) riddle in Mantuan dialect. Given in: Franco Agostini
& Nicola Alberto De Carlo; Intelligence Games; (As: Giochi della
Intelligenza; Mondadori, Milan, 1985);
Simon & Schuster, NY, 1987; p. 69.
Two fathers and two sons make three people. The discussion is a bit unclear as to the date of this riddle.
Book of Merry Riddles.
1629?
Tombstone in the church at Martham, Norfolk. 1730.
??NYS -- quoted in a letter from Judith Havens (Norwich, Norfolk) in
Challenging Centipede; The Guardian, section 2, (1 Dec 1994) 6.
"Here
Lyeth / The Body of Christ. / Burraway, who departed / this life ye 18 day / of
October, Anno Domini 1730 / Aged 59 years / And their Lyes / Alice, who by hir
Life / was my Sister, my Mistress, / My Mother, and my Wife. / Dyed Feb ye 12,
1729 / Aged 76 years."
This
was a response to a vague description of the epitaph in: Centipede; Famous last words; The Guardian,
section 2 (24 Nov 1994) 4. There it is
stated that Burraway was the result of an incestuous union between a man and
his daughter. The baby was sent away to
be brought up and years later happened to return to his native village where he
met a older woman and became her lover, then her husband! (Shades of Oedipus!) Consequently he was his own uncle,
step-father and brother-in-law! Cf
Cooper, above for the idea of double incest -- this seems to be a triple
incest.
If
Burraway's union with his sister/mother/wife produced a child, then it would
have only three grandparents and six great-grandparents (as is the case for the
offspring of any half-siblings).
Vyse. Tutor's
Guide. 1771?
Jackson. Rational
Amusement. 1821. Arithmetical Puzzles. No. 6, pp. 2 & 52. Father, mother, son, grandson, brother and
daughter comprise only 3 people in the situation of Alcuin/Bede 11a
when a couple has a son.
Curiosities for the Ingenious selected from The most
authentic Treasures of Nature, Science and Art, Biography, History, and General
Literature. 2nd ed., Thomas Boys,
London, 1822. Singular intermarriage,
p. 100. Man and daughter marry daughter
and father. "My father is my son,
and I am my mother's mother; / My sister is my daughter, and I'm grandmother to
my brother."
Richard Breen. Funny
Endings. Penny Publishing, UK, 1999, p.
25. Gives the following.
"Here,
beneath this stone, / Lie buried alone / The father and his
daughter, /
The
brother and his sister, / The man and his wife, /
And only two bodies. //
Work
it out...
Early
19th century, Erfunt Cemetery, Germany."
I suspect 'Erfunt'
is a misprint for 'Erfurt' and I'm a bit suspicious as to the authenticity of
this, but it's a good puzzle problem.
Judge Leicester King (1789-1856) of Akron, Ohio, and his son
married sisters, so he was his son's brother-in-law. BION-11.
Illustrated Boy's Own Treasury. 1860. Arithmetical and
Geometrical Problems, No. 32, pp. 430 & 434. Will involving 6 relatives who turn out to be just 3
due to the situation of Alcuin/Bede 11a.
Charades, Enigmas, and Riddles. 1862: prob. 40, pp. 138 & 144-145; 1865: prob. 584, pp. 110 & 157-158. "How can a man be his own
grandfather?" Mother and daughter
marry son and father. Mother and son
produce a son, Tom. Notes that perhaps
Tom is only his own grandfather-in-law.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 564-12, pp. 253 & 395. 25
relationships among only 7 people.
Answer is a couple, with their son and his wife, with their son and two
daughters. This omits a grandchild, so
there really ought to be 26 relationships here. Dudeney, AM 54, gives the same grouping, but
with a different list of 23 relationships.
However neither counts the relationships properly -- e.g. both
count 4 children and
2 sons and 2 daughters. I find
23 reasonable relationships -- Dudeney's 23 less 4 repeated children
plus 2 husbands and 2 wives that he omitted. Leske has the husbands and wives, but omits the grandmother and a
grandchild.
In
theory, n people can have
n(n-1) possible relationships,
but not all of these relationships have distinct names. E.g. in the above, the son of the first
couple is a son to both parents, so two distinct relationships are both denoted
by 'son'. However, in the classic
problem of man, son and grandson, we actually have 2 fathers, 2 sons, 1 grandfather and 1 grandson, giving a full 6 relationships among
just 3
people. Extending this to a
string of n generations gives the full
n(n-1) relationships among n
people. One might ask if one can
compact this a bit by using fewer generations for the n people. E.g., Leske's problem has 3
generations. So I pose the
following problem: for n
people in g generations, how many of the n(n-1)
relationships are distinctly named in English (where 'distinctly named'
is a bit vague!). I now realise that
the same relationship may have different names, indeed several different
names! See Alcuin, above, and Carroll,
below.
Mittenzwey.
1880.
E. S. Turner. Op.
cit. above, p. 109. [Retold in
his: Amazing Grace; Michael Joseph,
London, 1975; pp. 279-280.] The 7th Duke
of Marlborough described to the House of Lords, c1880, the supposedly real case
of a father and son marrying a daughter and mother. The son was his own grandfather and he became so confused that he
committed suicide. In a footnote,
Turner quotes a letter in the Welwyn News-Chronicle of 1949 from a man who
married his step-mother's sister, i.e. widower and son married sisters.
J. M. Letter: Genealogical puzzle. Knowledge 3 (6 Jul 1883) 13, item 865. +
Answer to genealogical puzzle in our last. Ibid. (13 Jul 1883) 29.
Two unrelated persons have the same brother. Editorial note to the Answer says there are several ways to solve
the puzzle -- how many?
Lewis Carroll. A
Tangled Tale. (1885) = Dover,
1965. The dinner party. Knot II: Eligible apartments. Pp. 7-8 & 84-85. (In the answers, this part of the Knot is
denoted §1. The dinner party.)
= Carroll-Wakeling II, prob. 41: Who's coming to dinner?, pp. 59
& 75. A man's: father's brother-in-law; brother's father-in-law; father-in-law's brother and
brother-in-law's father are all
the same person. This involves several
marriages between cousins.
But
brother-in-law denotes both sister's
husband and wife's brother! With an
earlier marriage between cousins, we can have this person being both the man's
father's wife's brother and father's sister's husband. His wife's brother's father is just his
father-in-law, but another marriage between cousins makes this person also the
man's brother-in-law's father-in-law.
E. W. Cole. Cole's
Fun Doctor. The Funniest Book in the
World. Routledge, London &
E. W. Cole, Melbourne, nd
[HPL dates it 1886 and gives the author's name as Arthur C. Cole]. P. 57:
A smart cut-out & Genealogy
are two stories of widow and daughter marrying son and father. = Alcuin 11b.
Lemon. 1890. How is this?, no. 725, pp. 83 &
123. 33 relatives who are only
8 people.
Hoffmann. 1893. Chap. IX, no. 42: The family party, pp. 321
& 328‑329 = Hoffmann‑Hordern, p. 214. A man is
his father's brother‑in‑law, his brother's father‑in‑law, his father‑in‑law's brother‑in‑law and
his brother‑in‑law's father‑in‑law!
C. C. Bombaugh.
Facts and Fancies for the Curious.
Lippincott, 1905, ??NYS. A Mr.
Harwood and John Cosick, both widowers, married each other's daughter, at
Durham in eastern Canada. Quoted
in: George Milburn; A Book of Puzzles
and Brainteasers; Little Blue Book No. 1103, Haldeman‑Julius, Girard,
Kansas, nd [1920s?], pp. 33‑34.
Pearson. 1907.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
26:6 (Oct 1911) 569. The mean
Duke. Same as Hoffmann.
Henry Edwards Huntington (1850-1927) married Arabella Duval
Huntington ( -1924), the widow of his
uncle Collis P. Huntington ( -1900) in
1913. Henry and Arabella were the same
age.
Dudeney. AM. 1917.
Several examples, including the following.
Ahrens.
A&N. 1918. Pp. 105‑122, esp. 111‑122. Describes an 11C report of a person with
only three grandparents, but it turns out to be erroneous. However, Cleopatra had only three
grandparents, since her parents were half‑siblings -- I had believed they
were full siblings which would give her only two grandparents. See also the 1730 entry about Christopher
Burraway.
For
convenience in the following, let n‑parents
denote one's ancestors n generations back, so 1‑parents are parents, 2‑parents are grandparents, 3‑parents are great‑grandparents,
etc. Normally one has 2n n‑parents.
Prince
Don Carlos of Spain (1545‑1568) had only
4 3‑parents, 6 4‑parents, 12 5‑parents and
20 6‑parents. Ahrens also gives more extended examples,
e.g. the 12 generations of ancestors of Kaiser Wilhelm II comprise only 1549
people instead of the expected
8190, and one person occurs in 70
places.
Smith. Number
Stories. 1919. Pp. 114‑116 & 142. Tartaglia's problem. Also a more complex version where 27
apparent people are only 7.
Ackermann.
1925. Pp. 92‑93. Three mothers, each with two daughters,
require only 7 beds.
Loyd Jr. SLAHP. 1928.
A puzzling estate, pp. 68 & 111.
Three fathers and three sons are only four people.
Collins. Fun with
Figures. 1928. He was his own grandfather, pp.
231-232. Claims to quote a Pittsburgh
newspaper story of a resident who committed suicide when he found he was his
own grandfather. Son and father married
widow and daughter.
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia
Verlagsbuchhandlung, Prague, 1937.
Haldeman-Julius.
1937. No. 88: Marriage problem,
pp. 11 & 25. How can a man have
married his widow's sister? (Also
entered under Deceased Wife's Sister.)
Depew. Cokesbury
Game Book. 1939.
Joseph Leeming.
Riddles, Riddles, Riddles.
Franklin Watts, 1953; Fawcett
Gold Medal, 1967.
Ripley's Believe It Or Not!
6th series, Pocket Books, NY, 1958.
P. 145. Jacob van Nissen, of
Zwolle, Holland, and his son married a girl and her mother.
W. Leslie Prout.
Think Again. Frederick Warne
& Co., London, 1958. Catch Quiz,
No. 6, pp. 12 & 115. "Said one
boy to another: "My mother's
sister is your sister's mother."
What relation were the two boys?"
Kathleen Rafferty.
Dell Pencil Puzzles & Word Games.
Dell, NY, 1975. Noodle Nudger 3,
pp. 106 & 127. "Mary's
husband's father-in-law is Mary's husband's brother's brother-in-law, and
Mary's sister-in-law is Mary's brother's stepmother. HOW COME?"
"Mary's father married her husband's sister."
Scot Morris. The
Book of Strange Facts and Useless Information.
Doubleday, NY, 1979, p. 98.
In order to give his divorced mother some benefit from his father's
estate, Robert Berston adopted his mother in 1967.
Patrick Donovan.
Peculiar People. Fontana,
1984. P. 76 reports that Dave Woodhouse
of Wolverhampton divorced his wife and married her mother in 1983 at a double
wedding where his ex-wife was also married to a new man.
Ripley's Believe It or Not! - Strange Coincidences. Tor (Tom Doherty Associates), NY, 1990, p.
21. George Clark Cheever, of Warsaw,
Indiana, three of his sons and one of his daughters all married siblings.
Marcia Ascher.
Ethnomathematics. Op. cit. in
4.B.10. 1991. Chapter Three: The logic of kinship relations, pp. 66-83. Gives a number of folk puzzles and
references.
Two
mothers and two daughters are three people (Brazil).
Two
brothers say "My brother's son is
buried there"; third brother
says "My brother's son is not
buried there" (Ireland).
Who
is the sister of my aunt, who is not my aunt? (Puerto Rico).
His
mother is my mother's mother-in-law (Russia).
Who
is my mother's brother's brother-in-law? (Wales).
"Smallweed" diary column. The Guardian (10 Apr 1993) 18. Bill Wyman of the Rolling Stones married and
later divorced Mandy Smith. Mandy's
mother Patsy is about to marry Stephen Wyman, Bill's son from his first
marriage.
Item on front of Society section of The Guardian (2 Apr
1997) 1. "A 46-year-old
Bedfordshire woman has married the father of an 18-year-old girl who eloped
with her husband."
David Singmaster.
Some years before I read a description of some English aristocrat which
said that he only had four grandparents while most people had eight. This was later corrected -- he had only four
great-grandparents while most people had eight. Of course this puzzled me for a bit and I worked out a reasonable
way this could have happened. I later
realised that there is another way it could have happened, though I think the
second method is slightly less likely.
I posed the problem of finding both ways as a Brain Jammer: Four
Great-Grandparents; Weekend Telegraph (12 Dec 1998) 21 & (2 Jan
1999) 17. At the time I didn't know of
any real examples, but Ahrens (1918) says Prince Don Carlos of Spain (1545‑1568)
had only four great-grandparents. I
also used this on a Puzzle Panel program.
When cousins marry, their children have six great-grandparents.
Mindgames. Frontiers
(a UK popular science magazine) No. 1 (Spring 1998) 107. "My wife's sister is also her
cousin. How can this be?" Solution is that her father married his dead
wife's sister and had another daughter.
9.E.1. THAT MAN'S FATHER IS MY FATHER'S SON, ETC.
New section -- I have just found the
17C example, but there must be other and older examples?? But see Proctor, 1883. The problem continues to perplex people --
see Fairon, 1992.
For the 'Blind beggar' type of
problem see: Boy's Own Book, 1828; Rowley, 1866; Rowley, 1875; Neil, 1880s;
Lemon, 1890; Clark, 1897; Home Book, 1941; Charlot, c1950s. Again one thinks this should be older.
Some of the items given above, could
go in here.
Tomé Pinheiro da Veiga.
Fastiginia o fastos geniales.
(This work is a chronicle of courtly life in Castilla, from 1601 to
1606. Translated & edited by
Narciso Alonso Cortés. Imprenta del
Colegio de Santiago, Valladolid, 1916, pp. 155b-156a, ??NYS.) French translation in: Augustin Redondo; Le
jeu de l'énigme dans l'Espagne du XVIIe siècle. Aspect ludique et subversion; IN: Les Jeux à
la Renaissance; Actes du XXIIIe Colloque International d'Études
Humanistes (Tours, 1980); J. Vrin, Paris, 1982; pp. 445-458, with the problem
being on p. 453. There are two
brothers, born of the same father and mother.
One is my uncle, but the other isn't.
How is this possible?
The Book of Merry Riddles.
London, 1629. ??NYS -- Santi 235
gives this and says it is reprinted in J. O. Halliwell, The literature of the
sixteenth and seventeenth centuries, London, 1851, pp. 67‑102,
??NYS, and as pp. 7-29 in Alois Brandl, Shakespeares Book of Merry Riddles und
die anderen Räthselbücher seiner Zeit, Jahrbuch der deutschen
Shakespeare-Gesellschaft 42 (1906) 1-64, ??NYS. Bryant, pp. 100-102, quotes from: A Booke of Merrie
Riddles, Robert Bird, London, 1631 and says it is also known as Prettie
Riddles. Santi 237 gives Booke of Merrie
Riddles, London, 1631, and says it is reprinted as pp. 53-63 in Brandl,
??NYS. Santi 307 gives The Booke of
Merry Riddles, London, 1660, reprinted by J. O. Halliwell in 1866 in an edition
of 25 copies, of which 15 were destroyed!, ??NYS. In Bryant, no. 275, pp. 102 & 336: "I know a child borne
by my mother, / naturall borne as other children be, / that is neither my
sister nor my brother. / Answer me shortly: what is he?"
Boy's Own Book.
Conundrums.
The Riddler.
1835. Conundrums Nos. 66 and 73,
p. 16, no answers in my copy. These are
identical to Nos. 70 and 77 in Boy's Own Book.
Boy's Own Book. 1843
(Paris): 437 & 441, no. 11. Woman
says: 'Your mother was my mother's only daughter.' = Boy's Treasury, 1844, pp. 425 & 429. = de Savigny, 1846, pp. 354 & 358, no.
8: 'sa mère à lui est la seule fille de sa mère à elle', which doesn't seem
quite right to me.
Child. Girl's Own Book. 1848: Enigma 46, p. 236; 1876: Enigma 37, p. 199. "His mother was my mother's only
child."
=
Fireside Amusements, 1850: No. 36, pp. 111 & 181; 1890: No. 21, p. 99.
Fireside Amusements.
Charades, Enigmas, and Riddles. 1860: prob. 176 & 177, pp. 21 & 44;
1862:
prob. 176 & 177, pp. 73 & 111.
Boy's Own Conjuring Book.
1860. P. 381. "I've no sister or brother; / You may
think I am wild; / But that man's mother / Was my mother's child."
Hugh Rowley. Puniana
or Thoughts Wise and Other-wise A New
Collection of the Best Riddles, Conundrums, Jokes, Sells, etc, etc. Chatto & Windus, London, 1866.
Hugh Rowley. More
Puniana; or, Thoughts Wise and Other-Why's.
Chatto & Windus, London, 1875.
[Richard A. Proctor]
Letters received and short answers.
Knowledge 3 (26 Oct 1883) 264. Answer to Harry.
"Sisters and brothers have I none
But that man's father is my father's son." Implies that the puzzle is not well
known. Recalls it being posed on a ship
and distracting all the passengers for a day.
E. W. Cole. Cole's
Fun Doctor. The Funniest Book in the
World. Routledge, London &
E. W. Cole, Melbourne, nd
[HPL dates it 1886 and gives the author's name as Arthur C. Cole]. P. 329: A riddle. "His mother was my mother's only child."
James Neil [= "A Literary Clergyman"]. Riddles:
And Something New About Them.
(Lang Neil & Co., London, nd [1880s?]; Simpkin Marshall & Co., London); Village Games, London, 1993.
General Riddles: Relationship, p. 28.
Lemon. 1890.
Fireside Amusements -- A Book of Indoor Games. Op. cit. in 7.L.1. 1890? P. 99, no. 21. "His mother was my mother's only
child."
William Crompton.
The odd half-hour. The Boy's Own
Paper 13 (No. 657) (15 Aug 1891) 731-732.
A true friend. "If your
uncle's sister is not your aunt, what relationship does she bear to you?"
Bennett Coll. Prove
it! The Idler 2 (1892-1893, probably
Dec 1892) 510-517. Man in front of a
portrait says "Sisters and
brothers have I none; That man's father
is my father's son." Says the portrait
is himself! Observes that this leads to
his father being his own son and being the father of his father. Describes the difficulties people have in
trying to see this answer [not surprisingly].
Various other solutions given:
grandfather, brother, uncle on the mother's side.
Hoffmann. 1893. Chap. IX, no. 25: The portrait, pp. 318
& 326 = Hoffmann-Hordern, p. 211. "Uncles and brothers have I none, But that man's father is
my father's son." He notes
"This venerable puzzle forms the subject of a humorous article, entitled
"Prove It," in a recent number of the Idler. Its most amusing feature is that the writer
has himself gone astray, ...."
[I'm not sure whether Coll has gone astray or is using the error to
generate humour??]
W. H. Howe. Everybody's
Book of Epitaphs Being for the Most Part What the Living Think of the
Dead. Saxon & Co., London, nd
[c1895] (reprinted by Pryor Publications, Whitstable, 1995). P. 165 has the following entry.
"In
Llanidan Churchyard, Anglesea:--
Here
lies the world's mother,
By
nature my Aunt -- sister to my mother,
My
grandmother -- mother to my mother,
My
great grandmother -- mother to my grandmother,
My
grandmother's daughter and her mother."
Could this be
a real case of 'I'm my own grandmother'??
Clark. Mental
Nuts.
Somerville Gibney.
So simple! The Boy's Own Paper
20 (No. 992) (15 Jan 1898) 252. 'That
very old catch -- "If Dick's father is Tom's son, What relation is Dick to Tom?"'
Dudeney. "The
Captain" puzzle corner. The
Captain 3:1 (Apr 1900) 1 & 90
& 3:3 (Jun 1900) 193
& 279. No. 3: Overheard in an
omnibus. "Was that your
father." "No, that
gentleman's mother was my mother's mother-in-law." Essentially the same as: AM; 1917; Prob. 53: Heard on the tube
railway, pp. 8 & 153; "That
gentleman's mother was my mother's mother-in-law, but he is not on speaking
terms with my papa."
Hummerston. Fun,
Mirth & Mystery. 1924. Grandfather's problems: The portrait, p.
68. "Sisters and brothers have I
none, But that man's father is my
father's son."
James Joyce.
Ulysses. (Dijon, 1922); Modern Library (Random House), NY, 1934,
apparently printed 1946. P. 692
(Gardner says the 1961 ed. has p. 708; this is about 4/5 of the way between the
start of Part III and Molly's soliloquy).
"Brothers and sisters had he none, Yet that man's father was his
grandfather's son." This is given
as a quotation, while Bloom is looking in a mirror -- otherwise it could be a
cousin. [Given in Bryant, no. 782, pp.
194 & 342.]
Streeter & Hoehn.
Op. cit. in 7.AE. Vol. 2, 1933,
p. 16, no. 10: "Brain twister".
"My son's father is your father's only child. What relative of yours am I?"
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia
Verlagsbuchhandlung, Prague, 1937.
Prob. 33, pp. 194 & 204.
'This man's mother is my mother's only child.'
Haldeman-Julius.
1937. No. 44: Portrait problem,
pp. 7 & 22. Woman points to a man's
portrait and says to her brother: "The man's mother was my mother's
mother-in-law." Answer is that she
is his daughter, but she might be his step-daughter.
McKay. Party
Night. 1940. No. 7, p. 175.
"Brothers and sisters have I none;
yet this man's father was my father's son."
Meyer. Big Fun
Book. 1940. No. 5, pp. 175 & 756.
"My father is the brother of your sister. What relative am I of yours?" Answer is nephew, but
son is also possible.
The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941.
John Henry Cutler.
Dr. Quizzler's Mind Teasers.
Greenberg, NY, 1944. ??NYS --
excerpted in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun
1992) 47 & 43, prob. 14, with additional comments in Ibid. 16:4 (No. 110)
(Aug 1992) 4 and 16:6 (No. 112) (Dec 1992) 4.
"What relation is a man to his mother's only brother's only
niece?" Answer is her brother, but
comments point out that she could be his cousin, i.e. his mother's sister's
daughter, or even a kind of cousin-in-law, i.e. his mother's brother's wife's
sibling's daughter.
Yvonne B. Charlot.
Conundrums of All Kinds.
Universal, London, nd [c1950?].
P. 77:
"If your aunt's brother is not your uncle, who is he?"
P. 82:
"What kin are those children to their own father who are not their own
father's sons?"
Hubert Phillips.
Party Games. Witherby, London,
1952. Chap. XIII, prob. 3: Photograph,
pp. 204 & 252‑253.
"Though
sons and brothers have I none,
Your
father was my father's son."
Solution says
this "is my own invention".
See Ascher in 9.E for some examples.
Iona & Peter Opie.
I Saw Esau: The Schoolchild's
Pocket Book. (Williams & Norgate,
London, 1947.) Revised edition, Walker
Books, London, 1992, ??NX. No. 42, p.
45, just gives the rhyme; illustration
on p. 44; answer on p. 144 just gives
the answer, with no historical comments.
Ripley's Believe It or Not!, 8th series. Pocket Books, 1962, p. 26. Jimmy Burnthet, of Emmerdale, England,
courted a girl for 40 years. She broke
off the engagement and married his nephew, whereupon he married her niece.
Philip Kaplan. More
Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 16, pp. 25 & 88. "The father of the person in the
portrait is my father's son, but I have no brothers or sons."
Ripley's Puzzles and Games.
1966.
Pat Fairon. Irish
Riddles. Appletree Press, Belfast,
1992; Chronicle Books,
San Francisco, 1992; pp. 27 &
60. "Brothers and sisters have I
none But this man's father Was my father's son." Answer is "Oneself".
9.E.2. IDENTICAL SIBLINGS WHO ARE NOT TWINS
Two siblings are born on the same
day to the same parents but are not twins.
New section. This must be older
than the example below.
Harold Hart. The
World's Best Puzzles. Op. cit. in
7.AS. 1943. The problem of the two students, pp. 4 & 50
Nicholas Falletta.
The Paradoxicon. Op. cit. in
8.H.1. 1983. Pp. 162‑163 relates that during World War II, Swedish Radio
announced there would be an unexpected civil defence exercise next week. Lennart Ekbom, a Swedish professor of
mathematics, noted the paradoxical nature of this and discussed it with his
students.
D. J. O'Connor.
Pragmatic paradoxes. Mind 57 (1948)
358‑359. Discusses several other
paradoxes, e.g. "I remember nothing", then the unexpected blackout
exercise.
L. Jonathan Cohen.
Mr. O'Connor's "Pragmatic paradoxes". Mind 59 (1950) 85‑87. Doesn't deal much with the unexpected
blackout.
Peter Alexander.
Pragmatic paradoxes. Mind 59
(1950) 536‑538. Also doesn't deal
much with the unexpected blackout.
Michael Scriven.
Paradoxical announcements. Mind
60 (1951) 403‑407. "A new
and powerful paradox has come to light."
Entirely concerned with the unexpected blackout and considers the case
of only two possible dates.
Max Black. Critical
Thinking. 1952. Op. cit. in 6.F.2. Prob. 1, pp. 156 & 433.
Gamow & Stern.
1958. The date of the
hanging. Pp. 23‑27.
M. Gardner. SA (Mar
63) = Unexpected, chap. 1, with an extensive historical addendum and
references.
Joseph S. Fulda. The
paradox of the surprise test. MG 75
(No. 474) (Dec 1991) 419-421.
I have just started to consider problems
where a number of statements are given and we know at least or at most some
number of them are lies.
Find correct answer in one question
from a truthteller or liar:
Goodman; Gardner; Harbin;
Rice; Doubleday - 2; Eldin.
See: Nozaki for a generalization.
Problem with three truthtellers or
liars and first one mumbles:
Rudin; Haldeman-Julius; Depew;
Catch-My-Pal; Kraitchik; Hart;
Leopold; Wickelgren.
Magician's Own Book.
1857. P. 216. A
lies 1/4 of the time; B lies 1/5
of the time; C lies
1/7 of the time. "What is the probability of an event
which A and B assert, and
C denies?" Answer is
140/143, but I get 2/3.
= Book of 500 Puzzles, 1859, p. 54.
Chas. G. Shaw.
Letter: The doctrine of
chances. Knowledge 7 (27 Feb 1885) 181,
item 1620. Says Whitaker's Almanac for
this year, under The Doctrine of Chances, gives the following problem with a
wrong answer. A lies
1/4 of the time; B
lies 1/5 of the time; C lies 1/6
of the time. What is the chance
of an event which A and
B assert, but C
denies? Whitaker and I get
(3/4)(4/5)(1/6) / [(3/4)(4/5)(1/6) + (1/4)(1/5)(5/6)] = 12/17,
but Shaw claims 19/24 by asserting that the probability of an
event when A and B testify to it ought to be 1 ‑ (1/4)(1/5) =
19/20 instead of (3/4)(4/5)
= 3/5, He then says this leads to
19/24 by modifying the above
formula, but I can't see how this can be done.
Lewis Carroll. Diary
entry for 27 May 1894. "I have
worked out in the last few days some curious problems on the plan of 'lying'
dilemma. E.g. 'A says B lies; B says C lies; C says A and B lie.'
Answer: 'A and C lie; B speaks
truly'. The problem is quoted in
Carroll-Gardner with his discussion of the result, pp. 22-23. Gardner says this was printed as an
anonymous leaflet in 1894.
Carroll-Wakeling. Prob. 9: Who's telling the truth?, pp. 11
& 65. Wakeling says "This is a
puzzle based on a piece of logic that appears in his diary.
The
Dodo says the Hatter tells lies.
The
Hatter says that the March Hare tells Lies.
The
March Hare says that both the Dodo and the Hatter tell lies.
Who
is telling the truth?"
In
a letter of 28 May 2003, Wakeling quotes the text from the Diary as above, but
continues with it: "And today 'A says
B says C says D lies; D says two lie
and one speaks true.' Answer: 'D lies;
the rest speak truly.' Wakeling adds
that the Diary entry for 2 Jun 1896 (not given in the Lancelyn Green edition)
says: "Finished the solution of
the hardest 'Truth-Problem' I have yet done", but Carroll gives no indication what it was.
Lewis Carroll. The
problem of the five liars. In his
unpublished Symbolic Logic, Part II. He
was working on this after Part I appeared in 1896 and he had some galley proofs
when he died in 1898. Published in
Lewis Carroll's Symbolic Logic, ed. by William Warren Bartley III; Clarkson N.
Potter, NY, 1977, pp. 352‑361, including facsimiles of several letters of
1896 to John Cook Wilson (not in Cohen).
Each of five people make two statements, e.g. A says "Either B or D
tells a truth and a lie; either C or E
tells two lies." When analysed,
one gets contradictions because a form of the Liar Paradox is embedded.
Carroll-Wakeling II.
c1890? Prob. 32: Bag containing
tickets, pp. 50 & 73. This is one
of the problems on undated sheets of paper that Carroll sent to Bartholomew
Price. Wakeling reproduces the MS.
A
bag contains 12 tickets, 3 marked 'A', 4 'B', 5 'C'. One is drawn in the presence of 12 witnesses of equal
credibility: three say it was 'A', four 'B', five 'C'. What is the chance that it was 'A'?
There is no
answer on the Carroll MS. Wakeling
gives an answer.
Let
the credibility of a witness be "a" when telling the truth. Hence, the credibility of a witness when
telling a lie is "1 - a".
If
it was A, then 3 tell the truth, and 9 lie; hence the credibility is 3 in
12, or 1 in 4.
Therefore
the chance that it is A, and no other, is:
3/12
x 1/4 + 4/12 x 3/4 + 5/12 x 3/4
= 5/8
I cannot see how
this last formula arises. Wakeling
writes that he was assisted in this problem by a friend who has since died, so
he does not know how the formula was obtained.
Assuming a is
the probability of a person telling the truth, this probability depends on what
the chosen ticket is and is not really determined by the information given --
e.g., if the ticket was A, then any value of a between 0
and 1 is possible. The
value 3/12 is an estimate of a, indeed the maximum likelihood estimate. If
k of the 12
people are telling the truth, I would take the situation as a binomial
distribution. There are BC (12, k)
ways to select them and the probability of having k
liars is then BC (12, k) ak
(1-a)12-k. Now it seems that
Bayes' Theorem is the most appropriate way to proceed. Our basic events can be denoted A, B, C
and their a priori probabilities are
3/12, 4/12, 5/12. Taking a = k/12,
the a posteriori probabilities
are proportional to
k/12 x BC
(12, k) (k/12)k ({12-k}/12)12-k, for
k = 3, 4, 5.
Dropping the common denominator of
1213, these
expressions are 6.904, 8.504,
10.1913 times 1012. Dividing by the total gives the a posteriori probabilities
of A, B, C as 27.0%, 33.2%,
39.8%.
This
is really a probability problem rather than a logical problem, but it
illustrates that the logical problem seems to have grown out of this sort of
probability problem.
A. C. D. Crommelin.
Problem given in an after-dinner speech, reported by Arthur Eddington in
1919. ??where, ??NYS -- quoted in: Philip Carter & Ken Russell; Classic
Puzzles; Sphere, London, 1990, pp. 50 & 120-121. Four persons who tell the truth once with probability 1/3.
If "A affirms that B denies that C
declares that D is a liar, what is the probability that D
was speaking the truth?"
Collins. Fun with
Figures. 1928. The evidence you now give, etc., etc., pp.
22-23. Three witnesses who tell the
truth 1/3, 1/5, 1/10 of the time. First two assert something which the third denies. What is the probability the assertion is
true? Asserts it is 9
to 8, which I also get.
Nelson Goodman. The
problem of the truth-tellers and liars.
Anonymous contribution to the Brainteasers column, The Boston Post (Jun
1929). ??NYS -- described in an undated
letter from Goodman to Martin Gardner, 1960s?, where he says he 'made it up out
of whole logical cloth' and submitted it to the paper.
H. A. Ripley. How
Good a Detective Are You? Frederick A.
Stokes, NY, 1934, prob. 22: An old Spanish custom. King will present the Princess's suitor a choice of two slips,
one marked 'win', the other 'lose'. The
king is determined to double-cross the suitor so he has both marked
'lose'. But the suitor realises this,
so when he picks a slip, he drops it in the fire and then asks the King to
reveal the other slip!
Rudin. 1936.
Haldeman-Julius.
1937. No. 5: Noblemen-huntsmen
problem, pp. 3-4 & 20. Noblemen
tell the truth and huntsmen lie. Same
as Rudin, no. 200. Says the problem was
sent in by J. C. Furnas.
Hubert Phillips.
Question Time, op. cit. in 5.U.
1937.
Depew. Cokesbury
Game Book. 1939. Mixed blood, p. 202. Same as Rudin 200.
Catch-My-Pal cards, issued with The Hotspur, a magazine
published by D. C. Thomson, Dundee, in 1939.
The cards are unnumbered, but the sixth in my set has three truthtellers
or liars being asked what they are. The
reply of the first is inaudible. The
second says: He says he's a truthteller and he is a truthteller and so am
I. The third says: I'm a truthteller
and they're liars. Solution finds they
are truthteller, truthteller, liar (TTL).
This assumes that all parts of the statements are true or false
simultaneously. If one assumes that a
conjunctive statement is a lie if any one part is a lie, then there are more
possibilities -- I get TTL, TLL, LLT, LLL.
M. Kraitchik.
Mathematical Recreations, 1943, op. cit. in 4.A.2, chap. 1, prob. 2, pp.
14‑15. Truthtellers and
liars. (Not in Math. des Jeux.) Same as Rudin 200, but doesn't give number
of liars so only determines which the second and third are.
Harold Hart. The
World's Best Puzzles. Op. cit. in
7.AS. 1943. The problem of the nobles and the slaves, pp. 11 & 51. Similar to Rudin, but doesn't say how many
liars, but the statements are more elaborate so all can be determined.
Leopold. At
Ease! 1943. Simpletons and liars, pp. 6-7 & 194. Similar to Rudin.
Hubert Phillips.
Something to Think About, op. cit. in 7.AD, 1945.
Hubert Phillips.
Hubert Phillips's Heptameron.
Eyre & Spottiswoode, London, 1945.
Day 1, prob. 25: Crazy island, pp. 18 & 231. Same as prob. 83 in Something to Think
About.
Leeming. 1945. Chap. 3, prob. 17: Which was the officer?,
pp. 25‑26 & 155‑156.
Two truthtellers and a liar.
Gardner. SA (Feb
1957) c= 1st Book, chap. 3, problem 4: The fork in the road. Truthteller or liar. The book version includes a number of
letters and comments. I have
photocopies from Gardner's files of letters from people who claim to have
invented this problem -- only one of these seemed reasonable -- cf Goodman
above. Other material ??NYR -- DO.
N. A. Longmore, proposer;
editorial solution. The oracle
of three gods. RMM 4 (Aug 1961) 47 &
5 (Oct 1961) 59. Truthteller,
liar and alternator.
Robert Harbin. Party
Lines. Op. cit. in 5.B.1. 1963.
The road to freedom, p. 30.
Truthteller or liar.
Charlie Rice.
Challenge! Op. cit. in 5.C. 1968.
Prob. 8, pp. 22-23 & 55.
Truthteller or liar.
F. W. Sinden. Logic
puzzles. In: R. P. Dilworth, et al., eds.; Puzzle Problems and Games Project
-- Final Report; Studies in Mathematics, vol. XVIII; School Mathematics Study
Group, Stanford, Calif., 1968; pp. 197‑201. The District Attorney, pp. 200‑201. Two truthtellers and a liar -- determine
which in two questions.
Doubleday - 2.
1971. Truth will out, pp.
151-152. Truthteller or liar.
Peter Eldin. Amaze
and Amuse Your Friends. Piccolo (Pan),
London, 1973. No. 34: Where am I?, pp.
79 & 106. You are on an island of
truthtellers or an island of liars.
Determine which in one question.
Wickelgren. How to
Solve Problems. Op. cit. in 5.O. 1974.
Pp. 36‑37. He uses 'truar'
for truthteller. From statements by
three truars or liars, you can deduce the number of each, though you can't tell
which is which!!
Rowan Barnes-Murphy.
Monstrous Mysteries. Piccolo,
1982. Tollimarsh Tower, pp. 14 &
57. Two monster guards, one truar and
one liar, and you have one question.
You discover one is asleep and the other says: "It doesn't matter
that he's asleep, he always tells people to do
A." Do you do A
or not A?
Shari Lewis.
Abracadabra! Magic and Other
Tricks. (World Almanac Publications,
NY, 1984); Puffin, 1985. Free choice, p. 22. Truthteller and liar have distributed
items A and B. You want to determine who has which item
with one question. You ask "Did
the liar take B?" If the person answers 'yes', he has item A;
if 'no', he has item B.
Akihiro Nozaki. How
to get three answers from a single yes‑no question. JRM 20:1 (1988) 59‑60. You have to ask a truthteller or a liar
which of three roads is correct. The
author's question results in neither being able to answer in the third case. He suggests extensions.
Ken Weber. More
Five-Minute Mysteries. Running Press,
Philadelphia, 1991. No. 13,
pp. 59-61 & 189-190. You have
gotten lost near the border between the truthtellers and the liars and you come
out in an open area where there is a border marker with a guard on each side,
but you cannot tell which side is which.
Further the guards are walking back and forth and exchanging positions,
so you are not sure if they are on their correct sides of the border or not. You can approach one of the guards and ask
one question to determine which side you are on. You ask "Are we in your country?" If he answers 'yes', you are in truthtelling
country.
Charles Babbage. On
the Economy of Machinery and Manufactures.
(1832, ??NYS); 4th ed., (1835),
reprinted by Augustus M. Kelley, NY, 1971. Section 348, p. 289. "... both parties are often led to
adopt arrangements ... at variance ... with the true interests of both."
Frederick Winslow Taylor.
The Principles of Scientific Management. (1911); Harper &
Brothers, NY, 1923. P. 10. Speaking of employers and employés, he says
"that perhaps the majority on either side do not believe that it is
possible so to arrange their mutual relations that their interests become
identical."
Merrill M. Flood & Melvin Dresher. c1950.
??NYS -- details. They
identified the paradox, but I have no reference to any publication.
Robert Axelrod. The
Evolution of Cooperation. Basic Books,
NY, 1986. p. 216, ??NYS. "The Prisoner's Dilemma game was
invented in about 1950 by Merrill Flood and Melvin Dresher, and formalised by
A. W. Tucker, shortly thereafter."
Keith Devlin. It's
only a game. The Guardian, second
section (17 Nov 1994) 12-13. Says
Tucker invented the dilemma in 1950.
Sylvia Nasar. Albert
W. Tucker, 89, pioneering mathematician.
New York Times (27 Jan 1995) ??
Asserts Tucker invented the dilemma when teaching game theory to
psychology students at Stanford in 1950.
Carl G. Hempel.
Studies in the logic of confirmation.
(Mid 1940s?). Reproduced
in: M. H. Foster & M. L.
Martin, eds.; Probability, Confirmation and Simplicity; Odyssey Press, NY,
1966, pp. 145‑183. ??NYS.
How do you
use a fallen signpost to find your way?
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
9:5 (Mar 1903) 490-491 & 10:1 (May 1903) 50-51. The sense of direction.
King. Best 100. 1927.
No. 22, pp. 14 & 44.
H. A. Ripley. How
Good a Detective Are You? Frederick A.
Stokes, NY, 1934, prob. 30: Class day.
Bernd Plage. Der
umgestürzte Wegweiser. Verlag von R.
Oldenbourg, München, 1944. Prob. 1: Der
umgestürzte Wegweiser, pp. 11-12 & 67.
(Kindly sent by Heinrich Hemme.)
John Paul Adams. We
Dare You to Solve This!. Op. cit. in
5.C. 1957? Prob. 180: On the right track?, pp. 67 & 121.
Jonathan Always.
Puzzles for Puzzlers. Tandem,
London, 1971. Prob. 84: A resourceful
motorist, pp. 41 & 87.
Martin Gardner asked me to look up
some of these items as he is doing a section on it in a book about Carroll
which will be much more detailed than the following, citing numerous other
discussions. This presumably refers to
the now-appeared Carroll-Gardner, 1996.
Carroll-Wakeling.
c1890? Prob. 16: Going out, pp.
20-21 & 67-68. This is on one of
the undated sheets Carroll sent to Bartholomew Price. Wakeling notes that it is an early version of the Barber shop
paradox and says that Mark Richards has pointed out a mistake in Carroll's
solution.
Lewis Carroll. Diary
entry for 31 Mar 1894. Says he has just
had a leaflet "A Disputed Point in Logic" printed containing the
problem that he and John Cook Wilson "have been arguing so
long." ??NYS -- quoted in
Carroll-Gardner. Gardner says the
pamphlet was revised in Apr 1894.
Lewis Carroll. A
logical paradox. Mind (NS) 3 (No. 11)
(Jul 1894) 436-438. Gardner says
Carroll reprinted this as a pamphlet.
Alfred Sidgwick.
"A logical paradox".
Mind (NS) 3 (No. 12) (Oct 1894) 582.
W. E. Johnson. A
logical paradox. Mind (NS) 3 (No. 12)
(Oct 1894) 583.
Lewis Carroll. Diary
entry for 21 Dec 1894. Not in Lancelyn
Green. Discusses the problem and seems
to recognise the distinction between material and strict implication. ??NYS -- quoted in Carroll-Gardner.
Alfred Sidgwick & W. E. Johnson. "Hypotheticals in a context". Mind (NS) 4 (No. 13) (Jan 1895)
143-144.
E. E. C. Jones.
Lewis Carroll's logical paradox (Mind, N.S., 3). Mind (NS) 14 (No. 53) (Jan 1905)
146-148.
W. [= John Cook Wilson, according to Gardner]. Lewis Carroll's logical paradox (Mind, N.S.,
3 and 53, P. 146). Mind (NS) 14 (No.
54) (Apr 1905) 292-293. He admits that
Carroll had been right all along.
R. B. Braithwaite.
Lewis Carroll as logician. MG 16
(No. 219) (Jul 1932) 174-178. Brief
discussion and solution of the paradox.
Warren Weaver. Lewis
Carroll: Mathematician. Op. cit. in
1. 1956. Discusses the paradox.
Alexander B. Morris's letter says the paradox is not real. Weaver's response discusses this and other
unpublished letters, saying he is not sure if the paradox is resolved.
Carroll-Gardner.
1996. Pp. 67-71 discusses this
in detail, citing a number of other references, including John Venn, but he
only gives the page numbers of an article in Symbolic Logic.
Lewis Carroll discusses this in his unpublished Symbolic
Logic, Part II. He was working on this
after Part I appeared in 1896 and he had some galley proofs when he died in
1898. Published in Lewis Carroll's
Symbolic Logic, ed. by William Warren Bartley III; Clarkson N. Potter, NY,
1977. Bartley includes all eight known
versions.
See also 7.S and 7.Y.
I will collect here some material on
physics toys in general.
Christian Ucke.
Physics toys for teaching. IN:
H. Kühnelt, ed.; Interdisciplinary Aspects of Physics Education; proc. of conf.
at Altmünster, Austria, 1989; World Scientific, Singapore, 1990, pp.
267-273. 1: Some new and not so well
known literature about physics toys gives 12 references. 2: Presenting a database about physics
toys. 3. Physics toy experiments with
PET bottles. 16 references at the end,
some repeating those in part 1.
10.A. OVERTAKING AND MEETING PROBLEMS
See Tropfke 588.
Note. Meeting problems include two pipe Cistern Problems, 7.H. Overtaking problems include Snail in Well
problems without end effect, 10.H, and Cisterns with one inlet and one outlet,
7.H. Many of the Indian versions
involve gaining or losing wealth rather than covering distance. Versions going around a circle or an island
are related to Conjunction of Planets, 7.P.6, and to problems of Clock Hands
meeting, 10.R.
In the 17, 18 and 19 C, this problem
was often discussed in relation to negative numbers as a change in the relative
values leads to a negative solution -- cf: Clairaut; Manning; Hutton,
1798?; Lacroix; De Morgan, 1831? & 1836;
NOTATION. There are five types of meeting (M) and overtaking (O) problems
which recur frequently with slight variations.
I have recently converted all problems to this notation and I hope I
have done it correctly. Tropfke 590
gives a more extended classification which includes motion on a right triangle
(see 6.BF.5) and on a circle (see 7.P.6, though some occur here) and
alternating motion (see 10.H), but doesn't distinguish between problems where
times are given and those where rates are given.
M-(a, b). Two travellers can cover a route in a, b (usually days). They start at opposite ends at the same time
toward each other. When do they
meet? This is identical to the cistern
problem (a, b) of 7.H.
Sometimes, the distance apart is given and the point of meeting is also
wanted. If the distance is, say 100, I will then say "with
D = 100". This
corresponds to asking how much each pipe contributes to a cistern of capacity D.
Sometimes, one starts later than the other. If, say the first starts
2 later, I will say "with the first delayed by 2".
This corresponds to opening one pipe later than the other. This is the version of Tropfke's I B a where times are given.
MR-(a, b; D). The same problem except that
a, b are the actual rates of the
two travellers and hence D must be given. This corresponds to a simple form of cistern problem which does
not have the characteristic feature of giving the times required to do the
entire task. This is the version of
Tropfke's I B a with rates given.
MR-(a, b; c, d; D). The same problem except that the travellers travel in arithmetic
progressions, so this gives:
a + a+b + ... + a+(n-1)b + c
+ c+d + ... + c+(n-1)d = D.
Hence MR-(a, 0; c, 0; D) =
MR-(a, c; D). The value of d is
often 0. One can interpret this as a cistern problem as for MR-(a, b; D), but it is even harder to imagine a pipe increasing its rate in
arithmetic progression that to imagine a traveller doing so. (An additional difficulty is that the
traveller is usually viewed discretely while a pipe ought to be viewed
continuously.) This is the version of
Tropfke's I B b with an arithmetic progression specified.
O-(a, b). Two travellers start from the same point at rates a, b,
with the slower starting some time
T before the faster, or they
start at the same time at rates a,
b, with the slower starting some
distance D ahead of the other. I.e.
the slower has a headstart of time
T or distance D. When does the faster overtake the slower? This corresponds to a cistern with rates
given, so that O-(a, b) with headstart D, which is the same as
MR-(a, -b; D), is a cistern
problem with one inlet and one outlet.
When a > b, then this corresponds to a full cistern of
size D, inlet rate a and outlet rate b. The case a < b
is most easily viewed by negating the amount done, which interchanges
inlet and outlet, and taking an empty cistern.
Hound and hare problems are basically of this form, with a headstart of
some distance, but usually with rates and distances complicatedly
expressed. This is Tropfke's I C a.
Sometimes the rates are not given explicitly, so I assume the first has
the headstart.
O-(a, b; c, d). Two travellers start from the same point at the same time, but in
arithmetic progressions. When do they
meet again? This gives us:
a + a+b + ... + a+(n-1)b = c
+ c+d + ... + c+(n-1)d. Some versions
of this are in 7.AF.
If the
first has a headstart of time T, then we either increase the first n
by T or decrease the second n by
T, depending on which number of
days is wanted. This is Tropfke's I C b
with an arithmetic progression specified. Sometimes the first has a headstart of distance D.
Occasionally it is the second that has the headstart which is denoted by
negative values of T or
D.
Snail in the well problems without
end effect (see 10.H) are special cases of meeting problems, usually MR-(a, 0; D) (Tropfke's I A c). When there are approaching animals, then it
may be MR(a, b; D) or O‑(a,
b) with headstart D
(Tropfke's I B c).
Hound and hare problems. Here one is often only given the ratio of
speeds, r, so one can determine where the hare is caught, but not when. In this case, the problem is O-(a, ra)
with some headstart and one asks for the distance to overtaking, but not
the time. See: Chiu Chang Suan Ching; Zhang Qiujian; Alcuin;
Fibonacci; Yang Hui; BR;
Bartoli;
Pseudo-dell'Abbaco; AR; The Treviso Arithmetic; Ulrich Wagner; HB.XI.22;
Calandri, c1485; Calandri,
1491; Pacioli; Tagliente;
Riese; Apianus; van Varenbraken; Cardan;
Buteo; Gori; Wingate/Kersey; Lauremberger; Les Amusemens; Euler;
Vyse; Hutton, c1780?; Bonnycastle (= Euler); King;
Hutton, 1798?; D. Adams,
1801; De Morgan, 1831?; Bourdon;
D. Adams, 1835; Hutton‑Rutherford;
Family Friend (& Illustrated Boy's Own Treasury); Anon: Treatise (1850); Brooks;
Clark; Haldeman-Julius (in
verse).
Cases where leaps differ
in both time and distance:
Pacioli?; Apian; Cardan;
Wingate/Kersey; Lauremberger; Euler; Bonnycastle; King;
Hutton, 1798? (= Lauremberger); De Morgan, 1831?;
Bourdon (= Lauremberger);
Brooks; Todhunter; Mittenzwey;
Clark (= Lauremberger);
General versions. Newton.
Versions with geometric
progressions. Chiu Chang Suan
Ching; della Francesca; Chuquet;
Pacioli; Cardan. See also 7.L.
Versions with sum of squares. Simpson.
Circular versions. Aryabhata(?); AR; Wingate/Kersey; Vyse;
Pike; Anon: Treatise (1850); Todhunter;
Perelman. See also 7.P.6, where
problems with more than two travellers in a circle occur.
Versions using negatives. Clairaut;
Manning.
Chiu Chang Suan Ching.
c‑150. (See also Vogel's
notes on pp. 126‑127.)
Chap. VI.
Chap. VII.
Zhang Qiujian. Zhang
Qiujian Suan Jing. Op. cit. in
7.E. 468, ??NYS. Analogous to above
hare and hound with values 37, 145, 23. (English in
Sanford 212, Mikami 41 and
H&S 74.)
Aryabhata. 499. Chap. II, v. 30-31, pp. 72-74. (Clark edition: pp. 40-42.)
Bakhshali MS.
c7C.
Bhaskara I.
629. Commentary to Aryabhata,
chap. II, v. 30-31. Sanskrit is on pp.
127-132; English version of the examples is on pp. 308-309.
V.
30, Ex. 1: 7x + 100 = 9x
+ 80.
V.
30, Ex. 2: 8x + 90 = 12x
+ 30.
V.
30, Ex. 3: 7x + 7 = 2x
+ 12.
V.
30, Ex. 4: 9x + 7 = 3x
+ 13.
V.
30, Ex. 5: 9x - 24 = 2
x + 18.
V.
31, Ex. 1: MR-(3/2, 5/4; 18).
V.
31, Ex. 2: O-(3/2, 2/3), D = 24.
Anania Schirakatzi (= Ananias of Shirak). Arithmetical problems. c640.
Translated by: P. Sahak
Kokian as: Des Anania von Schirak
arithmetische Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919)
112-117. See 7.E for description.
Mahavira. 850. Chap VI, v. 320‑327, pp. 177‑179.
Alcuin. 9C. Prob. 26: Propositio de campo et cursu canis
ac fuga leporis. Hound catching hare,
hound goes 9 to hare's 7, hare has
150 head start. =
O-(7, 9), D = 150. (H&S 72 gives Latin and
English.) The actual rates are not given,
only their ratio.
Sridhara. c900.
V. 65‑67(i),
ex. 81‑83, pp. 52‑53 & 95.
The verses give rules for various cases.
V. 96‑98,
ex. 111‑112, pp. 78‑79 & 96.
al‑Karkhi.
c1010. Sect. I, no. 5‑9,
p. 82.
Tabari. Miftāh
al-mu‘āmalāt. c1075. Pp. 103f.
??NYS -- quoted in Tropfke 593.
Ibn Ezra, Abraham (= ibn Ezra, Abraham ben Meir = ibn Ezra, Abu Ishaq Ibrahim al‑Majid). Sefer ha‑Mispar. c1163.
Translated by Moritz Silberberg as:
Das Buch der Zahl ein
hebräisch-arithmetisches Werk; J. Kauffmann, Frankfurt a. M., 1895. P. 56.
Brothers meeting, MR-(17, 19;
100). (H&S 72 gives English.) Silberberg's note 121 (p. 109) says a
similar problem occurs in Elia Misrachi, c1500.
Fibonacci.
1202. He has many examples. I give a selection.
Yang Hui.
Supplements to the Analysis of the Arithmetical Rules in the Nine
Sections (=?? Chiu Chang Suan Fa
Tsuan Lei). 1261. Repeats first problem of the Chiu Chang Suan
Ching.
BR. c1305.
Gherardi?. Liber
habaci. c1310.
Lucca 1754.
c1330. F. 59r, p. 134. Couriers,
M-(20, 30) with D = 200, though this is not used.
Munich 14684.
14C. Prob. VIII & XX, pp. 78
& 81. Discusses O‑(1, 1; k, 0).
Folkerts.
Aufgabensammlungen. 13-15C. 18 sources for O-(1, 1; c, 0). Cites AR
for extensions; Chiu Chang Suan Ching,
Alcuin, Fibonacci.
See Smith, op. cit. in 3.
Bartoli. Memoriale. c1420.
Prob. 28, f. 78r (= Sesiano, pp. 144 & 149-150. Fox is
121 (fox-)steps ahead of a
dog. 9
dog-steps = 13
fox-steps. He computes 9/13
of 121. Sesiano notes that this assumes the fox
stands still and determines how many dog-steps they are apart. The correct answer, which assumes that both
steps take the same time, is that the dog gains 4 fox-steps for every
9 steps he makes,
so he has to make
121 · 9/4 = 272 1/4
dog-steps, which are equal
to 121 · 13/4 = 393 1/4 =
121 + 272 1/4
fox-steps.
Pseudo-dell'Abbaco.
c1440.
AR. c1450. Several problems with arithmetic
progressions which I omit.
Benedetto da Firenze.
c1465. P. 67. O-(1, 1; 30, 0); O-(1, 2; 30, 0); O-(3,
3; 60, 0).
The Treviso Arithmetic = Larte de Labbacho. Op. cit. in 7.H. 1478.
Muscarello.
1478.
della Francesca.
Trattato. c1480. F. 40r (102). 25 + 25n = 1 + 2 + 4 + 8 + ... + 2n-1 = 2n
- 1. The exact solution is 7.785540889. He does linear interpolation on the seventh day, getting 7 73/103
= 7.7087378641. English in Jayawardene.
Ulrich Wagner. Das
Bamberger Rechenbuch, op. cit. in 7.G.1.
1483.
Chuquet. 1484. See also 7.L for a problem with geometric
progression.
Borghi.
Arithmetica. 1484.
Calandri.
Aritmetica. c1485.
HB.XI.22. 1488. P. 52 (Rath 247). (Rath says it is similar to Alcuin, but with different numbers.)
Calandri.
Arimethrica. 1491. F. 97v.
Pacioli. Summa. 1494.
PART II.
Blasius. 1513. F. F.iii.v: Decimasexta regula. O-(10, 13),
T = 9.
Köbel. 1514. ??NYS -- given in H&S 73. O-(10, 13),
T = 9.
Tagliente. Libro de
Abaco. (1515). 1541.
Ghaligai. Practica
D'Arithmetica. 1521.
Tonstall. De Arte
Supputandi. 1522.
Riese. Die
Coss. 1524.
Apianus. Kauffmanss
Rechnung. 1527.
van Varenbraken.
1532. ??NYS -- cited by the
editor of King, 1795, p. 154. Hound and
hare -- O-(12, 15), D = 200.
Cardan. Practica
Arithmetice. 1539. Chap. 66.
Buteo.
Logistica. 1559.
Baker. Well Spring
of Sciences. 1562? ??check if this in the Graves copy of the
1562/1568 ed.)
Gori. Libro di
arimetricha. 1571.
van Halle.
1568. ??NYS -- cited by the
editor of King, 1795, p. 155. O‑(11,
0; 1, 1).
Io. Baptiste Benedicti (= Giambattista Benedetti). Diversarum Speculationum Mathematicarum,
& Physicarum Liber. Turin, (1580),
Nicolai Bevilaquæ, Turin, 1585; (Venice, 1599). [Rara 364. Graves
141.f.16.] This has a number of
overtaking and meeting problems, but he makes diagrams showing the sums of the
arithmetic progressions involved.
Wingate/Kersey.
1678?.
Edward Cocker.
Arithmetic. Op. cit. in
7.R. 1678. Chap. 10, quest. 32.
1678, p. 181; 1715: p.
121; 1787: p. 106. This problem is attributed to Moor's or
More's Arith. cap. 8, qu. 7. 1678 states O-(40, 50),
T = 3, giving answers 12
days and 600 miles.
1715 states O-(40, 50), T = 3,
giving answers 32 days and
600 miles. 1787 states
O-(48, 50), T = 3, giving answers 12 days and 600
miles. I am surprised at the
misprints here.
Wells. 1698.
Peter Lauremberger (Petrus Laurembergus). Institutiones Arithmeticæ .... 4th ed., Joh. Lud. Gleditsch, Leipzig,
1698. P. 196, prob. 12.XII. Fox is
60 hare-leaps ahead of a
hound. She makes 9
leaps while the hound does
6, but 7 hare-leaps are as long
as 3
hound-leaps.
Isaac Newton.
Arithmetica Universalis, 1707.
??NYS. English version: Universal Arithmetic, translated by Mr.
Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James
Maguire and Theaker Wilder; W. Johnston, London, 1769. (De Morgan, in Rara, 652‑653, says
there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of
1720 and 1728, ??NYS.) Resolution of
Arithmetical Questions, Problem V, pp. 180‑184. Begins with MR(7/2, 8/3;
59) with second delayed by 1. Then "The same more generally"
does O(c/f, d/g) with second having a headstart of
distance e and either a headstart or delay of time h and MR(c/f, d/g; e) with second having either a headstart or delay of time h.
Then does example: O(13, 1) with second starting distance 90
ahead, but first delayed by
3 days. Then repeats original example from general
viewpoint.
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790).
Alexis-Claude Clairaut.
Élémens d'algèbre. 1746. ??NYS -- cited by Tom Henley. Discusses the relation between overtaking
and meeting problems and the use of negative rates or negative headstarts to
make them algebraically the same.
Les Amusemens.
1749.
Euler. Algebra. 1770.
I.IV.III: Questions for practice.
Vyse. Tutor's
Guide. 1771?
Dodson. Math.
Repository. 1775.
Charles Hutton. A
Complete Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2. [c1780?]
Bonnycastle.
Algebra. 1782.
Pike.
Arithmetic. 1788. P. 350, no. 17. Circle 268 in circumference. Two men start at ends of a diameter and go in the same direction
at rates 22/2 and 34/3. When and where do they meet? I. e.
O-(34/3, 22/6), D = 134.
Eadon. Repository. 1794.
John King, ed. John
King 1795 Arithmetical Book.
Published by the editor, who is the great-great-grandson of the 1795
writer, Twickenham, 1995.
Thomas Manning. An
Introduction to Arithmetic and Algebra.
2 vols., Nicholson, Lunn & Deighton, Cambridge, 1796 &
1798. Vol. 1, pp. 208-210. O-(5/2, 7/2), T = 2. Then considers
what would happen if the second traveller went slower than the first [more
simply, suppose the second traveller had the headstart, i.e. T is
negative]. This gives a negative
solution and he interprets this as that they must have met before the starting
time. He gives a general solution and
discussion of the problem. ??NX.
Hutton. A Course of
Mathematics. 1798?
D. Adams. Scholar's
Arithmetic. 1801.
Robert Goodacre. Op.
cit. in 7.Y. 1804. Miscellaneous Questions, no. 125, p. 205
& Key p. 269. A goes
4 mi/hr for 7
hrs each day. B starts a day later at 5
mi/hr for 8 hours each day, both starting at the same
time each morning. When does B
overtake?
Silvestre François Lacroix.
Élémens d'Algèbre, a l'Usage de l'École Centrale des
Quatre-Nations. 14th ed., Bachelier,
Paris, 1825. Sections 64-75, pp.
94-110 plus Addition, pp. 359-360.
Discusses general problems
MR-(a, b; D) and O-(a, b)
and considers negative solutions and what happens when the divisor is
zero!
Augustus De Morgan.
Arithmetic and Algebra. (1831
?). Reprinted as the second, separately
paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin &
Craddock, London, 1836. Art. 116, pp.
30-31. Hare is 80
hare-leaps ahead of a greyhound.
Hare makes 3 leaps for every 2 of the hound, but a
hound-leap is twice as long as a hare‑leap. Then considers a hound-leap as
n/m of a hare-leap. Takes
n/m = 4/3 and finds a negative
solution which he discusses. Takes n/m = 3/2
and finds division by zero which he interprets as the hound never
catching the hare.
Bourdon.
Algèbre. 7th ed., 1834.
D. Adams. New Arithmetic. 1835.
Augustus De Morgan. On The Study and Difficulties of Mathematics. First, separately paged, part of: Library of
Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836. P. 30 mentions O-(2, 3), T = 4 and
O-(a, b) with second delayed by time
T. Pp. 37-39 discusses the
general courier problems O-(m, n) with second having headstart of
distance D and MR-(m, n; D). He considers different signs and sizes,
getting six cases.
Unger. Arithmetische
Unterhaltungen. 1838. Pp. 135 & 258, nos. 515 & 516. MR‑(5/4, 3/5; 57) with first delayed by time 2½.
The second problem asks what delay of time for the first will make them
meet at the half-way point?
Hutton-Rutherford. A
Course of Mathematics. 1841?
Tate. Algebra Made
Easy. Op. cit. in 6.BF.3. 1848.
P. 46, no. 15. Man makes a
journey at 4 mph and returns at 3 mph, taking
21 hours in total. How far did he go?
Family Friend 1 (1849) 122 & 150. Arithmetical problems -- 1. "A hare starts 40
yards before a greyhound, and is not perceived by him till she has been
up 40
seconds: she gets away at the
rate of 10 miles an hour and the dog pursues her at the rate of 18
miles an hour: how long will the
course last, and what distance will the hare have run?" = Illustrated Boy's Own Treasury, 1860,
Prob. 4, pp. 427 & 431. = Hutton,
c1780?, prob. 68.
Anonymous. A
Treatise on Arithmetic in Theory and Practice: for the use of The Irish
National Schools. 3rd ed., 1850. Op. cit. in 7.H.
John Radford Young (1799-1885). Simple Arithmetic, Algebra, and the Elements of Euclid. IN: The Mathematical Sciences, Orr's Circle
of the Sciences series, Orr & Co., London, 1854. [An apparently earlier version is described in 7.H.]
Vinot. 1860. Art. LXIII: Problème du Renard et du
Lévrier. Fox is 72
(fox-)leaps in front of a greyhound.
Fox makes 9 leaps to hound's 5, but fox-leaps are 3/7
the size of hound-leaps.
Edward Brooks. The
Normal Mental Arithmetic A Thorough and
Complete Course by Analysis and Induction.
Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough
and Complete Course by Analysis and Induction;
Sower, Potts & Co., Philadelphia, 1873. Many examples. I mention
just one example.
1863
-- pp. 132-13, no. 19; 1873 -- p. 161,
no. 18. "E takes
60 steps before he is overtaken
by D;
how many steps does D take to catch E, provided E
takes 4 steps while
D takes 3,
and 5 of D's equal
8 of E's, and how far ahead
was E
when they started?"
Todhunter. Algebra,
5th ed. 1870. Several examples, including the following.
Mittenzwey.
1880. Prob. 78, pp. 15 &
66; 1895?: 85, pp. 19 & 69; 1917: 85, pp. 18 & 65. Hare is
50 leaps ahead and makes 6
leaps while the hound makes
5, but 7 houndleaps are as long
as 9 hare leaps. How leaps does the hare make before being
caught?
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E. No. 4, pp. 166 & 334. General problem: ax = b (n - x).
Solves for n ‑ x.
Clark. Mental
Nuts. 1897, no. 46; 1904, no. 62. The fox and the hound. Fox
is 60
(fox‑)leaps in front of a hound.
Hound takes 6 leaps to fox's 9, but the hound leaps
are 7/3 times as long as the fox's.
(= Lauremberger)
Dudeney. Weekly
Dispatch (17 May 1903) 13 & (14 Jun 1903) 16.
Perelman. 1937.
MCBF. At the cycle track, prob.
150, pp. 255-256. Circular track of
circumference 170m. When cyclists are going in opposite
directions, they meet every 10 sec;
when going the same direction, the faster passes the slower every 170 sec.
Haldeman-Julius.
1937. No. 109: Hare and hounds
problem, pp. 13 & 26. Gives a verse
version -- the only one I have seen -- and says the problem 'is about 150 years
old.'
As
I was walking o'er my forest grounds
Up
jumped a hare before my two greyhounds.
The
distance that she started up before
Was
fourscore rods, just, and no more.
My
dogs did fairly run
Unto
her 16 rods just 21.
Now
I would have you unto me declare
How
far they ran before they caught the hare.
This is O-(a, [21/16]a) with headstart D = 80.
C. Dudley Langford.
Note 1558: A graphical method of
solving problems on "Rate of Work" and similar problems. MG 25 (No. 267) (Dec 1941) 304-307. +
Note 2110: Addition to Note
1558: "Rate of Work"
problems. MG 34 (No. 307) (Feb 1950)
44. Uses a graph to show (a, b)
cistern problems as meeting problems.
Also solves problems
(A, x) in B
and (a, -b), the latter appearing as an overtaking
problem. The Addition gives a clearer
way of viewing (a, ‑b) problems as overtaking problems
Gamow & Stern.
1958. Pp. 9‑10, 59‑63. Elevator problem.
The linear form has an army of
length L moving with velocity
v. A rider goes at velocity V
from the rear to the front and then back to the rear, reaching the rear
when it has advanced d. How far,
D, does he go? Since one version gives an erroneous answer,
I will give the basic equations.
Let t be the time for the rider to reach the front of the army and T be
the time to get from the front back to the rear. We then have: Vt = vt +
L; v(t+T) = d; d = V(t-T); D = V(t+T).
Here may be insufficient equations to determine all the values (as also
occurs in 10.A.3), but the value of D can be found as D = L + Ö(L2 + d2). Note that
D/d = V/v. Other versions of the
problem can be solved, sometimes more simply, e.g. L = (V2 - v2)d/2vV; if
V = rv, then L = (r2-1)d/2r.
If the army has a width W,
and the rider goes across the moving army at each end, the situation is
more complex -- see the first example and Loyd.
J. Gale, proposer;
Joseph Edwards, Jr. & Mr. Coultherd, solvers. Question III. A Companion to the Gentleman's Diary; ... for the year 1798, pp.
59-60 & The Gentleman's Mathematical Companion, for the year 1799, pp.
16-17. Wagoner walking around his wagon
and team while it is travelling. L = 20
yd, W = 4 yd. He can walk V = 4
mph and he walks 74⅔ yd in his circumambulation (hence taking T = 7/11 min).
A complication arises as to how he goes crossways. The proposer says he turns at right angles
and passes 2 yd clear at the front and at the back. (He also says the walker passes 2 yd
clear on each side, but he never gives a width, taking the distance
between the two side paths as 4, which I have accounted for by taking W = 4.
In some cases, we can account for the
2 yd at each end by taking L = 24.)
How fast, v, is the team going? Both solvers get v =
2 mph, but I find neither is viewing the problem correctly and neither has
correctly formulated the problem he has described!
Edwards
starts in the middle, 2 in front of the horses, and says the the walker must go
2 to the left, 24 back and 2 to the right to get the the middle in back. He then says the time required is 28/(V+v).
But this assumes the crossways motion is at the same relative speed as
the lengthwise relative motion and seems definitely incorrect to me. Edwards then says that a similar argument
gives the time in the other direction as
28/(V-v). Setting the sum of
these equal to T does give
v = 2.
Coultherd
starts in the middle, 2 behind the wagon, but I will rephrase it to parallel
Edwards' solution. It takes the
man 2/V to go 2 to the left. During this time, the wagon moves ahead 2v/V, so he is now
only 2 - 2v/V in front. When he gets to
the back, he only needs to be 2 -
2v/V behind the wagon when he turns to
cross, so that he will be exactly 2 behind at the middle. So he must make a relative motion of 24 ‑ 4v/V at the relative velocity V + v.
Similarly, for the forward trip, he makes relative motion 24 + 4v/V
at relative velocity V - v. Adding the times for these to 8/V
for the crossways trips and setting equal to T, I get v = 2.056 mph, which seems to be the correct
answer. Coultherd forgets to account
for the 2 - 2v/V at the end of his first lengthwise trip and
confuses distance travelled at V and at
V-v, which simplifies his
algebra to the same final equation as Edwards.
I
am rather surprised at the basic errors in both solutions.
Clark. Mental
Nuts. 1897, no. 88; 1904, no. 98; 1916, no. 99. A West
Pointer. Column is 25
miles long. Courier goes from
the rear to the front and returns to the rear and sees that he is now where the
front of the column was when he started.
I.e. his trip takes the same time as the time the column moves 25
miles. How far did he go? Answer is
60 miles, 1876 feet, which is
correct. Here L = 25 = d, so D = 60.36.
Loyd. The courier
problem. Cyclopedia, 1914, pp. 315
& 382. (= MPSL2, prob. 146, pp. 103
& 167‑168, with solution method provided by Gardner -- Loyd only
gives the values "following the rule for solving puzzles of this
kind".) Army 50
miles long. Rider goes from back
to front to back of the army in the time it moves forward 50
miles. Here L = 50 = d, so D = 120.71 -- Loyd says "a little over 120
miles". Note this is the double of
Clark's problem.
Loyd
also extends to the case of a square army and Gardner gives a solution,
assuming the courier goes across the ends on an angle with velocity V,
so his crossways velocity is
w = Ö(V2-v2). The total time of his trip is then T = L/(V‑v) + L/(V+v)
+ 2W/w. Assuming that the army advances its own length in
this time, i.e. vT = L, and setting
x = V/v leads to L = L/(x‑1) + L/(x+1) + 2W/Ö(x2-1). Assuming
W = L simplifies the expression,
but it remains a fourth degree equation.
Gardner says the only relevant solution is x = 4.18112+.
The courier's distance is VT =
xvT = xL = 209.056+.
Abraham. 1933. Prob. 67 -- The column of troops, pp. 33
& 44 (19 & 116). Rider circling
army -- same as first part of Loyd.
Haldeman-Julius.
1937. No. 74: Train problem, pp.
10 & 24. Brakeman walks from rear
of a train going at V = 27 mph, to the front, thereby passing a point in
time T = 2½ minutes earlier than he would have. How long,
L, is the train? Though this seems like a problem of this
section, in fact it simply says the train takes 2½ minutes to pass a
given point and L = VT.
Perelman. 1937. MCBF.
Reconnaissance at sea, prob. 149 a & b, pp. 253-255. Squadron moving and a reconnaissance ship is
sent ahead. Part a
gives distance to go ahead and asks how long it will take. Part
b gives time for reconnaissance
and asks when the reconnaissance ship will turn back.
McKay. At Home
Tonight. 1940. Prob. 17: The orderly's ride, pp. 65 &
79. Same as first part of Loyd
with 1
mile army.
William R. Ransom.
Op. cit. in 6.M. 1955. An army courier, p. 103. Same as first part of Loyd with 25
mile army, but says it takes a day and asks how fast the courier rides.
G. J. S. Ross
& M. Westwood. Problems drive, 1960. Eureka 23 (Oct 1960) 23-25 & 26. Prob. G.
v = 15, d + L = 23, V = 60,
find L. At first I thought the result was wrong, but
the phrasing is a bit different than usual as the back of the army is 23
from where the front finishes.
Nathan Altshiller Court.
Mathematics in Fun and in Earnest.
Op. cit. in 5.B. 1961. Prob. m, pp. 189 & 191-192. Courier going from back to front and then
back again at three times the speed of the army. Where is he when he gets to the back again?
Philip Kaplan. More
Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 76, pp. 78 & 107. L = 100,
V = 3v, find d.
Julius Sumner Miller.
Millergrams. Ure Smith, Sydney,
1966. Prob. 25, pp. 25 & 70. Army is 3 mi long, officer starts at back,
goes to front and returns, reaching the back when it has advanced 4 miles. How far did he go? L = 3, d = 4, giving
D = 8 miles. He says: "you
can get the right answer by erroneous logic", but he doesn't explain how
to get the answer!
Birtwistle. Math.
Puzzles & Perplexities. 1971. Procession, pp. 43, 167 & 189. = Birtwistle; Calculator Puzzle Book; 1978;
prob. 55, pp. 38-39 & 99.
Procession 1½ miles long going at 2
mph. Marshall starts at head,
walks to the back and then forward, reaching his starting point when half the
army has passed. What is his
speed? He continues on the head and
returns to the same point. Where is the
end of the procession when he gets back?
New section
-- I haven't done much on this yet.
Kraitchik has several examples.
Mittenzwey.
1880. Prob. 75, pp. 14-15 &
66; 1895?: 82, pp. 19 & 69; 1917: 82, pp. 18 & 65. Daily trains crossing America in 7
days.
[Richard A. Proctor].
Letters received and short answers.
Knowledge 3 (26 Oct 1883) 264. Answer to Harry. Recalls
it being posed on by the captain's wife, Mrs Cargill, on the S. S.
Australasia, but no date is clear.
Trains going between New York and San Francisco taking 7
days. How many does one meet on
such a trip? Says he gave the wrong
answer!
[Richard A. Proctor].
Editorial gossip. Knowledge 3
(23 Nov 1883) 318. Gives a careful
answer to the problem stated above.
L. Carroll. A
Tangled Tale. (1885) = Dover,
1958.
Laisant. Op. cit. in
6.P.1. 1906. Chap. 49: Du Havre à New-York, pp. 123-125. Definitely asserts that this problem was
posed by Lucas at a meeting 'longtemps déjà'.
Boats leave every noon each way between le Havre and New York and take
exactly seven days to make the trip -- how many do they pass?
Pearson. 1907. Part II, no. 106, pp. 136 & 212‑213. Tube trains run every 2
minutes. How many are met in
a 30
min journey? Answer: 30.
Peano. Giochi. 1924.
Prob. 11, p. 4. Buses take 7 minutes
from the centre to the terminus. Buses
leave from the centre and from the terminus every minute. How many buses are met in going one way? Observes that it takes 14
buses to run such a service and so one bus meets the other 13.
McKay. At Home
Tonight. 1940. Prob. 14: Passing the trains, pp. 65 &
79. 12
hour journey and trains start from the other end every hour.
Sullivan.
Unusual. 1943. Prob. 24: Back in the days of gasoline,
tires, and motorists. 4½ hours from Chicago to Indianapolis, buses
leaving every hour [on the hour].
Anonymous. The
problems drive, 1956. Eureka 19 (Mar
1957) 12-14 & 19. No. 10. Circular line with trains departing each way
every 15 minutes, but the east bound trains take 2 hours for the circuit while
the west bound ones take 3 hours for a circuit. Starting after one leaves the station, how many trains does one
see?
Doubleday - 1.
1969. Prob. 36: Traveler's tale,
pp. 48 & 162. = Doubleday - 5, pp.
55-56. Trains going between Moscow and
Paris, taking seven days.
[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers.
1973. Op. cit. in 5.E. How many buses, pp. 7-8.
10.A.3. TIMES FROM MEETING TO FINISH GIVEN
Simpson.
Algebra. 1745. Section XI (misprinted IX in 1790), prob.
XLVI, pp. 110-111 (1790: prob. LIX, pp. 111‑112). Travellers set out from each of two cities
toward the other, at the same time.
After meeting, they take 4 and
9 hours to finish their
journeys. How long did they take? He gives a general solution -- if x is
the time before meeting and a, b are the times from meeting to finishing,
then x2 = ab. [I have seen a 20C version where only the
ratio of velocities is asked for -- indeed I used it in one of my puzzle
columns, before I knew that the times could be found.]
Dodson. Math.
Repository. 1775.
Ozanam‑Montucla.
1778. Supplement, prob. 42,
1778: 432; 1803: 425; 1814: 360;
1840: 186. Couriers set out
toward one another from cities 60 apart.
After meeting, they take 4 and
6 hours to reach their
destinations. What are their
velocities?
Thomas Grainger Hall.
The Elements of Algebra: Chiefly Intended for Schools, and the Junior
Classes in Colleges. Second Edition:
Altered and Enlarged. John W. Parker,
London, 1846. P. 136, ex. 38. Simpson's problem with cities of London and
York and times of 9 and 16.
T. Tate. Algebra
Made Easy. Op. cit. in 6.BF.3. 1848.
P. 89, no. 4. Same as Simpson.
Todhunter. Algebra,
5th ed. 1870. Examples XXIV, nos. 20 & 22, pp. 211-212 & 586.
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E. No. 132, pp. 306 & 347. Travellers between London and York reach
their destinations 25 &
36 hours after meeting. How long did each take?
W. W. Rouse Ball.
Elementary Algebra. CUP, 1890
[the 2nd ed. of 1897 is apparently identical except for minor changes at the
end of the Preface]. Ex. 5, p.
231. A
and B are 168 miles apart.
Trains leave each end for the other starting at the same time. They meet after 1 hour 52 minutes. The
train from A reaches B half an hour before the other reaches A.
Haldeman-Julius.
1937. No. 133: Racers' problem,
pp. 15 & 28. Racers on a circular
track start in opposite directions.
After meeting, they take 4 and
9 minutes to pass the starting
point, but they continue. When do they
meet each other at the starting point?
Nathan Altshiller Court.
Mathematics in Fun and in Earnest.
Op. cit. in 5.B. 1961. Prob. b, pp. 188-190. Trains start toward each other at 7
am; one takes 8
hours, the other takes 12. When do they meet?
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 37: Walking all the way, pp. 29 &
106-107. Similar to Simpson, but they
start at given times and the time of meeting is given and they get to their
destination at the same time. This is
thus the same as Simpson if it is considered with time reversed. The elapsed times to meeting are 25/12
and 49/12 hours.
Birtwistle.
Calculator Puzzle Book.
1978.
A man is usually met by a car at his
local train station, but he arrives
A minutes early and begins
walking home. The car meets him and
picks him up and they arrive home
B minutes early. How long was he walking? The car's trip is B/2 minutes shorter each
way, so the commuter is met B/2 minutes before his usual time and he has
been walking A - B/2 minutes.
New section -- there must be older examples.
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 155, pp. 61 & 151: Catching the
postman. Man driving 10 mph usually overtakes
the postman walking 4 mph at the same point every morning. One morning, the man is four minutes late
and he overtakes the postman half a mile beyond the usual point. Was the postman early or late, and by how
much? (Note that the man is four minutes
late starting, not in overtaking.)
Meyer. Big Fun
Book. 1940. No. 10, pp. 162-163 & 752.
A = 60, B = 10.
Harold Hart. The
World's Best Puzzles. Op. cit. in
7.AS. 1943. The problem of the commuter, pp. 8 & 50. A = 60,
B = 20.
The Little Puzzle Book.
Op. cit. in 5.D.5. 1955. Pp. 59-60: An easy problem. A = 60,
B = 20, he walks
at 4
mph. How fast does the chauffeur
drive?
Nathan Altshiller Court.
Mathematics in Fun and in Earnest.
Op. cit. in 5.B. 1961. Prob. f, pp. 189 & 191. A = 60,
B = 16. The time of arrival is
also given, but is not needed.
Liz Allen. Brain
Sharpeners. Op. cit. in 5.B. 1991.
The Commuter's tale, pp. 87-88 & 136. A = 60, B = 10.
Doubleday - 2 gives a typical
example. New section -- I have seen
other examples but didn't record them.
Gardner, in an article: My ten favorite brainteasers in Games (collected in Games Big Book of
Games, 1984, pp. 130-131) says this is one of his favorite problems. ??locate
Todhunter. Algebra,
5th ed. 1870. Examples XIII, no. 23, pp. 103 & 578. In racing a mile (= 1760 yds), A
gives B a headstart of 44 yd and wins by 51 sec;
but if A gives
B a headstart of 75 sec,
A loses by 88 yd.
Find the times each can run a mile.
Charles Pendlebury.
Arithmetic. Bell, London,
(1886), 6th ed, 1893. Section XXXV (b):
Races and games of skill, pp. 267-268, examples XXXV (b) & answers, p.
viii. Does an example: A
can give B 10
yards in 100 yards and
A can give C
15 yards in 100
yards. How much should B
give C in 150 yards?
He gives 11 similar problems. On pp. 363-364 & answers, part II, p. xix, he gives some
further problems, 45-48.
W. W. Rouse Ball.
Elementary Algebra. CUP, 1890
[the 2nd ed. of 1897 is apparently identical except for minor changes at the
end of the Preface].
Lewis Carroll. Letter
of 8 Apr 1897 to Enid Stevens. =
Carroll-Wakeling, prob. 25: Handicaps, pp. 33 & 71. A
loses 10 yards in every 100 against
B, while B
gains 10 yards in every 100 against
C. What handicaps should B
give A and C in a quarter mile (= 440 yards) race? Mentioned in Carroll-Gardner,
pp. 50-51 Cohen's comment is that
is that this is too ambiguous to have a precise answer, but such problems were
common in late Victorian times and Wakeling gives an answer (but no method)
which seems correct to me and in line with other problems of the time. However, Gardner does not comment on Cohen's
remark and does not give an answer.
Letting A, B, C denote the uniform speeds of the runners,
Wakeling and I would interpret the problem statements as: A = 90/100 · B and
B = 110/100 · C. The
ambiguity is that one could read the statements as saying A = 100/110 · B and B = 100/90 · C, but I think the first interpretation is more
natural. Either case permits a clear
answer. Cohen says it is clear that A = 90/100 · B, but is unsure whether B = 100/90 · C or
B = 110/100 · C, but
the first case makes C = A, which makes the problem much less interesting.
Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg,
1960 and says it was reset for the 49th ptg of 1944. Examination Papers XIV, prob. 11, p. 187. A runs at
12 1/2 mph and B runs at 12 7/16 mph. If A gives B ten yards headstart, when will he overtake him. Who is ahead when A has run a mile?
Hummerston. Fun,
Mirth & Mystery. 1924. Speed, Puzzle no. 18, pp. 49 & 174. Cook can run 100 yards in 12
sec. Brown can give Cook a 10
yd headstart and finish even with him.
Adams can give Brown a 5 yd headstart and finish even. Adams gives Cook a 15 yd headstart and Cook
loses 1/20 of a sec in a bad start.
Who wins and by how much? He
notes that one must assume uniform speeds.
Collins. Fun with
Figures. 1928. The 100-yard dash, pp. 24-26. = Hummerston with Adams, Brown, Cook replaced
by Bob, Jack, Bill.
Doubleday - 2.
1971. Running commentary, pp.
59. In a 100 yard dash, A
gives B a head start of 25 to make an even race
and B
gives C a head start of 20. How much head start
should A give C to make an even race?
10.A.6. DOUBLE CROSSING PROBLEMS
New section. I recall Loyd and/or Dudeney have
versions. Also, the problems in 10.A
with the faster meeting the slower on the return trip are related.
Mittenzwey.
1880. Prob. 296, pp. 54 &
105; 1895?: 326, pp. 57 & 106; 1917: 326, pp. 52 & 100. Field in the shape of a right triangle. Runners whose speeds are in the ratio 13 : 11 start at the right angled corner along different legs. They first meet at the midpoint of the
hypotenuse and then at 60m from the starting point. As far as I can see, this leads to a
quadratic.
Wood. Oddities. 1927.
Prob. 61: The errand boys, p. 47.
Alan goes from A to
B and back while Bob goes
from B
to A and back, both starting at the same time. They first cross at 720
from A and then at 400 from
B. How far is it between A
and B? Gives a general formula:
If the two crossings occur at distances
a, b, then the width is 3a - b.
He asserts that Alan goes faster, but the ratio of velocities is a/(2a-b)
which is 9/13 in this case.
Philip Kaplan. More
Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 73, pp. 76 & 106. Two swimmers in a pool, like above with
distances 40, 20.
New section. If a fast train, going at velocity V,
takes time T to overtake a slow train, going at velocity v,
and it takes time t for the trains to pass when meeting, then
the lengths of the trains cancel out and one gets V/v = (T + t)/(T - t).
Haldeman-Julius.
1937. No. 28: Speed problem, pp.
6 & 22. T = 35, t = 3,
V = 38.
New section. If I go at rate v1, I
arrive t1 too late; if I go at rate v2, I arrive t2 too soon.
I do not recall seeing anything like this before.
Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg,
1960 and says it was reset for the 49th ptg of 1944. Examination Papers XI, prob. 185, p. 185. If a man goes at 4 mph, he arrives 5 min
late; if he goes 5 mph, he arrives 10 min early. How far is he going?
The general solution is easily found to be D = v1v2(t1+t2)/(v2-v1). However, I wondered what was the correct
rate, v, so that I arrive on time.
This leads to the following.
(t1+t2)/v = t1/v2
+ t2/v1 or 1/v
= t1/(t1+t2)·1/v2 + t2/(t1+t2)·1/v1, so
v is a weighted harmonic mean
of v1 and
v2. On the other
hand, we find the time to arrive is t =
D/v =
v1/(v2-v1)·t1 + v2/(v2-v1)·t2, which is a weighted mean of t1 and t2.
The only type of problem that I can
recall that leads to similar means is the classic problem: If I make a journey at 30 mph and return at
20 mph, what is my average speed. In
general, if I travel di at speed
vi, what is my
average speed? Letting ti be the time at speed vi, we have
ti = di/vi, so the average speed is given by
(d1+d2)/v = t1
+ t2 = d1/v1 + d2/v2 or
1/v = d1/(d1+d2)·1/v1 + d2/(d1+d2)·1/v2, so
v is again a weighted harmonic
mean of v1 and
v2.
There are two trains d
apart, approaching at rates a,
b. A fly starts at one, flies to the
other, then back to the first, then back to the second, etc., flying at
rate c. How far does he go?
NOTATION. We denote this problem by
(a, b, c, d). For overtaking
problems, we let b be negative. I need to check for details.
Laisant. Op. cit. in
6.P.1. 1906. Chap. 53: Le chien et les deux voyagers, pp. 132-133. A
is 8 km ahead and goes at
4 km/hr. B
starts after him at 6 km/hr.
A dog starts at one and runs back and forth between them at 15
km/hr until they meet. I.e. (4, ‑6,
15, 8). Notes that the distance the dog
travels is independent of where he starts and that the travellers could be
meeting rather than overtaking.
Dudeney. Problem
464: Man and dog. Strand Mag. (Jul
1919). ??NX.
Dudeney. Problem
643: Baxter's dog. Strand Mag. (1924?).
??NX. A goes at rate 2 and has an hour's head
start. B goes at rate 4 and dog starts with him at rate 10.
I.e. (4, ‑2, 10, 2).
G. H. Hardy. Letter
of 5 Jan 1924 to M. Riesz. In: M. L. Cartwright. Manuscripts of Hardy, Littlewood, Marcel Riesz and
Titchmarsh. Bull. London Math. Soc. 14
(1982) 472‑532. (Letter is on p.
502, where it is identified as Add. MS. a. 275 33, presumably at Trinity
College.) Says it defeated Einstein,
Jeans, J. J. Thomson, etc. Fly between
cyclists (10, 10, 15, 20). "One thing only is necessary: you must
not know the formula for the sum of a geometrical progression. If you do, you will take 15-20
minutes: if not, 2 seconds."
Ackermann.
1925. Pp. 116‑117. Couple walking up a hill. Their dog, who is twice as fast as they are,
runs to the top and back to them continually.
(a, 0, 2a, d) -- he only says the couple start 1/4 of the way up the
hill.
Dudeney. Problem
754: The fly and the motor‑cars.
Strand Mag. (Jun 1925)
??NX. (?= PCP 72.)
H&S 53, 1927, says this is 'a modern problem'.
M. Kraitchik. La
Mathématique des Jeux, 1930, op. cit. in 4.A.2, chap. 2, prob. 17, p. 30. (15, 25, 100, 120). (Identified as from L'Echiquier, 1929, 20,
??NYS.) I can't find it in his
Mathematical Recreations.
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 181, pp. 71 & 156: Little Red
Riding Hood. Little Red Riding Hood
with her dog and Grandma set out at the same time to meet: (2, 2, 8, 6).
Dudeney. PCP. 1932.
Prob. 72: The fly and the motor‑cars, pp. 28 & 86. = 536; prob. 86: The fly and the cars, pp.
26 & 243. (50, 100, 150, 300).
Phillips. Week‑End. 1932.
Time tests of intelligence, no. 38, pp. 21 & 193. Fly between cyclists. (10, 15, 20, 60).
Abraham. 1933.
Streeter & Hoehn.
Op. cit. in 7.AE. Vol. 2, 1933,
p. 29: Aeroplane dilemma. Destroyer
going 25 overtaking battleship going
20 with headstart of 30.
Plane flies back and forth at
90. I.e. (25, ‑20, 90, 30).
Phillips.
Brush. 1936. Prob. G.21: The busy fly, pp. 20 &
87. Same as in Week‑End.
J. R. Evans. The
Junior Week‑End Book. Op. cit. in
6.AF. 1939. Prob. 37, pp. 265 & 270. Fly between cyclists,
(10, 10, 15, 20).
Haldeman-Julius.
1937.
McKay. At Home
Tonight. 1940. Prob. 20: The fluttered pigeon, pp. 66 &
80-81. Pigeon between walkers -- (3, 3, 21, 30). Gives solution by adding a GP and the easy solution.
Sullivan.
Unusual. 1943. Prob. 4: A busy bee. Bee between motorists -- (10, 10, 15, 20).
L. Lange. Another
encounter with geometric series. SSM 55
(1955) 472‑476. Studies the
series involved.
William R. Ransom.
Op. cit. in 6.M. 1955. The bicycles and the fly, pp. 22‑23. Studies the series.
Eugene Wigner. In
the film: John von Neumann, MAA, 1966,
he relates this as being posed by Max Born to von Neumann, involving a swallow
between bicyclists. He says it was a
popular problem in the 1920s.
Paul R. Halmos. The
legend of John von Neumann. AMM 80
(1973) 382-394. Gives the fly between
two cyclists puzzle and story on pp. 386-387.
David Singmaster.
The squashed fly -- (60, 40, 50, 100)
with fly starting on first train.
Used in several of my series.
Herbert R. Bailey. The
girl and the fly: A von Neumann Legend.
MS 24 (1991/92) 108-109. Cites
Halmos, but gives a version with a girl walking toward a wall. Finds the relevant series.
Yuri B. Chernyak & Robert S. Rose. The Chicken from Minsk. BasicBooks, NY, 1995.
10.C. LEWIS CARROLL'S MONKEY PROBLEM
A monkey and a barrel of equal
weight are on the ends of a rope over a pulley. The monkey starts to climb the rope -- what happens?
Stuart Dodgson Collingwood.
The Life and Letters of Lewis Carroll.
T. Fisher Unwin, London, 1898.
Pp. 317-318. Referring to Dec
1893, he says "Mr. Dodgson invented a new problem to puzzle his
mathematical friends with, which was called "The Monkey and Weight
Problem." A rope is supposed to be
hung over a wheel fixed to the roof of a building; at one end of the rope a
weight is fixed, which exactly counterbalances a monkey which is hanging on to
the other end. Suppose that the monkey
begins to climb the rope, what will be the result?"
Carroll. Letter of
19 Dec 1893 to Price. Discussed and
reproduced for the first time in Edward Wakeling: Lewis Carroll and the Bat;
ABMR (Antiquarian Book Monthly Review) 9:2 (No. 99) (Jul 1982) 252-259. Wakeling has sent me a copy of this. Carroll starts: "Many thanks for your
solution of the "Monkey & Weight" Problem -- It is the reverse
of the solution given me by Sampson", and discusses consequences of
Sampson's argument. As a postscript,
Carroll states: "I own to an inclination to believe that the weight
neither rises nor falls!"
Carroll. Diary entry
for 21 Dec 1893. In Roger Lancelyn
Green's edition, p. 505. Quoted in
Collingwood, above; in Carroll-Wakeling II, below; and slightly differently in
Wakeling's 1982 article, above.
"Got Prof. Clifton's answer to the "Monkey and Weight
Problem." It is very curious, the
different views taken by good mathematicians.
Price says the weight goes up, with increasing velocity. Clifton (and Harcourt) that it goes up,
at the same rate as the monkey; while Sampson says it goes down."
Carroll-Wakeling II, prob. 9: The monkey and weight problem,
pp. 15-16 & 66. Quotes the problem
and the Diary entry. Identifies the
people, who were all Oxford academics:
Robert Bellamy Clifton, Professor of Experimental Philosophy; Bartholomew "Bat" Prices -- cf
Carroll‑Wakeling in Common References;
Augustus Vernon Harcourt, Lee's Reader in Chemistry, at Christ
Church; Rev. Edward Francis Sampson,
assistant mathematical tutor to Carroll at Christ Church. Wakeling notes that most modern
mathematicians and scientists agree with Clifton and Harcourt.
Carroll. Letter of
23 Dec 1893 to Mrs. Price. Discussed
and partially reproduced in Wakeling's 1982 article, above. In a PS, he asks her to remind Price to
return Sampson's proof, summarises the various solutions received and concludes
the paragraph with "And my
present inclination is to believe that it goes neither up nor down!!!
Carroll-Collingwood.
1899. Pp. 267‑269
(Collins: 193-194). Repeats material
from The Life and Letters. Then
includes a letter from Arthur Brook, arguing that 'the weight remains
stationary'. Collingwood discusses it
and sides with Sampson.
Pearson. 1907. Part II, no. 11: A climbing monkey, pp. 9
& 188. Asserts that the weight goes
up, but the monkey does not!
Dudeney. Some much‑discussed
puzzles. Op. cit. in 2. 1908.
Cites the diary entry from Collingwood's Life and Brook's comment in the
Picture Book. Says mechanical devices
have been built.
Loyd. Lewis
Carroll's monkey puzzle. Cyclopedia, 1914,
pp. 44, 344‑345 (erroneous solution).
(= MPSL2, prob. 1, pp. 1‑2 & 121.)
William F. Rigge.
The climbing monkey. SSM 17
(1917) 821. (Refers to L'Astronomie
(Jul 1917) ??NYS.) Asserts that
the weight remains fixed and that he made a clockwork climber and demonstrated
this.
Wilbert A. Stevens.
The monkey climbs again. SSM 19
(1919) 815. Assert's Rigge's climber
was too light to overcome friction and that both should ascend together.
William F. Rigge.
The monkey stops climbing. SSM
20 (1920) 172‑173. Says he
increased the speed of his monkey and that now it and the weight go up
together.
Editors, proposers;
E. V. Huntington & L. M. Hoskins, solvers. Problem 2838. AMM 27
(1920) 273‑274 & 28 (1921) 399‑402. Proposal quotes Carroll and cites Collingwood
& SSM.
Ackermann.
1925. Pp. 1‑2. Says they would go up together, but the rope
moves to the side of the monkey and so the weight will rise faster. Cites Carroll.
Ernest K. Chapin.
Loc. cit. in 5.D.1. 1927.
Jerome S. Meyer.
Fun-to-do. Op. cit. in 5.C.
1948. Prob. 6: How good are you in
physics?, part 2, pp. 19-20 & 181-182.
Carroll's problem, with no reference to Carroll. Answer says the weight will go up.
Warren Weaver. Lewis
Carroll: Mathematician. Op. cit. in
1. 1956. Mentions the problem. A.
G. Samuelson's letter says a modern version has a mirror at the other end and
asks if the monkey can get away from his image. He says the monkey and the mirror will behave identically so he
cannot get away. Weaver's response is
that this is correct, though he was saying that the behaviour of the weight
cannot be known unless you know how the monkey climbs.
H. T. Croft
& S. Simons. Some little naggers. Eureka 23 (Oct 1960) 22 & 36. No. 5.
Answer says they go up together.
Carroll-Gardner.
1996. Pp. 23-24. Quotes the problem and gives the accepted
solution. Says there is a demonstration
at the Chicago Museum of Science and Industry.
10.D.1 MIRROR REVERSAL PARADOX
Why does a mirror reverse right and
left, but not up and down? This is a
perennial problem in Notes and Queries type columns.
Alice Raikes.
Letter. The Times (22 Jan
1932). ??NYS -- cited in: Michael Barsley; The Left Handed Book;
(Souvenir, London, 1966); Pan, 1989,
pp. 199‑200. Quoted in: Roger Lancelyn Green; Alice -- an excerpt
from his: Lewis Carroll, Bodley Head,
1960; IN: Robert Phillips, ed.; Aspects of Alice; (1971; Gollancz, London, 1972); Penguin, 1974, pp. 52-53. Quoted in:
Florence Becker Lennon; Escape through the looking-glass -- an excerpt
from her: Victoria Through the
Looking-Glass, 1971; IN: Robert Phillips, ibid., pp. 108-109.
Miss
Raikes was one of Carroll's girl friends.
She relates that Carroll put an orange in her right hand and then asked
her to stand in front of a mirror and say which hand the reflection had the
orange in. She said the left hand and
Carroll asked her to explain. She
finally said "If I was on the
other side of the glass, wouldn't the orange still be in my right
hand?" Carroll said this was the
best answer he had had and later said it was the idea for "Through the
Looking Glass". Green dates this
as Aug 1868 and says it took place when Carroll was visiting his uncle Skeffington
Lutwidge at his house in Onslow Square, London.
Dudeney. PCP. 1932.
Prob. 327: Two paradoxes, pp. 112‑113 & 526. = 536, prob. 526, pp. 216‑217
& 412.
Gardner. Left and
right handedness. SA (Mar 1958) = 1st
Book, chap. 16.
Eric Laithwaite. Why
Does a Glow-worm Glow? Beaver (Hamlyn),
London, 1977. Pp. 54‑56:
When you look into a mirror, your left hand becomes your right, so why doesn't
your head become your feet? He gives
the correct basic explanation but then introduces an interesting complication. If you hold a rotating wheel with its axis
parallel to the mirror, the image appears to be rotating in the opposite
direction to the real wheel. Now turn
the wheel so the axis is perpendicular to the mirror and the image now appears
to be rotating in the same direction as the real wheel!! [What do you see when you're halfway in the
turning process??]
Gardner. The
Ambidextrous Universe. 2nd ed., Pelican
(Penguin), 1982. Pp. 6‑9 & 22‑26
surveys the question and cites three 1970s serious(!) philosophical articles on
the question.
Richard L. Gregory.
Mirror reversal. IN: R. L.
Gregory, ed.; The Oxford Companion to the Mind; OUP, 1987, pp. 491-493. ??NX.
Gives the basic explanation, but seems unhappy with the perceptual
aspects. I would describe it as making
heavy weather of a simple problem.
Don Glass, ed. Why
You Can Never Get to the End of the Rainbow and Other Moments of Science. Indiana Univ Press, Bloomington, Indiana,
1993. (Adapted from scripts for the
radio series A Moment of Science, WFIU, Bloomington, Indiana.) A mirror riddle, pp. 32-33. States that a mirror usually reverses right
and left, but a reflecting lake (or a flat mirror) reverses up and down. Says that the explanation is that a mirror
reverses clockwise and anticlockwise, but this is basically inadequate.
In c1993, there was correspondence in the Answers column of
The Sunday Times. Collected in: The
Sunday Times Book of Answers; ed. by Christopher Lloyd, Times Books, London,
1993. (The column stared in Jan 1993,
but 70% of the book material did not appear in the paper.) Pp. 63-67.
One correspondent said he asked the question in New Scientist almost 20
years ago. One correspondent clearly
states it reverses front and back, not right and left. Another clearly notes that the appearance of
reversing right and left is due to the bilateral symmetry of our bodies, so we
turn around to consider the mirror image.
In c1994, there were several letters on the problem in The
Guardian's Notes & Queries. These
are reproduced in: Joseph Harker, ed.; Notes & Queries, Vol. 5., Fourth
Estate, London, 1994, pp. 178-180 and
in: Joseph Harker, ed.; The Weirdest Ever Notes & Queries; Fourth Estate,
London, 1997, pp. 140-142. R. Thomson
clearly sees that right and left are not reversed, but front and back are. Richard L. Gregory has a slightly confusing
letter but adds that the problem goes back to Plato and that he has given the
history and solution in his Odd Perceptions (Routledge, 1986, ??NYS) and in the
Oxford Companion to the Mind -- see above.
Erwin Brecher.
Surprising Science Puzzles.
Sterling, NY, 1995. The mirror
phenomenon, pp. 16 & 80.
"Why does a mirror reverse only the left and right sides but not up
and down?" Gives a nonsensical
answer: "Left and right are directional concepts while top and bottom, or
up and down, are positional concepts." and then follows with an
unreasonable analogy to walking over the North Pole.
Seckel, 2002, op. cit. in 6.AJ, fig. 17, pp. 26 &
44. Straightforward discussion of the
paradox.
Richard A. Proctor.
Our puzzles. Knowledge 10 (May 1887) 153 & (Jun 1887)
186-187. Prob. XXVIII: how to see
yourself properly -- use two mirrors at right angles. Prob. XXIX: in a fully mirrored room,
what do you see when you look into a corner?
-- yourself inverted. Prob. XXX: in the same room, how many images
of yourself do you see? -- 26.
These are oriental (Chinese or
perhaps Japanese) polished discs which cast reflections containing a
pattern. I first came across them in
one of R. Austin Freeman's detective stories and I was kindly brought one from
China a few years ago. Since about
1991, they have been made in and exported from China and are commercially
available. A fine example, with
explanation, is in the Museum of the History of Science in Oxford, but it is
not illuminated. Basically, the pattern
is hammered on the disc and this causes molecular changes which remain even
when the surface has been made apparently smooth. Apparently other methods of producing a difference in metallic
structure have been used. Sometimes the
pattern is also made in relief on the opposite side of the disc, and sometimes
a different pattern is made.
Peter Rasmussen and Wei Zhang have
sent a bundle of material on this, ??NYR.
Aignan Stanislas Julien.
Notice sur les miroirs magiques des Chinois et leur fabrication. CR, 1847?
Separately printed or extracted from the journal, Bachelier, Paris,
1847. 15pp. ??NYS -- seen in a dealer's catalogue.
John Timbs. Things
Not Generally Known, Familiarly
Explained. A Book for Old and Young
(spine says First Series and a note by a bookdealer on the flyleaf
says 2 vol.). Kent & Co., London, (1857?), 8th ed., 1859. Chinese magic mirrors, p. 114. Says the reflected pattern is in relief on
the other side of the disc. He quotes
an explanation given by 'Ou-tseu-hing' who lived between 1260 and 1341 and who
worked out the process by inspecting a broken mirror. He says that the disc with the relief pattern is made by casting,
in fine copper. The pattern is then
copied by engraving deeply on the smooth side and the removed parts are filled
with a rough copper, then the disc is fired, polished and tinned. The rough copper produces dark areas in the
reflection.
R. F. Hutchinson.
The Japanese magic mirror.
Knowledge 10 (Jun 1887) 186.
Says he has managed to fulfil a boyhood longing and obtain one. Describes the behaviour and asks for an
explanation.
J. Parnell. The
Japanese magic mirror. Knowledge 10
(Jul 1887) 207. Says he studied it some
20 years earlier and cites: The Reader
(1866); Nature (Jul 1877) and
a paper read to the Royal Society by Ayrton & Perry in Dec 1878 -- all ??NYS. Says the mirror is somewhat convex and the picture lines are
slightly flatter, so the reflection of the picture is brighter than of the
surrounding area.
The Japanese mirror.
Cassell's Magazine (Dec 1904) 159-160.
Short note in the Flotsam and Jetsam section. Says they are bronze covered with a mercury amalgam with a raised
pattern on the back, whose image is reflected by the front. Says it "is due to inequalities of
convexity on the face, caused by the pattern on the back dispersing the
sunlight more or less."
R. Austin Freeman.
The Surprising Adventures of Mr. Shuttlebury Cobb. Story VI: The magic mirror. The series first appeared in Pearson's Magazine: (1 Jun 1913) 438-448, (15 Jun 1913) 565-574, (1 Jul 1913) 748-757 and
Red Magazine: (15 Jul 1913)
??, (1 Aug 1913) ??, (15 Aug 1913). Collected as a book:
Hodder & Stoughton, London, 1927,
with Story VI on pp. 231-281.
"... product of Old Japan, ...." Says the device on the back is cast as a dark shape with a bright
halo. Says the design is formed by
chasing or hammering the lines, which makes them harder than the surrounding
bronze (or similar metal), so when polished, they project slightly and produce
an image in the reflection. Says there
is an Encyclopedia Britannica article on the subject, but it is not in my 1971
ed.
R. Austin Freeman.
The magic casket. Pearson's
Magazine (Oct 1926) 288-299, ??NYS.
Collected in: The Magic Casket;
Hodder & Stoughton, London, 1927 and reprinted numerous times (I have 5th
ptg, 1935), pp. 7-41. Collected
in: Dr. Thorndyke Omnibus (variously
titled); Hodder & Stoughton, London, 1929, pp. 398-425. Says the phenomenon was explained by
Sylvanus Thompson. Says the polished
lines project slightly and produce dark lines with a bright edges in the
reflection.
Will Dexter. Famous
Magic Secrets. Abbey Library, London,
nd [Intro. dated Nov 1955].
P. 56. Describing a visit
to the premises of The Magic Circle, he says:
"Here are Japanese Magic Mirrors -- a whole shelf of them. Made of a secret bronze alloy, ..., they
have a curious property. ... Why?
Well now, people have written books to explain this phenomenon,
.... Some other time we'll talk about
it...."
Clark. Mental
Nuts. 1897, no. 36; 1904, no. 44; 1916, no. 46. The
wheel. "Does the top go faster
than the bottom?" Answer is:
"Turning on ground, yes; on shaft, no."
See Laithwaite, 1977, in 10.D.1 for a combined wheel and
mirror paradox.
10.E.1. ARISTOTLE'S WHEEL PARADOX
Wheels of different sizes joined
concentrically and rolling on two tracks at different heights. At first, it appears that they each roll the
same distance and hence must have the same circumference!
Aristotle (attrib.).
Mechanical Problems. c-4C? In:
Aristotle -- Minor Works. Trans.
by W. S. Hett; Loeb Classical Library, 1936, pp. 329‑441. The wheel paradox is section 24, pp. 386‑395.
Heron (attrib.).
Mechanics (??*). c2C? Ed. by L. Nix & W. Schmidt. Heronis Opera, vol. II, Teubner,
Leipzig. Chap. 7. ??NYS.
(HGM II 347‑348.)
Cardan. Opus Novum
de Proportionibus Numerorum.
Henricpetrina, Basil, 1570, ??NYS.
= Opera Omnia, vol. IV, pp. 575-576.
Galileo. Discorsi e
Dimostrazione Matematiche intorno à Due Nuove Scienze Attenenti alla Mecanica
& i Movimenti Locali. Elzevirs,
Leiden, 1638. Trans. by S. Drake
as: Two New Sciences; Univ. of
Wisconsin Press, 1974; pp. 28‑34
& 55‑57. (English also
in: Struik, Source Book, pp. 198‑207.)
E. P. Northrop.
Riddles in Mathematics.
1944. 1944: 59-60; 1945: 56‑57; 1961: 63‑64. Footnote on p. 248 (1945: 229;
1961: 228) only dates it back to Galileo.
Israel Drabkin.
Aristotle's wheel: notes on the
history of a paradox. Osiris 9 (1950)
162-198.
10.E.2. ONE WHEEL ROLLING AROUND ANOTHER
How often does a wheel turn when it
is rolled around another? This is a
well-known astronomical phenomenon -- the solar day and the sidereal day are
different. This is closely related to
Section 10.E.3, qv.
Gardner. Some
mathematical models embedded in the solar system. SA (Apr 1970) = Circus, chap. 16, Solar system oddities. In 1806, a reader's letter to SA asked
"How many revolutions on its own axis will a wheel make in rolling once
round a fixed wheel of the same size?"
The editors replied "One", which started an enormous
correspondence. In vol. 18 (1868)
105-106, they printed a selection of letters, but more continued to come until
in Apr 1868, they announced they were dropping the topic, but would continue it
in a new monthly, The Wheel. This was
advertised as being available in the 23 May issue of SA. In Carroll-Gardner, p. 46, Gardner indicates
he has never seen a copy of The Wheel.
Clark. Mental
Nuts. 1897, no. 38; 1904, no. 46; 1916, no. 48. The cog
wheels. "Suppose two equal cog
wheels or coins (one stationary), how many turns will the other make revolving
around it?" Answer is: "Two full
turns."
Pearson. 1907. Part II, no. 58: The geared wheels, pp. 58
& 172. 10 tooth wheel turning
about a 40 tooth wheel.
Dudeney. AM. 1917.
Prob. 203: Concerning wheels, pp. 55 & 188.
McKay. Party
Night. 1940. No. 23, p. 181. Two equal
circles, one rolling around the other.
W. T. Williams & G. H. Savage. The Penguin Problems Book.
Penguin, 1940. No. 80: Revolving
coins, pp. 46 & 129. Equal coins,
then rolling coin of half the diameter.
E. P. Northrop.
Riddles in Mathematics.
1944. 1944: 55-57; 1945: 52‑54; 1961: 60‑62. Also considers movement of a slab on rollers.
Gardner. Coin
puzzles. SA (Feb 1966) =
Carnival, chap. 2, Penny Puzzles.
Gives the basic problem and the elegant generalization for rolling
around an arbitrary ring of coins of the same size.
This is closely related to Section 10.E.2, qv.
Gardner. Some
mathematical models embedded in the solar system. SA (Apr 1970) = Circus, chap. 16, Solar system oddities. He says there have been many, mostly
self-published, booklets arguing that the moon does not rotate and that several
are included in De Morgan's Budget of Paradoxes of 1872.
Jelinger Symons.
Letter: The moon has no rotary motion.
The Times (8 Apr 1856). ??NYS --
discussed by Edward Wakeling in his edition of Lewis Carroll's Diaries, because
Carroll responded and made an entry on 8 Apr.
Discussed in Carroll‑Gardner, pp. 45-46. Carroll "noticed for the first time the fact that though
[the moon] only goes 13 times round the earth in the course of the year, it
makes 14 revolutions round its own axis, the extra one being due to its motion
round the sun." There were
numerous responses to Symons' letter, and The Times printed seven of them on 9
Apr, but not Carroll's. Symons riposted
on 14 Apr, Carroll responded again, with a brief diary entry, and a reply
appeared on 15 Apr, signed E.B.D., but
this is not known to be a pseudonym of Carroll and if it were, Carroll would
most likely have recorded it in his diary.
Bubbenhall. Letter: A puzzle.
Knowledge 3 (9 Feb 1883) 91, item 719.
"A squirrel is sitting upon a post and a man is standing facing the
squirrel, the squirrel presently turns round and the man moves round with it,
always keeping face to face. When the
man has been round the post has he been round the squirrel?" Proctor was editor at the time -- could he
have written this letter??
Richard A. Proctor.
Editorial comment. Knowledge 3
(9 Mar 1883) 141-142. Says the hunter
does go round the squirrel and that the problem is purely verbal.
W. Smith.
Letter: The squirrel
puzzle. Knowledge 3 (4 May 1883) 268,
item 807. Disagrees with above, but not
very coherently. Proctor's comments do
not accept Smith's points. "In
what way does the expression going round an object imply seeing every side of
it? Suppose the man shut his eyes,
would that make any difference? Or,
suppose the man stood still and the squirrel turned round, so as to show him every
side -- would the stationary man have gone round the squirrel?"
Clark. Mental
Nuts. 1897, no. 22; 1904, no. 15; 1916, no. 21. The hunter
and the squirrel. Here the squirrel
always keeps on the opposite side of the tree from the hunter. The 1897 answer is: "No; he did
not." The 1904 ed extends this by:
"They travel on parallel lines and do not change their relative
position." and the 1916 abbreviates the answer as: "No; they travel
on parallel lines, don't change relative position."
Pearson. 1907. Part I: Round the monkey, p. 126. Says R. A. Proctor discussed this some years
ago in "Knowledge".
William James.
Pragmatism. NY, 1907. ??NYS.
Dudeney. Some much‑discussed
puzzles. Op. cit. in 2. 1908.
'The answer depends entirely on what you mean by "go around."'
Loyd.
Cyclopedia. 1914. The hunter and the squirrel, p. 61. c= SLAHP: The hunter and the squirrel,
p. 9. Loyd Jr. says it is a
"hundred‑year‑old question".
Collins. Fun with
Figures. 1928. Monkey doodles business, pp. 232-234. Monkey on a pole as in Bubbenhall. "It is really a matter of personal
opinion, ...." Quotes Proctor's
comments of 4 May 1883 with some minor changes.
Harriet Ventress Heald.
Op. cit. in 7.Z. 1941. Prob. 36, p. 17. Answer says it depends.
Robin Ault. On going
around squirrels in trees. JRM 10 (1977‑78)
15‑18. Cites James. Develops a measure such that if the hunter
and the squirrel move in concentric circles of radii a and b,
then the hunter goes
a/(a+b) around the squirrel and
the squirrel goes b/(a+b) around the hunter.
10.E.4. RAILWAY WHEELS PARADOX
New Section. Although I learned this years ago and have
used it as a problem, I don't recall seeing it in print before acquiring Clark
in Aug 2000.
Clark. Mental
Nuts. 1897, no. 49. Argument.
"The wheels of a locomotive are fixed fast on the axle. The outer rail in a curve is the
longest. How do the outer and inner
wheels keep even in rounding the curve?"
"By the bevel of the wheels and sliding." His 'bevel' refers to the fact that the
wheels are tapered, being smaller on the outside. Centrifugal force on a curve causes the locomotive to move toward
the outside of the curve so the outer wheel has a larger effective
circumference than the inner wheel.
David Singmaster.
Wheel trouble. Problem used as
Round the bend, Weekend Telegraph (26 Nov 1988) = Wheel trouble, Focus,
No. 9 (Aug 1993) 76-77 & 90.
Car
enthusiasts will know that the rear axle of a rear-wheel-drive vehicle has a
differential gear to allow the wheels to turn at different speeds. This is essential for turning because the
outside wheel travels along a circle of larger radius and hence goes further
and turns more than the inner wheel.
But a railroad car has rigid axles, with the wheels firmly attached at
each end. How can a railroad car go
around a bend?
One
must look closely at the wheels of a railroad car to see the answer. The wheels are tapered, with the larger part
toward the inside of the car, and the rails are slightly rounded. Consequently, when the car goes around a
bend, its inertia (generally called centrifugal force) causes it to move a bit
outward on the rails. The outer wheel
then rides out and up, giving it a larger radius, while the inner wheel moves
in and down, giving it a smaller radius.
The
diagram shows an exaggerated view of the wheels and axle.
/| |\
/ | | \
| | | |
| | | |
| | | |
| | |
|
| | | |
\ | | /
\| |/
Of course, the original floating body
problem is Archimedes' testing of Hieron's crown. I have only included a few examples of this -- it is fairly
widely available.
Archimedes. On
Floating Bodies, Book I. In: T. L. Heath, The Works of Archimedes, ...,
op. cit. in 6.AN, 1897 & 1912, pp. 253-262. On pp. 258-261, Heath describes how Archimedes probably analysed
Hieron's crown.
Marcus Vitruvius [Pollo].
De Architectura. c-20. Translated by Morris Hicky Morgan as:
Vitruvius The Ten Books on
Architecture; Harvard Univ. Press, 1914 = Dover, 1960. Pp. 253-254. Describes Hiero's crown problem and says Archimedes simply
measured the amount of water displaced by the crown and equal weights of gold
and silver.
E. J.
Dijksterhuis. Archimedes. Op. cit. in 6.S.1. 1956. Pp. 18-21 discusses
the problem, noting that the object was actually a wreath, stating that the
oldest known source is Vitruvius (late 1C) and giving several versions of the
method thought to have been used by Archimedes.
[I've recently read Vitruvius and his description of
Archimedes' work says he measured the displacements when the crown and equal
weights of gold and silver were placed in water. I'll expand on this when I find my photocopy.]
Isaac Disraeli.
Miscellanies of Literature.
Preface dated 1840 -- my copy is: New edition, revised; Ward, Lock, London, 1882 (date of
publisher's catalogue at end). P. 211
relates that Charles II, when dining with the Royal Society "on the
occasion of constituting them a Royal Society", asked what would happen if
one had two pails of water of equal weight and put fish in one of them --
"he wanted to know the reason why that pail with such addition, should not
weigh more than the other ...."
This produced numerous confused explanations until one member burst into
laughter and denied the fact. I find
the phrasing confusing -- it seems that he wanted to know why the pails
remained of equal weight. However,
there is a possible way that the pails would remain of equal weight -- if both
pails were full to the brim, then the insertion of fish would cause an equal
weight of water to overflow from the pail.
Disraeli is not specific about dates -- there are two basic dates of the
beginning of the Society. It was
founded on 28 Nov 1660 and it was chartered on 15 Jul 1662, with a second charter
on 22 Apr 1663. The 1662 date seems
most likely.
Dudeney?? Breakfast
Table Problems No. 334: Water and ice.
Daily Mail (3 & 4 Feb 1905) both p. 7. Ice in a full glass of water. "..., what volume of water will
overflow when the ice melts?"
Ackermann.
1925.
Dudeney. Problem
1060: Up or down? Strand Mag. (Jun?
1931). ??NX. Boat full of iron in a reservoir.
Perelman. FMP. c1935?
W. A. Bagley.
Paradox Pie. Op. cit. 6.BN. 1944.
John Henry Cutler.
Dr. Quizzler's Mind Teasers.
Greenberg, NY, 1944. ??NYS -- excerpted
in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun 1992) 47
& 43, prob. 7. Balance two bowls of
water on a scales. Add some goldfish to
one of them. Do the bowls still
balance?
W. T. Williams & G. H. Savage. The Third Penguin Problems Book.
Penguin, Harmondsworth, 1946.
Prob. 79: Aquatics, pp. 38 & 117.
Boat with iron weight in a bathtub.
J. De Grazia. Maths
is Fun. Allen & Unwin, London, 1949
(reprinted 1963). Chap. I,
prob. 7, Cobblestones and water level, pp. 12 & 111.
Gamow & Stern.
1958. The barge in the
lock. Pp. 104‑105. Barge full of iron in a lock.
H. T. Croft
& S. Simons. Some little naggers. Eureka 23 (Oct 1960) 22 & 36.
David Singmaster.
Any old iron? Barge full of iron
in the lock. What happens to the boat
when the iron is thrown into the lock?
Appeared as: Watertight
problem. The Weekend Telegraph (9 Jun
1990) XXIV & (16 Jun 1990) XXIV.
About 1994, there was some correspondence -- possibly in the
Guardian's Notes & Queries column -- about barges in a canal on a
viaduct. Apparently Telford's 1801
Pontcysyllte viaduct on the Shropshire Union Canal at Chirk, near Llangollen,
is 1007 ft long, but has a notice restricting the number of barges on it
to three, though a barge is about
70 ft long. Responses indicated that the reason for the
restriction may have been wave problems.
Erwin Brecher
& Mike Gerrard. Challenging Science Puzzles. Sterling, 1997. [Reprinted by Goodwill Publishing House, New Delhi, India, nd
[bought in early 2000]]. Pp. 33 &
73: Water level. Discusses the problem
of a boat in a pool and throwing a brick overboard. Asks what happens if the boat springs a leak and slowly sinks?
Simple problems of this type are
just variations of meeting problems -- see Vyse and Pike, etc. -- but they only
seem to date from the late 18C.
The comparison of up and down stream
versus across and back is the basis of the Michelson-Morley experiment and the
Lorentz-Fitzgerald contraction in relativity, so this idea must have been
pretty well known by about 1880, but the earliest puzzle example I have is
Chapin, 1927.
More recently, I posed a problem
involving travel uphill, downhill and on the level and I have now seen
Todhunter. I will add such problems here,
but I may make a separate subsection for them.
Vyse. Tutor's
Guide. 1771? Prob. 13, 1793: p. 186; 1799: p. 198 & Key p. 241. Two rowers who can row at 5
set out towards each other at points
34 apart on a river flowing 2½.
Though this appears to belong here, it is simply MR-(2½, 7½; 34).
Pike.
Arithmetic. 1788. P. 353, no. 33. Two approaching rowers, starting
18 apart and normally able to
row at rate 4, but the tide is flowing
at rate 1½. I. e. MR‑(2½, 5½; 18). (Sanford 218 says this is first published
version!)
D. Adams. New
Arithmetic. 1835. P. 244, no. 86. Two boats, with normal speeds
8 start to meet from 300
apart on a river flowing at rate
2. I. e. MR-(6, 10; 300).
I imagine this appears in many 19C texts. I have seen the following.
T. Tate. Algebra
Made Easy. Op. cit. in 6.BF.3. 1848.
Colenso. Op. cit. in
7.P.1. 1849.
Edward Brooks. The
Normal Mental Arithmetic A Thorough and
Complete Course by Analysis and Induction.
Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough
and Complete Course by Analysis and Induction;
Sower, Potts & Co., Philadelphia, 1873. Several examples of the following type.
1863
-- p. 136, no. 4; 1873 -- p. 146, no.
4. Steamboat goes 12
in still water; current is
4. Goes down river and returns
in 6
hours. How far did it go?
Todhunter. Algebra,
5th ed. 1870.
Colenso. Op. cit. in
7.H. These are from the (1864), 1871
material. Let V be the velocity of the
rower and T the velocity of the tide.
Horatio N. Robinson.
New Elementary Algebra: Containing the Rudiments of the Science for
Schools and Academies. Ivison,
Blakeman, Taylor & Co., New York, 1875.
Prob. 90, p. 305. Man can row 15
miles down river in 2½ hours, but requires 7½
to row back. What are the rates
of the rower and of the river?
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E. No. 89, pp. 301 & 347. Steamboat goes 10 in still water,
current is 4, boat goes down and returns in
10 hours -- how far did it go?
Briggs & Bryan.
The Tutorial Algebra, Part II. Op. cit. in 7.H.
1898. Exercises X, prob. 24, pp.
125 & 580. Stream goes 4.
Man rows up and back and takes
39 minutes longer than in still
water. With a second rower, they can
go 3/2
as fast as the single man. They
do the same trip in only 8 minutes longer than in still water.
W. P. Workman. The
Tutorial Arithmetic, op. cit. in 7.H.1.
1902. Section IX, art. 240,
example 3, p. 410 (= 416 in c1928 ed.).
Boat goes 4 mi in
20 min in still water and in 16
min with the tide. How long against
the tide?
Richard von Mises.
c1910. Described in: George
Pólya; Mathematical Methods in Science; (Studies in Mathematics Series, vol.
XI; School Mathematics Study Group, 1963); reprinted as New Mathematical
Library No. 26, MAA, 1977; Von Mises'
flight triangle, pp. 78-81. How can one
determine the airspeed of a plane from the ground speed when the wind is
unknown? "It was solved by Von
Mises some fifty years ago; this I well remember as I heard it from him at that
time."
Schubert. Op. cit.
in 7.H.4. 1913. Section 17, no. 99, pp. 64‑65 &
140. Steamship covers 60 km in 4
hours going upstream and in
3 hours going downstream.
Loyd.
Cyclopedia. 1914. Riding against the wind, pp. 199 &
365. = MPSL2, prob. 49, pp. 34
& 137. = SLAHP: Wind influence, pp.
38 & 97. Against and with the wind.
Loyd.
Cyclopedia. 1914. The Santos‑Dumont puzzle, pp. 202
& 366. Against and with the wind.
Peano. Giochi. 1924.
Prob. 14, pp. 4-5. Two ships
travel 6000 miles and return. The
first goes at 8 mph and returns at 12 mph; the second
goes 10 mph both ways. Which is
faster?
Ackermann.
1925. Pp. 77‑81. Determine speed of wind using sound echoes.
Ernest K. Chapin.
Loc. cit. in 5.D.1. 1927.
Harriet Ventress Heald.
Op. cit. in 7.Z. 1941. Prob. 48, pp. 21‑22. Current of
1 mph. Man rows up in 3 hours & back in 2
hours. How far did he go?
D. A. Hindman. Op.
cit. in 6.AF. 1955. Chap. 16, prob. 23: The floating hat, pp.
260‑263. Hat falls off rower
going upstream and is picked up on his return.
Gamow & Stern.
1958. Boat and bottle. Pp. 100‑102. Bottle falls off boat going upstream and is picked up later.
David Singmaster.
Racing along. Used in my puzzle
columns. Test track has one mile
stretches of level, uphill, level, downhill, then two mile stretches, then
three mile stretches, until a total of 60 miles. For optimum economy, a car goes
30 mph uphill, 60 mph
downhill and 40 mph on the level. How long does it take to cover the track?
Fuel
on the hill. Weekend Telegraph (17 Dec 1988) xv & (24 Dec 1988) x.
G&P
16 (Jul 1995) p. 26. (Solution never
appeared as this was the last issue.)
10.H. SNAIL CLIMBING OUT OF WELL
A snail is at the bottom of a well
which is D deep. He climbs A in
the day and slips back B in the night. How long does it take to get out? The earlier versions had serpents, snakes and lions. The 'end effect' is that when the snail gets
to within A of the top, he doesn't slip back. The earlier versions did not have this clear and thus are just
equivalent to meeting problems, MR-(A,
0; D) or MR-(A, B; D) or O-(A, B) with headstart D.
See 10.A: Bakhshali ff.
60r-60v; al-Karkhi no. 9(?); Fibonacci pp. 177-178 & 182 (S: 274
& 280).
It appears that the problem grows out of the usage of unit
fractions such as 1/2 - 1/3 to specify rates. At first this just meant
1/6 per day, but then it began
to be interpreted as going ahead 1/2 followed by retreating 1/3,
resulting in the 'end effect'.
The earliest to treat the end effect clearly seems to be c1350.
For convenience, let the net gain
per day be G = A - B. The solution is to take the least N such
that NG + A ³
D, i.e. N = é(D‑A)/Gù, then interpolate during the daytime of
the (N + 1)‑st day, getting
(D - NG)/A of the day time on
the (N + 1)-st day as the time of
meeting.
When we have approaching animals,
going +a, -b; +c, -d, set
A = a + c, B = b + d, so G
= a - b + c - d. If we are considering
meeting without an end termination such as the cat eating the mouse, and A ³ G, then one can have multiple meetings. The last night on which a meeting occurs during the retrogression
is the last M such that MG ³
D, i.e. M = ëD/Gû. Thus there will be 2 (M - N) + 1 meetings, though this is reduced by one if (D - A)/G
is an integer and by one if
D/G is an integer. (When both are integers, one reduces by
two.) Simple modifications deal with
the cases of negative B, say if a snail continues going up at night,
but at a different rate, and the case with
A < B.
Versions with two approaching
animals: Fibonacci, Muscarello,
Chuquet, Borghi, Pacioli,
Tagliente, Ghaligai, Buteo,
Tartaglia. Chuquet and Buteo
treat the end effect clearly and Pacioli and Tagliente almost get it.
More complex versions: Mahavira, Wood.
See Tropfke 588. Tropfke 589 classifies these as I A c (one person), I B c (two persons meeting), I C c (two persons overtaking).
Bakhshali MS.
c7C.
Bhaskara I.
629. Commentary to Aryabhata,
chap. II, v. 26-27. Sanskrit is on pp.
115-122; English version of the examples is on pp. 304-306. The material of interest is example 4. v
= + 1/2 - 1/5, D = 480.
Chaturveda. 860. Commentary to Brahma‑sphuta‑siddhanta,
chap. XII, section 1, v. 10. In
Colebrooke, pp. 283‑284. v = +
4/5 ‑ 1/60, D = 76,800,000.
Mahavira. 850. Chap. V, v. 24‑31, pp. 89‑90.
Sridhara. c900. Ex. 32‑33, pp. 24 & 93.
Tabari. Miftāh
al-mu‘āmalāt. c1075. Pp. 103f.
??NYS -- quoted in Tropfke 593.
No. 7. "A boat comes
forward 18 parasangs per day and goes backward 12 parasangs per day. It comes and goes for 40 days. How many days does it come and how many days
does it go?" This is not clear --
I wonder if 'days' in the last sentence should be 'parasangs' -- ??
Fibonacci.
1202.
Columbia Algorism.
c1350. Prob. 67, pp. 88‑89. Pigeon going down tower. Rate = ⅔ ‑ (⅓ +
¼), D = 10 (but Cowley says 50). These are alternatively day and night, and
the end effect is clearly treated to get
112 full days and one more
daytime. (Cowley 399.)
Pseudo-dell'Abbaco.
c1440. Prob. 191, pp. 151‑153
with fine plate on p. 152. Serpent in
well of depth 30. Goes
+2/3 in day and ‑1/5
at night. Gives simple answer
and then carefully analyses the end effect to get 63 full days plus 9/10
of daytime. Says that some solve
it erroneously. I have a colour slide
of this.
AR. c1450. Prob. 65, pp. 47, 177, 224. Tower
10 high. Dove flies up ⅔ in the day and
drops back ¼ + ⅓ at night.
Answer is 120 days, which ignores the end effect. Should be
112½, as in the Columbia
Algorism.
Muscarello.
1478.
Tommaso della Gazzaia.
Liber geometriae. Manuscript
C.III.23, Biblioteca Communali di Siena.
15C?? F. 169r. ??NYS -- quoted in Franci, op. cit. in 3.A,
p. 29. Serpent going +⅓ -¼
up a tower 30 high.
End effect ignored.
Chuquet. 1484. Mentioned on FHM 204 as 'the frog in the
well'. All these treat the end effect
clearly.
Borghi.
Arithmetica. 1484.
Calandri.
Arimethrica. 1491.
Pacioli. Summa. 1494.
Blasius. 1513. F. F.ii.r: Secunda regula. +50, -19
to reach 992. No end effect considered, even though the
rates are in the day and in the night.
Tagliente. Libro de
Abaco. (1515). 1541.
Prob. 121, f. 59r. Cat and mouse
on tree which is 26¾ tall.
Cat goes +1/2, -1/3 (misprinted 2/3);
mouse goes +1/4, -1/5. Seems to say they meet on the 120th
day, but it should be the
121st day.
Ghaligai. Practica
D'Arithmetica. 1521. Prob. 19, f. 64v. Two ants 100 apart going toward a pile of grain. Further goes +7, -4; other goes +5, -3
during day, night. How far is
the pile if they get there at the same time?
He says they take 100 days, but this ignores the end effect --
they meet at the end of the 99th daytime.
This is equivalent to a snail going
+2, -1 up a wall of 100,
(H&S 65 gives Italian and English.)
Riese.
Rechnung. 1522. 1544 ed. -- pp. 107‑109; 1574 ed. -- pp. 72v‑73r. The 1574 ed. calls it Schneckengang. +4⅔, ‑3¾, D = 32. Treats end effect clearly and says it was first done correctly by
Hansen Conrad, Probierer zu Eissleben (?= assayer at Eissleben). He discusses how to convince people of the
end effect.
Tonstall. De Arte
Supputandi. 1522. Quest. 36, p. 168. +70, -15 clearly stated
to be day and night, to go 4000, but he ignores the end effect.
Riese. Die
Coss. 1524. No. 142, p. 61. +4¼, ‑3⅓, D = 32.
Treats end effect properly. Says
the Nurmbergk Rechenmeister N. Kolberger got it wrong and that Hans
Conradtt got it right.
Giovanni Sfortunati.
Nuovo lume. Venice, 1545. F. 88r.
??NYS -- described by Franci, op. cit. in 3.A, p. 41. Solves a problem with end effect and says
that Borghi, Pacioli and Calandri have done it wrong.
Christoff Rudolph.
Künstlich rechnung mit der ziffer und mit den zalpfennigen ... Auffs new wiederumb fleissig ubersehen und
an vil arten gebessert. Nuremberg,
1553, 1561. ??NYS -- quoted by: Grosse; Historische Rechenbücher ...; op.
cit. in 7.H under Faulhaber, p. 28.
Von einem schnecken. Snail
climbing + 7, ‑2 to get out of a well 20
deep. End effect clearly
treated. [This is a revision of the
1526 ed., ??NYS, which may have the problem?]
J.
De Grazia; Maths is Fun; p. 12, attributes the problem to Christoff Rudolph
(1561).
Buteo.
Logistica. 1559. Prob. 33, pp. 234-237. Problem of ships with oscillating winds. They start
20000 apart. First goes
+1200 per day and -700
per night. Second starts on the
first night and goes +1400 per night and -600 per day. He first solves without end effect and then
end effect, getting 14 days +
12 hours of day + 10 2/7
hours of the night. (H&S 65
gives Latin & English.)
Gori. Libro di
arimetricha. F. 73r (p. 80). Height
50, rates +⅓, ‑ ¼. Notes end effect, but then forgets to add
the last day!
H&S 64 says examples with alternating motion are
in: Fibonacci (1202), Columbia Algorism (c1350), Borghi (1484), Calandri (1491),
Pacioli (1494?),
Tartaglia and Riese.
Also that examples with two animals approaching are in: Fibonacci (two ants), Borghi (1484, hawk & dove), Pacioli (1494?, both types) and
Tartaglia (both types).
Faulhaber. Op. cit.
in 7.H. 1614. ??NYS -- quoted by Grosse, loc. cit. in 7.H under Faulhaber, p.
120. No. 12, p. 212. Worm climbs
+¾, ‑⅓ up a
tree 100 high. Gives answer with
end effect.
Dilworth.
Schoolmaster's Assistant.
1743. P. 166, no. 91. +20, -15
to go 150. No end effect considered so answer is 30 days.
Walkingame. Tutor's
Assistant. 1751. 1777: p. 172, prob. 47; 1860, p. 180, prob. 46. Snail going
+8, -4 to get up a May pole 20
high. Answer is 4
days, presumably meaning 3 whole days and a daytime.
Vyse. Tutor's
Guide. 1771? Prob. 27, 1793: p. 40; 1799: pp. 43-44 & Key p. 39. Same as Walkingame, but answer is done step
by step and says 'the fourth Day at Night'.
D. Adams. Scholar's
Arithmetic. 1801. P. 209, no. 1. Frog in well +3, -2 to go
30. No answer.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions. No. 5, pp. 15 & 72. Snail going
+8, -4 to get up a maypole 20
high. Treats end effect properly but
states the answer is 4 days.
Family Friend 1 (1849) 150 & 178. Problems, arithmetical puzzles, &c.
-- 2. Snail on wall. +5, -4
to reach 20. End effect clearly treated. = The Illustrated Boy's Own Treasury, 1860,
Arithmetical and Geometrical Problems, No. 24, pp. 429 & 433.
Magician's Own Book.
1857. The industrious frog, p.
234. + 3, ‑ 2 up a well
30 deep. End effect treated. = Boy's Own Conjuring Book, 1860, p. 200.
Charades, Enigmas, and Riddles. 1860: prob. 24, pp. 59 & 63;
1862: prob. 25, pp. 135 & 1865: prob. 569, pp. 107 & 154. Snail going
+5, -4 up a wall 20
high. Answer is 16
days, which considers the end effect, but doesn't describe the final
daytime well.
Bachet-Labosne.
Problemes. 3rd ed., 1874. Supp. prob. II, 1884: 182. +9, -5
to cover 173. Gives simple answer and then considers end
effect.
Mittenzwey.
1880.
Lucas.
L'Arithmétique Amusante.
1895. Prob. XI: La ballade de
l'escargot rétrograde, pp. 25-26. +5, -2 going up a tree of
height 9. Also quoted in Laisant; op. cit. in 6.P.1; 1906; p. 125.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine
9:6 (Apr 1903) 544-546. "A frog
was trying to get up a slippery bank twelve feet high. In the first twelve hours he climbs eight
feet, but in the next twelve hours he loses four feet. How long will he be in reaching the
top?" "Why, having lost his
four feet, how could he get to the top at all?"
Clark (1897, 1904 & 1916), Pearson (1907), Loyd
(Cyclopedia, 1914), Ahrens (A&N,
1918), Loyd Jr. (SLAHP,
1928) all have versions.
Wood. Oddities. 1927.
M. Adams. Puzzles
That Everyone Can Do. 1931. Prob. 197, pp. 76 & 158: Seaside
problem. Deck-chair man earns £6 when
it is sunny but loses £3 when it rains.
The weather alternates, being sunny on the first day. When is he £60 ahead?
Stephen Leacock.
Model Memoirs and Other Sketches from Simple to Serious. John Lane, The Bodley Head, 1939, p.
290. In a sketch on quizzes, he has the
following.
"Sometime
we drop into straight mathematics, which has the same attraction as playing
with fire: for example: -- If a frog falls into a sand-pit twenty feet deep
and gets up the side in jumps two feet at a time, but slips back one foot on
the sand while taking his breath after each jump, how many jumps would it take
him to get out of the pit?
There,
be careful with it. Don't say you can
do it by algebra -- that's cheap stuff -- and anyway you can't."
Evelyn August. The
Black-Out Book. Op. cit. in 5.X.1. 1939.
The man who hoarded petrol, pp. 154 & 215. Tank holds 8 gallons, man puts in 2
gallons every day and 1 gallon leaks out every night; when is it full?
10.I. LIMITED MEANS OF TRANSPORT -- TWO MEN AND A BIKE, ETC.
The men have to get somewhere as
quickly as possible and have to share a vehicle which may be left to be picked
up or may return to pick up others.
Laisant. Op. cit. in
6.P.1. 1906.
Loyd. Tandem
puzzle. Cyclopedia, 1914, pp. 322 &
382 (erroneous solution). = MPSL2,
prob. 123, pp. 88 & 160‑161, with solution by Gardner. Three men and a tandem bike.
Loyd Jr. SLAHP. 1928.
A tandem for three, pp. 52 & 104.
Like Loyd's but with different data.
Dudeney. PCP. 1932.
Prob. 75: A question of transport, pp. 29 & 136. = 536, prob. 89, pp. 27 & 244. Twelve soldiers and a taxi which can take
four of them.
Haldeman-Julius.
1937. No. 64: Tandem for three
problem, pp. 9 & 24. Tom, Dick and
Harry can walk 3, 4, 5 mph.
They have a tandem bike and any one or two of them can ride it at 20 mph.
They want to go 43 1/3 miles as quickly as possible. Clearly the slowest boy should stay on the
bike at all times, so this is really two men and a taxi.
Gaston Boucheny.
Curiositiés & Récréations Mathématiques. Larousse, Paris, 1939, pp. 77‑78. Two men and a bike.
R. L. Goodstein.
Note 1797: Transport
problems. MG 29 (No. 283)
(Feb 1945) 16‑17. Graphical
technique to solve general problem of a company of men and a lorry.
William R. Ransom.
Op. cit. in 6.M. 1955. A ride and walk problem, pp. 108‑109. Same as Dudeney.
Karl Menninger.
Mathematics in Your World. Op.
cit. in 7.X. 1954?? A bit of 'hitch hiking', pp. 100‑101. Three men and a motorcyclist who can carry
one passenger. Graphical method, but he
neglects to consider that the passengers can be dropped off before the goal to
walk the remaining distance while the cyclist returns to pick up the others.
Doubleday - 2.
1971. Dead heat, pp. 79-80. Two men and a pony, but the pony is assumed
to stay where it is left, so this is like two men and a bike. If the pony could go back to meet the second
traveller, then this would be two men and a motorcyclist.
Jonathan Always.
Puzzles for Puzzlers. Tandem,
London, 1971. Prob. 103: Two men and a
bicycle, pp. 50 & 94. Journey of
25, men can walk at rates 3 and 4 or either can ride the bike at rate 7.
David Singmaster.
Symmetry saves the solution.
IN: Alfred S. Posamentier &
Wolfgang Schulz, eds.; The Art of Problem Solving: A Resource for the Mathematics
Teacher; Corwin Press, NY, 1996, pp. 273-286.
Men and a vehicle, pp. 282-285.
Uses appropriate variables to make the equations more symmetric and
hence easily solvable, even with different speeds.
Find the
resistance between two points in some network of unit resistors.
E. E. Brooks & A. W. Poyser. Magnetism and Electricity.
Longmans, Green & Co., London, 1920. Pp. 277‑279. ??NYS.
Anon. An electrical
problem. Eureka 14 (Oct 1951) 17. Infinite square lattice of one ohm resistors
along edges. Asks for resistance
from (0, 0) to (0, 1) and to
(1, 1). Solution to be in No.
15, but isn't there, nor in the next few issues.
Huseyin Demir, proposer;
C. W. Trigg, solver. Problem 407
-- Resistance in a cube. MM 33 (1959‑60)
225‑226 & 34 (1960‑61) 115‑116. All three inequivalent resistances for a
cubical network are found, i.e. the resistances between points which are
distance 1, 2, 3 apart.
Trigg says distance = 1 and
3 cases are in Brooks & Poyser.
Gardner. SA (Dec
1958) = 2nd Book, p. 22. Gives cube
solution and cites Trigg's reference to Brooks & Poyser.
B. van der Pol & H. Bremmer. Operational Calculus -- Based on the Two‑sided Laplace
Integral. 2nd ed., Cambridge Univ.
Press, 1959. ??NYS. 'Very last section' obtains formulae for the
resistance R(m, n) in the infinite plane network between (0, 0)
and (m, n). These are in terms of Chebyshev polynomials
or Bessel functions.
Albert A. Mullin & Derek Zave, independent
proposers; A. A. Jagers, solver. Problem E2620 -- Symmetrical networks with
one‑ohm resistors. AMM 83 (1976)
740 & 85 (1978) 117‑118.
All regular polyhedra and the n‑cube,
but only between furthest vertices.
(Editorial note says the cubical case first occurs in Coxeter's Regular
Polytopes, but I can't find it?? It
also erroneously says Gardner gives other solutions.)
D. C. Morley, proposer;
Friend H. Kierstead Jr., solver.
Problem 529 -- Hyper‑resistance.
JRM 9 (1976‑77) 211
& 10 (1977‑78) 223‑224. 4‑cube, resistance between points 3
apart.
David Singmaster, proposer;
Brian Barwell, solver. Problem
879 -- Hyper‑resistance II. JRM
12 (1979‑80) 220 & 13 (1980‑81) 229‑230. n‑cube, furthest vertices. What happens as n goes to infinity?
David Singmaster, proposer;
B. C. Rennie, partial solver.
Problem 79‑16 -- Resistances in an n‑dimensional cube.
SIAM Review 21 (1979) 559
& 22 (1980) 504‑508. In an
n‑cube, what are the resistances
R(n, i) between points i
apart? Results are only known
for i = 1, 2, 3, n‑2, n‑1,
n. Other solvers considered the
infinite n‑cubical lattice and
obtained a general result in terms of an
n‑fold integral, including
R(n, 1) = 1/n.
P. Taylor & C. Feather.
Problems drive 1981. Eureka 44
(Spring 1984) 13‑15 & 71. No.
1. Find all resistances for regular
polyhedral networks.
P. E. Trier. An
electrical resistance network and its mathematical undercurrents. Bull. Inst. Math. Appl. 21:3/4 (Mar/Apr
1985) 58‑60. Obtains a simple
form for R(m, n) (as defined under van der Pol) which
involves a double integral. He
evaluates this explicitly for small m,
n. The integral is the same as that of
the other solvers of my problem in SIAM Review.
P. E. Trier. An
electrical network -- some further undercurrents. Ibid. 22:1/2 (Jan/Feb 1986) 30‑31. Letter making a correction to the above and
citing several earlier works (McCrea & Whipple, 1940; Scraton, 1964; Hammersley, 1966 -- all ??NYS) and extensions: an explicit form for R(m, m),
the asymptotic value of R(m, m) and extensions to three and n
dimensions.
10.K. PROBLEM OF THE DATE LINE
A man who
circles the earth gains or loses a day.
I include
some related problems here. See also
6.AF.
E. John Holmyard.
Alchemy. Penguin, 1957, p. 119,
says Roger Bacon (1214-1294 (or 1292)) foresaw circumnavigation, but doesn't
indicate if he recognized the date problem.
Nicolas Oresme.
Traitié de l'espere. c1350. ??NYS -- described in: Cora E. Lutz; A fourteenth‑century
argument for an international date line; Yale University Library Gazette 47
(1973) 125‑131. Chap. 39. Three men.
One circles the world eastward in
12 days, another westward and
the third stays at home. He computes
their effective day lengths.
Lutz
describes the occurrence of the problem in other of Oresme's writings.
Quaestiones
supra speram (c1355), where the
travellers take 25 days and Oresme suggests "one ought to
assign a definite place where a change of the name of the day would be
made".
His
French translation of Aristotle's De
caelo et mundo as: Traitié du ciel et du monde (1377), where they take 9 days.
Kalendrier des Bergers; 1493. = The Kalendayr of the Shyppars (in a Scottish dialect);
1503. = The Shepherds' Kalendar;
R. Pynson, London, 1506. ??NYS. Described in: E. G. R. Taylor; The Mathematical Practitioners of
Tudor & Stuart England; (1954); CUP for the Inst. of Navigation, 1970; pp.
11‑12 & 311. Three friends,
one stays put, others circle the earth in opposite directions. When they meet, they disagree on what day it
is.
Antonio Pigafetta.
Magellan's Voyage: A Narrative Account of the First Circumnavigation
.... Translated and edited by R. A.
Skelton, 1969. Vol. I, pp. 147‑148. ??NYS -- quoted by Lutz. When they reached Cape Verde in 1522, a
landing party was told "it was Thursday, at which they were much amazed,
for to us it was Wednesday, and we knew not how we had fallen into error."
Cardan. Practica
Arithmetice. 1539. Chap. 66, section 34, ff. DD.iiii.v ‑
DD.v.r (p. 145). Discusses ship which
circles world three times to the west, and also mentions going east. (H&S 11 gives Latin. Sanford 214 thinks he was first to note the
problem.)
Cardan. De Rerum
Varietate. 1557, ??NYS. = Opera Omnia, vol. III, pp. 239-240. Liber XVII.
Navis qua orbem cirūambivit.
Describes Magellan's circumnavigation, giving dates and saying they had
lost a day.
van Etten.
1624. Prob. 91 (96), part IV
(7), p. 141 (231‑232). "How
can two twins, who are born at the same time and who die together, have seen a
different number of days?" The
English edition comments on a Christian, a Jew and a Saracen having their
Sabbaths on the same day.
"A Lover of the Mathematics." A Mathematical Miscellany in Four
Parts. 2nd ed., S. Fuller, Dublin,
1735. The First Part is: An Essay towards the Probable Solution of
the Forty five Surprising PARADOXES, in GORDON's Geography, so
the following must have appeared in Gordon.
Ozanam‑Montucla.
1778. Vol. III, prob. 5, 1778:
32-33; 1803: 34-35; 1814: 33-34; 1840: 415-416. Two
men, born & dying together, but one living one day, or even two days, more
than the other.
Philip Breslaw (attrib.).
Breslaw's Last Legacy.
1784? Op. cit. in 6.AF. 1795: pp. 36-38.
Geographical
Paradoxes.
Carlile.
Collection. 1793.
Jackson. Rational
Amusement. 1821. Geographical Paradoxes.
Endless Amusement II.
1826? Pp. 74-75: "How Two Men may be born on the same Day,
die at the same Moment, and yet one may have lived a Day, or even two Days more
than the other." Notes that this
can give three Thursdays in one week.
Carroll. The Rectory
Umbrella. c1845?? Difficulties: No. 1. In:
The Rectory Umbrella and Mischmasch; (Cassell, London, 1932), Dover,
1971, pp. 31‑33. Also in: Carroll-Collingwood, pp. 4‑5 (Collins:
11-12). Carroll-Gardner, pp. 81-82,
says he published a letter on this in the Illustrated London News of 18 Apr
1857. In 1860, he later gave this as a
lecture to the Ashmolean Society at Oxford.
Warren Weaver. Lewis
Carroll: Mathematician. Op. cit. in 1. 1956. Mentions Carroll's
interest in the problem.
J. Fisher. Where
does the day begin? The Magic of Lewis
Carroll. Op. cit. in 1, pp. 22‑24.
This discusses several of Carroll's
versions of the problem.
Vinot. 1860. Art. CXXIV: La semaine des trois jeudis, p.
145. How to have three Thursdays in a
week.
Jules Verne. Le tour
du monde en quatre-vingts jours (Around the World in Eighty Days). 1873.
Features the gain of a day by going eastward.
In 1883 and 1884, the Rome and Washington Conferences for
the Purpose of Fixing a Prime-Meridian and a Universal Day adopted the
Greenwich Meridian and the basic idea of time zones, which implied the acceptance
of the International Date Line. The
Philippines, having been colonized from the New World, had to skip a new to
conform with its Asian neighbours, but I don't know when this happened. When Alaska was purchased in 1867, it had to
drop 12 days to convert to the Gregorian calendar and then had to have one
eight-day week to conform with the rest of the New World.
Clark. Mental
Nuts. 1897, no. 89. Where is the day lost? Travel from San Francisco to Pekin at the
speed of the sun and one finds it is a day later. Answer says "New day begins at 180th meridian, which is
midway in the Pacific Ocean."
A. M. W. Downing.
Where the day changes. Knowledge
23 (May 1900) 100-101. Observes that
the precise location of the Date Line has not yet been fixed. Gives four versions on a map, the last two
of which only differ at a few islands.
F. &. V. Meynell.
The Week‑End Book. Op.
cit. in 7.E. 1924. 2nd. ed., prob. eight, pp. 276‑277; 5th? ed., prob. twelve, p. 410. What happens if you go around the world from
west to east in 24 hours, or in less than 24
hours? Suggests that in the
latter case, you get back before you started!
McKay. Party
Night. 1940. No. 36, p. 184. Two
airmen circle the earth in opposite directions, both taking 14
days. Which gets back home
first? Answer is that the eastbound one
gets back two days sooner because the
14 days are considered as days
viewed by the airmen.
A. P. Herbert.
Codd's Last Case and Other Misleading Cases. Methuen, London, 1952.
Chap. 14: In re Earl of Munsey:
Stewer v. Cobley -- The missing day case, pp. 72-83. (This probably appeared in Punch, about 7
Dec 1949.) Reprinted in: More Uncommon Law; Methuen, 1982, pp.
74-83. Lord Munsey left property to his
great-nephew "if he has attained the age of 21 before the date of my
death", otherwise the property went to Lord Munsey's brother. The great-nephew's birthday was 2 May. On 1 May, Lord Munsey was on a cruise around
the world and the ship crossed the Dateline at about noon, going westward, so
the Captain declared that the next day would be 3 May at midnight, but Lord
Munsey expired just after midnight. So
who inherits?
Jonathan Always.
Puzzles for Puzzlers. Tandem,
London, 1971. Prob. 2: Another birthday
poser, pp. 11 & 60. Four persons,
born on different dates, lived to be fifty years of age, but never saw their
fiftieth birthday. One was born on 29
Feb. "Another crossed the
International Date-line travelling eastward, so that he gained a day and never
arrived at his fiftieth birthday at all however long he lived." This confuses me -- crossing the Dateline
goes back a day, so it would seem his fiftieth birthday would simply be delayed
by at most a day. Always continues:
"The third crossed the International Date-line travelling westwards on the
eve of his fiftieth birthday, thus losing a day, a died a few hours
later." I think he has confused
things. If a man travels eastwards
across the Dateline on the day before his 50th birthday, then he goes back a
day and hasn't yet arrived at his 50th birthday, though he is now 50 years
old. If he dies before midnight, then
that fits the problem. If another man
travels westwards across the Dateline at midnight on the eve of his fiftieth
birthday, he goes ahead two days and really loses a day, namely his fiftieth
birthday, which he never saw. The
fourth person was blind!
Shakuntala Devi.
Puzzles to Puzzle You. Op. cit.
in 5.D.1. 1976. Prob. 138: The Sabbath day, pp. 86 &
134. In order for a Moslem, a Jew and a
Christian to have their Sabbath at the same time, send the Moslem around the
world to the west and the Christian around the world to the east. When they meet again they will all have the
same Sabbath day!
David Singmaster.
Letter [on the International Date Line]. Notes and Queries column, The Guardian, section 2 (20 Dec 1995)
7. Reprinted in Guardian Weekly (7 Jan
1996) 24. Sketches the history from
Oresme onward and notes the anomalies that Alaska and the Philippines
essentially crossed the Date Line.
Ed Barbeau. After
Math. Wall & Emerson, Toronto,
1995. Double Christmas, pp.
153-154. Straightforward problem
illustrating that flying east from Australia, one may return to the previous
day. Here the heroine gets 36 hours of
Christmas, in two disconnected parts!
[Indeed if one just crosses the date line from west to east at midnight,
one gets 48 hours of the same day.]
10.L. FALLING DOWN A HOLE THROUGH THE EARTH
The angular frequency of the
oscillation is Ö(g/R) where
g is the acceleration of gravity
at the surface and R is the radius of the earth. Taking a mean radius of 3956.665 mi
and g = 32.16 ft/sec2 gives half-period of 42.20 min = 42 min 12 sec.
Hesiod. The
Theogony. c-700. IN: The Homeric Hymns and Homerica,
translated by H. G. Evelyn White; Harvard Univ. Press, 1959, p. 131, lines
724-725. ??NYS -- information sent by
Andrew Simoson [email of 25 Feb 2003].
This claims that a brazen anvil dropped from the Earth's surface will
reach Tartarus in nine days.
Plutarch. Plutarch's
Moralia, vol. XII, translated by H. Cherniss & W. C. Helmbold. Harvard Univ. Press, 1957, pp. 65-67. ??NYS -- cited by Simoson. Plutarch notes that if the centre of the
earth is at a person's navel, then both his head and his feet are pointing
up. Simoson quotes Plutarch as:
"Not that ... masses ... falling through the ... earth stop when they
arrive at the centre, though nothing encounter or support them; and, if in
their downward motion the impetus should carry them past the centre, they swing
back again and return of themselves?"
Galileo. Dialogo ...
sopra i due Massimi Sistemi del Mondo ... (Dialogue Concerning the Two Chief
World Systems). Gio. Batista Landini,
Florence, 1632. Translated by Stillman
Drake; Univ. of Calif. Press, Berkeley, 1953; pp. 22‑23, 135‑136,
227 & 236. Asserts the object will
oscillate. No mention of air
resistance.
van Etten.
1624. Prob. 91 (88), part II
(2), p. 139 (220). Says a millstone
dropped down such a hole at 1 mile per minute will take more than 2½
days to reach the centre, where "it would hang in the air".
Ozanam. 1694. Prob. 7, corollary 3, 1696: 218; Prob. 7, Remark, 1708: 312. Prob. 7, part 8, corollary 3, 1725: vol. 2,
151-152. Considers falling down a well
to the centre of the earth but uses a hypothetical constant value of g.
Then considers a tube through the earth and says the object will
oscillate, but air resistance will slow it down to rest at the centre.
Euler. A Physical
Dissertation on Sound. 1727? ??NYS -- described by Truesdell in his Introduction
to Euler's Algebra, p. xiv. Annex
announces the solution of the problem of oscillation through a hole in the
earth.
Euler. Algebra. 1770.
I.III.X.501, p. 163. How far
would an object fall in a hour under constant
g as at the earth's surface? 39272
miles!
Euler. Letters to a
German Princess. ??NYS -- Simoson cites
an Arno Press, 1975, reprint of the 1833 ed.
Vol. 1, letter L, pp. 178-182.
Simoson quotes Euler as: "You will remember how Voltaire used to
laugh at the idea of a hole reaching to the centre of the earth, ... but there
is no harm is supposing it, in order to discover what would be the end
results."
Ozanam‑Montucla.
1778. Vol. IV, prob. 9, 1778:
41-42; 1803: 42-43; 1814: 34-35; 1840: 616-617. First
finds the time to reach the centre if
g is constant, namely 19 minutes. Then considers that gravity will decrease
and quotes a result of Newton to find the time to the centre is 21' 5" 12"'.
John Baines, proposer;
Wm. Rutherford; N. J. Andrew
& George Duckett; independent
solvers. Question (23). The Enigmatical Entertainer and Mathematical
Associate for the Year 1830; ....
Sherwood & Co., London. No.
III, 1829. This has two separate parts
with separate pagination. The second
part is The Mathematical Associate and the problem is on pp. 36-37 in the
Answers to the Questions Proposed Last Year.
"If a hole were bored through the earth, parallel to the equator,
in lat. 20o, and a heavy body let fall into it from the surface, it
is required to determine its velocity at any point of its descent, taking into
account the variation of gravity, but abstracting all resistance."
Lewis Carroll. Alice
in Wonderland. Macmillan, London,
1865. Chap. I, pp. 27-28 in Gardner's
Annotated Alice, below. "I wonder
if I shall fall right through the earth!"
Lewis Carroll.
Sylvie and Bruno Concluded.
Macmillan, London, 1893. Chap.
7, pp. 96‑112, esp. pp. 106‑108.
Discusses trains using straight holes, not through the centre. Cf Carroll-Gardner, pp. 7-8, where Gardner
notes that Carroll frequently uses the value 42 which is the half-period in
minutes.
Martin Gardner. The
Annotated Alice. Revised ed., Penguin,
1970. Chap. I, note 4 (to the line
given above), pp. 27‑28.
Describes Carroll's interest in the problem. Says it interested Plutarch, Bacon and Voltaire and that it had
been resolved by Galileo (see above).
Gardner also cites C. Flammarion, Strand Mag. 38 (1909) 348;
??NYS.
Clark. Mental
Nuts. 1897, no. 53. Matters of gravity. "Suppose you drill a hole through the
earth and drop an iron ball in it, where will the ball go?" Answer is "Centre of earth."
Pearson. 1907. Part II, no. 79: Dropped through the globe,
pp. 130 & 207. Says it will
oscillate, but air friction will cause it to come to rest at the centre.
Ackermann.
1925. Pp. 60‑61. Similar to Pearson.
Collins. Fun with
Figures. 1928. A hole through the earth, p. 203. Says it will oscillate like a pendulum and
if air is present, it will slow down and stop at the centre.
W. A. Bagley. Puzzle
Pie. Op. cit. in 5.D.5. 1944.
No. 46: Down Under, pp. 51-53.
Various discussions of what happens to a person or a cannonball falling
through the earth. Seems to think a man
would turn over so that he would be rightside up at the other side?? Says the oscillation will continue with
diminishing periods (presumably meaning amplitude) until the person is stuck at
the centre. A cannonball would burn up
from friction.
R. E. Green. A
problem & H. Martyn Cundy. A
solution. Classroom Note 178: Quicker round the bend! MG 52 (No. 382) (Dec 1968) 376‑380. Green notes the well known fact that the
time to fall through a straight frictionless hole is independent of its length
-- about 42 minutes. He asks what
path gives the least time? Cundy says
it is a straightforward application of the calculus of variations. He finds a solution for θ
as a function of r, in terms of
R, the radius of the earth,
and m,
the distance of closest approach of the curve to the centre. Also
m/R = 1 ‑ 2a/π,
where 2a is the central angle between the ends of the
tube. The straight through time is π Ö(R/g). The shortest time is
π Ö{(R2‑m2)/Rg}.
K. E. Bullen. The
earth and mathematics. MG 54 (No. 390)
(Dec 1970) 352‑353. A riposte to
Classroom Note 178, pointing out Saigey's result of c1890, that g
increases as you start down a hole because the interior of the earth is
denser than the surface. More recent
theoretical and practical work indicates
g is essentially constant for at
least the first 2000 km.
H. Lindgren.
Classroom Note 250: Quicker
round the bend (Classroom Note 178). MG
55 (No. 393) (Jun 1971) 319‑321.
Shows the optimal curve is a hypocycloid and rephrases the time
required. If d is the surface distance
between the ends and C is the earth's circumference, the minimal
time is Ö{d(C‑d)/Rg}. He cites 1953 and 1954 papers which treat
the problem in general.
Erwin Brecher.
Surprising Science Puzzles.
Sterling, NY, 1995. Hole through
the earth, pp. 20 & 88. Asks a
number of straightforward questions and then asks whether a ball will take more
or less time to fall through a hole in the moon. He says it will take about
53 min on the moon -- I get 54.14 min.
In 2000 or 2001, Tim Rowett asked me the following. If an apple could be dropped from a point on
the earth's orbit, but only attracted by the sun, how long would it take to
reach the sun? This is complicated by
gravity varying and I found it quite awkward to do, having to make some
approximations to get a time of 64.4 days.
After seeing Simoson's article, below, I asked him if his work would
deal with the problem and he pointed out that the formula for falling to the
centre of the earth considered as a point mass (p. 349, case 3) can be used if
one adapts the parameters appropriately.
After a little conversion, this is identical to the formula I
obtained. He computes 64.57 days,
or 63.89 days to the surface of the sun, the difference
being due to his using more accurate values for the astronomical constants.
Andrew J. Simoson.
The gravity of Hades. MM 75:5
(Dec 2002) 335-350. Considers the
problem for several models of the earth.
Cites Plutarch, Halley, Euler and discusses models from Greek mythology,
Dante, Hooke (a multi-layer, onion-like, earth), Halley (a hollow earth), etc. He finds the time to the centre is
about 21.2 min. He finds the time to the centre for a
constant acceleration model is about
19.0 min -- cf Ozanam and Ozanam-Montucla. For the best known results on the earth's mass distribution, he
finds about 19.2 min. If all the earth's mass is at the centre, he
gets about 15.0 min and observes this is the model of the the
earth's mass distribution which gives the fastest time (p. 349, case 3). He then poses a problem of falling down
the z-axis to a galaxy in the x-y plane, as Satan may have done when cast
out of heaven, thereby estimating the Miltonian distance between Heaven and
Hell.
When rotated, these objects stop and
then start turning in the opposite direction.
The word 'celt', with a soft 'c', so it sounds like 'selt', means a
stone hand axe, chisel or similar primitive implement -- see 1910 below. I once heard that the phenomenon was
discovered by anthropologists examining handaxes and that they used the spin as
a form of classification.
The OED entry for Celt is long and not definitive. The word 'celte' appears in the Vulgate
translation of the Bible and is understood to mean some kind of tool, but
others feel it is a miscopying -- it seems to be 'certe' in some manuscripts. By 1700, it was considered a proper Latin
word and was adopted by British archaeologists for primitive tools.
Chambers's Encyclopedia.
Revised edition, W. & R. Chambers, London & Edinburgh,
1885. Vol. II, p. 711. About a column on celts. "CELT (Lat. celtis (?), a
chisel), the name by which certain weapons or implements of the early
inhabitants of Western Europe are known among archaeologists."
G. T. Walker. J.
Walker says his investigations occur in old books on rotational mechanics in
the chapter on asymmetrical tops. ??NYS
G. T. Walker. On a
curious dynamical property of celts.
Proc. Camb. Philos. Soc. 8 (1892/95) 305‑306. (Meeting of 13 May 1895.) ??NX
G. T. Walker. On a
dynamical top. Quarterly J. Pure &
App. Mathematics 28 (1896) 175‑184.
(& diagrams??) ??NX
Harold Crabtree. An
Elementary Treatment of the Theory of Spinning Tops and Gyroscopic Motion. Longmans, Green & Co., London, 1909,
??NX; 2nd ed, 1914. Pp. 7 & 54 and Plate I. (I have a citation to these pages in the 1st
ed and the material is on the same pages in the 2nd ed. The Preface to the 2nd ed says changes occur
elsewhere.) Says the term comes from
Latin, celtis, a chisel, and uses it for the stone axes in general. Shows and discusses three examples. The first is an ordinary axe. The second is a two-way version -- when spun
in either direction, it will reverse.
The third is a one-way version -- when spun in one direction, it
continues, but when spun in the other direction, it reverses. Cites Walker, 1896. The plate has photos of the three examples.
Encyclopedia Britannica.
11th ed., 1910. Entry for
Celt. ??NYS -- quoted by Bürger,
below. "CELT, a word in common use
among British and French archeologists to describe the hatchets, adzes or
chisels of chipped or shaped stone used by primitive man. The word is variously derived from the Welsh
celit, a flintstone (that being the material of which the weapons are chiefly
made, though celts of basalt, felstone and jade are found); from being supposed to be the implement
peculiar to the Celtic peoples; or from
a Low Latin word celtis, a chisel. The
last derivation is more probably correct."
Andrew Gray.
Treatise of Gyrostatics and Rotational Motion. (1918); Dover, 1959. ??NYS.
Charles W. Sherburne (3409 Patton Ave., San Pedro,
California, 90731, USA). US Design
210,947 -- Scientific demonstration toy.
Filed: 12 Nov 1995; patented: 7
May 1968. 1p. This simply says it 'shows my new design' and there is no
indication of the skew curvature of the bottom. Sherburne has published material claiming that the rattleback
demonstrates the failure of Newton's laws and that it is the shape of Noah's
Ark!
Karl Magnus. The
stability of rotations of a non-symmetrical body on a horizontal surface. Festschrift Szabo, Berlin, 1971, pp.
19-23. ??NYS -- cited by Bürger. This determines which direction of rotation
is stable. I don't know if it deals
with 'both-way reversing' examples.
Jearl Walker. The
mysterious "rattleback": a stone that spins in one direction and then
reverses. SA 241:4 (Oct 1979) 144‑150. Reprinted with extra Note and recent
references in: Jearl Walker; Roundabout
-- The Physics of Rotation in the Everyday World; Freeman, NY, 1985; Chap. 6,
pp. 33‑38 & 66. Cites Crabtree
and G. T. Walker. Discusses work
of Nicholas A. Wheeler and of A. D. Moore.
Allan J. Boardman.
The mysterious celt. Fine
Woodworking (Jul/Aug 1985) 68‑69.
Describes how to make celts.
Hermann Bondi. The
rigid body dynamics of unidirectional spin.
Proc. Royal Soc. London A 405 (1986) 265‑274. Analyses the dynamics and shows that the
phenomenon occurs even without friction.
Only cites G. T. Walker, Quarterly J.
The Cavendish Laboratory has made a fine steel model with adjustable
weights which Bondi has seen make five reversals.
Wolfgang Bürger. A
Celtic rocking top. English version of
leaflet to accompany the plastic version of the celt distributed by Nixdorf
Computers. Nd [probably late
1980s]. Cites Walker, 1896, and quotes
Encyclopedia Britannica, 11th ed., 1910, for the term as quoted above. He conjectures that the spinning property
may have been discovered by an archaeologist and he gives a myth that such
spinning was used by ancient priests to determine guilt or innocence. (Frame-ups were common even then.) Until recent realization that these objects
were man-made, they were the subject of superstitions throughout the
world. He gives a short discussion of the
physics/geometry involved and says that since 1980, nine scientific papers have
tried to analyse the motion and that it was the subject of a German Jugend
forscht (Young Researchers) prize winning project in 1985.
The physics
of this is hard and I will only give some general articles.
Helene Sperl. German
Patent 63261 -- Wendekreisel. Patented:
7 Oct 1891; published: 12 Jul
1892. 1p + 1p diagrams. Several slightly diferent shapes. Diagram is reproduced by Kuypers & Ucke,
below.
Harold Crabtree. An
Elementary Treatment of the Theory of Spinning Tops and Gyroscopic Motion. Longmans, Green & Co., London, (1909,
??NX); 2nd ed, 1914. Pp. 4-5 describes some rising tops. One of these is a loaded sphere, which seems
to have the same inverting properties as the Tippee Top. Appendix IV: The rising of a spinning top,
pp. 145-158, is new in the 2nd ed and discusses the loaded sphere on pp.
155-156.
Gwen White. Antique
Toys and Their Background. Op. cit. in
5.A. 1971. P. 45. "An
interesting little top known as a 'tippe top' came in 1953 .... It was a great commercial success for the
British Indoor Pastimes Company."
D. G. Parkyn. The
inverting top. MG 40 (No. 334) (Dec
1956) 260‑265. Cites 4 papers in
1952.
Leslie Daiken.
Children's Toys Throughout the Ages.
Spring Books, London, 1963.
[This may be a reprint of an earlier publication??] P. 38.
"... the most recent craze, invented by a Swede. Made from plastic, and known as the Tippy
Tap [sic], this type will turn upside down and spin on its head!"
Jearl Walker.
Amateur Scientist columns. SA (Oct
1979) -- op. cit. in 10.M and The physics of spinning tops, including some
far‑out ones. SA 244:3 (Mar 1981)
134‑142. Reprinted with extra
notes in: Roundabout, op. cit. in 10.M,
chap. 6 & 7, pp. 33‑44 & 66, esp. pp. 37‑44. 15 references on p. 66, not including
Parkyn.
Wolfgang Bürger.
Elementary dynamics of simple mechanical toys. Mitteil. der Ges. f. Angew. Math. und Mechanik 2 (Jul 1980)
21-60. (Reproduced in:
Spielzeug-Physik; Bericht Nr. 98, Akademie für Lehrerfortbildung Dillingen,
1986, pp. 159-199.) Pp. 49‑52
(= 188-191) discusses the Tippe Top, noting that Fraülein Sperl's explanation
in her patent 'was far from being correct.'
Says the dynamics is difficult and cites 1952 & 1978 articles on it. J. L. Synge asserted that friction was not
necessary for the turning over motion, but Bürger shows it is essential.
Friedhelm Kuypers & Christian Ucke. Steh' auf Kreisel! Physik in unserer Zeit 25:5 (Sep 1994) cover &
214-215. The German names are
Stehaufkreisel and Kippkreisel.
Describes Sperl's patent and modern work on the mechanics involved --
"it is not so easy as we first believed".
10.N. SHIP'S LADDER IN RISING TIDE
Water is touching the bottom rung of
a rope ladder hanging over the side of a ship.
The tide is rising at a known rate.
How many rungs will be covered after some time?
Phillips.
Brush. 1936. Prob. O.1: The ship's ladder, pp. 49 &
106.
Haldeman-Julius.
1937. No. 42: The Queen Mary
problem, pp. 7 & 22.
Morley Adams. The
Children's Puzzle Book. Faber, London,
1940. Prob. 174: The ship's
ladder, pp. 55 & 78.
Shirley Cunningham.
The Pocket Entertainer.
Blakiston Co., Philadelphia, and Pocket Books, NY, 1942. Prob. VI: The rope ladder, pp. 72 & 222.
Harold Hart. The
World's Best Puzzles. Op. cit. in
7.AS. 1943. The rope ladder, pp. 31 & 59.
Leopold. At
Ease! 1943. Of time and tide, pp. 9-10 & 195.
Sullivan.
Unusual. 1943. Prob. 3: Time the tide.
Leeming. Op. cit. in
5.E. 1946. Chap. 3, prob. 40: Rising tide, pp. 36 & 162.
10.O. ERRONEOUS AVERAGING OF VELOCITIES
See also 7.Y which involves
erroneous averaging of unit costs.
There must be earlier examples than I have here.
H. A. Ripley. How
Good a Detective Are You? Frederick A.
Stokes, NY, 1934, prob. 42: Class day.
Average 10 mph and
50 mph over the same distance.
Dr. Th. Wolff. Die
lächelnde Sphinx. Academia
Verlagsbuchhandlung, Prague, 1937.
Prob. 17, pp. 190 & 201.
Going and returning at 100 km/hr versus going at 120 and
returning at 80.
Harriet Ventress Heald.
Op. cit. in 7.Z. 1941. Prob. 31, pp. 15‑16. Man goes
30 mph for a mile. How fast must he go for a second mile in
order to average 60 mph overall?
Sullivan.
Unusual. 1943. Prob. 22: Don't get caught trying it. If you are going 60 mph, how much faster
do you have to go to save a minute on each mile?
E. P. Northrop.
Riddles in Mathematics.
1944. 1944: 11-13; 1945: 10‑12; 1961: 20-22. Same as Heald with rates 15, 30.
Relates to airplane going with and against the wind.
Leeming. Op. cit. in
5.E. 1946. Chap. 5, prob. 20: Sixty miles per hour, pp. 59‑60 &
178. Does 30 mph for 10
miles. How fast for the
next 10 miles to average 60 mph?
Sullivan.
Unusual. 1947. Prob. 31: A problem without a title. Same as Heald.
Birtwistle.
Calculator Puzzle Book.
1978. Prob. 40: Motoring
problem, pp. 30 & 87-88. Same as
Heald with rates 15, 30.
Solomon (or The Preacher). c-960.
Hosea 12:7. "He
is a chapman, the balances of deceit are in his hand."
Aristotle.
Mechanical Questions.
c-340. ??NYS -- cited by van
Etten.
Pappus.
Collection. Book 3. c320.
??NYS - cited by W. Leybourn.
Muhammed.
Koran. c630. Translated by J. M. Rodwell, Everyman's
Library, J. M. Dent, 1909. Sura LXXXIII
-- Those who stint: 1-3. "Woe to
those who STINT the measure: Who when
they take by measure from others, exact the full; But when they mete to them or weigh to them, minish
--". (I saw the following version
on a British Museum label, erroneously attributed to Sura LXXX: "Woe be unto those who give short
measure or weight.")
Cardan. De
Subtilitate. 1550, Liber I, ??NYS. = Opera Omnia, vol. IV, p. 370: Modus
faciendi librā, que pondera rerum maiora quàm sunt[?? -- nearly
obliterated in the text I have seen] ostendat.
Describes a scale with arm divided
11 : 12.
John Wecker. Op.
cit. in 7.L.3. (1582), 1660. Book XVI -- Of the Secrets of Sciences:
Chap. 20 -- Of Secrets in Arithmetick: Fraud in Balances where things
heavier shall seem to be lighter, p. 293.
Says such fraud is mentioned by Aristotle.
van Etten.
1624. Prob. 54 (49), pp. 49‑50
(73‑74). Mentions Aristotle's
mechanical questions and cites Archimedes' law of the lever. Discusses example with arm lengths 12 : 11
W. Leybourn.
Pleasure with Profit. 1694. Tract. IV, pp. 2-3. Cites Solomon and Pappus' Collections, Book
3. Discusses arm lengths 11 : 10.
Ozanam. 1694. Prob. 4, 1696: 275-276 & fig. 131, plate
46. Prob. 4 & fig. 26, plate 14,
1708: 351‑352. Vol. II, prob. 7,
1725: 339‑340 & fig. 131, plate 46.
Vol. II, prob. 3, 1778: 4-5;
1803: 4-6; 1814: 3-5; 1840: 196-197. Construct a balance which is correct when empty, but gives
dishonest weight. This can be detected
by interchanging the contents of the two pans.
Hutton notes that the true weight is the geometric mean of the two
weights so obtained, and that this is close to the average of these two
values. Illustrates with weights 16
and 14. The figure is just a picture of a balance
and is not informative -- the same figure is also cited for various sets of
weights.
Vyse. Tutor's
Guide. 1771? Prob. 1, 1793: p. 303; 1799: p. 316 & Key p. 356. Cheese weighs 76 in one pan and 56
in the other. States the general
rule with no explanation.
Jackson. Rational
Amusement. 1821. Curious Arithmetical Questions. No. 8, pp. 16 & 72. Cheese weighs 16 on one side and 9 on
the other. Says the answer is the mean
proportional. = Illustrated Boy's Own
Treasury, 1860, prob. 12, pp. 428 & 431.
Julia de Fontenelle.
Nouveau Manuel Complet de Physique Amusante ou Nouvelles Récréations
Physiques .... Nouvelle Édition, Revue,
..., Par M. F. Malepeyre. Librairie
Encyclopédique de Roret, Paris, 1850.
Pp. 407-408 & fig. 146 on plate 4 (text erroneously says V): Balance
trompeuse. A bit like Ozanam, but
doesn't indicate the true weight is the geometric mean. Figure copied from Ozanam, 1725.
Magician's Own Book.
1857. The false scales, p.
251. Cheese weighs 9 on
one side and 16 on the other. Says the true weight is the mean proportional, hence 12
here. = Book of 500 Puzzles,
1859, p. 65. = Boy's Own Conjuring
Book, 1860, p. 223. Almost identical to
Jackson.
Hoffmann. 1893. Chap. IV, no. 94: The false scales, pp. 169
& 229 = Hoffmann-Hordern, p. 154. On one side a cheese weighs
9 and on the other it
weighs 16. Answer notes that the true weight is always the geometric
mean. Almost identical to Jackson and
Magician's Own Book.
Clark. Mental
Nuts. 1904, no. 66; 1916, no. 84. The grocer puzzled.
Weights of 8 and
18. Answer says to solve 8 : x :: x : 18.
Briggs & Bryan.
The Tutorial Algebra -- Part II.
Op. cit. in 7.H. 1898. Exercises X, pp. 125 & 580. Weighing one way gains an extra 11%
profit, but weighing the other way gives no profit at all. What is the legitimate profit?
Pearson. 1907. Part II, no. 72, pp. 128 & 205. Same problem as Hoffmann. Answer says to take the square root of 9 x 16.
Ernest K. Chapin.
Loc. cit. in 5.D.1. 1927. Prob. 1, p. 98 & Answers p. 10: The
druggist's balance. One arm is longer
than the other, but the shorter is weighted to give balance when the scales are
empty. He uses the two sides equally --
does he gain or lose (or come out even)?
If the lengths are L and
l, then balancing against
weight W will result in WL/l and
Wl/L equally often and the
arithmetic mean of these is greater than their geometric mean of W,
so the druggist is losing.
Answer only does the example with
L = 2l.
Loyd Jr. SLAHP. 1928.
The jeweler's puzzle, pp. 21‑22 & 90. More complex version.
Kraitchik. La
Mathématique des Jeux. Op. cit. in
4.A.2. 1930. Chap. II, prob. 26, p. 34.
= Mathematical Recreations; op. cit. in 4.A.2; 1943; Chap. 2, prob.
54, pp. 41‑42. Merchant has a
false balance. He weighs out two lots
by using first one side, then the other.
Is this fair on average?
Holding a bicycle upright, with the
pedals vertical, push the bottom pedal backward. What happens?
Pearson. 1907. Part II, no. 17: A cycle surprise, pp. 14
& 189.
Ernest K. Chapin.
Loc. cit. in 5.D.1. 1927. P. 99 & Answers p. 11. Pull the bottom pedal forward. What happens? What is the locus of the pedal in ordinary travel? Answer says it is a cycloid but it is
actually a curtate cycloid.
W. A. Bagley.
Paradox Pie. Op. cit. in
6.BN. 1944. No. 60: Another cycling problem, p. 46. Uses a tricycle -- which simplifies the experimental process.
David E. Daykin. The
bicycle problem. MM 45:1 (Jan 1972)
1. Short analysis of this 'old
problem'. No references.
These are questions as to when the
hands can meet, be opposite, be interchangeable, etc. There are also questions with fast and slow clocks. Many examples are in 19C arithmetic and
algebra books and in Loyd, Dudeney, etc.
I am somewhat surprised that my earliest examples are 1678? and 1725, as
clocks with minute hands appeared in the late 16C. The problems here are somewhat related to conjunction problems --
see 7.P.6.
See also 5.AC for digital clocks,
which are a combinatorial problem.
These are related to Section 7.P.6.
Wingate/Kersey.
1678?. Quest. 20, p. 490. Clock with an hour hand and a day hand, which
goes round once every 30 days. They are
together. When are they together again? In 30 days, the faster hand must pass the
slower 59 times, so the time between coincidences is 30/59 of a day.
Ozanam. 1725. Prob. 11, question 3, 1725: 76‑77. Prob. 2, 1778: 75-76; 1803: 77-78; 1814: 69; 1840: 37. When are the hands together? 1725 does it as a geometric progression,
like Achilles and the tortoise. 1778
adds the idea that there are 11 overtakings in 12 hours, but this does
not appear in the later eds.
Les Amusemens.
1749. Prob. 122, p. 264. When are clock hands together?
Vyse. Tutor's
Guide. 1771? Prob. 7, 1793: p. 304; 1799: p. 317 & Key pp. 357-358. When are the hands together between 5 and 6
o'clock?
Dodson. Math.
Repository. 1775.
Charles Hutton. A
Complete Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2. [c1780?] 1804: prob. 35,
p. 136. When are watch hands together between 4
and 5?
Pike.
Arithmetic. 1788. P. 352, no. 27. When are the clock hands together next after noon?
Bonnycastle.
Algebra. 1782.
Eadon.
Repository. 1794. P. 195, no. 13. When do the hands meet after noon?
Hutton. A Course of
Mathematics. 1798? Prob. 36,
1833: 223; 1857: 227. When do the hands next meet after noon?
Robert Goodacre. Op.
cit. in 7.Y. 1804. Miscellaneous Questions, no. 128, p.
205 & Key p. 270. When are
clock hands next together after 12:00?
Silvestre François Lacroix.
Élémens d'Algèbre, a l'Usage de l'École Centrale des
Quatre-Nations. 14th ed., Bachelier,
Paris, 1825. Section 82, ex. 6, p. 122. When are clock hands together after
noon? Notes that this is related to
overtaking problems.
Bourdon.
Algèbre. 7th ed., 1834. Art. 57, prob. 10, p. 85. When are clock hands together?
D. Adams. New
Arithmetic. 1835. P. 244, no. 83. When are hands next together after 12:00? Observes that the minute hand gains 11
spaces (i.e. hour marks) per hour on the hour hand.
Augustus De Morgan.
Examples of the Processes of Arithmetic and Algebra. Third, separately paged, part of: Library of
Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836. P. 90.
How much time elapses between times when the hands are together? Suppose one hand revolves in a
hours, the other in b hours -- how long between conjunctions? In the latter case, how long does it take
the faster hand to gain p/q of a whole revolution on the other?
Hutton-Rutherford. A
Course of Mathematics. 1841?
T. Tate. Algebra
Made Easy. Op. cit. in 6.BF.3. 1848.
Anonymous. A
Treatise on Arithmetic in Theory and Practice: for the use of The Irish
National Schools. 3rd ed., 1850. Op. cit. in 7.H.
Dana P. Colburn.
Arithmetic and Its Applications.
H, Cowperthwait & Co., Philadelphia, 1856. Miscellaneous Examples.
Todhunter. Algebra,
5th ed. 1870.
Colenso. Op. cit. in
7.H. These are from the (1864), 1871
material.
Problem 203.
Zeitschrift für mathematische und naturwissenschaftlichen Unterricht 15
(1874) 197-198. Interchanging clock
hands. Taken from Journ. élém., which
must be a French journal, but I don't recognise it. Shows that interchanging hands gives a legal appearance when the
hour hand is at 60k/143 minutes, k = 0, ..., 143. Copy sent by Heinrich Hemme.
William J. Milne.
The Inductive Algebra ....
1881. Op. cit. in 7.E.
Carroll-Wakeling.
1888 to 1898. Prob. 14:
Looking-glass time, pp. 17 & 66-67.
This was on one of the undated typed sheets Carroll sent to Bartholomew
Price. "A clock face has all the
hours indicated by the same mark, and both hands the same in length and
form. It is opposite to a
looking-glass. Find the time between 6
and 7 when the time as read direct and in the looking-glass shall be the
same." This seems to be the first
example of looking at the mirror image of a clock. This can also occur by reading a transparent clock from the wrong
side. Mentioned in Carroll-Gardner, p.
53. Wakeling has sent me a copy of the
typescript.
Anon. Prob. 80. Hobbies 31
(No. 795) (7 Jan 1911) 350
& (No. 798) (28 Jan 1911)
412. Three pendulum clocks, supposed to
have one second swings. One gains 5
minutes per day, the second loses ten minutes per day and the third is
correct. Assuming we start all three
pendula at the same time at the same end of their swing, when will they all be
together at this point again? Answer: 9
min 36 sec = 1/150 of a day.
Ernest K. Chapin.
Loc. cit. in 5.D.1. 1927. Prob. 5, p. 89 & Answers p. 8. Can the three hands make equal angles? [Last line of the answer has slipped into
the next column.]
W. B. Campbell, proposer;
W. E. Buker, solver. Problem
??. AMM 41 (Sep 1934) 447 &
42 (Feb 1935) 110-111.
Interchanging clock hands. ??NYS
- information sent by Heinrich Hemme.
Perelman. 1937. MCBF.
Interchanging the hands of a clock, prob. 141, pp. 235-238. Says the following was posed to Einstein by
his friend A. Moszkowski: at what times
can one exchange the hour and minute hands and get a valid position of the
hands? There are 143
solutions, of which 11 are the positions where the hands coincide,
which Perelman discusses as prob. 142, pp. 238-239.
Sullivan.
Unusual. 1943. Prob. 1: Watch and see. How many places do hands meet?
Anonymous. Problems
drive, 1957. 20 (Oct 1957) 14-17 &
29-30. No. 10. B reckons A's clock gains one minute per
day. C reckons B's clock gains one
minute per day. They synchronise
watches. Assuming rates are constant,
when do all three agree again?
Pierre Berloquin.
The Garden of the Sphinx, op. cit. in 5.N. 1981.
Jamie & Lea Poniachik.
Cómo Jugar y Divertirse con su Inteligencia; Juegos & Co. &
Zugarto Ediciones, Argentina & Spain, 1978 & 1996. Translated by Natalia M. Tizón as:
Hard-to-Solve Brainteasers. Ed. by Peter
Gordon. Sterling, NY, 1998. Pp. 14 & 70, prob. 18: What time is it
-- IV. Hour hand is on a minute mark
and the minute hand is on the previous mark.
??, proposer; Ken Greatrix & John Bull, separate
solvers, with editorial note by ADF [Tony Forbes]. Problem 168.1 -- Clock.
M500 168 (?? 1999) & 170 (Oct 1999) 11-14. Hour hand is 3" long; minute hand is 4". When are the ends 'travelling apart at the
fastest speed'? Bull notes that the
problem can be interpreted in two ways.
A: 'When is the relative velocity of the ends the greatest?', which
occurs when the hands are at 180o.
B: 'When is the distance between the ends increasing most rapidly?',
which occurs when the angle between the hands has cosine of 3/4.
Greatrix did the first case in general, finding the maximum occurs when
the cosine equals h/m, where
h, m are the lengths of the hour
and minute hand.
John Conway. Naming
and understanding the PENTAgonal polytopes; When are the three hands equally
spaced? 4pp, handwritten, Handout at
G4G5, 2002. Basically the hands are
never equally spaced, but they are close at
9:05:25 (and at the mirror image
time). When are they closest? This depends on how you measure
'closeness'. He finds there are just
five reasonable answers within a second of this time. There is a sixth answer but it is nearly two seconds away. The most interesting solution is at 9 hr 5 min 25/59 sec when the angle between the minute and second
hands is 120o, the angle between the second and hour hands
is 120o + ε and the angle between the hour and minute
hands is 120o - ε, where
ε = 10¢ 10² 10‴ ....
W. A. Bagley.
Paradox Pie. Op. cit. in
6.BN. 1944. No. 93: Better take a brolly, p. 61. Says the faster you go, the wetter you get.
A. Sutcliffe. Note
2271: A walk in the rain. MG 41 (1957) 271‑272.
C. O. Tuckey. Note
2384: A walk in the rain [Note
2721]. MG 43 (No. 344) (May 1959)
124‑125.
M. Scott.
Nature. ??NYS -- reported in
This England 1965-1968, p. 70.
"When walking into the rain one should lower the head and walk as
fast as possible. When the rain is
coming from behind one should either walk forward leaning backwards, or
backwards leaning forwards, at a deliberate pace."
David E. Bell. Note
60.21: Walk or run in the rain? MG 60 (No. 413) (Oct 1976) 206-208. "... keep pace with the wind if it is
from behind; otherwise, run for it."
D. R. Brown. Answer
to question. The Guardian (2 Apr 1993)
13. Cites Bell.
Yuri B. Chernyak & Robert S. Rose. The Chicken from Minsk. BasicBooks, NY, 1995.
Erwin Brecher
& Mike Gerrard. Challenging Science Puzzles. Sterling, 1997. [Reprinted by Goodwill Publishing House, New Delhi, India, nd
[bought in early 2000]]. Pp. 38 &
76‑77: Lingering in the Rain.
Starts with no wind and wearing a hat, but says he ran home and the rain
stopped when he got there. So the
wetness is due to moving into the rain and he says it makes no difference if
the rain continues, but since the rain stopped, walking would have been
drier. Then says that if he has no hat,
his head gets wetter the longer he is in the rain. Then says if there is a head wind, one gets wetter the longer one
is out, but if there is a tail wind, the best strategy is to move as fast as
the wind.
Fred Swithenbank. UK
Patent 11,801 -- An Improved Toy or Puzzle, applicable also as an Advertising
Medium. Applied: 21 May 1913; completed: 21 Nov 1913; accepted: 1 Jan 1914. 2pp + 1p diagrams. Shows two ball version and four ball round version.
I have acquired an example in wood with a celluloid(?) top
with the following printed on the sides.
'Zebra' Grate Polish. Get one
ball in each hole at the same time.
'Brasso' Metal Polish. Patent
No. 11801/13. I have seen others of similar
date advertising: "Swan" Pens
/ "Swan" Inks; Hoffmann
Roller Bearings. Both have the same
instruction and patent no.
Slocum says that he has the same Zebra/Brasso version and
another version from R. Journet, called Spoophem, with the same patent number
and that a round version "The Balls in the Hole Puzzle" is in
Gamage's 1913 catalogue. He cites
Maxwell's patent, below.
James Dalgety. R.
Journet & Company. A Brief History
of the Company & its Puzzles.
Published by the author, North Barrow, Somerset, 1989. P. 13 mentions the Spoophem puzzle, patented
in 1913.
Slocum.
Compendium. Shows 4 ball
Centrifugal Puzzle from Johnson Smith 1919 & 1929 catalogues. Shows 2 ball Spoophem Puzzle from the latter
catalogue.
Western Puzzle Works, 1926 Catalogue. No. 72: Four Ball Puzzle.
William R. Maxwell.
US Patent 1,765,019 -- Ball puzzle device. Filed: 6 Apr 1929;
patented: 17 Jun 1930. 2pp + 1 p
diagrams. Shows 2-ball and 4-ball
versions.
L. Davenport & Co. and Maskelyne's Mysteries. New Amusement Guide [a novelties
catalogue]. Davenport's, London, nd
[Davenport's identifies this as being in the period 1938-1942], p. 6, shows a
Boat Puzzle, which is a 2 ball version..
10.U. SHORTEST ROUTE VIA A WALL, ETC.
I have just added this. I imagine there are ancient versions of this
problem. See also 6.M and 6.BF.3 for
some problems which use the same reflection principle. Basically, this will cover two-dimensional
cases where the reflection process is very physical.
Ozanam‑Montucla.
1778. Prob. 24 & fig. 36,
plate 5, 1778: 305; 1803: 300-301; 1814: 256;
1840: 130. Run from A
to B, touching the wall CD.
Ozanam‑Montucla.
1778. Vol. II. Prob. 35: Du jeu de billard & figs. 34
& 35, plate 7, 1778: 58-62; 1803:
63-66; 1814: 52-55; 1840: 222-224. Use cue ball to hit another after hitting one or two cushions.
Birtwistle. Math.
Puzzles & Perplexities. 1971. Playground circuit, pp. 142-143. In a rectangular playground with a point
marked in it, what is the shortest route from the point to all four walls and
back to the point? He draws it and says
it's a parallelogram, but doesn't see that the length is twice the diagonal of
the rectangle, independently of the starting point.
10.V. PICK UP PUZZLES = PLUCK IT
I have recently added these. The puzzle comprises a ball which is about
half into a hole and the object is to remove it without moving the hole. It can be done with the fingers, but one can
also use the Bernoulli or Venturi effect!
Western Puzzle Works, 1926 Catalogue. No. 11: Pick up ball. Negro head.
S&B 140-141 gives an outline of
the ancient history. Vases which fill
from the bottom were found in Cyprus, by General Louis Palma di Cesnola, who
was American Consul there for ten years from about 1865. He discovered the Treasure of Curium. He shipped back a large amount which forms
the Cesnola Collection at the Metropolitan Museum of Art, NY. Such a vase is named an 'askos' by
archaeologists.
There are examples or descriptions
of such jugs in ‑5C Greece, in the works of Hero and Philo of Alexandria
in 1C, and in 13C France. From the 16C,
they are common throughout Europe and China, and the bottom filling 'Cadogan'
teapot goes back to perhaps c1000 in China.
I have now looked at some of the literature and have realised that this
topic covers a much larger range of types than I had initially thought. In Banu Musa, there are about a hundred
types. Some of these are common in
later works, e.g. van Etten, Ozanam, etc. have casks which pour different
liquids from the same spout. I don't
know if I will try to include all these later versions. Archaeologists refer to these as 'trick
vessels' or 'trick vases'.
A Tantalus, greedy, justice,
temperance or dribble cup has a siphon such that if it is filled above a
certain point, the siphon drains the cup, usually into the drinker's lap!
In response to an exchange on NOBNET
in Feb 1999, Peter Rasmussen kindly sent a page and a half of bibliography on
Chinese puzzle vessels. A number of his
entries are auction, sale or collection catalogues which apparently only show
one or two items, so I won't list them here, but I have included the more
general books at the end of this section.
Norman Sandfield has sent a draft
copy of his A Monograph on Chinese Ceramic Puzzle Vessels (Antique, Vintage and
Contemporary) Featuring a Classification System and Descriptive Inventory of
more than 60 different puzzle vessels, dated 20 Jan 2000, with 3pp of
bibliographical material and a 4pp version of this section, 25pp in all. He also sent further material: 3pp of time
line and a 7pp extended version of his bibliographical material incorporating
this section, dated 27 Jan 2000. He has
more recently sent An Annotated Bibliography on Chinese and Non-Chinese Puzzles
Vessels (Over 155 records) and Museums with Puzzle Vessels (over 30 Museums),
25pp, dated 4 Jun 2002. I will not try
to copy all these references. Most
Chinese examples are 'Cadogan' wine pots (= Magic Wine Pot) or 'Tantalus' Cups
(= Justice Cups -- this is the literal translation of the Chinese name). The Chinese did not adopt teapots until the
Ming Dynasty (1368-1644) and they then adapted the Cadogan style to
teapots. The Justice Cup traditionally
has a base to hold the spillage, sometimes inside it as a mystery to the
drinker. Sandfield feels there is a
difference between the ordinary Justice Cup and the Dribble Cup, but it may
just be a difference in the outlet hole size.
(The modern Dribble Glass has cut decorations, one or several of which
go through the glass so that it dribbles when you tilt it to drink, but this is
unrelated to the present topic.)
CYPRUS
General Louis Palma di Cesnola was the American Consul in
Cyprus for ten years from about 1865 (his commission was one of the last
documents signed by Lincoln). He
collected antiquities enthusiastically and discovered the Treasure of
Curium. He shipped a large amount back
to the Metropolitan Museum of Art in New York.
He describes his adventures and collecting in: Cyprus: Its Ancient
Cities, Tombs and Temples; New York, 1877; but there is no mention of puzzle
vessels there. He also wrote A Descriptive Atlas of the Cesnola
Collection of Cypriote Antiquities in the Metropolitan Museum of Art -- ??NYS,
vol. 2 appeared in 1894. Karageorghis
cites pl. XCV.813 for Myres 518.
John L. Myres.
Handbook of the Cesnola Collection of Antiquities from Cyprus. The Metropolitan Museum of Art, New York,
1914. Items 518, 519, 930, 931, pp.
67-68 & 113. The first two and the
last are illustrated with photos. The
first two are described as 'animal-headed vases' from the Early Iron Age (c‑1000,
with the second being later than the first) and their trickiness is only
mentioned in the index. The latter two
are in a group of Spout Vases with Modelled Head and in these two examples the
woman's head "blocks the apparent neck of the vase, and the real opening,
through which the vase is filled, is inside the foot, which communicates with a
long tube running up the middle of the inside of the vessel after the manner of
the neck of a modern safety ink-pot.
These trick vases are not common, ....
Both these examples are as early as the sixth century, and 931 may go
back to the seventh." (These are
centuries BC.) These were fairly common
at Kourion (= Curium) and at Marion, but rare in other parts of Cyprus.
Einar Gjerstad. The
Swedish Cyprus Expedition: Vol. IV, Part 2: The Cypro-geometric, Cypro-archaic
and Cypro-classical Periods. The
Expedition, Stockholm, 1948. This has
several sections - the drawing numbers refer to the section Pottery Types,
drawn by Bror Millberg, following p. 545, labelled Fig. I, Fig. II, .... The individual items on the page have two
numbers, one being its number on the page, e.g. 1), 2), ..., at
the upper left of the item; the other being the number within its category,
e.g. 1, 2, ..., below the item -- I will use the first. The sources of the pots drawn are in a
preliminary section with capital Roman numeral page numbers. The author is interested in pottery styles
and rarely gives dates or even periods.
There is a Relative Chronology for pottery on pp. 186-206 and an
Absolute Chronology on pp. 421‑427.
Date ranges below are deduced from these chronologies.
Pp.
52-53: White Painted II Ware (-950/-850), says that the askos, Fig. XV, 3, in
the Cyprus Museum, B. 1933, is a trick vessel.
As with several other entries, it says to see Fig. XXXVI, 9, which is
Myres 519, with a cross-sectional view showing how it works.
P.
60: Bichrome I Ware (-1050/-850), says that the askos, Fig. VIII, 24 (= Myres
518), at the Metropolitan, is a trick vase and says to see Fig. XXXVI, 9.
Pp.
62-64: Bichrome IV Ware (-600/-475), says that the askos, Fig. XXXVI, 9
(= Myres 519), at the Metropolitan, is a trick vase.
Pp.
69-71: Black-on-Red II (IV) Ware (-700/-475), says that 'the spout-jug with the
neck in the shape of a human protome', Fig. XXXIX, 15, in the BM (Brit. Mus.
Cat. Vases I:2, C 882), is a trick vase and says to see Fig. XXXVI, 9,
P.
73: Bichrome Red I (IV) Ware (-700/-600), says the spout-jug with human
protome, Fig. XLII, 5 (= Myres 931) in the Metropolitan is a trick jug
similar to the item above. He also says
that the askos, Fig. XLII, 9 in the BM (Corp. Vas. Ant., Great Britain 2, Brit,
Mus. 2, II Cc, Pl. 10.24. (Gr. Brit. 63) 'has a low foot, but apart from that
its shape is identical with the corresponding type of Black-on-Red II
(IV). It is a trick vase (cf. Fig.
XXXVI, 9)." However, the material
on Black-on-Red II (IV) is on pp. 69‑71 and it does not explicitly
say there that the askos, Fig. XXXIX, 19 in the BM (Corp. Vas. Ant., Great
Britain 2, Brit, Mus. 2, II Cc, Pl. 13.12. (Gr. Brit. 57)) is a trick vase, but
says the remaining items are similar to Bichrome IV Ware and that seems to
imply this is, like XXXVI, 9, a trick vase.
Vassos Karageorghis.
Ancient Art from Cyprus The
Cesnola Collection in the Metropolitan Museum of Art. The Museum, New York, 2000.
P. 80, #126: Trick vase in the form of an askos. = Myres 518
= Gjerstad Fig. VIII, 24. A bottom-filling
(Cadogan) pot with a goat's head from c-1050/c-950, with colour picture. He says it is Cypro-Geometric I, White
Painted I Ware and is a "trick vase".
Joseph Veach Noble.
Some trick Greek vases. Proc.
Amer. Philosophical Soc. 112:6 (Dec 1968) 371-378. He describes and illustrates a number of
types of trick vases from Athens: false
bottoms (-4C); dribble vases (which
dribble wine over the guest) (-5C); a
covered drinking cup which fills from the bottom like a Cadogan teapot (‑5C); vases with concealed contents which pour
when the host releases fingers from holes (-5C). In each case he cites museums and extended descriptions. One reference gives a list of covered cups.
Jasper Maskelyne.
White Magic. Stanley Paul,
London, nd [c1938], p. 110. Discusses
puzzle pitchers, saying "On the
site of a Greek temple at Athens, excavators a few years ago discovered a magic
pitcher which was famous in Greek legend in trials for witchcraft." He describes the standard hollow handled
pitcher.
Walter Gibson.
Secrets of the Great Magicians.
(Grosset & Dunlap, 1967);
Collins, Glasgow, 1976.
Describes several ancient devices based on siphons.
John Timbs. Things
Not Generally Known, Familiarly
Explained. A Book for Old and Young
(spine says First Series and a note by a bookdealer on the flyleaf
says 2 vol.). Kent & Co., London, (1857?), 8th ed., 1859. Hydrostatic wonders, p. 111, mentions some
classical examples as: "The magic
cup of Tantalus, which he could never drink, though the beverage rose to his
lips; the fountain in the island of Andros, which discharged wine for seven
days, and water during the rest of the year; the fountain of oile which burnt
out to welcome the return of Augustus from the Sicilian war; the empty urns
which, at the annual feast of Bacchus, filled themselves with wine, to the
astonishment of the assembled strangers; the glass tomb of Belus, which, after
being emptied by Xerxes, would never again be filled; the weeping statues of
the ancients; and the weeping virgin of modern times, whose tears were
uncourteously stopped by Peter the Great when he discovered the trick; and the
perpetual lamps of the magic temples, -- were all the obvious effects of
hydrostatical pressure. -- North
British Review, No. 5."
Banu Musa =
Banū Mūsā bin Shākir
(Sons of Moses), but largely the
work of Ahmad = Abu-l-Hasa
[the H
should have an underdot] Ahmad [the
h should have an underdot] ibn
Mūsa. Kitāb al-Hiyal
[the H
should have an underdot].
c870. Translated and annotated
by Donald R. Hill as: The
Book of Ingenious Devices; Reidel, 1979.
Describes 103 devices, most of which are trick vessels, as well as
fountains, etc. E.g. Model 1: "We wish to explain how a beaker is
made in which a quantity of wine is poured, and if [a measure] of wine or water
is added to it all its contents are discharged." This is a Tantalus cup.
Model 4: "We wish to make a
jar with an open outlet: if water is poured into it nothing issues from the
outlet, and if pouring is stopped the water issues from the outlet, and if
pouring is resumed [discharge] ceases again, and if pouring is stopped the
water discharges, and so on continuously." Models 12 - 15 have finger holes under the handle allowing the
pourer to produce various effects.
Sandfield says the earliest Chinese puzzle vessels are in
the Xian Museum and are dated to either the Song (951-960) or Northen Song
(960-1127 or 1279). (My Chinese
chronology has Song being 960-1279, with Northern Song being 960-1127 and
Southern Song being 1127-1279.)
Sandfield says they visited the Yaozhou Kiln Museum which apparently has
been making Celadon Magic Teapots ever since that time! [See his Annotated Bibliography items 6
& 7.]
Al-Jazari =
Badī‘al-Zamān Abū al-‘Izz Isma‘il ibn al-Razzāz
al-Jazarī. Kitāb fī
‘rifat al-hiyal (the h should have an underdot) al-handasiyya. c1204.
Translated and annotated by:
Donald R. Hill as: The Book of
Knowledge of Ingenious Mechanical Devices; Reidel, Dordrecht, 1974. Introduction, pp. 3‑12, and
Conclusion, pp. 279-280, include comparisons with other works. Not much of interest to us, except for the
following.
In the Historical Museum at Cologne is the 'goblet of
Albertus Magnus', which has a false bottom which allowed him to introduce
antimony and make the drink emetic. He
lived 1193-1280, so this might date from c1250. Described and illustrated in:
Edwin A. Dawes; The Great Illusionists; Chartwell Books, Secaucus, New
Jersey, 1979, p. 21.
Ashmolean Museum, Oxford, item 1921-202, presently in Room
4, in the case devoted to Decorative Techniques, item 14. Late 13C, different than the Exeter jug
below. Described, with photo and photo
of a cut-away model, in Crossley, below.
The same item is illustrated by a drawing in the following entry.
Jeremy Haslam.
Medieval Pottery in Britain.
Shire, Aylesbury, (1978); 2nd
ed., 1984. On pp. 19, under: Some
regional types; Oxford region; Thirteenth and fourteenth centuries, he
mentions, among characteristic items of the Oxford region in the late 13C,
"'puzzle‑jugs' decorated with applied scales, strips of red-firing
clay and stags' heads (13, 10)".
Fig. 13 on p. 47, Oxford, Fourteenth- and Fifteenth-Century
Vessels, item 10, is "Puzzle-jug, glazed green, with app. scales, red
strips, face masks and deer head over spout." The drawing is by the author from the above mentioned item in the
Ashmolean Museum, Oxford.
The Royal Albert Museum, Queen Street, Exeter, EX4 3RX, has
a notable example of a puzzle jug, c1300, from SW France or the Saintonge
region of west France, described as the finest piece of imported medieval
pottery in England. A postcard of it is
available from the Museum. Crossley,
below, says it is late 13C from the Saintonge and was excavated in Exeter in
1899. He gives a cross-sectional
drawing and cites a 1988 excavation report.
In fact, the puzzle aspect is quite simple -- the upper level is
connected to the base via the handle which leaves room for three levels of
decoration, showing some apparently unclothed bishops inside, then some ladies
leaning out of windows and then some musicians serenading outside! It is depicted and described in: John Allan
& Simon Timbs; Treasures of Ancient Devon; Devon Books (Devon County
Council), Tiverton, 1996, pp. 34-35.
A 16C German puzzle glass is shown and described in: Franz Sales Meyer. (Ornamentale Formenlehre; 1888.
Translated as: A Handbook of
Ornament; Batsford, 1894. Reprinted
as: Meyer's Handbook of Ornament; Omega
Books, London, 1987, pp. 351 & 353, plate 207 (on p. 351), fig. 10. "To those times also belonged:
Puzzle-glasses from which the liquor had to be sucked-out at the end of the
handle (fig. 10): ...." The figure
shows a tallish vessel apparently with a closed top through the hollow handle
runs to the bottom of the vessel and the bottom of the handle bends out to
provide a sucking point. The interior
tube is marked with dotted lines.
Cardan. De Rerum
Varietate. 1557, ??NYS. = Opera Omnia, vol. III, pp. 252. Liber XVIII. Urcei qui non se mergunt.
Apparently a vase which one fills with water and then pours wine out of.
The Percival David Foundation of Chinese Art, London, has a
late 16C Tantalus Cup (labelled Mazer Cup, item PDF A721) with a statue in the
middle.
Prévost. Clever and
Pleasant Inventions. (1584), 1998. Pp. 41-44.
A Tantalus cup.
John Bate. The
Mysteries of Nature and Art. In foure
severall parts. The first of Water
works. The second of Fire works. The third of Drawing, Washing, Limming,
Painting, and Engraving. The fourth of
sundry Experiments. (Ralph Mab [sic],
London, 1634.) The second Edition; with
many additions into every part. Ralph
Mabb, London, 1635. (3rd ed., Andrew Crooke,
London, 1654.) [BCB 20-22. Toole Stott 81-83. HPL [Bate] RBC has 2nd & 3rd eds.] The first part has several examples, notably the following.
P.
2. How to make a conceited pot, which
being filled with water, will of it selfe runne all out; but not being filled
will not run out.
P.
3. Another conceited Pot out of which
being first filled with wine and water, you may drinke pure wine apart, or
faire water apart, or els both together.
Louis L. Lipski.
Dated English Delftware: Tin-glazed Earthenware, 1600-1800. Sotheby Publications, London &
Harper & Row, Scranton, Pennsylvania, 1984. ??NYS -- cited and
quoted by Sandfield. Their earliest
upright puzzle jug is no. 1009, dated to 1653.
Edgar Gorer
& J. F. Blacker. Chinese Porcelain and Hard Stones,
Illustrated .... Quaritch, London,
1911. Vol. 1, plate 135 and preceding
page. ??NYS -- cited by Rasmussen and
quoted by Sandfield. Two illustrated
examples of puzzle jugs, i.e. 'how-to-pour puzzles', dated from the Kang-he (=
K'ang Hsi) period, 1662-1722. ??check
dates.
Ozanam. 1694. Probs. 14 & 26: 1696: 286 & 294
& figs. 137 & 138, plate 47 & fig 146, plate 49; 1708: 362-363 & 370-371 & figs. 33
& 34, plate 15 & fig. 44, plate 18.
Probs. 18 & 30: 1725: vol. 2: 389-390 & 404 & figs. 137
& 138, plate 47 & fig. 146, plate 49.
Probs. 4 & 5: 1790: vol. 4: 33-35 & figs. 5-7, plate 1; 1803: 34-36 & figs. 5-7, plate 1; 1840: 613‑614. Three forms of Tantalus cups. In 1790, it is called Tantalus and a figure
of Tantalus is put in the cup so that when the liquid approaches him, it runs
out. Some of these are such that they
pour out when the cup is tilted.
Patrick J. Donnelly.
Blanc de Chine: The Porcelain of Tehua in Fukien. London, 1969. Pp. 95, 345, Appendix 5: N86 & N69. ??NYS -- cited and quoted by Sandfield. Describes some tantalus cups, where the
siphon is concealed in a figure, dated 1675‑1725. The T'ao-ya says the figure represents Lu
Hung-chien (= Lu Hong Jian), a 16C author on porcelain whose name was Hsiang
Yuan-pien Tzu-ching and that the vessel is called a 'quiet mind' dish. Augustus the Strong (of Saxony?) had an
example before 1721. There is an
example in the Princessehof Museuem, Leeuwarden, The Netherlands. Other versions mentioned in Sandfield have
naked men or women in the cups.
The Percival David Foundation of Chinese Art, London, has a
'peach shaped wine pot', i.e. a Cadogan pot, from the Kangxi period, 1662-1722,
item PDF 826.
The Victoria and Albert Museum, London, has an example from
the K'ang Hsi period (1622‑1722).
??NYS -- cited and quoted by Sandfield.
??check dates.
Alberti. 1747. Art 31, pp. ?? (131-132) & fig. 57 on
plate XVI, opp. p. ?? (130). Vase pours
different liquors.
Tissandier.
Récréations Scientifiques.
1880?; 2nd ed., 1881, pp.
327-328 describes 'vases trompeurs', with illustrations on pp. 324-325. Says they were popular in the 18C and
earlier and the illustrations are of examples in the Musée de la Manufacture de
Sèvres. Not in 5th ed., 1888.
Tissandier. Popular
Scientific Recreations. 1890? Pp. 65-67, with illustrations on pp. 67-69,
describes "vases of Tantalus" -- cups which cannot be filled too full
-- and then gives the material from the 2nd French ed., 1881.
The Ashmolean Museum, Oxford, Room 47 has a fine example of
a Lambeth puzzle jug dated 1745.
The Rijksmuseum, Amsterdam, Room 255 has a fine example of a
Delft puzzle jug dated 1768.
Henry Pierre Fourest.
Delftware: Faience Production at Delft.
(Translated from La Faiences de Delft.)
Rizzoli, NY, 1980. P. 136, no.
130. ??NYS - cited and quoted by
Sandfield. Shows a Delft puzzle jug
from second half of 18C. In a footnote,
the author says: "This very old common form was copied from the Chinese
for the East India Company."
Sandfield says the earliest English Cadogan teapots were
made at Swindon, Yorkshire, after 1806 and they became popular after 1810.
Badcock. Philosophical
Recreations, or, Winter Amusements.
[1820]. Pp. 130-131, no. 195
& Frontispiece fig. 10: Hydraulic Experiment called Tantalus's Cup.
Rational Recreations.
1824. Experiment 66, pp.
122-123. Tantalus cup.
The Boy's Own Book.
1828: 446. Tantalus's cup. With statue of Tantalus inside.
An exhibit in the Bramah Tea & Coffee Museum, London,
says Mrs Cadogan brought the first examples, of what became Cadogan teapots, to
England about 1830, but see above.
Young Man's Book.
1839. Pp. 172-173. Tantalus's Cup.
The Secret Out.
1859.
G. J. Monson-Fitzjohn.
Drinking Vessels of Bygone Days
from the Neolithic age to the Georgian period. Herbert Jenkins, London, 1927.
On
pp. 42-43, he discusses and illustrates fuddling cups and pot crowns. He says fuddling cups, tygs and puzzle pots
dates from Elizabethan times. Says
fuddling cups have three to six cups and there is an example with six in the
British Museum, dated 1790, with the inscriptions: "My friend is he that
loves me well, but who he is I cannot tell". Says Drake, Frobisher and Raleigh used or owned examples.
A
pot crown has four cups mounted on a ring, with four tubes rising to a central
spout. A maiden wore this on her head
and the object was to drain the cups by sucking on the spout, which is not a
puzzle, but the maiden could tilt her head to prevent anyone but her chosen
suitor managing to drink.
On
p. 49-50, he describes tygs and posset pots as simple large drinking vessels.
On
p. 58, he illustrates and describes an ordinary puzzle jug. An example in the BM bears the following
verse: "Here, gentlemen, come try
your skill / I'll hold a wager if you will
/ That you don't drink this
liquor all / Without you spill or let some fall." He says there is a good example in the
Unicorn Hotel, Ripon,. and that an example, dated 1775, has the following: "God save the King I say /
God bless the King I pray / God save the King."
On
pp. 103-104, he describes and illustrates the 17C 'Milkmaid Cup' of the
Vintners' Company. When inverted, the
milkmaid's skirt is a large cup. A
smaller second cup is supported on pivots between her upraised arms. When inverted, this second cup is also
filled and the drinker must drain the larger cup without spilling any from the
swinging cup.
Robert Crossley. The
circulatory systems of puzzle jugs.
English Ceramic Circle Transactions 15:1 (1993) 73-98, front cover &
plate V. Starts with medieval examples
from late 13C -- see above. Says the
traditional jug with sucking spouts is first known from late 14C England,
though it probably derives from Italy but no early examples are known
there. Classifies puzzle jugs into 10
groups, subdivided into 20 types and into 40 variations. Some of these are only known from one
example, while others were commercially produced and many examples survive. 64 B&W illustrations, 4 colour
illustrations on the cover and plate V.
Many of the illustrations show the circulation systems by cut-away
drawings or photos of cut-away models, followed by photos of actual jugs. 72 references, some citing several sources
or items. The Glaisher collection at
the Fitzwilliam Museum, Cambridge, provided the most examples -- 13. (I wrote to Crossley in about 1998, but had
no answer and the Secretary of the English Ceramic Circle sent me a note about
a year later saying he had died.)
In
the same issue of the journal, pp. 45, 48 & plate IIb show 6 examples of
Cadogan teapots.
The Glaisher Collection of pottery in the Fitzwilliam
Museum, Trumpington Street, Cambridge, CB2 1RB; tel: 01223-332900, has the largest
number of puzzle vessels on display that I know of. I found 22 on display, including one specially made for
Glaisher. James Whitbread Lee Glaisher
(1848-1928) was a mathematician of some note at Trinity College,
Cambridge. His collection included over
3000 items. Many of the items on
display are described in Crossley's article, but Crossley mentions six items of
the Glaisher Collection which are not on display. I have prepared a list of the items on display and the further
six items.
The
collection includes several 'tygs', which are large cups with several
handles. Some of these are specifically
described as puzzle vessels. In some
other cases one cannot see if a tyg is a puzzle tyg or not, but these cases are
all included in Crossley. [Crossley,
pp. 92-93.]
The
collection includes some 'fuddling cups' which are multiple interconnected
small cups which either spill on you if you don't use them correctly or cause
you to drink several cupfuls instead of one -- when this is not expected, you get
fuddled! One example has ten cups in a
triangular array. [Crossley, pp. 91-92]
mentions these briefly. The collection
is described in the following.
Bernard Rackham.
Catalogue of the Glaisher Collection of Pottery & Porcelain in the
Fitzwilliam Museum, Cambridge, 2 vol., CUP, 1935; Reprinted: Antique Collectors' Club, Woodbridge (Suffolk),
1987. ??NYS -- found in BLC.
J. F. Blacker. Chats
on Oriental China. T. Fisher Unwin,
Ltd., London, 1908, 406 pp. See pp. 272‑273. Cadogan pots. (??NYS -- from Rasmussen.)
R. L. Hobson.
Chinese Pottery and Porcelain.
Funk & Wagnalls, New York, 1915.
Reprinted by Dover, 1976. See
pp. 276, 278. Tantalus cups, Cadogan
pots. (??NYS -- from Rasmussen.)
R. L. Hobson & A. L. Hetherington. The Art of the Chinese Potter from the Han
Dynasty to the end of the Ming. Benn,
London & Knopf, New York, 1923, 20[sic ??] pp. Reprinted as: The Art of the Chinese Potter:
An Illustrated Survey, Dover, 1982, 137 pp.
See Plates 128 & 149.
Tantalus cups. Cadogan pots. (??NYS -- from Rasmussen.)
R. L. Hobson; Bernard Rackham & William King. Chinese Ceramics in Private
Collections. Halton & Truscott
Smith, London, 1931, 201 pp. See pp.
109‑110. Tantalus cups. (??NYS -- from Rasmussen.)
D. F. Lunsingh Scheurleer.
Chinese Export Porcelain: Chine de Commande. Pitman, New York, 1974, 256 pp.
See pp. 94‑95, 215, plate 105.
Puzzle jugs. (??NYS -- from
Rasmussen.)
Miriam Godofsky. The
Cadogan pot. The Wedgewoodian (Oct
1982) 141-142. ??NYS -- cited and
quoted by Sandfield. Three
illustrations and mentions of several in museums.
Tassos N. Petris.
Samos History - Art - Folklore -
Modern Life. Toubis, Athens, 1983,
pp. 32 & 66-68] says the local potters at Mavratzaioi, Samos, were
still making puzzle jugs (maskara bardak) and tantalus cups (dikia koupa), but
"the secrets of making these vessels are now known to only a few, and it
must be regarded as a dying trade."
C. J. A. Jörg.
Interaction in Ceramics: Oriental Porcelain & Delftware. Hong Kong Museum of Art, Hong Kong, 1984,
218 pp. See pp. 78‑79 and 162‑163,
nos. 36 and 115. Puzzle jugs. (??NYS -- from Rasmussen.)
Nob Yoshigahara.
Puzzlart. Tokyo, 1992. Puzzle Collection, pp. 50-57 shows many
examples and a photo of Laurie Brokenshire, all in colour.
Lynn Pan. True to
Form: A Celebration of the Art of the Chinese Craftsman. FormAsia Books Ltd., Hong Kong, 1995, 148
pp. See p. 18. Cadogan pots. (??NYS -- from Rasmussen.)
Fang Jing Pei.
Treasures of the Chinese Scholar.
Weatherhill, New York, 1997, 165 pp.
See pp. 116‑117. Cadogan
pots. (??NYS -- from Rasmussen.)
Rik van Grol.
Puzzling China. CFF 47 (1998??)
27-29. ??NYR -- cited and quoted by
Sandfield.
Franz de Vreugd.
Oriental puzzle vessels. CFF 49
(Jun 1999) 18-20. ??NYR-- cited by
Sandfield. 12 puzzle vessels in colour.
10.X. HOW FAR DOES A PHONOGRAPH NEEDLE TRAVEL?
New section. I have just seen a recent version of this
and decided it ought to be entered.
There must be examples back to the early part of this century.
Meyer. Big Fun
Book. 1940. No. 2, pp. 173 & 755.
Young World.
c1960. P. 29. "How many grooves are there in a long
playing record? Just one long
one." [Actually there are two!]
The Diagram Group.
The Family Book of Puzzles. The
Leisure Circle Ltd., Wembley, Middlesex, 1984.
Problem 95, with Solution at the back of the book.
10.Y. DOUBLE CONE ROLLING UPHILL
W. Leybourn.
Pleasure with Profit. 1694. Tract. III, pp. 12-13: "A Mechanical
Paradox: or, A New and Diverting Experiment.
Whereby a Heavy Body shall by its own Weight move up a sloping
Ascent. Written by J. P." Nice drawing. No indication of who J. P. is and he is not one of the publishers
or the additional author.
Andrew Q. Morton.
Science in the 18th Century. The
King George III Collection. Science
Museum, London, 1993. P. 33 shows the
example in the George III Gallery and says it is c1750 and that the idea was
invented at the end of the 17C and was popular in 18C lectures on mechanics.
Henk J. M. Bos.
Descriptive Catalogue Mechanical
Instruments in the Utrecht University Museum.
Utrecht University Museum, 1968, pp. 35-37. Describes several examples, saying they are described in the
classic experimental mechanics texts of the early 18C, citing 's Gravesande,
Desaguliers, Nollet, Musschenbroek.
Item M 5 was bought in 1755.
The
Museum also has two oscillatory versions where the cone seems to roll over a
peak and down the other side, then back again, .... Item M 7 is first mentioned in an inventory of 1816.
Ozanam-Montucla.
1778. Vol. II, prob. 26: 1790: 45-46 & figs. 22-24, plate 5; 1803: 49-50 & figs. 22-24, plate 5; 1840: 216-217.
Catel.
Kunst-Cabinet. 1790. Der bergangehende Kegel, p. 12 & fig. 4
on plate I.
Bestemeier.
1801. Item 40 -- Der Berggehende
Kegel. Copies most of Catel. Diagram is copied from Catel, but less well
done.
Gardner D. Hiscox.
Mechanical Appliances Mechanical
Movements and Novelties of Construction.
A second volume to accompany his previous Mechanical Movements, Powers and
Devices. Norman W. Henley Publishing
Co, NY, (1904), 2nd ed., 1910. Item
931, p. 372, is an attempt to use this device as a perpetual motion by having
the rails diverging where it goes uphill and parallel where it goes downhill in
alternate sections. Patented in 1829!
Magician's Own Book (UK version). 1871. A double cone
ascending a slope by its own weight, pp. 136-137. The picture of the cone looks more like an octahedron!
Will Goldston.
Tricks & Illusions for
amateur and professional conjurers.
Routledge & Dutton,
9th ed (revised), nd [1920s?], pp. 28-29: The magic cone.
This is formed by slotting two discs
on radial lines and fitting them together.
When the distance between centres is
Ö2 times the radius,
then the centre of gravity at two obvious positions is at the same height and
the object rolls rather smoothly with a 'wobbly' motion. The basic result is that the centre of
gravity stays at constant height as it rolls.
Similar shapes are included here.
Gardner D. Hiscox.
Mechanical Appliances Mechanical
Movements and Novelties of Construction.
A second volume to accompany his previous Mechanical Movements, Powers
and Devices. Norman W. Henley
Publishing Co, NY, (1904), 2nd ed., 1910.
Item 355: The pantanemone, p. 144.
Take a disc and cut it along a diameter. Twist one half by 90o about the diameter perpendicular
to the cut, so the planes of the semi-circles are at right angles, looking bit like the Wobbler. Mount the pieces on an axle at 45o
to each plane -- in practice this requires some guy wires between the
pieces. He asserts that this rotates in
a wind from any direction, except perpendicular to the axle, so it can be used
as a stationary windmill, giving 60 days more work per year than conventional
windmills.
A. T. Stewart.
Two-circle roller. Amer. J.
Physics 34:2 (Feb 1966) 166-167. Shows
the basic result and gets the height of the centre of gravity for any
spacing. Shows examples made from
slotted plastic discs. I learned of
Stewart's work in 1992 from Mike Berry.
In a letter of 27 Oct 1992 to Berry, he says the problem is difficult
and little known. It took him "a
dozen pages of cumbersome trigonometry".
He offered $5 to anyone who could derive the result on one page and only
two people ever collected.
Anthea Alley.
Rocking Toy 1969. This is an example of the wobbler made from
48" diameter disks, 1" thick, with holes about 12" diameter in
the middle. This is shown, with no
text, in a photograph on p. 108 of:
Jasia Reichardt, ed.; Play Orbit [catalogue of an exhibition at the ICA,
London, and elsewhere in 1969-1970]; Studio International, 1969.
Paul Schatz.
Rhythmusforschung und Technik.
Verlag Freies Geistesleben, Stuttgart, 1975. In this he considers a number of related ideas, but his object --
the Oloid -- has the distance between centres being the radius. This does not roll smoothly on the level,
but he would roll it down a slope -- see Müller.
Paul Schatz. Swiss
patent 500,000. (??NYS -- cited in
Müller.)
Georg Müller, ed. Phänomena
-- Eine Dokumentation zur Ausstellung über Phänomene und Rätsel der Umwelt an
der Seepromenade Zürichhorn, 12 Mai - 4 November 1984. Zürcher Forum, Zürich, 1984. P. 79 shows the Oloid rolling down a slope.
David Singmaster and Frederick Flowerday. The wobbler. Eureka 50 (1990) 74-78.
Flowerday developed the idea in the 1980s, but I can't recall if he
developed it from Schatz's work. This
paper proves the basic result and discusses unsolved problems such as the path
of the centre of gravity. For the
Oloid, the distance between the two contact points is of constant length. Stewart's paper was not known when this was
written.
Christian Ucke & Hans-Joachim Schlichting. Wobbler, Torkler oder
Zwei-Scheiben-Roller. Physik in unserer
Zeit 25:3 (1994) 127-128. Short
description, mentioning Flowerday and citing Stewart and Schatz. Says they are sold in Germany and
Switzerland as Wobbler or Go-On. Shows
a version using slotted plastic discs called Rondi. Instructions on how to make one.
Christoph Engelhardt & Christian Ucke. Zwei-Scheiben-Roller. Preprint of 3 May 1994, to appear in
Mathematisch-Naturwissenschaftlicher Unterricht (1995). Basic description and derivation of the
basic result. Shows this also works for
elliptical discs! Attempts to find the
path of the centre of gravity.
Singmaster & Flowerday was not known when this was written.
This deals with determining the
probability of the various faces of a die which is not a regular
polyhedron. The immediate approach is a
simple geometric model -- the probability of a face should be proportional to
the solid angle subtended by that face viewed from the centroid. However, this fails to agree with reality
and a number of authors have attempted to explain the real situation by more
complex modellings of the physical situation.
Note that a coin or cylinder is a non-regular die, so the phenomenon of
a coin landing on edge is included here.
I have included everything I know of here as there has been considerable
interest in this problem in recent years.
Ricky Jay recently asked about the history of loaded or mis-spotted dice
and we would appreciate references on this -- it is commonly asserted that
mis-spotted dice occurred in the 11C, but I cannot relocate my source for this
and there may not have been very specific -- cf Jay, 2000.
Scott Beach.
Musicdotes. Ten Speed Press,
Berkeley, California, 1997, p. 77. Says
Jeremiah Clarke (c1674-1707), the organist of St. Paul's Cathedral and a
composer best known for the Trumpet Voluntary (properly the Prince of Denmark's
March, long credited to Purcell) became enamoured of a lady above his station
and was so despondent that he decided to commit suicide. Being somewhat indecisive, he threw a coin
to determine whether to hang himself or drown himself. It landed on the ground and stuck on
edge! Failing to recognise this clear
sign, he went home and shot himself!
(The text is given as a Gleaning: A loss of certainty, submitted by me,
in MG 66 (No. 436) (1982) 154.)
P. S. de Laplace.
Théorie Analytique des Probabilités.
3rd ed, Courcier, Paris, 1820.
P. 359 remarks that determining the probabilities for a cuboid is beyond
analytic techniques.
J. D. Roberts. A
theory of biased dice. Eureka 18 (Nov
1955) 8-11. Deals with slightly
non-cubical or slightly weighted dice (e.g. due to the varying number of
pips). He changes the lengths by ε
and ignores terms of order higher than first order. He uses the simple geometric theory.
J. R. Probert-Jones.
Letter to the Editor. Eureka 19
(Mar 1957) 17-18. Says Roberts' article
is the first to treat the problem quantitatively. States that about 1900, Weldon made two extensive trials on
dice. Weldon's work is not known to
have been published, but Pearson cited his results of 26,306 throws of six dice
in the paper in which he introduced the
χ2 test. The analysis showed the results would occur
with fair dice about once in
20,000 trials, but using
Roberts' estimate of the bias of dice, the result would occur about once in
every two trials. Edgeworth describes
Weldon's other trials, which were a bit more elaborate. Assuming fair dice, the probability of the
results was .0012, while Roberts's estimate leads to
probability .067. Both trials show that fairness of ordinary
dice is not reasonable and Roberts' estimates are reasonable.
L. E. Maistrov.
Probability Theory. A Historical
Sketch. Academic, 1974. ??NYS -- Heilbronner says he measured
ancient dice at Moscow and Leningrad, finding them quite irregular -- the worst
cases having ratios of edge lengths as great as 1.2 and 1.3.
Robert A. Gibbs, proposer;
P. Merkey & Martin Berman, independent solvers. Problem 1011 -- An old dice problem. MM 50:2 (Mar 1977) 99 &
51 (1978) 308. Editor says it is
an old problem which might be of interest to new readers. Can one can load a pair of dice so each
value has equal probability of occurring.
I.e. each of 2, 3, ..., 12 has probability 1/11. Solution notes
refer to Problem E 925, AMM 57 (1951)
191-192 and E. J. Dudewicz & R. E. Dann; Equally likely dice sums do not
exist; Amer. Statistician 26 (1972) 41-42,
both ??NYS.
Scot Morris. The
Book of Strange Facts and Useless Information.
Doubleday, 1979, p. 105.
Says 6 is the most common face to appear on an ordinary die because the
markings are indentations in the material, making the six side the lightest and
hence most likely to come up. He says
that this was first noticed by ESP researchers who initially thought it was an ESP
effect. The effect is quite small and
requires a large number of trials to be observable. (I asked Scot Morris for the source of this information -- he
couldn't recall but suspected it came from Martin Gardner. Can anyone provide the source?)
Frank Budden. Note
64.17: Throwing non-cubical dice. Math.
Gaz. 64 (No. 429) (Oct 1980) 196-198.
He had a stock of 15mm square rod and cut it to varying lengths. His students then threw these many times to
obtain experimental values for the probability of side versus end.
David Singmaster.
Theoretical probabilities for a cuboidal die. Math. Gaz. 65 (No. 433) (Oct 1981) 208-210. Gives the simple geometric approach and
compares the predictions with the experimental values obtained by Budden's
students and finds they differ widely.
Correspondence
with Frank Budden led to his applying the theory to a coin and this gives
probabilities of landing on edge of
8.1% for a UK 10p coin and 7.4%
for a US quarter. [And 9.5%
for a US nickel.]
Trevor Truran.
Playroom: The problem of the five-sided die. The Gamer 2 (Sep/Oct 1981) 16 & 4 (Jan/Feb 1982) 32. Presents Pete Fayers' question about a fair
five-sided die and responses, including mine.
This considered a square pyramid and wanted to determine the shape which
would be fair.
Eugene M. Levin.
Experiments with loaded dice.
Am. J. Physics 51:2 (1983) 149-152.
Studies loaded cubes. Seeks for
formulae using the activation energies, i.e. the energies required to roll from
a face to an adjacent face, and inserts them into an exponential. One of his formulae shows fair agreement
with experiment.
E. Heilbronner.
Crooked dice. JRM 17:3 (1984-5)
177-183. He considers cuboidal
dice. He says he could find no earlier
material on the problem in the literature.
He did extensive experiments, a la Budden. He gives two formulae for the probabilities using somewhat
physical concepts. Taking r as
the ratio of the variable length to the length of the other two edges, he
thinks the experimental data looks like a bit of the normal distribution and
tries a formulae of the form exp(-ar2). He then tries other formulae, based on the
heights of the centres of gravity, finding that if R is the ratio of the
energies required to tilt from one side to another, then exp(-aR)
gives a good fit.
Frank H. Berkshire. The
'stochastic' dynamics of coins and irregular dice. Typescript of his presentation to BAAS meeting at Strathclyde,
1985. Notes that a small change in r
near the cubical case, i.e. r =
1, gives a change about 3.4
times as great in the probabilities.
Observes that the probability of a coin landing on edge depends greatly
on how one starts it - e.g. standing it on edge and spinning it makes it much
more likely that it will end up on edge.
Says professional dice have edge
3/4" with tolerance of 1/5000 " and that the pips are filled
flush to the surface with paint of the same density as the cube. Further, the edges are true, rather than
rounded as for ordinary dice. These
carry a serial number and a casino monogram and are regularly changed. Describes various methods of making crooked
dice, citing Frank Garcia; Marked Cards and Loaded Dice; Prentice Hall, 1962,
and John Scarne; Scarne on Dice; Stackpole Books, 1974. Studies cuboidal dice, citing Budden and
Singmaster. Develops a dynamical model
based on the potential wells about each face.
This fits Budden's data reasonably well, especially for small values
of r.
But for a cylinder, it essentially reduces to the simple geometric
model. He then develops a more
complicated dynamical model which gives the probability of a 10p coin landing
on edge as about 10-8.
John Soares. Loaded
Dice. (Taylor Publishing, Dallas,
1985); Star (W. H. Allen), 1988. On p. 49, He says "Aristotle wrote a
scholarly essay on how to cheat at dice."
In an Afterword by George Joseph, pp. 243-247, on p. 246, he describes
'flats': "Certain sides of the dice are slightly larger (Flat) 1/5,000th
to 1/10,000th of an inch."
David Singmaster. On
cuboidal dice. Written in response to
the cited article by Heilbronner and submitted to JRM in 1986 but never
used. The experimental data of Budden
and Heilbronner are compared and found to agree. The geometric formula and Heilbronner's empirical formulae are
compared and it is found that Heilbronner's second formulae gives the best fit
so far.
I
had a letter in response from Heilbronner at some point, but it is buried in my
office.
Joseph B. Keller.
The probability of heads. AMM
93:3 (Mar 1986) 191-197. Considers the
dynamics of a thin coin and shows that if the initial values of velocity and
angular velocity are large, then the probability of one side approaches 1/2.
One can estimate the initial velocity from the amount of bounce -- he
finds about 8 ft/sec. Persi Diaconis examined coins with a
stroboscope to determine values of the angular velocity, getting an average
of 76π rad/sec. He considers other devices, e.g. roulette
wheels, and cites earlier dynamically based work on these lines.
Frank H. Berkshire.
The die is cast. Chaotic
dynamics for gamblers. Copy of his
OHP's for a talk, Jun 1987. Similar to
his 1985 talk.
J. M. Sharpey-Schafer.
Letter: On edge. The Guardian
(20 Jul 1989) 31. An OU course asks
students to toss a coin 100 times and verify that the distribution is about 50
: 50. He tried it 1000 times and the
coin once landed on edge.
D. Kershaw. Letter:
Spin probables. The Guardian (10 Aug
1989) 29. Responding to the previous
letter, he says the probability that a tossed coin will land on edge is zero,
but this does not mean it is impossible.
A. W. Rowe.
Letter. The Guardian (17 Aug
1989) ?? Asserts that saying the
probability of landing on edge is zero admits 'to using an over-simplified
mathematics model'.
K. Robin McLean.
Dungeons, dragons and dice. MG
74 (No. 469) (Oct 1990) 243-256. Considers
isohedral polyhedra and shows that there are 18 basic types and two infinite
sets, namely the duals of the 5 regular and 13 Archimedean solids and the sets
of prisms and antiprisms. Then notes
that unbiased dice can be made in other shapes, e.g. triangular prisms, but
that the probabilities are not obvious, citing Budden and Singmaster, and
describing how the probabilities can change with differing throwing processes.
Joe Keller, in an email of 24 Feb 1992, says Frederick
Mosteller experimented with cylinders landing on edge 'some time ago', probably
in the early 1970s. He cut up an old
broom handle and had students throw them.
He proposed the basic geometric theory.
Keller says Persi Diaconis proposed the cuboidal problem to him
c1976. Keller developed a theory based
on energy loses in rolling about edges.
Diaconis made some cuboidal die and students threw them each 1000
times. The experimental results
differed both from Diaconis' theory (presumably the geometric theory) and
Keller's theory.
Hermann Bondi. The
dropping of a cylinder. Eur. J. Phys.
14 (1993) 136-140. Considers a
cylindrical die, e.g. a coin. Considers
the process in three cases: inelastic, perfectly rough planes; smooth plane, for which an intermediate case
gives the geometric probabilities;
imperfectly elastic impacts.
In late 1996 through early 1997, there was considerable
interest in this topic on NOBNET due to James Dalgety and Dick Hess describing
the problem for a cubo-octahedron. I
gave some of the above information in reply.
Edward Taylor Pegg, Junior.
A Complete List of Fair Dice.
Thesis for MSc in Applied Mathematics, Dept. of Mathematics, Univ. of
Colorado at Colorado Springs, 1997.
40pp. He started with the
question of finding a fair five-sided die and tried to find one in the form of
a triangular prism as well as a square pyramid. He considers the simple geometric model and notes it cannot work
as some polyhedra have unstable faces!
Asserts that an isohedron is obviously fair. [A polyhedron is isohedral if any face can be mapped onto any
other by a symmetry (i.e. a rotation or a rotation and reflection) of the
polyhedron.] Notes that the method of
throwing a die affects its fairness and says this shows that isohedra are the
only fair dice. He lists all the
isohedra -- he says there are 24 of them and the two infinite families of dual
prisms (the bipyramids) and dual antiprisms (the trapezohedra), but I only see
23 of them in his lists). He explicitly
describes a fair but unsymmetrical cardboard tetrahedron, but it seems he has
to cut a circular hole in one face or weight the face. Develops an energy state model and finds
that the probability of a US nickel landing on edge is 3/10,000, but then revises his model and
gets 15/100,000,000 -- Budden's 1981 method gives 9.5%.
20 references, but he was unaware of the above literature and the
references are mostly to polyhedra and their groups. Pegg also sent a graph of Frank Budden's data versus the
mathematical model - presumably one of Pegg's models.
[A. D. [Tony] Forbes.]
Problem 171.1 - Cylinder. M500
171 (Dec 1999) 9. Asks for the shape of
cylinder such that the probability of landing on the side is 50%.
David Singmaster.
Re: Problem 171.1 -- Cylinder.
M500 173 (Apr 2000) 19. Sketch
of history of the problem.
David Singmaster.
Non-regular Dice. M500 174 (Jun
2000) 12-15. The material of this
section up through 1997, though I have since amended some of it. With a prefatory note by ADF that he had no
idea the problem had any history when he posed it.
Gordon Alabaster.
Re: Problem 171.1 -- Cylinder.
M500 174 (Jun 2000) 16-17. Gives
a simple physical argument that 50% should occur when the cross-section is
square.
Colin Davies. Re:
Problem 171.1 -- Cylinder. M500 176
(Oct 2000) 22. Differs with Alabaster's
analysis and notes the the way in which the cylinder is tossed has a major
effect.
Ricky Jay. The story
of dice. The New Yorker (11 Dec 2000)
90-95. A set of 24 loaded or mismarked
dice was found in a container, dated late 15C, in the Thames in 1984,
apparently ditched by an early Tudor gambler to avoid being caught out; these
are now in the London Museum. Two 19C
histories of gambling state that loaded dice were discovered at Pompeii or
Herculaneum, but neither gives a reference or details of the dice and they give
different sites. The earliest
discussion of false dice in English is in Roger Ascham's Toxophilus of
1545. The Mirror of Saxony, a 13C legal
compendium, specifies that users of false dice could have their hands cut off
and the makers of false dice could have their eyes put out.
10.AB. BICYCLE TRACK PROBLEMS.
There are three different problems
involved here.
First, does the front wheel wear out
more rapidly than the rear one? Why?
Second, ignoring the first point, can
you determine from bicycle tracks which one is front and which is rear?
Third, can you determine which way
the bicycle was going?
This
section was inspired by running across several modern items and recalling
Doyle's article.
I have recently read that the front wheel of a bicycle wears
out faster than the rear wheel because the front wheel travels further -- on a
curve, front wheels travel in an arc of larger radius than rear wheels, and
even in fairly straight travel, the front wheel oscillates a bit to each side
of the line of travel. In addition,
front wheels are often turned when the vehicle is at rest. However, I cannot relocate my source of
this, though I recall that the answer simply said the front wheel travels
farther with no further explanation.
Yuri B. Chernyak & Robert M. Rose. The Chicken from Minsk. BasicBooks (HarperCollins), NY, 1995. Asks why the front tires on a car wear out
faster than the rear tires and says that proper turning requires the front
wheels to turn by different amounts and that this, with some other undiscussed
reasons, leads to the front wheels being set slightly out of parallel, which
causes the extra wear. The solution
concludes: "The perfect suspension, which would turn the wheels at exactly
the proper angles, has yet to be devised."
(On the other hand, a cross-country cyclist recently told me
that his rear tire wears out faster.)
See Gerrard & Brecher in Section 6.Y for a somewhat
related problem.
From the fact that the front wheel makes a more sinuous path,
or that it is the outer track on a curve, or that the rear track goes over the
front track, or, perhaps, that it makes a shallower track, we can tell which of
the two tracks is the front wheel, but Doyle, below, does not refer to this
point.
A. Conan Doyle. The
Adventure of the Priory School. In
this, Holmes says he can tell which way a bicycle was going from its tracks on
a pathway.
A. Conan Doyle. The
truth about Sherlock Holmes. The
National Weekly (= Collier's Weekly) (29 Dec 1923). Reprinted in: The Final Adventures of Sherlock Holmes; ed. by
Peter Haining; W. H. Allen, London, 1981, pp. 27-40 in the PB ed., esp.
p. 38. See also: The Uncollected
Sherlock Holmes; ed. by Richard Lancelyn Green; Penguin, 1983, pp. 305-315,
esp. pp. 313-314, which gives a longer version of the article that appeared as
Sidelights on Sherlock Holmes; Strand Mag. (Jan 1924) and is basically a part
of a chapter in Doyle's autobiography which he was writing in 1923.
Doyle
recalls the bicycle track episode in "The Adventure of the Priory
School" and says that a number of letters objected to this, so he went out
and tried it, finding that he couldn't tell on the level, but that "on an
undulating moor the wheels make a much deeper impression uphill and a more
shallow one downhill; so Holmes was justified of his wisdom after
all."
Ruth Thomson
& Judy Hindley. Tracking & Trailing. Usborne Spy Guides, 1978, Usborne, London,
pp. 44-45. This says: "The front
wheel of a cycle makes a loopy track as the cyclist turns it from side to side
to keep his balance. As he goes faster
he turns it less, so the loops are flatter.
The narrow end of the loops point in the direction where the cyclist is
heading." When I first read this,
I thought that one could tell the direction from the fact that the loops get
flatter as the cycle goes downhill, but the track going uphill will look
similar - the cycle travels faster at the bottom then at the top. I am not convinced that 'the narrow end of
the loops' works -- see my analysis below.
Joseph D. E. Konhauser, Dan Velleman & Stan Wagon. Which Way Did the Bicycle Go? ...and Other
Intriguing Mathematical Mysteries. MAA,
Dolciani Math. Expos. 18, 1996, prob. 1, pp. 1 & 63-64. This is a careful treatment of determining
which way the bicycle was going from the geometry of the tracks in general, but
I have found there is a much simpler solution in ordinary cases.
Consider
when the bicycle is going essentially straight and begins to turn. Both wheels move off the straight route onto
curves, so the front wheel will have gone a bit further (namely the distance
between the axles) along the straight route than the rear one did, so the outer
track, which is made by the front wheel, has a short straight section at the
beginning of the turn. When the bicycle
completes its turn and both wheels are now going straight, the front wheel is
the same distance ahead, so the rear wheel makes a bit of a straight track
before meeting the track of the front wheel.
So the inner track, made by the rear wheel, has a short straight section
at the end of the turn. Knowing this,
one can tell which way the cycle was going from examination of one end of a
turn, provided the track is distinct enough.
This is a mechanism commonly used in
pan balances but if one extends part of it outward, then it exhibits the
paradoxical behaviour that the position of a weight doesn't affect the
equilibrium, apparently in violation of the law of the lever. Imagine a rectangle with pivoted
corners. Let the long edges be
horizontal and the short edges be vertical.
Attach the midpoints of the long edges to an upright, so these can
pivot. As the rectangle pivots the
short ends will remain vertical. Now
attach horizontal rods to these ends.
As the rectangle pivots, these remain horizontal. If you hang equal weights on these rods, the
whole thing balances, regardless of where the weights are positioned on these
rods.
Nouvelle maniere de Balance inventée par M. de Roberval, Professor
Royal des Mathématiques dans l'Université de Paris. Journal des Sçavans (10 Feb 1670). ??NYS - cited and described in: Henk J. M. Bos; Descriptive
Catalogue Mechanical Instruments in the
Utrecht University Museum; Utrecht University Museum, 1968, pp. 37-38.
New section. The usual question is: which is heavier -- a
pound of feathers or a pound of gold?
This has the trick answer dependent on the different pounds used to
weigh these materials. However, I have
recently seen the 1850 & 1930 items below and decided to add this section.
Fireside Amusements.
1850: No. 108, pp. 138 & 187?;
1890: No. 77, p. 114.
"Which is heaviest -- a pound of lead, or a pound of feathers?" "Both the same."
Clark. Mental
Nuts. 1897, no. 69; 1904, no. 79; 1916, no. 75. Which
weighs the most? "A pound of
feathers or a pound of gold."
Answer: "Feathers, 7000 grains; gold, 5760." (The editions vary slightly.)
Heinrich Voggenreiter.
Deutsches Spielbuch Sechster
Teil: Heimspiele. Ludwig Voggenreiter,
Potsdam, 1930. P. 108: Was ist
schwerer? 'Which is heavier, a pound of
feathers or a pound of lead.' There is
no answer or explanation. 'Schwer' has
several other related meanings, especially 'strong' and 'difficult'.
Hummerston. Fun,
Mirth & Mystery. 1924. Problem, Puzzle no. 63, pp. 148 &
182. "If one pound of potatoes
balances with a box containing one pound of gold What is the weight of the box?" Correctly notes that Avoirdupois and Troy ounces are different,
so one has to go to grains of which there are
5760 in a Troy pound (1 Troy oz
= 480 grains) and 7000 in an Avoirdupois pound (1 Avoirdupois oz =
437.5 grains). So the box must
weigh 1240 grains = 2.58333... oz = 2 oz 11 pennyweight 16
grain Troy = 2.83429... oz = 2
oz 13 dr 9 17/32 grain Avoirdupois.
New section.
Doubleday - 1.
1969. Prob. 10, Over the limit,
pp. 17 & 157-158. = Doubleday - 4,
pp. 21‑22. A man of weight
145 finds three balls, each weighing 2 and wishes to carry them across a bridge
over a ravine. But the bridge can only
carry a weight of 150. How does he do
it? Solution says to juggle the balls,
so the man is only holding at most two balls at a time. He says he was once challenged about this by
someone who claimed that air pressure and humidity invalidated the method.
David Singmaster.
Problem proposal 78.B. MG 78
(No. 481) (Mar 1994) 112. Shows that
the above is a fatal delusion, as the average force, on the juggler, of a ball
being juggled is its weight.
Many of the puzzles described here
have the common characteristic that a loop of string is entangled in some
object and the entangled string has to be worked through a number of holes in
order to remove the string or to release a ring, etc.
In 11.I, the end of the loop is
worked through holes until it can be looped around the other end of the string
which has an obstructive object on it.
Alternatively, the loop can be passed round the object containing the
holes. Which is easier depends on the
relative sizes of the two objects involved.
In 11.A, the other end of the loop
is inaccessible and the end of the loop is then passed around the object, which
is equivalent to passing it over the other end of the loop.
In 11.E, the other end of the loop
is inaccessible and the end of the loop must be partly passed over the
deformable object to allow the obstruction to pass through the deformed
object. 11.I, 11.A and 11.E are thus all
based on the reef or square knot and topologically equivalent. Some of the trick purses in 11.F use this
idea.
In 11.B, the basic process is
obscured by using people so that one does not readily see the necessary
path. 11.H are 11.H.1 are essentially
the same as this, both in their topology and their obscuring process. The wire puzzle called The United Hearts,
Cupid's Bow, etc. is isomorphic to this.
In 11.C, the basic process is
obscured by using a flexible object which is deformed to act as the loop.
In 11.F, the basic process is again
obscured, this time by the fact that the holes do not appear to be part of the
puzzle and by the fact that one does not remove the loop, but instead a ring is
released.
11.D is somewhat similar, but the
process of moving the end of the loop is quite different and the object is to
move objects along the string, so this is basically a different type of puzzle.
7.M.5 is in this general category,
but the systematic binary pattern of disentanglement makes it quite different
from the items below.
James Dalgety has shown me some
examples of Puzzle Boxes from John Jaques & Son, London, c1900, which
contain many of these puzzles. The
decorative features on these are very similar to those in Hoffmann's illustrations. It is clear that Hoffmann (or his artist)
drew from such examples (Jaques are not known to have published any
illustrations of these puzzles), so Jaques must have been producing them in the
1880s. I will note 'drawing based on
Jaques' puzzle' in the entry for Hoffmann for such puzzles.
Ozanam. 1725. Vol. IV, prob. 35, p. 437 & fig. 42,
plate 18 (error for 13) (15).
Minguet. 1733. Pp. 108-109 (1755: 76-77; 1822: 83-84 &
127-128; 1864: 72-73 & 107-108).
Somewhat similar to Ozanam.
Alberti. 1747. Art. 35, p. 209 (110) and fig. 43,
plate XII, opposite p. 212 (110). Taken
from Ozanam.
Manuel des Sorciers.
1825. Pp. 210-211, art. 25.
de Savigny.
Livre des Écoliers. 1846. P. 265: Le nœud des ciseaux.
The Sociable.
1858. Prob. 40: The scissors
entangled, pp. 298 & 316.
"This is an old but a capital puzzle." Says the ends are held in the hand, but
figure shows them tied to a post. =
Book of 500 Puzzles, 1859, prob. 40, pp. 16 & 34. = The Secret Out, 1859, pp. 238‑239:
The Disentangled Scissors, but says the ends 'are held by the hand or tied
firmly to a post ...', and with a diagram for the solution. See Magician's Own Book (UK version) for a
clearer version. = Wehman, New Book of
200 Puzzles, 1908, p. 44.
Indoor & Outdoor.
c1859. Part II, p. 129, prob. 9:
The scissors entangled. Almost
identical to The Sociable, but the figure omits the post and the problem
statement starts with 56. -- apparently the problem number in the source
from which this was taken.
Magician's Own Book (UK version). 1871. The liberated
prisoner, pp. 211-212. Shows a prisoner
chained in this manner, but the diagram is too small to really see what is
going on. Then says it is equivalent to
the scissors problem, which is clearly drawn and much bigger than in The
Sociable. The explanation is clearer
than in The Sociable.
Tissandier.
Récréations Scientifiques.
1880? 2nd ed., 1881, gives a
brief unlabelled description on pp. 330-331, with figure copied from Ozanam on
p. 328.
5th
ed., 1888, La cordelette et les ciseaux, p. 259. Based on Ozanam, copying the diagram.
The
index of the English ed. has a reference to this, but the relevant pages 775‑776
have become the title for the Supplement!
This is included on p. 84 of Marvels of Invention -- cf Tissandier in
Common References.
Anonymous. Social
Entertainer and Tricks (thus on spine, but running title inside is New Book of
Tricks). Apparently a compilation with
advertisements for Johnson Smith (Detroit, Michigan) products, c1890? P. 38b: The scissors trick.
Hoffmann. 1893. Chap. X, no. 46: The entangled scissors, pp.
355 & 393 = Hoffmann‑Hordern, p. 254.
A. Murray. Tricks
with string. The Boy's Own Paper 17 or
18?? (1894??) 526-527. Well drawn.
Devant. Tricks for
Everyone. Op. cit. in 4.A.1. 1910.
The scissors trick, pp. 35-36.
Simple version.
M. Adams. Indoor
Games. 1912. The tailor's scissors, pp. 28‑30.
Williams. Home
Entertainments. 1914. The entangled scissors, pp. 111-112.
Hummerston. Fun,
Mirth & Mystery. 1924. The entangled scissors, p. 127.
Collins. Book of
Puzzles. 1927. The dressmaker's puzzle, pp. 21-22.
J. F. Orrin. Easy
Magic for Evening Parties. Op. cit. in
7.Q.2. 1930s?? The scissors puzzle, pp. 36-37.
11.B. TWO PEOPLE JOINED BY ROPES AT WRISTS
This is isomorphic to 11.K.8. See von Hartwig, Goldston and Svengarro for
one person versions.
Ozanam. 1725. Vol. IV, prob. 38, p. 438 & fig. 45,
plate 18 (error for 13) (15).
Minguet. 1733. Pp. 110-111 (1755: 77-78; 1822: 129-130;
1864: 108-109). Similar to Ozanam.
Alberti. 1747. Art. 38, p. 212 (111) and fig. 46, plate
XII, opposite p. 212 (110). Taken from
Ozanam.
Family Friend 2 (1850) 267 & 353. Practical Puzzle -- No. IX. = Illustrated Boy's Own Treasury, 1860,
Practical Puzzles, No. 37, pp. 402 & 442.
Parlour Pastime, 1857.
= Indoor & Outdoor, c1859, Part 1.
= Parlour Pastimes, 1868.
Mechanical puzzles, no. 10, pp. 180-181 (1868: 191-192).
Magician's Own Book.
1857. No. 11: The handcuffs, p.
11. = The Secret Out, 1859, pp. 248‑250,
but with a few changes of words, a diagram for the solution and more elegant
drawing. = Boy's Own Conjuring Book,
1860, p. 23.
Leske. Illustriertes
Spielbuch für Mädchen. 1864? Prob. 177, p. 96: Die verschlungenen
Schnüre.
Anonymous. Every Little
Boy's Book A Complete Cyclopædia of in
and outdoor games with and without toys, domestic pets, conjuring, shows,
riddles, etc. With two hundred and
fifty illustrations. Routledge, London,
nd. HPL gives c1850, but the text is
identical to Every Boy's Book, whose first edition was 1856, and which has not
yet been entered. In 4.A.1, I've
guessed this book may be c1868. Pp.
360-361: The handcuffs.
Magician's Own Book (UK version). 1871. The prisoner's
release, pp. 209-211. Adds that one can
also intertwine the two cords in the form of a square or reef knot which allows
a simpler disentanglement.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 4, no. 13:
The handcuffs, p. 97.
Cassell's.
1881. P. 94: The prisoners'
release puzzle. = Manson, 1911, 143-144
Tissandier.
Récréations Scientifiques.
1880? 2nd ed., 1881, brief
unlabelled description on p. 330 with figure copied from Ozanam on p.
328.
5th
ed., 1888, Les deux prisonniers, pp. 257-258.
Based on Ozanam, copying the diagram.
The
index of the English ed. has a reference to this, but the relevant pages 775‑776
have become the title for the Supplement!
This is included on p. 84 of Marvels of Invention -- cf Tissandier in
Common References.
Anonymous. Social Entertainer
and Tricks (thus on spine, but running title inside is New Book of
Tricks). Apparently a compilation with
advertisements for Johnson Smith (Detroit, Michigan) products, c1890? P. 14a: Slipping the bonds.
Richard von Hartwig.
UK Patent 3859 -- A New Game or Puzzle.
Applied: 27 Feb 1892; accepted:
2 Apr 1892. 1p + 1p
diagrams. One man, the other loop being
tied to a tree.
Hoffmann. 1893. Chap. X, no. 36: Silken fetters, pp. 349‑350
& 390 = Hoffmann-Hordern, pp. 246-247.
A. Murray. Tricks
with string. The Boy's Own Paper 17 or
18?? (1894??) 526-527.
M. Adams. Indoor
Games. 1912. Release the prisoners, pp. 28‑29.
Will Goldston. The
Young Conjuror. [1912 -- BMC]; 2nd ed., Will Goldston Ltd, London, nd [1919
-- NUC], Vol. 1, pp. 34-39: Three Malay
rope tricks. No. two is the
present section, with one person having his wrists tied and a string looped
around and held by another person.
Goldston thanks E. R. Bartrum for the text and illustrations.
Prof. Svengarro.
Book of Tricks and Magic. I.
& M. Ottenheimer, Baltimore, 1913.
Rope trick, p. 15. As in
Goldston, with wrists tied by a handkerchief and then a rope looped around it.
Williams. Home
Entertainments. 1914. The looped chains, pp. 109-110. Jailer tries to secure prisoner by this
method.
J. F. Orrin. Easy
Magic for Evening Parties. Op. cit. in
7.Q.2. 1930s?? The magic release (no. 1), pp. 26-27.
McKay. Party
Night. 1940. How did it get there?, p. 150.
This is an alternate method which gives a 'knot' between the two strings. It is most easily described from the undone
state and the second loop is most easily visualised as a ring. Form a bight in the string and pass it
through the ring, then pass it under the loop around one wrist, over the hand
and back under the loop. This leaves
the bight around the wrist below the loop.
Now just lift it off the hand and the string will be knotted to the
ring.
11.C. TWO BALLS ON STRING THROUGH LEATHER HOLE AND STRAP = CHERRIES PUZZLE
The basic version has a leather
strap with two longish cuts allowing the central part to flex away from the
rest of the strap. There is a hole or
two holes at the bottom of the strap. A
string with balls at each end comes through the hole(s) and around the central
part. The balls are larger than the
hole(s), but the central part can be brought through the hole(s) to form a loop
big enough to pass a ball through. (I
have just noted that two hole versions occur, but I haven't checked all
versions.)
The balls were often called cherries
and even drawn as such. I wonder if the
puzzle originally used a pair of joined-together cherries??
An equivalent version has a slit in
a card (sometimes tubular), producing a thin part on which hangs the string
with two balls with a ring or cylinder about the double string. A variation of this has a doubled paper or
leather object such as a pair of boots attached at the tops, with a ring or
just a paper loop or annulus around it.
The key to these versions is folding the card so the thin bit can be
brought through the ring, cylinder or loop.
A version with names like Key, Heart
and Arrow has a card heart with slits and a card arrow as the doubled object
and the key acting as the ring.
SEE: Girl's Own Book; Secret
Out; Magician's Own Book (UK);
Pacioli. De
Viribus. c1500.
Schwenter.
1636. Part 10, exercise 30, p.
411. Version using a card.
Witgeest. Het
Natuurlyk Tover-Boek. 1686.
Prob. 18, pp.
160-161. Elaborate card version.
Prob. 19, pp.
162-163. Cherries, with two holes.
Ozanam. 1725.
Minguet. 1733. Pp. 112-113 (1755: 79; 1822: 131-132; 1864:
110-111). Cherries version. Similar to Ozanam, prob. 33. Two holes and one hole are both shown.
Alberti. 1747. Loc. cit. in 11.A.
Catel.
Kunst-Cabinet. 1790. Die verbundenen Kirschen, pp. 13-14 &
fig. 18 on plate I.
Bestelmeier.
1801. Item 273: Der
verschlungenen Kirschen. Copies part of
Catel's text.
Manuel des Sorciers.
1825. Pp. 182-183: Le jeu des
cerises.
The Boy's Own Book.
Child. Girl's Own
Book. Heart, dart, and key. 1833: 138-139; 1839: 122-123;
1842: 203‑204. A
variation of the card version with the key as the ring. Cf: Secret Out: Magician's Own Book (UK).
Nuts to Crack II (1833), no. 92. The card puzzle. Almost
identical to Boy's Own Book.
Crambrook.
1843. P. 4, no. 16: Cherry Cheat
Puzzle. Check??
Family Friend 2 (1850) 208 & 239. Practical Puzzle, No. VII. Repeated as Puzzle 10 -- The button puzzle
in (1855) 339 with solution in (1856) 28 .
= The Illustrated Boy's Own Treasury, 1860, Practical Puzzles, No. 43,
pp. 403 & 442. Identical to Magician's
Own Book, prob. 11. Cherries puzzle
using buttons.
Magician's Own Book.
1857.
Book of 500 Puzzles.
1859.
The Secret Out.
1859. Key, Heart, and Arrow, pp.
390-391. As in Girl's Own Book, but
with much better pictures and clearer text.
Cf Magician's Own Book (UK).
Boy's Own Conjuring Book.
1860.
Magician's Own Book (UK version). 1871.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 10:
The button puzzle, pp. 29 & 31.
Hanky Panky.
1872. The undetachable cylinder,
pp. 125‑126. Card version.
Martin Appleton Wright.
UK Patent 7002 -- Improved Advertisement Cards. Dated: 28 Apr 1884 and 30 Apr 1884. 1p + 2pp diagrams. Card versions.
Gaston Tissandier.
Jeux et Jouets du jeune age
Choix de récréations amusantes
& instructives. Ill. by Albert
Tissandier. G. Masson, Paris, nd
[c1890]. No. 10-11: Le problème des
cerises, pp. 40-41, with elegant coloured plate.
William Crompton.
The odd half-hour. The Boy's Own
Paper 13 (No. 657) (15 Aug 1891) 731-732.
The slippery buttons.
Handy Book for Boys and Girls. Op. cit. in 6.F.3.
1892. Pp. 57-59: The boot
puzzle. Card version with a pair of
boots.
Tom Tit, vol. 3.
1893. Le jeu de la fève, pp.
225-226. = K, no. 171: The bean trick,
pp. 394‑395. Card version,
using a bean pod to make all the parts.
Hoffmann. 1893. Chap. II, no. 24: The ball and three
strings, pp. 34‑35 & 59‑60 = Hoffmann-Hordern, pp.
38-39, with photos. This is a more
complex puzzle, but based on the same principle. A string goes around other strings, through a ball and then has
ends separately knotted, so you have to bring the other strings through the
ball in order to release the string.
Drawing based on Jaques' puzzle.
Photos on p. 39 show The Three-String Ball Puzzle, with instructions, by
Jaques & Son, in its original state and with the strings through the ball,
1870-1900.
Benson. 1904. The ball and strings puzzle, p. 217. As in Hoffmann.
Anon. Triangular
string puzzle. Hobbies 31 (No. 793) (24
Dec 1910) 302-303. Same as Hoffmann,
with the strings inside a triangular prism
Williams. Home
Entertainments. 1914. String and button puzzle, p. 112. Cherries puzzle using buttons.
Hummerston. Fun,
Mirth & Mystery. 1924. The buttons puzzle, Puzzle no. 73, pp. 160
& 183.
Collins. Book of
Puzzles. 1927. The bachelor's button puzzle, pp. 22-23.
A. B. Nordmann. One
Hundred More Parlour Tricks and Problems.
Wells, Gardner, Darton & Co., London, nd [1927 -- BMC]. No. 88: The button trick, pp. 80-82. This is a card version, but with a string
and buttons. The card has two long parallel
cuts, as in the leather version, but it then has two slots off to the side that
the string goes through. One has to
fold the card to superimpose these two slots, then fold it again to bring the
thin strip to these slots.
See S&B, p. 114. Slocum and Gebhardt have pointed out that
there are two approaches to this problem, particularly in the two ball case,
depending on how the ends of the string are attached to the board. If the string is passed through a hole and a
knot is tied in the string, rather than tying the string to the board, then one
can partially undo the central loop by passing it through the end hole and
around the end of the string -- repeating with a slight change completely
undoes the central loop. This approach
makes the problem really a form of 11.I.
The only example in print that I have seen is in Hobbies, 1910. However, I think that in most cases with a
knot in the string, the indicated size of the hole in the board is too small to
permit a loop to pass through and this method is not possible.
I have recently acquired an example
of the Waterloo Puzzle, produced by Jaques, c1900, and seen other examples at
James Dalgety's. This has seven holes
in a bone plank. There are four small
holes at the corners and three larger holes on the midline, say A, B, C in
order. Two strings run lengthwise
between pairs of end holes, but one dips into the centre hole B. The small holes seem too small to admit
another string, and the dip makes a very small loop on the other side. (The other string has some slack in it and
probably could also be brought through to make a double dip.) A loop starts around both upper strings,
goes down through hole A, through a ring, through the small dip (which is
barely big enough for two strings), through another ring and up through hole C
and around both strings. I cannot
decide if the string has been broken and then erroneously restrung. Dalgety has provided a copy of the
instructions which say the object is to remove the rings. The instructions also state that the loop has
to be passed through each of the end holes to pass around the ends of the
strings -- at present, there is not enough slack to do this and it would be
difficult to get that much string into the small holes.
Pacioli. De
Viribus. c1500. Ff. 206v-207r, Part 2, Capitulo. CI.
Do(cumento) un altro filo pur in .3. fori in la stecca con unambra. per sacca
far le andare' tutte in una (Another string also through three holes in the
stick with one bead per loop, make them go onto one (loop)). = Peirani 283‑284. The problem titles vary between the actual
problem and the Table of Contents and the latter shows that 'unambra' should be
'una ambra' -- Peirani has given it as un'ambra. Sacca means pocket or bay or inlet and it seems clear he means a
loop which has that sort of shape.
Ambra is amber, but seems to mean an amber bead here. Sadly there is no picture though the text
refers to one.
Schwenter.
1636. Part 10, exercise 27,
pp. 408-410. With two rings.
Witgeest. Het
Natuurlyk Tover-Boek. 1686. Prob. 43, pp. 33-34. Clearly taken from Schwenter.
Ozanam. 1725. Vol. IV, prob. 40, pp. 439‑440 &
fig. 47, plate 14 (16). Le Sigillum
Salomonis, ou Sceau de Salomon -- version with four rings.
Alberti. 1747. Art. 40, pp. 214 (112) and fig. 48, plate
XIII, opposite p. 214 (111). Il
Sigillum Salomonis, o Sigillo Salomone -- version with 4 rings. Taken from Ozanam.
Catel.
Kunst-Cabinet. 1790. Die Salomonsringe, pp. 14-15 & fig. 24
on plate I. Version with 4 rings. Describes how to solve it.
Bestelmeier.
1801. Item 214: Die Salomons‑Ringe. Version with 3 rings. Brief text.
Boy's Own Book. 1843
(Paris): 438 & 442, no. 17: The bead puzzle. "This puzzle may be procured at many toy-shops." = Boy's Treasury, 1844, pp. 426 & 430,
no. 14.
Family Friend 3 (1850) 30 & 61. Practical puzzle -- No. XI. Love's Puzzle with two hearts. = The Sociable, 1858, Prob. 24: Love's
Puzzle, pp. 294 & 310. = Book of
500 Puzzles, 1859, prob. 24, pp. 12 & 28.
Magician's Own Book. 1857. Prob. 37: The string
and balls puzzle, pp. 277-278 & 301.
Two balls. = Boy's Own Conjuring
Book, 1860, prob. 36, pp. 240-241 & 265.
= Wehman, New Book of 200 Puzzles, 1908, p. 9.
Book of 500 Puzzles.
1859.
The Illustrated Boy's Own Treasury. 1860.
Practical Puzzles, No. 5: Love's puzzle, pp. 396 & 436. Identical with Family Friend.
Magician's Own Book (UK version). 1871. The puzzle of
Cupid, p. 227. Two hearts. Diagram is hard to make out.
Elliott. Within‑Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 11:
The string and balls, pp. 29 & 31.
Cassell's.
1881. P. 90: The string and
balls puzzle. = Manson, 1911, p.
147. Version with 2 balls.
Hoffmann. 1893. Chap. II, no. 13: The two balls, pp. 27‑28
& 53‑54 = Hoffmann-Hordern, p. 28, with photo. Photo on p. 28 shows an ivory example,
1850-1900. Hordern Collection,
p. 22, shows a different example, apparently in ivory, 1870-1900.
Burnett Fallow. An
ingenious bead puzzle. The Boy's Own
Paper 15 (No. 755) (1 Jul 1893) 638.
Shows two loop version but notes it can be extended.
Anon. Hexagonal wood
puzzle. Hobbies 31 (No. 791) (10 Dec 1910)
261-262 & 279. Two loop version,
but asks to undo the central loop from the wood, as discussed in the
introduction to this section.
Collins. Book of
Puzzles. 1927. The string and ball puzzle, pp. 23-24.
James Dalgety. Email
of 3 Sep 1999. Reports that his father
saw the puzzle in use by Inuits in the Canadian Arctic or Greenland in c1930,
but his family lost the walrus ivory (or bone) and leather examples that his
father brought back. Also that a collection
of topological puzzles from near Lake Tanganyika, gathered in the 1920s and now
in the Horniman Museum, London, does not contain an example of Solomon's
Seal.
R. P. Lelong.
Casse-tête guerzé. Notes
Africaines 22 (Apr 1944) 1. ??NYS --
cited and described by Béart. Says M.
Gienger found the variant with an extra ring encircling both loops in the
forest of the Ivory Coast in 1940, named
kpala kpala powa [body of a toucan]
or kpa kpa powa [body of a
parrot].
Paul Niewenglowski.
Bulletin de l'IFAN [Institut Français d'Afrique Noire] 14:1 (Jan
1952). ??NYS -- cited and described by
Béart. Describes his invention of an
interesting, rather simpler, variant as a result of seeing a standard version
from Béart.
Charles Béart. Jeux
et jouets de l'ouest africain. Tome
I. Mémoires de l'Institut Français
d'Afrique Noire, No. 42. IFAN, Dakar,
Senegal, 1955. Pp. 413-418 discusses
and carefully illustrates several versions.
The standard version, but with several beads on one loop, is called pèn
and is common in the forests of Guinea and Ivory Coast. Describes variants of Gienger/Lelong and
Niewenglowski.
Ch'ung‑En Yü.
Ingenious Ring Puzzle Book.
1958. Op. cit. in 7.M.1. P. 28 shows a version, called Double Coin
Ring Puzzle.
Fred Grunfield.
Games of the World. Ballantine,
NY, 1975. On p. 267, he calls this
"African String Game", but gives no reference. Probably based on Béart.
Pieter van Delft & Jack Botermans. Creative Puzzles of the World. Abrams, New York, 1978. African ball puzzles. "It was once used in magic rites by
tribes living in the jungles of the Ivory Coast. The puzzle is still used for amusement in this part of Africa,
not only by the people who inhabit the remote outlying areas but also by city
dwellers. ... The puzzles were not restricted to this part of Africa. Variations may be found in Guinea, and some
... were made in China." No
reference given, but I suspect it must come from Béart, although this is not
listed in their bibliography. My thanks
to Mark Peters for the reference to van Delft and Botermans.
See S&B, p. 114. I have seen it claimed that the phrase 'to
buttonhole someone' derives from the use of this.
Devant. Tricks for
Everyone. Op. cit. in 4.A.1. 1910.
The flexible pencil, pp. 13-15.
No history.
Will Goldston. More
Tricks and Puzzles without Mechanical Apparatus. Op. cit. in 6.AK.
1910?. The pencil, loop and
buttonhole, pp. 69‑71.
W. P. Eaton. Loc.
cit. in 1. 1911. Gives Loyd's narration of the invention of
this for John A. McCall, President of the New York Life Insurance Co.
A. C. White. Sam
Loyd and His Chess Problems. 1913. Op. cit. in 1. P. 103. Quotes from
Eaton.
A. B. Nordmann. One
Hundred More Parlour Tricks and Problems.
Wells, Gardner, Darton & Co., London, nd [1927 -- BMC]. No. 87: Latch key trick, pp. 79-80 &
111. This is the only version I have
seen using something other than a pencil.
It has the advantage that a key has a loop at the end to tie the loop of
string to, but the buttonhole will have to be large!
Rohrbough. Puzzle
Craft. 1932. The Buttonholer, p. 4 (= p. 4 of 1940s?).
Abraham. 1933. Prob. 165 -- Pencil and buttonhole, pp. 77
(49).
J. F. Orrin. Easy
Magic for Evening Parties. Op. cit. in
7.Q.2. 1930s?? Looping the loop, pp. 34-36. No mention of Loyd.
Slocum. Compendium. Shows Magic Coat Pencil from Johnson Smith
1937 catalogue.
Depew. Cokesbury
Game Book. 1939. Lapel needle, p. 167. No mention of Loyd.
"Willane".
Willane's Wizardry. Academy of
Recorded Crafts, Arts and Sciences, Croydon, 1947. The looped pencil, pp. 10-11.
Gardner. SA (Nov
1971) = Wheels, Chap. 12.
This could be
combined into 11.I.
Catel.
Kunst-Cabinet. 1790. Das einfache Ringspiel, p. 13 & fig. 39
on plate II.
Bestelmeier.
1801. Item 199: Das einfache
Ringspiel. Copies part of Catel's text.
The Boy's Own Book.
The scale and ring puzzle. 1828:
424‑425; 1828‑2: 429; 1829 (US): 220-221; 1843 (Paris): 437 & 442, no. 13; 1855: 575;
1868: 677-678. In the 1843
(Paris), the engraved heading of the section, p. 435, shows the puzzle. Boy's Treasury, 1844, omits the puzzle, but
copies this engraving on p. 424. = de
Savigny, 1846, pp. 354 & 358, no. 10: Le problème de la balance et de
l'anneau.
Nuts to Crack III (1834), no. 82. The scale and ring puzzle.
Almost identical to Boy's Own Book.
Crambrook.
1843. P. 4, no. 8: Imperial
Scale. Check??
See Bogesen, 6.W.2, for an actual example, mid 19C.
Magician's Own Book.
1857. Prob. 15: The scale and
ring puzzle, pp. 270-271 & 295. Identical
to Boy's Own Book, but with an illogical break between puzzle and
solution. = Book of 500 Puzzles,
1859, pp. 84-85 & 109. = Boy's Own
Conjuring Book, 1860, prob. 14, pp. 232‑233 & 258. = Magician's Own Book (UK version), 1871,
pp. 229‑230. c= Wehman, New
Book of 200 Puzzles, 1908, p. 23, which omits the solution, which is not really
needed.
F. Chasemore. Loc.
cit. in 6.W.5. 1891. Item 1: The balance puzzle, p. 571.
Hoffmann. 1893. Chap. II, no. 19: The imperial scale, pp. 31‑32
& 56‑57 = Hoffmann‑Hordern, pp. 32-33, with photo. Drawing based on Jaques' puzzle. Photo on p. 32 shows The Imperial Scale
Puzzle, with instructions, by M.M.M.M., London, in the "St. Dutius"
Series of Novelties, 1870-1900. Hordern
Collection, p. 27, shows another version without instructions.
William Hollins. UK
Patent 21,097 -- An Improved Puzzle.
Applied: 23 Sep 1896;
accepted: 5 Dec 1896. 1p +
1p diagrams. Identical to all the
above!!
van Etten.
1624. Prob. 60 (55), p. 55 &
fig. opp. p. 33 (p. 80).
Schwenter.
1636. Part 15, exercise 18, p,
545. Almost identical to van Etten.
Witgeest. Het
Natuurlyk Tover-Boek. 1686. Prob. 33, pp. 25-26. Similar to van Etten, but with a more
decorative purse.
Ozanam. 1725. Vol. IV.
Alberti. 1747.
Catel.
Kunst-Cabinet. 1790.
Bestelmeier.
1801. Item 387: Ein
Zauberbeutel. Copies Catel's
Jägertasche. Like Ozanam, prob. 42.
Child. Girl's Own
Book. 1833: 219-220; 1839: 198-199; 1842: 316. The miser's
purse, apparently a trick stitch as in Ozanam, prob. 39 ??
Crambrook.
1843. P. 3. The Miser's Purse; The Ring Purse. These
might be trick items??
The Sociable.
1858. Prob. 15: Puzzle purse,
pp. 291 & 307. Like Ozanam, 1725,
prob. 52 but with a second similar locking mechanism. = Book of 500 Puzzles, 1859, prob. 15, pp. 9 & 25. = The
Secret Out, 1859, pp. 372-373: The Magic Purse.
The Illustrated Boy's Own Treasury. 1860.
Practical Puzzles, No. 44: Puzzle purse, pp. 403 & 442‑443. Identical to The Sociable.
Peck & Snyder.
1886.
Hoffmann. 1893. Chap. II, no. 42: Puzzle purse, pp. 43 &
68 = Hoffmann-Hordern, p. 54. This
is Ozanam's prob. 32 and finally made it clear to me how it worked.
Burnett Fallow. How
to make the "B.O.P." puzzle purse.
The Boy's Own Paper 16 (No. 771) (21 Oct 1893) 43-45. As in Ozanam's no. 32.
Jaques puzzle boxes of c1900 included The Secret Purse,
based on the trick stitching of Ozanam prob. 39.
Will Goldston.
Tricks & Illusions for
amateur and professional conjurers.
Routledge & Dutton,
9th ed (revised), nd [1920s?], pp. 136-137. Uses the trick stitching of Ozanam prob. 39 to make a sack large
enough to hold a person, who easily escapes from it behind a screen and then
tightens up the stitching before coming out.
Davenport's catalogue, op. cit. in 10.T, c1940, p. 3. Tantalizing pants. They are based on a trick stitch like Ozanam prob. 39. Described as the very latest novelty. Remnants of this stock, made in Japan, are
still available from Davenport's as Puzzle pants.
In 1996, Harriet Hall gave me an example of Ozanam's prob.
32, made from deerskin, labelled:
Genuine Indian Handicraft Native
Renaissance II Deseronto, Ontario Made in Canada. She obtained it from Puzzletts in Seattle.
11.H. REMOVING WAISTCOAT WITHOUT REMOVING COAT
Philip Breslaw (attrib.). Breslaw's Last Legacy. 1784?
Op. cit. in 6.AF. 1795: 128-129:
'To pull off any Person's Shirt, without undressing him, or having Occasion for
a Confederate.'
Manuel des Sorciers.
1825. Pp. 132-133, art. 9:
Enlever la chemise à quelqu'un sans le déshabiller.
The Boy's Own Book.
1828: 362. To pull off a
person's shirt without undressing him.
Magician's Own Book (UK version). 1871. To take a man's
vest off without removing his coat, pp. 239-240. The fact that this English book uses the American term 'vest'
makes me suspect this is taken from a US source.
Cassell's.
1881. P. 95: To take a man's
waistcoat off without removing his coat.
= Manson, 1911, pp. 148-149: Waistcoat puzzle.
Anonymous. Social
Entertainer and Tricks (thus on spine, but running title inside is New Book of
Tricks). Apparently a compilation with
advertisements for Johnson Smith (Detroit, Michigan) products, c1890? P. 78: How to remove a man's shirt without
taking off his coat or vest.
H. D. Northrop.
Popular Pastimes. 1901. No. 20: The vest puzzle, pp. 69 &
74-75. "How can you take a man's
vest off with out removing his coat?"
Solution is = Cassell's.
Devant. Tricks for
Everyone. Op. cit. in 4.A.1. 1910.
The waistcoat trick, pp. 106-109.
Hugh Shaw. The
Helping Hand: How to take a man's waistcoat off without removing his coat. Hobbies 32 (No. 821) (8 Jul 1911) 303.
Games and Tricks -- to make the Party "Go". Supplement to "Pearson's Weekly",
Nov. 7th, no year indicated [1930s??].
Removing a waistcoat without the coat, p. 7.
Foulsham's New Party Book.
Foulsham, London, nd [1950s?].
P. 49: Removing a boy's waistcoat without taking off his coat.
Alfred C. Gilbert.
Gilbert Knots and Splices with
Rope-Tying Tricks Explains methods of
knot tying and reveals rope tricks made famous by great artists. A. C. Gilbert Co, New Haven, Connecticut,
1920, pp. 79-82. Turning vest inside
out without removing coat and with the wrists securely tied.
Gardner.
MM&M. 1956.
11.H.1. REMOVING LOOP FROM ARM
Family Friend 3 (1850) 341 & 351. Practical puzzle. No. XXII. Loop around arm
with hand in waist coat pocket. Remove
the loop without moving the hand. Posed
in verse. = Magician's Own Book,
1857, prob. 4: The endless string, pp. 267 & 292. = Book of 500 Puzzles, 1859, pp. 81 & 106. = Boy's Own Conjuring Book, 1860, prob. 5,
pp. 229 & 255, but this has a different picture. = Illustrated Boy's Own Treasury, 1860, no. 20, pp. 399 &
438, with slightly abbreviated answer.
Magician's Own Book (UK version). 1871. The string without
an end, pp. 204-205. Only shows a loop
of string and the problem is a bit unclearly set. Solution is similar to Family Friend.
J. F. Orrin. Easy
Magic for Evening Parties. Op. cit. in
7.Q.2. 1930s?? Another magic release (no. 4), pp.
30-31.
McKay. Party
Night. 1940. Removing the string, pp. 147-148. As in Family Friend.
Notes that many people will fail to do it because they put their hand in
their trousers pocket!
Gardner.
MM&M. 1956. The puzzling loop, p. 86 & fig. 41 on p.
90. Coat off, loop around arm with hand
in waistcoat pocket. Loop must be big
enough to pass round his chest.
11.I. HEART AND BALL PUZZLE AND OTHER LOOP PUZZLES
The ALLIANCE or VICTORIA puzzle has
two boards, each with two holes, and a length of string with a largish loop at
each end. To disentangle, a loop has to
be worked along the string and taken around the entire board at the other end
(or, equivalently, around the first board).
One can also use a large single loop of string. There are also versions with three holes in
each board. One can also have one board
with the free ends of the string tied to something else. See most entries below.
The HEART AND BALL puzzle has a
heart-shaped piece with several holes in it and a loop of string which goes
through the holes and then loops around itself. The other end of the loop has a ball on both strings of the loop
and a large knot to prevent it coming free.
One has to work the first end of the loop along the string until it
comes out along the free end of the loop when it can be passed around the free
end and worked back to release the entire string and hence the ball. A simple version of this has just three
holes in a straight line on a board and is often sold with the Alliance puzzle.
SEE: Boy's Own Book; Nuts to
Crack; Crambrook; Family Friend, 1850; Illustrated Boy's Own Treasury; Magician's Own Book (UK); Hoffmann;
Benson; Slocum; Collins;
The BOARD AND BALL puzzle has two
holes in a board. A string goes through
both holes, crosses itself and then each end comes back through its hole and is
terminated with a bead and a knot. A
largish bead is located on one of the loops through the holes. To remove, this loop is put through the
other hole and passed over the end and brought back, which allows everything to
come apart. The alternate approach to
the Solomon's Seal described at 11.D is a variant of this approach.
SEE: Family Friend, 1858;
Secret Out; Illustrated Boy's
Own Treasury; Magician's Own Book (UK); Anon: Social ...;
The CHINESE LADDER has several crossbars
with holes and a long string winding through the holes and through intermediate
discs, with a needle attached to the end of the string. One threads the needle back through all the
holes, keeping the end of the loop from pulling through, then threads back
through all the ladder holes, avoiding all the disc holes and again keeping the
loop from pulling through. When
completed, pulling the end causes all the discs to fall off, but the string is
still in its original place on the ladder.
Making another pass through the discs but not the holes has the effect
that when you pull the string free, all the discs are on it.
SEE: Adams & Co.;
Hoffmann; Benson; Slocum, Compendium, c1890; Williams;
Collins.
Pacioli. De
Viribus. c1500.
Cardan. De Rerum
Varietate. 1557, ??NYS. = Opera Omnia, vol. III, pp. 245-246. Liber XIII.
Ludicrum primum & Ludicrum secondum. The first is the Alliance or Victoria puzzle, with one diagram
showing it apart. Second is the version
with three holes in each piece, with one diagram, again showing it apart.
Prévost. Clever and
Pleasant Inventions. (1584), 1998.
Schwenter.
1636. Part 10, exercise 29, pp.
410-411. Two hole version.
John Wecker. Op.
cit. in 7.L.3. 1660. Book XVIII -- Of the Secrets of Sports: The first
Pastime and the second Pastime, p. 338.
Taken from and attributed to Cardan, with the same diagrams. (I don't know if this material appeared in
the 1582 ed.??)
Witgeest. Het
Natuurlyk Tover-Boek. 1686. Prob. 44, pp. 35-36 is a two hole version taken
from Schwenter.
Ozanam. 1725. Vol. IV, prob. 31, p. 435 & fig. 37,
plate 11 (13). This is like the
Alliance puzzle in Hoffmann, below.
Alberti. 1747. Art. 31, pp. 205-206 (108‑109) &
fig. 38, plate X, opp. p. 206 (between pp. 108 & 109). Taken from Ozanam.
Catel.
Kunst-Cabinet. 1790. Das Scheibenspiel, p. 15 & fig. 23 on
plate I. The Alliance puzzle, looking
much like one of Ozanam's figures.
Bestelmeier.
1801. Item 291: Die
verschlungenen Bretchen. Alliance
puzzle. The diagram is very obscure and
the description is too brief to be certain, but I now see it is copied from
part of Catel's text, so this is definitely the Alliance.
Boy's Own Book. The
heart and ball puzzle. 1828: 424; 1828‑2: 428; 1829 (US): 220; 1843 (Paris): 438 & 442, no. 16; 1855: 574‑575; 1868: 675-676. = Wehman, New Book of 200 Puzzles, 1908, p.
18. In the 1843 (Paris), the engraved
heading of the section, p. 435, shows the puzzle. Boy's Treasury, 1844, omits the puzzle, but copies this engraving
on p. 424. = de Savigny, 1846, pp. 355
& 358, no. 12: Problème du cœur et de la boule.
Nuts to Crack II (1833), no. 91. The heart and ball puzzle.
Almost identical to Boy's Own Book.
Crambrook.
1843. P. 4, no. 9: Heart and
Ball Puzzle. Check??
Family Friend 3 (1850) 300 & 331. Practical puzzle, No. XX. Drawing shows the artist didn't understand
the puzzle at all. = Magician's Own
Book, 1857, prob. 16: The heart puzzle, pp. 271-272 & 295. = Book of 500 Puzzles, 1859, pp. 85-86
& 109. = Boy's Own Conjuring Book,
1860, prob. 15, pp. 233 & 258.
Family Friend (Dec 1858) 359. Practical puzzles -- 3.
Board and ball, but no name given to it. I don't have the solution.
The Secret Out.
1859.
Illustrated Boy's Own Treasury. 1860.
Adams & Co., Boston.
Advertisement in The Holiday Journal of Parlor Plays and Pastimes, Fall
1868. Details?? -- photocopy sent by
Slocum. P. 6: Chinese Ladder
Puzzle. Made of ivory and silk. Ornamentally carved. Probably the same as Hoffmann 17, below --
??
Magician's Own Book (UK version). 1871.
Anonymous. Social
Entertainer and Tricks (thus on spine, but running title inside is New Book of
Tricks). Apparently a compilation with
advertisements for Johnson Smith (Detroit, Michigan) products, c1890? P. 68: The board and ball. Same as Family Friend, 1858.
Hoffmann. 1893. Chap. II.
A. Murray. Some
simple puzzles. The Boy's Own Paper 17
or 18?? (1894??) 46. Well drawn.
Benson. 1904.
Slocum.
Compendium. Shows Chinese ladder
from Joseph Bland catalogue, c1890, and heart and string puzzle from Johnson
Smith 1929 catalogue.
Williams. Home
Entertainments. 1914. Another string and buttons puzzle, pp.
112-114. Chinese ladder. Get all the buttons out of the ladder and
onto the thread.
"Toymaker".
The New-century Cross Puzzle.
Work, No. 1394 (4 Dec 1915) 158 (or 153??). Like half of a Victoria or Alliance Puzzle with two holes in the
arms of a cross.
"Toymaker".
The "Wheel of Fate" Puzzle.
Work, No. 1467, (28 Apr 1917) no page number on the photocopy from
Slocum -- ?? (= Hoffmann, pp. 28-29.)
Collins. Book of
Puzzles. 1927.
Charles Béart. Jeux
et jouets de l'ouest africain. Tome
I. Mémoires de l'Institut Français
d'Afrique Noire, No. 42. IFAN, Dakar,
Senegal, 1955. Pp. 418-419. Describes and illustrates a one-board,
two-hole, version called djibilibi or jibilibi from Fonta-Djallon.
Lorraine L. Larison.
The Möbius band in Roman mosaics.
Amer. Scientist 61 (1973) 544‑547. Describes and illustrates a Roman mosaic in the Museum of Pagan
Art, Arles, France, which has a band with five twists. No date given.
At the Möbius Conference at Oxford, in 1990. it was stated
that the strip appears in Listing's notes for 1858, apparently a few months
before it appears in Möbius's notes.
Walter Purkert. Die
Mathematik an der Universität Leipzig von ihrer Gründung bis zum zweiten
Drittel des 19. Jahrhunderts. In: H. Beckert & H. Schumann, eds.; 100
Jahre Mathematisches Seminar der Karl‑Marx‑Universität Leipzig; VEB
Deutscher Verlag der Wissenschaften, Berlin, 1981; pp. 9‑39. On p. 31, he says that Möbius's Nachlass
shows that he discovered the strip in 1858.
Chris Pritchard: Aspects of the life and work of Peter
Guthrie Tait, pp. 77-88 IN: James Clerk
Maxwell Commemorative Booklet; James Clerk Maxwell Foundation, Edinburgh,
1999. On p. 81, he says that Listing
found the Strip in Jul 1858 and brought it to public attention in 1861, while
Möbius found it around Sep 1858 but did not publish until 1865.
J. B. Listing. Der
Census räumlicher Complexe. Abh. der
Ges. der Wiss. zu Göttingen 10 (1861) 97-180.
This appeared as a separate book in 1862. ??NYS -- cited by M. Kline, p. 1164.
A. F. Möbius. Über
die Bestimmung des Inhaltes eines Polyeders.
Königlich Sächsischen Ges. der Wiss. zu Leipzig 17 (1865) 31-68. = Gesammelte Werke, Leipzig, 1885-1887, vol.
2, pp. 473-512. ??NYS -- cited by M.
Kline, p. 1165. He also considers
multitwisted bands.
Tissandier.
Récréations Scientifiques. 5th
ed., 1888, Les anneaux de papier, pp. 272-273.
Illustration by Poyet. Shows and
describes rings with 0, 1, 2 twists.
Not in 2nd ed., 1881. I didn't
see whether this was in the 1883 ed.
=
Popular Scientific Recreations; [c1890]; Supplement: The paper rings,
pp. 867‑869.
Gardner,
MM&M, 1956, p. 70 says the earliest magic version he has found is the
"1882 enlarged edition of" Tissandier. This may be Popular Scientific Recreations, but I don't see any
date in it and the Supplement contains material that was not in the 1883 French
ed. -- cf. comments in Common References.
P. G. Tait.
Listing's Topologie.
Philosophical Mag. (5) 17 (No. 103) (Jan 1884) 30‑46 &
plate opp. p. 80. This is based on
Listing's Vorstudien zur Topologie (1847) and Der Census räumlicher Complexe
(1861). Section 8, pp. 37-38, discusses
strips with twists, noting that an odd number of half-twists gives one side and
one edge. If the odd number m of
half-twists is greater than 1, then cutting it down the middle gives a
knotted band with 2m half-twists. He
says this was the basis of a pamphlet which was popular in Vienna a few years
ago, which showed how to tie a knot in a closed loop. Pritchard [op. cit. above, p. 83] says Tait originally did not
credit either Listing or Möbius for these results, and in an article in Nature
in 1883, he noted that these properties were common currency among conjurers
for some time as alluded to in Listing's Vorstudien.
Anon. [presumably prepared by the editor, Richard A.
Proctor]. Trick with paper bands. Knowledge 11 (Jan 1888) 67-68. Short description, based on La Nature, i.e.
Tissandier, with copy of the illustration, omitting Poyet's name.
J. B. Bartlett. A
glimpse of the "Fourth Dimension".
The Boy's Own Paper 12 (No. 588) (19 Apr 1890) 462. Simple description.
Lewis Carroll.
Letter of Jun 1890 to Princess
Alice. Not in Cohen. In 1890, Carroll met Princess Alice (whose
father, Prince Leopold had been a student at Christ Church and had been
enamoured of the original Alice, then aged 18, but the Queen prevented such a
marriage), then age 6 and became friends.
This letter has the plan of a Möbius strip. This letter was advertised for sale by Quaritch's at the 2001
Antiquarian Book Fair in London.
Carroll refers to it in his letter of 12 Aug 1890 to R. H. Collins and
Cohen's note quotes Sylvie and Bruno Concluded, qv below, and explains the
object.
Tom Tit, vol. 3. 1893. See entry in 6.Q for
a singly-twisted band.
Lewis Carroll.
Sylvie and Bruno Concluded.
Macmillan, 1893. Chap. 7, pp. 96‑112,
esp., pp. 99‑105. Discusses
Möbius band ("puzzle of the Paper Ring"), Klein bottle and projective
plane. Quoted, with extended discussion
in John Fisher; The Magic of Lewis Carroll; op. cit. in 1; pp. 230-234. Cf Carroll-Gardner, pp. 6-7
Lucas.
L'Arithmétique Amusante.
1895. Note IV: Section II: No. 3: Les hélices paradromes,
pp. 222-223. Attributes the ideas to
Listing's Vorstudien zur Topologie, 1848.
Says this is the basis of a lengthy memoir sold at Vienna some years ago
showing how one can make a knot in a closed loop -- cf Tait. Gives basic results, including cutting in
half.
Herr Meyer.
Puzzles. The Boy's Own Paper 19
(No. 937) (26 Dec 1896) 206. No. 3:
Paper ring puzzle. Asks what happens
when you cut into halves or thirds after one or two or more twists.
Somerville Gibney.
So simple! The hexagon, the
enlarged ring, and the handcuffs. The
Boy's Own Paper 20 (No. 1012) (4 Jun 1898) 573-574. Cuts Möbius band in half twice, then does the same with doubly
twisted band.
Hoffmann. Later
Magic. (Routledge, London, 1903); Dover, NY, 1979. The Afghan bands, pp. 471‑473. Cuts various strips in half.
Gardner,
MM&M, 1956, p. 71, says this is the first usage of the name Afghan Bands
that he has found.
C. H. Bullivant. The
Drawing Room Entertainer. C. Arthur
Pearson, London, 1903. Paper rings, p.
48. Cuts various rings in half.
Dudeney. Cutting-out
paper puzzles. Cassell's Magazine ??
(Dec 1909) 187-191 & 233-235.
Calls it "paradromic ring" and says it is due to Listing,
1847. Probably based on Lucas.
Devant. Tricks for
Everyone. Op. cit. in 4.A.1. 1910.
Curious paper patterns, pp. 20-21.
Cutting rings with 0, 1, 2 half-twists in half.
Anon. [H. W. R.]
Games and Amusements. Ward, Lock
& Co., London, nd [c1910??]. The
mysterious paper bands, p. 128.
Describes bands with 1, 2,
3 twists and cutting them in half.
Williams. Home
Entertainments. 1914. The magic paper rings, pp. 104-106. Rings with
0, 1, 2 half-twists to be
cut down the middle. Good diagram.
Lee de Forest. US
Patent 1,442,682 -- Endless Sound Record and Mechanism Therefor. Filed: 5 Oct 1920; patented: 16 Jan 1923. 3pp
+ 2pp diagrams. No references and no
mention of any previous forms.
Hummerston. Fun,
Mirth & Mystery. 1924. The magic rings, p. 91. 0, 1, 2
half-twists, each cut in half.
William Hazlett Upson.
A. Botts and the Moebius strip.
Saturday Evening Post (?? 1945).
Reprinted in: Clifton Fadiman,
ed.; Fantasia Mathematica; Simon & Schuster, NY, 1958; pp. 155-170.
Owen H. Harris. US
Patent 2,479,929 -- Abrasive Belt.
Applied: 19 Mar 1949;
patented: 23 Aug 1949. 2pp
+ 1p diagrams. Same comments as on de
Forest.
William Hazlett Upson.
Paul Bunyan versus the conveyor belt.
??, 1949. Reprinted in: Clifton Fadiman, ed.; The Mathematical
Magpie; Simon & Schuster, NY, 1962; pp. 33‑35.
Gardner.
MM&M. 1956. The Afghan Bands, pp. 70-73 & figs.
9-14, pp. 74-77. Describes several
usages as a magic effect. Cites
Tissandier and Hoffman, cf. above.
James O. Trinkle. US
Patent 2,784,834 -- Conveyor for Hot Material.
Applied: 22 Jul 1952; patented:
12 Mar 1957. 2pp + 1p diagrams. Same comments as on de Forest, except that 5
references are mentioned in the file, but not in the patent itself.
James W. Jacobs. US
Patent 3,302,795 -- Self-cleaning Filter.
Filed: 30 Aug 1963; patented: 7
Feb 1967. 2pp + 2pp diagrams. "This invention relates to dry cleaning
apparatus and more particularly to a self-cleaning filter element comprised of
an endless belt having a half twist therein." The diagram does not show the twist very clearly. Same comments as on de Forest, except that
the Examiner cites three patents.
Making resistors with math.
Time (25 Sep 1964) 49. Richard
L. Davis of Sandia Laboratories has made a Möbius strip noninductive
resistor. This has metal foil on both
sides of a nonconductive Möbius strip with connections opposite to one
another. The current flows equally both
ways and passes through itself. He
found the inductance as low as he had hoped, but he is not entirely clear why it
works! Gardner, below, describes this
and also cites Electronics Illustrated (Nov 1969) 76f, ??NYS.
Jean J. Pedersen.
Dressing up mathematics. MTr 60
(Feb 1968) 118-122. Describes garments
with two sides and one edge or one side and one edge or two sides and no
edges!!
Gardner. The Möbius
strip. SA (Dec 1968) = Magic Show, chap.
9. Describes the above mentioned
patents and inventions and numerous stories and works of art using the idea.
Ross H. Casey. US
Patent 4,161,270 -- Continuous Loop Stuffer Cartridge having Improved Moebius
Loop Tensioning System. Filed: 15 Jul
1977; patented: 17 Jul 1979. 2pp + 3pp diagrams. This is actually only for an improvement in
the idea. "Typically, a
cubically-shaped wire form or a plurality of guides are used to effect a
Moebius twist in the continuous loop.
The invention includes an improved Moebius loop device and tensioner in
the form of an easily constructed planar triangular-shaped device to effect a
Moebius twist and tensioning in a continuous loop." Basically the loop folds around three edges
of a triangle. Cites several earlier
patents, but Joe Malkevich says none of them relate to the Möbius idea.
In the early 1990s, Tim Rowett found a German making a
stainless steel strip with a double twist in it which could be manoeuvred into
a double Möbius strip which appeared to be cut in half through the thickness of
the strip and which sprang apart when released. In fact it can also be seen as the result of an ordinary cutting
of a Möbius strip in half. I cannot
recall seeing this behaviour described anywhere, though I imagine it is well
known. The process is shown clearly in
the following.
Jean-Pierre Petit.
Gémellité Cosmique. Text for the
month of Juin. Mathematical
Calendar: Tous les mois sont maths! for
1990 produced by Editions du Choix, Bréançon, 1989.
Scot Morris. The
Next Book of Omni Games. Op. cit. in
7.E. 1991. Pp. 53-54 describes Jacobs' 1963 patent. He says that David M. Walba and colleagues
at Univ. of Colorado have synthesised "the first molecular Möbius
strip", a molecule called trisTHYME and that they have managed to cut it
down the middle!!
Wire and string puzzles are
difficult to describe. Only a few were
illustrated before 1900. S&B, p.
90, says they first appeared in the 1880s (when wire became common), though
some are older -- see 7.M.1, 7.M.5, 11.A, 11.C, 11.D, 11.F, 11.I, 11.K.7 and
possibly 11.B, 11.E, 11.H. Ch'ung-En
Yü's Ingenious Ring Puzzle Book (op. cit. in 7.M.1) implies that wire and ring
puzzles, besides the Chinese Rings, were popular in the Sung Dynasty (960-1279)
but I have no confirmation of this assertion.
See the entry under Stewart Culin in 7.M.1 for a vague reference to
Japanese ring puzzles called Chiye No
Wa. This section will cover the various
later versions, but without trying to describe them in detail. I have separated the Horseshoes (11.K.7) and
the Caught Heart (11.K.8), as they are so common.
Wire puzzles were included in
general puzzle boxes by 1893 -- see the ad at the end of Hoffmann mentioned in
11.I. By 1912, they were being sold in boxes of just wire puzzles. Six boxes of wire puzzles are offered in
Bartl's c1920 Zauberkatalog, p. 305.
Wire puzzles are a major component of the Western Puzzle Works, 1926
Catalogue.
See S&B, pp. 88‑115.
Peck & Snyder.
1886. P. 245, No. A -- The
puzzle brain links. 11 interlocked
links. Not in Slocum's Compendium.
Slocum.
Compendium. Shows Egyptian
Mystery from Joseph Bland's catalogue, c1890.
Herr Meyer. An
improved ring puzzle. In: Hutchison; op. cit. in 5.A; 1891; chap. 70,
section III, pp. 573‑574.
Folding ring on loop on loop on loop on bar.
Hoffmann. 1893. Chap. VIII: Wire puzzles, pp. 302‑314
= Hoffmann-Hordern, pp. 198-206, with six photos.
Montgomery Ward & Co.
Catalog No 57, Spring & Summer, 1895. Facsimile by Dover, 1969, ??NX. P. 237, item 25479, is a collection of 8
wire puzzles. The Chilian Puzzle is
shown. Cf 1903.
Ernest C. Fincham.
Street Toys. Strand Mag. 10
(1895) 765-773. Shows: bicycle puzzle (a circular version of
Hoffmann No. 2); "Ally
Sloper" puzzle (a caught heart, cf 11.K.8); gridiron puzzle (basically an elaborated version of the caught
heart, cf 11.K.8); magnetic puzzle (involves
placing three needles by use of a magnet);
key puzzle (a cast puzzle of two interlocked keys); oriental ring (ordinary four strand puzzle
ring, cf 11.K.4).
H. F. Hobden. Wire
puzzles and how to make them. The Boy's
Own Paper 19 (No. 945) (13 Feb 1896) 332-333.
Montgomery Ward & Co.
Catalogue. 1903. Reproduced in: Joseph J. Schroeder, Jr.; The
Wonderful World of Toys, Games & Dolls
1860··1930; DBI Books, Northfield, Illinois, 1977?, p. 118. Shows one puzzle (Chilian Puzzle) from a
set. Cf 1895. Slocum's Compendium also shows this.
Benson. 1904. Chap. VIII: Wire puzzles, pp. 236‑240.
M.
Adams. Indoor Games. 1912.
Pp. 337‑341.
Bartl. c1920. Vexier- und Geduldspiele, pp. 305-312, shows
43 wire puzzles.
Western Puzzle Works, 1926 Catalogue. Shows about 65 wire puzzles.
Collins. Book of
Puzzles. 1927. Some good metal puzzles, pp. 52-53. Shows:
The devil's ring; The
teaser; The three-in-one rings; The link and rings; The union puzzle. Pp. 44-54 cover other wire puzzles: The Twin Hearts (= Hoffmann no. 1, 11.K.8); The eternal triangle (= Hoffmann no.
2); The bead and spiral (= Hoffmann no.
8); The snake and ring
(= Hoffmann no. 3, 11.K.1);
The great seven-ring puzzle (= Chinese rings, 7.M.1); The eastern question (= Hoffmann no. 10,
Double witch key, 11.K.6).
Ch'ung-En Yü.
Ingenious Ring Puzzle Book.
1958. Op. cit. in 7.M.1.
Richard I. Hess.
Compendium of Over 7000 Wire Puzzles. 5th ed., published by the author, Rancho Palos Verdes, Mar 1991,
259pp. 7090 wire, entanglement and cast iron puzzles classified in 17
categories. There is an 53 page index
and then 205 pages of reduced photocopies of the actual puzzles. The pictures come out almost as good as
drawings. Some of the more obvious
combinations do not have pictures. The
naming of the puzzles has a certain poetry about it. A118: Type 2 trapeze with
baffle, horseshoes, lock and 1-ring key.
A226: Cascaded double compound
trapeze with baffled heart. B053: Hong Kong house with semicircles (Hard
2-story). C226: Bug (3-ring) with triple cross. D105:
Triple Finnish diddle. (Previous
editions: 2nd ed, 1982, c500
puzzles; 3rd ed, 1985, c1400 puzzles; 4th ed, 1988, c2600 puzzles.)
11.K.1. RING AND SPRING PUZZLE
A ring is on
a spring with sealed ends.
I have seen this illustrated in Wizard's Guide, a catalogue
of magic apparatus by W. J. Judd, 1882.
??NX.
Montgomery Ward & Co.
Catalogue. 1886. Reproduced in: Joseph J. Schroeder, Jr.; The
Wonderful World of Toys, Games & Dolls
1860··1930; DBI Books, Northfield, Illinois, 1977?, p. 24. Slocum's Compendium also shows this.
Peck & Snyder.
1886. P. 162: Universal pocket
puzzle.
Hoffmann. 1893. Chap. VIII, no. 3: The snake and ring (=
ring and spring puzzle) = Hoffmann-Hordern, p. 199. Photo on p. 199 shows L'Anneau Prisonnier,
with box, 1891-1905. Hordern
Collection, p. 80, shows the same, dated 1880-1905.
Benson. 1904. Chap.
VIII: Wire puzzles, pp. 236‑240.
The spiral and ring. (=
Hoffmann's no. 3.)
I have seen a French example, called L'anneaux prisonnier,
dated 1900-1920, but see above.
Slocum.
Compendium. Shows many later
examples: 1913, 1915, 1915, 1919, etc.
Bartl. c1920.
Western Puzzle Works, 1926 Catalogue. No. 9: Down and Out.
Collins. Book of
Puzzles. 1927. The snake and ring puzzle, pp. 48-49.
11.K.2. STRING AND SPRING PUZZLE
A loop of string goes through the
spring which has a few turns and long tails so the string doesn't come off
obviously.
Western Puzzle Works, 1926 Catalogue. No. 1877: Loop and Chain.
Slocum.
Compendium. Shows Magic Chain
from Johnson Smith 1929 catalogue.
11.K.3. MAGIC CHAIN = TUMBLE RINGS
One holds the two top rings and
releases the upper one which appears to drop to the bottom.
A magic chain. The
Boy's Own Paper 12 (No. 581) (1 Mar 1890) 351.
Good picture by Poyet, so this ought to be in Tissandier or Tom Tit, but
I haven't seen it.
Der Gute Kamerad.
Kolumbus‑Eier. 1890. ??NYS, but reproduced in Edi Lanners' 1976
edition, translated as: Columbus' Egg;
Paddington Press, London, 1978; The magic chain, pp. 176‑177, with good
illustration on p. 177 -- an enlargement of Poyet's picture with his name
removed.
Bartl. c1920. P. 308, no. 71: Konsilkette.
Western Puzzle Works, 1926 Catalogue.
Davenport's catalogue, op. cit. in 10.T, c1940, pp. 8 &
29, calls it The Wizard's Chain.
Gardner. SA (Aug
1962). = Unexpected, chap. 13. Calls it "tumble rings". Describes the chain with a good diagram for
making your own, but gives no indication of its history.
William Jones. Finger-Ring Lore. Chatto & Windus, London, 1877. Pp. 313-321 discusses gemmel or gemmow rings with two or three
parts. He cites Herrick -- "a ring
of jimmals", quotes Dryden's play
Don Sebastian describing a two part ring,
describes a three-part jointed ring,
describes a two part ring excavated in 1800 (medieval??), describes 'a plain geemel wedding-ring'
given by the Prince Regent to Mrs. Fitzherbert, describes the five-link wedding ring of Lady Catherine Gray, illustrates a 15C gemmel ring with a head of
Lucretia of the type mentioned in Twelfth Night II.v, illustrates Sir Thomas Gresham's betrothal ring of 1544, describes and illustrates several other
examples -- 16C, 13C. On pp. 321-322,
he mentions an exhibition of some puzzle rings by Rev. John Beck at a meeting
of the Archaeological Institute in Mar 1863.
These had 7, 4 and 9 parts.
Ernest C. Fincham.
Street Toys. Strand Mag. 10
(1895) 765-773. Shows: Oriental ring (ordinary four strand puzzle
ring).
Dudeney. Great
puzzle crazes. Op. cit. in 2. 1904.
Says he believes it is of Indian origin.
George Frederick Kunz.
Rings for the Finger. J. B.
Lippincott, Philadelphia, 1917;
reprinted by Dover, NY, 1973 (with the two colour plates done in
B&W).
Pp.
218‑233 discusses puzzle rings.
The earliest forms were gimmel rings which had two or three parts which
could be separated for use at betrothal, with the parts rejoined at the wedding
and given to the bride. These are known
from the 16C -- e.g. the plate facing p. 219 shows a 16C German example from
the BM and Sir Thomas Gresham's betrothal ring (c1540) is similar. P. 219 gives a quote from Dryden's play Don
Sebastian which describes a two part ring.
The plates opp. pp. 220, 221 & 230 show other examples including a
three part one from 17C.
Kunz
says "the so‑called 'puzzle ring' ... was derived from the
East." The plate facing p. 220
shows examples of a three part ring and the common four part Middle Eastern
ring, in gold, from the 17C in the BM.
The
plate opp. p. 233 shows a six part gold betrothal ring from the Albert Figdor
collection, Vienna. This forms a simple
chain of six rings.
Stewart Culin.
Korean Games. Op. cit. in
4.B.5. Section XX: Ryou‑Kaik‑Tjyo
-- Delay Guest Instrument (Ring Puzzle), pp. 31‑32. Says there are many Japanese ring puzzles,
called Chiye No Wa, and shows one which seems to be 10 rings linked in a
chain -- possibly the simple type of puzzle ring??
Slocum.
Compendium. Shows The Lady's and
Gentleman's "Wonderful Ring" from Joseph Bland's catalogue,
c1890. This has four parts which form a
simple chain. The Compendium also shows
Puzzle Ring from Johnson Smith 1929 catalogue, which is the classic Turkish or
Middle Eastern four part ring.
These are plates with holes and
perhaps raised sections. An open ring
must be removed by working it from hole to hole.
S&B 92 says a version was sold by Hamleys in 1879 and
appeared as The Boston In-and-Out Puzzle in 1880-1885 and as The Queen's
Jubilee Puzzle in 1887.
Peck & Snyder.
1886. P. 250: no. 188 -- The
order of stupids.
Hoffmann. 1893. Chap. X, pp. 353-354 & 392
= Hoffmann-Hordern, pp. 251-252, with photos.
Western Puzzle Works, 1926 Catalogue.
Dudeney. Great
puzzle crazes. Op. cit. in 2. 1904.
"... the 'Conjurer's Medal,' that came out some years ago
...." Medal with five holes.
11.K.6. INTERLOCKED NAILS, HOOKS, HORNS, ETC.
I have just added this. The puzzle has two interlocked objects. One type has nails bent around a 270o twist -- see S&B 96-97.
A variation, called Wishbone Puzzle, has longer tails -- see S&B
97. A variation has one of the nails
made longer with twists at each end, sometimes called Tangle Twister -- see
S&B 96. A variation has two
circular bits with tails, sometimes called Double Witch Key -- see S&B 97
& 102. Another type has two
S-shaped pieces or two J-shaped pieces (hooks) -- see S&B 95.
Hoffmann. 1893. Chap. VIII, no. 10: The Eastern question, p.
307 = Hoffmann-Hordern, pp. 204-205.
Interlocked circular bits with tails
= Double witch key. Photo on p.
205 shows an example, with instructions, by Hamley Bros., 1890‑1900. Hordern Collection, p. 86, shows two
examples of Question Romaine, with their boxes, by W. S. and R. T., both Paris,
1890-1910.
Burnett Fallow. How
to make an ingenious link puzzle. The
Boy's Own Paper 16 (No. 777) (2 Dec 1893) 143.
= Hoffmann no. 10.
Benson. 1904. Chap.
VIII: Wire puzzles, pp. 236‑240.
The two crooks. (= Hoffmann no.
10.)
Walter S. Jenkins.
US Patent 969,481 -- Puzzle.
Filed: 16 Mar 1908; patented: 6
Sep 1910. 3pp + 1p diagrams. Interlocked nails.
Bartl. c1920.
Western Puzzle Works, 1926 Catalogue.
Collins. Book of
Puzzles. 1927. The Eastern question, pp. 53-54. As in Hoffmann, with very similar diagram.
S&B 95 calls the interlocked hooks, Loop the Loop or The
Devil's Keys, and shows it, but gives no date.
S&B 96-97
Ch'ung-En Yü.
Ingenious Ring Puzzle Book.
1958. Op. cit. in 7.M.1. P. 21 shows the interlocked nails puzzle.
See S&B
99.
Magician's Own Book (UK version). 1871. The ring and
wire-loop puzzle, p. 113. With
elongated horseshoe parts.
Hoffmann. 1893. Chap. VIII: Wire puzzles, no. 6: The double
bow and ring (= Horseshoes puzzle) = Hoffmann-Hordern p. 201. Photo on p. 201 shows an undated
example. Hordern Collection, p. 82,
shows two versions, La Question du Divorce and Les Anneaux Diaboliques, dated
1880-1905.
H. F. Hobden. Wire
puzzles and how to make them. The Boy's
Own Paper 19 (No. 945) (13 Feb 1896) 332-333. Hourglass and ring.
Benson. 1904. Chap.
VIII: Wire puzzles, pp. 236‑240.
The bow and ring puzzle.
(= Hoffmann no. 6.)
M. Adams. Indoor
Games. 1912. Pp. 337‑341. The
hourglass and ring. (= Hoffmann 6.)
Bartl. c1920.
Ch'ung-En Yü.
Ingenious Ring Puzzle Book.
1958. Op. cit. in 7.M.1. Pp. 18-19 discusses the "horseshoes
puzzle" which is called "Jade Interlocked Ring Puzzle" and is
"the simplest and easiest puzzle of the Incomplete Ring Type". Yü then discusses more complex versions of
the puzzle.
This puzzle
is isomorphic to 11.B.
Hoffmann. 1893. Chap. VIII: Wire puzzles, pp. 302‑314. No. 1: The united hearts.
Ernest C. Fincham.
Street Toys. Strand Mag. 10
(1895) 765-773. Shows: "Ally Sloper" puzzle (a Caught
heart) and Gridiron puzzle (basically
an elaborated version of the Caught heart).
H. F. Hobden. Wire
puzzles and how to make them. The Boy's
Own Paper 19 (No. 945) (13 Feb 1896) 332-333.
Cupid's bow (similar to Hoffmann 1).
Benson. 1904. Chap.
VIII: Wire puzzles, pp. 236‑240.
The two hearts. (= Hoffmann no.
1.)
M. Adams. Indoor
Games. 1912. Pp. 337‑341.
Bartl. c1920. P. 308, no. 65: Vexierherz.
Collins. Book of Puzzles. 1927.
The twin hearts puzzle, pp. 44-45.
As in Hoffmann, with similar diagram.
Ch'ung-En Yü.
Ingenious Ring Puzzle Book.
1958. Op. cit. in 7.M.1. P. 26 shows a single heart, similar to Hoffmann
1, but with a doubled second part, called Recessed Handle Ring Puzzle.
11.L. JACOB'S LADDER AND OTHER HINGING DEVICES
I am now realising that a number of
the objects in 6.D, namely the tetraflexagons and the trick or magic books, are
just extended forms of the Jacob's ladder.
See also Engel in 6.X. The
Chinese wallet has two boards with this kind of hinging so it can open on
either side, giving different effects.
Pacioli. De
Viribus. c1500. Ff. 229r - 229v, Part 2, Capitolo. CXXXII. Do(cumento).
del solazo puerile ditto bugie (On the childish recreation called
deception?). = Peirani 316‑317. Uses two tablets and three leather
straps. Describes how to use it to
catch a straw. His references show that
it is called 'calamita di legno' (calamity of wood (or magnet of wood --
depending on whether the Italian is
calamità or calamita) in Italian.
Bernardino Luini.
"A Boy with a Toy" or "Cherub with a Game of
Patience". Proby Collection, Elton
Hall, Peterborough, Cambridgeshire. The
painting is 15" by 13" (38 cm by 33 cm). Luini was a fairly well known Lombard
follower of Leonardo, born c1470 and last known in 1533. I have found no indication of the date of
the work, but the middle of his working life is c1510. The figure is reproduced in: Angela Ottino
della Chiesa; Bernardino Luini; Electa Editrice, Milan, 1980, item 59,
but the quality is not good. The
painting is described in: G. C. Williamson; Bernardino Luini; George
Bell and Sons, London, 1900, pp. 104-105.
Williamson says the tapes holding the boards together are red and are
apparently holding a straw, but he doesn't seem to recognize the object.
F.
Bartolozzi made a nice engraving of this in 1795, attributing the painting to
da Vinci, as was generally believed at the time. James Dalgety has an example and I have a photo of it.
Gustav Pauli.
Ausstellung von Gemälden der Lombardischen Schule im Burlington Fine
Arts Club London April - Juni 1898. (Schluss). Zeitschrift
für bildene Kunst (NF) 10 (1898-1899) 149.
The Luini was on display and the author describes the toy as a
'Taschenspielerstückchen', a little juggler's trick -- but recall that juggler
was long a synonym for magician -- with two boards which allow one to vanish
the straw.
Bernardino Licino (attrib.). "Portrait of a Man with a Puzzle". Picture Gallery, Hampton Court Palace,
London, Richmond upon Thames, London.
Licinio was a Venetian painter born about 1485 and last known in
1549. The painting is described and
illustrated (in B&W) in John Shearman; The Early Italian Pictures in the
Collection of Her Majesty The Queen; CUP, 1983, item 141, plate 124. It is very similar to another portrait known
to be by Licinio and dated 1524, so this is probably c1524 and hence a bit
later than the Luini. Shearman cites
the Luini painting and Pauli's notice of it.
The description says the binding tapes are red, as in the Luini, and
both show something like a straw being trapped in the wallet, which suggests
some connection between the two pictures, though it may just be that this toy
was then being produced in or imported to North Italy and was customarily made
with red tapes. On the toy is an
inscription: Carpendo Carperis Ipse (roughly: Snapping snaps the snapper), but
Shearman says it definitely appears to be an addition, though its paint is not
noticeably newer than the rest of the painting. Shearman says the toy comprises 'three or more rectangles ...',
though both paintings clearly show just two pieces. My thanks to Peter Hajek who reported seeing this in an email of
22 May 1998
Prévost. Clever and
Pleasant Inventions. (1584), 1998. Pp. 136-140. Chinese wallet.
Schwenter.
1636, Part 15, exercise 27, pp.
551-552. "Ein einmaul zu
machen." Chinese wallet. I can't find Einmaul in my dictionaries.
Witgeest. Het
Natuurlyk Tover-Boek. 1686. Prob. 66, pp. 49-50. Chinese wallet.
Peter Haining.
Moveable Books An Illustrated
History. New English Library,
1979. Pp. 12‑13 shows The
Unique Click Tablets of c1790 which is a Jacob's Ladder of six pieces with a two
part picture glued to each face, described as 'possibly unique'. The pictures are uncoloured and similar to
those occurring in chapbooks and/or hornbooks of the period. (Thanks to Christopher Holtom for this
information.)
de Savigny.
Livre des Écoliers. 1846. P. 272: Les deux planchettes. Shows 3½ parts and indicates one can
continue beyond four. "Tous les
écoliers connaissent le jeu des planchettes ...."
Leslie Daiken.
Children's Toys Throughout the Ages.
Spring Books, London, 1963.
Plate 6 on p. 24 shows "Hand‑operated game of changing
pictures, c. 1850" which clearly shows the "Jacob's ladder"
hinging with four parts.
An example with nine blocks with coloured pictures, possibly
home-made, mid 19C, was advertised by a book dealer in 2003.
Edward Hordern's collection has an example with four panels
from c1854.
Hanky Panky.
1872. The magic pocket-book (Die
zwei magichen [sic] Brieftaschen), pp. 270-272. This seems to be a form of this mechanism similar to the Chinese
wallet.
H. F. Hobden. Jacob's
ladder, and how to make it. The Boy's
Own Paper 12 (No. 592) (17 May 1890) 526.
Says he doesn't know why it is called Jacob's ladder, but that it has
been popular "for a number of years". Suggests at least 7 blocks, preferably 12.
Hutchison. Op. cit.
in 5.A. 1891. Chap. 71: Jacob's ladder, and other contrivances. Section I -- Jacob's ladder, pp. 583‑585. Shows it clearly. Suggests use of 12 parts, or at least 7.
Axel O. Sodergren.
US Patent 778,282 -- Folding Picture-Album. Applied: 28 Mar 1904;
patented: 27 Dec 1904. 2pp + 2pp
diagrams. Ordinary Jacob's Ladder used
to display pictures.
Davenport's catalogue, op. cit. in 10.T, c1940, pp. 8 &
21, shows the two part version called The Wonderful Magic Book, now often sold as
the Chinese Wallet.
Norman F. Rutherford.
US Patent 2,245,875 -- Toy.
Applied: 4 Dec 1939; patented:
17 June 1941. 4pp + 7pp diagrams. "... a development of and improvement
upon the toy known for years under the name of Jacob's ladder." Starts with two block versions, then shows 8
blocks formed into a ring and a number of shapes it can form. Then considers three blocks in an L with
hinges in two directions, using with square or hexagonal blocks. Then considers five blocks in a W, or closed
into a four block ring and the seven block extension of this idea, closed into
a six block ring, and the nine block version closed into an eight block
ring. then considers the six hexagon
version in a ring. All in all, a pretty
direct ancestor of Rubik's Magic puzzle.
A six piece burr with identical flat
notched pieces and no key piece is sometimes assembled by forcing together
(perhaps after steaming) to make an unopenable money box. Cf 6.W.5.
Vesta boxes were small boxes to hold matches and were popular from c1850
to c1920. Some of these had trick
openings and form a distinct class of puzzle boxes.
Catel.
Kunst-Cabinet. 1790. Das Vexierkästchen, p. 21 & fig. 21 on
plate I. Figure just shows a box. Text says the cover is "made like a
see-saw" and one presses firmly on one side and the other lifts up.
Bestelmeier.
1801. Item 208: Eine Kästchen,
welches man ohne das Geheimniss zu wissen nicht öfnen kan.
Jerry Slocum, compiler.
Match box puzzles and related trick boxes from catalogs, books and
patents. Jan 1993. 12pp, with 25 entries during 1843-1912, some
entries being lists of many items.
Includes Crambrook, Taylor, Peck & Snyder (only items 35 & 122),
Hoffmann (items 30-39), below.
Crambrook.
1843. P. 3 lists 24 types of Puzzling
Boxes. ??
The 'Ne Plus Ultra' match or snuff box has a lid which opens
by pressing on one edge. Slocum, above,
says the earliest known example is in silver, produced by Alfred Taylor,
silversmiths of Birmingham, and hall-marked 1864.
Roger Fresco-Corbu.
Vesta Boxes. Antique Pocket
Guides, Lutterworth Press, Guildford, Surrey, 1983. He mentions several trick boxes.
Peck & Snyder.
1886.
Montgomery Ward & Co.
Catalogue. 1889. Reproduced in: Joseph J. Schroeder, Jr.; The
Wonderful World of Toys, Games & Dolls
1860··1930; DBI Books, Northfield, Illinois, 1977?, p. 27. Puzzle Locomotive Savings Bank. Does not open until it is full of money.
Marshall Field & Co.
Catalogue. 1892. Reproduced in: Joseph J. Schroeder, Jr.; The
Wonderful World of Toys, Games & Dolls
1860··1930; DBI Books, Northfield, Illinois, 1977?, p. 66.
Hoffmann. 1893. Chap. 2, pp. 37-68 = Hoffmann-Hordern,
pp. 44-54, with photos, shows several items which could be included here.
Carl P. Stirn.
1894 Trade Price List Toys, ....
Reproduced in: Ronald S. Barlow, ed.; The Great American 1879-1945
Antique Toy Bazaar; Windmill Publishing, El Cajon, California,
1998. The Magic Bank for dimes, on p.
99 of Barlow. I believe this is a
cylindrical device which will not open until it is filled with dimes -- I have
several examples.
James Scott. Chinese
puzzles, tricks, and traps. Strand Mag.
20 (No. 120) (Dec 1900) 715‑720.
Figure 7 shows an object that looks like a three-piece burr but which is
a puzzle box made of cardboard or thin wood.
It opens by sliding one stick and then pressing its end, when its sides
are seen to be hinged and they flex outward.
Figures 8‑10 show an ivory globular trinket casket, which has
three orthogonal rods crossing in the centre.
These have to be turned and slid in the right order to open the box.
Montgomery Ward & Co.
Catalogue. 1903. Reproduced in: Joseph J. Schroeder, Jr.; The
Wonderful World of Toys, Games & Dolls
1860··1930; DBI Books, Northfield, Illinois, 1977?, p. 120.
Butler Brothers (now City Products Corp.). Catalogue.
1914. Reproduced in: Joseph J.
Schroeder, Jr.; The Wonderful World of Toys, Games & Dolls 1860··1930; DBI Books, Northfield, Illinois,
1977?, p. 179. "Gem" Pocket
Money Banks. Types which open when 20
nickels or 50 dimes are inserted.
M. Adams. Indoor
Games. 1912. A puzzle matchbox, pp. 260-261.
Simply concealed slide.
Cecil Henry Bullivant.
Every Boy's Book of Hobbies. T.
C. & E. C. Jack, London, nd [c1912].
Bartl. c1920. P. 306, no. 25 is very similar to Hoffmann's
Psycho match-box.
Butler Brothers.
[Toy catalogue], 1921.
Reproduced in: Ronald S. Barlow, ed.; The Great American 1879-1945
Antique Toy Bazaar; Windmill Publishing, El Cajon, California,
1998. "Gem" Pocket Savings
Banks, on p. 186 of Barlow. One of
these may be the same as the item mentioned under Stirn, above.
11.N. THREE KNIVES MAKE A SUPPORT
New section. The pattern of interlocking knives involved
is also used in basketry, woven fences, latticework, etc. Indeed, if they are beams with notches, this
gives frameworks which can roof a space much wider than any available beam,
e.g. Wren's ceiling of the Sheldonian Theatre, Oxford, and this is used in
puzzle pot stands.
Four knives versions: Magician's Own Book; Hoffmann; Blyth;
Collins; Bile Beans; Doubleday;
Pacioli. De
Viribus. c1500. Ff. 228r - 228v, Part 2, Capitolo. CXXIX. Do(cumento).
atozzare .iij tagli de coltelli insiemi (Join together three blades of
knives). = Peirani 315. Pacioli says this was shown to him on 1 Apr
1509 (Peirani has misread this as isog)
by "due dorotea veneti et u perulo 1509 ad primo aprile ebreo". Peirani transcribes u
as un but Dario Uri thinks it is the initial of Perulo's given
name. I wonder if 'dorotea' might refer
to some occupation, e.g. nuns at S. Dorothy's Convent. In Vienna, the Dorotheum is a huge public
auction house where estates are auctioned off.
The word 'ebreo' means 'Hebrew', but I cannot see what it refers to.
Cardan. De
Subtilitate, Book 17, 1550. = Op. Omnia
III, p. 629.
Prévost. Clever and
Pleasant Inventions. (1584), 1998. PP. 19-20.
van Etten.
1624. Prob. 6 (6), p. 7 (15‑16). Henrion's 1630 Notte cites Cardan.
Hunt. 1631
(1651). Pp. 278-279 (270-271).
Schwenter.
1636. Part 15, exercise 7, pp.
536-537. Three knives.
John Wecker. Op.
cit. in 7.L.3. 1660. Book XVIII -- Of the Secrets of Sports:
Sticks that mutually support one the other, p. 345. Taken from and attributed to Cardan, with very similar
diagram. (I don't know if this material
appeared in the 1582 ed.??)
Witgeest. Het
Natuurlyk Tover-Boek. 1686. Prob. 74, pp. 56-57. Three pipes shown. Text refers to sticks, etc.
Ozanam. 1694. Prob. 16: 1696: 287-288 & fig. 140,
plate 47. Prob. 16 & fig. 35, plate
15, 1708: 364. Prob. 20 & fig. 140,
plate 47 (45), 1725: vol. 2, 392-393.
Prob. 47 & fig. 50, plate 11, 1778: vol. 2, 87; 1803: vol. 2, 93-94; 1814: vol. 2, 76; Prob. 46, 1840: 235.
Badcock.
Philosophical Recreations, or, Winter Amusements. [1820].
P. 15, no. 25 & Frontispiece fig. 2: To place three sticks, or tobacco
pipes, upon a table in such a manner, that they may appear to be unsupported by
any thing but themselves.
Rational Recreations.
1824. Feat 16, p. 64. Three knives or tobacco pipes.
Endless Amusement II.
1826?
Boy's Own Book.
1828.
Nuts to Crack II (1833).
Julia de Fontenelle.
Nouveau Manuel Complet de Physique Amusante ou Nouvelles Récréations
Physiques .... Nouvelle Édition, Revue,
..., Par M. F. Malepeyre. Librairie
Encyclopédique de Roret, Paris, 1850.
P. 408 & fig. 147 on plate 4: Disposer trois batons .... Figure copied from Ozanam, 1725.
Magician's Own Book.
1857.
Anonymous. Every
Little Boy's Book A Complete Cyclopædia
of in and outdoor games with and without toys, domestic pets, conjuring, shows,
riddles, etc. With two hundred and
fifty illustrations. Routledge, London,
nd. HPL gives c1850, but the text is
clearly derived from Every Boy's Book, whose first edition was 1856. The material here is in the 1856 ed of Every
Boy's Book (with J. G. Wood as unnamed editor), not yet entered, and later
editions, but with different text and pictures. These are essentially the same as the versions in Boy's Own Book,
1843 (Paris).
P.
339: The tobacco pipe jug stand.
P.
351: Bridge of knives.
Magician's Own Book (UK version). 1871.
Tissandier.
Récréations Scientifiques.
1883? Not in the 2nd ed.,
1881. I didn't see if these were in the
3rd ed., 1883. 5th ed., 1888. Illustrations by Poyet.
Tissandier. Popular
Scientific Recreations;
Supplement. 1890? Pp. 798-799. To poise a tumbler upon three sticks, each one of which has one
end in the air. The water-bottle and
the three knives. = (Beeton's) Boy's
Own Magazine 3:6 (Jun 1889) 249-251.
Der Gute Kamerad.
Kolumbus‑Eier. 1890. ??NYS, but reproduced in Edi Lanners' 1976
edition, translated as: Columbus' Egg;
Paddington Press, London, 1978. Uses
Poyet's illustrations.
Hoffmann. 1893. Chap. X, pp. 350-351 & 390-391 = Hoffmann/Hordern pp. 248-249, with
photo.
Blyth. Match-Stick
Magic. 1921.
Collins. Book of
Puzzles. 1927. The bridge builder's puzzle, pp. 41-42. Three and four match bridges over goblets.
The Bile Beans Puzzle Book.
1933. No. 15. Use four matches to support a wine glass
above four wine glasses.
Ripley's Believe It or Not!, 16th series, 1971, unpaginated,
next to last page, shows six strips forming the sides of a hexagon and then
extending, so each strip goes over,
under, over, under, four other strips,
forming six versions of the three knife configuration.
Doubleday - 2.
1971. What a corker!, pp.
17-18. Four matches have to support a
cork off the table, and the match heads must not touch the table.
New section.
The Borromean rings occur as part of
the coat of arms of the Borromeo family, who were counts of the area north of
Milan since the 15C. The Golfo Borromeo
and the Borromean Islands are in Lago Maggiore, off the town of Stresa. In the 16 and 17C, the Counts of Borromeo
built a baroque palace and gardens on the main island, Isola Bella. The Borromean rings can be seen in many
places in the palace and gardens, including the sides of the flower pots! Although the Rings have been described as a
symbol of the Trinity, I don't know how they came to be part of the Borromean
crest, though the guide book describes some of the other features of the
crest. (Thanks to Alan and Philippa
Collins for the information and loan of the guide book.) Perhaps the most famous member of the family
was San Carlo Borromeo (1538-1584), Archbishop of Milan and a leader of the
Counter-Reformation, but he does not seem to have used the rings in his crest.
Clarence Hornung, ed.
Traditional Japanese Crest Designs.
Dover, 1986. On plates 10, 20,
24 & 39 are examples of Borromean rings.
On plate 10, the outer parts of the rings are split open. On plates 20 and 23, the Borromean rings are
in the centre of an extended pattern.
On plate 39, three extra rings are added, giving the Borromean ring
pattern four times. These designs have
no descriptions and the only dating is in the Publisher's Note which says such
designs were common in the 12C-17C.
Claude Humbert, ed.
1000 Ornamental Designs for Artists and Craftspeople. Ill. by Geneviève Durand. [As: Ornamental Design; Office du Livre,
Fribourg, Switzerland, 1970.] Dover,
2000. Design 260 on p. 73 is like the
pattern on plate 10 of Hornung, but more opened. It is only identified as from Japan.
Pietro Canetta.
Albero Genealogico Storico
Biografico della nobile Famiglia Borromeo.
1903. This is available at: www.verbanensia.org/biographica/Canetta%20‑%20albero%20genealogico.htm
. This says it is copied from a
manuscript of the archivist Pietro Canetta, with a footnote: Il Bandello, p.
243, vol. VIII. I suspect that this
refers to a publication of the MS.
This
simply says that the three rings represent the three houses of Sforza, Visconti
and Borromeo which are joined by marriages.
[Thanks to Dario Uri [email of 17 Jul 2001] for this source].
Takao Hayashi has investigated the Borromean Rings in Japan
and the following material comes from him.
Dictionary of Representative Crests. Nihon Seishi Monshō Sōran (A
Comprehensive Survey of Names and Crests in Japan), Special issue of Rekishi
Dokuhon (Readings in History), Shin
Jinbutsu Oraisha, Tokyo, 1989, pp. 271-484.
Photocopies of relevant pages kindly sent by Takao Hayashi.
The
Japanese term for the pattern is 'mitsu-wa-chigai mon' (three-rings-cross
crest).
Crest
3077 is called 'mitsu-kumi-kana-wa mon' (three-cross-metal-rings crest) and is
a clear picture of the Borromean Rings with thin rings.
Crest
3083 = 3930 is the Borromean Rings with moderately thick rings and an extra
rounded triangular design behind it.
Crest
3114 is the example on Hornung's plate 10.
Crests
3638-3644 are Borromean Rings with open rings, almost identical to the example
in Humbert.
Crest
3850 is the Borromean Rings with the rings being wiggly diamond shapes.
Crest
3926 is called 'mitsu-wa-chigai mon' (three-rings-cross crest) and is a clear
picture with moderately thick rings.
The
pattern usually occurs in a group of interlinked rings patterns called
'wa-chigai mon' (rings-cross crest), starting with two interlocked rings in
various surrounds (crests 3921-3925), then going to three rings (crests
3926-3931, all containing the Borromean Rings) and then more rings, but without
the Borromean property, though some are partially Borromean in that some of the
rings are not really interlocked and will be released when another ring is
removed. 3931 is the example on
Hornung's plate 10 without the surrounding circle. 3939 is on Hornung's plate 39.
Crests
4058 and 4060 are Borromean Rings where the rings are hexagons.
Crest
4284 is the Borromean Rings where the rings are squares.
Hayashi says the patterns may have developed from a linear
pattern of interlocked or overlapping circles used as early as the 8C or
9C. But the Borromean Rings do not
appear before the Edo Period (1603-1867).
Kansei Chōshū Shokafu. This is a book of family records compiled in 1799-1812. ??NYS -- information supplied by
Hayashi. Three families use the
Borromean Rings as a crest: the Tanabes
(from c1630); the Fushikis (from
c1700); the Kobayashis (from c1700).
P. G. Tait. Letter
to J. C. Maxwell in Jun 1877. Quoted on
p. 81 of Chris Pritchard: Aspects of the life and work of Peter Guthrie Tait,
pp. 77-88 IN: James Clerk Maxwell
Commemorative Booklet; James Clerk Maxwell Foundation, Edinburgh, 1999. Tait says he is confused about the diagram
which he draws, but does not name, of the Borromean rings: "This is
neither Knot nor Link. What
is it?"
Birtwistle. Math.
Puzzles & Perplexities. 1971.
New section. I know of earlier examples from perhaps the
1950s, but the problem must be much older.
A monk starts at dawn and walks to
top of a mountain to meditate. Next
day, at dawn, he walks down. Show that
he is at some point at the same time that he was there on the previous
day. This can be approached in two
ways.
Let
D be the distance to the top of
the mountain and let d(t) be his position at time t on
the second day minus his position at time t on the first day. Then
d(dawn) = D while d(evening) = -D, so at some time t, we must have d(t) = 0.
Equivalently, draw the graphs of his
position at time t on both days. The first day's graph starts at
0 at dawn and goes up to D at
evening, while the second day's graph begins at D at dawn and goes down
to 0
at evening. The two graphs must
cross.
This is an application of either the
Intermediate Value Theorem of basic real analysis or of the Jordan Curve Theorem
of topology.
Ivan Morris. The
Lonely Monk and Other Puzzles.
(Probably first published by Bodley Head.) Little, Brown and Co., 1970.
(Later combined into the Ivan Morris Puzzle Book, Penguin, 1972.) Prob. 1, pp. 14-15 & 91. Monk leaves his mountaintop retreat to go
down to the village at 5:00 one morning and starts back at 5:00 the next
morning. Is there always a place which
he is at at the same time each day? A
note says the problem is based on an idea of Arthur Koestler.
Yuri B. Chernyak & Robert S. Rose. The Chicken from Minsk. BasicBooks, NY, 1995. Chap. 5, prob. 9: Another triumph of central
planning, pp. 43-45 & 129-130. This
is a complicated version of the problem.
There are two roads from A to
B, such that for each point on
the first road, there is a point at most 20 m away of the second road. This is verified by sending two cars along
the two roads, attached by a 20 m phone line.
Is it possible to start trucks of width 22 m simultaneously from A
to B on one road and from B
to A on the other road? The
solution uses a graphical method, plotting the distance from A of
the first vehicle going along the first road, versus the distance from A of
the second vehicle, going along the second road. If the distance along the longer road is D,
the verification cars give a graph starting at (0, 0) and ending at (D, D),
while the trucks give a graph starting at (0, D) and going to (D, 0).
11.Q. TURNING AN INNER TUBE INSIDE OUT
New section.
Gardner. SA (Jan 1958) c= 1st Book, Chap. 14: Fallacies. Says that the process was illustrated in SA
(Jan 1950) and a New Jersey engineer sent in an inner tube which had been
turned inside out. Gardner then
describes and illustrates painting rings in both directions, one inside, the
other outside so they are interlinked at the beginning. He then draws the inner tube apparently
inverted, with the rings unlinked and asks for the resolution of this paradox. The solution is given in the Addendum:
"the reversal changes the 'grain,' so to speak, of the torus. As a result, the two rings exchange places
and remain linked." Several
readers made examples using parts of socks.
Victor Serebriakoff.
(A Second Mensa Puzzle Book.
Muller, London, 1985.) Later
combined with A Mensa Puzzle Book, 1982, Muller, London, as: The Mensa Puzzle
Book, Treasure Press, London, 1991. (I
have not seen the earlier version.)
Problem P13: The Yonklowitz diamond, pp. 164-166, & answer A22, pp.
237-238. After some preliminaries, he
asks three questions.
A) Can you pull an inner tube inside out
through a small hole?
B) What is the resulting shape?
C) If you draw circles in the two directions,
one inside and one outside so they are linked at the beginning, are they linked
at the end?
His
answers are: A) yes; B) the result has the same shape; C) the circles become unlinked. The last two are wrong. The shape changes -- effectively the two
radii change roles. When this is
carefully seen, the two circles are seen to still be linked.
This topic has a vast literature and
I intended to omit it as it is very difficult to summarise. However, I have recently acquired the books
of Haddon which are an excellent source, so I have included here just a few
books with extensive bibliographies which will lead the reader into the
literature. I have just learned
of:
International String Figure
Association, PO Box 5134, Pasadena, California, 91117, USA.
Web: http://members.iquest.net/~webweavers/isfa.htm . They publish Bulletin of the
International String Figure Association (formerly without the
International). [SA (Jun 1998)
77.]
String figures seem to be relatively
late in getting to Europe. The OED
cites 1768, 1823, 1824, 1867, 1887 for Cat's Cradle. Magician's Own Book and Secret Out are the earliest examples I
have where the process is explicitly described, but I have only started looking
for such material.
Abraham Tucker. The
Light of Nature Pursued. 7 vols. Vol. I, (1768); reprinted 1852,
p. 388. ??NYS -- quoted in the
OED. "An ingenious play they call
cat's cradle; ...." This is the
first citation in the OED.
Charles Lamb. Essays
of Elia: Christ's Hospital. 1823, p.
326. ??NYS -- quoted in the OED. "Weaving those ingenious parentheses
called cat-cradles." A popular
book says Lamb is describing his school days of 1782.
Child. Girl's Own
Book. Cat's Cradle. 1833: 76;
1839, 63; 1842: 57. "It is impossible to describe how this
is done; but every little girl will find some friend kind enough to teach her."
Fireside Amusements.
1850: No. 31, p. 95; 1890: No.
31, p. 71. As a forfeit, we find:
"Make a good cat's cradle."
Magician's Own Book.
1857. No. 9: The old man and his
chair, pp. 8-10.
The Secret Out.
1859. The Old Man and his Chair,
pp. 231-236. Like Magician's Own Book,
but many more drawings and more detailed explanation.
Caroline Furness Jayne.
String Figures. Scribner's,
1906, ??NYS. Retitled: String Figures
and How to Make Them; Dover, 1962. 97
figures given with instructions;
another 134 figures are pictured without instructions. 55
references.
Kathleen Haddon [Mrs. O. H. T. Rishbeth]. Cat's Cradles from Many Lands. Longmans, Green and Co., 1911. 16 + 95 pp, 50 string figures and 12 tricks,
14 items of bibliography.
Kathleen Haddon [Mrs. O. H. T. Rishbeth]. Artists in String. String Figures: Their Regional Distribution and Social
Significance. Methuen, 1930. Discusses string figures in their cultural
setting, describing five cultures and some of their figures. 41
string figures given with instructions, and some variants. P. 149 says "descriptions of over eight
hundred figures have been published and many more have been
collected." The Appendix on p. 151
gives the numbers of figures classified by type of objects represented and
location, with a total of 1605 figures plus 101 tricks. Two bibliographies, totalling 116
items. (An abbreviated version,
called String Games for Beginners, containing
28 of the figures and omitting
the cultural discussions, was printed by Heffers in 1934 and has been in print
since then, recently from John Adams Toys.)
Alex Johnston Abraham.
String Figures. Reference
Publications, Algonac, Michigan, 1988.
31 figures given with
instructions plus a chapter on Cat's Cradle.
156 references.
New section,
inspired by finding Moore.
Mr. X [cf 4.A.1].
His Pages. The Royal Magazine 26
(Jul-Oct 1911). Each issue offers
puzzle penknives as prizes.
Simon Moore.
Penknives and other folding
knives. Shire Album 223, Shire,
1988. Pp. 5-6 describe and illustrate
several examples which he says became popular in the early 17C. Most of these are of the type where the two
sides of the handle have to be separated, then one side turns 360o
to bring the blade out 180o and bring the handle back together. There are a few 18C and 19C examples, but
they were basically superseded by the spring-back knife in the late 17C. On p. 10, he says the 'lockback' knives,
with a mechanism to prevent accidental closure, were made from the late 18C.
A
mid 17C example has lugs between the two sides of the handle which are locked
by a pin, whose head is concealed by a false rivet near the pivot. When the rivet is moved, the pin can be
removed.
An example
dated 23 Oct 1699 has two dials which have to be set correctly to release the
locking lugs.