5.       COMBINATORIAL RECREATIONS

 

          7.AZ is actually combinatorial rather than arithmetical and I may shift it.

 

          5.A.    THE 15 PUZZLE, ETC.

 

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.

 

                    GENERAL

 

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?

 

L. E. 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.

S&B, pp. 126‑129, shows several versions of the puzzle.

 

                    EARLY ALPHABETIC VERSIONS

 

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.

 

                    LOYD

 

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.

 

                    THE 15 PUZZLE

 

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 90^o, 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.

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".

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 90^o 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 Hordern's book of Hoffmann puzzles, p. 74.)  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 Mondesir.  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 90^o 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. 

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.

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 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 90^o or by inverting the 6 and the 9.

Hoffmann.  1893.  Chap IV, no. 69: The "Fifteen" or "Boss" puzzles, pp. 161‑162 & 217‑218.  (Hordern, p. 74, has a photo of a version by Cremer.)  "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.  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."  (Surprisingly, Hordern doesn't mention this.)

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.  Side 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.  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....'  [I suspect the author's name is a pseudonym, but have not found this confirmed anywhere.  On pp. 132-138, he discusses the Diaphote Hoax, from a Pennsylvania daily newspaper of 10 Feb 1880, which features:  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.]

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.

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 Schofiled'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 90^o 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. 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  A_n  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 (Altenberg 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.

 

          5.A.1. NON‑SQUARE PIECES

 

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.

 

          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.

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.  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 aroud 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 havinf vertical connections.  The pieces have two markings: three colours and three shapes.  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.

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.

 

          5.A.3. ROLLING PIECE PUZZLES

 

          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 2870 _ 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.

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 120^o 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 Ko_niowski.  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. 

 

          5.A.4. PANEX PUZZLE

 

          Invented by Toshio Akanuma (??SP).  Manufactured by Tricks Co., Japan.  Described in Hordern, pp. 144-145 & 220, E35.  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 [c1985].  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 £ 7  by computer search.  X(6) = 881,  compared to the bounds of  796  and  881.  For  n = 10, the bounds are  27,564  and  31,537.

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.  Sketchs 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.

 

          5.B.    CROSSING PROBLEMS

 

See MUS I 1-13, Tropfke 658 and also 5.N.

 

Wolf, goat and cabbages:  Alcuin,  Abbot Albert,  Columbia Algorism,  Munich 14684,  Chuquet,  Tartaglia,  Merry Riddles,  van Etten,  Ozanam,  Dilworth,  Wingate/Dodson,  Jackson,  Endless Amusement II,  Boy's Own Book,  Nuts to Crack,  Taylor; The Riddler,  Child,  Magician's Own Book,  Book of 500 Puzzles,  Boy's Own Conjuring Book,  Secret Out,  Kamp,  Berg,  Lemon,  Hoffmann,  Brandreth Puzzle Book,  Carroll,  King,  Stein,  Zaslavsky,  Ascher

Extension to four items:  Gori,  Phillips,  Adams,  Gibbs,  Ascher

Adults and children:  Alcuin,  Kamp,  Hoffmann,  Parker?,  Gibbs

Three jealous husbands:  Alcuin,  Abbot Albert,  Columbia Algorism,  Munich 14684,  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,  Secret Out,  Lemon,  Hoffmann,  Fourrey,  Mr. X,  Loyd,  Williams,  Clark,  Goodstein,  O'Beirne,  Doubleday,  Allen, 

Four or more jealous husbands:  Pacioli,  Filicaia,  Tartaglia,  Bachet,  Delannoy,  Ball,  Carroll,  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,  Schubert,  Arbiter,  H&S,  Abraham,  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.  c1370. 

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.

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.  Prob. 61: De 3 mariti et 3 mogli gelosi.  3  couples.  Says that  4  or  5  couples requires a  3  person boat.

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.

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.  Tony Augarde; The Oxford Guide to Word Games; OUP, 1984; p. 6 says wolf, goat, cabbage appears in this.

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.  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.)

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."

Minguét.  Engaños.  1733.  Pp. 158-159 (1755: 114-115; 1822: 175-176).  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.

Problem 7: Three jealous husbands.  (Dilworth cites Wingate for this _ but this is in Kersey's additions _ cf Wingate/Kersey, 1678? above.)

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.

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 artihmetic 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?

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.

The Secret Out.  Op. cit. in 4.A.1.  1871?.  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 _ letters of 1873 & 1878 _ see below at 1899.

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 send me a xerox copy 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.

Mittenzwey.  1879?  Prob. 255, pp. 42 & 90.  Three kings and three servants who will rob the kings if they outnumber them _ i.e. 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 _ ??)

Hoffmann.  1893.  Chap. IV, pp. 157‑158 & 211‑213.

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.

No. 57: The wolf, the goat, and the cabbages.  (Hordern has a photo on p. 72 of "La Chevre et le Chou".  S&B, p. 134, shows 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.

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 Fonetenays' 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.

L. Carroll.  The Lewis Carroll Picture Book, S. D. Collingwood, ed.  T. Fisher Unwin, London, 1899.  I have a reprint, apparently in reduced format, Collins, c1900.  = Diversions and Digressions of Lewis Carroll, Dover, 1961.  = The Unknown Lewis Carroll, Dover, 1961(?).  (The pagination of the main text is the same in both reprints and agrees with some citations I have, but is quite different than the Collins.  I will indicate the Collins pages separately.) 

Letter of 22 Jan 1878 to Jessie.  Pp. 205-207 (Collins: 150).  Fox, goose and corn.

Letter of 15 Mar 1873 to Helen.  Pp. 212-215 (Collins: 154-155).  Fox, goose and corn. 

P. 317 (Collins: 231 or 232 (missing in my copy)).  Four couples _ only posed, no solution.

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.

Mr. X.  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.

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 misionaries 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 additonal 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.

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.

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.)

Phillips.  Brush.  1936.  Prob. L.2: Crossing the Limpopo, pp. 39‑40 & 98.  Same as in Week‑End, 1932.

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.

Nathan Altshiller Court.  Mathematics in Fun and in Earnest.  Mentor (New American Library), NY, 1961.  [John Fauvel has 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 - II.  1971.  Family outing, pp. 49-50.  Three couples, but one man has quarreled with the other men and his wife has quarreled 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.

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.  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.

 

          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, otherwise they go too fast.  This is often attributed to Carroll.

 

L. Carroll.  Op. cit. in 5.B, 1899, p. 318 (Collins: 232-233 (232 is lacking in my copy)).  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??

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.  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.

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.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), Adams, Rice

 

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.  The twelve coin problem.  Eureka 13 (1950) 7,20.  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. Bollabas, 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.

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.

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.1  RANKING 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 ‑ 2^k  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.

 

          5.D.   MEASURING PROBLEMS

 

          5.D.1. JUGS & BOTTLES

 

          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-(11,4,3;9): McKay

II-( ¥,5,3;1): Wood,  Serebriakoff,  Diagram Group

II-( ¥,5,3;4): Chuquet,  Wood

II-( ¥,7,4;5): Meyer,  Stein,  Brandes

 

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,  Lawrence

Versions with 4 jugs:  Tartaglia,  Anon (1961),  O'Beirne.

 

Abbot Albert.  c1240.  Prob. 4, p. 333.  I-(8,5,3)  _ one solution.

Columbia Algorism.  c1370.  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.

Dell'Abbaco.  c1370.  Prob. 66, p.62.  I-(8,5,3)  _ one solution.  "This problem is of little utility ...."  I have a colour slide of this.

Munich 14684.  14C.  Prob. XVIII  &  XXIX, pp. 80  &  83.  I-(8,5,3).

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. 

Prob. 53: A partire una botte de vino fra doi.  I-(8,5,3). 

Prob. 54: A partire un altra botte tra doi.  I-(12,7,5).

Prob. 55: De doi altri sotili divisioni de botti come se dira.  I‑(10,6,4)  and  I-(12,8,4).  These are impossible but only the heading is given and it makes no indication of this _ perhaps some trick method is used

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).

Yoshida (Shichibei) K_y_ (= Mitsuyoshi Yoshida) (1598-1672).  Jink_‑ki.  2nd ed., 1634 or 1641??.  ??NYS  The recreational problems are discussed in Kazuo Shimodaira; Recreative Problems on "Jing_ki", 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 Jing_ki and Jink_ki 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 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.)

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.

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.

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. 

Boy's Treasury.  1844.  Puzzles and paradoxes, no. 7, pp. 425 & 429.  I‑(8,5,3).  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.

Boy's Own Book.  The can of ale.  1855: 395;  1868: 432.  I‑(8,5,3).  One solution.

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, ...."

The Secret Out.  Op. cit. in 4.A.1.  1871?.  To divide equally eight pints of wine ..., pp. 12-13.

Kamp.  Op. cit. in 5.B.  1877.  No. 17, p. 326:  I-(8,5,3).

Bachet-Labosne.  1874.  For details, see Bachet, 1612.  Labosne adds a consideration of the general case which seems to be the first such.

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 Mag. 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).

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.

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.  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.

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.

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)

P. A. 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 measureable.  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.

L. Redei & A. Renyi.  On the representation of  1, 2, ..., N  by differences (in Russian).  Mat. Sbornik 66 (NS 24) (1949) 385‑389.

John Leech.  On the representation of  1, 2, ..., n  by differences.  J. London Math. Soc. 31 (1956) 160‑169.  Improves Redei & Renyi'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.

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.

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.

 

          5.D.5. MEASURE HALF A BARREL

 

          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.

 

          5.E.    EULER CIRCUITS AND MAZES

 

          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.

          Non-crossing Euler circuits:  Endless Amusement II,  Bellew,  Carroll,  Meyer,  Willson,  Scott,  Singmaster.

          K_n  denotes the complete graph on  n  vertices.

          Trick solutions:  Tom Tit,  Dudeney (1913),  Houdini,  Loyd Jr,  Ripley,  Meyer,  Leeming,  Adams,  Gibson,  Anon. (1986).

 

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 colleague Jeremy Wyndham has become interested in the seven bridges problem and has made inquiries which have 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  K_2n+1  and Euler's criterion.  Discusses square with diagonals.

Endless Amusement II.  1826?  Prob. 34, p. 211.  Pattern of two overlapping squares can be traced without crossings.

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  K_2n+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 to be traced without crossings.

Lewis Carroll.  Stuart Dodgson Collingwood; The Life and Letters of Lewis Carroll; T. Fisher Unwin, Lndon, (Dec 1898), 2nd ed., Jan 1899, p. 370, says:  "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."  Carroll mentions this in a letter of 22 Aug 1869 _ "Have you succeeded in drawing the three squares?" _ which is given on the same page of Collingwood.  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 here.

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 it, 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 dominos and  K_2n+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.  1892.  Chap. X, no. 9: Single‑stroke figures, pp. 338 & 375.  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.

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.

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', 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 this seems to be considering the central crossing as a further vertex.  I have now written a program to compute 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.  I have also computed 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.

 

          5.E.1. MAZES

 

          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 began in 1980 and issue 27 appeared in 1996.

          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 '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 17C, all mazes were unicursal, i.e. labyrinths.  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, 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.

          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.

          In the 17C, non-unicursal puzzle mazes developed.  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, but the map has 40.] 

          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.)  It has dead ends and islands, 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 centre on an island, so that the 'hand on wall' rule will not solve it.  The fourth Earl planted one of these at Chevening, Kent, in c1820 and it is extant though not open to the public.  It is additionally interesting in that it has no dead ends.  [Fisher, p. 71, with photo on p. 72 and diagram on p. 73.]

          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.]

          The first 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, Randall 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 Muhammad ibn ’Ahmad 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.

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.

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.

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.

 

          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) = k^n‑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  2^f(n) solutions,  where  f(n) = 2^n-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 the Second Gathering for Gardner, Atlanta, 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.

 

          5.E.2.a.         PANTACTIC SQUARES

 

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.]

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 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.  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 (it's not clear, but I think only one-sided ??).  He obtained US Patent 3,677,549 for it on 18 Jul 1972 (??NYS) and the version produced by Orda Ltd. was reviewed in G&P 54 (Nov 1976) (??NYS).  So it seems clear that Moscovich had the idea before Bouwkamp's version was published.

 

          5.F.    HAMILTONIAN CIRCUITS

 

          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, Zurich, 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.

G. 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.

 

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.

 

 

Rudrata.  K_vy_lank_ra.  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 was‑Sûlî wa ghair‑huma [Book of the Chess; extracts from the works of al‑'Adlî, as‑Sûlî and others].  Copied by Abû Ishâq Ibrâhîm ibn al‑Mubârak ibn ‘Alî al‑Mudhahhab al‑Baghdâdî.  Murray 171‑172 says it is MS ‘Abd‑al‑Hamid 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‑shatranj by Abû’l‑‘Abbâs Ahmad 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 inital 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 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, which 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 previous one.  [This is equivalent to the octagram puzzle of 5.R.6.]

"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 ibn Ibrâhîm al‑Hakîm.  Nuzhat al‑arbâb al‑‘aqûl fî’sh‑shatranj 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 Rudrata, 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 ibn Yahya ibn Abî Hajala at‑Tilimsâni alH‑anbalî.  Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shatranj (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 on the  3 x 3  board.  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.  Murray 337, 673 (CB236) & 690 and Wieber 481 show these problems.

Risâlahi 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.]

Sifat mal ‘ûb al‑faras fî gamî abyât aš‑šatran_.  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.)

Bhatta N_lakantha.  Bhagavantabh_skara.  17C.  End of 5th book.  ??NYS, described by Murray 63‑66.  The author gives three tours, in the poetic form of Rudrata, 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 belief 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 180^o 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 180^o symmetry, two halves, and other size boards.

La Corsa del Cavallo per tutt'i scacchi dello scacchiere.  Lelio della Volpe, Bologna, 1766.  11pp.  [Lelio della Volpe is sometimes given as the author.]  ??NYS _ 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.  He says it gives 38 tours.

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.

Paul de Hijo [= Abbe 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!

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.

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  10^23±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 Jellis.  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.

 

          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, with photo of the board.)

For a long time, the only known example of the game 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 are lacking.  Dalgety obtained another board in 1998, again without the pegs and instructions.

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.

Pearson.  1907.  Part III, no. 60: The open door, pp. 60 & 130.  Prisoner is 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.

C. W. Ceram.  Gods, Graves and Scholars.  2nd ed., Gollancz, London, 1971, p. 25.  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 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.

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  11487  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.    CONNECTION PROBLEMS

 

          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.

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.

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 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;  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).

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.  The coloured cubes problem.  Eureka 9 (1947) 9‑11.  General graphical solution 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.

 

          5.H.2. MACMAHON PIECES

 

          Haubrich's 1995 survey, in 5.H.4, includes MacMahon puzzles as one class.

 

P. A. MacMahon & J. R. J. Jocelyn.  UK Patent 3927 _ Appliances to be used in Playing a New Class of Games.  Applied 29 Feb 1892;  complete application 28 Nov 1892;  accepted 28 Jan 1893 [though the patent is headed A.D. 1892].  5pp + 2 plates.  Describes the 24 triangles with four types of edge and mentions other numbers of edge types.  Describes various games and  puzzles.

F. H. Richards.  US Patent 331,652.  1895.  For MacMahon 4-coloured triangles.  Cited by Gardner in Magic Show, but there must be some confusion as the patent number corresponds to 1885 and this patent is not in Marcel Gillen's compendium of US puzzle patents _ ??  ??NYS

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, ??NYS  Is this a confusion with the above patent??

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.  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 bünte Wurfel _ MacMahon's problem.  Teubner, Leizig, 1934, 128pp.  ??NYS.

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.

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.

 

          5.H.3. PATH FORMING PUZZLES

 

          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.

 

Hoffmann.  1893.  Chap. III, No. 18: The endless chain, pp. 99-100 & 131.  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 piece are rectangles of width one.

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 almost identical to Hoffmann.

See Haubrich in 5.H.4.

 

          5.H.4. OTHER AND GENERAL

 

          These all have coloured edges unless specified.  See S&B, p. 36, for examples.

 

Edwin L. Thurston.  US Patent 487,797 _ Puzzle.  Applied 30 Sep 1890;  patented 13 Dec 1892.  3pp + 3pp diagrams.  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.  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.  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.

Angus K. Rankin.  US Patent 1,006,878 _ Puzzle.  Applied 3 Feb 1911;  patented 24 Oct 1911.  2pp + 1p diagrams.  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.

Le Berger Malin.  France, c1915.  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.

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.  Aserts 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.  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'.

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]

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??)

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 seven most common matching rules:  Heads and Tails;  Edge Matching (i.e. MacMahon);  Path Matching;  Corner Matching;  Corner Dismatching;  Jig-Saw-Like;  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.

 

          5.I.     LATIN SQUARES AND EULER SQUARES

 

          This topic ties in with certain tournament problems but I have not covered them.

 

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 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‘_rif.  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 rearrangment 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 17^th of Ian

                              1709/8 in the 22^th 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.  I don't know how the acrostic is physically laid out.

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.   (I have a reference for this to the 1723 edition.)

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.

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.

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.

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.

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.

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.

 

          5.I.1.  EIGHT QUEENS PROBLEM

 

          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 (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.)

E. 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.

E. 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.

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.

No. 24: No two in a row.  Eight queens.

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.  This is a solution of the eight queens problem, cut into four quadrants and jumbled.  The goal is to reconstruct the solution.

                    Hordern, p. 94, and S&B, p. 37, show a version of this called  Jeu des Manifestants  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 soluiton:  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.

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 - II.  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.

 

          5.J.     SQUARED SQUARES, ETC.

 

          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 Moro_'s work.  Moro_ later found the first example of Sprague but did not publish it.

Z. Moro_.  O rozk_adach prostok_tów na kwadraty (In Polish) (On the dissection of a rectangle into squares).  Przegl_d Matematyczno‑Fizyczny (Warsaw) 3 (1925) 152‑153.  Decomposes rectangles into 9 and 10 unequal squares.  (Translation provided by A. M_kowski, 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 & M. Dehn.  Einführung in der analytische Geometrie der Ebene und des Raumes.  2nd ed., Appendix VI: Ungelöste Probleme der Analytischen Geometrie, pp. 402‑411.  Springer, Berlin, 1931.  ??NYS

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 Moro_'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.

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 fo 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.

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 [probably C. 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.

 

          5.J.1.  MRS PERKINS'S QUILT

 

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.

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.

 

          5.J.2.  CUBING THE CUBE

 

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.

 

          5.K.    DERANGEMENTS

 

          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.  Ganita-Bh_rat_ (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??);  2nd ed., (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)), Jombert & Quillau, 1714.  [The 1714 may differ from the 1713 ??]  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.)

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 derrangement 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.

 

          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.

          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

 

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.

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. 

 

          5.K.3. CAYLEY'S MOUSETRAP

 

          This is a solitaire 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.

 

          5.L.    MÉNAGE PROBLEM

 

          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 arangement.  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 geenrating 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  K_a  contains a monochromatic  K_a  or  K_b.

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 Competiton 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  K_n  contains a  K_a  of the first colour or a  K_b  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 nottion.

 

Alcuin.  9C.  Prob. 52: Propositio de homine patrefamilias.  Wants to get supplies from  A  to  B.  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??)  He is trying to get the most to the other side, so this is different than the 20C versions.

Pacioli.  De Viribus.  c1500.  Probs. 49‑52.  Agostini only describes Prob. 49 in some detail.

Prob. 49: De doi a portar pome ch' più n'avanza.  One has 90 apples to transport 30 miles, but one eats one apple per mile and one can carry at most 30 apples.  (So these are similar to Alcuin.)

Prob. 50: De 3 navi per 30 gabelle 90 mesure.

Prob. 51: De portar 100 perle 10 miglia lontano 10 per volta et ogni miglio lascia una.

Prob. 52: El medesimo con più avanzo per altro modo.

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.

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.

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 below.

Gamow & Stern.  1958.  Refueling.  Pp. 114‑115.

Martin Gardner, SA (May  &  June 1959) c= 2nd Book, chap. 14, prob. 1.  (The book gives extensive references which were not in SA.)

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,  ....

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.  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)]

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 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.

 

          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.  Barbeau, 1995, notes that moving to  BWBWBWBW  is a different problem, requiring an extra move.  I had not noticed this difference before _ indeed I had it the wrong way round in the heading of this section.  I must check to see if this occurs earlier.

 

Genjun Nakane (= H_jiku 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.  La Nature 15 (1887, 2nd sem) 10-11.  ??NYS _ cited by Ahrens.

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.

          No. 19: The "Four and Four" puzzle.

          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...  = Graham, 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.

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.

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 & dimes, pp. 1 & 25.  Usual version with 8 coins.  Solution has blanks at the opposite end to where they began.

No. 2: Nickles & 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.

Walter Gibson.  Big Book of Magic for All Ages.  Kaye & Ward, Kingswood, Surrey, 1982. 

Six cents at a time, p. 117.  Uses pennies and nickles.  .....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.

 

          5.P.1. SHUNTING PUZZLES

 

          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.

 

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.

Mr. X.  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.  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.

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.

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 - II.  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.

 

          5.P.2. TAQUIN

 

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

 

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, Zurich, 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 passed 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.]

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.

 

          5.R.    JUMPING PIECE GAMES

 

          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.

 

          5.R.1. PEG SOLITAIRE

 

          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 Remond 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??);  2nd ed., (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)), Jombert & Quillau, 1714.  [The 1714 may differ from the 1713 ??]  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 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. Kosi_ski & 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. 

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  45^o  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.

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.

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??

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.

 

          5.R.1.a.        TRIANGULAR VERSION

 

          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.

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.

 

          5.R.1.b.        OTHER SHAPES

 

          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.]

 

          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 intercahnge 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.  1879?  Prob. 267 & 268, pp. 44 & 91‑92.  Problem with  3 & 3  and  4 & 4  brown and white horses.

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 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.

No. 17: The "Right and Left" puzzle.  Three and three.  Hordern, p. 77, shows a Jumping Frogs puzzle.

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.

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.

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.

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, particulary in the United States and France" and they show an example of the period labelled The Sphinx and Pyramid Puzzle _ An Egyptian Novelty.

See Harbin in 5.R.4 for a 1963 example.

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.

 

          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.  Mentions that it is produced by Messrs Heywood, as below.  Solution in 52 moves, which he believes is minimal.

John Heywood, Manchester, produced a version called 'The English Sixteen Puzzle', undated, but by 1893 as Hoffmann cites it.

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.

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 arrangment.  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.  (1973);  Tempo Books (Grosset & Dunlap), NY, 1978.  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.

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.  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.

 

          5.R.5. FOX AND GEESE, ETC.

 

          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.

Shackley Marmion.  A Fine Companion (a play).  1633.  II, v.  "Let him sit in the shop ... and play at fox and geese with the foreman."  Earliest English occurrence of fox-and-geese.  ??NYS _ 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.

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 solitatire 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).  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; A Fine Companion (a play); 1633; II,v, for the oldest English occurrence of fox-and-geese.  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.  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.

 

          5.R.6. OCTAGRAM PUZZLE

 

          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.  This is equivalent to the 7 knights problem mentioned in 5.F.1 under King's Library MS 13 (c1275);  Shihâbaddîn at‑Tilimsâni (c1370)  and Bonus Socius (MS of c1530).

          Versions with different numbers of points.

 5 points:  Rohrbough.

 7 points:  Meyer.

 9 points:  Dudeney.

10 points:  Bell & Cornelius; 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.

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.

Endless Amusement II.  1826?  Prob. 28, pp. 203-204.

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.

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, no. 13, pp. 267 & 280‑281.  Basic octagram puzzle.

Chap. VI, no. 14: The "Okto" puzzle, pp. 268 & 281.  Here the counters and points are coloured.

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.

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.

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.

 

          5.R.7. PASSING OVER COUNTERS

 

          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. 

          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.  I have only seen two other instances of this form.  One could also permit passing over one pile, which is solvable for any even number  ³ 4.

          Double Five Puzzle and Singmaster & Abbott deal with the problem in a circle and with piles to be left in specific locations.

          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. 4 is "Analysis of the Essay of Games".  F. 4.v 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  D^r 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.

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.

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.

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.

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.

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.

Sullivan.  Unusual.  1947.  Prob. 39: On the line.  Ten 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.

 

          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.

Mr. X.  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.

Clark.  Mental Nuts.  1904: no. 14: The chain question.  1916: no. 59: The chain puzzle.  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.

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.  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.

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.

 

          5.T.    DIVIDING A CAKE FAIRLY

 

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.

 

          5.U.   PIGEONHOLE RECREATIONS

 

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.

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.

Sullivan.  Unusual.  1943.  Prob. 18: In a dark room.  Picking shoes and socks to get pairs.

Doubleday - II.  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.]

 

          5.V.    THINK‑A‑DOT, ETC.

 

          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: [email protected]; 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.

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.  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.

 

          5.W.1.          BOILING EGGS

 

          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:  Gooding (1994) in 5.X.1.

 

          5.X.1. COUNTING TRIANGLES

 

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: Pharoah's seal, pp. 75 & 174.  Isoceles right triangles in a square pattern with some diagonals.

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 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, not paginated.  Triangular pattern with  2  triangles on a side, with the three altitudes drawn.

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.

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.

Doubleday - II.  1971.  Count down, pp. 127-128.  How many triangles in the pentagram (i.e. a pentagon with all its diagonals)?

Gyles Brandreth.  Brandreth's Bedroom Book.  Eyre Methuen, London, 1973.  Triangular, pp. 27 & 63.  Count triangles in an irregular pattern.

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.  Consider a pentagram and draw lines from each star point through the centre to the opposite crossing point.  How many triangles?

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 amd 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-II.

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-II.

Problem 116, with Solution at the back of the book.  Count the triangles in a 'butterfly' pattern.

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 counitng 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 Silke and Mogens Esrom Larsen.  Silke 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.

 

          5.X.2. COUNTING RECTANGLES OR SQUARES

 

          I have just seen Adams.  There are probably earlier examples of these types of problems.

 

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.

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?

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.

 

          5.X.3. COUNTING HEXAGONS

 

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.

 

          5.X.4. COUNTING CIRCLES

 

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 C

                                               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.  Nothing asks for the number of ways of reading the inscription.  I get  4 BC(16,10)  =  32032  ways.

 

                                                           eineRnhoJsesJohnRenie

                                                           ineRnhoJseiesJohnReni

                                                           neRnhoJseiliesJohnRen

                                                            eRnhoJseileliesJohnRe

                                                            RnhoJseilereliesJohnR

                                                             nhoJseilerereliesJohn

                                                             hoJseilereHereliesJoh

                                                             nhoJseilerereliesJohn

                                                            RnhoJseilereliesJohnR

                                                            eRnhoJseileliesJohnRe

                                                           neRnhoJseiliesJohnRen

                                                           ineRnhoJseiesJohnReni

                                                           eineRnhoJsesJohnRenie

 

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 receved 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  P_n(3)  where P_n(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  P_n(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.

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.  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.

          New section.  Though these results seem like they must be old, the ideas seem to have originated with the eight queens problem, c1850, 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.

          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. Jellis.  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.

 

          5.Z.1. KINGS

 

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.

 

          5.Z.2. QUEENS

 

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, M^r 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 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.  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.

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.

 

          5.Z.3. BISHOPS

 

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.

 

          5.Z.4. KNIGHTS

 

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.

 

          5.Z.5. ROOKS

 

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.

 

          5.Z.6. MIXTURES

 

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.

 

          5.AA. CARD SHUFFLING

 

          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 the Second Gathering for Gardner in Atlanta, Jan 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.  Op. cit. in 4.A.1.  1871?  Permutation table, pp. 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).

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  2^e º ±1 (mod 4n+1),  which is  12  for  n = 26.

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 (Jun 1957) 70‑71, (Jul 1957) 78-79, (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, so perhaps Gardner had written Ibidem by mistake.

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.)

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.  Pi Mu Epsilon J. 6 (1975) 85-92.  ??NYS _ cited by Herstin & Kaplansky.

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  2^a-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.

          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 car goes 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.

 

          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.

 

          5.AE.  REVERSING CUPS

 

          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.

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.