Sources page biographical material

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. The diagram is just an 8-cycle and is the same as the knight's connections on the 3 x 3 board, so the octagram puzzle is equivalent to the 7 knights problem mentioned in 5.F.1. Further, the 4 knights problem of 5.F.1 has the same 8-cycle, with men at alternate points of it.

Versions with different numbers of points.

5 points: Rohrbough.

7 points: Mittenzwey; Meyer.

9 points: Dudeney.

10 points: Bell & Cornelius; Hoffmann; Cohen; Williams; Toymaker; Rohrbough; Putnam.

13 points: Berkeley & Rowland.
Bell & Cornelius. Board Games Round the World. Op. cit. in 4.B.1. 1988. Pentalpha, p. 15. Says that a pentagram board occurs at Kurna, Egypt, c-1400 and that the solitaire game of Pentalpha is played in Crete. This has 9 men to be placed on the vertices and the intersections of the pentagram. Each man must be placed on a vacant point, then slid ahead two positions along one straight line. The intermediate point may be occupied, but the ending point must be unoccupied. Unfortunately we don't know if the Egyptian board was used for this game.

Pacioli. De Viribus. c1500.

Ff. 112r - 113v. .C(apitolo). LXVIII. D(e). cita ch' a .8. porti ch' cosa convi(e)ne arepararli (Chap. 68. Of a city with 8 gates which admits of division ??). = Peirani 158-160. Octagram puzzle with a complex story about a city with 8 gates and 7 disputing factions to be placed at the gates.

F. IVv. = Peirani 8. The Index gives the above as Problem 83. Problem 82: De .8. donne ch' sonno aun ballo et de .7. giovini quali con loro sa con pagnano (Of 8 ladies who are at a ball and of 7 youths who accompany them).

Schwenter. 1636. Part 2, exercise 36, pp. 149-150. Octagram.

Witgeest. Het Natuurlyk Tover-Boek. 1686. Prob. 4, pp. 224-225. Octagram, taken from Schwenter.

Les Amusemens. 1749. P. xxxiii.

Catel. Kunst-Cabinet. 1790. Das Achteck, pp. 12-13 & fig. 36 on plate II. The rules are not clearly described.

Bestelmeier. 1801. Item 290: Das Achteckspiel. Text copies part of Catel.

Charles Babbage. The Philosophy of Analysis -- unpublished collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more details. Ff. 131-133 are an analysis of the heptagram puzzle.

Rational Recreations. 1824. Feat 34, pp. 161-164. Octagram.

Endless Amusement II. 1826? Prob. 28, pp. 203-204. = New Sphinx, c1840, pp. 137-138.

Nuts to Crack IV (1835), no. 194 -- part of a long section called Tricks upon Travellers.

Family Friend (Dec 1858) 359. Practical puzzles -- 1. I don't have the answer.

The Boy's Own Magazine 3 (1857) 159 & 192. Puzzle of the points.

Illustrated Boy's Own Treasury. 1860. Practical Puzzles No. 6, pp. 396 & 436.

The Secret Out. 1859. To Place Seven Counters upon an Eight-Pointed Star, pp. 373-374.

J. J. Cohen, New York. Star puzzle. Advertising card for Star Soap, Schultz & Co., Zanesville, Ohio, Copyright May 1887. Reproduced in: Bert Hochberg; As advertised Puzzles from the collection of Will Shortz; Games Magazine 17:1 (No. 113) 10-13, on p. 11. Identical to pentalpha - see Bell & Cornelius above.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles No. IX: The reversi puzzle, pp. 8-10. Version with 13 cards in a circle and one can move ahead by any number of steps. If there are x cards and one moves ahead s steps, then x and s must have no common factor.

Hoffmann. 1893.

Chap. VI, pp. 267-268 & 280-281 = Hoffmann-Hordern, pp. 180-181, with photos.

No. 13. No name. Basic octagram puzzle. Photo on p. 181 shows: The Seven Puzzles, by W. & T. Darton, dated 1806-1811; a Tunbridge ware version dated 1825-1840; and Jeu de Zig Zag, by M. D., Paris, 1891-1900.

No. 14. The "Okto" puzzle, pp. 268 & 281. Here the counters and points are coloured. Photo on p. 181 of The "Okto" Puzzle by McGaw, Stevenson & Orr, Ltd. for John Stewart, dated 1880-1895.

Chap. X, no. 8: Crossette, pp. 337 & 374-375 = Hoffmann-Hordern, pp. 229-230, with photo. 10 counters in a circle. Start anywhere and move ahead three. Photo on p. 230 shows The Mystic Seven, a seven counter version, by the Lord Roberts Workshops, 1914-1920.

Mittenzwey. 1895? Prob. 329, pp. 58 & 106; 1917: 329, pp. 52 & 101. Heptagram.

Dudeney. Problem 58: A wreath puzzle. Tit Bits 33 (6 & 27 Nov 1897) 99 & 153. Complex nonagram puzzle involving moves in either direction and producing the original word again.

Clark. Mental Nuts. 1897, no. 54; 1904, no. 80; 1916, no. 69. A little puzzle. Usual octagram.

Benson. 1904. The eight points puzzle, pp. 250 251. c= Hoffmann, no. 13.

Slocum. Compendium. Shows the "Octo" Star Puzzle from Gamage's 1913 catalogue.

Williams. Home Entertainments. 1914.

Crossette, pp. 115-116. Ten points, advancing three places.

Eight points puzzle, pp. 120-121. Usual octagram.

"Toymaker". Top in Hole Puzzle. Work (23 Dec 1916) 200. 10 holes and one has to move to the third position and reverse the top in that hole.

Blyth. Match-Stick Magic. 1921. Crossing the points, pp. 83-84.

Hummerston. Fun, Mirth & Mystery. 1924.

The sacred seven, Puzzle no. 5, pp. 26 & 173. Octagram puzzle on the outer points of the diagram shown in 5.A.

The four rabbits, Puzzle no. 6, pp. 26 & 173. Using the octagram shown in 5.A, put black counters on locations 1 and 2 and white counters on 7 and 8. The object is to interchange the colours. This is like the 4 knights problem except the corresponding 8-cycle has men at positions 1, 2, 5, 6. He counts a sequence of steps by the same man as a move and hence solves it in 6 moves (comprising 16 steps).

Will Blyth. Money Magic. C. Arthur Pearson, London, 1926. Turning the tails, pp. 66-69. 8 coins in a circle, tails up. Count from a tail four ahead and reverse that coin. Get 7 heads up. Counting four ahead means that if you start at 1, you count 1, 2, 3, 4 and reverse 4.

King. Best 100. 1927. No. 64, pp. 26-27 & 54.

Rohrbough. Puzzle Craft. 1932.

Count 4, p. 6. 10 points on a circle, moving ahead 3. (= Rohrbough; Brain Resters and Testers; c1935, p. 21.)

Star Puzzle, p. 8 (= p. 10 of 1940s?). Consider the pentagram with its internal vertices. First puzzle is Pentalpha. Second is to place a counter and move ahead three positions. The object is to get four counters on the points, which is the same as the pentagram puzzle, moving one position.

Jerome S. Meyer. Fun-to-do. Op. cit. in 5.C. 1948. Prob. 18: Odd man out, pp. 27 & 184. Version with 7 positions in a circle and 6 men where one must place a man and then move him three places ahead.

Putnam. Puzzle Fun. 1978. No. 63: Ten card turnover, pp. 11 & 35. Ten face down cards in a circle. Mark a card, count ahead three and turnover.

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 than given here. In describing solutions, 4/1 means move the 4th piece on top of the 1st piece.

There are trick solutions where a counter moves to a vacated space or even lands between two spaces. See: Mittenzwey; Haldeman-Julius; Hemme.

Berkeley & Rowland give a problem where each move must pass a counter over two piles. This makes the problem easier and it is solvable for any even number of counters  6, but it gives more solutions. See: Berkeley & Rowland; Wood; Indoor Tricks & Games; Putnam; Doubleday - 1.

One could also permit passing over one pile, which is solvable for any even number  4.

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

Mittenzwey, Lucas and Putnam consider making piles of three by passing over 3, etc.

Kanchusen. Wakoku Chiekurabe. 1727. Pp. 38-39. Jukkoku-futatsu-koshi (Ten stones jumping over two). Ten counters, one solution.

Charles Babbage. The Philosophy of Analysis -- unpublished collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more details. F. 4r is "Analysis of the Essay of Games". F. 4v has "The question of the shillings passing at each time over two or a certain number 8 is the least number Any number being given and any law of transit 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: 4/1 7/3 5/9 2/6 8/10. = New Sphinx, c1840, pp. 135-135.

Nuts to Crack II (1833), no. 122. 10 counters, identical to Endless Amusement II.

Nuts to Crack V (1836), no. 68. Trick of the eight sovereigns. Usual form.

Young Man's Book. 1839. P. 234. Ingenious Problem. 10 halfpence. Identical to Endless Amusement II.

Family Friend 2 (1850) 178 & 209. Practical Puzzle, No. VI. Usual form with eight counters or coins. One solution.

Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles, no. 2, p. 176 (1868: 187). Passing over coins. Gives two symmetric solutions.

Magician's Own Book. 1857. Prob. 34: The counter puzzle, pp. 277 & 300. Identical to Book of 500 Puzzles, prob. 34.

The Sociable. 1858. Prob. 16: Problem of money, pp. 291-292 & 308. Start with 10 half dimes, says to pass over one, but solution has passing over two. One solution. = Book of 500 Puzzles, 1859, prob. 16, pp. 9-10 & 26.

Book of 500 Puzzles. 1859.

Prob. 16: Problem of money, pp. 9-10 & 26. As in The Sociable.

Prob. 34: The counter puzzle, pp. 91 & 114. Eight counters, two solutions given. Identical to Magician's Own Book.

The Secret Out. 1859. The Crowning Puzzle, p. 386. 'Crowning' is here derived from the idea of crowning in draughts or checkers. One solution: 4/1 6/9 8/3 2/5 7/10.

Boy's Own Conjuring Book. 1860.

Prob. 33: The counter puzzle, pp. 240 & 264. Identical to Magician's Own Book, prob. 34.

The puzzling halfpence, p. 342. Almost identical to The Sociable, prob. 16, with half-dimes replaced by halfpence.

Illustrated Boy's Own Treasury. 1860. Prob. 17, pp. 398 & 438. Same as prob. 34 in Magician's Own Book but only gives one solution.

Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 593, part 6, pp. 299-411: Sechs Knacknüsse. 10 counters, one solution.

Hanky Panky. 1872. Counter puzzle, p. 132. Gives two solutions for 8 counters and one for 10 counters.

Kamp. Op. cit. in 5.B. 1877. No. 12, p. 325.

Mittenzwey. 1880. Prob. 235-238, pp. 43-44 & 93-94; 1895?: 262-266, pp. 47-48 & 95-96; 1917: 262-266, pp. 43-44 & 91.

235 (262). Usual problem, with 10 counters. Two solutions.

--- (263). Added in 1895? Same with 8 counters. Two solutions.

236 (264). 12 numbered counters in a circle. Pass over two to leave six piles of two on the first six positions. Solution is misprinted in all editions!

237 (265). 12 counters in a circle. Pass over three to leave six piles of two, except the last move goes over six. The solution allows landing counters on vacated locations!

238 (266). 15 counters in a row. Pass over 3 to leave five piles of three. The solution allows landing a counter on a vacated location and landing a counter between two locations!!

Lucas. RM2. 1883. Les huit pions, pp. 139-140. Solves for 8, 10, 12, ... counters. Says Delannoy has generalized to the problem of mp counters to be formed into m piles (m  4) of p by passing over p counters.

[More generally, using one of Berkeley & Rowland's variations (see below), one can ask when the following problem is solvable: form a line of n = kp counters into k piles of p by passing over q [counters or piles]. Does q have to be  p?]

Double Five Puzzle. c1890. ??NYS -- described by Slocum from his example. 10 counters in a circle, but the final piles must alternate with gaps, e.g. the final piles are at the even positions. This is also solvable for 4, 8, 12, 16, ..., and I conjectured it was only solvable for 4n or 10 counters -- it is easy to see there is no solution for 2 or 6 counters and my computer gave no solutions for 14 or 18. For 4, 8, 10 or 12 counters, one can also leave the final piles in consecutive locations, but there is no such solution for 6, 14, 16 or 18 counters. See Singmaster & Abbott, 1992/93, for the resolution of these conjectures.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles.

No. VII: The halma puzzle, pp. 6-7. Arrange the first ten cards of a suit in a row so that passing over two cards leaves five piles whose cards total 11 and are in the odd places. Arrangement is 7,6,3,4,5,2,1,8,9,10. Move 2 to 9, 4 to 7, 8 to 3, 6 to 5, 10 to 1.

No. VIII: Another version, p. 7. With the cards in order and passing over two piles, leave five piles of two. But this is so easy, he adds that one wants to leave as low a total as possible on the tops of the piles. He moves 7 to 10, 6 to 3, 4 to 9, 1 to 5, 2 to 8, leaving a total of 20.

[However, this is not minimal -- there are six solutions leaving 18 exposed, e.g. 1 to 4, 3 to 6, 7 to 10, 5 to 9, 2 to 8. For 6 cards, the minimum is 6, achieved once; for 8 cards, the minimum is 11, achieved 3 times; for 12 cards, the minimum is 27, achieved 10 times. For the more usual case of passing over two cards, the minimum for 8 cards is 15, achieved twice; for 10 cards, the minimum is 22, achieved 4 times, e.g. by 7 to 10, 5 to 2, 3 to 8, 1 to 4, 6 to 9; for 12 cards, the minimum is 31, achieved 6 times; for 14 cards, the minimum is 42, achieved 8 times. For passing over one pile, the minimum for 4 cards is 3, achieved once; the minimum for 6 cards is 7, achieved twice; the minimum for 8 cards is 13, achieved 3 times; for 10 cards, the minimum is 21, achieved 4 times; the minimum for 12 cards is 31, achieved 5 times. Maxima are obtained by taking mirror images of the minimal solutions.]

Puzzles with draughtsmen. The Boy's Own Paper 17 or 18?? (1894??) 751. 8 men, passing over two men each time. Notes that it can be extended to any even number of counters.

Clark. Mental Nuts. 1897, no. 63: Toothpicks; 1904, no. 83 & 1916, no. 70: Place 8 toothpicks in a row. One solution.

Parlour Games for Everybody. John Leng, Dundee & London, nd [1903 -- BLC], p. 32: The five pairs. 10 counter version, one solution.

Wehman. New Book of 200 Puzzles. 1908. P. 15: The counter puzzle and Problem of money. 8 and 10 counter versions, the latter using pennies. Two and one solutions.

Ahrens. MUS I. 1910. Pp. 15-17. Essentially repeats Lucas.

Manson. 1911. Decimal game, pp. 253-254. Ten rings on pegs. "Children are frequently seen playing the game out of doors with pebbles or other convenient articles."

Blyth. Match-Stick Magic. 1921. Straights and crosses, pp. 85-87. 10 matchsticks, one must pass over two of them. Two solutions, both starting with 4 to 1.

Hummerston. Fun, Mirth & Mystery. 1924.

The pairing puzzle, Puzzle no. 8, pp. 27 & 173. Essential 8 counters in a circle, with four in a row being white, the other four being black. Moving only the whites, and passing over two, form four piles of two.

Pairing the pennies, Puzzle no. 39, pp. 102 & 178. Ten pennies, one solution.

Will Blyth. Money Magic. C. Arthur Pearson, London, 1926. Marrying the coins, pp. 113 115. Ten coins or eight coins, passing over 2. Gives two solutions for 10, not noting that the case of 10 is immediately reduced to 8. Says there are several solutions for 8 and gives two.

Wood. Oddities. 1927. Prob. 45: Fish in the basket, pp. 39-40. 12 fish in baskets in a circle. Move a fish over two baskets, continuing moving in the same direction, to get get two fish in each of six baskets, in the fewest number of circuits.

Rudin. 1936. No. 121, pp. 43 & 103. 10 matches. Two solutions.

J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Straights and crosses, pp. 105-106. As in Blyth, 1926.

Indoor Tricks and Games. Success Publishing, London, nd [1930s??].

How to pair the pennies, p. 4. 8 pennies, one solution.

The ten rings. p. 4. 10 rings, passing over two piles, one solution.

Haldeman-Julius. 1937.

No. 91: Jumping pennies, pp. 11 & 25. Six pennies to be formed into two piles of three by jumping over three pennies each time. Solution has a trick move. Jump 1 to 5, 6 to 3 and 2 to 1/5, which gives the position: 6/3 4 2/1/5. He then says: "No. 4 jumps over 5, 1 and 2 -- then jumps back over 5, 1 and 2 and lands upon 3 and 6, ...." Since the rules are not clear about where a jumping piece can land, the trick move can be viewed as a legitimate jump to the vacant 6 position, then a legitimate move over the 2/1/5 pile and the now vacant 4 position onto the 6/3 pile. If the pennies are considered as a cycle, this trick is not needed.

No. 148: Half dimes, pp. 16 & 143. 10 half dimes, passing over one dime (i.e. two counters).

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

Doubleday - 1. 1969. Prob. 75: Money moves, pp. 91 & 170. Ten pennies. Jump over two piles. Says there are several solutions and gives one, which sometimes jumps over three or four pennies.

Putnam. Puzzle Fun. 1978.

No. 26: Pile up the coins, pp. 7 & 31. 12 in a row. Make four piles of three, passing over three coins each time.

No. 27: Pile 'em up again, pp. 7 & 31. 16 in a row. Make four piles of four, passing over four or fewer each time.

No. 60: Coin assembly, pp. 11 & 35. Ten in a row, passing over two each time.

No. 61: Alternative coin assembly, pp. 11 & 35. Ten in a row, passing over two piles each time.

David Singmaster, proposer; H. L. Abbott, solver. Problem 1767. CM 18:7 (1992) 207 & 19:6 (1993). Solves the general version of the Double Five Puzzle, which the proposer had not solved. One can leave the counters on even numbered locations if and only if the number of counters is a multiple of 4 or a multiple of 10. One can leave the counters in consecutive locations if and only if the number of counters is 4, 8, 10 or 12.

Heinrich Hemme. Email of 25 Feb 1999. Points out that the rules in the usual version should say that the counter must land on a pile of a single coin. This would eliminate the trick solutions given by Mittenzwey and Haldeman-Julius. Hemme says that without this rule, the problem is easy and can be solved for 4 and 6 counters!
5.S. CHAIN CUTTING AND REJOINING
The basic problem is to minimise the cost or effort of reforming a chain from some fragments.
Loyd. Problem 25: A brace of puzzles -- No. 25: The chain puzzle. Tit Bits 31 (27 Mar 1897) & 32 (17 Apr 1897) 41. 13 lengths: 5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 12. (Not in the Cyclopedia.)

Loyd. Problem 42: The blacksmith puzzle. Tit Bits 32 (10 & 31 Jul & 21 Aug 1897) 273, 327 & 385. Complex problem involving 10 pieces of lengths from 3 to 23 to be joined.

Clark. Mental Nuts. 1897, no. 7 & 1904, no. 14: The chain question; 1916, no. 59: The chain puzzle. 5 pieces of 3 links to make into a single length.

Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 9:4 (Feb 1903) 390-391 & 9:5 (Mar 1903) 490-491. The five chains. 5 pieces of 3 links to make into a single length.

Pearson. 1907. Part II, no. 67, pp. 128 & 205. 5 pieces of 3 links to make into a single length.

Dudeney. The world's best puzzles. Op. cit. in 2. 1908. He attributes such puzzles to Loyd (Tit Bits prob. 25) and gives that problem.

Cecil H. Bullivant. Home Fun. T. C. & E. C. Jack, London, 1910. Part VI, Chap IV, No. 9: The broken chain, pp. 518 & 522. 5 3 link pieces into an open chain.

Loyd. The missing link. Cyclopedia, 1914, pp. 222 (no solution) (c= MPSL2, prob. 25, pp. 19 & 129). 6 5 link pieces into a loop.

Loyd. The necklace puzzle. Cyclopedia, 1914, pp. 48 & 345 (= MPSL1, prob. 47, pp. 45 46 & 138). 12 pieces, with large and small links which must alternate.

D. E. Smith. Number Stories. 1919. Pp. 119 & 143 144. 5 pieces of 3 links to make into one length.

Hummerston. Fun, Mirth & Mystery. 1924. Q.E.D. -- The broken chain, Puzzle no. 38, pp. 99 & 178. Pieces of lengths 2, 2, 3, 3, 4, 4, 6 to make into a closed loop.

Ackermann. 1925. Pp. 85 86. Identical to the Loyd example cited by Dudeney.

Dudeney. MP. 1926. Prob. 212: A chain puzzle, pp. 96 & 181 (= 536, prob. 513, pp. 211 212 & 408). 13 pieces, with large and small links which must alternate.

King. Best 100. 1927. No. 7, pp. 9 & 40. 5 pieces of three links to make into one length.

William P. Keasby. The Big Trick and Puzzle Book. Whitman Publishing, Racine, Wisconsin, 1929. A linking problem, pp. 161 & 207. 6 pieces comprising 2, 4, 4, 5, 5, 6 links to be made into one length.
5.S.1. USING CHAIN LINKS TO PAY FOR A ROOM
The landlord agrees to accept one link per day and the owner wants to minimise the number of links he has to cut. The solution depends on whether the chain is closed in a cycle or open at the ends. Some weighing problems in 7.L.2.c and 7.L.3 are phrased in terms of making daily payment, but these are like having the chain already in pieces. See the Fibonacci in 7.L.2.c.

New section. I recall that there are older versions.

Rupert T. Gould. The Stargazer Talks. Geoffrey Bles, London, 1944. A Few Puzzles -- write up of a BBC talk on 10 Jan 1939, pp. 106-113. 63 link chain with three cuts. On p. 106, he says he believes it is quite modern -- he first heard it in 1935. On p. 113, he adds a postscript that he now believes it first appeared in John O'London's Weekly (16 Mar 1935) ??NYS.

Anonymous. Problems drive, 1958. Eureka 21 (Oct 1958) 14-16 & 30. No. 3. Man has closed chain of 182 links and wants to stay 182 days. What is the minimum number of links to be opened?

Birtwistle. Math. Puzzles & Perplexities. 1971. Pp. 13-16. Begins with seven link open-ended bracelet. Then how big a bracelet can be dealt with using only two cuts? Gets 23. Then does general case, getting n + (n+1)(2ⁿ⁺¹ - 1).

Angela Fox Dunn. Second Book of Mathematical Bafflers. Dover, 1983. Selected from Litton's Problematical Recreations, which appeared in 1959 1971. Prob. 26, pp. 28 & 176. 23 link case.

Howson. Op. cit. in 5.R.4. 1988. Prob. 30. Says a 23 link chain need only be cut twice, giving lengths 1, 1, 3, 6, 12, which make all values up to 23. Asks for three cuts in a 63 link chain and the maximum length chain one can deal with in n cuts.
5.T. DIVIDING A CAKE FAIRLY
Mittenzwey. 1880. Prob. 200, pp. 37 & 89; 1895?: 225, pp. 41 & 91; 1917: 225, pp. 38 & 88. Family of 4 adults and 4 children. With three cuts, divide a cake so the adults and the children get equal pieces. He makes two perpendicular diametrical cuts and then a circular cut around the middle. He seems to mean the adults get equal pieces and the children get equal pieces, not necessarily the same. But if the circular cut is at 2/2 of the radius, then the areas are all equal. Not clear where this should go -- also entered in 5.Q.

B. Knaster. Sur le problème du partage pragmatique de H. Steinhaus. Annales de la Société Polonaise de Mathématique 19 (1946) 228 230. Says Steinhaus proposed the problem in a 1944 letter to Knaster. Outlines the Banach & Knaster method of one cutting 1/n and each being allowed to diminish it -- last diminisher takes the piece. Also shows that if the valuations are different, then everyone can get > 1/n in his measure. Gives Banach's abstract formulations.

H. Steinhaus. Remarques sur le partage pragmatique. Ibid., 230 231. Says the problem isn't solved for irrational people and that Banach & Knaster's method can form a game.

H. Steinhaus. The problem of fair division. Econometrica 16:1 (Jan 1948) 101 104. This is a report of a paper given on 17 Sep. Gives Banach & Knaster's method.

H. Steinhaus. Sur la division pragmatique. (With English summary) Econometrica 17 (Supplement) (1949) 315 319. Gives Banach & Knaster's method.

Max Black. Critical Thinking. Prentice Hall, Englewood Cliffs, (1946, ??NYS), 2nd ed., 1952. Prob. 12, pp. 12 & 432. Raises the question but only suggests combining two persons.

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.

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

H. Phillips. News Chronicle "Quiz" No. 3: Natural History. News Chronicle, London, 1946. Pp. 22 & 43. 12 blue, 9 red and 6 green marbles in a bag. Choose enough to have three of one colour and two of another colour.

H. Phillips. News Chronicle "Quiz" No. 4: Current Affairs. News Chronicle, London, 1946. Pp. 17 & 40. 6 yellow, 5 blue and 2 red marbles in a bag. Choose enough to have three of the same colour.

L. Moser, proposer; D. J. Newman, solver. Problem 4300 -- The identity as a product of successive elements. AMM 55 (1948) 369 & 57 (1950) 47. n elements from a group of order n have a a subinterval with product = 1.

Doubleday - 2. 1971.

In the dark, pp. 145-146. How many socks do you have to pick from a drawer of white and black socks to get two pairs (possibly different)?

Lucky dip, pp. 147-148. How many socks do you have to pick from a drawer of with many white and black socks to get nine pairs (possibly different)? Gives the general answer 2n+1 for n pairs. [Many means that the drawer contains more than n pairs.]

Doubleday - 3. 1972. In the dark, pp. 35-36. Four sweaters and 5, 12, 4, 9 socks of the same colours as the sweaters. Lights go out. He can only find two of the sweaters. How many socks must he bring down into the light to be sure of having a pair matching one of the sweaters?

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: [phone]; email: [e-mail]; web: http://www.edmundscientific.com). But he later wrote that investigation of the manuals of DifiComp, one of their other products, reveals that there appears to be no connection. E.S.R. = Education Science Research. The inventors of DigiComp, as listed in the patent for it, are: Irving J. Lieberman, William H, Duerig and Charles D. Hogan, all of Montclair, and they were the founders of the company. The DigiComp manuals say Think A Dot was later invented by John Weisbacker. There is a website devoted to DigiComp which contains this material and/or pointers to related sites and has a DigiComp emulator: http://members.aol.com/digicomp1/DigiComp.html . www.yahoo.com has a Yahoo club called Friends of DigiComp. There is another website with the DigiComp manual: http://galena.tj.edu.inter.net/digicomp/ .

E.S.R. Instructions, 8pp, nd -- but box says ©1965. No patent number anywhere but leaflet says the name Think-A-Dot is trademarked.

E.S.R., Inc. Corporation. US trademark registration no. 822,770. Filed: 8 Dec 1965; registered: 24 Jan 1967. First used 23 Aug 1965. Expired. The US Patent and Trademark Office website entry says the owner is the company and gives no information about the inventor(s). The name has been registered for a computer game on 23 Jul 2002.

Benjamin L. Schwartz. Mathematical theory of Think A Dot. MM 40:4 (Sep 1967) 187 193. Shows there are two classes of patterns and that one can transform any pattern into any other pattern in the same class in at most 15 drops.

Ray Hemmings. Apparatus Review: Think a Dot. MTg 40 (1967) 45.

Sidney Kravitz. Additional mathematical theory of Think A Dot. JRM 1:4 (Oct 1968) 247 250. Considers problems of making ball emerge from one side and of viewing only the back of the game.

Owen Storer. A think about Think a dot. MTg 45 (Winter 1968) 50 55. Gives an exercise to show that any possible transformation can be achieved in at most 9 drops.

T. H. O'Beirne. Letter: Think a dot. MTg 48 (Autumn 1969) 13. Proves Steiner's (Storer?? - check) assertion about 9 drops and gives an optimal algorithm.

John A. Beidler. Think-A-Dot revisited. MM 46:3 (May 1973) 128-136. Answers a question of Schwarz by use of automata theory. Characterizes all minimal sequences. Suggests some generalized versions of the puzzle.

Hans Dockhorn. Bob's binary boxes. CFF 32 (Aug 1993) 4-6. Bob Kootstra makes boxes with the same sort of T-shaped switch present in Think-A-Dot, but with just one entrance. One switch with two exits is the simplest case. Kootstra makes a box with three switches and four exits along the bottom, and the successive balls come out of the exits in cyclic sequence. Using a reset connection between switches, he also makes a two switch, three exit, box.

Boob Kootstra. Box seven. CFF 32 (Aug 1993) 7. Says he has managed to design and make boxes with 5, 6, 7, 8 exits, again with successive balls coming out the exits in cyclic order, but he cannot see any general method nor a way to obtain solutions with a minimal number of movable parts (switches and reset levers). Further his design for 7 exits is awkward and the design of an optimal box for seven is posed as a contest problem.

5.W. MAKING THREE PIECES OF TOAST
This involves an old fashioned toaster which does one side of two pieces at a time. An alternative version is frying steaks or hamburgers on a grill which holds two objects, assuming each side has to be cooked the same length of time. The problem is probably older than these examples.
Sullivan. Unusual. 1943. Prob. 7: For the busy housewife.

J. E. Littlewood. A Mathematician's Miscellany. Op. cit. in 5.C. 1953. P. 4 (26). Mentions problem and solution.

Simon Dresner. Science World Book of Brain Teasers. 1962. Op. cit. in 5.B.1. Prob. 40: Minute toast, pp. 18 & 93.

D. St. P. Barnard. 50 Daily Telegraph Brain Twisters. 1985. Op. cit. in 4.A.4. Prob. 5: Well done, pp. 16, 80, 103 104. Grilling three steaks on a grill which only holds two. He complicates the problem in two ways: a) each side takes a minute to season before cooking; b) the steaks want to be cooked 4, 3, 2 minutes per side.

Edward Sitarski. When do we eat? CM 27:2 (Mar 2001) 133-135. Hamburgers which require time T per side. After showing that three hamburgers take 3T, he asks how long it will take to cook H hamburgers. Easily shows that it can be done in HT, except for H = 1, which takes 2T. Then remarks that this is an easy version of a scheduling problem -- in reality, the hamburgers would have different numbers of sides, there would be several grills and each hamburger would have different parts requiring different grills, but in a particular order!
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: Young World (c1960); Gooding (1994) in 5.X.1.

5.X.1. COUNTING TRIANGLES
Counting triangles in a pattern is always fraught with difficulties, so I have written a program to do this, but I haven't checked all the examples here.
Pearson. 1907. Part II.

No. 74: A triangle of triangles, p. 74. Triangular array with four on a side, but with all the altitudes also drawn. Gets 653 triangles of various shapes.

No. 75: Pharaoh's seal, pp. 75 & 174. Isosceles right triangles in a square pattern with some diagonals.

Anon. Prob. 76. Hobbies 31 (No. 791) (10 Dec 1910) 256 & (No. 794) (31 Dec 1910) 318. Make as many triangles as possible with six matches. From the solution, it seems that the tetrahedron was expected with four triangles, but many submitted the figure of a triangle with its altitudes drawn, but only one solver noted that this figure contains 16 triangles! However, if the altitudes are displaced to give an interior triangle, I find 17 triangles!!

Loyd. Cyclopedia. 1914. King Solomon's seal, pp. 284 & 378. = MPSL2, No. 142, pp. 100 & 165 c= SLAHP: Various triangles, pp. 25 & 91. How many triangles in the triangular pattern with 4 on a side? Loyd Sr. has this embedded in a larger triangle.

Collins. Book of Puzzles. 1927. The swarm of triangles, pp. 97-98. Same as Pearson No. 74. He says there are 653 triangles and that starting with 5 on a side gives 1196 and 10,000 on a side gives 6,992,965,420,382. When I gave August's problem in the Weekend Telegraph, F. R. Gill wrote that this puzzle with 5 on a side was given out as a competition problem by a furniture shop in north Lancashire in the late 1930s, with a three piece suite as a prize for the first correct solution.

Evelyn August. The Black-Out Book. Harrap, London, 1939. The eternal triangle, pp. 64 & 213. Take a triangle, ABC, with midpoints a, b, c, opposite A, B, C. Take a point d between a and B. Draw Aa, ab, bc, ca, bd, cd. How many triangles? Answer is given as 24, but I (and my program) find 27 and others have confirmed this.

Anon. Test your eyes. Mathematical Pie 7 (Oct 1952) 51. Reproduced in: Bernard Atkin, ed.; Slices of Mathematical Pie; Math. Assoc., Leicester, 1991, pp. 15 & 71 (not paginated - I count the TP as p. 1). Triangular pattern with 2 triangles on a side, with the three altitudes drawn. Answer is 47 'obtained by systematic counting'. This is correct. Cf Hancox, 1978.

W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. How many triangles, pp. 43 & 130. Take a pentagon and draw the pentagram inside it. In the interior pentagon, draw another pentagram. How many triangles are there? Answer is 85.

Young World. c1960. P. 57: One for Pythagoras. Consider a L-tromino. Draw all the midlines to form 12 unit squares. Or take a 4 x 4 square array and remove a 2 x 2 array from a corner. Now draw the two main diagonals of the 4 x 4 square - except half of one diagonal would be outside our figure. How many triangles and how many squares are present? Gives correct answers of 26 & 17.

J. Halsall. An interesting series. MG 46 (No. 355) (Feb 1962) 55 56. Larsen (below) says he seems to be the first to count the triangles in the triangular pattern with n on a side, but he does not give any proof.

Although there are few references before this point, the puzzle idea was pretty well known and occurs regularly. E.g. in the children's puzzle books of Norman Pulsford which start c1965, he gives various irregular patterns and asks for the number of triangles or squares.

J. E. Brider. A mathematical adventure. MTg (1966) 17 21. Correct derivation for the number of triangles in a triangle. This seems to be the first paper after Halsall but is not in Larsen.

G. A. Briggs. Puzzle and Humour Book. Published by the author, Ilkley, 1966. Prob. 2/12, pp. 23 & 75. Consider an isosceles right triangle with legs along the axes from (0,0) to (4,0) and (0,4). Draw the horizontals and verticals through the integer lattice points, except that the lines through (1,1) only go from the legs to this point and stop. Draw the diagonals through even-integral lattice points, e.g. from (2,0) to (0,2). How many triangles. Says he found 27, but his secretary then found 29. I find 29.

Ripley's Puzzles and Games. 1966. Pp. 72-73 have several problems of counting triangles.

Item 3. Consider a Star of David with the diameters of its inner hexagon drawn. How many triangles are in it? Answer: 20, which I agree with.

Item 4. Consider a 3 x 3 array of squares with their diagonals drawn. How many triangles are there? Answer: 150, however, there are only 124.

Item 5. Consider five squares, with their midlines and diagonals drawn, formed into a Greek cross. How many triangles are there? Answer: 104, but there are 120.

Doubleday - 2. 1971. Count down, pp. 127-128. How many triangles in the pentagram (i.e. a pentagon with all its diagonals)? He says 35.

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

[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers. 1973. Op. cit. in 5.E. An unusual star, pp. 49-50. Consider a pentagram and draw lines from each star point through the centre to the opposite crossing point. How many triangles? They say 110.

[Henry] Joseph and Lenore Scott. Master Mind Pencil Puzzles. 1973. Op. cit. in 5.R.4.

Diamonds are forever, pp. 35-36. Hexagon with Star of David inside and another Star of David in the centre of that one. How many triangles? Answer is 76.

Count the triangles, pp. 55-56. Ordinary Greek cross of five squares, with all the diagonals and midlines of the five squares drawn. How many triangles> Answer is 104.

C. P. Chalmers. Note 3353: More triangles. MG 58 (No. 403) (Mar 1974) 52 54. How many triangles are determined by N points lying on M lines? (Not in Larsen.)

Nicola Davies. The 2nd Target Book of Fun and Games. Target (Universal-Tandem), London, 1974. Squares and triangles, pp. 18 & 119. Consider a chessboard of 4 x 4 cells. Draw all the diagonals, except the two main ones. How many squares and how many triangles?

Shakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 136: The triangles, pp. 85 & 133. How many triangles in a Star of David made of 12 equilateral triangles?

Michael Holt. Figure It Out -- Book Two. Granada, London, 1978. Prob. 67, unpaginated. How many triangles in a Star of David made of 12 equilateral triangles?

Putnam. Puzzle Fun. 1978. No. 91: Counting triangles, pp. 12 & 37. Same as Doubleday - 2.

D. J. Hancox, D. J. Number Puzzles For all The Family. Stanley Thornes, London, 1978.

Puzzle 8, pp. 2 & 47. Draw a line with five points on it, say A, B, C, D, E, making four segments. Connect all these points to a point F on one side of the line and to a point G on the other side of the line, with FCG collinear. How many triangles are there? Answer is 24, which is correct.

Puzzle 53, pp. 24 & 54. Same as Anon.; Test Your Eyes, 1952. Answer is 36, but there are 47.

The Diagram Group. The Family Book of Puzzles. The Leisure Circle Ltd., Wembley, Middlesex, 1984.

Problem 40, with Solution at the back of the book. Same as Doubleday - 2.

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

Sue Macy. Mad Math. The Best of DynaMath Puzzles. Scholastic, 1987. (Taken from Scholastic's DynaMath magazine.) Shape Up, pp. 5 & 56.

Take a triangle, trisect one edge and join the points of trisection to the opposite vertex. How many triangles? [More generally, if one has n points on a line and joins them all to a vertex, there are 1 + 2 + ... + n-1 = n(n-1)/2 triangles.]

Take a triangle, join up the midpoints of the edges, giving four smaller triangles, and draw one altitude of the original triangle. How many triangles?

1980 Celebration of Chinese New Year Contest Problem No. 5; solution by Leroy F. Meyers. CM 17 (1991) 2 & 18 (1992) 272-273. n x n array of squares with all diagonals drawn. Find the number of isosceles right triangles. [Has this also been done in half the diagram? That is, how many isosceles right triangles are in the isosceles right triangle with legs going from (0,0) to (n,0) and (0,n) with all verticals, horizontals and diagonals through integral points drawn?]

Mogens Esrom Larsen. The eternal triangle -- a history of a counting problem. Preprint, 1988. Surveys the history from Halsall on. The problem was proposed at least five times from 1962 and solved at least ten times. I have sent him the earlier references.

Marjorie Newman. The Christmas Puzzle Book. Hippo (Scholastic Publications), London, 1990. Star time, pp. 26 & 117. Consider a Star of David formed from 12 triangles, but each of the six inner triangles is subdivided into 4 triangles. How many triangles in this pattern? Answer is 'at least 50'. I find 58.

Erick Gooding. Polygon counting. Mathematical Pie No. 131 (Spring 1994) 1038 & Notes, pp. 1-2. Consider the pentagram, i.e. the pentagon with its diagonals drawn. How many triangles, quadrilaterals and pentagons are there? Gets 35, 25, 92, with some uncertainty whether the last number is correct.

When F. R. Gill (See Pearson and Collins above) mentioned the problem of counting the triangles in the figure with all the altitudes drawn, I decided to try to count them myself for the figure with N intervals on each side. The theoretical counting soon gets really messy and I adapted my program for counting triangles in a figure (developed to verify the number found for August's problem). However, the number of points involved soon got larger than my simple Basic could handle and I rewrote the program for this special case, getting the answers of 653 and 1196 and continuing to N = 22. I expected the answers to be like those for the simpler triangle counting problem so that there would be separate polynomials for the odd and even cases, or perhaps for different cases (mod 3 or 4 or 6 or 12 or ??). However, no such pattern appeared for moduli 2, 3, 4 and I did not get enough data to check modulus 6 or higher. I communicated this to Torsten Sillke and Mogens Esrom Larsen. Sillke has replied with a detailed answer showing that the relevant modulus is 60! I haven't checked through his work yet to see if this is an empirical result or he has done the theoretical counting.

Heather Dickson, Heather, ed. Mind-Bending Challenging Optical Puzzles. Lagoon Books, London, 1999, pp. 40 & 91. Gives the version m = n = 4 of the following. I have seen other versions of this elsewhere, but I found the general solution on 4 Jul 2001 and am submitting it as a problem to AMM.

Consider a triangle ABC. Subdivide the side AB into m parts by inserting m 1 additional points. Connect these points to C. Subdivide the side AC into n parts by inserting n-1 additional points and connect them to B. How many triangles are in this pattern? The number is [m²n + mn²]/2. When m = n, we get n³, but I cannot see any simple geometric interpretation for this.
5.X.2. COUNTING RECTANGLES OR SQUARES
I have just seen M. Adams. There are probably earlier examples of these types of problems.
Anon. Prob. 63. Hobbies 30 (No. 778) (10 Sep 1910) 488 & 31 (No. 781) (1 Oct 1910) 2. How many rectangles on a 4 x 4 chessboard? Solution says 100, which is correct, but then says they are of 17 different types -- I can only get 16 types.

Blyth. Match-Stick Magic. 1921. Counting the squares, p. 47. Count the squares on a 4 x 4 chessboard made of matches with an extra unit square around the central point. The extra unit square gives 5 additional squares beyond the usual 1 + 4 + 9 + 16.

King. Best 100. 1927. No. 9, pp. 10 & 40. = Foulsham's, no. 5, pp. 6 & 10. 4 x 4 board with some diagonals yielding one extra square.

Loyd Jr. SLAHP. 1928. How many rectangles?, pp. 80 & 117. Asks for the number of squares and rectangles on a 4 x 4 board (i.e. a 5 x 5 lattice of points). Says answers are 1 + 4 + 9 + 16 and (1 + 2 + 3 + 4)² and that these generalise to any size of board.