Sources page biographical material

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 The Philosophy of Analysis -- unpublished collection of MSS in the BM as Add.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!