aa r X i v : . [ m a t h . C O ] D ec Triangular Peg Solitaire Unlimited
George I. Bell
December 2004 [email protected] Abstract
Triangular peg solitaire is a well-known one-person game or puzzle. When one peg capturesmany pegs consecutively, this is called a sweep. We investigate whether the game can end ina dramatic fashion, with one peg sweeping all remaining pegs off the board. For triangularboards of side 6 and 8 (with 21 and 36 holes, respectively) the geometrically longest sweepcan occur as the final move in a game. On larger triangular boards, we demonstrate how toconstruct solutions that finish with arbitrarily long sweeps. We also consider the problemof finding solutions that minimize the total number of moves (where a move is one or moreconsecutive jumps by the same peg).
Peg solitaire on a 15-hole triangular board is an old puzzle but it remains popular. In theUnited States, one can find these puzzles at tables in Cracker Barrel R (cid:13) restaurants. The15-hole puzzle is also amenable to exhaustive computer search and this is a common pro-gramming assignment for computer science classes.Here we consider peg solitaire on an (equilateral) triangular board with n holes on each side.This board will be referred to as Triangle ( n ) and can be conveniently presented on an arrayof hexagons (Figure 1). The Triangle ( n ) board has T ( n ) = n ( n + 1) / T ( n ) isthe n ’th triangular number. The notation used to identify the holes in the board is shownin Figure 1—it differs from that given in Beasley [1] and is based on the system for squarelattice boards. A nice property of this notation is that the top corner hole is always “a1”.The rules of the game are simple: start from a board with a peg in every hole but one, calledthe starting vacancy . Then jump one peg over another into an empty hole, removing thejumped peg from the board. The goal is to choose a sequence of jumps to finish with onepeg (the coordinate of this final peg is called the finishing location ). This general problem Original version at http://gpj.connectfree.co.uk/gpjr.htm
Converted to L A TEX by the author with some modifications to the text, November 2007. he Games and Puzzles Journal—Issue 36, November-December 2004 a1a2 b2a3 b3 c3a4 b4 c4 d4a5 b5 c5 d5 e5a6 b6 c6 d6 e6 f6 Figure 1: The general triangular board with hole coordinates.of going from a board position with one peg missing to a board position with one peg willbe referred to as a peg solitaire problem . The special case where the starting vacancyand finishing location are the same is referred to as a complement problem . GPJ jumps because we start with 14 pegs and finish with 1. However, when the same peg jumps one ormore pegs consecutively, we call this one move . Given a particular peg solitaire problem,what is the solution with the least number of moves? While this question has historicallyplayed an important role in peg solitaire on the standard 33-hole cross-shaped board, it hasbarely been considered for triangular boards. The following terminology is used in referringto moves involving multiple jumps: when a peg removes i pegs in a single move, we refer toit as a sweep , or more specifically, an i -sweep .After attempting a peg solitaire problem, many people get the idea to try to solve it backwards from the final peg. What is not so obvious is that this is exactly the same as the originalgame, where the concepts of “hole” and “peg” are interchanged. In fact the solution to anypeg solitaire problem really contains two solutions: the original (“forward” solution) plusthis “backward” solution, where the individual jumps are executed in the same direction, butin reverse order, and the starting vacancy and finishing location are swapped. An importantobservation is that when an i -sweep is reversed, we must reverse the individual jumps, and itbecomes i single jump moves. In other words, sweeps in forward solutions cannot be sweepsin reversed solutions (and vice versa).Backward play is hard to comprehend, because our brain does not easily interchange theconcepts of “hole” and “peg”. It is easier to understand “backward play” by realizing thatit is the same as forward play from the complement of the current board position (thecomplement of a board position is where we replace every hole by a peg, and vice versa).This leads to a theorem used extensively in the remainder of this paper. Suppose we havea board position B and we wonder if it could appear during a solution to a peg solitaireproblem. he Games and Puzzles Journal—Issue 36, November-December 2004 Forward/Backward Theorem : Board position B can appear during a solution to a pegsolitaire problem if and only if both1. B can be reduced to a single peg (the finishing location) using peg solitaire jumps.2. The complement of B can be reduced to a single peg (the starting location) using pegsolitaire jumps.A proof of this theorem just boils down to the equivalence of playing the game forward orbackward. It may seem obvious, but is key to understanding how to reach complex sweeppositions.A practical problem that arises is finding a triangular board to play on. Fifteen hole Trian-gle (5) boards are common, but cannot be easily extended to larger triangular boards. Thebest board I have found is a Chinese Checkers set, which allows one to play on boards aslarge as
Triangle (13). To help follow the arguments below, I recommend playing out thesolutions on a Chinese Checkers set. It is particularly helpful to see solutions played forwardand backward.
Triangular boards support the longest sweeps of any peg solitaire board. This is becausefrom any board location the total number of possible jumps is even . If the board size n is odd, there exist sweeps that jump into or over every single location on the board. Suchsweeps are maximal in the sense that they are the longest sweep geometrically possible onthe board. The figure below shows examples of the first four maximal sweepsFigure 2: Maximal Sweeps on Triangle ( n ), where n = 3, 5, 7 and 9. These sweeps are shownstarting and ending at a1, but can begin and end at other board locations.The length of this sweep is 3 T (( n − /
2) = 3( n − /
8. Related geometrical tours on atriangular lattice have appeared in GPJ he Games and Puzzles Journal—Issue 36, November-December 2004 / / The same maximal sweep on an odd triangular board is also maximal on the even board onesize larger, because the added row cannot be reached to extend the sweep. However, theadded row can make it possible to reach the sweep position during a peg solitaire game. Onthe
Triangle (6) and
Triangle (8) boards this can be worked out by hand, Figure 3 shows thisprocess on
Triangle (6): (a)(b)
Figure 3: Constructing a solitaire solution that finishes with the 9-sweep to a
1. (a) Backward:playing from the complement of the sweep pattern to c5. (b) Forward: The completedsolution from the c5 vacancy ending with a 9-sweep to a1.The forward solution in Figure 3b has only 9 moves. In the final section of this paper,we prove that it’s impossible to solve this problem in fewer than 9 moves. Thus, besidescontaining a maximal length sweep, the solution of Figure 3b solves the problem in theminimum number of moves. This solution was discovered before 1975 by Harry O. Davis [3].There are three problems on
Triangle (6) that can contain a 9-sweep:1. Vacate c5, play to finish at a1 with the last move a 9-sweep (Figure 3).2. Vacate c5, play to finish at a4 with the last move a 9-sweep. he Games and Puzzles Journal—Issue 36, November-December 2004
53. Vacate c5, play to finish at a4 with the second to the last move a 9-sweep.The reader is invited to solve problems 2 and 3. As in Figure 3, the trick is to figure outwhat the final sweep must look like, and then solve backwards (playing forward from thecomplement of the sweep position). Then exactly reverse the jumps in your “backward”solution and you will reproduce the sweep position.On
Triangle (8), the 18-sweep can also be reached. There are three problems on this boardthat can contain the 18-sweep:1. Vacate c5, play to finish at a1 with the last move an 18-sweep (Figure 4).2. Vacate b6 or e6, play to finish at c8 with the last move an 18-sweep.3. Vacate c5, play to finish at b6 with the second to the last move an 18-sweep.These problems are more difficult than those on
Triangle (6), but are still quite reasonable towork out by hand. Figure 4 shows the solution to the first problem, played forward. Thissolution was discovered, as usual, by attempting to play backward from the complement ofthe sweep position. This solution contains 15-moves, but it is possible to solve this problemin 14 moves (without the 18-sweep), so this solution does not have the minimum number ofmoves.Figure 4: A 15-move solution to problem
Triangle (8) (finishing with an 18-sweep).Note: more than one move is sometimes shown between board snapshots.On
Triangle (10), a computational search for a peg solitaire solution containing a maximal30-sweep has come up empty (although a solution was found which ends with a 29-sweep).
Triangle (12) does not appear to have a maximal 45-sweep solution either. I haven’t checkedall possible configurations of this 45-sweep, but my program has shown the most obviouscandidates can’t be reached from a single vacancy start. It appears that the
Triangle (8)board is the largest triangular board for which a solution to a peg solitaire problem cancontain a maximal sweep. he Games and Puzzles Journal—Issue 36, November-December 2004 Although maximal sweeps appear not to be attainable in peg solitaire problems on largetriangular boards, it turns out sweeps of only slightly reduced length are possible. A veryspecial solution was discovered by hand on the 78-hole
Triangle (12) board. This solutionfinishes with a 42-sweep when run forward, and is shown in Figure 5. (a)(b) moves1&2 moves3&4 moves5&6 moves7&8move9 move10 move11
Figure 5: Building a solitaire solution that finishes with the 42-sweep (a) Forward: thefinishing 42-sweep (from a1 to a3) (b) Backward: showing how to reduce the complement ofthe sweep pattern to one peg.This solution is remarkable because it can be extended to even larger triangular boards.We can extend the final sweep to cover the bottom of
Triangle (14) and the complement canstill be reduced to a single peg. We do this by keeping moves along the bottom row thesame, while extending other moves vertically. In particular moves 1 & 2 end at the sameboard locations but begin from the bottom row of the board. The U-shaped moves 9 & 10become vertically elongated, but have the same starting and ending board locations. Finallyadditional moves need to be added after move 11 to reduce the remaining pattern of pegsto a single survivor (which does not end up at the same location as in Figure 5b). We cancontinue stepping the board size up by 2 and still reduce the complement to a single peg upto
Triangle (20). However if we try this on
Triangle (22), we find we can no longer reduce theremaining pattern to a single peg. he Games and Puzzles Journal—Issue 36, November-December 2004 original problem on Triangle (12). This key insight suggests that we may be able to performan inductive step, and extend the board indefinitely . This is in fact possible, and will bedescribed below. In effect we can construct solutions to peg solitaire problems on triangularboards of arbitrarily large size. Not only that, these solutions finish with arbitrarily longsweep moves! A Triangle(12) B Triangle(12)
Figure 6: Fitting the inductive components together to make a long finishing sweep on
Triangle (24).To complete the inductive step, we add another solution for
Triangle (12) underneath thefirst one to obtain a solution on the 300-hole
Triangle (24) board. Figure 6 shows the overallgeometry of the solution; we use “component A ” (Figure 5) in the upper part of the boardand “component B ” (Figure 7) in the lower part of the board. In the reversed solution, theremaining (white) areas of the board are cleared by extensions of the moves to clear A , andthe final move snakes down vertically through both A and B to the bottom of the board.The final sweep pattern will be close to the maximal sweep, but component A contains a“defect” in this sweep pattern, and so must component B . Figuring out exactly where toplace this defect in component B is critical to making everything work out. At first I triedputting the defect symmetrically over the y -axis of the board, but this never seems to workout. What did work was to put it slightly off center.While component A can be considered a solution itself on Triangle (12), component B isinherently tied to the board above it, for there must be moves which link the two boards.Figure 7 shows how to solve component B , which looks very similar to component A . Note,however, that the U-shaped moves actually go in the opposite direction from before. Theentire solution is constructed to enable the final move to pass vertically down the board.The final sweep pattern always starts at a1 and finishes at a3. The initial vacancy, orfinishing location for the reversed solution is always directly under a3 on the lowest row thathas a hole in vertical alignment with a3 (on Triangle (24), this starting vacancy is at k23).The entire solution on
Triangle (24), when executed in the forward direction, has a final move he Games and Puzzles Journal—Issue 36, November-December 2004 B . Backward: showing how to reduce the complement ofthe sweep pattern to one peg. The top two holes (in green) are part of the board above.which is a 191-sweep (Figure 8):Figure 8: Forward: the finishing 191-sweep on Triangle(24). This sweep pattern is themaximal 198-sweep with two “defects”.By stacking additional copies of component B under the diagram of Figure 6, we can extendthis process indefinitely. The net result is that on Triangle (12 i ), we can construct a solutionto a solitaire problem that finishes with a sweep of length 54 i − i + 1 (this is 4 i − i + 6 i holes, so asymptoticallythis finishing sweep removes 3 / i + 19 i − i + 14 i −
3. Table 1 he Games and Puzzles Journal—Issue 36, November-December 2004 i Board (
Triangle (12) 78 42 55.3% 292
Triangle (24) 300 191 64.1% 973
Triangle (36) 666 448 67.5% 20110
Triangle (120) 7,260 5,271 72.6% 1,937Table 1: Statistics on long sweep solutions on
Triangle (12 i ).Although this construction technique has given us solutions on boards with sides a multipleof 12, it is not hard to extend it to Triangle ( n ) for any even n ≥
12. The way to do this isusing the same technique we used to extend component A on Triangle (12) to
Triangle (14),
Triangle (16), etc.
Given a board and a (solvable) peg solitaire problem, what is the least number of moves thatcan solve it? While this question has been important in the history of the standard 33-holecross-shaped board, it has not received much attention on triangular boards. Informally,this question can be rephrased: “What is the solution which involves touching the smallestnumber of pegs?”In 1966, Harry O. Davis studied short solutions on the
Triangle (5) board analytically [2]. Hewas able to find minimal solutions “for all starting locations” and prove that his solutionswere the shortest possible. In particular, he found a 10 move solution to the a1-complement,and proved that the problem could not be solved in less than 10 moves.I have now completed exhaustive computer calculations on all peg solitaire problems onboards up to
Triangle (7), and all complement problems on
Triangle (8). Table 2 summarizesmy results, for more information with examples and complete lists of shortest solutions see[4].In Table 2, we count only distinct peg solitaire problems that cannot be reduced to anotherproblem by means of rotation or reflection of the board. For example, on the
Triangle (5)board, Table 2 indicates that there are 17 distinct peg solitaire problems, but only 12 aresolvable. Of these 12 solvable problems, two can be done in a minimum of 9 moves, six in 10moves and four in 11 moves. Surprisingly, over half the problems on the
Triangle (6) boardcan be solved in 9 moves, so on average, it’s possible to solve problems on this board in fewer moves than for
Triangle (5)! he Games and Puzzles Journal—Issue 36, November-December 2004
Triangle (5) 15 17 12 2 6 4 - - - -
Triangle (6) 21 29 29 16 11 2 - - - -
Triangle (7) 28 27 27 - - - 19 8 - -
Triangle (8) 36 80 80 - - - - 1 † † † Table 2: Summary of minimal solution lengths (found by computational search) on triangularboards of side 5–8 ( † - complement problems only).There are 80 different problems on Triangle (8), I have only run the 8 complement problemsplus a few non-complement problems. Among the complement problems, only one can beaccomplished in 13 moves, the a7-complement (Figure 9) [due to board symmetry, thisproblem is equivalent to the complement problem at 5 other board locations: a2, b2, b8,g7 or g8]. The computer run also determined that this 13-move solution is unique in thesense that any other 13-move solution to the a7-complement must contain the exact sameset of jumps , although not necessarily in the same order. I have found several other 13-movesolutions to non-complement problems on the
Triangle (8) board.Figure 9: The minimal 13-move solution to the a7-complement on
Triangle (8). Note: morethan one move is sometimes shown between board snapshots.
It is quite difficult to determine the shortest solution on boards larger than
Triangle (8).However, we can obtain a lower bound on the length of the shortest solution by dividingthe board into “Merson Regions” (named after Robin Merson who first used this concept in1962 [1, p. 203]). The shape of a region is chosen such that when it is completely filled withpegs, there is no way to remove a peg in the region without a move that originates in theregion. On the edge of the board the regions can be corners or two consecutive holes, but in he Games and Puzzles Journal—Issue 36, November-December 2004
R = 9 R = 13 R = 18
Figure 10: Dividing
Triangle (6),
Triangle (8) and
Triangle (10) into “Merson Regions” ( R isthe number of regions).Any region that starts out full must have at least one move starting from inside it. Sincethe starting position has every hole filled by a peg except one, all regions start full exceptpossibly the region that contains the starting location. If we start in a corner, then thisregion starts out empty but is filled by the first move, hence there still must be a move outof this corner region. We can summarize these results as follows: If R is the number ofMerson Regions, then1. If the starting vacancy is a corner, or is not in any region (white board locations inFigure 10), then any solution to the peg solitaire problem (no matter where it finishes)has at least R moves.2. If the starting vacancy is in a region (but not a corner), then any solution to the pegsolitaire problem (no matter where it finishes) has at least R − Triangle (6), this proves that any solution to “Vacate c5, finish at a1” willtake at least 9 moves. On
Triangle (8), this analysis indicates that the a7-complement musttake at least 12 moves. It can be seen that the 13-move solution in Figure 9 contains one“extra” move above this minimum, in that there are two moves out of the central (blue)hexagon region. Therefore, 13 moves is not proved the shortest possible by this method,although exhaustive computer search indicates no 12-move solution exists.As the triangular board becomes very large, the number of interior hexagons eventuallydominates the number of regions, because this is the only region count growing quadratically.We can then “tile” the board with these hexagons without leaving any gaps (except nearthe edge of the board). So no solution can be shorter than the number of holes in the boarddivided by 7. No matter how large the triangular board, we cannot hope to find a solutionwhich does better than a 7-sweep averaged over all moves. Of course, averaging anywherenear this would be quite remarkable—note that the 13-move solution in Figure 9 removesan average of only 34 / ≈ . Triangle (12 i ). This upper bound is a solution oflength 18 i + 14 i − i + 6 i . As i increases, this shows that there exist he Games and Puzzles Journal—Issue 36, November-December 2004 References [1] J. Beasley,
The Ins and Outs of Peg Solitaire , Oxford Univ. Press, Oxford, New York, 1992.[2] Letter to Martin Gardner, 16 June 1966, now preserved in the Bodleian Library. Information com-municated by J. D. Beasley.[3] M. Gardner, Penny Puzzles, in
Mathematical Carnival , 12–26, Alfred A. Knopf, Inc., 1975 (reprintof an original article which appeared in
Scientific American , February 1966).[4]