Peg Solitaire
These pages were created and are maintained by George Bell (gibell@comcast.net) Last Modified April 12th, 2009 Copyright © 2009 by George I. Bell	To triangular peg solitaire To the peg solitaire army
We can merely mention bean-bags, peg-boards, size and form boards, as some apparatus found useful for the purpose of amusing and instructing the weak-minded. Albutt's "System of Medicine", 1899, VIII, 246 ([B3])

Table of Contents:

• Introduction
• Preliminaries
• Boards: English, 6x6, French, Diamond41, Diamond32, Diamond13 (Hoppers),
  Wiegleb, 9x9, Antique English, Rectangular, Generalized Cross
• Online Games and Puzzles
• Solution Catalogs
• Computational Search Techniques
• How Hard Is Peg Solitaire?
• References

Introduction

The "Central Game" A carefully selected sequence of jumps The starting board position The target board position

The basic game consists of a cross-shaped board, usually made of wood, together with a set of pegs, or more commonly marbles. To start the game, one fills the board with pegs except for the central hole. A jump consists of jumping one peg over another into an empty spot in the board, removing the peg jumped over from the board. Diagonal jumps are not allowed (in the standard version of the game). The goal is to choose a sequence of jumps and finish with as few pegs as possible, ideally a single peg in the center.

It seems almost everybody has run into this puzzle at some point. Boards range from drilled planks using golf tees, to beautifully crafted hardwood boards with indentations for marbles, including a nice rim around the edge for storage of marbles as they are removed. Computer versions of the game are also common (see my Javascript games below).

After unsuccessful attempts at a "manual solution", many people (myself included) try to write a simple computer program to solve it. Even if there are only 4-8 jumps available at each board position, this can lead to a prohibitively large number of possible jump sequences after only 15 jumps. This game is much older than the computer, and it is remarkable how much was proved about this game without the aid of computers.

Instead of trying to memorize a computer solution, or a YouTube video, the best way to remember a solution is to understand "block removals" or "packages", sequences of moves that remove a whole block of pegs but leave the rest of the board untouched. Once you understand and have mastered the "3-removal", "6-removal" and "L-move", the central game rapidly goes from frustrating to being quite simple (see [B1], [B3] or [W2] for details). This is similar to Rubik's cube in that you memorize a specific sequence of moves that place you closer to your goal but do not mess up the rest of the board.

After the central game above has been solved, what then? We can try to define the "best possible" solution. All solutions have exactly 31 jumps, because we start with 32 pegs and one is lost with each jump. However, when the same peg jumps one or more pegs, we call this one move. The question is, what is the solution with the least number of moves? This question may be rephrased more informally as: what is the solution that involves touching the smallest number of pegs? This question turns out to be much harder to answer than simply finding a solution to the central game. The following terminology is useful in referring to moves involving multiple jumps: when a peg removes n pegs in a single move, we refer to it as a sweep, or more specifically, an n-sweep. A sweep that begins and ends at the same board location is called a loop.

This web site will not show you how to play peg solitaire. It documents my own efforts to answer difficult or unsolved questions about the game, often involving boards different from than the standard board. Most of the rest of this page concerns the search (by computer) for solutions to certain problems with the least number of moves (called minimal length solutions).

The boards on this page have holes based on a square lattice; each hole has (at most) 4 neighbors. It is also possible to play on a triangular lattice, where each hole has (at most) 6 neighbors. I now have a separate page for Triangular Peg Solitaire. If you want some tips and solutions for the 15-hole triangular board, check out my page on tips and solutions for the Cracker Barrel Puzzle.

Preliminaries

Board Notation and Identification

We need some kind of notation to identify board locations. On the cross-shaped 33-hole board the most common notation is to label the board columns a-g (left to right) and the rows 1-7 (top to bottom). This notation works well for any board that fits inside a 7x7 square, and is referred to as "Standard 7x7 Notation". An alternative notation where the holes are numbered consecutively is not recommended because it must be changed every time the board shape changes.

c1 d1 e1 c2 d2 e2 a3 b3 c3 d3 e3 f3 g3 a4 b4 c4 d4 e4 f4 g4 a5 b5 c5 d5 e5 f5 g5 c6 d6 e6 c7 d7 e7	y= 3       y= 2       y= 1         y= 0           y=-1         y=-2       y=-3       x= -3 -2 -1 0 1 2 3
Standard 7x7 Notation [used by Beasley]	The bulky but general alternative: Cartesian Coordinates

For larger boards, this coordinate system must be extended. "Standard 9x9 Notation" is the obvious extension, where the columns are a-i and the rows 1-9 (the most interesting boards have an odd width, and that is why we go from 7x7 to 9x9). An unfortunate aspect of this notational switch is that the central hole goes from being called "d4" to "e5". An alternative notation is Cartesian Coordinates, where the center of the board is (0,0).

In standard 7x7 notation, to refer to a jump we simply list the starting and ending board locations separated by a dash, i.e. "e5-e3" for one of the jumps available above, and "e5-e3-c3-a3" for the 3-sweep move "e5-e3, e3-c3, c3-a3".

Any hole on the board that cannot be jumped over is called a corner hole. The standard board above has 8 corners: c1, e1, a3, g3, a5, g5, c7 and e7. The number and geometry of the corners is an important aspect of any peg solitaire board. We will call a board gapless if for any two board locations in the same row or column, all the intervening points are also on the board. All of the boards on this page are gapless, and this concept is mainly useful to exclude pathological boards with interior holes or gaps along the edge.

A board is called rectangular-symmetric if it is unchanged when reflected about the x- or y-axes, and rotationally-symmetric if it is unchanged by any 90 degree rotation. A board is called square-symmetric if it is both rectangular-symmetric and rotationally-symmetric. Finally, a square-symmetric board with a unique central hole is called odd because its width is odd, and it also has an odd total number of holes. We see that the board above is odd, square-symmetric and gapless.

Null-Class Boards

Consider a diagonal labeling of the board holes (in two ways) as shown below:

1 2 3 2 3 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 3 1 2 1 2 3

4 5 6 6 4 5 6 4 5 6 4 5 6 5 6 4 5 6 4 5 4 5 6 4 5 6 4 5 6 4 4 5 6

Given a board position B, let N_i(B) be the number of pegs in the cells marked i, and T(B) be the total number of pegs on the board. Now consider what happens to these functions after a solitaire jump is executed. The total number of pegs T always goes down by one, while one of N₁, N₂ or N₃ increases by one while the other two decrease by one (and similarly for N₄, N₅, N₆). Therefore the evenness or oddness of the differences T-N_i does not change as the game is played. This partitions the set of all possible boards into sixteen position classes depending on the parity (even or oddness) of the six numbers: (T-N₁,T-N₂,T-N₃,T- N₄,T-N₅,T-N₆). As you play solitaire you cannot leave the position class that you start in. [You might think there should be 2⁶ or 64 position classes. However the numbers N_i are not independent, because T = N₁+N₂+N₃ = N₄+N₅+N₆, and this implies that among (T-N₁,T- N₂,T-N₃), there are either zero or two odd parities (and similarly for the other half) and the number of position classes is reduced to 2⁴ or 16].

At the standard starting position with only the center hole vacant, you can easily check that N₁=N₃=N₄=N₆ =11, N₂=N₅=10, and the total number of pegs T=32. Therefore the 6 starting parities (T-N_i), i=1,2, ... 6 and therefore the position class of the board is (Odd, Even, Odd, Odd, Even, Odd). All board positions reachable from this starting position must be in this same position class. The position class of a board position with only a single peg is easy to calculate, it is odd on all diagonals except for the two the peg is in. Hence we see that the only possible finishing locations for the game must be the intersections between diagonals 2 and 5, or the board locations (0,0)=d4, (3,0)=g4, (0,3)=d1, (-3,0)=a4 or (0,-3)=d7.

Consider the board position with every peg filled. Then T=33, N_i=11 for all i, and all six parities are Even. The empty board with no pegs also lies in the same position class: (Even,Even,Even,Even,Even,Even). This is the defining property of a null-class board: any board position and its complement are always in the same position class (the complement of a board position is the board position where each peg is replaced by a hole and vice versa). It is important to realize that null-class boards are special, and not all boards are null-class boards.

By using the parity argument above, one can easily prove the following two facts about null-class boards:

1. Only on a null-class board is it possible to solve the complement problem.
2. Consider a null-class board with an initial vacancy at (x₀,y₀). Then we can only finish with a single peg at board positions (x₁,y₁) where the differences x₁-x₀ and y₁-y₀ are multiples of 3.

From a practical standpoint, how do we determine whether a particular board is null-class or not? The most obvious technique is to label the board in the above fashion and count the number of 1's, 2's, 3's through 6's. If these six numbers have the same parity (all odd or all even) then the board is null-class, otherwise it is not. A more clever technique is to apply local transformations to the full board that do not change the position class we are in, and try to reduce it to the empty board. A simple class of very useful transformations is to take the complement of any three consecutive board locations (vertically or horizontally). A solitaire move itself is such a transformation, but there are others, such as removing three pegs in a row, or replacing two pegs separated by a hole by a peg in the hole. Using this technique we can discover which position class any pattern of pegs is in, and which finishing holes are possible.

Boards

The English "Standard" Board (33 holes)

A 33-hole board from India, 1830 © 2003 puzzlemuseum.com

The English Board has more going for it than a pleasing shape. It is the smallest square-symmetric, gapless board on which the central game is solvable (see [P3]). In fact, every complement problem is solvable on this board.

Ernest Bergholt found an 18-move solution to the central game in 1912, and this was proven analytically to be minimal length by John Beasley in 1964 [B1]. Since then the problem of finding minimal length solutions has been attacked by several people using a computer, and minimal length solutions to all single vacancy to single survivor problems have been found.

If diagonal jumps are allowed, what is the shortest solution to the central game? This problem was unresolved until recently. Using a computer, it is not so easy to find the minimal length solution because the number of moves from each board position is approximately double. However I finally have been able to complete the calculation, to show that the minimal length solution has 15 moves [P4].

A web page on the shortest solution when diagonal jumps are allowed.

A page with example computer calculations by level for the English 33-hole board.

The 6x6 Board (36 holes)

This board does have one unusual property: it is possible for any peg on the board to reach some corner. No matter where the final peg is to be left, it is possible to arrange things so that the last four moves start from the four corners. If this board is easy for you try finding solutions with this property. You can see several of these corner finish solutions in the solution catalog. It is not always possible to have a minimal length solution with this corner finish property.

The French Board (37 holes)

In Winning Ways For Your Mathematical Plays [B3] this board is referred to as "the Continental Board". Unfortunately, there is no standard name by which these boards are known.

If you are solving a problem by hand, the best technique to use (on any board) is to decompose it into block removals. Note that the French board decomposes nicely into L and 3-removals as shown in the figure on the right. In order to solve a particular problem, one can modify this diagram with appropriate starting and ending moves, as shown in the figures on the left.

In each diagram, the location of the starting hole is marked by an "S", and the finishing hole by an "F". After dividing the board up into block removals, one must also check that the catalyst needed by each block removal is present. The sequencing of the block removals is indicated by the numbering. If you don't understand how to turn such a diagram into a solution, read the introduction to block removals at [W2].

If the game begins with the center peg missing, this board position is in the position class of the empty board, so it is not possible to finish with one peg. An elegant modification of the rules which allows for a d4 to d4 solution is known as Cremers' Key [W16]. According to [W16], this trick was invented by Frans Cremers, a retired teacher from Aalter, Belgium. Play proceeds from the center as normally, but the player is allowed to replace the central peg once when the hole is unoccupied. This puts the board position in the correct position class, and a solution can be obtained. Click here to see a center to center solution using Cremers' Key in 20 moves (the shortest possible).

Note that Cremers' key works not only from the center start, but from ANY start! In the general case one is allowed to replace the peg at the starting hole once. The goal is to finish with one peg, NOT at the starting location, but ALWAYS in the center. The results of [P3] show that Cremers' Key will put the board in the position class of the center finish on any gapless, odd, square-symmetric board that is not null-class.

Another way to make the central game solvable is to allow diagonal jumps. In this case the central game can be solved in as little as 13 moves. Another interesting puzzle is to begin with pegs at all locations except for the central 9 holes {c3, c4, c5, d3, d4, d5, e3, e4, e5}, and try to play to the complement of this position with only the center 9 holes occupied. Allowing diagonal jumps, this "big central game" is solvable, but the same problem is not solvable on the 33-hole English board. The shortest possible solution to the "big central game" has 13 moves, just like the normal central game.

A web page on the shortest solution when diagonal jumps are allowed

The 41-Hole Diamond Board

This board is not easy to decompose into block removals. In his book, Beasley [B1] shows how to accomplish the task using more complex block removals to deal with the corners.

One can try to obtain a center to center solution using the Cremers' Key [W16] rule modification, but it fails on this board. Replacing the center peg does give one a board position in the correct position class, but using the resource count shown below you can easily prove that it is not possible to finish with one peg.

This board is a member of a general class known as a draughtsboard. Such boards are obtained by taking any square board, and labeling the holes alternately as on a chess or checkers board. Then the board is rotated 45 degrees and the black squares define the holes of the draughtsboard. The 41-hole diamond board can be so obtained starting from a 9x9 square board.

The 32-Hole Diamond Board

This board has a similar size as the standard English board, but it has 14 corners rather than 8. In 1941, B. M. Stewart showed that all 25 single vacancy to single survivor problems were solvable on this board. My program has found that all of them are solvable in 17-19 moves, with only the d1 (top hole) complement requiring 19 moves. The central game can be solved in 18 moves, but no fewer.

The 13-Hole Diamond Board ("Hoppers" board)

In 2007, ThinkFun repackaged this game under the name "Cool Moves" with a completely new card set.

A web page with strategy tips for this board.

Wiegleb's Board (45 holes)

The proof of the impossibility of the (4,0) or e1 complement is not a simple matter, a brief outline of a proof is given in Beasley's book [B1]. I have found an alternate proof by integer programming techniques, but no short, simple proof has been found. A computer proof by exhaustive search is possible, but time consuming. It feels like you are smashing a peanut with a sledgehammer. This problem is several orders of magnitude harder than any problem on the English Board.

Problems on this board can be solved most easily using block removals as shown on the figures to the right. The numbering shows the ordering of the block removals. Numbers subdivided a, b, c are block removals that must be interleaved, in other words part b begins before part a is finished. To see the sequence of moves to solve the central game this way see the solution given in [W3]. Beasley [B1] gives block removal diagrams for one other complement problem on this board.

The longest sweep that can appear in any solution to a single vacancy to single survivor problem has length 16, one such sweep is shown on the board above left. There are three problems on this board for which a 16-sweep can occur, these make for interesting problems to solve by hand:

1. Vacate g4 (or equivalently d7), play to finish at d4 with the last move a 16-sweep (the above board shows the last move).
2. Vacate g5, play to finish at d2 with the last move a 16-sweep.
3. Vacate g6, play to finish at d3 with the second to the last move a 16-sweep. The 16-sweep begins and ends at d2.

The solution catalog shows the shortest length solution to all 35 solvable problems on this board, an effort which took about 3 months of CPU time on a 1 GHz Pentium PC. All of them are solvable in 20-23 moves, with only the (3,0) or e2 complement requiring 23 moves. There are a few problems on this board with unique minimal length solutions, at least up to symmetry and move order. The central game can be solved in 22 moves, but no fewer.

A web page of computational results for this board

The 9x9 Board (81 holes)

This board also represents a significant computational barrier. On the standard 33-hole board, it is possible to answer any question using a computer in a reasonable amount of time. On that board, the total number of distinct board positions is 2³³, about 8.6 billion. However only 1/16'th of these board positions lie in the same position class as the starting position, so the number of board positions reachable from any single vacancy start is at most 2²⁹, about 540 million. From an initial vacancy at c3 [or (1,1)] one can calculate that 264 million board positions are reachable (and somewhat less from any other starting location). If we use 33 bits per board, storing all 264 million board positions requires about 1 Gigabyte of hard drive space, not that large a number by today's standards.

On the 9x9 board the total number of distinct board positions is 2⁸¹, which is greater than 10²⁴. Even if we assume 1/32 of them are reachable from a single vacancy start, this is still 7x10²² board positions, which would require 7x10¹⁴ gigabytes of storage (assuming 81 bits per board). By using symmetry and other ideas (for example we probably don't need to compute all positions reachable anyway) we may be able to reduce this by 2-4 orders of magnitude, reducing our storage requirements to a mere hundred billion gigabytes! These numbers give a good idea how much harder this board is to deal with computationally compared to the standard 33-hole board.

A web page on the 9x9 board.

The Antique English Board (67 holes)

For some reason two additional holes have also been added to the top row of this board (breaking the square symmetry, as well as no longer being gapless). The holes outlined in green appear to have some special significance-- it may have been that some special set of rules applied to this board. Robert Reid has emailed me that this board is in David Parlett's Oxford History of Board Games (p. 192), and that the markings are for a Fox and Geese variant called "Siege". He doubts this board was ever used to play peg solitaire.

It is not hard to show that this board shares the same problems as the 37-hole French board. Namely, if one begins with a full board with only the center peg missing, it is impossible to play to one peg. This is true for the 67-hole board as well as the square-symmetric 65-hole version where the two holes on the ends of the top row are removed.

Rectangular Boards

Generalized Cross Boards

I looked into these boards and discovered there are exactly 12 generalized cross boards where the complement problem is solvable at every board location. Of these 12, the only board with square symmetry is the standard 33-hole board. Two are rectangular-symmetric (top row of figure to the left), three have rectangular symmetry about only one axis, two have diagonal symmetry (second row) and the remaining four have no symmetries at all [W1].

A web page on Generalized Cross Boards (not available online).

Triangular Boards

There is now a separate web page for Triangular Solitaire.

Online Puzzles

Here is a list of the games on this site. These are all written in JavaScript, and you must have JavaScript activated in your browser to view them. Works best with Internet Explorer.

• GPJ Puzzles produced for Issue #28 of The Games and Puzzles Journal [W1] (2003).
• G4G7 Puzzles created for the seventh Gathering for Gardner (2006).
• Ultimate Triangular Peg Solitaire ( http://tinyurl.com/2v53ya , 2006) You can play on a triangular board of any size. If you get stumped, watch the computer solve the problem.
• "Never Lose" Triangular Peg Solitaire ( http://tinyurl.com/5rh2rx , 2008) This version has the ability to point out all jumps that lead to a dead end, which tends to make the puzzle seem trivial!

Solution Catalogs

Given a board and a (solvable) single vacancy to single survivor problem, there is a minimum number of moves that can solve it. These solutions have an elegant look to them and they tend to be extremely hard to find by hand.

The table below shows a list of boards, together with some statistics about each. If you click on a board, you will see another table listing all single vacancy to single survivor problems solvable on that board, together with information about these solutions. These boards are all square-symmetric, and we only list unique single vacancy to single survivor problems. In other words, if one problem can be obtained from another by rotation and/or reflection, only one will be listed. If you keep clicking on the tables, you can view diagrams of minimal length solutions. The solution catalog shown may be incomplete. Only those with a check mark before them can display all solutions. It is labor intensive producing these and they may not be complete for some time.

Some column heads you will see that require explanation:

• Number of Problems: This is the number of different solvable single vacancy to single survivor problems on this board (not including problems equivalent by symmetry).
• Longest Sweep: The first column is the longest sweep possible in any single vacancy to single survivor problem, regardless of solution length. The second "Longest Sweep" column is the longest sweep possible in a minimal length solution.
• Solution Lengths: This is the range in the number of moves in the minimal length solutions, over all problems.
• Time to Calculate: This is the amount of CPU time that was needed to calculate the underlying table (find all minimal solutions to all problems).

Board Name [click to see catalog]	Number of Holes	Null-Class?	Number of Problems	Longest Sweep (any problem)	Minimal Length Solution Properties
					Solution Lengths	Longest Sweep	Time to Calculate
Diamond32	32	Yes	35	8 or 9	17-19	8	2 hours
English	33	Yes	21	9  †	15-19	8	2 hours
6x6	36	Yes	21	10 ‡	15-16	10	6 hours
French	37	No	10	9 †	20-21	9	24 hours
Diamond41	41	No	4	9	26	9	3 hours
Wiegleb's	45	Yes	35	16  ‡	20-23	14	3 months
Table Footnotes: (†) From John Beasley's book [B1]. (‡) This is the longest sweep geometrically possible on this board, and at least one such sweep can be realized as the final move to a single vacancy to single survivor problem.

The information in the above table has been calculated by other authors in the case of the English, French and 6x6 boards. My results have been checked against their results. To my knowledge, nobody before has calculated minimal length solutions for the 41-hole diamond and 45-hole Wiegleb's board.

Computational Search Techniques

A good introductory reference which has a nice progression of problems is the recent book by Koetke [B4]. This book starts out with small boards that are easy to solve, and discusses the problems encountered as larger boards are considered. It also has example programs in java. This is one of the few peg solitaire references that contains a lot of detail about solving the game computationally.

If you try to solve a peg solitaire problem in in inefficient manner your program can take forever to run, even on the standard 33-hole board. For example the most obvious technique is to store the sequence of moves (or jumps) and try to exhaustively go through all possible sequences. Because there are a large number of move sequences that result in the same board position, such an algorithm is extremely inefficient. One solution is to use a hash table or some other means to store board positions seen previously so you do not have to investigate them farther.

A similar technique is to only keep track of the set of boards at each level in the tree, rather than the moves. I call this a search by levels. This is much faster than a straight search over move sequences and can solve any problem on the 33-hole board relatively quickly. For boards larger than this, additional techniques are needed.

I have used four ideas to speed up a search by levels:

1. If the problem is symmetric, there is no need to search over solutions that are really the same. On the 33-hole English board this can help a lot for some problems (for example the central game or d4 complement) but not on others (for example the c1 complement).
2. The use of resource counts or Pagoda functions as explained below (see also [B1] or [B3]). Whether this is useful computationally depends on the board as well as the problem. It is most helpful on the 41-hole Diamond Board.
3. Search forward from the starting position and also backwards from the desired final position, looking for matching board positions in the middle. This is known as a bidirectional search. A trade-off is that the programming is more complex.
4. A* search. This works particularly well when diagonal jumps are allowed. For details see [P4].

Finally, one should be careful not to confuse the computer's failure to find a solution with a proof that no solution exists. These calculations are complex and lengthy, particularly when resource counts, symmetry and forward/backward calculations are all being used. Logical bugs in the code can easily prevent the computer from finding a solution, and much testing is required to make sure the results are reasonable. For example, no program that reports "no 17 move solution to the English board central game" should be trusted unless it can find the Bergholt solution in 18 moves. You can also test your program by reproducing these tables.

For more details on computational search techniques, see my paper [P4].

Resource Counts

The most useful resource counts generally have negative values in corners (in fact, it is not hard to show that a resource count can only have a negative value at a corner). If a certain move leaves the board with a value that is less than that of our final position, we know that this move cannot possibly lead to a solution. A very useful resource count on the 41-hole diamond board is:

					-1
				-1	1	-1
			-1	1	0	1	-1
		-1	1	0	1	0	1	-1
	-1	1	0	1	0	1	0	1	-1
		-1	1	0	1	0	1	-1
			-1	1	0	1	-1
				-1	1	-1
					-1

This resource count provides a constraint on board positions that can appear in any single vacancy to single survivor problem. One can show that for any single vacancy to single survivor problem on this board, after the first move and before the last move, this resource count must always evaluate to zero. This is a powerful constraint on the jumps available from any board position (that can lead to a solution). No solution can contain a jump over a peg in a hole marked "1" (white squares) unless that jump starts at the edge of the board. This constraint can be added to a computer program, and with it this board is not much harder to find minimal solutions (on a computer) than the 33-hole English board.

The above resource constraint can be applied on the French board (by ignoring any value that is not on that board) and it is still somewhat useful. It can also be used on the English board, but with only a very minor speed increase.

On Wiegleb's board a moderately useful resource count (for computations) has a "-1" at the eight corners and "+1" at the 12 interior locations: d2, f2, b4, d4, f4, h4, b6, d6, f6, h6, d8 and f8 (and zeros everywhere else).

How Hard Is Peg Solitaire?

© 1996 John Robinson

Consider a rather arbitrary board without any interior holes (gapless). This board might be quite large, and in the simplest case could be square. We will try to quantify the difficulty of three different problems:

1. Given an arbitrary pattern of pegs, determine if it can be reduced to a single peg. In this case the specific shape of the board is not important, we could even consider the board to be infinite.
2. Given a complement problem on this board, find any solution to it. Here the shape of the board is very important. The central game on the 33-hole board is an example of such a problem.
3. Given a complement problem on this board, find the shortest solution to it (minimum number of moves).

It was shown in 1990 [P1] that problem #1 is NP Complete. In practical terms this means that any algorithm to solve this problem will have a run time that increases exponentially with the problem size.

Like problem #1, problem #2 also asks if you can reduce a pattern of pegs to a single peg. Does this mean problem #2 is NP Complete also? No it does not, because a complement problem does not start from an arbitrary pattern of pegs. The starting position has every hole occupied, except for the starting vacancy, hence it is a very regular pattern. Of course the shape of the board could still be somewhat complicated, but as a gapless board it is only the edge that may be complicated. It is still unknown how hard problem #2 is to solve.

Clearly if the board is not null-class, there is no solution to problem #2 or #3 (we can also check in problem #1 if the configuration of pegs is in the position class of one peg or not). It is very easy to check if a board is null-class, so we don't need to require this. As the difficult case we may as well assume all boards from now on are null-class.

Suppose we consider problem #2 on an arbitrary rectangular null-class board (with both edges at least 4). Finding a solution to a complement problem on such a board, I claim, is actually very easy, and can probably be solved in linear time. Why? Because this problem is easy to solve by hand using block removals. You simply visually identify blocks of pegs you need to remove and make sure the required catalyst is present and that after the block removal you don't strand any pegs too far away from the core bunch. You also need to be careful near the edge of the board. This logic could be programmed, and would result in a computer algorithm that could solve complement problems on rectangular boards extremely quickly.

Even boards which are "not too different" from a rectangular board should also be easy. It is rather difficult to quantify "not too different", but basically any gapless null-class board that doesn't have any tight spaces can be solved using block removals. The standard 33-hole board should fit into this category. It easier to complete this argument on triangular boards, and I now have a simple algorithm that can solve any single vacancy problem on a triangular board of arbitrary size.

In summary, problem #1 has been proven NP Complete. I believe problem #2 is not very hard on "well behaved" boards, which include at least rectangular boards (and certainly triangular boards). Problem #3 is very difficult on any board with more than about 50 holes, and I believe no algorithm can find shortest solutions in polynomial time.

Peg Solitaire References

Books

Papers

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Copyright © 2009 by George I. Bell

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