Detail of a French board marketed as Solitaire Di Venezia, solid ebony board with individually hand-blown glass marbles.
These pages were created and are maintained by George Bell (gibell@comcast.net) Last Modified November 4th, 2024 Copyright © 2006–2024 by George I. Bell 2016 snapshot of this site on recmath.org	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]

A 1697 engraving by C-A Berey of the Princess Soubise beside a French solitaire board.

Table of Contents:

• Introduction
• Preliminaries: Position Class, Where can the final peg be? Short Solutions, Playing Backwards
• Boards: English, French, 6x6, Diamond41, Diamond32, Hoppers,
    Wiegleb, 8x8, 9x9, Rectangular, Siege, Generalized Cross, Kralenspel
• Generalized Boards: Gridless, Toroidal, 3D
• Online Games and Puzzles
• Solution Catalogs
• Computational Search Techniques
• How Hard Is Peg Solitaire?
• References

Introduction

This one-person game (or puzzle) first appeared in France in the late 17th century. We know this because the puzzle is depicted in several art works of the period, most notably the engraving by Claude Auguste Berey shown on the right. The first mention of the game in print was in the French literary magazine Mercure Galant in 1697. Remarkably, this article escaped the notice of game historians for hundreds of years, and it came to light only in 2014, shortly after Mercure Galant became available on the internet. You can read more about this discovery in John Beasley's historical update [P10].

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, often made of wood, together with a set of pegs (or alternatively 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 everyone (in my generation) 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 jumps 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 (for details, jump ahead to here, or see [B1], [B3], [W2] or [W20]).

Lithographed Tin board with plastic pegs in two colors, © 2015 puzzlemuseum.com

After the central game above has been solved, what then? We can try to find the "shortest" solution. All solutions contain 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 is more difficult than finding a solution to the central game, and we will return to it when discussing the English board.

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. Below we will also discuss gridless boards.

James Dalgety [W13] is a puzzle collector who owns many peg solitaire boards. On his web page he discusses design faults of peg solitaire boards, which I include here for future puzzle designers. Common design faults include:

• Not enough room for the fingers between the pegs or balls.
• No space, or not enough space, in which to place the captured pieces.
• Difficult to lift full board off the table.
• Difficult to place pegs into the holes
• Nowhere to put the pegs when not in use.

Preliminaries

Board Notation and Identification

We mention quickly other possible notations: first, one can simply number the holes consecutively. This notation is quite common, the main problem with it is that the numbering changes every time the board shape changes. John Conway invented an alphabetic notation which mirrors the symmetry of the board [B3]. The most general notation is 2D Cartesian coordinates, with the origin at the center of the board (to preserve symmetry), or at the lower left corner (to avoid negative coordinates). This notation (see below, right) is bulky and is primarily used in computer programs.

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 [B1]) The bulky but general alternative: Cartesian coordinates

For larger boards, "Standard 7x7 notation" 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 "d4" to "e5".

In standard 7x7 notation, to refer to a jump we simply list the starting and ending holes separated by a dash, i.e. "e5-e3" for one of the jumps available on the left, and "e5-e3-c3-a3" is shorthand for the move consisting of the jumps "e5-e3, e3-c3, c3-a3".

A hand carved board with clay marbles, dating from the early 1900's (photo courtesy St. John Stimson).

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. By requiring that the board be gapless we exclude pathological boards with missing interior holes or gaps along the edge. For an example of a board which is not gapless see Dot Solitaire 4714.

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.

Complement Problems and Single Vacancy to Single Survivor Problems

The general peg solitaire problem is to play from a full board with one peg missing to a board position where only one peg remains. These problems are called single vacancy to single survivor problems, sometimes abbreviated SVSS. The special problem where the initial vacancy and survivor are the same board location is called the (single vacancy) complement problem.

Position Class and 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

This English board designed by Michael Graves has a novel spiral marble trap.

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 hole filled by a peg. 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.

A French board marketed as Solitaire Di Venezia, solid ebony board with individually hand blown glass marbles.

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. This is referred to by John Conway et. al. as "the rule of three" [B3].

From a practical standpoint, how do we determine whether a 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 jump 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.

Where can the final peg be?

Figuring out possible finishing holes from the starting position.

Figuring out possible finishing holes from a failed game.

The transformations in the last section are useful to find all possible finishing holes from any configuration of pegs on any board (null-class or not). Consider, for example, the peg solitaire puzzle to the right, from the iPhone App Puzzlium [W23]. In order to determine where the final peg can be, I mentally erase lines of three pegs (denoted by the red bars). This leaves only one peg, so the only possible finishing holes are the two marked "F". The finishing peg in the "F" hole must never be captured, because no other peg can finish there.

We can use this technique not just at the start, but at any time during a solution. One can attempt to solve the puzzle, and failing that, observe carefully the final pattern of pegs. Using the same transformations, reduce it to a single peg, which will give the possible finishing holes. The simplest case is two pegs a distance 2 apart, in which case the finish must be the hole between them (or any hole a multiple of 3 away in any direction). Here is a summary of local transformations for use in determining finish holes:

1. Three pegs together in a row (or column) can be removed.
2. Two pegs in the same row (or column) a distance 2 apart can be replaced by one peg between them.
3. Any peg can be moved 3 holes along any column or row. If it lands on another peg, both are removed.
4. Any solitaire jump can be executed (forward or backward).

When determining the finishing hole, one can also ignore the board during the process, and not worry when your transformation results in a peg off the board. Of course, in the end a finishing hole can only be one that is on the board. Can you determine the possible finishing holes from the failed game shown on the right?

How can the knowledge of finishing holes help solve a puzzle? You can keep track of the pegs which can finish, being careful not to remove all of them. In addition, on cramped boards you may be able to figure out the final jumps, and aim for such a board position. On the first Puzzlium board shown, this is particularly useful.

What happens if the local transformations fail to reach a one peg state? This can only happen if all pegs can be removed, leaving an empty board. This shows the original puzzle is not solvable to one peg! Puzzlium [W23] has internal checks which prevent users from entering unsolvable peg solitaire puzzles.

Short Solutions

On the left we see a board position on Wiegleb's board from which a remarkable 16-loop can be executed. This loop move fits on the English Board, however on Wiegleb's board this board position can actually be reached in a single vacancy to single survivor game. An interesting puzzle which can be solved by hand is to find a solution with a 16-loop.

Playing Backwards

Backward play is hard to comprehend, because our brain does not easily interchange the concepts of "hole" and "peg". It is hard to shake the perception that a hole is the absence of a peg. It is much easier to understand "backward play" by realizing that it is the same as forward play from the complement of the current board position.

For example, suppose we want to solve the puzzle mentioned above on Wiegleb's board. Namely, we want to find a solution beginning with one peg missing, and ending with a spectacular 16-loop. Playing forward, how can we possibly know which 27 jumps might end at the complex 17-peg pattern shown above? It seems hopeless, even impossible; but now we know that playing the game backwards from the 16-loop position is the same as playing the game forward from the complement of this position. So we begin from the complement of the 16-loop position (shown on the left). If we can solve this board to one peg (and this is possible), then the 16-sweep is ours! We then begin with a board filled except for where this solution ended, and play the jump sequence in reverse. This will magically reproduce the 16-loop board position! We will return to this puzzle later. In [B3] this technique is called the "time reversal trick", and it can be a powerful tool for solving certain problems.

Boards

The English "Standard" Board (33 holes)

A 33-hole board from India, ca. 1830 © 2015 puzzlemuseum.com

Over the years, literally hundreds of versions of this board have been produced under brand names such as Puzzle Pegs, Hi-Q, and Classical Solitaire. Even today, if you google peg solitaire board you will find dozens of English boards to choose from. The most expensive boards, with exotic marbles, usually include the four extra holes of the French Board.

Screen capture of "Pegged".

The computer puzzle "Pegged" was included in Microsoft Windows 3.0 (1990), one of eight puzzles in the first Microsoft Entertainment Pack.

Why is the English board so popular? It is the smallest square-symmetric, gapless board on which the central game is solvable (see [P3]). In fact, it is the smallest such board on which every complement problem is solvable.

Ernest Bergholt found an 18-move solution to the central game in 1912. In 1964, John Beasley proved that there is no shorter solution to the central game [B1]. Since then the problem of finding minimal length solutions has been attacked by many people using a computer, and minimal length solutions to all single vacancy to single survivor problems have been found. In 2012, Joseph Barker and Richard Korf [P8] applied advanced search techniques to this problem—their search algorithm can find Bergholt's solution using less than 3 seconds of CPU time! They can find shortest solutions to all 21 SVSS problems using less than 2 minutes of CPU time [P8].

Over the years, there have been various attempts to describe a solution to the central game that is easy to remember. Bergholt's 18-move solution is shortest, but tends to be difficult to recall. Nonetheless, a specialized "Wolstenholme notation" has been invented to assist in recalling it [W19]. Another solution, called "Jabberwockey", is given in Martin Gardner's book [B2]. This solution has some nice symmetry properties, which make it easier to remember, as well as faster because at a certain point one can capture pegs with both hands simultaneously. Click here to see a diagram of this solution (paired jumps are shown in red). Another solution where the board passes through seven positions with diagonal symmetry can be seen here.

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

My personal favorite easy to remember solution to the central game requires that you memorize the L-move (see [W2] under L-Package). Start with d2-d4, f3-d3, and now focus in on the (blue) L-shaped region defined by L1={e4, e3, e2, e1, d1, c1} (see the diagram on the left). The L-move consists of three moves (four jumps) whose net effect is to complement or reverse the state of the pegs in L1: e1-e3, e4-e2, c1-e1-e3 (or you can do e1-e3, c1-e1, e4-e2, e1-e3 if you find this easier to remember). We now do the L-move on the other three arms of the cross. First, e6-e4 to open up (orange) L2={d5, e5, f5, g5, g4, g3}, then the second L-move: g5-e5, d5-f5, g3-g5-e5. Next, b5-d5 to open up (purple) L3={c4, c5, c6, c7, d7, e7}, then the third L-move: c7-c5, c4-c6, e7-c7-c5. Finally, c2-c4 to open up (red) L4={d3, c3, b3, a3, a4, a5}, then the last L-move: a3-c3, d3-b3, a5-a3-c3.

This leaves you in the heart-shaped board position shown on the upper right. Notice the six jump loop starting from d5: d5-b5-b3-d3-f3-f5-d5. This leaves the board with a "T" shaped configuration of pegs on the right which can be solved (by inspection): d4-f4, d6-d4, c4-e4, f4-d4. The animation on the right shows the whole solution, or see this diagram, or this WikiHow Page [W21]. Note that the screen capture of "Pegged" (above left), is actually one frame of a 2007 YouTube video by Kevin Paul Scarrott. The video shows the same solution, but reflected about the x-axis.

If diagonal jumps are allowed, what is the shortest solution to the central game? In Beasley's book [B1] he gives a 16-move solution, and remarks that it is not known if this is the shortest. In 2006, I was able to complete the exhaustive search, and the shortest solution has 15 moves [P4].

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

A web page with sample computer calculations for the English 33-hole board.

Symmetric English Puzzles, a collection of 279 symmetric positions which can be solved to one peg in the center.

A French board made by C. Jeandin, France, ca 1890. Photo courtesy the Slocum Collection.

The French Board (37 holes)

This board is not a null-class board, therefore no complement problem is solvable on it. Using the position class theory, one can prove that if you begin from a filled board with only the center vacant, it is impossible to finish with one peg. To show this, simply compute the six parities of the starting board position. All six parities are even, and therefore all six parities will always be even. But any one peg position has four odd parities and two even parities, so no single peg position can be reached.

When solving by hand, many of us record the first solution we find, preserving a moment of triumph. Things were not so different 300 years ago! On the left we see the oldest published solution to a peg solitaire puzzle, from the 1697 edition of the French literary magazine Mercure Galant. In this solution, the holes are numbered 1 to 37, and the problem to be solved is "vacate 1, finish at 3" (in our notation "vacate c1, finish at e1"). The solution translates as: "Solitaire Table: Starting with the third peg, take 2 with 3, 6 with 12, 7 with 8" (in our notation "e1-c1, d3-d1, f2-d2"). The complete diagram of this 300 year old solution can be found here.

Solitaire was very popular in France at the start of the 18th century. Below we show a shield or escutcheon bearing a French peg solitaire board (see [P10]). This is part of a French coat of arms registered in 1709. Peg solitaire is possibly the only puzzle to have appeared in a traditional coat of arms.

There are ten single vacancy to single survivor problems solvable on this board, and each can be solved in 20 or 21 moves. 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 on the right. In order to solve a particular problem, one can modify this diagram with appropriate starting and ending moves, as shown 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. Click on each diagram to show the full solution played out. 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 interesting modification of the rules which allows for a d4 to d4 solution is known as Cremers' Key [W16]. According to [W16], this concept was invented around 1998 by Frans Cremers, a retired teacher from Aalter, Belgium. Starting with the center vacant, play proceeds as usual, except that 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. The shortest solution to the central game using Cremers' Key has 20 moves.

An escutcheon, ca. 1709 "Jaque Chavillot" [P10].

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 reason why this works is clear if you understand position classes. On the French board, the board with every hole filled by a peg is in the same position class as the board with one peg in the center. Therefore, when you replace a peg at the starting hole, what you are doing is changing the position class to that of the full board, so that finishing in the center is now possible. 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. One interesting puzzle with diagonal jumps 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, the shortest possible solution has 13 moves. One reason the "big central game" is interesting is that using it one can prove this board is universal [P4].

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

Symmetric French Puzzles, a collection of 428 symmetric positions which can be solved to one peg in the center.

The 6x6 Board (36 holes)

The 6x6 board supports some remarkably short solutions. John W. Harris and Harry O. Davis studied this board in the 1960's, and found that most single vacancy to single survivor problems can be solved in 15 moves, with a few cases requiring 16 moves [W1].

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. Click here to see an elegant 15-move solution with this property, due to John Harris [W1]. Here is another one I found. It is not always possible to have a minimal length solution with this corner finish property.

A double vacancy complement problem is a puzzle where two pegs are removed at the start, and your goal is to finish with two pegs in the original vacancies. On the 6x6 board, all 93 double vacancy complement problems are solvable (unlike the English 33-hole board, where four double vacancy complement problems are not solvable [B1, p. 106-9]). It is tedious to find solutions for the 93 cases, but all are solvable (here is a sample solution). The 6x6 board is not the smallest board where all double vacancy complement problems are solvable, the 6x4 rectangular board also has this property.

The 41-Hole Diamond Board

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.

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

Several interesting variations to the 41-Hole Diamond Board have been proposed. Around 1882, H.-A.-H. Hermary proposed removing the leftmost and rightmost holes from the board, giving Hermary's 39-Hole board, shown on the right. This board is null-class, rectangular-symmetric, and most (but not all) complement problems are solvable [B1]. The central game on Hermary's 39-hole board can be solved in a minimum of 23 moves.

In 1894, A. Huber removed 4 holes to produce Huber's 37-Hole board, shown on the left. This board is null-class and square-symmetric—the central vacancy and other complement problems are solvable. The central game on Huber's 37-hole board can be solved in a minimum of 20 moves. We note that the complement problem at the "tip of the arm" is not solvable (proved in [P3]).

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 SVSS 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 d4-complement can be solved in a minimum of 18 moves.

A classic problem on a checkers board is to begin with 24 pegs along the edges with the 4x4 center section empty. It is not hard to show that this board position is in the position class of the empty board, hence it cannot be solved to one peg [W26].

Hoppers Board

Cool Moves, 2007

In "Hoppers", Nob Yoshigahara came up with a whole set of challenge cards for this board. He also had the idea of including a special red frog which must be preserved and perform the final jump. The original name for Nob's version was "Marsh Madness".

If you rotate this board 45 degrees, it can be seen as a 13 hole Diamond Board, or draughtsboard created from a 5x5 square board, with the addition of diagonal jumps along both diagonals. Without diagonal jumps, no single vacancy to single survivor problem on this board is solvable. However with the diagonal jumps many problems become solvable, and its small size makes solutions easier to work out. An interesting advanced problem is to try to solve the central game on this board in as few as 7 moves.

The Hoppers Boad can be extended by adding 4 holes to the outside and new jumps into and out of the new holes as shown on the left. With each iteration the board area doubles, the left figure shows 13-hole (Hoppers or Great 13), 17-hole (Great-17) and 21-hole (Great-21) boards. The 17-hole board is the same as Breitenbach's board but has eight additional jumps added.

A web page with strategy tips for this board.

Wiegleb's Board (45 holes)

A Soli2 board.

There are 36 different SVSS problems on Wiegleb's Board, all are solvable except for the (4,0) or e1 complement problem. A proof that the (4,0) or e1-complement is unsolvable is not easy. An outline of a proof is given in Beasley's book [B1], but filling in the details is non-trivial (I asked John Beasley about this, and he agrees that filling in the details is not easy). I have found an alternate proof by integer programming techniques [P3]. A computer proof by exhaustive search is possible, but time consuming.

Trying to solve the (4,0) or e1-complement problem is several orders of magnitude harder than any problem on the English board. If you have a fast peg solitaire solver try testing it on this problem. If you want to try a problem that actually has a solution, try any other complement problem on this board, or try the (4,0) or d1-complement problem on the 3232 Board.

Problems on this board can be solved most easily using block removals as shown on the figures to the left. 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. Click on each diagram to see the full solution played out. Beasley [B1] gives a block removal diagram for one other complement problem on this board.

Hand-made Wiegleb's Board in painted plywood, 16 mm marbles, DIY Puzzles.

The longest sweep possible on Wiegleb's board has length 16—one possibility is shown on the board on the upper left. Because it begins and ends at the same place, it is a 16-loop. This sweep also fits on the English board, but it cannot appear during a single vacancy to single survivor problem (proved in Beasley's book [B1]). In 2005, I discovered that on Wiegleb's board the 16-loop can be reached [P2]! There are three SVSS problems for which a 16-loop can occur. Here is a web page that you can print out to try these problems by hand:

1. Vacate g4 (or equivalently d7), play to finish at d4 with the last move a 16-loop (the above board shows the last move).
2. Vacate g5, play to finish at d2 with the last move a 16-loop.
3. Vacate d6 or g6, play to finish at d3 with the second to the last move a 16-loop. The 16-loop begins and ends at d2.

No other combination of starting and finishing holes can contain a 16-sweep, except of course rotations or reflections of the above three problems. The three problems above can be solved in a minimum of 22, 24, and 23 total moves, respectively. For solutions, see [P2] or this web page.

The solution catalog shows the shortest length solution to all 35 solvable problems on this board, an effort in 2004 which required 3 months of CPU time on a 1 GHz Pentium PC. The results can be found in our paper [P2], and in 2012 were confirmed by Barker and Korf [P8]. All the problems are solvable in a minimum of 20-23 moves, with only the (3,0) or e2 complement requiring 23 moves. The central game can be solved in 22 moves, but no fewer. There are a few problems on this board with unique minimal length solutions, up to symmetry and move order. For example, there is a unique 20 move solution from c4 to i4.

A web page of computational results for Wiegleb's Board

A web page on 16-loops on Wiegleb's Board (print it out and try them yourself)

The 8x8 Board (64 holes)

Using the same technique [W1] as for the 6x6 board, it is easy to prove that the solution to any SVSS problem must have at least 24 moves. In 1986, John Harris found a 25-move solution by hand [W1]; finding a shorter solution is a difficult computational task, but in 2014 I found a 24-move solution (and here is another).

The longest finishing sweep on this board has length 21. It is not difficult to find a solution finishing with a 21-sweep—this is a fun exercise to work out by hand. As usual, you must begin from the complement of the sweep position (one possibility is shown on the right).

The 9x9 Board (81 holes)

On this board, what is the shortest solution to the central game? This is currently an open question. In 2004, Alain Maye found a solution by hand in 34 moves. It is likely that the shortest solution has around 30 moves, by analogy with the 10x8 board, which is of similar size. Curiously, the odd side length makes the techniques used on the 10x8 board much less powerful.

An interesting challenge is to demonstrate that the 32-loops shown to the left can be reached from single vacancy starts. Some of them can also be the final move to a complement problem. It is also possible to find a solution to the central game ending with a 30-loop.

A web page on the 9x9 board

Rectangular Boards and Even-Even Boards

Rectangular boards where both n and m are even are more interesting than you might think, we call them "even-even boards". Why are even-even boards interesting? By analogy with the proof [W1] on the 6x6 board, it is easy to prove that the solution to any SVSS problem on an even-even board must have at least (n/2+1)(m/2+1)-1 = nm/4+(n+m)/2 moves. Therefore, if one can find a solution of this length, it must be the shortest possible. This may not sound very exciting, but the remarkable thing is that it can be rather easy to find solutions of this length. Even-even boards are interesting because it is surprisingly easy to find the shortest solutions.

The grid to the left summarizes my results on even-even boards. The green squares denote null-class boards, otherwise the square is red. Square boards appear along the diagonal. The tiny 4x4 board has a minimum solution length of 8 moves, an 8 move solution does not exist, although 9 moves is possible. For many larger even-even boards, the minimum solution length seems to be attainable.

On the 6x4 board it is not difficult to find 11-move solutions, and we have already mentioned that the 6x6 board has 15-move solutions. On the 8x6 board, I have found 19-move solutions, and on the 8x8 square board, 24-move solutions. All of these are automatically the shortest possible, by the above argument.

In 2014, I found a 29-move solution on the 10x8 board. In 2015, I found a 34-move solution on the 12x8 board! This is the largest board (96 holes) for which a solution having minimal length is known. Finding these shortest solutions on such large boards requires advanced search techniques, I use A* search based on the number of filled Merson regions [B1].

67-hole Siege Board, ca. 1880

The "Siege" Board (67 holes)

Generalized Cross Boards

I investigated these boards in 2003 and found 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 have rectangular symmetry (top row of figure to the left), three have mirror symmetry about one axis, two have diagonal symmetry (second row) and the remaining four have no symmetries at all.

c1 d1 e1 c2 d2 e2 c3 d3 e3 a4 b4 c4 d4 e4 f4 g4 a5 b5 c5 d5 e5 f5 g5 a6 b6 c6 d6 e6 f6 g6 c7 d7 e7 c8 d8 e8 c9 d9 e9

Of particular interest is the 39-hole board in the upper right with alternating arm lengths 3,2,3,2. Our hole notation for this board is shown to the right. A very hard problem is to solve the d1-complement problem. The solution to this complement problem is unique up to symmetry and move order [P2]. This is an unusual property for a peg solitaire solution and makes this puzzle hard to solve, either by hand or using a computer. Try this problem yourself using this online puzzle. Here are four challenges of increasing difficulty on the "3232 Board".

1. Find a solution to the central game (d5-complement).
2. Find a solution to the a4-complement ending with the 12-loop: a4-c4-c2-e2-e4-g4-g6-e6-e8-c8-c6-a6-a4.
3. Vacate a4, and mark the man at g4. Play to finish at a4 with the marked man sweeping off 13 in the final move.
4. Find a solution to the d1-complement.

A web page on Generalized Cross Boards

Three challenges on the 3232 Board (print it and try them yourself)

A Solo Board. Photo courtesy Jaap Scherphuis [W20].

Kralenspel or Solo Board

The Solo board is not null-class. For detailed analysis of this board see this web page on Jaap's Puzzle Page [W20].

Triangular Boards

There is now a separate web page for Triangular Solitaire.

Generalized Boards

Jumping rules leading to a universal board indicate a fundamental change in the game—the position classes have "collapsed" into a single set, really the whole concept of a position class becomes meaningless. Since every board position is in the same position class, all boards are null-class (this term has also become meaningless). A universal board might sound like a good thing, but it usually means that the puzzle is too complicated.

Consider, for example, the English Board. From a randomly chosen board position (probability of 1/2 to have a peg in any hole) there are an average of 9.5 jumps available. If diagonal jumps are allowed, the board becomes universal, with an average of 17.0 jumps available. This means that the game tree expands almost twice as fast. For example after 5 jumps under normal rules about (9.5)⁵ ~ 77,000 board positions should be reachable. With diagonal jumps this number is 17⁵, which is greater than one million.

CE = C9 "Center Empty"

C9 = CE "Center Nine"

How can we prove that a board under certain jumping rules is universal? Computer solvers can demonstrate this. However, there is a clever analytical trick that may work. Consider the French Board under diagonal jumping rules. Suppose we can prove these two propositions:

1. The "Big Central Game" is solvable. In other words, it is possible to play between CE and C9 (shown on the right). Diagonal jumps (as well as orthogonal jumps) are required.
2. We can play from C9 to a one peg finish, anywhere on the board. Because of symmetry, it suffices to show that we can finish with one peg at eight board locations: c1, d1, b2, c2, d2, c3, d3, d4.

To apply this trick in general, we need to find a board position like C9. It must have the following properties:

1. It is possible to play from the complement of C9 (CE) to finish at C9.
2. It is possible to play from C9 to a one-peg finish anywhere on the board.

Gridless Boards

Our derivation of null-class does not extend to gridless boards; however, the concept can sometimes be extended using "invariant parity counts", as explained in Beasley's book [B1, p. 63-5 and p. 241-2]. Several examples of invariant parity counts are given below.

Star Jump, by Creative Crafthouse

Hyper Solitaire, photo courtesy the Slocum collection

Clock Solitaire, 2017

Solomon, by Kadon Enterprises

Clock Solitaire, photo courtesy St. John Stimson

Breitenbach's Board

The most common gridless board seems to be Star Jump (shown to the left), a 10-hole board sold by Creative Crafthouse, it has also been sold under the names Penta, Star Solitaire, and Star Trekker. One marble is removed at the start and this puzzle uses normal peg solitaire rules, with jumps allowed along the lines on the board. As usual, the goal is to finish with one peg.

Let N₁ be the number of pegs in the inner ring. A peg solitaire jump can never change the parity of N₁. If we begin from a vacancy in the inner ring, then N₁=4 (even). The only one peg positions with N₁ even have N₁=0 with one peg in the outer ring. Thus, if we start in the inner ring, we can only finish in the outer ring (and vice-versa). It turns out all such problems are solvable. Star jump can be extended to a 16-hole board with 5-fold symmetry. One such board is given in Beasley's book [B1, p. 242-3], another option is shown in this blog [W24].

Solomon was invented by Martin Gardner around 1970 as a two player game, and can also be used for peg solitaire. This 19-hole board in the shape of a hexagram can be purchased from Kadon Enterprises.

Some gridless boards have no corners, a property which we have not seen to this point. Hyper Solitaire is a 33-hole English board which has been warped so that jumps can be made "around the corners". Although the number of possible jumps has increased from 76 to 108, Hyper Solitaire is not universal, as shown by the following argument: consider the 12 hole set R marked in red on the photo to the left. We note that there is no jump which can add or remove one peg from R (pegs can only be removed from R in pairs). Therefore, the parity of the number of pegs in the set R can never change. Thus if the initial vacancy is in R, you can only finish in R, and if you start outside R, you can only finish outside R.

Round Solitaire is a modification of Hyper Solitaire invented in 2009 by Tetsuro Kawahara, he removed the outer ring of 12 holes. The central game is solvable on this board in a minimum of 8 moves. Round Solitaire is not universal [P10], this can be demonstrated using the set R of the previous paragraph.

The brass board with clay marbles on the right was probably made in the early 1900's, the engraving seems to read "H.S.H. (PROT)ECTED". Another copy can be found in item #3663 of the Jerry Slocum collection [W22]. It is not known if this board was meant for playing peg solitaire, and this seems unlikely to me because the marbles are so close together. If viewed as a peg solitaire board, the jumping rules are not obvious. It can be interpreted as Hexagon(3) with additional jumps around the outer ring, giving 66 possible jumps. With this 66 jump set, the board becomes universal. John Beasley has investigated possible jumping rules and has found a certain minimal set of 42 jumps which make a nice peg solitaire board. He calls it "Clock Solitaire" [W18]. For an analysis of this board see [P11].

The 17-hole board to the right was patented in 1899 by W. C. Breitenbach but since then has been ignored. It is actually quite a nice gridless board, I call it "Breitenbach's Board". The usual coloring argument (see [P11]) proves that the starting and ending holes must be the same color. The central game is solvable in a minimum of 6 moves. This board is related to Clock Solitaire, although the board has been "squared" and the number of spokes increased from 6 to 8. My solving program has found that if you add the eight jumps around the corners (in the outer ring only), this board becomes universal.

The Great boards show a way to create an increasing sequence of gridless boards, all with square symmetry. The Great 13 board is the same as Hoppers, the Great 17 board is similar to Breitenbach's Board (and was also patented by him in 1899). My solving program has found that the Great 17 board is not universal. The Great 21 board board is universal, and probably all larger Great boards are also universal.

This is not an exhaustive list of gridless boards, more can be found in Beasley's book [B1], or his more recent update article [P10].

Toroidal Solitaire

Legal jumps in 4x4 toroidal solitaire
a1   b1   c1   d1  a2 b2 c2 d2 a3 b3 c3 d3 a4 b4 c4 d4 Notation	An 8-loop?
A 5-move solution to the a1-complement!

A toroidal chess board!

Consider, for example, the 4x4 board. Under normal peg solitaire rules, this board isn't very interesting, the only SVSS problems which are solvable have the form: vacate a2, finish at a3 (or d3). Under toroidal jumping rules, the board becomes universal. The symmetry of the board is also very different: there are no corners or edges—all holes are equivalent. The red graphic to the left shows a curious 8-loop on the 4x4 toroidal board. Unfortunately, it cannot be reached from any single vacancy start. My program has found that all SVSS problems on this board can be solved in a minimum of 5 moves (see the example solution to the left). It is remarkable that the first move can be a double jump!

The reader may enjoy demonstrating that the central game on 3x3 toroidal solitaire is solvable. In fact, it is difficult to lose this game, is there any dead end? An analysis of n by m rectangular boards under toroidal jumping rules reveals that if n and m are both multiples of 3 the board is still null-class and the "rule of three" still applies. If both n and m are not multiples of 3, the board is probably universal. A simple extension of the rule of three gives this, because by wrapping around the board any pair of holes can be considered to have coordinates a multiple of 3 apart (when n and m are not multiples of 3). If one of n and m is a multiple of 3 and the other is not, then we have an intermediate case where the rule of three applies in only one dimension (the one divisible by 3). My program has found that the shortest solution to the complement problem on the 5x5 toroidal board has 8 moves, and on the 6x6 toroidal board, 11 moves.

Starting from any rectangular board, we can also identify opposite edges in other orientations to play peg solitaire on a cylinder, Möbius strip, or Klein bottle. It is not clear if these geometries have any more to offer than the torus.

A proof that the 4x4 toroidal board is universal.

Solidaire (3D solitaire), and beyond

Online Puzzles

"Pegged", an early computer version of peg solitaire.

"Never Lose" Triangle(5) (2014).

Most computer versions of peg solitaire have the ability to take back moves, all the way back to the beginning if necessary. In my opinion, this tends to make a puzzle seem easier, compared with solving on a mechanical board. Resetting the board is also trivial for a computer puzzle. Computer versions of peg solitaire can also include demos or solutions, and they can point out bad or good jumps.

Here are links to online peg solitaire puzzles I have created. You must have JavaScript activated in your browser to play them.

Square grid peg solitaire:

• GPJ Puzzles produced for Issue #28 of The Games and Puzzles Journal [W1] (2003).
• G4G7 Puzzles created for the seventh Gathering for Gardner (2006).
• Symmetric English Puzzles Solve 279 symmetric puzzles on the English board (2016). The hardest one? Also difficult.
• Symmetric French Puzzles Solve 428 symmetric puzzles on the French board (2016).
• Symmetric Square (6x6) Puzzles Solve 414 symmetric puzzles on the Square 6x6 board (2016).
• Difficult English Puzzles Solve 51 difficult puzzles on the English board (2016).
• Difficult French Puzzles Solve 197 difficult puzzles on the French board (2016).

Triangular peg solitaire:

• Ultimate Triangular Peg Solitaire (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 (2014). This version has the ability to point out all jumps that lead to a dead end, which tends to make the puzzle seem trivial! First released in 2008, the 2014 version includes Truncated Triangle boards.
• Symmetric Hexagonal Puzzles Solve 335 symmetric puzzles on the 37-hole hexagon board (2016).
• Difficult Hexagonal Puzzles Solve 40 symmetric puzzles on the 37-hole hexagon board (2016).

I have also designed the (hexagonal) levels of a free online peg solitaire game. You will need to download Shockwave to use it. Thanks to Rob Gordon of Article19 for the GUI.

• Shockwave Peg Solitaire (2006)

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 square-symmetric (except for Diamond32) 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.

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. In 2012, Barker and Korf [P8] confirmed my results for the 41-hole diamond and 45-hole Wiegleb's board.

Computational Search Techniques

Read the story behind the creation of this "very limited edition" board.

A good introductory reference which has a nice progression of problems is the 2007 book by Koetke [B4]. This book starts out with small boards that are easy to solve, and discusses the problems encountered for larger boards. 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 an 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 (generally using a depth-first search). Because there are a large number of move sequences that result in the same board position, such an algorithm is rather inefficient. Somewhat surprisingly, this inefficient algorithm may still quickly find a solution to the central game, but we seek algorithms which perform well even in the worst case. One significant improvement is to use a hash table or some other means to store board positions seen previously so you do not have to investigate them further. This results in a very speedy algorithm, the only problem is what to do when the hash table fills all memory.

A related 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 technique can solve any problem on the 33-hole board relatively quickly. For boards larger than this, additional techniques are needed.

I have used five 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 very 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. See [P8].
4. A* search. For examples of this algorithm applied to peg solitaire see [P4] and [P8].
5. This algorithm is easy to parallelize. For example, with 32 processors each can go through all boards at the current level, but only store 1/32 of the resulting board positions.

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 details on computational search techniques see [P4] and [P8].

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), see a diagram here.

How Hard Is Peg Solitaire?

"Culture", a vertically oriented board with hanging pegs © 2013 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 problem #1 will have a run time that increases exponentially with the board size. A challenging problem to solve by hand is shown on the left, it has only 13 pegs! Here is a board position which can be reduced to a single peg, but only if you can find the correct sequence of 12 jumps. A computer search found this board position, which has the property that only one opening jump leads to a solution, the other 15 jumps are dead ends. Can you find the correct opening jump?

While this 13-peg puzzle is difficult to solve by hand, a computer can easily solve it using exhaustive earch. However, imagine a similar problem with, say, 50 pegs.

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. Clearly if the board is not null-class, there is no solution to problem #2 or #3. 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. In 2012, Joseph Barker and Richard Korf wrote a paper on the subject of searching for short solutions [P8].

Peg Solitaire References

Books

Papers

Web References

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