English 33-hole Board Example

This page shows the results of a calculation by levels both forward and backward for the central game on the standard 33-hole board. This is done first by moves, which can be used to find the shortest solution, and then by jumps, which gives all possible board positions that can be reached. As you will see this allows us to view a peg solitaire problem as a problem in set theory (although the sets may be quite large). Next we consider a player who always makes random jumps, and compute the probabilities that the game will finish at n pegs. Finally, a discussion on counting and classification of minimal length solutions.

Calculation by Moves

If we are looking for the shortest solution, the calculation by levels needs to be done by moves rather than jumps (a move is one or more jumps by the same peg). See below for the same calculation by jumps.

First a few definitions:

  1. Let Bn be the set of board positions that can be reached from the starting position in n moves, but no fewer.
  2. Let B-n bet the set of board positions from which the finishing board position can be reached in n moves, but no fewer.
  3. The complement board position is obtained by replacing every peg by a hole (i.e. removing it), and replacing every hole by a peg. For any board position b, the complement board position will be denoted b'. Similarly, for a set of board positions S, S' is the set of complement board positions. In other words b'∈S' if and only if b∈S.
  4. |Bn| is the number of elements in the set Bn.
The task of computing Bn or B-n are quite similar, because backward play from the final board position is equivalent to forward play from the complement of the final position. However there is some subtlety around the nature of multi-jump moves when the direction of play is reversed.

For the central game on the 33-hole English Board, the following table shows the size of the sets Bn and B-n. The 8-fold symmetry of the problem was taken into account, but no resource count was used. Because of the symmetry of the problem, we really should refer to equivalence classes of board positions (this is important to realize when calculating set intersections). However it simplifies our thinking if we select one representative board position from each equivalence class, so the sets Bn are sets of board positions.

n (level) |Bn| Ratio Pegs |B-n| Ratio Pegs
0 1   32 1   1
1 1 1.00 31 1 1.00 2
2 2 2.00 30 16 16.0 3-10
3 9 4.50 28-29 979 61.2 4-19
4 47 5.22 27-28 14,115 14.4 5-21
5 263 5.60 25-27 126,902 8.99 6-22
6 1,452 5.52 23-26 632,451 4.98 7-23
7 7,772 5.35 20-25 1,954,152 3.09 8-25
8 38,150 4.91 16-24 3,770,028 1.93 9-26
9 164,297 4.31 14-23 4,949,879 1.31 10-26
10 585,567 3.56 12-22 4,796,624 0.97 11-27
11 1,641,177 2.80 10-21 3,576,434 0.75 13-28
12 3,426,496 2.09 8-20 2,113,413 0.59 14-28
13 5,111,387 1.49 5-19 1,003,420 0.48 15-29
14 5,327,135 1.04 5-18 387,982 0.39 16-30
15 3,909,899 0.73 2-17 115,922 0.30 17-30
16 2,093,768 0.54 2-16 28,134 0.24 19-31
17 843,917 0.40 1-15 4,589 0.16 20-31
18 255,301 0.30 1-14 604 0.13 22-32
19 57,579 0.23 3-13 37 0.06 22-29
20 10,037 0.17 4-12 5 0.14 26-29
21 1,268 0.13 5-11 0 0.00  
22 148 0.12 6-10
23 15 0.10 7-9
24 0 0.00  
Total 23,475,688 ~130 MB 23,475,688 ~130 MB
Table 1:
"Ratio" is |Bn|/|Bn-1|, or |B-n|/|B-(n-1)|.
"Pegs" is the range in the number of pegs for all boards on this level.

For a forward/backward search, only the green entries need to be calculated,
only about 2% of the calculation time for this entire table (4-6 hours).
See a similar table for the Wiegleb's Central Game

Note that the total number of boards in Bn or B-n over all n are identical. This actually provides a good check that the algorithm is working properly, why must it be the case? If a specific board pattern b lies in some Bn, then by definition this board position is reachable from the central vacancy in n moves. Therefore it is possible to play from b' to the final state, but not necessarily in n moves, so b' must lie in some B-m. In fact,
.

Bergholt's 18 move solution can be recovered from B18, B-18, or the intersection of any Bn and B-m where n+m=18. In an optimal search algorithm, one would clearly want to choose n to minimize the sum of the time to compute Bn plus the time to compute Bn-18. This occurs at about n=11, corresponding the green region of the above table, with total computation time on my machine about 8.5 minutes. This compares with over 2 hours for going all the way either forward or backward, so a significant time savings is realized by calculating both forward and backward.

The above table also contains some general results:

  1. Any board position that can be reached from the central vacancy, can be reached in 23 moves or less.
  2. Any board position that can be reduced to a single peg in the center can in 20 moves or less.
I was curious to see what strange board positions fall into the extreme categories, and a few examples are shown below:

   
 
           
             
           
 
     
     
     
       
         
       
     
     
   
     
       
         
           
 
     
   
 
           
         
       
     
     
   
 
             
       
       
     
     
23 moves are required to reach these positions from the central vacancy. The positions with 9 pegs (the right three) can only be reached by using single jump moves. The rightmost board is also the only one (of 15 total) that can appear in a solution to the central game.
     
 
 
   
   
 
 
       
 
         
 
 
   
 
 
These positions can be reduced to a single peg at the center in a minimum of 20 moves. None of these board positions can appear in a solution to the central game. How do we know this? Because the complement cannot be reduced to a single peg.

One common peg solitaire puzzle is to take a given "shape" of pegs and attempt to reduce it to a single peg, usually in the center of the board. Once this "reduction problem" has been solved, we then can try to solve it in the smallest number of moves. If you try to solve the problem using a computer, checking each problem individually can take quite some time. Calculating the sets B-m is a much more efficient way to proceed. Note that (by definition) every board pattern in B-m can be reduced to one peg in m moves, and the set contains every possible pattern that can be so reduced. In a single (2 hour) calculation, we therefore have solved every conceivable reduction problem! The only constraint is that the jumps must all be made on the 33-hole board. Some of these problems can be solved in fewer moves if one allows jumps to holes that are not on this board, and there exist patterns of pegs on the 33-hole board that cannot be reduced to a single peg by using jumps on the board, but can be reduced to a single peg if jumps off this board are allowed.

Calculation by Jumps

It is also possible to calculate a table similar to that above by going one jump at a time rather than one move. This is simpler to program, and allows us to describe all board positions that can occur in a solution to the central game (or any peg solitaire problem, at least in theory).
  1. Let Sn be the set of board positions that can be reached from the starting board position in exactly n jumps.
  2. Let S-n be the set of board positions from which the finishing board position can be reached in exactly n jumps.
  3. Let Wn be the subset of Sn that can be reduced to the finishing board position (the "Winning positions").
  4. Let N be the total number of jumps in a solution (31 for the single vacancy to single survivor problems on the English Board).
This calculation is quite a bit simpler, because all elements of Sn have the same number of pegs. By definition, we have Wn = Sn ∩ Sn-N.

If the final board position is the complement of the initial board position, then in addition we have:

  1. Sn = (S-n)'
  2. Wn = Sn ∩ (SN-n)'.
  3. Wn = (WN-n)'.
The following table shows the sizes of these sets calculated again for the central game on the 33-hole English Board. Although this table cannot be used to find the shortest solution to the central game, it can be used to find all solutions, or at least all board positions that can appear during a solution to the central game.

n (level) |Sn| Ratio Pegs |Wn| % Winning
0 1   32 1 100%
1 1 1.00 31 1 100%
2 2 2.0 30 2 100%
3 8 4.0 29 8 100%
4 39 4.88 28 38 97.4%
5 171 4.38 27 164 95.9%
6 719 4.20 26 635 88.3%
7 2,757 3.83 25 2,089 75.8%
8 9,751 3.54 24 6,174 63.3%
9 31,312 3.21 23 16,020 51.2%
10 89,927 2.87 22 35,749 39.8%
11 229,614 2.55 21 68,326 29.8%
12 517,854 2.26 20 112,788 21.8%
13 1,022,224 1.97 19 162,319 15.9%
14 1,753,737 1.72 18 204,992 11.7%
15 2,598,215 1.48 17 230,230 8.9%
16 3,312,423 1.27 16 230,230 8.9%
17 3,626,632 1.09 15 204,992 5.7%
18 3,413,313 0.94 14 162,319 4.8%
19 2,765,623 0.81 13 112,788 4.1%
20 1,930,324 0.70 12 68,326 3.5%
21 1,160,977 0.60 11 35,749 3.1%
22 600,372 0.52 10 16,020 2.7%
23 265,865 0.44 9 6,174 2.3%
24 100,565 0.38 8 2,089 2.1%
25 32,250 0.32 7 635 2.0%
26 8,688 0.27 6 164 1.9%
27 1,917 0.22 5 38 2.0%
28 348 0.18 4 8 2.3%
29 50 0.14 3 2 4.0%
30 7 0.14 2 1 14.3%
31 2 0.29 1 1 50.0%
Total 23,475,688 ~130 MB 1,679,072 7.2%
Table 2:
"Ratio" is |Sn|/|Sn-1|.
"Pegs" is the number of pegs for all boards on this level.
See a similar table for the Wiegleb's Central Game

Table 2 agrees with that calculated by "Durango Bill" [W9]. Note also that the total |Sn| over all n is the same as the total in the first table, calculated by move.

I have entered the (finite) sequences |Sn| and |Wn| into The On-Line Encyclopedia of Integer Sequences as A112737 and A112738.

If we take the union of Wn over n=0,2, ..., 15, we obtain a special set X of 839,536 board positions. What is so special about this set? If b(0) is any board position, let b(1), ... b(7) be the board positions obtained by all rotations and/or reflections of this board position, and b(8), ... b(15) be the complements of the first 8 board positions. Then b(0) can occur during a solution to the central game if and only if some b(i) is in X. This set X can be used to create a peg solitaire computer game with the ability to point out all winning moves from the current board position. If we use 33 bits to store each board position, then the set X requires around 4 MB of storage, hence can be easily held in memory.

Can the game be solved by jumping randomly?

We can simulate the game by assuming that the player always selects a jump at random. To be precise, the player counts the total number of jumps available from the current board position and selects one at random. How likely is it that such a player will be able to reduce the board to one peg? This can be calculated by playing a lot of random games and tabulating the results, a "Monte Carlo simulation". Alternatively we can recalculate Table 2, associating a real number with each board position that is the probability that it can occur in a random game. The exact probability of each board on a level can be calculated from the boards on the previous level. This method has the advantage that board positions with very small probability are calculated exactly.

The symmetry of the board must be carefully considered in this calculation. As before we consider board positions that are identical via reflection and/or rotation as the same board position. For example after the first jump there are four possible board positions, however since these are all rotations of each other we consider this as only one board position, which occurs with probability 1. Things become more complicated later in the game, because two completely different jump sequences may result in board positions that are exact 90-degree rotations of each other (for example), which by our accounting are the same board position.

Looking at the last column of Table 2, you may conclude that if our random player makes 15 jumps, there is an 8.9% chance that the board is still in a "winning position" (can be reduced to one peg assuming perfect play). However this type of reasoning is incorrect because it assumes each board position is equally likely, which is actually far from true. In fact a random player is much less likely to remain in a winning board position, and after 15 random jumps the chance that the board is in a winning position is less than 1%.

Table 3 below shows where a randomly played game is likely to end. A "dead end" is a board position from which no jump is possible (game over), and we show the number of dead ends, and the probability that the game ends after exactly n jumps. The probability that the central vacancy start finishes with one peg (at either the center d4, or d2, b4, f4 or d6) is 2.6978E-08, or about 1 in 37 million games. The probability that the final peg ends up in the center is exactly half this, or 1 chance in 74 million (even on the final jump, our random player has a 50% chance of going the wrong direction).

n (jump) Pegs left Dead ends Probability that the
game ends after n jumps
6 26 1 7.4074E-03 exactly 1 in 135
7 25 0 0 never
8 24 0 0 never
9 23 0 0 never
10 22 1 9.7435E-06 1 in 102,633
11 21 1 6.8999E-05 1 in 14,493
12 20 0 0 never
13 19 5 5.7230E-05 1 in 17,473
14 18 10 1.0757E-04 1 in 9,296
15 17 7 2.6607E-04 1 in 3,758
16 16 27 1.7353E-04 1 in 5,763
17 15 47 6.6371E-04 1 in 1,507
18 14 121 1.4401E-03 1 in 694
19 13 373 4.1355E-03 1 in 242
20 12 925 9.8602E-03 1 in 101
21 11 1972 2.5485E-02 1 in 39
22 10 3346 7.6543E-02 1 in 13
23 9 4356 1.5044E-01 1 in 6.6
24 8 4256 1.9958E-01 1 in 5.0
25 7 3054 2.4665E-01 1 in 4.1
26 6 1715 1.6347E-01 1 in 6.1
27 5 665 7.9883E-02 1 in 13
28 4 182 3.0672E-02 1 in 33
29 3 39 3.0107E-03 1 in 332
30 2 6 7.7690E-05 1 in 12,872
31 1 2 2.6978E-08 1 in 37.07 million
Total   21,111 1.0000E+00  
Table 3:
How the game will end if a player selects jumps at random.
Calculated exact values from level calculations,
probabilities confirmed by Monte Carlo simulation.

Our random player is expected to finish most often with 7 pegs, and with a mean of 7.64 pegs. If we compare these results with those of "average" human players, we find that most are able to do quite a bit better than this. The reason, of course, is that human players do not select jumps at random. Most people will try to clear out the arms without stranding a peg near the edge of the board, and a lot of people can get down to 2 or 3 pegs after 10 attempts or less.

While a modern computer can simulate a million games in a matter of seconds, such a brute force technique expends an awful lot of effort to find a single solution, and will fail entirely for larger boards. Making random jumps is actually one of the worst strategies to adopt in solving the central game.

Finally, it is interesting to find the most and least likely ending board positions for a randomly played game:

 
       
 
 
     
   
             
       
             
 
     
The most likely finish to a random game, also the shortest possible game (6 jumps). One out of every 135 randomly played games ends at this board position. What six jumps are needed to reach this board position? The least likely finish to a random game. One out of every 70.4 billion randomly played games ends at this board position. Can you figure out how to reach this board position? Hint: Where must the pegs at d2, b4, d4 and f4 have started from?

Finding and Counting Minimal Length Solutions

When we calculate by moves, we can find the shortest solution to a particular peg solitaire problem. However, the technique does not directly give us an actual solution (only its length). This section outlines my techniques for finding and counting these solutions.

The Solution Graph

The first step is to construct the solution graph; this in fact enables us to calculate all solutions of minimal length. It turns out that all that follows is very easy computationally and can be calculated in a few seconds once the level sets Bn and B-m are calculated (this assumes the number of solutions is not extremely large, however). In what follows we again use the central game for the English board as an illustrative example.

The nodes of the solution graph are the board positions along minimal length solutions, and the links joining them are the moves between these board positions. As a result of our calculation by levels, we already have some of these board positions, namely those in the intersection between the forward and backward level sets (B11 and B-7 in our example). To recover all the board positions in any minimal length solution, it is just a matter of tracing these board positions both backward and forward in the level sets.

The solution graph for the English central game is shown below. The left node (B) is the board position with the central vacancy, while the right node (E1) is the final board position with one peg in the center. This graph was produced by Sidney Cadot. He uses a different notation for moves and board locations (he numbers the holes consecutively 1-33).
Note that Sidney has arranged the graph horizontally so that positions with the same number of pegs lie on a vertical line. I would align boards vertically by level, which would show that they all contain the same number of moves. The figure below shows one of the possible 18-move solutions to the central game (first found by Ernest Bergholt in 1912).

Two Ways to Count Solutions

The number of solutions (counting move order) is simply the number of ways to traverse the solution graph. There is a simple and quick algorithm for computing this. Label the starting node with a "1". Then at each level, go through all the nodes, and label each node with the sum of the labels along incident edges. You have to be careful to count loop moves as either duplicated links or multipliers (the above graph shows the loop move as a double link). Each label counts the number of ways to traverse the graph to that point, and the label on the final node is the number of ways to traverse the entire graph, 936 in this case.

936 sounds like quite a few solutions, but most of these solutions have the exact same moves, just in a different order. Can we quantify this? If you look at the solution graph, you can see that there are really only two different paths to take. If you don't count the order of moves, there are really only two solutions. The top route in the solution graph is taken by the above solution, while the lower route is that taken in the solution below. As you can see, the difference between these solutions is quite subtle.

To count solutions regardless of move order, a similar, although somewhat more complex algorithm is used. At each node, instead of having a single number to keep track of, I use a sequence of lists, each of which is a set of moves that can used to reach that node, not counting order. These lists are sorted (in some known, but arbitrary order) so we can easily check for duplicates. Going through the tree in the same fashion, the number of solutions regardless of move order is the number of lists at the terminal node, and each list gives a possible solution.

Also, when all is done you have lists of solution moves, but you no longer know how to order the moves to make them solutions! Figuring out an order that will make it work can be non-trivial, so I found that at each node in the tree I wanted to store BOTH a sorted move list and the original (sequential) move list. This way at the end one can go back to the original list and recover each solution.

By the way it is also possible to use the same technique for solutions calculated by jump. In other words, we consider solutions of any length and consider them as sequences of jumps (not moves). The number of solutions to the central game regardless of jump order is extremely large and the memory on my PC fills up before this calculation is complete. However, I have been able to complete this calculation on smaller boards, particularly the triangular peg solitaire boards. In this case, the counting is a bit strange, for example a solution to a complement problem and the jumps executed in the exact reverse order are considered the same solution. This oddity doesn't happen when counting minimal solutions, because when such a solution is reversed it is almost never minimal length. But some counter-intuitive behavior can still occur, as shown in the next section.

An Additional Subtlety and Solution Classification

The solution catalog originally had solutions classified by the final sweep move. Then for the French Board, the program came up with the following solutions:


These two solutions have the first nine moves identical, but after that they appear to be different, and they finish very differently. However a careful check will show they actually contain exactly the same moves just in a different order. Thus, according to our accounting method, they are the same solution, even though their endings are completely different. Therefore, when grouping solutions regardless of move order, there is no longer a unique finishing move. The situation above is a rather unusual occurence and is not true for the vast majority of cases (usually solutions with the same moves differ only in the middle), but must be taken into account for the general case. Although there is no unique final move, there is a definite longest final sweep length for any solution no matter what order it's moves are taken in (3 in the above example). Therefore, I now classify solutions according to three numbers (A,B,C):

  1. Length of the longest sweep
  2. Length of the second longest sweep
  3. Length of the (longest possible) final sweep
For example the solution above is classified as (5,5,3) [in the French Board catalog under Vacate (-1,0) Finish (-2,0)].

Created in 2007. Last modified August 17th, 2011

Copyright © 2011 by George I. Bell

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