# Wiegleb's 45-hole Board Results

 d1 e1 f1 d2 e2 f2 d3 e3 f3 a4 b4 c4 d4 e4 f4 g4 h4 i4 a5 b5 c5 d5 e5 f5 g5 h5 i5 a6 b6 c6 d6 e6 f6 g6 h6 i6 d7 e7 f7 d8 e8 f8 d9 e9 f9
 -1 0 -1 1 0 1 0 0 0 -1 1 0 1 0 1 0 1 -1 0 0 0 0 0 0 0 0 0 -1 1 0 1 0 1 0 1 -1 0 0 0 1 0 1 -1 0 -1
Standard 9x9 Notation The useful resource count

We consider first the central game (e5 to e5) on this board. We calculate by levels both forward and backward. 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. We then look at the shortest possible game beginning with the center e5 empty. We then consider the longest possible sweep that can appear in the solution to any single vacancy to single survivor problem. We give solutions to the three special problems both backwards and forwards.

## 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.

For the central game on the 45-hole Wiegleb's 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, plus the resource count at the top of this page. For the forward problem, we can ignore any board position where this resource count evaluates to less than 0 (this has no effect until n=7). For the backward problem, we can ignore any board position where this resource count evaluates to more than 4 (this has no effect until n=2). Without this resource count the board counts at the final level would be approximately twice as large.

n (level) |Bn| Ratio Pegs |B-n| Ratio Pegs
0 1   44 1   1
1 1 1.00 43 2 2.00 2-3
2 3 3.00 42 78 39.0 3-11
3 15 5.00 40-41 5,929 76.0 4-21
4 87 5.80 38-40 193,261 32.6 5-23
5 564 6.48 36-39 3,228,718 16.7 6-25
6 3,786 6.71 34-38 32,916,284 10.2 7-27
7 25,210 6.66 31-37 217,782,198 6.62 8-29
8 161,708 6.41 29-36 956,482,597 4.39 9-31
9 957,967 5.92 26-35 Not Calculated
10 5,045,265 5.27 24-34
11 23,010,765 4.56 21-33
12 89,501,077 3.89 18-32
13 294,359,870 3.29 16-31
14 813,471,187 2.76 14-30
Total 1,226,537,506 ~7 GB 1,210,609,068 ~7 GB
"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.

Calculating the entire table took about 4 days of CPU time
(on a 1 GHz Pentium with 512 MB of RAM) and 14 GB of disk space
(plus another 50 GB of temporary disk space).
See a similar table for the English Board Central Game

The first intersection that is nonempty is B14 and B-8, which has exactly 7 board positions in the intersection. These correspond to 3 unique 22-move solutions (regardless of move order) that all look very similar to the one shown below.

## 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 (43 for the single vacancy to single survivor problems on Wiegleb's 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 45-hole Wiegleb's Board. A constraint used here is that the above resource count (up to jump 22) must be greater than or equal to 1. This does not eliminate any solutions, because when calculating backwards, it is impossible to keep the resource count at 0 for more than 12 jumps. In other words, for any solution to the central game, at the intersection point S21, the resource count cannot be at 0. At jump 21, the resource count must in fact be 1, 2 or 3.

n (level) |Sn| Ratio Pegs |Wn| % Winning Split
0 1   44 1 100% 1
1 1 1.00 43 1 100% 1
2 3 3.00 42 3 100% 1
3 11 3.67 41 10 90.9% 1
4 60 5.46 40 54 90.0% 1
5 297 4.95 39 236 79.5% 1
6 1,427 4.81 38 900 63.1% 1
7 6,459 4.53 37 3,007 46.6% 1
8 27,317 4.23 36 9,056 33.2% 1
9 106,347 3.89 35 24,990 23.5% 1
10 379,537 3.57 34 64,182 16.9% 1
11 1,238,520 3,26 33 154,345 12.5% 1
12 3,702,227 2.99 32 348,705 9.4% 1
13 10,160,129 2.74 31 741,102 7.3% 3
14 25,647,378 2.52 30 1,483,185 5.8% 5
15 59,620,492 2.33 29 2,788,600 4.7% 11
16 127,737,457 2.14 28 4,898,948 3.8% 29
17 252,239,569 1.98 27 7,981,238 3.2% 53
18 458,623,402 1.82 26 11,958,747 2.6% 127
19 766,145,054 1.67 25 16,344,138 2.1% 211
20 1,172,139,707 1.53 24 20,224,817 1.7% 367
21 1,635,783,432 1.40 23 22,532,441 1.4% 499
22 2,073,430,928 1.27 22 = |W21| 1.1% 701
Total 6,586,989,754 ~37 GB 89,558,705 2.0%
Table 2:
"Ratio" is |Sn|/|Sn-1|.
"Pegs" is the number of pegs for all boards on this level.
"Split" is the number of subproblems this level was split into.

Calculating the entire table took about 4 weeks of CPU time
(on a 1 GHz Pentium with 512 MB of RAM) and 38 GB of disk space
(plus another 100 GB of temporary disk space).
See a similar table for the English Board Central Game

If we take the union of Wn over n=0,2, ..., 21, we obtain a special set X of 89,558,705 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.

## The Shortest Possible Game

### No Jumps Possible

If you begin from the central vacancy, how quickly can you reach a position from which no jump is possible? It is not hard on the 33-hole English board to find a dead end in only 6 jumps. If these 6 jumps are executed on Wiegleb's board, further jumps are still possible, and the shortest possible game is not easy to find. Exhaustive computer search has found that the answer is 13 jumps with a solution shown below.

### One Peg Finish Impossible

A different question is how quickly one can reach a board position from which a solution (one peg finish) is no longer possible. For the central game on the 33-hole English board, a well-known sequence of four jumps ([B1], [B3]) leads to a board position that cannot be solved to one peg. This same sequence of jumps on Wiegleb's board also leads to an unsolvable position. However we can see from the above table that there is some combination of three jumps that leads to an unsolvable position. This combination is: e3-e5, e6-e4, e8-e6, the shortest possible "dead end" for the central game.

These three moves put the board in the position shown above. It is difficult to show (analytically or computationally) that a one peg finish cannot be reached from this 41-peg board position. I have only managed a computational "proof".

## The Longest Possible Sweep

Rather surprisingly, it is possible to finish a single vacancy to single survivor problem with a sweep of length 16. Here are the three problems which can contain a 16-loop, with solutions for the backwards problem (starting from the complement of the board position before the 16-loop).
1. Vacate g4, play to finish at d4 with the last move a 16-loop.
2. Vacate g5, play to finish at d2 with the last move a 16-loop.
3. Vacate g6, play to finish at d3 with the second to the last move a 16-loop. The 16-loop begins and ends at d2.
These now solve the original problems by reversing the sequence of jumps. By rearranging the order of the jumps, the shortest possible solutions contain 22, 24 and 23 moves, respectively, with sample solutions given by: