Wiegleb's 45hole 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:

Let B_{n} be the set of board positions that can be reached
from the starting position in n moves, but no fewer.

Let B_{n} bet the set of board positions from which the
finishing board position can be reached in n moves, but no fewer.

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.

B_{n} is the number of elements in the set B_{n}.
For the central game on the 45hole Wiegleb's Board,
the following table shows the size of the sets
B_{n} and B_{n}.
The 8fold 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) 
B_{n} 
Ratio 
Pegs 
B_{n} 
Ratio 
Pegs 
0 
1 

44 
1 

1 
1 
1 
1.00 
43 
2 
2.00 
23 
2 
3 
3.00 
42 
78 
39.0 
311 
3 
15 
5.00 
4041 
5,929 
76.0 
421 
4 
87 
5.80 
3840 
193,261 
32.6 
523 
5 
564 
6.48 
3639 
3,228,718 
16.7 
625 
6 
3,786 
6.71 
3438 
32,916,284 
10.2 
727 
7 
25,210 
6.66 
3137 
217,782,198 
6.62 
829 
8 
161,708 
6.41 
2936 
956,482,597 
4.39 
931 
9 
957,967 
5.92 
2635 
Not Calculated 
10 
5,045,265 
5.27 
2434 
11 
23,010,765 
4.56 
2133 
12 
89,501,077 
3.89 
1832 
13 
294,359,870 
3.29 
1631 
14 
813,471,187 
2.76 
1430 
Total 
1,226,537,506 
~7 GB 
1,210,609,068 
~7 GB 
"Ratio" is B_{n}/B_{n1}, or
B_{n}/B_{(n1)}.
"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 B_{14} and
B_{8}, which has exactly 7 board positions in the intersection.
These correspond to 3 unique 22move 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).

Let S_{n} be the set of board positions that can be reached
from the starting board position in exactly n jumps.

Let S_{n} be the set of board positions from which the
finishing board position can be reached in exactly n jumps.

Let W_{n} be the subset of S_{n} that can be reduced
to the finishing board position (the "Winning positions").

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
S_{n} have the same number of pegs.
By definition, we have
W_{n} = S_{n} ∩ S_{nN}.
If the final board position is the complement of the initial board position,
then in addition we have:

S_{n} = (S_{n})'

W_{n} = S_{n} ∩ (S_{Nn})'.

W_{n} = (W_{Nn})'.
The following table shows the sizes of these sets calculated again
for the central game on the 45hole 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 S_{21}, the resource count cannot be at 0.
At jump 21, the resource count must in fact be 1, 2 or 3.
n (level) 
S_{n} 
Ratio 
Pegs 
W_{n} 
% 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 
= W_{21} 
1.1% 
701 
Total 
6,586,989,754 
~37 GB 
89,558,705 
2.0% 

Table 2:
"Ratio" is S_{n}/S_{n1}.
"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 W_{n} 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 33hole 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 33hole English board, a wellknown 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: e3e5, e6e4, e8e6, 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 41peg 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 16sweep, with solutions for the backwards problem
(starting from the complement
of the board position before the 16sweep).
 Vacate g4, play to finish at d4 with the last move a 16sweep.
 Vacate g5, play to finish at d2 with the last move a 16sweep.
 Vacate g6, play to finish at d3 with the second to the last move a 16sweep.
The 16sweep 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:
Last modified November 22nd, 2014
Copyright © 2014 by George I. Bell
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