9x9 Peg Solitaire |

Last Modified August 24, 2005 Copyright © 2005 by George I. Bell Extended version of this page [Not available online] |

This web page documents some interesting properties and solutions for peg solitaire on the square 9x9 board.

- The Central Game: Inductive Technique, Shortest Solution, Symmetry

It is unfortunate that a standard chess board is just slightly too small to play 9x9 peg solitaire on. You can play the game on a go board, however the best board is really a computer. On a computer game you can easily backtrack and record move sequences, and taking the complement of a board position is trivial once a suitable button has been programmed. Many of the solutions below were discovered by hand using a Javascript program which I modified from a version by JC Meyrignac. Unfortunately this game is rather hacked up and I don't think anyone else would find it useful if they can't program in Javascript.

In 1962, Robin Merson found an elegant argument which gives a lower bound
for the length of a solution on the 6x6 board.
This argument can be generalized to any square (or even rectangular) null class
board, but on an *n* x *n* board the bound is not very tight
if *n* is odd.
The complement problem from a corner must use at least
*n*/2+1)²*n* is odd one must round *n*/2 down to the nearest integer.
If the problem does not begin in a corner the bound is one less.
On the 6x6 board this gives a lower bound of 15 for all problems that do not begin
at a corner. In fact one can come up with solutions in 15 moves, so one
immediately has a proof that they are optimal.

On the 9x9 board the lower bound indicates that any solution must have at least 24 moves. The best solution I have seen was constructed by Alain Maye, by hand, and has 34 moves. It is likely it can be done in fewer than 34 moves, but I doubt the minimum length solution is under 28 moves.

The solution below answers this question. After 8 jumps (or 6 moves) the board position becomes square symmetric (shown in red). The next 60 jumps come in sets of 4 moves that are rotational copies of one another, so every 4 moves you pass through a position with rotational symmetry (shown in green). Then the final 11 jumps (or 6 moves) finish at the center. The final solution has 8+60+11=79 jumps (or 72 moves) and passes through 16 positions with rotational symmetry, 5 of them being square symmetric. It is not possible to go through more than 16 positions with rotational symmetry, because one cannot be reached in under 8 jumps from either end.