Conway's Soldiers or the checker-jumping problem is a one-person mathematical game or puzzle devised and analyzed by mathematician John Horton Conway in 1961. A variant of peg solitaire, it takes place on an infinite checkerboard. The board is divided by a horizontal line that extends indefinitely. Above the line are empty cells and below the line are an arbitrary number of game pieces, or "soldiers". As in peg solitaire, a move consists of one soldier jumping over an adjacent soldier into an empty cell, vertically or horizontally (but not diagonally), and removing the soldier which was jumped over. The goal of the puzzle is to place a soldier as far above the horizontal line as possible.
Conway proved that, regardless of the strategy used, there is no finite sequence of moves that will allow a soldier to advance more than four rows above the horizontal line. His argument uses a carefully chosen weighting of cells (involving the golden ratio), and he proved that the total weight can only decrease or remain constant. This argument has been reproduced in a number of popular math books.
Simon Tatham and Gareth Taylor have shown that the fifth row can be reached via an infinite series of moves. If diagonal jumps are allowed, the 8th row can be reached, but not the 9th row. In the n-dimensional version of the game, the highest row that can be reached is
3n-2
3n-1
Define
\varphi=
\sqrt{5 | |
-1}{2} |
≈ 0.6180339887\ldots
\varphi
\varphi2=1-\varphi
Let the target square be labeled with the value
\varphi0=1
\varphin
n
\varphi1+\varphi2
When a soldier jumps over another soldier, there are three cases to consider:
\varphin
n
\varphin-1
\varphin-2-\varphin-1-\varphin=\varphin-2(1-\varphi-\varphi2)=0
\varphin-\varphin-1-\varphin=-\varphin-1
\varphin+2-\varphin+1-\varphin=\varphin(\varphi2-\varphi-1)=-2\varphin+1
So, no jump will ever increase the configuration's total score.
Consider now a starting configuration where only one infinite horizontal line is completely filled with soldiers.
If this horizontal line of soldiers is immediately below the target square, then the score of the configuration is
\varphi+2\varphi2+2\varphi3+\ldots
\varphi2+2\varphi3+2\varphi4+\ldots=\varphi(\varphi+2\varphi2+2\varphi3+\ldots)
\varphi2(\varphi+2\varphi2+2\varphi3+\ldots)
Consider the full starting configuration, where soldiers fill the whole half-plane below the red line. This configuration's score is the sum of the scores of the individual lines. Therefore, if the target square is immediately above the red line, the score is
S1=(\varphi+2(\varphi2+\varphi3+\varphi4\ldots))(1+\varphi+\varphi2+\varphi3+\ldots)
At this point, observe another interesting property of
\varphi
infty | |
\sum | |
n=2 |
\varphin=1
S1=(\varphi+2)(1+\varphi+1)=(\varphi+2)2=5+3\varphi ≈ 6.85410\ldots
If the target square is in the second row above the red line, every soldier is one space further from the target square, and so the score is
S2=\varphiS1=3+2\varphi
Similarly:
S3=\varphiS2=2+\varphi
S4=\varphiS3=1+\varphi
S5=\varphiS4=1
When a soldier reaches the target square after some finite number of moves, the ending configuration has score
E=\varphi0+\epsilon
\varphi0
\epsilon
Thus, we have shown that when the target square is in the fifth row above the infinite half-plane of soldiers, the starting configuration's score is exactly
S5=1
E=1+\epsilon
S5\geE
\BbbZd