## Answers

The second game in Figure 52 is not possible. Zero must have played first and last, but X had a win before the final move, so the last move would not have been made. In the third game, X could have completed a win if his first two moves had been on either side, therefore the first two moves must have been diagonally opposite, and his final move in the top right corner.

These two problems are so easily solved that I will add here a difficult one that involves what chess players call retrograde analysis. Figure 55 shows the pattern after two perfect players have agreed to a draw. Your task is to determine the first and last moves. If you can't solve it, you will find the solution in the Journal of Recreational Mathematics, Vol. 11, No. 1, 1978, page 70. The problem had been earlier posed in the same journal by Les Marvin.

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What were the first and last moves?

In Silverman's first problem, X can always win, regardless of whether he plays first or second. Assume that the cells are numbered (left to right, top to bottom) from 1 to 9. Here is Silverman's proof:

If X begins, he takes 1. O must take 5, otherwise X can get three of his marks in a row by standard ticktacktoe strategy. X2 forces 03, then X4 forces 07, which completes three O's in a line, giving X the win.

If 0 starts the game, he has a choice of corner, side or center opening. If he opens at the center (5), X responds with 1. If the move is 02, X7 forces 04, then X9 forces 08, which loses. If O's second move is 3, X4 forces 07, which also loses. If O's second move is 6, X7 forces 0 to lose at 4. If O's second move is 9, X2 forces 03, then X4 forces O to lose at 7. All other lines of play are symmetrically equivalent.

If 0 opens at the side, say at 4, X5 will win. As before, there are four basically different continuing lines of play: (1) 01, X3, 07 (loses), (2) 02, X3, 07, X9, Ol (loses), (3) 03, X9, Ol, X8, 02 (loses), (4) 06, X3, 07, X9, 01 (loses).

A corner opening by 0, say at 1, is met with X5, which leads again to four basically different continuations: (1) 02, X7, 03 (loses), (2) 03, X8, 02 (loses), (3) 06, X8, 02, X7, 03 (loses), (4) 09, X2, 08, X3, 07 (loses).

When this game is played on a four-by-four field (X winning if there are four of either mark in a row, 0 winning if the final position is drawn), the play is so enormously more complex, Silverman informs me, that it has not yet been fully analyzed.

0 wins Silverman's go-moku problem by playing 01 [see Figure 56]. X2 is forced, 03 forces X4, 05 forces X6, then 07 creates an open-end diagonal row of four O's, which X cannot block. If X plays at either end, 0 wins by playing at the opposite end. As Silverman points out in his book, 0 wins only by counterattacking. He loses quickly if he plays defensively by trying to block X's open-end diagonal row of three.

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Solution to the go-moku problem

### Solution to the go-moku problem

Note that when X plays on the cell marked 2 it creates a fork. This is permitted, however, because the move is forced. It is the only way to prevent 0 from winning on the next move.

chapter 9

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