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Another puzzle/ next question/ SPOILER I hope)_

Posted By: Bob Koca
Date: Sunday, 16 May 2010, at 2:03 a.m.

In Response To: Another puzzle/ next question (Bob Koca)

I think that the player who goes second can force a win no matter the value of N. Note that it is possible to place (N-1)^2 X's on the board without making X in a row and then by pigeonhole principle the next X must make N in a row. The player who goes 2nd can force that configuration (unless the other player commits suicide earlier than that by making N in a row when it was not forced). The strategy is to scan the row and columns and find which has the fewest amount of X's. Of those choices find one which fills in a row or column with minimum number of X's. For example suppose that you have found that columns 1, 3, and 8 have only 1 X and all other columns and rows have more than one X already. Of those three columns find which open space to play in so that the row you choose has a minimum number of X's.

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