Snapping a 6-by-8 chocolate bar into squares

Question

You have a rectangular chocolate bar scored into a 6 by 8 grid of small squares (48 squares total). Two players alternate turns. A move is to pick up any single piece and snap it in a straight line along one groove, splitting it into two pieces. The player who cannot move (because every piece is already a single square) loses.

With perfect play, does the first or the second player win? Does your strategy even matter?

Accepted Answer

Strategy is irrelevant, the winner is fixed in advance, and it is the first player. Count the breaks. Start with 1 piece. Every single snap turns one piece into two, so each move raises the piece count by exactly one. The game ends when there are 48 single squares. To go from 1 piece to 48 pieces you need exactly 48 - 1 = \mathbf{47} snaps, no matter where or in what order anyone breaks. There is no way to make more or fewer. Follow the parity. The players alternate, so the first player makes breaks number 1, 3, 5, \dots and the second player makes 2, 4, 6, \dots. Since 47 is odd, the first p…

Snapping a 6-by-8 chocolate bar into squares

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