2975 -- Nim

Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or more stones from any single pile. Play ends when all the stones have been removed, at which point the last player to have moved is declared the winner. Given a position in Nim, your task is to determine how many winning moves there are in that position.

A position in Nim is called “losing” if the first player to move from that position would lose if both sides played perfectly. A “winning move,” then, is a move that leaves the game in a losing position. There is a famous theorem that classifies all losing positions. Suppose a Nim position contains n piles having k₁, k₂, …, k_n stones respectively; in such a position, there are k₁ + k₂ + … + k_n possible moves. We write each k_i in binary (base 2). Then, the Nim position is losing if and only if, among all the k_i’s, there are an even number of 1’s in each digit position. In other words, the Nim position is losing if and only if the xor of the k_i’s is 0.

Consider the position with three piles given by k₁ = 7, k₂ = 11, and k₃ = 13. In binary, these values are as follows:

 111
1011
1101

There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, suppose k₃ were changed to be 12. Then, there would be exactly two 1’s in each digit position, and thus, the Nim position would become losing. Since a winning move is any move that leaves the game in a losing position, it follows that removing one stone from the third pile is a winning move when k₁ = 7, k₂ = 11, and k₃ = 13. In fact, there are exactly three winning moves from this position: namely removing one stone from any of the three piles.