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 removingone or more stonesfrom 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 containsnpiles havingk1,k2, …,knstones respectively; in such a position, there arek1+k2+ … +knpossible moves. We write eachkiin binary (base 2). Then, the Nim position is losing if and only if, among all theki’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 thexorof theki’s is 0.
Consider the position with three piles given byk1= 7,k2= 11, andk3= 13. In binary, these values are as follows:
11110111101
There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, supposek3were 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 whenk1= 7,k2= 11, andk3= 13. In fact, there are exactly three winning moves from this position: namely removing one stone from any of the three piles.
，从此便踏上征途，也许会孤独一程。
