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Consider a modified version of standard nim game. In this game, there are K piles of stones.There are two players.In each turn a player can remove exactly 1 stone from any of the piles.The winner is who removes the last stone from any of the pile. Note: It is not necessary to remove all stones from all piles.

Example:

Lets say distribution of piles is: (1,1).Clearly 1st player will win. For (2,2) 1st player have to remove 1 stone from any of the one pile.Then 2nd player removes remaining one stone and wins.

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Suppose there are $x$ stones in total. If any pile have one stone, then the first player wins straight away.

Otherwise, consider the parity of $x-2K$ (or x). If it's odd, then the first player wins. If it's even, then the second player wins. The reason is obvious: every pile must remain at least two stones, otherwise whoever takes one loses instantly.

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