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There are some piles of stones.

Two players move in turn. One can remove a stone from a pile or merge two piles in a move. The player that removes the last stone wins.

With the number of stones in each pile given, how to know who will win?

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  • $\begingroup$ In fact, I got a conclusion, but I don't know how to prove it. Suppose there are $p$ piles, $k$ piles that contain only one stone, and a total of $n$ stones. If $n \le p+1$, the first player lose when $k\bmod{3}=0$. If $n>p+1$ the first player lose when $p+n$ is odd and $k$ is even. $\endgroup$ – Tang Xianghao Jul 6 '13 at 12:31
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Hint: Only by taking a pile which consists of a single stone can a player change the parity (see Brian's hint). Is there a strategy for a player who is winning on parity which avoids either player taking a single stone until the last move?

What happens when the initial position has a single pile of size 1?

What happens when there are several piles of size 1 in the initial position?

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