# Winning strategy - nim variation

i was reading about different variations of nim game and i'm trying to find winning strategy to one of them: There are n empty places on the circle. Two players are placing their "coins: on empty places. They can put their "coin" only on empty places which aren't next to oponnent "coin". In one turn each player can place only one "coin". Player who can't make any move lose. Who's got the winning strategy for each n.
I think that for even n second player will win, because he can always place his "coins" symetrical to the first player. But what about odd n?
sorry for my english.

• Can you structure your question, because I don't reall understand the game. – hlapointe Nov 1 '15 at 21:33

The first person will always win for odd $n$.
Your logic for why the second person will always win for even $n$ is correct. Now consider two players, Al and Bob, who are playing with odd $n$. Al makes the first move. Now, the game is the same as if Bob was making the first move in a game with even $n$, with the added restriction that he can't make his move next to where Al already has a coin. No matter what Bob does, Al can just pretend Al's the second player in a game with even $n$ and mirror Bob's moves, and so Al will always win with odd $n$. This is no problem because there's an added restriction to Bob, not to Al, and so Al will always be able to mirror Bob's moves.