Consider the following game. It is one of the many variants of Nim, also known a

Question

Consider the following game. It is one of the many variants of Nim, also known as the Marienbad game. Initially there is a heap of matches on the table between two players. The first player may remove as many matches as he likes, except that he must take at least one and he must leave at least one. There must therefore be at least two matches in the initial heap. Thereafter, each player in turn must remove at least one match and at most twice the number of matches his opponent just took. The player who removes the last match wins. There are no draws.

Suppose we change the rules of the game above so that the player who is forced to take the last match loses. This is the misère version of the game. Suppose also that the first player must take at least one match and that he must leave at least two. Among the initial positions with three to eight matches, which are now winning positions for the first player?

Explanation / Answer

Ans:There are two as the rule is reverse to actual game rule.

1) If m=1 and n=1, or m=n=1 then the player 1 has a winning strategy.

2) If mn, then player 1 has a winning strategy. (If m not equal to n then player 1 will be having the winning strategy).

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