Combinatorial Game Theory
who wins, and by how much
Generator

The same position, two conventions, two winners

The same position, two conventions, two winners
The same position, two conventions, two winners. Three-player Nim with the last counter winning. The two columns differ only in what a player does when they cannot win themselves, which is a question the rules do not answer — and the answer decides who wins.

Three-player Nim with the last counter winning. The two columns differ only in what a player does when they cannot win themselves, which is a question the rules do not answer — and the answer decides who wins.

1 essay calls three-player. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

The same position, two conventions, two winners. Three-player Nim with the last counter winning. The two columns differ only in what a player does when they cannot win themselves, which is a question the rules do not answer — and the answer decides who wins. Where it stops

Three players and no answer

Every theorem here is about two players, and the reason is not convenience. With two players the game is zero-sum, so 'play well' needs no further explanation. Add a third and the winner of a Nim position becomes a fact about the convention: two reasonable ones disagree on 56 of the 71 positions swept. The one question no convention touches — can a player force a win against the other two together — is answered 'nobody' in 65 of the 71.

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