Combinatorial Game Theory
who wins, and by how much
Generator

Hydras, and how long each takes to kill

Hydras, and how long each takes to kill
Hydras, and how long each takes to kill. Six small hydras with the ordinal the termination proof assigns to each and the exact number of chops it takes to finish it. Two of them are not finished here: the fight is guaranteed to end and the machine runs out of memory long before it does, which is the gap between a termination proof and a bound.

Six small hydras with the ordinal the termination proof assigns to each and the exact number of chops it takes to finish it. Two of them are not finished here: the fight is guaranteed to end and the machine runs out of memory long before it does, which is the gap between a termination proof and a bound.

3 essays call hydra-tree. 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.

Hydras, and how long each takes to kill. Six small hydras with the ordinal the termination proof assigns to each and the exact number of chops it takes to finish it. Two of them are not finished here: the fight is guaranteed to end and the machine runs out of memory long before it does, which is the gap between a termination proof and a bound. Where it stops

It ends, and nothing says when

The recursion this site runs needs every line of play to reach a position with no moves, and the condition is usually met by an obvious decreasing quantity. The hydra meets it with no such quantity anywhere: the tree grows at nearly every step and the fight ends regardless, because the only thing that decreases is an ordinal. A four-node hydra dies in twenty chops; one level deeper and 279 chops reach forty thousand nodes with no end in sight.

 Two ways to be certain and ignorant at once. Ten positions from two games that both terminate for reasons no bound comes out of. Sylver Coinage's proof counts something that goes down and can be counted; the hydra's counts an ordinal, which cannot, and the last column shows what that difference is worth. Where it stops

Two ways to end with no bound

Sylver Coinage and the hydra are both guaranteed to finish and neither will say when. The difference is that one of them carries its own bound: every move in Sylver removes at least one gap, the gaps can be counted in a moment, and over ten openings the longest play uses every single one. The hydra has no decreasing quantity a solver can hold — three hydras of five nodes each take seven chops, twenty-one, and a number past two hundred and seventy-nine that this machine never reaches.

 Two ways to be certain and ignorant at once. Ten positions from two games that both terminate for reasons no bound comes out of. Sylver Coinage's proof counts something that goes down and can be counted; the hydra's counts an ordinal, which cannot, and the last column shows what that difference is worth. Where it stops

Which games end at which level

Between a game that ends within a computable bound and one that ends with no bound at all there are levels, each corresponding to a strength of induction. This site's games sit at three of them, and which level a game is at is decided by exhibiting its termination measure and checking that every move lowers it.

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