Combinatorial Game Theory
who wins, and by how much
Concept

Ordinal — where it appears

A number for counting past every finite stage, used here to prove a process ends when no quantity in the picture decreases. A number in it is not a count of anything a solver can hold, which is why the proof it supplies comes with no bound.

Named by 5 essays across 3 fields — each of them below, with the objects they name alongside it.

The days this site can compute, and the ones it cannot. Zero on the first day, ±1 on the second, and thereafter the simplest number in every remaining gap — the construction run by the game recursion, which produces only fractions with a power of two underneath however long it goes on. Below it, three objects the same recursion reaches when the stopping rule is removed, each written with its option set and the exact reason this site's machinery cannot hold it. They are named rather than drawn, which is the honest half of a figure-first collection.

The recursion this site cannot run

Remove the stopping condition from the construction and it reaches ω, its reciprocal, and one third — none of which this site's evaluator can represent, because it interns a position from a finite list of options. The figure draws what it computes and names what it cannot, which is where the boundary belongs.

values · Numbers 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.

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.

limits · Termination 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.

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.

limits · Termination 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.

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.

limits · Termination A game every play of which ends, and no round settles. A game whose first move chooses how long the game will be, cut off at several sizes. Every play of it is finite and no position is drawn, so the fourth outcome class has nothing to do with what goes wrong. What goes wrong is the round counter: the opening is a loss, a loss settles only when the last of its options is known, and there is no last option. Cut the game off larger and the round grows, so no number in the column is the answer for the untruncated game — and the induction that labels it has to run past every finite stage.

Every play ends and no round settles

Take the finiteness hypothesis away carefully — not by adding a cycle, which has already been priced twice, but by adding infinitely many positions to a game every play of which still ends. Nothing is drawn, every line finishes, and the round the opening settles in grows with every cut: two, four, six, eight, twelve, sixteen, and no number in the column is the answer.

history · Determinacy

Named alongside it

The objects these essays reach for when they reach for this one.

TerminationInductionExhaustive searchSylver CoinageBackward inductionBoundFixed length gameGame lengthHydraProofRecursionSurreal

All concepts