Combinatorial Game Theory
who wins, and by how much
Concept

Backward induction — where it appears

Working out a game from its ends, labelling each position from the labels of everything it leads to. It is the same recursion every value on this site is computed by, run over a finite graph rather than a tree.

Named by 8 essays across 3 fields — each of them below, with the objects they name alongside it.

Backward induction on a game that ends, one round at a time. Zermelo's argument as it actually runs. Round zero is the positions where the player to move has no move at all, which is the only thing the procedure knows without being told; each later round is what those settle. Anything still unlabelled when nothing more can be deduced has no label and never will — and on a game with a cycle in it, that leftover is exactly the set of drawn positions. The theorem is a statement about this procedure terminating, and it names the winner of nothing.

The first theorem, and the winner it declines to name

Zermelo proved in 1913 that a finite game with no chance and no hidden information is decided before anybody sits down — every position is a win for one side or a draw, and which one is settled already. The proof is a labelling procedure, and watching it run shows exactly how little it says.

history · Determinacy 1 ko point, no ko rule. A ko fight drawn as a position graph and labelled by retrograde analysis: blue edges are Black's captures, red are White's, and each position carries the verdict for whichever side is to move. Positions the propagation never reaches are drawn — neither player can force a win and the game does not end — and they appear only where the rules permit a repetition.

The rule that makes Go a finite game

A ko is a point in Go where a capture can be recaptured for ever, and every set of rules forbids it. That prohibition is not etiquette or tidiness — it is the hypothesis that puts Go inside the class of games every theorem on this site is about, and removing it removes the values.

applied · Go 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 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.

limits · Multiplayer 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 A fortress, and the counter that gives it a label. A pawn ending where the defender's king shuffles for ever and the attacker needs time. Down the rows, how many moves of preparation the breakthrough needs; across the columns, how many moves the rule allows before declaring a draw. With no breakthrough the position is drawn whatever the rule says, and drawn as a residue the backward induction never reaches. With a breakthrough and no rule the attacker wins despite the cycle. Where the march is longer than the counter allows, the rule turns a won position into a drawn one.

A position with no value, and the rule that gives it one

A fortress is a cycle in the position graph, so the recursion defining a value has nowhere to bottom out and the propagation never reaches it. Chess has a rule for that — count fifty moves and call it drawn — and the rule does not merely tidy the theory up. On eleven cells of the sweep it takes away a win.

applied · Chess How long a win takes, against how long the argument allows. Ordinary impartial games with the size of their position graphs, the number of rounds the backward labelling takes, and the number of moves the longest win actually lasts. The round a position settles in is the length of the play from it, which is computed here a second way so the two must agree. The rounds are a handful and the positions are many, which is the gap Zermelo's 1913 paper is about — his question was how many moves a forced win needs, and the answer he could prove was the size of the whole graph.

The paper was about how long

Zermelo's 1913 paper is remembered for a theorem it proves in passing. The question it actually asks is how many moves a forced win takes, the answer it can prove is the size of the whole position graph, and the round counter in the procedure is the real answer — a quantity nobody named for another forty years.

history · Determinacy Two solutions to one set of equations. The winning condition written as a single predicate and solved twice: once as the least solution of its own equations and once as the greatest. The least says Left can force a win; the greatest says Left cannot be forced to lose, which admits the positions where Left can keep the game going for ever. On a graph with no cycle in it the two coincide and the equations determine an answer. Where they differ, the difference is exactly the set the backward propagation never reaches — so a draw is not a leftover of the algorithm, it is the equations failing to have one answer.

The gap between two answers

A draw is usually described as what the backward labelling never reached, which makes it sound like a shortfall of the algorithm. Written as one predicate the winning condition is an equation, the equation is monotone, and it has a least solution and a greatest one — and the set the two disagree about is exactly the drawn set, on every game checked.

history · Determinacy 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.

TerminationDeterminacyOutcome classPosition graphDrawExhaustive searchInductionLoopyNormal playRetrograde analysisBoundChess

All concepts