Combinatorial Game Theory
who wins, and by how much
Concept

Alternation — where it appears

Players moving in turn, which is what separates a game from a puzzle and what makes the winner's certificate a strategy. Replace it with an auction and a position gets a number in the unit interval instead of an outcome class.

Named by 16 essays across 3 fields — each of them below, with the objects they name alongside it.

Every quantifier is a move. A quantified boolean formula with its quantifiers drawn as turns: an existential is a choice by the player to move, a universal a choice by the opponent. The same formula is put through the reduction to Generalized Geography and the two answers are checked against each other, so the prefix of quantifiers and the game beside it are one claim.

A puzzle asks once, a game asks alternately

Quantifier alternation is the whole difference between a puzzle and a game. One chooser is an existential and its answer is a witness somebody can check; two choosers taking turns is a prefix of alternating quantifiers, and the witness stops being an assignment and becomes a strategy.

complexity · Alternation A winning strategy on 3×3, drawn whole. The whole of one player's winning strategy on a small Domineering board: their own move at each of their turns, and every reply the opponent has at each of theirs. The strategy branches only where the loser chooses. Its size is what somebody would have to be handed to check the claim that this player wins, and it is far larger than the claim itself.

"Left wins" has no short proof

A complete solution of Nim on heaps of 7, 11 and 13 is 480 table entries. A winning strategy for the same position — one move of the winner's at each of their turns, and an answer to every reply — has 56,167,022 nodes in it. The answer is smaller than the proof by a factor of a hundred thousand.

complexity · Complexity What the auction can and cannot see. Values under both conventions. The Richman value is the share of the money the second player needs; a half means the position itself decides nothing and whoever has more money wins. Every infinitesimal on the list, and zero with them, comes out at a half.

Nobody has to move

Every convention here rests on one sentence nobody examines — the players move alternately. Replace it with an auction and a position stops having an outcome class and starts having a number: the share of the money the second player needs. The 22 values born by day two collapse to seven of those numbers, eight of them landing on exactly a half; the new number respects the game order on all 179 comparable pairs, and is not determined by the parts under addition on 14 of 49.

limits · Bidding A shuttle and a loop, judged by what the play returns to. A three-node loopy game drawn as a graph, with Left's moves in blue, Right's in red and position a marked. Beside it, each position-and-mover pair under the backward labelling and under the rule that a never-ending play goes to Left when it returns to a infinitely often. Four pairs are drawn by the labelling; the new rule gives two to Left and two to Right and leaves the decided pairs as they were.

What the play keeps coming back to

A draw is what the backward labelling never reaches, and handing every never-ending play to one player turns the draws into wins wholesale. Judge an infinite play instead by what it keeps returning to, and every draw gets a winner of its own: over the 262,144 three-node games, 15,432 send some of their draws to one player and some to the other, which no wholesale rule can do. Finding those winners takes a fixed point inside a fixed point.

history · Determinacy A hub with two spokes, and the bit of memory it needs. A three-node loopy game in which Left, at a hub, chooses between two spokes and Right must return from either. Left wins a never-ending play that passes through both spokes infinitely often. With one bit of memory recording which spoke is owed, Left wins from the hub; with a strategy that depends only on the position, Left always takes the same spoke and loses. Three position-and-mover pairs change hands.

One bit of memory

Judge an infinite play by whether one position keeps recurring and every winner can play from a table of one move per position, with nothing remembered. Ask for two positions to keep recurring and that stops being true. At a hub with two spokes a player has to alternate, and a table cannot alternate: over every three-node game, 49,487 position-and-mover pairs are won with one bit of memory and lost without it.

history · Determinacy Which conditions make a winner remember. Every condition on which set of three positions a never-ending play keeps returning to, grouped by two properties of the condition alone, against the arenas swept. 32 of 128 conditions have an arena Left wins and cannot win from a table of one move per position; 19 of those are closed under union.

The rule decides who has to remember

Whether a winner needs memory is a property of the winning condition and not of the board, and the property everybody reaches for is the wrong one. Of the 128 conditions on which of three positions a play keeps returning to, 32 demand memory and 19 of those are closed under union. What separates them is measured two independent ways and the two agree on all 128: a condition needs no memory exactly when it can be rewritten as a number on each position.

history · Determinacy Two ways to search on, one position. A 4 × 5 Domineering position with Right to move, which Right wins, beside what deepening says at each depth when it declines to guess where the counts of placements are close. Searching on one move at a time, depths 0 and 1 agree on the wrong verdict; searching on two moves at a time, the search stops at depth 3 with the right one.

Search on in pairs of moves

Deepening until two depths agree gives a proved verdict, and searching on where the counts are close gives a better one; put together the obvious way, they stop on a wrong verdict at 3,231 positions of 4 × 5 Domineering. A guess one move past the cut has the other player to move and flatters the wrong side. Searching on two moves at a time keeps the proof, and the window that suits it is one-sided — but however it is widened, the certificate gets cheaper only by turning into the search that finishes, and on four boards it never gets below it.

complexity · Search 6 turns in strict alternation, priced. One quantifier prefix taken apart into its blocks, with the size of a winning strategy computed a term at a time. A choice made after k of the opponent's turns has to be written down once for each of the 2^k lines the opponent can produce, so the total depends on where the opponent's turns sit and not merely on how many there are.

Twelve turns, and three different prices

The earlier essay prices a universal quantifier at a doubling and leaves it there. Twelve turns with six of them the opponent's cost 6, 63 or 384 decisions to write down, depending on nothing but the order the turns come in — and the cheap arrangements are cheap for only one of the two players. What a claim costs is the number of times the choosing changes hands.

complexity · Alternation How many turns are choices. Every position of each game with both sides to move, classified by whether the turn is a choice at all: no move, exactly one move, several moves that all lead to the same verdict, and several that do not. Only the last is a turn at which the alternation is doing any work.

Eleven moves and one decision

A prefix has one quantifier a turn, so a game of eleven moves is eleven alternations. Counted on the boards themselves, a Toads and Frogs strip of eleven moves has twenty-six turns with exactly one move available and one turn anywhere at which the choice changes the answer; a Clobber board has a hundred and fourteen turns and none. Nim, the game everybody calls solved, decides at four turns in five.

complexity · Alternation A win is proved by one move and a loss by all of them. The smallest proof of each position's verdict, averaged by verdict. At a node the mover wins the proof takes the cheapest single option; at a node the mover loses it has to answer every option, which is the existential and universal quantifiers of the prefix showing up as two different objects.

Proving a loss means answering everything

A win is established by one move and a loss by every move, so the two verdicts are certified by objects of different shapes. Measured over every position of four games, a loss costs between 1.07 and 2.31 times a win — a small constant, never an exponential. The obvious explanation is the branching and it is wrong: Nim answers six options at a losing turn and pays 2.18, not six.

complexity · Alternation What the opponent's choosing is worth. Every position answered twice: against an opponent who searches, and against one following a fixed rule with no search in it. Only a loss can change, so the share is taken over the losses. The spread between games is the measurement — in one of them nearly every loss is recovered and in another none is.

The opponent stops choosing

Replace one player by a rule with no search in it and the question has one chooser left, which is a puzzle rather than a game. Nim recovers five of its six lost positions that way, and six of seven on three heaps of five. Domineering recovers six of a hundred and twenty-two while the fixed rule throws away a winning move eighty-eight times, and one Clobber board recovers none at all — because on that board no rule can misplay.

complexity · Alternation A turn is not a bit. The number of turns a game lasts, beside the number of quantified bits those turns amount to. Each ply is measured over the positions actually reachable at it rather than along one line, and the bits are the logarithm of the branching, which is what a quantifier prefix would need one of.

A turn is not a bit

The prefix a game is read as gives each player one quantifier a turn, and a turn on a board is a choice among however many moves there are. Nim on heaps of 3, 4 and 5 lasts twelve moves and carries 23.6 bits of choice; a Toads and Frogs strip lasts eleven and carries two. Corrected for that, the model predicts a strategy 539 times too large on one board and 67 times too small on another, and the two failures have different causes.

complexity · Alternation The money played out, and it never mattered. The bidding rule played move by move with a countable pool of chips, at every way of splitting it. The verdict is constant across the splits and opposite under the two ways of resolving equal bids, so what settles these positions is the tie-break rather than the money.

The auction never gets to the money

The critical fraction is computed and never played. Played out with a countable pool of chips — twelve positions, four pool sizes, every split of the chips, every bid answered — the verdict does not move with the money on a single one of the forty-eight sweeps, and the rule for equal bids settles all forty-eight. The reason is one line long: declining every auction wins, and bidding nothing declines.

limits · Bidding The same number, from a rule that needs no tie-break. The number computed twice: once as the critical share of a pot under the auction, and once as the probability that Left wins when a fair coin decides who moves at each turn. They agree on every position, and only the second derivation survives being played out.

A coin needs no tie-break

The same recursion has a second derivation: a fair coin decides who moves at each turn, a player whose turn it is with no move has lost, and both play to win. Written from those rules it comes out identical on every position — and it needs no rule for equal bids, because there are no bids. The number is a probability, it belongs to Left rather than Right, and the empty position is the one where the coin decides everything.

limits · Bidding A position Left always wins, and not always. Values grouped by the outcome class alternating play assigns them, with the range of probabilities the coin gives Left inside each class. A class that alternating play calls a win for Left every time holds no position the coin makes certain.

Left always wins, and loses more often than not

Alternating play answers with one of four classes and the coin answers with a chance, and the two do not have to agree. Over the twenty-two values born by day two they never disagree and the margin is exactly nothing — the lowest chance on a position Left wins whoever moves is a half. Over the 1,474 born by day three, seven of them sit at seven sixteenths, and seven mirror them on the other side.

limits · Bidding The coin's move is often a blunder. Positions where Left has a choice and at least one option wins under alternating play, with how often the option maximising Left's chance under random turns is an option that loses the alternating game outright.

The best chance is the wrong move

Maximising a probability and denying an opponent a reply are different objectives, and on 189 of the 904 day-three positions where Left has a choice and a winning move, the option the coin prefers is one that loses the alternating game outright. The smallest case is two options and one line of arithmetic: five eighths beats a half, and a half is the move that wins.

limits · Bidding

Named alongside it

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

Exhaustive searchStrategyDeterminacyOutcome classComplexityCertificateDecisionNormal playCounterexampleCountingDomineeringClobber

All concepts