Combinatorial Game Theory
who wins, and by how much
Concept

Strategy stealing — where it appears

An argument that the first player wins, by showing an extra move can never hurt, which names no move and yields none. It settles Hex and Chomp and supplies nothing playable, which is the clearest gap in the subject between knowing and doing.

Named by 12 essays across 3 fields — each of them below, with the objects they name alongside it.

Hex on 3 × 3, with every winning opening found. A rhombic Hex board with each cell marked according to whether taking it first wins. Left joins the top edge to the bottom and Right joins left to right; a filled board is always a win for exactly one of them, so the search needs no draw test. Strategy stealing proves that a winning opening exists without exhibiting one — these are the ones exhaustive search finds, on a board small enough for exhaustive search to finish.

The theorem that names a winner and no move

Strategy stealing proves that the first player wins Hex and wins Chomp, on every board, in about four lines. It exhibits no move, contains nothing a move could be extracted from, and is not going to. The moves have to come from somewhere else, and where they come from runs out almost immediately.

applied · Strategy stealing Three things the word “solved” is used for. The three standard senses of a solved game, priced on positions this solver can settle completely. Ultra-weak names the winner; weak supplies a strategy from the opening; strong supplies one from every position. They differ by orders of magnitude, and a claim that a game is solved is nearly useless until it says which of the three it means.

Three different claims are all called solved

Hex is solved in the sense that the first player provably wins, by an argument that names no move whatever. Nim is solved in the sense that a formula gives the right move from any position at any size. Between them sit strategies for one opening, and databases of a few billion positions. The word covers all four.

complexity · Complexity 7 symmetries, and the one that is a strategy. 4 games and 7 candidate symmetries, each tested by playing the strategy out against every opponent line rather than by argument. A pairing strategy needs a map that fixes the start, is an involution, and carries one player's moves to the other's — and the last condition is where most of these fail.

The strategy that is a symmetry

A pairing strategy is a symmetry of the board that turns one player's moves into the other's, and it wins without computing anything. Tested by playing it out rather than argued, it wins one of seven candidate symmetries across four games — exactly the Cram boards with both sides even, which is exactly where no domino is its own image.

impartial · Pairing A thousand shapes, and twelve pairings. Cram on every connected shape of at most eight squares, with the search for a symmetry that answers each of the opponent’s moves. Every pairing found is a second-player win, most shapes have no involution at all, and the strategy accounts for a sixth of the second-player wins there are.

Looking for the symmetry

Answering every move with its mirror image wins Cram on a board with both sides even, which is the argument everybody meets. Asked of every connected shape of at most eight squares instead of of thirteen rectangles, it wins twelve — and accounts for a sixth of the second-player wins there are, because 852 of the 1,042 shapes have no symmetry to answer with in the first place.

impartial · Pairing One test in front of a search. Five Cram boards solved with and without a check for a reachable pairing. A 4 × 5 board takes 17,348 node expansions without it and one with it.

A check in front of a search

The rung below found a pairing one move away on 288 of the 767 even first-player shapes, and asked what a solver that tested for one before recursing would save on a real game. On an even Cram board it saves nearly the whole search — a 4 × 5 board takes 17,348 node expansions without the check and one with it — and the depth profile shows why that number flatters: the check settles every winning position at the opening and at the last two moves, and about one in ten in between.

impartial · Pairing The Bridg-It board of size 3, both players at once. A Bridg-It board of size 3: blue dots in 4 rows of 3, red dots in 3 rows of 4, interleaved. Every bridge blue can usefully build is drawn in blue and every bridge red can usefully build in red, and each blue bridge crosses exactly one red one. Blue's switching graph and its planar dual have the same numbers of points and links, because the dual is red's board turned a quarter.

Cut is Short on another graph

Everything proved about the switching game is proved from Short's side, and Cut appears only as the player whose moves get enumerated. On a graph drawn without crossings Cut does not need a theory of its own: deleting a link is securing the link that crosses it in the dual, so Cut's game is Short's game on a different graph. Bridg-It is the board that is its own dual — one link short of two trees at every size, which is why its first player wins.

applied · Switching Chomp to 12 × 8: one needle on every bar but one. Every Chomp rectangle up to 12 columns by 8 rows, with the number of winning opening moves in each cell, found by search. All but one have exactly one; the 10 × 8 bar has 2. Cells in blue belong to the families whose winning move can be stated in a sentence — a single row, two rows, or a square; cells in gold are found only by searching.

Where the needle has a sentence

Strategy stealing proves the first player wins every Chomp bar and names no square to take. On two families the square can be said in a sentence — a square bar and a bar two rows deep — and in both the sentence is a pairing that names every later move too. Three rows deep the needle wanders, and the observation that every bar has exactly one needle survives ninety-four rectangles and fails on the ninety-fifth.

applied · Strategy stealing Hex on 3 × 4: the nearer edges win whoever starts. Two copies of a Hex board of 3 rows and 4 columns. On the left each cell is coloured by whether Down, joining top to bottom, wins by taking it first: all 12 do. On the right each cell is coloured by whether Across, joining left to right, wins by taking it first: none do. Down's edges are one row nearer together than Across's, and Down wins whoever moves first.

A board one column wider

Strategy stealing proves the first player wins Hex, and it needs three things: no draws, an extra stone never hurting, and rules that treat the two players alike. Add one column to the board and the third goes. The player whose edges are now nearer together wins whoever moves first — and does it with a table of pairs that names every reply, checked against every line to a board of twenty cells.

applied · Strategy stealing The positions that pair their gaps off, and who loses them. Every Sylver Coinage position with at most sixteen unnameable numbers, counted by genus, split into the symmetric and pseudo-symmetric semigroups — the irreducible ones — and the rest, with the positions lost for the player to move in each. Of 584 irreducible positions exactly one is lost, the single position whose only gap is 1; the remaining 11,185 positions include 1,405 losses.

Every move closes the largest gap

A census of Sylver Coinage by genus finds a parity that nearly decides the game and asks whether any known property of a numerical semigroup predicts the outcome. One does, completely: a semigroup whose gaps pair off around the largest one is never lost for the player to move — none of 583 up to genus sixteen. The reason is strategy stealing, and it is the same reason the top-right square decides Chomp: every move from such a position closes the largest gap.

applied · Sylver Two replies to the first stone. The 4 by 5 Hex board after Across's first stone in row 1, column 1, with every empty cell labelled by the weight of Across's unblocked chains through it. The potential answers in the heaviest cell, row 2, column 4; the pairing, which wins this board for Down, answers in row 1, column 2.

A potential that names every move

Strategy stealing names no move, and the pairings that do name moves need a board with the right symmetry. The Erdős–Selfridge potential needs neither: Down, moving second in Hex, takes the empty cell through which Across's unfinished chains weigh most. Its guarantee reaches only boards two rows deep. It wins far past the guarantee — on every board of three rows that Down can win — and then, on a four-by-five board that a table of pairs wins for Down with certainty, it answers Across's first stone in a different cell and loses along the bottom edge.

applied · Strategy stealing The gaps of ⟨5, 7, 9, 11⟩, and which of them the pairing removes. A Sylver Coinage position drawn as the numerical semigroup it is, with every legal move marked by what a single pass says about it. Gold squares are the numbers already named; plain squares are sums of them; magenta squares are the gaps, which are the legal moves. Under each gap is "struck" when the position that move reaches has its own gaps paired around its own largest gap — a position the opponent wins, so the move loses — and "wins" when the search says the move wins.

The pairing removes moves it cannot name

Symmetric positions were settled by an argument that names a winner and no move. Turned on the moves instead, the same one-pass test strikes off 27,215 of the 159,728 moves in the census and not one of the 21,234 winning ones — a quarter of a full search — and still names nothing. On 583 paired positions nine arithmetic descriptions of the winning gap reach at most 123, and 367 of those positions have exactly one winning move.

applied · Sylver An edge bonus, on every board it could help. The Erdős–Selfridge potential for Hex with the chains along the outer rows weighted more heavily, on six boards. No bonus wins the four-by-five board the plain potential loses, and the bonus costs Down 4 boards it was already holding.

The winning reply is the fourth choice

The repair proposed for the potential was to weigh an edge chain more heavily. Fifty-five weightings later, none holds the four-by-five board, and an edge bonus costs Down four boards it was already holding. The reason is not the numbers: over 393,660 turns of the pairing that does hold that board, the potential would take the same cell 26.1% of the time, and the winning cell is its 3.7th choice on average and as low as its seventeenth.

applied · Strategy stealing

Named alongside it

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

Exhaustive searchSymmetryCertificateCounterexamplePairing strategyStrategySolved gameImpartialCramEnumerationHeuristicInvariant

All concepts