Combinatorial Game Theory
who wins, and by how much
Series

Strategy stealing — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · applied
  2. 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.

    part 2 · applied
  3. 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.

    part 3 · applied
  4. 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.

    part 4 · applied
  5. 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.

    part 5 · applied

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