Domineering as a combinatorial game. #
We define the game of Domineering, played on a chessboard of arbitrary shape (possibly even disconnected). Left moves by placing a domino vertically, while Right moves by placing a domino horizontally.
This is only a fragment of a full development; in order to successfully analyse positions we would need some more theorems. Most importantly, we need a general statement that allows us to discard irrelevant moves. Specifically to domineering, we need the fact that disjoint parts of the chessboard give sums of games.
The equivalence (x, y) ↦ (x, y+1)
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- SetTheory.PGame.Domineering.shiftUp = (Equiv.refl ℤ).prodCongr (Equiv.addRight 1)
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The equivalence (x, y) ↦ (x+1, y)
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A Domineering board is an arbitrary finite subset of ℤ × ℤ
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Left can play anywhere that a square and the square below it are open.
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- SetTheory.PGame.Domineering.left b = b ∩ Finset.map SetTheory.PGame.Domineering.shiftUp.toEmbedding b
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Right can play anywhere that a square and the square to the left are open.
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- SetTheory.PGame.Domineering.right b = b ∩ Finset.map SetTheory.PGame.Domineering.shiftRight.toEmbedding b
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After Left moves, two vertically adjacent squares are removed from the board.
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- SetTheory.PGame.Domineering.moveLeft b m = (Finset.erase b m).erase (m.1, m.2 - 1)
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After Left moves, two horizontally adjacent squares are removed from the board.
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- SetTheory.PGame.Domineering.moveRight b m = (Finset.erase b m).erase (m.1 - 1, m.2)
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The instance describing allowed moves on a Domineering board.
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- One or more equations did not get rendered due to their size.
Construct a pre-game from a Domineering board.
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All games of Domineering are short, because each move removes two squares.
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The Domineering board with two squares arranged vertically, in which Left has the only move.
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- SetTheory.PGame.domineering.one = SetTheory.PGame.domineering [(0, 0), (0, 1)].toFinset
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The L
shaped Domineering board, in which Left is exactly half a move ahead.
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- SetTheory.PGame.domineering.L = SetTheory.PGame.domineering [(0, 2), (0, 1), (0, 0), (1, 0)].toFinset