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♟️ The Mutilated Chessboard

A beautiful impossibility proof using a colouring argument

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♟️ The Setup
Topics: proof by contradiction, colouring arguments, combinatorics
Take a standard 8×8 chessboard (64 squares, alternating black and white).

Remove the two squares at diagonally opposite corners (e.g. top-left and bottom-right). You are left with 62 squares.

A domino covers exactly 2 adjacent squares (horizontally or vertically).

Challenge: Can you tile the 62 remaining squares perfectly with 31 dominoes — with no gaps and no overlaps?
Key insight: Every domino covers exactly one black square and one white square — because it covers two adjacent squares, which are always different colours.

Therefore, 31 dominoes would cover exactly 31 black and 31 white squares.

But the two removed corners are both the same colour (both white, or both black — opposite corners share colour on a standard board). So the mutilated board has 30 of one colour and 32 of the other. An exact tiling is impossible.
📝 Questions
1
A standard chessboard has 64 squares. How many are white? How many are black? 1 mark
2
Look at the chessboard diagram. The two removed corners are marked in red. What colour are they both? 1 mark
3
After removing the two corners, how many white squares remain? How many black squares remain? 2 marks
4
A single domino covers two adjacent squares. What colour are two adjacent squares always? Explain why. 2 marks
5
If 31 dominoes are placed on the board, how many white squares and how many black squares must they cover in total? 2 marks
6
Using your answers to questions 3 and 5, prove that it is impossible to tile the mutilated chessboard with 31 dominoes. Write a clear, logical argument. 4 marks
7
Extension: Suppose instead of the two corners, you removed one white square and one black square from anywhere on the board. Would 31 dominoes now be able to tile the board? Justify your answer. 4 marks
8
Extension: The colouring argument is a type of "invariant" proof. What is an invariant, and can you think of another puzzle where a colouring or parity argument proves impossibility? 4 marks

Answer Key

♟️ Mutilated Chessboard

1.32 white, 32 black.
2.Both are white (top-left and bottom-right corners on a standard board are the same colour).
3.30 white squares and 32 black squares remain.
4.Two adjacent squares are always different colours — the board alternates colours in every direction, so any two touching squares are one black and one white.
5.31 dominoes must cover exactly 31 white and 31 black squares.
6.The mutilated board has 30 white and 32 black squares. But any valid tiling with 31 dominoes must cover 31 white and 31 black squares (since each domino covers one of each). 31 ≠ 30 and 31 ≠ 32. Therefore no tiling is possible.
7.Not necessarily guaranteed, but the colouring argument no longer rules it out — removing one of each colour restores 31 white and 31 black squares, making tiling potentially possible. (In practice, whether it's achievable depends on which squares are removed.)
8.An invariant is a property that remains unchanged under any allowed operation. Examples: in a 15-puzzle (sliding tiles), the parity of permutations is invariant — certain configurations are unreachable; in a coin-flipping puzzle, the parity of face-up coins may be invariant.

Answer Key

Mutilated Chessboard

1.32 white, 32 black.
2.Both white.
3.30 white, 32 black.
4.Always different colours — alternating pattern.
5.31 white, 31 black.
6.Tiling needs 31W+31B but board has 30W+32B → impossible.
7.Colouring argument no longer rules it out (31W+31B restored). Still depends on position.
8.An invariant is unchanged under allowed moves. E.g. 15-puzzle permutation parity.
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