The Puzzle: You have a standard 8x8 Chessboard.
- You remove two squares from opposite corners (e.g., A1 and H8).
- You have 31 Dominoes, and each domino covers exactly two adjacent squares. Is it possible to cover the remaining 62 squares using these 31 dominoes?
1. The Strategy: Color Parity
A standard chessboard has 32 White squares and 32 Black squares.
- Opposite corners on a chessboard are always the same color.
- If you remove two opposite corners (say, both Black), the board now has:
- 32 White squares.
- 30 Black squares.
2. The Contradiction
Every single domino, no matter how it is placed, must cover one White square and one Black square.
- To cover 31 dominoes, you would need exactly 31 White and 31 Black squares.
- Since our board has 32 White and 30 Black, it is impossible to cover it perfectly.
Interview-Focused Questions
Q: What is the ‘Invariant’ in this puzzle?
A: The invariant is that a domino always consumes one square of each color. This parity check is a powerful tool in discrete mathematics to prove that a specific state is unreachable.
Q: Could you solve this for a 3D cube?
A: Yes. If you have a 4×4×4 cube and remove two opposite corners, the same checkerboard logic applies (3D parity). If the number of “Black” and “White” cells doesn’t match, you can’t fill it with 1×2 blocks.
Key Takeaway
This is a lesson in Proof by Contradiction. If the necessary conditions (equal color counts) are not met, the task is impossible.
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