A rook's tour from corner to opposite corner

Question

A rook starts on a1 and moves one square at a time (up, down, left or right). It must visit every square of the 8 by 8 board exactly once and finish on h8, the opposite corner. 1. Is this possible? 2. Is it possible to finish on h1 instead (the corner along the same edge)? 3. Does the answer change for a king, which can also move diagonally?

Accepted Answer

Finishing on h8 is impossible; finishing on h1 is possible. A king can do both. The parity argument. Visiting all 64 squares exactly once takes exactly 63 moves. Each one-square rook move changes the colour of the square (adjacent squares always differ in colour). After 63 moves, an odd number, the rook is on the opposite colour from where it started. But a1 and h8 are the same colour (both dark). So the tour cannot end on h8. h1 is fine. The corners a1 and h1 are different colours, so parity permits it, and a tour exists, snaking up and down the columns: go up column a, across to column b, d…

A rook's tour from corner to opposite corner

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