(related to Problem: The Two Rooks)

The second player can always win, but to ensure his doing so he must always place his rook, at the start and on every subsequent move, on the same diagonal as his opponent's rook. He can then force his opponent into a corner and win. Supposing the diagram to represent the positions of the rooks at the start, then, if Black played first, White might have placed his rook at $A$ and won next move. Any square on that diagonal from $A$ to $H$ will win, but the best play is always to restrict the moves of the opposing rook as much as possible. If White played first, then Black should have placed his rook at $B$ ($F$ would not be so good, as it gives White more scope); then if White goes to $C,$ Black moves to $D;$ White to $E,$ Black to $F;$ White to $G,$ Black to $C;$ White to $H,$ Black to $I;$ and Black must win next move. If at any time Black had failed to move on to the same diagonal as White, then White could take Black's diagonal and win.


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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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