Solution

(related to Problem: The Rookery)

The answer involves the little point that in the final position the numbered rooks must be in numerical order in the direction contrary to that in which they appear in the original diagram, otherwise, it cannot be solved. Play the rooks in the following order of their numbers. As there is never more than one square to which a rook can move (except on the final move), the notation is obvious — $5, 6, 7, 5, 6,$ $4, 3, 6, 4, 7, 5,$ $4, 7, 3, 6, 7, 3,$ $5, 4, 3, 1, 8,$ $3, 4, 5, 6,$ $7, 1, 8, 2, 1,$ and rook takes bishop, checkmate. These are the fewest possible moves — thirty-two. The Black king's moves are all forced, and need not be given.

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References

Project Gutenberg

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

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