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On a 15x15 chessboard there are 15 rooks that do not attack each other (via ordinary rook moves). Then each of the rooks makes one move like that of a knight.
Is it possible that after all this is done, the 15 rooks still do not attack each other (via ordinary rook moves)?
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asked Feb 23, 2016 at 20:05
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Since there is (initially and finally) a rook in each row and column, we can say the sum of all rooks' X and Y positions must be equal to 2 $\times$ (15 + 14 + ... + 1) = 120.
A knight's move will increment the rook's (X + Y) by +3, +1, -1, or -3, all of which are odd numbers.
The sum of 15 odd numbers (15 knight moves) is an odd number.
Since we are performing 15 moves, we are adding an odd number to the sum of the rooks' X and Y positions.
The final sum of the rooks' X and Y positions cannot be equal to the initial sum since performing 15 knight moves adds a non-zero number to that sum.
Since the initial sum is 120, the final sum cannot be 120 and therefore can't be a legal (non-attacking) position.
This proof can be extended to any $N\times N$ board where $N$ is odd.
answered Feb 23, 2016 at 20:47
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Yes, in at least three cases that I can think of:
The board wraps itself - i.e. from the top row, move 1 up, and you are now on the bottom row. In that case, all of the rooks make an identical move, and they are still safe.
You said all the rooks make one move. You didn't say, that they make ONLY one move. So, the rooks simple move out, then back to their original position.
The rooks move like knights. Presumably they also take like knights. In that case, I think the solution is obvious ;).
answered Feb 24, 2016 at 12:05