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 Mutually disagreeing queens. (Posted on 2006-05-19)
In how many ways can two queens be placed on a chessboard such that they are mutually attacking each other?

Show how to solve this problem quickly with only pencil and paper (not even a calculator).

 No Solution Yet Submitted by Jer Rating: 4.0000 (1 votes)

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 solution | Comment 2 of 5 |

Each queen is capable of orthogonal (rook-type) moves and diagonal (bishop-type) moves.  From any given position on the board, there are 14 places the queen might go via a rook-type move.  So if this were a question asked about rooks, the answer would be easy: 64 possible positions for the first rook, with 14 positions each for the second rook. Then 64 * 14 = 896; but then we'd need to divide by 2 as the identities of the rooks don't count (which one was first and which second), so that would be 448.

That's for rooks.  But it also counts for the queens' orthogonal relative positions.

The number of diagonal (bishop-type) moves available varies by location. From the outer edges of the board (28 squares like this) there are 7 bishop moves available from any given square. From the next-inner ring of squares (20 of these), there are 9 bishop moves available from each square. Going further inward, the next ring (12 squares) has 11 bishop moves available from each square.  From the 4 center squares, there are 13 bishop moves available from each.  That's 28*7 + 20*9 + 12*11 + 4*13.  Again, after we do that we have to divide the total by two as we don't care about the identities of the bishops or queens.  One could do the arithmetic with pencil and paper, but hey, the calculator's here.  The total comes out to 560, and when divided by two, that's 280.

So, adding the rook-type and bishop-type pairs of positions, that's 728.

 Posted by Charlie on 2006-05-19 10:24:32

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