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Mutually disagreeing queens. (Posted on 2006-05-19) Difficulty: 2 of 5
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 Queen v Queen - Solution "depends" | Comment 5 of 6 |

Without considering rotations, interchange of colors, etc., there are 1456 ways:

For each of 64 squares, each attacks 7 others in the same row, and 7 others in the same column (64 * 14 = 896).

Considering only diagonals, depending on ring:

Each of the 28 squares forming the outermost perimeter attacks 7 other squares (28 * 7 = 196)

Each of the 20 squares one removed from the outer ring attacks 9 other squares (20 * 9 = 180)

Each of the 12 squares two removed from the outer ring atacks 11 other squares (12 * 11 = 132)

Each of the 4 central squares attacks 13 other squares (4 * 13 = 52)

Adding all these gives 1456.  If you consider an exact interchange of white/black as the "same" way, this would divide by 2 (giving 728).  If  you considered the geometric relation of the two squares the "same" with rotations of the board, this would divide again by 4 giving 182.  Other topological equivalences could be considered.  Without a strict definition of a "way"  any of these are possible answers.

 


  Posted by badger on 2007-10-11 14:53:40
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