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The maximal number of knights of a single color that can be placed on a chessboard without any knight being one knights-move away from any other is 32; such a position may easily be achieved by placing all 32 knights on squares of the same color.

If one uses both black pieces and white pieces, it's possible to place 24 knights of each color (48 total) on the board without any knight being attacked by a knight of the opposite color. One simple approach is to place 24 white knights on the first three ranks, and 24 black knights on the last three. That position is but one of many, which can accommodate 24 knights of each color.

Is it possible to either place 24 knights of one color and more than 24 of the other or prove the impossibility of doing so? Note that if one may arbitrarily choose the number of knights of each color, one could easily place 62 knights (61 white and one black) or, for that matter, 64 (all white). To be of interest, I think it is possible to rank solutions by the number of pieces in the "shorter" army and then by the number of pieces in the other.

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  • $\begingroup$ I think the answer is actually proving that its not possible more than actually placing the pieces $\endgroup$ – skv Oct 16 '14 at 4:16
  • $\begingroup$ @skv, indeed. it's not possible $\endgroup$ – Rafe Oct 16 '14 at 5:15
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Ok let me stick my neck out, I would love to be proved wrong on this, but there are only a few ways (3 unique ways) 24 Knights of each side can be placed on the board i.e. without attacking any of the other coloured pieces

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Other combinations of swapping colours and rotating the boards could be achieved.

And even theoretically 48 squares are occupied by knights, leaving just 16, given the fact that a knight in the edge attacks 4 squares and in the center attacks 8, it is logical to understand that 2 and 4 of these have to fall in "enemy" territory so to say. so even ensuring an overlap of 2 knights per square, we are talking of 8 squares per colour being "attacked" over and above the squares occupied by their own forces, for both sides this is 16 squares.

Hence it is not possible to place more than 24 knights per side

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  • $\begingroup$ You are right, but I meant those in the "middle corner" as in edge :) I will edit $\endgroup$ – skv Oct 16 '14 at 8:51
  • $\begingroup$ There is at least one other possibility (See my edit) $\endgroup$ – M.Herzkamp Oct 16 '14 at 12:55
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    $\begingroup$ @M.Herzkamp appreciate it and accepted the edit, but I think its your answer :), I mean I did not come up with that thought $\endgroup$ – skv Oct 16 '14 at 12:59
  • $\begingroup$ It does not improve the score :( So it just aims to complete your answer. And now it looks ugly ;) $\endgroup$ – M.Herzkamp Oct 16 '14 at 13:07
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    $\begingroup$ Oh, I just realized, this can be rotated as well, so you have four symmetric solutions like this. $\endgroup$ – M.Herzkamp Oct 16 '14 at 13:14

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