# Black queens on $n\times n$ board

I conjecture that the maximum number of black queens -- queens placed on black squares -- not threatening each other that can be placed on an $$n\times n$$ chessboard cannot exceed $$\frac{2n}{3}$$. I have not programmed this question, but have a 'lower half triangle' proof for $$n=8$$ (reducible to one case) and it seems that there is a similar exhaustive proof for $$n=12$$ (involving $$79$$ cases). A better approach is needed for (dis)proving this conjecture, would first like to know whether this is a known result.

Note: if this conjecture holds, it implies at once that the ratio of the numbers $$B, W$$ of black and white queens on an n x n chessboard (with $$B + W = n$$) is between $$1/2$$ and $$2$$. (On a $$8\times 8$$ chessboard always $$B = W = 4$$, on a $$12\times 12$$ chessboard a 'knight placement' easily yields an example of $$B = 8, W = 4$$.)

Your conjecture is false. Consider $$n=25$$. You claim there can be at most $$\frac{2n}{3}\approx 16.6667$$ black queens. Here is a configuration with $$17$$ black queens.
• The first counterexample is at $n=10$: Seven queens Mar 6 at 11:53
• @DanielMathias I am assuming the top left corner is black. Is there a counter example for $n<25$? Mar 6 at 12:07
• For even $n$, the board is symmetrical: One major diagonal is black. Just flip the solution in the link. Mar 6 at 12:10
• Counterexamples exist for all $n>9$. Eight queens on 11x11 (White diagonals, the solution can be shifted vertically for black diagonals.) Mar 6 at 12:20