# Eight queens problem, wondering about the non-unique solutions

I've done the code that generates all the solutions. But know I am suppose to filter out any redundant solutions based on symmetry and rotations. I have code for vertical symmetry, horizontal symmetry, rotation 90,180 and 270. The bit that remains is removing symmetry about the the diagonals. / and \ of the board. I guess I can rotate it 90 degress instead of making code for both diagonals. I want to implement the diagonal "\". I drew up some points.
From bottom part to the upper part.
(1,3) -> (6,8)
(2,1) -> (8,7)

From top part to bottom part.
(4,6)->(3,5)
(5,6)->(3,4)

I'm not sure what formulas would do this for me. And what about chess pieces on the diagonal it self, they would just stay put I guess?

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My code gives 23 "unique" answers without filtering for the diagonal ones. –  Algific Sep 11 '11 at 12:43
Maybe you could simply only count solutions where one queen is on one of the squares $a_1, a_2, a_3, a_4$? And then maybe delete all solutions which are already covered by symmetries, e.g. have a queen on $d_1$, or $c_1$ (if there isn't one on $a_4$), or $b_1$ (if there isn't one on $a_3, a_4$). –  TMM Sep 11 '11 at 13:42
I need it to work with a n*n board. –  Algific Sep 11 '11 at 13:50
$(a,b)\to(9-b,9-a){}{}{}{}{}{}$