# 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 $90^\circ$ 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)$I'm not sure what formulas would do this for me. And what about chess pieces on the diagonal it self, they would just stay put I guess? - My code gives 23 "unique" answers without filtering for the diagonal ones. – Algific Sep 11 '11 at 12:43 I need it to work with a n*n board. – Algific Sep 11 '11 at 13:50 It would be better if this question describes the "eight queens problem" in more detail, or at least contains a link to a description of the problem elsewhere. At least one reason for this is that readers of this question want to know what you're asking about. – Quinn Culver Jun 20 '15 at 6:36 ## 1 Answer$(a,b)\to(9-b,9-a){}{}{}{}{}{}\$

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Now I just feel stupid haha. How would it be if the diagonal went the other way? "/"? – Algific Sep 11 '11 at 13:50
@Algific: for the diagonal "/" it is (a, b) -> (b, a). – Jiri Sep 11 '11 at 14:17