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Contest: the 9 Queens Problem. Put 9 queens and 1 or 2 pawns such that queens do not see each other. Send your solution before Feb 29, and win a book![All Comments] [Add Comment or Rating]

Hans Bodlaender wrote on Mon, Jan 12, 2004 08:43 AM UTC:

The rules are modified as follows:
<p>
Solutions that arrive from now on should not have a pawn on a centre
square, that is: d4, d5, e4, and e5.
<p>
The solutions that arrived earlier (three so far) keep running for the
prize; new entries hence have to be essentially different from the posted
one. I hope this is fair enough.
<p>
For those interested in this type of puzzles, I would be willing to run a
similar contest later (still have about two books which I can give away
for prizes.)
<p>
Please try to refrain posting hints or solutions to this puzzle before
the
deadline. In the meantime, you may try to find out <em>how many</em>
essentially different solutions there are. I've read that the original 8
queens problem has 12 essentially different solutions. (Solutions are
essentially different if you cannot get one from another by mirroring
and/or rotating the board.)