|Perl: the Markov chain saw|
(e.g., for $N = 11: [ a1, b3, c5, d7, e9, f11, g2, h4, i6, j8, k10 ])
This is equivalent to starting in the bottom-left corner of the board and moving right one square and up two squares (wrapping when you hit the edge) N-1 times.
If the regex picks the first space not-already-capturable in each column (From brief inspection, it appears to do so -- It finds an equivalent solution for odd-N), this is the first solution it will find. In the even-N case, this process will not leave any not-already-capturable squares in the last column on the first pass, so it must then backtrack.
Update: Whoa there, Ben. I spoke way too soon. The above solution only applies when N is odd AND not divisible by 3. (So, for N = (1,5) mod 6)
More update: I was wrong about most of the analysis, too. This won't be the first solution found.
In reply to Re: Re: Re: The N-queens problem using pure regexes