In article <[EMAIL PROTECTED]>, Talin <[EMAIL PROTECTED]> wrote:
> I'm sure I am not the first person to do this, but I wanted to share > this: a generator which returns all permutations of a list: > > def permute( lst ): > if len( lst ) == 1: > yield lst > else: > head = lst[:1] > for x in permute( lst[1:] ): > yield head + x > yield x + head > return You're right that you're not the first person to do this: Many others have also posted incorrect permutation generators. Have you tried your code on some simple test cases? list(permute([1, 2, 3])) ==> [[1, 2, 3], [2, 3, 1], [1, 3, 2], [3, 2, 1]] Notably absent from this list are [2, 1, 3] and [2, 3, 1]. The problem gets worse with longer lists. The basic problem is that x needs to be able to occur in ALL positions, not just the beginning and the end. Cheers, -M -- Michael J. Fromberger | Lecturer, Dept. of Computer Science http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA -- http://mail.python.org/mailman/listinfo/python-list
