On Wed, 14 Mar 2007 19:25:46 +1100, Steven D'Aprano wrote: > Now that I've seen your _partition() function, I'm quite impressed. It is > still brute-force, but it avoids generating all 50**5 non-unique lists. By > my count, it only has to throw away 11,294 lists to generate the 2611 > unique lists it returns.
Ignore that count. I don't know where I got 11,294 from, the correct figure is 10,358. Oh, I seem to have found a bug in the _partitions() function. I believe it is missing some of the partitions. >>> list(_partitions(25, 24)) [(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2)] >>> list(_partitions(25, 25)) [] -- Steven D'Aprano -- http://mail.python.org/mailman/listinfo/python-list
