[EMAIL PROTECTED] wrote the following on 02/28/2008 12:36 AM: > On Feb 27, 8:47 pm, Michael Robertson <[EMAIL PROTECTED]> wrote: > Your only casualty here is all the zeroes in (4,0,0,..). You don't > want to swap k_2 and k_3 in (4,0,0). (Is that what permutation > means?)
Correct. Though by 'permutation', I meant 'permutations with repetitions'---so the algorithm would have handled that. > >> In addition to having to generate permutations for each integer >> partition, I'd have to ignore all integer partitions which had more than >> k parts...this seemed like a bad way to go (bad as in horribly inefficient). >> >> Better ideas are appreciated. > > Running time on my recursive was K** 2* N, counting the copy, I > think. sum( 1..i )== i( i+ 1 )/ 2, O( i** 2 ). My iterative was > slower, K** 3* N, unless you save a K or N in the small length of K > early on, I think. Did anyone take this course that can tell me? Thanks again for your efforts here. This particular problem didn't appear in any course I took...certainly similar problems did. -- http://mail.python.org/mailman/listinfo/python-list
