Its not clear what you mean by "permutation", row? column? both? In any case, sorting is faster than trying all permutations.
On Thursday, February 27, 2014 8:25:24 PM UTC+1, Keivan Monfared wrote: > > In a problem that I am solving I generate matrices and put them in a list. > The list gets long as if a matrix is a solution to my problem, all of its > permutations are, so I want to check to see if a matrix that I find is > permutation of another matrix which is already in the list. Are there any > fast ways to do this? Anything faster than checking all permutations of > that matrix? > > If it helps, all my matrices are 0-1 matrices. > -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/groups/opt_out.
