The easiest way is probably to store your matrices in a some sort of "normal form": permute the rows and columns so that one of the smallest entries in your matrix is in the (1,1) position and then permute rows 2,3, ... so that the entries in column one are weakly increasing from top to bottom. Now permute columns 2,3,... so that the smallest possible entry is in position (2,2) subject to the constraint that the entries in column one do not change. Continue in a suitable fashion. Then two matrices will be permutation equivalent if and only if they have the same normal form.
For your example matrices below the normal form would be [[0,1,1],[1,0,1],[1,1,1]]. For 0-1 matrices, however, I wouldn't be surprised if the sets of row and column sums determine the orbit uniquely. Andrew On Friday, 28 February 2014 06:25:24 UTC+11, 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 sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/groups/opt_out.
