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Jul 27 at 9:18 comment added Ihromant @fedja yes, It's according to reading like a page in a book. You "read" two incidence matrices of same size left to right, top to bottom until you find 1 in one position and 0 in second. Then first matrix is lexicographically bigger. In example above "output" is bigger due to 1 on 3rd position in "output" and 0 on 3rd position in "input".
Jul 27 at 2:42 comment added fedja Could you, please, remind us what is the lexicographic order on $m\times n$ $0-1$ matrices? Is it according to reading left to right top to bottom like a page in a book, or something else?
Jul 20 at 10:17 comment added joro I am permuting for a toy implementation. The problem can be formulated to integer program, which is faster than permuting, but still of exponential complexity.
Jul 20 at 10:08 comment added Ihromant It could be related. Still, while looked at your code I spotted that you are iterating over permutations(n). If n>10 it leads to unacceptable running time. Still, nauty runs pretty fast on non-regular graph. I assume that there is (or could be) algorithm which runs with approximately similar time.
Jul 20 at 9:55 comment added joro Is this related to Complete graph invariant based on integer programming?
Jul 20 at 9:53 comment added Ihromant @DanielWeber Yes, I understand that. Still, graph canonical form (I reimplemented nauty's approach from arxiv.org/abs/1301.1493 Practical graph isomorphism) produces not lexicographically largest incidence matrix.
Jul 20 at 2:29 comment added Daniel Weber This is similar to lexicographically smallest graph canonization, which is known to be NP-hard, but for bipartite graphs. I think there's a good chance it's NP-hard as well
S Jul 20 at 1:35 history suggested J. W. Tanner CC BY-SA 4.0
Corrected spelling in the title
Jul 19 at 23:27 review Suggested edits
S Jul 20 at 1:35
Jul 19 at 22:25 history asked Ihromant CC BY-SA 4.0