Revision history [back]

Thank you for your help, I wasn't able to find the documentation for is_isomorphic yesterday, now I finally got it. In fact, I don't even need to do the edge labeling, when I do WW = BipartiteGraph(Z,weighted=True), the numbers in the matrix are treated as weights, so I can get the result I want by just doing WW.is_isomorphic(XX,edge_labels=True).

In fact, I am interested in a more general isomorphism test. For example, in X = Matrix([(1,1,2,2),(1,1,2,3),(1,2,2,1)]), I not only want to be able to treat swapping rows and columns as equivalent (which equivalent to change the labeling of nodes on the left and right of the biparitite graph separately), I can also do bijections on each rows. So for row one of X, (1,1,2,2) can be replaced with (2,2,1,1), for row two of X, (1,1,2,3) can be replaced with (1,1,3,2) (2,2,3,1) (2,2,1,3) (3,3,1,2) (3,3,2,1). I am wondering if there exist algorithms to do that directly? Right now the only way I can think of is to first list all combinations of bijections, in this example. there are 262 = 24 equivalent cases, then for each of them, I do a isomorphism test with the original matrix, if at least one graph out of 24 is isomorphism with the original graph, I consider them equivalent. However, this approach will become intractable even the size of the matrix is not that big.

Thank you for your help, I wasn't able to find the documentation for is_isomorphic yesterday, now I finally got it. In fact, I don't even need to do the edge labeling, when I do WW = BipartiteGraph(Z,weighted=True), the numbers in the matrix are treated as weights, so I can get the result I want by just doing WW.is_isomorphic(XX,edge_labels=True).

In fact, I am interested in a more general isomorphism test. For example, in X = Matrix([(1,1,2,2),(1,1,2,3),(1,2,2,1)]), I not only want to be able to treat swapping rows and columns as equivalent (which equivalent to change the labeling of nodes on the left and right of the biparitite graph separately), I can also do bijections on each rows. So for row one of X, (1,1,2,2) can be replaced with (2,2,1,1), for row two of X, (1,1,2,3) can be replaced with (1,1,3,2) (2,2,3,1) (2,2,1,3) (3,3,1,2) (3,3,2,1). I am wondering if there exist algorithms to do that directly? Right now the only way I can think of is to first list all combinations of bijections, in this example. there are 262 * 6 * 2 = 24 equivalent cases, then for each of them, I do a isomorphism test with the original matrix, if at least one graph out of 24 is isomorphism with the original graph, I consider them equivalent. However, this approach will become intractable even the size of the matrix is not that big.