> The obvious answer: if the vertex labels are not all distinct They are all distinct.
> The answer if you care about fast computation: Define == as strict identity > of pair(Hasse diagram, labels), not isomorphism up to permutation. Provide a > separate is_isomorphic() method for the latter. I am not talking of isomorphism test. Two posets defined on the same sets of points, and such that x<y in one iif x<y in the other must be declared equal by Sage. Nathann -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-devel@googlegroups.com. Visit this group at http://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
