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Mar 7 at 23:00 comment added David Roberson You could ask about asymmetric graphs (graphs whose only automorphism is the identity) instead of non-regular graphs. I think this might be an open problem.
Mar 7 at 12:44 comment added Eauriel That makes sense, thank you for a clarification!
Mar 7 at 12:02 comment added Gordon Royle If you could determine if two non-regular graphs were isomorphic in polynomial time, then you could do the same for regular graphs - stick a pendant edge onto a vertex of one graph, then do the same for each vertex in turn of the second graph. Then after $n$ runs of the non-regular algorithm you would have an answer.
S Mar 7 at 11:27 review First questions
Mar 7 at 13:17
S Mar 7 at 11:27 history asked Eauriel CC BY-SA 4.0