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Aug 15, 2021 at 22:06 comment added Algernon I have assumed that $G$ is simple, but I believe the same holds if you allow multiple edges. The proof goes through with minor adaptations (the induction starts at $n=1$, etc.)
Aug 15, 2021 at 20:11 comment added Sanket Biswas Thanks for this. I have just one question, did you assume $G$ to be a simple bipartite graph, or did you assume it to be any bipartite graph?
Aug 6, 2021 at 17:39 history answered Algernon CC BY-SA 4.0