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Dec 15, 2013 at 18:15 vote accept hbm
Dec 15, 2013 at 14:38 comment added joro This method appears to work for the other question too. Found $G$ on $12$ vertices, 3-connected, non-hamiltonian. Replacing with a triangle gave a solution to the other problem.
Dec 15, 2013 at 7:27 history answered bof CC BY-SA 3.0