Is a 2-edge-connected graph necessarily Hamiltonian? I would think you could write any 3-edge-connected component of the graph as a union of cycles and construct a Hamiltonian cycle across the cut set that way. Anybody have a counter example? I'm having trouble convincing myself this is true.
2026-03-27 23:31:20.1774654280
2-edge-connectivity and Hamiltonian Cycles
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Consider the complete bipartite graph $K_{n,n+1}$. This is certainly 2-edge-connected (in fact, I believe it is $n$-edge-connected) but it does not have a Hamiltonian cycle by a parity argument: any cycle must alternate between the two parts of the bipartition, but one of the parts has more vertices than the other.