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Tutte graph is a counterexample for the Tait's conjecture stating that all cubic graphs are Hamiltonian. For the non-hamiltonian graphs - is it true that all vertices of any such graph can be covered by a set of cycles?

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    $\begingroup$ Each vertex is on a cycle, so the answer to your question would be yes. But this is probably not what you mean. Please clarify. $\endgroup$ – Leen Droogendijk May 21 '14 at 6:46
  • $\begingroup$ Tait's conjecture says that every planar 3-connected cubic graph is Hamiltonian. In every at least 2-connected graph, each vertex is on a cycle, and so yes, the graph can be covered with a set of cycles. If you don't restrict to at least 2-connected graphs, there are cubic planar graphs with vertices not participating in any cycles. $\endgroup$ – Perry Elliott-Iverson May 21 '14 at 12:51
  • $\begingroup$ Sorry, I meant to say "covered by a set of mutually non-overlapping cycles". $\endgroup$ – Leonid May 21 '14 at 13:54
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Yes, each cubic polyhedral graph can be covered by a set of mutually disjoint cycles. This is a trivial consequence of Petersen's theorem that states that every bridgeless, cubic graph has a 1-factor. Since a polyhedral graph is bridgeless, it has a 1-factor. Removing this 1-factor leaves a regular graph of degree 2, which is a union of disjoint cycles.

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