Does having two edge-disjoint Hamiltonian paths imply having a Hamiltonian circuit?

Question

Suppose a simple graph G has two edge-disjoint Hamiltonian paths, can we construct a Hamiltonian circuit in G? If no, can anyone provide a counterexample?

Answer 1

Let $P_1$ and $P_2$ be paths with vertex set $V=V(P_1)=V(P_2)=\{v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8,v_9,v_{10},v_{11}\}$ and edge sets $E(P_1)=\{v_1v_2,v_2v_3,v_3v_4,v_4v_5,v_5v_6,v_6v_7,v_7v_8,v_8v_9,v_9v_{10},v_{10}v_{11}\}$ and $E(P_2)=\{v_1v_3,v_3v_5,v_5v_2,v_2v_4,v_4v_6,v_6v_8,v_8v_{10},v_{10}v_7,v_7v_9,v_9v_{11}\}.$

In the graph $G=P_1\cup P_2,\ P_1$ and $P_2$ are edge-disjoint Hamiltonian paths with the same endpoints. The graph $G$ has no Hamiltonian circuit, since $G-v_6$ is disconnected.

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