I know that if a graph is Eulerian then there exists an Eulerian cycle that contains all edges of the graph. I also know that if a graph is Hamiltonian then there exists a Hamiltonian cycle that contains all vertices of the graph.

It is easy for me to observe that a Hamiltonian graph may not be Eulerian (because may exist edges not contained in the Hamiltonian cycle). However, I'm a bit confused about the other direction. Is all Eulerian graphs also Hamiltonian?


It is not the case that every Eulerian graph is also Hamiltonian. It is required that a Hamiltonian cycle visits each vertex of the graph exactly once and that an Eulerian circuit traverses each edge exactly once without regard to how many times a given vertex is visited. Take as an example the following graph:

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It's easy to find an Eulerian circuit, but there is no Hamiltonian cycle because the center vertex is the only way one can get from the left triangle to the right.

  • $\begingroup$ Thank you. The "...without regard to how many times a given vertex is visited" was missing for me. $\endgroup$ – Pedro Alves Nov 28 '17 at 17:15

Counterexample: the complete bipartite graph $K_{2,4}$ is Eulerian but not Hamiltonian.

  • $\begingroup$ Can you clarify this? I don't see how to draw an Eulerian cycle (no repeated edges or vertex) on $K_{2,4}$. $\endgroup$ – Pedro Alves Nov 28 '17 at 17:05
  • 1
    $\begingroup$ @PedroAlves, $K_{2n,2m}$ has no odd vertices, so it has an Eulerian cycle. The statement in the question ("that contains all edges of the graph") is the correct definition. $\endgroup$ – Peter Taylor Nov 29 '17 at 12:55

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