# Is every Eulerian graph also Hamiltonian?

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?

Counterexample: the complete bipartite graph $K_{2,4}$ is Eulerian but not Hamiltonian.
• Can you clarify this? I don't see how to draw an Eulerian cycle (no repeated edges or vertex) on $K_{2,4}$. – Pedro Alves Nov 28 '17 at 17:05
• @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. – Peter Taylor Nov 29 '17 at 12:55