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A colleague of mine claims that there exists one, but I can't figure how an eulerian graph can be disconnected, since you have to visit all the graph vertices in the cycle...

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It can’t be disconnected. Neither can a graph with a Hamilton circuit. – Brian M. Scott Apr 11 '12 at 1:26
Maybe you should ask your colleague to exhibit the graph. – Rahul Apr 11 '12 at 1:27
It might be a trick question with a silly answer like "empty graph". Depending on the definition the empty graph might be a disconnected one, e.g. if disconnected is defined as $\forall v \in V.\ \exists x, y \in V.\ x$ and $y$ are not connected. I do know that such definition does not make sense, just giving an example. – dtldarek Apr 11 '12 at 1:40
@dtldarek, good point! There are good reasons to consider the empty graph disconnected. But I have no idea if it qualifies as Eulerian, Hamiltonian, or bipartite. – Rahul Apr 11 '12 at 2:02

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It's possible that your colleague is using a non-standard definition by saying that the disjoint union of eulerian/hamiltonian connected components is itself eulerian/hamiltonian. If this is the case, then simply considering one connected component of the graph will suffice, and so the problem reduces to showing a connected, hamiltonian, eulerian, bipartite graph. There are many such examples, the cycle on four vertices being the smallest nontrivial example.

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But the cycle of four vertices isn't a bipartite graph, isn't it ? – Pacane Apr 11 '12 at 3:31
Sure it is. ${}$ – Gerry Myerson Apr 11 '12 at 4:17
@Pacane Label the vertices in order around the cycle $A$, $B$, $C$, and $D$. The sets $\{A, C\}$ and $\{B, D\}$ are a bipartition. – Austin Mohr Apr 11 '12 at 12:53

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