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...
Asked
Active
Viewed 249 times
1
-
1It can’t be disconnected. Neither can a graph with a Hamilton circuit. – Brian M. Scott Apr 11 '12 at 01:26
-
Maybe you should ask your colleague to exhibit the graph. – Apr 11 '12 at 01:27
-
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 01:40
-
@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. – Apr 11 '12 at 02:02
1 Answers
2
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.
Austin Mohr
- 25,662
-
-
-
@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