eulerian-pathgraph theorygraph-connectivityhamiltonian-path
If a graph $G$ is Eulerian then I can show that the line graph of $G$ is Eulerian but is the converse true ?
If $G$ is Hamiltonian then I can show that the line graph of $ G$ is Hamiltonian but is it Hamiltonian connected ?
If $G$ contains a dominating circuit then the line graph of $G$ is Hamiltonian but does it contain a dominating circuit ?
The complete bipartite graph $K_{2,4}$ has an Eulerian circuit, but is non-Hamiltonian (in fact, it doesn't even contain a Hamiltonian path).
Any Hamiltonian path would alternate colors (and there's not enough blue vertices). Since every vertex has even degree, the graph has an Eulerian circuit.
Recall that an Eulerian path exists iff there are exactly zero or two odd vertices. Since $v_0$, $v_2$, $v_4$, and $v_5$ have odd degree, there is no Eulerian path in the first graph.
It is clear from inspection that the first graph admits a Hamiltonian path but no Hamiltonian cycle (since $\operatorname{deg}v_0=1$).
The other two graphs posted each have exactly two odd vertices, and so admit an Eulerian path but not an Eulerian circuit.
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