0
$\begingroup$

If $G = (V,E)$ is a finite, connected, simple, undirected graph, is there a Hamiltonian path in the line graph $L(G)$ of $G$?

$\endgroup$

1 Answer 1

5
$\begingroup$

No. Consider the graph $G = (V, E)$ with $$V = \{0,1,2,3,4,5,6\}\quad \text{and} \quad E = \{\{0,1\}, \{1, 2\}, \{0,3\}, \{3, 4\}, \{0,5\}, \{5, 6\}\}.$$ Note that its line-graph has three vertices of degree 1. Therefore $L(G)$ has no Hamiltonian path.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .