# If a Graph have Eulerian Cycle and Hamiltonian Path, does it mean that the Graph have Hamiltonian Cycle?

Let $$G$$ be a Graph that have Eulerian Cycle and Hamiltonian Path, does it mean that $$G$$ must have Hamiltonian Cycle?

I tried to find a counter example but I always got stuck since when I notice an Eulerian Cycle I always find a Hamiltonian Cycle. I know that if $$G$$ have Eulerian Cycle then all the degrees are even, and we visited all the edges exactly one but it doesn't necessary means that we have visited all the vertices exactly once.

I will be happy if someone can help by getting a counter example or a proof if its true.

Take two cycles sharing one vertex. The resulting graph looks like a bowtie (at least for two $$3$$-cycles – MathWorld calls it the butterfly graph and it has $$5$$ vertices) and clearly has a Hamiltonian path and Eulerian cycle, but no Hamiltonian cycle.
• @mm8511 The new graph has a cut-vertex; that vertex has degree $4$ and all other ones have degree $2$. Having a cut-vertex is by itself enough to preclude a Hamiltonian cycle. May 26, 2021 at 15:27