About the proof of a graph is not Hamiltonian.

Question

Given the following graph:

https://i.stack.imgur.com/fg2Q9.png

Is this graph Hamiltonian or not?

The answer is no. What I tried to prove is by using the fact that: "if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit." Following that we know, edges 11,6,4,12 must be in any circuit and the edges 17,18,19 and 9,15 must be in any circuit. Knowing that, what should I do to conclude that the graph is not Hamiltonian?

Answer 1

Edges $18$ and $19$ are incident on vertex $m$ and must be in any Hamiltonian circuit. But then none of the other edges from vertex $m$ are in the circuit, therefore the circuit can't contain any edges connecting vertices $m, g, n$ to the rest of the graph.

Answer 2

The graph has drawn has a cut-vertex. Indeed removing $m$ breaks the graph into two components. If $G$ has a Hamiltonian circuit it cannot have any cut vertices [make sure you see why for yourself].

Answer 3

Let G be any graph. If G is Hamiltonian, then κ(G) ≥ 2.

Since κ(G) = 1, with κ-set = {m}, then G must NOT be Hamiltonian.