I googled it and found:
If G = (V,E) has n ≥ 3 vertices and every vertex has degree ≥ n/2 then G has a Hamilton circuit.
But my question is if the degree of each vertex is 2 or more, then the graph can also have a Hamiltonian Cycle.
example:-
1---->2
2---->3
3---->4
4---->5
5---->6
6---->7
7---->8
8---->1
suppose the graph is undirected...
in the above example degree of each vertex is 2, so the graph will have a Hamiltonian cycle.
Then, what does the above-quoted text make sense?
Attach 2 triangular graphs by a single vertex:
The vertices on the sides have degrees of 2-2-2-2, the one in the middle has a degree of 4.
n=5
in this case and vertices would need a degree>=2.5
(so, 3 in practice) in order to contain a Hamiltonian circle for sure.And the for sure is the important part here: while you may find graphs which do not fulfill the
>=n/2
requirement and still contain a Hamiltonian circle, you can't find a graph which fulfills the requirement and doesn't contain a Hamiltonian circle.