A graph G with number of vertices greater and equal than three i.e. (n≥3) is a…

Correct criterion: For a simple graph with n ≥ 3 vertices, if every vertex has degree at least n/2, then the graph is Hamiltonian (Dirac's theorem).

Proof sketch:

• Assume every vertex has degree ≥ n/2. Let C be a longest cycle in the graph.

• If C contains all vertices, it is a Hamiltonian cycle and we are done. Otherwise pick a vertex x not on C.

• Vertex x has at least n/2 neighbors on C. These neighbors divide C into many segments, so two neighbors of x on C are not consecutive along C. By connecting x into C between such nonconsecutive neighbors, one can form a longer cycle that includes x, contradicting the maximality of C.

• This contradiction implies C must include all vertices; hence the graph is Hamiltonian.

Remarks: The condition is sufficient but not necessary. Also note the maximum possible degree in an n-vertex simple graph is n−1, so suggestions like degree equal to n or multiples of n are impossible.