If a graph (G) has no loops or parallel edges, and if the number of vertices…

Concept

A stated condition on a simple graph G (n ≥ 3, no loops or parallel edges) is a genuine sufficient condition for G to be Hamiltonian only if EVERY graph satisfying it has a Hamiltonian cycle — one graph that meets the condition yet has no Hamiltonian cycle is enough to disprove it, no matter how many other graphs meeting the same condition ARE Hamiltonian. Two classical theorems fix the real thresholds here: Dirac's theorem (deg(v) ≥ n/2 for every vertex v guarantees a Hamiltonian cycle) and Ore's theorem (deg(v) + deg(w) ≥ n for every pair of non-adjacent vertices v, w guarantees a Hamiltonian cycle).

Application

• Condition (ii) — deg(v) + deg(w) ≥ n for every non-adjacent pair — is exactly Ore's theorem's hypothesis, so it is sufficient: every simple graph on n ≥ 3 vertices meeting it has a Hamiltonian cycle.

• Condition (i) only requires deg(v) ≥ n/3, which is weaker than the n/2 bound Dirac's theorem actually needs, so it does NOT guarantee a Hamiltonian cycle. Counterexample: for n = 6, two disjoint triangles give every vertex degree 2 = n/3, but the graph is disconnected and has no Hamiltonian cycle. This does not conflict with a hexagon (the 6-cycle C₆): a hexagon also has every vertex of degree 2 = n/3 and IS Hamiltonian (it is itself the Hamiltonian cycle) — but a sufficiency claim must hold for every graph meeting the condition, not just for the ones that happen to work. Since the two-disjoint-triangles graph meets condition (i) yet fails, (i) alone still cannot be relied on, even though the hexagon example meets it and succeeds.

• Condition (iii) requires E(G) ≥ (1/3)(n − 1)(n − 2) + 2, a smaller (looser) bound than the edge count actually known to force Hamiltonicity. Counterexample: the complete graph on n − 1 vertices together with one isolated vertex has (n − 1)(n − 2)/2 edges — at or above (iii)'s bound for n ≥ 5 — yet the isolated vertex means it has no Hamiltonian cycle.

Cross-check

Dirac's theorem is in fact a special case of Ore's theorem (if every vertex has degree ≥ n/2, any non-adjacent pair automatically has degree sum ≥ n), so (ii) sits exactly on the strongest degree-based guarantee among the three options, while (i) and (iii) both use bounds weaker than their respective real thresholds and both admit the explicit counterexamples above.

So among (i), (ii), (iii), only condition (ii) guarantees that G is Hamiltonian — the correct answer is (ii) only.