Questions: Hamiltonian Cycles: Dirac and Ore Conditions

Question 1

A graph has 10 vertices and every vertex has degree 4. A student concludes: 'Dirac's threshold is n/2 = 5, and every vertex has degree only 4, so the graph is not Hamiltonian.' Is this reasoning correct?

• A. Yes — the Dirac degree condition is not met, so the graph cannot have a Hamiltonian cycle
• B. No — Dirac's condition is sufficient but not necessary; the graph might still be Hamiltonian even though the condition fails
• C. No — Dirac's threshold for 10 vertices is actually 4, not 5, so the condition is satisfied
• D. Yes — degree 4 in a 10-vertex graph is always too low for any Hamiltonian cycle to exist

Answer: (B)

This is the central misconception. Dirac's theorem is a one-way guarantee: if the degree condition is satisfied, a Hamiltonian cycle must exist. But failing the condition tells you nothing — the graph might still be Hamiltonian. A simple cycle (Cₙ) on 10 vertices has every vertex with degree 2 and is trivially Hamiltonian. The student incorrectly treats a sufficient condition as if it were necessary.

Question 2

Ore's theorem is strictly more broadly applicable than Dirac's theorem because:

• A. Ore requires every vertex to have degree ≥ n/2 individually, while Dirac only requires non-adjacent pairs to have high degree
• B. Every graph satisfying Dirac also satisfies Ore, but not every graph satisfying Ore satisfies Dirac
• C. Ore's condition applies only to bipartite graphs, which Dirac cannot handle
• D. Ore uses algebraic conditions while Dirac uses purely combinatorial ones, making Ore more general

Answer: (B)

Ore's condition requires that for every non-adjacent pair u, v: deg(u) + deg(v) ≥ n. If every vertex individually has degree ≥ n/2 (Dirac), then any non-adjacent pair sums to at least n (satisfying Ore). So Dirac implies Ore — every Dirac graph is also an Ore graph. But Ore can hold when individual vertices have low degree, as long as every low-degree vertex is adjacent to every other low-degree vertex, so some Ore graphs are not Dirac graphs.

Question 3

A cycle graph Cₙ on n ≥ 3 vertices is Hamiltonian despite having every vertex with degree 2, far below the Dirac threshold of n/2.

• T. True
• F. False

Answer: True

The n-cycle is a Hamiltonian cycle by construction — it visits every vertex exactly once and returns to the start. Yet every vertex has degree 2, which is far below Dirac's n/2 threshold for n ≥ 5. This is one of the simplest examples showing that Dirac and Ore conditions are not necessary for Hamiltonicity.

Question 4

If a graph fails Ore's condition for some non-adjacent pair (u, v) where deg(u) + deg(v) < n, then the graph can seldom have a Hamiltonian cycle.

• T. True
• F. False

Answer: False

Ore's theorem is a sufficient condition, not necessary. Failing it means only that Ore's guarantee doesn't apply — not that no Hamiltonian cycle exists. A graph can fail Ore's condition for some pair and still be Hamiltonian by some other structural reason. The converse of a sufficient condition is not generally true.

Question 5

Why is the fact that Dirac and Ore conditions are sufficient but not necessary important for understanding the computational difficulty of detecting Hamiltonian cycles?

Model answer: If these conditions were necessary and sufficient, you could check Hamiltonicity in polynomial time: compute degrees, verify the condition, done. But because they are only sufficient, they identify a restricted class of graphs (dense enough ones) where the hard problem becomes easy — but leave the general case untouched. Deciding Hamiltonicity in an arbitrary graph is NP-hard: no polynomial-time algorithm is known. Dirac and Ore carve out well-structured regions where the guarantee holds, but failing those conditions leaves you with no general shortcut. The contrast with Eulerian circuits — which have a clean necessary-and-sufficient degree condition and are easy to detect — makes the hardness of the Hamiltonian case especially striking.

This distinction between sufficient-only and necessary-and-sufficient conditions directly maps to algorithmic difficulty. A sufficient condition is a fast path for a subset of cases; a necessary-and-sufficient condition is a complete decision procedure. The absence of a necessary-and-sufficient condition for Hamiltonicity is not an accident — it reflects the NP-hardness of the problem.