- 2073 Magh · 2+2+2+2 marks
Euler and Hamilton circuits with conditions
Short answer
An Euler circuit traverses every edge exactly once; exists iff all vertices have even degree. A Hamilton circuit visits every vertex exactly once and returns to start; no simple necessary and sufficient condition exists, but Dirac's theorem (n≥3, all degrees ≥ n/2) is sufficient. Example: bowtie has Euler circuit, pentagon has Hamilton circuit.
Explain the Euler circuit and Hamilton circuit with example. State the necessary and the sufficient conditions for them.
Euler circuit
A closed walk that uses every edge exactly once. Example: the bowtie with edges ab, bc, ca, cd, de, ec has the Euler circuit c → a → b → c → d → e → c.
a d
|\ /|
| \ / |
| \ / |
| \ / |
| c |
| / \ |
| / \ |
| / \ |
b e (edges: ab, ac, bc, cd, ce, de)
(Bowtie: triangles abc and cde sharing vertex c.)
Conditions: a connected graph has an Euler circuit if and only if every vertex has even degree. It has an Euler path (not a circuit) if and only if exactly two vertices have odd degree.
Hamilton circuit
A cycle that visits every vertex exactly once and returns to the start. Example: in the pentagon a–b–c–d–e–a, the circuit a, b, c, d, e, a visits all five vertices.
a
/ \
e b
| |
d---c
Conditions: there is no known simple necessary and sufficient condition.
- Necessary: the graph is connected, every vertex has degree ≥ 2, and it has no cut vertex.
- Sufficient (Dirac): a simple graph with n ≥ 3 vertices in which every vertex has degree ≥ n/2 has a Hamilton circuit.
- Sufficient (Ore): if a simple graph with n ≥ 3 vertices has deg(u) + deg(v) ≥ n for every pair of non-adjacent vertices, it is Hamiltonian.
| Point | Euler circuit | Hamilton circuit |
|---|---|---|
| Visits | all edges once | all vertices once |
| Test | all degrees even (exact) | only sufficient conditions known |
| Example with only this | bowtie (not Hamiltonian) | K₄ (degrees 3, not Eulerian) |
✓ Answer checked by Claude Opus 5.5 (high effort)
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