Proof of Dirac's theorem on Hamiltonian cycles from Ore's theorem
Builds on GraphDegree of a vertexHamiltonian graphOre's theorem
Proof
We deduce the theorem from Ore's theorem.
- Let be a finite simple graph with vertices, and suppose that for every vertex , where is the degree.
- Let and be two distinct vertices of that are not adjacent.
- By Step 1, .
- So satisfies the hypothesis of Ore's theorem: it has vertices, and every pair of distinct nonadjacent vertices has degree sum at least .
- By Ore's theorem, is Hamiltonian.
Idea
Dirac's condition bounds each degree from below by . Adding two such bounds gives Ore's condition on pairs of nonadjacent vertices. So Dirac's theorem is the special case of Ore's theorem where every vertex is "rich" on its own.
Reading path
To understand this page from the ground up, read these 14 pages in order. The foundations come first. The path takes the proof Proof of Ore's theorem.
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- Complete graph example
- Graph definition
- Neighbourhood definition
- Walk definition
- Degree of a vertex definition
- Path definition
- Trail definition
- Circuit definition
- Cycle definition
- Hamiltonian cycle definition
- Hamiltonian graph definition
- Proof of Ore's theorem proof
- Ore's theorem statement
- Proof of Dirac's theorem on Hamiltonian cycles from Ore's theorem proof
Where this leads
1 page builds on this one.
Further reading
- Web https://en.wikipedia.org/wiki/Ore's_theorem, Related results
About this page
| Tag | X8WAMEEJ4D |
|---|---|
| URI | m2://X8WAMEEJ4D |
| Kind | proof |
| Writing | W0 stub |
| MSC | 05C45 |
| Proves | Dirac's theorem on Hamiltonian cycles |
| Uses | Graph, Degree of a vertex, Hamiltonian graph, Ore's theorem |
| Changed | 4 November 2026 |
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