Ore's Theorem

Theorem
Let $G = \struct {V, E}$ be a simple graph of order $n \ge 3$.

Let $G$ be an Ore graph, that is:


 * For each pair of non-adjacent vertices $u, v \in V$:
 * $(1): \quad \deg u + \deg v \ge n$

Then $G$ is Hamiltonian.

Proof
From Ore Graph is Connected it is not necessary to demonstrate that $G$ is connected.

it were possible to construct a graph that fulfils condition $(1)$ which is not Hamiltonian.

For a given $n \ge 3$, let $G$ be the graph with the most possible edges such that $G$ is non-Hamiltonian which satisfies $(1)$.

Although it does not contain a Hamilton cycle, $G$ has to contain a Hamiltonian path $\tuple {v_1, v_2, \ldots, v_n}$.

Otherwise it would be possible to add further edges to $G$ without making $G$ Hamiltonian.

Since $G$ is not Hamiltonian, $v_1$ is not adjacent to $v_n$, otherwise $\tuple {v_1, v_2, \ldots, v_n, v_1}$ would be a Hamilton cycle.

By $(1)$, we have that:
 * $\deg v_1 + \deg v_n \ge n$

By the Pigeonhole Principle, for some $i$ such that $2 \le i \le n - 1$, $v_i$ is adjacent to $v_1$, and $v_{i - 1}$ is adjacent to $v_n$.

But the cycle $\tuple {v_1, v_2, \ldots, v_{i - 1}, v_n, v_{n - 1}, \ldots, v_i, v_1}$ is then a Hamilton cycle.

So $G$ is Hamiltonian after all.

Also see

 * Hamiltonian Graph is not necessarily Ore Graph, demonstrating that the converse of Ore's Theorem does not hold.


 * Dirac's Theorem