Let be a graph with vertices. We denote by the degree of a vertex in, i.e. the number of incident edges in to. Then, Ore's theorem states that if then is Hamiltonian.
Proof
It is equivalent to show that every non-Hamiltonian graph does not obey condition '. Accordingly, let be a graph on vertices that is not Hamiltonian, and let be formed from by adding edges one at a time that do not create a Hamiltonian cycle, until no more edges can be added. Let and be any two non-adjacent vertices in. Then adding edge to would create at least one new Hamiltonian cycle, and the edges other than in such a cycle must form a Hamiltonian path in with and. For each index in the range, consider the two possible edges in from to and from to. At most one of these two edges can be present in, for otherwise the cycle would be a Hamiltonian cycle. Thus, the total number of edges incident to either or is at most equal to the number of choices of, which is. Therefore, does not obey property ', which requires that this total number of edges be greater than or equal to. Since the vertex degrees in are at most equal to the degrees in, it follows that also does not obey property .
Algorithm
describes the following simple algorithm for constructing a Hamiltonian cycle in a graph meeting Ore's condition.
Arrange the vertices arbitrarily into a cycle, ignoring adjacencies in the graph.
While the cycle contains two consecutive vertices vi and vi + 1 that are not adjacent in the graph, perform the following two steps:
*Search for an index j such that the four vertices vi, vi + 1, vj, and vj + 1 are all distinct and such that the graph contains edges from vi to vj and from vj + 1 to vi + 1
*Reverse the part of the cycle between vi + 1 and vj.
Each step increases the number of consecutive pairs in the cycle that are adjacent in the graph, by one or two pairs, so the outer loop can only happen at most n times before the algorithm terminates, where n is the number of vertices in the given graph. By an argument similar to the one in the proof of the theorem, the desired index j must exist, or else the nonadjacent vertices vi and vi + 1 would have too small a total degree. Finding i and j, and reversing part of the cycle, can all be accomplished in time O. Therefore, the total time for the algorithm is O, matching the number of edges in the input graph.
Related results
Ore's theorem is a generalization of Dirac's theorem that, when each vertex has degree at least, the graph is Hamiltonian. For, if a graph meets Dirac's condition, then clearly each pair of vertices has degrees adding to at least. In turn Ore's theorem is generalized by the Bondy–Chvátal theorem. One may define a closure operation on a graph in which, whenever two nonadjacent vertices have degrees adding to at least, one adds an edge connecting them; if a graph meets the conditions of Ore's theorem, its closure is a complete graph. The Bondy–Chvátal theorem states that a graph is Hamiltonian if and only if its closure is Hamiltonian; since the complete graph is Hamiltonian, Ore's theorem is an immediate consequence. found a version of Ore's theorem that applies to directed graphs. Suppose a digraph G has the property that, for every two vertices u and v, either there is an edge from u to v or the outdegree of uplus the indegree of v equals or exceeds the number of vertices in G. Then, according to Woodall's theorem, G contains a directed Hamiltonian cycle. Ore's theorem may be obtained from Woodall by replacing every edge in a given undirected graph by a pair of directed edges. A closely related theorem by states that an n-vertex strongly connected digraph with the property that, for every two nonadjacent vertices u and v, the total number of edges incident to u or v is at least 2n − 1 must be Hamiltonian. Ore's theorem may also be strengthened to give a stronger conclusion than Hamiltonicity as a consequence of the degree condition in the theorem. Specifically, every graph satisfying the conditions of Ore's theorem is either a regularcomplete bipartite graph or is pancyclic.
