Definition

A graph is Hamilton if there exists a closed walk that visits every vertex exactly once.

Although the definition of a Hamiltonian graph is extremely similar to an Eulerian graph, it is much harder to determine whether a graph is Hamiltonian or not: doing so is an NP-complete problem.

Examples

Proposition

The Petersen graph is not Hamiltonian.

Proof

Suppose the Petersen graph was Hamiltonian, and let denote the Hamiltonian cycle, as we have drawn below:

The Petersen graph has 15 edges, and the Hamiltonian cycle uses 10 of them, so there must be 5 edges left in the Petersen graph that are not in the Hamiltonian cycle. The proof works by considering the possibilities for these 5 extra edges.

Let us consider the extra edge incident to ; the others are equivalent. Since the Petersen graph has no loops or multiple edges, this edge cannot go from to itself, or to or .

Since the Petersen graph has no triangles or 4 cycles, this edge cannot “skip” 1 or 2 vertices in the Hamiltonian cycle. More specifically, we cannot we cannot have an edge from to or , as these make triangles, as in the dashed edges below:

And we cannot have edges to or to , as these make 4 cycles:

Thus, the only possibility for these edges is that they “skip” 4 or 5 vertices (skipping more than 5 vertices is the same as skipping less vertices in the other direction). So for instance, can only be connected to or with its extra edge.

We now claim that at least one of these extra edges must “skip” 4 vertices. If not, then every extra edge would skip 5 vertices. Since each vertex has degree 3, there must be one extra edge through each of these vertices, connecting to the opposite vertex. But this configuration has many four cycles – for instance, . Since the Petersen graph does not have any 4 cycles, we see this cannot occur:

Relabeling our edges if necesary, we can make the extra edge that skips the 4 cycle be the edge . Now, consider the extra edge at . This cannot skip 5, because that would make it adjacent to , which already has its extra edge. Thus, it must skip 4, and be adjacent to either or . But since each of these are adjacent to , it would create a 4-cycle: either , or .

We have drawn the edge from to in solid red; any of the dashed edges from create a configuration not found in the Petersen graph:

Partial results on Hamiltonian graphs

Although there is no complete characterisation of Hamiltonian graphs, there are several nice sufficient conditions for a graph to be Hamiltonian;

Ore’s Theorem

Let be a simple graph with vertices such that for any two nonadjacent vertices and , we have . Then is Hamiltonian.

Proof

We will give a proof by contradiction. First, suppose that were a counterexample. Adding edges to preserves the condition on the degrees, and so we may continually add edges to until adding any more would create a Hamiltonian cycle. We will thus assume that is a counterexample with the maximal set of edges.

The benefit of this is that such a must have a non-closed path that visits every vertex exactly once. To see this, add one more edge to – the resulting graph then has a Hamiltonian cycle, which must pass through , deleting give the desired path.

Let be the path visiting every vertex of . We will complete the proof by showing there exists an so that is adjacent to and is adjacent to . This completes the proof because is a Hamiltonian path. We illustrate this below, with and :

The existence of such an follows from the pigeon hole principle. Since and are not adjacent by supposition, the sum of their degrees is at least . Two edges incident to them are accounted for , and , so there must be at least other edges incident to or . Since is simple, any such edge out of has possibilities: . Similarly, there are possibilities for an edge adjacent to . We pair these edges up into the possibilities we want, creating bins. Since there are edges we need to distribute among these bins, one of the bins must contain at least two, which is exactly what we needed to show.