Definition

A graph $\Gamma$ 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 $v_0-v_1-v_2-v_3-\cdots-v_{9}-v_0$ 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 $v_0$; the others are equivalent. Since the Petersen graph has no loops or multiple edges, this edge cannot go from $v_0$ to itself, or to $v_1$ or $v_9$.

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 $v_0$ to $v_2$ or $v_8$, as these make triangles, as in the dashed edges below:

And we cannot have edges $v_0$ to $v_3$ or to $v_7$, 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, $v_0$ can only be connected to $v_4, v_5$ or $v_6$ 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, $v_0-v_1-v_6-v_5-v_0$. 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 $v_0-v_4$. Now, consider the extra edge at $v_5$. This cannot skip 5, because that would make it adjacent to $v_0$, which already has its extra edge. Thus, it must skip 4, and be adjacent to either $v_9$ or $v_1$. But since each of these are adjacent to $v_0$, it would create a 4-cycle: either $v_0-v_4-v_5-v_9-v_0$, or $v_0-v_4-v_5-v_1-v_0$.

We have drawn the edge from $v_0$ to $v_4$ in solid red; any of the dashed edges from $v_5$ 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 $\Gamma$ be a simple graph with $n$ vertices such that for any two nonadjacent vertices $v$ and $w$, we have $\text{deg}(v)+\deg{w}\geq n$. Then $\Gamma$ is Hamiltonian.

Proof

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

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

Let $v_1-v_2-\dots v_{n-1}-v_n$ be the path visiting every vertex of $\Gamma$. We will complete the proof by showing there exists an $i\in 3,\cdots, n-1$ so that $v_1$ is adjacent to $v_i$ and $v_n$ is adjacent to $v_{i-1}$. This completes the proof because $v_1-v_2-\cdots v_{i-1}-v_n-v_{n-1}-v_{n-2}-\dots v_i-v_1$ is a Hamiltonian path. We illustrate this below, with $n=9$ and $i=5$:

The existence of such an $i$ follows from the pigeon hole principle. Since $v_1$ and $v_n$ are not adjacent by supposition, the sum of their degrees is at least $n$. Two edges incident to them are accounted for $v_1-v_2$, and $v_{n-1}-v_{n-2}$, so there must be at least $n-2$ other edges incident to $v_1$ or $v_{n}$. Since $\Gamma$ is simple, any such edge out of $v_1$ has $n-3$ possibilities: $v_3, v_4, \dots, v_{n-1}$. Similarly, there are $n-3$ possibilities for an edge adjacent to $v_n$ – $v_2,v_3,\dots, v_{n-2}$. We pair these edges up into the possibilities we want, creating $n-3$ bins. Since there are $n-2$ 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.