I have been digging on the internet for the proof of this theorem for
the last couple of days without success. The result was published by Sir
Crispin Nash-Williams as *Valency Sequences which force graphs to have
Hamiltonian Circuits*. Interim Rep, University of Waterloo Res Rep.,
Waterloo, Ontario, 1969. However, this old paper is unavailable online
but I have a proof in some lecture notes from my class, that I want to
share here.

**Theorem.** Let [latex]G=(V, E)[/latex] be an [latex]n[/latex]-regular
graph with [latex]|V| = 2n + 1[/latex]. Then, [latex]G[/latex] is Hamiltonian.

**Proof.** We first remark that [latex]n[/latex] must be even, since
\$\$\sum_{x \in V} d(x) = n(2n + 1) = 2|E|\$\$ We might try to apply
Dirac’s theorem, which would give us a Hamiltonian cycle if [latex]
\forall x \in V, d(x) \geq \frac{|V|}{2}[/latex]. But in the current
case, [latex]\forall x \in V, d(x) = n \< \frac{2n+1}{2}[/latex].

So we force Dirac by adding an extra vertex [latex]w[/latex] and connecting it to all [latex] x \in V [/latex]. In this new graph [latex]G’[/latex], [latex]d(x) = n + 1 \forall x \in V[/latex] and [latex]d(w) = 2n + 1[/latex]. Therefore we have a Hamiltonian cycle that passes through [latex]w[/latex] and in which, [latex]w[/latex] is adjacent to two vertices [latex]x[/latex] and [latex]y \in V[/latex]. Therefore this cycle induces a Hamiltonian path in [latex]G[/latex]: \$\$P = [x = v_0, v_1, …, v_{2n-1}, v_{2n}=y] \$\$

Suppose that [latex]G[/latex] is not Hamiltonian. It follows that if [latex] v_0v_i \in E [/latex], then [latex] v_{i-1}v_{2n} \notin E[/latex] and also that if [latex] v_0v_i \notin E [/latex], then [latex] v_{i-1}v_{2n} \in E[/latex].

We have two cases. If [latex]v_0[/latex] is adjacent to [latex]v_1, …, v_n[/latex] then it follows that [latex]v_{2n}[/latex] is adjacent to [latex]v_n, v_{n+1}, …, v_{2n-1}[/latex], since it cannot be adjacent to any [latex]v_i, i \< n[/latex] without creating a Hamiltonian cycle. But in this case, in the graph induced by the first half [latex]G[\{v_0, v_1, … v_n\}][/latex], [latex]v_n[/latex] cannot be adjacent to all the others, since in [latex]G[/latex] it has degree [latex]n[/latex] and it already has [latex]2[/latex] outgoing edges. So there is at least one vertex [latex]v_i, i \< n[/latex] that isn’t adjacent to it, which means [latex]v_i[/latex] is adjacent to some [latex]v_j, j > n[/latex], thus forming a Hamiltonian cycle.

In the second case, we have a vertex [latex]v_i, 2 \leq i \leq 2n - 1[/latex] such that [latex]v_0v_i \notin E[/latex] and [latex]v_0v_{i+1} \in E[/latex]. This also means that [latex]v_{i-1}v_{2n} \in E[/latex].

We therefore have a cycle of length [latex]2n[/latex] in [latex]G[/latex] that excludes [latex]v_i[/latex]. Let’s rename this cycle [latex]C=[y_1, y_2, …, y_{2n}, y_1][/latex] and [latex]v_i=y_0[/latex].

[latex]y_0[/latex] cannot be adjacent to two consecutive vertices [latex]y_i[/latex] and [latex]y_{i+1}[/latex] because this will give a Hamiltonian cycle. But we know that [latex]deg(y_0) = n[/latex]. It follows that it’s adjacent to all of the even or odd numbered vertices. We assume the latter, without loss of generality. Let [latex]2k[/latex] be some even index. Notice that we have [latex]\{y_0y_{2k-1}, y_0y_{2k+1}\} \subset E[/latex] and we can follow the cycle [latex]C[/latex] from [latex]y_{2k+1}[/latex] all the way back to [latex]y_{2n-1}[/latex] giving us a new cycle [latex]C’ = [y_1, y_2, …, y_{2n-1}, y_0, y_{2k+1}, …, y_{2n}, y_1][/latex] also of length [latex]2n[/latex]. So by repeating the same reasoning for every even vertex, by placing it in the middle and building a cycle around it, it follows that every even vertex is adjacent to all the odd vertices. But there are [latex]n+1[/latex] even indices, so it follows that the degree of any odd vertex is at least [latex]n+1[/latex], contradicting the initial conditions of the theorem. [latex]\square[/latex]

## Comments !