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J. Adrian Bondy and Vaˇsek Chv´atal [34] observed that the proof of Ore’s theorem (Theorem 6.1) neither uses nor needs the full strength of the require-ment that the degree sum of each pair of nonadjacent vertices is at least the order of the graph being considered. Initially, Bondy and Chv´atal made the following observation.

Theorem 6.4 Let u and v be nonadjacent vertices in a graph G of order n such that deg u + deg v ≥ n. Then G + uv is Hamiltonian if and only if G is Hamiltonian.

Proof. Certainly, if G is Hamiltonian, then G + uv is Hamiltonian. For the converse, suppose that G + uv is Hamiltonian but G is not. Then every tonian cycle in G + uv contains the edge uv, implying that G contains a Hamil-tonian u − v path. We can now proceed exactly as in the proof of Theorem 6.1 to produce a contradiction.

The preceding result inspired a definition. Let G be a graph of order n. The closure CL(G) of G is the graph obtained from G by recursively joining pairs of nonadjacent vertices whose degree sum is at least n (in the resulting graph at each stage) until no such pair remains. A graph G and its closure are shown in Figure 6.4.

First, we show that the closure is a well-defined operation on graphs, that is, the same graph is obtained regardless of the order in which edges are added.

6.2. SUFFICIENT CONDITIONS FOR HAMILTONICITY 131

Figure 6.4: Constructing the closure of a graph

Theorem 6.5 Let G be a graph of order n. If G1 and G2 are graphs obtained by recursively joining pairs of nonadjacent vertices whose degree sum is at least n until no such pair remains, then G1= G2. are nonadjacent vertices of H. Since deg_Hx + deg_Hy ≥ n, it follows that deg_G2x + deg_G2y ≥ n, which produces a contradiction.

Repeated application of Theorem 6.4 gives us the following result.

Theorem 6.6 A graph is Hamiltonian if and only if its closure is Hamiltonian.

Since each complete graph with at least three vertices is Hamiltonian, we obtain the following sufficient condition for a graph to be Hamiltonian due to Bondy and Chv´atal [34].

Theorem 6.7 Let G be a graph with at least three vertices. If CL(G) is complete, then G is Hamiltonian.

If a graph G satisfies the conditions of Theorem 6.1, then CL(G) is complete and so, by Theorem 6.7, G is Hamiltonian. Thus, Ore’s theorem is an immediate corollary of Theorem 6.7 (although chronologically it preceded the theorem of Bondy and Chv´atal by several years). In fact, many sufficient conditions for a graph to be Hamiltonian based on the degrees of the vertices of a graph can be deduced from Theorem 6.7. The following result of Chv´atal [54] is an example of one of the strongest of these.

Theorem 6.8 Let G be a graph of order n ≥ 3, the degrees d_iof whose vertices satisfy d₁≤ d₂≤ · · · ≤ d_n. If there is no integer k < n/2 for which d_k ≤ k and d_n−k≤ n − k − 1, then G is Hamiltonian.

Proof. Let H = CL(G). We show that H is complete which, by Theorem 6.7, implies that G is Hamiltonian. Assume, to the contrary, that H is not complete.

Let u and w be nonadjacent vertices of H for which deg_Hu + deg_Hw is as large as possible. Since u and w are nonadjacent vertices of H, it follows that deg_Hu+

deg_Hw ≤ n − 1. Assume, without loss of generality, that deg_Hu ≤ deg_Hw.

Thus, if k = deg_Hu, we have that k ≤ (n − 1)/2 < n/2 and deg_Hw ≤ n − 1 − k.

Let W denote the vertices other than w that are not adjacent to w in H. Then

|W | = n − 1 − deg_Hw ≥ k. Also, by the choice of u and w, every vertex v ∈ W satisfies

deg_Gv ≤ deg_Hv ≤ deg_Hu = k.

Thus, G has at least k vertices of degree at most k and so d_k ≤ k. Similarly, let U denote the vertices other than u that are not adjacent to u in H. Then

|U | = n − 1 − deg_Hu = n − k − 1. Every vertex v ∈ U satisfies deg_Gv ≤ deg_Hv ≤ deg_Hw ≤ n − 1 − k, implying that dn−k−1≤ n − k − 1. However,

deg_Gu ≤ deg_Hu ≤ deg_Hw ≤ n − 1 − k,

so dn−k≤ n − k − 1. This, however, contradicts the hypothesis of the theorem.

Thus, H = CL(G) is complete.

All of the sufficient conditions presented thus far for a graph G to be Hamil-tonian involve the degrees of the vertices of G. In fact, if G has order n, then each of these conditions requires some of the vertices of G to have degree at least n/2. In the case of regular graphs, however, this situation can be im-proved. Bill Jackson [134] showed that every 2-connected r-regular graph of order at most 3r is Hamiltonian. As we are about to see, the Petersen graph shows that 3r cannot be replaced by 3r + 1. We mentioned earlier that the girth (the length of a smallest cycle) of the Petersen graph P is 5 and its circumfer-ence (the length of a longest cycle) is 9. In fact, P − v contains a 9-cycle for every vertex v of P . We now verify that the Petersen graph has circumference 9 by showing that it is not Hamiltonian.

Theorem 6.9 The Petersen graph is not Hamiltonian.

Proof. Assume, to the contrary, that the Petersen graph P is Hamiltonian.

Then P has a Hamiltonian cycle

C = (v₁, v₂, . . . , v₁₀, v₁).

Since P is cubic, v₁ is adjacent to exactly one of the vertices v₃, v₄, . . . , v₉. However, since P contains neither a 3-cycle nor a 4-cycle, v₁ is adjacent to exactly one of v₅, v₆ and v₇. Because of the symmetry of v₅ and v₇, we may assume that v₁ is adjacent either to v₅or to v₆.

6.2. SUFFICIENT CONDITIONS FOR HAMILTONICITY 133 Case 1. v1 is adjacent to v5 in P . Then v10 is adjacent to exactly one of v4, v5and v6, which results in a 4-cycle, a 3-cycle or a 4-cycle, respectively, each of which is impossible.

Case 2. v1 is adjacent to v6 in P . Again, v10 is adjacent to exactly one of v4, v5 and v6. Since P does not contain a 3-cycle or a 4-cycle, v10 must be adjacent to v4. However, then this returns us to Case 1, where v1 and v5 are replaced by v10 and v4, respectively.

The next sufficient condition for a graph G to be Hamiltonian does not involve the degrees of the vertices of G but, rather, involves the cardinality of sets of pairwise nonadjacent vertices of a graph and its connectivity. A set U of vertices in a graph G is independent if no two vertices in U are adjacent.

(Some refer to an independent set of vertices as a stable set.) The maximum number of vertices in an independent set of vertices of G is called the vertex independence number or, more simply, the independence number of G and is denoted by α(G). For example, α(Kr,s) = max{r, s}, α(Cn) =n

2 and α(Kn) = 1. Vaˇsek Chv´atal and Paul Erd˝os [58] showed that if G is a graph of order at least 3 whose connectivity is at least as large as its independence number, then G must be Hamiltonian.

Theorem 6.10 Let G be a graph of order at least 3. If κ(G) ≥ α(G), then G is Hamiltonian.

Proof. If α(G) = 1, then G is complete and therefore Hamiltonian. Hence, we may assume that α(G) = k ≥ 2. Since κ(G) ≥ 2, it follows that G is 2-connected and so G contains a cycle by Theorem 4.15. Let C be a longest cycle in G. By Theorem 4.15, C contains at least k vertices. We show that C is a Hamiltonian cycle. Assume, to the contrary, that C is not a Hamiltonian cycle. Then there is some vertex w of G that does not lie on C. Since G is k-connected, it follows with the aid of Corollary 4.13 that G contains k paths P1, P2, · · · , Pk such that Pi is a w − vipath where viis the only vertex of Pi on C and such that the paths are pairwise-disjoint except for w.

In some cyclic ordering of the vertices of C, let ui be the vertex that follows vi on C for each i (1 ≤ i ≤ k). No vertex ui is adjacent to w, for otherwise, replacing the edge viui by Pi and wui produces a cycle whose length exceeds that of C. Let S = {w, u1, u2, . . . , uk}. Since |S| = k + 1 > α(G) and wui ∈/ E(G) for each i (1 ≤ i ≤ k), there are distinct integers r and s such that 1 ≤ r, s ≤ k and urus∈ E(G). Replacing the edges urvr and usvs by the edge urus and the paths Pr and Psproduces a cycle that is longer than C. This is a contradiction.

For the non-Hamiltonian Petersen graph P , κ(P ) = 3 and α(P ) = 4; so, as anticipated by Theorem 6.10, κ(P ) < α(P ). On the other hand, κ(P ) ≥ α(P ) − 1 and, consequently, P has a Hamiltonian path (see Exercise 13).

6.3 Toughness of Graphs

While we have described several sufficient conditions for a graph to be Hamil-tonian, there are also useful necessary conditions. Certainly, every Hamiltonian graph is connected. Since every pair u, v of distinct vertices of a Hamiltonian graph G lies on a Hamiltonian cycle of G, it follows that G contains at least two internally disjoint u − v paths. By Theorem 4.11, every Hamiltonian graph is 2-connected. Consequently, if S is a subset of V (G) with |S| = 1, then G − S consists of a single component, that is k(G − S) ≤ |S|, where, recall, that k(G − S) is the number of components in G − S. The next result generalizes this observation.