An Ore-type condition for hamiltonicity in tough graphs

Let $G$ be a $t$-tough graph on $n\ge 3$ vertices for some $t>0$. It was shown by Bauer et al. in 1995 that if the minimum degree of $G$ is greater than $\frac{n}{t+1}-1$, then $G$ is hamiltonian. In terms of Ore-type hamiltonicity conditions, the problem was only studied when $t$ is between 1 and 2. In this paper, we show that if the degree sum of any two nonadjacent vertices of $G$ is greater than $\frac{2n}{t+1}+t-2$, then $G$ is hamiltonian.


Introduction 2 Introduction
We consider only simple graphs. Let G be a graph. Denote by V (G) and E(G) the vertex set and edge set of G, respectively. Let v ∈ V (G), S ⊆ V (G), and H ⊆ G. Then Throughout this paper, if not specified, we will assume t to be a nonnegative real number. The number of components of a graph G is denoted by c(G). The graph G is said to be t-tough if |S| ≥ t · c(G − S) for each S ⊆ V (G) with c(G − S) ≥ 2. The toughness τ (G) is the largest real number t for which G is t-tough, or is ∞ if G is complete. This concept, a measure of graph connectivity and "resilience" under removal of vertices, was introduced by Chvátal [6] in 1973. It is easy to see that if G has a hamiltonian cycle then G is 1-tough. Conversely, Chvátal [6] conjectured that there exists a constant t 0 such that every t 0 -tough graph is hamiltonian. Bauer, Broersma and Veldman [1] have constructed t-tough graphs that are not hamiltonian for all t < 9 4 , so t 0 must be at least 9 4 if Chvátal's toughness conjecture is true.
Chvátal's toughness conjecture has been verified when restricted to a number of graph classes [2], including planar graphs, claw-free graphs, co-comparability graphs, and chordal graphs. In general, the conjecture is still wide open. In finding hamiltonian cycles in graphs, sufficient conditions such as Dirac-type and Ore-type conditions are the most classical ones.
Analogous to Dirac's Theorem, Bauer, Broersma, van den Heuvel, and Veldman [4] proved the following result by incorporating the toughness of the graph. Theorem 2.3 (Bauer et al. [4]). Let G be a t-tough graph on n ≥ 3 vertices. If δ(G) > n t+1 − 1, then G is hamiltonian.
A natural question here is whether we can find an Ore-type condition involving the toughness of G that generalizes Theorem 2.3. Various theorems were proved prior to Theorem 2.3 by only taking τ (G) between 1 and 2. Jung in 1978 [8] showed that if G is a 1-tough graph on n ≥ 11 vertices with σ 2 (G) ≥ n − 4, then G is hamiltonian. In 1991, Bauer, Chen, and Lasser [3] showed that the degree bound in Jung's Theorem can be slightly lowered if τ (G) > 1. The result states that if G is a graph on n ≥ 30 vertices with τ (G) > 1 and σ 2 (G) ≥ n − 7, then G is hamiltonian. In 1989/90, Bauer, Veldman, Morgana, and Schmeichel [5] showed that if G is a 2-tough graph on n ≥ 3 vertices with σ 2 (G) ≥ 2n 3 , then G is hamiltonian (a consequence of Corollary 16 from [5]). In this paper, we obtain the following result, which provides an Ore-type condition involving τ (G) that guarantees a hamiltonian cycle in a graph.
In fact, we believe that the following stronger statement might be true. Conjecture 1. Let G be a t-tough graph on n ≥ 3 vertices. If σ 2 (G) > 2n t+1 − 2, then G is hamiltonian.
Considering both toughness and degree sum conditions such as in Theorem 1 and Conjecture 1 is an approach to investigate Chvátal's toughness conjecture while the conjecture remains open. However, in light of the conjecture, those results might only be relevant for some small values of t.
For odd integers n ≥ 3, the complete bipartite graph G := K n−1 2 , n+1 2 is n−1 n+1 -tough and satisfies σ 2 (G) = n − 1 = 2n However, G is not hamiltonian. Thus, if true, the degree sum condition in Conjecture 1 would be best possible. In fact, for odd integers n ≥ 3, any graph from the family H = {H n−1 is any graph on n−1 2 vertices} is an extremal graph, where "+" represents the join of two graphs. In light of the results mentioned in the paragraph right above Theorem 1 and Chvátal's toughness conjecture, it suggests that t-tough non-hamiltonian graphs G with σ 2 (G) = 2n t+1 − 2 exist only when t < 1. Furthermore, by looking at the non-hamiltonian t-tough graphs G with t < 1 and δ(G) = n t+1 − 1, which are exactly the graphs in the family H, it suggests that when t < 1, any non-hamiltonian t-tough graph G with t < 1 and σ 2 (G) = 2n t+1 − 2 belongs to the family H. So we propose the following conjecture.
In attempting to prove Conjecture 1 by contradiction, the most difficult case to deal with is when G has a cycle C of length n − 1 and G − V (C) is just a single vertex component H. It seems very hard to deduce any nontrivial property of G using the σ 2 (G) and toughness conditions. However, by adding t to the σ 2 (G) bound, vertices in V (C) \ N C (H) can be shown to have degree bigger than n t+1 +t−1. This degree condition allows us to find |N C (H)| disjoint subgraphs each of order t + 2 such that there is no edge between any two of them. Then we get to use the toughness condition to give a smaller upper bound on |N C (H)| (|N C (H)| ≤ n 2(t+1) − 1 2 ), which plays a key role in the proof of Theorem 1. Therefore, it might require a completely different approach to confirm Conjecture 1.
The remainder of this paper is organized as follows: in Section 2, we introduce some notation and preliminary results, and in Section 3, we prove Theorem 1.

Preliminary results
Let G be a graph and λ a positive integer. Following [11], a cycle C of G is a D λ -cycle if every component of G−V (C) has order less than λ. Clearly, a D 1 -cycle is just a hamiltonian cycle. We denote by c λ (G) the number of components of G with order at least λ, and write c 1 (G) just as c(G). Two subgraphs H 1 and H 2 of G are remote if they are disjoint and there is no edge of G joining a vertex of H 1 and a vertex of H 2 . For a subgraph H of G, let d G (H) = |N G (H)| be the degree of H in G. We denote by δ λ (G) the minimum degree of a connected subgraph of order λ in G. Again δ 1 (G) is just δ(G).
Let C be an oriented cycle, and we assume that the orientation is clockwise throughout the rest of this paper. For x ∈ V (C), denote the immediate successor of x on C by x + and the immediate predecessor of x on C by x − . For u, v ∈ V (C), u ⇀ Cv denotes the segment of C starting at u, following C in the orientation, and ending at v. Likewise, u ↼ Cv is the opposite segment of C with endpoints as u and v. Let dist⇀ C (u, v) denote the length of the path u ⇀ Cv. For any vertex u ∈ V (C) and any positive integer k, define to be the set of k consecutive successors of u and the set of k consecutive predecessors of u, respectively. A chord of C is an edge uv with u, v ∈ V (C) and uv ∈ E(C). Two chords ux and vy that do not share any endvertices of C are crossing if the four vertices u, x, v, y appear along ⇀ C in the order u, v, x, y or u, y, x, v. Hereafter, all cycles under consideration are oriented.
A path P connecting two vertices u and v is called a (u, v)-path, and we write uP v or vP u in order to specify the two endvertices of P . Let uP v and xQy be two paths. If vx is an edge, we write uP vxQy as the concatenation of P and Q through the edge vx.
has a component of order λ. The result below with d G (H) replaced by δ λ (G) was proved in [4].
Proof. Let k = d G (H), which equals the total number of neighbors of vertices of H on C.
We assume the k neighbors are v 1 , . . . , v k and appear in the same order along If such a vertex w * i exists, let L * v i (λ) be the union of the vertex set V (v + i ⇀ Cw * i ) and all those vertex sets of graphs in To prove Lemma 1, it suffices to show that L * v 1 (λ), . . . , L * v k (λ) and H are pairwise remote.
Since H has order λ and no vertex of H is adjacent in G to any internal vertex of v i ⇀ Cv i+1 , it follows that each component of contradicting the choice of C. The argument above verifies Statement (a).
For Statement (b), assume to the contrary that Since there is no edge between any two components of (v j , y) minimum such that xy ∈ E(G). By this choice of x and y, it follows The following lemma provides a way of extending a cycle C provided that the vertices outside C have many neighbors on C. The proof follows from Lemma 2 and is very similar to the proof of Lemma 10 in [10].
Lemma 3. Let t ≥ 1 and G be an n-vertex t-tough graph, and let C be a non-hamiltonian

Proof of Theorem 1
We may assume that G is not a complete graph. Thus G is 2⌈t⌉-connected as it is t-tough. Suppose to the contrary that G is not hamiltonian. By Theorem 2.3, we have δ(G) ≤ n t+1 − 1. Since δ(G) ≥ 2⌈t⌉, we get n ≥ (t + 1)(2⌈t⌉ + 1).
Since t > 1 and G is not complete, G is 2-connected and so contains cycles. We choose λ ≥ 0 to be a smallest integer such that G admits no D λ -cycle but a D λ+1 -cycle. Then we choose C to be a longest D λ+1 -cycle such that c λ (G − V (C)) is minimum. As G is not hamiltonian, we have λ ≥ 1. Thus V (G) \ V (C) = ∅. Since λ is taken to be minimum, G − V (C) has a component H of order λ. Let W = N C (V (H)) and ω = |W |.
Since G is a connected t-tough graph, it follows that ω ≥ 2⌈t⌉. On the other hand, Lemma 1 implies that ω ≤ n t+λ − 1.
Proof. Assume to the contrary that λ+ω > n t+1 . If λ = 1, then H has only one vertex and ω > n t+1 − 1. By Lemma 3, we can find a cycle C ′ with V (C ′ ) = V (C)∪ V (H), contradicting the choice of C. Thus λ ≥ 2. Since 2t ≤ ω ≤ n t+λ − 1 ≤ n t+2 − 1, we have n ≥ (t + 2)(2t + 1). By Lemma 1, we have reaching a contradiction. Since H is the only component of G − V (C), every vertex v ∈ V (C) \ W is only adjacent in G to vertices on C. As d G (u) ≤ n t+1 − 1 for any u ∈ V (H) by Claim 2, using Equation (1) allows us to construct the vertex sets L + u (t + 2) for each u ∈ W . For notation simplicity, we use L + u for L + u (t + 2).

Claim 4. For any two distinct vertices
Proof. Let u * ∈ N H (u), v * ∈ N H (v) and P be a (u * , v * )-path of H. For the first part of the statement, it suffices to show that when we arrange the vertices of W along ⇀ C, for any two consecutive vertices u and v from the arrangement, we have dist⇀ For the second part of the statement, we assume to the contrary that E G (L + u , L + v ) = ∅. Applying the first part, we know that dist⇀ (v, y) minimum such that xy ∈ E(G). By this choice of x and y, Proof. Assume otherwise that ω > n 2(t+1) − 1 2 . By Claim 4, for any two distinct u, v ∈ W , G[L + u ] and G[L + v ] are remote, and G[L + u ] and H are remote. Thus in G, there are ω + 1 pairwise remote subgraphs. By the definition, G[L + u ] has order t + 2 for each u ∈ W . Let contradicting the toughness of G.
By Claim 3 and Claim 6, we have We will now explore the neighborhood of vertices from W + := {w ∈ V (C) : w − ∈ W }, and show that some vertices from the neighborhood have similar properties as those in W + . By Claim 1, we know that |W | ≥ 3 and so |W + | ≥ 3. Equation (3) allows us to construct the vertex sets L − x ( 0.25n t+1 + 2) for each x ∈ N C (W + ). For notation simplicity, we use L − x for L − x ( 0.25n t+1 + 2). Note that the statement below is not true in general if we replace L − x ( 0.25n t+1 + 2) by L + x ( 0.25n t+1 + 2).
Claim 7. Let u ∈ W + and x ∈ N C (u). Then (2) Let v ∈ W + and y ∈ N C (v) such that ux and vy are two crossing chords of C. Then dist⇀ C (x, y) ≥ 0.25n t+1 + 3.
Proof. For Statement (1), suppose to the contrary that there exists z ∈ W such that z ∈ L − x . Then dist⇀ , and P * be a (u * , z * )-path of H. Then C * = z ↼ Cux ⇀ Cu − u * P * z * z is a cycle. As H is complete of order λ (2), suppose to the contrary that dist⇀ C (x, y) ≤ 0.25n t+1 + 2. We assume without loss of generality that u, v, x, y appear in this order along For two distinct vertices x, y ∈ N C (W + ), we say x and y form a crossing if there exist distinct vertices u, v ∈ W + such that ux and vy are crossing chords of C. By Claim 7 (2), there are at least 0.5n t+1 + 2 vertices between x and y along ⇀ C for any two x, y ∈ N C (W + ) such that x and y form a crossing. Our goal below is to find at least n 2(t+1) vertices from N C (W + ) such that there are at least 0.5n t+1 + 2 vertices between any two of them along ⇀ C. Then we will reach a contradiction by showing that |V (C)| ≥ n. Define Let u ∈ W + and p = deg G (u, B) for some positive integer p, and let We may assume that x 1 , x 2 , . . . , x p appear in the same order along ⇀ C. We separate those vertices according to vertices of W . By Claim 7(1), we have L − x i ∩ W = ∅ for each i ∈ [1, p]. Therefore for some integer q ≥ 1, we assume that x 1 , . . . , x p are grouped into q sets . Furthermore, we may assume that the number q of sets with the property above is minimum.
Claim 8. For each i ∈ [1, q], B i has at least |B i |/2 vertices such that the distance between any two of them on C is at least 0.5n t+1 + 3.
Proof. We partition B i into two subsets according to whether or not vertices in N G (x) ∩ By the Pigeonhole Principle, we have |B i1 | ≥ |B i |/2 or |B i2 | ≥ |B i |/2. We show that any two distinct vertices from B i1 or from B i2 have distance at least 0.5n t+1 + 3 between them on t+1 + 3 by (4), we have dist⇀ C (x a , x b ) ≥ 0.5n t+1 + 3. Thus we assume a < b. Suppose now that x a , x b ∈ B i2 . If a > b i−1 + 1 or a = b i−1 + 1 but x a ∈ W , then ). Then the four vertices u, v, x a , x b appear in the order x a , x b , u, v along ⇀ C and so ux a and vx b are crossing chords of C. By Claim 7(2), we have dist⇀ C (x a , x b ) ≥ 0.5n t+1 + 3. Thus we assume a = b i−1 + 1 and x a ∈ W . By Claim 7(1), we know that dist⇀ C (x a , x a+1 ) ≥ 0.5n t+1 +3 and so dist⇀ C (x a , x b ) ≥ 0.5n t+1 + 3.