An Ore-type condition for hamiltonicity in tough graphs and the extremal examples

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, and recently the author proved a general result. The result states 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. It was conjectured in the same paper that the ``$+t$"in the bound $\frac{2n}{t+1}+t-2$ can be removed. Here we confirm the conjecture. The result generalizes the result by Bauer, Broersma, van den Heuvel, and Veldman. Furthermore, we characterize all $t$-tough graphs $G$ on $n\ge 3$ vertices for which $\sigma_2(G) = \frac{2n}{t+1}-2$ but $G$ is non-hamiltonian.


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 N G (v) denotes the set of neighbors of v in G, d G (v) := |N G (v)| is the degree of v in G, and δ(G) \ S, and we write N G (H) for N G (V (H)).Let N H (v) = N G (v)∩V (H) and N H (S) = N G (S)∩V (H).Again, we write N H (R) for N H (V (R)) for any subgraph R of G.We use G[S] and G − S to denote the subgraphs of G induced by S and V (G) \ S, respectively.For notational simplicity we write G − x for G − {x}.Let V 1 , V 2 ⊆ V (G) be two disjoint vertex sets.Then E G (V 1 , V 2 ) is the set of edges in G with one endvertex in V 1 and the other endvertex in V 2 .For two integers a and b, let [a, b] = {i ∈ Z : a ≤ i ≤ b}.
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 was introduced by Chvátal [7] in 1973.It is easy to see that if G has a hamiltonian cycle then G is 1-tough.Conversely, Chvátal [7] 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.
Analogous to Dirac's Theorem, Bauer, Broersma, van den Heuvel, and Veldman [4] proved the following result by incorporating the toughness of the graph.
A natural question here is whether we can find an Ore-type condition involving the toughness of G that generalizes Theorem 1.3.Various theorems were proved prior to Theorem 1.3 by only taking τ (G) between 1 and 2 [10,3,5].Let G be a t-tough graph on n ≥ 3 vertices.The author showed in [14] that if σ 2 (G) > 2n t+1 +t−2, then G is hamiltonian.It was also conjectured in [14] that σ 2 (G) > 2n t+1 − 2 is the right bound.In this paper, we confirm the conjecture.For any odd integer n ≥ 3, the complete bipartite graph G : However, G is not hamiltonian.Thus, the degree sum condition that σ 2 (G) > 2n t+1 − 2 is best possible for a t-tough graph on at least three vertices to be hamiltonian.In fact, for any odd integers n ≥ 3, any graph from the family is any graph on n−1 2 vertices} is an extremal graph, where "+" represents the join of two graphs.We also show that H is the only family of extremal graphs.
Theorem 1.Let G be a t-tough graph on n ≥ 3 vertices.Then the following statements hold.
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 be a positive integer.Following [18], a cycle C of G is a D λcycle 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 with 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).

Lemma 1 ([16]
).Let t > 0 and G be a non-complete n-vertex t-tough graph.Then |W | ≤ n t+1 for every independent set W in G.For any vertex u ∈ V (C) and any positive integer k, define

Denote by
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 1 and is very similar to the proof of Lemma 10 in [16]: if we assume instead that C cannot be extended by including x, then N + C (x) ∪ {x} is an independent set in G.
Lemma 2. Let t > 0 and G be an n-vertex t-tough graph, and let C be a non-hamiltonian cycle of G.
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.

Lemma 3 ([14]
).Let G be a t-tough 2-connected graph of order n.Suppose G has a D s+1cycle but no D s -cycle for some integer s The lemma below is the key to get rid of the "+t" in the lower bound 2n t+1 + t − 2 on σ 2 (G) for guaranteeing the existence of a hamiltonian cycle [14].
Lemma 4. Let G be a t-tough 2-connected graph of order n.Suppose that G has a D λ+1cycle but no D λ -cycle for some integer λ ≥ 1.Let C be a cycle of G. Then G − V (C) has a component H with order at least λ such that deg G (x, C) ≤ n t+1 − λ for some x ∈ V (H).
Proof.Since G has no D λ -cycle, it is clear that G − V (C) has a component of order at least λ.We suppose to the contrary that for each component H with order at least λ of G− V (C) and each x ∈ V (H), we have deg G (x, C) > n t+1 − λ.Among all cycles C ′ of G that satisfy the two conditions below, we may assume that C is one that minimizes c p (G− V (C)) prior to minimizing c q (G − V (C)) for any p ≥ λ and any q with q < p.
(1) each component of G − V (C) either has order at most λ − 1, or (2) the component H has order at least λ such that for each x ∈ V (H), we have deg We take a component H with order at least λ and assume that N C (H) has size k for some integer k ≥ 2, and that the k neighbors are v 1 , . . ., v k and appear in the same order along ⇀ C. Note that k > n t+1 − λ by our assumption.For each i ∈ [1, k], and each ) and all those vertex sets of graphs in We will show that we can make the following assumptions: With Assumptions (a) and (b), we can reach a contradiction as follows: note that and so k ≤ n t+λ − 1 ≤ n t+1 − λ.This gives a contradiction to k > n t+1 − λ.Thus we are only left to show Assumptions (a) and (b).We show that if any one of the assumptions is violated, then we can decrease c p (G − V (C)) for some p ≥ λ.
Thus we may just assume j = i + 1, where the index is taken modulo k.
Note that every component of G − V (C) not having any vertex joining to a vertex from

. Those components automatically satisfy
Conditions ( 1) and ( 2) as listed in the beginning of this proof.Vertices in v , and the component has order at most λ − 1 by the assumption that either has order at most λ − 1 or is a component of order at least λ such that each vertex from the component has in G more than n t+1 − λ neighbors on C 1 .However, For Assumption (b), suppose it is false.Then there exist distinct i, j ] are not remote.By the definition of remote subgraphs, we have either In order to achieve a contradiction, we first show the following general claim, call it Claim ( * ).
. By this choice of y and z, it follows that and Assumption (a).By the assumption that Cz − are respectively contained in distinct components of G − V (C 1 ) that each of order at most λ − 1.By the same reasoning as in proving Assumption (a), we know that each component of G − V (C 1 ) has order at most λ − 1 or is a component such that each vertex from the component has in contains at most λ − 1 vertices and they are proper subsets of V (v By the choices of v and u that for any Cu − have order at most λ − 1.By the same reasoning as in proving Assumption (a), we know that each component of G − V (C 1 ) has order at most λ − 1 or is a component such that each vertex from the component has in G more than n t+1 − λ neighbors on ] are remote, contradicting our assumption.Thus Assumption (b) holds.

Proof of Theorem 1
We may assume that G is not a complete graph.Thus G is ⌈2t⌉-connected as it is t-tough.Suppose to the contrary that G is not hamiltonian.Claim 1.We may assume that G is 2-connected.
Proof.Since t > 0, G is connected.Assume to the contrary that G has a cutvertex x.By considering the degree sum of two vertices respectively from two components of G − x, we know that σ 2 (G) ≤ n − 1.On the other hand, G has a cutvertex implies t ≤ 1 2 and so Also as G is 2-connected, G contains cycles.Let λ ≥ 0 be the 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 that minimizes c p (G − V (C)) prior to minimizing c q (G − V (C)) for any p, q ∈ [1, λ] with p > q.As G is not hamiltonian, we have λ Since G is a connected t-tough graph, it follows that ω ≥ ⌈2t⌉.On the other hand, Lemma 3 implies that ω ≤ n t+λ − 1.
Proof.If λ = 1, then the assertion holds by ω ≤ n t+λ − 1.Thus we assume λ ≥ 2 and assume to the contrary that λ + ω ≥ n t+1 .Then we have n ≤ (λ + ω)(t + 1).By Lemma 3, we have n ≥ (λ + t)(ω + 1).Thus we have (λ + t)(ω + 1) ≤ (λ + ω)(t + 1), which implies λω + λ + tω + t ≤ λt + λ + tω + ω and so (λ − 1)ω ≤ (λ − 1)t.Since λ ≥ 2, we get ω ≤ t, a contradiction to ω ≥ 2t.Note that the argument above for λ ≥ 2 holds for all components of G − V (C) as Lemma 3 holds for all components of G − V (C).Thus we assume λ = 1.We get the same contradiction as above if σ 2 (G) > 2n t+1 − 2 or λ + ω < n t+1 .Thus we have σ 2 (G) = 2n t+1 − 2 and ω = n t+1 − 1 by Claim 2. Then H and H * each contains only one vertex, say x and y, respectively.We first claim that the vertex y is adjacent in G to at most one vertex from W + .For otherwise, suppose there are distinct We then claim that the set W + is an independent set in G.For otherwise, suppose there We construct the vertex sets L + u for each u ∈ W as follows: Proof.We only show Claim 4(a), as the proof for Claim 4(b) follows the same argument by just using the strict inequality.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⇀ This shows a contradiction to Lemma 4.
For the second part of the statement, we assume to the contrary that . By this choice of x and y, it follows that E G (V (u that each is of order less than n t+1 − ω but at least one of them has order at least λ.
This shows a contradiction to Lemma 4.
By Claim 5, Theorem 1(a) holds.In the rest of the proof, we show Theorem 1(b).Let Since u + ∈ L + u for each u ∈ W , Claim 4 implies that W * is an independent set in G.
Claim 6.Every vertex in V (G) \ W * is adjacent in G to at least two vertices from W * .
Proof.Suppose to the contrary that there exists x ∈ V (G) \ W * such that x is adjacent in G to at most one vertex from < t, a contradiction.
Claim 7.For every v ∈ W + , we have deg G (v, C) = n t+1 − 1 and v is not adjacent in G to any two consecutive vertices on C. Our goal is to show that N C (W + ) = N C (H).To do so, we investigate how vertices in N C (W + ) are located along ⇀ C. We start with some definitions.A chord of C is an edge uv with u, v ∈ V (C) and uv ∈ E(C).Two chords ux and vy of C that do not share any endvertices 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.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. Claim 8.For any two distinct x, y ∈ N C (W + ) with xy ∈ E(C), it follows that x and y do not form any crossing.
Proof.Suppose to the contrary that for some distinct x, y ∈ N C (W + ) with xy ∈ E(C), the two vertices x and y form a crossing.Let u, v ∈ W + such that yu, yv ∈ E(G).Assume, without loss of generality, that the four vertices u, v, x, y appear in the order u, v, x, y along Then by our selection of x and y, we know that each internal vertex of x ⇀ Cy is adjacent in G to a vertex from W + \ {v}.Applying Claim 8, x + does not form a crossing with x, and so x + forms a crossing with y.Similarly, x ++ does not form a crossing with x + , and so forms a crossing with y.Continuing this argument for all the internal vertices of x ++ ⇀ Cy, we know that y − forms a crossing with y, a contradiction to Claim 8.
We assume that the ω neighbors of the vertex from V (H) on C are v 1 , . . ., v ω and they appear in the same order along

⇀C
an orientation of C. 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 − .We use N + C (x) to denote the set of immediate predecessors for vertices from N C (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.
has a component of order λ.The result below with d G (H) replaced by δ λ (G) and H replaced by any component of G − V (C) with order λ was proved in [4, Corollary 7(a)].
contradicting the choice of C. Therefore we have Assumption (a), which implies that the vertex w * i exists for each i ∈ [1, k].
contradicting the choice of C. Thus Claim ( * ) holds.Now let us get back to prove Assumption (b) by contradiction.Assume first that L * we then further choose v closest to v i and u closest to v j along ⇀ C with the property.Thus for any w Repeating exactly the same argument for |V (H * )| + |N C (H * )| as in the proof of Claim 2 leads to a contradiction.
for such pairs of u and v. Assume to the contrary that there are distinct u, v ∈ W with V (u + ⇀ Cv − ) ∩ W = ∅ and dist⇀ C (u, v) < n t+1 − ω + 1.Let C * = u ↼ Cvv * P u * u.Since H has order λ and V (u Again, since G has no D λ -cycle, it follows that deg G (v, C * ) = n t+1 − 1 and v is not adjacent in G to any two consecutive vertices on C * .The claim follows as deg G (v, C) = deg G (v, C * ) and two neighbors of v that are consecutive on C will also be consecutive on C * .

⇀C.Claim 9 .
Let V (H) = {w}.Then ux ↼ Cvy ⇀ Cu − wv − ↼ Cu is a hamiltonian cycle of G, a contradiction to our assumption that G is not hamiltonian.For any vertex v ∈ W + and any two distinct x, y ∈ N C (v), x ⇀ Cy contains a vertex from W + .Proof.By Claim 7, x ⇀ Cy has at least three vertices.Suppose to the contrary that x ⇀ Cy contains no vertex from W + .We furthermore choose x and y so that x ⇀ Cy contains no other vertex from N C (v) \ {x, y}.Assume that the three vertices v, x, y appear in the order v, x, y along ⇀ C. By Claim 6, each internal vertex of x ⇀ Cy is adjacent in G to a vertex from W + .

Proof. Since x ⇀↼⇀
Cy contains a vertex from W + for any two distinct x, y ∈ N C (v) by Claim 9, it follows that no I i can contain more than one vertex from N C (v).Since deg G (v, C) = ω = |W + | by Claim 7 and {I 1 , . . ., I ω } is a partition of V (C), the Pigeon-hole Principle implies that each I i contains exactly one vertex from N C (v).Assume to the contrary that NC (v) = W .Let i ∈ [1, ω] be the index such that dist⇀ C (v, v i ) is largest and vv i ∈ E(G).Note that the index i exists since v − ∈ W and vv − ∈ E(G).In particular, every vertex u ∈ W ∩ V (v + i ⇀Cv) is adjacent to v by the choice of i.Let z be the vertex in N C (v) ∩ I i−1 .We prove the four subclaims below.Let V (H) = {x} in the rest arguments.Claim A: z = v − i .Proof of Claim A. Suppose otherwise that z = v − i .Then by Claim 6, z + is adjacent in G to at least two vertices from W + .By Claim 8, N C (z + )∩ W + ⊆ V (v + i ⇀ Cv).Thus z + is adjacent in G to a vertex from W + ∩ V (v + i ⇀ Cv − ) as z is the only neighbor of v from I i−1 in G.By repeating this procedure for all the vertices from V (z ++ ⇀ Cv − i ) iteratively, we conclude that v − i is adjacent in G to a vertex u ∈ W + ∩ V (v + i ⇀ Cv − ).As v + i v i ∈ E(G) and v i v − i ∈ E(C), Claim 7 implies that v + i is not adjacent in G to v − i .Thus we have u ∈ {v + i , v}.However, since u − v ∈ E(G) by our choice of the index i, the cycle xv − Cv i x is in G longer than C, a contradiction.Thus z must be v − i .Claim B:v i+1 = v − .by Claim 6.If u = v + i , then the cycle xv i ⇀ Cu − x is in G longer than C, acontradiction.Thus we have u = v + i .We consider the cycleC * = v + ⇀ Cv − i v i xv − ↼ Cv + i v + in G.Note that we have V (C * ) = V (G) \ {v}.Then since the length of C * is equal to the length of C, we can apply Claim D to C * .However, v − , x, v i , v − i are four consecutive vertices on C * appearing in the order v− , x, v i , v − i and v − , v − i ∈ N C * (v),showing that C * does not satisfy Claim D, a contradiction.This completes the proof of Claim 10.Claim 10 implies that N C (W * ) = W . Thus every vertex from W * is adjacent in G to every vertex from W . Therefore t ≤ τ (G) ≤ |W | |W * | as W * is an independent set in G. Consequently, |W | ≥ t|W * | = tn t+1 and so W = V (G) \ W * by noticing |W * | = n t+1 .Thus G contains a spanning complete bipartite graph between W * and W .On the other hand, since |W + | = |W | = n t+1 − 1 and V (G) = W * ∪ W = (W + ∪ V (H)) ∪ W , we know that 2( n t+1 − 1) + 1 = n and so t = n−1 n+1 .Thus |W | = n−1 2 and |W * | = n−1 2 + 1 = n+1 2 .Therefore, G ∈ H.The proof of Theorem 1 is now complete.