Some sufficient conditions on hamilton graphs with toughness

Abstract

Let $G$ be a graph, $\omega(G)$ denote the number of components of a graph $G$, and $t$ is real number. If $\omega(G-S)\ge2 \Rightarrow |S|\ge t\omega(G-S)$ holds for each set $S$ of vertex set $(G)$ of $G$, $G$ be said to be $t$-tough. The $toughness$ of $G$ is the maximum value of $t$ for which $G$ is $t$-tough, denoted with $\tau(G)$. The graph $G$ is called Hamilton graph if it has a cycle which contains all vertices of $G$. Chv$\acute{a}$tal and other scholars have studied the relation between toughness conditions to the existence of cycle structures. In this paper, we first establish some sufficient conditions for a graph with toughness to be Hamiltonian in terms of the edge number, the spectral radius and the signless Laplacian spectral radius of the graph.

1. Introduction

Let G = [V(G), E(G)] be a finite simple undirected graph with vertex set V(G) = {v₁, v₂, …, v_n} and edge set E(G). Write by m = |E(G)| the number of edges and n = |V(G)| the number of vertices of the graph G, respectively. The set of neighbors of a vertex v in graph G is denoted by N_G(v). Let v_i ∈ V(G), we denote by d_i = d_vi = d_G(v_i) = |N_G(v_i)| the degree of v_i. Denote by δ(G) [Δ(G)] or simply δ (Δ) the minimum (maximum) degree of G. Let (d₁, d₂, …, d_n) be a nondecreasing degree sequence of G, that is, d₁ ≤ d₂ ≤ ⋯ ≤ d_n. For convenience, we use (0x0,1x1,⋯,kxk,⋯,ΔxΔ) (0x0,1x1,⋯,kxk,⋯,ΔxΔ) to denote the degree sequence of G, where x_k is the number of vertices of degree k in the graph G. We denote a bipartite graph with bipartition (X, Y) by using G[X, Y]. We denote the cycle and the complete graph on n vertices by using C_n and K_n, respectively. We use K_{m, n} to denote a complete bipartite graph with two parts having m, n vertices, respectively. Let G and H be two disjoint graphs. We denote by G + H the disjoint union of G and H, which is a graph with vertex set V(G) ∪ V(H) and edge set E(G) ∪ E(H). If G₁ = G₂ = … = G_k, we denote G₁ + G₂ + ⋯ + G_k by kG₁. We denote by G ∨ H the join of G and H, which is a graph obtained from the disjoint union of G and H by adding edges joining every vertex of G to every vertex of H. Let K_{n − 1} + v denote the complete graph K_{n − 1} together with an isolated vertex v. Other undefined symbols reference can be seen in Bondy and Murty (1982) and Bauer et al. (2006).

The adjacency matrix of G is A(G) = (a_ij), where a_ij = 1 if v_i and v_j are adjacent in G and a_ij = 0 otherwise. Let D(G) be the degree diagonal matrix of G, i.e., D(G) = diag{d_G(v₁), d_G(v₂), …, d_G(v_n)}. The matrix Q(G) = D(G) + A(G) is called the signless Laplacian matrix of G. The largest eigenvalue of A(G), denoted by μ(G), is called to be the spectral radius of G. The largest eigenvalue of Q(G), denoted by q(G), is called to be the signless Laplacian spectral radius of G.

A cycle (path) containing every vertex of a graph is called a Hamilton cycle (path) of the graph. Graph G is called a Hamilton graph if it has a Hamilton cycle, and then we also call G Hamiltonian. The number of components of G is denoted by c(G). Let t be a positive real number. A connected graph G is t-tough if tc(G − S) ≤ |S| for every vertex cut S of V(G). The toughness of G is the largest value of t for which G is t-tough, denoted by τ(G). If G is a complete graph, take τ(K_n) = ∞ for all n ≥ 1. If G is not a complete graph, τ(G)=min{|S|c(G−S):S⊆V(G),c(G−S)≥2} τ(G)=min{|S|c(G-S):S⊆V(G),c(G-S)≥2} , where the minimum is taken over all cut sets of vertices in G. Obviously, a t-tough graph is s-tough for all s < t.

On the one hand, more than 40 years ago, Chvátal (1973) introduced the concept of toughness. From then on a lot of research has been obtained, mainly relating to the relationship between toughness conditions and the existence of cyclic structures. Historically, most of the research was based on some conjectures in Chvátal (1973). D. Bauer etc. (Bauer et al., 1991, 1995a,b, 1999), surveyed results on toughness and its relationship to cycle structure. If we want to know more about the Hamiltonian problems related to toughness, we can refer to Bauer et al. (2006) and Huang et al. (2022). On the other hand, the problem of determining whether a graph is Hamiltonian is an NP-complete problem. In recent years, the study of Hamiltonian problem using spectrum graph theory has received extensive attention, and some meaningful results are obtained, such as Fiedler and Nikiforov (2010), Zhou (2010), Lu et al. (2012), Yu and Fan (2013), Liu et al. (2015), Li and Ning (2016), Feng et al. (2017), Zhou et al. (2018), and Yu et al. (2019). We would naturally think of what a t-tough graph is Hamiltonian when adding to other conditions. Inspired by the above results, in this paper, we establish some sufficient conditions that a graph with toughness is Hamiltonian based on the number of edges, spectral radius, and signless Laplacian spectral radius of the graph.

2. Preliminary

At the beginning of this section, we first give some definitions. Let G be a graph on n vertices. A vector X ∈ Rⁿ is called to be defined on the vertex set V(G) of the graph G, if there is a one-to-one mapping φ from vertex set V(G) of the graph to the components of the vector X; simply written X_u = φ(u).

When μ is an eigenvalue of the adjacency matrix A(G) corresponding to the eigenvector X if and only if X ≠ 0,

μXv=∑w∈NG(v)Xw,for each vertex v∈V(G). (2.1) μXv=∑w∈NG(v)Xw,for each vertex v∈V(G). (2.1)

The Equation (2.1) is called the characteristic equation of G.

When q is an eigenvalue of signless Laplacian matrix Q(G) corresponding to the eigenvector X if and only if X ≠ 0,

[q−dG(v)]Xv=∑w∈NG(v)Xw,for each vertex v∈V(G). (2.2) [q−dG(v)]Xv=∑w∈NG(v)Xw,for each vertex v∈V(G). (2.2)

The Equation (2.2) is called the signless Laplacian characteristic equation of G.

Lemma 2.1. Hoàng (1995) let t ∈ {1, 2, 3} and G be a t-tough graph with a non-decreasing degree sequence d₁ ≤ d₂ ≤ ⋯ ≤ d_n. If for all integers k with t≤k<n2 t≤k<n2 , d_k ≤ k implies d_n−k+t ≥ n − k, then G has a Hamilton cycle.

Lemma 2.2. Yuan (1988) let G be a connected graph with n vertices and m edges. Then

μ(G)≤2m−n+1,−−−−−−−−−−√ μ(G)≤2m-n+1,

and the equality holds if and only if G = K_n or G = K_1,n−1.

Lemma 2.3. Yu and Fan (2013) let G be a graph with n vertices and m edges. Then

q(G)≤2mn−1+n−2. q(G)≤2mn-1+n-2.

If G is connected, the equality holds if and only if G = K_1,n−1 or G = K_n. Otherwise, the equality holds if and only if G = K_n−1+v.

Lemma 2.4. Hoàng (1995) every Hamiltonian graph is 1-tough.

Lemma 2.5. Let the graph G be not a complete graph with minimum degree δ(G), and G is t-tough, then δ(G) ≥ 2t.

Proof Let δ(G) = d_v, S = N_G(v), then |S| = |N_G(v)| = δ(G). We can get c(G − S) ≥ 2. Because G is not a complete graph, by

τ(G)=min{|S|c(G−S):S⊆V(G),c(G−S)≥2}, τ(G)=min{|S|c(G-S):S⊆V(G),c(G-S)≥2},

then

t≤τ(G)≤|S|c(G−S)≤δ(G)2, t≤τ(G)≤|S|c(G-S)≤δ(G)2,

thus, we can get δ(G) ≥ 2t.

The proof is completed.■

Lemma 2.6. Jung (1978) let G be a graph without a Hamiltonian cycle and at least 11 vertices. Then

(i) there exist two non-adjacent vertices x, y such that d(x) + d(y) ≤ |V(G)| − 5 or

(ii) there exist for some t ≥ 1 vertices x₁, x₂, …, x_t such that G − x₁ − ⋯ − x_t has at least t + 1 components.

Corollary 2.7. Let t ∈ {1, 2, 3}, and G be a t-tough graph without a Hamiltonian cycle with at least 11 vertices. Then there exist two non-adjacent vertices x, y such that d(x) + d(y) ≤ |V(G)| − 5.

Proof Because G has no Hamiltonian cycle, G is not a complete graph. If there exist for some s ≥ 1 vertices x₁, x₂, …, x_s such that G − x₁ − ⋯ −x_s has at least s + 1 components, by

τ(G)=min{|S|c(G−S):S⊆V(G),c(G−S)≥2}, τ(G)=min{|S|c(G-S):S⊆V(G),c(G-S)≥2},

then

t≤τ(G)≤ss+1<1, t≤τ(G)≤ss+1<1,

a contradiction.

The proof is completed.■

The closure of a graph G, denoting by Cn(G) Cn(G) , is the graph obtained from G by recursively joining pairs of nonadjacent vertices whose degree sum is at least n until no such pair remains, refer to Bondy and Chvátal (1976).

Lemma 2.8. Bondy and Chvátal (1976) a graph G is Hamiltonian if and only if Cn(G) Cn(G) is Hamiltonian.

3. Main results

Theorem 3.1. Let G be a t-tough (t ∈ {1, 2, 3}) and simple connected graph with n(≥ 8t) vertices and m edges. If

m≥(n−2t2)+3t2, (3.1) m≥(n−2t2)+3t2, (3.1)

then

(i) G is Hamiltonian when t ∈ {1, 2} and n ≥ 8t.

(ii) G is Hamiltonian when t = 3 and n > 9t.

Proof Suppose that G is not a Hamilton graph. By Lemma 2.1, there exists a positive integer k for t≤k<n2 t≤k<n2 and d_k ≤ k, such that d_{n − k + t} ≤ n − k − 1. Then we have

2m=∑i=1kdi+∑i=k+1n−k+tdi+∑i=n−k+t+1ndi≤k2+(n−2k+t)(n−k−1)+(k−t)(n−1)=n2−n+3k2+(1−2n−t)k=2⎛⎝⎜n−2t2⎞⎠⎟+6t2−(k−2t)(2n−3k−5t−1), 2m=∑i=1kdi+∑i=k+1n-k+tdi+∑i=n-k+t+1ndi≤k2+(n-2k+t)(n-k-1)+(k-t)(n-1)=n2-n+3k2+(1-2n-t)k=2(n-2t2)+6t2-(k-2t)(2n-3k-5t-1),

thus

m≤(n−2t2)+3t2−(k−2t)(2n−3k−5t−1)2. (3.2) m≤(n−2t2)+3t2−(k−2t)(2n−3k−5t−1)2. (3.2)

Since ⎛⎝⎜n−2t2⎞⎠⎟+3t2≤m≤⎛⎝⎜n−2t2⎞⎠⎟+3t2−(k−2t)(2n−3k−5t−1)2 (n-2t2)+3t2≤m≤(n-2t2)+3t2-(k-2t)(2n-3k-5t-1)2 , thus (k − 2t)(2n − 3k − 5t − 1) ≤ 0. Next, we discuss three cases.

Case 1 t = 1.

In this case, n ≥ 8t = 8, (k − 2)(2n − 3k − 6) ≤ 0. By Lemma 2.5, δ (G)≥ 2t = 2. Since δ(G) ≤ d_k ≤ k, then k ≥ 2.

Case 1.1 (k − 2)(2n − 3k − 6) = 0, i.e., k = 2; or k ≠ 2 and 2n − 3k − 6 = 0.

Case 1.1.1 k = 2.

In this case, G is a graph with d₂ ≤ 2, d_{n − 1} ≤ n−3, d_n ≤ n − 1, and we have (n−2)(n−3)+6≤∑i=1ndi≤(n−2)(n−3)+6 (n−2)(n−3)+6≤∑i=1ndi≤(n−2)(n−3)+6 by (3.1) and (3.2). Thus, all inequalities of (3.2) become equality. In this time, G must be with degree sequence [2², (n − 3)^{n − 3}, n − 1].

If two 2-degree vertices of G are non-adjacent, G must be the graph K₁ ∨ (K_{n − 3} − uv + wu + zv), where u, v ∈ V(K_{n − 3}), w, z ∉ V(K_{n − 3}), is Hamiltonian, a contraction.

If two 2-degree vertices of G are adjacent, G must be the graph K₁ ∨ (K₂ + K_{n − 3}), but τ[K1∨(K2+Kn−3)]≤12, τ[K1∨(K2+Kn-3)]≤12, a contraction.

Case 1.1.2 k ≠ 2 and 2n − 3k − 6 = 0.

In this case, we can get n ≤ 11 because k<n2 k<n2 , and hence n = 9, k = 4. Then d₄ ≤ 4, d₆ ≤ 4, d₉ ≤ 8, and we have 48≤∑i=19di≤48 48≤∑i=19di≤48 by (3.1) and (3.2). Thus, all inequalities of (3.2) become equality. G must be with degree sequence (4⁶, 8³), and G = K₃ ∨ (3K₂) is Hamiltonian, a contradiction.

Case 1.2 (k − 2)(2n − 3k − 6) < 0.

In this case, k ≥ 3 and 2n − 3k − 6 < 0. Since k<n2 k<n2 , we have n ≥ 2k + 1. From these results, we have 4k + 2 ≤ 2n ≤ 3k + 5, that is k ≤ 3. Thus, we have k = 3, n = 7. A contradiction with known condition n ≥ 8.

Case 2 t = 2.

In this case, (k − 4)(2n − 3k − 11) ≤ 0. By Lemma 2.5, δ(G)≥ 2t = 4. Since δ(G) ≤ d_k ≤ k, then k ≥ 4.

Case 2.1 (k − 4)(2n − 3k − 11) = 0, i.e., k = 4; or k ≠ 4 and 2n − 3k − 11 = 0.

Case 2.1.1 k = 4.

In this case, G is a graph with d₄ ≤ 4, d_{n − 2} ≤ n − 5, d_n ≤ n − 1, and we have (n−4)(n−5)+24≤∑i=1ndi≤(n−4)(n−5)+24 (n−4)(n−5)+24≤∑i=1ndi≤(n−4)(n−5)+24 by (3.1) and (3.2). Thus, all inequalities of (3.2) become equality. During this time, G is the graph with degree sequence (4⁴, (n − 5)^{n − 6}, (n − 1)²), and n ≥ 8t = 16 for t = 2. Let S is the set containing four 4 − degree vertices of G, and there exist two non-adjacent vertices in S by Corollary 2.7.

By Lemma 2.8, Cn(G) Cn(G) is also not Hamiltonian. According to the definition of Cn(G) Cn(G) , all points except the vertices of S form a complete graph K_{n − 4} in Cn(G) Cn(G) . Let us discuss graph Cn(G) Cn(G) . If one vertex of the K_{n − 4} is adjacent to one vertex of S, it must be adjacent to all vertices of S. Moreover, there are at least 2 vertices of the K_{n − 4} adjacent to all vertices of S because there exist two non-adjacent vertices in S by Corollary 2.7.

If there are 2 vertices of the K_{n − 4} adjacent to all vertices of S, then Cn(G)⊇K2∨(Kn−6+K2,2) Cn(G)⊇K2∨(Kn-6+K2,2) . Since K₂ ∨ (K_{n − 6} + K_2,2) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are 3 vertices of the K_{n − 4} adjacent to all vertices of S, then Cn(G)⊇K3∨(Kn−7+2K2) Cn(G)⊇K3∨(Kn-7+2K2) . Since K₃ ∨ (K_{n − 7} + 2K₂) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are 4 vertices of the K_{n − 4} adjacent to all vertices of S, then Cn(G)⊇K4∨(Kn−8+4K1) Cn(G)⊇K4∨(Kn-8+4K1) . Since K₄ ∨ (K_{n − 8} + 4K₁) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are more than 4 vertices of the K_{n − 4} adjacent to all vertices of S, Cn(G)⊇Ki∨(Kn−4−i+4K1)(i>4) Cn(G)⊇Ki∨(Kn-4-i+4K1)(i>4) . Since K_i ∨ (K_{n − 4 − i} + 4K₁) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

Case 2.1.2 k ≠ 4 and 2n − 3k − 11 = 0.

In this case, we can get 16 ≤ n ≤ 21 because k<n2 k<n2 and 16 = 8t ≤ n, hence n = 16, k = 7. Then d₇ ≤ 7, d₁₁ ≤ 8, d₁₆ ≤ 15, and we have 156≤∑i=116di≤156 156≤∑i=116di≤156 by (3.1) and (3.2). Thus, ∑i=116di=156 ∑i=116di=156 , and all inequalities of (3.2) become equality. The corresponding permissible graphic sequence is (7⁷, 8⁴, 15⁵). Then there are no two vertices x, y that are not adjacent such that d(x) + d(y) ≤ ∣V(G)∣ − 5. By Corollary 2.7, we can get G is Hamiltonian, a contradiction.

Case 2.2 (k − 4)(2n − 3k − 11) < 0.

In this case, k ≥ 5 and 2n − 3k − 11 < 0. Since k<n2 k<n2 , we have n ≥ 2k + 1. From these results, we have 4k + 2 ≤ 2n ≤ 3k + 10, that is k ≤ 8. Thus, we have 16 = 8t ≤ n ≤ 17.

When n = 16, we have k = 7 from 4k + 2 ≤ 2n ≤ 3k + 10. At this time 2n − 3k − 11 = 0, which contradicts to 2n − 3k − 11 < 0.

When n = 17, we have k = 8 from 4k + 2 ≤ 2n ≤ 3k + 10. In this case, d₈ ≤ 8, d₁₁ ≤ 8, d₁₇ ≤ 16. We have 180≤∑i=117di≤184 180≤∑i=117di≤184 by (3.1) and (3.2). Then, ∑i=117di=184 ∑i=117di=184 or ∑i=117di=182 ∑i=117di=182 or ∑i=117di=180 ∑i=117di=180 .

When ∑i=117di=184 ∑i=117di=184 or ∑i=117di=182 ∑i=117di=182 . The corresponding permissible degree sequence and its properties are as follows in Table 1. By Corollary 2.7, we can get G is Hamiltonian, a contradiction.

TABLE 1

Table 1. The degree sequence of G and its properties.

When ∑i=117di=180 ∑i=117di=180 , The corresponding permissible degree sequence and its properties are shown in the follows in Table 2.

TABLE 2

Table 2. The degree sequence of G and its properties.

From Table 2, we can find that:

(1) for degree sequence (4,8¹⁰,16⁶) and (5,7,8⁹,16⁶). They are not graphic, a contradiction.

(2) for degree sequence (6²,8⁹,16⁶). If the corresponding graphs are not Hamiltonian, there must exist two non-adjacent 6-degree vertices by Corollary 2.7, and the corresponding graphs are isomorphic to K₆ ∨ (C₉ + 2K₁) or K₆ ∨ (C₄ + C₅ + 2K₁) or K₆ ∨ (C₆ + K₃ + 2K₁). We can find these graphs are Hamiltonian, a contraction.

(3) the other degree sequence except (4,8¹⁰,16⁶), (5,7,8⁹,16⁶), and (6²,8⁹,16⁶) in Table 2, there are no two vertices x, y that are not adjacent such that d(x) + d(y) ≤ ∣V(G)∣ − 5. By Corollary 2.7, we can get that G is Hamiltonian, a contradiction.

Case 3 t = 3.

In this case, (k − 6)(2n − 3k − 16) ≤ 0. By Lemma 2.5, δ(G) ≥ 2t = 6. Since δ ≤ d_k ≤ k, then k ≥ 6.

Case 3.1 (k − 6)(2n − 3k − 16) = 0, i.e., k − 6 = 0 or k ≠ 6 and 2n − 3k − 16 = 0.

Case 3.1.1 k = 6.

In this case, G is a graph with d₆ ≤ 6, d_{n − 3} ≤ n − 7, d_n ≤ n − 1. We have (n−6)(n−7)+54≤∑i=1ndi≤(n−6)(n−7)+54 (n-6)(n-7)+54≤∑i=1ndi≤(n-6)(n-7)+54 by (3.1) and (3.2), thus ∑i=1ndi=(n−6)(n−7)+54 ∑i=1ndi=(n-6)(n-7)+54 . During this time, we have the corresponding permissible degree sequence of G is (6⁶, (n − 7)^{n − 9}, (n − 1)³), and we have n > 9t = 27 because t = 3. Let S is the set containing six 6 − degree vertices of G.

By Lemma 2.8, Cn(G) Cn(G) is also not Hamiltonian. According to the definition of Cn(G) Cn(G) , all points except the vertices of S form a complete graph K_{n − 6} in Cn(G) Cn(G) . Let us discuss the graph Cn(G) Cn(G) . If one vertex of the K_{n − 6} is adjacent to one vertex of S, it must be adjacent to all vertices of S. Moreover, there are at least 2 vertices of the K_{n − 6} adjacent to all vertices of S because there exist two non-adjacent vertices in S by Corollary 2.7.

If there are 2 vertices of the K_{n − 6} adjacent to all vertices of S, then Cn(G)⊇K2∨(Kn−8+C6) Cn(G)⊇K2∨(Kn-8+C6) . Since K₂ ∨ (K_{n − 8} + C₆) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are 3 vertices of the K_{n − 6} adjacent to all vertices of S, then Cn(G)⊇K3∨(Kn−9+C6) Cn(G)⊇K3∨(Kn-9+C6) . Since K₃ ∨ (K_{n − 9} + C₆) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are 4 vertices of the K_{n − 6} adjacent to all vertices of S, then Cn(G)⊇K4∨(Kn−10+C6) Cn(G)⊇K4∨(Kn-10+C6) or Cn(G)⊇K4∨(Kn−10+2C3) Cn(G)⊇K4∨(Kn-10+2C3) . Since K₄ ∨ (K_{n − 10} + C₆) and K₄ ∨ (K_{n − 10} + 2C₃) are Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are 5 vertices of the K_{n − 6} adjacent to all vertices of S, then Cn(G)⊇K5∨(Kn−11+3K2) Cn(G)⊇K5∨(Kn-11+3K2) . Since K₅ ∨ (K_{n − 11} + 3K₂) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

If there are 6 vertices of the K_{n − 6} adjacent to all vertices of S, then Cn(G)⊇K6∨(Kn−12+6K1) Cn(G)⊇K6∨(Kn-12+6K1) , so C_n(G) is Hamiltonian, a contradiction.

If there are more than 6 vertices of the K_{n − 6} adjacent to all vertices of S, Cn(G)⊇Ki∨(Kn−6−i+6K1)(i>6) Cn(G)⊇Ki∨(Kn-6-i+6K1)(i>6) . Since K_i ∨ (K_{n − 6 − i} + 6K₁)(i > 6) is Hamiltonian, then Cn(G) Cn(G) is Hamiltonian, a contradiction.

Case 3.1.2 k ≠ 6 and 2n − 3k − 16 = 0.

In this case, we can get 28 ≤ n ≤ 31 because k<n2 k<n2 and n > 9t = 27. Since 2n − 3k − 16 = 0, then n = 29, k = 14. So d₁₄ ≤ 14, d₁₈ ≤ 14, d₂₉ ≤ 28. We have 560≤∑i=129di≤560 560≤∑i=129di≤560 by (3.1) and (3.2), thus ∑i=129di=560 ∑i=129di=560 . The corresponding permissible degree sequence is (14¹⁸, 28¹¹). For this degree sequence, we can get that there are no two vertices x, y that are not adjacent such that d(x) + d(y) ≤∣ V(G) ∣ − 5. By Corollary 2.7, we can get G is Hamiltonian, a contradiction.

Case 3.2 (k − 6)(2n − 3k − 16) < 0.

In this case, k ≥ 7 and 2n − 3k − 16 < 0. Since k<n2 k<n2 , we have n ≥ 2k + 1. From these results, we have 4k + 2 ≤ 2n ≤ 3k + 15, that is k ≤ 13. Thus, we have n ≤ 27. A contradiction with known condition n > 9t = 27. ■

Theorem 3.2 Let G be a t − tough(t ∈ {1, 2, 3}) and simple connected graph with n vertices and m edges. If

μ(G)≥n2−4tn−2n+10t2+2t−1−−−−−−−−−−−−−−−−−−−−−−−−√, μ(G)≥n2-4tn-2n+10t2+2t-1,

then

(i) G is Hamiltonian when t ∈ {1, 2} and n ≥ 8t.

(ii) G is Hamiltonian when t = 3 and n > 9t.

Proof Suppose that G is not Hamiltonian. Since K_n is Hamiltonian, then G ≠ K_n. By theorem conditions, we can get n ≥ 8 when t ∈ {1, 2, 3}, thus τ(K1,n−1)≤1n−1≤17<1 τ(K1,n-1)≤1n-1≤17<1 . Thus, G ≠ K_{1, n − 1}.

By Lemma 2.2,

n2−4tn−2n+10t2+2t−1−−−−−−−−−−−−−−−−−−−−−−−−√≤μ(G)<2m−n+1−−−−−−−−−−√, n2-4tn-2n+10t2+2t-1≤μ(G)<2m-n+1,

then

m>⎛⎝⎜n−2t2⎞⎠⎟+3t2. m>(n-2t2)+3t2.

By Theorem 3.1, we get G is Hamiltonian, a contradiction. ■

Theorem 3.3 Let G be a t − tough(t ∈ {1, 2, 3}) and simple connected graph with n vertices and m edges. If

q(G)≥n2+10t2−4nt+2t−n−2+(n−1)(n−2)n−1, q(G)≥n2+10t2-4nt+2t-n-2+(n-1)(n-2)n-1,

then

(i) G is Hamiltonian when t ∈ {1, 2} and n ≥ 8t.

(ii) G is Hamiltonian when t = 3 and n > 9t.

Proof Suppose that G is not a Hamilton graph. Since K_n is Hamiltonian, then G ≠ K_n and G ≠ K_{n − 1} + v. By theorem conditions, we can get n ≥ 8 when t ∈ {1, 2, 3}, thus τ(G)≤1n−1≤17<1 τ(G)≤1n-1≤17<1 . So, G ≠ K_n, G ≠ K_{1, n − 1}.

By Lemma 2.3,

n2+10t2−4nt+2t−n−2+(n−1)(n−2)n−1≤q(G)<2mn−1+n−2, n2+10t2-4nt+2t-n-2+(n-1)(n-2)n-1≤q(G)<2mn-1+n-2,

then

m>⎛⎝⎜n−2t2⎞⎠⎟+3t2. m>(n-2t2)+3t2.

By Theorem 3.1, we get G is Hamiltonian, a contradiction.

■

4. Concluding remarks

We suggest the following general problems.

Problem 1 Let G be a t − tough(t ∈ {1, 2, 3}) and simple connected graph with n vertices and m edges. If

m≥⎛⎝⎜n−2t2⎞⎠⎟+3t2, m≥(n-2t2)+3t2,

then

(i) when t ∈ {1, 2} and n ≥ 8t, G is pancyclic.

(ii) when t = 3 and n > 9t, G is pancyclic.

Problem 2 Let G be a t − tough(t ∈ {1, 2, 3}) and simple connected graph with n vertices and m edges. If

μ(G)≥n2−4tn−2n+10t2+2t−1−−−−−−−−−−−−−−−−−−−−−−−−√, μ(G)≥n2-4tn-2n+10t2+2t-1,

then

(i) when t ∈ {1, 2} and n ≥ 8t, G is pancyclic.

(ii) when t = 3 and n > 9t, G is pancyclic.

Problem 3 Let G be a t − tough(t ∈ {1, 2, 3}) and simple connected graph with n vertices and m edges. If

q(G)≥n2+10t2−4nt+2t−n−2+(n−1)(n−2)n−1, q(G)≥n2+10t2-4nt+2t-n-2+(n-1)(n-2)n-1,

then

(i) when t ∈ {1, 2} and n ≥ 8t, G is pancyclic.

(ii) when t = 3 and n > 9t, G is pancyclic.

Data availability statement

The original contributions presented in the study are included in the article/supplementary material, further inquiries can be directed to the corresponding author.

Author contributions

GY provide writing ideas and master the thesis as a whole. GC contributed significantly to analysis and manuscript preparation. TY performed the data analyses and wrote the manuscript. HX helped perform the analysis with constructive discussions. All authors contributed to the article and approved the submitted version.

Funding

This study was supported by the National Natural Science Foundation of China (No. 11871077), the NSF of Anhui Province (1808085MA04 and 1908085MC62), the NSF of Anhui Provincial Department of Education (KJ2020A0894 and KJ2021A0650), Graduate Scientific Research Project of Anhui Provincial Department of Education (YJS20210515), Research and Innovation Team of Hefei Preschool Education College (KCTD202001), and Graduate offline course graph theory (2021aqnuxskc02).

Conflict of interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

Publisher's note

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