Hamiltonian cycles in tough $(P_2\cup P_3)$-free graphs

TL;DR

This paper proves that every 15-tough $(P_2rac12;P_3)$-free graph with at least three vertices contains a Hamiltonian cycle, advancing understanding of toughness conditions for Hamiltonicity.

Contribution

It establishes a new toughness threshold for Hamiltonicity specifically in $(P_2rac12;P_3)$-free graphs, a class previously not fully characterized.

Findings

15-tough $(P_2rac12;P_3)$-free graphs are Hamiltonian

Provides a specific toughness bound for Hamiltonicity in this graph class

Advances the toughness-Hamiltonicity conjecture in restricted graph classes

Abstract

Let $t>0$ be a real number and $G$ be a graph. We say $G$ is $t$-tough if for every cutset $S$ of $G$, the ratio of $|S|$ to the number of components of $G-S$ is at least $t$. Determining toughness is an NP-hard problem for arbitrary graphs. The Toughness Conjecture of Chv\'atal, stating that there exists a constant $t_0$ such that every $t_0$-tough graph with at least three vertices is hamiltonian, is still open in general. A graph is called $(P_2\cup P_3)$-free if it does not contain any induced subgraph isomorphic to $P_2\cup P_3$, the union of two vertex-disjoint paths of order 2 and 3, respectively. In this paper, we show that every 15-tough $(P_2\cup P_3)$-free graph with at least three vertices is hamiltonian.

Equations

$$C^{\prime }=(C-\{uw\})\cup \{ux,xw\}$$

$$|W|>\frac{n}{t+1}.$$

$$C^{\prime }=u^{+}\stackrel{\rightharpoonup }{C}wxu\stackrel{\leftharpoonup }{C}{w}^{+}{u}^{+}$$

$$\frac{|S|}{c(G-S)}=\frac{|S|}{|W|}<\frac{\frac{tn}{t+1}}{\frac{n}{t+1}}=t,$$

$$C^{\prime }=(C-\{uw\})\cup \{ux,zw\}\cup P$$

$$u\in N_G(x)\cap V(C)$$

$$W_x$$

$$W_z$$

$$|W_x|=|N_G(x)\cap V(C)|>\frac{4.5n}{16},\text{mbox}and|W_z|=|N_G(z)\cap V(C)|>\frac{4.5n}{16}.$$

$$C^{\prime }=u^{+}\stackrel{\rightharpoonup }{C}wzPxu\stackrel{\leftharpoonup }{C}{w}^{+}{u}^{+}$$

$$E_G(W_x,W_z)=\varnothing .$$

$$N_G(x)\cap V(C)$$

$$(N_G(x)\cap V(C))\cap W_x=\varnothing ,\text{mbox}and(N_G(z)\cap V(C))\cap W_z=\varnothing .$$

$$W_{xz}=W_x\cap W_z.$$

$$n$$

$$S_0$$

$$S_2$$

$$\frac{n-|S|}{c(G-S)}$$

$$Q_1$$

$$Q_2$$

$$Q_3$$

$$Q_i\cap Q_j=\varnothing ,i\ne j,i,j=1,2,3.$$

$$W$$

$$\frac{|N_G(W_1)\cap S_2|}{c(G-(N_G(W_1)\cap S_2))}<4<15.$$

$$M$$

$$C=u_1{\stackrel{\rightharpoonup }{C}}_1{v}_1{x}_1{u}_2{\stackrel{\rightharpoonup }{C}}_2{v}_2\cdots u_{t-1}{\stackrel{\rightharpoonup }{C}}_{t-1}{v}_{t-1}{x}_{t-1}{u}_t{\stackrel{\rightharpoonup }{C}}_t{v}_t{x}_t{u}_1$$

$$G_1=G[V(D_1)\cup Q_3]\text{mbox}and{G}_2=G[V(D_2)\cup Q_2].$$

$$C=x_1{P}_1{y}_1{\stackrel{\rightharpoonup }{C}}_2{y}_2{P}_2{x}_2{\stackrel{\leftharpoonup }{C}}_1{x}_1$$

$$C=x_{1}x{x}_2{\stackrel{\rightharpoonup }{C}}_2{y}_{2}y{y}_1{\stackrel{\leftharpoonup }{C}}_1{x}_1$$

$$|V(C)|$$

$$n-|S|-\frac{2|S|}{15}\ge \frac{3n}{20}$$

$$|V(D_1)|+|V(D_2)|\ge n-|S|-\frac{|S|}{15}\ge \frac{n}{5}.$$

Full text

Hamiltonian cycles in tough $(P_{2}\cup P_{3})$-free graphs

Songling Shan

Illinois State University, Normal, IL 61790

sshan12@ilstu.edu

(8 January 2019)

Abstract. Let $t>0$ be a real number and $G$ be a graph. We say $G$ is $t$-tough if for every cutset $S$ of $G$, the ratio of $|S|$ to the number of components of $G-S$ is at least $t$. Determining toughness is an NP-hard problem for arbitrary graphs. The Toughness Conjecture of Chvátal, stating that there exists a constant $t_{0}$ such that every $t_{0}$-tough graph with at least three vertices is hamiltonian, is still open in general. A graph is called $(P_{2}\cup P_{3})$-free if it does not contain any induced subgraph isomorphic to $P_{2}\cup P_{3}$, the union of two vertex-disjoint paths of order 2 and 3, respectively. In this paper, we show that every 15-tough $(P_{2}\cup P_{3})$-free graph with at least three vertices is hamiltonian.

Keywords. Toughness; Hamiltonian cycle; $(P_{2}\cup P_{3})$-free graph

1 Introduction

Graphs considered in this paper are simple, undirected, and finite. Let $G$ be a graph. Denote by $V(G)$ and $E(G)$ the vertex set and edge set of $G$, respectively. For $v\in V(G)$, $N_{G}(v)$ denotes the set of neighbors of $v$ in $G$. For $S\subseteq V(G)$ and $x\in V(G)$, define $deg_{G}(x,S)=|N_{G}(x)\cap S|$. If $H\subseteq G$, we simply write $deg_{G}(x,H)$ for $deg_{G}(x,V(H))$. We skip the subscript $G$ if the graph in consideration is clear from the context. Let $S\subseteq V(G)$. Then the subgraph induced on $V(G)-S$ is denoted by $G-S$. For notational simplicity, we write $G-x$ for $G-\{x\}$. If $uv\in E(G)$ is an edge, we write $u\sim v$. Let $V_{1},V_{2}\subseteq V(G)$ be two disjoint vertex sets. Then $E_{G}(V_{1},V_{2})$ is the set of edges of $G$ with one end in $V_{1}$ and the other end in $V_{2}$.

The number of components of $G$ is denoted by $c(G)$. Let $t\geq 0$ be a real number. The graph is said to be $t$-tough if $|S|\geq t\cdot c(G-S)$ for each $S\subseteq V(G)$ with $c(G-S)\geq 2$. The toughness $\tau(G)$ is the largest real number $t$ for which $G$ is $t$-tough, or is $\infty$ if $G$ is complete. This concept, a measure of graph connectivity and “resilience” under removal of vertices, 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 (Chvátal’s toughness conjecture). Bauer, Broersma and Veldman [2] have constructed $t$-tough graphs that are not hamiltonian for all $t<\frac{9}{4}$, so $t_{0}$ must be at least $\frac{9}{4}$.

There are a number of papers on Chvátal’s toughness conjecture, and it has been verified when restricted to a number of graph classes [3], including planar graphs, claw-free graphs, co-comparability graphs, and chordal graphs. A graph $G$ is called $2K_{2}$-free if it does not contain two independent edges as an induced subgraph. Recently, Broersma, Patel and Pyatkin [5] proved that every 25-tough $2K_{2}$-free graph on at least three vertices is hamiltonian, and the author of this paper improved the required toughness in this result from 25 to 3 [13].

Let $P_{\ell}$ denote a path on $\ell$-vertices. A graph is $(P_{2}\cup P_{3})$-free if it does not contain any induced copy of $P_{2}\cup P_{3}$, the disjoint union of $P_{2}$ and $P_{3}$. In this paper, we confirm Chvátal’s toughness conjecture for the class of $(P_{2}\cup P_{3})$-free graphs, a superclass of $2K_{2}$-free graphs.

Theorem 1**.**

Let $G$ be a $15$-tough $(P_{2}\cup P_{3})$-free graph with at least three vertices. Then $G$ is hamiltonian.

In [10] it was shown that every 3/2-tough split graph on at least three vertices is hamiltonian. And the authors constructed a sequence $\{G_{n}\}_{n=1}^{\infty}$ of split graphs with no 2-factor and $\tau(G_{n})\rightarrow 3/2$. So $3/2$ is the best possible toughness for split graphs to be hamiltonian. Since split graphs are $(P_{2}\cup P_{3})$-free, we cannot decrease the bound in Theorem 1 below 3/2. Although it is certain that 15-tough is not optimal, we are not sure about the “best possible” toughness for giving a hamiltonian cycle in a $(P_{2}\cup P_{3})$-free graph.

The class of $2K_{2}$-free graphs is well studied, for instance, see [5, 6, 8, 9, 11, 12]. It is a superclass of split graphs, where the vertices can be partitioned into a clique and an independent set. One can also easily check that every cochordal graph (i.e., a graph that is the complement of a chordal graph) is $2K_{2}$-free and so the class of $2K_{2}$-free graphs is at least as rich as the class of chordal graphs. By the definition, the class of $(P_{2}\cup P_{3})$-free graphs is a superclass of $2K_{2}$-free graphs but with much more complicated structures than graphs that are $2K_{2}$-free. The proof techniques used in [5] and [13] for showing that certain tough $2K_{2}$-free graphs are hamiltonian seem to be not applicable for $(P_{2}\cup P_{3})$-free graphs. The proof approach used in this paper for showing Theorem 1 is new and more general and reveals some structural properties of $(P_{2}\cup P_{3})$-free graphs.

2 Proof of Theorem 1

We start this section with some definitions. Let $G$ be a graph and $S\subseteq V(G)$ a cutset of $G$, and let $D$ be a component of $G-S$. For a vertex $x\in S$, we say that $x$* is adjacent to $D$* if $x$ is adjacent in $G$ to a vertex of $D$. We call $D$ a clique component of $G-S$ if $V(D)$ is a clique in $G$. We call $D$ a trivial component of $G-S$ if $D$ has only one vertex.

A star-matching is a set of vertex-disjoint copies of stars. The vertices of degree at least 2 in a star-matching are called the centers of the star-matching. In particular, if all the stars in a star-matching are isomorphic to $K_{1,t}$, where $t\geq 1$ is an integer, we call the star-matching a $K_{1,t}$-matching. For a star-matching $M$, we denote by $V(M)$ the set of vertices covered by $M$.

Let $C$ be an oriented cycle. For $x\in V(C)$, denote the successor of $x$ on $C$ by $x^{+}$ and the predecessor of $x$ on $C$ by $x^{-}$. For $u,v\in V(C)$, $u\overset{\rightharpoonup}{C}v$ denotes the portion of $C$ starting at $u$, following $C$ in the orientation, and ending at $v$. Likewise, $u\overset{\leftharpoonup}{C}v$ is the opposite portion of $C$ with endpoints as $u$ and $v$. We assume all cycles in consideration afterwards are oriented. A path $P$ connecting two vertices $u$ and $v$ is called a $(u,v)$-path, and we write $uPv$ or $vPu$ in specifying the two endvertices of $P$. Let $uPv$ and $xQy$ be two paths. If $vx$ is an edge, we write $uPvxQy$ as the concatenation of $P$ and $Q$ through the edge $vx$.

Lemma 2.1** ([1], Theorem 2.10).**

Let $G$ be a bipartite graph with partite sets $X$ and $Y$, and let $f$ be a function from $X$ to the set of positive integers. If for every $S\subseteq X$, it holds that $|N_{G}(S)|\geq\sum\limits_{v\in S}f(v)$ , then $G$ has a subgraph $H$ such that $X\subseteq V(H)$, $d_{H}(u)=f(u)$ for every $u\in X$, and $d_{H}(v)=1$ for every $v\in Y\cap V(H)$.

Lemma 2.2** (Bauer et al. [4]).**

Let $t>0$ be real and $G$ be a $t$-tough $n$-vertex graph ($n\geq 3$) with $\delta(G)>\frac{n}{t+1}-1$. Then $G$ is hamiltonian.

Lemmas 2.3 and 2.4 below are consequences of $(P_{2}\cup P_{3})$-freeness.

Lemma 2.3**.**

Let $G$ be a $(P_{2}\cup P_{3})$-free graph and $S\subseteq V(G)$ a cutset of $G$. If $G-S$ has a component that is not a clique component, then all other components of $G-S$ are trivial. Consequently, if $G-S$ has at least two nontrivial components, then all components of $G-S$ are clique components.

Lemma 2.4**.**

Let $G$ be a $(P_{2}\cup P_{3})$-free graph and $S\subseteq V(G)$ a cutset of $G$, and let $x\in S$. Suppose that $x$ is adjacent to exactly one component $D$ of $G-S$, and $G-S$ has a nontrivial component to which $x$ is not adjacent, then $x$ is adjacent in $G$ to all vertices of $D$.

Lemma 2.5**.**

Let $G$ be a connected $(P_{2}\cup P_{3})$-free graph and $S\subseteq V(G)$ a cutset of $G$ such that each vertex in $S$ is adjacent to at least two components of $G-S$. Then each of the following statement holds.

1. (i)

For every nontrivial clique component $D\subseteq G-S$ and for every vertex $x\in S$, $x$ is adjacent to $D$. 2. (ii)

For every nontrivial clique component $D\subseteq G-S$ and for every vertex $x\in S$, if $x$ is adjacent to at least three components, then $x$ is adjacent in $G$ to at least $|V(D)|-1$ vertices of $D$. 3. (iii)

Let $D_{1}$ and $D_{2}$ be two nontrivial clique components of $G-S$. Then for every vertex $x\in S$, either $x$ is adjacent in $G$ to at least $|V(D_{i})|-1$ vertices of each $D_{i}$, or $x$ is adjacent in $G$ to all vertices of one of $D_{i}$, $i=1,2$.

Proof. For (i), let $w_{1}$ and $w_{2}$ be two neighbors of $x$ in $G$ respectively from two distinct components of $G-S$. Then $w_{1}xw_{2}$ is an induced $P_{3}$. Now for every edge $uv\in E(D)$, we must have that $\{w_{1},w_{2}\}\cap\{u,v\}\neq\emptyset$ or $x$ is adjacent in $G$ to $u$ or $v$, by the $(P_{2}\cup P_{3})$-freeness. Therefore, $x$ is adjacent to $D$. For (ii), let $x\in S$ and $D$ be a nontrivial clique component of $G-S$. Since $x$ is adjacent to at least three components, there exists $u,w$, respectively from two components of $G-S$ that are distinct from $D$ such that $x\sim u$ and $x\sim w$ in $G$. Thus, $uxw$ is an induced $P_{3}$ in $G$. Furthermore, since $u,w\in V(G)-S-V(D)$, $E_{G}(\{u,w\},V(D))=\emptyset$. Thus, by the $(P_{2}\cup P_{3})$-freeness assumption, for every edge in $D$, $x$ is adjacent to at least one endvertex of that edge. This, together with the fact that $D$ is a clique, we know that $x$ is adjacent at least $|V(D)|-1$ vertices in $D$. For (iii), assume to the contrary that the statement does not hold. By symmetry, we assume that there exists $uv\in E(D_{1})$ such that $x\not\sim u,v$ in $G$, and there exists $w\in V(D_{2})$ such that $x\not\sim w$ in $G$. Let $y\in V(D_{2})\cap N_{G}(x)$ that exists by Lemma 2.5 (i). Then $uv\cup xyw$ is an induced $P_{2}\cup P_{3}$, giving a contradiction. ∎

Lemma 2.6**.**

Let $t\geq 1$ and $G$ be an $n$-vertex $t$-tough graph, and let $C$ be a non-hamiltonian cycle of $G$. If $x\in V(G)-V(C)$ satisfies that $deg(x,C)>\frac{n}{t+1}$, then $G$ has a cycle $C^{\prime}$ such that $V(C^{\prime})=V(C)\cup\{x\}$.

Proof. It is clear that if $x$ is adjacent to two consecutive vertices $u,w$ on $C$, then

[TABLE]

is a cycle with the desired property. So we assume that for any $u,w\in N_{G}(x)\cap V(C)$, $uw\not\in E(C)$. Let $W=\{u^{+}\,|\,u\in N_{G}(x)\cap V(C)\}$ be the set of the successors of the neighbors of $x$ on $C$. Because there is a one-to-one correspondence between $W$ and $N_{G}(x)\cap V(C)$, by the assumption that $deg(x,C)>\frac{n}{t+1}$, we know that

[TABLE]

If there exist $u^{+},w^{+}\in W$ with $u,w\in N_{G}(x)\cap V(C)$ such that $u^{+}\sim w^{+}$ in $G$, then

[TABLE]

is a desired cycle. Therefore, we assume that $W$ is an independent set in $G$. Let $S=V(G)-W$. Then $c(G-S)=|W|$, as $W$ is an independent set in $G$. However, by (1),

[TABLE]

showing a contradiction to the toughness of $G$. ∎

Lemma 2.7**.**

Let $G$ be an $n$-vertex $15$-tough $(P_{2}\cup P_{3})$-free graph, and let $C$ be a non-hamiltonian cycle of $G$. Let $P\subseteq G-V(C)$ be an $(x,z)$-path. If both $x$ and $z$ are adjacent to more than $\frac{4.5n}{16}$ vertices in $V(C)$, then $G$ has a cycle $C^{\prime}$ such that $V(C^{\prime})=V(C)\cup V(P)$.

Proof. It is clear that if $x$ is adjacent to a vertex $u$ on $C$ and $z$ is adjacent to a vertex $w$ on $C$ such that $uw\in E(C)$, then

[TABLE]

is a cycle with the desired property. So we assume that

[TABLE]

Let

[TABLE]

Clearly,

[TABLE]

If there exist $u^{+}\in W_{x}$ and $w^{+}\in W_{z}$ with $u\in N_{G}(x)\cap V(C)$ and $w\in N_{G}(z)\cap V(C)$ such that $u^{+}\sim w^{+}$ in $G$, then

[TABLE]

is a desired cycle. Therefore, we assume that

[TABLE]

We further claim that

[TABLE]

By symmetry, we only show that no two vertices in $N_{G}(x)\cap V(C)$ are consecutive on $C$.

Assume to the contrary that there exists a path $v_{1}v_{2}\cdots v_{\ell}\subseteq C$ with $\ell\geq 2$ such that for each $i$ with $1\leq i\leq\ell$, $v_{i}\in N_{G}(x)\cap V(C)$, $v_{1}^{-}\not\in N_{G}(x)\cap V(C)$, and $v_{\ell}^{+}\not\in N_{G}(x)\cap V(C)$. Note that such vertices $v_{1}$ and $v_{\ell}$ exist by the assumption in (2). We assume that there exist $w_{1},w_{2}\in W_{z}$ such that $w_{1}\sim w_{2}$ in $G$. For otherwise, letting $S=V(G)-W_{z}$ gives that $c(G-S)=|W_{z}|$, and consequently, $|S|/c(G-S)=|S|/|W_{z}|<15$ by (3).

Then $xv_{\ell}v_{\ell}^{+}$ is an induced $P_{3}$ in $G$. Consider the edge $w_{1}w_{2}$. By the assumption in (2), $x\not\sim w_{1},w_{2}$ in $G$ (otherwise, $w_{1}^{-}w_{1}\in E(C)$ or $w_{2}^{-}w_{2}\in E(C)$ with $w_{1}^{-},w_{2}^{-}\in N_{G}(z)\cap V(C)$), and by the assumption in (4), $v_{\ell}^{+}\not\sim w_{1},w_{2}$ in $G$. Thus, $v_{\ell}\sim w_{1}$ or $v_{\ell}\sim w_{2}$ in $G$ by the $(P_{2}\cup P_{3})$-freeness assumption. However, $v_{\ell}=v_{\ell-1}^{+}\in W_{x}$, showing a contradiction to (4).

Therefore, by (5),

[TABLE]

Let

[TABLE]

By the assumption in (4), $W_{xz}$ is an independent set in $G$. By the toughness of $G$, we know that $|W_{xz}|\leq\frac{n}{16}$. Therefore, $|N_{G}(x)\cap N_{G}(z)\cap V(C)|\leq\frac{n}{16}$. These, together with (3) and (6), imply that

[TABLE]

showing a contradiction. ∎

Lemma 2.8**.**

Let $G$ be an $n$-vertex $15$-tough $(P_{2}\cup P_{3})$-free graph with $n\geq 31$, and let $S\subseteq V(G)$ be a cutset of $G$ with $|S|\leq\frac{3n}{4}$. Assume that $G-S$ has at least two nontrivial clique components, and that for every edge $uv\in E(G)$, $d(u)+d(v)\geq|S|$. Then $G$ has a hamiltonian cycle.

Proof. By Lemma 2.3, every component of $G-S$ is a clique component. If there exists $x\in S$ such that $x$ is adjacent to exactly one component, say $D$ of $G-S$, then we move $x$ from $S$ into $D$. By Lemma 2.4, every component of $G-(S-\{x\})$ is still a clique component. We move out all such vertex $x$ from $S$ iteratively and denote the remaining vertices in $S$ by $S_{1}$. Note that $S_{1}\neq\emptyset$, since $G$ is a connected graph and $S$ is a cutset of $G$. Also, $c(G-S)=c(G-S_{1})$ and $G-S_{1}$ has at least two nontrivial components. By Lemma 2.3, every component of $G-S_{1}$ is a clique component. Let

[TABLE]

Note that $S_{2}=S_{1}-S_{0}$.

Since $G-S_{1}$ has a nontrivial component that has no edge going to $S_{0}$, the $(P_{2}\cup P_{3})$-freeness of $G$ implies that $G[S_{0}]$ consists of vertex-disjoint complete subgraphs of $G$. Thus $S_{2}$ is a cutset of $G$ with components consisting those from $G-S_{1}$ and $G[S_{0}]$. Also, all components of $G-S_{2}$ are clique components in which at least two of them are nontrivial. By the toughness of $G$, $|S_{2}|\geq 15c(G-S_{2})$.

We will construct a hamiltonian cycle in $G$ through two steps: (1) combing spanning cycles from every clique component of $G-S_{2}$ that has at least three vertices into a single cycle $C$, and (2) insterting remaining vertices in $V(G)-V(C)$ into $C$ to obtain a hamiltonian cycle of $G$.

Suppose that $G-S_{2}$ has exactly $h$ clique components $D_{1},\cdots,D_{h}$ with $|V(D_{1})|\geq|V(D_{2})|\geq\cdots\geq|V(D_{h})|\geq 1$, and that the first $t$ ($0\leq t\leq h$) of them are components that contain at least three vertices. Since $G-S_{2}$ has at least two nontrivial components, both $D_{1}$ and $D_{2}$ are nontrivial.

Claim 1**.**

The component $D_{1}$ contains at least 5 vertices.

Proof: Since $|S|\leq\frac{3n}{4}$, $n\geq\frac{4|S|}{3}$. Also, $c(G-S)\leq\frac{|S|}{15}$ by $\tau(G)\geq 15$. Therefore, a largest component of $G-S$ contains at least

[TABLE]

vertices. $\blacksquare$

Let

[TABLE]

By Lemma 2.5 (i) and the definition of $Q_{1}$, we know that if $Q_{1}\neq\emptyset$, then every vertex in $Q_{1}$ is adjacent to at least three components of $G-S_{2}$. By Lemma 2.5 (ii), we get the following claim.

Claim 2**.**

Suppose that $Q_{1}\neq\emptyset$. Then for every $x\in Q_{1}$ and for every nontrivial component $D$ of $G-S$, $x$ is adjacent to at least $|V(D)|-1$ vertices of $D$.

Claim 3**.**

Suppose that $Q_{2}\neq\emptyset$. Then for every $x\in Q_{2}$, $x$ is adjacent to all vertices of $D_{2}$ and $Q_{2}$ is a clique in $G$.

Proof: Note that both $D_{1}$ and $D_{2}$ are nontrivial components of $G-S_{2}$. Since $|V(D_{1})|\geq 5$ by Claim 1, $x$ is not adjacent to at least three vertices of $D_{1}$ by the definition of $Q_{2}$. Therefore, $x$ is adjacent to all vertices of $D_{2}$ by Lemma 2.5 (iii). For the second part, suppose to the contrary that there exist $x,y\in Q_{2}$ such that $x\not\sim y$ in $G$. Let $w\in V(D_{2})$. Then $w\sim x$ and $w\sim y$ in $G$ by the first part of this claim. Thus, we find an induced $P_{3}=xwy$. Since $E_{G}(\{w\},V(D_{1}))=\emptyset$, the $(P_{2}\cup P_{3})$-freeness implies that for every edge in $D_{1}$, at least one of $x$ and $y$ is adjacent to at least one endpoint of the edge. Since $D_{1}$ is complete, by Pigeonhole Principle, one of $x$ and $y$ is adjacent to at least $\frac{|V(D_{1})|-1}{2}$ vertices of $D_{1}$. This gives a contradiction to the assumption that $x,y\in Q_{2}$. $\blacksquare$

Similarly, we have the following result.

Claim 4**.**

Suppose that $Q_{3}\neq\emptyset$. Then for every $x\in Q_{3}$, $x$ is adjacent to all vertices of $D_{1}$ and $Q_{3}$ is a clique in $G$.

By Claims 2 to 4, we have that

[TABLE]

Define

[TABLE]

If $W\neq\emptyset$, we claim that there is a $K_{1,2}$-matching $M$ between $W$ and $S_{2}$ such that every vertex in $W$ is the center of a $K_{1,2}$-star. For otherwise, by Theorem 2.1, there exists $W_{1}\subseteq W$ such that $2|W_{1}|>|N_{G}(W_{1})\cap S_{2}|$. By the definition of $W$, we then see that $G-(N_{G}(W_{1})\cap S_{2})$ has at least $|W_{1}|/2+1\geq 2$ components (vertices in $S_{2}-(N_{G}(W_{1})\cap S_{2})$ form at least one component of $G-(N_{G}(W_{1})\cap S_{2})$ that is disjoint from those containing vertices from $W_{1}$), implying that

[TABLE]

This gives a contradiction to the toughness.

[TABLE]

Claim 5**.**

If $M\neq\emptyset$, then for every $x\in V(M)\cap S_{2}$, $x\in Q_{1}$. Consequently, for every nontrivial component $D$ of $G-S$, $x$ is adjacent to at least $|V(D)|-1$ vertices of $D$.

Proof: If $G-S_{2}$ has at least three nontrivial components, then every vertex of $S_{2}$ is adjacent to all those nontrivial components by Lemma 2.5 (i). Therefore, $S_{2}=Q_{1}$ by the definition of $Q_{1}$. In particularly, $x\in Q_{1}$ for $x\in V(M)\cap S_{2}$. Hence, we assume that $G-S_{2}$ has exactly two nontrivial components, which are $D_{1}$ and $D_{2}$. This assumption implies that $|V(D_{3})|\leq 1$. Consequently, $|V(D_{3})|=1$ since $M\neq\emptyset$. Then for every $x\in V(M)\cap S_{2}$, $x$ is adjacent to both $D_{1}$ and $D_{2}$ by Lemma 2.5 (i). Also $x$ is adjacent to a trivial component of $G-S_{2}$. Thus $x\in Q_{1}$. The second part of Claim 5 is a consequence of Claim 2. $\blacksquare$

Claim 6**.**

There is a cycle $C$ in $G-V(M)$ with at least $\frac{3n}{20}$ vertices such that $C$ contains all vertices from every $D_{i}$, $i=1,2,\cdots,t$, and $Q_{2}\cup Q_{3}\subseteq V(C)$.

Proof: Suppose first that $G-S_{2}$ has at least three nontrivial components. Then by Lemma 2.5 (i), every vertex of $S_{2}$ is adjacent to all those nontrivial components of $G-S_{2}$. Consequently, $S_{2}=Q_{1}$ and $Q_{2}=Q_{3}=\emptyset$. Therefore, for every $x\in S_{2}$ and every $D_{i}$, $x$ is adjacent to at least $|V(D_{i})|-1$ vertices of $D_{i}$ by Claim 2.

Let $x_{1},\cdots,x_{t}$ be $t$ distinct vertices in $S_{2}-V(M)$. (By the toughness of $G$, $|S_{2}|\geq 15c(G-S_{2})$. Since $|V(M)\cap S_{2}|\leq 4c(G-S_{2})$, we have enough vertices in $S_{2}-V(M)$ to pick.) Let $C_{i}$ be a hamiltonian cycle of $D_{i}$, and let $u_{i},v_{i}\in V(C_{i})$ with $u_{i}v_{i}\in E(C_{i})$ such that for $i=1,2,\cdots,t-1$, $x_{i}\sim v_{i},u_{i+1}$, and $x_{t}\sim u_{1},v_{t}$ in $G$. Then

[TABLE]

is a cycle that contains all vertices from each $D_{i}$ and the vertices $x_{1},\cdots,x_{t}$ from $S_{2}-V(M)$. Also $Q_{2}\cup Q_{3}\subseteq V(C)$ trivially as $Q_{2}=Q_{3}=\emptyset$.

So we assume that $G-S_{2}$ has exactly two nontrivial clique components, which are $D_{1}$ and $D_{2}$, call this assumption ($\ast$). Let

[TABLE]

By Claims 3 and 4, we know that both $G_{1}$ and $G_{2}$ are complete subgraphs of $G$.

If $t=1$ and $Q_{2}=\emptyset$, we let $C$ be a hamiltonian cycle of $G_{1}$. Clearly, $Q_{2}\cup Q_{3}\subseteq V(C)$.

Thus, we assume that $t\geq 2$ or $Q_{2}\neq\emptyset$. Since $D_{1}$ has at least three vertices by Claim 1, $G_{1}$ contains at least three vertices. Note that $|V(D_{2})|\geq 2$ by the assumption that $G-S_{2}$ has at least two nontrivial components and $D_{2}$ is one of them. Thus, $G_{2}$ contains at least three vertices either by $t\geq 2$ or $Q_{2}\neq\emptyset$.

If there are two disjoint edges between $G_{1}$ and $G_{2}$, then $G[V(G_{1})\cup V(G_{2})]$ has a hamiltonian cycle $C$. Thus, we may assume, without loss of generality, that there is either no edge between $G_{1}$ and $G_{2}$ or all edges between $G_{1}$ and $G_{2}$ are incident to only a single vertex, say in $G_{1}$.

If $c(G-S_{2})=2$, then $M=\emptyset$ by the definitions of $W$ and $M$. Since $G[V(G_{1})\cup V(G_{2})]$ has a cutvertex or is disconnected, the toughness of $G$ implies that $|S_{2}-Q_{2}-Q_{3}|\geq 29$. In addition, there are vertex-disjoint paths $P_{1}$ and $P_{2}$ connecting $G_{1}$ and $G_{2}$ in $G$ such that each $P_{i}$ only has exactly one of its endvertices in $G_{1}$ and $G_{2}$. Let $V(P_{i})\cap V(G_{1})=\{x_{i}\}$ and $V(P_{i})\cap V(G_{2})=\{y_{i}\}$, $i=1,2$. Let $C_{1}$ be a hamiltonian cycle in $G_{1}$ such that $x_{1}x_{2}\in E(C_{1})$, and $C_{2}$ be a hamiltonian cycle in $G_{2}$ such that $y_{1}y_{2}\in E(C_{2})$. Then

[TABLE]

is a cycle that contains all vertices in clique components of $G-S_{2}$ that contain at least three vertices and the vertices from $P_{1}$ and $P_{2}$. Also $Q_{2}\cup Q_{3}\subseteq V(C)$ by the construction of $C$.

So we assume that $c(G-S_{2})\geq 3$. Since $D_{1}$ and $D_{2}$ are the only nontrivial components of $G-S_{2}$ by assumption ($\ast$), the assumption that $c(G-S_{2})\geq 3$ implies that $|Q_{1}|\geq 15(c(G-S_{2})-2)$. Thus $|Q_{1}-V(M)|\geq 15(c(G-S_{2})-2)-2(c(G-S_{2})-2)\geq 13$, since each trivial component of $G-S$ uses exactly two vertices from $S_{2}\cap V(M)$. Hence, we can find two vertices $x,y\in Q_{1}-V(M)=S_{2}-Q_{2}-Q_{3}-V(M)$ such that both $x$ and $y$ are adjacent to at least $|V(D_{1})|-1$ vertices of $D_{1}$, and at least $|V(D_{2})|-1$ vertices of $D_{2}$ by Claim 5. We claim that $x$ is adjacent to at least two vertices of $G_{2}$. This is clear if $x$ is adjacent to at least two vertices of $D_{2}$. So we assume that $x$ is adjacent to only one vertex of $D_{2}$. Let $w$ be the neighbor of $x$ from $D_{2}$, and let $w_{1}\in V(D_{2})-\{w\}$, $w_{2}\in V(G_{2})-V(D_{2})$. By this choice, $x\not\sim w_{1},w_{2}$ in $G$. Note that $w_{1}$ is not adjacent to any vertex of $D_{1}$, and $w_{2}$ is adjacent to less than $\frac{|V(D_{1})|-1}{2}$ vertices of $D_{1}$. Therefore, we can find a vertex $w^{*}\in V(D_{1})$ such that $w_{1},w_{2}\not\sim w^{*}$ in $G$ and $x\sim w^{*}$ in $G$. By the choice of $x$, there is a vertex $w^{\prime}\in V(G)-S-V(D_{1})-V(D_{2})$ such that $x\sim w^{\prime}$ in $G$. However, $w_{1}w_{2}\cup w^{*}xw^{\prime}$ is an induced $P_{2}\cup P_{3}$. This gives a contradiction. Since $D_{1}$ has at least 5 vertices, both $x$ and $y$ have at least four neighbors in $D_{1}$. Thus we can select distinct vertices $x_{1},y_{1}\in V(G_{1})$ and $x_{2},y_{2}\in V(G_{2})$ such that $x\sim x_{1},x_{2}$ and $y\sim y_{1},y_{2}$ in $G$.

Let $C_{1}$ be a hamiltonian cycle of $G_{1}$ such that $x_{1}y_{1}\in E(C_{1})$, and let $C_{2}$ be a hamiltonian cycle of $G_{2}$ such that $x_{2}y_{2}\in E(C_{2})$. Then

[TABLE]

is a cycle that contains all vertices in clique components of $G-S_{2}$ that contain at least three vertices and the vertices $x$ and $y$. Furthermore, $Q_{2}\cup Q_{3}\subseteq V(C)$.

Since for each $i$, $1\leq i\leq t$, $V(D_{i})\subseteq V(C)$ and $\bigcup_{1\leq i\leq t}V(D_{i})\subseteq V(G)-S-W$, we have that

[TABLE]

$\blacksquare$

Claim 7**.**

Let $C$ be the cycle defined in Claim 6. For any $x\in S_{2}-V(C)$, $x$ has more than $\frac{n}{16}$ neighbors on $C$.

Proof: Note that every vertex in $S_{2}$ is adjacent to at least two components of $G-S_{2}$. If $G-S_{2}$ has at least three nontrivial clique components, then Lemma 2.5 (ii) implies that for every $x\in S_{2}$, and for every nontrivial clique component $D$ of $G-S_{2}$, $x$ is adjacent to at least $|V(D)|-1$ vertices of $D$. By the toughness of $G$ and the assumption that $|S|\leq\frac{3n}{4}$, $x$ has at least

[TABLE]

neighbors in the union of the nontrivial clique components of $G-S_{2}$ that contain at least three vertices. Therefore, $x$ has more than $\frac{n}{16}$ neighbors on $C$.

So we assume that $G-S_{2}$ has exactly two nontrivial clique components. Since $S_{2}-V(C)\subseteq S_{2}-Q_{2}-Q_{3}$ (recall that $Q_{2}\cup Q_{3}\subseteq V(C)$), we know that $x$ is adjacent to at least $\frac{|V(D_{1})|-1}{2}$ vertices of $D_{1}$, and is adjacent to at least $\frac{|V(D_{2})|-1}{2}$ vertices of $D_{2}$. Note that

[TABLE]

Since $C$ contains all vertices from $D_{1}\cup D_{2}$, we conclude that $x$ is adjacent to at least $\frac{n}{10}-1>\frac{n}{16}$ (by $n\geq 31$) neighbors on $C$. $\blacksquare$

By Claim 7, and by applying Lemma 2.6 for $C$ and vertices in $S_{2}-V(C)-V(M)$ iteratively, we get a longer cycle $C^{\prime}$ such that $V(C^{\prime})=V(C)\cup(S_{2}-V(C)-V(M))$. Note also that

[TABLE]

Recall that for every $x\in S_{2}-V(C^{\prime})=S_{2}\cap V(M)$, $x$ is adjacent to at least $|V(D_{i})|-1$ vertices in each $D_{i}$, $i=1,2,\cdots,t$ by Claim 5. Therefore, if $|S|\leq\frac{7n}{12}$, then $x$ has in $G$ at least

[TABLE]

neighbors on $C^{\prime}$. Then applying Lemma 2.7 for $C^{\prime}$ and every path in $M$ iteratively, we obtain a hamiltonian cycle in $G$. Hence we

[TABLE]

Claim 8**.**

For any two $K_{1,2}$-stars $x_{1}u_{1}y_{1},x_{2}u_{2}y_{2}\in M$, if $u_{1}u_{2}$ is a 2-vertex component of $G-S_{2}$ and $|S|>\frac{7n}{12}$, then at least one of $u_{1}$ and $u_{2}$ has more than $\frac{n}{16}$ neighbors on $C^{\prime}$.

Proof: For otherwise, since $u_{i}$ is adjacent to exactly one vertex in $V(M)\cap(V(G)-S_{2})$, and $|V(M)\cap S_{2}|\leq 4|V(M)\cap(V(G)-S_{2})|\leq\frac{4|S|}{15}$,

[TABLE]

show a contradiction to the assumption that for every edge $uv\in E(G)$, $d_{G}(u)+d_{G}(v)\geq|S|$. $\blacksquare$

Let

[TABLE]

Take $uwv\in M_{1}$, note that $u,v\in S_{2}$ and $w\in V(G)-S_{2}$. By the definition of $M_{1}$, $deg(w,C^{\prime})>\frac{n}{16}$. By Claim 7, $deg(u,C^{\prime})>\frac{n}{16}$ and $deg(v,C^{\prime})>\frac{n}{16}$. Now applying Lemma 2.6 for $C^{\prime}$ and every path in $M_{1}$ iteratively, we get a longer cycle $C^{*}$ such that $V(C^{*})=V(C^{\prime})\cup V(M_{1})$.

By the toughness of $G$, $G-S_{2}$ has at most $\frac{|S|}{15}$ components in total. Particularly, $G-S_{2}$ has at most $\frac{|S|}{15}$ components that have at most two vertices in total. By Claim 8, we know that for every 2-vertex component $uv$ of $G-S_{2}$, at least one of $u$ or $v$ has more than $\frac{n}{16}$ neighbors on $C^{\prime}$. Therefore, at least one of the two $K_{1,2}$-stars centered, respectively, at $u$ and $v$ is contained in $M_{1}$. In other words, there is at most one $K_{1,2}$-star from $M_{2}$ that centers at a vertex from a same component of $G-S_{2}$. Therefore,

[TABLE]

By the definition of $M_{2}$ and by the assumption that for any $uv\in E(G)$, $d_{G}(u)+d_{G}(v)\geq|S|$, we know that for any path $xwy\in M_{2}$, where $x,y\in V(G)-S_{2}$ and $w\in S_{2}$, we have that $d_{G}(x)+d_{G}(w)\geq|S|$. Therefore, the number of neighbors that $x$ has in $G$ on $C^{*}$ is at least

[TABLE]

Similarly, the vertex $y$ has in $G$ at least $\frac{4.5n}{16}$ neighbors on $C^{*}$. Now applying Lemma 2.7 for $C^{*}$ and every path in $M_{2}$ iteratively gives a hamiltonian cycle in $G$. ∎

Proof of Theorem 1.

Since $G$ is 15-tough, it is 30-connected, and consequently, $\delta(G)\geq 30$. By Lemma 2.2, we may assume that

[TABLE]

We consider two case to finish the proof.

**Case 1: For every edge $e=uv\in E(G)$, **$d_{G}(u)+d_{G}(v)>\frac{3n}{4}$.

Denote by

[TABLE]

By the assumption of Case 1, we know that $V_{1}$ is an independent set in $G$. Therefore,

[TABLE]

by $\tau(G)\geq 15$.

Since $G$ is 15-tough, Lemma 2.1 implies that $G$ has a $K_{1,2}$-matching $M$ with all vertices in $V_{1}$ as the centers of the $K_{1,2}$-matching. Let $V_{2}$ be the set of the vertices contained in $M$. By (12), we have that

[TABLE]

Denote by $G_{1}=G-V_{2}$. Then by the definitions of $V_{1},V_{2}$ and (13), we get that

[TABLE]

We first assume that $G_{1}$ has a hamiltonian cycle $C$. For every copy of $K_{1,2}$, say $xyz\in M$, by (15),

[TABLE]

Let

[TABLE]

By (16), applying Lemma 2.6 with respect to $C$ and every vertex in $M_{1}$ iteratively, we get a longer cycle $C^{*}$ such that $V(C^{*})=V(C)\cup V(M_{1})$.

By the definition of $M_{2}$ and by the assumption that for any $uv\in E(G)$, $d_{G}(u)+d_{G}(v)>\frac{3n}{4}$, we know that for any path $xwy\in M_{2}$, where $x,y\in V_{2}-V_{1}$ and $w\in V_{1}$, we have that $d_{G}(x)+d_{G}(w)>\frac{3n}{4}$. Therefore, the number of neighbors that $x$ has in $G$ on $C^{*}$ is at least

[TABLE]

Similarly, the vertex $y$ has in $G$ at least $\frac{4.5n}{16}$ neighbors on $C^{*}$. Now applying Lemma 2.7 for $C^{*}$ and every path in $M_{2}$ iteratively gives a hamiltonian cycle in $G$.

Hence we assume that $G$ does not have a hamiltonian cycle. By Lemma 2.2 and (14), we know that $\tau(G_{1})<7$. Therefore, there exists $S_{1}\subseteq V(G_{1})$ such that

[TABLE]

If $|S_{1}|\geq\frac{3n}{16}$, then we get that $c(G_{1}-S_{1})=c(G-(S_{1}\cup V_{2}))$, and thus by (13),

[TABLE]

showing a contradiction to $\tau(G)\geq 15$. So we assume that $|S_{1}|<\frac{3n}{16}$, thus $|S_{1}|\leq\lfloor\frac{3n}{16}\rfloor$. As $\delta(G_{1})\geq\lfloor\frac{3n}{16}\rfloor+1$ by (14), we know that each component of $G_{1}$ contains at least

[TABLE]

vertices. By Lemma 2.3, we know that every component of $G_{1}-S_{1}$ is a clique component. Let $S=S_{1}\cup V_{2}$. We then see that all components of $G-S$ are nontrivial. Also, $|S|<\frac{6n}{16}<\frac{3n}{4}$ since $|S_{1}|<\frac{3n}{16}$ and $|V_{2}|\leq\frac{3n}{16}$ by (13). Furthermore, by the assumption of Case 1, for every edge $uv\in E(G)$, $d_{G}(u)+d_{G}(v)>\frac{3n}{4}>|S|$. Now we can apply Lemma 2.8 on $G$ and $S$ to find a hamiltonian cycle in $G$.

**Case 2: There exists an edge $e=uv\in E(G)$ such that **$d_{G}(u)+d_{G}(v)\leq\frac{3n}{4}$.

Let

[TABLE]

such that $d_{G}(u)+d_{G}(v)$ is smallest among all the degree sums of two adjacent vertices in $G$.

By the assumption of this case and the choice of $S$, we know that

[TABLE]

By the definition of $S$, $c(G-S)\geq 2$ and $uv$ is one of the components of $G-S$. Since $\tau(G)\geq 15$, and $|V(G)-S-\{u,v\}|\geq n-|S|\geq\frac{|S|}{3}=\frac{5|S|}{15}$, $G-S-\{u,v\}$ has a component with at least 5 vertices. This, together with the fact that $uv$ is one of the components of $G-S$, Lemma 2.3 implies that every component of $G-S$ is a clique component, and $G-S$ has at least two nontrivial components. Again Lemma 2.8 implies that $G$ has a hamiltonian cycle. ∎