Spanning k-Ended Tree in 2-Connected Graph

Abstract

Win proved a very famous conclusion that states the graph G with connectivity $\kappa \left(G\right)$, independence number $\alpha \left(G\right)$ and $\alpha \left(G\right)\le \kappa \left(G\right)+k-1$$(k\ge 2)$ contains a spanning k-ended tree. This means that there exists a spanning tree with at most k leaves. In this paper, we strengthen the Win theorem to the following: Let G be a simple 2-connected graph such that $\left|V\right(G\left)\right|\ge 2\kappa \left(G\right)+k$, $\alpha \left(G\right)\le \kappa \left(G\right)+k$$(k\ge 2)$ and the number of maximum independent sets of cardinality $\kappa +k$ is at most $n-2\kappa -k+1$. Then, either G contains a spanning k-ended tree or a subgraph of $K_{\kappa }\vee ((k+\kappa -1)K_1\cup K_{n-2\kappa -k+1})$.

1. Introduction

Notation regarding graph theory is not covered in this paper. We refer the reader to [1]. Let $G=\left(V\right(G),E(G\left)\right)$ be a graph satisfying vertex set $V\left(G\right)$ and edge set $E\left(G\right)$. We denote the set of vertices adjacent to v in G as $N\left(v\right)$. We write $N\left(X\right)={\displaystyle \bigcup _{x\in X}}N\left(x\right)$ for $X\subseteq V\left(G\right)$. We also denote the subgraph of G induced by S as $G\left[S\right]$ for $S\subseteq V\left(G\right)$. Let $H_1$ and $H_2$ be two subgraphs of G which vertex disjoint, and P be a path of G. A path $xPy$ in G with end vertices $x,y\in V\left(G\right)$ is called a path from $H_1$ to $H_2$ if $V\left(xPy\right)\cap V\left(H_1\right)=\left\{x\right\}$ and $V\left(xPy\right)\cap V\left(H_2\right)=\left\{y\right\}$. $(x,U)$-path is a path from $\left\{x\right\}$ to a vertex set U. We write an $(x,U)$-fan of width k for $F\subseteq G$ if F is a union of $(x,U)$-paths $P_1,P_2,\dots ,P_k$, where $V\left(P_i\right)\cap V\left(P_j\right)=\left\{x\right\}$ for $i\ne j$. Let $G_1$ and $G_2$ be two subgraphs of G. We denote by $x{G}_1$$(G_{1}x$, respectively) the Hamilton path of $G[\left\{x\right\}\cup V\left(G_1\right)]$, which starts at x (terminates at x, respectively). We denote by $x{G}_{1}y$ the Hamilton path of $G[V\left(G_1\right)\cup \{x,y\}]$, which starts at x and terminates at y. We denote by $G_{1}x{G}_2$ the Hamilton path of $G[V\left(G_1\right)\cup \left\{x\right\}\cup V\left(G_2\right)]$. A nontrivial graph, G, is considered k-connected if the maximum number of pairwise internally disjoint $xy$-paths for any two distinct vertices, x and y, is greater than or equal to k. A trivial graph is considered 0-connected or 1-connected, but it is not considered k-connected for any k greater than 1. The connectivity, $\kappa \left(G\right)$, of G is defined as the maximum value of k for which G is k-connected.

If a graph contains a Hamilton path, then the graph is said traceable, and if a graph contains a Hamilton cycle, then the graph is said hamiltonian. The sufficient conditions under which a graph can be traceable involving connectivity ($\kappa \left(G\right)$) and independence number ($\alpha \left(G\right)$) were given by Chvátal and Erdős in 1972.

Theorem 1.

(Chvátal and Erdős, [2]) If a graph G with $\left|V\right(G\left)\right|\ge 3$ satisfies the conditions $\alpha \left(G\right)\le \kappa \left(G\right)$, $\alpha \left(G\right)\le \kappa \left(G\right)+1$, respectively, then G is Hamiltonian and traceable, respectively.

Theorem 1 has been extended in various directions, as documented in previous studies [3,4,5,6,7,8]. For recent results, see [9,10,11,12]. Fouquet and Jolivet [13] conjecture whether a graph’s circumference can have a best possible lower bound when its independence number exceeds its connectivity. This has been proved by Suil O et al.

Theorem 2.

(Suil O et al., [14]) If G is a simple graph such that $\left|V\right(G\left)\right|=n$ and $\alpha \left(G\right)\ge \kappa \left(G\right)$, then G contains a cycle with length of at least ${\displaystyle \frac{\kappa \left(G\right)(n-\alpha (G)-\kappa (G\left)\right)}{\alpha \left(G\right)}}$.

The number of maximum independent sets of H for a subgraph $H\subseteq G$ is denoted by $m\left(H\right)$. In their study [15], Chen et al. presented the following theorem that generalizes Theorem 1 by bound $m\left(G\right)$. Specifically, the authors demonstrated that expanding the independence number (i.e., $\alpha \left(G\right)\le \kappa \left(G\right)+2$) slightly and bounding m(G) does not alter the traceability of G. It is worth noting that $K_s$ represents a complete graph with s vertices, while $\overline{K_s}$ is the complement of $K_s$. Additionally, the join $G\vee H$ of disjoint graphs G and H is obtained by joining each vertex of G to each vertex of H in the graph $G+H$.

In the following, we construct three graphs which are excluded. Let $H_i\left(k_i\right)$ be a copy of ${\overline{K}}_{k_i}$ where $i=1,2$. The graph $F_0(k_1,k_2)$ is defined as $(H_1\left(k_1\right)\vee H_2\left(k_2\right))\cup K_{n-k_1-k_2}\cup M_1\left(k_2\right)$, where $n-k_1-k_2\ge k_2$ and $M_1\left(k_2\right)$ is a matching of cardinality $k_2$ between $H_2\left(k_2\right)$ and $K_{n-k_1-k_2}$. If $n-k_1-k_2\ge k_2$, then $F_{11}(k_1,k_2)$ is obtained from $F_0(k_1,k_2)$ by joining exactly two (nonadjacent) vertices of $H_2\left(k_2\right))$ or by joining all vertices of $V\left(K_{n-k_1-k_2}\right)\backslash V\left(M_1\left(k_2\right)\right)$ and some fixed vertex $w_0\in H_2\left(k_2\right)$. Let $F_{00}(k_1,k_2)$ be the graph $(H_1\left(k_1\right)\vee H_2\left(k_2\right))\cup K_{n-k_1-k_2}\cup M_2\left(k_2\right)$, where $n-k_1-k_2\le k_2$ and $M_2\left(k_2\right)$ is a matching of cardinality $n-k_1-k_2$ between $K_{n-k_1-k_2}$ and $H_2\left(k_2\right)$. Define the graph $F_2(k_1,k_2)=K_{k_2}\vee (\overline{K_{k_1}}\cup K_{n-k_1-k_2})$; see Figure 1.

Figure 1. $F_0(k_1,k_2)$, $F_{00}(k_1,k_2)$ and $F_2(k_1,k_2)$.

Theorem 3.

(Chen et al., [15]) Let G be a 2-connected graph with $\left|V\left(G\right)\right|\ge 2{\kappa }^2\left(G\right)$, $\kappa \left(G\right)=\kappa$, $\alpha \left(G\right)\le \kappa +1$ and $m\left(G\right)\le n-2\kappa$. Then, either G is Hamiltonian or $F_{11}(\kappa ,\kappa )\subseteq G\subseteq F_2(\kappa ,\kappa )$, where $F_{11}(\kappa ,\kappa )$ and $F_2(\kappa ,\kappa )$ are two graphs defined above.

Theorem 4.

(Chen et al., [15]) Let G be a connected graph with $\left|V\left(G\right)\right|\ge 2{\kappa }^2\left(G\right)$, $\kappa \left(G\right)=\kappa$, $\alpha \left(G\right)\le \kappa +2$ and $m\left(G\right)\le n-2\kappa -1$. Then either G is traceable or $F_{11}(\kappa +1,\kappa )\subseteq G\subseteq K_{\kappa }\vee ((\kappa +1)K_1\cup K_{n-2\kappa -1})$, where $F_{11}(\kappa +1,\kappa )$ is the graph defined above.

A Hamilton path is viewed as a spanning tree with exactly two leaves. This perspective allows for the generalization of sufficient conditions for a graph to be traceable to those for a spanning tree with at most k leaves. A tree is called a k-ended tree if it has at most k leaves. Our focus now shifts to spanning k-ended trees. Clearly, if $s\le t$, then a spanning s-ended tree is also a spanning t-ended tree. Theorem 1 demonstrates that each graph G such that $\alpha \left(G\right)\le \kappa \left(G\right)+1$ is traceable. In [16], Win proved the following theorem, which generalizes Theorem 1.

Theorem 5.

(Win, [16]) Let G be a connected graph and let $k\ge 2$ be an integer. If $\alpha \left(G\right)\le \kappa \left(G\right)+k-1$, then G contains a spanning k-ended tree.

In [17], Lei et al. extend Theorem 5 in cases when $\kappa \left(G\right)=1$ to the following direction.

Theorem 6.

(Lei et al., [17]) Let $k\ge 3$ and G be a connect graph with $\left|V\right(G\left)\right|\ge 2k+2$, $\alpha \left(G\right)\le 1+k$ and $m\left(G\right)\le n-2k-2$. Then G contains a spanning k-ended tree.

In [15], Chen et al. generalize Theorem 1 by bound $m\left(G\right)$. The authors demonstrated that expanding the independence number (i.e., $\alpha \left(G\right)\le \kappa \left(G\right)+2$) slightly and bounding $m\left(G\right)$ does not alter the traceability of G.

In this paper, our focus will be on the existence of spanning k-ended tree. We will work on extending Theorem 5 to a more general case. A natural question is whether expanding the independence number can alter the existence of the spanning k-ended tree. In the following section, we introduce the k-ended system, which is an important tool for studying the k-ended tree.

k-Ended System

If there exists a set of paths and cycles where the elements are pairwise vertex-disjoint, we refer to it as a system. This system is often viewed as a subgraph. Let $\mathcal{S}$ be a system in a graph. For $S\in \mathcal{S}$, we put $f\left(S\right)=2$ if S is a path of order at least 3 and $f\left(S\right)=1$ otherwise (i.e., S is a vertex, an edge or a cycle). We write $V\left(\mathcal{S}\right)={\displaystyle \bigcup _{S\in \mathcal{S}}}V\left(S\right)$ and $f\left(\mathcal{S}\right)={\displaystyle \sum _{S\in \mathcal{S}}}f\left(S\right)$. If $f\left(\mathcal{S}\right)\le k$, $\mathcal{S}$ is called a k-ended system. Moreover, we call $\mathcal{S}$ a spanning k-ended system of G, if $V\left(\mathcal{S}\right)=V\left(G\right)$. Let

$${\mathcal{S}}_P=\{S\in \mathcal{S}:f\left(S\right)=2\},{\mathcal{S}}_C=\{S\in \mathcal{S}:f\left(S\right)=1\}.$$

Then,

$$\mathcal{S}={\mathcal{S}}_P\cup {\mathcal{S}}_C,V\left(\mathcal{S}\right)=\bigcup _{S\in \mathcal{S}}V\left(S\right).$$

Additionally, $V\left({\mathcal{S}}_P\right)$ and $V\left({\mathcal{S}}_C\right)$ can be defined in a similar manner. We use $\left|\mathcal{S}\right|$, $|{\mathcal{S}}_P|$ and $|{\mathcal{S}}_C|$ to represent the number of elements in $\mathcal{S}$, ${\mathcal{S}}_P$ and ${\mathcal{S}}_C$, respectively. For each $S\in \mathcal{S}$, we assign an orientation denoted by the symbol <, where $x<y$ if x precedes y in the orientation. Let $\overrightarrow{S}$ be the orientation of $S\in \mathcal{S}$ and let $\overleftarrow{S}$ be the reverse orientation of $\overrightarrow{S}$ for $S\in \mathcal{S}$. By assigning an orientation to each $S\in \mathcal{S}$, we identify $\mathcal{S}$ as a system with an orientation, where each element is ordered relative to the others.

Let $\mathcal{S}$ be a system with a defined orientation. For any $P\in {\mathcal{S}}_P$, we define $v_L\left(P\right)$ and $v_R\left(P\right)$ as the two end-vertices of P such that $v_L\left(P\right)<v_R\left(P\right)$. Additionally, for each $C\in {\mathcal{S}}_C$, we select an arbitrary vertex $v_C$ within C. These definitions will be used in subsequent analyses. Then define

$$End\left({\mathcal{S}}_P\right)=\bigcup _{P\in {\mathcal{S}}_P}\{v_L\left(P\right),v_R\left(P\right)\},End\left({\mathcal{S}}_C\right)=\bigcup _{C\in {\mathcal{S}}_C}\left\{v_C\right\},End\left(\mathcal{S}\right)=End\left({\mathcal{S}}_P\right)\cup End\left({\mathcal{S}}_C\right).$$

For $S\in \mathcal{S}$ and $x\in V\left(S\right)$, we write the first, second and ith predecessor (successor, respectively) of x as $x^{-}$, $x^{--}$ and $x^{i-}$ ($x^{+}$, $x^{++}$ and $x^{i+}$, respectively). For convenience, we write $x=x^{+}=x^{-}$ for $K_1=x$ and $y=x^{+}=x^{-}$ for $K_2=xy$.

For $P\in {\mathcal{S}}_P$, if $x=v_R\left(P\right)$$(x=v_L\left(P\right)$, respectively), we have only the predecessor of $v_R\left(P\right)$ (successor of $v_L\left(P\right)$, respectively). For $\{x,y\}\subseteq V\left(P\right)$, we denote by the section $P(x,y)$ a path $x^{+}{x}^{2+}{x}^{3+}...x^{s+}(=y^{-})$ of consecutive vertices of P and denote by the section $P[x,y]$ a path $x{x}^{+}{x}^{2+}...x^{s+}(=y)$ of consecutive vertices of P. Moreover, if $x=y$, then the section $P[x,y]$ is trivial.

The following lemma illustrates the importance of k-ended systems for spanning k-ended trees.

Lemma 1.

(Win, [16]) Let $k\ge 2$ be an integer and let G be a connected simple graph. If G contains a spanning k-ended system, then G also contains a spanning k-ended tree.

A k-ended system $\mathcal{S}$ in G is considered a maximal k-ended system if there is no other k-ended system $\widehat{\mathcal{S}}$ in G satisfying $V\left(\mathcal{S}\right)\subset V\left(\widehat{\mathcal{S}}\right)$. The following lemma presents some useful properties of k-ended systems. It is important to note that two distinct elements of $\mathcal{S}$ are connected by a path in $G-V\left(\mathcal{S}\right)$ if there exists a path in G whose end-vertices are in elements of $\mathcal{S}$ and whose inner vertices are not all contained in $V\left(\mathcal{S}\right)$. It is worth mentioning that a path may not have any inner vertex.

Lemma 2.

(Akiyama and Kano, [18]) Let $k\ge 2$ be an integer and G be a connected simple graph. Assume that G does not contain a spanning k-ended system and let $\mathcal{S}$ be a maximal k-ended system of G satisfying the cardinality of the maximum value of ${\mathcal{S}}_P$ subject to the maximum value of $V\left(\mathcal{S}\right)$. Then the following characterizations are true.

(i)

There is no path connecting two distinct elements of ${\mathcal{S}}_C$ whose inner vertices are in $V\left(G\right)\backslash V\left(\mathcal{S}\right)$.

(ii)

There is no path connecting an element of ${\mathcal{S}}_C$ and one end-vertex of an element of ${\mathcal{S}}_P$ whose inner vertices are in $V\left(G\right)\backslash V\left(\mathcal{S}\right)$.

(iii)

There is no path connecting an end-vertex of an element of ${\mathcal{S}}_P$ and an end-vertex of another element of ${\mathcal{S}}_P$ whose inner vertices are in $V\left(G\right)\backslash V\left(\mathcal{S}\right)$.

(iv)

There are no two internally disjoint paths $Q_1$ and $Q_2$ connecting two distinct elements of ${\mathcal{S}}_C$ whose inner vertices are in $V\left(G\right)\backslash V\left(\mathcal{S}\right)$ with $|V\left(Q_1\right)\cap V\left(Q_2\right)|=1$.

2. Methods

In this paper, our focus will be on the existence of spanning k-ended tree. We will work on extending Theorem 5 to a more general case. We tried to prove that it does not change the existence of spanning k-ended tree if we expand the independent numbers a little bit and bound m(G). The proof will follow an approach similar to Theorem 6, but with additional considerations for the increased connectivity of the graph. Our proof follows a method of contradiction. We primarily utilize the crucial tool of the maximal k-ended system, as mentioned above, to derive contradictions. The subsequent section is the crucial property of the maximal k-ended system which we obtained. This property plays a pivotal role in our proof.

Important Properties of Maximal k-Ended System Based on Theorem 7

In this section, for convenience, we assume the following: Let $k\ge 2$ and G be a graph with $\left|V\right(G\left)\right|>2\kappa \left(G\right)+k$, $\kappa \left(G\right)=\kappa \ge 2$, $\alpha \left(G\right)=\kappa +k$ and $m\left(G\right)\le n-2\kappa -k+1$. Suppose that there is no spanning k-ended system in G and let $\mathcal{S}$ be a k-ended system of G satisfying the following:

(I)

The cardinality of the set $V\left(\mathcal{S}\right)$ is maximized.

(II)

The cardinality of ${\mathcal{S}}_P$ is maximized subject to condition $\left(I\right)$.

Then $\mathcal{S}$ is a set of subgraphs of G satisfying the hypothesis of Lemma 2. Let $H=G-V\left(\mathcal{S}\right)$. Then $\left|V\right(H\left)\right|\ge 1$. Let $w\in \left(V\right(G)-V(\mathcal{S})$. The following lemma is easily obtained from the selection of $\mathcal{S}$ and we omit the proof.

Lemma 3.

The following characterizations are true.

(1)

For any $P\in {\mathcal{S}}_P$, $v_L\left(P\right)$ and $v_R\left(P\right)$ are not in $N\left(H\right)$.

(2)

For any $C\in {\mathcal{S}}_C$ such that $C\cong K_1$, $N\left(H\right)\cap V\left(C\right)=\varnothing$.

By the Fan Lemma, there exists a $(w,V(\mathcal{S}\left)\right)$-fan $\mathcal{L}$ with width $\kappa$. For $S\in \mathcal{S}$ with $V\left(S\right)\cap V\left(\mathcal{L}\right)\ne \varnothing$, let $V\left(S\right)\cap V\left(\mathcal{L}\right)=\{u_{S,1},\cdots ,u_{S,t_S}\}$ (where $u_{S,1},\cdots ,u_{S,t_S}$ are the vertices of S along the direction of S) and $L_{S,i}$ be the path of $\mathcal{L}$ between w and $u_{S,i}$. Then $U={\bigcup }_{S\in \mathcal{S}}\{V\left(S\right)\cap V\left(\mathcal{L}\right)\}$ and $L_S=\{L_{S,1},\cdots ,L_{S,t_S}\}$ is the set of paths between w and S. Denote $U^{+}=\{u^{+}:u\in U\}$ and $U^{-}=\{u^{-}:u\in U\}$. By Lemma 3(1),(2), $U^{+}$ and $U^{-}$ are well defined and hence $|U^{+}|=|U^{-}|=|U|$.

The proof of the following lemmas can be easily obtained from the choice of $\mathcal{S}$ and we will omit it.

Lemma 4.

A graph G cannot have a $k^{\prime }$-ended system $\mathcal{T}$ that includes all vertices in a k-ended system $\mathcal{S}$, where $k^{\prime }<k$.

Lemma 5.

The following characterizations are true:

(1)

Both $U^{+}$ and $End\left(\mathcal{S}\right)$ are independent sets of G.

(2)

$N\left(w\right)\cap U^{+}=\varnothing$.

(3)

$\mid \{C\in {\mathcal{S}}_C:V\left(C\right)\cap U\ne \varnothing \}\mid \le 1$.

(4)

Let $x\in U^{+}\cap V\left(P\right)$, where $P\in {\mathcal{S}}_P$. Then $N\left(x\right)\cap (End\left({\mathcal{S}}_P\right)\backslash \left\{v_R\left(P\right)\right\})=\varnothing$. Furthermore, if $x\ne u_{P,t_P}^{+}$, then $N\left(x\right)\cap End\left({\mathcal{S}}_P\right)=\varnothing$.

Lemma 6.

Let $v_1\in End\left(\mathcal{S}\right)$ and $S_1\in \mathcal{S}$ with $v_1\in V\left(S_1\right)$. Then the following statements are true:

(1)

$[{(N\left(v_1\right)\cap V\left(S\right))}^{-}\cup {(N\left(v_1\right)\cap V\left(S\right))}^{+}]\cap N\left(v\right)=\varnothing$ for any $v\in V({\mathcal{S}}_{\mathcal{C}}-\{S,S_1\})\cup End(\mathcal{S}-\{S,S_1\})$ and $S\in \mathcal{S}-\left\{S_1\right\}$.

(2)

If $v_1=v_L\left(P\right)$ ($v_1=v_R\left(P\right)$, respectively), then ${(N\left(v_1\right)\cap V\left(S_1\right))}^{-}\cap N\left(v\right)=\varnothing$ ($N\left(v_1\right)\cap {(N\left(v\right)\cap V\left(S_1\right))}^{-}=\varnothing$, respectively) for any $v\in V\left({\mathcal{S}}_C\right)\cup V\left(H\right)\cup (End\left(\mathcal{S}\right)\backslash \left\{v_1\right\})$.

Let Y be an independent set of G with size $k+\kappa$. Then the following lemma holds.

Lemma 7.

Let $S^{\prime }$ belong to $V\left(G\right)$ which satisfies $S^{\prime }\cap Y$ having precisely one vertex, denoted as z, i.e, $S^{\prime }\cap Y=\left\{z\right\}$. If $N\left(x\right)\cap Y=\left\{z\right\}$ for each $x\in S^{\prime }\backslash \left\{z\right\}$, then $G\left[S^{\prime }\right]$ forms a clique.

Proof of Lemma 7.

We begin by assuming the opposite and using a proof by contradiction. Suppose that $x_1{x}_2\notin E\left(G\right)$ for some pair of vertices $x_1$ and $x_2$ in $S^{\prime }$, where $x_1\ne x_2$. Then, $(Y\backslash \left\{z\right\})\cup \{x_1,x_2\}$ forms an independent set of G with a size of $k+\kappa +1$. This contradicts the fact that $\alpha \left(G\right)=k+\kappa$. Hence, $G\left[S^{\prime }\right]$ is a clique. □

For convenience, suppose that x is an element of $V\left(P\right)$. For each $P\in {\mathcal{S}}_P$, we define $\overrightarrow{\mathcal{Q}}(x,P)$ as follows:

$$\begin{array}{c}\hfill \overrightarrow{\mathcal{Q}}(x,P)=\left\{\begin{array}{cc}{v}_R\left(P\right),\hfill & \mathrm{if}x=v_R\left(P\right),\hfill \\ x{v}_R\left(P\right),\hfill & \mathrm{if}{x}^{+}=v_R\left(P\right),\hfill \\ x\overrightarrow{P}{v}_R\left(P\right)x,\hfill & \mathrm{if}x{v}_R\left(P\right)\in E\left(G\right)(x\ne v_R\left(P\right),x^{+}\ne v_R\left(P\right)).\hfill \end{array}\right.\end{array}$$

Similarly, we define $\overleftarrow{\mathcal{Q}}(x,P)$ as follows:

$$\begin{array}{c}\hfill \overleftarrow{\mathcal{Q}}(x,P)=\left\{\begin{array}{cc}x=v_L\left(P\right),\hfill & \mathrm{if}x=v_L\left(P\right),\hfill \\ x{v}_L\left(P\right),\hfill & \mathrm{if}{x}^{-}=v_L\left(P\right),\hfill \\ x\overleftarrow{P}{v}_L\left(P\right)x,\hfill & \mathrm{if}x{v}_L\left(P\right)\in E\left(G\right)(x\ne v_L\left(P\right),x^{-}\ne v_L\left(P\right)).\hfill \end{array}\right.\end{array}$$

Therefore, $f\left(\overrightarrow{\mathcal{Q}}(x,P)\right)=1$ and $f\left(\overleftarrow{\mathcal{Q}}(x,P)\right)=1$. We say that $C\left(G_0\right)$ is a spanning subgraph of $G_0$ satisfying $f\left(C\left(G_0\right)\right)=1$ if $G_0\subseteq G$.

Some properties of $\mathcal{S}$ are described in the following lemmas, as proved in Appendix B.

Lemma 8.

$U\subseteq V\left({\mathcal{S}}_P\right)$ and $\mid {\mathcal{S}}_P\mid =1$.

By Lemma 8, ${\mathcal{S}}_P=\left\{P\right\}$ (say). Then $U=V\left(P\right)\cap V\left(\mathcal{L}\right)=\{u_{P,1},\cdots ,u_{P,t_P}\}$ and $t_P=\kappa$. For convenience, denote that $U=\{u_1,u_2,\cdots ,u_{\kappa }\}$ and

$$\begin{array}{c}\hfill \mathcal{X}=\left\{\begin{array}{cc}End\left(\mathcal{S}\right)\cup U^{+}\cup \left\{w\right\},\hfill & \mathrm{if}End\left(\mathcal{S}\right)\cap U^{+}\ne \varnothing ,\hfill \\ (End\left(\mathcal{S}\right)\cup U^{+}\cup \left\{w\right\})\backslash \left\{u_{\kappa }^{+}\right\},\hfill & \mathrm{if}End\left(\mathcal{S}\right)\cap U^{+}=\varnothing .\hfill \end{array}\right.\end{array}$$

Lemma 9.

The following statements hold.

(1)

$N\left(u_{\kappa }^{+}\right)\cap End\left({\mathcal{S}}_C\right)\ne \varnothing$ or $f\left(\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)\right)=1$ and $N\left(u_1^{-}\right)\cap End\left({\mathcal{S}}_C\right)\ne \varnothing$ or $f(\overleftarrow{\mathcal{Q}}(u_1^{-},P\left)\right)=1$.

(2)

If $f\left(\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)\right)\ne 1$ and $f\left(\overleftarrow{\mathcal{Q}}(u_1^{-},P)\right)\ne 1$, then there exist at least two elements C, $C^{\prime }\in {\mathcal{S}}_C$ such that $u_{\kappa }^{+}{v}_{C^{\prime }}\in E\left(G\right)$, $u_1^{-}{v}_C\in E\left(G\right)$.

(3)

$\mathcal{X}$ forms an independent set of G with size $k+\kappa$.

(4)

If $N\left(u_{\kappa }^{+}\right)\cap End\left({\mathcal{S}}_C\right)\ne \varnothing$ and $u_{\kappa }^{2+}\ne v_R\left(P\right)$, then $u_{\kappa }^{2+}{v}_R\left(P\right)\in E\left(G\right)$.

(5)

Let $y\in V\left(P\right)$ satisfy $\left|V\right(P[y^{+},v_R\left(P\right)]\left)\right|\ge 1$, $y{v}_R\left(P\right)\in E\left(G\right)$ and $V\left(P[y,v_R\left(P\right)]\right)\cap U=\varnothing$. Then $G\left[V\left(P[y^{+},v_R\left(P\right)]\right)\right]$ forms a clique. Additionally, if the intersection of $N\left(y\right)$ and $\mathcal{X}$ is $\left\{v_R\left(P\right)\right\}$, then the graph $G\left[V\left(P[y,v_R\left(P\right)]\right)\right]$ forms a clique.

(6)

$G\left[V\right(C\left)\right]$ forms a clique for any $C\in {\mathcal{S}}_C$. Furthermore, $N\left(x\right)\cap \mathcal{X}=\left\{v_C\right\}$ for any $x\in V\left(C\right)\backslash \left\{v_C\right\}$.

3. Results and Discussion

In [17], the authors provide a novel extension by imposing a limit on the maximum number of independent sets, although the limit is not sharp. Note that G has no spanning $k_1+1-k_2$-ended tree for each $G\in \{F_0(k_1,k_2),F_{00}(k_1,k_2),F_2(k_1,k_2)\}$. In this paper, we extend Theorem 5 to the case where $\kappa \left(G\right)\ge 2$ and the bound on the number of maximum independent sets is already sharp.

Theorem 7.

Let $k\ge 2$ and G be a graph with $\left|V\right(G\left)\right|\ge 2\kappa \left(G\right)+k$, $\kappa \left(G\right)=\kappa \ge 2$, $\alpha \left(G\right)\le \kappa +k$ and $m\left(G\right)\le n-2\kappa -k+1$. Then G contains a spanning k-ended tree, unless either $F_0(k+\kappa -1,\kappa )\subseteq G\subseteq F_2(k+\kappa -1,\kappa )$ for $n>3\kappa +k-1$, or $F_{00}(k+\kappa -1,\kappa )\subseteq G\subseteq F_2(k+\kappa -1,\kappa )$ for $2\kappa +k\le n\le 3\kappa +k-1.$

Note that a spanning tree having exactly two leaves is called a Hamilton path. Then, we can immediately obtain the following result.

Corollary 1.

Let G be a graph with $\left|V\right(G\left)\right|\ge 2\kappa \left(G\right)+2$, $\kappa \left(G\right)=\kappa \ge 2$, $\alpha \left(G\right)\le \kappa +2$ and $m\left(G\right)\le n-2\kappa -1$. Then G is traceable, unless either $F_0(\kappa +1,\kappa )\subseteq G\subseteq F_2(\kappa +1,\kappa )$ for $n>3\kappa +1$, or $F_{00}(\kappa +1,\kappa )\subseteq G\subseteq F_2(\kappa +1,\kappa )$ for $2\kappa +2\le n\le 3\kappa +1.$

In the case of 2-connected, the bounds of $\left|V\right(G\left)\right|$ can do better. Clearly, Corollary 1 improves the result of Theorem 4. It demonstrated that expanding the independence number slightly and bounding $m\left(G\right)$ also does not alter the traceability in highly connected graphs.

If $F_0(k+\kappa -1,\kappa )\subseteq G\subseteq F_2(k+\kappa -1,\kappa )$ for $n>3\kappa +k-1$, or $F_{00}(k+\kappa -1,\kappa )\subseteq G\subseteq F_2(k+\kappa -1,\kappa )$ for $2\kappa +k\le n\le 3\kappa +k-1$, then $m\left(G\right)=n-2\kappa -k+1$. Hence, we can obtain the following result immediately.

Corollary 2.

Let $k\ge 2$ and G be a graph of order $n\ge 2\kappa \left(G\right)+k$ such that $\kappa \left(G\right)=\kappa \ge 2$, $\alpha \left(G\right)\le \kappa +k$ and $m\left(G\right)\le n-2\kappa -k$. Then G contains a spanning k-ended tree.

4. Proof of Theorem 7

In this section, we employ the same terminology and notation in Section 3.

Proof of Theorem 7.

Let $k\ge 2$ and G be a graph with $\left|V\right(G\left)\right|>2\kappa \left(G\right)+k$, $\kappa \left(G\right)=\kappa \ge 2$, $\alpha \left(G\right)\le \kappa +k$ and $m\left(G\right)\le n-2\kappa -k+1$. We begin by assuming the opposite and using a proof by contradiction. Suppose that G does not have a spanning k-ended tree. This assumption, along with Theorem 5, implies the following equation:

$$\alpha \left(G\right)=\kappa \left(G\right)+k.$$

(1)

Thus, by Lemma 1, G cannot have a spanning k-ended system. We select a maximal k-ended system $\mathcal{S}$ of G that satisfies conditions (I) and (II) outlined in Section 3. Define $H=G-V\left(\mathcal{S}\right)$. Clearly $\left|V\right(H\left)\right|\ge 1$. Let $w\in \left(V\right(G)-V(\mathcal{S}\left)\right)$.

We will show that $F_0(k+\kappa -1,\kappa )\subseteq G\subseteq F_2(k+\kappa -1,\kappa )$ for $n>3\kappa +k-1$, or $F_{00}(k+\kappa -1,\kappa )\subseteq G\subseteq F_2(k+\kappa -1,\kappa )$ for $2\kappa +k\le n\le 3\kappa +k-1.$

Fact 1.

$m\left(G\right)\ge n-2\kappa -k+1$.

Proof of Fact 1.

We consider a $(w,V(\mathcal{S}\left)\right)$-fan $\mathcal{L}$ in Section 3. By Lemma 8, we choose $U=\{u_1,u_2,\cdots ,u_{\kappa }\}$ and ${\mathcal{S}}_P=\left\{P\right\}$ in Section 3.

We consider a $(w,V(\mathcal{S}\left)\right)$-fan $\mathcal{L}$ in Section 3. By Lemma 8, we choose $U=\{u_1,u_2,\cdots ,u_{\kappa }\}$ and ${\mathcal{S}}_P=\left\{P\right\}$ in Section 3.

Claim 1.

$N\left(u_{\kappa }^{+}\right)\cap V\left({\mathcal{S}}_C\right)=\varnothing$ and $N\left(u_1^{-}\right)\cap V\left({\mathcal{S}}_C\right)=\varnothing$.

Proof of Claim 1.

Using symmetry, we can focus on proving that $N\left(u_{\kappa }^{+}\right)\cap V\left({\mathcal{S}}_C\right)=\varnothing$. By contradiction, suppose that $N\left(u_{\kappa }^{+}\right)\cap V\left({\mathcal{S}}_C\right)\ne \varnothing$; say $v\in N\left(u_{\kappa }^{+}\right)\cap V\left({\mathcal{S}}_C\right)$ and $v\in V\left(C^{\prime }\right)$. By Lemma 2$\left(ii\right)$, $u_{\kappa }^{+}\ne v_R\left(P\right)$.

Denote

$$\begin{array}{c}\hfill A=\left\{\begin{array}{cc}V\left(P[u_{\kappa }^{3+},v_R\left(P\right)]\right),\hfill & \mathrm{if}{u}_{\kappa }^{+}{v}_R\left(P\right)\notin E\left(G\right),\hfill \\ V\left(P[u_{\kappa }^{2+},v_R\left(P\right)]\right),\hfill & \mathrm{if}{u}_{\kappa }^{+}{v}_R\left(P\right)\in E\left(G\right).\hfill \end{array}\right.\end{array}$$

If $u_{\kappa }^{+}{v}_R\left(P\right)\notin E\left(G\right)$, then, by Lemma 9(4), $u_{\kappa }^{2+}{v}_R\left(P\right)\in E\left(G\right)$. Therefore, by Lemma 9(5), $G\left[V\left(P[u_{\kappa }^{3+},v_R\left(P\right)]\right)\right]$ forms a clique. If $u_{\kappa }^{+}{v}_R\left(P\right)\in E\left(G\right)$, then, according to Lemma 9(5), $G\left[V\right(P[u_{\kappa }^{2+},$$v_R\left(P\right)\left]\right)]$ forms a clique. Hence,

$$G\left[A\right]formsaclique.$$

(2)

As G is a connected graph, $N\left(A\right)\cap V(G-A)\ne \varnothing$. For $y\in N\left(A\right)\cap V(G-A)$, there exists a vertex $x\in A$ with $xy\in E\left(G\right)$. We will show that

$$y\in \{u_{\kappa }^{+},u_{\kappa }^{2+}\}.$$

(3)

By contradiction, suppose that $y\notin \{u_{\kappa }^{+},u_{\kappa }^{2+}\}$. By Lemma 6(2) and (2), $y\notin V\left({\mathcal{S}}_C\right)\cup V\left(H\right)$. We will examine the following two scenarios to reach a contradiction:

• Suppose that $y\in V\left(P[v_L\left(P\right),u_1^{-}]\right)$. By Lemma 9(1), $f\left(\overleftarrow{\mathcal{Q}}(u_1^{-},P)\right)=1$ or $N\left(u_1^{-}\right)\cap V\left({\mathcal{S}}_C\right)\ne \varnothing$. We will show that $u_1^{2-}{v}_L\left(P\right)\in E\left(G\right)$. If $f\left(\overleftarrow{\mathcal{Q}}(u_1^{-},P)\right)=1$, then, by symmetry and Lemma 9(5), $u_1^{2-}{v}_L\left(P\right)\in E\left(G\right)$. If $f\left(\overleftarrow{\mathcal{Q}}(u_1^{-},P)\right)\ne 1$, then, by Lemma 9(1), $N\left(u_1^{-}\right)\cap V\left({\mathcal{S}}_C\right)\ne \varnothing$. By symmetry and Lemma 9(4), $u_1^{2-}{v}_L\left(P\right)\in E\left(G\right)$. Then, by symmetry and Lemma 9(5) and (2), either the set of vertices of the subgraph $C^{\prime }{u}_{\kappa }^{+}\overrightarrow{P}{x}^{-}{v}_R\left(P\right)\overleftarrow{P}xy\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }w$ and $\overleftarrow{\mathcal{Q}}(y^{-},P)$ is equal to $V(P\cup C^{\prime })\cup \left\{w\right\}$, which contradicts (I); or $C^{\prime }{u}_{\kappa }^{+}\overrightarrow{P}{x}^{-}{v}_R\left(P\right)\overleftarrow{P}x{v}_L\left(P\right)\overrightarrow{P}{u}_{\kappa }$ in G is equal to $V(P\cup C^{\prime })$, which contradicts Lemma 4.
• Assume that $y\in V\left(P[u_1,u_{\kappa }]\right)$. Then, by Lemma 9(1)(2) and (2), the set of vertices of the subgraph

  $$\begin{array}{c}\hfill Q_1=\left\{\begin{array}{cc}{C}^{\prime }{u}_{\kappa }^{+}\overrightarrow{P}{x}^{-}{v}_R\left(P\right)\overleftarrow{P}xy\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }w{L}_1{u}_1\overrightarrow{P}{y}^{-},\hfill & \mathrm{if}y\in V\left(P(u_1,u_{\kappa })\right),\hfill \\ C^{\prime }{u}_{\kappa }^{+}\overrightarrow{P}{x}^{-}{v}_R\left(P\right)\overleftarrow{P}x{u}_1\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }w,\hfill & \mathrm{if}y=u_1,\hfill \\ C^{\prime }{u}_{\kappa }^{+}\overrightarrow{P}{x}^{-}{v}_R\left(P\right)\overleftarrow{P}x{u}_{\kappa }\overleftarrow{P}{u}_1{L}_{1}w,\hfill & \mathrm{if}y=u_{\kappa },\hfill \end{array}\right.\end{array}$$

  and

  $$\begin{array}{c}\hfill Q_2=\left\{\begin{array}{cc}C{u}_{Q,1}^{-}\overleftarrow{Q}{v}_L\left(Q\right),\hfill & \mathrm{if}f\left(\overleftarrow{\mathcal{Q}}(u_{Q,1}^{-},Q)\right)\ne 1,\hfill \\ \overleftarrow{\mathcal{Q}}(u_{Q,1}^{-},Q),\hfill & \mathrm{if}f\left(\overleftarrow{\mathcal{Q}}(u_{Q,1}^{-},Q)\right)=1.\hfill \end{array}\right.\end{array}$$

  is equal to $V(P\cup C\cup C^{\prime })\cup \left\{w\right\}$ or $V(P\cup C^{\prime })\cup \left\{w\right\}$, which contradicts (I).

This contradiction shows that (3) holds.

If $y=u_{\kappa }^{+}$ and $u_{\kappa }^{+}{v}_R\left(P\right)\notin E\left(G\right)$, then, $u_{\kappa }^{2+}{v}_R\left(P\right)\in E\left(G\right)$. By (2), $G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right]$ has a cycle $C_{\kappa }=v_R\left(P\right)\overleftarrow{P}x{u}_{\kappa }^{+}\overrightarrow{P}{x}^{-}{v}_R\left(P\right)$. By (2), we structure a new path $P^{\prime }$ such that $P^{\prime }=v_L\left(P\right)\overrightarrow{P}{u}_{\kappa }^{+}x\overrightarrow{P}{v}_R\left(P\right)x^{-}\overleftarrow{P}{u}_{\kappa }^{2+}$ by rearranging the order of the vertices in P. Then $v_R\left(P^{\prime }\right)=u_{\kappa }^{2+}$. It is easy to verify that $G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right]\cong G\left[V\left(P^{\prime }[u_{\kappa }^{+},v_R\left(P^{\prime }\right)]\right)\right]$. Note that $u_{\kappa }^{+}{v}_R\left(P^{\prime }\right)\in E\left(G\right)$. By Lemma 9(5), the subgraph $G\left[V\left(P^{\prime }[x,v_R\left(P^{\prime }\right)]\right)\right]$ forms a clique. Let $A^{\prime }=V\left(P^{\prime }[u_{\kappa }^{+},v_R\left(P^{\prime }\right)]\right)\backslash \left\{u_{\kappa }^{+}\right\}=V\left(P^{\prime }[x,v_R\left(P^{\prime }\right)]\right)]$. Since G is connected, $N\left(A^{\prime }\right)\cap V(G-A^{\prime })\ne \varnothing$. For $y^{\prime }\in N\left(A^{\prime }\right)\cap V(G-A^{\prime })$, there exists a vertex $x^{\prime }\in A$ with $x^{\prime }{y}^{\prime }\in E\left(G\right)$. By the proof of $\left(3\right)$, $y^{\prime }=u_{\kappa }^{+}$. That means $|N\left(A^{\prime }\right)\cap V(G-A^{\prime })|=1$, contradicting $|N\left(A^{\prime }\right)\cap V(G-A^{\prime })|\ge \kappa \ge 2$. Therefore, by (3), we have either $y=u_{\kappa }^{2+}$ and $u_{\kappa }^{+}{v}_R\left(P\right)\notin E\left(G\right)$ or $y=u_{\kappa }^{+}$ and $u_{\kappa }^{+}{v}_R\left(P\right)\in E\left(G\right)$. Then, $\left|N\right(A)\cap V(G-A\left)\right|=1$, contradicting $\left|N\right(A)\cap V(G-A\left)\right|\ge \kappa \ge 2$. This contradiction indicates that Claim 1 is true. □

According to Claim 1 and Lemma 9(1), $f\left(\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)\right)=1$ and $f\left(\overleftarrow{\mathcal{Q}}(u_1^{-},P)\right)=1$. Denote

$$\begin{array}{c}\hfill \mathcal{X}=\left\{\begin{array}{cc}End\left(\mathcal{S}\right)\cup U^{+}\cup \left\{w\right\},\hfill & \mathrm{if}End\left(\mathcal{S}\right)\cap U^{+}\ne \varnothing ,\hfill \\ (End\left(\mathcal{S}\right)\cup U^{+}\cup \left\{w\right\})\backslash \left\{u_{\kappa }^{+}\right\},\hfill & otherwise(i.e.,\mathrm{if}End\left(\mathcal{S}\right)\cap U^{+}=\varnothing ).\hfill \end{array}\right.\end{array}$$

By Lemma 9(3), $\mathcal{X}$ is an independent set of G with size $\kappa +k$. Thus,

$$N\left(v\right)\cap \mathcal{X}\ne \varnothing foranyv\in V\left(G\right)\backslash \mathcal{X}.$$

(4)

Claim 2.

$G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right]$ and $G\left[V\left(P[v_L\left(P\right),u_1^{-}]\right)\right]$ form cliques.

Proof of Claim 2.

By virtue of symmetry, we may restrict our consideration to prove that $G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right]$ forms a clique. As $G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right]$ is connected, we can assume that $\left|V\right(P[u_{\kappa }^{+},v_R\left(P\right)]\left)\right|\ge 3$. According to Lemma 5(1) (4) and Claim 1, $N\left(u_{\kappa }^{+}\right)\cap (\mathcal{X}\backslash \left\{v_R\left(P\right)\right\})=\varnothing$. By (4), $N\left(u_{\kappa }^{+}\right)\cap \mathcal{X}=\left\{v_R\left(P\right)\right\}$. By Lemma 9(5), $G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right]$ forms a clique. □

Claim 3.

$N\left(V\left(H\right)\right)\cap V\left({\mathcal{S}}_C\right)=\varnothing$.

Proof of Claim 3.

By contradiction, suppose that $N\left(V\left(H\right)\right)\cap V\left({\mathcal{S}}_C\right)\ne \varnothing$; say $x\in N\left(V\right(H\left)\right)\cap V\left(C\right)$ for some $C\in {\mathcal{S}}_C$. This implies that there is a vertex $v\in V\left(H\right)$ with $e=vx$. By Lemma 8, $v\ne w$.

We will show that

$$v\notin V\left(\mathcal{L}\right).$$

(5)

Suppose, by way of contradiction, that $v\in V\left(L_{i_0}\right)$ for some $i_0\in \{1,\cdots ,\kappa \}$. Suppose that $v\in V\left(L_1\right)\cup V\left(L_{\kappa }\right)$. By symmetry, we may only think of $v\in V\left(L_1\right)$. Then, by Claim 2, $v_R\left(P\right)\overleftarrow{P}{u}_1{L}_{1}vC$ and $\overleftarrow{\mathcal{Q}}(u_1^{-},P)$ cover $V\left(P\right)\cup V\left(C\right)\cup \left\{v\right\}$, contradicting (I). Therefore, $v\in V\left(L_{i_0}\right)$ for some $i_0\in \{2,\cdots ,\kappa -1\}$. Then, by Claim 2, the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }w{L}_{i_0}vC$ and $\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ in G is equal to $V\left(P\right)\cup V\left(C\right)\cup \{w,v\}$, which again contradicts (I). Thus, we have shown that (5) holds.

Next, we will prove that

$$vw\in E\left(G\right).$$

(6)

By contradiction, suppose that $vw\notin E\left(G\right)$. Note that $x\in V\left(C\right)$. We consider the neighbourhood of the vertex $x^{+}$. According to Lemma 2$\left(i\right)\left(ii\right)$, $N\left(x^{+}\right)\cap (End\left(\mathcal{S}\right)\backslash \left\{v_C\right\})=\varnothing$. If $x^{+}$ and $u_i^{+}\in V\left(P\right)$ for some $i\in \{1,\cdots ,\kappa -1\}$ are adjacent in G, then, by Claim 2 and (5), $v_L\left(P\right)\overrightarrow{P}{u}_i{L}_{i}w{L}_{\kappa }{u}_{\kappa }\overleftarrow{P}{u}_i^{+}Cv$ and $\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ in G covers $V\left(P\right)\cup V\left(C\right)\cup \{w,v\}$, contradicting (I). Hence, $N\left(x^{+}\right)\cap (U^{+}\backslash \left\{u_{\kappa }^{+}\right\})=\varnothing$. If v and $u_i^{+}\in V\left(P\right)$ for some $i\in \{1,\cdots ,\kappa -1\}$ are adjacent in G, then, by Claim 2 and (5), $v_L\left(P\right)\overrightarrow{P}{u}_i{L}_{i}w{L}_{\kappa }{u}_{\kappa }\overleftarrow{P}{u}_i^{+}vC$ and $\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ in G covers $V\left(P\right)\cup V\left(C\right)\cup \{w,v\}$, contradicting (I). Hence, $N\left(v\right)\cap (U^{+}\backslash \left\{u_{\kappa }^{+}\right\})=\varnothing$. Note that $vw\notin E\left(G\right)$. Therefore, by Lemma 2$\left(i\right)\left(ii\right)$, $\{v_L\left(P\right),v_R\left(P\right),x^{+},w,v\}\cup (U^{+}\backslash \left\{u_{\kappa }^{+}\right\})\cup (End\left({\mathcal{S}}_C\right)\backslash \left\{v_C\right\})$ forms an independent set of size $\kappa +k+1$. This contradicts the fact that $\alpha \left(G\right)=\kappa +k$ and thus establishes that (6) holds.

Then, by Claim 2, (5) and (6), the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }wvC$ and $\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ is equal to $V(P\cup C)\cup \{w,v\}$. This contradicts (I) and establishes that $N\left(V\right(H\left)\right)\cap V\left(C\right)=\varnothing$. □

Claim 4.

$N\left(V\right(C\left)\right)\cap \left(V\right(G)\backslash V(C\left)\right)=U$for each element $C\in {\mathcal{S}}_C$.

Proof of Claim 4.

Since G is connected, $N\left(V\right(C\left)\right)\cap \left(V\right(G)\backslash V(C\left)\right)\ne \varnothing$ for any $C\in {\mathcal{S}}_C$. For $z\in N\left(V\right(C\left)\right)\cap \left(V\right(G)\backslash V(C\left)\right)$, there exists a vertex $x\in V\left(C\right)$ with $xz\in E\left(G\right)$. According to Lemma 2$\left(i\right),\left(ii\right)$, $z\notin V({\mathcal{S}}_C\backslash \left\{C\right\})$. By Claim 3, $z\notin V\left(H\right)$. This implies that

$$z\in V\left(P\right).$$

(7)

Next, we will show that $z\in U$. By contradiction, suppose that $z\notin U$. By Lemma 6(2) and Claim 2, z does not belong to $V\left(P(u_{\kappa }^{+},v_R\left(P\right)]\right)\cup V\left(P[v_L\left(P\right),u_1^{-})\right)$. To arrive at a contradiction, we will examine the following three scenarios using (7):

• Suppose that $z\in \{u_{\kappa }^{+},u_1^{-}\}$. Then $N\left(u_{\kappa }^{+}\right)\cap V\left({\mathcal{S}}_C\right)\ne \varnothing$ or $N\left(u_1^{-}\right)\cap V\left({\mathcal{S}}_C\right)\ne \varnothing$, contradicting Claim 1.
• Suppose that $z\in U^{+}\backslash \left\{u_{\kappa }^{+}\right\}$ or $U^{-}\backslash \left\{u_1^{-}\right\}$. By symmetry, we consider that $z\in U^{+}\backslash \left\{u_{\kappa }^{+}\right\}$ say $z=u_i^{+}$ for some $i\in \{1,\cdots ,\kappa -1\}$. Then, by Claim 2, $C{u}_i^{+}\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }w{L}_i{u}_i\overleftarrow{P}{v}_L\left(P\right)$ and $\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ cover $V(P\cup C)\cup \left\{w\right\}$, contradicting (I).
• Suppose that $z\in V\left(P[u_i^{2+},u_{i+1}^{2-}]\right)$ for some $i\in \{1,\cdots ,\kappa -1\}$. We consider the neighbourhood of the vertex $z^{+}$. We claim that $N\left(z^{+}\right)\cap \mathcal{X}=\left\{v_C\right\}$. Suppose otherwise that there exists a vertex $y\in N\left(z^{+}\right)\cap \mathcal{X}$ such that $y\ne v_C$. By Lemma 6(1), $y\notin End\left({\mathcal{S}}_C\right)\backslash \left\{v_C\right\}$. According to the definition of $\mathcal{X}$, we will examine the following two scenarios to reach a contradiction.

  –

  Assume that $y\in U^{+}\cap \mathcal{X}$; say $y=u_j^{+}$ for some $j\in \{1,\cdots ,\kappa -1\}$. If $j>i$, then, by Claim 2, the set of vertices of the subgraph $v_R\left(P\right)\overleftarrow{P}{u}_j^{+}{z}^{+}\overrightarrow{P}{u}_j{L}_{j}w{L}_1{u}_1\overrightarrow{P}zC$ and $\overleftarrow{\mathcal{Q}}(u_1^{-},P)$ is equal to $V(P\cup C)\cup \left\{w\right\}$, which contradicts (I). If $j\le i$, then, by Claim 2, the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}{u}_j{L}_{j}w{u}_{\kappa }\overleftarrow{P}{z}^{+}{u}_j^{+}\overrightarrow{P}zC\cup \overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ is equal to $V(P\cup C)\cup \left\{w\right\}$, which again contradicts (I).

  –

  Assume that $y\in End\left({\mathcal{S}}_P\right)$. By Lemma 6(2), $y=v_R\left(P\right)$. Then, by Claim 2, the set of vertices of the subgraph $u_{\kappa }^{-}\overleftarrow{P}{z}^{+}{v}_R\left(P\right)\overleftarrow{P}{u}_{\kappa }{L}_{\kappa }w{L}_1{u}_1\overrightarrow{P}zC\cup \overleftarrow{\mathcal{Q}}(u_1^{-},P)$ is equal to $V(P\cup C)\cup \left\{w\right\}$, which contradicts (I).

This contradiction establishes that $N\left(z^{+}\right)\cap \mathcal{X}\subseteq \left\{v_C\right\}$. By $\left(4\right)$, $N\left(z^{+}\right)\cap \mathcal{X}=\left\{v_C\right\}$. If $x\ne v_C$ and $\left|V\right(C\left)\right|>1$ or $\left|V\right(C\left)\right|=1$, then, by Lemma 9(6), the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}zC{z}^{+}\overrightarrow{P}{v}_R\left(P\right)$ is equal to $V(P\cup C)$, which contradicts Lemma 4. If $x=v_C$ and $\left|V\right(C\left)\right|>1$, then, according to Lemma 9(6), $N\left(v_C^{+}\right)\cap \mathcal{X}=\left\{v_C\right\}$. Note that $N\left(z^{+}\right)\cap \mathcal{X}=\left\{v_C\right\}$. If $z^{+}{v}_C^{+}\notin E\left(G\right)$, then, by Lemma 9(6), $(\mathcal{X}\backslash \left\{v_C\right\})\cup \{z^{+},v_C^{+}\}$ would be an independent set of cardinality $\kappa +k+1$, contradicting (1). Therefore, $z^{+}{v}_C^{+}\in E\left(G\right)$. Then the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}zC{z}^{+}\overrightarrow{P}{v}_R\left(P\right)$ is equal to $V(P\cup C)$, which contradicts Lemma 4.

This contradiction establishes that $z\in U$. Since $\left|N\right(V\left(C\right))\cap (V\left(G\right)\backslash V\left(C\right)\left)\right|\ge \kappa$ and $\left|U\right|=\kappa$, $N\left(V\right(C\left)\right)\cap \left(V\right(G)\backslash V(C\left)\right)=U$ for any element $C\in {\mathcal{S}}_C$. □

Claim 5.

Let $C\in S_C$ with $\left|V\right(C\left)\right|>1$. For any two disjoint vertices $u_i$, $u_j\in U$, there exist two disjoint vertices v, $v^{\prime }\in V\left(C\right)$ such that $u_{i}v$, $u_j{v}^{\prime }\in E\left(G\right)$.

Proof of Claim 5.

We establish Claim 5 by contradiction. Suppose that either $N\left(u_{i_0}\right)\cap V\left(C\right)=N\left(u_{j_0}\right)\cap V\left(C\right)=\varnothing$ or $N\left(u_{i_0}\right)\cap V\left(C\right)=N\left(u_{j_0}\right)\cap V\left(C\right)=\left\{v\right\}$ and $v\in V\left(C\right)$ for some $u_{i_0}$, $u_{j_0}\in U$.

If $N\left(u_{i_0}\right)\cap V\left(C\right)=N\left(u_{j_0}\right)\cap V\left(C\right)=\varnothing$, then $N\left(V\right(C\left)\right)\cap \left(V\right(G)\backslash V(C\left)\right)\ne U$, contradicting Claim 4. Now suppose that $N\left(u_{i_0}\right)\cap V\left(C\right)=N\left(u_{j_0}\right)\cap V\left(C\right)=\left\{v\right\}$. Let $\widehat{U}=(U\backslash \{u_{i_0},u_{j_0}\})\cup \left\{v\right\}$ and $\widehat{C}=C-v$. Since $\left|V\right(C\left)\right|>1$, $\widehat{C}\ne \varnothing$. Then, by hypothesis and Claim 4, $N\left(V\left(\widehat{C}\right)\right)\cap (V\left(G\right)\backslash V\left(\widehat{C}\right))\subseteq \widehat{U}$. However, $|\widehat{U}|=\kappa -1$, contradicting the hypothesis that G is $\kappa$-connected. These contradictions establish that Claim 5 is true. □

Claim 6.

$G\left[V\right(H\left)\right]$ forms a clique.

Proof of Claim 6.

We will only focus on the case where $\left|V\right(H\left)\right|\ge 2$. For every vertex $v\in V\left(H\right)\backslash \left\{w\right\}$, $N\left(v\right)\cap \mathcal{X}\ne \varnothing$. We assume that there is at least one vertex $x\in N\left(v\right)\cap \mathcal{X}$ with $x\ne w$. By Claim 3, x is not an element of $End\left({\mathcal{S}}_C\right)$. By Lemma 3(1), $x\notin End\left({\mathcal{S}}_P\right)$. Then, $x\in \mathcal{X}\cap U^{+}$; say $x=u_i^{+}$ for some $i\in \{1,\cdots ,\kappa -1\}$. Then, by Claims 2, 4 and 5, there exist a path Q and $\overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ cover $V\left(P\right)\cup V\left(C\right)\cup \left\{v\right\}$, see Figure 2, contradicting (I). This contradiction shows that $N\left(v\right)\cap \mathcal{X}\subseteq \left\{w\right\}$. By (4), $N\left(v\right)\cap \mathcal{X}=\left\{w\right\}$ for every vertex $v\in V\left(H\right)\backslash \left\{w\right\}$. Let $S^{\prime }=V\left(H\right)$. Then, according to Lemma 7, $G\left[V\right(H\left)\right]$ forms a clique. □

Figure 2. $v{u}_i^{+}\in E\left(G\right)$$(i\ne \kappa )$.

Denote $A_1=V\left(P[v_L\left(P\right),u_1^{-}]\right)$ and $A_2=V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)$.

Claim 7.

The following two statements are true.

(1)

$N\left(A_i\right)\cap (G-A_i)=U$for$i\in \{1,2\}$;

(2)

$N\left(V\right(H\left)\right)\cap V\left(\mathcal{S}\right)=U$.

Proof of Claim 7.

We will prove the first statement. By symmetry, we have only proved that $N\left(A_2\right)\cap (G-A_2)=U$. Let $C^{*}=G\left[A_2\right]$. By Claim 2(1), $f\left(C^{*}\right)=1$. We pick an element $C\in {\mathcal{S}}_C$, by Claim 4, a new path $Q=v_L\left(P\right)\overrightarrow{P}{u}_{\kappa }C$ would be obtained. We structure a new system ${\mathcal{S}}^{*}$ such that ${\mathcal{S}}^{*}={\mathcal{S}}_C^{*}\cup {\mathcal{S}}_P^{*}$, ${\mathcal{S}}_P^{*}=\left\{Q\right\}$ and ${\mathcal{S}}_C^{*}=({\mathcal{S}}_C\backslash \left\{C\right\})\cup \left\{C^{*}\right\}$. It is easy to verify that $V\left({\mathcal{S}}^{*}\right)=V\left(\mathcal{S}\right)$, $|{\mathcal{S}}_C^{*}|=|{\mathcal{S}}_C|$ and $|{\mathcal{S}}_P^{*}|=|{\mathcal{S}}_P|$. Hence, ${\mathcal{S}}^{*}$ is also a k-ended system satisfying (I), (II). Then, by Claim 4, $N\left(A_2\right)\cap (G-A_2)=U$.

Next, we need to prove the second statement. The proof here is similar to Claim 4. (For details, see Appendix A.) □

Claim 8.

Suppose that $\left|V\right(H\left)\right|>1$. For any two disjoint vertices $u_i$, $u_j\in U$, there exist two disjoint vertices v, $v^{\prime }\in V\left(H\right)$ such that $u_{i}v$, $u_j{v}^{\prime }\in E\left(G\right)$.

Proof of Claim 8.

The proof here is similar to Claim 5. (For details, see Appendix A.) □

Claim 9.

$u_{i+1}^{-}{u}_i^{+}\in E\left(G\right)$ for each $i\in \{1,\cdots ,\kappa -1\}$.

Proof of Claim 9.

Since $G\left[V\left(P[u_i^{+},u_{i+1}^{-}]\right)\right]$ is connected, we only need to focus on the case where $\left|V\right(P[u_i^{+},u_{i+1}^{-}]\left)\right|\ge 3$. By contradiction, suppose that $u_{i_0+1}^{-}{u}_{i_0}^{+}\notin E\left(G\right)$ for some $i_0\in \{1,\cdots ,\kappa -1\}$. By (4), there exists at least one vertex $y\in N\left(u_{i_0+1}^{-}\right)\cap \mathcal{X}$ satisfying $y\ne u_{i_0}^{+}$. By Claim 4, $y\notin End\left({\mathcal{S}}_C\right)$. We will examine the following two scenarios to reach a contradiction, based on the definition of $\mathcal{X}$:

• Assume that $y\in \mathcal{X}\cap (U^{+}\backslash \left\{u_{i_0}^{+}\right\})$; say $y=u_j^{+}$ for some $j\in \{1,\cdots ,\kappa -1\}\backslash \left\{i_0\right\}$. If $j>i_0$, then, by Claim 4, the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}{u}_{i_0+1}^{-}{u}_j^{+}\overrightarrow{P}{u}_{\kappa }{L}_{\kappa }w{L}_{i_0+1}{u}_{i_0+1}\overrightarrow{P}{u}_{j}C\cup \overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ is equal to $V(P\cup C)\cup \left\{w\right\}$, which contradicts (I). If $j<i_0$, then, by Claims 4 and 5, the set of vertices of the subgraph $v_L\left(P\right)\overrightarrow{P}{u}_j{L}_{j}w{L}_{\kappa }{u}_{\kappa }\overleftarrow{P}{u}_{i_0+1}C{u}_{i_0}\overrightarrow{P}{u}_{i_0+1}^{-}{u}_j^{+}\overrightarrow{P}{u}_{i_0}^{-}\cup \overrightarrow{\mathcal{Q}}(u_{\kappa }^{+},P)$ is equal to $V(P\cup C)\cup \left\{w\right\}$, which again contradicts (I).
• Assume that $y\in End\left({\mathcal{S}}_P\right)$. If $y=v_L\left(P\right)$, then $v_R\left(P\right)\overleftarrow{P}{u}_{i_0+1}{L}_{i_0+1}w{L}_{i_0}{u}_{i_0}\overleftarrow{P}{v}_L\left(P\right)u_{i_0+1}^{-}\overleftarrow{P}{u}_{i_0}^{+}$ covers $V\left(P\right)\cup \left\{w\right\}$, which contradicts (I). Therefore, $y=v_R\left(P\right)$. Then $v_L\left(P\right)\overrightarrow{P}{u}_{i_0+1}^{-}{v}_R\left(P\right)\overleftarrow{P}{u}_{i_0+1}$$L_{i_0+1}w$ covers $V\left(P\right)\cup \left\{w\right\}$, which again contradicts (I).

This contradiction demonstrates that $N\left(u_{i+1}^{-}\right)\cap \mathcal{X}\subseteq \left\{u_i^{+}\right\}$. By (4), $N\left(u_{i+1}^{-}\right)\cap \mathcal{X}=\left\{u_i^{+}\right\}$. □

By Claim 9, it holds that $u_{i+1}^{-}{u}_i^{+}\in E\left(G\right)$ for every $i\in \{1,\cdots ,\kappa -1\}$. Let $C_i=G\left[E\right(P[u_i^{+},u_{i+1}^{-}]\cup \left\{u_i^{+}{u}_{i+1}^{-}\right\}\left)\right]$ for every $i\in \{1,\cdots ,\kappa -1\}$.

Claim 10.

For each section$P[u_i^{+},u_{i+1}^{-}]$, the following two statements are true.

• $G\left[V\left(P[u_i^{+},u_{i+1}^{-}]\right)\right]$forms a clique;
• $N\left(V\left(P[u_i^{+},u_{i+1}^{-}]\right)\right)\cap (V\left(G\right)\backslash V\left(P[u_i^{+},u_{i+1}^{-}]\right))=U$.

Proof of Claim 10.

We pick an element $C\in {\mathcal{S}}_C$; by Claims 4 and 5, a new path $Q_i=v_L\left(P\right)\overrightarrow{P}{u}_{i}C{u}_{i+1}$$\overrightarrow{P}{v}_R\left(P\right)$ would be obtained. We structure a new system ${\mathcal{S}}_i$ such that ${\mathcal{S}}_i={\mathcal{S}}_{iC}\cup {\mathcal{S}}_{iP}$, ${\mathcal{S}}_{iP}=\left\{Q_i\right\}$ and ${\mathcal{S}}_{iC}=({\mathcal{S}}_C\backslash \left\{C\right\})\cup \left\{C_i\right\}$. It is easy to verify that $V\left({\mathcal{S}}_i\right)=V\left(\mathcal{S}\right)$, $|{\mathcal{S}}_{iC}|=|{\mathcal{S}}_C|$ and $|{\mathcal{S}}_{iP}|=|{\mathcal{S}}_P|$. Hence, ${\mathcal{S}}_i$ is also a k-ended system satisfying (I), (II). According to Lemma 9(6), $G\left[V\left(C_i\right)\right]$ forms a clique; by Claim 4, $N\left(V\left(C_i\right)\right)\cap (V\left(G\right)\backslash V\left(C_i\right))=U$. Claim 10 is proved. □

By Claims 2–10 and Lemma 9(6), $\omega (G-U)=k+\kappa$ and every component of $G-U$ forms a clique. Then,

$$\begin{gathered} m\left(G\right)=\left|V\left(H\right)\right|·|V\left(P[v_L\left(P\right),u_1^{-}]\right)|·|V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)|·{\displaystyle \prod _{i=1}^{\kappa -1}}|V\left(P[u_i^{+},u_{i+1}^{-}]\right)|·{\displaystyle \prod _{C\in {\mathcal{S}}_C}} \\ \left|V\left(C\right)\right|\ge \underset{\kappa +k}{\underbrace{1·1·\cdots ·1·1}}·[n-2\kappa -k+1]=n-2\kappa -k+1 \end{gathered}$$. This completes the proof of Fact 1. □

Fact 2.

$\left|V\right(H\left)\right|=1$ and $m\left(G\right)=n-2\kappa -k+1$.

Proof of Fact 2.

By Fact 1 and the condition of Theorem 5, $m\left(G\right)=n-2\kappa -k+1$. Then, we will show that $\left|V\right(H\left)\right|=1$.

By contradiction, suppose that $\left|V\right(H\left)\right|\ge 2$. Then $\left|V\right(P[u_i^{+},u_{i+1}^{-}]\left)\right|=1$ for any $i\in \{1,\cdots ,\kappa -1\}$. Otherwise, $m\left(G\right)>n-2\kappa -k+1$, contradicting $m\left(G\right)=n-2\kappa -k+1$. Let $x=V\left(P[u_{i_0}^{+},u_{i_0+1}^{-}]\right)$ for some $i_0\in \{1,\cdots ,\kappa -1\}$. By Claims 7(2) and 8, $v_L\left(P\right)\overrightarrow{P}{u}_{i_0}H{u}_{i_0+1}\overrightarrow{P}{v}_R\left(P\right)$ in G cover $\left(V\right(P)\backslash \{x\left\}\right)\cup V\left(H\right)$, which contradicts (I). This contradiction shows that $\left|V\right(H\left)\right|=1$. □

Denote

$$\begin{array}{c}\hfill J\left(r\right)=\left\{\begin{array}{cc}G\left[V\left(P[u_i^{+},u_{i+1}^{-}]\right)\right],\hfill & \mathrm{if}r\in U^{+}\backslash \left\{u_{\kappa }^{+}\right\},sayr=u_i^{+},\hfill \\ G\left[V\left(P[v_L\left(P\right),u_1^{-}]\right)\right],\hfill & \mathrm{if}r=v_L\left(P\right),\hfill \\ G\left[V\left(P[u_{\kappa }^{+},v_R\left(P\right)]\right)\right],\hfill & \mathrm{if}r=v_R\left(P\right),\hfill \\ G\left[V\right(C\left)\right].\hfill & \mathrm{if}r=v_C.\hfill \end{array}\right.\end{array}$$

Finally, we need to prove that G is isomorphic to one of those graphs F with $F_0(\kappa +k-1,\kappa )\subseteq F\subseteq F_2(\kappa +k-1,\kappa )$ or $F_{00}(\kappa +k-1,\kappa )\subseteq F\subseteq F_2(\kappa +k-1,\kappa )$. Denote $R=End\left(\mathcal{S}\right)\cup (U^{+}\backslash \left\{u_{\kappa }^{+}\right\})$. By Fact 2, $\left|V\right(\mathcal{S}\left)\right|=n-1$ and there exists at most one vertex $r_0\in R$ such that $\left|J\right(r_0\left)\right|\ge 2$. Then $|J\left(r_0\right)|=n-2\kappa -k+1=m\left(G\right)$. Let $W_1\left(G\right)=\left\{w\right\}\cup (R\backslash \left\{r_0\right\})$. It follows that $W_1\left(G\right)$ is an independent set of G with a cardinality of $k+\kappa -1$. Additionally, $W_1\left(G\right)\cup \left\{x\right\}$ is a maximum independent set of G for any vertex $x\in J\left(r_0\right)$. By Claims 4, 7 and 10, $y\in R\backslash \left\{r_0\right\}$ is not adjacent to any vertex in $J\left(r_0\right)\cup \left\{w\right\}$; it should be adjacent to $u_i$ for all $i\in \{1,\cdots ,\kappa \}$. Now let $H_1=G\left[W_1\left(G\right)\right]$ and $H_2=G\left[U\right]$. This implies that $F_0(\kappa +k-1,\kappa )\subseteq G\subseteq F_2(\kappa +k-1,\kappa )$ and $n>3\kappa +k-1$ or $F_{00}(\kappa +k-1,\kappa )\subseteq G\subseteq F_2(\kappa +k-1,\kappa )$ and $2\kappa +k\le n\le 3\kappa +k-1$ (note that $J\left(r_0\right)\cong K_{n-2\kappa -k+1}$), which completes the proof of Theorem 7. □

5. Conclusions

We demonstrats that it does not change the existence of spanning k-ended tree if we expand the independent numbers a little bit and bound $m\left(G\right)$. Therefore, we generalize Theorem 5 and the bound on the number of maximum independent sets is already sharp. Note that a Hamilton path is viewed as a spanning tree with exactly two leaves; in other words, a Hamilton path is a spanning 2-ended tree. Hence, our results extend Theorem 4, which has significant implications for traceability and the existence of spanning trees. Moreover, we extend Theorem 5 to the case where $\kappa \left(G\right)\ge 2$. This extension has important implications for the study of independent sets in highly connected graphs.

The proof of the results is currently too complex and difficult. We hope to find a more clever and concise proof technique for Theorem 7 in the future.