#### Abstract

A graph G is called -free if G contains no as an induced subgraph. A tree with at most m leaves is called an m-ended tree. Let be the minimum degree sum of k independent vertices in G. In this paper, it is shown that every connected -free graph G contains a spanning 6-ended tree if .

#### 1. Introduction

We will begin with some basic definitions, notations, and terminologies used by Bondy and Murty . The graph generally means a simple undirected graph G with a vertex set and an edge set . We write for the order of G. For any , let be the set of vertices adjacent to in G and . For a nonempty subset , we write

For an integer i, we define and . Let T be a tree. A vertex of degree one in T is called a leaf of T, and a vertex of degree at least three is called a branch vertex. If are vertices of T, the path in T connecting u and is unique and is denoted by . Define , and . The distance between u and in T is denoted by .

A subset is called an independent set of G if no two vertices of A are adjacent in G. The independence number of G, , is the order of the maximum independent set of G. For an integer k, we define

For an integer , a graph G is called -free if G contains no as an induced subgraph. A tree with at most m leaves is called an m-ended tree. If T is an m-ended tree in G such that is as large as possible, we call T the maximum m-ended tree of G. There are several well-known conditions ensuring that a graph G contains a spanning m-ended tree (see the survey paper ). In 1979, Win  obtained a sufficient condition related to an independent number for k-connected graphs, which confirms the conjecture of Las Vergnas .

Theorem 1 (Win ). Let G be a k-connected graph, and let . If , then G has a spanning m-ended tree.

There are many results on the degree sum conditions for a graph to contain a spanning tree with a bounded number of leaves or branch vertices. Among them are the following theorems.

Theorem 2 (Broersma and Tuinstra ). Let G be a connected graph with n vertices, and let . If , then G has an m-ended tree.

Theorem 3 (Gargano et al. ). Let k be a nonnegative integer, and let G be a connected claw-free graph. If , then G has a spanning tree with at most k branch vertices.

Theorem 4 (Kano et al. ). Let k be a nonnegative integer, and let G be a connected claw-free graph. If , then G has a spanning tree with at most leaves.

Theorem 5 (Kyaw ). Every connected -free graph with contains a spanning tree with at most 3 leaves.

Theorem 6 (Kyaw ). Let G be a connected -free graph.(1)If , then G has a Hamiltonian path.(2)If for an integer , then G has a spanning tree with at most k leaves.

Theorem 7 (Chen et al. ). Let G be a connected -free graph with n vertices. If , then G contains a spanning 4-ended tree.

In this paper, we further consider -free graphs and generalize Theorem 7 as follows.

Theorem 8. Let G be a connected -free graph. If , then G contains a spanning 6-ended tree.

Before proving Theorem 8, we first give an example to show that the condition in Theorem 8 is sharp.

Let m be a positive integer, and let G be the graph obtained from a triangle and 7 copies of by joining x to each vertex of , y to each vertex of , and z to each vertex of (see Figure 1). It is easy to see that , but each spanning tree of G has at least 7 leaves. Therefore, the condition in Theorem 8 is sharp.

#### 2. Proof of Theorem 8

In this section, we will prove Theorem 8. First, we prove the following two lemmas.

Lemma 1. Let G be a connected graph with no spanning 6-ended tree, and let T be a maximum 7-ended tree of G. Then, G does not contain a 6-ended tree such that .

Proof of Lemma 1. By way of contradiction, suppose that there is a 6-ended tree in G such that . Since G has no spanning 6-ended tree, is not a spanning tree. Hence, . As G is connected, there are two vertices such that and . Noting that has at most 7 leaves, we conclude that is a 7-ended tree, contrary to the maximality of T. Hence, Lemma 1 is true.

Lemma 2. Let T be a tree with n vertices. For an integer , if T has at most k leaves, then it has at most branch vertices.

Proof of Lemma 2. Assume that T has x vertices of degree two, y branch vertices, and z leaves. Then, and . Hence, . This completes the proof of Lemma 2.
Based on the above two lemmas, we can now prove the main result of this paper.

Proof of Theorem 8. Let G be a connected -free graph. Suppose, by contradiction, that G does not contain a spanning 6-ended tree. As G is connected, there is a spanning tree T in G. By our hypothesis, i.e., G does not contain a spanning 6-ended tree, the number of leaves in T is at least 7. If the number of leaves in T is 7, then we have found a tree with exactly 7 leaves. If the number of leaves in T is more than 7, then we can delete some of the leaves, and hence, there always exists a 7-ended tree in G. We choose a maximum 7-ended tree T. By Lemma 1, T has exactly 7 leaves. Denote by the set of leaves of T, and let be the set of branch vertices of T. From Lemma 2, we can conclude that . By the maximality of T, we have . For , let be the unique component of containing , and let be the only neighbor of S in . Denote by the only neighbor of in S. For convenience, we let . Note that . For , let be the predecessor of x and be the successor of x. If , let be the longest path between the branch vertices. Without loss of generality, we may assume that and . For convenience, denote by the predecessor of and the successor of for all . Set . If for some and , then we call the branch vertex for .

Claim 1. For all with , if , then .

Proof of Claim 1. By way of contradiction, assume Claim 1 is false. SetThen, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 1.
It follows from Claim 1 that U is an independent set. Since G is -free, .

Claim 2. For , .

Proof of Claim 2. By way of contradiction, assume that there exists a vertex for some . Then, there exist three distinct integers such that . By Claim 1, we can derive that and . As , . Set . Then, is a 6-ended tree with , which is a contradiction. This completes the proof of Claim 2.
By an argument similar to the proof of Claim 2, we can derive that , .

Claim 3. for all .

Proof of Claim 3. Suppose, on the contrary, that there are two vertices for some . Without loss of generality, assume that . By Claim 1, . Assume that for some . Since , there exists such that .
If , then . Then, is a 6-ended tree with a vertex set , which contradicts Lemma 1. Hence, . Since G is -free, it can be derived that . Set . Then, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 3.

Claim 4. If , then for all .

Proof of Claim 4. By way of contradiction, assume that Claim 4 is false. Then, there is a vertex x such that for some with . By Claim 1, . Set . Then, is a 6-ended tree with , which contradicts Lemma 1. This completes the proof of Claim 4.

Claim 5. For all , , , , and are pairwise disjoint, where .

Proof of Claim 5. It can be easily concluded that . It follows from Claim 1 that . This completes the proof of Claim 5.
By Claim 5, we haveThis together with Claim 4 implies that

Claim 6. If for some and , then and for some .

Proof of Claim 6. By way of contradiction, assume that or . SetThen, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 6.
For every vertex , defineFor every vertex set , defineIf , we consider the following two cases.

Case 1. .
In this case, . We choose the maximum T with 7 leaves such that(C1) is as large as possible.(C2) is as small as possible, subject to (C1).Without loss of generality, we may assume that , , and . Hence, and . SetLet be the family of graphs shown in Figure 2. Then, in Case 1, .
The following claim plays a crucial role in the proof of Case 1.

Claim 7. Let be two consecutive vertices in such that , where and . Let be the branch vertex for , where . If G has no 6-ended tree, then and .

Proof of Claim 7. By way of contradiction, assume that . Set . Then, is a 6-ended tree with , which is contrary to Lemma 1. Similarly, . This completes the proof of Claim 7.

Claim 8. Let be the branch vertex for for some and . Then, . Specifically, for all and for all .

Proof of Claim 8. Suppose, for contradiction, that for some . Set . Hence, is a 6-ended tree with , contradicting Lemma 1. Hence, . Likewise, . This completes the proof of Claim 8.

Claim 9. If for some , then .

Proof of Claim 9. By way of contradiction, suppose and . Hence, there exist two integers with such that . By Claim 8, we can see that is not the branch vertex for or . This together with , by applying Claim 7, implies that . This and Claim 8 imply that , which is a contradiction. Hence, , and similarly, if , then .
The proof of Case 1 has been divided into 6 subcases.

Subcase 1. .
Without loss of generality, we may assume that , .
We choose T such that(C3) is as large as possible, subject to (C1) and (C2).

Subclaim 1. For every vertex .

Proof of Subclaim 1. By way of contradiction, suppose that there is a vertex such that . Then, there exist three distinct integers such that . By (C2), we can see that . If , set . Hence, is a 6-ended tree with , which is contrary to Lemma 1. Thus, . Since G is -free, there exists such that or , contradicting Claim 7. This completes the proof of Subclaim 1.
Set . If for every vertex , then . Otherwise, suppose that there are s vertices in satisfying for all . Define , , and . Therefore, .
For every , . It follows from Claim 7 and (C2) that . Hence, . Since , . Thus, . Therefore,

Subclaim 2.

Proof of Subclaim 2. If , by (C2), , say . Then, . Otherwise, assume that and for some . Then by (C2), , that is, . If , the result holds. Hence, we assume that . Since , by Claims 4 and 6, for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. It follows from (5) that

Subclaim 3. .

Proof of Subclaim 3. If there are three distinct integers in such that , then since G is -free, then , which contradicts (C2).
If for , by (C2), . Without loss of generality, set . Since , by (C2) and Claims 1 and 6, we have . Also by Claim 6, . Moreover, for all .
If for some , set . Then by Claim 7, . If there is a vertex such that , then . Otherwise, for every , . By Claim 7, . Specifically, . By applying Claim 7, we get a contradiction. Hence, . If for some , then . Suppose, for contradiction, and . Then by (C3), . Set , which contradicts (C2). This together with (5) implies that the result holds.
It follows from Subclaims 2-3 thatwhich implies that , which is a contradiction. This completes the proof of Subcase 1.

Claim 10. If and there is a vertex , then .

Proof of Claim 10. Assume that there is a vertex such that . Then, there exist three distinct integers such that . Set . Then, is a maximum tree with 7 leaves. Since are branch vertices, by (C2), and . This implies that we can find a tree such that . This completes the proof of Claim 10.
Thus, in the following subcases, we suppose that, for every vertex , .
If for every vertex , then . Otherwise, suppose that there are s vertices in P satisfying for all . Define , , and . Therefore, .
It follows from Claim 7, Claim 8, and (C2) that . Since , by applying Claim 7 with and (C2), we can derive that either or there exists some such that for all . In the latter case, combining with , we can conclude that contradict Claim 7. Thus, , and hence, , where .
If , similar arguments apply to and then . Thus,If , by Claim 7, we can deduce that for all . Hence,In the following, if , we choose T such that(C4) is as large as possible, subject to (C1) and (C2).

Subcase 2. .
Without loss of generality, we may assume that and . If , by (C2), .
Since , there are three integers in , say , such that or for some . Hence, and for all .(i) . In this case, and for all . If , by Claim 6 and (C2), we can derive that . Hence, .(ii) . Then, , since otherwise is a 6-ended tree with , which contradicts Lemma 1. If , then assume that and for some . By (C2), . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Thus, . Hence, .(iii). Then by similar arguments, . If , then we claim that . If , then the above claim is true. Hence, . Assume and . By similar arguments, we have . Thus, . Since , by Claims 1, 4, and 6, , where . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence, . Above all, we haveIf there are three distinct integers such that , then . Since , we have for some , which contradicts Claim 7.
From (C4), for all . If , then .
If , there is an integer such that . By symmetry, set . Then, and . Otherwise, if , set . If , then . Set . Then, is a 6-ended tree with , which contradicts Lemma 1. Moreover, . This together with the proof of (14) implies that .
In either case, we can getwhich implies that , which is a contradiction. This completes the proof of Subcase 2.

Subcase 3. .
In this case, T satisfies and . Without loss of generality, we may assume that and .
If , then . Otherwise, we can find a tree satisfying Subcase 2. Analysis similar to that of Subcase 2 shows thatIf for some , then set . Then, . If , the result is true. Hence, . Suppose, for contradiction, for some . By (C4), we have . Since G is -free, for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence, if for some , then .
If for some , then set . Then, and , that is, .
If for some , then set . Since , for some , say . Hence, . By similar arguments, we can derive that if , then . Hence,Above all, we haveThis together with (20) shows thatwhich implies that , which is a contradiction. This completes the proof of Subcase 3.

Subcase 4. .
In this case, T satisfies and . Without loss of generality, we may assume that , .

Subclaim 4. for all . Similarly, for all and for all .

Proof of Subclaim 4. By way of contradiction, suppose that for some . Set . Then, , which is a contradiction to Subcase 2. Similarly, suppose that or for some . Set or . Then, , which is a contradiction to Subcase 3. This completes the proof of Subclaim 4.

Subclaim 5. .

Proof of Subclaim 5. It follows from (C4) thatIt , then . By way of contradiction, assume that and . By (25), . Set . Then, , which contradicts (C2).
If for , then . Hence, . By Claim 7, . Moreover, . Otherwise, assume that . Then, . Set . is a 6-ended tree with , which is contrary to Lemma 1. Hence, . By a similar argument as in the proof of (10), we haveThus, . This completes the proof of Subclaim 5.

Subclaim 6. .

Proof of Subclaim 6. From (C4), we haveIf such that , then we assert that . By way of contradiction, assume that and