Discrete Dynamics in Nature and Society

Volume 2015 (2015), Article ID 476182, 5 pages

http://dx.doi.org/10.1155/2015/476182

## Spanning 3-Ended Trees in Almost Claw-Free Graphs

College of Science, Liaoning University of Technology, Jinzhou 121001, China

Received 10 November 2015; Accepted 3 December 2015

Academic Editor: Chenguang Yang

Copyright © 2015 Xiaodong Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We prove that if is a -connected () almost claw-free graph of order and , then contains a spanning 3-ended tree, where is an independent set of with .

#### 1. Introduction

In this paper, only finite and simple graphs are considered. We refer to [1] for notation and terminology not defined here. If , as well as every edge of incident with the vertices in , then is a* dominating set*. We use to denote the* domination number* of a graph , where . If a graph has no subgraph, then is* claw-free*. The vertex of degree 3 in is called* claw center* of a claw. Let and . If, for any vertex , and the set of all claw centers in subgraphs of is an independent set, then is* almost claw-free*. A claw-free graph is almost claw-free.

and for some vertex , and . We use to denote the minimum degree sum of all the independent sets with order in . or denotes a path with a positive orientation from to with end vertices , . For a path , , let denote the subpath with end vertices , with positive orientation and denote the subpath with end vertices , with negative orientation. Let denote the number of components of a graph .

As we know, graph theory focuses on graphs composed by vertices and edges. The vertices in a graph are considered as discrete points which is usually discussed in control problems. Then there are a lot of results using graph theory to solve control problems [2–8].

A spanning tree in a graph with no more than leaves is called a spanning -ended tree of . There are many results about the properties of spanning trees [9–13]. Kyaw [14, 15] gave some degree sum conditions for -free graphs to contain spanning -ended trees.

Theorem 1 (see [14]). *If is a connected -free graph and , then contains spanning 3-ended trees.*

Theorem 2 (see [15]). *If is a connected -free graph, then the following properties hold:*(i)* contains a hamiltonian path if .*(ii)* contains spanning -ended trees if for an integer .*

*On the other hand, Kano et al. [16] obtained sharp sufficient conditions for claw-free graphs to contain spanning -ended trees.*

*Theorem 3 (see [16]). If is a connected claw-free graph of order and (), then contains spanning -ended trees with the maximum degree at most 3.*

*Recently, Chen et al. [17] gave some degree sum conditions for -connected -free graphs to contain spanning 3-ended trees.*

*Theorem 4 (see [17]). If is a -connected -free graph of order and (), then contains spanning 3-ended trees.*

*Inspired by Theorems 4 and 5, in this paper, we further explore sufficient conditions for -connected almost claw-free graphs to contain spanning 3-ended trees which holds for claw-free graphs.*

*Theorem 5. If is a -connected almost claw-free graph of order and (), then contains spanning 3-ended trees.*

*Obviously, there are a lot of almost claw-free graphs containing subgraphs, so to some extent Theorem 5 is a generalization of Theorem 4.*

*2. Preliminaries*

*We need the following result given by Ryjáček [18] to prove Theorem 5.*

*Lemma 6 (see [18]). for any claw center in an almost claw-free graph.*

*We mainly use the definition and properties of insertible vertex defined in [16] to prove Theorem 5.*

*Suppose that is a connected nonhamiltonian graph and is longest cycle in with counterclockwise direction as positive orientation. Assume that is a component of and such that are labeled in order along the positive direction of Let , , and . A vertex in is an insertible vertex if has two consecutive neighbors and in .*

*Chen and Schelp [19] proposed the following two results.*

*Lemma 7 (see [19]). For each contains a noninsertible vertex.*

*Assume that is the first noninsertible vertex in for each .*

*Lemma 8 (see [19]). Let , with . Then(a)there is no path in such that ,(b)for any vertex in , if , then . By symmetry, for any vertex in , if , then ,(c)for any vertex in , if , then . By symmetry, for any vertex in , if , then .*

*Suppose, for some , and is the second noninsertible vertex in . Then Chen et al. [17] gave the following result.*

*Lemma 9 (see [17]). Let , and . Then the following properties hold:(a)There does not exist a path in such that .(b)For every vertex , if , then ; similarly, for every , if , then .(c)For every vertex , if , then ; by symmetry, for any vertex in , if , then .*

*3. Proof of Theorem 5*

*3. Proof of Theorem 5**To the contrary, suppose that satisfies the conditions of Theorem 5 and any spanning tree in contains more than 3 leaves. Let be longest path in such that satisfies the following two conditions:(T1) is minimum.(T2) is minimum such that is the first vertex in with , subject to (T1).*

*Suppose that is a component in , and with in order along the positive direction of . Let denote an independent set in .*

*Let denote a graph with , . Then is a maximal cycle in . We define the counterclockwise orientation as the positive direction of . Let denote the segment for , and . By Lemma 7, let denote the first noninsertible vertex in , for , and . By Lemma 8(a), is an independent set.*

* can be divided into disjoint intervals with and . We call the intervals -segments. If , then ; that is, if , then . By the definition of -segment, for any -segment , there exists such that (subscripts expressed modulo ).*

*Claim 1. * and for any .

*Proof. *Suppose that is an insertible vertex such that , where . If , then we can get a path , a contradiction to (T2). If , then let , a contradiction to (T2). Thus . Suppose , for some . Obviously, . Since and is an insertible vertex, a contradiction.

*Claim 2. *For any vertex , if is claw-free, then .

*Proof. *Suppose that is in some -segment with , , and with . Then by Lemma 8(c), . Since , or . Similarly, or . Obviously, at least one vertex in is not in . Without loss of generality, suppose and . Then by is an insertible vertex, a contradiction.

*Claim 3. * for any vertex , and if , then is a claw center.

*Proof. *Without loss of generality, suppose that is in -segment and . If , then . Suppose and , where are in order along the positive direction of Then , for . For some , suppose with . Then is a claw center. By Lemma 6, suppose , are two distinct domination vertices of . Then , are claw-free and at least two vertices in are incident with or . Without loss of generality, suppose . Then , and by Lemma 8(c). Suppose containing , . Obviously, at least one vertex in is not in . Without loss of generality, suppose . Since is a noninsertible vertex and , . Thus , a contradiction. If , then by Claim 2, is a claw center.

*Claim 4. *For any -segment not containing , contains at most one vertex with , and .

*Proof. *If , then there exists a -segment with . Suppose and in -segment , then by and . Thus if , then . Without loss of generality, suppose , , where are in order along the positive direction of . By Claim 3, suppose that is the first vertex in with , , and , where . Then , and, by Claim 3, is a claw center. Then is claw-free, and, by Claim 2, . Thus if , then we are done.

Suppose and , . Since contains at least one edge in . Since is a noninsertible vertex and , if or , a contradiction to Lemma 8(b) since . Thus , and . Since is a claw center, is claw-free and, by Claim 2, . Thus if , then we are done.

Suppose and , . Since and , by Lemma 8(b) . Then . Since or . If , then . Then, by Lemma 8(b), and since . Then , a contradiction to the noninsertablity of . Thus , and then is claw-free. By Claim 2, Thus if , then we are done. If , then, proceeding in the above manners to the set , we can get that is claw-free for any vertex in , and then, by Claim 2, . It follows that has exactly one vertex with for . Thus the claim holds.

*Claim 5. *Suppose that the -segment contains . Then for any vertex .

*Proof. *If , then and . Suppose . Then, by Claim 1, . By Lemma 8(b), and then for any vertex .

*Claim 6. *For any vertex in , .

*Proof. *Suppose , where are distinct vertices, . Since is a claw center. By Lemma 6, suppose , are the two distinct domination vertices in . Then , and , are claw-free. Since , . Suppose , are both in . Then at least two vertices of are adjacent to or . Without loss of generality, suppose , and then , a contradiction. Thus one vertex of , is in , and the other vertex is in . Without loss of generality, suppose , . Then . If , then , a contradiction, and, by the preceding proof, there is exactly one vertex of adjacent to . Thus . Without loss of generality, suppose . Since , . Then we can get a longer cycle than , a contradiction.

*Claim 7. *For any vertex in , if and , then and , for any vertex , where is the -segment containing .

*Proof. *Suppose and , . Since is a claw center. Assume and , . If , then, by Lemma 8(a), for any vertex . Thus if with , then . Suppose , . By the proof of Claim 4, is claw-free for any vertex in , which contradicts that is a claw center. It follows that .

*Claim 8. *For any vertex in , if , then .

*Proof. *Without loss of generality, we consider Suppose with and , where , are two distinct vertices. By Claim 3, is a claw center. By Lemma 6, suppose , are the two distinct domination vertices in . Then , are claw-free and . Thus . Without loss of generality, suppose . If , then , a contradiction to Claim 2. Suppose . By Claim 2, , can not both be incident with or . Without loss of generality, suppose . Obviously, . Suppose , . If , then , a contradiction. Suppose . Then, by , . Obviously, all the vertices in can be inserted into . Let denote the path by the above inserting process with and denote the path connecting and with internal vertices in . Then we can get a cycle longer than , a contradiction.

*Claim 9. *Consider the following: and .

*Proof. *Obviously, and . Then . By Claim 5, . By Claim 4, for . Thus + .

Obviously, . Then = . Without loss of generality, we consider . Let denote the -segment containing . By Claim 4, for any -segment in . Obviously, . By Claim 6, .

Suppose . Then is the first vertex in and . By Lemma 8(a), . Thus and + .

Suppose . If , then ≤ . If , then, by Claim 7, . Thus + .

Suppose . Then, by Claim 8, . Thus = .

It follows that . Similarly, for any , . By Claim 5, . Thus .

*Claim 10. * and is hamiltonian-connected for each component of .

*Proof. * from Lemma 8(a), for , and then . By the connectedness of , . If , then we get an independent set of order at least . By Claim 9, , a contradiction to . Suppose and . Then we can get an independent set of order . By Claim 9, . Then = , a contradiction to by . Thus .

Suppose that is not hamiltonian-connected. Then by Ore’s theorem [20], suppose with and . We can get an independent set with order . By Claim 9, ≤ , a contradiction to by .

*If contains only one component , then, by Claim 10, there is a spanning 3-ended tree in . Thus we only consider the case that contains at least two components and suppose is a component in .*

*Claim 11. *Consider the following: for some .

*Proof. *Suppose for any . Then . Let , . Then we get an independent set of order . By Claim 10, and, by Claim 9, . Thus = + , a contradiction to .

*By Claim 11, we assume for some . By Lemma 8(a), for .*

*Claim 12. * contains a second noninsertible vertex .

*Proof. *Suppose contains only one noninsertible vertex . Then we can get a path such that by inserting all the vertices in to . Suppose . Then we get a cycle . Let . Then , a contradiction to (T1). Suppose . If , then is a vertex cut of with order , a contradiction to Claim 10. Thus . By Claim 10, there is a hamiltonian path of . Thus we can get a cycle longer than , a contradiction.

*Now, we complete the proof of Theorem 5. By Claim 12 and Lemma 9, is an independent set. Then, by the preceding proof, has the same properties as . By Lemma 9, , for any two distinct vertices . Thus . By Claim 9, . Let , . Then is an independent set of order in . Thus = . By Claim 10, , . Thus ≤ , a contradiction to . It follows that Theorem 5 holds.*

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments*

*This research is supported by the National Natural Science Foundation of China (Grant nos. 11426125 and 61473139), Program for Liaoning Excellent Talents in University LR2014016, and the Educational Commission of Liaoning Province (Grant no. L2014239).*

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