The Existence of Spanning Ended System on Claw-Free Graphs

Abstract

We prove that every connected claw-free graph contains a spanning -ended system if and only if contains a spanning -ended system, where denotes Ryjáček closure of .

1. Introduction

Graph theory focuses on graphs composed of vertices and edges. The vertices in a graph are considered as discrete points usually discussed in control problems. Then there are a lot of results using graph theory to solve control and other application problems [1–13].

We consider only finite and simple graphs. For notation and terminology not defined here we refer to [14]. For a subgraph of a graph , denotes the induced subgraph by , and denotes the induced subgraph by for For , let denote the set of vertices adjacent to , and . denotes a path with end vertices , and a positive orientation from to For a path , , let denote the subpath from to with positive orientation and denote the subpath from to with negative orientation. Similarly, for a cycle with a given direction, we can define , with In the paper, we define clockwise as the positive direction of a cycle. We use to denote a complete graph with order , and if , then it is trivial vertex. A tree with at most leaves is called -ended tree.

A graph is called claw-free if it does not contain induced subgraph. For a vertex , let denote the graph with and , and then is called the local completion of at For a graph , , if is connected, then is locally connected; if is a complete induced subgraph of , then is simplicial; if is locally connected, but not simplicial, then is eligible.

Ryjáček [15] proposed a closure operation on a claw-free graph by joining all nonadjacent pairs of vertices in the neighbourhood of every eligible vertex till there is no eligible vertex, and then we get the closure Ryjáček also gave the following result, which is considered a useful tool to research on the Hamiltonian properties of claw-free graphs.

Theorem 1 (Ryjáček [15]). If is a connected claw-free graph, then is Hamiltonian if and only if is Hamiltonian.
Actually, there are a lot of results which present that and have many common properties.

Theorem 2 (Brandt et al. [13]). A claw-free graph is traceable if and only if is traceable.

Theorem 3 (Ryjáček et al. [16]). Let be a claw-free graph. If contains 2 factors with components, then contains 2 factors with at most components.

A tree with at most leaves is called -ended tree. Win [17] provided sufficient conditions for a graph to contain spanning -ended trees by spanning -ended system. A system of a graph which contains paths, cycles, and trivial vertices is defined by a function as follows:

A system is called -ended system if Moreover, if , then is called a spanning -ended system of Obviously, if contains a spanning -ended system, then contains a spanning -ended tree. It follows that if a graph contains no spanning -ended tree, then it contains no spanning -ended systems.

In this paper, we prove that can preserve the existence of spanning -ended system of

Theorem 4. A claw-free graph contains a spanning -ended system if and only if contains a spanning -ended system.

2. Proof of Theorem 4

We divide a -ended system of a graph into two sets and and let

For every component , we take one vertex For every path , let and denote the two end vertices of We define

For a spanning -ended system of a graph , if there is no spanning -ended system with , then we call the system minimum spanning -ended system. Obviously, for a minimum spanning -ended system of , is an independent set of with

In order to prove Theorem 4, we only need to prove that the following result holds.

Theorem 5. Let be a claw-free graph with Then contains a spanning -ended system for any vertex if and only if contains a spanning -ended system.

Proof. Obviously, the sufficiency holds and we only need to prove the necessity. Assume contains a spanning -ended system which satisfies the following properties. (T1) is a minimum spanning -ended system of .(T2) is minimum, subject to (T1).(T3) is minimum, subject to (T1) and (T2).(T4)If contains , then contains as many vertices in as possible, where subject to (T1), (T2), and (T3).(T5)If contains , then contains as many vertices in as possible, where subject to (T1), (T2), and (T3). If , then is a spanning -ended system in , and we are done. Thus we assume

Claim 1. For , if , then

Proof. To the contrary, suppose , , and with Then Suppose , Since , or Without loss of generality, suppose and Then contains two paths and Replacing and by and , then contains a spanning -ended system with less edge than , a contradiction to (T2).
Suppose , If , then contains a path with Replacing and by , contains a spanning -ended system, a contradiction to (T1). Thus Since , or Without loss of generality, suppose Then contains a path with Replacing and by , then contains a spanning -ended system, a contradiction to (T1).

Claim 2. For , if , then

Proof. Since , Suppose , Assume to the contrary Suppose , Since , or Without loss of generality, assume and Then contains a path with Replacing , by , then contains a spanning -ended system, a contradiction.
Suppose , . If , then contains a cycle with Replacing by , then contains a spanning -ended system, a contradiction. If , then by the preceding proof or Without loss of generality, assume Then contains a cycle with Replacing , by , then contains a spanning -ended system, a contradiction.
Since is a disjoint system, we can get the following two results by Claims 1 and 2.

Claim 3. For , if , then

Claim 4. For , if , then
Now we prove the case that , for Then, by Claim 3, Suppose , and then we can get the following results.

Claim 5. Consider the following: .

Proof. Suppose, to the contrary, , where , , , , , and are labeled in order along the positive orientation of Since , , or Without loss of generality, assume Then contains a path with and , a contradiction to (T2).

Claim 6. Consider the following: .

Proof. Suppose, to the contrary, by Claim 5 and , where , , , and are labeled in order along the positive orientation of By the proof of Claim 5, , By Claim 1, Assume Then ; otherwise contains a path with and , a contradiction to (T2) by Claim 3. If , then contains a path with and , a contradiction to (T2) by Claim 3. Thus , a contradiction. By similar proof, we can prove that Claim 6 holds if
By Claim 6, we assume that , where , are labeled in order along the positive orientation of , and without loss of generality assume Since is eligible, there exists at least one path in connecting and Suppose is the shortest path in connecting and Since is claw-free, Assume ,

Claim 7. Consider the following: .

Proof. To the contrary, suppose , where If , then contains a path and a path Replacing and by and , then contains a spanning -ended system with no edge in , a contradiction to (T2). Similarly, Thus If , then contains a path and a path Replacing and by and , contains a spanning -ended system with , a contradiction to (T2). Thus Similarly, If , then contains a path and a path Replacing and by and , contains a spanning -ended system with such that contains more vertices in than , a contradiction to (T4). Thus . ; otherwise , a contradiction. ; otherwise contains a path with and , a contradiction to (T2). Then by . or by and If , then contains two paths and Replacing and by and , contains a spanning -ended system with no edge in , a contradiction to (T2). If , contains two paths and Replacing and by and , then contains a spanning -ended system, a contradiction. Using a similar proof, we can get a contradiction if with Thus
By Claim 1 without loss of generality, in the following proof, assume Then , since is claw-free and by (T2).

Claim 8. Consider the following: .

Proof. To the contrary, suppose By Claim 7, without loss of generality, assume If , then, replacing by , contains a spanning -ended system with no edge in , a contradiction to (T2). Thus If , then, replacing by , contains a spanning -ended system with no edge in , a contradiction to (T2). Similarly, By , By , or If , then, replacing by , contains a spanning -ended system with no edge in , a contradiction to (T2). Thus Replacing by path , contains a spanning -ended system with no edge in , a contradiction to (T2).

Claim 9. can be transformed to a path such that , , and

Proof. By Claim 7, without loss of generality, assume Since , If , then If , then If , then
By Claim 8, , and then Suppose By Claim 9, replace by By the proof of Claim 7, By the proof of Claim 9, contains a spanning -ended system with no edge in , a contradiction to (T2). It follows that Theorem 4 holds and then Theorem 5 holds.

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

The research was supported by NSFC, Tian Yuan Special Foundation 11426125, and Educational Commission of Liaoning Province L2014239.