MPE Mathematical Problems in Engineering 1563-5147 1024-123X Hindawi Publishing Corporation 10.1155/2016/6041841 6041841 Research Article The Existence of Spanning Ended System on Claw-Free Graphs http://orcid.org/0000-0001-6494-7082 Chen Xiaodong 1 http://orcid.org/0000-0002-9563-0616 Xu Meijin 1 http://orcid.org/0000-0002-1771-6342 Liu Yanjun 1 Meng Xiangyu College of Science Liaoning University of Technology Jinzhou 121001 China lnit.edu.cn 2016 2092016 2016 27 06 2016 25 08 2016 2016 Copyright © 2016 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.

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

NSFC Tian Yuan Special Foundation 11426125 Educational Commission of Liaoning Province L2014239
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 .

We consider only finite and simple graphs. For notation and terminology not defined here we refer to . For a subgraph H of a graph G, G-H denotes the induced subgraph by V(G)-V(H), and G[S] denotes the induced subgraph by S for SV(G). For vV(G), let N(v) denote the set of vertices adjacent to v, and N[v]=N(v){v}. P[a,b] denotes a path with end vertices a, b and a positive orientation from a to b. For a path P[a,b], x,yV(P), let xPy denote the subpath from x to y with positive orientation and yP-x denote the subpath from y to x with negative orientation. Similarly, for a cycle C with a given direction, we can define C[a,b], C-[a,b] with a,bV(C). In the paper, we define clockwise as the positive direction of a cycle. We use Km to denote a complete graph with order m, and if m=1, then it is trivial vertex. A tree with at most k leaves is called k-ended tree.

A graph is called claw-free if it does not contain K1,3 induced subgraph. For a vertex xV(G), let Gx denote the graph with V(Gx)=V(G) and E(Gx)=E(G){uv:u,vN(x)}, and then Gx is called the local completion of G at x. For a graph G, xV(G), if G[N(x)] is connected, then x is locally connected; if G[N(x)] is a complete induced subgraph of G, then x is simplicial; if x is locally connected, but not simplicial, then x is eligible.

Ryjáček  proposed a closure operation on a claw-free graph G 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 cl(G). 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 [<xref ref-type="bibr" rid="B15">15</xref>]).

If G is a connected claw-free graph, then cl(G) is Hamiltonian if and only if G is Hamiltonian.

Actually, there are a lot of results which present that cl(G) and G have many common properties.

Theorem 2 (Brandt et al. [<xref ref-type="bibr" rid="B12">13</xref>]).

A claw-free graph G is traceable if and only if cl(G) is traceable.

Theorem 3 (Ryjáček et al. [<xref ref-type="bibr" rid="B16">16</xref>]).

Let G be a claw-free graph. If cl(G) contains 2 factors with k components, then G contains 2 factors with at most k components.

A tree with at most k leaves is called k-ended tree. Win  provided sufficient conditions for a graph to contain spanning k-ended trees by spanning k-ended system. A system of a graph which contains paths, cycles, and trivial vertices is defined by a function f(α) as follows: (1)fα=1ifαisK1,K2,oracycle,2ifαisapathoforderatleast3.

A system S is called k-ended system if αSf(α)k. Moreover, if V(S)=V(G), then S is called a spanning k-ended system of G. Obviously, if G contains a spanning k-ended system, then G contains a spanning k-ended tree. It follows that if a graph contains no spanning k-ended tree, then it contains no spanning k-ended systems.

In this paper, we prove that cl(G) can preserve the existence of spanning k-ended system of G.

Theorem 4.

A claw-free graph G contains a spanning k-ended system if and only if cl(G) contains a spanning k-ended system.

2. Proof of Theorem <xref ref-type="statement" rid="thm4">4</xref>

We divide a k-ended system S of a graph G into two sets S1 and S2 and let(2)S1=αS:fα=1,S2=αS:fα=2.

For every component CS1, we take one vertex xCV(C). For every path PS2, let xP and yP denote the two end vertices of P. We define(3)EndS1=CS1xC,EndS2=PS2xP,yP,EndS=EndS1EndS2.

For a spanning t-ended system S of a graph G, if there is no spanning s-ended system with s<t, then we call the system minimum spanning t-ended system. Obviously, for a minimum spanning t-ended system S of G, End(S) is an independent set of G with EndS=t.

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

Theorem 5.

Let G be a claw-free graph with δ(G)2. Then Gx contains a spanning k-ended system for any vertex xV(G) if and only if G contains a spanning k-ended system.

Proof.

Obviously, the sufficiency holds and we only need to prove the necessity. Assume Gx contains a spanning k-ended system S which satisfies the following properties.

S is a minimum spanning k-ended system of Gx.

E(Gx)-E(G) is minimum, subject to (T1).

|S2| is minimum, subject to (T1) and (T2).

If P contains x, then P contains as many vertices in N(x) as possible, where PS2 subject to (T1), (T2), and (T3).

If C contains x, then C contains as many vertices in N(x) as possible, where CS1 subject to (T1), (T2), and (T3).

If E(S)-E(G)=, then S is a spanning k-ended system in G, and we are done. Thus we assume |E(S)-E(G)|1.

Claim 1.

For PS2, if E(P)-E(G), then xV(P).

Proof.

To the contrary, suppose xV(P), P=aPb, and uvE(P)-E(G) with v=u+. Then u,vN(x). Suppose xcP1d, P1S2. Since G[x,u,v,x+(x-)]K1,3, ux+(ux-)E(G) or vx+(vx-)E(G). Without loss of generality, suppose xd and ux+E(G). Then Gx contains two paths P2=cP1xvPb and P3=aPux+P1d. Replacing P and P1 by P2 and P3, then Gx contains a spanning k-ended system with less edge than E(Gx)-E(G), a contradiction to (T2).

Suppose xV(C), CS1. If C={u}, then Gx contains a path P=aPuxvPb with V(P)=V(C)V(P). Replacing P and C by P, Gx contains a spanning (k-1)-ended system, a contradiction to (T1). Thus VC2. Since G[x,u,v,x+]K1,3, ux+E(G) or vx+E(G). Without loss of generality, suppose ux+E(G). Then Gx contains a path P=aPuC[x+,x]vPb with V(P)=V(P)V(C). Replacing P and C by P, then Gx contains a spanning (k-1)-ended system, a contradiction to (T1).

Claim 2.

For CS1, if E(C)-E(G), then xV(C).

Proof.

Since E(C)-E(G), |V(C)|2. Suppose uvE(C)-E(G), v=u+. Assume to the contrary xV(C). Suppose xaPb, PS2. Since G[x,x+(x-),u,v]K1,3, x+u(x-u)E(G) or x+v(x-v)E(G). Without loss of generality, assume xb and x+uE(G). Then Gx contains a path P=bP-x+C-[u,v]xP-a with V(P)=V(P)V(C). Replacing P, C by P, then Gx contains a spanning (k-1)-ended system, a contradiction.

Suppose xV(C), CS1-{C}. If V(C)={x}, then Gx contains a cycle C=C[v,u]xv with V(C)=V(C)V(C). Replacing C,C by C, then Gx contains a spanning (k-1)-ended system, a contradiction. If |VC|2, then by the preceding proof x+uE(G) or x+vE(G). Without loss of generality, assume x+uE(G). Then Gx contains a cycle C=xCv,uC[x+,x] with V(C)=V(C)V(C). Replacing C, C by C, then Gx contains a spanning (k-1)-ended system, a contradiction.

Since S is a disjoint system, we can get the following two results by Claims 1 and 2.

Claim 3.

For PS2, if E(P)-E(G), then E(S)-E(G)E(P).

Claim 4.

For CS1, if E(C)-E(G), then E(S)-E(G)E(C).

Now we prove the case that E(P)-E(G), for PS2. Then, by Claim 3, E(S)-E(G)E(P). Suppose P=aPb, and then we can get the following results.

Claim 5.

Consider the following: EP-EG2.

Proof.

Suppose, to the contrary, u1v1,u2v2,u3v3E(P)-E(G), where u1, v1, u2, v2, u3, and v3 are labeled in order along the positive orientation of P. Since G[x,u1,v1, v2]K1,3, u1v2E(G) or v1v2E(G). Without loss of generality, assume u1v2E(G). Then Gx contains a path P=aPu1v2Pu3v1Pu2v3Pb with V(P)=V(P) and |EP-E(G)|<|E(P)-E(G)|, a contradiction to (T2).

Claim 6.

Consider the following: EP-EG=1.

Proof.

Suppose, to the contrary, |E(P)-E(G)|=2 by Claim 5 and u1v1,u2v2E(P)-E(G), where u1, v1, u2, and v2 are labeled in order along the positive orientation of P. By the proof of Claim 5, u1v2,v1u2E(G), v1v2E(G). By Claim 1, xV(P). Assume xP[a,u1]. Then x+v1E(G); otherwise Gx contains a path P=aPxu2P-v1x+Pu1v2Pb with V(P)=V(P) and |E(P)-E(G)|=1, a contradiction to (T2) by Claim 3. If x+v2E(G), then Gx contains a path P=aPxu2P-v1u1P-x+v2Pb with V(P)=V(P) and |E(P)-E(G)|<|E(P)-E(G)|, a contradiction to (T2) by Claim 3. Thus G[x,x+,v1,v2]=K1,3, a contradiction. By similar proof, we can prove that Claim 6 holds if xP[v1,u2]P[v2,b].

By Claim 6, we assume that E(P)-E(G)={u1v1}, where u1, v1 are labeled in order along the positive orientation of P, and without loss of generality assume xP[a,u1]. Since x is eligible, there exists at least one path in N(x) connecting u1 and v1. Suppose P0 is the shortest path in N(x) connecting u1 and v1. Since G is claw-free, 3VP04. Assume yV(P0), u1yE(G).

Claim 7.

Consider the following: yV(P).

Proof.

To the contrary, suppose yV(P), where P=P[c,d]S2-{P}. If y=c, then Gx contains a path P1=aPu1yPd and a path P2=v1Pb. Replacing P and P by P1 and P2, then Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), a contradiction to (T2). Similarly, yd. Thus y{c,d}. If y-v1E(G), then Gx contains a path P1=aPu1yPd and a path P2=cPy-v1Pb. Replacing P and P by P1 and P2, Gx contains a spanning k-ended system S with E(Gx)-E(G)=, a contradiction to (T2). Thus y-v1E(G). Similarly, v1y+E(G). If y-y+E(G), then Gx contains a path P1=aPu1yv1Pb and a path P2=cPy-y+Pd. Replacing P and P by P1 and P2, Gx contains a spanning k-ended system S with E(Gx)-E(G)E(P1) such that P1 contains more vertices in N(x) than P, a contradiction to (T4). Thus y-y+E(G). yv1E(G); otherwise G[y,v1,y-,y+]=K1,3, a contradiction. x+v1E(G); otherwise Gx contains a path P1=aPxu1P-x+v1Pb with V(P)=V(P1) and E(P1)-E(G)=, a contradiction to (T2). Then x+yE(G) by G[x,x+,y,v1]K1,3. y-x+E(G) or y+x+E(G) by G[y,y-,y+,x+]K1,3 and y-y+E(G). If x+y-E(G), then Gx contains two paths P1=aPxv1Pb and P2=cPy-x+Pu1yPd. Replacing P and P by P1 and P2, Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), a contradiction to (T2). If x+y+E(G), Gx contains two paths P1=aPxv1Pb and P2=cPyu1P-x+y+Pd. Replacing P and P by P1 and P2, then Gx contains a spanning k-1-ended system, a contradiction. Using a similar proof, we can get a contradiction if y1V(C) with CS1. Thus yV(P).

By Claim 1 without loss of generality, in the following proof, assume xP[a,u1). Then x+u1,x-v1E(G), x-x+,x-u1,x+v1E(G) since G is claw-free and by (T2).

Claim 8.

Consider the following: v1yE(G).

Proof.

To the contrary, suppose v1yE(G). By Claim 7, without loss of generality, assume yP[a,u1]. If y=a, then, replacing P by P1=u1P-yv1Pb, Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), a contradiction to (T2). Thus ya. If y-u1E(G), then, replacing P by P1=aPy-u1P-yv1Pb, Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), a contradiction to (T2). Similarly, y-y+E(G). By G[y,y-,y+,u1]K1,3, y+u1E(G). By G[y,y-,y+,x]K1,3, y-xE(G) or y+xE(G). If y-xE(G), then, replacing P by P1=aPxy-P-x+u1P-yv1Pb, Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), a contradiction to (T2). Thus y+xE(G). Replacing P by path P1=aPxy+Pux+PyvPb, Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), a contradiction to (T2).

Claim 9.

P can be transformed to a path P1 such that V(P1)=V(P), v1yE(P1), and E(Gx)-E(G)={v1y}.

Proof.

By Claim 7, without loss of generality, assume yP(x,u1). Since G[y,y-,y+,x]K1,3, y-y+,y-x,y+xE(G). If y-y+E(G), then P1=aPy-y+Pu1yv1Pb. If y-xE(G), then P1=aPxy-P-x+u1P-yv1Pb. If y+xE(G), then P1=aPxy+Pu1x+Pyv1Pb.

By Claim 8, v1yE(G), and then VP0=4. Suppose P0=u1yzv1. By Claim 9, replace P by P1. By the proof of Claim 7, zP1. By the proof of Claim 9, Gx contains a spanning k-ended system with no edge in E(Gx)-E(G), 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.

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