We prove that if G is a k-connected (k≥2) almost claw-free graph of order n and σk+3(G)≥n+2k-2, then G contains a spanning 3-ended tree, where σk(G)=min{∑v∈Sdeg(v):S is an independent set of G with S=k}.

1. Introduction

In this paper, only finite and simple graphs are considered. We refer to [1] for notation and terminology not defined here. If B⊆V(G), as well as every edge of G incident with the vertices in B, then B is a dominating set. We use γ(G) to denote the domination number of a graph G, where γ(G)=min{S:S is a dominating set of G}. If a graph G has no K1,3 subgraph, then G is claw-free. The vertex of degree 3 in K1,3 is called claw center of a claw. Let N(v)={u:uv∈E(G)} and N[v]=N(v)∪{v}. If, for any vertex v∈V(G), γ(N(v))≤2 and the set of all claw centers in K1,3 subgraphs of G is an independent set, then G is almost claw-free. A claw-free graph is almost claw-free.

NH(S)={v:v∈V(H) and uv∈E(G) for some vertex u∈V(S)}, and dH(S)=NHS. We use σk(G) to denote the minimum degree sum of all the independent sets with order k in G. aPb or P[a,b] denotes a path with a positive orientation from x to y with end vertices a, b. For a path P[a,b], x,y∈V(P), let xPy denote the subpath with end vertices x, y with positive orientation and yP-x denote the subpath with end vertices y, x with negative orientation. Let w(G) denote the number of components of a graph G.

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 G with no more than k leaves is called a spanning k-ended tree of G. There are many results about the properties of spanning trees [9–13]. Kyaw [14, 15] gave some degree sum conditions for K1,4-free graphs to contain spanning k-ended trees.

Theorem 1 (see [<xref ref-type="bibr" rid="B2">14</xref>]).

If G is a connected K1,4-free graph and σ4(G)≥G-1, then G contains spanning 3-ended trees.

Theorem 2 (see [<xref ref-type="bibr" rid="B4">15</xref>]).

If G is a connected K1,4-free graph, then the following properties hold:

G contains a hamiltonian path if σ3(G)≥G.

G contains spanning k-ended trees if σk+1(G)≥G-k/2 for an integer k≥3.

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

Theorem 3 (see [<xref ref-type="bibr" rid="B6">16</xref>]).

If G is a connected claw-free graph of order n and σk+1(G)≥n-k (k≥2), then G contains spanning k-ended trees with the maximum degree at most 3.

Recently, Chen et al. [17] gave some degree sum conditions for k-connected K1,4-free graphs to contain spanning 3-ended trees.

Theorem 4 (see [<xref ref-type="bibr" rid="B5">17</xref>]).

If G is a k-connected K1,4-free graph of order n and σk+3(G)≥n+2k-2 (k≥2), then G contains spanning 3-ended trees.

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

Theorem 5.

If G is a k-connected almost claw-free graph of order n and σk+3(G)≥n+2k-2 (k≥2), then G contains spanning 3-ended trees.

Obviously, there are a lot of almost claw-free graphs containing K1,4 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 [<xref ref-type="bibr" rid="B8">18</xref>]).

γ(N(v))=2 for any claw center v 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 G is a connected nonhamiltonian graph and C is longest cycle in G with counterclockwise direction as positive orientation. Assume that R is a component of G-C and NC(R)={u1,u2,…,um} such that u1,u2,…,um are labeled in order along the positive direction of C. Let Sj=C(uj,uj+1], 1≤j≤m-1, and Sm=C(um,u1]. A vertex u in Sj is an insertible vertex if u has two consecutive neighbors v and v+ in C-Sj.

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

Lemma 7 (see [<xref ref-type="bibr" rid="B3">19</xref>]).

For each Sj,Sj-{vj+1} contains a noninsertible vertex.

Assume that vj is the first noninsertible vertex in Sj-{uj+1} for each j∈[1,m].

Lemma 8 (see [<xref ref-type="bibr" rid="B3">19</xref>]).

Let xi∈C[ui+,vi], xj∈C[uj+,vj] with 1≤i<j≤m. Then

there is no path P[xi,xj] in G such that P[xi,xj]∩V(C)={xi,xj},

for any vertex u in C[xi+,xj-], if uxi∈E(G), then u-xj∉E(G). By symmetry, for any vertex u in C[xj+,xi-], if uxj∈E(G), then u-xi∉E(G),

for any vertex u in C[xi,xj], if uxi,uxj∈E(G), then u-u+∉E(G). By symmetry, for any vertex u in C[xj,xi], if uxi,uxj∈E(G), then u-u+∉E(G).

Suppose, for some i∈[1,m], N(vi)∩V(G-C-R)≠∅ and vi′ is the second noninsertible vertex in Si-{ui+1}. Then Chen et al. [17] gave the following result.

Lemma 9 (see [<xref ref-type="bibr" rid="B5">17</xref>]).

Let 1≤i<j≤m, xi∈C[vi+,vi′] and xj∈C[uj+,vj]. Then the following properties hold:

There does not exist a path P[xi,xj] in G such that P[xi,xj]∩V(C)={xi,xj}.

For every vertex u∈C[xi+,xj-], if uxi∈E(G), then u-xj∉E(G); similarly, for every u∈C[xj+,xi-], if uxj∈E(G), then u-xi∉E(G).

For every vertex u∈C[xi,xj], if uxi,uxj∈E(G), then u-u+∉E(G); by symmetry, for any vertex u in C[xj,xi], if uxi,uxj∈E(G), then u-u+∉E(G).

3. Proof of Theorem <xref ref-type="statement" rid="thm1.5">5</xref>

To the contrary, suppose that G satisfies the conditions of Theorem 5 and any spanning tree in G contains more than 3 leaves. Let P=P[x,y] be longest path in G such that P satisfies the following two conditions:

w(G-P) is minimum.

|P[x,u1]| is minimum such that u1 is the first vertex in P with N(u1)∩V(G-P)≠∅, subject to (T1).

Suppose that R is a component in G-P, and {u1,…,um}=NP(R) with u1,…,um in order along the positive direction of P. Let RI denote an independent set in R.

Let G′ denote a graph with V(G′)=V(G)∪{u0}, E(G′)=E(G)∪{u0u:u∈V(G)}. Then C=u0P[x,y]u0 is a maximal cycle in G′. We define the counterclockwise orientation as the positive direction of C. Let Si denote the segment C(ui,ui+1] for 0≤i≤m-1, and Sm=C(um,u0]. By Lemma 7, let vi denote the first noninsertible vertex in Si, for i∈[0,m], and U={v0,v1,…,vm}. By Lemma 8(a), U is an independent set.

C can be divided into disjoint intervals S=C[a,b] with a,b+∉N(U) and C[a+,b]⊆N(U). We call the intervals U-segments. If a=b, then Ca+,b=∅; that is, if S=1, then dU(S)=0. By the definition of U-segment, for any U-segment S, there exists l∈[0,m] such that S⊆C[vl,vl+1-] (subscripts expressed modulo m+1).

Claim 1.

x=v0 and y∉N(vi) for any i∈[0,m-1].

Proof.

Suppose that x is an insertible vertex such that xu,xu+∈E(G), where u,u+∈C-S0. If u≠y, then we can get a path P′=P[x+,u]xP[u+,y], a contradiction to (T2). If u=y, then let P′=P[x+,y]x, a contradiction to (T2). Thus x=v0. Suppose viy∈E(G), for some i∈[0,m-1]. Obviously, u0=y+. Since yvi,y+vi∈E(G) and y∈C-Si,vi is an insertible vertex, a contradiction.

Claim 2.

For any vertex u∈V(P), if N[u] is claw-free, then dU(u)≤1.

Proof.

Suppose that u is in some U-segment S with S⊆C[vi,vi+1-], i∈[0,m], and vi1,vi2∈NU(u) with 0≤i1<i2≤m. Then by Lemma 8(c), u-u+∉E(G). Since G[u,u-,u+,vi1]≠K1,3, vi1u-∈E(G) or vi1u+∈E(G). Similarly, vi2u-∈E(G) or vi2u+∈E(G). Obviously, at least one vertex in {vi1,vi2} is not in C[vi,vi+1-]. Without loss of generality, suppose vi1∉C[vi,vi+1-] and vi1u-∈E(G). Then by vi1u-,vi1u∈E(G),vi1 is an insertible vertex, a contradiction.

Claim 3.

dU(u)≤2 for any vertex u∈V(P), and if dU(u)=2, then u is a claw center.

Proof.

Without loss of generality, suppose that u is in U-segment S and S⊆C[v0,v1-]. If S=1, then dU(u)=0. Suppose S≥2 and S={x0,x1,x2,…,xh}, where x0,x1,x2,…,xh are in order along the positive direction of C. Then x0∉N(U), xi∈N(U) for i∈[1,h]. For some i∈[1,h], suppose vi1,vi2,vi3∈NU(xi) with 0≤i1<i2<i3≤m. Then xi is a claw center. By Lemma 6, suppose y1, y2 are two distinct domination vertices of N(xi). Then N[y1], N[y2] are claw-free and at least two vertices in {vi1,vi2,vi3} are incident with y1 or y2. Without loss of generality, suppose vi1y1,vi2y1∈E(G). Then y1∈V(P), and y1-y1+∉E(G) by Lemma 8(c). Suppose Sj=C(uj,uj+1] containing y1, 0≤j≤m. Obviously, at least one vertex in {vi1,vi2} is not in Sj. Without loss of generality, suppose vi1∉Sj. Since vi1 is a noninsertible vertex and vi1y1∈E(G), y1-vi1,y1+vi1∉E(G). Thus G[y1,y1-,y1+,vi1]=K1,3, a contradiction. If dU(u)=2, then by Claim 2, u is a claw center.

Claim 4.

For any U-segment S not containing y, S contains at most one vertex u with dU(u)=2, and dU(S)≤S.

Proof.

If y∉N(U), then there exists a U-segment S with S={y,u0}. Suppose y∈N(U) and y in U-segment S, then u0∈S by u0=y+ and u0∈N(U). Thus if y∉S, then u0∉S0. Without loss of generality, suppose S={x0,x1,x2,…,xh}⊆C[vi,vi+1-], 0≤i≤m, where x0,x1,x2,…,xh are in order along the positive direction of C. By Claim 3, suppose that xj is the first vertex in S with dU(xj)=2, 1≤j≤h, and {vi1,vi2}=NU(xj), where 0≤i1<i2≤m. Then vi2≠vi, and, by Claim 3, xj is a claw center. Then N[xj+] is claw-free, and, by Claim 2, dU(xj+)≤1. Thus if j≤h≤j+1, then we are done.

Suppose h>j+1 and NU(xj+1)={vi3}, i3∈[0,m]. Since G[xj+1,xj+2,xj,vi3]≠K1,3,E(G) contains at least one edge in {xjvi3,xj+2vi3,xjxj+2}. Since vi3 is a noninsertible vertex and vi3xj+1∈E(G), vi3=vi if xjvi3 or xj+2vi3∈E(G), a contradiction to Lemma 8(b) since xjvi1,xjvi2∈E(G). Thus vi3≠vi, and xjxj+2∈E(G). Since xj is a claw center, N[xj+2] is claw-free and, by Claim 2, dU(xj+2)≤1. Thus if h=j+2, then we are done.

Suppose h>j+2 and {vi4}=NU(xj+2), i4∈[0,m]. Since xj+1vi3,xj+2vi4∈E(G) and vi3≠vi, by Lemma 8(b) vi4≠vi. Then vi4xj+3∉E(G). Since G[xj+2,xj,xj+3,vi4]≠K1,3,vi4xj or xjxj+3∈E(G). If vi4xj∈E(G), then vi4∈{vi1,vi2}. Then, by Lemma 8(b), vi3=vi4 and vi4=vi2 since xj+1vi3,xj+2vi4∈E(G). Then vi2xj+1,vi2xj+2∈E(G), a contradiction to the noninsertablity of vi2. Thus xjxj+3∈E(G), and then N[xj+3] is claw-free. By Claim 2, dU(xj+3)≤1. Thus if h=j+3, then we are done. If h>j+3, then, proceeding in the above manners to the set L={xj+4,…,xh}, we can get that N[u] is claw-free for any vertex u in L, and then, by Claim 2, dU(u)≤1. It follows that S has exactly one vertex xj with dU(xj)=2 for j∈[1,h]. Thus the claim holds.

Claim 5.

Suppose that the U-segment S0 contains y. Then dU(u)≤1 for any vertex u∈S0-{u0}.

Proof.

If y∉N(U), then S0={y,u0} and dU(y)=0. Suppose y∈N(U). Then, by Claim 1, NU(y)={vt}. By Lemma 8(b), NU(u)⊆{vt} and then dU(u)≤1 for any vertex u∈S0-{u0}.

Claim 6.

For any vertex ui(1≤i≤m) in NP(R)={u1,…,um}, dRI(ui)≤2.

Proof.

Suppose z1,z2,z3∈NRI(ui), where z1,z2,z3 are distinct vertices, i∈[1,m]. Since G[ui,z1,z2,z3]=K1,3,ui is a claw center. By Lemma 6, suppose y1, y2 are the two distinct domination vertices in N(ui). Then ui-,ui+∈N(y1)∪N(y2), and N[y1], N[y2] are claw-free. Since ui-,ui+∉NP(R), {y1,y2}∩V(C)≠∅. Suppose y1, y2 are both in V(C). Then at least two vertices of z1,z2,z3 are adjacent to y1 or y2. Without loss of generality, suppose z1,z2∈N(y1), and then G[y1,z1,z2,y1+]=K1,3, a contradiction. Thus one vertex of y1, y2 is in V(C), and the other vertex is in V(R). Without loss of generality, suppose y1∈V(C), y2∈V(R). Then ui+y1∈E(G). If z1,z2,z3∈N(y2), then G[y2,z1,z2,z3]=K1,3, a contradiction, and, by the preceding proof, there is exactly one vertex of z1,z2,z3 adjacent to y1. Thus y1∈NP(R)-{ui}. Without loss of generality, suppose z1∈N(y1). Since G[y1,y1-,y1+,z1]≠K1,3, y1-y1+∈E(G). Then we can get a longer cycle C′=y1z1C-[ui,y1+]C-[y1-,ui+]y1 than C, a contradiction.

Claim 7.

For any vertex ui(1≤i≤m) in NP(R)={u1,…,um}, if dU(ui)=1 and dRI(ui)=2, then dU(u)≤1 and dU(S)=S-1, for any vertex u∈S, where S is the U-segment containing ui.

Proof.

Suppose dU(ui)=1 and NRI(ui)={z1,z2}, i∈[1,m]. Since G[ui,ui-,z1,z2]=K1,3,ui is a claw center. Assume S={x1,x2,…,xh} and xj=ui, 2≤j≤h. If h>j, then, by Lemma 8(a), NU(u)={vi} for any vertex u∈{xj+1,…,xh}. Thus if u∈S with dU(u)=2, then u∈{x2,x3,…,xj-1}. Suppose dU(xp)=2, p∈[2,j-1]. By the proof of Claim 4, N[u] is claw-free for any vertex u in {xp+1,…,xh}, which contradicts that xj is a claw center. It follows that dU(S)=|S|-1.

Claim 8.

For any vertex ui(1≤i≤m) in NP(R)={u1,…,um}, if dU(ui)=2, then dRI(ui)≤1.

Proof.

Without loss of generality, we consider u2. Suppose {vi,vj}=NU(u2) with 0≤i<j≤m and z1,z2∈NRI(u2), where z1, z2 are two distinct vertices. By Claim 3, u2 is a claw center. By Lemma 6, suppose y1, y2 are the two distinct domination vertices in N(u2). Then N[y1], N[y2] are claw-free and vi,vj∈N(y1)∪N(y2). Thus {y1,y2}∩V(C)≠∅. Without loss of generality, suppose y1∈V(C). If y2∈V(R), then viy1,vjy1∈E(G), a contradiction to Claim 2. Suppose y2∈V(C). By Claim 2, vi, vj can not both be incident with y1 or y2. Without loss of generality, suppose z1y1,z2y2,viy1,vjy2∈E(G). Obviously, y1∈NP(R)-{u2}. Suppose y1=uj, j∈{1,3,…,m}. If vi≠vj-1, then G[y1,y1+,vi,z1]=K1,3, a contradiction. Suppose vi=vj-1. Then, by G[y1,y1+,vi,z1]≠K1,3, y1+vi∈E(G). Obviously, all the vertices in C(uj-1,vj-1-) can be inserted into C[y1,uj-1]. Let y1P1uj-1 denote the path by the above inserting process with V(P1)=V(C) and uj-1PRy1 denote the path connecting uj-1 and y1 with internal vertices in R. Then we can get a cycle C′=y1-P1-vj-1y1+P1uj-1PRy1 longer than C, a contradiction.

Claim 9.

Consider the following: ∑i=0mdP(vi)≤P-1 and ∑i=0mdP(vi)+∑z∈RIdP(z)≤P+m-1.

Proof.

Obviously, V(P)=⋃i=0m-1V(P[vi,vi+1-])∪V(P[vm,y]) and ∑i=0mdP(vi)=dU(P). Then ∑i=0mdP(vi)=∑i=0m-1dU(P[vi,vi+1-])+dU(P[vm,y]). By Claim 5, dU(P[vm,y])≤|P[vm,y]|-1. By Claim 4, dU(P[vi,vi+1-])≤|P[vi,vi+1-]| for 1≤i≤m-1. Thus ∑i=0mdP(vi)≤∑i=0m-1|P[vi,vi+1-]| + |P[vm,y]|-1=|P|-1.

Obviously, NP(RI)⊆{u1,u2,…,um}. Then ∑i=0mdP(vi)+∑z∈RIdP(z)=dU(P)+∑z∈RIdNP(RI)(z) = ∑i=0m-1(dU(P[vi,vi+1-])+∑z∈RIdui(z))+dU(P[vm,y]). Without loss of generality, we consider P[v1,v2-]. Let Su2 denote the U-segment containing u2. By Claim 4, dU(S)≤|S| for any U-segment in P[v1,v2-]. Obviously, ∑z∈RIdu2(z)=dRI(u2). By Claim 6, dRI(u2)≤2.

Suppose dU(u2)=0. Then u2 is the first vertex in Su2 and Su2=C[u2,v2). By Lemma 8(a), NU(Su2)⊆{v2}. Thus dU(Su2)=Su2-1 and dU(P[v1,v2-])+∑z∈RIdu2(z)=∑S≠Su2dU(S)+dU(Su2)+dRI(u2)≤∑S≠Su2S + (Su2-1)+2=Pv1,v2-+1.

Suppose dU(u2)=1. If dRI(u2)≤1, then dU(P[v1,v2-])+∑z∈RIdu2(z)=∑S≠Su2dU(S)+dU(Su2)+dRI(u2) ≤ ∑S≠Su2S+Su2+1=Pv1,v2-+1. If dRI(u2)=2, then, by Claim 7, dU(Su2)=Su2-1. Thus dU(P[v1,v2-])+∑z∈RIdu2(z)=∑S≠Su2dU(S)+dU(Su2)+dRI(u2)≤∑S≠Su2S+(Su2-1) + 2=Pv1,v2-+1.

It follows that dU(P[v1,v2-])+∑z∈RIdu2(z)≤Pv1,v2-+1. Similarly, for any i∈[1,m-1], dU(P[vi,vi+1-])≤Pvi,vi+1-+1. By Claim 5, dU(P[vm,y])≤|P[vm,y]|-1. Thus dU(P)+∑z∈RIdP(z)≤∑i=0m-1(Pvi,vi+1-+1)+Pvm,y-1≤|P|+m-1.

Claim 10.

dP(R)=k and R is hamiltonian-connected for each component R of G-P.

Proof.

NG-P(vi)∩NG-P(vj)=∅ from Lemma 8(a), for 0≤i≠j≤m, and then ∑i=0mdG-P(vi)≤n-P-R. By the connectedness of G, m≥k. If m≥k+2, then we get an independent set {v0,v1,…,vm} of order at least k+3. By Claim 9, ∑i=0md(vi)=∑i=0mdG-P(vi)+∑i=0mdP(vi)≤(n-P-R)+(|P|-1)=n-1-R, a contradiction to σk+3(G)≥n+2k-2. Suppose m=k+1 and z∈V(R). Then we can get an independent set {z,v0,v1,…,vm} of order k+3. By Claim 9, dP(z)+∑i=0k+1dP(vi)≤(k+1)-1+P=P+k. Then d(z)+∑i=0k+1d(vi)=∑i=0k+1dG-P(vi)+dR(z)+∑i=0k+1dP(vi)+dP(z)≤(n-|P|-|R|)+(|R|-1)+(|P|+k) = n+k-1, a contradiction to σk+3(G)≥n+2k-2 by k≥2. Thus m=k.

Suppose that R is not hamiltonian-connected. Then by Ore’s theorem [20], suppose z1,z2∈V(R) with z1z2∉E(G) and dR(z1)+dR(z2)≤R. We can get an independent set {z1,z2,v0,v1,…,vk} with order k+3. By Claim 9, d(z1)+d(z2)+∑i=0kd(vi)=dR(z1)+dR(z2)+∑i=0kdG-P(vi)+(∑i=0kdP(vi)+dP(z1)+dP(z2)) ≤ |R|+(n-|P|-|R|)+(|P|+k-1)=n+k-1, a contradiction to σk+3(G)≥n+2k-2 by k≥2.

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

Claim 11.

Consider the following: NR′(vi)≠∅ for some i∈[1,k].

Proof.

Suppose NR′(vi)=∅ for any i∈[1,k]. Then ∑i=0kdG-P(vi)≤n-|P|-|R|-|R′|. Let z1∈V(R), z2∈V(R′). Then we get an independent set {z1,z2,v0,v1,…,vk} of order k+3. By Claim 10, dP(z2)≤k and, by Claim 9, ∑i=0kd(vi)+dP(z1)≤|P|+k-1. Thus ∑i=0kd(vi)+d(z1)+d(z2) = dP(z2)+dR(z1)+dR′(z2)+(∑i=0kdP(vi)+dP(z1))+∑i=0kdG-P(vi)≤k+(|R|-1)+(|R′|-1) + (|P|+k-1)+(n-|P|-|R|-|R′|)=n+2k-3, a contradiction to σk+3(G)≥n+2k-2.

By Claim 11, we assume NR′(vi)≠∅ for some i∈[1,k]. By Lemma 8(a), NR′(vj)=∅ for j∈[0,i-1]∪[i+1,k].

Claim 12.

Si-{ui+1} contains a second noninsertible vertex vi′.

Proof.

Suppose Si-{ui+1} contains only one noninsertible vertex vi. Then we can get a path P1[ui+1,ui] such that V(P1)=V(C)-{vi} by inserting all the vertices in Si-{vi} to C[ui+1,ui]. Suppose |V(R)|={u}. Then we get a cycle C′=P1[ui+1,ui]uui+1. Let P′=V(C′)-{u0}. Then w(G-P′)<w(G-P), a contradiction to (T1). Suppose |V(R)|≥2. If NR(ui)∪NR(ui+1)={z}, then NP(R)∪{z}-{ui,ui+1} is a vertex cut of G with order k-1, a contradiction to Claim 10. Thus NRui∪NRui+1≥2. By Claim 10, there is a hamiltonian path uiP2ui+1 of R∪{ui,ui+1}. Thus we can get a cycle C1=ui+1P1uiP2ui+1 longer than C, a contradiction.

Now, we complete the proof of Theorem 5. By Claim 12 and Lemma 9, U′={v0,…,vi-1,vi′,vi+1,…,vk} is an independent set. Then, by the preceding proof, U′ has the same properties as U. By Lemma 9, NG-P(vi)∩NG-P(vj)=∅, for any two distinct vertices vi,vj∈U′. Thus ∑v∈U′dG-P(v)≤n-|P|-|R|-|R′|. By Claim 9, ∑v∈U′dP(v)≤|P|-1. Let z1∈V(R), z2∈V(R′). Then U′∪{z1,z2} is an independent set of order k+3 in G. Thus ∑v∈U′d(v)=∑v∈U′dP(v)+∑v∈U′dG-P(v)≤(|P|-1)+(n-|P|-|R|-|R′|) = n-1-|R|-|R′|. By Claim 10, d(z1)≤|R|-1+k, d(z2)≤|R′|-1+k. Thus d(z1)+d(z2)+∑v∈U′d(v) ≤ (|R|-1+k)+(|R′|-1+k)+(n-1-|R|-|R′|)=n+2k-3, a contradiction to σk+3(G)≥n+2k-2. It follows that Theorem 5 holds.

Conflict of Interests

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

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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