DDNS Discrete Dynamics in Nature and Society 1607-887X 1026-0226 Hindawi Publishing Corporation 10.1155/2015/476182 476182 Research Article Spanning 3-Ended Trees in Almost Claw-Free Graphs Chen Xiaodong 1 http://orcid.org/0000-0002-9563-0616 Xu Meijin 1 Liu Yanjun 1 Yang Chenguang College of Science Liaoning University of Technology Jinzhou 121001 China lnit.edu.cn 2015 21122015 2015 10 11 2015 03 12 2015 2015 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.

We prove that if G is a k -connected ( k 2 ) almost claw-free graph of order n and σ k + 3 ( G ) n + 2 k - 2 , then G contains a spanning 3-ended tree, where σ k ( G ) = min { v S deg ( 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  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 K 1,3 subgraph, then G is claw-free. The vertex of degree 3 in K 1,3 is called claw center of a claw. Let N ( v ) = { u : u v 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 K 1,3 subgraphs of G is an independent set, then G is almost claw-free. A claw-free graph is almost claw-free.

N H ( S ) = { v : v V ( H ) and u v E ( G ) for some vertex u V ( S ) } , and d H ( S ) = N H S . We use σ k ( G ) to denote the minimum degree sum of all the independent sets with order k in G . a P b 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 x P y denote the subpath with end vertices x , y with positive orientation and y P - 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 .

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 . Kyaw [14, 15] gave some degree sum conditions for K 1,4 -free graphs to contain spanning k -ended trees.

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

If G is a connected K 1,4 -free graph and σ 4 ( 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 K 1,4 -free graph, then the following properties hold:

G contains a hamiltonian path if σ 3 ( G ) G .

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

On the other hand, Kano et al.  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 σ k + 1 ( G ) n - k ( k 2 ), then G contains spanning k -ended trees with the maximum degree at most 3.

Recently, Chen et al.  gave some degree sum conditions for k -connected K 1,4 -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 K 1,4 -free graph of order n and σ k + 3 ( G ) n + 2 k - 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 σ k + 3 ( G ) n + 2 k - 2 ( k 2 ), then G contains spanning 3-ended trees.

Obviously, there are a lot of almost claw-free graphs containing K 1,4 subgraphs, so to some extent Theorem 5 is a generalization of Theorem 4.

2. Preliminaries

We need the following result given by Ryjáček  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  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 N C ( R ) = { u 1 , u 2 , , u m } such that u 1 , u 2 , , u m are labeled in order along the positive direction of C . Let S j = C ( u j , u j + 1 ] , 1 j m - 1 , and S m = C ( u m , u 1 ] . A vertex u in S j is an insertible vertex if u has two consecutive neighbors v and v + in C - S j .

Chen and Schelp  proposed the following two results.

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

For each S j , S j - { v j + 1 } contains a noninsertible vertex.

Assume that v j is the first noninsertible vertex in S j - { u j + 1 } for each j [ 1 , m ] .

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

Let x i C [ u i + , v i ] , x j C [ u j + , v j ] with 1 i < j m . Then

there is no path P [ x i , x j ] in G such that P [ x i , x j ] V ( C ) = { x i , x j } ,

for any vertex u in C [ x i + , x j - ] , if u x i E ( G ) , then u - x j E ( G ) . By symmetry, for any vertex u in C [ x j + , x i - ] , if u x j E ( G ) , then u - x i E ( G ) ,

for any vertex u in C [ x i , x j ] , if u x i , u x j E ( G ) , then u - u + E ( G ) . By symmetry, for any vertex u in C [ x j , x i ] , if u x i , u x j E ( G ) , then u - u + E ( G ) .

Suppose, for some i [ 1 , m ] , N ( v i ) V ( G - C - R ) and v i is the second noninsertible vertex in S i - { u i + 1 } . Then Chen et al.  gave the following result.

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

Let 1 i < j m , x i C [ v i + , v i ] and x j C [ u j + , v j ] . Then the following properties hold:

There does not exist a path P [ x i , x j ] in G such that P [ x i , x j ] V ( C ) = { x i , x j } .

For every vertex u C [ x i + , x j - ] , if u x i E ( G ) , then u - x j E ( G ) ; similarly, for every u C [ x j + , x i - ] , if u x j E ( G ) , then u - x i E ( G ) .

For every vertex u C [ x i , x j ] , if u x i , u x j E ( G ) , then u - u + E ( G ) ; by symmetry, for any vertex u in C [ x j , x i ] , if u x i , u x j 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 , u 1 ] | is minimum such that u 1 is the first vertex in P with N ( u 1 ) V ( G - P ) , subject to (T1).

Suppose that R is a component in G - P , and { u 1 , , u m } = N P ( R ) with u 1 , , u m in order along the positive direction of P . Let R I denote an independent set in R .

Let G denote a graph with V ( G ) = V ( G ) { u 0 } , E ( G ) = E ( G ) { u 0 u : u V ( G ) } . Then C = u 0 P [ x , y ] u 0 is a maximal cycle in G . We define the counterclockwise orientation as the positive direction of C . Let S i denote the segment C ( u i , u i + 1 ] for 0 i m - 1 , and S m = C ( u m , u 0 ] . By Lemma 7, let v i denote the first noninsertible vertex in S i , for i [ 0 , m ] , and U = { v 0 , v 1 , , v m } . 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 C a + , b = ; that is, if S = 1 , then d U ( S ) = 0 . By the definition of U -segment, for any U -segment S , there exists l [ 0 , m ] such that S C [ v l , v l + 1 - ] (subscripts expressed modulo m + 1 ).

Claim 1.

x = v 0 and y N ( v i ) for any i [ 0 , m - 1 ] .

Proof.

Suppose that x is an insertible vertex such that x u , x u + E ( G ) , where u , u + C - S 0 . If u y , then we can get a path P = P [ x + , u ] x P [ u + , y ] , a contradiction to (T2). If u = y , then let P = P [ x + , y ] x , a contradiction to (T2). Thus x = v 0 . Suppose v i y E ( G ) , for some i [ 0 , m - 1 ] . Obviously, u 0 = y + . Since y v i , y + v i E ( G ) and y C - S i , v i is an insertible vertex, a contradiction.

Claim 2.

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

Proof.

Suppose that u is in some U -segment S with S C [ v i , v i + 1 - ] , i [ 0 , m ] , and v i 1 , v i 2 N U ( u ) with 0 i 1 < i 2 m . Then by Lemma 8(c), u - u + E ( G ) . Since G [ u , u - , u + , v i 1 ] K 1,3 , v i 1 u - E ( G ) or v i 1 u + E ( G ) . Similarly, v i 2 u - E ( G ) or v i 2 u + E ( G ) . Obviously, at least one vertex in { v i 1 , v i 2 } is not in C [ v i , v i + 1 - ] . Without loss of generality, suppose v i 1 C [ v i , v i + 1 - ] and v i 1 u - E ( G ) . Then by v i 1 u - , v i 1 u E ( G ) , v i 1 is an insertible vertex, a contradiction.

Claim 3.

d U ( u ) 2 for any vertex u V ( P ) , and if d U ( u ) = 2 , then u is a claw center.

Proof.

Without loss of generality, suppose that u is in U -segment S and S C [ v 0 , v 1 - ] . If S = 1 , then d U ( u ) = 0 . Suppose S 2 and S = { x 0 , x 1 , x 2 , , x h } , where x 0 , x 1 , x 2 , , x h are in order along the positive direction of C . Then x 0 N ( U ) , x i N ( U ) for i [ 1 , h ] . For some i [ 1 , h ] , suppose v i 1 , v i 2 , v i 3 N U ( x i ) with 0 i 1 < i 2 < i 3 m . Then x i is a claw center. By Lemma 6, suppose y 1 , y 2 are two distinct domination vertices of N ( x i ) . Then N [ y 1 ] , N [ y 2 ] are claw-free and at least two vertices in { v i 1 , v i 2 , v i 3 } are incident with y 1 or y 2 . Without loss of generality, suppose v i 1 y 1 , v i 2 y 1 E ( G ) . Then y 1 V ( P ) , and y 1 - y 1 + E ( G ) by Lemma 8(c). Suppose S j = C ( u j , u j + 1 ] containing y 1 , 0 j m . Obviously, at least one vertex in { v i 1 , v i 2 } is not in S j . Without loss of generality, suppose v i 1 S j . Since v i 1 is a noninsertible vertex and v i 1 y 1 E ( G ) , y 1 - v i 1 , y 1 + v i 1 E ( G ) . Thus G [ y 1 , y 1 - , y 1 + , v i 1 ] = K 1,3 , a contradiction. If d U ( 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 d U ( u ) = 2 , and d U ( S ) S .

Proof.

If y N ( U ) , then there exists a U -segment S with S = { y , u 0 } . Suppose y N ( U ) and y in U -segment S , then u 0 S by u 0 = y + and u 0 N ( U ) . Thus if y S , then u 0 S 0 . Without loss of generality, suppose S = { x 0 , x 1 , x 2 , , x h } C [ v i , v i + 1 - ] , 0 i m , where x 0 , x 1 , x 2 , , x h are in order along the positive direction of C . By Claim 3, suppose that x j is the first vertex in S with d U ( x j ) = 2 , 1 j h , and { v i 1 , v i 2 } = N U ( x j ) , where 0 i 1 < i 2 m . Then v i 2 v i , and, by Claim 3, x j is a claw center. Then N [ x j + ] is claw-free, and, by Claim 2, d U ( x j + ) 1 . Thus if j h j + 1 , then we are done.

Suppose h > j + 1 and N U ( x j + 1 ) = { v i 3 } , i 3 [ 0 , m ] . Since G [ x j + 1 , x j + 2 , x j , v i 3 ] K 1,3 , E ( G ) contains at least one edge in { x j v i 3 , x j + 2 v i 3 , x j x j + 2 } . Since v i 3 is a noninsertible vertex and v i 3 x j + 1 E ( G ) , v i 3 = v i if x j v i 3 or x j + 2 v i 3 E ( G ) , a contradiction to Lemma 8(b) since x j v i 1 , x j v i 2 E ( G ) . Thus v i 3 v i , and x j x j + 2 E ( G ) . Since x j is a claw center, N [ x j + 2 ] is claw-free and, by Claim 2, d U ( x j + 2 ) 1 . Thus if h = j + 2 , then we are done.

Suppose h > j + 2 and { v i 4 } = N U ( x j + 2 ) , i 4 [ 0 , m ] . Since x j + 1 v i 3 , x j + 2 v i 4 E ( G ) and v i 3 v i , by Lemma 8(b) v i 4 v i . Then v i 4 x j + 3 E ( G ) . Since G [ x j + 2 , x j , x j + 3 , v i 4 ] K 1,3 , v i 4 x j or x j x j + 3 E ( G ) . If v i 4 x j E ( G ) , then v i 4 { v i 1 , v i 2 } . Then, by Lemma 8(b), v i 3 = v i 4 and v i 4 = v i 2 since x j + 1 v i 3 , x j + 2 v i 4 E ( G ) . Then v i 2 x j + 1 , v i 2 x j + 2 E ( G ) , a contradiction to the noninsertablity of v i 2 . Thus x j x j + 3 E ( G ) , and then N [ x j + 3 ] is claw-free. By Claim 2, d U ( x j + 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 = { x j + 4 , , x h } , we can get that N [ u ] is claw-free for any vertex u in L , and then, by Claim 2, d U ( u ) 1 . It follows that S has exactly one vertex x j with d U ( x j ) = 2 for j [ 1 , h ] . Thus the claim holds.

Claim 5.

Suppose that the U -segment S 0 contains y . Then d U ( u ) 1 for any vertex u S 0 - { u 0 } .

Proof.

If y N ( U ) , then S 0 = { y , u 0 } and d U ( y ) = 0 . Suppose y N ( U ) . Then, by Claim 1, N U ( y ) = { v t } . By Lemma 8(b), N U ( u ) { v t } and then d U ( u ) 1 for any vertex u S 0 - { u 0 } .

Claim 6.

For any vertex u i ( 1 i m ) in N P ( R ) = { u 1 , , u m } , d R I ( u i ) 2 .

Proof.

Suppose z 1 , z 2 , z 3 N R I ( u i ) , where z 1 , z 2 , z 3 are distinct vertices, i [ 1 , m ] . Since G [ u i , z 1 , z 2 , z 3 ] = K 1,3 , u i is a claw center. By Lemma 6, suppose y 1 , y 2 are the two distinct domination vertices in N ( u i ) . Then u i - , u i + N ( y 1 ) N ( y 2 ) , and N [ y 1 ] , N [ y 2 ] are claw-free. Since u i - , u i + N P ( R ) , { y 1 , y 2 } V ( C ) . Suppose y 1 , y 2 are both in V ( C ) . Then at least two vertices of z 1 , z 2 , z 3 are adjacent to y 1 or y 2 . Without loss of generality, suppose z 1 , z 2 N ( y 1 ) , and then G [ y 1 , z 1 , z 2 , y 1 + ] = K 1,3 , a contradiction. Thus one vertex of y 1 , y 2 is in V ( C ) , and the other vertex is in V ( R ) . Without loss of generality, suppose y 1 V ( C ) , y 2 V ( R ) . Then u i + y 1 E ( G ) . If z 1 , z 2 , z 3 N ( y 2 ) , then G [ y 2 , z 1 , z 2 , z 3 ] = K 1,3 , a contradiction, and, by the preceding proof, there is exactly one vertex of z 1 , z 2 , z 3 adjacent to y 1 . Thus y 1 N P ( R ) - { u i } . Without loss of generality, suppose z 1 N ( y 1 ) . Since G [ y 1 , y 1 - , y 1 + , z 1 ] K 1,3 , y 1 - y 1 + E ( G ) . Then we can get a longer cycle C = y 1 z 1 C - [ u i , y 1 + ] C - [ y 1 - , u i + ] y 1 than C , a contradiction.

Claim 7.

For any vertex u i ( 1 i m ) in N P ( R ) = { u 1 , , u m } , if d U ( u i ) = 1 and d R I ( u i ) = 2 , then d U ( u ) 1 and d U ( S ) = S - 1 , for any vertex u S , where S is the U -segment containing u i .

Proof.

Suppose d U ( u i ) = 1 and N R I ( u i ) = { z 1 , z 2 } , i [ 1 , m ] . Since G [ u i , u i - , z 1 , z 2 ] = K 1,3 , u i is a claw center. Assume S = { x 1 , x 2 , , x h } and x j = u i , 2 j h . If h > j , then, by Lemma 8(a), N U ( u ) = { v i } for any vertex u { x j + 1 , , x h } . Thus if u S with d U ( u ) = 2 , then u { x 2 , x 3 , , x j - 1 } . Suppose d U ( x p ) = 2 , p [ 2 , j - 1 ] . By the proof of Claim 4, N [ u ] is claw-free for any vertex u in { x p + 1 , , x h } , which contradicts that x j is a claw center. It follows that d U ( S ) = | S | - 1 .

Claim 8.

For any vertex u i ( 1 i m ) in N P ( R ) = { u 1 , , u m } , if d U ( u i ) = 2 , then d R I ( u i ) 1 .

Proof.

Without loss of generality, we consider u 2 . Suppose { v i , v j } = N U ( u 2 ) with 0 i < j m and z 1 , z 2 N R I ( u 2 ) , where z 1 , z 2 are two distinct vertices. By Claim 3, u 2 is a claw center. By Lemma 6, suppose y 1 , y 2 are the two distinct domination vertices in N ( u 2 ) . Then N [ y 1 ] , N [ y 2 ] are claw-free and v i , v j N ( y 1 ) N ( y 2 ) . Thus { y 1 , y 2 } V ( C ) . Without loss of generality, suppose y 1 V ( C ) . If y 2 V ( R ) , then v i y 1 , v j y 1 E ( G ) , a contradiction to Claim 2. Suppose y 2 V ( C ) . By Claim 2, v i , v j can not both be incident with y 1 or y 2 . Without loss of generality, suppose z 1 y 1 , z 2 y 2 , v i y 1 , v j y 2 E ( G ) . Obviously, y 1 N P ( R ) - { u 2 } . Suppose y 1 = u j , j { 1,3 , , m } . If v i v j - 1 , then G [ y 1 , y 1 + , v i , z 1 ] = K 1,3 , a contradiction. Suppose v i = v j - 1 . Then, by G [ y 1 , y 1 + , v i , z 1 ] K 1,3 , y 1 + v i E ( G ) . Obviously, all the vertices in C ( u j - 1 , v j - 1 - ) can be inserted into C [ y 1 , u j - 1 ] . Let y 1 P 1 u j - 1 denote the path by the above inserting process with V ( P 1 ) = V ( C ) and u j - 1 P R y 1 denote the path connecting u j - 1 and y 1 with internal vertices in R . Then we can get a cycle C = y 1 - P 1 - v j - 1 y 1 + P 1 u j - 1 P R y 1 longer than C , a contradiction.

Claim 9.

Consider the following:    i = 0 m d P ( v i ) P - 1 and i = 0 m d P ( v i ) + z R I d P ( z ) P + m - 1 .

Proof.

Obviously, V ( P ) = i = 0 m - 1 V ( P [ v i , v i + 1 - ] ) V ( P [ v m , y ] ) and i = 0 m d P ( v i ) = d U ( P ) . Then i = 0 m d P ( v i ) = i = 0 m - 1 d U ( P [ v i , v i + 1 - ] ) + d U ( P [ v m , y ] ) . By Claim 5, d U ( P [ v m , y ] ) | P [ v m , y ] | - 1 . By Claim 4, d U ( P [ v i , v i + 1 - ] ) | P [ v i , v i + 1 - ] | for 1 i m - 1 . Thus i = 0 m d P ( v i ) i = 0 m - 1 | P [ v i , v i + 1 - ] | + | P [ v m , y ] | - 1 = | P | - 1 .

Obviously, N P ( R I ) { u 1 , u 2 , , u m } . Then i = 0 m d P ( v i ) + z R I d P ( z ) = d U ( P ) + z R I d N P ( R I ) ( z ) = i = 0 m - 1 ( d U ( P [ v i , v i + 1 - ] ) + z R I d u i ( z ) ) + d U ( P [ v m , y ] ) . Without loss of generality, we consider P [ v 1 , v 2 - ] . Let S u 2 denote the U -segment containing u 2 . By Claim 4, d U ( S ) | S | for any U -segment in P [ v 1 , v 2 - ] . Obviously, z R I d u 2 ( z ) = d R I ( u 2 ) . By Claim 6, d R I ( u 2 ) 2 .

Suppose d U ( u 2 ) = 0 . Then u 2 is the first vertex in S u 2 and S u 2 = C [ u 2 , v 2 ) . By Lemma 8(a), N U ( S u 2 ) { v 2 } . Thus d U ( S u 2 ) = S u 2 - 1 and d U ( P [ v 1 , v 2 - ] ) + z R I d u 2 ( z ) = S S u 2 d U ( S ) + d U ( S u 2 ) + d R I ( u 2 ) S S u 2 S + ( S u 2 - 1 ) + 2 = P v 1 , v 2 - + 1 .

Suppose d U ( u 2 ) = 1 . If d R I ( u 2 ) 1 , then d U ( P [ v 1 , v 2 - ] ) + z R I d u 2 ( z ) = S S u 2 d U ( S ) + d U ( S u 2 ) + d R I ( u 2 ) S S u 2 S + S u 2 + 1 = P v 1 , v 2 - + 1 . If d R I ( u 2 ) = 2 , then, by Claim 7, d U ( S u 2 ) = S u 2 - 1 . Thus d U ( P [ v 1 , v 2 - ] ) + z R I d u 2 ( z ) = S S u 2 d U ( S ) + d U ( S u 2 ) + d R I ( u 2 ) S S u 2 S + ( S u 2 - 1 ) + 2 = P v 1 , v 2 - + 1 .

Suppose d U ( u 2 ) = 2 . Then, by Claim 8, d R I ( u 2 ) 1 . Thus d U ( P [ v 1 , v 2 - ] ) + z R I d u 2 ( z ) = S S u 2 d U ( S ) + d U ( S u 2 ) + d R I ( u 2 ) S S u 2 | S | + | S u 2 | + 1 = | P [ v 1 , v 2 - ] | + 1 .

It follows that d U ( P [ v 1 , v 2 - ] ) + z R I d u 2 ( z ) P v 1 , v 2 - + 1 . Similarly, for any i [ 1 , m - 1 ] , d U ( P [ v i , v i + 1 - ] ) P v i , v i + 1 - + 1 . By Claim 5, d U ( P [ v m , y ] ) | P [ v m , y ] | - 1 . Thus d U ( P ) + z R I d P ( z ) i = 0 m - 1 ( P v i , v i + 1 - + 1 ) + P v m , y - 1 | P | + m - 1 .

Claim 10.

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

Proof.

N G - P ( v i ) N G - P ( v j ) = from Lemma 8(a), for 0 i j m , and then i = 0 m d G - P ( v i ) n - P - R . By the connectedness of G , m k . If m k + 2 , then we get an independent set { v 0 , v 1 , , v m } of order at least k + 3 . By Claim 9, i = 0 m d ( v i ) = i = 0 m d G - P ( v i ) + i = 0 m d P ( v i ) ( n - P - R ) + ( | P | - 1 ) = n - 1 - R , a contradiction to σ k + 3 ( G ) n + 2 k - 2 . Suppose m = k + 1 and z V ( R ) . Then we can get an independent set { z , v 0 , v 1 , , v m } of order k + 3 . By Claim 9, d P ( z ) + i = 0 k + 1 d P ( v i ) ( k + 1 ) - 1 + P = P + k . Then d ( z ) + i = 0 k + 1 d ( v i ) = i = 0 k + 1 d G - P ( v i ) + d R ( z ) + i = 0 k + 1 d P ( v i ) + d P ( z ) ( n - | P | - | R | ) + ( | R | - 1 ) + ( | P | + k ) = n + k - 1 , a contradiction to σ k + 3 ( G ) n + 2 k - 2 by k 2 . Thus m = k .

Suppose that R is not hamiltonian-connected. Then by Ore’s theorem , suppose z 1 , z 2 V ( R ) with z 1 z 2 E ( G ) and d R ( z 1 ) + d R ( z 2 ) R . We can get an independent set { z 1 , z 2 , v 0 , v 1 , , v k } with order k + 3 . By Claim 9, d ( z 1 ) + d ( z 2 ) + i = 0 k d ( v i ) = d R ( z 1 ) + d R ( z 2 ) + i = 0 k d G - P ( v i ) + ( i = 0 k d P ( v i ) + d P ( z 1 ) + d P ( z 2 ) ) | R | + ( n - | P | - | R | ) + ( | P | + k - 1 ) = n + k - 1 , a contradiction to σ k + 3 ( G ) n + 2 k - 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: N R ( v i ) for some i [ 1 , k ] .

Proof.

Suppose N R ( v i ) = for any i [ 1 , k ] . Then i = 0 k d G - P ( v i ) n - | P | - | R | - | R | . Let z 1 V ( R ) , z 2 V ( R ) . Then we get an independent set { z 1 , z 2 , v 0 , v 1 , , v k } of order k + 3 . By Claim 10, d P ( z 2 ) k and, by Claim 9, i = 0 k d ( v i ) + d P ( z 1 ) | P | + k - 1 . Thus i = 0 k d ( v i ) + d ( z 1 ) + d ( z 2 ) = d P ( z 2 ) + d R ( z 1 ) + d R ( z 2 ) + ( i = 0 k d P ( v i ) + d P ( z 1 ) ) + i = 0 k d G - P ( v i ) k + ( | R | - 1 ) + ( | R | - 1 ) + ( | P | + k - 1 ) + ( n - | P | - | R | - | R | ) = n + 2 k - 3 , a contradiction to σ k + 3 ( G ) n + 2 k - 2 .

By Claim 11, we assume N R ( v i ) for some i [ 1 , k ] . By Lemma 8(a), N R ( v j ) = for j [ 0 , i - 1 ] [ i + 1 , k ] .

Claim 12.

S i - { u i + 1 } contains a second noninsertible vertex v i .

Proof.

Suppose S i - { u i + 1 } contains only one noninsertible vertex v i . Then we can get a path P 1 [ u i + 1 , u i ] such that V ( P 1 ) = V ( C ) - { v i } by inserting all the vertices in S i - { v i } to C [ u i + 1 , u i ] . Suppose | V ( R ) | = { u } . Then we get a cycle C = P 1 [ u i + 1 , u i ] u u i + 1 . Let P = V ( C ) - { u 0 } . Then w ( G - P ) < w ( G - P ) , a contradiction to (T1). Suppose | V ( R ) | 2 . If N R ( u i ) N R ( u i + 1 ) = { z } , then N P ( R ) { z } - { u i , u i + 1 } is a vertex cut of G with order k - 1 , a contradiction to Claim 10. Thus N R u i N R u i + 1 2 . By Claim 10, there is a hamiltonian path u i P 2 u i + 1 of R { u i , u i + 1 } . Thus we can get a cycle C 1 = u i + 1 P 1 u i P 2 u i + 1 longer than C , a contradiction.

Now, we complete the proof of Theorem 5. By Claim 12 and Lemma 9, U = { v 0 , , v i - 1 , v i , v i + 1 , , v k } is an independent set. Then, by the preceding proof, U has the same properties as U . By Lemma 9, N G - P ( v i ) N G - P ( v j ) = , for any two distinct vertices v i , v j U . Thus v U d G - P ( v ) n - | P | - | R | - | R | . By Claim 9, v U d P ( v ) | P | - 1 . Let z 1 V ( R ) , z 2 V ( R ) . Then U { z 1 , z 2 } is an independent set of order k + 3 in G . Thus v U d ( v ) = v U d P ( v ) + v U d G - P ( v ) ( | P | - 1 ) + ( n - | P | - | R | - | R | ) = n - 1 - | R | - | R | . By Claim 10, d ( z 1 ) | R | - 1 + k , d ( z 2 ) | R | - 1 + k . Thus d ( z 1 ) + d ( z 2 ) + v U d ( v ) ( | R | - 1 + k ) + ( | R | - 1 + k ) + ( n - 1 - | R | - | R | ) = n + 2 k - 3 , a contradiction to σ k + 3 ( G ) n + 2 k - 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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