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Guidong Yu 1 and Miaolin Ye 1 and Gaixiang Cai 1 and Jinde Cao 2, 3
Academic Editor:Francesco Soldovieri
1, School of Mathematics & Computation Sciences, Anqing Normal College, Anqing 246011, China
2, Department of Mathematics, Southeast University, Nanjing, Jiangsu 210096, China
3, Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia
Received 23 January 2014; Revised 7 June 2014; Accepted 16 June 2014; 30 June 2014
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.
1. Introduction
Let a graph, G=(V,E) be a simple graph of order n with vertex set V={v1 ,v2 ,...,vn } and edge set E . Denote by e(G)...=|E| the number of edges of the graph G . Write by Kn a complete graph of order n , On an empty graph of order n (without edges), and Kn,m a complete bipartite graph with two parts having n , m vertices, respectively. The graph G is said to be Hamiltonian, if it has a Hamiltonian cycle which is a cycle of order n contained in G . The graph G is said to be traceable if it has a Hamiltonian path which is a path of order n contained in G . The problem of deciding whether a graph is Hamiltonian is Hamiltonian problem, which is one of the most difficult classical problems in graph theory. Indeed, it is NP-complete problem.
The adjacency matrix of G is defined to be a matrix A(G)=[aij ]n×n , where aij =1 if vi is adjacent to vj and aij =0 otherwise. The largest eigenvalue of A(G) is called to be the spectral radius of G , which is denoted by μ(G) . The degree matrix of G is written by D(G)=diag...(dG (v1 ),dG (v2 ),...,dG (vn )) , where dG (vi ) (i=1,2,...,n) denotes the degree of the vertex vi in the graph G . The signless Laplacian matrix of G is defined by Q(G)=D(G)+A(G) . The largest eigenvalue of Q(G) is called to be the signless Laplacian spectral radius of G , which is denoted by q(G) .
Recently, using spectral graph theory to study the Hamiltonian problem has received a lot of attention. Some spectral conditions for a graph to be Hamiltonian or traceable have been given in [1-6]. In this paper, we still study the Hamiltonicity of a graph. Firstly, we present a signless Laplacian spectral radius condition for a bipartite graph to be Hamiltonian in Section 2. Secondly, we give some signless Laplacian spectral radius conditions for a graph to be traceable or Hamilton-connected in Section 3 and Section 4, respectively.
2. Signless Laplacian Spectral Radius in Hamiltonian Bipartite Graphs
The definition of the closure of a balanced bipartite graph can be found in [7, 8]. For a positive integer k , the k-closure of a balanced bipartite graph GBPT ...=(X,Y;E) , where |X|=|Y| , written by Ck (GBPT ) , is a graph obtained from GBPT by successively joining pairs of nonadjacent vertices x∈X and y∈Y , whose degree sum is at least k , until no such pairs remain. By the definition of the Ck (GBPT ) , we have that dCk (GBPT ) (x)+dCk (GBPT ) (y)...4;k-1 for any pair of nonadjacent vertices x∈X and y∈Y of Ck (GBPT ) .
Lemma 1 (see [9]).
Let GBPT =(X,Y;E) be a connected balanced bipartite graph, where |X|=|Y|=r...5;2 . Then, GBPT is Hamiltonian if and only if Cr+1 (GBPT ) is Hamiltonian.
For a graph G , write Z(G)...=∑uv∈E(G) ...(dG (u)+dG (v))=∑u∈V(G) ...dG2 (u) , and let Δ(G) be maximum degree of G . A regular graph is a graph for which every vertex in the graph has the same degree. A semi-regular graph is a bipartite graph for which every vertex in the same partite set has the same degree.
Lemma 2 (see [2]).
Let G be a graph with at least one edge. Then, [figure omitted; refer to PDF] if and only if G is regular or semi-regular.
Let M be a Hermitian matrix of order n , and let λ1 (M)...5;λ2 (M)...5;......5;λn (M) be the eigenvalues of M .
Lemma 3 (see [10]).
Let B and C be Hermitian matrices of order n , 1...4;i , j...4;n . Then, [figure omitted; refer to PDF] if i+j...4;n+1 .
Lemma 4.
Let G be a graph. Then, [figure omitted; refer to PDF]
Proof.
Because Q(G)=A(G)+D(G) , by Lemma 3, [figure omitted; refer to PDF] We notice that λ1 (A(G))=μ(G) , λ1 (D(G))=Δ(G) , and λ1 (Q(G))=q(G) . So, the result follows.
Let GBPT =(X,Y;E) be a bipartite graph, the quasi-complement of GBPT is denoted by GBPT* ...=(X,Y;E[variant prime] ) , where E[variant prime] ={xy:x∈X,y∈Y,xy∉E} .
Theorem 5.
Let GBPT =(X,Y;E) be a connected balanced bipartite graph, where |X|=|Y|=r...5;2 . If [figure omitted; refer to PDF] then GBPT is Hamiltonian.
Proof.
Suppose that GBPT is not Hamiltonian. Then, HBPT ...=Cr+1 (GBPT ) is not Hamiltonian too by Lemma 1, and therefore, HBPT is not Kr,r . Thus, there exists a vertex x∈X and a vertex y∈Y such that xy∉E(HBPT ) . We find that dHBPT (x)+dHBPT (y)...4;r for any pair of nonadjacent vertices x∈X and y∈Y in HBPT . So, [figure omitted; refer to PDF] for any pair of adjacent vertices x∈X , y∈Y in HBPT* . Hence, [figure omitted; refer to PDF]
By Lemma 2, we have that [figure omitted; refer to PDF]
As HBPT* is a subgraph of GBPT* , by Perron-Frobenius theorem, [figure omitted; refer to PDF]
Thus, by (5), (8), and (9), we have that [figure omitted; refer to PDF] a contradiction.
Li [4] has given a sufficient condition for a bipartite graph to be Hamiltonian as follows.
Theorem 6 (see [4]).
Let GBPT =(X,Y;E) be a connected balanced bipartite graph, where |X|=|Y|=r...5;2 . If [figure omitted; refer to PDF] then GBPT is Hamiltonian.
Remark 7.
We now compare Theorems 5 and 6. If μ(GBPT* )...4;(r-2)/2 and Δ(GBPT* )<r-(r-2)/2 , we have that q(GBPT* )<r by Lemma 4. Hence Theorem 5 improves Theorem 6 when Δ(GBPT* )<r-(r-2)/2 . For example, let GBPT be a regular connected balanced bipartite graph with degree (r+1)/2 , where r is odd and |X|=|Y|=r...5;6 . Then, its quasi-complement GBPT* is a regular graph with degrees (r-1)/2 , μ(GBPT* )=(r-1)/2 , and q(GBPT* )=r-1 . GBPT satisfies the condition of Theorems 5, and hence, it is Hamiltonian. But it does not satisfy the condition of Theorem 6.
3. Signless Laplacian Spectral Radius in Traceable Graphs
Write Kn-1 +v for Kn-1 together with an isolated vertex. Let G=(V(G),E(G)) and H=(V(H),E(H)) be two disjoint graphs. The disjoint union of G and H , denoted by G∪H , is the graph with vertex set V(G)∪V(H) and edge set E(G)∪E(H) . If G1 [congruent with]...[congruent with]Gk , we write kG1 for G1 ∪...∪Gk . The join of G and H , denoted by G⋁H , is the graph obtained from G∪H by adding edges joining every vertex of G to every vertex of H .
Lemma 8 (see [3]).
Let G be a connected graph of order n...5;4 . If [figure omitted; refer to PDF] then G is traceable unless G[congruent with]K1 ⋁(Kn-3 ∪2K1 ) , K2 ⋁(3K1 ∪K2 ) , or K4 ⋁(6K1 ) .
Let G be a graph containing a vertex v . Denote mG (v)=m(v)=(1/dG (v))∑u∈NG (v) ...dG (u) if dG (v)>0 and mG (v)=0 otherwise, where NG (v) or simply N(v) denotes the neighborhood of v in G .
Lemma 9 (see [11]).
Let G be a graph of order n . Then, [figure omitted; refer to PDF] with equality if and only if G⊇K1,n-1 or G=Kn-1 +v .
Lemma 10 (see [12]).
Let G be a connected graph. Then [figure omitted; refer to PDF] with equality if and only if G is a regular graph or a semi-regular graph.
In fact, if G is disconnected, there exists a component H of G such that [figure omitted; refer to PDF] So the inequality (14) also holds when G is a disconnected graph. By Lemmas 9 and 10, we have the following result; also see [13].
Corollary 11.
Let G be a graph of order n . Then, [figure omitted; refer to PDF] If G is connected, then the equality in (16) holds if and only if G=K1,n-1 or G=Kn . Otherwise, the equality in (16) holds if and only if G=Kn-1 +v .
Given a graph G of order n , a vector X∈Rn is called a function defined on G , if there is a 1-1 map [straight phi] from V(G) to the entries of X , simply written as Xu =[straight phi](u) for each u∈V(G) , Xu is also called the value of u given by X . If X is an eigenvector of Q(G) corresponding to the eigenvalue q , then X is defined naturally on G ; that is, Xu is the entry of X corresponding to the vertex u . One can find that [figure omitted; refer to PDF] where NG (v) denotes the neighborhood of v in G . The equation (17) is called (q,X) -eigenequation of G .
Theorem 12.
Let G be a connected graph of order n...5;4 . If [figure omitted; refer to PDF] then G is traceable.
Proof.
By Corollary 11 and (18), we have [figure omitted; refer to PDF]
Suppose that G is non-traceable. Then, by Lemma 8 and (19), G[congruent with]K1 ⋁(Kn-3 ∪2K1 ) , K2 ⋁(3K1 ∪K2 ) , or K4 ⋁(6K1 ) .
If G[congruent with]K1 ⋁(Kn-3 ∪2K1 ) , let X=(X1 ,X2 ,...,Xn )T be the eigenvector of Q(G) corresponding to eigenvalue q(G) . By (18), we know that q(G)...0;1 , n-4 . Thus, by (17), all vertices of degree 1 have the same values given by X , say X1 ; all vertices of degree n-3 have the same values by X , say X2 . Denote by X3 the value of the vertex of degree n-1 given by X . Also, by (17), we have [figure omitted; refer to PDF] Transform (20) into a matrix equation (B-q(G)I)X[variant prime] =0 , where X[variant prime] =(X1 ,X2 ,X3 )T and [figure omitted; refer to PDF]
Thus, q(G) is the largest root of the following equation: [figure omitted; refer to PDF] Let f(x)=x3 +(-3n+7)x2 +(2n2 -7n)x-2n2 +14n-24 ; then f[variant prime] (x)=3x2 +2(-3n+7)x+2n2 -7n . Let f[variant prime] (x)=0 ; we have two values x1 and x2 , such that f[variant prime] (x1 )=f[variant prime] (x2 )=0 , where [figure omitted; refer to PDF]
Hence, f(x) is strictly increasing with respect to x for x>x2 .
Because f(2(n-3))=2n2 -17n+33>0 and (2(n-2)2 +4)/(n-1)>2(n-3)>x2 , we have that f((2(n-2)2 +4)/(n-1))>0 , which implies that q(G)<(2(n-2)2 +4)/(n-1) .
If G[congruent with]K2 ⋁(3K1 ∪K2 ) , let X=(X1 ,X2 ,...,X7)T be the eigenvector of Q(G) corresponding to eigenvalue q(G) . By (18), we know that q(G)...0;2,5 . Thus, by (17), three vertices of degree 2 have the same values given by X , say X1 ; two vertices of degree 3 have the same values, say X2 ; two vertices of degree 6 have the same values, say X3 . Also, by (17), we have [figure omitted; refer to PDF] Transform (24) into a matrix equation (B-q(G)I)X[variant prime] =0 , where X[variant prime] =(X1 ,X2 ,X3 )T and [figure omitted; refer to PDF]
Thus, q(G) is the largest root of the following equation: [figure omitted; refer to PDF] Let g(x)=x3 -13x2 +40x-24 ; we can easily get that g(x) is strictly increasing with respect to x for x>20/3 .
Consider g(9)=12>0 , which implies that q(G)<9 .
If G[congruent with]K4 ⋁(6K1 ) , we easily calculate q(G)=8+210<44/3 .
Thus, in either case, we have a contradiction.
Lu et al. [3] have given a sufficient condition for a graph to be traceable as follows.
Theorem 13 (see [3]).
Let G be a connected graph of order n...5;5 . If [figure omitted; refer to PDF] then G is traceable.
Example 14.
There are graphs to which Theorem 12 may apply but Theorem 13 may not. Let G=(Kr ∪Kr )⋁K1 of order n...=2r+1 , where r...5;4 . Surely, the graph G is traceable. By a little computation, μ(G) is the largest root of the polynomial f(x)=x[x-(r-1)]-2r and q(G) is the largest root of the polynomial g(x)=[x-(2r-1)](x-2r)-2r . Hence, [figure omitted; refer to PDF] So, we can apply Theorem 12 but not Theorem 13 for G to be traceable.
4. Signless Laplacian Spectral Radius in Hamilton-Connected Graphs
For a graph G of order n , Erdös and Gallai [14] prove that if [figure omitted; refer to PDF] for any pair of nonadjacent vertices u and v , then G is Hamilton-connected.
The idea for the closure of a graph can be found in [7]. For a positive integer k , the k -closure of a graph G=(V,E) , denoted by Ck (G) , is a graph obtained from G by successively joining pairs of nonadjacent vertices u∈V and v∈V , whose degree sum is at least k until no such pairs remain. By the definition of the k -closure of G , we have that dCk (G) (u)+dCk (G) (v)...4;k-1 for any pair of nonadjacent vertices u∈V and v∈V of Ck (G) .
Lemma 15 (see [7]).
Let G be a graph of order n . Then, G is Hamilton-connected if and only if Cn+1 (G) is Hamilton-connected.
Lemma 16.
Let G be a simple graph with degree sequence (dG (v1 ),dG (v2 ),...,dG (vn )) , where dG (v1 )...4;dG (v2 )...4;......4;dG (vn ) and n...5;3 . Suppose that there is no integer k...4;n/2 such that dG (vk-1 )...4;k and dG (vn-k )...4;n-k . Then, G is Hamilton-connected.
Proof.
Let H-=Cn+1 (G) be the (n+1) -closure of G . Next, we will prove that H- is a complete graph; then the result follows according to (29). To the contrary, suppose that H- is not a complete graph, and let u and v be two nonadjacent vertices in H- with [figure omitted; refer to PDF] and dH- (u)+dH- (v) being as large as possible. By the definition of Cn+1 (G) , we have [figure omitted; refer to PDF] Denote by S the set of vertices in V\{v} which are nonadjacent to v in H- . Denote by T the set of vertices in V\{u} which are nonadjacent to u in H- . Then, [figure omitted; refer to PDF] Furthermore, by dH- (u)+dH- (v) being as large as possible, each vertex in S has degree at most dH- (u) and each vertex in T∪{u} has degree at most dH- (v) . Let k...=dH- (u) . According to (31) and (32), we have that |S|=n-1-dH- (v)...5;dH- (u)-1=k-1 , |T|+1=n-1-dH- (u)+1=n-dH- (u)=n-k . Then H- has at least k-1 vertices of degree not exceeding k and at least n-k vertices of degree not exceeding n-k . Because G is a spanning subgraph of H- , the same is true for G ; that is, dG (vk-1 )...4;k and dG (vn-k )...4;n-k . Because k...4;n/2 by (30) and (31), this is contrary to the hypothesis. So we have that the (n+1) -closure H- of G is indeed complete graph and hence that G is Hamilton-connected by (29).
We write Kn-1 +e+e[variant prime] for Kn-1 together with a vertex joining two vertices of Kn-1 by edges e,e[variant prime] , respectively.
Lemma 17.
Let G be a connected graph of order n...5;6 . If [figure omitted; refer to PDF] then G is Hamilton-connected unless G[congruent with]Kn-1 +e+e[variant prime] or G[congruent with]O3 ⋁K3 .
Proof.
Suppose that G is not a Hamilton-connected graph with degree sequence (dG (v1 ),dG (v2 ),...,dG (vn )) , where dG (v1 )...4;dG (v2 )...4;......4;dG (vn ) and n...5;6 . By Lemma 16, there is integer k...4;n/2 such that dG (vk-1 )...4;k and dG (vn-k )...4;n-k . Since G is connected, k...5;2 . Thus, [figure omitted; refer to PDF]
Because 2...4;k...4;n/2 , (9-2n)/6...4;k-(2n+3)/6...4;(n-3)/6 . Thus, if n...5;6 , [figure omitted; refer to PDF] Since e(G)...5;(n-1)(n-2)/2+2 , then all inequalities in the above argument should be equalities. From the last equality in (35), we have k=2 or k=3 and n=6 . If k=2 , by the equality in (34), G is a graph with dG (v1 )=2,dG (v2 )=dG (v3 )=...=dG (vn-2 )=n-2,dG (vn-1 )=dG (vn )=n-1 , which implies G[congruent with]Kn-1 +e+e[variant prime] . If k=3 and n=6 , by the equality in (34), G is a graph with dG (v1 )=dG (v2 )=dG (v3 )=3,dG (v4 )=dG (v5 )=dG (v6 )=5 , which implies G[congruent with]O3 ⋁K3 .
Theorem 18.
Let G be a connected graph of order n...5;6 . If [figure omitted; refer to PDF] then G is Hamilton-connected.
Proof.
By Corollary 11 and (36), we have [figure omitted; refer to PDF]
Suppose that G is not Hamilton-connected. Then, by Lemma 17 and (37), G[congruent with]Kn-1 +e+e[variant prime] or G[congruent with]O3 ⋁K3 .
If G[congruent with]Kn-1 +e+e[variant prime] . Let X=(X1 ,X2 ,...,Xn)T be the eigenvector of Q(G) corresponding to the eigenvalue q(G) . By (36), we know that q(G)...0;n-3 and q(G)...0;n-2 . Thus, by (17), all vertices of degree n-2 have the same values given by X , say X1 , and all vertices of degree n-1 have the same values, say X2 . Denote by X3 the value of the vertex of degree 2 given by X . Also, by (17), we have [figure omitted; refer to PDF] Transform (38) into a matrix equation (B-q(G)I)X[variant prime] =0 , where X[variant prime] =(X1 ,X2 ,X3 )T and [figure omitted; refer to PDF]
Thus, q(G) is the largest root of following equation: [figure omitted; refer to PDF]
Let f(x)=x3 +(4-3n)x2 +(2n2 -2n-8)x-4n2 +20n-24 ; then f[variant prime] (x)=3x2 +2(4-3n)x+2n2 -2n-8 . Let f[variant prime] (x)=0 ; we have two values x1 and x2 , such that f[variant prime] (x1 )=f[variant prime] (x2 )=0 , where [figure omitted; refer to PDF]
Hence, f(x) is strictly increasing with respect to x for x>x2 .
Consider f(2(n-2)+4/(n-1))=4(n-3)2 (n2 -3n+6)/(n-1)3 >0 and 2(n-2)+4/(n-1)>x2 , which implies that q(G)<2(n-2)+4/(n-1) .
If G[congruent with]O3 ⋁K3 . We can calculate that q(G)=5+13<8.8=2(6-2)+4/(6-1) . Thus, in either case, we have a contradiction.
Acknowledgments
This work is jointly supported by the National Natural Science Foundation of China under Grant nos. 11071001 and 11071002, the Natural Science Foundation of Anhui Province of China under Grant no. 11040606M14, the Natural Science Foundation of Department of Education of Anhui Province of China under Grant nos. KJ2011A195 and KJ2013A196, and the Young Scientific Research Foundation of Anqing Normal College, no. KJ201307.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Copyright © 2014 Guidong Yu et al. Guidong Yu 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 establish some signless Laplacian spectral radius conditions for a graph to be Hamiltonian or traceable or Hamilton-connected.
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