Journal of Applied Mathematics

Journal of Applied Mathematics / 2014 / Article

Research Article | Open Access

Volume 2014 |Article ID 282053 | 6 pages | https://doi.org/10.1155/2014/282053

Signless Laplacian Spectral Conditions for Hamiltonicity of Graphs

Academic Editor: Francesco Soldovieri
Received23 Jan 2014
Revised07 Jun 2014
Accepted16 Jun 2014
Published30 Jun 2014

Abstract

We establish some signless Laplacian spectral radius conditions for a graph to be Hamiltonian or traceable or Hamilton-connected.

1. Introduction

Let a graph, be a simple graph of order with vertex set and edge set . Denote by the number of edges of the graph . Write by a complete graph of order , an empty graph of order (without edges), and a complete bipartite graph with two parts having , vertices, respectively. The graph is said to be Hamiltonian, if it has a Hamiltonian cycle which is a cycle of order contained in . The graph is said to be traceable if it has a Hamiltonian path which is a path of order contained in . 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 is defined to be a matrix , where if is adjacent to and otherwise. The largest eigenvalue of is called to be the spectral radius of , which is denoted by . The degree matrix of is written by , where denotes the degree of the vertex in the graph . The signless Laplacian matrix of is defined by . The largest eigenvalue of is called to be the signless Laplacian spectral radius of , which is denoted by .

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 [16]. 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 , the -closure of a balanced bipartite graph , where , written by , is a graph obtained from by successively joining pairs of nonadjacent vertices and , whose degree sum is at least , until no such pairs remain. By the definition of the , we have that for any pair of nonadjacent vertices and of .

Lemma 1 (see [9]). Let be a connected balanced bipartite graph, where . Then, is Hamiltonian if and only if is Hamiltonian.

For a graph , write , and let be maximum degree of . 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 be a graph with at least one edge. Then, if and only if is regular or semi-regular.

Let be a Hermitian matrix of order , and let be the eigenvalues of .

Lemma 3 (see [10]). Let and be Hermitian matrices of order , , . Then, if .

Lemma 4. Let be a graph. Then,

Proof. Because , by Lemma 3, We notice that , , and . So, the result follows.

Let be a bipartite graph, the quasi-complement of is denoted by , where .

Theorem 5. Let be a connected balanced bipartite graph, where . If then is Hamiltonian.

Proof. Suppose that is not Hamiltonian. Then, is not Hamiltonian too by Lemma 1, and therefore, is not . Thus, there exists a vertex and a vertex such that . We find that for any pair of nonadjacent vertices and in . So, for any pair of adjacent vertices , in . Hence,
By Lemma 2, we have that
As is a subgraph of , by Perron-Frobenius theorem,
Thus, by (5), (8), and (9), we have that a contradiction.

Li [4] has given a sufficient condition for a bipartite graph to be Hamiltonian as follows.

Theorem 6 (see [4]). Let be a connected balanced bipartite graph, where . If then is Hamiltonian.

Remark 7. We now compare Theorems 5 and 6. If and , we have that by Lemma 4. Hence Theorem 5 improves Theorem 6 when . For example, let be a regular connected balanced bipartite graph with degree , where is odd and . Then, its quasi-complement is a regular graph with degrees , , and . 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 for together with an isolated vertex. Let and be two disjoint graphs. The disjoint union of and , denoted by , is the graph with vertex set and edge set . If , we write for . The join of and , denoted by , is the graph obtained from by adding edges joining every vertex of to every vertex of .

Lemma 8 (see [3]). Let be a connected graph of order . If then G is traceable unless , , or .

Let be a graph containing a vertex . Denote if and otherwise, where or simply denotes the neighborhood of in .

Lemma 9 (see [11]). Let be a graph of order . Then, with equality if and only if or .

Lemma 10 (see [12]). Let G be a connected graph. Then with equality if and only if is a regular graph or a semi-regular graph.

In fact, if is disconnected, there exists a component of such that So the inequality (14) also holds when is a disconnected graph. By Lemmas 9 and 10, we have the following result; also see [13].

Corollary 11. Let be a graph of order . Then, If is connected, then the equality in (16) holds if and only if or . Otherwise, the equality in (16) holds if and only if .

Given a graph of order , a vector is called a function defined on , if there is a 1-1 map from to the entries of , simply written as for each , is also called the value of given by . If is an eigenvector of corresponding to the eigenvalue , then is defined naturally on ; that is, is the entry of corresponding to the vertex . One can find that where denotes the neighborhood of in . The equation (17) is called -eigenequation of .

Theorem 12. Let be a connected graph of order . If then is traceable.

Proof. By Corollary 11 and (18), we have
Suppose that is non-traceable. Then, by Lemma 8 and (19), , , or .
If , let be the eigenvector of corresponding to eigenvalue . By (18), we know that , . Thus, by (17), all vertices of degree have the same values given by , say ; all vertices of degree have the same values by , say . Denote by the value of the vertex of degree given by . Also, by (17), we have Transform (20) into a matrix equation , where and
Thus, is the largest root of the following equation: Let ; then . Let ; we have two values and , such that , where
Hence, is strictly increasing with respect to for .
Because and , we have that , which implies that .
If , let be the eigenvector of corresponding to eigenvalue . By (18), we know that . Thus, by (17), three vertices of degree have the same values given by , say ; two vertices of degree have the same values, say ; two vertices of degree have the same values, say . Also, by (17), we have Transform (24) into a matrix equation , where and
Thus, is the largest root of the following equation: Let ; we can easily get that is strictly increasing with respect to for .
Consider , which implies that .
If , we easily calculate .
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 be a connected graph of order . If then is traceable.

Example 14. There are graphs to which Theorem 12 may apply but Theorem 13 may not. Let of order , where . Surely, the graph is traceable. By a little computation, is the largest root of the polynomial and is the largest root of the polynomial . Hence, So, we can apply Theorem 12 but not Theorem 13 for to be traceable.

4. Signless Laplacian Spectral Radius in Hamilton-Connected Graphs

For a graph of order , Erdös and Gallai [14] prove that if for any pair of nonadjacent vertices and , then is Hamilton-connected.

The idea for the closure of a graph can be found in [7]. For a positive integer , the -closure of a graph , denoted by , is a graph obtained from by successively joining pairs of nonadjacent vertices and , whose degree sum is at least until no such pairs remain. By the definition of the -closure of , we have that for any pair of nonadjacent vertices and of .

Lemma 15 (see [7]). Let be a graph of order . Then, is Hamilton-connected if and only if is Hamilton-connected.

Lemma 16. Let be a simple graph with degree sequence , where and . Suppose that there is no integer such that and . Then, is Hamilton-connected.

Proof. Let be the -closure of . Next, we will prove that is a complete graph; then the result follows according to (29). To the contrary, suppose that is not a complete graph, and let and be two nonadjacent vertices in with and being as large as possible. By the definition of , we have Denote by the set of vertices in which are nonadjacent to in . Denote by the set of vertices in which are nonadjacent to in . Then, Furthermore, by being as large as possible, each vertex in has degree at most and each vertex in has degree at most . Let . According to (31) and (32), we have that , . Then has at least vertices of degree not exceeding and at least vertices of degree not exceeding . Because is a spanning subgraph of , the same is true for ; that is, and . Because by (30) and (31), this is contrary to the hypothesis. So we have that the -closure of is indeed complete graph and hence that is Hamilton-connected by (29).

We write for together with a vertex joining two vertices of by edges , respectively.

Lemma 17. Let be a connected graph of order . If then is Hamilton-connected unless or .

Proof. Suppose that is not a Hamilton-connected graph with degree sequence , where and . By Lemma 16, there is integer such that and . Since is connected, . Thus,
Because , . Thus, if , Since , then all inequalities in the above argument should be equalities. From the last equality in (35), we have or and . If , by the equality in (34), is a graph with , which implies . If and , by the equality in (34), is a graph with , which implies .

Theorem 18. Let be a connected graph of order . If then is Hamilton-connected.

Proof. By Corollary 11 and (36), we have
Suppose that is not Hamilton-connected. Then, by Lemma 17 and (37), or .
If . Let be the eigenvector of corresponding to the eigenvalue . By (36), we know that and . Thus, by (17), all vertices of degree have the same values given by , say , and all vertices of degree have the same values, say . Denote by the value of the vertex of degree given by . Also, by (17), we have Transform (38) into a matrix equation , where and
Thus, is the largest root of following equation:
Let ; then . Let ; we have two values and , such that , where
Hence, is strictly increasing with respect to for .
Consider and , which implies that .
If . We can calculate that . Thus, in either case, we have a contradiction.

Conflict of Interests

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

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.

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Copyright © 2014 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.

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