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Spectral radius and Hamiltonian graphs Mei Lu a,∗,1 , Huiqing Liu b,2 , Feng Tian c,3 a b c

Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China School of Mathematics and Computer Science, Hubei University, Wuhan 430062, China Institute of Systems Science, Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences, Beijing 100080, China

ARTICLE INFO

ABSTRACT

Article history: Received 10 August 2011 Accepted 14 May 2012 Available online 6 June 2012

In the paper, we will present two sufficient conditions for a bipartite graph to be Hamiltonian and a graph to be traceable, respectively. © 2012 Elsevier Inc. All rights reserved.

Submitted by R.A. Brualdi AMS classification: 05C50 15A18 Keywords: Graph Hamiltonian bipartite graph Traceable graph Adjacent matrix Spectral radius

1. Introduction Let G = (V , E ) be a simple undirected graph with vertex set V (G) = {v1 , v2 , · · · , vn }. For vi ∈ V , the degree of vi , written by di , is the number of edges incident with vi and let (d1 , d2 , · · · , dn ) be the degree sequence of G, where d1 ≤ d2 ≤ · · · ≤ dn . We use G to denote the complement of G. Let Kn,n−1 = (X , Y ; E ) be a complete bipartite graph with |X | = n and |Y | = n − 1. Denoted Kn,n−1 + e the bipartite graph obtained from Kn,n−1 by adding a pendent edge to one of vertices in X. Let G and H be two disjoint graphs. The disjoint union of G and H, denoted G + H, is the graph with vertex V (G) ∪ V (E ) ∗ Corresponding author. 1 2 3

E-mail addresses: [email protected] (M. Lu), [email protected] (H. Liu), [email protected] (F. Tian). Partially supported by NNSFC (Nos. 10971114, 10990011). Partially supported by NNSFC (Nos. 10971114, 11171097), NSF of Hubei Provincial Department of Education (No. T200901). Partially supported by NNSFC (No. 11071115).

0024-3795/$ - see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.05.021

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and edge set E (G) ∪ E (H ). If G1 ∼ = ··· ∼ = Gk , we write kG1 for G1 + · · · + Gk . The join of G and H, denoted G ∨ H, is the graph obtained from disjoint union of G and H by adding edges joining every vertex of G to every vertex of H. Let A(G) be the adjacency matrix of G and let µ1 ≥ µ2 ≥ · · · ≥ µn be the eigenvalues of A. When G is connected, A(G) is irreducible and so by Perron–Frobenius Theorem, µ1 is simple. The eigenvalue µ1 (also call the spectral radius of G) has received a great deal of attention (see, for example [1,5,7,9– 12,14,16,17]). In Section 2, we will give a sufficient condition for a bipartite graph to be Hamiltonian. In section 3, we will present a sufficient condition for a graph to be traceable. Other spectral conditions for Hamiltonian cycles or Hamiltonian paths have been given in [3,6,8,13,15]. 2. Spectral radius in Hamiltonian bipartite graphs In this section, we will consider the bipartite graphs. We first cite the following result about Hamiltonian bipartite graphs. Lemma 2.1 [2]. Let G := G[X , Y ; E ] be a bipartite graph, where |X | = |Y | = n ≥ 2, with degree sequence (d1 , d2 , · · · , d2n ), where d1 ≤ d2 ≤ · · · ≤ d2n . Suppose that there is no integer k ≤ n/2 such that dk ≤ k and dn ≤ n − k. Then G is Hamiltonian. Lemma 2.2. Let G

:= G[X , Y ; E ] be a connected bipartite graph, where |X | = |Y | = n ≥ 2. If

|E (G)| ≥ n2 − n + 1, then G is Hamiltonian unless G

∼ = Kn,n−1 + e.

Proof. Suppose G is a connected non-Hamiltonian bipartite graph with degree sequence (d1 , d2 , · · · , d2n ), where d1 ≤ d2 ≤ · · · ≤ d2n and n ≥ 2. By Lemma 2.1, there is k ≤ n/2 such that dk ≤ k and dn ≤ n − k. Thus

|E (G)| =

2n 1

2 i=1

di

≤

k2

+ (n − k)2 + n2 2

= n − n + 1 − (k − 1)(n − k − 1) 2

≤ n2 − n + 1.

(1)

Since |E (G)| ≥ n − n + 1, we have |E (G)| = n − n + 1. Then all inequalities in the above argument should be equalities. From (1) and k ≤ n/2, we have k = 1. That is, G is a bipartite graph of size n2 − n + 1 with d1 = 1, d2 = · · · = dn = n − 1 and dn+1 = · · · = d2n = n. Thus G ∼ = Kn,n−1 + e. 2

2

Lemma 2.3 [4]. Let G d(x) + d(y)

:= G[X , Y ; E ] be a bipartite graph, where |X | = |Y | = n ≥ 2. If

≥n+1

for every pair of nonadjacent vertices x

∈ X and y ∈ Y , then G is Hamiltonian.

Let G := G[X , Y ; E ] be a bipartite graph, where |X | = |Y | = n ≥ 2. Perform the following operation on G: if there are two nonadjacent vertices x ∈ X and y ∈ Y with dG (x) + dG (y) ≥ n + 1, add the edge xy to E (G). The closure of G is the graph obtained from G by successively applying this operation as long as possible. Then the closure of G is well defined and we have the following result from Lemma 2.3. Theorem 2.4 [4]. Let G := G[X , Y ; E ] be a bipartite graph, where |X | Hamiltonian if and only if its closure is Hamiltonian.

= |Y | = n ≥ 2. Then G is

Let G := G[X , Y ; E ] be a bipartite graph. We construct a new bipartite graph G∗ = [X , Y ; E ], which called the quasi-complement of G, from G as follows: V (G∗ ) = V (G) and for x ∈ X, y ∈ Y , xy ∈ E (G∗ ) if and only if xy ∈ E (G). Then we have the following result. The main idea of the proof comes from [6].

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Theorem 2.5. Let G := G[X , Y ; E ] be a bipartite graph and G∗ the quasi-complement of G, where |X | |Y | = n ≥ 2. Let µ(G∗ ) be the spectral radius of G∗ . If

µ(G∗ ) ≤

=

√

n − 1,

then G is Hamiltonian unless G

∼ = Kn,n−1 + e.

Proof. For short, denoted H the closure of G and H ∗ the quasi-complement of H. Suppose G is nonHamiltonian. Then H is non-Hamiltonian from Theorem 2.4. By the construction of H, dH (x) + dH (y) ≤ n for every pair of nonadjacent vertices x ∈ X and y ∈ Y . Thus dH ∗ (x) + dH ∗ (y)

= n − dH (x) + n − dH (y) ≥ n

for every edge xy ∈ E (H ∗ ). So (dH ∗ (x) + dH ∗ (y)) xy∈E (H ∗ )

On the other hand, 2 dH ∗ (v)

=

v∈V (H )

≥ n|E (H ∗ )|.

(dH ∗ (x) + dH ∗ (y)) ≥ n|E (H ∗ )|.

xy∈E (H ∗ )

Using the inequality of Hofmeister [9], we have 2 ∗ nµ2 (H ∗ ) ≥ dH ∗ (v) ≥ n|E (H )|. v∈V (H )

Since H ∗

⊆ G∗ ,

µ(H ∗ ) ≤ µ(G∗ ) ≤

√

n − 1.

Thus n(n − 1)

≥ nµ2 (H ∗ ) ≥ n|E (H ∗ )|,

which implies |E (H ∗ )|

≤ n − 1. Hence

|E (H )| = n2 − |E (H ∗ )| ≥ n2 − n + 1. Since H is non-Hamiltonian, we have G ∼ = Kn,n−1 + e from Lemma 2.2. If G = H, we are done, so we can assume that G is a proper subgraph of H. Thus G∗ contains a connected proper subgraph K1,n−1 . By the Perron-Frobenius theorem, we have

µ(G∗ ) > µ(K1,n−1 ) =

√

n − 1,

a contradiction. From Theorem 2.5, we easily have the following corollary. Corolary 2.6. Let G := G[X , Y ; E ] be a 2-connected bipartite graph, where |X | = |Y | = n ≥ 2. Let µ(G∗ ) be the spectral radius of G∗ . If √ µ(G∗ ) ≤ n − 1, then G is Hamiltonian.

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3. Spectral radius in traceable graphs A sufficient conditions for a graph to be traceable in terms of its vertex degree was given in [2]. Theorem 3.1 [2]. Let G be a nontrivial graph with degree sequence (d1 , d2 , · · · , dn ), where d1 ≤ d2 ≤ · · · ≤ dn and n ≥ 4. Suppose that there is no integer k < (n + 1)/2 such that dk ≤ k − 1 and dn−k+1 ≤ n − k − 1. Then G is traceable. By Theorem 3.1, we have the following result. Lemma 3.2. Let G be a connected graph of order n

|E (G)| ≥

(n − 2)(n − 3) 2

then G is traceable unless G

≥ 4. If

+ 2,

∼ = K1 ∨ (Kn−3 + 2K1 ), or K2 ∨ (3K1 + K2 ), or K4 ∨ (6K1 ).

Proof. Suppose that G is a non-traceable graph with degree sequence (d1 , d2 , · · · , dn ), where d1 ≤ 1 d2 ≤ · · · ≤ dn and n ≥ 4. By Theorem 3.1, there is k < n+ such that dk ≤ k−1 and dn−k+1 ≤ n−k−1. 2 Since G is connected, k ≥ 2. Thus

|E (G)| = ≤ = ≤

n 1

di 2 i=1 1 (k(k − 1) + (n − 2k + 1)(n − k − 1) + (k − 1)(n − 1)) 2 (n − 2)(n − 3) (k − 2)(2n − 3k − 5) +2− 2 2 (n − 2)(n − 3) + 2. 2 (n−2)(n−3)

(2)

(3)

(n−2)(n−3)

Since |E (G)| ≥ + 2, |E (G)| = + 2. Then all inequalities in the above argument 2 2 should be equalities. From (3), we have k = 2 or 2n = 3k + 5. If k = 2, then G is a graph with d1 = d2 = 1, d3 = · · · = dn−1 = n − 3 and dn = n − 1, which implies G ∼ = K1 ∨ (Kn−3 + 2K1 ). If 2n = 3k + 5, then n ≤ 10 as k < (n + 1)/2, and hence n = 7, k = 3, or n = 10, k = 5. By (2), G is a graph of order 7 with d1 = d2 = d3 = 2, d4 = d5 = 3, d6 = d7 = 6, or G is a graph of order 10 with d1 = · · · = d6 = 4, d7 = · · · = d10 = 9, which implies G ∼ = K2 ∨ (3K1 + K2 ), or G ∼ = K4 ∨ (6K1 ). Theorem 3.3 [10]. Let G be a connected graph of order n. Then µ1 ≤ 2|E (G)| − n + 1 and the equality holds if and only if G

∼ = Kn or G ∼ = K1,n−1 .

Now we have the following theorem. Theorem 3.4. Let G be a connected graph of order n µ1 ≥ (n − 3)2 + 2,

≥ 5. If (4)

then G is traceable. Proof. By Theorem 3.3 and (4), we have 2|E (G)|

≥ µ21 + n − 1 ≥ (n − 3)2 + 2 + n − 1 = (n − 2)(n − 3) + 4.

(5)

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Suppose that G is non-traceable. Then by Lemma 3.2 and (5), G ∼ = K1 ∨ (Kn−3 + 2K1 ), or G ∼ = K2 ∨ (3K1 + K2 ), or G ∼ = K4 ∨ (6K1 ). If G ∼ = K1 ∨ (Kn−3 + 2K1 ), we let X = (x1 , x2 . . . , xn )t be the eigenvector corresponding to µ1 (G), where xi (1 ≤ i ≤ 2) corresponds to the vertex of degree 1, xi (3 ≤ i ≤ n − 1) corresponds to the vertex of degree n − 3 and xn corresponds to the vertex of degree n − 1. Then by µ1 X = A(G)X, we have ⎧ ⎪ ⎪ x1 = x2 , x3 = · · · = xn−1 , ⎪ ⎪ ⎪ ⎪ ⎨x = µ x , n 1 1 ⎪ ⎪ x + ( n − 4)x3 = µ1 x3 , ⎪ n ⎪ ⎪ ⎪ ⎩ 2x1 + (n − 3)x3 = µ1 xn . Thus

µ31 − (n − 4)µ21 − (n − 1)µ1 + 2(n − 4) = 0. Let f (x)

= x3 − (n − 4)x2 − (n − 1)x + 2(n − 4). Then

f ( (n − 3)2

+ 2) = (n − 3)(n − 4)

(n − 3)2 + 2 − (n − 3) > 0

≥ 5, which implies µ1 (G) < (n − 3)2 + 2. If G ∼ = K2 ∨ (3K1 + K2 ), then µ1 (G) = 3.9095 < √ √ 18 = (7 − 3)2 + 2. If G ∼ = K4 ∨ (6K1 ), then µ1 (G) = 6.6234 < 51 = (10 − 3)2 + 2. Thus, for n

in either case, we have a contradiction.

Acknowledgments Many thanks to the anonymous referee for his/her many helpful comments and suggestions, which have considerably improved the presentation of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17]

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