On the Signless Laplacian Spectral Radius of Cacti

A cactus is a connected graph in which any two cycles have at most one vertex in common. We determine the unique graphs with maximum signless Laplacian spectral radius in the class of cacti with given number of cycles (cut edges, respectively) as well as in the class of cacti with perfect matchings and given number of cycles.


INTRODUCTION
LL graphs in this paper are simple.Let G be a graph on n vertices with vertex set V(G) and edge set E(G).Let A(G) be the adjacency matrix of G.The matrix Q(G) = D(G) + A(G) is called the signless Laplacian matrix of G, where D(G) is a diagonal matrix of (vertex) degrees of G.It is well known that Q(G) is a semi-definite matrix and thus its eigenvalues are all nonnegative.The signless Laplacian spectral radius of G is the largest eigenvalue of Q(G), denoted by q(G).For more matrices associated to a graph, see the book of Janežič et al. [1] If G is connected, then Q(G) is irreducible and by the Perron-Frobenius theorem, q(G) has multiplicity one and there exists a unique positive unit eigenvector corresponding to q(G), which is the Perron vector of Q(G).][5][6][7][8][9][10] A cactus is a connected graph in which every edge appears in at most one cycle, see, e.g.Ref. [11].Note that trees and unicyclic graphs are cacti.The (adjacency) spectral radius and least eigenvalue of a cactus have been studied to some extent, [12][13][14][15] and the distance spectral radius of a cactus was also studied. [16]Li and Zhang [7] determined the unique graphs with maximum signless Laplacian spectral radius in the class of cacti with given numbers of vertices and pendant vertices, and in the class of cacti with perfect matching and given number of vertices, respectively.
be the class of all cacti on n vertices with k cycles.
For 0 ≤ k ≤ n -3, let ( , ) n k F be the class of all cacti on n vertices with k cut edges.
For 0 ≤ k ≤ n -1, let ( , ) n k G be the class of all cacti on 2n vertices with perfect matchings and k cycles.
In this paper, we determine the unique cacti with maximum signless Laplacian spectral radius in ( , ) The spectral radius was, long ago, put forward as a measure of molecular branching, [17] while the Laplacian spectral radius was used for describing the shape and folding of DNA molecules. [18]It is well known that the Laplacian and signless Laplacian spectra of bipartite graphs coincide.Thus, chemically interesting cases for signless Laplacian spectral radius are the fullerenes, fluoranthenes and other non-alternant conjugated species.

PRELIMINARIES
Let G be a connected graph with V(G) = {v1,...,vn}.The Perron vector of Q(G) is the column vector x = ( Denote by Cn, Pn and Sn the cycle, the path and the star on n vertices, respectively. For a graph G with u ∈ V(G), NG(u) denotes the set of neighbors of u in G, and the degree of The following two lemmas were proved in Refs.[19]  and [20], respectively.
Lemma 2.2.Let G be a connected graph and e = uv a non-pendant edge of G. Suppose that NG(u) ∩ NG(v) = ∅.Let G' be the graph obtained from G -{uv} by identifying u and v into u, and adding a new pendant edge at u.Then, q(G') > q(G).

Lemma 2.3. Let H be a proper subgraph of a connected graph G. Then, q(G) > q(H).
For n ≥ 3, k ≥ 0, and 0 ≤ p ≤ (n -2k -1)/2, let Gn,k,p be an n-vertex graph obtained by identifying a vertex of each of k triangles, a vertex of each of t paths P2, and a terminal vertex of each of p paths P3, where t = n -2k -2p -1, see Figure 1.
Lemma 2.4.For n ≥ 3, k ≥ 0, and 0 ≤ p ≤ (n -2k -1)/2, q(Gn,k,p) is the largest root of the equation f(x) = 0, where f(x) = x 5 -(n -p + 6)x 4 + (6n -6p + 10)x 3 -(10n + 4k -9p + 3)x 2 + (3n + 12k)x -4k.In particular, q(Gn,k,0) is the largest root of the equation x 3 -(n + 3)x 2 + 3nx -4k = 0, and for odd n, Proof.The characteristic polynomial of Q(Gn,k,p) was given in the proof of Lemma 4 in Ref. [8], from which the first part follows. Let , which, together with the fact that q(Gn,k,0) ≥ q(P3) = 3, implies that q(Gn,k,0) is the largest root of the equation g For a cactus G with at least one cycle, the deletion of all edges of G on cycles results in a forest.A nontrivial connected component of such a forest containing a unique vertex v on some cycle is said to be a branch of G at v. If G is a tree, we also call it a branch at one of its vertices.

RADIUS OF GRAPHS IN
, where 0 ≤ k ≤ [(n -1)/2] and n ≥ 3.Then, q(G) ≤ q(Gn,k) with equality if and only if G ≌ Gn,k, where q(Gn,k) is the largest root of the equation Proof.By Lemma 2.4, q(Gn,k) is the largest root of the equation with maximum signless Laplacian spectral radius.We need only to show that G ≌ Gn,k.
Let x be the Perron vector of Q(G).Suppose that k ≥ 1 and there is a cycle of length at least 4. Let v1 ...vrv1 be such a cycle with length r ≥ 4. Without loss of generality assume that . By Lemma 2.1, we have q(G 0 ) > q(G), a contradiction.Thus, if k ≥ 1, then all cycles of G are triangles.If n = 3, then k = 1 and thus the result follows.So let n ≥ 4.
Claim 1.If k ≥ 2, then any two triangles of G have one common vertex.
Suppose that there are two disjoint triangles T1 and T2 in G.Then, there exists a unique shortest path v1 ...vs joining them, where Since G is a cactus, any path joining T1 and T2 starts from v1 and ends in vs.We may assume that xv1 ≥ xvs.Let w1 and w2 the neighbors of vs in T2.
Claim 2. If k ≥ 3, then any three triangles of G have exactly one common vertex.
Suppose that there are three triangles T1, T2 and T3 in G such that they have no common vertex.By Claim 1, the common vertices of T1 and T2, T2 and T3, and T1 and T3 induce x is maximum among vertices with degree at least 3 on the unique triangle, and if k = 0, choose v0 to be a non-pendant vertex.
Claim 3.For k ≥ 1, there is no branch of G at a vertex different from v0.
Suppose that there is a branch , and by Lemma 2.1, we have q(G') > q(G), a contradiction.Thus, Claim 3 follows.
If there is a branch T of G at v0 (possibly k = 0 and G = T), then T is a star with center v0.Otherwise, suppose that there is a path v0v1 ...vs in T, where s ≥ 2. Let G' be the graph obtained from G by deleting the edge v0v1, identifying v0 and v1 into v0, and adding a new pendant edge to v0.

RADIUS OF GRAPHS IN
In view of the proofs of Lemmas 2.6 and 2.7 in Ref. [7], we deduce the following two lemmas.
Lemma 4.1.Let Y be a connected graph with u0 ∈ V(Y ).For m ≥ 5, let G1 be the graph obtained by identifying u0 and a vertex of Cm, G2 be the graph obtained by identifying u0 and a vertex of each of (m -1)/2 triangles for odd m, and G3 be the graph obtained by identifying u0, a vertex of each of (m -4)/2 triangles, and a vertex of one quadrangle for even m, see Figure 2.Then, q(G1) < q(G2) if m is odd, and q(G1) < q(G3) if m is even.
Lemma 4.2.Let Y be a connected graph with u0 ∈ V (Y ).Let G4 be the graph obtained by identifying u0 and a vertex of each of two quadrangles, and G5 be the graph obtained by identifying u0 and a vertex of each of three triangles, see Figure 3.Then, q(G4) < q(G5).
G ) and x be the Perron vector of Q.By Lemma 2.3, q ≥ q(C4) = 4.
The result is trivial for n = 3 and for n = 4 with k = 0. Suppose n ≥ 4 and (n,k) ≠ (4,0).By Lemma 2.3 and the fact that q(Cs) = 4 for s ≥ 3, we have G ≇ Cn.
Since k ≤ n -3, G contains at least one cycle.By similar arguments as in Claims 1 and 2 in the proof of Theorem 3.1, if there are at least two cycles, then all cycles of G have exactly one common vertex, denoted by v0.If there is exactly one cycle in G, then choose v0 such that the corresponding entry of the Perron vector of Q(G) is maximum among vertices with degree at least 3 on the unique cycle.By similar arguments as in Claim 3 in the proof of Theorem 3.1, there is no branch of G at a vertex different from v0.If there is a branch T of G at v0, then by Lemma 2.2, T is a star with center v0.
If G contains a cycle of length at least 5, then by Lemma 4.1, we may have a graph G' ∈ ( , ) n k F (of the form G2 or G3) such that q(G) < q(G'), a contradiction.Thus, any cycle of G has length 3 or 4. If G contains two quadrangles, then by Lemma 4.2, we may have a graph G' ∈ ( , ) n k F (of the form G5) such that q(G) < q(G'), a contradiction.Thus, G has at most one quadrangle.Thus, G ≌ Gn,(n -k -1)/2 if n -k is odd and G ≌ G'n,(n -k -2)/2 if nk is even.The following lemma is a particular case of Theorem 4.2 in Ref. [21].

RADIUS OF GRAPHS IN
Lemma 5.1.Let G be a graph on at least two vertices with ∆i = ∆i(G) for i = 1,2.Then, which can be considered as a function defined on V(G) which maps vertex vi to i v x for i = 1,2,...,n.For an edge subset F of a graph G, G -F denotes the graph obtained from G by deleting the edges in F, while for A an edge subset F' of the complement of G, G + F' denotes the graph obtained from G by adding the edges from F'.