On Maximum Signless Laplacian Estrada Index of Graphs with Given Parameters II

Recently Ayyaswamy [1] have introduced a novel concept of the signless Laplacian Estrada index (after here $SLEE$) associated with a graph $G$. After works, we have identified the unique graph with maximum $SLEE$ with a given parameter such as: number of cut vertices, (vertex) connectivity and edge connectivity. In this paper we continue out characterization for two further parameters; diameter and number of cut vertices.

Recently, Ayyaswamy [1] developed the innovative notion of the signless Laplacian Estrada index as He also established lower and upper bounds for SLEE in terms of the number of vertices and edges. Grone and Merris [17,18] proved that for a bipartite graph G, SLEE(G) = LEE(G).
Previousely in [8], we characterized the unique graphs with maximum SLEE among the set of all graphs with given number of cut edges, pendent vertices, (vertex) connectivity and edge connectivity. In this paper, we continue our research by characterizing the unique graph according to two further parameters: diameter and number of cut vertices.

Preliminaries and Lemmas
In this section, we first introduce basic definitions, notations and concepts used thorough this paper and restate some proved results found in [3,8]. Then, we prove some needful propositions for proving the main result of the next section.
, and e 1 , e 2 , . . . , e k ∈ E(G) such that the vertices v i and v i+1 are (not necessarily distinct) end points of edge e i , for any i = 1, 2, . . . , k. If v 1 = v k+1 , then we say W is a closed semi-edge walk.
By following [8], we denote The k-th signless Laplacian spectral moment of the graph G by T k (G) , i.e., For a graph G, the signless Laplacian spectral moment T k is equal to the number of closed semi-edge walks of length k.
Let G and G ′ be two graphs, and x, y ∈ V (G), and x ′ , y ′ ∈ V (G ′ ). Denote by SW k (G; x, y) the set of all semi-edge walks of length k in graph G, which are begining at vetex x, and ending at vertex y. For convenience, we use SW k (G; x, x) instead of SW k (G; x), and set SW k (G) = x∈V (G) SW k (G; x). Thus, by Theorem 2.2, we have T k (G) = |SW k (G)|. Note that, by Taylor expansions, we have Lemma 2.4. [8] Let G be a graph and v, u, w 1 , w 2 , . . . , w r ∈ V (G). suppose that E v = {e 1 = vw 1 , . . . , e r = vw r } and E u = {e ′ 1 = uw 1 , . . . , e ′ r = uw r } are subsets of edges of the complement of G.
For a vertex x and an edge e, let SW k (G; x, [e]) be the set of all closed semi-edge walks of length k in the graph G starting at vetex x and containing the edge e.
Lemma 2.5. Let G be a graph and H = G + e, such that e = uv ∈ E(G).
Proof . We know that for each z ∈ {u, v}, and k ≥ 0, . We can uniquely decompose W to W = W 1 eW 2 e . . . eW r , such that W i ∈ SW k i (G; x, y), where x, y ∈ {u, v}, and k i ≥ 0, and 1 ≤ i ≤ r. Note that W i is a semi-edge walk in G and does not contain e, Thus the decomposition is unique. For each W i excactly one of the following cases occurs: In Note that if there exists k 0 such that |SW k 0 (G; v)| < |SW k 0 (G; u)|, then f k 0 is not surjective. Thus h k 0 is not a surjection, and we have By a similar method, we prove the following statement: . W decomposes uniquely to W 1 eW 2 e . . . eW r , where W i is a semi-edge walk of length k i in G. Three cases will be considered as follows for W 1 : If 1 < i ≤ r, then three cases will be considered as follows for W i : The secound part of the lemma is clear.

The graph with maximum SLEE with given diameter
is the maximum eccentricity of the vertices, whereas the radius r(G) is the minimum eccentricity. Also, x is a central vertex if e(x) = r(G) and a diametral path is a shortest path between two vertices whose distance is equal to d(G). For convenience, let us denote ⌈ d 2 ⌉ by d where is the smallest integer number greater than d 2 . It is obvious that K n is the unique graph with diameter 1. Also, the path on n vertices P n , is the unique graph with diameter n − 1. Furthermore, K n − e is the graph with maximum SLEE with diameter 2, where e is an edge of K n .
Let n > 4, and 2 < d < n − 1, and 1 ≤ j ≤ d. We denote by H d,j , the set of all graphs H d,j , constructed from K n−1−d and For example, all graphs H 4,2 with n = 7 are shown in Fig.1. Lemma 3.2. Let n > 4, and 2 < d < n − 1, and 2 ≤ j ≤ d. If H j ∈ H d,j , then either H j ∈ H d,j−1 , or there exists a graph, say H j−1 ∈ H d,j−1 , such that SLEE(H j ) < SLEE(H j−1 ).
is the set of vertices that are adjacent to v i . To facilitate the understanding of the proof, we divide the argument into two parts. We first discuss about N K (v d−j ) and then proceed to N K (v d+j ). Note that if d is odd, j = 2, and N K (v d+2 ) = ∅, then by renaming the vertices of P d+1 such that v i changes to v d−i we conclude that H j ∈ H d,j−1 . Let H j ∈ H d,j−1 . Therefore either at least one of the vertex subsets N K (v d−j ) or N K (v d+j ) is not empty, or d is odd and j = 2 and N K (v d+2 ) is not empty. If , it is enough to prove the following statements: We start the prove of (1) by the following claim: , where e = yz, and k ≥ 0. We can decompose W to W = W 1 W 2 W 3 , where W 1 and W 3 are as long as possible and consisting of just the vertices v 0 , v 1 , . . . , y, and edges in {v t v t+1 : i obtains from W i , for i = 1, 3, by replacing each vertex v t by v a , and each edge v t v t+1 by v a v a−1 , and each edge yx by zx, where x ∈ N K (y), and a = 2 d − 2j − t + 3 (In fact, the distance between v t and y is equal to the distance between v a and z in P d+1 ). It is easy to check that the map f ′ k : . Now, the claim follows from lemma 2.5. Let f k : SW k (H ′ j ; y) → SW k (H ′ j ; z) be an injection, for each k ≥ 0. If W ∈ SW k (H ′ j ; v), then W can be decomposed to W = W 1 W 2 W 3 , where W 2 ∈ SW k 2 (H ′ j ; y) is as long as possible . Let W ′ i obtain form W i , for each i = 1, 3, by replacing each vertex v t by v a , and each edge v t v t+1 by v a v a−1 , where a = 2 d − 2j − t + 3. The map g k : (1). By a similar method used above, we prove the statement (2). First, we claim that: x, w) is as long as possible, where w ∈ N K (y), and W 2 ∈ SW k 2 (H ′ j − e; w, y). Suppose that W ′ 2 obtains from W 2 by replacing each vertex v t by v a , and the edge wy by wz, and each edge v t v t+1 by v a v a−1 , where a = 2 d − 2j − t + 3. One can easily check that the map h ′ k : x, z). Now, the claim follows from lemma 2.6.
x, y) is as long as possible, and W 2 ∈ SW k 2 (H ′ j ; y, v). Let W ′ 2 obtain from W 2 by replacing each vertex v t by v a , and replacing each is not empty. By repeating the above discussion The following theorem is our main result of this section, which is determined the unique graph with maximum SLEE among the set of all unicyclic graphs with diameter d, where 2 < d < n − 1.  Proof . Suppose that G is a graph, having maximum SLEE with diameter d. Let P d+1 = v 0 v 1 . . . v d be a diameterical path in G, and H be the graph obtained from G by adding some edges such that: (a) x is adjacent with exactly 3 vertices of P d+1 in H, ) is a complete graph on n − 1 − d vertices. By lemma 3.1, such a graph H exists. Obviously, we have H ∈ H d,j , for some j, 1 ≤ j ≤ d, and SLEE(G) ≤ SLEE(H), with equality if and only if G = H.
If j > 1, then by lemma 3.2, we may get a sequence of graphs, say H d,j−1 , H d,j−2 , . . . , H d,1 , such that for each t, H d,t ∈ H d,t , and with equalities hold, if and only if the graphs are equal. since the diameter of H d,1 is d, and G has the maximal SLEE among the set of all graphs with diameter d, hence SLEE(G) = SLEE(H d,1 ) which imlplies that G = H d,1 , as expected.

The graph with maximum SLEE with given number of cut vertices
A cut vertex of a graph is a vertex whose removal increases the number of components of the graph. Let G be a connected graph and x be a vertex of G, a block of G is defined to be a maximal subgraph without cut vertices. A pendent path at x in a graph G is a path in which no vertex other than x is incident with any edge of G outside the path, where deg G (x) ≥ 3. In particular, we consider a vertex x as a pendent path at x of length zero in G only when x is neither a pendent vertex nor a cut vertex of G. Let G and H be two vertex-disjoint connected graphs, such that x ∈ V (G) and y ∈ V (H). We denote the coalescence of G and H by G(x) • H(y), which is obtained by identifying the vertex x of G with the vertex y of H. Lemma 4.1. Let H 1 and H 2 be two graphs and P s = y 0 y 1 . . . y s−1 be a path on s vertices, and u ∈ V (H 2 ) and xy ∈ E(H 1 ), such that x = y.
is the set of vertices of H 1 that are adjacent to y .
Let 0 ≤ r ≤ n − 2. Suppose that G r n is the graph obtained from K n−r by attaching n − r pendent path of orders n 1 , n 2 , . . . , n n−r to its vertices, where each vertex of K n−r has exactly one pendent path and | n i − n j |≤ 1 for 1 ≤ i, j ≤ n − r. More precisely, each pendent path is of order ⌊ r n−r ⌋ or ⌊ r n−r ⌋ + 1. For example, the graphs G r 6 with r = 0, 1, 2, 3, 4 are shown in Fig.4 . Theorem 4.2. If 0 ≤ r ≤ n − 2, then G r n is the unique graph with maximum SLEE among all graphs on n vetices with r cut vertices.
Proof . Since P n = G n−2 n is the uniqe graph with n − 2 cut vertices, the case r = n − 2 is obviouse. If r = 0, then by lemma 2.3, K n = G 0 n is the unique graph on n vertices with maximum SLEE. Let 1 ≤ r ≤ n − 3, and G be a graph with maximum SLEE among all graphs on n vertices with r cut vertices. First, we prove that G is connected. Otherwise, if G is not connected and x is a cut vertex of G, then x is also a cut vertex of a component, say G 1 of G. Let G 2 be another component of G. If G 2 has a cut vertex, say y, then set G ′ = G + {xy}. If G 2 has no cut vertex, then suppose that G ′ is the graph obtained from G by attaching x to each vertex of G 2 . It is easy to check that in both cases, G ′ is a graph with r cut vertex and SLEE(G) < SLEE(G ′ ), a contradiction. Thus G is connected.
By lemma 2.3, every block of G is complete. Let x be a cut vertex contained in at least 3 blocks, say B 1 , B 2 and B 3 . Suppose that, B 1 and B 3 will be disjointed if the vertex x is removed. Let G ′ be the graph obtained from G by attaching each vertex of B 1 to each vertex of B 2 . Obviously, G ′ has r cut vertex and by lemma 2.3, SLEE(G) < SLEE(G ′ ), a contradiction. Thus, each cut vertex of G is contained in exactly two blocks. Suppose that G has at least one block with at least 3 vertices. Otherwise, since each block of G has 2 vertices, G is a tree with maximum degree 2. Thus G ∼ = P n , and r = n − 2, a contradiction. Let P s be a pendent path with minimum length in G at x. Obviously, x lies in a block of G, say B, with at least 3 vertices. Note that if s = 1, then x is not a cut vertex. For each y ∈ V (B), let H y be the component of G−E(B) which is containing y. Obviously, H x = P s . Let y ∈ V (B) such that y = x. Let H be the component of G− E(H x )∪E(H y ) containing y. We have G ∼ = H(x)•H x (x) (y)•H y (y). Suppose that H y is not a path. Since P s has minimal length, there is a pendent path on at least s vertices at a vertex in H y , say z, where z = y. Thus H y contains a path on at least s + 2 vertices with an end vertex y. Note that since H y is not a path, we can choose some vertices of H y and construct the path of length at least s + 2 with an end vertex y. By lemma 4.1, we may get another graph on n vertices with r cut vertices, which has larger SLEE, a contradiction. Therefore, H y is a pendent path, say P t at y. By the choice of P s , we have t ≥ s. If t ≥ s + 2, then by lemma 4.1, we may obtain another graph on n vertices with r cut vertices, which has a larger SLEE that G, a contradiction. Therefore, for each y ∈ V (B), H y ∼ = P s or P s+1 . Hence G ∼ = G r n .