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On Maximum $SLEE$ of Graphs with Given Parameters II
\hspace{2ex} $\arrowvert$\hspace{2ex}
Ramin Nasiri et al.}


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\begin{document}

\begin{frontmatter}
% Write your paper title here
\papertitle{On Maximum Signless Laplacian Estrada Index of Graphs with Given Parameters II}

%% use optional labels to link authors explicitly to addresses. The label may be more than one, use comma to separate

%\author[label1,label]{Edy Tri Baskoro${}^1$}

\author[label1]{Ramin Nasiri}
\author[label1]{Hamid Reza Ellahi}
\author[label2]{Gholam Hossein Fath-Tabar}
\author[label1]{Ahmad Gholami}

\address[label1]{\small Department of Mathematics, Faculty of Science, University of Qom, Qom 37161-46611,	I. R. Iran}
\address[label2]{Department of Mathematics, Statistics and Computer Science, Faculty of Science, University of Kashan, Kashan 87317-51167, I. R. Iran

\vspace*{2.5ex} 
 {\normalfont R.Nasiri.82@gmail.com, H.R.Ellahi@stu.qom.ac.ir, Fathtabar@kashanu.ac.ir, A.Gholami@qom.ac.ir}
 }

\begin{abstract}
The signless Laplacian Estrada index of a graph $G$ is defined as $SLEE(G)=\sum^{n}_{i=1}e^{q_i}$ where $q_1, q_2, \ldots, q_n$ are the eigenvalues of the signless Laplacian matrix of $G$. 
Following the previous work in which we have identified the unique graphs with maximum signless Laplacian Estrada index with each of the given parameters, namely, number of cut edges,  pendent vertices, (vertex) connectivity, and edge connectivity, in this paper we continue our characterization for two further parameters: diameter and number of cut vertices.

\let\thefootnote\relax\footnotetext{Received: xx xxxxx 20xx,\quad
  Accepted: xx xxxxx 20xx.\\[3ex]
  }
  
\end{abstract}

\begin{keyword}
% Separate keyword by \sep
Estrada index \sep signless Laplacian Estrada index \sep extremal graph \sep diameter \sep cut vertex

% Write the classification number
Mathematics Subject Classification : 05C12, 05C35, 05C50

\end{keyword}

\end{frontmatter}

%% Main text
\section{Introduction}
Let $G = (V, E)$ be a simple, finite, and undirected graph with vertex set $V(G)$, the edge set $E(G)$, and $|V(G)|=n$. The adjacency matrix $\mathbf{A}=\mathbf{A}(G)=[ a_{ij}]$ of $G$ is the binary matrix, where the element $a_{ij}$ is equal to 1 if vertices $i$ and $j$ are adjacent, and 0 otherwise. The matrices $\mathbf{L}=\mathbf{D}-\mathbf{A}$ and $\mathbf{Q}=\mathbf{D}+\mathbf{A}$, where $\mathbf{D} = diag(d_1,d_2,\ldots,d_n)$ is the diagonal matrix of vertex degrees, are known as the \emph{Laplacian matrix} and \emph{signless Laplacian matrix} of $G$. The spectrum of $\mathbf{Q}$ is denoted by $(q_1, q_2, \ldots, q_n)$.

Spectral graph theory is the study of properties of a graph in relationship to the eigenvalues of matrix $\mathbf{M}$ associated with the graph. This theory is called \emph{$\mathbf{M}$-theory} (see [5--7]
%\cite{Cv1,Cv2,Cv3}
).
The spectral graph theory is widely used in various fields such as
physics, chemistry, computer sciences, and mathematics (see [2--5, 12--16] %\cite{BrPa,ChTr,CoSi,Cv1,Huc,Kas,Kle,MoPo,Nasiri}
).
For some special graphs, $\mathbf{Q}$-theory is equivalent to other theories. 
For example, for regular graphs $\mathbf{Q}$-theory is equivalent to $\mathbf{A}$-theory and $\mathbf{L}$-theory or the matrices $\mathbf{L}$ and $\mathbf{Q}$ are similar if and
only if $G$ is a bipartite graph \cite{Cvetkovic01,Dam,Grone01}.
For studying various graph properties, some evidence is presented that the positive semi-definite matrix $\mathbf{Q}$ might be better suited than the other graph matrices (see  \cite{Dam}).

For a graph $G$, Ayyaswamy et al. \cite{Ayyaswamy01} introduced the innovative notion of the \emph{signless Laplacian Estrada index} as
$$SLEE(G)=\sum^{n}_{i=1}e^{q_i}.$$
They also established lower and upper bounds for $SLEE$ in terms of the number of vertices and edges. Previously in \cite{Elahi01}, we investigated 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 determining the unique graphs with maximum $SLEE$ according to two further parameters: diameter and number of cut vertices.
Our main results are the following two theorems: 
\begin{theorem} \label{t0}
	If $G$ has maximum $SLEE$ with diameter $d$, $2<d<n-1$, then
	$G\cong H_{d,1}$, (see Figure \ref{fig:1}). 
\end{theorem}
%---------figure----------------
\begin{figure}[ht]
	\center
	\begin{tikzpicture}[thick, scale=1.5, x={(0:1cm)}, y={(90:0.7cm)}]
	\foreach \i  in {0,...,6} 
	{
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	\path [fill=gray!30, opacity=0.3] (2.1,1.9)--(3)--(3.9,1.9);
	\path [fill=gray!30, opacity=0.3] (2.2,1.7)--(4)--(3.9,2);
	\draw [ opacity=0.75] (2.1,1.9)--(3)--(3.9,1.9);
	\draw [ opacity=0.75] (2.1,2)--(2)--(3.8,1.7);
	\draw [ opacity=0.75] (2.2,1.7)--(4)--(3.9,2);
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	\path [fill=white!100, opacity=1]  (3,2) ellipse (0.9 and 0.6)node {$K^{}_{n-d-1}$};
	\filldraw (0) circle (0pt) node [anchor=north]{$v^{}_{0}$};
	\filldraw (1) circle (0pt) node [anchor=north]{$v^{}_{1}$};
	\filldraw (2) circle (0pt) node [anchor=north]{$v^{}_{\ddd-1}$};
	\filldraw (3) circle (0pt) node [anchor=north]{$v^{}_{\ddd}$};
	\filldraw (4) circle (0pt) node [anchor=north]{$v^{}_{\ddd+1}$};
	\filldraw (5) circle (0pt) node [anchor=north]{$v^{}_{d-1}$};
	\filldraw (6) circle (0pt) node [anchor=north]{$v^{}_{d}$};
	\filldraw [ opacity=0.75](0) circle (1.5pt);
	\filldraw [ opacity=0.75](1) circle (1.5pt);
	\filldraw [ opacity=0.75](2) circle (1.5pt);
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	\draw [ opacity=0.75](2)--(4);
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	\draw [ opacity=0.6, dashed] (4)--(5);
	\end{tikzpicture}
	%\includegraphics[width =0.5  \textwidth]{p2.eps}
	\caption{Graph $H_{d,1}$.}
	\label{fig:1}
\end{figure}
%-----------------------------------------------------------
\begin{theorem} \label{t1}
	If $G$ has maximum $SLEE$ on $n$ vertices with $r$ cut vertices, $0\leq r\leq n-2$, then $G\cong G^r_n$, where $G^r_n$ is the graph obtained from $K_{n-r}$ by attaching $n-r$ pendent path of orders $n_1, n_2, \ldots, n_{n-r}$ to its vertices; such that each vertex of $K_{n-r}$ has exactly one pendent path and also $\mid n_i-n_j\mid \leq 1$ for $1\leq i,j\leq n-r$. 
	More precisely, each pendent path is of order $\lfloor \frac{r}{n-r} \rfloor$ or $\lfloor \frac{r}{n-r} \rfloor+ 1$. 
	For example, the graphs $G^r_6$ with $r=0,1,2,3,4$ are shown in Figure \ref{fig:2}.
\end{theorem}
%==============================
\begin{figure}[ht]
	\center
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	\foreach \i in {1,...,5}
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			\draw (\i)--(\j);
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	\begin{scope}[shift={(3, 0)}]
	\foreach \i in {1,...,6}
	{
		\coordinate (\i) at (\i*60-90:1);
		\filldraw (\i) circle (1.5 pt);
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	\foreach \i in {1,...,4}
	{
		\foreach \j in {2,...,5}
		{
			\draw (\i)--(\j);
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	\draw (1)--(6);
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	\foreach \i in {1,...,6}
	{
		\coordinate (\i) at (\i*60-90:1);
		\filldraw (\i) circle (1.5 pt);
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	\draw (0,-1.5) circle (0pt) node [anchor=north] {$G^{2}_{6}$};
	\foreach \i in {1,2,4,5}
	{
		\foreach \j in {2,4,5}
		{
			\draw (\i)--(\j);
		}
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	\draw (1)--(6);
	\draw (3)--(4);
	\end{scope}
	\begin{scope}[shift={(9, 0)}]
	\foreach \i in {1,...,6}
	{
		\coordinate (\i) at (\i*60-90:1);
		\filldraw (\i) circle (1.5 pt);
	}
	\draw (0,-1.5) circle (0pt) node [anchor=north] {$G^{3}_{6}$};	
	\draw (1)--(3)--(5)--(1);		
	\draw (1)--(6);
	\draw (3)--(2);
	\draw (5)--(4);
	\end{scope}
	\begin{scope}[shift={(12, 0)}]
	\foreach \i in {1,...,6}
	{
		\coordinate (\i) at (\i*60-90:1);
		\filldraw (\i) circle (1.5 pt);
	}
	\draw (0,-1.5) circle (0pt) node [anchor=north] {$G^{4}_{6}$};	
	\draw (2)--(3)--(4)--(5)--(6)--(1);		
	\end{scope}
	\end{tikzpicture}
	\caption{The graphs $G^r_6$ with $r=0,1,2,3,4$ .}
	\label{fig:2}
\end{figure}
%================================

\section{Preliminaries and Lemmas}

In this section, initially, basic definitions, notations, and concepts used in the study are introduced and some findings proved in \cite{Cvetkovic01, Elahi01} are restated as well. Then, relevant propositions required to prove the results reported in the
next sections are given and proved. 



%===========================================
\begin{definition}\cite{Cvetkovic01}
	A \emph{semi-edge walk} of length $k$ in graph $G$ is an alternating sequence 
	$W=v_1 e_1 v_2 e_2 \ldots  v_k e_k v_{k+1}$, where $v_1, v_2, \ldots , v_k, v_{k+1}\in V(G)$, and $e_1, e_2, \ldots , e_k\in E(G)$ such that  the vertices $v_i$ and $v_{i+1}$ are (not necessarily distinct)  end points of edge $e_i$, for any $i=1, 2, \ldots , k$.
	If $v_1=v_{k+1}$, then we say $W$ is a \emph{closed semi-edge walk}.
\end{definition}
%===========================================

By following \cite{Elahi01}, we denote the $k$-th signless Laplacian spectral moment of the graph $G$ by $T_k(G)$ , i.e., $T_k(G)=\sum^{n}_{i=1}q^{k}_{i}$.
%===========================================
\begin{theorem} \cite{Cvetkovic01}\label{theo22}
	For a graph $G$, the signless Laplacian spectral moment $T_k$ is equal to the number of closed semi-edge walks of length $k$.
\end{theorem} 
%===========================================
Note that, by Taylor expansions, we have
\begin{equation*}\label{eq001}
	SLEE(G)=\sum_{k\geq 0}\frac{T_k(G)}{k!}.
\end{equation*}
Bearing this relation in mind, one can find that for two $n$-vertex graphs $G$ and $H$, if  $T_k(G)\geq T_k(H)$ for all $k\geq 0$, then $SLEE(G)\geq  SLEE(H)$. So, to compare the signless Laplacian Estrada indices of two graphs, we can compare their signless Laplacian spectral moments.

By $(G;v,u)\qs(G';v',u')$ we mean $ |SW_k (G;v,u)|\leq|SW_k(G'; v', u')|$, for any $k\geq 0$.\\
 Moreover, if $(G;v,u)\qs(G';v',u')$ and there exists some $k_0$ such that we have $|SW_{k_0}(G;v,u)|<|SW_{k_0} (G';v',u')|$, then we write $(G;v,u)\s(G';v',u')$. Let $SW_k(G;v)=SW_k(G;v,v)$.
Similarly, we may define $(G;v)\qs(G';v')$ and $(G;v)\s(G';v')$.
By the above notations, we have 
$$T_k(G)=\sum_{v\in V(G)}|SW_k (G;v)|.$$
The notation $G\qs G'$ means that $T_k(G)\leq T_k(G')$, for each $k\geq 0$. If $G\qs G'$ and for some $k_0$, $T_{k_0}(G)< T_{k_0}(G')$, then we use the notation $G\s G'$.
Also, if $T_k(G)=T_k(G')$, for each $k\geq 0$, then we write $G=_s G'$.
%===========================================
\begin{lemma} \cite{Elahi01}\label{lem01}
	Let $G$ be a graph. If an edge $e$ does not belong to $E(G)$, then $G\s G+e$, thus $SLEE(G)< SLEE(G+e)$.
\end{lemma}
%===========================================
\begin{lemma} \cite{Elahi01}\label{lem02}
	Let $G$ be a graph and $v, u, w_1, w_2, \ldots , w_r\in V(G)$.
	Suppose that
	$E_v=\{e_1=vw_1, \ldots , e_r=vw_r\}$ and
	$E_u=\{e'_1=uw_1, \ldots ,  e'_r=uw_r\}$ are subsets of edges of the complement of $G$.
	Let $G_u=G+E_u$ and $G_v=G+E_v$.
	If $(G;v)\s (G;u)$ and $(G;w_i,v)\qs (G;w_i,u)$ for each $i=1,2,\ldots,r$,
	then $G_v\s G_u$, thus $SLEE(G_v)<SLEE(G_u)$.
\end{lemma}
For a vertex $v$ and an edge $e$, let $SW_k(G;v, [e])$ be the set of all closed semi-edge walks of length $k$ in the graph $G$,  starting at vertex $v$ and containing  $e$. 
%===========================================
\begin{lemma}\label{lem03}
	Let $G$ be a graph and $H=G+e$, such that $e=uv\in E(\overline{G})$. If $(G;v)\qs (G;u)$, then $(H;v)\qs (H;u)$. Moreover, if $(G;v)\s (G;u)$, then $(H;v)\s (H;u)$.
\end{lemma}
\begin{proof}
	We know that for each $z\in \{u,v\}$ and $k\geq 0$,
	$$|SW_k(H;z)|=|SW_k(G;z)|+|SW_k(H;z,[e])|.$$
	Since $(G;v)\qs (G;u)$, $|SW_k(G;v)|\leq |SW_k(G;u)|$, for each $k\geq 0$. Thus, there is a bijection $f_k:SW_k(G;v)\to A_k\ssq SW_k(G;u)$, for each $k\geq 0$.
	\\
	It is enough to show that $|SW_k(H;v,[e])|\leq |SW_k(H;u,[e])|$, for each $k\geq 0$. Let $W\in SW_k(H;v,[e])$. We can uniquely decompose $W$ to $W=W_1eW_2e\dots eW_r$, such that $W_i\in SW_{k_i}(G;x,y)$, where $x,y\in \{u,v\}$, $k_i\geq 0$ and $1\leq i\leq r$. Note that $W_i$ is a semi-edge walk in $G$ and does not contain $e$, thus the decomposition is unique. For each $W_i$ , exactly one of the following cases occurs:
	\begin{enumerate}
		\item[1)] $W_i\in SW_{k_i}(G;v,v)$. In this case, we set $h(W_i)=f_{k_i}(W_i)$. Thus, $h(W_i)\in A_{k_i}\ssq SW_{k_i}(G;u,u)$.
		\item[2)] $W_i\in A_{k_i}\ssq SW_{k_i}(G;u,u)$. In this case, set $h(W_i)=f^{-1}_{k_i}(W_i)\in SW_{k_i}(G;v,v)$.
		\item[3)] $W_i\in SW_{k_i}(G;u,u)\setminus A_{k_i}$, or $W_i\in SW_{k_i}(G;u,v)$, or $W_i\in SW_{k_i}(G;v,u)$. In these cases, let $h$ fix $W_i$, i.e. $h(W_i)=W_i$.
	\end{enumerate}
	Now, it is easy to show that the map $h:SW_k(H;v,[e])\to SW_k(H;u,[e])$ by the rule $h_{k}(W)=h_{k}(W_1eW_2e\ldots W_r)=h(W_1)eh(W_2)e\ldots eh(W_r)$ is an injection. 
	
	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 
	$$|SW_{k_0}(H;v,[e])|<|SW_{k_0}(G;u,[e])|$$
	which implies that $(H;v)\s(H;u)$.
\end{proof}

\begin{lemma}\label{lem04}
	Let $G$ be a graph and $H=G+e$, such that $e=uv\in E(\overline{G})$ and $(G;v)\qs (G;u)$.
	If there exists a vertex $x\in V(G)$ such that $(G;x,v)\\
	\qs (G;x,u)$, then $(H;x,v)\qs (H;x,u)$.
	Moreover, if $(G;v)\s (G;u)$ or $(G;x,v)\s (G;x,u)$, then $(H;x,v)\s (H;x,u)$.
\end{lemma}
\begin{proof}
	Since $(G;v)\qs (G;u)$, there is a bijection $f_k:SW_k(G;v)\to A_k\ssq SW_k(G;u)$, for each $k\geq 0$. Similarly, since $(G;x,v)\qs (G;x,u)$, there is a bijection \\
	$g_k:SW_k(G;x,v) \to B_k\ssq SW_k(G;x,u)$, for each $k\geq 0$.
	It is obvious that for each $k\geq 0$,
	$$|SW_k(H;x,z)|=|SW_k(G;x,z)|+|SW_k(H;x,z,[e])|$$
	where $z\in \{v,u\}$. It is enough to show that for each $k\geq 0$,
	$$|SW_k(H;x,v,[e])|\leq |SW_k(H;x,u,[e])|.$$
	Let $W\in SW_k(H;x,v,[e])$.
	$W$ can be decomposed uniquely to $W_1eW_2e\ldots eW_r$, where $W_i$ is a semi-edge walk of length $k_i$ in $G$.
	Three cases will be considered as follows for $W_1$:
	\begin{enumerate}
		\item[1)] If $W_1\in SW_{k_1}(G;x,v)$, set $h_1(W_1)=g^{}_{k_1}(W_1)\in B_{k_1}\ssq SW_{k_1}(G;x,u)$.
		\item[2)] If $W_1\in B_{k_1}\ssq SW_{k_1}(G;x,u)$, set $h_1(W_1)=g^{-1}_{k_1}(W_1)\in SW_{k_1}(G;x,v)$.
		\item[3)] If $W_1\in SW_{k_1}(G;x,u)\setminus B_{k_1}$, set $h_1(W_1)=W_1$.
	\end{enumerate}
	If $1<i\leq r$, then three cases will be considered as follows for $W_i$:
	\begin{enumerate}
		\item[1)] If $W_i\in SW_{k_i}(G;v)$, then set $h_i(W_i)=f_{k_i}(W_i)\in A_{k_i}$.
		\item[2)] If $W_i\in A_{k_i}\ssq SW_{k_i}(G;u)$, then set $h_i(W_i)=f^{-1}_{k_i}(W_i)\in SW_{k_i}(G;v)$.
		\item[3)] If $W_i\in SW_{k_i}(G;u)\setminus A_{k_i}$, $W_i\in SW_{k_i}(G;v,u)$ or $W_i\in SW_{k_i}(G;u,v)$, then set $h_i(W_i)=W_i$.
	\end{enumerate}
	One can  easily see that the map $h_k:SW_k(H;x,v,[e])\to SW_k(H;x,u,[e])$ by the rule $h_{k}(W)=h_{k}(W_1eW_2e\ldots W_r)=h_1(W_1)eh_2(W_2)e\ldots eh_r(W_r)$ is injective.
	
	The second part of the lemma is clear.
\end{proof}
%================================

\section{The proof of Theorem \ref{t0}}

For $x\in V(G)$, the \emph{eccentricity} $e(x)$ of $x$ is defined as $e(x)=max\lbrace d(x,y) : y\in V(G)\rbrace$.
The \emph{diameter} $d(G)$ is the maximum eccentricity over all vertices, whereas the \emph{radius} $r(G)$ is the minimum eccentricity. Also, $x$ is a \emph{central vertex} if  $e(x)=r(G)$ and a \emph{diametrical path} is a shortest path between  two vertices whose distance is equal to $d(G)$.
For the sake of convenience, we denote $\lceil \frac{d}{2}\rceil$ by $\ddd$, which is the smallest integer number greater than $\frac{d}{2}$.
\\
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 the greatest signless Laplacian spectral moments, and so the maximum $SLEE$, with diameter $2$, where $e$ is an edge of $K_n$. 

%===========================================
\begin{lemma}\label{lem05}
	Let $G$ be a graph with diameter $d$ and $P_{d+1}=v_0v_1\ldots v_d$ be a diametrical path in $G$. If $d\geq 2$ and $x\in V(G)\setminus V(P_{d+1})$, then $x$ has at most $3$ neighbors in  $V(P_{d+1})$.
\end{lemma}
\begin{proof}
	Suppose that  $x$ has neighbors $v_{i_1},v_{i_2},\ldots , v_{i_{r}}$ in $P_{d+1}$, where $r>3$ and $i_1<i_2<\ldots<i_r$. Since $i_r-i_1>2$, the path $P'=v_0v_1\ldots v_{i_1}xv_{i_r}v_{i_r+1}\ldots v_d$ from $v_0$ to $v_d$ has a length $d-i_r+i_1+2<d$, which  is a contradiction.
\end{proof}
%===========================================
Let $n>4$, $2<d<n-1$, and $1\leq j\leq \ddd$.
We denote by $\mathcal{H}_{d,j}$ the set of  all graphs $H_{d,j}$; each of these members are constructed from $K_{n-1-d}$ and $P_{d+1}=v_0v_1\ldots v_d$ by attaching each vertex of $K_{n-d-1}$ to exactly 3 vertices of $P_{d+1}$, such that  for each $x\in V(K_{n-d-1})$ there exists an index $i$, where $\ddd-j\leq i\leq \ddd +j-2$, and $x$ is attached to $v_i$, $v_{i+1}$, and $v_{i+2}$.
Therefore, none of vertices $v_i$, where $0\leq i<\ddd-j$ or $\ddd +j<i\leq d$, has a neighbor in  $K_{n-d-1}$.
Note that $v_{\ddd}$  is a central vertex of the path $P_{d+1}$. For example, all graphs $H_{4,2}$ with $n=7$  are shown in Figure \ref{fig:3}.
%===========================================
\begin{figure}[ht]
	\center
	\begin{tikzpicture}[thick, scale=1, x={(0:1cm)}, y={(90:0.8cm)}]
	\foreach \i in {0,1,2}
	{
		\foreach \j in {0,1,2}
		{
			\begin{scope}[shift={(\i*5, -2.5*\j)}]
			\foreach \k in {0,1,2,3, 4}
			{
				\coordinate (\k) at (\k,0);
			}
			\coordinate (5) at (1,1);
			\coordinate (6) at (2,1);
			\foreach \k in {0,1, 2,3,4}
			{
				\filldraw (\k) circle (1.5pt) node [anchor=north]{$v^{}_\k$};
			}
			\filldraw (5) circle (1.5pt);
			\filldraw (6) circle (1.5pt);
			\draw (0)--(1)--(2)--(3)--(4);
			\draw (5)--(6);
			\foreach \k in {0,1,2}
			{
				\draw (5)-- +(\j+\k-1,-1)	;
				\draw (6)-- +(\i+\k-2,-1)	;		
			}
			\end{scope}
		}
	}
	
	\end{tikzpicture}
	\caption{All graphs $H_{4,2}$ with $n=7$.}
	\label{fig:3}
\end{figure}
%==============================
\begin{lemma}\label{lem06}
	Let $n>4$, $2<d<n-1$, and $2\leq j\leq \ddd $. If $H_j\in \mathcal{H}_{d,j}$, then either $H_j\in \mathcal{H}_{d,j-1}$, or there exists a graph, say $H_{j-1}\in \mathcal{H}_{d,j-1}$, such that $H_j\s H_{j-1}$, resulting $SLEE(H_j)<SLEE(H_{j-1})$. 
\end{lemma}
%-------------------------------------------------------------------------------------
\begin{proof}
	Let $H_j\in\mathcal{H}_{d,j}$ and $N_K(v_i)=N(v_i)\cap V(K_{n-1-d})$, where $0\leq i\leq d$ and $N(v_i)$ is the set
	of vertices that are adjacent to $v_i$.
	For a better understanding of the proof, our argument is divided into two parts; that is, first we discuss $N_K(v^{}_{\ddd-j})$ and then, proceed to  $N_K(v^{}_{\ddd+j})$.  
	Note that if $d$ is odd, $j=2$,
	and $N_K(v^{}_{\ddd+2})= \emptyset$, then by renaming the vertices of $P_{d+1}$ such that $v_i$ changes to $v^{}_{d-i}$,
	we conclude that $H_j\in \mathcal{H}_{d,j-1}$.
	Let $H_j\not\in \mathcal{H}_{d,j-1}$. Therefore, either at least one of the  subsets $N_K(v^{}_{\ddd-j})$ or $N_K(v^{}_{\ddd+j})$ is not empty, or $d$ is odd and $j=2$ and  $N_K(v^{}_{\ddd+2})$  is not empty.  
	
	If $N_K(v^{}_{\ddd-j})=\emptyset$, then we set $H'_{j-1}=H_j$.
	In this case, we have $H'_{j-1}=_s H_j$.
	Let $N_K(v^{}_{\ddd-j})\neq \emptyset$. For convenience, suppose that $v=v^{}_{\ddd-j}$, $y=v^{}_{\ddd-j+1}$, $z=v^{}_{\ddd-j+2}$, and $u=v^{}_{\ddd-j+3}$.
	By the definition of $\mathcal{H}_{d,j}$, it is obvious that $N_K(v)\ssq N_K(y)\ssq N_K(z)$ and $N_K(v)\cap N_K(u)=\emptyset$.
	
	Let $E=\{vx:x\in N_K(v)\}$, $E'=\{ux:x\in N_K(v)\}$, $H'_j=H_j-E$, and $H'_{j-1}=H'_j+E'$.
	
	By lemma \ref{lem02}, in order to show that $H_j\s H'_{j-1}$, it is enough to  prove the following statements:
	\begin{enumerate}
		\item[1)] $(H'_j;v)\s (H'_j;u)$.
		\item[2)] $(H'_j;x,v)\qs (H'_j;x,u)$, for each $x\in N_K(v)$.
	\end{enumerate}
	In order to prove (1), we begin with the following claim: \\
	{\bf Claim.} $(H'_j;y)\qs (H'_j;z)$:
	
	To prove the claim, let $W\in SW_k(H'_j-e;y)$, where $e=yz$ and $k\geq 0$.
	We can decompose $W$ to $W=W_1W_2W_3$, where $W_1$ and $W_3$ are as long as possible and consist of just the vertices $v_0,v_1,\ldots,y$ and edges in $\{v_tv_{t+1}: 0\leq t\leq \ddd-j\}\cup \{yx:x\in N_K(y)\}$;
	where also $W_2\in SW_{k_2}(H'_j-e;x,w)$, such that $x,w\in N_K(y)\ssq N_K(z)$.
	Suppose that $W'_i$ is obtained from $W_i$, for $i=1,3$, by replacing each vertex $v_t$ with $v_a$,  each edge $v_tv_{t+1}$ with $v_av_{a-1}$, and each edge $yx$ with $zx$; where $x\in N_K(y)$ and $a=2\ddd-2j-t+3$ (in fact the distance between vertices $v_t$ and $y$ is equal to the distance between vertices $v_a$ and $z$ in $P_{d+1}$).
	\\
	It is easy to show that the map $f'_k:SW_k(H'_j-e;y)\to SW_k(H'_j-e;z)$, defined by the rule $f'_k(W_1W_2W_3)=W'_1W_2W'_3$, is injective; thus, $(H'_j-e;y)\qs (H'_j-e;z)$. Now, the claim follows from lemma \ref{lem03}.
	
	For each $k\geq 0$, let $f_k:SW_k(H'_j;y)\to SW_k(H'_j;z)$ be an injection. If $W\in SW_k(H'_j;v)$, then $W$ can be decomposed to $W=W_1W_2W_3$, where $W_2\in SW_{k_2}(H'_j;y)$ is as long as possible . Let $W'_i$ be obtained form $W_i$ for each $i=1,3$; by replacing each vertex $v_t$ with $v_a$ and each edge $v_tv_{t+1}$ with $v_av_{a-1}$, where  $a=2\ddd-2j-t+3$.
	The map $g_k:SW_k(H'_j;v)\to SW_k(H'_j;u)$, defined by the rule $g_k(W_1W_2W_3)=W'_1f_{k_2}(W_{k_2})W'_3$, is injective.
	Note that if $j>2$ or $d$ is even, then the path $v_0v_1\ldots v$ is a proper subgraph of the path $v_dv_{d-1}\ldots u$. Also, if $d$ is odd and $j=2$, then  $N_K(u)\neq \emptyset$, which implies that $deg^{}_{H'_j}(v)=2<deg^{}_{H'_j}(u)$. Therefore, $(H'_j;v)\s (H'_j;u)$, which is (1).
	
	We use a similar procedure to prove statement (2): First, we claim that:\\
	{\bf Claim.} For each $x\in N_K(v)$, $(H'_j;x,y)\qs (H'_j;x,z)$.
	
	In order to prove the claim, let $x\in N_K(v)$ and $W\in SW_k(H'_j-e;x,y)$, where $e=yz$. We can decompose $W$ to $W=W_1W_2$, such that $W_1\in SW_{k_1}(H'_j-e;x,w)$ is as long as possible, where $w\in N_K(y)$ and $W_2\in SW_{k_2}(H'_j-e;w,y)$. Suppose that $W'_2$ is obtained from $W_2$ by replacing each vertex $v_t$ with $v_a$, the edge $wy$ with $wz$ and each edge $v_tv_{t+1}$ with $v_av_{a-1}$, where  $a=2\ddd-2j-t+3$.
	\\
	One can easily see that the map $h'_k:SW_k(H'_j-e;x,y)\to SW_k(H'_j-e;x,z)$, defined by the rule $h'_k(W_1W_2)=W_1W'_2$, is injective. 
	Resulting, $(H'_j-e;x,y)\qs (H'_j-e;x,z)$. Now, the claim is obtained from lemma \ref{lem04}.
	
	Consider $h_k:SW_k(H'_j;x,y)\to SW_k(H'_j;x,z)$ is an injective map for each $k\geq 0$. Let $W\in SW_k(H'_j;x,v)$. We can decompose $W$ to $W=W_1W_2$, where $W_1\in SW_{k_1}(H'_j;x,y)$ and is as long as possible; where also $W_2\in SW_{k_2}(H'_j;y,v)$. Let $W'_2$ be obtained from $W_2$ by replacing each vertex $v_t$ with $v_a$ and each edge $v_tv_{t+1}$ with $v_av_{a-1}$, where $a=2\ddd-2j-t+3$.
	It is elementary to show that the map $l_k:SW_k(H'_j:x,v)\to SW_k(H'_j;x,u)$, defined by the rule $l_k(W_1W_2)=h_{k_1}(W_1)W'_2$, is an injection.
	Thus, $(H'_j;x,v)\qs (H'_j;x,u)$ for each $x\in N_K(v)$, from which statement (2) follows. 
	\\
	Now, by the above discussion and lemma  \ref{lem02}, we have
	$H_j\qs H'_{j-1}$,  with equality if and only if $H'_{j-1}\cong H_j$.
	The first part of the argument ends here.
	
	If  $N_K(v^{}_{\ddd+j})$ is empty or $d$ is odd and $j=2$, then $H'_{j-1}\in \mathcal{H}_{d,j-1}$. In this case, set $H_{j-1}=H'_{j-1}$ and of course $H_{j-1}=_s H'_{j-1}$.
	Let $H'_{j-1}\not\in \mathcal{H}_{d,j-1}$, then $N_K(v^{}_{\ddd+j})$ is not empty. 
	By repeating the above discussion for $v=v^{}_{\ddd+j}$, $y=v^{}_{\ddd+j-1}$, $z=v^{}_{\ddd+j-2}$, and $u=v^{}_{\ddd+j-3}$, we get the graph $H_{j-1}=H'_{j-1}-E+E'$, such that $H_{j-1}\in \mathcal{H}_{d,j-1}$ and  $H'_{j-1}\s H_{j-1}$.
	Therefore,$$H_j\qs H'_{j-1}\qs H_{j-1}\in \mathcal{H}_{d,j-1},$$
	with equalities holding, if and only if graphs are isomorphic.
\end{proof}
%===========================================

Now, we may prove the main result of this section.\\[3mm]
{\bf Proof  of Theorem \ref{t0}.}
Suppose that $G$ is a graph, having the greatest signless Laplacian spectral moments, and so the maximum $SLEE$, with diameter $d$.
Let $P_{d+1}=v_0v_1\ldots v_d$ be a diametrical path in $G$, and $H$ be the graph obtained from $G$ by adding some edges such that:
\begin{enumerate}
	\item[(a)] For each $x\in V(G)\setminus V(P_{d+1})$, $x$ is a neighbor of exactly 3 vertices of $P_{d+1}$ in $H$, say $v_i$, $v_{i+1}$ and $v_{i+2}$.
	\item[(b)] $H-V(P_{d+1})$ is a complete graph on $n-1-d$ vertices.
\end{enumerate}
By lemma \ref{lem05}, such a graph $H$ exists. Obviously, we have $H\in \mathcal{H}_{d,j}$ for some $j$, where $1\leq j\leq \ddd$ and $G\qs H$, with equality if and only if $G\cong H$.

If $j>1$, then by lemma \ref{lem06}, we may get a sequence of graphs, say \\
$H_{d,j-1}, H_{d,j-2}, \ldots , H_{d,1}$, such that for each $t$, $H_{d,t}\in \mathcal{H}_{d,t}$ and 
$$G\qs H\qs H_{d,j-1}\qs H_{d,j-2}\qs \ldots \qs H_{d,1},$$
with equality if and only if the graphs are isomorphic.
Since the diameter of $H_{d,1}$ is $d$ and $G$ has the greatest signless Laplacian spectral moments among the set of all graphs with diameter $d$,  $G=_s H_{d,1}$ which implies that $G\cong H_{d,1}$, as expected.
\hfill{$\square$}

\section{The proof of Theorem \ref{t1}}

A \emph{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 \emph{block} of $G$ is defined to be a maximal subgraph without cut vertices.\\
A \emph{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 $deg^{}_G(x)\geq3$.
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 $x \in V (G)$
and $y \in V (H)$. We denote the \emph{coalescence} of $G$ and $H$ by $G(x) \circ H(y)$, which
is obtained by identifying the vertex $x$ of $G$ with the vertex $y$ of $H$.

\begin{lemma}\label{lem07}
	Let $H_1$ and $H_2$ be two graphs, $P_s=y_0y_1\ldots y_{s-1}$ be a path on $s$ vertices, $u\in V(H_2)$, and $xy\in E(H_1)$ such that $x\neq y$.
	Let $G=\big(H_1(y)\circ P_s(y_0)\big)(x)\circ H_2(u)$. 
	If $H_2$ contains a path $Q_{s+2}=ux_1x_2\ldots x_{s+1}$, then $G\s G-E_y+E_{x_1}$ --  thus $SLEE(G)< SLEE(G-E_y+E_{x_1})$ --
	where $E_y=\big\{yw:w\in N_{H_1}(y)\setminus \{x\}\big\}$, $E_{x_1}=\big\{x_1w:w\in N_{H_1}(y)\setminus \{x\}\big\}$, and $N_{H_1}(y)$ is the set
	of vertices of $H_1$ that are neighbors of $y$ (see Figure \ref{fig:4}).
\end{lemma}
\begin{figure}[h]
	\center
	\begin{tikzpicture}[thick, scale=0.92, x={(0:1.1cm)}, y={(90:1.1cm)}]
	\foreach \sh/\tx in {0/G,1/G-E_{y}+E_{x_1}}
	{
		\begin{scope}[shift={(6*\sh+6, 0)}]
		\foreach \i in {1,2,3,4}
		{
			\coordinate (w\i) at ([shift={(2.6+\sh, -0.8)}]0:\i*0.35);				
		}		
		\foreach \i in {1,2,3}
		{
			\coordinate (x\i) at ([shift={(5, 0)}]135:\i*0.85);		
		}	
		\foreach \i in {0,...,5}
		{
			\coordinate (\i) at (\i,0);
		}	
		\path[shift={(\sh+3, 0)}, rotate=90, fill=gray!20, opacity=0.5] (0,0) arc (45:315:0.8 and 0.7); 
		\draw [shift={(\sh+3, 0)}, rotate=90,](0,0) arc (45:315:0.8 and 0.7); 
		
		\path [shift={(5, 0)}, fill=gray!20, opacity=0.5, rotate=-60](0,0) arc (15:310:1.5 and 0.825); 
		\draw [shift={(5, 0)}, rotate=-60](0,0) arc (15:310:1.5 and 0.825); 
		
		\draw (5)--(x1);
		\draw [dashed] (x1)--(x2);
		\draw (x2)--(x3);
		\filldraw (x1) circle (1.5pt) node[anchor=west]{$x_{2}$};
		\filldraw (x2) circle (1.5pt) node[anchor=west]{$x_{s}$};
		\filldraw (x3) circle (1.5pt) node[anchor=west]{$x_{s+1}$};	
		\filldraw (0) circle (1.5pt) node[anchor=south]{$y_{s-1}$};
		\filldraw (1) circle (1.5pt) node[anchor=south]{$y_{s-2}$};
		\filldraw (2) circle (1.5pt) node[anchor=south]{$y_{1}$};
		\filldraw (3) circle (1.5pt) node[anchor=south]{$y$};
		\filldraw (4) circle (1.5pt) node[anchor=south]{$x$};
		\filldraw (5) circle (1.5pt) node[anchor=west]{$x_{1}$};
		\draw (0)--(1);
		\draw [dashed] (1)--(2);
		\draw (2)--(5);
		\foreach \i[evaluate=\k as \k using int(2*\sh+3)] in {1,2,4}
		{
			\filldraw (w\i) circle (1.5pt);	
			\draw (w\i)--(\k);
		}
		\draw (w3) circle (0pt) node {$\dots$};
		\draw (2.5,-2) circle (0pt) node {$\tx$};
		\end{scope}
	}
	
	\end{tikzpicture}
	\caption{An illustration of graphs in Lemma \ref{lem07}.}
	\label{fig:4}
\end{figure}
\begin{proof}
	Let $G'=G-E_y$. By lemma \ref{lem02}, it is enough to show that $(G';y)\s(G';x_1)$ and $(G';w,y)\qs(G';w,x_1)$ for each $w\in  N_{H_1}(y)\setminus \{x\}$.
	Let $P'_{s+1}=xy_0y_1\ldots y_{s-1}$, $A_k=SW_k(G';y)\setminus SW_k(P'_{s+1};y)$, and $B_k=SW_k(G';x_1)\setminus SW_k(Q_{s+2};x_1)$.
	Since $P'_{s+1}$ is a proper subgraph of $Q_{s+2}$, it is easy to show that
	$|SW_k(p'_{s+1};y)|\leq |SW_k(Q_{s+2};x_1)|$ and inequality is strict for some $k=k_0\geq s$.
	\\
	Let $W\in A_k$. 
	We can decompose $W$ to $W_1W_2W_3$, such that $W_2\in SW_{k_2}(G';x)$ and is as long as possible; also $W_1\in SW_{k_1}(G';y,x), W_3\in SW_{k_3}(G';x,y)$ and $k=k_1+k_2+k_3$.
	Let $W'_j$ be obtained from $W_j$ by replacing each $y_i$ with $x_{i+1}$, where $j=1,3$ and $i=0,1,\ldots , s-1$.
	The map $f:A_k\to B_k$, defined by the rule $f(W_1W_2W_3)=W'_1W_2W'_3$, is injective, and thus, $|A_k|\leq |B_k|$. 
	Therefore, $|SW_k(G';y)|\leq |SW_k(G';x_1)|$ and for some $k=k_0$ the inequality is strict.
	Hence, $(G';y)\s (G';x_1)$.
	
	Let $w\in  N_{H_1}(y)\setminus \{x\}$ and $W\in SW_k(G';w,y)$.
	We can decompose $W$ uniquely to $W_1W_2$, such that $W_1\in SW_{k_1}(G';w,x)$ is as long as possible.
	Let $W'_2$ be obtained from $W_2$ by replacing each $y_i$ with $x_{i+1}$, where $W_2\in SW_{k_2}(G';x,y)$,
	$k=k_1+k_2$, and $i=0,1,\ldots , s-1$.\\
	The map $g_{w,k}:SW_k(G';w,y)\to SW_k(G';w,x_1)$, defined by the rule\\  $g_{w,k} (W_1W_2) = W_1W'_2$, is injective.
	Thus, $|SW_k(G';w,y)|\leq |SW_k(G';w,x_1)|$ for each $k$.
	Therefore,
	$(G';w,y)\qs (G';w,x_1)$ for each $w\in  N_{H_1}(y)\setminus \{x\}$.
\end{proof}

Now, we get to the most important proof of this section.\\[3mm]
{\bf Proof  of Theorem \ref{t1}.}
Since $P_n=G_n^{n-2}$ is the unique graph with $n-2$ cut vertices, the case $r=n-2$ is obvious.
If $r=0$, then by lemma \ref{lem01}, $K_n=G_n^0$ is the unique graph on $n$ vertices with the greatest signless Laplacian spectral moments, and also maximum $SLEE$.
Let $1\leq r\leq n-3$ and $G$ be a graph with the greatest signless Laplacian spectral moments among all graphs on $n$ vertices with $r$ cut vertices.
\\
First, we prove that $G$ is connected,
for 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 show that in both cases $G'$ is a graph with $r$ cut vertices and $G\s G'$, a contradiction. Thus, $G$ is connected.

By lemma \ref{lem01}, every block of $G$ is complete.
Let $x$ be a cut vertex contained in at least $3$ blocks, say $B_1$, $B_2$ and $B_3$.
Assume that $B_1$ and $B_3$ are 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 vertices, and by lemma \ref{lem01} $G\s G'$, 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, $G\cong P_n$ 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 $y\in V(B)$, let $H_y$ be the component of $G-E(B)$ which contains $y$. 
Obviously, $H_x=P_s$.
Let $y\in V(B)$ such that $y\neq x$.
Let $H$ be the component of $G-\big(E(H_x)\cup E(H_y)\big)$ containing $y$.
We have $G\cong \big(H(x)\circ H_x(x)\big)(y)\circ H_y(y)$.\\
Suppose that $H_y$ is not a path. 
Since $P_s$ has minimum length, there is a pendent path on at least $s$ vertices at a vertex in $H_y$, say $z$, where $z\neq y$.
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 \ref{lem07}, we may get another graph on $n$ vertices with $r$ cut vertices, which has greater signless Laplacian spectral moments, a contradiction.
Therefore,  $H_y$ is a pendent path, say $P_t$ at $y$.
Bearing in mind the choice of $P_s$, we have $t\geq s$.
If $t\geq s+2$, then by lemma \ref{lem07}, we can obtain another graph on $n$ vertices with $r$ cut vertices, which has greater signless Laplacian spectral moments than $G$, a contradiction. 
Therefore, for each $y\in V(B)$, $H_y\cong P_s$ or $P_{s+1}$.
Hence, $G\cong G_n^r$.
\hfill{$\square$}

\section{Concluding Remarks}

In this paper and \cite{Elahi01}, we have studied the $\mathbf{Q}$-spectral moments and signless Laplacian Estrada index ($SLEE$)  of graphs. more precisely, we have determined graphs with greatest $\mathbf{Q}$-spectral moments, thus maximum $SLEE$, through the set of all $n$-vertex graphs with a given parameter, namely, the number of cut edges, cut vertices, pendent vertices, (vertex) connectivity, edge connectivity, and diameter.

It would be of interest to investigate the behavior of these quantities on other classes of graphs such as chemical, $c$-cyclic, cactus graphs and linear polymers. 
Also, one might continue our work, by considering some other given parameters and finding their corresponding extremal graphs.


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\end{document}

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