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\fancyhead[CE]{M. Gholamnia Taleshani, M.Taghidoost Laskukalayeh and A. Abbasi} 
\fancyhead[CO]{ON DIMENSION OF SOME FINITE TOTAL GRAPHS}



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{\noindent Journal of Mathematical Extension \\
Journal Pre-proof}\\
ISSN: 1735-8299\\
URL: http://www.ijmex.com\\
Original Research Paper\\
‎\vspace*{9mm}
‎
\begin{center}

{\Large \bf 
On dimension of some finite Total graphs\\}
%{\bf Do You Have a Subtitle? \\ If so, Write It Here} 


\let\thefootnote\relax\footnote{\scriptsize Received: January 2008; Accepted: February 2009 }

{\bf M. Gholamnia Taleshani $^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\
\vspace{2mm} {\small  University of Guilan} \vspace{2mm}

{\bf M. Taghidoost Laskukalayeh\vspace*{-2mm}\\
\vspace{2mm} {\small  University of Guilan} \vspace{2mm}

{\bf  A. Abbasi}\vspace*{-2mm}\\
\vspace{2mm} {\small  University of Guilan} \vspace{2mm}
}
\end{center}

\vspace{4mm}


{\footnotesize
\begin{quotation}
{\noindent \bf Abstract.} In this paper we decompose $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$,
a graph with vertises $Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha})$ and two distinct vertices $x$ and $y$ are adjacent if and only if there exists $i$, $1 \leq i \leq \alpha$, such that $p_i \mid x, y$. We obtain dimension, edge metric dimension, strong metric dimension, and fractional metric dimension for it.
\end{quotation}
\begin{quotation}
\noindent{\bf AMS Subject Classification:} 05C12; 05C51; 05C75; 05C90.

\noindent{\bf Keywords and Phrases:} distance, resolving set,metric dimension, Total graph.
\end{quotation}}
\section{Introduction}

The concept of \textit{metric dimension} of a general metric space was introduced in 1953 by Blumenthal \cite{Blumenthal}. About twenty years later, it was applied by Slater \cite{Slater} who introduced the concept of \textit{locating set} of a graph. Independently Harary and Melter\cite{11} introduced the same concept as the \textit{resolving sets} and calculated the metric dimension of a tree graph. Since then, it has been frequently used in graph theory, chemistry, biology, robotics and many other disciplines.

Let $G = (V, E)$ be a simple, finite, undirected and connected graph. For graph theoretic terminology we refer to \cite{harary1}. We say that a vertex $u \in V$ distinguishes (determines or recognizes) two vertices $x, y \in V$ if $d(u, x) \neq d(u,y)$, where $d(x, y)$ represents the length of a shortest $x-y$-path in $G$. A \textit{metric generator} for $G$ is a set $\mathtt{B} \subseteq V$ with the property that, for each pair of vertices $x, y \in V$, there exists a vertex $u\in \mathtt{B}$ that distinguishes $x$ and $y$. If for some metric generator $ \mathtt{A} \subseteq V$, we have that $\vert \mathtt{A} \vert= min\{\vert \mathtt{B}\vert : \mathtt{B}$ is a metric generator for $G\}$, we say that $ \mathtt{A} $ is a \textit{metric basis} for $G$ and $dim(G) = \vert \mathtt{A} \vert$, is the \textit{metric dimension} of $G$. 

A set $\mathtt{L} \subseteq V$ is said to be a \textit{local metric generator} for $G$ if for each pair of vertices $x, y\in V$ such that $uv \in E$, there exists a vertex $u \in \mathtt{L}$ that distinguishes $u$ and $v$. If for some local metric generator $\mathtt{M} \subseteq V$, we have that $\vert \mathtt{M} \vert = min \{\vert \mathtt{L} \vert : \mathtt{L} $ is a local metric generator for $G\}$, then we say that $M$ is a \textit{local metric basis} for $G$ and $dim_{\ell}(G) = \vert \mathtt{M} \vert$, is the \textit{local metric dimension} of $G$.  The concept of adjacency generator was introduced by Jannesari and Omoomi \cite{omoomi} as a tool to study the metric dimension of lexicographic product of graphs. An \textit{adjacency generator} for $G$ is a set $\mathtt{B} \subseteq V$ such that for each $x, y \in V - \mathtt{B}$ there exists $b \in \mathtt{B}$ such that $b$ is adjacent to exactly one of $x$ and $y$. An adjancency which has the minimum cardinality among all adjacency generators of $G$ is called an \textit{adjacency basis} of $G$, and its cardinality is said to be the \textit{adjacency dimension} of $G$, denoted by $dim_{a}(G)$. 
The concepts of \textit{local adjacency generator}, \textit{local adjacency basis} and \textit{local adjacency dimension} are defined analogously, and the local adjacency dimension of a graph $G$ is denoted by $dim_{a,\ell}(G)$.

The distance between the vertex $v$ and the edge $e$ is defined as $d(e, v)=min \lbrace d(u, v), d(w, v) \rbrace$, where $e = uw$. A vertex $w \in V$ distinguishes two edges $e_{1}, e_{2} \in E$ if $d(w, e_{1}) \neq d(w, e_{2})$. A nonempty set $\mathtt{S} \subseteq V$ is an \textit{edge metric generator} for $G$ if any two edges of $G$ are distinguished by some vertex of $S$. An edge metric generator with the smallest possible cardinality is called an \textit{edge metric basis} for $G$, and its cardinality is the \textit{edge metric dimension}, which is denoted by $dim_e (G)$.

A set $\mathtt{S}$ of vertices of $G$ is a \textit{mixed metric generator} if any two elements (vertices or edges) of $G$ are distinguished by some vertex of $\mathtt{S}$. The smallest cardinality of a \textit{mixed metric generator} for $G$ is called the \textit{mixed metric dimension} and is denoted by $dim_m (G)$. A mixed metric basis for $G$ is a mixed metric generator for $G$ of cardinality $dim_m (G)$. It immediately follows that $dim_m (G) \geq max \{dim(G), dim_e (G) \}$ .

For any two vertices $u$ and $v$ of $G$, the interval $I[u, v]$ is defined as the collection of all vertices that belong to some shortest $u-v$-path. A vertex $w$ strongly distinguishes $u$ and $v$ if $v \in I[u, w]$ or $u \in I[v, w]$. A set $\mathtt{S}$ of vertices in a connected graph $G$ is a \textit{strong metric generator} for $G$ if every two vertices of $G$ are strongly distinguished by some vertex of $\mathtt{S}$. The smallest cardinality of a strong  metric generator of $G$ is called \textit{strong metric dimension} and is denoted by $dim_s (G)$.

For any two vertices $x$ and $y$ of $G$, $R \lbrace x, y \rbrace$ denotes the set of vertices $z$ such that $d(x, z) \neq d(y, z)$. In this view, a metric generating of $G$ is a subset $W$ of $V$ such that $W \cap R \lbrace x, y \rbrace \neq \emptyset$ for any two distinct vertices $x$ and $y$ of $G$. 
Let $f : V(G) \longrightarrow [0, 1]$ be a real valued function. For $W \subseteq V$, denote $f(W) = \sum_{v \in W} f (v)$. We call $f$ a \textit{resolving function} of $G$ if $f (R \lbrace x, y \rbrace) \geq 1$ for any two distinct vertices $x$ and $y$ of $G$. The \textit{fractional metric dimension}, denoted by $dim_f (G)$, is given by $dim_f (G) = min \lbrace |g| : g$ is a resolving function of $G \rbrace$, where $|g| = g(V(G))$.

By definition of the different variants of generators, the following inequalities hold for any graph $G$. For more details see \cite{omoomi}.
\begin{itemize} 
	\item $dim(G) \leq dim_{a}(G)$,
	\item $dim_{l}(G) \leq dim(G) \leq  dim_{l}(G) + dim_{a,l}(G^{c})$,
	\item $dim_{l}(G) \leq dim_{a,l}(G)$,
	\item $dim_{a,l}(G) \leq dim_{a}(G)$,
\end{itemize}

For a non-zero commutative ring $R$, let $Z(R)$ be the set of zero-divisors of $R$. Anderson and Badawi in \cite{anderson} introduced the total graph of $R$, denoted by $\Gamma(R)$, as an undirected graph with all elements of $R$ as vertices, and for distinct $x, y \in R$, the vertices $x$ and $y$ are adjacent if and only if $x+y \in Z(R)$. They also studied the induced subgraph $\Gamma(Z(R))$ of $\Gamma(R)$ with vertices $Z(R)$.

In this paper, we let $P =\{ p_{1}, p_{2}, \ldots, p_{\alpha} \}$ as a set of odd prime numbers and consider the total graph $\Gamma(\mathbb{Z}_{p_1p_2\ldots p_\alpha})$ and its induced subgraph $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ . We define the graph $\Gamma^{\alpha} (Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$ as a subgraph of $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$, with all elements of $Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha})$ as vertices, and two distinct vertices $x$ and $y$ are adjacent if and only if there exists $i$, $1 \leq i \leq \alpha$, such that $p_i \mid x, y$. We find a decomposition for this graph, and investigate the various dimensions.

\section{decomposition}
In this section we first decompose the subgraph $\Gamma^{2}(Z(\mathbb{Z}_{pq}))$, where $p, q$ are two distinct odd prime numbers and then generalize it into $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$.
We know that $Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}})$ is not an ideal of $\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}$, and $2 \notin Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}})$, so by \cite{anderson}, $\Gamma(Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$ is a connected graph with diameter two.

\begin{remark} \label{0000}
	It is known by Euler's formula that $\vert Z(\mathbb{Z}_{n})\vert= n-\phi(n)$ such that $\phi(n)=n \Pi_{i=1}^{\alpha}(1-\frac{1}{p_i})$. So we have $$\vert Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha})\vert= p_1p_2\ldots p_\alpha-\prod\limits_{i=1}^{\alpha}(p_i-1),$$ which is the number of vertices of $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$.
\end{remark}

\begin{remark}
	One can see that $\Gamma^{2}(Z(\mathbb{Z}_{pq}))$ is the graph $\Gamma(Z(\mathbb{Z}_{pq}))$. For $\alpha \geq 3$, $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$ and $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ are distinct. It is easy to see that $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$ is a subgraph of $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ because for any two adjacent vertices $x, y$ in $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$, there exists $i$, $1 \leq i \leq \alpha$, such that $p_i \mid x, y$. It leads to $p_i \mid x+y$. So, $x+y \in Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}})$, and $x \sim y$ in $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$. Since the logical proposition $p_i \mid x, y \Rightarrow p_i \mid x+y$, is not reversible, there exist adjacent vertices in $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ which are nonadjacent in $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_{1} p_{2} \ldots p_{\alpha}}))$. 
\end{remark}

\begin{example}
	Consider the graphs $\Gamma(Z(\mathbb{Z}_{105}))$ and $\Gamma^{3}(Z(\mathbb{Z}_{105}))$ with $P =\{ 3, 5, 7 \}$ . It is clear that the number of edges in $\Gamma(Z(\mathbb{Z}_{105}))$ is more than $\Gamma^{3}(Z(\mathbb{Z}_{105}))$ . For example, $3 \sim 7$, $33 \sim 77$ in $\Gamma(Z(\mathbb{Z}_{105}))$ since $3 + 7, 33 + 77 \in Z(\mathbb{Z}_{105})$. But $3 \nsim 7$, $33 \nsim 77$ in $\Gamma^{3}(Z(\mathbb{Z}_{105}))$, since there isn't any common prime factor in $P$. 
\end{example}

\begin{lemma} \label{21}
	The following facts hold in $\Gamma^{2}(Z(\mathbb{Z}_{pq}))$: 
	\item[(i)] If $u=kp$ where $1 \leq k \leq q-1$, then $deg(u)=q-1$.
	\item[(ii)] If $v=kq$ where $1 \leq k \leq p-1$, then $deg(v)=p-1$.
	\item[(iii)] $deg(0)=p+q-2$.
\end{lemma}
\begin{proof}
	It is clear that $u=kp$ is adjacent to zero and all of the other multiples of $p$. Similarly, $v=kq$ is adjacent to zero and all of the other multiples of $q$. Obviously, $u$ and $v$ are not adjacent.  
\end{proof}

\begin{definition}
	A \textit{decomposition} of a graph $G$ is a list of subgraphs $G_{1},G_{2}, \ldots ,G_{r}$ such that each edge appears in exactly one subgraph in the list. In this terminology, we say that $G$ is decomposed by $G_{1},G_{2}, \ldots ,G_{r}$ and show it by $G=G_{1}+G_{2}+ \cdots +G_{r}$.
\end{definition}

\begin{theorem} \label{39}
	$\Gamma^{2} (Z(\mathbb{Z}_{pq}))$ has the following decomposition;
	\begin{center}
		$\Gamma^{2} (Z(\mathbb{Z}_{pq}))=K_p + K_q$.
	\end{center}
\end{theorem}
\begin{proof}
	In $\Gamma^{2} (Z(\mathbb{Z}_{pq}))$ we have $q$ vertices of the form $v=kp$ which induce $K_q$ and $p$ vertices of the form $v=kq$ which induce $K_p$, by Lemma \ref{21}. It is clear that there is no other adjacency and zero is the only common vertex of them.
\end{proof}

\begin{example}
\end{example}
\begin{figure}[!ht]
	\begin{center}
		\begin{tikzpicture}
			\coordinate (1a) at (0,0) ;
			\coordinate (1b) at (1,1.5) ;
			\coordinate (1c) at (1,-1.5) ;
			\coordinate (1d) at (3,1) ;
			\coordinate (1r) at (3,-1) ;
			\coordinate (1s) at (-1,1.5) ;
			\coordinate (1t) at (-1,-1.5) ;
			\coordinate (1f) at (-3,2) ;
			\coordinate (1g) at (-3,-2) ;
			\coordinate (1h) at (-5,1) ;
			\coordinate (1i) at (-5,-1) ;
			
			\fill (1a) circle (0.5 mm) node [above,] {$0$};
			\fill (1b) circle (0.5 mm)  node[above,] {$28$};
			\fill (1c) circle (0.5 mm)  node [below,] {$7$};
			\fill (1d) circle (0.5 mm)  node[right,] {$21$};
			\fill (1r) circle (0.5 mm)  node[below,] {$14$};
			\fill (1s) circle (0.5 mm)  node[above,] {$5$};
			\fill (1t) circle (0.5 mm)  node [below,] {$30$};
			\fill (1f) circle (0.5 mm)  node [above,] {$10$ };
			\fill (1g) circle (0.5 mm)  node[below,]{$25$};
			\fill (1h) circle (0.5 mm)  node [left,] {$15$} ;
			\fill (1i) circle (0.5 mm)  node [left,] {$20$};
			
			\draw [thin] (1a) -- (1b) ;
			\draw [thin] (1a) -- (1c) ;
			\draw [thin] (1a) -- (1d) ;
			\draw [thin] (1a) -- (1r) ;
			\draw [thin] (1a) -- (1s) ;
			\draw [thin] (1a) -- (1t) ;
			\draw [thin] (1a) -- (1f) ;
			\draw [thin] (1a) -- (1g) ;
			\draw [thin] (1a) -- (1h) ;
			\draw [thin] (1a) -- (1i) ;
			\draw [thin] (1b) -- (1c) ;
			\draw [thin] (1b) -- (1d) ;
			\draw [thin] (1b) -- (1r) ;
			\draw [thin] (1c) -- (1d) ;
			\draw [thin] (1c) -- (1r) ;
			\draw [thin] (1d) -- (1r) ;
			\draw [thin] (1s) -- (1t) ;
			\draw [thin] (1s) -- (1f) ;
			\draw [thin] (1s) -- (1g) ;
			\draw [thin] (1s) -- (1h) ;
			\draw [thin] (1s) -- (1i) ;
			\draw [thin] (1t) -- (1f) ;
			\draw [thin] (1t) -- (1g) ;
			\draw [thin] (1t) -- (1h) ;
			\draw [thin] (1t) -- (1i) ;
			\draw [thin] (1f) -- (1g) ;
			\draw [thin] (1f) -- (1h) ;
			\draw [thin] (1f) -- (1i) ;
			\draw [thin] (1g) -- (1h) ;
			\draw [thin] (1g) -- (1i) ;
			\draw [thin] (1h) -- (1i) ;
		\end{tikzpicture} 
		\caption{$Z(\Gamma^{2} (\mathbb{Z}_{35}))=K_{7}+K_{5}$}
	\end{center}
\end{figure}
\begin{theorem}\label{d2}
	$\Gamma^{3} (Z(\mathbb{Z}_{pqr}))=K_{pq}+ K_{pr}+K_{qr}-(K_p+ K_q+K_r).$
\end{theorem}
\begin{proof}
	The set of zerodivisors of $\mathbb{Z}_{pqr}$ is classified in multiples of $p, q$ and $r$. It is clear that $qr$ elements of them are multiples of $p$, $pr$ elements are multiples of $q$ and $pq$ elements are multiples of $r$. Hence, the edges of $G$  can be decomposed into three complete graphs $K_{qr}, K_{pr}$ and $K_{pq}$.
	
	There are also some other edges formed by common multiples of $p, q$ and $r$. Since there are $r$ common multiples of $p$ and $q$,
	$p$ common multiples of $q$ and $r$ and $q$ common multiples of $p$ and $r$, which form $K_r, K_p$ and $K_q$, respectively, in order to avoid considering repeated edges, we omit $K_r, K_p$ and $K_q$ from the decomposition.
\end{proof}
\begin{figure}[!ht]
	\begin{center}
		\begin{tikzpicture}
			
			\draw (-1.5,0) circle [radius=2.5] node {};
			\draw (1.5,0) circle [radius=2.5] node {};
			\coordinate (A) at (-3,-2) ;
			\coordinate (B) at (0,-2) ;
			\coordinate (C) at (3,-2) ;
			\coordinate (D) at (0,-5) ;
			\coordinate (E) at (0,0) ;
			\coordinate (F) at (-1.5,-2) ;
			\coordinate (G) at (1.5,-2) ;
			\coordinate (H) at (4,0) ;
			\coordinate (I) at (-4,0) ;
			\coordinate (J) at (0,-5) ;
			
			\fill (B) circle (0.5mm) node [below = 1mm]  {$0$};
			\fill (E) node []  {$K_{q}$};
			\fill (F) node []  {$K_{p}$};
			\fill (G) node []  {$K_{r}$};
			\fill (H) node [right = 1mm]  {$K_{qr}$};
			\fill (I)  node [left = 1mm]  {$K_{pq}$};
			\fill (J)  node [below = 1mm]  {$K_{pr}$};
			
			\draw (A) .. controls (-1.5,-1) ..(B);
			
			\draw (B) .. controls (1.5,-1) ..(C);
			
			\draw (A) to [bend right] (D);
			\draw (D) to [bend right] (C);
			%\draw (A) .. controls (0,-6) ..(C);
		\end{tikzpicture}
		\caption{Decomposition of $\Gamma^{3} (Z(\mathbb{Z}_{pqr}))$}\label{}
	\end{center}
\end{figure}

\begin{theorem}\label{d3}
	$\Gamma^{4} (Z(\mathbb{Z}_{pqrs}))=K_{pqr}+ K_{pqs}+K_{prs}+K_{qrs}-(K_{pq}+ K_{pr}+K_{ps}+K_{qr}+ K_{qs}+K_{rs})+K_p+ K_q+K_r+K_s.$
\end{theorem}
\begin{proof}
	Similar to proof of theorem \ref{d2}, multiples of $p, q, r$ and $s$ form the complete graphs $K_{qrs}, K_{prs}, K_{pqs}$ and $K_{pqr}$, respectively. Notice that there are $\binom{4}{2}$ common multiples of every two factors of $p, q, r$ and $s$ which are counted once in above complete graphs. So, we should omit the edges of $K_{pq}+ K_{pr}+K_{ps}+K_{qr}+ K_{qs}+K_{rs}$. Since the edges formed by common multiples of every three factors of $p, q, r$ and $s$ are missed by the procedure of deletion, we have to add the complete graphs $K_p+ K_q+K_r+K_s$.
\end{proof}
\begin{figure}[!ht]
	\begin{center}
		\begin{tikzpicture}
			
			\coordinate (A) at (0,0) ;
			\coordinate (B) at (-2,2) ;
			\coordinate (C) at (2,2) ;
			\coordinate (D) at (2,-2) ;
			\coordinate (E) at (-2,-2) ;
			\coordinate (F) at (-4.3,0) ;
			\coordinate (G) at (0,4.3) ;
			\coordinate (H) at (4.3,0) ;
			\coordinate (I) at (0,-4.3) ;
			\coordinate (J) at (-4,4) ;
			\coordinate (K) at (4,4) ;
			\coordinate (L) at (4,-4) ;
			\coordinate (M) at (-4,-4) ;
			\coordinate (a) at (-5.5,2) ;
			\coordinate (b) at (-2.5,5.5) ;
			\coordinate (g) at (5.5,2) ;
			\coordinate (h) at (2.5,5.5) ;
			\coordinate (c) at (-5.5,-2) ;
			\coordinate (d) at (-2.5,-5.5) ;
			\coordinate (e) at (5.5,-2) ;
			\coordinate (f) at (2.5,-5.5) ;
			
			\coordinate (T) at (-1,-1) ;
			\coordinate (U) at (-1,1) ;
			\coordinate (V) at (1,1) ;
			\coordinate (W) at (1,-1) ;
			\coordinate (O) at (-2.5,0) ;
			\coordinate (X) at (0,2.5) ;
			\coordinate (Z) at (2.5,0) ;
			\coordinate (Y) at (0,-2.5) ;
			\coordinate (S) at (-2.7,-2.7) ;
			\coordinate (P) at (-2.7,2.7) ;
			\coordinate (Q) at (2.7,2.7) ;
			\coordinate (R) at (2.7,-2.7) ;
			\coordinate (m) at (-3.6,-3.6) ;
			\coordinate (n) at (-3.6,3.6) ;
			\coordinate (k) at (3.6,3.6) ;
			\coordinate (l) at (3.6,-3.6) ;
			
			\fill (T) node []  {$K_{p}$};
			\fill (U) node []  {$K_{q}$};
			\fill (V) node []  {$K_{r}$};
			\fill (W) node []  {$K_{s}$};
			\fill (O) node []  {$K_{pq}$};
			\fill (X) node []  {$K_{qr}$};
			\fill (Z) node []  {$K_{rs}$};
			\fill (Y) node []  {$K_{ps}$};
			\fill (S) node []  {$K_{pqs}$};
			\fill (P) node []  {$K_{pqr}$};
			\fill (Q) node []  {$K_{qrs}$};
			\fill (R) node []  {$K_{prs}$};
			\fill (m) node []  {$K_{qs}$};
			\fill (n) node []  {$K_{pr}$};
			\fill (k) node []  {$K_{qs}$};
			\fill (l) node []  {$K_{pr}$};
			
			\fill (A) circle (0.5mm) node [below = 1mm]{$0$};
			\draw [dashed](b) .. controls (-3,3) ..(a);
			
			\draw [dashed](g) .. controls (3,3) ..(h);
			
			\draw [dashed](c) .. controls (-3,-3) ..(d);
			
			\draw [dashed](e) .. controls (3,-3) ..(f);
			
			
			\draw (B) to [bend left] (A);
			\draw (B) to [bend right] (A);
			\draw (A) to [bend left] (C);
			\draw (A) to [bend right] (C);
			\draw (A) to [bend left] (D);
			\draw (A) to [bend right] (D);
			\draw (A) to [bend left] (E);
			\draw (A) to [bend right] (E);
			\draw (F) to [bend left] (B);
			\draw (F) to [bend right] (E);
			\draw (G) to [bend left] (C);
			\draw (G) to [bend right] (B);
			\draw (H) to [bend left] (D);
			\draw (H) to [bend right] (C);
			\draw (I) to [bend left] (E);
			\draw (I) to [bend right] (D);
			\draw (J) to [bend left] (G);
			\draw (J) to [bend right] (F);
			\draw (K) to [bend left] (H);
			\draw (K) to [bend right] (G);
			\draw (L) to [bend left] (I);
			\draw (L) to [bend right] (H);
			\draw (M) to [bend left] (F);
			\draw (M) to [bend right] (I);
		\end{tikzpicture}
		\caption{Decomposition of $\Gamma^{4} (Z(\mathbb{Z}_{pqrs}))$}
	\end{center}
\end{figure}

In general, we have the following theorem.
\begin{theorem}\label{tf}
	$$\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))=\sum\limits_{\substack{i=1}}^{\alpha} K_{\prod\limits_{\substack{j=1\\ j\neq i}}^{\alpha}p_j}-\sum\limits_{\substack{i,j=1\\ i<j}}^{\alpha} K_{\prod\limits_{\substack{k=1\\ k\neq i,j}}^{\alpha}p_k}+\sum\limits_{\substack{i,j,k=1\\ i<j<k}}^{\alpha} K_{\prod\limits_{\substack{l=1\\ l\neq i,j,k}}^{\alpha}p_l}-\ldots+(-1)^{\alpha}\sum\limits_{i=1}^{\alpha}K_{p_i}.$$
\end{theorem}
\begin{proof}
	The proof follows by inclusion-exclusion principle and induction on $\alpha$ due to the proofs of Theorems \ref {d2}, \ref {d3}.
\end{proof}
\begin{corollary}
	The clique number of $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ is $\max\lbrace \prod\limits_{\substack{j=1\\ j\neq i}}^{\alpha}p_j \mid 1\leq i\leq \alpha\rbrace$. In the case $p_1 < p_2 < \ldots < p_\alpha$, the clique number is $p_2 p_3 \ldots p_\alpha$.
\end{corollary}
\begin{proof}
	According to Theorem \ref{tf}, it is clear.
\end{proof}

\begin{corollary}
	$diam(\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))) = 2.$ 
\end{corollary}
\begin{proof}
	The decomposition in Theorem \ref{tf} shows that $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ is not a complete graph. Since $\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ is a subgaph of $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$, $diam(\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))) = 2$.  
\end{proof}

\section{Twin equivalence classes of $\Gamma(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$}
For a vertex $u$, the \textit{open neighborhood} of $u$ in $G$ is $N(u) = \lbrace v \in V  \mid uv \in E \rbrace$ and the \textit{closed neighborhood} of $u$ is $N[u] = N(u) \cup \{u\}$. Two vertices $u, v$ are \textit{true twins} of $G$ if $N[u] = N[v]$. They are \textit{false twins} if $N(u) = N(v)$; and \textit{twins} if they are any of the previous. Define a relation $\equiv$ on $V (G)$ by $u \equiv v$ if and only if $u = v$ or $u, v$ are twins. By Lemma 2.6 in \cite{hernando}, $\equiv$ is an equivalence relation. It is not difficult to see that the equivalence classes of the true-twin relations are cliques and those of the false-twin relations are independent sets.
There are three possibilities for each twin equivalence class $U$:
\begin{itemize}
	\item[(a)]$U$ is a singleton set, or
	\item[(b)]$|U|> 1$ and $N(u) = N(v)$ for any $u, v \in  U$, or
	\item[(c)]$|U|> 1$ and $N[u] = N[v]$ for any $u, v \in  U$.
\end{itemize}
We will refer to the type $(c)$ as the true twin equivalence classes.

Consider the equivalence relation $\equiv$. For each vertex $v \in V (G)$, let $v^{*}$ be the set of vertices of $G$ that are equivalent to $v$ under $\equiv$. Let $\{ v^{*}_1, . . . , v^{*}_k \}$ be the partition of $V (G)$ induced by $\equiv$, where each $v_i$ is a representative of the set $v^{*}_i$. The \textit{twin graph} of $G$, denoted by $G^{*}$, is the graph with vertex set $V (G^{*}) := \{ v^{*}_1, . . . , v^{*}_k \}$, where $v^{*}_i v^{*}_j \in E(G^{*})$ if and only if $v_i v_j \in E(G)$. By Lemma 2.10 in \cite{hernando}, one can see that this definition is independent of the choice of representatives.

Note that in $\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ the vertices can be classified in multiples and common multiples of $p_i$'s. In the next, we show that this partition forms equivalence classes. Also, we obtain the number of the equivalence classes by counting the ways of selecting common multiples of $p_i$'s.


\begin{notation}\label{r}
	For any nonempty subset $S\subseteq P$, let $A_{S}=\{x \in Z(\mathbb{Z}_{p_1p_2 \ldots p_\alpha});\quad p \mid x \Longleftrightarrow p \in S\}$. Set $\mathcal{A}=\{A_{S}; S\subseteq P\}$ and for all $ 1\leq i \leq \alpha$, $\mathcal{A}_{i}=\{A_{S};\quad S\subseteq P, |S|=i\}$. 
\end{notation}
One sees that $|\mathcal{A}_{i}|=\binom{\alpha}{i}$. In the next theorem we show that for all $\emptyset\neq S\subseteq P$, $A_{S}$ is an equivalence class.


\begin{theorem}\label{2p}
	The number of twin equivalence classes of $\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ is $2^{\alpha}-1$.
\end{theorem} 
\begin{proof}
	Let $S=\{p_{i_{1}},\ldots, p_{i_{s}}\} \subseteq P$. We show that $N[x]=N[y]$ for every $x, y \in A_{S} \in \mathcal{A}_s$. Suppose that $z \in N[x]$. There exists $p_i$ such that $p_i \mid x$ , $p_i \mid y$. If $p_i \in S$, then $p_i = p_{i_{l}}$, for some $1 \leq l \leq s$. So, $p_{i_{l}} \mid x$ and $p_{i_{l}} \mid z$. Since $p_{i_{l}} \mid y$, it leads to $z \sim y$. Therefore, $z \in N[y]$. If $p_i \notin S$, then $p_{i_{l}} \nmid y$. So, $y \notin A_S$, which contradicts to the assumption. Thus, $A_S$ is an equivalence class. 
	Note that $p \mid  0$ for all $p \in P$. So, $A_{P}$ is the zero singleton class.
	
	By assumption, we have $\binom{\alpha}{s}$ sets of $A_S$'s.  Therefore, the number of the equivalence classes is equal to $\sum\limits_{i=1}^{\alpha} \binom{\alpha}{i} = 2^{\alpha}-1$.
\end{proof}
\begin{figure}[!ht]
	\begin{center}
	\begin{tikzpicture}
		
		\draw (-1.2,0) circle [radius=2] node {};
		\draw (1.2,0) circle [radius=2] node {};
		\coordinate (A) at (-2.27,-1.7) ;
		\coordinate (B) at (0,-1.6) ;
		\coordinate (C) at (2.27,-1.7) ;
		\coordinate (D) at (0,-4) ;
		\coordinate (E) at (0,0) ;
		\coordinate (F) at (-1.1,-1.6) ;
		\coordinate (G) at (1.1,-1.6) ;
		\coordinate (H) at (2,0) ;
		\coordinate (I) at (-2,0) ;
		\coordinate (J) at (0,-3.2) ;
		\coordinate (B1) at (0,-1.9) ;
		\fill (B1) node [] {$0$};
		\fill (E) node []  {$A_{\{p,r\}}$};
		\fill (F) node []  {$A_{\{q,r\}}$};
		\fill (G) node []  {$A_{\{p,q\}}$};
		\fill (H)  node []  {$A_{\{p\}}$};
		\fill (I)  node []  {$A_{\{r\}}$};
		\fill (J) node []  {$A_{\{q\}}$};
		\fill (B) circle (0.5 mm) node [] {};
		\draw (A) .. controls (-1.1,-1) ..(B);
		\draw (B) .. controls (1.1,-1) ..(C);
		\draw (A) to [bend right] (D);
		\draw (D) to [bend right] (C);
	\end{tikzpicture}
	\caption{ Equivalence classes of $\Gamma^{3}(Z(\mathbb{Z}_{pqr}))$}
	\end{center}
\end{figure}


\begin{figure}[!ht]
	\begin{center}
	\begin{tikzpicture}
		
		\coordinate (A) at (0,0) ;
		\coordinate (B) at (-2,2) ;
		\coordinate (C) at (2,2) ;
		\coordinate (D) at (2,-2) ;
		\coordinate (E) at (-2,-2) ;
		\coordinate (F) at (-4.3,0) ;
		\coordinate (G) at (0,4.3) ;
		\coordinate (H) at (4.3,0) ;
		\coordinate (I) at (0,-4.3) ;
		\coordinate (J) at (-4,4) ;
		\coordinate (K) at (4,4) ;
		\coordinate (L) at (4,-4) ;
		\coordinate (M) at (-4,-4) ;
		\coordinate (a) at (-5.5,2) ;
		\coordinate (b) at (-2.5,5.5) ;
		\coordinate (g) at (5.5,2) ;
		\coordinate (h) at (2.5,5.5) ;
		\coordinate (c) at (-5.5,-2) ;
		\coordinate (d) at (-2.5,-5.5) ;
		\coordinate (e) at (5.5,-2) ;
		\coordinate (f) at (2.5,-5.5) ;
		
		\coordinate (T) at (-1,-1) ;
		\coordinate (U) at (-1,1) ;
		\coordinate (V) at (1,1) ;
		\coordinate (W) at (1,-1) ;
		\coordinate (O) at (-2.5,0) ;
		\coordinate (X) at (0,2.5) ;
		\coordinate (Z) at (2.5,0) ;
		\coordinate (Y) at (0,-2.5) ;
		\coordinate (S) at (-2.7,-2.7) ;
		\coordinate (P) at (-2.7,2.7) ;
		\coordinate (Q) at (2.7,2.7) ;
		\coordinate (R) at (2.7,-2.7) ;
		\coordinate (m) at (-3.6,-3.6) ;
		\coordinate (n) at (-3.6,3.6) ;
		\coordinate (k) at (3.6,3.6) ;
		\coordinate (l) at (3.6,-3.6) ;
		
		\fill (T) node []  {$A_{\{q,r,s\}}$};
		\fill (U) node []  {$A_{\{p,r,s\}}$};
		\fill (V) node []  {$A_{\{p,q,s\}}$};
		\fill (W) node []  {$A_{\{p,q,r\}}$};
		\fill (O) node []  {$A_{\{r,s\}}$};
		\fill (X) node []  {$A_{\{p,s\}}$};
		\fill (Z) node []  {$A_{\{p,q\}}$};
		\fill (Y) node []  {$A_{\{q,r\}}$};
		\fill (S) node []  {$A_{\{r\}}$};
		\fill (P) node []  {$A_{\{s\}}$};
		\fill (Q) node []  {$A_{\{p\}}$};
		\fill (R) node []  {$A_{\{q\}}$};
		\fill (m) node []  {$A_{\{p,r\}}$};
		\fill (n) node []  {$A_{\{q,s\}}$};
		\fill (k) node []  {$A_{\{p,r\}}$};
		\fill (l) node []  {$A_{\{q,s\}}$};
		
		\fill (A) circle (0.5mm) node [below = 1mm]{$0$};
		\draw [dashed](b) .. controls (-3,3) ..(a);
		\draw [dashed](g) .. controls (3,3) ..(h);
		\draw [dashed](c) .. controls (-3,-3) ..(d);
		\draw [dashed](e) .. controls (3,-3) ..(f);
		
		\draw (B) to [bend left] (A);
		\draw (B) to [bend right] (A);
		\draw (A) to [bend left] (C);
		\draw (A) to [bend right] (C);
		\draw (A) to [bend left] (D);
		\draw (A) to [bend right] (D);
		\draw (A) to [bend left] (E);
		\draw (A) to [bend right] (E);
		\draw (F) to [bend left] (B);
		\draw (F) to [bend right] (E);
		\draw (G) to [bend left] (C);
		\draw (G) to [bend right] (B);
		\draw (H) to [bend left] (D);
		\draw (H) to [bend right] (C);
		\draw (I) to [bend left] (E);
		\draw (I) to [bend right] (D);
		\draw (J) to [bend left] (G);
		\draw (J) to [bend right] (F);
		\draw (K) to [bend left] (H);
		\draw (K) to [bend right] (G);
		\draw (L) to [bend left] (I);
		\draw (L) to [bend right] (H);
		\draw (M) to [bend left] (F);
		\draw (M) to [bend right] (I);
	\end{tikzpicture}
	\caption{ Equivalence classes of $\Gamma^{4}(Z(\mathbb{Z}_{pqrs}))$}
	\end{center}
\end{figure}


\begin{lemma}\label{l2q}
	Let $S$ and $T$ be two nonempty subsets of $P =\{p_1, \ldots, p_\alpha\}$. For any $v\in A_{S}$ and $w\in A_{T}$;
	\begin{itemize}
		\item[(i)] If $S \cap T\neq\emptyset$, then $d(v,w)=1$;
		\item[(ii)] If $S \cap T=\emptyset$, then $d(v,w)=2$.
	\end{itemize}
\end{lemma}
\begin{proof}
	(i) If $p_{l} \in S \cap T$, then $p_{l}\mid v$ and $p_{l}\mid w$. So, $v \sim w$.
	(ii) Let $S=\lbrace  p_{i_{1}},\ldots, p_{i_{s}}\rbrace$ and $T=\lbrace  p_{j_{1}},\ldots, p_{j_{t}}\rbrace$.
	Since the intersection is empty, $p_{i_{l}} \nmid w$ for any $1\leq l \leq s$, and $p_{j_{r}} \nmid v$ for any $1\leq r \leq t$. 
	If $d(v,w)=1$, then there exists $p_k \in P$ such that $p_k \mid v$ and $p_k \mid w$. By the assumption, $p_k \notin S \cup T$. It means that there is no prime factor in $P$ which aliquots both $u$ and $v$. So, $v \nsim w$. Hence, $d(v,w)=2$.
\end{proof}

We know that a subset $S \subseteq V$ is an \textit{independent set} in $G$ if no two vertices in $S$ are adjacent. \textit{The independence number} of $G$ is the maximum size of all independent sets of vertices, denoted by $\alpha (G)$.

\begin{theorem}
	The independence number of $\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ is $\alpha$.
\end{theorem}
\begin{proof}
	Consider the equivalence classes $\mathcal{A}_{1}=\lbrace A_{\lbrace p_{i} \rbrace} ; \quad 1 \leq i \leq \alpha \rbrace$. Let $I=\lbrace v_1,v_2, \ldots, v_\alpha \rbrace$ such that $v_i \in A_{\lbrace p_{i} \rbrace}$. For any two distinct vertices $v_i, v_j \in I$, by part (ii) of Lemma \ref{l2q}, $v_i$ and $v_j$ are not adjacent. Therefore, $I$ is an independent set. Suppose that $u \notin I$. If $u\in \mathcal{A}_{1}$, then $u \in A_{\lbrace p_{_{t}} \rbrace}$ for some $1 \leq t \leq \alpha $, so $p_{t}\mid u$. Since $p_{t} \mid v_{t}$, it leads to $u$ is adjacent to $v_t$. Let $u\notin \mathcal{A}_{1}$, then there is an equivalence class $A_{S}$ such that $u\in A_{S}$. Clearly, $A_{S} \notin \mathcal{A}_{1}$, $|S| \geq 2$, and $\lbrace p_{_{t}} \rbrace \cap S \neq \emptyset$  for some $1 \leq t \leq \alpha$. Let $p_r \in S$ such that $p_r \neq p_t$, then $p_r \mid u$ and $p_t \mid u$. So, $u \sim v_r$  and $u \sim v_t$. Hence, $I$ is the maximal independent set.
\end{proof}
\section{dimension}
In this section we obtain some types of dimensions for the graph $G=\Gamma^{\alpha}(Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$. First, we note that determining whether a given set $\mathtt{B}$ of vertices of $G$ is a metric generating set of $G$, one need only investigate the pairs of vertices in $V(G) - \mathtt{B}$, since $u \in \mathtt{B}$ is the only vertex of $G$ whose distance from $u$ is $0$. 

\begin{theorem} \label{2o}
	\cite{omoomi2} If $G^{*}$ is the twin graph of $G$, then $dim(G) \geq n(G) - n(G^{*})$. 
\end{theorem}

\begin{theorem}
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$. Then $dim(G)= n(G)-2^{\alpha}+1$.
\end{theorem}

\begin{proof}
	By Theorems \ref{2o} and \ref{2p}, $dim(G) \geq n(G) - 2^{\alpha} + 1$. 
	Set $R$ as a set of representative vertices of equivalence classes. By Theorem \ref{2p}, $|R|=2^{\alpha}-1$. We show that $\mathtt{M} = V(G) - R$ is a metric basis for $G$. For any two distinct vertices $x, y \in \mathtt{M}$, $d(x,x)=0$ and $d(x,y)=1$ or $d(x,y)=2$. So, $d(x,x)\neq d(x,y)$. 
	
	Let $u,v\in R$ such that $u,v \neq 0$, and suppose $S$ and $T$ be two subset of $P$ such that $u\in A_{S}$ and $v\in A_{T}$. There exists $L\subseteq P$ such that $L \cap S\neq \emptyset$ and $L \cap T= \emptyset$. So, for every vertex $x \in A_{L}$, $x \notin R$; $d(x,u)=1$ and $d(x,v)=2$, by Lemma \ref{l2q}. 
	
	Again, let $u,v \in R$ and $u = 0$. There exists $L\subseteq P$ 
	such that $L \cap S\neq \emptyset$ and $L \cap T= \emptyset$. Then for all $x\in A_{L}$, $d(x,v)=2$ and $d(x,u)=1$; since zero is adjacent to all vertices. Hence, $\mathtt{M}$ is a metric basis and $dim(G)\leq n(G)-2^{\alpha}+1$. 
\end{proof}

\begin{corollary}
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$, Then $dim_{a}(G)= n(G)-2^{\alpha}+1$.
\end{corollary}
\begin{proof}
	Since $diam(G) = 2$, it is easy to see that  $dim_{a}(G)= dim(G) = n(G)-2^{\alpha}+1$.
\end{proof}

\begin{lemma}\label{l10}
	Let $y$ and $z$ be true twins. If $e, f \in E(G)$ such that $e=xy$ and $f=xz$, then at least one of $y$ and $z$ belongs to an edge metric basis of $G$.
\end{lemma}
\begin{proof}
	Let $x \in A_{S}$ for a subset $S$ of $P$. Consider two edges $e=xy$ and $f=xz$. It is clear that $d(e, x)=d(f, x)=0$. So, $x$ doesn't distinguish $e$ and $f$. Now for $v \neq x, y , z$, consider the following cases.
	
	Case 1: Let $y, z \in A_{S}$, if $v \in A_{S}$, then $d(e, v)=d(f, v)=1$. If $v\notin A_{S}$, then $v\in A_{T}$ for some $T\subseteq P$ and by Lemma \ref{l2q}, $d(e,v)=d(f,v)=1$ or $d(e,v)=d(f,v)=2$.
	\begin{center}
		$d(e, v) = d(f, v)= \Bigg\{ \begin{array}{cc}
			1 & \:  S \cap T\neq\emptyset \\ 
			2 & \:  S \cap T=\emptyset \\
		\end{array}$
	\end{center}
	Case 2: If $y, z \notin A_{S}$, since $y$ and $z$ are twin vertices, then by Lemma \ref{l2q}, $y, z\in A_T$ for some $T\subseteq P$ such that $S \cap T \neq \emptyset$. If $v\in A_{S\cap T}$, then $v$ is adjacent to $x, y$ and $z$. So, $d(e,v)=d(f,v)=1$.
	If $v\in A_{S}$ or $v\in A_{T}$, then $v\in A_{T}$ is adjacent to $x$ or $y, z$, respectively. Therefore, $d(e,v)=d(f,v)=1$. If $v\in A_{P\setminus \{S\cup T\}}$, then $d(e,v)=d(f,v)=2$, by Lemma \ref{l2q}.
	\begin{center}
		$d(e,v) = d(f,v)= \Bigg\{ \begin{array}{cc}
			1 & \: v \in A_{S \cup T} \\ 
			2 & \: v \in A_{P \setminus (S \cup T)}
		\end{array}$
	\end{center}
	So, any vertex $v \neq x, y , z$ doesn't distinguish $e$ and $f$. By the first part of the proof, $x$ doesn't distinguish $e$ and $f$. Thus, at least one of $y$ and $z$ must be in an edge metric basis $\mathtt{E}$.
\end{proof}

\begin{lemma} \label{eee}
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$ and $S \subseteq P$. Then every edge metric basis $\mathtt{E}$ of $G$ contains at least  $|A_{S}|-1$ vertices of the equivalence class $A_{S}$.
\end{lemma}
\begin{proof}
	Let $S \subseteq P$ and consider the true twin equivalence class $A_{S}$ of $G$. Let $x, y, z$ be three vertices of $A_{S}$ and consider two edges $e=xy$ and $f=xz$. Then by Lemma \ref{l10}, at least one of $y$ and $z$ must be in an edge metric basis $\mathtt{E}$. This argument can be repeated for any pair of such edges.
\end{proof}

\begin{theorem}
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$. Then $dim_{e}(G)= n(G) - 2$.
\end{theorem}
\begin{proof}
	For any edge metric basis $\mathtt{E}$ of $G$, by Theorem \ref{2p} and Lemma \ref{eee}, $| \mathtt{E} | \geq n(G) - 2^{\alpha} + 1$. If $\mathtt{E} = V(G) - \lbrace u, v, w \rbrace$ such that $u \in A_{U}$, $v \in A_{V}$, and $w \in A_{W}$  where $A_{U}$, $A_{V}$, $A_{W}$ are three distinct equivalence classes associated with $U, V, W \subseteq P$.
	
	Let $u, v, w \neq 0$, then for two edges $e=ou$, $f=ov$, $d(e,0)=d(f,0)=0$, and $d(e,x)=d(f,x)=1$ for all $x \in \mathtt{E}$. Thus, 
	$\mathtt{E}$ does not distinguish $e$ and $f$.
	If $u = 0$, $\mathtt{E} = V(G) - \lbrace 0, v, w \rbrace$, then the edges $e=0w$, $f=0v$ have distance one to all vertices in $\mathtt{E}$, which is a contradiction to edge metric basis of $\mathtt{E}$. By this argument, there is no edge metric basis of size $n - 3$. Hence, $dim_{e}(G) \geq n - 2$.
	
	Let $\mathtt{E^{'}}$ contains at least one of the vertices $v$ and $w$. Suppose that $w \in \mathtt{E^{'}}$ and $\mathtt{E^{'}} = V(G) - \lbrace 0, v \rbrace$. We show that $\mathtt{E^{'}}$ is an edge metric basis.
	According to the structure of the graph, for any pair edges $e$ and $f$ we have the following cases. In each case, we show that there is an $x \in \mathtt{E^{'}}$ which distinguishes $e$ and $f$.
	
	Case 1: Let $e=xy$, $f=zt$, such that $x, y, z, t \notin \lbrace 0, v \rbrace $. If $e, f$ belong to an equivalence class $A_S$, then $d(e, x)=0$ and $d(f, x)=1$. If $e, f$ belong to distinct equivalence classes $A_S$ , $A_T$, respectively, then $d(e, x)=0$. Also, for $T \cap S \neq \emptyset$, $d(f, x)=1$ and $T \cap S = \emptyset$, $d(f, x)=2$.
	
	Case 2: Let $e=0v$, $f=0x$, then $d(f, x)=0$ and $d(e, x)=1$; since zero is adjacent to all vertices.
	
	Case 3: Let  $e=0x$, $f=0y$, then $d(e, x)=0$ and $d(f, x)=1$ or $2$; by Lemma \ref{l2q}.
	
	Case 4: Let $e=0v$, $f=xy$, then $d(f, x)=0$ and $d(e, x)=1$; since zero is adjacent to all vertices.
	
	Case 5: Let  $e=0v$, $f=vx$, then $d(f, x)=0$, $d(e, x)=1$; since $x \in A_V$, or $x \in A_S$ for some $S \subseteq P$ such that $V \cap S \neq \emptyset $.
	
	Case 6: Let $e=vx$, $f=yz$, then $d(e, x)=0$, $d(f, x)=1$ or $2$; by Lemma \ref{l2q}.
	
	Case 7: Let $e=vx$, $f=vy$, then $d(e, x)=0$, $d(f, x)=1$; since $y \in A_V$, or $y \in A_T$ for some $T \subseteq P$ such that $V \cap T \neq \emptyset $.  
	
	Case 8: Let $e=0x$, $f=yz$, then $d(e, x)=0$, $d(f, x)=1$ or $2$; by Lemma \ref{l2q}.
	
	In each case, $\mathtt{E^{'}}$ is an edge metric basis of $G$. Thus, $dim_{e}(G)\leq n -2$. 
\end{proof}

\begin{definition} A vertex $u \in N(v)$ is said to be a \textit{maximal neighbour} of $v$ if $N[v] \subseteq N[u]$.
\end{definition}
\begin{theorem} (See \cite{Kelenc}) \label{5}
	Let $G$ be a connected graph of order $n$. Then $dim_m (G) = n$ if and only if every vertex of the graph $G$ has a maximal neighbour.
\end{theorem}

\begin{theorem} 
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$. Then $dim_{m}(G)= n(G)$.
\end{theorem}
\begin{proof}
	By the structure of the graph $G$, every vertex $v$ belongs to a twin equivalence class of $A_{S}$ for some $S\subseteq P$. So, each vertex has a maximal neighbour, and the result follows by Theorem \ref{5}.
\end{proof}

\begin{definition} $X \subseteq V$ is called a \textit{twin-free clique} in $G$ if the subgraph induced by $X$ is a clique and for every $u, v \in X$ it follows $N[u] \neq N[v]$. We say that the \textit{twin-free clique number} of $G$, denoted by $\bar \omega (G)$, is the maximum cardinality among all twin-free cliques in $G$.
\end{definition}

\begin{theorem} (See \cite{yero2}) \label{3}
	Let $H$ be a connected graph of order $n \geq 2$. Then $dim_s (H) \leq n - \bar \omega (H)$.
	Moreover, if $H$ has diameter two, then $dim_s (H) = n - \bar \omega (H)$.
\end{theorem}

\begin{remark} \label{rrrr}
	Let $G^{*}$ be the twin graph of $G$ and $\chi$ be a maximal clique in $G^{*}$. Then it induces a clique in $G$ which is twin-free, i.e. $\omega (G^{*}) \leq \bar \omega (G)$.
	Let $\Theta$ be a maximal twin-free clique in $G$, i.e. every vertex is belong to a twin equivalence class $A_S$ for some $S \subseteq P$. Then it induces a clique in $G^{*}$. So, $\bar \omega (G) \leq \omega (G^{*})$. Hence, $\bar \omega (G)  = \omega (G^{*})$.
\end{remark}

\begin{theorem}
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$, then $dim_s (G) = n - 2^{\alpha -1}$.
\end{theorem}
\begin{proof}
	By remark \ref{rrrr}, $\bar \omega (G)  = \omega (G^{*})$.
	If $\alpha =2$, according to decomposition in Theorem \ref{39}, $G ^{*}$ is isomorphic to $P_3$. So, $\omega (G ^{*}) = 2$.
	
	For $\alpha \geq 3$, we claim that $\omega (G^{*}) = 2^{\alpha -1}$. Without lose of generality, let $T=\lbrace p_1 \rbrace \subseteq P$ and $\mathtt{S^{'}}=\lbrace S \subseteq P; \quad p_1 \in S, |S| \geq 2 \rbrace$. Consider the equivalence classes $\mathcal{B} = \lbrace A_{S} ; \quad S \in \mathtt{S^{'}} \rbrace \cup \lbrace A_{T^{c}} \rbrace$. Let $X$ be the set of representative vertices of the equivalence classes $\mathcal{B}$.
	It is clear that $| \mathtt{S^{'}} | = 2^{\alpha -1} - 1$. So, $|X| = 2^{\alpha  - 1}$.For all $S \in \mathtt{S^{'}}$, $S \cap T^{c}\neq \emptyset$. So, by Lemma \ref{l2q}, all the vertices of $X$ are adjacent. It means that the graph induced by $X$ is the complete graph $K_{2^{\alpha -1}}$.
	
	Now, suppose $R \subseteq P$ such that $R \notin \mathtt{S^{'}}$. If $R =T$, then the representative vertex of $A_R$ is not adjacent to the representative vertex of $A_{T^{c}}$. If $R \neq T$, then $R^{c}=S$, for some $S \in \mathtt{S^{'}}$. So, by Lemma \ref{l2q}, the representative vertex of $A_{R}$ is not adjacent to at least one vertex of the clique $K_{2^{\alpha - 1}}$. Hence, $K_{2^{\alpha - 1}}$ is a maximal clique of $G^{*}$.
\end{proof}
%_________________________________________________________________
\begin{lemma} \label{20}
	For any twin vertices $x, y$ of a connected graph $G$, $R \lbrace x, y \rbrace = \lbrace x, y \rbrace$. 
\end{lemma}
\begin{proof}
	Let $z \in R \lbrace x, y \rbrace - \lbrace x, y \rbrace $, then $d(x, z) \neq d(y, z)$. So, $z \notin N(x) \cap N(y)$. Since $G$ is connected and $x, y$ are twins, $d(x, z) = d(y, z)$, which is a contradiction.
\end{proof}

\begin{theorem} ( See \cite{yero3}) \label{ffff}
	Let $G$ be a connected graph of order at least two. Then $dim_f (G) =\frac{|V (G)|}{2}$ if and only if there exists a bijection $\alpha:V (G) \longrightarrow V (G)$ such that $\alpha (v) \neq v$ and $\vert R \lbrace v, \alpha (v) \rbrace \vert = 2$ for all $v \in V (G)$.
\end{theorem}

\begin{theorem}
	Let $G=\Gamma^{\alpha} (Z(\mathbb{Z}_{p_1p_2\ldots p_\alpha}))$. Then $dim_f (G) =\frac{n(G)}{2}$.
\end{theorem}
\begin{proof}
	Let $\alpha: V (G) \longrightarrow V (G)$ such that $\alpha (x) = x +\prod\limits_{p_i \mid x} p_i$. Then it is easy to check that $\alpha$ is a bijection which takes any vertex $x$ to it's twin and $\alpha (x) \neq x$. Moreover, by Lemma \ref{20}, $R \lbrace x, \alpha (x) \rbrace = \lbrace x, \alpha (x) \rbrace$, and the result follows by Theorem \ref{ffff}.
\end{proof}

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{\small

\noindent{\bf Mona Gholamnia Taleshani}

\noindent Department of Mathematics

\noindent Faculty of Mathematical Sciences

\noindent Rasht, Iran

\noindent E-mail: m.gholamniai@gmail.com}\\

{\small
\noindent{\bf   Mozhgan Taghidoost Laskukalayeh}

	\noindent Department of Mathematics
	
\noindent Faculty of Mathematical Sciences

\noindent Rasht, Iran

\noindent E-mail: mtaghidoust@phd.guilan.ac.ir}\\
{\small
	
	\noindent{\bf  Ahmad Abbasi }
	
	\noindent Department of Mathematics
	
	\noindent Associate Professor of Mathematics
	
	\noindent Faculty of Mathematical Sciences
	
	\noindent Rasht, Iran
		
	\noindent E-mail: aabbasi@guilan.ac.ir}\\


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