Very cost effective bipartition in Gamma(Z_n)

TL;DR

This paper explores the conditions under which the zero-divisor graph of Z_n and related graphs have very cost effective bipartitions, focusing on cases where n is a product of distinct primes, and extends the analysis to line and total graphs.

Contribution

It introduces the concept of very cost effective bipartitions in zero-divisor graphs of Z_n and investigates their existence for various n, including line and total graphs.

Findings

Characterization of very cost effective bipartitions for Gamma(Z_n) when n is a product of distinct primes.

Conditions under which N(Z_n) and Omega(Z_n) graphs have very cost effective bipartitions.

Results on very cost effective bipartitions of line and total graphs of Gamma(Z_n).

Abstract

Let Z_n be the finite commutative ring of residue classes modulo n and Gamma(Z_n) be its zero-divisor graph. The nilradical graph and non-nilradical graph of Z_n are denoted by N(Z_n) and Omega(Z_n) respectively. In 2012, Haynes et al. [5] introduced the concept of very cost effective graph. For a graph G = (V,E) and a set of vertices S subset of V, a vertex v in S is said to be very cost effective if it is adjacent to more vertices in V\S than in S. A bipartition Pi = {S, V\S} is called very cost effective if both S and V\S are very cost effective sets [5,6]. In this paper, we investigate the very cost effective bipartition of Gamma(Z_n), where n = p_1 p_2 ... p_m, here all p_i's are distinct primes. In addition, we discuss the cases in which N(Z_n) and Omega(Z_n) graphs have very cost effective bipartition for different n. Finally, we derive some results for very cost effective bipartition of the Line graph and Total graph of Gamma(Z_n), denoted by L(Gamma(Z_n)) and T(Gamma(Z_n)) respectively.

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Full text

††copyright: ©2009: American Mathematical Society

1. INTRODUCTION

Let $R=\mathbb{Z}_{n}$ be the finite commutative ring of residue classes $modulo~{}n$ with identity$(1\neq 0)$ and $\Gamma(\mathbb{Z}_{n})$ be its zero-divisor graph. The study of zero-divisor graphs of commutative rings reveals interesting relation between ring theory and graph theory. Because, algebraic tools help to understand graphs properties and vice versa. An element $z(\neq 0)\in R$ is said to be a zero-divisor if there exists a non-zero $r\in R$ such that $rz=0$. The set of zero-divisors is denoted by $Z(R)$ and zero-divisor graph of $R$, $\Gamma(R)$, is the graph whose vertices are the zero-divisors of $R$ and its two vertices are connected by an edge if and only if their product is [math]. In 1988, the concept of zero-divisor graph of a commutative ring was introduced by I. Beck [2] in context of coloring of rings and were redefined in 1999 by D. F Anderson and P. Livingston in[1].

For a graph $G=(V,E)$, the open neighborhood of a vertex $u\in V$ is the set $N(u)=\{v~{}|~{}uv\in E\}$, and the closed neighborhood of $u$ is the set $N[u]=N(u)\cup\{u\}$. In the same fashion, the open neighborhood of a set $S\subseteq V$ is the set $N(S)=\cup_{u\in S}N(u)$, and the closed neighborhood is the set $N[S]=N(S)\cup S$ respectively. The order of the open neighborhood of a vertex $u\in V$ is denoted by $|N(u)|$. The degree of a vertex $v$ in a graph $G$, denoted by $deg(v)$, is $|N(v)|$. For basic definitions and results on graph we refer [4].

In 2012, the concept of cost effective and very cost effective sets in graphs were introduced by Haynes et al.[5] and further studied in [6] for various graphs. A vertex $v$ in a set $S$ is said to be cost effective if it is adjacent to at least as many vertices in $V\setminus S$ as in $S$, that is, $|N(v)\cap S|\leq|N(v)\cap V\setminus S|$. A Vertex $v$ in a set $S$ is very cost effective if it is adjacent to more vertices in $V\setminus S$ than in $S$, that is, $|N(v)\cap S|<|N(v)\cap V\setminus S|$. A set $S$ is (very) cost effective if every vertex $v\in S$ is (very) cost effective. Moreover, very cost effective bipartition were also introduced in [5]. A bipartition $\pi={\{S,V\setminus S\}}$ is called cost effective if each of $S$ and $V\setminus S$ is cost effective, and $\pi$ is very cost effective if each of $S$ and $V\setminus S$ is very cost effective. Graphs that have a (very) cost effective bipartition are called (very) cost effective graphs. It was shown in [5] that every connected, non-trivial graph is cost effective. Also, they observed in [6] that all bipartite graphs with no isolated vertices are very cost effective.

A line graph $L(G)$ of a simple graph $G$ is obtained by associating a vertex with each edge of the graph $G$ and connecting two vertices by an edge if and only if the corresponding edges of $G$ have a vertex in common. The total graph $T(G)$ of the graph $G$ has a vertex for each edge and each vertex of $G$ and an edge in $T(G)$ for every edge-edge, vertex-edge, and vertex-vertex adjacency in $G$. Here, we illustrate an example of a Zero divisor graph and its line graph over ring $\mathbb{Z}_{16}$:

[TABLE]

Figure 1 $\varGamma(\mathbb{Z}_{16})$

[TABLE]

Figure 2 $L(\varGamma(\mathbb{Z}_{16}))$

2. Bipartition in $\Gamma(\mathbb{Z}_{n})$ and $L(\Gamma(\mathbb{Z}_{n}))$

Let $\pi={\{R,B\}}$ be a bipartition of the graph $G$. If $\pi$ is a very cost effective bipartition of $G$, then we say that $G$ is very cost effective under $\pi$.

Theorem 2.1**.**

*If $n=p_{1}.p_{2}...p_{m},m\geq 1$ and $p_{1}<p_{2}<...<p_{m}$ are primes, then $\Gamma(\mathbb{Z}_{n})$ is a very cost effective graph.

Proof.

Let $n=p_{1}.p_{2}...p_{m},m\geq 1$ for distinct primes $p_{1},p_{2},...,p_{m}$. Then all the zero divisor elements of $\mathbb{Z}_{n}$ are $p_{1},2p_{1},...,(p_{2}...p_{m}-1)p_{1};p_{2},2p_{2},...,(p_{1}p_{3}\\ ....p_{m}-1)p_{2};...,p_{m},2p_{m},...,(p_{1}p_{2}...p_{m-1}-1)p_{m}$. Let $\pi={\{R,B\}}$ be a bipartition of vertices in $\Gamma(\mathbb{Z}_{n})$. Suppose the set $R$ contains all the elements which are multiple of $p_{m}$ and the other set $B$ contains rest of elements. Then $B$ is an independent set because it does not contain any element which is multiple of $p_{m}$. Now, we take $u\in B$. Then there exists at least one element $v\in R$ such that $uv=0$. Therefore, $|N(u)\cap B|=0$ and $|N(u)\cap R|\geq 1$, for all $u\in B$. Hence, all the elements in set $B$ are very cost effective and thus the set $B$ is very cost effective set.

Now, we divide $R$ into two sets $R_{1}$ and $R_{2}$. $R_{1}$ is the subset of $R$ containing multiple of $p_{m}$ as well as some of $p_{i}^{\prime}s$ (not all at a time), for $i=1,...,m-1$ and $R_{2}$ is containing those multiples of $p_{m}$ which are not multiple of any $p_{i}^{\prime}s$, $i=1,...,m-1$. Then element $v\in R_{2}$ is not adjacent to any element in $R$. But, these elements are adjacent to the elements of set $B$, so $|N(v)\cap R|=0$ and $|N(v)\cap B|\geq 1$. Again, if $u\in R_{1}$, then number of vertices adjacent to $u$ in $R$ is $\lfloor\frac{p_{1}.p_{2}...p_{m-1}-1}{\Pi p_{i}}\rfloor=M$, where $\Pi p_{i}$ is the product of those $p_{i}^{\prime}s$ $i=1,...,m-1$ which are not available in $u$ and number of vertices which are adjacent to $u$ in $B$ is $((\Pi p_{i})p_{m}-1-M)$ where $\Pi p_{i}$ is the product of those $p_{i}^{\prime}s$ $i=1,...,m-1$ which are available in $u$. Since $|N(u)\cap R|=M$ and $|N(u)\cap B|>M$. So $|N(u)\cap R|<|N(u)\cap B|$. Therefore, the set $B$ is also very cost effective and the partition $\pi={\{R,B\}}$ is very cost effective bipartition. Hence, the graph $\Gamma(\mathbb{Z}_{n})$ is very cost effective graph. ∎

Corollary 2.2**.**

$\Gamma(\mathbb{Z}_{n})$* is very cost effective graph, for $n=pq$, where $p,q$ are distinct primes.*

Proof.

For $n=pq$, there are two independent sets of vertices in $\Gamma(\mathbb{Z}_{n})$. One set contains multiple of $p$ and other multiple of $q$ respectively. Therefore, the graph $\Gamma(\mathbb{Z}_{n})$ is a complete bipartite graph. Since every bipartite graph without isolated vertices is very cost effective. Thus, $\Gamma(\mathbb{Z}_{n})$ is very cost effective graph.

∎

Theorem 2.3**.**

*Let $p$ and $q$ be distinct primes and $n$ a positive integer.

$(i)$ If $n=p^{2}q$, $p,q\geq 2$, then $\Gamma(\mathbb{Z}_{n})$ is very cost effective.

$(ii)$ If $n=p^{2}q^{2}$, $p,q\geq 3$ and $p<q$, then $\Gamma(\mathbb{Z}_{n})$ is very cost effective.

Proof.

$(i)$ Let $n=p^{2}q$, where $p,q$ be distinct primes. Certainly, zero divisor elements of $\mathbb{Z}_{n}$ are either multiples of $p$ or multiples of $q$ or multiples of $pq$. Let $R$ be the set of vertices contains all those elements which are multiple of $q$ and $B$ contains all those elements which are multiple of $p$ but not $q$. Let $\pi={\{R,B\}}$ be the bipartition of $V(\Gamma(\mathbb{Z}_{n}))$. Then $B$ is an independent set. Now, take $u\in B$, then $|N(u)\cap B|=0$ and $|N(u)\cap R|\geq 1$. In the set $R$, take $v\in R$. If $v$ is multiple of $pq$, then $|N(v)\cap R|=p-2$ and $|N(v)\cap B|>p-2$. Again, if $v$ is a multiple of $q$ only, then $|N(v)\cap R|=0$ and $|N(v)\cap B|\geq 1$. Hence, from both the condition $|N(v)\cap R|<|N(v)\cap B|$. Therefore, the sets $R$ and $B$ are very cost effective and thus, the graph $\Gamma(\mathbb{Z}_{n})$ is very cost effective.

$(ii)$ Let $n=p^{2}q^{2}$, where $p,q$ are distinct odd primes. Then the zero divisor elements are either multiple of $p$ or $q$ or both. Now, take a bipartition $\pi={\{R,B\}}$ in such a way that set $R=R_{1}\cup R_{2}\cup R_{3}$ where $R_{1}=\{v\in V(\Gamma(\mathbb{Z}_{p^{2}q^{2}})):p^{2}\mid v\}$, $R_{2}=\{v\in V(\Gamma(\mathbb{Z}_{p^{2}q^{2}})):p\mid v~{}and~{}q\notdivides v\}$ and $R_{3}=\{v\in V(\Gamma(\mathbb{Z}_{p^{2}q^{2}})):pq\mid v~{}and~{}p^{2}\notdivides v~{}and~{}q^{2}\notdivides v\}$. Here, $R_{3}$ contains $\frac{q(p-2)+1}{2}$ number of vertices. Now, set $B=B_{1}\cup B_{2}\cup B_{3}$ where $B_{1}=\{v\in V(\Gamma(\mathbb{Z}_{p^{2}q^{2}})):q^{2}\mid v\}$, $B_{2}=\{v\in V(\Gamma(\mathbb{Z}_{p^{2}q^{2}})):q\mid v~{}and~{}p\notdivides v\}$ and $B_{3}=\{v\in V(\Gamma(\mathbb{Z}_{p^{2}q^{2}})):pq\mid v~{}and~{}p^{2}\notdivides v~{}and~{}q^{2}\notdivides v\}$. Here, $B_{3}$ contains $\frac{p(q-2)+1}{2}$ number of vertices. Let $u\in R$. If $u$ is not a multiple of $q$, then $|N(u)\cap R|=0$ and $|N(u)\cap B|\geq 1$. Again, if $u$ is a multiple of $p$ as well as $q$, then $|N(u)\cap R|=\frac{pq-3}{2}$ and $|N(u)\cap B|\geq\frac{pq-1}{2}$. So $|N(u)\cap R|<|N(u)\cap B|$ and the set $R$ is very cost effective.

Similarly, set $B$ is also very cost effective and the bipartition $\pi$ is very cost effective bipartition. Thus, the graph $\Gamma(\mathbb{Z}_{n})$ is very cost effective. ∎

Theorem 2.4**.**

$L(\Gamma(\mathbb{Z}_{n}))$* is very cost effective, Where $n=pq$, $p<q$ and $p,q$ are primes.

Proof.

If $n=pq$, then zero divisor graph $\Gamma(\mathbb{Z}_{n})$ is a complete bipartite graph. So, there are two independent sets of vertices in which each vertex of a set is adjacent to every vertex of the other set. We draw the line graph of $\Gamma(\mathbb{Z}_{n})$ with $(p-1)(q-1)$ vertices. This line graph is $(p+q-4)$ regular graph. Let $[u_{i},v_{j}]\in V(L(\Gamma(\mathbb{Z}_{n})))$, where $u_{i}^{\prime}s$ are multiple of $p$ $i.e$ $u_{i}=p.i$, $1\leq i\leq q-1$ and $v_{j}^{\prime}s$ are multiple of q $i.e$ $v_{j}=q.j$, $1\leq j\leq p-1$. Now, Let $\pi={\{R,B\}}$ be a bipartition of vertices in $L(\Gamma(\mathbb{Z}_{n}))$. Now we shall prove that $\pi$ is very cost effective bipartition. Since each set $R$ and $B$ contain $\frac{(p-1)(q-1)}{2}$ vertices. Therefore, set $R$ contains $\{[u_{1},v_{k}],[u_{2},v_{L}],[u_{3},v_{k}],[u_{4},v_{L}],...,[u_{q-2},v_{k}],[u_{q-1},v_{L}]\}$ where $1\leq k\leq\frac{p-1}{2}$ and $\frac{p+1}{2}\leq L\leq p-1$ and $B$ contains $\{[u_{1},v_{L}],[u_{2},v_{k}],[u_{3},v_{L}],[u_{4},v_{k}],...,$ $[u_{q-2},v_{L}],[u_{q-1},v_{k}]\}$ where $1\leq k\leq\frac{p-1}{2}$ and $\frac{p+1}{2}\leq L\leq p-1$. Now, take $[u_{i},v_{j}]\in R$. Then $|N([u_{i},v_{j}])\cap R|=\frac{p+q}{2}-3$ and $|N([u_{i},v_{j}])\cap B|=\frac{p+q}{2}-1$. Therefore, $|N([u_{i},v_{j}])\cap R|<|N([u_{i},v_{j}])\cap B|$. Hence, $R$ is very cost effective set. Similarly, if we take $[u_{i},v_{j}]\in B$, then $|N([u_{i},v_{j}])\cap B|<|N([u_{i},v_{j}])\cap R|$. Thus, the line graph $L(\Gamma(\mathbb{Z}_{n}))$ is very cost effective graph.

∎

3. Bipartition in $N(\mathbb{Z}_{n})$ and $\Omega(\mathbb{Z}_{n})$

In 2008, Bishop et al. [3] introduced the concept of Nilradical graph and Non-Nilradical graph and further some work appeared in [7]. They defined these graphs as follows:

Definition 3.1**.**

[3] The nilradical graph, denoted $N(R)$, is the graph whose vertices are the nonzero nilpotent elements of $R$ and two vertices are connected by an edge if and only if their product is [math].

Definition 3.2**.**

[3] The non-nilradical graph, denoted $\Omega(R)$, is the graph whose vertices are the non-nilpotent zero-divisors of $R$ and where two vertices are connected by an edge if and only if their product is [math].

Fig 1 is serve an example of $N(\mathbb{Z}_{16})$. Also the example of $\Omega(\mathbb{Z}_{1}8)$ is given below:

[TABLE]

Figure 3 $\Omega(\mathbb{Z}_{18})$

Theorem 3.3**.**

*Let $p,q$ be distinct primes and $n$, a positive integer.

$(i)$ If $n=p^{2}$ and $p>2$, then $N(\mathbb{Z}_{n})$ is very cost effective graph.

$(ii)$ If $n=p^{2}q^{2}$ and $p,q\geq 2$, then $N(\mathbb{Z}_{n})$ is very cost effective graph.

$(iii)$ If $n=p^{3}$ and $p\geq 2$, then $N(\mathbb{Z}_{n})$ is very cost effective graph.

$(iv)$ If $n=p^{2}q$ and $p>2$, then $N(\mathbb{Z}_{n})$ is very cost effective graph.

Proof.

$(i)$ Let $n=p^{2}$, where $p$ is a prime number and $p>2$. Then the nilpotent elements in $\mathbb{Z}_{n}$ are $p,2p,...,(p-1)p$. So, the number of nilpotent elements is $p-1$ and every element is adjacent to the other element. So, these $(p-1)$ elements forms a complete graph. Since $p$ is prime, then $p-1$ is even and every complete graph of even order is very cost effective. Thus, $N(\mathbb{Z}_{n})$ is very cost effective graph.

$(ii)$ Let $n=p^{2}q^{2}$, where $p,q$ are distinct primes. Then the nilpotent elements in $\mathbb{Z}_{n}$ are multiple of $pq$ and number of nilpotent elements are $pq-1$. Since all the nilpotent elements are multiple of $pq$, so every vertex is adjacent to the all other vertices in $N(\mathbb{Z}_{n})$. Therefore, $(pq-1)$ elements form a complete graph with $pq-1$ vertices. Since $p$ and $q$ are prime number and $p,q>2$, so $pq-1$ is even number. Hence, the graph of $N(\mathbb{Z}_{n})$ is very cost effective.

$(iii)$ If $n=p^{3}$, where $p$ is prime number, then all the nilpotent elements are multiple of $p$ and total number of nilpotent elements are $p^{2}-1$. Now, we take bipartition $\pi={\{R,B\}}$ of vertex set in $N(\mathbb{Z}_{n})$ such that $R$ contains only those elements which are divisible by $p$ but not by $p^{2}$ and the set $B$ contains those elements which are divisible by only $p^{2}$. Then the set $R$ has $p(p-1)$ elements and the set $B$ has $p-1$ elements. Now, set $R$ is independent set and in $B$, each vertex is adjacent to every other vertices in $B$. Since all the elements of $R$ is adjacent to all elements of set $B$. Therefore, $R$ is very cost effective set. Now, take $u\in B$, then $|N(u)\cap B|=p-2$ and $|N(u)\cap R|=p(p-1)$. Since $|N(u)\cap B|<|N(u)\cap R|$, so set $B$ is very cost effective. Hence, $\pi={\{R,B\}}$ is very cost effective bipartition and graph $N(\mathbb{Z}_{n})$ is very cost effective.

$(iv)$ Let $n=p^{2}q$, where $p$ and $q$ are distinct prime number and $p\neq q$. Then the nilpotent elements of $N(\mathbb{Z}_{n})$ are multiples of $pq$ and the number of nilpotent elements are $p-1$. These $p-1$ elements are connected to each other so these $p-1$ vertices forms a complete graph. Since $p$ is an odd prime so $p-1$ is even and complete graph of even number is very cost effective. Hence, $N(\mathbb{Z}_{n})$ is very cost effective graph.

∎

Theorem 3.4**.**

*If $p$ and $q$ are distinct prime number and $n$ is a positive integer, then $\Omega(\mathbb{Z}_{n})$ is not very cost effective graph, where $n=p^{2}q$.

Proof.

Let $n=p^{2}q$, where $p$ and $q$ be a distinct primes. Then the non-nilradical elements are all zero-divisors which are not divisible by $pq$. Since these elements are not adjacent to themselves but the vertices which are multiple of $p^{2}$ is adjacent to multiple of $q$. Here, $p$ is also another vertex which is not adjacent to any other vertices. Therefore, $p$ is an isolated vertices. Hence $\Omega(\mathbb{Z}_{n})$ is not very cost effective graph.

∎

Theorem 3.5**.**

*Let $p_{1},p_{2},...,p_{m}$ be distinct prime and $n=p_{1}.p_{2}...p_{m},m\geq 1$. Then $\Omega(\mathbb{Z}_{n})$ is very cost effective graph.

Proof.

Here, $\Omega(\mathbb{Z}_{n})=\Gamma(\mathbb{Z}_{n})$. So $\Omega(\mathbb{Z}_{n})$ is very cost effective graph.

∎

4. Bipartition in $T(\Gamma(\mathbb{Z}_{n}))$

In this section, we have studied the very cost effective properties of $T(\Gamma(\mathbb{Z}_{n}))$ for $n=2p$ and $n=pq$.

Theorem 4.1**.**

Let $n=2p$, p be an odd prime. Then the total graph $T(\Gamma(\mathbb{Z}_{n}))$ is not very cost effective.

Proof.

If $n=2p$, then $\Gamma(\mathbb{Z}_{n})$ is a star graph with $p$ vertices. Now, total graph of $\Gamma(\mathbb{Z}_{n})$ is a graph with $2p-1$ vertices in which $p-1$ vertices have degree two, $p-1$ vertices have degree $p$ and one vertex which is itself $p$ has degree $2p-2$ respectively. In order to prove $T(\Gamma(\mathbb{Z}_{n}))$ is not very cost effective, let $T(\Gamma(\mathbb{Z}_{n}))$ be a very cost effective graph. Then it has a very cost effective bipartition $\pi={\{R,B\}}$. Take $u\in R$ and suppose $u$ is a vertex whose degree is $2$. Then $|N(u)\cap R|=0$ and $|N(u)\cap B|=2$. Now, the vertex $p\in B$ because $u$ is adjacent to the vertex $p$. So, all the vertices whose degree are $2$ belong to the set $R$ because these vertices are adjacent to vertex $p$. Also, rest of the $p-1$ vertices which are of the form $[2,p],[2.2,p],...,[2(p-1),p]$ will be in $B$. If any one of these vertices belongs to $R$, then $R$ is not very cost effective set. So, take $[u_{i},p]\in B$, where $u_{i}=2.i$, $1\leq i\leq p-1$. Then $|N([u_{i},p])\cap B|=p-1$ and $|N([u_{i},p])\cap R|=1$. Hence, the set $B$ is not very cost effective set, which contradicts our assumption. Thus, the graph $T(\Gamma(\mathbb{Z}_{n}))$ is not very cost effective.

∎

Theorem 4.2**.**

Let $n=pq$, $p,q\geq 3$ and $p<q$ are primes. Then total graph $T(\Gamma(\mathbb{Z}_{n}))$ is very cost effective.

Proof.

If $n=pq$, then $\Gamma({Z}_{n})$ is a bipartite graph with $p+q-2$ vertices and $(p-1)(q-1)$ edges. Now, in total graph $T(\Gamma(\mathbb{Z}_{n}))$, there is $pq-1$ vertices in which $q-1$ vertices are multiples of $p$ and each has degree $2(p-1)$. Also $p-1$ vertices which are multiples of $q$ have degree $2(q-1)$. The vertices $[u_{i},v_{j}]\in V(T(\Gamma(\mathbb{Z}_{n}))$, $u_{i}=p.i$, $1\leq i\leq q-1$ and $v_{j}=q.j$, $1\leq j\leq p-1$ have degree $p+q-2$. Suppose set $R$ contains the element $\{p,2p,...,(q-1)p,[u_{1},v_{k}],[u_{2},v_{L}],[u_{3},v_{k}],...,[u_{q-1},v_{L}]\}$ and $B$ contains $\{q,2q,...,(p-1)q,[u_{1},v_{L}],[u_{2},v_{k}],[u_{3},v_{L}],...,[u_{q-1},v_{k}]\}$ where $1\leq k\leq\frac{p-1}{2}$, $\frac{p+1}{2}\leq L\leq p-1$. Take bipartition $\pi={\{R,B\}}$. We have to show that this bipartition is very cost effective.

Now, consider $u_{i}\in R$, then $|N(u_{i})\cap R|=p-2$ and $|N(u_{i})\cap B|=p$. Also, for $[u_{i},v_{j}]\in R$, we have $|N([u_{i},v_{j}])\cap R|=\frac{p+q-4}{2}$ and $|N([u_{i},v_{j}])\cap B|=\frac{p+q}{2}$. So, for every vertex $v\in R$, $|N(v)\cap R|<|N(v)\cap B|$ and the set $R$ is very cost effective set. In the set $B$, take $v_{j}\in B$, then $|N(v_{j})\cap B|=q-2$ and $|N(v_{j})\cap R|=q$. Again, take $[u_{i},v_{j}]\in B$, we have $|N([u_{i},v_{j}])\cap B|=\frac{p+q-4}{2}$ and $|N([u_{i},v_{j}])\cap R|=\frac{p+q}{2}$. Therefore, $|N(v)\cap B|<|N(v)\cap R|$ for every vertex $v$ in $B$. So set $B$ is also very cost effective. Hence, the bipartition $\pi$ is very cost effective bipartition and the total graph $T(\Gamma(\mathbb{Z}_{n}))$ is very cost effective graph.

∎