#  Construction of Linear Codes from the Unit Graph $G(\mathbb{Z}_{n})$

Dr. Rupali S. Jain  Associate Professor, School of Mathematical Sciences,
S.R.T.M. University, Nanded  Dr. B. Surendranath Reddy  Assistant Professor,
School of Mathematical Sciences, S.R.T.M. University, Nanded  Mr. Wajid M.
Shaikh  Research Scholar, School of Mathematical Sciences, S.R.T.M.
University, Nanded

###### Abstract

In this paper, we consider the unit graph $G(\mathbb{Z}_{n})$, where
$n=p_{1}^{n_{1}}\text{ or }p_{1}^{n_{1}}p_{2}^{n_{2}}\text{ or
}p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}$ and $p_{1},p_{2},p_{3}$ are distinct
primes. For any prime $q$, we construct $q$-ary linear codes from the
incidence matrix of the unit graph $G(\mathbb{Z}_{n})$ with their parameters.
We also prove that the dual of the constructed codes have minimum distance
either 3 or 4. Lastly, we stated two conjectures on diameter of unit graph
$G(\mathbb{Z}_{n})$ and linear codes constructed from the incidence matrix of
the unit graph $G(\mathbb{Z}_{n})$ for any integer $n$.

##  1 Introduction

In 1990, the unit graph was first introduced by Grimaldi, R. P.[14] for
$\displaystyle\mathbb{Z}_{n}$, is the graph with no loops and parallel edges
and $\displaystyle x,y\in\mathbb{Z}_{n}$ are adjacent if and only if $x+y$ is
a unit in $\mathbb{Z}_{n}$.

In 2010, Fish, W., Key, J. D., $\&$ Mwambene, E.[7], constructed linear codes
form incidence matrices of line graphs of Hamming graphs. Key, J. D., $\&$
Rodrigues, B. G. [8] constructed codes from lattice graphs and examine their
decoding techniques using permutation decoding. Similar type of results were
examined by several researcher [9, 10]. In 2013, Dankelmann, P., Key, J. D.,
$\&$ Rodrigues, B. G. [3] gave the generalization of relationship between
parameters of connected graphs and codes generated from incidence matrices of
graphs and obtained upper bounds for minimum distance of their dual codes.
Also several researchers carried out their research in construction of linear
codes from adjacency matrices of some special graphs, as mentioned in [11, 12,
13].

Recently, in 2021, Annamalai, N., $\&$ Durairajan, C.[1], constructed linear
codes from the incidence matrices of unit graph $\displaystyle
G(\mathbb{Z}_{p})$ and $\displaystyle G(\mathbb{Z}_{2p})$, for odd prime $p$.
In this paper, we generalise their work by constructing linear codes from the
incidence matrices of unit graph $\displaystyle G(\mathbb{Z}_{n})$, where
$n=p_{1}^{n_{1}}\text{ or }p_{1}^{n_{1}}p_{2}^{n_{2}}\text{ or
}p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}$ and $p_{1},p_{2},p_{3}$ are distinct
primes. We also find the parameters for these codes and their dual codes.
Finally we conclude by stating two conjectures.

##  2 Preliminaries

In this section, we recall definitions and results related to unit graphs and
linear codes. Let $\mathbb{Z}_{n}$ denote the ring of integers modulo $n$.
Here, we denote units and non-units of $\displaystyle\mathbb{Z}_{n}$ by
$\displaystyle U(\mathbb{Z}_{n})$ and $\displaystyle N_{U}(\mathbb{Z}_{n})$
respectively.

######  Definition 2.1.

[18][Linear Code] Let $\displaystyle\mathbb{F}_{q}$ represents the finite
field with $q$ elements. A linear code $C_{q}$ of length $n$ is a subspace of
$\displaystyle\mathbb{F}^{n}_{q}$ and it is called $q$-ary linear code.
Dimension of linear code $C_{q}$ is the dimension of $C_{q}$ as a vector space
over field $\displaystyle\mathbb{F}_{q}$ and is denoted by
$\displaystyle\text{dim}(C_{q})$.

######  Definition 2.2.

[18][Dual of code] Let $C_{q}$ be a linear code of length $n$ over
$\displaystyle\mathbb{F}_{q}$. Then dual of code $C_{q}$ is the orthogonal
compliment of the subspace $C_{q}$ in $\displaystyle\mathbb{F}^{n}_{q}$ and is
denoted by $\displaystyle C^{\perp}_{q}$.

######  Theorem 2.3.

[18] Let $C_{q}$ be a $q$-ary code of length $n$ over a field
$\displaystyle\mathbb{F}_{q}$. Then $\displaystyle C^{\perp}_{q}$ is a linear
code of length $n$ and
$\displaystyle\text{dim}(C_{q}^{\perp})=n-\text{dim}(C_{q})$.

######  Definition 2.4.

  1. 1.

Let $x$ and $y$ be vectors in $\displaystyle\mathbb{F}^{n}_{q}$. Then Hamming
distance of $x$ and $y$, denoted by $\displaystyle d_{C}(x,y)$, is defined to
be the number of places at which $x$ and $y$ differ.

  2. 2.

Let $x$ be a vector in $\displaystyle\mathbb{F}^{n}_{q}$. Then Hamming weight
of $x$ is defined to be the number of non-zero coordinates in $x$ and is
denoted by $\text{wt}(x)$. Clearly, $\displaystyle\text{wt}(x-y)=d_{C}(x,y)$.

######  Definition 2.5.

[18][Minimum Hamming weight] Let $C_{q}$ be a linear code. Then minimum
Hamming weight of $C_{q}$, denoted by $\text{wt}(C_{q})$, is defined as
$\displaystyle\text{wt}(C_{q})=\text{min}\\{\text{wt}(x)\ |\ x\in C_{q}\ \&\
x\neq 0\\}$.

######  Remark 2.6.

[18][Minimum Hamming distance] Let $C_{q}$ be a linear code. The minimum
Hamming distance of code $C_{q}$, denoted by $d(C_{q})$, is defined as
$\displaystyle d(C_{q})=\text{min}\\{d_{C}(x,y)\ |\ x,y\in C_{q}\ \&\ x\neq
y\\}$. Note that $d(C_{q})=\text{wt}(C_{q})$.

######  Remark 2.7.

[18] A $q$-ary linear code $C_{q}$ of length $n$, dimension $k$ and minimum
distance $d$ is called $[n,k,d]_{q}$ linear code.

######  Definition 2.8.

[18] A generator matrix of linear code $C_{q}$ is a matrix $H$ whose rows form
a basis for $C_{q}$ and a generator matrix $\displaystyle H^{\perp}$ of linear
code $C^{\perp}_{q}$ is called parity-check matrix of $C_{q}$.

Let $\displaystyle G=(V,E)$ be a graph with vertex set $V$ and edge set $E$.
For any $x,y\in V$, $[x,y]$ denote the edge between $x$ and $y$ and if
$[x,y]\in E$ then we call $x$ is adjacent to $y$. A graph is called simple if
it does not have loops and parallel edges. A complete graph is a simple graph
in which any distinct pair of vertices is joined by an edge. If the vertex set
$V$ of $G$ can be partitioned into two non-empty subsets $W_{1}$ and $W_{2}$
such that each edge in $G$ has one end in $W_{1}$ and one end in $W_{2}$ then
$G$ is called bipartite. A complete bipartite graph is a simple bipartite
graph $G$, with bipartition $V=W_{1}\cup W_{2}$, in which every vertex in
$W_{1}$ is joined to every vertex in $W_{2}$.

An edge $e$ of graph $G$ is said to be incident with the vertex $x$, if $x$ is
the end vertex of $e$. In this case we also say that $x$ is incident with $e$.
The degree of vertex $x$ is the number of edges of $G$ incident with $x$ and
it is denoted by $\text{deg}(x)$. If $x\in V$ and
$\text{deg}(x)\leq\text{deg}(y)$ for all $y\in V$ then $\text{deg}(x)$ is
called minimum degree of $G$ and it is denoted by $\delta(G)$. If for some
positive integer $k$, $\text{deg}(x)=k$ for every vertex of the graph of $G$,
then $G$ is called $k$-regular graph. A vertex $x$ is said to be connected to
a vertex $y$ in a graph $G$ if there is path in $G$ from $x$ to $y$. A graph
is called connected if every two vertices are connected. A nontrivial closed
trail in graph $G$ is called cycle if its origin and internal vertices are
distinct. A cycle of length $k$, i.e. with $k$ edges is called $k$-cycle. The
girth of graph $G$, is denoted by $g_{r}(G)$, is the length of the shortest
cycle contained in $G$.

######  Definition 2.9.

The distance between two vertices $x$ and $y$, denoted by $d(x,y)$, is the
length of a shortest path from $x$ to $y$. The diameter of a graph $G$ is
denoted by $\text{diam}(G)$, is the maximum distance between any two vertices
in $G$. i.e. $\displaystyle\text{diam}(G)=\text{Max}\\{d(x,y)\ |\ x,y\in
V\\}$.

######  Definition 2.10.

[19] Let $G$ be a simple graph. The edge connectivity of $G$, denoted by
$\lambda(G)$, is the smallest number of edges in $G$ whose deletion from $G$
either leaves a disconnected graph or an empty graph.

######  Definition 2.11.

[2] Let $R$ be a ring with nonzero identity. The unit graph of $R$, denoted by
$G(R)$, is a graph with vertex set as $R$ and two distinct vertices $x$ and
$y$ are adjacent if and only if $x+y$ is a unit of $R$.

######  Theorem 2.12.

[4] Let $\displaystyle G=(V,E)$ be a connected graph with vertex set $V$.
If $\text{diam}(G)\leq 2$ then edge connectivity of $G$ is
$\lambda(G)=\delta(G)$.

######  Theorem 2.13.

[5] Let $\displaystyle G=(V,E)$ be a connected bipartite graph, if
$\displaystyle\text{diam}(G)\leq 3$ then edge connectivity of $G$ is
$\displaystyle\lambda(G)=\delta(G)$.

######  Theorem 2.14.

[2] Let $\mathcal{R}$ be a finite ring. Then the following statements hold for
the unit graph of $\mathcal{R}$

  1. 1.

If $\displaystyle 2\notin U(\mathcal{R})$, then the unit graph
$G(\mathcal{R})$ is a $|U(\mathcal{R})|$-regular graph.

  2. 2.

If $\displaystyle 2\in U(\mathcal{R})$, then for every $x\in U(\mathcal{R})$
we have $\displaystyle\text{deg}(x)=|U(\mathcal{R})|-1$ and for every
$\displaystyle x\in N_{U}(\mathcal{R})$ we have
$\displaystyle\text{deg}(x)=|U(\mathcal{R})|$.

######  Theorem 2.15.

[15] Let $R$ be a ring. Then

  1. 1.

If $\displaystyle|U(R)|=2$ then $g_{r}(G)\in\\{3,4,6\\}$.

  2. 2.

If $\displaystyle|U(R)|\geq 3$ then $g_{r}(G)\in\\{3,4\\}$.

######  Theorem 2.16.

[3] Let $G=(V,E)$ be a connected graph and let $H$ be a
$\displaystyle|V|\times|E|$ incidence matrix for $G$. Then binary code
generated by $H$ is
$\displaystyle C_{2}(H)=\displaystyle[|E|,|V|-1,\lambda(G)]_{2}$.

######  Theorem 2.17.

[3] Let $G=(V,E)$ be a connected bipartite graph and let $H$ be a
$\displaystyle|V|\times|E|$ incidence matrix for $G$, and $q$ be an odd prime.
Then $q$-ary code generated by $H$ is $\displaystyle
C_{q}=[|E|,|V|-1,\lambda(G)]_{q}$.

######  Theorem 2.18.

[3] Let $G$ be a connected graph with girth $g_{r}(G)$ and even girth
$g_{r}(G)_{e}$. Let $H$ be an incidence matrix for $G$, $C=C_{q}(H)$, where
$q$ is any prime, and $\displaystyle d^{\perp}$ is the minimum distance of
$\displaystyle C_{q}^{\perp}$. If $q=2$ or $g_{r}(G)$ is even then
$\displaystyle d^{\perp}=g_{r}(G)$.

##  3 Construction of Linear Codes from the Incidence Matrix of a Unit Graph
$G(\mathbb{Z}_{p^{n}})$

In this section, we construct binary and $q$-ary linear codes $C_{2}$ and
$C_{q}$ generated from the incidence matrix of the unit graph $\displaystyle
G(\mathbb{Z}_{p^{n}})$, where $p$ is any prime number and $n\in\mathbb{N}$. We
also examine the dual codes $C^{\perp}_{2}$ and $C^{\perp}_{q}$ with their
parameters.

Let $\displaystyle\mathbb{Z}_{p^{n}}$ denote the ring. Then $\displaystyle
U(\mathbb{Z}_{p^{n}})=\\{x\in\mathbb{Z}_{p^{n}}\ |\
\text{g.c.d}(x,p^{n})=1\\}$ and $\displaystyle
N_{U}(\mathbb{Z}_{p^{n}})=\\{x\in\mathbb{Z}_{p^{n}}\ |\ x=\alpha
p\\}\neq\phi$. Let $G(\mathbb{Z}_{p^{n}})$ be a unit graph with vertex set
$\displaystyle V=\mathbb{Z}_{p^{n}}$ and $x,y\in\mathbb{Z}_{p^{n}}$ are
adjacent if and only if $x+y\in U(\mathbb{Z}_{p^{n}})$.

######  Theorem 3.1.

Let $\displaystyle G(\mathbb{Z}_{p^{n}})$ be a unit graph and $p$ is an odd
prime. Then the graph $\displaystyle G(\mathbb{Z}_{p^{n}})$ is connected with
$\displaystyle|V|=p^{n}$ and
$\displaystyle|E|=\frac{(p^{n}-1)\phi(p^{n})}{2}$.

###### Proof.

Clearly, $\displaystyle|V|=p^{n}$. Note that for every $\displaystyle x\in
U(\mathbb{Z}_{p^{n}})$ and $\displaystyle y\in N_{U}(\mathbb{Z}_{p^{n}})$,
$[x,y]$ is an edge in $G(\mathbb{Z}_{p^{n}})$. Suppose this is not true which
gives $x+y$ is not a unit, this implies $\displaystyle p|x+y$. But
$\displaystyle y\in N_{U}(\mathbb{Z}_{p^{n}})$ i.e. $y=\alpha p$ this implies
$p|x$ which is a contradiction. Hence $\displaystyle G(\mathbb{Z}_{p^{n}})$ is
connected graph.

Since $p$ is odd prime, $2\in U(\mathbb{Z}_{p^{n}})$. By theorem (2.14), we
have $\displaystyle\text{deg}(x)=|U(\mathbb{Z}_{p^{n}})|-1=\phi(p^{n})-1,\
\forall\ x\in U(\mathbb{Z}_{p^{n}})$ and
$\displaystyle\text{deg}(x)=\phi(p^{n}),\ \forall\ x\in
N_{U}(\mathbb{Z}_{p^{n}})$. Now

$\displaystyle|E|$
$\displaystyle=\frac{\sum_{x\in\mathbb{Z}_{p^{n}}}\text{deg}(x)}{2}$
$\displaystyle=\frac{\sum_{x\in U(\mathbb{Z}_{p^{n}})}\text{deg}(x)+\sum_{x\in
N_{U}(\mathbb{Z}_{p^{n}})}\text{deg}(x)}{2}$
$\displaystyle=\frac{\phi(p^{n})[\phi(p^{n})-1]+[p^{n}-\phi(p^{n})]\phi(p^{n})}{2}$
$\displaystyle=\frac{\phi(p^{n})[\phi(p^{n})-1+p^{n}-\phi(p^{n})]}{2}$
$\displaystyle|E|$ $\displaystyle=\frac{\phi(p^{n})[p^{n}-1]}{2}$

∎

######  Theorem 3.2.

Let $\displaystyle G(\mathbb{Z}_{p^{n}})$ be a unit graph and $p$ is an odd
prime. Then edge connectivity of $\displaystyle G(\mathbb{Z}_{n})$ is
$\displaystyle\lambda(G(\mathbb{Z}_{p^{n}}))=\displaystyle\phi(p^{n})-1$.

###### Proof.

First we show that $\text{diam}(G(\mathbb{Z}_{p^{n}}))\leq 2$. For any
$\displaystyle x,y\in\mathbb{Z}_{p^{n}}$ we have following cases,
Case I: Let $\displaystyle x,y\in U(\mathbb{Z}_{p^{n}})$. Then there exist
$\displaystyle z\in N_{U}(\mathbb{Z}_{p^{n}})$ such that $x$ and $y$ adjacent
to $z$. Hence $d(x,y)\leq 2$.
Case II: If $\displaystyle x,y\in N_{U}(\mathbb{Z}_{p^{n}})$ then for any
$z\in U(\mathbb{Z}_{p^{n}})$, we have $[x,z]$ and $[z,y]$ are edges in
$\displaystyle G(\mathbb{Z}_{p^{n}})$, which gives $d(x,y)\leq 2$
Case III: If $\displaystyle x\in U(\mathbb{Z}_{p^{n}})$ and $\displaystyle
y\in N_{U}(\mathbb{Z}_{p^{n}})$ then $x$ is adjacent to $y$, which gives
$d(x,y)=1$.
We get $d(x,y)\leq 2$ for all $\displaystyle x,y\in\mathbb{Z}_{p^{n}}$ which
gives $\text{diam}(G(\mathbb{Z}_{p^{n}}))\leq 2$.
Now by Theorem 2.12, we have
$\displaystyle\lambda(G(\mathbb{Z}_{p^{n}}))=\delta(G(\mathbb{Z}_{p^{n}}))=\phi(p^{n})-1$.
∎

######  Corollary 3.3.

Let $\displaystyle G(\mathbb{Z}_{p^{n}})$ be a unit graph, where $p$ is an odd
prime and $\displaystyle p^{n}\neq 3$. Then $\displaystyle
g_{r}(G(\mathbb{Z}_{p^{n}}))=3$.

###### Proof.

If $n=1$ then $p\geq 3$. Note that $\displaystyle 1,2\in\mathbb{Z}_{p}$ and
$1$ is adjacent to $2$. Since, $\displaystyle 0\in\mathbb{Z}_{p}$ which gives
$1$ and $2$ are adjacent to $0$. Hence, we get cycle of length 3, which
conclude our result. If $n>1$ then for any $\displaystyle x\in
U(\mathbb{Z}_{p^{n}})$ we have $\displaystyle x+p\in U(\mathbb{Z}_{p^{n}})$.
Also $x$ is adjacent to $x+p$. Note that $x$ and $x+p$ are adjacent to $p$.
Hence, we get cycle of length $3$, which implies
$g_{r}(G(\mathbb{Z}_{p^{n}}))=3$. ∎

######  Lemma 3.4.

Let $\displaystyle G(\mathbb{Z}_{2^{n}})$ be a unit graph. Then $x$ and $y$
are adjacent if and only if $x\in U(\mathbb{Z}_{2^{n}})$ and $y\in
N_{U}(\mathbb{Z}_{2^{n}})$.

###### Proof.

Let $G(\mathbb{Z}_{2^{n}})$ be a unit graph. Suppose that, $x$ and $y$ are
adjacent for some $x,y\in\mathbb{Z}_{2^{n}}$. That is $[x,y]$ is an edge in
$G(\mathbb{Z}_{2^{n}})$, which implies $x+y\in U(\mathbb{Z}_{2^{n}})$, hence
$x+y=2k+1$ for some integer $k$. Assume that $\displaystyle x,y\in
U(\mathbb{Z}_{2^{n}})$ which imlies $x=2k_{1}+1$ and $y=2k_{2}+1$ for some
integer $k_{1}$ and $k_{2}$. From this, we have $\displaystyle 2|x+y$, which
gives a contradiction to our assumption. Similarly, if $\displaystyle x,y\in
N_{U}(\mathbb{Z}_{p^{n}})$, then $\displaystyle 2|x+y$, which is again a
contradiction. Hence $\displaystyle x\in U(\mathbb{Z}_{p^{n}})$ and
$\displaystyle y\in N_{U}(\mathbb{Z}_{2^{n}})$.
Conversely, let $\displaystyle x\in U(\mathbb{Z}_{2^{n}})$ and $\displaystyle
y\in N_{U}(\mathbb{Z}_{2^{n}})$. Then $\displaystyle x=2k_{1}$ and
$\displaystyle y=2k_{1}+1$ this implies $2\nmid x+y$ and hence $[x,y]$ is an
edge in $G(\mathbb{Z}_{2^{n}})$. ∎

######  Corollary 3.5.

Let $\displaystyle G(\mathbb{Z}_{2^{n}})$ be a unit graph. Then $\displaystyle
G(\mathbb{Z}_{2^{n}})$ is complete bipartite graph with bipartition
$\displaystyle W_{1}=\\{x\in\mathbb{Z}_{2^{n}}\ |\ x=2\alpha\ \text{for some
}\ \alpha\in\mathbb{Z}\\}$ and $\displaystyle W_{2}=\\{x\in\mathbb{Z}_{2^{n}}\
|\ x=2\alpha+1\ \text{for some }\ \alpha\in\mathbb{Z}\\}$.

######  Theorem 3.6.

Let $\displaystyle G(\mathbb{Z}_{2^{n}})$ be a unit graph. Then

  1. 1.

$\displaystyle|V|=2^{n}$ and $\displaystyle|E|=2^{2(n-1)}$.

  2. 2.

$\displaystyle\lambda(G(\mathbb{Z}_{2^{n}}))=\phi(2^{n})$.

###### Proof.

  1. 1.

By the definition of unit graph $\displaystyle G(\mathbb{Z}_{2^{n}})$, we have
$\displaystyle|V|=2^{n}$. By Corollary (3.5), $\displaystyle
G(\mathbb{Z}_{2^{n}})$ is complete bipartite graph, which implies
$\displaystyle|E|=2^{n-1}2^{n-1}=2^{2(n-1)}$.

  2. 2.

From Corollary (3.5), $\displaystyle G(\mathbb{Z}_{2^{n}})$ is complete
bipartite graph and hence
$\displaystyle\lambda(G(\mathbb{Z}_{2^{n}}))=\phi(2^{n})$.

∎

######  Theorem 3.7.

Let $\displaystyle G(\mathbb{Z}_{p^{n}})$ be a unit graph and $H$ be a
$|V|\times|E|$ incidence matrix of $G(\mathbb{Z}_{p^{n}})$.

  1. 1.

If $p$ is an odd prime then binary code generated by $H$ is a
$\displaystyle
C_{2}(H)=\left[\frac{\phi(p^{n})(p^{n}-1)}{2},p^{n}-1,\phi(p^{n})-1\right]_{2}$
code over the finite field $\displaystyle\mathbb{F}_{2}$.

  2. 2.

If $p=2$ then for any odd prime $q$, the $q$-ary code generated by $H$ is a
$\displaystyle C_{q}(H)=[2^{2(n-1)},2^{n}-1,2^{(n-1)}]_{q}$ code over the
finite field $\displaystyle\mathbb{F}_{q}$.

###### Proof.

  1. 1.

Let $G(\mathbb{Z}_{p^{n}})$ be a unit graph, where $p$ is an odd prime and $H$
be an incidence matrix of $G(\mathbb{Z}_{p^{n}})$. By Theorem (3.1),
$G(\mathbb{Z}_{p^{n}})$ is connected graph and hence by Theorem (2.16), binary
code generated by $H$ is
$C_{2}(H)=[|E|,|V|-1,\lambda(G(\mathbb{Z}_{p^{n}}))]_{2}$. By theorem (3.1)
and (3.2), we get $\displaystyle|E|=\frac{\phi(p^{n})(p^{n}-1)}{2},\
|V|=p^{n}$ and the edge connectivity of $\displaystyle G(\mathbb{Z}_{p^{n}})$
is $\lambda(G(\mathbb{Z}_{p^{n}}))=\phi(p^{n})-1$. Hence we get $\displaystyle
C_{2}(H)=\left[\frac{\phi(p^{n})(p^{n}-1)}{2},p^{n}-1,\phi(p^{n})-1\right]_{2}$.

  2. 2.

Let $p=2$. Then by Corollary (3.5), $\displaystyle G(\mathbb{Z}_{2^{n}})$ is a
complete bipartite graph, which implies $G(\mathbb{Z}_{2^{n}})$ is connected
bipartite graph and hence by Theorem (2.16) and (2.17), for any prime $q$,
$q$-ary code generated by $H$ is
$C_{q}(H)=[|E|,|V|-1,\lambda(G(\mathbb{Z}_{2^{n}}))]_{q}$. Using Theorem
(3.6), we get $\displaystyle|E|=2^{2(n-1)},\ |V|=2^{n}$ and the edge
connectivity of $\displaystyle G(\mathbb{Z}_{2^{n}})$ is
$\lambda(G(\mathbb{Z}_{2^{n}}))=2^{n-1}$. Hence $\displaystyle
C_{q}(H)=[2^{2(n-1)},2^{n}-1,2^{n-1}]_{q}$.

∎

######  Corollary 3.8.

Let $C_{q}(H)$ and $C_{2}(H)$ denote the codes generated by incidence matrix
of $\displaystyle G(\mathbb{Z}_{2^{n}})$ and $\displaystyle
G(\mathbb{Z}_{p^{n}})$. Then

  1. 1.

Dual of code $C_{2}(H)$ is $\displaystyle
C^{\perp}_{2}=\left[\frac{(p^{n}-1)\phi(p^{n})}{2},\frac{(p^{n}-1)[\phi(p^{n})-2]}{2},3\right]_{2}$,
where $p^{n}\neq 3$.

  2. 2.

Dual of code $C_{q}(H)$ is $\displaystyle
C^{\perp}_{q}=\left[2^{2(n-1)},2^{n}(2^{n-2}-1)+1,4\right]_{q}$.

###### Proof.

  1. 1.

It follows from Theorems (2.3) & (2.18) and Corollary (3.3).

  2. 2.

Since $\displaystyle G(\mathbb{Z}_{2^{n}})$ is complete bipartite graph, we
have $\displaystyle g_{r}(G)=4$ and hence result follows from Theorems (2.3)
and (2.18).

∎

##  4 Construction of Linear Codes from the Incidence Matrix of a Unit Graph
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$

In this section, we construct binary and $q$-ary linear codes from the
incidence matrix of the unit graphs $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, where $p_{1}$ and $p_{2}$ are
distinct primes and $\displaystyle n_{1},n_{2}\in\mathbb{N}$. We also discuss
the parameters their dual codes.

######  Lemma 4.1.

Let $\displaystyle G(\mathbb{Z}_{n})$ be a unit graph. If $2\in
N_{U}(\mathbb{Z}_{n})$, then $\displaystyle G(\mathbb{Z}_{n})$ is bipartite.

###### Proof.

We have $\displaystyle W_{1}\cup W_{2}=\mathbb{Z}_{n}$, where $\displaystyle
W_{1}=\\{x\in\mathbb{Z}_{n}\ |\ x=2\alpha\ \text{for some
}\alpha\in\mathbb{Z}\\}$ and $\displaystyle W_{2}=\\{x\in\mathbb{Z}_{n}\ |\
x=2\alpha+1\ \text{for some }\alpha\in\mathbb{Z}\\}$. Now for any
$\displaystyle x,y\in W_{1}$, we have $\displaystyle\text{gcd}(x+y,n)\geq 2$,
which implies that $x$ is not adjacent to $y$. Similarly, for any
$\displaystyle x,y\in W_{2}$, we have $\displaystyle\text{gcd}(x+y,n)\geq 2$.
Hence $\displaystyle G(\mathbb{Z}_{n})=G(W_{1}\cup W_{2})$ is a bipartite
graph. ∎

######  Theorem 4.2.

Let $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ be a unit
graph, where $\displaystyle p_{1}$ and $\displaystyle p_{2}$ are distinct odd
primes. Then $G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ is a connected graph
and $\displaystyle\text{diam}(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))\leq
2$.

###### Proof.

We can rewrite

$\displaystyle\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}$
$\displaystyle=U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})\cup
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$
$\displaystyle=U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})\cup N_{p_{1}}\cup
N_{p_{2}}\cup N_{p_{1}p_{2}}$

where

$\displaystyle N_{p_{1}}$
$\displaystyle=\\{x\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}\ |\ x=\alpha
p_{1}\ \&\ p_{2}\nmid\alpha\\}$ $\displaystyle N_{p_{2}}$
$\displaystyle=\\{x\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}\ |\ x=\alpha
p_{2}\ \&\ p_{1}\nmid\alpha\\}$ $\displaystyle N_{p_{1}p_{2}}$
$\displaystyle=\\{x\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}\ |\ x=\alpha
p_{1}p_{2}\\}$

For $\displaystyle x,y\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}$, we have
following cases
Case I: If $\displaystyle x,y\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$,
then $\displaystyle[x,0]$ and $\displaystyle[0,y]$ are edges in $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, which gives $d(x,y)\leq 2$.
Case II: If $\displaystyle x\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$
and $\displaystyle y\in N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, then
we have the following possibilities for $y$:
(a) If $\displaystyle y\in N_{p_{1}}$ ,then $y=\alpha p_{1}$. Suppose that
$\displaystyle[x,y]=[x,\alpha p_{1}]$ is not an edge in $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ which implies $\displaystyle
p_{2}\mid x+\alpha p_{1}$ i.e. $x+\alpha p_{1}=\beta p_{2}$. Note that
$p_{1}\nmid\beta$, hence $\displaystyle\beta p_{2}\in N_{p_{2}}$. Now we claim
that $\displaystyle[x,\beta p_{2}]$ is an edge in $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Suppose this is no true. Then
$p_{1}\mid x+\beta p_{2}$, we get $\displaystyle p_{1}\mid 2x+\alpha p_{1}$,
i.e. $\displaystyle p_{1}\mid x$ which is a contradiction. Hence, we get $x$
is adjacent to $\beta p_{2}$. Note that $\displaystyle[\beta p_{2},\alpha
p_{1}]$ is an edge in $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. i.e. $\displaystyle[x,\beta
p_{2}]$ and $\displaystyle[\beta p_{2},\alpha p_{1}]$ are edges in
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Hence $d(x,y)\leq
2$.
(b) If $\displaystyle y\in N_{p_{2}}$, then $y=\alpha p_{2}$. By following the
above procedure, we get $d(x,y)\leq 2$.
(c) If $\displaystyle y\in N_{p_{1}p_{2}}$, then $y=\alpha p_{1}p_{2}$. Since
$x$ is a unit, we get $\displaystyle[x,\alpha p_{1}p_{2}]$ is an edge in
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. For if this not
hold, then we have $\displaystyle p_{1}\mid x+\alpha p_{1}p_{2}$ i.e.
$p_{1}\mid x$ which is not possible. Similarly if $\displaystyle p_{2}\mid
x+\alpha p_{1}p_{2}$ then $p_{2}\mid x$. Hence $d(x,y)=1\leq 2$.
Case III: If $\displaystyle x,y\in
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ then, consider the following
possibilities
(a) If $\displaystyle x,y\in N_{p_{1}}$ then $x=\alpha p_{1}$ and
$\displaystyle y=\alpha^{\prime}p_{1}$. Since, $\displaystyle p_{2}\in
N_{p_{2}}$ which implies $N_{p_{2}}\neq\phi$. Then for any $\displaystyle\beta
p_{2}\in N_{p_{2}}$,. $\displaystyle[\alpha p_{1},\beta p_{2}]$ is an edges in
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Similarly, we can
prove that $\displaystyle[\beta p_{2},\alpha^{\prime}p_{1}]$ is an edge in
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. i.e.
$\displaystyle[x,\beta p_{2}]$ and $\displaystyle[\beta p_{2},y]$ are edges in
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Hence, we have
$\displaystyle d(x,y)\leq 2$.
(b) If $\displaystyle x,y\in N_{p_{2}}$, then $d(x,y)\leq 2$ as in similar to
case (a).
(c) If $\displaystyle x,y\in N_{p_{1}p_{2}}$, then $x=\alpha p_{1}p_{2}$ and
$\displaystyle y=\alpha^{\prime}p_{1}p_{2}$. Note that $\displaystyle
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})\neq\phi$. Then for any
$\displaystyle z\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, we have
$\displaystyle[\alpha p_{1}p_{2},z]$ and
$\displaystyle[z,\alpha^{\prime}p_{1}p_{2}]$ are edges in $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Hence $\displaystyle d(x,y)\leq
2$.
(d)If $\displaystyle x\in N_{p_{1}}$ and $\displaystyle y\in N_{p_{2}}$, then
$x$ is adjacent to $y$, hence $\displaystyle d(x,y)=1\leq 2$.
(e) If $\displaystyle x\in N_{p_{1}}$ and $\displaystyle y\in N_{p_{1}p_{2}}$,
then $x=\alpha p_{1}\ \&\ y=\alpha^{\prime}p_{1}p_{2}$. Now, let $\beta
p_{2}\in N_{2}$ then $\displaystyle z=\alpha p_{1}+\beta p_{2}\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, from this we have
$[z,\alpha^{\prime}p_{1}p_{2}]$ is an edge. Also $\displaystyle[\alpha
p_{1},z]$ is an edge in $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Hence $d(x,y)\leq 2$.
(f) If $\displaystyle x\in N_{p_{2}}$ and $\displaystyle y\in N_{p_{1}p_{2}}$,
then using the above procedure, we can show that $d(x,y)\leq 2$.
Hence, $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ is connected
graph and $d(x,y)\leq 2$, for all $\displaystyle
x,y\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}$, which gives
$\text{diam}(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))\leq 2$. ∎

######  Corollary 4.3.

Let $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ be a unit
graph, where $\displaystyle p_{1}$ and $\displaystyle p_{2}$ are odd primes.
Then $\displaystyle g_{r}(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))=3$.

###### Proof.

Note that, $p_{1}$ is adjacent to $p_{2}$ and $p_{1}+p_{2}$ is adjacent to
both $p_{1}\ \&\ p_{2}$, from this we conclude the result. ∎

######  Theorem 4.4.

Let $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ be a unit
graph, where $\displaystyle p_{1}$ and $p_{2}$ are odd distinct primes. Then

  1. 1.

$\displaystyle|V|=p^{n_{1}}_{1}p^{n_{2}}_{2}$ and
$\displaystyle|E|=\frac{(p^{n_{1}}_{1}p^{n_{2}}_{2}-1)\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})}{2}$.

  2. 2.

$\displaystyle\lambda(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\delta(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})-1$.

###### Proof.

  1. 1.

Here $\displaystyle|V|=p^{n_{1}}_{1}p^{n_{2}}_{2}$. Since $\displaystyle 2\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, by Theorem (2.14),
$\displaystyle\text{deg}(x)=\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})-1$ for all
$\displaystyle x\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$ and
$\displaystyle\text{deg}(x)=\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})$ for all
$\displaystyle x\in N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Thus,

$\displaystyle|E|$ $\displaystyle=\frac{\sum_{x\in V}\text{deg}(x)}{2}$
$\displaystyle=\frac{\sum_{x\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})}\text{deg}(x)+\sum_{x\in
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})}\text{deg}(x)}{2}$
$\displaystyle=\frac{\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})(\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})-1)+(p^{n_{1}}_{1}p^{n_{2}}_{2}-\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}))\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})}{2}$
$\displaystyle|E|$
$\displaystyle=\frac{(p^{n_{1}}_{1}p^{n_{2}}_{2}-1)\phi(p^{n_{1}}_{1}p^{n_{2}}_{2})}{2}.$

  2. 2.

It follows from Theorem (4.2) and Theorem (2.12).

∎

######  Theorem 4.5.

Let $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})=(V,E)$ be a unit graph,
where $p_{1}$ is an odd prime. Then $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ is a connected graph and
$\displaystyle\text{diam}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))\leq 3$.

###### Proof.

Let
$\displaystyle\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}=U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})\cup
N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$.Then

$\displaystyle N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$
$\displaystyle=N_{2}\cup N_{p_{1}}\cup N_{2p_{1}}$ $\displaystyle\text{where
}N_{2}$ $\displaystyle=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}\ |\ x=2\alpha\
\&\ p_{1}\nmid\alpha\\}$ $\displaystyle N_{p_{1}}$
$\displaystyle=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}\ |\ x=p_{1}\alpha\ \&\
2\nmid\alpha\\}$ $\displaystyle N_{2p_{1}}$
$\displaystyle=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}\ |\ x=2\alpha p_{1}\\}$

Now for any $x,y\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}$, we have the following
cases:
Case I: If $\displaystyle x,y\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, then
$\displaystyle[x,0]$ and $\displaystyle[0,y]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$. Hence $\displaystyle d(x,y)=1$.
Case II: If $\displaystyle x\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ and
$\displaystyle y\in N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, then we have the
following possibilities for $y$,
(a) If $y\in N_{2}$, then $y=2\alpha$ for some integer $\alpha$. Suppose $x$
is not adjacent to $y$ which gives $\displaystyle p_{1}\mid x+2\alpha$. Since
$p_{1}\in N_{P_{1}}$, which implies $N_{p_{1}}\neq\phi$. Let $\beta p_{1}\in
N_{p_{1}}$. Then $\displaystyle z=2\alpha+\beta p_{1}\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$. Then $[z,y]$ is an edge. Since
$\displaystyle x,z\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, $x$ and $y$ are
adjacent to $0$. Hence, we get $\displaystyle[x,0],\ [0,z]$ and
$\displaystyle[z,y]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, which gives $d(x,y)\leq 3$.
(b) If $\displaystyle y\in N_{p_{1}}$, then $y=\alpha p_{1}$. Note that $x$ is
not adjacent to $y$. Since $\displaystyle 2\in N_{2}$ which implies
$N_{2}\neq\phi$. Now for any, $z=2\beta\in N_{2}$, $y$ is adjacent to $z$. Now
it is enough to show that $x$ is adjacent to $\displaystyle 2\beta$ or
$\displaystyle 2\beta$, suppose $x$ is not adjacent to $\displaystyle
2\beta\in N_{2}$ and $\displaystyle-2\beta\in N_{2}$, which implies
$\displaystyle p_{1}\mid x+2\beta$ and $\displaystyle p_{1}\mid x-2\beta$
which gives $p_{1}\mid 4\beta$, hence $p_{1}\mid\beta$, which is contradiction
to choice of $\displaystyle\beta$. Hence, $\displaystyle[x,2\beta]$ or
$\displaystyle[x,-2\beta]$ is edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, which implies $d(x,y)=2$.
(c) If $\displaystyle y\in N_{2p_{1}}$, then $y=2\alpha p_{1}$. Since
$\displaystyle x\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, $x$ is adjacent to
$y$, if this not hold then either $\displaystyle p_{1}\mid x+y$ or
$\displaystyle 2\mid x+y$. If $\displaystyle p_{1}\mid x+2\alpha p_{1}$, then
$p_{1}\mid x$ which is a contradiction. In similar way, If $\displaystyle
2\mid x+2\alpha p_{1}$, then $\displaystyle 2\mid x$ which is also
contradiction to $\displaystyle x\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$.
Hence $\displaystyle[x,y]$ is an edge in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ which gives, $\displaystyle d(x,y)=1\leq
3$.
Case III: If $\displaystyle x,y\in N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$,
then we have the following possibilities for $x$ and $y$:
(a) If $\displaystyle x,y\in N_{2}$, then $\displaystyle x=2\alpha$ and
$\displaystyle y=2\alpha^{\prime}$. For any $\displaystyle z=\beta p_{1}\in
N_{p_{1}}$, we have $[x,z]$ and $[z,y]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$. Hence, we have $\displaystyle d(x,y)=2$.
(b) If $\displaystyle x,y\in N_{p_{1}}$, then from above procedure for any
$\displaystyle z\in N_{2}$, we have $[x,z]$ and $[z,y]$ are edges in
$\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$. Hence we get
$\displaystyle d(x,y)=2\leq 3$.
(c) If $\displaystyle x,y\in N_{2p_{1}}$, then for any $\displaystyle z\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, we have $\displaystyle[x,z]$ and
$\displaystyle[z,y]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, which gives $\displaystyle d(x,y)=2$.
(d) If $\displaystyle x\in N_{2}$ and $\displaystyle y\in N_{p_{1}}$ or vice
versa, then $x$ is adjacent to $y$, which implies $d(x,y)=1$.
(e)If $\displaystyle x\in N_{2}$ and $y\in N_{2p_{1}}$, then $x=2\alpha$ and
$\displaystyle y=2\alpha^{\prime}p_{1}$. Now for any $z=\beta p_{1}\in
N_{p_{1}}$, we have $\displaystyle x+z\in
N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ and $x$ is adjacent to $x+z$. Since
$\displaystyle x+z\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, we have $x+z$ is
adjacent to $y$. Hence, we get $[x,x+z]$ and $[x+z,y]$ are edges in
$\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, which gives $d(x,y)=2\leq
3$.
(f) If $\displaystyle x\in N_{p_{1}}$ and $y\in N_{2p_{1}}$, then $x=\alpha
p_{1}$ and $\displaystyle y=2\alpha^{\prime}p_{1}$. Now for any $z=2\beta\in
N_{2}$, we have $x$ is adjacent to $z$ and $z$ is adjacent to $x+z$ both
follows from above part. Note that $\displaystyle x+z\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$. Hence, we have $x+z$ is adjacent to $y$,
from this, we get $\displaystyle[x,z],\ [z,x+z]$ and $\displaystyle[x+z,y]$
are edges in $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$, which give
$d(x,y)\leq 3$.
Hence, $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ is connected graph
and $\displaystyle d(x,y)\leq 3$, for all $\displaystyle
x,y\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}$. Therefore, we have
$\displaystyle\text{diam}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))\leq 3$. ∎

######  Corollary 4.6.

Let $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ be a unit graph, where
$p_{1}$ is an odd prime.

  1. 1.

If $\displaystyle 2^{m}p^{n_{1}}_{1}\neq 6$ then $\displaystyle
g_{r}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))=4$

  2. 2.

If $\displaystyle 2^{m}p^{n_{1}}_{1}=6$ then $\displaystyle
g_{r}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))=6$

###### Proof.

Proof follows from Theorems (2.15) and (4.1). ∎

######  Theorem 4.7.

Let $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ be a unit graph, where
$\displaystyle p_{1}$ is an odd prime. Then

  1. 1.

$\displaystyle|V|=2^{m}p^{n_{1}}_{1}$ and
$\displaystyle|E|=2^{m-1}p^{n_{1}}_{1}\phi(2^{m}p^{n_{1}}_{1})$.

  2. 2.

$\displaystyle\lambda(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))=\delta(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))=\phi(2^{m}p^{n_{1}}_{1})$.

###### Proof.

  1. 1.

From the definition of unit graph, we have
$\displaystyle|V|=2^{m}p^{n_{1}}_{1}$. Since $\displaystyle 2\in
N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ from Theorem (2.14), we get
$\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ is
$\displaystyle\phi(2^{m}p^{n_{1}}_{1})$-regular graph and hence, we have
$\displaystyle|E|=\frac{2^{m}p^{n_{1}}_{1}\phi(2^{m}p^{n_{1}}_{1})}{2}=2^{m-1}p^{n_{1}}_{1}\phi(2^{m}p^{n_{1}}_{1})$.

  2. 2.

From Lemma (4.1) and Theorems (4.4) & (2.13), we get
$\displaystyle\lambda(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))=\delta(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))=\phi(2^{m}p^{n_{1}}_{1})$.

∎

######  Theorem 4.8.

Let $\displaystyle G(\mathbb{Z}_{n})$ be a unit graph, where
$n=p^{n_{1}}_{1}p^{n_{2}}_{2}$ and $p_{1}\ \&\ p_{2}$ are distinct primes. Let
$H$ be a $|V|\times|E|$ incidence matrix of $G(\mathbb{Z}_{n})$.

  1. 1.

If $\displaystyle 2\in U(\mathbb{Z}_{n})$, then binary code generated by $H$
is a $\displaystyle
C_{2}(H)=\left[\frac{(n-1)\phi(n)}{2},n-1,\phi(n)-1\right]_{2}$ code over
finite field $\mathbb{F}_{2}$.

  2. 2.

If $\displaystyle 2\in N_{U}(\mathbb{Z}_{n})$, then for any odd prime $q$, the
$q$-ary code generated by $H$ is a $\displaystyle
C_{q}(H)=\left[\frac{n\phi(n)}{2},n-1,\phi(n)\right]_{q}$ code over finite
field $\mathbb{F}_{q}$.

###### Proof.

  1. 1.

If $\displaystyle 2\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, then
$p_{1}$ and $p_{2}$ are odd primes. By Theorem (4.2), $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})=(V,E)$ is a connected graph and
hence by Theorem (2.16), binary code generated by $H$ is $\displaystyle
C_{2}(H)=[|E|,|V|-1,\lambda(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))]_{2}$.
Now from Theorem (4.4), we get $\displaystyle|E|=\frac{(n-1)\phi(n)}{2},\\\
|V|-1=n-1$ and
$\displaystyle\lambda(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\phi(n)-1$.

  2. 2.

If $\displaystyle 2\in N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$, then
either $p_{1}=2$ or $p_{2}=2$. By Theorem (4.5) and Lemma (4.1),
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})=(V,E)$ is a
connected bipartite graph and hence by Theorems (2.16) and (2.17), $q$-ary
code generated by $H$ is $\displaystyle
C_{q}(H)=[|E|,|V|-1,\lambda(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}}))]_{q}$. Now
using Theorem (4.7), we conclude the result.

∎

######  Corollary 4.9.

Let $C_{q}(H)$ and $C_{2}(H)$ denote the linear codes generated from incidence
matrices of $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}})$ and
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Then

  1. 1.

Dual of code $C_{2}$ is $\displaystyle
C^{\perp}_{2}=\left[\frac{(n-1)\phi(n)}{2},\frac{(n-1)[\phi(n)-2]}{2},3\right]_{2}$,
where $\displaystyle n=p^{n_{1}}_{1}p^{n_{2}}_{2}$.

  2. 2.

Dual of code $C_{q}$ is $\displaystyle
C^{\perp}_{q}=\left[\frac{n\phi(n)}{2},\frac{n(\phi(n)-2)+2}{2},4\right]_{q}$,
where $\displaystyle n=2^{m}p^{n_{1}}_{1}\neq 6$.

###### Proof.

  1. 1.

From Theorem (2.3), $\text{diam}(\displaystyle
C^{\perp}_{2})=\frac{(n-1)[\phi(n)-2]}{2}$.
By Theorem (2.18) and Corollary (4.3), $\displaystyle d(C^{\perp}_{2})=3$.

  2. 2.

Proof follows from Theorems (2.3) & (2.18) and Corollary (4.6).

∎

##  5 Construction of Linear Codes from the Incidence Matrix of a Unit Graph
$G(\mathbb{Z}_{p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}})$

In this section, we extend the result for three distinct primes $p_{1}$,
$p_{2}$ and $p_{3}$.

######  Theorem 5.1.

Let $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ be
a unit graph, where $\displaystyle p_{1},\ p_{2}$ and $\displaystyle p_{3}$
are distinct odd primes. Then $\displaystyle
G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ is connected graph
and
$\displaystyle\text{diam}(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}))\leq
2$.

###### Proof.

Let
$\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}=U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})\cup
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Also we can
express
$N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})=\displaystyle\bigcup_{i=1}^{3}N_{p_{i}}\bigcup_{i\neq
j}N_{p_{i}p_{j}}\bigcup N_{p_{1}p_{2}p_{3}}$,
where $\displaystyle
N_{p_{i}}=\left\\{x\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}\ |\
x=\alpha p_{i}\ \&\ p_{j}\nmid\alpha,\ \forall\ i\neq j\right\\}$,
$\displaystyle
N_{p_{i}p_{j}}=\left\\{x\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}\
|\ x=\alpha p_{i}p_{j}\ \&\ p_{k}\nmid\alpha,\ \ k\neq i,j\right\\}$
and $\displaystyle
N_{p_{1}p_{2}p_{3}}=\left\\{x\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}\
|\ x=\alpha p_{1}p_{2}p_{3}\right\\}$. Now for any $\displaystyle
x,y\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}$ We have the
following cases:
Case I: If $\displaystyle x,y\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$, then
$\displaystyle[x,0]$ and $\displaystyle[0,y]$ are edges. Hence, $d(x,y)\leq
2$.
Case II: If $\displaystyle x\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ and $\displaystyle
y\in N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$, then we
have the following possibilities for $y$:
(a) If $\displaystyle y\in N_{p_{i}}$, then $y=\alpha p_{i}$. Suppose $x$ and
$y$ are not adjacent then $\displaystyle p_{j}\mid x+\alpha p_{i}$ for $i\neq
j$ which implies $x+\alpha p_{i}=\beta p_{j}$. Note that $\displaystyle
p_{i}\nmid\beta$, hence $\beta p_{j}p_{k}\in N_{p_{j}p_{k}}$ and $-\beta
p_{j}p_{k}\in N_{p_{j}p_{k}}$ for $i\neq j,k$. Now our claim is either
$\displaystyle[x,\beta p_{j}p_{k}]$ or $[x,-\beta p_{j}p_{k}]$ is edge in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Suppose both not
hold which implies $\displaystyle p_{i}\mid x+\beta p_{j}p_{k}$ and
$\displaystyle p_{i}\mid x-\beta p_{j}p_{k}$, from this we get $\displaystyle
p_{i}\mid 2x$, this implies $p_{i}\mid x$ which is a contradiction. Now we
prove $y$ is adjacent to $\beta p_{j}p_{k}$, suppose this is not true, which
implies $\displaystyle p_{i}\mid\alpha p_{i}+\beta p_{j}p_{k}$, from this we
have, $p_{i}\mid\beta$ which is a contradiction. Similar contradiction occurs,
when we assuming $\displaystyle p_{j}\mid\alpha p_{i}+\beta p_{j}p_{k}$ and
$\displaystyle p_{k}\mid\alpha p_{i}+\beta p_{j}p_{k}$. Hence
$\displaystyle[\alpha p_{i},\beta p_{j}p_{k}]$ is an edge. In similar way, we
can show that $\displaystyle[\alpha p_{i},-\beta p_{j}p_{k}]$ is an edge in
$\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$.
Hence, there exist $\displaystyle z\in N_{p_{j}p_{k}}$ such that
$\displaystyle[x,z]$ and $\displaystyle[z,y]$ are edges in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$, from this we
concluded that $\displaystyle d(x,y)\leq 2$.
(b) If $\displaystyle y\in N_{p_{i}p_{j}}$, then $y=\alpha p_{i}p_{j}$.
Suppose $x$ and $y$ are not adjacent then $\displaystyle p_{k}\mid x+\alpha
p_{i}p_{j}$ for $k\neq i,j$, which implies $x+\alpha p_{i}p_{j}=\beta p_{k}$.
Note that $\displaystyle p_{i}\nmid\beta$ and $\displaystyle p_{j}\nmid\beta$
which gives $\beta p_{k}\in N_{p_{k}}$. Now we prove that
$\displaystyle[x,\beta p_{k}]$ is an edge, suppose this is not an edge, which
implies $\displaystyle p_{i}\mid x+\beta p_{k}$ for some $i\neq k$. Hence
$\displaystyle p_{i}\mid x+x+\alpha p_{i}p_{j}$ this implies $p_{i}\mid 2x$,
which is a contradiction. $\displaystyle[x,\beta p_{k}]$ is an edge and using
procedure in $(a)$, we can show that $\displaystyle[\beta p_{k},\alpha
p_{i}p_{j}]=[\beta p_{k},y]$. Hence $d(x,y)\leq 2$.
(c) If $\displaystyle y\in N_{p_{1}p_{2}p_{3}}$, then $\displaystyle[x,\alpha
p_{1}p_{2}p_{3}]$ is an edge in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Hence $d(x,y)=1$.
Case III: If $\displaystyle x,y\in
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$, then we have the
following possibilities:
(a) If $\displaystyle x,y\in N_{p_{i}}$, then $x=\alpha p_{i}$ and
$\displaystyle y=\alpha^{\prime}p_{i}$. Since $N_{p_{j}p_{k}}\neq\phi$. For
any $\displaystyle\beta p_{j}p_{k}\in N_{p_{j}p_{k}}$ for $i\neq j,k$, we have
$\displaystyle[x,\beta p_{j}p_{k}]$ and $\displaystyle[\beta p_{j}p_{k},y]$
are edges in $G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Hence,
$d(x,y)\leq 2$.
(b) If $\displaystyle x,y\in N_{p_{i}p_{j}}$, then $x=\alpha p_{i}p_{j}$ and
$\displaystyle y=\alpha^{\prime}p_{i}p_{j}$. Let $\displaystyle\beta p_{k}\in
N_{p_{k}}$ for $k\neq i,j$. Then $\displaystyle[x,\beta p_{k}]$ and
$\displaystyle[\beta p_{k},y]$ are edges in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Hence, $d(x,y)\leq
2$.
(c) $\displaystyle x,y\in N_{p_{1}p_{2}p_{3}}$, then for any $\displaystyle
z\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$,
$\displaystyle[x,z]$ and $\displaystyle[z,x]$ are edges in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Hence, $d(x,y)\leq
2$.
(d) If $\displaystyle x\in N_{p_{i}}$ and $\displaystyle y\in N_{p_{j}}$ for
$i\neq j$, then $\displaystyle x=\alpha p_{i}$ and $y=\alpha^{\prime}p_{j}$.
Assume that $\displaystyle[x,y]$ is not an edge, which gives $\displaystyle
p_{k}\mid x+y$, for $k\neq i,j$. Now our claim is that $\displaystyle[x+y,x]$
and $\displaystyle[y,x+y]$ are edges. Suppose $x+y$ is not adajcent to $x$,
hence $p_{m}\mid 2x+y$ for some $\displaystyle m=1,2,3$. If $p_{i}\mid 2x+y$
i.e. $p_{i}\mid 2\alpha p_{i}+\alpha^{\prime}p_{j}$, then
$p_{i}\mid\alpha^{\prime}$, which is a contradiction. Similarly, if $p_{j}\mid
2x+y$, then $p_{j}\mid\alpha$ and if $p_{k}\mid 2x+y$ i.e. $p_{k}\mid x+y+x$,
then $p_{k}\mid x$, since $\displaystyle p_{k}\mid x+y$. Hence,
$\displaystyle[x+y,x]$ is an edge in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. In similar way, we
can show that $x+y$ is adjacent to $y$, which gives $\displaystyle d(x,y)\leq
2$.
(e) If $\displaystyle x\in N_{p_{i}p_{j}}$ and $\displaystyle y\in N_{p_{j}}$,
then $\displaystyle x=\alpha p_{i}p_{j}$ and $\displaystyle
y=\alpha^{\prime}p_{j}$. Note that for any $\displaystyle z=\beta p_{k}$, for
$k\neq i,j$, $x$ is adjacent to $z$. Now assuming that $y=\alpha p_{j}\in
N_{p_{j}}$ and $\displaystyle\beta p_{k}\in N_{p_{k}}$ are not adjacent, then
$\displaystyle p_{i}\mid\alpha p_{j}+\beta p_{k}$. Since $\displaystyle\beta
p_{k}\in N_{p_{k}}$, which implies $-\beta p_{k}\in N_{p_{k}}$ this gives
$\displaystyle[y,-\beta p_{k}]$ is an edge. Hence, $y$ is adjacent to either
$\displaystyle\beta p_{k}$ or $\displaystyle-\beta p_{k}$, which gives
$d(x,y)\leq 2$.
(f) If $\displaystyle x\in N_{p_{i}p_{j}}$ and $\displaystyle y\in
N_{p_{j}p_{k}}$ for $\displaystyle i\neq k$, then $\displaystyle x=\alpha
p_{i}p_{j}$ and $\displaystyle y=\alpha^{\prime}p_{j}p_{k}$. Now for any
$\displaystyle\beta p_{i}\in N_{p_{i}}$, we have $\displaystyle z=(\beta
p_{i}+\alpha^{\prime}p_{j})p_{k}+\alpha p_{i}p_{j}\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ such that
$\displaystyle[x,z]$ and $\displaystyle[z,y]$ are edges in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. To see this,
suppose $x$ is not adjacent to $z$ which implies $\displaystyle
p_{i}\mid\alpha p_{i}p_{j}+(\beta p_{i}+\alpha^{\prime}p_{j})p_{k}+\alpha
p_{i}p_{j}$, we get $p_{i}\mid\alpha^{\prime}$ which is not possible.
Similarly , if $\displaystyle p_{j}\mid\alpha p_{i}p_{j}+(\beta
p_{i}+\alpha^{\prime}p_{j})p_{k}+\alpha p_{i}p_{j}$, then $p_{j}\mid\beta$ and
if $\displaystyle p_{k}\mid\alpha p_{i}p_{j}+(\beta
p_{i}+\alpha^{\prime}p_{j})p_{k}+\alpha p_{i}p_{j}$, then $p_{k}\mid 2\alpha$
both are not possible, which gives $x$ is adjacent to $z$. Hence, we get
$d(x,y)\leq 2$.
(g) If $\displaystyle x\in N_{p_{i}p_{j}}$ and $\displaystyle y\in N_{p_{k}}$
for $k\neq i,j$, then $\displaystyle[x,y]$ is an edge in
$G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ gives $\displaystyle
d(x,y)=1$.
Hence, $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$
is connected graph and $d(x,y)\leq 2$, for all $\displaystyle
x,y\in\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}$, which gives,
$\displaystyle\text{diam}(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}))\leq
2$. ∎

######  Corollary 5.2.

Let $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ be
a unit graph, where $\displaystyle p_{1},p_{2}$ and $\displaystyle p_{3}$ are
odd primes. Then $\displaystyle
g_{r}(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}))=3$.

###### Proof.

The result follows from $p_{1}$ is adjacent to $p_{2}p_{3}$ and
$p_{1}+p_{2}p_{3}$ is adjacent to both $p_{1}\ \&\ p_{2}p_{3}$. ∎

######  Theorem 5.3.

Let $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ be
a unit graph, where $\displaystyle p_{1},p_{2}$ and $p_{3}$ are odd primes.
Then

  1. 1.

$\displaystyle|V|=p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}$ and
$\displaystyle|E|=\frac{(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}-1)\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})}{2}$.

  2. 2.

$\displaystyle\lambda(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}))=\delta(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}))=\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})-1$.

###### Proof.

  1. 1.

Clearly, $\displaystyle|V|=p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}$. Since
$\displaystyle 2\in U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$
by Theorem (2.14)
$\displaystyle\text{deg}(x)=\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})-1$
for all $\displaystyle x\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$ and
$\displaystyle\text{deg}(x)=\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})$ for
all $\displaystyle x\in
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$. Thus,

$\displaystyle|E|$ $\displaystyle=\frac{\sum_{x\in V}\text{deg}(x)}{2}$
$\displaystyle=\frac{\sum_{x\in
U(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})}\text{deg}(x)+\sum_{x\in
N_{U}(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})}\text{deg}(x)}{2}$
$\displaystyle=\frac{\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})(\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})-1)+(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}-\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}))\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})}{2}$
$\displaystyle|E|$
$\displaystyle=\frac{(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}-1)\phi(p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3})}{2}.$

  2. 2.

It follows from Theorem (5.1) and Theorem (2.12).

∎

######  Theorem 5.4.

Let $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ be a unit
graph, where $p_{1}$ and $p_{2}$ are odd primes. Then $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ is connected graph and
$\displaystyle\text{diam}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))\leq
3$.

###### Proof.

Let
$\displaystyle\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}=U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})\cup
N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ and
$\displaystyle N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})=N_{2}\cup
N_{p_{1}}\cup N_{p_{2}}\cup N_{2p_{1}}\cup N_{2p_{2}}\cup N_{p_{1}p_{2}}\cup
N_{2p_{1}p_{2}}$
where $\displaystyle
N_{2}=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}\ |\ x=2\alpha\ \&\
p_{1}\nmid\alpha\ \&\ p_{2}\nmid\alpha\\}$
$\displaystyle N_{p_{i}}=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}\
|\ x=\alpha p_{i}\ \&\ 2\nmid\alpha\ \&\ p_{j}\nmid\alpha,\ i\neq j\\}$
$\displaystyle N_{2p_{i}}=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}\
|\ x=2\alpha p_{i}\ \&\ p_{j}\nmid\alpha,\ i\neq j\\}\\\
N_{p_{1}p_{2}}=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}\ |\
x=\alpha p_{1}p_{2}\ \&\ 2\nmid\alpha\\}$
and $N_{2p_{1}p_{2}}=\\{x\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}\ |\
x=\alpha 2p_{1}p_{2}\\}$.
Now for any $x,y\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}$, we have the
following cases:
Case I: If $\displaystyle x,y\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, then $x$ and $y$ adjacent to
$0$, which gives $\displaystyle d(x,y)=1\leq 3$.
Case II: If $\displaystyle x\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ and $\displaystyle y\in
N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, then we have the
following possibilities for $y$:
(a) If $\displaystyle y\in N_{2}$, then $\displaystyle y=2\alpha$. Suppose $x$
is not adjacent to $y$. For any $\displaystyle z=\beta p_{1}p_{2}$, we have
$\displaystyle y+z\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Now
our claim is that, $\displaystyle[y,y+z]$ is an edge in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$. If this is not true, then
$\displaystyle 2\mid 2y+z$, which gives $2\mid\beta$. Similarly, if
$\displaystyle p_{i}\mid 4\alpha+\beta p_{1}p_{2}$, then $p_{i}\mid\alpha$ for
$i=1,2$ in both cases we get contradiction. Since, $\displaystyle y+z,x\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, which gives $[y+z,0]$ and
$[0,x]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Hence, we get $d(x,y)\leq
3$.
(b) If $y\in N_{p_{i}}$, for $i=1,2$, then $x$ is not adjacent to $y$. First
we show that $x$ is adjacent to some $z=2\beta p_{j}\in N_{2p_{j}}$ for $i\neq
j$. Suppose $x$ is not adjacent to $z=2\beta p_{j}$, which implies
$\displaystyle p_{i}\mid x+z$. Note that $\displaystyle p_{j}\nmid x+z$ and
$\displaystyle 2\nmid x+z$. Since $\displaystyle z=2\beta p_{j}\in
N_{2p_{j}}$, which gives $-2\beta p_{j}\in N_{2p_{j}}$. If $\displaystyle
p_{i}\mid x-2\beta p_{j}$, then $\displaystyle p_{i}\mid\beta$, which is not
possible. Hence, there exits $\displaystyle z\in N_{2p_{j}}$ such that $x$ is
adjacent to $z$. Clearly, $y$ is adjacent to $z$, for all $\displaystyle z\in
N_{2p_{j}}$. Hence, we get $\displaystyle d(x,y)=2$.
(c) If $\displaystyle y\in N_{2p_{i}}$, then $y=2\alpha p_{i}$, for $i=1,2$.
Suppose $x$ is not adjacent to $y$. For any $\displaystyle z=\beta p_{j}\in
N_{p_{j}}$ for $i\neq j$, we have $\displaystyle y+z\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Now $y$ is adjacent to
$y+z$, suppose not then we have following possibilities. If $\displaystyle
p_{i}\mid 2y+z$, then $p_{i}\mid\beta$ and if $\displaystyle p_{j}\mid 2y+z$,
then $\displaystyle p_{j}\mid\alpha$. Similarly, if $\displaystyle 2\mid
2y+z$, then $\displaystyle 2\mid\beta$, which are not possible. Since,
$\displaystyle y+z,x\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, we
have $y+z$ and $x$ is adjacent to $0$. Hence, we get $[y,y+z],[y+z,0]$ and
$[0,x]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, which implies $d(x,y)\leq
3$.
(d) If $\displaystyle y\in N_{2p_{1}p_{2}}$, then $x$ is adjacent to $y$,
which gives $d(x,y)=1$.
Case III: If $\displaystyle x,y\in
N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, then we have the
following possibilities:
(a) If $\displaystyle x,y\in N_{2}$, then for any $\displaystyle z\in
N_{p_{1}p_{2}}$, we have $x,y$ are adjacent to $z$. Hence, we get $d(x,y)\leq
3$.
(b) If $\displaystyle x\in N_{2}$ and $\displaystyle y\in N_{p_{i}}$, then
$x=2\alpha$ and $y=\alpha^{\prime}p_{i}$. Suppose $x$ is not adjacent to $y$,
then $\displaystyle p_{j}\mid x+y$, for $i\neq j$ i.e. $p_{j}\mid
2\alpha+\alpha^{\prime}p_{i}$, which gives $2\alpha+\alpha^{\prime}p_{i}=\beta
p_{j}$. Note that $\displaystyle 2\nmid\beta$ and $\displaystyle
p_{i}\nmid\beta$ this implies $\beta p_{j}\in N_{p_{j}}$ and
$\displaystyle\beta p_{i}p_{j}\in N_{p_{i}p_{j}}$. First we prove that
$\displaystyle y=\alpha^{\prime}p_{i}$ is adjacent to $\displaystyle
z=\alpha^{\prime}p_{i}+\beta p_{j}$, suppose this is not true, then we have,
$\displaystyle 2\mid z+y$ i.e. $2\mid 2\alpha^{\prime}p_{i}+\beta p_{j}$,
which gives $2\mid\beta$ which is a contradiction. Similar, contradiction
occurs, when $\displaystyle p_{i}\mid y+z$ and $\displaystyle p_{j}\mid y+z$.
Now we prove that $z$ is adjacent to $u=\beta p_{i}p_{j}\in N_{p_{i}p_{j}}$,
suppose it is not holds, which implies $\displaystyle 2\mid z+u$ i.e.
$2\mid\alpha^{\prime}p_{i}+\beta p_{j}+\beta p_{i}p_{j}$, we get
$\displaystyle 2\mid\alpha^{\prime}p_{i}+2\alpha+\alpha^{\prime}p_{i}+\beta
p_{i}$, which implies $2\mid\beta$ which is not possible. Similar
contradiction occurs, when $\displaystyle p_{i}\mid u+z$ and $\displaystyle
p_{j}\mid u+z$. Finally, observe that $u=\beta p_{i}p_{j}$ is adjacent to
$x=2\alpha$. Hence, we get $[y,y+z],\ [y+z,u]$ and $[u,x]$ are edges in
$\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, which implies
that $d(x,y)\leq 3$.
(c) If $\displaystyle x\in N_{2}$ and $\displaystyle y\in N_{2p_{i}}$, for
$i=1,2$ then $x=2\alpha$ and $y=2\alpha^{\prime}p_{i}$. First we prove that
$x$ is adjacent to some, $z\in N_{p_{j}}$, for $i\neq j$. Let $\beta p_{j}\in
N_{p_{2}}$ and assume that $x$ is not adjacent to $\beta p_{j}$, which implies
that $\displaystyle p_{i}\mid x+\beta p_{j}$. Note that $\displaystyle\beta
p_{j}\in N_{p_{2}}$, which gives $\displaystyle-\beta p_{j}\in N_{p_{j}}$ and
if $x$ is not adjacent to $\displaystyle-\beta p_{j}$, then $\displaystyle
p_{i}\mid x-\beta p_{j}$, this implies $\displaystyle p_{i}\mid x$, which is a
contradiction. Hence, we get $\displaystyle z\in N_{p_{j}}$ such that $x$ is
adjacent to $z$. Note that $\displaystyle y=2\alpha^{\prime}p_{i}$ is adjacent
to every $z$ in $\displaystyle N_{p_{j}}$. Hence, we get $d(x,y)\leq 3$.
(d) If $\displaystyle x\in N_{2}$ and $\displaystyle y\in N_{p_{1}p_{2}}$,
then $x$ is adjacent to $y$, which gives $d(x,y)=1$.
(e) If $\displaystyle x\in N_{2}$ and $\displaystyle y\in N_{2p_{1}p_{2}}$,
then $x=2\alpha$ and $y=2\alpha^{\prime}p_{1}p_{2}$. Note that for any
$\displaystyle z=\beta p_{1}p_{1}$, we have $\displaystyle x+z\in
U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$. Then $x$ is adjacent to
$x+z$. Since $x+z$ is unit, $x+z$ is adjacent $y$. Hence, we have $[x,x+z]$
and $[x+z,y]$ are edges in $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, which gives $d(x,y)\leq 3$.
(f) If $\displaystyle x,y\in N_{P_{i}}$, then $x=\alpha p_{i}$ and
$y=\alpha^{\prime}p_{i}$ for $i=1,2$. Note that, for any $\displaystyle x\in
N_{p_{i}}$ and $\displaystyle z\in N_{2p_{j}}$, for $i\neq j$, we have $x$ is
adjacent to $z$, which gives $d(x,y)\leq 2$.
(g) If $\displaystyle x\in N_{p_{i}}$ and $\displaystyle y\in N_{p_{j}}$, for
$i\neq j$ then $x=\alpha p_{i}$ and $y=\alpha^{\prime}p_{j}$. Note that $x$ is
not adjacent to $y$, we prove that $x+y$ is adjacent to $x$ and $y$. Assume
that $x+y$ is not adjacent to $x$, which gives $\displaystyle 2\mid 2x+y$, we
get $2\mid\alpha^{\prime}$ which is not possible and if $\displaystyle
p_{i}\mid 2x+y$, then $p_{i}\mid\alpha^{\prime}$ which is a contradiction.
Similar contradiction occurs, if $\displaystyle p_{j}\mid 2x+y$. Hence we get
$x$ is adjacent to $x+y$. Similarly, we can show that $y$ is adjacent to
$x+y$, which gives $d(x,y)\leq 3$.
(h) If $\displaystyle x\in N_{p_{i}}$ and $\displaystyle y\in N_{2p_{i}}$,
then $x=\alpha p_{i}$ and $y=2\alpha^{\prime}p_{i}$. Let $\displaystyle\beta
p_{j}\in N_{p_{j}}$, for $i\neq j$. Then $\displaystyle z=2(\beta p_{j}+\alpha
p_{i})p_{j}\in N_{2p_{j}}$, we already proved $x$ is adjacent to every
$\displaystyle z\in N_{2p_{j}}$. Now we prove that $\displaystyle z$ is
adjacent to $\displaystyle z+x$ and $\displaystyle z-x$. Suppose this is not
hold, then $\displaystyle 2\mid 2z+x$ i.e. $2\mid\alpha p_{i}$, which gives
$2\mid\alpha$ or if $\displaystyle p_{i}\mid 2z+x$ i.e. $p_{i}\mid 4(\beta
p_{j}+\alpha p_{i})p_{j}+\alpha p_{i}$, then $p_{i}\mid\beta$ or if
$\displaystyle p_{j}\mid 2z+x$ i.e. $p_{j}\mid 4(\beta p_{j}+\alpha
p_{i})p_{j}+\alpha p_{i}$, then $p_{j}\mid\alpha$ which are not possible.
Similarly, we can prove that $z-x$ is adjacent to $z$. Now we claim that
either $z+x$ or $z-x$ is adjacent to $y$, assume that both are not true, then
only possibility is $\displaystyle p_{j}\mid z+x+y$ and $\displaystyle
p_{j}\mid z-x+y$, which gives $\displaystyle p_{j}\mid z+y$. Hence, we get
$p_{j}\mid\alpha^{\prime}$, which is a contradiction. Hence, either $z+x$ or
$z-x$ is adjacent to $y$, which gives $d(x,y)\leq 3$.
(i) If $\displaystyle x\in N_{p_{i}}$ and $\displaystyle y\in N_{2p_{j}}$ for
$i\neq j$, then $x$ is adjacent to $y$, which gives $d(x,y)=1$.
(j) If $\displaystyle x\in N_{p_{i}}$ and $\displaystyle y\in N_{p_{1}p_{2}}$,
for $i=1,2$ then $x=\alpha p_{i}$ and $y=\alpha^{\prime}p_{1}p_{2}$. We know
that $y$ is adjacent to every $z=2\beta\in N_{2}$. It is enough to prove that
$x$ is adjacent to some $\displaystyle 2\beta\in N_{2}$. If $\displaystyle
z=2\beta\in N_{2}$ is not adjacent to $x=\alpha p_{i}$, then $p_{j}\mid z+x$,
for $i\neq j$. Now note that $\displaystyle-z=-2\beta\in N_{2}$ and if $x$ is
not adjacent to $-z$ then $\displaystyle p_{j}\mid x-z$ this implies
$\displaystyle p_{j}\mid 2x$ which gives $p_{j}\mid\alpha$ which is a
contradiction. Hence we get either $z$ or $-z$ is adjacent to $x$. Hence,
$d(x,y)\leq 3$.
(k) If $\displaystyle x\in N_{p_{i}}$ and $\displaystyle y\in
N_{2p_{1}p_{2}}$, for $i=1,2$, then using the procedure in $(e)$, we can show
that $d(x,y)\leq 3$.
(l) If $\displaystyle x,y\in N_{2p_{i}}$, for $i=1,2$ then for any
$\displaystyle z\in N_{p_{j}}$, for $i\neq j$, we have $x$ and $y$ are
adjacent to $z$. Hence, $d(x,y)\leq 3$.
(m) If $\displaystyle x\in N_{2p_{1}}$ and $\displaystyle y\in N_{2p_{2}}$,
then $x=2\alpha p_{1}$ and $y=2\alpha^{\prime}p_{2}$ . Note that for any
$\displaystyle\beta p_{2}\in N_{p_{2}}$, we have $\displaystyle x+\beta p_{2}$
is adjacent to $x$. Now it is enough to prove that, $y$ is adjacent to
$x+\beta p_{2}$, for some $\displaystyle\beta p_{2}\in N_{p_{2}}$. Let
$\displaystyle\beta p_{2}\in N_{p_{2}}$. Then $\displaystyle-\beta p_{2}\in
N_{p_{2}}$, if $y$ is not adjacent to $x+\beta p_{2}$ and $x-\beta p_{2}$,
then $\displaystyle p_{2}\nmid x\pm\beta p_{2}+y$ and $\displaystyle 2\nmid
x\pm\beta p_{2}+y$. Hence, only possibility is that $\displaystyle p_{1}\mid
x+\beta p_{2}+y$ and $\displaystyle p_{1}\mid x-\beta p_{2}+y$, which implies
$p_{1}\mid x+y$, from this we have $p_{1}\mid\alpha^{\prime}$, which is a
contradiction. Hence, either $x+\beta p_{2}$ or $x-\beta p_{2}$ is adjacent to
$y$, which gives $d(x,y)\leq 3$.
(n) If $\displaystyle x\in N_{2p_{i}}$, for $i=1,2$ and $\displaystyle y\in
N_{p_{1}p_{2}}$, then $x=2\alpha p_{i}$ and $y=\alpha^{\prime}p_{1}p_{2}$. For
any $\displaystyle 2\beta\in N_{2}$, we have $y$ is adjacent to $2\beta$, and
$2\beta+y$ is adjacent to $2\beta$. Now we prove that for some $z\in N_{2}$,
$x$ is adjacent to $y+z$. Let $\displaystyle 2\beta\in N_{2}$. Then
$\displaystyle-2\beta\in N_{2}$ and assume that $y+2\beta$ and $y-2\beta$ are
not adjacent to $x$. Note that $\displaystyle p_{i}\nmid y+2\beta+x$ and
$\displaystyle 2\nmid y+2\beta+x$, hence the only possibility is
$\displaystyle p_{j}\mid y+2\beta+x$ for $i\neq j$. Similarly from above we
get $\displaystyle p_{j}\mid y-2\beta+x$ this implies $p_{j}\mid y+x$, since
$p_{j}\in\\{p_{1},p_{2}\\}$, we get $p_{j}\mid\alpha$ which is a
contradiction, which gives either $y+2\beta$ is adjacent to $x$ or $y-2\beta$
is adjacent to $x$. From this, we conclude that $d(x,y)\leq 3$.
(o) If $\displaystyle x\in N_{2p_{i}}$ and $\displaystyle y\in
N_{2p_{1}p_{2}}$, for $i=1,2$ then using the procedure in $(e)$ we can show
that $d(x,y)\leq 3$.
(p) If $\displaystyle x,y\in N_{p_{1}p_{2}}$, then for any $\displaystyle z\in
N_{2}$ we have $x$ and $y$ are adjacent to $z$. Hence we get $d(x,y)\leq 3$.
(q) If $\displaystyle x\in N_{p_{1}p_{2}}$ and $\displaystyle y\in
N_{2p_{1}p_{2}}$, then using the procedure in $(e)$ we can show that
$d(x,y)\leq 3$.
(r) If $\displaystyle x,y\in N_{2p_{1}p_{2}}$, then for any $\displaystyle
z\in U(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, we have $x$ and $y$ are
adjacent to $z$, which gives $d(x,y)\leq 3$.
Hence, $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ is
connected graph and $d(x,y)\leq 3$, for all $\displaystyle
x,y\in\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}$, which gives
$\displaystyle\text{diam}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))\leq
3$. ∎

######  Corollary 5.5.

Let $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ be a unit
graph, where $\displaystyle p_{1}$ and $\displaystyle p_{2}$ are distinct odd
primes. Then $\displaystyle
g_{r}(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))=4$.

###### Proof.

Proof follows from Theorem (2.15) and Lemma (4.1). ∎

######  Theorem 5.6.

Let $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ be a unit
graph, where $\displaystyle p_{1}$ and $\displaystyle p_{2}$ are odd primes.
Then

  1. 1.

$\displaystyle|V|=2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}$ and
$\displaystyle|E|=\frac{(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})\phi(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})}{2}$

  2. 2.

$\displaystyle\lambda(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\delta(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\phi(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})$.

###### Proof.

  1. 1.

Clearly, $\displaystyle|V|=2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}$. Since,
$\displaystyle 2\in N_{U}(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$, from
Theorem (2.14), we get $\displaystyle
G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$ is
$\displaystyle\phi(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})$-regular graph and hence
$\displaystyle|E|=\frac{(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})\phi(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})}{2}$.

  2. 2.

From Lemma (4.1) and Theorems (5.4) and (2.13), we get
$\displaystyle\lambda(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\delta(G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}}))=\phi(2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2})$.

∎

######  Theorem 5.7.

Let $\displaystyle G(\mathbb{Z}_{n})$ be a unit graph, where
$n=p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}$ and $p_{1},\ p_{2}\ \&\ p_{3}$ are
distinct primes. Let $H$ be a $|V|\times|E|$ incidence matrix of
$G(\mathbb{Z}_{n})$.

  1. 1.

If $\displaystyle 2\in U(\mathbb{Z}_{n})$, then binary code generated by $H$
is a $\displaystyle
C_{2}(H)=\left[\frac{(n-1)\phi(n)}{2},n-1,\phi(n)-1\right]_{2}$ code over
finite field $\mathbb{F}_{2}$.

  2. 2.

If $\displaystyle 2\in N_{U}(\mathbb{Z}_{n})$, then for any odd prime $q$, the
$q$-ary code generated by $H$ is a $\displaystyle
C_{q}(H)=\left[\frac{n\phi(n)}{2},n-1,\phi(n)\right]_{q}$ code over finite
field $\mathbb{F}_{q}$.

###### Proof.

  1. 1.

Let $\displaystyle 2\in U(\mathbb{Z}_{n})$ then $p_{1},p_{2}$ and $p_{3}$ are
odd primes. By Theorem (5.1), $\displaystyle G(\mathbb{Z}_{n})=(V,E)$ is a
connected graph and hence by Theorem (2.16) binary code generated by $H$ is
$\displaystyle
C_{2}(H)=[|E|,|V|-1,\lambda(G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}}))]_{2}$.
Now from Theorem (5.3), we get $\displaystyle|E|=\frac{(n-1)\phi(n)}{2},\
|V|-1=n-1$ and $\displaystyle\lambda(G(\mathbb{Z}_{n}))=\phi(n)-1$.

  2. 2.

Let $\displaystyle 2\in N_{U}(\mathbb{Z}_{n})$. Then we have $p_{i}=2$, for
some $i$. By Theorem (5.6) and Lemma (4.1), $\displaystyle
G(\mathbb{Z}_{n})=(V,E)$ is a connected bipartite graph and hence by Theorems
(2.16) and (2.17), $q$-ary code generated by $H$ is $\displaystyle
C_{q}(H)=[|E|,|V|-1,\lambda(G(\mathbb{Z}_{n}))]_{q}$. Now from Theorem (5.6),
the result follows.

∎

######  Corollary 5.8.

Let $C_{q}(H)$ and $C_{2}(H)$ denote the linear codes generated from incidence
matrices of $\displaystyle G(\mathbb{Z}_{2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}})$
and $\displaystyle G(\mathbb{Z}_{p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}})$.
Then

  1. 1.

Dual of code $C_{2}$ is $\displaystyle
C^{\perp}_{2}=\left[\frac{(n-1)\phi(n)}{2},\frac{(n-1)[\phi(n)-2]}{2},3\right]_{2}$,
where $\displaystyle n=p^{n_{1}}_{1}p^{n_{2}}_{2}p^{n_{3}}_{3}$.

  2. 2.

Dual of code $C_{q}$ is $\displaystyle
C^{\perp}_{q}=\left[\frac{n\phi(n)}{2},\frac{n(\phi(n)-2)+2}{2},4\right]_{q}$,
where $\displaystyle n=2^{m}p^{n_{1}}_{1}p^{n_{2}}_{2}$.

###### Proof.

  1. 1.

From Theorem (2.3), $\text{diam}(\displaystyle
C^{\perp}_{2})=\frac{(n-1)[\phi(n)-2]}{2}$. By Theorem (2.18) and Corollary
(5.2), $\displaystyle d(C^{\perp}_{2})=3$.

  2. 2.

Proof follows from Theorems (2.3) & (2.18) and Corollary (5.5).

∎

######  Theorem 5.9.

Let $G(\mathbb{Z}_{n})$ be a unit graph, where $n$ is any positive integer.

  1. 1.

If $2\in U(\mathbb{Z}_{n})$, then $\displaystyle|E|=\frac{(n-1)\phi(n)}{2}$.

  2. 2.

If $2\in N_{U}(\mathbb{Z}_{n})$, then $\displaystyle|E|=\frac{n\phi(n)}{2}$.

###### Proof.

  1. 1.

Let $\displaystyle 2\in U(\mathbb{Z}_{n})$. By Theorem (2.14),
$\displaystyle\text{deg}(x)=\phi(n)-1$ for all $\displaystyle x\in
U(\mathbb{Z}_{n})$ and $\displaystyle\text{deg}(x)=\phi(n)$ for all
$\displaystyle x\in N_{U}(\mathbb{Z}_{n})$. Thus,

$\displaystyle|E|$ $\displaystyle=\frac{\sum_{x\in V}\text{deg}(x)}{2}$
$\displaystyle=\frac{\sum_{x\in U(\mathbb{Z}_{n})}\text{deg}(x)+\sum_{x\in
N_{U}(\mathbb{Z}_{n})}\text{deg}(x)}{2}$
$\displaystyle=\frac{\phi(n)(\phi(n)-1)+(n-\phi(n))\phi(n)}{2}$
$\displaystyle|E|$ $\displaystyle=\frac{(n-1)\phi(n)}{2}.$

  2. 2.

Let $\displaystyle 2\in N_{U}(\mathbb{Z}_{n})$. Then from Theorem (2.14), we
get $\displaystyle G(\mathbb{Z}_{n})$ is $\displaystyle\phi(n)$-regular graph.
Hence, $\displaystyle|E|=\frac{n\phi(n)}{2}$.

∎

Bases on our results for unit graphs, in Theorems (3.2), (3.3), (4.2), (4.5),
(5.1) and (5.4), we state following conjectures for any natural number $n$:
Conjecture I: Let $G(\mathbb{Z}_{n})$ be a unit graph. Then
$G(\mathbb{Z}_{n})$ is connected graph and

  1. 1.

If $2\in U(\mathbb{Z}_{n})$, then $\text{diam}(G(\mathbb{Z}_{n}))\leq 2$.

  2. 2.

If $2\in N_{U}(\mathbb{Z}_{n})$, then $\text{diam}(G(\mathbb{Z}_{n}))\leq 3$.

Conjecture II: Let $\displaystyle G(\mathbb{Z}_{n})$ be a unit graph and $H$
be a $|V|\times|E|$ incidence matrix of $G(\mathbb{Z}_{n})$.

  1. 1.

If $\displaystyle 2\in U(\mathbb{Z}_{n})$, then binary code generated by $H$
is a $\displaystyle
C_{2}(H)=\left[\frac{(n-1)\phi(n)}{2},n-1,\phi(n)-1\right]_{2}$ code over
finite field $\mathbb{F}_{2}$.

  2. 2.

If $\displaystyle 2\in N_{U}(\mathbb{Z}_{n})$, then for any odd prime $q$, the
$q$-ary code generated by $H$ is a $\displaystyle
C_{q}(H)=\left[\frac{n\phi(n)}{2},n-1,\phi(n)\right]_{q}$ code over finite
field $\mathbb{F}_{q}$.

##  6 Conclusion

In this paper, we generate $q$-ary linear codes, for any prime $q$, from
incidence matrices of unit graphs $\displaystyle G(\mathbb{Z}_{n})$. Moreover,
we found parameters and dual of constructed codes over finite filed
$\displaystyle\mathbb{F}_{q}$. In this article, we consider $n$ as product of
power of three distinct primes. Examine the permutation decoding techniques,
covering radius of constructed codes and one can construct linear codes from
unit graph over different commutative rings is the further scope to work.

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