# An Algebraic Construction Technique for Codes over Hurwitz Integers

Ramazan DURAN
Department of Mathematics
Afyon Kocatepe University
Afyonkarahisar, TR 03200
& Murat GUZELTEPE
Department of Mathematics
Sakarya University
Sakarya, TR 54187

###### Abstract

Let $\alpha$ be a prime Hurwitz integer. $\mathcal{H}_{\alpha}$, which is the
set of residual class with respect to related modulo function in the rings of
Hurwitz integers, is a subset of $\mathcal{H},$ which is the set of all
Hurwitz integers. We consider left congruent modulo $\alpha$ and, the domain
of related modulo function in this study is $\mathbb{Z}_{N(\alpha)},$ which is
residual class ring of ordinary integers with $N(\alpha)$ elements, which is
the norm of prime Hurwitz integer $\alpha.$ In this study, we present an
algebraic construction technique, which is a modulo function formed depending
on two modulo operations, for codes over Hurwitz integers. Thereby, we obtain
the residue class rings of Hurwitz integers with $N(\alpha)$ size. In
addition, we present some results for mathematical notations used in two
modulo functions, and for the algebraic construction technique formed
depending upon two modulo functions. Moreover, we presented graphs obtained by
graph layout methods, such as spring, high-dimensional, and spiral embedding,
for the set of the residual class obtained with respect to the related modulo
function in the rings of Hurwitz integers.

_Keywords_ Algebraic construction $\cdot$ average energy $\cdot$ block code
$\cdot$ code rate $\cdot$ graph

##  1 Introduction

The first study on codes over high-dimensional algebraic structures such as
Gauss integers, Einstein-Jacobi integers, quaternions, Lipschitz integers, and
Hurwitz integers was got by Huber. In [1], Huber showed how block codes over
Gaussian integers could be used for coding over two-dimensional algebraic
structure. The set of Gaussian integers that denoted by $\mathbb{Z}[i]$ is
shown by
$\mathbb{Z}[i]=\\{\alpha=\alpha_{1}+\alpha_{2}i:\alpha_{1},\alpha_{2}\in\mathbb{Z}\
\text{where}\ i^{2}=-1\\}.$ In this study, we use $\mathcal{G}$ notation
instead of $\mathbb{Z}[i].$ $\alpha_{1}$ is the real part, and $\alpha_{2}i$
is the imaginary part. The conjugate of Gaussian integer $\alpha$ is
$\overline{\alpha}=\alpha_{1}-\alpha_{2}i.$ The norm of Gaussian integer
$\alpha$ is $N(\alpha)=\alpha\overline{\alpha}=\alpha^{2}_{1}+\alpha^{2}_{2}.$
A Gaussian integer is called a prime Gaussian integer if its norm is a prime
integer. In [1], Huber considered Gaussian integers such that $p\equiv 1\mod
N(\alpha)$ where $p$ is a prime integer and, presented a technique known as
the modulo function to construct block codes over Gaussian integers, a two-
dimensional vector space. In this way, Huber constructed one Mannheim error-
correcting (OMEC) codes over Gaussian integers fields. The modulo function
$\mu:\mathcal{G}\rightarrow\mathcal{G}_{\alpha}$ is defined according to

$\mu(\xi)=\xi\mod\alpha=\eta=\xi-\left[\frac{\xi\overline{\alpha}}{\alpha\overline{\alpha}}\right]\alpha$
(1)

where $\xi\in\mathcal{G}$ and $\alpha$ is a prime Gaussian integer [1]. Here
$\mathcal{G}_{\alpha}$ is the set of residual class with respect to modulo
$\alpha.$ We give the definition of mathematical notation $[\cdot]$ in the
following section. In a similar technique, in [2], Huber showed how block
codes over Eisenstein-Jacobi integers could be used for coding over two-
dimensional vector space. In [3], Freudenberger et al. presented new coding
techniques for codes over Gaussian integers by using the modulo function in
[1].

Quaternions, a four-dimensional algebraic structure, are a number system that
expands over complex numbers. They form
$\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ where
$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{R}$ such that
$i^{2}=j^{2}=k^{2}=-1,$ and $ij=-ji=k,$ $jk=-kj=i,$ $ki=-ik=j.$
$\alpha_{1},\alpha_{2},\alpha_{3},$ and $\alpha_{4}$ are the components of
quaternion $\alpha$. In [4], Ozen and Guzeltepe studied codes over some finite
fields by using quaternion integers, which have the commutative property. In
[5], Ozen and Guzeltepe studied cyclic codes over some finite quaternion
integer rings. They considered the quaternion integers which have the
commutative property. In [6], Shah and Rasool established that; over
quaternion integers, for a given $n$ length cyclic code there exists a cyclic
code of length $2n-1.$ A lipschitz integer is a quaternion whose components
are all integers. In [7], Freudenberger et al. presented new block codes over
Lipschitz integers. They consider primitive Lipschitz integers, which is the
greatest common divisor of components is one. A quaternion is called a Hurwitz
integer if $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ are either in
$\mathbb{Z}$ or in $\mathbb{Z}+\frac{1}{2}.$ Also, Hurwitz integers forms a
subring of the ring of all quaternions. If the norm of a Hurwitz integer is a
prime integer, then it is called a prime Hurwitz integer. In this study, we
consider prime Hurwitz integers. In [8], Ozen and Guzeltepe obtained new
classes of linear codes over Hurwitz integers. They considered Hurwitz
integers, which have the commutative property. In [9], Rohweder et al.
presented a new algebraic construction technique for codes over Hurwitz
integers that is inherently accompanied by a respective modulo operation. They
were considered that the domain of related modulo function in [9] is
$\mathbb{Z}_{N(\alpha)}\times\mathbb{Z}_{N(\alpha)}.$ Therefore, they
constructed new sets of residual class, which has $N^{2}(\alpha)$ elements,
with respect to the related modulo technique in the rings of Hurwitz integers,
using the primitive Lipschitz integers. Other studies on codes over Hurwitz
integers were presented in papers to [10], [11], and [12].

The presentation of our results is organized as follows: In thenext section,
we give the necessary fundamental definitions used throughout this paper. In
section 3, we present an algebraic construction technique, which is a modulo
function formed depending upon two modulo functions such that the domain of it
is $\mathbb{Z}_{N(\alpha)}.$ Thereby, we construct new set of residual class
with $N(\alpha)$ elements, with respect to the related modulo function in the
rings of Hurwitz integers. In section 4, we give some of the natural results
for mathematical notations in two modulo functions and the modulo function. In
section 5, we give examples for the algebraic construction technique given in
this study. In section 6, we investigate the code rates of some codes over
Hurwitz integers. In section 7, we give the points and graph on $2D$ (two-
dimensional) and $3D$ (three-dimensional) the elements of the residual class
obtained with respect to the related modulo function in prime Hurwitz integers
using graph layout methods such as spring, high-dimensional, and spiral
embedding. We conclude this paper with section 8.

##  2 Preliminaries

We begin with some basic definitions.

###### Definition 2.1

The Hamilton quaternion algebra over $R$, denoted by $\mathbb{H}(\mathbb{R})$,
is the associative unital algebra given by the following presentation:

  * i.

$\mathbb{H}(\mathbb{R})$ is free $R-$module over the symbols $1,$ $i,$ $j,$
$k;$ that is,
$\mathbb{H}(\mathbb{R})=\\{\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k:\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{R}\\};$

  * ii.

$1$ is the multiplicative unit;

  * iii.

$i^{2}=j^{2}=k^{2}=-1;$

  * iv.

$ij=-ji=k;$ $jk=-kj=i;$ $ki=-ik=j$ [13].

Note that, $\alpha_{1}$ is the real part,
$\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ is imaginary part, and
$\alpha_{1},\alpha_{2},\alpha_{3},$ and $\alpha_{4}$ are the components. If
$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then $\alpha$ is
called a quaternion integer. The commutative property of multiplication does
not hold over quaternion integers, in general. The set of quaternion integers
which have the commutative property that denoted by $\mathcal{H}^{c}$ is shown
by
$\mathcal{H}^{c}=\\{\alpha_{1}+\alpha_{2}(i+j+k):\alpha_{1},\alpha_{2}\in\mathbb{Z}\\}.$
In the other words, if the imaginary parts of quaternion integers are parallel
to each other, then their product is commutative.

###### Definition 2.2

$\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ is called a Hurwitz
integer just if either
$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}$ or
$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2}.$ The
set of all Hurwitz integers that denoted by $\mathcal{H}$ is shown by

$\begin{array}[]{lll}\mathcal{H}&=&\left\\{{\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k:\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}\
or\
\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2}}\right\\}\hfill\\\
&=&\mathcal{H}(\mathbb{Z})\bigcup\mathcal{H}(\mathbb{Z}+\frac{1}{2}).\end{array}$
(2)

###### Example 2.1

  * i.

$\pm 1\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k$ is not a Hurwitz integer.

  * ii.

$\pm\frac{1}{2}\pm\frac{1}{2}j$ is not a Hurwitz integer.

  * iii.

$\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k$ is a Hurwitz
integer.

The ring of Hurwitz integers forms a subring of the ring of all quaternions
because of closed under multiplication and addition. The conjugate of $\alpha$
is $\overline{\alpha}=\alpha_{1}-\alpha_{2}i-\alpha_{3}j-\alpha_{4}k,$ the
norm of $\alpha$ is
$N\left(\alpha\right)=\alpha\cdot\overline{\alpha}=\alpha_{1}^{2}+\alpha_{2}^{2}+\alpha_{3}^{2}+\alpha_{4}^{2},$
and the inverse of $\alpha$ is
${\alpha^{-1}}=\frac{{\overline{\alpha}}}{{N\left(\alpha\right)}}$ where
$N(\alpha)\neq 0.$

###### Definition 2.3

Let $\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ be a Hurwitz
integers such that $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}.$
$\alpha$ is called a primitive Hurwitz integer just if the greatest common
divisor ($\mathrm{gcd}$) of $\alpha_{1},\alpha_{2},\alpha_{3},$ and
$\alpha_{4}$ is one, namely,
$(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})=1.$

###### Definition 2.4

Let $\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ be a Hurwitz
integers such that
$\alpha_{1}=\beta_{1}+\frac{1}{2},\alpha_{2}=\beta_{2}+\frac{1}{2},\alpha_{3}=\beta_{2}+\frac{1}{2},\alpha_{4}=\beta_{4}+\frac{1}{2}$
where $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z}.$ $\alpha$ is
called a primitive Hurwitz integer just if the greatest common divisor
($\mathrm{gcd}$) of $2\beta_{1}+1,2\beta_{2}+1,2\beta_{3}+1,$ and
$2\beta_{4}+1$ is one, namely,
$(2\beta_{1}+1,2\beta_{2}+1,2\beta_{3}+1,2\beta_{3}+1)=1.$

###### Definition 2.5

$\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ is called a prime
Hurwitz integer if its norm is a prime integer.

From definition 2.3, definition 2.4 and definition 2.5, we say that every
prime Hurwitz integer is a primitive Hurwitz integer but converse may not be
true. Note that we consider prime Hurwitz integers in this study.

###### Definition 2.6

Let $\alpha$ be a Hurwitz integer. If there exists $\lambda\in\mathcal{H}$
such that $q_{1}-q_{2}=\lambda\pi,$ then $q_{1},q_{2}\in\mathcal{H}$ are said
to be right congruent modulo $\alpha.$ This relation is denoted by
$q_{1}\equiv_{r}q_{2}.$ This relation $q_{1}\equiv_{r}q_{2}$ is an equivalence
relation. The elements in the right ideal

$\langle\alpha\rangle=\\{\lambda\alpha:\lambda\in\mathcal{H}\\}$ (3)

define a normal subgroup of the additive group of the ring $\mathcal{H}.$ The
set of cosets to $\langle\alpha\rangle$ in $\mathcal{H}$ defines Abelian group
denoted by $\mathcal{H}_{\alpha}=\mathcal{H}\diagup\langle\alpha\rangle.$
Analogous results are valid for left congruences modulo $\alpha$ [8].

Note that we use left congruences modulo $\alpha$ in this study. The following
section is going to present the algebraic construction technique, formed
depending on two modulo functions, used to construct block codes of
$N(\alpha)$ size over Hurwitz integers. However, we close this section by
giving the definitions of two mathematical notations used in the modulo
functions before going to the next section.

###### Definition 2.7

A notation for rounding to the nearest integer is denoted by
$\lfloor\cdot\rceil.$ It is defined as rounding a rational number to the
integer closest to it. All components in a quaternion are separately rounding
to the integer closest to it.

###### Definition 2.8

A notation for rounding to the nearest half-integer is denoted by
$\lfloor\lfloor\cdot\rceil\rceil.$ It is defined as rounding a rational number
to the half-integer closest to it. All components in a quaternion are
separately rounding to the half-integer closest to it [11].

Note that if the component is negative or zero, then the rounding is done in
the direction up. Otherwise the rounding is done in the direction down in this
study.

###### Example 2.2

  * i.

$\lfloor\frac{3}{2}\rceil=1,$ $\lfloor-\frac{1}{2}\rceil=0,$
$\lfloor-\frac{3}{2}\rceil=-1,$ and so on.

  * ii.

$\lfloor\lfloor 0\rceil\rceil=\frac{1}{2}$ where $0\in\mathbb{Z},$
$\lfloor\lfloor-1\rceil\rceil=-\frac{1}{2},$ $\lfloor\lfloor
1\rceil\rceil=\frac{1}{2},$ and so on.

  * iii.

$\lfloor\lfloor
0\rceil\rceil=\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k$ where $0$ is
a quaternion.

##  3 Algebraic Construction Technique

This section presents an algebraic construction technique formed depending on
two modulo functions used to construct codes of $N(\alpha)$ size over Hurwitz
integers.

###### Definition 3.1

Let $\alpha$ be a prime Hurwitz integer. The modulo function
$\mu:\mathbb{Z}_{N(\alpha)}\rightarrow\mathcal{H}_{\alpha}$ is defined by

$\mu_{\alpha}(z)=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}$
(4)

where

$\mu_{\alpha}^{(1)}(z)=z\mod\alpha=z-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil$
(5)

and

$\mu_{\alpha}^{(2)}(z)=z\mod\alpha=z-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil$
(6)

such that

$\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}=\begin{cases}\mu_{\alpha}^{(1)}(z),&\mbox{if
}N(\mu_{\alpha}^{(1)}(z))<N(\mu_{\alpha}^{(2)}(z))\\\
\mu_{\alpha}^{(1)}(z),&\mbox{if
}N(\mu_{\alpha}^{(1)}(z))=N(\mu_{\alpha}^{(2)}(z))\mbox{ where z is an even
integer}\\\ \mu_{\alpha}^{(2)}(z),&\mbox{otherwise}.\end{cases}$ (7)

where $z\in\mathbb{Z}_{N(\alpha)}.$ Here $\mathbb{Z}_{N(\alpha)}$ is the well-
known residual class ring of ordinary integers with $N(\alpha)$ elements,
$\mathcal{H}_{\alpha}$ is the residual class ring of prime Hurwitz integer
$\alpha,$ and $\mu_{\alpha}^{(1)}(z)$ and $\mu_{\alpha}^{(2)}(z)$ is given
remainder of $z(z\in\mathbb{Z}_{N(\alpha)})$ with respect to modulo functions
given in eq. 5 and eq. 6, respectively. The set of quotient ring of the
Hurwitz integers that denoted by $\mathcal{H}_{\alpha}$ is shown by

$\mathcal{H}_{\alpha}=\\{\mu_{\alpha}(z)=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}|z\in\mathbb{Z}_{N(\pi)}\\}.$
(8)

###### Proposition 3.1

Let $\alpha$ be a prime Hurwitz integer, and $z=0.$ Then

  * i.

$\mu_{\alpha}^{(1)}(0)=0,$

  * ii.

$\mu_{\alpha}^{(2)}(0)\equiv 0\mod\alpha.$

###### Proof 3.1

Let $\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ be a prime Hurwitz
integer, and $z=0.$

$(i.)$ The proof can be easily seen from eq. 5.

$(ii.)$ From eq. 6,

$\begin{array}[]{lll}\mu_{\alpha}^{(2)}(0)&=&0-\alpha\lfloor\lfloor\frac{\overline{\alpha}0}{N(\alpha)}\rceil\rceil\\\
&=&-(\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k)\lfloor\lfloor\frac{(\alpha_{1}-\alpha_{2}i-\alpha_{3}j-\alpha_{4}k)0}{N(\alpha)}\rceil\rceil\\\
&=&-(\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k)\lfloor\lfloor
0\rceil\rceil\\\
&=&-(\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k)(\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k)\\\
&=&-\frac{\alpha_{1}}{2}-\frac{\alpha_{1}}{2}i-\frac{\alpha_{1}}{2}j-\frac{\alpha_{1}}{2}k-\frac{\alpha_{2}}{2}i+\frac{\alpha_{2}}{2}-\frac{\alpha_{2}}{2}k+\frac{\alpha_{2}}{2}j\\\
&&-\frac{\alpha_{3}}{2}j+\frac{\alpha_{3}}{2}k+\frac{\alpha_{3}}{2}-\frac{\alpha_{3}}{2}i-\frac{\alpha_{4}}{2}k-\frac{\alpha_{4}}{2}j+\frac{\alpha_{4}}{2}i+\frac{\alpha_{4}}{2}\\\
&=&\frac{-\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}}{2}+\frac{-\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{4}}{2}i+\frac{-\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{4}}{2}j+\frac{-\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4}}{2}k\\\
&\equiv&0\mod\alpha.\end{array}$ (9)

This completes the proof.

The set $\mathcal{H}_{\alpha}$ contains $N(\alpha)$ elements. The modulo
function $\mu$ in definition 2.8 defines a bijective mapping from
$\mathbb{Z}_{N(\alpha)}$ into $\mathcal{H}_{\alpha}.$ In other words, there is
a ring isomorphism between $\mathbb{Z}_{N(\alpha)}$ and
$\mathcal{H}_{\alpha}.$ With the following theorems, we first show that the
modulo function $\mu$ in definition 2.8 is a ring homomorphism, and then we
show that it is a ring isomorphism.

###### Theorem 3.1

Let $\alpha$ be a prime Hurwitz integer, and
$z_{1},z_{2}\in\mathbb{Z}_{N(\alpha)}.$ The modulo function
$\mu:\mathbb{Z}_{N(\alpha)}\rightarrow\mathcal{H}_{\alpha}$ is a ring
homomorphism.

###### Proof 3.2

Let $\alpha$ be a prime Hurwitz integer. The modulo function
$\mu:\mathbb{Z}_{N(\alpha)}\rightarrow\mathcal{H}_{\alpha}$ by

$\mu_{\alpha}(z)=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}$
(10)

where

$\mu_{\alpha}^{(1)}(z)=z\mod\alpha=z-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil$
(11)

and

$\mu_{\alpha}^{(2)}(z)=z\mod\alpha=z-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil$
(12)

such that

$\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}=\begin{cases}\mu_{\alpha}^{(1)}(z),&\mbox{if
}N(\mu_{\alpha}^{(1)}(z))<N(\mu_{\alpha}^{(2)}(z))\\\
\mu_{\alpha}^{(1)}(z),&\mbox{if
}N(\mu_{\alpha}^{(1)}(z))=N(\mu_{\alpha}^{(2)}(z))\mbox{ where z is an even
integer}\\\ \mu_{\alpha}^{(2)}(z),&\mbox{otherwise}.\end{cases}$ (13)

where $z\in\mathbb{Z}_{N(\alpha)}.$ From eq. 10,
$\mu_{\alpha}(z_{1})=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z_{1}),\mu_{\alpha}^{(2)}(z_{1})\\},$
and
$\mu_{\alpha}(z_{2})=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z_{2}),\mu_{\alpha}^{(2)}(z_{2})\\}$
where $z_{1},z_{2}\in\mathbb{Z}_{N(\alpha)}.$ If
$z_{1},z_{2}\in\mathbb{Z}_{N(\alpha)},$ then there are four probable case with
respect to eq. 10;

  * i.

$\mu_{\alpha}(z_{1})=\mu^{(1)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(1)}_{\alpha}(z_{2})$ or,

  * ii.

$\mu_{\alpha}(z_{1})=\mu^{(1)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(2)}_{\alpha}(z_{2})$ or,

  * iii.

$\mu_{\alpha}(z_{1})=\mu^{(2)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(1)}_{\alpha}(z_{2})$ or,

  * iv.

$\mu_{\alpha}(z_{1})=\mu^{(2)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(2)}_{\alpha}(z_{2}).$

Let us show that
$\mu_{\alpha}(z_{1}+z_{2})=\mu_{\alpha}(z_{1})+\mu_{\alpha}(z_{2}).$

$(i.)$ Let $\mu_{\alpha}(z_{1})=\mu^{(1)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(1)}_{\alpha}(z_{2}).$ Then
$\mu_{\alpha}(z_{1}+z_{2})=\mu^{(1)}_{\alpha}(z_{1})+\mu^{(1)}_{\alpha}(z_{2}).$
From eq. 11,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(1)}_{\alpha}(z_{2})&=&z_{1}-\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil+z_{2}-\alpha\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\\\
&=&z_{1}+z_{2}-\alpha(\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil+\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil).\end{array}$
(14)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil=\lambda_{1}$ and
$\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil=\lambda_{2}.$ Hereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(1)}_{\alpha}(z_{2})&=&z_{1}+z_{2}-\alpha(\lambda_{1}+\lambda_{2}).\end{array}$
(15)

Let $\lambda_{1}+\lambda_{2}=\lambda$ where $\lambda\in\mathcal{H}.$ Thereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(1)}_{\alpha}(z_{2})&=&z_{1}+z_{2}-\alpha\lambda.\end{array}$
(16)

Since $\mu_{\alpha}(z_{1}+z_{2})=(z_{1}+z_{2})\mod\alpha,$ then there exists
$\exists\beta\in\mathcal{H}$ such that
$\mu_{\alpha}(z_{1}+z_{2})=z_{1}+z_{2}-\alpha\beta.$ So, we have

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(1)}_{\alpha}(z_{2})&=&\mu_{\alpha}(z_{1}+z_{2}).\end{array}$
(17)

$(ii.)$ Let $\mu_{\alpha}(z_{1})=\mu^{(1)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(2)}_{\alpha}(z_{2}).$ Then
$\mu_{\alpha}(z_{1}+z_{2})=\mu^{(1)}_{\alpha}(z_{1})+\mu^{(2)}_{\alpha}(z_{2}).$
From eq. 11, and eq. 12,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(2)}_{\alpha}(z_{2})&=&z_{1}-\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil+z_{2}-\alpha\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil\\\
&=&z_{1}+z_{2}-\alpha(\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil+\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil).\end{array}$
(18)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil=\lambda_{1}$ and
$\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil=\lambda_{2}.$
Hereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(2)}_{\alpha}(z_{2})&=&z_{1}+z_{2}-\alpha(\lambda_{1}+\lambda_{2}).\end{array}$
(19)

Let $\lambda_{1}+\lambda_{2}=\lambda$ where $\lambda\in\mathcal{H}.$ Thereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(2)}_{\alpha}(z_{2})&=&z_{1}+z_{2}-\alpha\lambda.\end{array}$
(20)

Since $\mu_{\alpha}(z_{1}+z_{2})=(z_{1}+z_{2})\mod\alpha,$ then there exists
$\exists\beta\in\mathcal{H}$ such that
$\mu_{\alpha}(z_{1}+z_{2})=z_{1}+z_{2}-\alpha\beta.$ So, we have

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})+\mu^{(2)}_{\alpha}(z_{2})&=&\mu_{\alpha}(z_{1}+z_{2}).\end{array}$
(21)

Similarly, we can also show cases in the $(iii.)$ and $(iv.).$ Consequently,

$\mu_{\alpha}(z_{1}+z_{2})=\mu_{\alpha}(z_{1})+\mu_{\alpha}(z_{2}).$ (22)

On the other hand, let us show that
$\mu_{\alpha}(z_{1}z_{2})=\mu_{\alpha}(z_{1})\mu_{\alpha}(z_{2}).$

$(i.)$ Let $\mu_{\alpha}(z_{1})=\mu^{(1)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(1)}_{\alpha}(z_{2}).$ Then
$\mu_{\alpha}(z_{1}z_{2})=\mu^{(1)}_{\alpha}(z_{1})\mu^{(1)}_{\alpha}(z_{2}).$
From eq. 11,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(1)}_{\alpha}(z_{2})&=&(z_{1}-\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil)(z_{2}-\alpha\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil)\\\
&=&z_{1}z_{2}-z_{1}\alpha\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil-
z_{2}\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil+\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil\alpha\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil.\end{array}$
(23)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil=\lambda_{1}$ and
$\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil=\lambda_{2}.$ Hereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(1)}_{\alpha}(z_{2})&=&z_{1}z_{2}-z_{1}\alpha\lambda_{2}-z_{2}\alpha\lambda_{1}+\alpha\lambda_{1}\alpha\lambda_{2}\\\
&=&z_{1}z_{2}-\alpha(z_{1}\lambda_{2}+z_{2}\lambda_{1}+\lambda_{1}\alpha\lambda_{2}).\end{array}$
(24)

Let $z_{1}\lambda_{2}+z_{2}\lambda_{1}+\lambda_{1}\alpha\lambda_{2}=\lambda$
where $\lambda\in\mathcal{H}.$ Thereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(1)}_{\alpha}(z_{2})&=&z_{1}z_{2}-\alpha\lambda.\end{array}$
(25)

Since $\mu_{\alpha}(z_{1}z_{2})=(z_{1}z_{2})\mod\alpha,$ then there exists
$\exists\beta\in\mathcal{H}$ such that
$\mu_{\alpha}(z_{1}z_{2})=z_{1}z_{2}-\alpha\beta.$ So, we have

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(1)}_{\alpha}(z_{2})&=&\mu_{\alpha}(z_{1}z_{2}).\end{array}$
(26)

$(ii.)$ Let $\mu_{\alpha}(z_{1})=\mu^{(1)}_{\alpha}(z_{1})$ and
$\mu_{\alpha}(z_{2})=\mu^{(2)}_{\alpha}(z_{2}).$ Then
$\mu_{\alpha}(z_{1}z_{2})=\mu^{(1)}_{\alpha}(z_{1})\mu^{(2)}_{\alpha}(z_{2}).$
From eq. 11, and eq. 12,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(2)}_{\alpha}(z_{2})&=&(z_{1}-\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil)(z_{2}-\alpha\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil)\\\
&=&z_{1}z_{2}-z_{1}\alpha\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil-
z_{2}\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil+\alpha\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil\alpha\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil.\end{array}$
(27)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\frac{\overline{\alpha}z_{1}}{N(\alpha)}\rceil=\lambda_{1}$ and
$\lfloor\lfloor\frac{\overline{\alpha}z_{2}}{N(\alpha)}\rceil\rceil=\lambda_{2}.$
Hereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(2)}_{\alpha}(z_{2})&=&z_{1}z_{2}-z_{1}\alpha\lambda_{2}-z_{2}\alpha\lambda_{1}+\alpha\lambda_{1}\alpha\lambda_{2}\\\
&=&z_{1}z_{2}-\alpha(z_{1}\lambda_{2}+z_{2}\lambda_{1}+\lambda_{1}\alpha\lambda_{2}).\end{array}$
(28)

Let $z_{1}\lambda_{2}+z_{2}\lambda_{1}+\lambda_{1}\alpha\lambda_{2}=\lambda$
where $\lambda\in\mathcal{H}.$ Thereby,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(2)}_{\alpha}(z_{2})&=&z_{1}z_{2}-\alpha\lambda.\end{array}$
(29)

Since $\mu_{\alpha}(z_{1}z_{2})=(z_{1}z_{2})\mod\alpha,$ then there exists
$\exists\beta\in\mathcal{H}$ such that
$\mu_{\alpha}(z_{1}z_{2})=z_{1}z_{2}-\alpha\beta.$ So, we have

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(z_{1})\mu^{(2)}_{\alpha}(z_{2})&=&\mu_{\alpha}(z_{1}z_{2}).\end{array}$
(30)

Similarly, we can also show cases in the $(iii.)$ and $(iv.).$ Consequently,

$\mu_{\alpha}(z_{1}z_{2})=\mu_{\alpha}(z_{1})\mu_{\alpha}(z_{2}).$ (31)

Namely, the modulo function
$\mu:\mathbb{Z}_{N(\alpha)}\rightarrow\mathcal{H}_{\alpha}$ is a ring
homomorphism. This completes this proof.

###### Theorem 3.2

The modulo function
$\mu:\mathbb{Z}_{N(\alpha)}\rightarrow\mathcal{H}_{\alpha}$ is a ring
isomorphism ring. Namely, $\mathbb{Z}_{N(\alpha)}\cong\mathcal{H}_{\alpha}.$

###### Proof 3.3

Let $\alpha$ be a prime Hurwitz integer and,
$z_{1},z_{2}\in\mathbb{Z}_{N(\alpha)}.$ The modulo function
$\mu:\mathbb{Z}_{N(\alpha)}\rightarrow\mathcal{H}_{\alpha}:$

$\mu_{\alpha}(z)=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}$
(32)

where

$\mu_{\alpha}^{(1)}(z)=z\mod\alpha=z-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil$
(33)

and

$\mu_{\alpha}^{(2)}(z)=z\mod\alpha=z-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil$
(34)

such that

$\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}=\begin{cases}\mu_{\alpha}^{(1)}(z),&\mbox{if
}N(\mu_{\alpha}^{(1)}(z))<N(\mu_{\alpha}^{(2)}(z))\\\
\mu_{\alpha}^{(1)}(z),&\mbox{if
}N(\mu_{\alpha}^{(1)}(z))=N(\mu_{\alpha}^{(2)}(z))\mbox{ where z is an even
integer}\\\ \mu_{\alpha}^{(2)}(z),&\mbox{otherwise}.\end{cases}$ (35)

where $z\in\mathbb{Z}_{N(\alpha)}.$

According to theorem 3.1, the modulo function $\mu$ is a ring homomorphism. We
should show that it is a bijective ring homomorphism, i.e., a ring
isomorphism. This mapping is a surjective ring homomorphism because of
$\text{Im}\mu=\\{\mu_{\alpha}(z)=\mathrm{min}\\{\mu_{\alpha}^{(1)}(z),\mu_{\alpha}^{(2)}(z)\\}:z\in\mathbb{Z}_{N(\alpha)}\\}=\mathcal{H}_{\alpha}.$
If $z=0,$ then

$\begin{array}[]{lll}\mu_{\alpha}(0)&=&\mathrm{min}\\{\mu_{\alpha}^{(1)}(0),\mu_{\alpha}^{(2)}(0)\\}.\end{array}$
(36)

From proposition 3.1,

$\begin{array}[]{lll}\mu_{\alpha}(0)&=&0.\end{array}$ (37)

If $z\neq 0,$ then $\mu_{\alpha}(z)$ is to greater than or equal to $1.$
Hereby, this mapping is an injective ring homomorphism because of
$Ker\mu=\\{z\in\mathbb{Z}_{N(\alpha)}:\mu_{\alpha}(z)=0\\}=\\{z\in\mathbb{Z}_{N(\alpha)}:z=0\\}=\\{0\\}.$
So, $\mu$ function is a ring isomorphism since it is both a surjective ring
homomorphism and an injective ring homomorphism, i.e.
${\mathbb{Z}_{N(\alpha)}}\cong{\mathcal{H}_{\alpha}}.$ This completes the
proof.

###### Definition 3.2

Let $\mathbb{F}_{q}$ be a finite field with $q$ elements. $C$ is called a code
if $C$ is a nonempty subset of $\mathbb{F}_{q}^{n}.$ An element of $C$ is
called a codeword in $C.$ A linear code $C$ of length $n$ over
$\mathbb{F}_{q}$ is defined to be a subspace of $\mathbb{F}_{q}^{n}.$

###### Definition 3.3

A code $C$ of length $n$ is a subset of the direct product $\mathcal{H}^{n}$
of $n$ copies of $\mathcal{H}.$ In each of the cases we consider,
$\mathcal{H}$ is an Abelian group, and thus the same is true for
$\mathcal{H}^{n}.$ A code $C$ is a group code if it is a subgroup of
$\mathcal{H}^{n},$ or equivalently as $\mathcal{H}^{n}$ is a finite group,

$c,c^{{}^{\prime}}\in C\Rightarrow c-c^{{}^{\prime}}\in C.$

In the case when $\mathcal{H}$ is a finite field, and thus $\mathcal{H}^{n}$
is a vector space of dimension $n$ over $\mathcal{H},$ then a linear code is a
subspace $C$ of $\mathcal{H}^{n}.$ Here we say that a code $C$ in
$\mathcal{H}^{n}$ is an $(n,k)-code$ if the size of $C$ is equal to
$|\mathcal{H}|^{k}$ (The case $k=0$ is of less interest, and thus left aside)
[12].

###### Definition 3.4

The average energy of $\mathcal{H}_{\alpha}$ that denoted by
$\mathcal{E}_{\pi}$ is computed by

$\mathcal{E}_{\pi}=\frac{1}{N(\alpha)}\sum_{z=0}^{N(\alpha)-1}N(\mu_{\alpha}(z))$
(38)

where $z\in\mathbb{Z}_{N(\alpha)}$.

##  4 Some Results

This section presents some results related to the modulo function $\mu$
defined in definition 3.1 and, the mathematical notations given in eq. 5 and
eq. 6. $\mathcal{H}^{(1)}_{\alpha}$ is the set of residual class with respect
to eq. 5, $\mathcal{H}^{(2)}_{\alpha}$ is the set of residual class with
respect to eq. 6, and $\mathcal{H}_{\alpha}$ is the set of residual class with
respect to the modulo function $\mu$ defined in definition 3.1. $2+i,$ $3+2j,$
and $4i+k$ are prime Hurwitz integers that have two-components, and so on.
$3+i+j,$ $3+2j+2k,$ and $4i+3j+2k$ are prime Hurwitz integers that have three-
components, and so on.

###### Corollary 4.1

Let $\alpha$ be a prime Hurwitz integer that have two-components. Then, we
have

$\mathcal{H}_{\alpha}=\mathcal{H}^{(1)}_{\alpha}.$ (39)

###### Proposition 4.1

Let $\beta=\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k$ be a Hurwitz integer.
If $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z},$ then,

$\lfloor\beta\rceil=\beta,$ (40)

and

$\lfloor\lfloor\beta\rceil\rceil=\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k.$
(41)

If $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z}+\frac{1}{2},$ then,

$\lfloor\beta\rceil=\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k,$
(42)

and

$\lfloor\lfloor\beta\rceil\rceil=\beta.$ (43)

###### Proof 4.1

Let $\beta=\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k$ be a Hurwitz integer.
If $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z},$ then,

$\begin{array}[]{lll}\lfloor\beta\rceil&=&\lfloor\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\rceil\\\
&=&\lfloor\beta_{1}\rceil+\lfloor\beta_{2}\rceil i+\lfloor\beta_{3}\rceil
j+\lfloor\beta_{4}\rceil k\\\ &=&\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\\\
&=&\beta.\end{array}$ (44)

On the other hand,

$\begin{array}[]{lll}\lfloor\lfloor\beta\rceil\rceil&=&\lfloor\lfloor\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\rceil\rceil\\\
&=&\lfloor\lfloor\beta_{1}\rceil\rceil+\lfloor\lfloor\beta_{2}\rceil\rceil
i+\lfloor\lfloor\beta_{3}\rceil\rceil j+\lfloor\lfloor\beta_{4}\rceil\rceil
k.\end{array}$ (45)

Since the property of rounding notation, then,

$\begin{array}[]{lll}\lfloor\lfloor\beta\rceil\rceil&=&\beta_{1}\pm\frac{1}{2}+(\beta_{2}\pm\frac{1}{2})i+(\beta_{3}\pm\frac{1}{2})j+(\beta_{4}\pm\frac{1}{2})k\\\
&=&\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\\\
&=&\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\\\
\end{array}$ (46)

If $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z}+\frac{1}{2},$ then,

$\begin{array}[]{lll}\lfloor\beta\rceil&=&\lfloor\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\rceil\\\
&=&\lfloor\beta_{1}\rceil+\lfloor\beta_{2}\rceil i+\lfloor\beta_{3}\rceil
j+\lfloor\beta_{4}\rceil k.\end{array}$ (47)

Since the property of rounding notation, then,

$\begin{array}[]{lll}\begin{array}[]{lll}\lfloor\beta\rceil&=&\beta_{1}\pm\frac{1}{2}+(\beta_{2}\pm\frac{1}{2})i+(\beta_{3}\pm\frac{1}{2})j+(\beta_{4}\pm\frac{1}{2})k\\\
&=&\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\\\
&=&\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k.\end{array}\end{array}$
(48)

On the other hand,

$\begin{array}[]{lll}\lfloor\lfloor\beta\rceil\rceil&=&\lfloor\lfloor\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\rceil\rceil\\\
&=&\lfloor\lfloor\beta_{1}\rceil\rceil+\lfloor\lfloor\beta_{2}\rceil\rceil
i+\lfloor\lfloor\beta_{3}\rceil\rceil j+\lfloor\lfloor\beta_{4}\rceil\rceil
k\\\ &=&\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k\\\ &=&\beta.\end{array}$
(49)

This completes the proof.

###### Corollary 4.2

Let $\beta=\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k$ and
$\pi=\pi_{1}+\pi_{2}i+\pi_{3}j+\pi_{4}k$ be Hurwitz integers. If
$\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z},$ and
$\pi_{1},\pi_{2},\pi_{3},\pi_{4}\in\mathbb{Z},$ then,

$\lfloor\pi\pm\beta\rceil=\lfloor\pi\rceil\pm\lfloor\beta\rceil=\pi\pm\beta,$
(50)

and

$\lfloor\lfloor\pi\pm\beta\rceil\rceil=\left(\pi\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\right)\pm\left(\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\right).$
(51)

If $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z},$ and
$\pi_{1},\pi_{2},\pi_{3},\pi_{4}\in\mathbb{Z}+\frac{1}{2},$ then,

$\lfloor\pi\pm\beta\rceil=\lfloor\pi\rceil\pm\lfloor\beta\rceil=\pi\pm\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k,$
(52)

and

$\lfloor\lfloor\pi\pm\beta\rceil\rceil=\lfloor\lfloor\pi\rceil\rceil\pm\lfloor\lfloor\beta\rceil\rceil=\pi\pm\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k.$
(53)

If $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{Z}+\frac{1}{2},$ and
$\pi_{1},\pi_{2},\pi_{3},\pi_{4}\in\mathbb{Z}+\frac{1}{2},$ then,

$\lfloor\pi\pm\beta\rceil=\lfloor\pi\rceil\pm\lfloor\beta\rceil=\left(\pi\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\right)\pm\left(\beta\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\right),$
(54)

and

$\lfloor\lfloor\pi\pm\beta\rceil\rceil=\lfloor\lfloor\pi\rceil\rceil\pm\lfloor\lfloor\beta\rceil\rceil=\pi\pm\beta.$
(55)

###### Proof 4.2

The proof can be easily seen from proposition 4.1.

###### Proposition 4.2

Let $\beta=\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k$ and
$\pi=\pi_{1}+\pi_{2}i+\pi_{3}j+\pi_{4}k$ be Hurwitz integers. We consider
$\pi_{1}$ and $\beta_{1}.$ If $\pi_{1},\beta_{1}\in\mathbb{Z},$ then,

$\lfloor\pi_{1}\beta_{1}\rceil=\pi_{1}\beta_{1},$ (56)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pi_{1}\beta_{1}\pm\frac{1}{2}.$
(57)

Let $\pi_{1}\in\mathbb{Z}$ and $\beta_{1}\in\mathbb{Z}+\frac{1}{2}.$ If
$\pi_{1}$ is an even integer, then,

$\lfloor\pi_{1}\beta_{1}\rceil=\pi_{1}\beta_{1},$ (58)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pi_{1}\beta_{1}\pm\frac{1}{2}.$
(59)

If $\pi_{1}$ is an odd integer, then,

$\lfloor\pi_{1}\beta_{1}\rceil=\pi_{1}\beta_{1}\pm\frac{1}{2},$ (60)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pi_{1}\beta_{1}.$ (61)

Let $\pi_{1},\beta_{1}\in\mathbb{Z}+\frac{1}{2}$ such that
$\pi=\frac{\lambda_{1}}{2}$ and $\beta=\frac{\lambda_{2}}{2}$ where
$\lambda_{1}$ and $\lambda_{2}$ are odd integers. If
$\lambda_{1}\lambda_{2}\equiv 1\mod 4,$ then,

$\lfloor\pi_{1}\beta_{1}\rceil=\pm\pi_{1}\beta_{1}\mp\frac{1}{4},$ (62)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pm\pi_{1}\beta_{1}\pm\frac{1}{4}.$
(63)

If $\lambda_{1}\lambda_{2}\equiv 3\mod 4,$ then,

$\lfloor\pi_{1}\beta_{1}\rceil=\pm\pi_{1}\beta_{1}\pm\frac{1}{4},$ (64)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pm\pi_{1}\beta_{1}\mp\frac{1}{4}.$
(65)

###### Proof 4.3

Let $\beta=\beta_{1}+\beta_{2}i+\beta_{3}j+\beta_{4}k$ and
$\pi=\pi_{1}+\pi_{2}i+\pi_{3}j+\pi_{4}k$ be Hurwitz integers. We consider
$\pi_{1}$ and $\beta_{1}.$ Let $\pi_{1},\beta_{1}\in\mathbb{Z}.$
$\pi_{1}\beta_{1}\in\mathbb{Z}$ because of $\pi_{1},\beta_{1}\in\mathbb{Z}.$
So, from the property of round notations,

$\lfloor\pi_{1}\beta_{1}\rceil=\pi_{1}\beta_{1},$ (66)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pi_{1}\beta_{1}\pm\frac{1}{2}.$
(67)

Let $\pi_{1}\in\mathbb{Z}$ and $\beta_{1}\in\mathbb{Z}+\frac{1}{2}.$
$\pi_{1}\beta_{1}\in\mathbb{Z}$ because of $\pi_{1}\in\mathbb{Z},$
$\beta_{1}\in\mathbb{Z}+\frac{1}{2},$ and $\pi_{1}$ is an even integer. So,
from the property of round notations,

$\lfloor\pi_{1}\beta_{1}\rceil=\pi_{1}\beta_{1},$ (68)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pi_{1}\beta_{1}\pm\frac{1}{2}.$
(69)

$\pi_{1}\beta_{1}\in\mathbb{Z}+\frac{1}{2}$ because of $\pi_{1}\in\mathbb{Z},$
$\beta_{1}\in\mathbb{Z}+\frac{1}{2},$ and $\pi_{1}$ is an odd integer. So,
from the property of round notations,

$\lfloor\pi_{1}\beta_{1}\rceil=\pi_{1}\beta_{1}\pm\frac{1}{2},$ (70)

and

$\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil=\pi_{1}\beta_{1}.$ (71)

Let $\pi_{1},\beta_{1}\in\mathbb{Z}+\frac{1}{2}$ and, let
$\pi_{1}=\frac{\lambda_{1}}{2}$ and $\beta_{1}=\frac{\lambda_{2}}{2}$ where
$\lambda_{1}$ and $\lambda_{2}$ are odd integers. If
$1\equiv\lambda_{1}\lambda_{2}\mod 4,$ then, $\lambda_{1}\lambda_{2}=4k+1$
where $k\in\mathbb{Z}.$ Hereby,

$\begin{array}[]{lll}\lfloor\pi_{1}\beta_{1}\rceil&=&\lfloor\pm\frac{\lambda_{1}\lambda_{2}}{4}\rceil\\\
&=&\lfloor\pm\frac{4k+1}{4}\rceil.\end{array}$ (72)

$\lfloor\frac{4k+1}{4}\rceil=\pm k$ because of the property of round notation,
and $k<\lfloor\frac{4k+1}{4}\rceil<\frac{2k+1}{2}.$. Hereby,

$\begin{array}[]{lll}\lfloor\pi_{1}\beta_{1}\rceil&=&\pm k\end{array}$ (73)

Since
$k=\frac{\lambda_{1}\lambda_{2}}{4}-\frac{1}{4}=\pi_{1}\beta_{1}-\frac{1}{4},$
then

$\begin{array}[]{lll}\lfloor\pi_{1}\beta_{1}\rceil&=&\pm\pi_{1}\beta_{1}\mp\frac{1}{4}.\end{array}$
(74)

On the other hand,

$\begin{array}[]{lll}\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil&=&\lfloor\lfloor\pm\frac{\lambda_{1}\lambda_{2}}{4}\rceil\rceil\\\
&=&\lfloor\lfloor\pm\frac{4k+1}{4}\rceil\rceil\\\ \end{array}$ (75)

$\pm\lfloor\lfloor\frac{4k+1}{4}\rceil\rceil=\pm\frac{2k+1}{2}$ because of the
property of round notation, and
$k<\lfloor\lfloor\frac{4k+1}{4}\rceil\rceil<\frac{2k+1}{2}.$ Hereby,

$\begin{array}[]{lll}\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil&=&\pm\frac{2k+1}{2}\\\
&=&\pm k\pm\frac{1}{2}.\end{array}$ (76)

Since
$k=\frac{\lambda_{1}\lambda_{2}}{4}-\frac{1}{4}=\pi_{1}\beta_{1}-\frac{1}{4},$
then

$\begin{array}[]{lll}\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil&=&\pm\pi_{1}\beta_{1}\mp\frac{1}{4}\pm\frac{1}{2}\\\
&=&\pm\pi_{1}\beta_{1}\pm\frac{1}{4}.\end{array}$ (77)

If $3\equiv\lambda_{1}\lambda_{2}\mod 4,$ then, $\lambda_{1}\lambda_{2}=4k+3$
where $k\in\mathbb{Z}.$ Hereby,

$\begin{array}[]{lll}\lfloor\pi_{1}\beta_{1}\rceil&=&\lfloor\pm\frac{\lambda_{1}\lambda_{2}}{4}\rceil\\\
&=&\lfloor\pm\frac{4k+3}{4}\rceil.\end{array}$ (78)

$\lfloor\frac{4k+3}{4}\rceil=\pm k\pm 1$ because of the property of round
notation, and $\frac{2k+1}{2}<\lfloor\frac{4k+3}{4}\rceil<k+1.$ Hereby,

$\begin{array}[]{lll}\lfloor\pi_{1}\beta_{1}\rceil&=&\pm k\pm 1.\end{array}$
(79)

Since
$k=\frac{\lambda_{1}\lambda{2}}{4}-\frac{3}{4}=\pi_{1}\beta_{1}-\frac{3}{4},$
then

$\begin{array}[]{lll}\lfloor\pi_{1}\beta_{1}\rceil&=&\pm\pi_{1}\beta_{1}\mp\frac{3}{4}\pm
1\\\ &=&\pm\pi_{1}\beta_{1}\pm\frac{1}{4}.\end{array}$ (80)

On the other hand,

$\begin{array}[]{lll}\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil&=&\lfloor\lfloor\pm\frac{\lambda_{1}}{4}\rceil\rceil\\\
&=&\lfloor\lfloor\pm\frac{4k+3}{4}\rceil\rceil\\\ \end{array}$ (81)

$\pm\lfloor\lfloor\frac{4k+3}{4}\rceil\rceil=\pm\frac{2k+1}{2}$ because of the
property of round notation, and
$\frac{2k+1}{2}<\lfloor\lfloor\frac{4k+3}{4}\rceil\rceil<k+1.$ Hereby,

$\begin{array}[]{lll}\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil&=&\pm\frac{2k+1}{2}\\\
&=&\pm k\pm\frac{1}{2}.\end{array}$ (82)

Since
$k=\frac{\lambda_{1}\lambda_{2}}{4}-\frac{3}{4}=\pi_{1}\beta_{1}-\frac{3}{4},$
then

$\begin{array}[]{lll}\lfloor\lfloor\pi_{1}\beta_{1}\rceil\rceil&=&\pm\pi_{1}\beta_{1}\mp\frac{3}{4}\pm\frac{1}{2}\\\
&=&\pm\pi_{1}\beta_{1}\mp\frac{1}{4}.\end{array}$ (83)

This completes the proof.

###### Proposition 4.3

Let $\alpha$ be a prime Hurwitz integer. Then,

$N(\mu_{\alpha}(0))=N(\alpha).$ (84)

###### Proof 4.4

Let $\alpha$ be a prime Hurwitz integer. From eq. 9, we know that

$\mu_{\alpha}(0)=\frac{-\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}}{2}+\frac{-\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{4}}{2}i+\frac{-\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{4}}{2}j+\frac{-\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4}}{2}k.$
(85)

Hereby,

$\begin{array}[]{lll}N(\mu_{\alpha}(0))&=&N(\frac{-\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}}{2}+\frac{-\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{4}}{2}i+\frac{-\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{4}}{2}j+\frac{-\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4}}{2}k)\\\
&=&(\frac{-\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}}{2})^{2}+(\frac{-\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{4}}{2})^{2}+(\frac{-\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{4}}{2})^{2}+(\frac{-\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4}}{2})^{2}\\\
&=&\frac{\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4}+2\alpha_{3}\alpha_{4}+2\alpha_{2}\alpha_{3}+2\alpha_{2}\alpha_{4}-2\alpha_{1}\alpha_{2}-2\alpha_{1}\alpha_{3}-2\alpha_{1}\alpha_{4}}{4}\\\
&&+\frac{\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4}-2\alpha_{3}\alpha_{4}+2\alpha_{2}\alpha_{3}-2\alpha_{2}\alpha_{4}+2\alpha_{1}\alpha_{2}+2\alpha_{1}\alpha_{3}-2\alpha_{1}\alpha_{4}}{4}\\\
&&+\frac{\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4}+2\alpha_{3}\alpha_{4}-2\alpha_{2}\alpha_{3}-2\alpha_{2}\alpha_{4}-2\alpha_{1}\alpha_{2}+2\alpha_{1}\alpha_{3}+2\alpha_{1}\alpha_{4}}{4}\\\
&&+\frac{\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4}-2\alpha_{3}\alpha_{4}-2\alpha_{2}\alpha_{3}+2\alpha_{2}\alpha_{4}+2\alpha_{1}\alpha_{2}-2\alpha_{1}\alpha_{3}+2\alpha_{1}\alpha_{4}}{4}\\\
&=&\frac{4(\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4})}{4}\\\
\end{array}$ (86)

Since $N(\alpha)=\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4},$
then,

$\begin{array}[]{lll}N(\mu_{\alpha}(0))&=&\alpha^{2}_{1}+\alpha^{2}_{2}+\alpha^{2}_{3}+\alpha^{2}_{4}\\\
&=&N(\alpha).\end{array}$ (87)

This completes the proof.

###### Proposition 4.4

Let $\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ be a prime Hurwitz
integer. If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then,

$\mu^{(1)}_{\alpha}(N(\alpha))=0,$ (88)

and

$\mu^{(2)}_{\alpha}(N(\alpha))\equiv 0\mod\alpha.$ (89)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2},$
then

$\mu^{(1)}_{\alpha}(N(\alpha))\equiv 0\mod\alpha,$ (90)

and

$\mu^{(2)}_{\alpha}(N(\alpha))=0.$ (91)

###### Proof 4.5

Let $\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ be a prime Hurwitz
integer. If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then,
from eq. 5,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\overline{\alpha}\rceil.\end{array}$ (92)

From eq. 40,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\overline{\alpha}\\\
&=&N(\alpha)-N(\alpha)\\\ &=&0.\end{array}$ (93)

From eq. 6,

$\begin{array}[]{lll}\mu^{(2)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\lfloor\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\overline{\alpha}\rceil\rceil.\end{array}$
(94)

From eq. 41,

$\begin{array}[]{lll}\mu^{(2)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\left(\overline{\alpha}+\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k\right)\\\
&=&N(\alpha)-N(\alpha)-\alpha\left(\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k\right)\\\
&=&-\alpha\left(\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k\right)\\\
&\equiv&0\mod\alpha\end{array}$ (95)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2},$
then, from eq. 5,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\overline{\alpha}\rceil.\end{array}$ (96)

From eq. 42,

$\begin{array}[]{lll}\mu^{(1)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\left(\overline{\alpha}+\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k\right)\\\
&=&N(\alpha)-N(\alpha)-\alpha\left(\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k\right)\\\
&=&-\alpha\left(\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k\right)\\\
&\equiv&0\mod\alpha.\end{array}$ (97)

From eq. 6,

$\begin{array}[]{lll}\mu^{(2)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\lfloor\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\overline{\alpha}\rceil\rceil.\end{array}$
(98)

From eq. 43,

$\begin{array}[]{lll}\mu^{(2)}_{\alpha}(N(\alpha))&=&N(\alpha)-\alpha\overline{\alpha}\\\
&=&N(\alpha)-N(\alpha)\\\ &=&0.\end{array}$ (99)

This completes the proof.

###### Proposition 4.5

Let $\alpha=\alpha_{1}+\alpha_{2}i+\alpha_{3}j+\alpha_{4}k$ be a prime Hurwitz
integer. Then,

$\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)\equiv 0\mod\alpha$ (100)

where $z\in\mathbb{Z}_{N(\alpha)}.$

###### Proof 4.6

Let $\alpha$ be a prime Hurwitz integer. Firstly, we suppose that
$\mu_{\alpha}(z)=\mu^{(1)}_{\alpha}(z)$ and
$\mu_{\alpha}(N(\alpha)-z)=\mu^{(1)}_{\alpha}(N(\alpha)-z)$ where
$z\in\mathbb{Z}_{N(\alpha)}.$ Then,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&\mu^{(1)}_{\alpha}(z)+\mu^{(1)}_{\alpha}(N(\alpha)-z).\end{array}$
(101)

From eq. 5,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&z-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil+N(\alpha)-z-\alpha\lfloor\frac{\overline{\alpha}(N(\alpha)-z)}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil-\alpha\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil-\alpha\lfloor\frac{\overline{\alpha}(-z)}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil-\alpha\lfloor\overline{\alpha}\rceil+\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\overline{\alpha}\rceil.\end{array}$ (102)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then, from eq.
92 and eq. 93,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&0\\\
&\equiv&0\mod\alpha.\end{array}$ (103)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2},$
then, from eq. 96 and eq. 97,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&\equiv&0\mod\alpha.\end{array}$
(104)

Secondly, we suppose that $\mu_{\alpha}(z)=\mu^{(1)}_{\alpha}(z)$ and
$\mu_{\alpha}(N(\alpha)-z)=\mu^{(2)}_{\alpha}(N(\alpha)-z)$ where
$z\in\mathbb{Z}_{N(\alpha)}.$ Then,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&\mu^{(1)}_{\alpha}(z)+\mu^{(2)}_{\alpha}(N(\alpha)-z).\end{array}$
(105)

From eq. 5 and eq. 6,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&z-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil+N(\alpha)-z-\alpha\lfloor\lfloor\frac{\overline{\alpha}(N(\alpha)-z)}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil-\alpha\lfloor\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil\rceil-\alpha\lfloor\lfloor\frac{\overline{\alpha}(-z)}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil-\alpha\lfloor\lfloor\overline{\alpha}\rceil\rceil+\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil.\end{array}$
(106)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then, from eq.
94 and eq. 95,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil+\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil.\end{array}$
(107)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil=\lambda_{1}$ and
$\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil=\lambda_{2}.$
Hereby,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lambda_{1}+\alpha\lambda_{2}\\\
&=&\alpha(-\lambda_{1}+\lambda_{2})\\\ &\equiv&0\mod\alpha.\end{array}$ (108)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2},$
then, from eq. 98 and eq. 99,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil+\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil.\end{array}$
(109)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil=\lambda_{1}$ and
$\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil=\lambda_{2}.$
Hereby,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lambda_{1}+\alpha\lambda_{2}\\\
&=&\alpha(-\lambda_{1}+\lambda_{2})\\\ &\equiv&0\mod\alpha.\end{array}$ (110)

Thirdly, we suppose that $\mu_{\alpha}(z)=\mu^{(2)}_{\alpha}(z)$ and
$\mu_{\alpha}(N(\alpha)-z)=\mu^{(1)}_{\alpha}(N(\alpha)-z)$ where
$z\in\mathbb{Z}_{N(\alpha)}.$ Then,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&\mu^{(2)}_{\alpha}(z)+\mu^{(1)}_{\alpha}(N(\alpha)-z).\end{array}$
(111)

From eq. 6 and eq. 5,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&z-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil+N(\alpha)-z-\alpha\lfloor\frac{\overline{\alpha}(N(\alpha)-z)}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil-\alpha\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil-\alpha\lfloor\frac{\overline{\alpha}(-z)}{N(\alpha)}\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil-\alpha\lfloor\overline{\alpha}\rceil+\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil.\end{array}$
(112)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then, from eq.
92 and eq. 93,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil+\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil.\end{array}$
(113)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil=\lambda_{1}$
and $\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil=\lambda_{2}.$
Hereby,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lambda_{1}+\alpha\lambda_{2}\\\
&=&\alpha(-\lambda_{1}+\lambda_{2})\\\ &\equiv&0\mod\alpha.\end{array}$ (114)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2},$
then, from eq. 96 and eq. 97,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil+\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil.\end{array}$
(115)

There exist $\lambda_{1},\lambda_{2}\in\mathcal{H}$ such that
$\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil=\lambda_{1}$
and $\alpha\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil=\lambda_{2}.$
Hereby,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&-\alpha\lambda_{1}+\alpha\lambda_{2}\\\
&=&\alpha(-\lambda_{1}+\lambda_{2})\\\ &\equiv&0\mod\alpha.\end{array}$ (116)

Lastly, we suppose that $\mu_{\alpha}(z)=\mu^{(2)}_{\alpha}(z)$ and
$\mu_{\alpha}(N(\alpha)-z)=\mu^{(2)}_{\alpha}(N(\alpha)-z)$ where
$z\in\mathbb{Z}_{N(\alpha)}.$ Then,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&\mu^{(2)}_{\alpha}(z)+\mu^{(12)}_{\alpha}(N(\alpha)-z).\end{array}$
(117)

From eq. 5,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&z-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil+N(\alpha)-z-\alpha\lfloor\lfloor\frac{\overline{\alpha}(N(\alpha)-z)}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil-\alpha\lfloor\lfloor\frac{\overline{\alpha}N(\alpha)}{N(\alpha)}\rceil\rceil-\alpha\lfloor\lfloor\frac{\overline{\alpha}(-z)}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil-\alpha\lfloor\lfloor\overline{\alpha}\rceil\rceil+\alpha\lfloor\lfloor\frac{\overline{\alpha}z}{N(\alpha)}\rceil\rceil\\\
&=&N(\alpha)-\alpha\lfloor\lfloor\overline{\alpha}\rceil\rceil.\end{array}$
(118)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z},$ then, from eq.
94 and eq. 95,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&=&0\\\
&\equiv&0\mod\alpha.\end{array}$ (119)

If $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{Z}+\frac{1}{2},$
then, from eq. 98 and eq. 99,

$\begin{array}[]{lll}\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)&\equiv&0\mod\alpha.\end{array}$
(120)

Consequently, $\mu_{\alpha}(z)+\mu_{\alpha}(N(\alpha)-z)\equiv 0\mod\alpha.$
This completes the proof.

##  5 Examples

###### Example 5.1

Let $\alpha=\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k.$ $\alpha$ is a
prime Hurwitz integer because of $N(\alpha)=13.$ From eq. 5,

$\mathcal{H}^{(1)}_{\alpha}=\left\\{\begin{gathered}\mu^{(1)}_{\alpha}(0)=0,\mu^{(1)}_{\alpha}(1)=1,\mu^{(1)}_{\alpha}(2)=2,\mu^{(1)}_{\alpha}(3)=\frac{1}{2}-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k,\hfill\\\
\mu^{(1)}_{\alpha}(4)=\frac{3}{2}-\frac{3}{2}i-\frac{3}{2}j-\frac{3}{2}k,\mu^{(1)}_{\alpha}(5)=-2+i+j+k,\mu^{(1)}_{\alpha}(6)=-1+i+j+k,\hfill\\\
\mu^{(1)}_{\alpha}(7)=i+j+k,\mu^{(1)}_{\alpha}(8)=-\frac{3}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\mu^{(1)}_{\alpha}(9)=-\frac{1}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\hfill\\\
\mu^{(1)}_{\alpha}(10)=\frac{1}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\mu^{(1)}_{\alpha}(11)=\frac{3}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\hfill\\\
\mu^{(1)}_{\alpha}(12)=\frac{5}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k\hfill\end{gathered}\right\\}.$
(121)

From eq. 6,

$\mathcal{H}^{(2)}_{\alpha}=\left\\{\begin{gathered}\mu^{(2)}_{\alpha}(0)=1-2i-2j-2k,\mu^{(2)}_{\alpha}(1)=-\frac{5}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(2)=-\frac{3}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\mu^{(2)}_{\alpha}(3)=-\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(4)=\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\mu^{(2)}_{\alpha}(5)=\frac{3}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(6)=-i-j-k,\mu^{(2)}_{\alpha}(7)=1-i-j-k,\mu^{(2)}_{\alpha}(8)=2-i-j-k,\hfill\\\
\mu^{(2)}_{\alpha}(9)=-\frac{3}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k,\mu^{(2)}_{\alpha}(10)=-\frac{1}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(11)=-2,\mu^{(2)}_{\alpha}(12)=-1\hfill\end{gathered}\right\\}.$
(122)

With respect to eq. 7

$\mathcal{H}_{\alpha}=\left\\{\begin{gathered}\mu_{\alpha}(0)=\mu^{(1)}_{\alpha}(0)=0,\mu_{\alpha}(1)=\mu^{(1)}_{\alpha}(1)=1,\mu_{\alpha}(2)=\mu^{(2)}_{\alpha}(2)=-\frac{3}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\hfill\\\
\mu_{\alpha}(3)=\mu^{(2)}_{\alpha}(3)=-\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\mu_{\alpha}(4)=\mu^{(2)}_{\alpha}(4)=\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\hfill\\\
\mu_{\alpha}(5)=\mu^{(2)}_{\alpha}(5)=\frac{3}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k,\mu_{\alpha}(6)=\mu^{(2)}_{\alpha}(6)=-i-j-k,\hfill\\\
\mu_{\alpha}(7)=\mu^{(1)}_{\alpha}(7)=i+j+k,\mu_{\alpha}(8)=\mu^{(1)}_{\alpha}(8)=-\frac{3}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\hfill\\\
\mu_{\alpha}(9)=\mu^{(1)}_{\alpha}(9)=-\frac{1}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\mu_{\alpha}(10)=\mu^{(1)}_{\alpha}(10)=\frac{1}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\hfill\\\
\mu_{\alpha}(11)=\mu^{(1)}_{\alpha}(11)=\frac{3}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k,\mu_{\alpha}(12)=\mu^{(2)}_{\alpha}(12)=-1\hfill\end{gathered}\right\\}.$
(123)

From eq. 38, the average energy of
$\mathcal{H}_{\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k}$ is

$\mathcal{E}_{\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k}=\frac{24}{13}=1.8462.$
(124)

###### Example 5.2

Let $\alpha=3+2i.$ $\alpha$ is a prime Hurwitz integer because of
$N(\alpha)=13.$ From eq. 5,

$\mathcal{H}^{(1)}_{\alpha}=\left\\{\begin{gathered}\mu^{(1)}_{\alpha}(0)=0,\mu^{(1)}_{\alpha}(1)=1,\mu^{(1)}_{\alpha}(2)=2,\mu^{(1)}_{\alpha}(3)=-2i,\mu^{(1)}_{\alpha}(4)=-1+i,\mu^{(1)}_{\alpha}(5)=i,\hfill\\\
\mu^{(1)}_{\alpha}(6)=1+i,\mu^{(1)}_{\alpha}(7)=-1-i,\mu^{(1)}_{\alpha}(8)=-i,\mu^{(1)}_{\alpha}(9)=1-i,\mu^{(1)}_{\alpha}(10)=2i,\hfill\\\
\mu^{(1)}_{\alpha}(11)=-2,\mu^{(1)}_{\alpha}(12)=-1\hfill\end{gathered}\right\\}.$
(125)

From eq. 6,

$\mathcal{H}^{(2)}_{\alpha}=\left\\{\begin{gathered}\mu^{(2)}_{\alpha}(0)=-\frac{1}{2}-\frac{5}{2}i-\frac{1}{2}j-\frac{5}{2}k,\mu^{(2)}_{\alpha}(1)=-\frac{3}{2}+\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(2)=-\frac{1}{2}+\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\mu^{(2)}_{\alpha}(3)=\frac{1}{2}+\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(4)=\frac{3}{2}+\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\mu^{(2)}_{\alpha}(5)=-\frac{1}{2}-\frac{3}{2}i-\frac{1}{2}j-\frac{5}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(6)=\frac{1}{2}-\frac{3}{2}i-\frac{1}{2}j-\frac{5}{2}k,\mu^{(2)}_{\alpha}(7)=-\frac{1}{2}+\frac{3}{2}i-\frac{1}{2}j-\frac{5}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(8)=\frac{1}{2}+\frac{3}{2}i-\frac{1}{2}j-\frac{5}{2}k,\mu^{(2)}_{\alpha}(9)=-\frac{3}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(10)=-\frac{1}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\mu^{(2)}_{\alpha}(11)=\frac{1}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(12)=\frac{3}{2}-\frac{1}{2}i-\frac{1}{2}j-\frac{5}{2}k\hfill\end{gathered}\right\\}.$
(126)

With respect to eq. 7, and corollary 4.1

$\mathcal{H}_{\alpha}=\mathcal{H}^{(1)}_{\alpha}$ (127)

From eq. 38, the average energy of $\mathcal{H}_{3+2i}$ is

$\mathcal{E}_{3+2i}=\frac{28}{13}=2.1539.$ (128)

The average energy for the transmitted signal, considering the sets of
residual class with the same cardinality, the average energy of
$\mathcal{H}_{\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k}$ is smaller
than the average energy of $\mathcal{H}_{3+2i}.$

###### Example 5.3

Let $\alpha=3+i+j.$ $\alpha$ is a prime Hurwitz integer because of
$N(\alpha)=11.$ From eq. 5,

$\mathcal{H}^{(1)}_{\alpha}=\left\\{\begin{gathered}\mu^{(1)}_{\alpha}(0)=0,\mu^{(1)}_{\alpha}(1)=1,\mu^{(1)}_{\alpha}(2)=-1-i-j,\mu^{(1)}_{\alpha}(3)=-i-j,\mu^{(1)}_{\alpha}(4)=1-i-j,\hfill\\\
\mu^{(1)}_{\alpha}(5)=2-i-j,\mu^{(1)}_{\alpha}(6)=-2+i+j,\mu^{(1)}_{\alpha}(7)=-1+i+j,\mu^{(1)}_{\alpha}(8)=i+j,\hfill\\\
\mu^{(1)}_{\alpha}(9)=1+i+j,\mu^{(1)}_{\alpha}(10)=-1,\hfill\end{gathered}\right\\}.$
(129)

From eq. 6,

$\mathcal{H}^{(2)}_{\alpha}=\left\\{\begin{gathered}\mu^{(2)}_{\alpha}(0)=-\frac{1}{2}-\frac{5}{2}i-\frac{3}{2}j-\frac{3}{2}k,\mu^{(2)}_{\alpha}(1)=-\frac{3}{2}+\frac{1}{2}i+\frac{3}{2}j-\frac{3}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(2)=-\frac{1}{2}+\frac{1}{2}i+\frac{3}{2}j-\frac{3}{2}k,\mu^{(2)}_{\alpha}(3)=\frac{1}{2}+\frac{1}{2}i+\frac{3}{2}j-\frac{3}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(4)=-\frac{3}{2}-\frac{1}{2}i+\frac{1}{2}j-\frac{3}{2}k,\mu^{(2)}_{\alpha}(5)=-\frac{1}{2}-\frac{1}{2}i+\frac{1}{2}j-\frac{3}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(6)=\frac{1}{2}-\frac{1}{2}i+\frac{1}{2}j-\frac{3}{2}k,\mu^{(2)}_{\alpha}(7)=\frac{3}{2}-\frac{1}{2}i+\frac{1}{2}j+\frac{3}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(8)=-\frac{1}{2}-\frac{3}{2}i-\frac{1}{2}j-\frac{3}{2}k,\mu^{(2)}_{\alpha}(9)=\frac{1}{2}-\frac{3}{2}i-\frac{1}{2}j-\frac{3}{2}k,\hfill\\\
\mu^{(2)}_{\alpha}(10)=\frac{3}{2}-\frac{3}{2}i-\frac{1}{2}j-\frac{3}{2}k,\hfill\end{gathered}\right\\}.$
(130)

With respect to eq. 7

$\mathcal{H}_{\alpha}=\left\\{\begin{gathered}\mu_{\alpha}(0)=\mu^{(1)}_{\alpha}(0)=0,\mu_{\alpha}(1)=\mu^{(1)}_{\alpha}(1)=1,\mu_{\alpha}(2)=\mu^{(1)}_{\alpha}(2)=-1-i-j,\hfill\\\
\mu_{\alpha}(3)=\mu^{(1)}_{\alpha}(3)=-i-j,\mu_{\alpha}(4)=\mu^{(1)}_{\alpha}(4)=1-i-j,\hfill\\\
\mu_{\alpha}(5)=\mu^{(2)}_{\alpha}(5)=-\frac{1}{2}-\frac{1}{2}i+\frac{1}{2}j-\frac{3}{2}k,\mu_{\alpha}(6)=\mu^{(2)}_{\alpha}(6)=\frac{1}{2}-\frac{1}{2}i+\frac{1}{2}j-\frac{3}{2}k,\hfill\\\
\mu_{\alpha}(7)=\mu^{(1)}_{\alpha}(7)=-1+i+j,\mu_{\alpha}(8)=\mu^{(1)}_{\alpha}(8)=i+j,\hfill\\\
\mu_{\alpha}(9)=\mu^{(1)}_{\alpha}(9)=1+i+j,\mu_{\alpha}(10)=\mu^{(1)}_{\alpha}(10)=-1,\hfill\end{gathered}\right\\}.$
(131)

From eq. 38, the average energy of $\mathcal{H}_{3+i+j}$ is

$\mathcal{E}_{3+i+j}=\frac{24}{11}=2.1818.$ (132)

Table 1: For $N\leq 50,$ the Average Energies of Prime Hurwitz Integers such that $N=6k+1$ where $k\in\mathbb{Z}^{+}.$ N | Hurwitz Integers ($\mathbb{Z}$) | Hurwitz Integers ($\mathbb{Z}+\frac{1}{2}$) | Average Energy
---|---|---|---
$7$ | $2+i+j+k$ | $\frac{3}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{1}{2}k$ | $0.8571$
|  | $\frac{5}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k$ |
$13$ | $2+2i+2j+k$ | $\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ | $1.8462$
|  | $\frac{7}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k$ |
$19$ | $4+i+j+k$ | $\frac{5}{2}+\frac{5}{2}i+\frac{5}{2}j+\frac{1}{2}k$ | $2.5263$
|  | $\frac{7}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ |
$31$ | $3+3i+3j+2k$ | $\frac{7}{2}+\frac{5}{2}i+\frac{5}{2}j+\frac{5}{2}k$ | $4.2581$
|  | $\frac{11}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k$ |
$37$ | $5+2i+2j+2k$ | $\frac{7}{2}+\frac{7}{2}i+\frac{7}{2}j+\frac{1}{2}k$ | $5.0270$
|  | $\frac{11}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ |
$43$ | $4+3i+3j+3k$ | $\frac{7}{2}+\frac{7}{2}i+\frac{7}{2}j+\frac{5}{2}k$ | $6.0000$
|  | $\frac{13}{2}+\frac{1}{2}i+\frac{1}{2}j+\frac{1}{2}k$ |

Table I is presented the average energies of prime Hurwitz integers such that
$N=6k+1\leq 50$ where $k\in\mathbb{Z}^{+}.$

##  6 The Code Rate for Codes over Hurwitz Integers

Let $\beta$ be a Hurwitz integer. $\beta$ is a unit if and only if
$N(\beta)=1.$ There are precisely $24$ units in $\mathcal{H}.$ The set of
units in $\mathcal{H}$ that denoted by $\mathcal{U}_{\mathcal{H}}$ is shown by

$\mathcal{U}_{\mathcal{H}}=\left\\{\pm 1,\pm i,\pm j,\pm k,\pm 1\pm i\pm j\pm
k,\pm\frac{1}{2}\pm\frac{1}{2}i\pm\frac{1}{2}j\pm\frac{1}{2}k\right\\}.$ (133)

In this section, we give the code rate of a code $C$ over $\mathcal{H}.$ Note
that $\mathcal{H}_{\alpha}$ has $N(\alpha)$ elements. Let $N(\alpha)=p$ be a
prime integer such that $p\equiv 1\mod 24.$ A code $C$ over $\mathcal{H}$ has
length $n=\frac{p^{k}-1}{24}.$ Here $k$ is the dimension of a code $C$ over
$\mathcal{H}.$ Let $|C|=|\mathcal{H}_{\alpha}|=N(\alpha)=p.$ If
$|C|=N(\alpha)=p,$ then $k=\log_{p}|C|=\log_{p}p=1.$ The coding rate of a code
$C$ over $\mathcal{H}$ is computed by

$R=\frac{k}{n}=\frac{24k}{p^{k}-1}.$ (134)

In this study, we consider $k=1.$

###### Example 6.1

For $p\leq 100,$ $p=73$ and $p=97$ are prime integers such that $p\equiv 1\mod
24.$

  * i.

For $N(\alpha)=p=73,$ $\alpha$ is a prime Hurwitz integer. The length of a
code $C$ over $\mathcal{H}$ is $n=\frac{72}{24}=3.$ The rate of a code $C$
over $\mathcal{H}$ is $R=\frac{24}{72}=\frac{1}{3}.$ $C$ is a $(3,1)-$ code
over $\mathcal{H}$ because of $n=3$ and $k=1.$

  * ii.

For $N(\alpha)=p=97,$ $\alpha$ is a prime Hurwitz integer. The length of a
code $C$ over $\mathcal{H}$ is $n=\frac{96}{24}=4.$ The rate of a code $C$
over $\mathcal{H}$ is $R=\frac{24}{96}=\frac{1}{4}.$ $C$ is a $(4,1)-$ code
over $\mathcal{H}$ because of $n=4$ and $k=1.$

##  7 Graph Layout Methods

Graphs are used to show the relationship between items, in general. Graph
drawing enables visualization of these relationships. The usefulness of the
visual representation depends upon whether the drawing is aesthetic. While
there are no strict criteria for aesthetic drawing, it is generally agreed
that such a drawing has minimal edge crossing and even spacing between
vertices. Two popular straight-edge drawing algorithms, the spring embedding,
and spring-electrical embedding work by minimizing the energy of physical
models of the graph. The high-dimensional embedding method, on the other hand,
embeds a graph in high-dimensional space and then projects it back to two or
three-dimensional space. Random, circular, and spiral embedding do not utilize
connectivity information for laying out a graph. In this study, we do not
consider random embedding, and circular embedding. We consider spring, the
high-dimensional, and spiral embedding methods, in this study. The spring
embedding algorithm assigns force between each pair of nodes. When two nodes
are too close together, a repelling force comes into effect. When two nodes
are too far apart, they are subject to an attractive force. This scenario can
be illustrated by linking the vertices with springs, hence the name "spring
embedding". In the high-dimensional embedding method, a graph is embedded in
high-dimensional space and then projected back to two or three-dimensional
space. The high-dimensional embedding method tends to be very fast but its
results are often of lower quality than force-directed algorithms. We use the
Wolfram Mathematica $10.2$ program, which has algorithms used for
layered/hierarchical drawing of directed graphs and, for drawing trees. These
algorithms are implemented via four functions: "GraphPlot," "GraphPlot3D,"
"LayeredGraphPlot," and "TreePlot". In this study, we use two of these
algorithms for graph drawing, i.e., "GraphPlot," and "GraphPlot3D". You can
find more details on the wolfram.com.

###### Definition 7.1

Spring Embedding The spring embedding is a graph-drawing technique to position
vertices of a graph so that they minimize the mechanical energy when each edge
corresponds to a spring. The spring embedding is typically used to lay out
regular structured graphs. You can find more details on the wolfram.com.

###### Definition 7.2

High Dimensional Embedding The high-dimensional embedding is a graph-drawing
technique to position vertices of a graph in a high-dimensional space, and
then project back to two- or three-dimensional space. The high-dimensional
embedding is typically used for fast layout of graphs. You can find more
details on the wolfram.com.

###### Definition 7.3

Spiral Embedding The spiral embedding is a graph-drawing technique to position
vertices of a graph on a 3D spiral projected to 2D. The spiral embedding is
typically used to lay out path graphs. You can find more details on the
wolfram.com.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 1: The Spring Embedding of $3+2i$ on Two or Three-Dimensional
Coordinate System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 2: The Spring Embedding of
$\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ on Two or Three-
Dimensional Coordinate System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 3: The High-Dimensional Embedding of $3+2i$ on Two or Three-Dimensional
Coordinate System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 4: The High-Dimensional Embedding of
$\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ on Two or Three-
Dimensional Coordinate System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 5: The Spiral Embedding of $3+2i$ on Two or Three-Dimensional
Coordinate System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 6: The Spiral Embedding of
$\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ on Two or Three-
Dimensional Coordinate System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 7: Points and Graph of $3+2i$ on Two or Three-Dimensional Coordinate
System.

(a) Points on three dimensional coordinate system

(b) Graph in a three-dimensional box

(c) Points on two-dimensional coordinate system

(d) Graph on two-dimensional coordinate system

Figure 8: Points and Graph of
$\frac{5}{2}+\frac{3}{2}i+\frac{3}{2}j+\frac{3}{2}k$ on Two or Three-
Dimensional Coordinate System.

##  8 Conclusion

In this study, we presented an algebraic construction technique, a modulo
function formed depending on two modulo operations, used to construct code
constructions over Hurwitz integers. Also, we presented some results for two
mathematical notations and, for the algebraic construction technique in this
study. In addition, we obtained some new block codes over Hurwitz integers
with respect to the modulo function defined in the definition 3.1 (See Section
5). Moreover, we obtained $(3,1)-$code, and $(4,1)-$code (See Section 6).
Lastly, we presented graphs of the residue class obtained with respect to the
modulo function defined in the definition 3.1 in the ring of Hurwitz integers,
particularly prime Hurwitz integers.

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