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\markboth{M.\,G\"{u}zeltepe and A.\,Alt{\i}nel}{Perfect
1-error-correcting Hurwitz weight codes}
\title[Perfect 1-error-correcting Hurwitz weight codes]{Perfect 1-error-correcting Hurwitz weight codes\footnote{The work was supported by T\"{U}B\.{I}TAK (The
Scientific and Technical Research Council of Turkey) under project
number 116F318.} }


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\author[M.\,G\"{u}zeltepe, A.\,Alt{\i}nel]{Murat G\"{u}zeltepe\corrauth   and Alev Alt{\i}nel}
\address{\ Department of Mathematics, Sakarya University, TR-54\,187 Sakarya, Turkey\\   }
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\emails{{\tt mguzeltepe@sakarya.edu.tr}\,\,(M.\,G\"{u}zeltepe),
{\tt alevaltinell@gmail.com} (A.\,Alt{\i}nel) }
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%%%%% Begin Abstract %%%%%%%%%%%
\begin{abstract}
Let $\pi$ be a Hurwitz prime and $p=\pi\pi^\star$. In this paper,
we construct perfect 1-error-correcting codes in
$\mathcal{H}_\pi^n$ for every prime number $p>3$, where
$\mathcal{H}$ denotes the set of Hurwitz integers.
%{\it Mathematical Communications}.
\end{abstract}
%%%%% end %%%%%%%%%%%

%%%%% Keywords %%%%%%%%%%%
\keywords{block codes, Hurwitz distance, perfect code}

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\ams{94B05, 94B60}

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\maketitle

%%%% Start %%%%%%






\section{Introduction}
\label{}

Since perfect codes play an important role in coding theory, both
for theoretical and practical reasons, many authors have focused
on these codes for many years. The first perfect codes which are
binary codes were defined by Hamming in \cite{hamming}. These
perfect codes were null-spaces of matrices $H$ and subspaces of
vector spaces $\mathbb{Z}_2^n$. Vasil'ev constructed the first
non-linear perfect 1-error correcting binary code in
\cite{vasilev}. Lindstr\"{o}m, 1969, and independently
Sch\"{o}nheim, 1968, generalized Vasil'ev's construction to the
$q-$ary case in \cite{lind, sch}. In 1977, mixed perfect codes
were given by Heden in \cite{heden1}. These codes were not equal
to any linear code. As for more recent perfect codes, a
generalization of perfect Lee-error-correcting codes and perfect
1-error-correcting Lipschitz weight codes were presented by Heden
and G\"{u}zeltepe \cite{biz2015,biz2016}.




Let $S$ be the direct product of $n$ copies of the finite field
$\mathbb{F}_q$ with $q$ elements and let $C\subset S$. If there is
a unique word $e=(0,0,\ldots ,\epsilon _i ,0,\ldots ,0)$ of weight
one to every word $u=(u_1, \ldots ,u_n)\in S$ such that $u+e \in
C$, then $C$ is a perfect 1-error-correcting code in the Hamming
metric, where $\epsilon _i \in \mathbb{F}_q$. There are some
generalizations of the error-correcting model introduced by
Hamming to the cases, when the error $\epsilon _i$ belongs to a
subset $\mathcal{E}$ of the alphabet used to form the codewords,
for example, ${\cal E} = \left\{ { \mp 1} \right\}$ and base ring
$\mathbb{Z}_n$. These codes are called \emph{perfect
1-error-correcting codes} in the Lee metric \cite{lee}. In a general
case, when $\mathcal{E}$ can be any given subset of the alphabet
in use, we call this kind of codes perfect
1-$\mathcal{E}$-error-correcting codes. More information can be
found in \cite{biz2015}.










\section{Preliminaries}

 The first codes constructed over Hurwitz integers were presented by G\"{u}zeltepe in
\cite{guz}. The set of Hurwitz intergers is a subset of
quaternions. It is a noncommutative ring. The set $H(\mathbb{Z})$
called integer quaternions or Lipschitz integers over the integers
$\mathbb{Z}$ is defined as

\[H(\mathbb{Z}) = \left\{ {{a_0} + {a_1}{e_1} + {a_2}{e_2} +
{a_3}{e_3}|\,{a_0},{a_1},{a_2},{a_3} \in \mathbb{Z}} \right\},\]
where
\[e_1^2 = e_2^2 = e_3^2 =  - 1, \] and \[{e_1}{e_2} =  - {e_2}{e_1}
= {e_3},\,\,\,{e_2}{e_3} =  - {e_3}{e_2} = {e_1},\,\,\,{e_3}{e_1}
= - {e_1}{e_3} = {e_2}.\]


 The set ${\cal H}= H(\mathbb{Z}) \cup H(\mathbb{Z} +
 \frac{1}{2})$ is the set of Hurwitz integers. A Hurwitz prime is an
 element $\pi  = {a_0} + {a_1}{e_1} + {a_2}{e_2} + {a_3}{e_3} \in
 \mathcal{H}$ such that $p = \pi {\pi ^\star} = ({a_0} + {a_1}{e_1} + {a_2}{e_2} + {a_3}{e_3})({a_0} - {a_1}{e_1} - {a_2}{e_2} - {a_3}{e_3}) = a_0^2 + a_1^2 + a_2^2 +
 a_3^2$ is an odd prime in $\mathbb{Z}$. The norm of $\pi$ is
 defined by $N(\pi)=\pi \pi^\star$.



 The elements in the right ideal
\[\left\langle \pi  \right\rangle  = \left\{ {\pi \delta
|\,\,\delta  \in \mathcal{H}} \right\}\] define a normal subgroup
of the additive group of the ring $\mathcal{H}$. The set of cosets
to $\left\langle \pi  \right\rangle $ in $\mathcal{H}$ defines an
Abelian group   denoted as $\mathcal{H}_\pi = {{\cal H}
\mathord{\left/
 {\vphantom {{\cal H} {\left\langle \pi  \right\rangle }}} \right.
 \kern-\nulldelimiterspace} {\left\langle \pi  \right\rangle }}$.

 Let
 \[
 \mathcal{E}=\{\pm1,\pm e_1,\pm e_2,\pm e_3,
\frac{1}{2}(\pm1\pm e_1\pm e_2\pm e_3)\}
\]
and let $\mathcal{E}_\pi$ denotes the family of cosets to
$\left\langle \pi  \right\rangle $ containing the elements of
$\mathcal{E}$. We define the distance between the words
$u=(u_1,u_2,\ldots ,u_n)$ and $v=(v_1,v_2,\ldots ,v_n)$ in
$\mathcal{H}_\pi ^n$, $d(u,v) = 1$, if there exists a $j \in
\left\{ {1,2, \ldots ,n} \right\}$ and an $\epsilon $ in $
\mathcal{E}_\pi$ such that ${u_j} = {v_j} + \epsilon $ and
$u_i=v_i$ for $i\ne j$. A perfect 1-error-correcting Hurwitz
weight code of length $n$ is a subset $C$ of $\mathcal{H}^n$, such
that every element in ${{\mathcal H}^n} \backslash C$ is at
distance one from exactly one word of $C$.

The following theorem was presented in \cite{biz2014}.
\begin{theorem}
If $\pi$ is an odd Hurwitz prime, then the size of
$\mathcal{H}_\pi $ is equal to $N(\pi)^2$.

\end{theorem}
Here note that if the norm of a Hurwitz integer $q$ is an odd integer,
then the element $q$ is called an odd Hurwitz integer.



\section{Notations and some lemmas}

The symbol $\left[  \cdot  \right]$ denotes the rounding to the
closest integer. For example, $\left[ {\frac{6}{{11}}} \right] =
1$. For Hurwitz integers we can define a rounding operation as
follows: \emph{The rounding of Hurwitz integers:}
\[\left[ {{a_0} + {a_1}{e_1} + {a_2}{e_2} + {a_3}{e_3}} \right] =
\left[ {{a_0}} \right] + \left[ {{a_1}} \right]{e_1} + \left[
{{a_2}} \right]{e_2} + \left[ {{a_3}} \right]{e_3}.\]


Also, the symbol $\left[ {\left[  \cdot  \right]} \right]$ denotes
the rounding to the closest number in ${H}(\mathbb{Z}+\frac{1}{2})$.
For example,
\[\left[ {\left[ {{ {6 \over {11}}}} \right]} \right] = \frac{1}{2},\,\,\left[ {
\left[ {{ {{ - 3} \over {11}}}} \right]} \right] =  -
\frac{1}{2},\,\,\left[ {\left[ {{   {{ - 1} \over {11}}}}
\right]} \right] =  - \frac{1}{2},\,\,\,\left[ {\left[
{{ {{12} \over {11}}}} \right]} \right] = \frac{3}{2}.\]

In case $\left[ {\left[ a \right]} \right]$, $a\in \mathbb{Z}$,
one can choose any   of the closest numbers, for example,
$\left[ {\left[ \frac{0}{11} \right]} \right]=\frac{1}{2}$,
$\left[ {\left[ \frac{0}{11} \right]} \right]=-\frac{1}{2}$.





\begin{proposition} Let $\pi$ be a prime in $H(\mathbb{Z})$ and
let $q$ be an element in Hurwitz integers $\mathcal{H}$. Then
there exist $\gamma  \in H(\mathbb{Z})$ and $\delta _1  \in
\mathcal{H}$ such that \[q = \pi \gamma  + {\delta
_1}\;and\;N({\delta _1}) < N(\pi ).\]

\end{proposition}


\begin{proof} Let $ \widetilde q =\pi ^\star
q=s_0+s_1e_1+s_2e_2+s_3e_3$. Then there exists $\gamma \in
H(\mathbb{Z})$, such that \[\begin{array}{c} \gamma  = \left[
{\frac{{{s_0}}}{{N(\pi )}}} \right] + \left[
{\frac{{{s_1}}}{{N(\pi )}}} \right]{e_1} + \left[
{\frac{{{s_2}}}{{N(\pi )}}} \right]{e_2} + \left[
{\frac{{{s_3}}}{{N(\pi )}}} \right]{e_3} = {r_0} + {r_1}{e_1} +
{r_2}{e_2} + {r_3}{e_3}
\end{array}\]
and $N(\widetilde q -\gamma N(\pi ))<N(\pi) ^2$. Indeed,
\[N(\widetilde q -\gamma
N(\pi))=t_0^2+t_1^2+t_2^2+t_3^2<4 (\frac{N(\pi
)}{2})^2=N(\pi)^2,\] where $t_i=s_i-N(\pi)r_i$ such that $\left|
{{t_i}} \right| < \frac{N(\pi )}{2}$. Hence we obtain that
\[N(\pi)N(\pi ^{\star})\!=\!N(\pi)^2\!>N(\widetilde q -  N(\pi)
\gamma)\!=\!N(\pi ^\star q- \pi ^\star \pi  \gamma)\!=\!N(\pi)N(q- \pi
\gamma)\!=\!N(\pi)N(\delta_1).\] This completes the proof.

\end{proof}





\begin{proposition} Let $\pi$ be a prime in $H(\mathbb{Z})$ and
let $q$ be an element in Hurwitz integers $\mathcal{H}$. Then
there exist $\gamma  \in H(\mathbb{Z}+\frac{1}{2})$ and $\delta
_2  \in \mathcal{H}$ such that
\[q=\pi \gamma +\delta _2 \ \rm{{and}}\ N(\delta _2)<N(\pi).\]
\end{proposition}


\begin{proof} Let $ \widetilde q =\pi ^\star
q=s_0+s_1e_1+s_2e_2+s_3e_3$. Then there exists $\gamma \in
H(\mathbb{Z}+\frac{1}{2})$, such that \[\gamma  = \left[ {\left[
{\frac{{{s_0}}}{{N(\pi )}}} \right]} \right] + \left[ {\left[
{\frac{{{s_1}}}{{N(\pi )}}} \right]} \right]{e_1} + \left[ {\left[
{\frac{{{s_2}}}{{N(\pi )}}} \right]} \right]{e_2} + \left[ {\left[
{\frac{{{s_3}}}{{N(\pi )}}} \right]} \right]{e_3} = {r_0} +
{r_1}{e_1} + {r_2}{e_2} + {r_3}{e_3}\]
and $N(\widetilde q -\gamma N(\pi ))<N(\pi) ^2$. The remainder of
the proof is similar to the proof of Proposition 1.

\end{proof}

We now define some sets as follows \[{A_i} = \left\{ {q \in
H\left( \mathbb Z \right):N\left( q \right) = i} \right\}.\]

Let $\pi$ be a prime in $H(\mathbb{Z})$ and let $q \in A_i$. From
now on, $w$ will denote $\frac{1}{2}(1+e_1+e_2+e_3)$. Using
Proposition 1 and Proposition 2, we get

\[\begin{array}{*{20}{c}}
{q = \pi {\gamma _1} + {\delta _1}},\ \ \mathrm{with} \ \  N(\delta _1)<N(\pi) \\
\end{array}\]
and
\[\begin{array}{*{20}{c}}
{q = \pi {\gamma _2} + {\delta _2}}, \ \ \mathrm{with} \ \  N(\delta _2)<N(\pi),\\
\end{array}\] respectively. Under the above conditions, we define

\[{A_{{i_1}}} = \left\{ q \in {A_i}:\,\,
\ N({\delta _1}) \ne N(\delta _2)\,\  \mathrm{and} \ \ N({\delta
_1}),N({\delta _2}) \ge i \right\}\]
and
\[{A_{{i_2}}} = \left\{ q \in {A_i}:\,\,  N({\delta _1}) = N(\delta _2)=i \right\}.\]
It is obvious that \[A_i=A_{{i_1}}\cup A_{{i_2}} \ \ \mbox{and} \ \
{A_{{i_1}}} \cap {A_{{i_2}}} = \emptyset .\]

Note that for any $q\in A_i$, if $N({\delta _1})< i$ or $N({\delta
_2})<i$, then we obtain that $A_{i_{1}}=A_{i_{2}}=
\emptyset$.


\begin{proposition} For all $i$, $\left| {{A_i}} \right| = 24k,$ where $
k \in \mathbb{Z}$.



\end{proposition}

\begin{proof} It is clear that $A_1=\mathcal{E}$ and $\left| \mathcal{E} \right| =
24$. Let $q_1\in A_s$, $1\le s \le i$; then $N(q_1)=s$. If $\mu
\in \mathcal{E}$, then $q_1 \mu \in A_s$ since $N(q_1 \mu
)=N(q_1)N(\mu )=s$. Recall that the norm $N$ is a multiplicative
norm. Hence, we get $\left| q_1\mathcal{E} \right| = 24.$ Let
$q_2\notin q_1\mathcal{E},  q_2 \in A_s$. Then $\left|
q_2\mathcal{E} \right| = 24.$ In this way, let us assume that the
number of distinct elements in $A_s$ is $k_1$. Hence, we obtain
\[{A_s} = {q_1}{\cal E} \cup {q_2}{\cal E} \cup  \cdots  \cup
{q_{k_1}}\mathcal{E}\  \text{and, therefore}\ \left| A_s \right| =
24{k_1}.
\] This completes the proof.

\end{proof}





\section{Construction of necessary partitions}




Under the above conditions, we now give a partition of
$\mathcal{H}_\pi$.





\begin{definition} Let $\pi$ be a prime in $\mathcal{H}$ with
$N(\pi)=p$, where $p>3$ is a prime integer. We define the set
${P_{{i_1}}}$ as

\[
P_{i_1} =A_{i_1}.\]

Let $q\in A_{i_2}$. Let us assume that $q=\pi \gamma_1 + \delta
_1$, $q=\pi \gamma_2 + \delta _2 $, where $\delta _1$ and $\delta
_2$ were defined in Proposition 1 and Proposition 2, respectively.
Then the set $P_{{i_2}}$ is constructed by $\delta _1\in
P_{{i_2}},\delta _2\notin P_{{i_2}}$; in other words,

\[{P_{{i_2}}}\mathop  \subset \limits_ {\ne}  {A_{{i_2}}} = \left\{ q \in {A_{{i_2}}}: \ q=\delta _1, \,\, N(\delta_1)=N(\delta_2)=i
 \right\} - \delta _2\mathcal{E}.\]
Then, the partition of $\mathcal{H}_\pi$ is obtained by

\[{P_i} = {P_{{i_1}}} \cup {P_{{i_2}}}.\]
\end{definition}

If $P_j=\emptyset$, then $P_j$ is not in the partition; that is,
the index $i$ is not ordered. For example, in Example 2, $P_4$ does
not appear in the partition.


\begin{example}  Let $\pi =3+e_1+e_2$. Then
\[\begin{array}{l}
{P_1} =P_{1_{1}}\cup P_{1_{2}}= \mathcal{E}\cup \emptyset =\mathcal{E},\\[1ex]
{P_2} = P_{2_{1}}\cup P_{2_{2}}= (1+e_1)\mathcal{E}\cup \emptyset
=\left( {1 + e_1} \right)\mathcal{E}
\end{array}\] and

\[{P_3} = P_{3_{1}}\cup P_{3_{2}},\]
where $P_{3_{1}}=A_{3_{1}}=(1+e_1+e_2)\mathcal{E}\cup
(1+e_1+e_3)\mathcal{E}$,


\[\begin{array}{rcl}
{P_{{3_2}}} &=& {A_{{3_2}}} - \left( {\dfrac{1}{2}\left( {-1 + {e_1} - {e_2} - 3{e_3}} \right)} \right)\mathcal{E}\\
& =& \left\{ \begin{array}{l}
\begin{array}{*{20}{l}}
{ \pm \left( {1 + {e_2} + {e_3}} \right), \pm \left( {1 + {e_1} - {e_2}} \right), \pm \left( {1 - {e_1} - {e_3}} \right), \pm \left( {{e_1} + {e_2} - {e_3}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} - \dfrac{{{e_1}}}{2} - \dfrac{{3{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} - \dfrac{{3{e_1}}}{2} + \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} + \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2} + \dfrac{{3{e_3}}}{2}} \right), \pm \left( {\dfrac{3}{2} + \dfrac{{{e_1}}}{2} + \dfrac{{{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{3}{2} - \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} - \dfrac{{{e_1}}}{2} + \dfrac{{3{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} + \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2}
- \dfrac{{3{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} +
\dfrac{{3{e_1}}}{2} + \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}}
\right),}
\end{array}\\[0.5ex]
\begin{array}{*{20}{l}}
{ \pm \left( {{e_1} + {e_2} + {e_3}} \right), \pm \left( {1 - {e_2} + {e_3}} \right), \pm \left( {1 + {e_1} - {e_3}} \right), \pm \left( {1 - {e_1} + {e_2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{3}{2} - \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} - \dfrac{{{e_1}}}{2} + \dfrac{{{e_2}}}{2} + \dfrac{{3{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} + \dfrac{{3{e_1}}}{2} - \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} + \dfrac{{{e_1}}}{2} + \dfrac{{3{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{3}{2} + \dfrac{{{e_1}}}{2} + \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} + \dfrac{{{e_1}}}{2} - \dfrac{{3{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} - \dfrac{{{e_1}}}{2} + \dfrac{{{e_2}}}{2}
- \dfrac{{3{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} -
\dfrac{{3{e_1}}}{2} - \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}}
\right),}
\end{array}
\end{array} \right\}\\[0.5ex]
&&-\left( {\dfrac{1}{2}\left( {-1 + {e_1} - {e_2} - 3{e_3}} \right)}
\right)\mathcal{E}\\[1ex]
&=&\left\{ {\begin{array}{*{20}{l}}
{ \pm \left( {1 + {e_2} + {e_3}} \right), \pm \left( {1 + {e_1} - {e_2}} \right), \pm \left( {1 - {e_1} - {e_3}} \right), \pm \left( {{e_1} + {e_2} - {e_3}} \right),}\\
{ \pm \left( {\dfrac{1}{2} - \dfrac{{{e_1}}}{2} - \dfrac{{3{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} - \dfrac{{3{e_1}}}{2} + \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} + \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2} + \dfrac{{3{e_3}}}{2}} \right), \pm \left( {\dfrac{3}{2} + \dfrac{{{e_1}}}{2} + \dfrac{{{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{3}{2} - \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} - \dfrac{{{e_1}}}{2} + \dfrac{{3{e_2}}}{2} - \dfrac{{{e_3}}}{2}} \right),}\\[0.5ex]
{ \pm \left( {\dfrac{1}{2} + \dfrac{{{e_1}}}{2} - \dfrac{{{e_2}}}{2}
- \dfrac{{3{e_3}}}{2}} \right), \pm \left( {\dfrac{1}{2} +
\dfrac{{3{e_1}}}{2} + \dfrac{{{e_2}}}{2} + \dfrac{{{e_3}}}{2}}
\right),}
\end{array}} \right\}\\
&=&{(1+e_2+e_3)} \ \mathcal{E}\cup
(e_1+e_2+e_3)\mathcal{E}-({\dfrac{1}{2}\left( {-1 + {e_1} - {e_2} -
3{e_3}} \right)})\mathcal{E}
\end{array}\]

Hence we obtain the partition of $H_{3+e_1+e_2}$ as $P_1,P_2,P_3$.
The sizes of $P_1,P_2,P_3$ are $24,24,72$, respectively. Also
${{\cal E}_\pi } = \left\{ {{P_1},{P_2},{P_3}} \right\}$.

In order to assist the reader in understanding the method, we give
some specific examples.


\begin{itemize}
    \item Take the element $q=1+e_1+e_2 \in A_3$. Using Proposition 1 and  Proposition 2, we obtain
  \begin{align*}
{\delta _1} &= q\\
{\delta _2} &=  \frac{1}{2} - \frac{1}{2}{e_1} -
\frac{3}{2}{e_2} + \frac{3}{2}{e_3}
\end{align*}
It is obvious that $N(\delta _1 )\ne N(\delta _2)$. So $1+e_1+e_2
\in P_{3_1}$.

    \item Take the element $q=1+e_2+e_3 \in A_3$. Using Proposition 1 and  Proposition 2, we obtain
  \begin{align*}
{\delta _1} &= q\\
{\delta _2} &=  -\frac{1}{2} + \frac{1}{2}{e_1} -
\frac{1}{2}{e_2} - \frac{3}{2}{e_3}
\end{align*} It is obvious that $N(\delta _1 )=N(\delta _2
)$. So $1+e_2+e_3 \in P_{3_2}$. It should also be noted that if an
element $q_1$
$$q_1\in (-\frac{1}{2} + \frac{1}{2}{e_1} - \frac{1}{2}{e_2} -
\frac{3}{2}{e_3})\mathcal{E}=\delta _2 \mathcal{E},$$ then $q_1
\notin P_{3_2}$.
    \item Take the element $q=\frac{1}{2} + \frac{3}{2}{e_1}
    +\frac{1}{2}{e_2} + \frac{1}{2}{e_3} \in A_3$. Using Proposition 1 and
Proposition 2, we obtain
   \begin{align*}
{\delta _1} &= q\\
{\delta _2}& =  -{e_1} -{e_2} -{e_3}
\end{align*} It is obvious that $N(\delta _1 )=N(\delta _2
)$. So $\frac{1}{2} + \frac{3}{2}{e_1}
    +\frac{1}{2}{e_2} + \frac{1}{2}{e_3}\in P_{3_2}$.
\end{itemize}


\end{example}



\begin{example}  Let $\pi =2+2e_1+2e_2+e_3$. Then
 \begin{align*}
{P_1} &=P_{1_{1}}\cup P_{1_{2}}= \mathcal{E}\cup \emptyset =\mathcal{E},\\
{P_2}& = P_{2_{1}}\cup P_{2_{2}}= (1+e_1)\mathcal{E}\cup \emptyset
=\left( {1 + e_1} \right)\mathcal{E},\\
 {P_{3_1}}& = \left( {1 + {e_1} + {e_2}} \right)\mathcal{E} \cup \left( {1 + {e_1} + {e_3}}
\right)\mathcal{E} \cup \left( {1 + {e_2} + {e_3}}
\right)\mathcal{E} \cup \left( {{e_1} + {e_2} + {e_3}}
\right)\mathcal{E}, \\
P_{3_2}&=\emptyset,\\
{P_3}& = P_{3_{1}}\cup P_{3_{2}},
\end{align*}
$P_4=\emptyset$ since either $N(\delta _1) <4$ or $N(\delta _2)
<4$ for any element $q\in A_4$. For example, let $q=2\in A_4$.
Then we get $\delta _1 =q$, $\delta _2 =-\frac{3}{2} +
\frac{1}{2}{e_1}
    -\frac{1}{2}{e_2} + \frac{1}{2}{e_3}$. It is obvious that
 $N(\delta _2)=3 <4$. So $2\notin P_4$.

\noindent $P_5=(2+e_1)\mathcal{E}$.





Note that if $q\in A_5, q\notin P_5$,  then we obtain that
either $N(\delta _1) <5$ or $N(\delta _2) <5$. For example, let
$q=2+e_2$. Then we get $\delta _1 =q=2+e_2$, $\delta _2
=-\frac{3}{2} + \frac{1}{2}{e_1}
    +\frac{1}{2}{e_2} + \frac{1}{2}{e_3}$. It is obvious that
 $N(\delta _2)=3 <5$. So $2+e_2\notin P_5$.

\bigskip

 Hence we obtain the partition of
$H_{2+2e_1+2e_2+e_3}$ as $P_1,P_2,P_3,P_5$. The sizes of
$P_1,P_2,P_3,P_5$ are $24,24,96,24$, respectively. Also
${{\mathcal E}_\pi } = \left\{ {{P_1},{P_2},{P_3},{P_5}}
\right\}$.


\end{example}

Observe that $P_1=\mathcal{E}$, $P_2=(1+e_1)\mathcal{E}, P_3$
appear in all partitions. $P_3$ are not fixed. $P_4$ did not
appear in all partitions.



Proof of the following theorem is straightforward from the
partition of $\mathcal{H}_\pi$.


\begin{theorem} Assume that $\pi$ is a prime in $H(\mathbb{Z})$
with $N(\pi)> 3$. Let the partition of $\mathcal{H}_\pi$ be
$P_{i_1}=\mathcal{E},P_{i_2}=(1+e_1)\mathcal{E},P_{i_3},\ldots ,
P_{i_n}$. Let $g_1=1\in P_{i_1}=\mathcal{E},g_2=(1+e_1)\in
P_{i_2}=(1+e_1)\mathcal{E},g_3\in P_{i_3},\ldots ,g_n\in P_{i_n}
$. Then the null-space $C$ of the matrix

\[H = \left( {\begin{array}{*{20}{c}} {1,}&{1 +
{e_1},}&{{g_3},}&{{g_4},}&{ \cdots ,}&{{g_n}}
\end{array}} \right)\]
is a perfect 1-error-correcting Hurwitz weight code with the
length $n = \left| {{\mathcal{E}_\pi }} \right|$ in
$\mathcal{H}_\pi^n$, in the metric defined by $\mathcal{E}\pi $.


\end{theorem}


\begin{example} Let $\pi =2+2e_1+2e_2+e_3$. The null-space of the
following matrix $H$ \[H = \left( {\begin{array}{*{20}{c}} {1,}&{1
+ {e_1},}&{1 + {e_1} + {e_2},}&{2 + {e_1}}
\end{array}} \right)\]
is a perfect 1-error-correcting Hurwitz weight code with the
length $n = \left| {{\mathcal{E}_\pi }} \right|=4$. After a
transmission, let the received vector $r = \left(
{\begin{array}{*{20}{c}} { - 1 - {e_1},}&{1 + {e_1},}&{1 +
{e_2},}&1
\end{array}} \right)$. Then the syndrome of $r$ is ž
\begin{align*}
S &= H{r^{tr}} = \left( {\begin{array}{*{20}{c}} {1,}&{1 +
{e_1},}&{1 + {e_1} + {e_2},}&{2 + {e_1}}
\end{array}} \right){\left( {\begin{array}{*{20}{c}}
{ - 1 - {e_1},}&{1 + {e_1},}&{1 + {e_2},}&1
\end{array}} \right)^{tr}}\\
 &= {e_1}(1 + {e_1}).
\end{align*}

This shows that the error has appeared in the second coordinate
position since  $1 + {e_1} \in P_2$. The value of the error is
$e_1$. Hence we obtain the corrected vector as
\[\begin{array}{l} c = r - e = \left( {\begin{array}{*{20}{c}} { -
1 - {e_1},}&{1 + {e_1},}&{1 + {e_2},}&1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{0,}&{{e_1},}&{0,}&0
\end{array}} \right)
 = \left( {\begin{array}{*{20}{c}}
{ - 1 - {e_1},}&{1,}&{1 + {e_2},}&1
\end{array}} \right).
\end{array}\]

\end{example}

\begin{theorem} Assume that $\pi$ is a prime in $H(\mathbb{Z})$
with $N(\pi)> 3$. Let the partition of $\mathcal{H}_\pi$ be
\[\begin{array}{rcl}
{P_{{i_1}}} &=& {\cal E},\\[0.5ex]
{P_{{i_2}}} &=& (1 + {e_1}){\cal E},\\[0.5ex]
{P_{{i_3}}}&=& {d_{11}}{\cal E} \cup {d_{12}}{\cal E} \cup , \ldots , \cup {d_{1{n_1}}}{\cal E},\\[0.5ex]
{P_{{i_4}}} &=& {d_{21}}{\cal E} \cup {d_{22}}{\cal E} \cup , \ldots , \cup {d_{2{n_2}}}{\cal E},\\[0.5ex]
 &\vdots& \\[0.5ex]
{P_{{i_m}}} &=& {d_{m1}}{\cal E} \cup {d_{m2}}{\cal E} \cup , \ldots
, \cup {d_{m{n_m}}}{\cal E}.
\end{array}\]
Then the null-space $C$ of the matrix
\[H = \left( {1,\,1 + {e_1},\,{d_{11}},\, \ldots ,{d_{1{n_1}}},{d_{21}}, \ldots {d_{2{n_2}}}, \ldots ,{d_{m1}}, \ldots {d_{m{n_m}}}} \right)\]
is a perfect 1-error-correcting Hurwitz weight code with the
length $n = \frac{{{N(\pi)^2} - 1}}{{24}}$ in $\mathcal{H}_\pi^n$,
in a metric that it is defined as follows:
\begin{itemize}
\item[]
the distance between the words $u=(u_1,u_2,\ldots ,u_n)$ and
$v=(v_1,v_2,\ldots ,v_n)$ in $\mathcal{H}_\pi ^n$, $d(u,v) = 1$,
if there exist a $j \in \left\{ {1,2, \ldots ,n} \right\}$ and an
$\epsilon \in d_{rs} \mathcal{E}$ such that ${u_j} = {v_j} +
\epsilon $ and $u_i=v_i$ for $i\ne j$.
\end{itemize}A perfect
1-error-correcting Hurwitz weight code of length $n$ is a subset
$C$ of $\mathcal{H}^n$ such that every element in ${{\mathcal
H}^n} \backslash C$ is at distance one from exactly one word of
$C$.

\end{theorem}








\begin{example} Let $\pi =2+2e_1+2e_2+e_3$.
Then the partition given in Example 2 is
\[
\begin{array}{*{20}{rcl}}
{P_1}& =& {\cal E},\\[0.5ex]
{P_2} & =& \left( {1 + {e_1}} \right){\cal E},\\[0.5ex]
{P_3}& =& \left( {1 + {e_1} + {e_2}} \right){\cal E} \cup \left( {1
+ {e_1} + {e_3}} \right){\cal E} \cup \left( {1 + {e_2} + {e_3}}
\right){\cal E} \cup \left( {{e_1} + {e_2} + {e_3}} \right){\cal
E},\\[0.5ex]
{P_4}& =& \emptyset ,\\[0.5ex]
{P_5} & =& \left( {2 + {e_1}} \right){\cal E},
\end{array}\]
that is, $d_{11}=1 + {e_1} + {e_2},d_{12}=1 + {e_1} + {e_3},d_{13}=1 + {e_2} + {e_3},d_{14}=e_1 + {e_1} + {e_3},$ $d_{31}=2+ {e_1}.$
The null-space of the following matrix $H$
\[H = \left( {\begin{array}{*{20}{c}}
{1,}&{1 + {e_1},}&{1 + {e_1} + {e_2},}&{1 + {e_1} + {e_3},}&{1 +
{e_2} + {e_3},}&{{e_1} + {e_2} + {e_3},}&{2 + {e_1}}
\end{array}} \right)\]
is a perfect 1-error-correcting Hurwitz weight code with the
length $n = \frac{{N\left( {{\pi ^2}} \right) - 1}}{{24}} $ $=
\frac{{169 - 1}}{{24}} = 7$.

The perfect 1-error-correcting code $C$ is the null-space in
$\mathcal{H}_\pi ^7$ of $H$, the number of codewords of $C$ is
$|C| = 169^6$. The number of errors for which the error is at
distance one from exactly one word of $C$ is $7\cdot24\cdot
169^6$. Considering the well-known sphere packing bound, the code $C$
is perfect since it satisfies the bound, that is, \[\left|
{{H_{2+2e_1+2e_2+e_3} }} \right |^7 = {169^7} = 7 \cdot 24 \cdot
{169^6} + {169^6}.\]


\end{example}








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\end{document}