Covering radius of repetition codes over \(F_{2}+vF_{2}+v^2F_2\) with \(v^3=1\)

1. Introduction

For more than a decade, codes over finite rings have gotten much attention of researchers due the definition of Gray map [1, 2, 3]. Particularly, codes over the ring \(F_{2}+vF_{2}+v^2F_2\) have been extensively studied [4, 5].

The covering radius is very interesting topic in coding theory. First time, the covering radius of binary linear codes were studied by Helleseth et al., [6]. Furthermore, covering radius of linear codes over \(F_2+uF_2\) with \(u^2=0\) was determined by using Lee distance, Chinese Euclidean distance, and Bachoc distance [7, 8, 9].

Recently, the covering radius of codes over \(Z_4\) have been determined by using lee weight and Chinese Euclidean lee weight [10, 11]. In [11], Manoj et al., introduced new reduction and torsion codes, repetition codes for octonary codes and determined its covering radius. Chatouh et al., [12], Pandian et al., [13] gave the upper and lower bounds on the covering radius of some classes of codes over rings \(Z_2\times Z_4\), \(Z_2\times R\) with \(R=F2+vF2, v^2=v\) respectively. Panchanathan et al., in [14] studied bounds on covering radius for various repetition codes with respect to different and similar length over \(F_2+uF_2+u^2F_2 \) with \(u^3=0\) using Lee weight and generalized Lee weight.

The goal of this paper is to investigate the covering radius of repetition codes over the finite ring \(F_2+vF_2+v^2F_2\) with \(v^3=1\).

2. Preliminaries

2.1. The overview of the ring \(M=F_2+uF_2+u^2F_2\) with \(u^3=1\)

Some basic information about the finite ring \(M\) have been recalled as follows:
Let \(F_2=\{0, 1\}\) be the binary Galois field of two element. Further, let \(M\) is commutative ring with characteristic 2 and 8 elements which is given by \(M=\{0, 1, v, v^2, 1+v, 1+v^2, v+v^2, 1+v+v^2 | v^3=1 \}\) [4]. The elements \(1,v,v^2\) are units of \(M\) and \(\{0, 1+v, 1+v^2, v+v^2, 1+v+v^2 \}\) is the set of Zero-divisors of \(M\).
The \(M\)-submodule \(C\) of \(M^{n}\) is called a linear code having length \(n\) over \(M\), and its elements are known as codewords. The Hamming weight \(w_{H}(c)\) of a non-zero codeword \(c=(c_1, c_2, \cdots, c_n)\) of \(C\) is the number of non-zero coordinates of element \(c\), and the Hamming distance between two codewords \(x\) and \(y\) \(\in M^n\) which is denoted by \(d_H(x,y)\) is equal to \(w_H(x-y)\).
For any \(m=a+bv+cv^{2} \in\) \(M\), the Gray map \(g\) from \(M\) to \(F_{2}^{3}\) is defined as \(g(m)=(a+b+c, a+b, a+c)\). The Lee weight \(w_{L}\) for an element \(m \in\) \(M\) is given by \(w_{L}(m)=w_{H}(g(m))\). Now, the Lee weight of the element of \(M\) are given as:
\(\begin{cases} w_{L}(0)=0, w_{L}(1)=3,\\ w_{L}(v)=w_{L}(v^2)=w_{L}(v+v^2)=2,\;\; \text{and}\\ w_{L}(1+v)=w_{L}(1+v^2)=w_{L}(1+v+v^2)=1.\end{cases}\)
Lee weight of a codeword \(c=(c_1, c_2, \cdots, c_n)\) of a linear code \(C\) is defined as \(w_{L}(c)=\sum_{i=1}^{n}w_{L}(c_i)\), and Lee distance between two codewords \(m_1,m_2\) of \(C\), (with \((m_1,m_2)\neq 0\)) is denoted by \(d_L(m_1, m_2)\) and defined as \(d_L(m_1, m_2)=w_L(m_1-m_2)\).

2.2. Covering radius of linear codes

Definition 1. Let \(C\) be a linear code over \(M\) of length \(n\). The covering radius of \(C\) with respect distance \(d\), where \(d \in \{d_H, d_L\},\) is defined as: $$r_d(C)= max\{d(x,C) | x\in M^n\}.$$

Proposition 1. [15] Let \(C\) be a code generated by \(G\) over \(M\), \(C^{'}\) be a code generated by \(G^{'}\) over \(M\) and \(C^{''}\) be a code generated by \(\begin{pmatrix} 0 & G^{'} \\ G^{''} & A \end{pmatrix}\), then \(r_d(C^{''})\leqslant r_d(C)+r_d(C^{''}).\) Moreover, if \(B\) be the concatenation of \(C\) and \(C^{'}\) then \(r_d(B)\geqslant r_d(C)+r_d(C^{''}).\)

3. Main results

3.1. Covering radius of repetition codes

In this section we represents the covering radius of repetition code over \(F_q\), as well as over \(M\).
Let \(F_{q}\) be a finite field of \(q\) element. Let \(C\) be a repetition code over \(F_q\) of length \(n\) defined by \(C=\left\{(a,a,\cdots,a) | a \in F_{q}\right\}\) with \(G\) its generator matrix \(G\)= \(\begin{pmatrix} a & a & \cdots & a \end{pmatrix}\). Hence \(C\) is \([n,1,n]\)-code over \(M\), and according to [16], the covering radius of \(C\) is equal to \(\left\lfloor\dfrac{n(q-1)}{q} \right\rfloor\), and the covering radius of \([n(q-1), 1, n(q-1)]\)-codes over \(F_q\) with generator matrix \(\begin{pmatrix} 1, 1,\cdots,1 \arrowvert & a_2, a_2,\cdots,a_2\arrowvert& \cdots \arrowvert& a_{q-1}, a_{q-1},\cdots, a_{q-1} \end{pmatrix}\) is equal to \( \left\lfloor\dfrac{n(q-1)^2}{q}\right\rfloor\).

Theorem 1. Let \(C_{1}\) be unit repetition code of generator matrix \(G_{1}=\begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix}\) and \(C_{2}\) be zero-divisor repetition code with generator matrix \(G_{2}=\begin{pmatrix} 1+v & 1+v & \cdots & 1+v \end{pmatrix}\) and \(C_{3}\) be zero-divisor repetition code with generator matrix \(G_{3}= \begin{pmatrix} 1+v+v^2 & 1+v+v^2 & \cdots & 1+v+v^2 \end{pmatrix}\). Then we have

• i) \(C_{1}\) is \([n, 8, d_{H}=n,d_{L}=3n]\)-code over \(M\) and \(r_{d_L}(C_1)= \frac{3n}{2}\),
• ii) \(C_{2}\) is \([n, 4, d_{H}=n,d_{L}=n]\)-code over \(M\) and \(4\left\lfloor \dfrac{n}{4} \right\rfloor \leq r_{d_L}(C_2)\leq 2n\),
• iii) \(C_3\) is \(\left[n, 2, d_{H}=n,d_{L}=n\right]\)-code over \(M\) and \(\lfloor \frac{n}{2}\rfloor \leq r_{d_L}(C_3) \leq \frac{5n}{2}\).

Proof.

• i)Since \(C_1=\{ a.(1,1,\cdots,1) | a \in M \}\) is linear code of length \(n\) and \(8\) elements, \(d_{H}(C_1)= min\{w_H(c)| c \in C, c\neq (0,0,\cdots,0) \}=n\) and \(d_L(C_1)=nw_L(1)=3n\). So \(C_1\) is \([n,\lvert C_1\arrowvert =8, n, 3n ]\)-code over \(M\). Now, we will prove that \(r_{d_L}(C_1)=\frac{3n}{2}\). Let \(x \in M^n\), and
  \(\begin{cases}w_0\; \text{be the number of coordinate has}\;0\; \text{component},\\ w_1\; \text{the number of coordinate has }\;1\; \text{component,} \\ w_2\;\text{ the number of coordinate has}\;v\;\text{ component,}\\ w_3\; \text{is the number of coordinate has}\; v^2\;\text{ component,}\\ w_4\;\text{the number of coordinate has}\;1+v\;\text{component,}\\ w_5\;\text{the number of coordinate has}\;1+v^2\;\text{component,}\\ w_6\;\text{is the number of coordinate has}\;v+v^2\;\text{component, and}\\ w_7\;\text{be the number of coordinate has }\;1+v+v^2\;\text{component.} \end{cases}\) We also have \begin{align*}\sum_{i=0}^{7}w_i=n\end{align*} and \(\begin{cases}w_L(x)=w_0w_L(0) + w_1w_L(1)+ w_2w_L(v) +w_3w_L(v^2)+ w_4w_L(1+v)+w_5w_L(1+v^2)+ w_6w_L(v+v^2)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;+ w_7w_L(1+v+v^2)= (w_4+ w_5+ w_7)+ 2(w_2+ w_3+ w_6)+ 3w_1.\\ d_L(x,00...0)=(w_4+ w_5+ w_7)+ 2(w_2+ w_3+ w_6)+ 3w_1=n-w_0+2w_1+ w_2+ w_3+ w_6,\\ d_L(x,11...1)=w_L(x-11...1)= n+2w_0-w_1+w_4+w_5+w_7.\end{cases}\)
  And by the same way, we have \(\begin{cases} d_L(x,v v...v)= n+w_0-w_2+w_3+2w_4+w_6,\\ d_L(x,v^2 v^2...v^2)= n+w_0+w_2-w_3+2w_5+w_6,\\ d_L(x,1+v 1+v...1+v)= n+w_1+2w_2-w_4+w_5+w_7,\\ d_L(x,1+v^2 1+v^2...1+v^2)= n+w_1+2w_3++w_4-w_5+w_7,\\ d_L(x,v+v^2 v+v^2...v+v^2)= n+w_0+w_2+w_3-w_6+2w_7,\\ d_L(x,1+v+v^2 1+v+v^2...1+v+v^2)= n+w_1+w_4 +w_5+2w_6-w_7.\end{cases}\) So, $$d_L(x,C_1)=min\{d_L(x,y)| y \in C_1 \}\leqslant \dfrac{1}{8}(8n+ 4n)=\dfrac{12n}{8}=\dfrac{3n}{2}$$ implies $$d_L(x,C_1)\leq \dfrac{3n}{2}\;\;\forall x \in M^n,$$ hence $$r_{d_L}(C_1) \leq \dfrac{3n}{2}.$$ To prove the reverse inequality, take \( x=\left[\overset{m}{ \widehat{0 0\cdots 0}}| \overset{m}{ \widehat{1 1\cdots 1}}| \overset{m}{\widehat{v v \cdots v}} |\overset{m}{\widehat{v^2 v^2 \cdots v^2}} |\overset{m}{ \widehat{1+v 1+v \cdots 1+v}} |\overset{m}{\widehat{1+v^2 1+v^2 \cdots 1+v^2}} \right.\)
  \(\left.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;|\overset{m}{\widehat{v+v^2 v+v^2...v+v^2}} |\overset{mn-7m}{\widehat{1+v+v^2 1+v+v^2 \cdots 1+v+v^2}}\right]\in M^n \), with \( m=\lfloor\frac{n}{8}\rfloor .\)
  So, we have \(\begin{cases}d_L(x,00\cdots 0)=2n-4m,\\ d_L(x,1 1 \cdots 1)=n+4m,\\ d_L(x,v v \cdots v)=n+4m ,\\ d_L(x,v^2 v^2\cdots v^2)=n+4m,\\ d_L(x,v+1 v+1 \cdots v+1)= 2n-4m,\\ d_L(x, v^2+1 v^2+1 \cdots v^2+1)=2n-4m,\\ d_L(x,v+v^2 v+v^2\cdots v+v^2)=3n-12m,\\ d_L(x,1+v+v^2 1+v+v^2\cdots 1+v+v^2)=n+4m,\\ d_L(x,C_1)=min\{n+4m, 2n-4m, 3n-12m\}= n+4m.\end{cases}\) Then \(r_{d_L}(C_1) \geq n+4m\geq n+\frac{n}{2}=\frac{3n}{2}\). Hence, \(r_{d_L}(C_1)=\frac{3n}{2}\).
• ii) Since \(C_2=\{ b.(1+v,1+v,\cdots,1+v) | b \in M \}=\{(0,0,\cdots,0), (1+v,1+v,\cdots,1+v), (1+v^2,1+v^2,\cdots,1+v^2), (v+v^2,v+v^2,\cdots,v+v^2)\}\), \(d_H(C_2)=n\) and \(d_L(C_2)=n\). So \(C_2\) is \([n,\lvert C_2\arrowvert =4,n,n]\)-code over \(M\). Now, we prove \(4\left\lfloor \dfrac{n}{4} \right\rfloor \leq r_{d_L}(C_2)\leq 2n\). To prove \(r_{d_L}(C_2)\leq 2n\), suppose \(x \in M^n\) then \(\begin{cases}d_L(x,(0,0,\cdots,0))=n-w_0+2w_1+ w_2+ w_3+ w_6,\\ d_L(x,(1+v,1+v,\cdots,1+v))=n+w_1+2w_2-w_4+w_5+w_7,\\ d_L(x,(1+v^2,1+v^2,\cdots,1+v^2))= n+w_1+2w_3+w_4-w_5+ w_7,\;\text{ and}\\ d_L(x,(v+v^2,v+v^2,\cdots,v+v^2))=n+w_0+w_2+w_3-w_6+2w_7.\end{cases}\) Now, since \(w_1+w_2+w_3+w_7 \leq n\), so \(d_L(x,C_2)=min\{n-w_0+2w_1+ w_2+ w_3+ w_6, n+w_1+2w_2-w_4+w_5+w_7\), \( n+w_1+2w_3+w_4-w_5+ w_7 \), \( n+w_0+w_2+w_3-w_6+2w_7 \}\leq \frac{1}{4}( 4n +4(w_1+w_2+w_3+w_7))\leq 2n\). To prove \(4\left\lfloor \dfrac{n}{4} \right\rfloor \leq r_{d_L}(C_2)\), suppose $$ x=\left[\overset {\lfloor \frac{n}{4}\rfloor }{\widehat{0 0\cdots0}}|\overset{\lfloor \frac{n}{4}\rfloor}{\widehat{1+v 1+v\cdots1+v}} | \overset{\lfloor \frac{n}{4}\rfloor }{ \widehat{1+v^2 1+v^2\cdots1+v^2}}|\overset{n-3\lfloor \frac{n}{4}\rfloor}{\widehat{v+v^2 v+v^2\cdots v+v^2}}\right]\in M^n,$$ then \(\begin{cases}d_L(x,(0,0,\cdots,0))= 2n-4\lfloor \frac{n}{4}\rfloor,\\ d_L(x,(1+v,1+v,\cdots,1+v))=d_L(x,(1+v^2,1+v^2,\cdots,1+v^2))=n,\; \text{and}\\ d_L(x,(v+v^2,v+v^2,\cdots,v+v^2))=4\lfloor \frac{n}{4}\rfloor.\end{cases}\) Then \(d_L(x,C_2)=min\{ n, 2n-4\lfloor \frac{n}{4}\rfloor, 4\lfloor \frac{n}{4}\rfloor \}= 4\lfloor \frac{n}{4}\rfloor\) implies \(r_{d_L}(C_2) \geq 4\lfloor \frac{n}{4}\rfloor \). Hence we achieved our desired result.
• iii) Since \(C_3\) is linear code with generator matrix \(G_{3}= \begin{pmatrix} 1+v+v^2 & 1+v+v^2 & \cdots & 1+v+v^2 \end{pmatrix}\), then \(C_3=\{ c.(1+v+v^2,1+v+v^2,\cdots,1+v+v^2) / c \in M \}=\{(0,0,\cdots,0), (1+v+v^2,1+v+v^2,\cdots,1+v+v^2)\} \), \(d_H(C_3)=n \), and \(d_L(C_3)= n.w_L(1+v+v^2)=n\). Hence \(C_3\) is \([n, 2, d_H=n, d_L=n] \). Now we prove that \(\lfloor \frac{n}{2}\rfloor \leq r_{d_L}(C_3) \leq \frac{5n}{2}\). To prove \(r_{d_L}(C_3) \leq \frac{5n}{2}\), suppose \(x \in M^n \) then \(\begin{cases}d_L(x,(0,0,\cdots,0))= n-w_0+2w_1+ w_2+ w_3+ w_6,\; \text{and}\\ d_L(x,(1+v+v^2, 1+v+v^2, \cdots,1+v+v^2))=n+w_1+w_4 +w_5+2w_6-w_7.\end{cases}\) Now, because \(w_0+w_1+w_6+w_7 \leq n\), so \(d_L(x,C_3) \leq \frac{1}{2}(3n-2w_0+2w_1+2w_6-2w_7)\) implies \(d_L(x,C_3) \leq \dfrac{5n}{2}\) \(\forall x \in M^n\). Hence \(r_{d_L(C_3)} \leq \dfrac{5n}{2}\). It remains to prove \(\lfloor \frac{n}{2}\rfloor \leq r_{d_L}(C_3)\). For this suppose \(x=[\overset{\lfloor \frac{n}{2}\rfloor}{\widehat{0 0\cdots0}} |\overset{\lfloor \frac{n}{2}\rfloor}{\widehat{1+v+v^2 1+v+v^2 \cdots 1+v+v^2}}]\). Then \(\begin{cases}d_L(x,(0,0,\cdots,0))= \lfloor \frac{n}{2}\rfloor\; \text{and}\\ d_L(x,(1+v+v^2, 1+v+v^2, \cdots,1+v+v^2))= \lfloor \frac{n}{2}\rfloor. \end{cases}\) So \(d_L(x,C_3)= \lfloor \frac{n}{2}\rfloor \) and by the definition of covering radius of code we have \(r_{d_L(C_3)} \geq \lfloor \frac{n}{2}\rfloor \).

3.2. Covering radius of block repetition code over \(M\)

Theorem 2. Let \(Rp_{U_s}^{3n}\) be a block repetition code of length \(3n\) with generator matrix \(G_{U_s} =\begin{pmatrix} \overset{n}{\widehat{1 1 \cdots 1}} & \overset{n }{\widehat{v v \cdots v}} & \overset{n }{\widehat{v^2 v^2 \cdots v^2 }} \end{pmatrix}\) and \(Rp_{U_d}^{n_1+n_2+n_3}\) be a block repetition code of length \(n_1+n_2+n_3\) with generator matrix \( G_{U_d}= \begin{pmatrix} V{n_1 }{\widehat{1 1 \cdots 1}} & \overset{n_2 }{ \widehat{v v \cdots v}} & \overset{n_3}{\widehat{v^2 v^2 \cdots v^2 }} \end{pmatrix}\). Then

• i) the code \(Rp_{U_s}^{3n}\) is \([3n,8,d_H=3n,d_L=3n]\)-code over \(M\) with \(r_{d_L}(Rp_{U_s}^{3n})=\frac{9n}{2}\), and
• ii) the code \(Rp_{U_d}^{n_1+n_2+n_3}\) is \([n_1+n_2+n_3,8,d_H=n_1+n_2+n_3, d_L=n_1+n_2+n_3]\)-code over \(M\) with \(r_{d_L}(Rp_{U_d}^{n_1+n_2+n_3})=\frac{3}{2}(n_1+n_2+n_3)\).

Proof.

• i) Let \(x \in M^3n\) such that \(x=(x_1,x_2,x_3)\) with \(x_i, 1\leq i \leq 3\) is vector of \(n\) coordinates with \((s_0,t_0,w_0)\) is the number of coordinate have \(0\) element in \((x_1,x_2,x_3)\) respectively, and \(\begin{cases}(s_1,t_1,w_1)\;\text{ is}\;1\;\text{ times in}\;\;x,\\ (s_2,t_2,w_2)\;\text{ is }\;\;v\;\text{ times in }\;\;x,\\ (s_3,t_3,w_3)\;\text{ is }\;v^2\;\text{ times in }\;x,\\ (s_4,t_4,w_4)\;\text{ is }\;1+v\;\text{ times in }\;x,\\ (s_5,t_5,w_5)\;\text{ is }\;1+v^2\;\text{ times in }\;x,\\ (s_6,t_6,w_6)\;\text{ is }\;v+v^2\;\text{ times in }\;x, \;\text{ and }\\ (s_7,t_7,w_7)\;\text{ is }\;1+v+v^2\;\text{ times in }\;x.\end{cases}\) Then we have $$\sum_{i=0}^{7}s_i=\sum_{j=0}^{7}t_j=\sum_{k=0}^{7}w_i=n.$$ Let \(c_{i,i=0\cdots 7} \in \{ \alpha \begin{pmatrix} 1 1 \cdots 1 & v v \cdots v & v^2 v^2 \cdots v^2 \end{pmatrix} \;\text{such that}\; \alpha \in M \}\). Then we have \(\begin{cases} d_L(x,c_0)=3n-s_0+2s_1+s_2+s_3+s_6-t_0+2t_1+t_2+t_3+t_6-w_0+2w_1+w_2+w_3+w_6,\\ d_L(x,c_1)=3n+2s_0-s_1+s_4+s_5+s_7+t_0-t_2+t_3+2t_4+t_6+w_0+w_2-w_3+2w_5+w_6,\\ d_L(x,c_2)=3n+s_0-s_2+s_3+2s_4+s_6+t_0+t_2-t_3+2t_5+t_6+2w_0-w_1+w_4+w_5+w_7,\\ d_L(x,c_3)=3n+2s_0-s_1+s_4+s_5+s_7+t_0-t_2+t_3+2t_4+t_6+w_0+w_2-w_3+2w_5+w_6,\\ d_L(x,c_4)=3n+s_1+2s_2-s_4+s_5+s_7+t_0+t_2+t_3-t_6+2t_7+w_1+2w_3+w_4-w_5+w_7,\\ d_L(x,c_5)=3n+s_1+2s_3+s_4-s_5+s_7+t_1+2t_2-t_4+t_5+t_7+w_0+w_2+w_3-w_6+2w_7,\\ d_L(x,c_6)=3n+s_0+s_2+s_3-s_6+2s_7+t_1+2t_3+t_4-t_5+t_7+w_1+2w_2-w_4+w_5+w_7,\\ d_L(x,c_7)=3n+s_1+s_4+s_5+2s_6-s_7+t_1+t_4+t_5+2t_6-t_7+w_1+2w_2+w_5+2w_6-w_7.\end{cases}\) Therefore \(d_L(x,Rp_{U_s}^{3n})=min\{ d_L(x,c_i), i=0..7\} \leq \frac{1}{8}(24n+4(\sum_{i=0}^{7} s_i)+4(\sum_{i=0}^{7} t_i)+4(\sum_{i=0}^{7} w_i))\) and hence \(d_L(x,Rp_{U_s}^{3n})\leq \frac{9n}{2}\) \(\forall x \in M^{3n}\). In other hand by using Preposition 1 we have \(r_{d_L}(Rp_{U_s}^{3n}) \geq 3r_{d_L}(C_1)=\dfrac{9n}{2}\).
• ii) The prove is similar to i) and left for the readers.

Theorem 3. Let \(C\) be a block repetition code of length \(2n\) generated by \(G=\begin{pmatrix} \overset{n }{\widehat{1+v 1+v \cdots 1+v}} & \overset{n}{\widehat{1+v+v^2 1+v+v^2 \cdots 1+v+v^2 }} \end{pmatrix}\). Then the code \(C\) is \([2n,8,d_H=n,d_L=n]\)-code over \(M^{2n}\) and \(4\lfloor\frac{n}{4} \rfloor+\lfloor \frac{n}{2}\rfloor \leq r_{d_L}(C) \leq 4n.\)

Proof. By Proposition 1, we have \(r_{d_L} \geq r_{d_L}(C_2)+r_{d_L}(C_3) \geq 4\lfloor\frac{n}{4} \rfloor+\lfloor \frac{n}{2}\rfloor\). So \(r_{d_L} \geq 4\lfloor\frac{n}{4} \rfloor+\lfloor \frac{n}{2}\rfloor\). Let \(x \in M^{2n}\) such that \(x=(x_1|x_2)\) with \(m_i, i=0..7\) is the number of coordinates have \((0, 1, v, v^2, 1+v, 1+v^2, v+v^2, 1+v+v^2)\) respectively in \(x_1\) and \(l_i, i=0..7\) is the number of coordinates in \(x_2\) have \((0, 1, v, v^2, 1+v, 1+v^2, v+v^2, 1+v+v^2)\) respectively. We conclude that \(\sum_{i=0}^{7}m_i=\sum_{i=0}^{7}l_i=n\). Now let \(c_{i,i=0..7} \in \{ \alpha \begin{pmatrix} 1+v 1+v \cdots 1+v & 1+v+v^2 1+v+v^2 \cdots 1+v+v^2 \end{pmatrix} \;\text{such that}\; \alpha \in M \}\), then we have \(\begin{cases}d_L(x,c_0)=2n-m_0+2m_1+m_2+m_3+m_6-l_0+2l_1+l_2+l_3+l_6,\\ d_L(x,c_1)=2n+m_1+2m_2-m_4+m_5+m_7+l_1+l_4+l_5+2l_6-l_7,\\ d_L(x,c_2)=2n+m_0+m_2+m_3-m_6+2m_7+l_1+l_4+l_5+2l_6-l_7,\\ d_L(x,c_3)=2n+m_1+2m_3+m_4-m_5+m_7+l_1+l_4+l_5+2l_6-l_7,\\ d_L(x,c_4)=2n+m_1+2m_3+m_4-m_5+m_7-l_0+2l_1+l_2+l_3+l_6,\\ d_L(x,c_5)=2n+m_0+m_2+m_3-m_6+2m_7-l_0+2l_1+l_2+l_3+l_6,\\ d_L(x,c_6)=2n+m_1+2m_2-m_4+m_5+m_7-l_0+2l_1+l_2+l_3+l_6,\\ d_L(x,c_7)=2n-m_0+2m_1+m_2+m_3+m_6+l_1+l_4+l_5+2l_6-l_7.\end{cases}\)
By definition of Lee distance of \(C,\) we have \(d_L(x,C)=min\{ d_L(x,c_i), i=0..7\} \leq 4n\) for any \(x\) in \(M^{2n}\) which implies that \( r_{d_L}(C) \leq 4n\). Hence the prove is complete.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflict of Interests

The authors declare no conflict of interest.