Finding a Generator Matrix of a Multidimensional Cyclic Code

TL;DR

This paper extends a method for constructing generator matrices from two-dimensional cyclic codes to more complex multicyclic codes, enabling the formation of a basis and the subsequent creation of a generator matrix.

Contribution

The paper generalizes Sepasdar's method to find a basis and generator matrix for multicyclic codes, broadening the applicability of the technique.

Findings

Successfully generalized the method to multicyclic codes

Provided a procedure to construct a generator matrix from the basis

Enhanced understanding of multicyclic code structure

Abstract

We generalize Sepasdar's method for finding a generator matrix of two-dimensional cyclic codes to find an independent subset of a general multicyclic code, which may form a basis of the code as a vector subspace. A generator matrix can be then constructed from this basis.

Equations

$$x_i^{{\rho }_i}=1,$$

$$x_i^m=x_i^{m\mathrm{mod}{\rho }_i}\text{for}m\in N\text{and}i=1,\dots ,s,$$

$${\mathcal{G}}_s=Z/{\rho }_{1}Z\times \dots \times Z/{\rho }_{s}Z,$$

$$Z/{\rho }_{i}Z=\{0,1,\dots ,{\rho }_i-1\}.$$

$$f(x_1,\dots ,x_s)=\sum _{({\alpha }_1,\dots ,{\alpha }_s)\in {\mathcal{G}}_s}{f}_{({\alpha }_1,\dots ,{\alpha }_s)}{x}_1^{{\alpha }_1}\cdots x_s^{{\alpha }_s}.$$

$$f(x)=\sum _{\alpha \in {\mathcal{G}}_s}{f}_{\alpha }{x}^{\alpha },$$

$$x^{\alpha }=x_1^{{\alpha }_1}\cdots x_s^{{\alpha }_s},$$

$$\alpha \mathrm{mod}\rho =({\alpha }_1\mathrm{mod}{\rho }_1,\dots ,{\alpha }_s\mathrm{mod}{\rho }_s)\in {\mathcal{G}}_s,$$

$$x^{\rho }=1\text{and}{x}^{\alpha }=x^{\alpha \mathrm{mod}\rho }.$$

$$\alpha {⩽}_{+}\beta ⟺{\alpha }_i⩽{\beta }_i\text{for}i=1,\dots ,s,$$

$$\alpha {<}_{\mathrm{lex}}\beta ⟺\text{for the first index i such thatαi=βi, one has αi<βi}.$$

$${\alpha }^{(1)}<\cdots {\alpha }^{(i)}<\cdots <{\alpha }^{(n)}$$

$$f(x)=f_{{\alpha }^{(1)}}{x}^{{\alpha }^{(1)}}+\cdots +f_{{\alpha }^{(i)}}{x}^{{\alpha }^{(1)}}+\cdots +f_{{\alpha }^{({\rho }_s)}}{x}^{{\alpha }^{(n)}}.$$

$$\mathrm{deg}f=\underset{{⩽}_{\mathrm{lex}}}{\max }\{{\alpha }^{(i)}|f_{{\alpha }^{(i)}}\ne 0\}.$$

$$f(x_1,\dots ,x_s)g(x_1,\dots ,x_s)$$

$$=\sum _{\alpha }\sum _{\beta }{f}_{\alpha }{g}_{\beta }{x}^{(\alpha +\beta )}$$

$$\mathrm{deg}(fg)=\underset{{⩽}_{\mathrm{lex}}}{\max }(\alpha +\beta )=\mathrm{deg}f+\mathrm{deg}g.$$

$$S=F_q[X_1,\dots ,X_s]/⟨X_1^{{\rho }_1}-1,\dots ,X_{s-1}^{{\rho }_{s-1}}-1⟩=F_q[x_1,\dots ,x_{s-1}],$$

$$\mathfrak{B}=\{{\mathfrak{p}}_1^{(0)},\dots ,{\mathfrak{p}}_{r_1}^{(0)},{\mathfrak{p}}_1^{(1)},\dots ,{\mathfrak{p}}_{r_1}^{(1)},\dots ,{\mathfrak{p}}_1^{(i)},\dots ,{\mathfrak{p}}_{r_i}^{(i)},\dots ,{\mathfrak{p}}_1^{({\rho }_s-1)},\dots ,{\mathfrak{p}}_{r_{{\rho }_s-1}}^{({\rho }_s-1)}\}$$

$$f(x_1,\dots ,x_s)=\sum _{i=0}^{r_{{\rho }_s-1}}\sum _{j=1}^{r_j}{q}_j^{(i)}(x_1,\dots ,x_{s-1}){\mathfrak{p}}_j^{(i)}(x_1,\dots ,x_s).$$

$${\mathfrak{B}}^{\prime }=\{x_1^{{\alpha }_1}\cdots x_{s-1}^{{\alpha }_{s-1}}\mathfrak{p}|({\alpha }_1,\dots ,{\alpha }_{s-1}){⩽}_{+}({\rho }_1,\dots ,{\rho }_{s-1})\text{and}\mathfrak{p}\in \mathfrak{B}\}$$

$$q_j^{(i)}(x_1,\dots ,x_{s-1})=\sum _{({\alpha }_1,\dots ,{\alpha }_{s-1}){⩽}_{+}({\rho }_1,\dots ,{\rho }_{s-1})}{q}_{j{\alpha }_1,\dots ,{\alpha }_{s-1}}^{(i)}{x}_1^{{\alpha }_1}\cdots x_{s-1}^{{\alpha }_{s-1}},$$

$${\mathfrak{p}}_0(x_1,\dots ,x_s),...,{\mathfrak{p}}_{{\rho }_s-1}(x_1,\dots ,x_s),$$

$${\mathfrak{p}}_k(x_1,\dots ,x_s)=\sum _{h=k}^{{\rho }_{s-1}}{p}_h^h(x_1,\dots ,x_{s-1})x_s^h,$$

$$\sum _{k=0}^{{\rho }_{s-1}}{l}_k(x_1,\dots ,x_{s-1}){\mathfrak{p}}_k(x_1,\dots ,x_{s-1})=0⟹$$

$$\text{for}k=0\dots ,{\rho }_{s-1}.$$

$$\sum _{k=0}^{{\rho }_{s-1}}{l}_k(x_1,\dots ,x_{s-1}){\mathfrak{p}}_k(x_1,\dots ,x_s)=0.$$

$$l_0(x_1,\dots ,x_{s-1})p_0(x_1,\dots ,x_{s-1})=0.$$

$$\begin{array}{rl}\mathrm{deg}(l_0{p}_0)=& \mathrm{deg}(l_0)+\mathrm{deg}(p_0)>0.\end{array}$$

$$B=$$

$$\cup$$

$$\dots$$

Full text

rator matrix of two-dimensional cyclic codes to find an independent subset of a general multicyclic code, which may form a basis of the code as a vector subspace. A generator matrix can be then constructed from this basis.

MSC(2010): primary: 13F20, 16D25; Secondary: 94B60

Keywords: quotient ring, lexicographic order, ideal basis, multicyclic code, generator matrix

$*$ Corresponding author

1. Introduction

Sepasdar, in [2] presented a method to find a generator matrix of two dimensional skew cyclic Codes. Then, Sepasdar and Khashyarmanesh, in [3] gave a method to find a generator matrix of some class of two-dimensional cyclic codes. Finally, Sepasdar, in [4], found a method to construct a generator matrix for general two-dimensional cyclic codes. In this paper, we will generalize this Sepasdar’s method for a general multicyclic code. Our method uses an ideal basis of the code whose construction was presented by Lalasoa et al. in [1].

In section 2 of this paper, we recall the notations used in [1] and the mathematical tools we will need, including two orderings : the partial ordering “$\leqslant_{+}$” and the well ordering “$\operatorname{\leqslant_{\rm lex}}$”. This latter allows to define degrees of polynomials in the quotient-ring with a special property, given by Proposition 3.2.

In section 3, we present our results. Proposition 3.1 gives an idea of how a basis of the multicyclic code, considered as a vector space will look like. The main result is Theorem 3.3, which allows the construction of an independent subset of the code. Once a basis is found, one can then construct a generator matrix by forming the matrix whose rows are the coefficients of the polynomials of the basis.

2. Notations and Preliminaries

We briefly recall the notations which are used in [1]. We denote the quotient ring $\operatorname{\mathbb{F}}_{q}[X_{1},\ldots,X_{s}]/\langle X_{1}^{\rho_{1}}-1,\ldots,X_{s}^{\rho_{s}}-1\rangle$ by $\operatorname{\mathbb{F}}_{q}[x_{1},\ldots,x_{s}]$, where $\operatorname{\mathbb{F}}_{q}$ is the finite field with $q$ element, $x_{i}$ the residue class of $X_{i}$ modulo the ideal $\langle X_{1}^{\rho_{1}}-1,\ldots,X_{s}^{\rho_{s}}-1\rangle$. We have

[TABLE]

so that

[TABLE]

where $m\mod\rho_{i}$ is the remainder of $m$ by the euclidean division of $m$ by $\rho_{i}$.

The additive product group $\mathcal{G}_{s}$ is defined by

[TABLE]

with

[TABLE]

An element of $\operatorname{\mathbb{F}}_{q}[x_{1},\ldots,x_{s}]$ is of the form

[TABLE]

For sake of simplicity, we denote $(\alpha_{1},\ldots,\alpha_{s})\in\mathcal{G}_{s}$ or, more generally, $(\alpha_{1},\ldots,\alpha_{s})\in\operatorname{\mathbb{N}}^{s}$ by $\alpha$. Then (2.3) can then be written as a

[TABLE]

where

[TABLE]

and we may omit the set $\mathcal{G}_{s}$. For $\alpha\in\operatorname{\mathbb{N}}^{s}$, we also adopt the notation

[TABLE]

where $\rho=(\rho_{1},\ldots,\rho_{s})$. Equations (2.1) and (2.2) are then “generalized” to the following:

[TABLE]

The set $\operatorname{\mathbb{N}}^{s}$, and therefore also he product group $\mathcal{G}_{s}$ is provided with two orders : a partial ordering $\leqslant_{+}$ defined by

[TABLE]

and a well ordering $\operatorname{\leqslant_{\rm lex}}$ (the “lexicographical ordering’), defined by

[TABLE]

Put $n=\rho_{1}\cdots\rho_{s}$. We then may write $\mathcal{G}_{s}=\{\alpha^{(1)},\ldots,\alpha^{(i)},\ldots,\alpha^{(n)}\}$ with

[TABLE]

and the polynomial $f(x)$ in (2.4) can be written as

[TABLE]

If $f(x)$ is non-zero, we may define its degree, denoted $\deg f(x)$ or simply $\deg f$ as

[TABLE]

(Note that it is the usual definition of the degree of a multivariate polynomial). However, due to equations (2.6), for two polynomials $f$ and $g$ of $\operatorname{\mathbb{F}}_{q}[x_{1},\ldots,x_{s}]$, the equality $\deg(fg)=\deg f+\deg g$ does not necessarily hold. The following proposition gives a sufficient condition for this property.

Proposition 2.1**.**

If $f$ and $g$ are non-zero elements of $\operatorname{\mathbb{F}}_{q}[x_{1},\ldots,x_{s}]$ such that $\deg f+\deg g<_{+}\rho$, then $\deg(fg)=\deg f+\deg g$.

Proof.

Write $f(x_{1},\ldots,x_{s})=\sum_{\alpha}f_{\alpha}x^{\alpha}$ and $g(x_{1},\ldots,x_{s})=\sum_{\beta}g_{\beta}x^{\beta}$. Then, using the second equation of (2.6), we have

[TABLE]

since $\alpha+\beta\leqslant_{+}\deg f+\deg g<_{+}\rho=(\rho_{1},\ldots,\rho_{s})$ for all $\alpha$ and $\beta$. Thus

[TABLE]

∎

All the previous results are also true for the quotient ring

[TABLE]

with $s-1$ variables, where $x_{i}$ is the residue class of $x_{i}$ modulo the ideal $\langle X_{1}^{\rho_{1}}-1,\ldots,X_{s-1}^{\rho_{s-1}}-1\rangle$. Note that we have used the same notation $x_{i}$, because the residue class of $x_{i}$ modulo the ideal $\langle X_{1}^{\rho_{1}}-1,\ldots,X_{s-1}^{\rho_{s-1}}-1\rangle$ may be identified with its class modulo the ideal $\langle X_{1}^{\rho_{1}}-1,\ldots,X_{s}^{\rho_{s}}-1\rangle$, (cf. Proposition 2.2, [1]).

A* multicyclic code* is an ideal of $R$.

Let $I$ be a non-zero ideal of $R$ and

[TABLE]

the basis of $I$, found by Lalasoa et al. by the method in [1]. Then an element $f(x_{1},\ldots,x_{s})\in R$ may be written as

[TABLE]

Note that in (2.11), the coefficients of the polynomials in $\mathfrak{B}$ are polynomials in $S$.

3. Results

Our aim in this section is to construct a basis of $I$, as an $\operatorname{\mathbb{F}}_{q}$-vector subspace of $R$ (an $\operatorname{\mathbb{F}}_{q}$-basis), from the ideal basis $\mathfrak{B}$ of $I$, in (2.10).

Proposition 3.1**.**

The set

[TABLE]

is a generating set of $I$, as an $\operatorname{\mathbb{F}}_{q}$-vector space.

Proof.

It suffices to use (2.11) and write

[TABLE]

where $q^{(i)}_{j\alpha_{1},\ldots,\alpha_{s-1}}\in\operatorname{\mathbb{F}}_{q}$. Then the polynomial $f$ is written as a linear combination of elements of $\mathfrak{B}^{\prime}$, with coefficients in $\operatorname{\mathbb{F}}_{q}$. ∎

The set $\mathfrak{B}^{\prime}$ in 3.1 may be too large to be an $\operatorname{\mathbb{F}}_{q}$- basis of $I$. In other words, the elements of $\mathfrak{B}$ may be linearly dependent. If this is the case, an $\operatorname{\mathbb{F}}_{q}$ - basis $B$ of $I$ should be then extracted from $\mathfrak{B}^{\prime}$.

We will find linearly independent elements of $\mathfrak{B}^{\prime}$ and check whether they form an $\operatorname{\mathbb{F}}_{q}$ -base of $I$.

According to the notations in (2.10), we choose polynomials

[TABLE]

where $\mathfrak{p}_{k}\in\{\mathfrak{p}_{1}^{(k)},\ldots,\mathfrak{p}^{(k)}_{r_{\rho_{s}-1}}\}\subset I_{k}$. Let $p_{k}(x_{1},\ldots,x_{s-1})\in S$ be the coefficient of $\mathfrak{p}_{k}$ with respect to $x_{s}^{k}$ and $a_{k}=\deg p_{k}$, where the degree is defined as in (2.9), but, now, in the quotient ring $S$. We have

[TABLE]

with $p_{h}^{h}\in S$ and $p_{k}^{k}=p_{k}$.

Proposition 3.2**.**

Let $l_{0}(x_{1},\ldots,x_{s-1}),\ldots,\l_{\rho_{s-1}}(x_{1},\ldots,x_{s-1})$ be polynomials in $\operatorname{\mathbb{F}}_{q}[x_{1},\ldots,x_{s-1}]$ such that $\deg(l_{k})<_{+}[(\rho_{1},\ldots,\rho_{s-1})-(a_{k})]$. Then

[TABLE]

Proof.

Let $l_{0}(x_{1},\ldots,x_{s-1}),\ldots,l_{\rho_{s-1}}(x_{1},\ldots,x_{s-1})$ be polynomials in

$\operatorname{\mathbb{F}}_{q}[x_{1},\ldots,x_{s-1}]$ which verify the hypothesis of the proposition, such that

[TABLE]

Then

[TABLE]

Supose that $l_{0}\neq 0$. By taking the degrees, we have, by Proposition 3.2,

[TABLE]

But this is impossible for a non-zero polynomial. It follows that $l_{0}=0$ and using (3.1), the same reasoning can be applied step by step to show that $l_{i}$ for $i=1,\ldots,\rho_{s-1}$. ∎

Theorem 3.3**.**

With the previous notations, let $B$ be the set

[TABLE]

*Then

$(1)$ The elements of $B$ are $\operatorname{\mathbb{F}}_{q}$-linearly independent.

$(2)$ If $|B|={\rm log_{q}}|I|$, then $B$ is an $\operatorname{\mathbb{F}}_{q}$-basis of $I$ (where $|I|$ is the cardinality of $I$).*

Proof.

(1) We construct the finite sequence of numbers

[TABLE]

for $k=0,\ldots,\rho_{s}-1$. Now, let $(\alpha_{j}^{k})_{1\leqslant j\leqslant N_{k}}$ be sequences of elements of $\operatorname{\mathbb{F}}_{q}$ such that

[TABLE]

(this is a linear combination of elements of $B$ which equals to zero). By taking

[TABLE]

for $k=0,\ldots,\rho_{s}-1$, equation (3.3) becomes

[TABLE]

By Proposition 3.2, we have $l_{k}(x_{1},\ldots,x_{s-1})=0$ for $k=0,\ldots,\rho_{s-1}$, i.e. $\alpha_{j}^{k}=0$ for $j=1,\ldots,N_{k}$.

(2) The ring $R$ is isomorphic to a subspace of $\operatorname{\mathbb{F}}_{q}^{n}$, by the mapping

[TABLE]

where $n=\prod_{i=1}^{s}{\rho_{i}}$. Thus, $I$ may be identified with a subspace of $\operatorname{\mathbb{F}}_{q}^{n}$, and it is known that in this case, $\dim I={\rm log_{q}}|I|$. The elements of the set $B$ being linearly independent, it follows that $B$ is a basis of $I$, when its when its cardinality equals to ${\rm log_{q}}|I|$. ∎

For an $\operatorname{\mathbb{F}}_{q}$-basis $B=\{g_{1}(x),\ldots,g_{l}(x)\}$ of $I$, where, according to (2.8)

[TABLE]

A generator matrix for $I$, as a multicyclic code is then

[TABLE]

where $\operatorname{\mathbb{F}}_{q}^{l,n}$ is the set of matrices with $l$ rows and $n$ columns and entries in $\operatorname{\mathbb{F}}_{q}$. In other words, $G$ is the matrix whose rows are the coefficients of the elements of $B$.

Acknowledgments

The authors would like to thank the referee for careful reading.