On Double Cyclic Codes and Applications to DNA Codes

Abstract

In this paper, we investigate the structure of double cyclic codes of length $(\alpha ,\beta )$ over $\mathfrak{R}={\mathbb{F}}_4+\varrho {\mathbb{F}}_4,{\varrho }^2=0$ and give a minimal spanning set of double cyclic codes. Moreover, we extend this study to construct DNA codes from $\mathfrak{R}$-double cyclic codes. Furthermore, we explore reversible and reversible-complement double cyclic codes over $\mathfrak{R}$ and present a necessary and sufficient condition for separable $\mathfrak{R}$-double cyclic codes to be reversible and reversible-complement $\mathfrak{R}$-double cyclic codes. Also, we study reversible and reversible-complement double cyclic codes over $\mathfrak{R}$ for non-separable case and construct some optimal $\mathfrak{R}$-double cyclic codes, and several DNA codes to support our results.

1. Introduction

Behind every trustworthy communication system, there is coding theory playing a crucial role. Coding theory primarily deals with the efficient transmission of information and the achievement of reliable communication and storage of data. Since the remarkable work of Shannon [1], coding theory has evolved significantly, and codes, especially linear codes over finite fields, have been intensively studied throughout the literature [2]. Linear codes are vital in error detection and correction during information transmission and play an important role in improving data storage reliability. In 1994, Hammons et al. [3] demonstrated that several nonlinear codes over ${\mathbb{Z}}_2$ can be viewed as the Gray images of linear codes over ${\mathbb{Z}}_4$. Since then, due to their simple encoding and decoding procedures, linear codes over finite rings have been extensively investigated. Among the class of linear codes, cyclic codes over finite rings are the most studied class because of their rich algebraic structure and convenient polynomial representation; for greater insight one can see [4,5,6,7,8,9,10]. Furthermore, reversible cyclic codes are important because they simplify both the encoding and decoding processes. In addition, reversibility is an essential property when studying DNA codes. DNA codes are crucial in coding theory, as many combinatorial problems have been addressed using DNA computing, such as the maximal clique problem [11] and the Hamiltonian path problem [12]. Mansuripur et al. [13] demonstrated the use of DNA codes for data storage media, and Adleman and coauthors [14] successfully cracked the Data Encryption Standard (DES) cryptosystem using DNA computing techniques. Consequently, efficient error-correcting codes have been constructed by modeling the structure of DNA, and codes with properties analogous to DNA sequences have also been employed to better understand DNA. The linear construction of DNA codes was first studied by Gaborit and King [15]. Abualrub et al. [16] investigated DNA codes over the finite field with four elements. Later, Siap et al. [17] discussed DNA codes over the ring ${\mathbb{F}}_2\left[u\right]/\langle u^2-1\rangle$ of four elements. DNA codes over the ring ${\mathbb{F}}_2\left[u\right]/\langle u^4-1\rangle$ with sixteen elements were then studied by Yildiz and Siap [18]. Guenda and Gulliver [19] examined cyclic DNA codes over the ring ${\mathbb{F}}_2+u{\mathbb{F}}_2,u^2=0$. Subsequently, Mostafanasab and Darani [20] explored cyclic DNA codes over the ring $R=F_2+u{F}_2+u^2{F}_2$, $u^3=0$.

In [21], Borges et al. introduced a new class of linear codes called double cyclic codes; this is a generalization of cyclic codes. The authors studied the structure of double cyclic codes over ${\mathbb{Z}}_2$ and their duals as submodules of ${\mathbb{Z}}_2\left[x\right]$ module ${\displaystyle \frac{{\mathbb{Z}}_2\left[x\right]}{\langle x^r-1\rangle }}\times {\displaystyle \frac{{\mathbb{Z}}_2\left[x\right]}{\langle x^s-1\rangle }}$, where r and s are any non-negative integers. Additionally, they also compare ${\mathbb{Z}}_2$-double cyclic codes with other families of cyclic codes. After that, Gao et al. [22] extended the study of double cyclic codes over the ring of integer modulo 4. Double cyclic codes over a chain ring ${\mathbb{F}}_q+u{\mathbb{F}}_q+u^2{\mathbb{F}}_q,u^3=0$ are studied by Yao and Shi [23]. Later, Gao and Hou [24] showed that ${\mathbb{Z}}_4$-double cyclic codes are asymptotically good. Patanker [25] studied the reversibility of double cyclic codes over ${\mathbb{Z}}_2$. In [26], Kanlaya and Klin-Eam extended this study to double cyclic codes over the ring ${\mathbb{F}}_2+u{\mathbb{F}}_2$ for the construction of DNA codes.

Motivated by these works, we study double cyclic codes over the ring $\mathfrak{R}={\mathbb{F}}_4+\varrho {\mathbb{F}}_4,{\varrho }^2=0$. We determine the algebraic structure of $\mathfrak{R}$-double cyclic codes. We show that an $\mathfrak{R}$-double cyclic code $\mathfrak{C}$ of length ($\alpha$,$\beta$) is an $\mathfrak{R}\left[x\right]$-submodule of ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$, where $\alpha$ and $\beta$ both are odd, and $\mathfrak{C}$ has the form $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle ,$ where ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are monic polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. We also find a minimal spanning set of $\mathfrak{R}$-double cyclic code of length ($\alpha$,$\beta$). Moreover, we explore reversibility of these codes and present a necessary and sufficient condition for double cyclic codes of length ($\alpha$,$\beta$) over $\mathfrak{R}$ to be reversible $\mathfrak{R}$-double cyclic codes. Further, we study reversible-complement $\mathfrak{R}$-double cyclic code of length ($\alpha$,$\beta$). We construct several optimal double cyclic codes over $\mathfrak{R}$, and using the methodology of Kanlaya and Klin-Eam [26], we construct some DNA codes from reversible-complement $\mathfrak{R}$-double cyclic codes. The following remarks should be emphasized.

• Similar to the work [22,23], this paper presents the results on another type of ring. Compared to the works [21,25], this paper presents reversible double cyclic codes over a finite chain ring. Therefore, this paper will improve and generalize the codes in [21,25].
• The Gray images of double cyclic codes over $\mathfrak{R}$ are linear codes over ${\mathbb{F}}_4$, yielding codes with improved parameters compared to the linear codes over ${\mathbb{F}}_2$.
• In contrast to [26], we construct DNA codes over the complete set of DNA double base pairs, allowing the construction of DNA codes with improved parameters.

This paper is organized as follows: In Section 2, some preliminaries are discussed. In Section 3, we explore the structure of $\mathfrak{R}$-double cyclic codes and we find a minimal spanning set for double cyclic codes of length ($\alpha$,$\beta$) over $\mathfrak{R}$. In Section 4, we study DNA codes obtained from double cyclic codes over $\mathfrak{R}$. For this, we divide this section into two subsections. In the first section, we discuss the reversibility of separable and non-separable $\mathfrak{R}$-double cyclic codes and present a necessary and sufficient condition for the reversibility of separable codes of arbitrary length ($\alpha$,$\beta$), where $\alpha$ and $\beta$ are odd positive integers. In the latter section, we consider reversible-complement $\mathfrak{R}$-double cyclic codes for the separable and non-separable cases and present a necessary and sufficient condition for separable $\mathfrak{R}$-double cyclic codes of arbitrary length ($\alpha$,$\beta$), where $\alpha$ and $\beta$ are odd positive integers, to be reversible-complement $\mathfrak{R}$-double cyclic codes. Finally, Section 5 concludes the paper.

2. Preliminaries

Let ${\mathbb{F}}_4=\{0,1,w,w^2=1+w\}$ be a finite field of four elements. Consider the set $\mathfrak{R}={\mathbb{F}}_4+\varrho {\mathbb{F}}_4,{\varrho }^2=0.$ Then, $\mathfrak{R}$ is a finite commutative ring of 16 elements with characteristic 2, endowed with usual addition and multiplication with the condition ${\varrho }^2=0$. The finite field ${\mathbb{F}}_4$ is a subring of $\mathfrak{R}$; therefore, the factorization of $x^n-1$ over ${\mathbb{F}}_4$ is still valid over $\mathfrak{R}$. The ring $\mathfrak{R}$ is a local chain ring with maximal ideal $\langle \varrho \rangle$. Recall that a linear code $\mathfrak{C}\subseteq {\mathfrak{R}}^{\mathfrak{n}}$ of length n over $\mathfrak{R}$ is an $\mathfrak{R}$-submodule of $\mathfrak{R}$-module ${\mathfrak{R}}^{\mathfrak{n}}$. A linear code $\mathfrak{C}$ is called cyclic if $\mathfrak{C}$ is closed under cyclic shift, i.e., for

$$(c_0,c_1,\dots ,c_{n-1})\in \mathfrak{C}\Rightarrow (c_{n-1},c_0,\dots ,c_{n-2})\in \mathfrak{C},\forall (c_0,c_1,\dots ,c_{n-1})\in \mathfrak{C}.$$

For any $b=(b_0,b_1,\dots ,b_{n-1})\in {\mathfrak{R}}^n$, we can identify b by a unique polynomial $b\left(x\right)=b_0+b_{1}x+\dots +b_{n-1}{x}^{n-1}$ in $\mathcal{R}={\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^n-1\rangle }}$. Then, a linear code $\mathfrak{C}$ of length n over $\mathfrak{R}$ is cyclic if and only if $\mathfrak{C}$ is an ideal in $\mathcal{R}$. The Gray map $\mathsf{\Gamma }:\mathfrak{R}\to {\mathbb{F}}_4^2$ is defined as $\mathsf{\Gamma }(a+\varrho b)=(b,a+b),$ where $a+\varrho b\in \mathfrak{R}$. This Gray map can naturally be extended to $\mathsf{\Gamma }:{\mathfrak{R}}^n\to {\mathbb{F}}_4^{2n}$. Then, $\mathsf{\Gamma }$ is a distance-preserving ${\mathbb{F}}_4$-linear map transforming Lee distance to Hamming distance.

Lemma 1.

If $\mathfrak{C}$ is an $\mathfrak{R}$-linear $[n,k,d]$ code, then $\mathsf{\Gamma }\left(\mathfrak{C}\right)$ is an ${\mathbb{F}}_4$-linear $[2n,k,d]$ code.

The structure of cyclic codes over $\mathfrak{R}$ is well studied by Prakash and coauthors [9]. A cyclic code of length n over $\mathfrak{R}$ is given by the following result:

Theorem 1

([9]). “Let $\mathfrak{C}$ be a cyclic code of length n over $\mathfrak{R}$. Then,

(1)

If n is odd then, ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^n-1\rangle }}$ is a principal ideal ring and $\mathfrak{C}=\langle \xi \left(x\right),\varrho \rho \left(x\right)\rangle =\langle \xi \left(x\right)+\varrho \rho \left(x\right)\rangle$, where $\xi \left(x\right),\rho \left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$ such that $\rho \left(x\right)\left|\xi \left(x\right)\right|(x^n-1)$.

(2)

If n is even then,

(a)

$\mathfrak{C}=\langle \xi \left(x\right)+\varrho t\left(x\right),\varrho \rho \left(x\right)\rangle$, where $\xi \left(x\right),\rho \left(x\right)$ and $t\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$ such that $\rho \left(x\right)\left|\xi \left(x\right)\right|(x^n-1)$, $\xi \left(x\right)+\varrho t\left(x\right)|(x^n-1)$, $\rho \left(x\right)|t\left(x\right)\left({\displaystyle \frac{x^n-1}{\xi \left(x\right)}}\right)$ and $deg\left(\xi \right(x\left)\right)>deg\left(\rho \right(x\left)\right)>deg\left(t\right(x\left)\right)$.

(b)

$\mathfrak{C}=\langle \xi \left(x\right)+\varrho t\left(x\right)\rangle$, where $\xi \left(x\right),t\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$ such that $\xi \left(x\right)|(x^n-1)$, $(\xi \left(x\right)+\varrho t\left(x\right))|(x^n-1)$, $\xi \left(x\right)|t\left(x\right)\left({\displaystyle \frac{x^n-1}{\xi \left(x\right)}}\right)$ and $\xi \left(x\right)=\rho \left(x\right)$”.

Let $\alpha ,\beta \in \mathbb{N}\cup \left\{0\right\}$ and $n=\alpha +\beta$. We consider a partition of n coordinates of each n-tuple of ${\mathfrak{R}}^{\mathfrak{n}}$ into two sets of sizes $\alpha$, $\beta$. Then, a subset $\mathfrak{C}$ of ${\mathfrak{R}}^{\mathfrak{n}}$ can be seen as a subset of ${\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$. Following Borges et al. [21], for any $s\in \mathfrak{R}$ and $r=(r_0,r_1,\dots ,r_{\alpha -1}|r_0^{\prime },r_1^{\prime },\dots ,r_{\beta -1}^{\prime })\in {\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$, we define a scalar multiplication as follows:

$$s\ast (r_0,r_1,\dots ,r_{\alpha -1}|r_0^{\prime },r_1^{\prime },\dots ,r_{\beta -1}^{\prime })=(s{r}_0,s{r}_1,\dots ,s{r}_{\alpha -1}|s{r}_0^{\prime },s{r}_1^{\prime },\dots ,s{r}_{\beta -1}^{\prime }).$$

Under this scalar multiplication, ${\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$ is an $\mathfrak{R}$-module. An $\mathfrak{R}$-linear code $\mathfrak{C}$ of length $n=\alpha +\beta$ can be viewed as an $\mathfrak{R}$-submodule of $\mathfrak{R}$-module ${\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$. For any vector $r=(r_0,r_1,\dots ,r_{\alpha -1}|r_0^{\prime },r_1^{\prime },\dots ,r_{\beta -1}^{\prime })\in {\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$, the double cyclic shift $\sigma$ of r is defined as follows:

$$\sigma \left(r\right)=(r_{\alpha -1},r_0,\dots ,r_{\alpha -2}|r_{\beta -1}^{\prime },r_0^{\prime },\dots ,r_{\beta -2}^{\prime }).$$

Definition 1.

“A linear code $\mathfrak{C}$ of length $n=\alpha +\beta$ over $\mathfrak{R}$ is said to be double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$ if $\sigma \left(r\right)\in \mathfrak{C}$ for all $r\in \mathfrak{C}$”.

We can identify any element $d=\left(r\right|r^{\prime })=(r_0,r_1,\dots ,r_{\alpha -1}|r_0^{\prime },r_1^{\prime },\dots ,r_{\beta -1}^{\prime })\in {\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$ by a unique element $d\left(x\right)$ in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$ as follows:

$$d\left(x\right)=\left(r\left(x\right)\right|r^{\prime }\left(x\right))=(r_0+r_{1}x+\dots +r_{\alpha -1}{x}^{\alpha -1}|r_0^{\prime }+r_1^{\prime }x+\dots +r_{\beta -1}^{\prime }{x}^{\beta -1}).$$

This provides a one-to-one correspondence between ${\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$ and ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$. The ring ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$ is an $\mathfrak{R}\left[x\right]$-module with respect to usual addition and multiplication ∗ defined in (1), for any $p\left(x\right)\in \mathfrak{R}\left[x\right]$ and $\left(r\left(x\right)\right|r^{\prime }\left(x\right))=d\left(x\right)\in {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$,

$$p\left(x\right)\ast d\left(x\right)=p\left(x\right)\ast \left(r\left(x\right)\right|r^{\prime }\left(x\right))=\left(p\left(x\right)r\left(x\right)\right|p\left(x\right)r^{\prime }\left(x\right)),$$

(1)

where multiplication $p\left(x\right)r\left(x\right)$ is performed under $mod(x^{\alpha }-1)$ and $p\left(x\right)r^{\prime }\left(x\right)$ is performed under $mod(x^{\beta }-1)$. Moreover, for $d\left(x\right)\in {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$, the multiplication $x\ast d\left(x\right)$ gives double cyclic shift of $d\in {\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$. Hence, we have the following proposition:

Proposition 1.

A linear code $\mathfrak{C}$ of length $n=\alpha +\beta$ over $\mathfrak{R}$ is a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$ if and only if $\mathfrak{C}$ is an $\mathfrak{R}\left[x\right]$-submodule of ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$.

From now on throughout this paper, we consider both $\alpha$ and $\beta$ to be odd.

3. Double Cyclic Codes

In this section, we discuss the structure of double cyclic codes of length ($\alpha$,$\beta$) over $\mathfrak{R}$. Let $\mathfrak{C}$ be an $\mathfrak{R}$-double cyclic code of length $(\alpha ,\beta )$; ${\mathfrak{C}}_{\alpha }$ and ${\mathfrak{C}}_{\beta }$ are the coordinate projections of $\mathfrak{C}$ on $\alpha$ coordinates first and $\beta$ coordinates last. Then, $\mathfrak{C}$ is said to be separable if $\mathfrak{C}={\mathfrak{C}}_{\alpha }\times {\mathfrak{C}}_{\beta }$. Let $\mathfrak{C}$ be an $\mathfrak{R}\left[x\right]$-submodule of ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$; consider the following maps defined by the following:

$$\begin{array}{cc}\hfill {\varphi }_{\alpha }:\mathfrak{C}& \to {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }},\hfill \\ \hfill {\varphi }_{\alpha }(r\left(x\right)& |r^{\prime }\left(x\right))=r\left(x\right);\hfill \end{array}$$

for all $\left(r\left(x\right)\right|r^{\prime }\left(x\right))\in \mathfrak{C}$, and

$$\begin{array}{cc}\hfill {\varphi }_{\beta }:\mathfrak{C}& \to {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }},\hfill \\ \hfill {\varphi }_{\beta }(r\left(x\right)& |r^{\prime }\left(x\right))=r^{\prime }\left(x\right);\hfill \end{array}$$

for all $\left(r\left(x\right)\right|r^{\prime }\left(x\right))\in \mathfrak{C}$. Then, ${\varphi }_{\alpha }$ and ${\varphi }_{\beta }$ are $\mathfrak{R}\left[x\right]$- module homomorphisms. Let us now find the structure of double cyclic code of length ($\alpha$,$\beta$) over $\mathfrak{R}$.

Theorem 2.

Let $\mathfrak{C}$ be an $\mathfrak{R}$-double cyclic code of length (α,β). Then,

$$\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle ,$$

where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$.

Proof. 

Consider the homomorphism ${\varphi }_{\beta }$; the kernel of ${\varphi }_{\beta }$ is given by $ker\left({\varphi }_{\beta }\right)=\{\left(f\left(x\right)\right|0)\in \mathfrak{C}|f\left(x\right)\in {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\}$. Consider the set $J=\{a\left(x\right)\in {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}|\left(a\left(x\right)\right|0)\in ker({\varphi }_{\beta })\}$. Clearly, J is also ideal in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}$, and the homomorphic image ${\varphi }_{\beta }\left(\mathfrak{C}\right)$ of $\mathfrak{C}$ under ${\varphi }_{\beta }$ is ideal in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$. Thus, from Theorem 1, $J=\langle {\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)\rangle$ and ${\varphi }_{\beta }\left(\mathfrak{C}\right)=\langle {\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)\rangle$, where ${\rho }_i\left(x\right),{\xi }_i\left(x\right)$ are polynomials over ${\mathbb{F}}_4$ such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. Hence, we have $ker\left({\varphi }_{\beta }\right)=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\rangle$ and ${\varphi }_{\beta }\left(\mathfrak{C}\right)=\langle {\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)\rangle$. Thus, $\mathfrak{C}$ is given by pullback along ${\varphi }_{\beta }$ as follows:

$$\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle ,$$

for some $ß\left(x\right)\in \mathfrak{R}\left[x\right]$.    □

Lemma 2.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$; then, we may assume that $deg(ß\left(x\right))<deg({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$.

Proof. 

Suppose that $deg(ß\left(x\right))\ge deg({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$. Let $j=deg(ß\left(x\right))-deg({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$ and $\mathsf{𝒞}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)-x^j({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$; then, $\mathsf{𝒞}\subseteq \mathfrak{C}$. However, we also have

$$\begin{array}{cc}\hfill (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=& (ß\left(x\right)-x^j({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))+x^j\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0).\hfill \end{array}$$

Thus, $\mathfrak{C}\subseteq \mathsf{𝒞}$; hence, $\mathfrak{C}=\mathsf{𝒞}$. Therefore, $deg(ß(x\left)\right)$ can be reduced in $\mathfrak{C}$ so that we may assume $deg(ß\left(x\right))<deg({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$.    □

Lemma 3.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be an $\mathfrak{R}$-double cyclic code of length $(\alpha ,\beta )$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$; then

(i)

${\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)$ divides ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)$,

(ii)

${\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)$ divides $\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}ß\left(x\right)$.

Proof. 

In the proof of Theorem 2, we see that $ker\left({\varphi }_{\beta }\right)=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\rangle$. This is because $(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\in \mathfrak{C}$, ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=({\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)|0)\in \mathfrak{C}$ and ${\varphi }_{\beta }({\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)|0)=0$; hence, $({\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)|0)\in ker\left({\varphi }_{\beta }\right)=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\rangle$. This implies that ${\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)$ divides ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)$. Similarly we can prove (ii).    □

Corollary 1.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$. If ${\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)$ is coprime with ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}$ then, $ß\left(x\right)=0$.

Lemma 3 tells us that if an $\mathfrak{R}$-double cyclic code has only one generator of the form $\langle (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$, then we have $(x^{\alpha }-1)|{\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. Thus, we have the following result:

Proposition 2.

Let $\mathfrak{C}$ be a double cyclic code of length (α,β) over $\mathfrak{R}$. Then, we can classify $\mathfrak{C}$ as follows:

(i)

$\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\rangle$ with ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$;

(ii)

$\mathfrak{C}=\langle (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ with $(x^{\alpha }-1)|{\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$;

(iii)

$\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ with ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$, $deg(ß\left(x\right))<deg({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))$, $({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$.

Theorem 3.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. Then, $\mathfrak{C}$ is separable if and only if $ß\left(x\right)=0$.

Proof. 

Consider that $ß\left(x\right)=0$. Then, $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),\left(0\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$, it follows that ${\varphi }_{\alpha }\left(\mathfrak{C}\right)=\langle {\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)\rangle$ and ${\varphi }_{\beta }\left(\mathfrak{C}\right)=\langle {\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)\rangle$. Thus, we observe that $\mathfrak{C}={\varphi }_{\alpha }\left(\mathfrak{C}\right)\times {\varphi }_{\beta }\left(\mathfrak{C}\right)={\mathfrak{C}}_{\alpha }\times {\mathfrak{C}}_{\beta }$.

Conversely, suppose that $\mathfrak{C}$ is separable and ${\mathfrak{C}}^{\prime }=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),\left(0\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle .$ Then, $\mathfrak{C}$ can be viewed as ${\varphi }_{\alpha }\left(\mathfrak{C}\right)\times {\varphi }_{\beta }\left(\mathfrak{C}\right)$. $ß\left(x\right)\in {\varphi }_{\alpha }\left(\mathfrak{C}\right)$ and $0\in {\varphi }_{\beta }\left(\mathfrak{C}\right)$; thus, $(ß\left(x\right)|0)\in {\varphi }_{\alpha }\left(\mathfrak{C}\right)\times {\varphi }_{\beta }\left(\mathfrak{C}\right)$ and ${\mathfrak{C}}^{\prime }\subseteq \mathfrak{C}$. Therefore, $(ß\left(x\right)|0)=\mu \left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+\lambda \left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)),$ for some $\mu \left(x\right),\lambda \left(x\right)\in \mathfrak{R}\left[x\right]$. Hence, we have

$$ß\left(x\right)\equiv \mu \left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+\lambda \left(x\right)ß\left(x\right)mod(x^{\alpha }-1),$$

(2)

and

$$0\equiv \lambda \left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))mod(x^{\beta }-1).$$

(3)

Equation (3) implies that $\lambda \left(x\right)=0$ or $\lambda \left(x\right)={\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{k}_1\left(x\right)$ or $\lambda \left(x\right)=\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}{k}_2\left(x\right)$ for some $k_1\left(x\right),k_2\left(x\right)\in \mathfrak{R}\left[x\right]$.

If $\lambda \left(x\right)=0$, then from 2, $ß\left(x\right)\equiv \mu \left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$$mod(x^{\alpha }-1)$. We obtain

$$\begin{array}{cc}\hfill (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=& \left(\mu \left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\hfill \\ \hfill =& \mu \left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+\left(0\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\in {\mathfrak{C}}^{\prime }.\hfill \end{array}$$

This means $\mathfrak{C}\subseteq {\mathfrak{C}}^{\prime }$; hence, $\mathfrak{C}={\mathfrak{C}}^{\prime }$. If $\lambda \left(x\right)={\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{k}_1\left(x\right)$, then, from (2),

$$\begin{array}{cc}\hfill ß\left(x\right)\equiv & \mu \left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+{\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{k}_1\left(x\right)ß\left(x\right)mod(x^{\alpha }-1).\hfill \end{array}$$

From Lemma 3, we have ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{k}_1\left(x\right)ß\left(x\right)={\mu }_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$ for some ${\mu }_1\left(x\right)\in \mathfrak{R}\left[x\right]$. This means $ß\left(x\right)\equiv (\mu \left(x\right)+{\mu }_1\left(x\right))({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))mod(x^{\alpha }-1)$. Therefore, we obtain

$$\begin{array}{cc}\hfill (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=& \left((\mu \left(x\right)+{\mu }_1\left(x\right))({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\hfill \\ \hfill =& (\mu \left(x\right)+{\mu }_1\left(x\right))\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+\left(0\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\in {\mathfrak{C}}^{\prime }.\hfill \end{array}$$

Thus, $\mathfrak{C}={\mathfrak{C}}^{\prime }$. Similarly, if $\lambda \left(x\right)=\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}{k}_2\left(x\right)$, we can show that $\mathfrak{C}={\mathfrak{C}}^{\prime }$.    □

Corollary 2.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle ,$ be an $\mathfrak{R}$-double cyclic code of length $(\alpha ,\beta )$ as in Theorem 2. If ${\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)$ and ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}$ are coprime in $\mathfrak{R}\left[x\right]$, then $\mathfrak{C}$ is separable.

Proof. 

The result is obvious from Theorem 3 and Corollary 1.    □

Let us now find a minimal spanning set for an $\mathfrak{R}$-double cyclic code of length ($\alpha$,$\beta$) as an $\mathfrak{R}$-module.

Theorem 4.

Suppose $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ is a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. Also, $x^{\alpha }-1={\xi }_1\left(x\right)h_1\left(x\right)$, $x^{\beta }-1={\xi }_2\left(x\right)h_2\left(x\right)$ with $deg\left({\xi }_1\left(x\right)\right)=t_1,deg\left({\rho }_1\left(x\right)\right)=t_2,deg\left({\xi }_2\left(x\right)\right)=s_1,deg\left({\rho }_2\left(x\right)\right)=s_2$. Consider the sets

$$\begin{array}{cc}\hfill {\mathit{F}}_1=& \bigcup _{i=0}^{\alpha -t_1-1}\left\{x^i\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\right\},\hfill \\ \hfill {\mathit{F}}_2=& \bigcup _{i=0}^{t_1-t_2-1}\left\{x^i\ast \left(\varrho h_1\left(x\right){\rho }_1\left(x\right)\right|0)\right\},\hfill \\ \hfill {\mathit{F}}_3=& \bigcup _{i=0}^{\beta -s_1-1}\left\{x^i\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\right\},\hfill \\ \hfill {\mathit{F}}_4=& \bigcup _{i=0}^{s_1-s_2-1}\left\{x^i\ast (h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right))\right\}.\hfill \end{array}$$

Then, $\mathit{F}={\mathit{F}}_1\cup {\mathit{F}}_2\cup {\mathit{F}}_3\cup {\mathit{F}}_4$ forms a minimal spanning set for $\mathfrak{C}$ as an $\mathfrak{R}$-module. Moreover, $\mathfrak{C}$ has $4^{2\alpha +2\beta -t_1-t_2-s_1-s_2}$ codewords.

Proof. 

Let $c\left(x\right)\in \mathfrak{C}$; then, there exist polynomials $p\left(x\right),q\left(x\right)\in \mathfrak{R}\left[x\right]$ such that

$$c\left(x\right)=p\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+q\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)).$$

If $deg\left(p\left(x\right)\right)\le \alpha -t_1-1$, then we have $p\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\in Span\left({\mathit{F}}_1\right)$. Otherwise, by the division algorithm, we have

$$p\left(x\right)=q_1\left(x\right)h_1\left(x\right)+r_1\left(x\right),$$

where $q_1\left(x\right),r_1\left(x\right)\in \mathfrak{R}\left[x\right]$ and $r_1\left(x\right)=0$ or $deg\left(r_1\left(x\right)\right)\le \alpha -t_1-1$. Therefore, we have

$$\begin{array}{cc}\hfill p\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)=& (q_1\left(x\right)h_1\left(x\right)+r_1\left(x\right))\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\hfill \\ \hfill =& q_1\left(x\right)\ast \left(\varrho h_1\left(x\right){\rho }_1\left(x\right)\right|0)+r_1\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0).\hfill \end{array}$$

If $deg\left(q_1\left(x\right)\right)\le t_1-t_2-1$, then $q_1\left(x\right)\ast \left(\varrho h_1\left(x\right){\rho }_1\left(x\right)\right|0)\in Span\left({\mathit{F}}_2\right)$. Otherwise, by the division algorithm,

$$q_1\left(x\right)={\displaystyle \frac{x^{\alpha }-1}{h_1\left(x\right){\rho }_1\left(x\right)}}{q}_2\left(x\right)+r_2\left(x\right),$$

where $q_2\left(x\right),r_2\left(x\right)\in \mathfrak{R}\left[x\right]$ and $r_2\left(x\right)=0$ or $deg\left(r_2\left(x\right)\right)\le t_1-t_2-1$. Thus, we have

$$\begin{array}{cc}\hfill q_1\left(x\right)\ast \left(\varrho h_1\left(x\right){\rho }_1\left(x\right)\right|0)=& \left({\displaystyle \frac{x^{\alpha }-1}{h_1\left(x\right){\rho }_1\left(x\right)}}{q}_2\left(x\right)+r_2\left(x\right)\right)\ast \left(\varrho h_1\left(x\right){\rho }_1\left(x\right)\right|0)\hfill \\ \hfill =& r_2\left(x\right)\ast \left(\varrho h_1\left(x\right){\rho }_1\left(x\right)\right|0)\in Span\left({\mathit{F}}_2\right).\hfill \end{array}$$

Therefore, $p\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\in Span({\mathit{F}}_1\cup {\mathit{F}}_2)$.

Now, if $deg\left(q\left(x\right)\right)\le \beta -s_1-1$, then $q\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\in Span\left({\mathit{F}}_3\right)$. Otherwise, from the division algorithm,

$$q\left(x\right)=h_2\left(x\right)q_3\left(x\right)+r_3\left(x\right),$$

where $q_3\left(x\right),r_3\left(x\right)\in \mathfrak{R}\left[x\right]$ and $r_3\left(x\right)=0$ or $deg\left(r_3\left(x\right)\right)\le \beta -s_1-1$. Therefore, we have

$$\begin{array}{cc}\hfill q\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=& (h_2\left(x\right)q_3\left(x\right)+r_3\left(x\right))\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\hfill \\ \hfill =& q_3\left(x\right)\ast (h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right))+r_3\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)).\hfill \end{array}$$

If $deg\left(q_3\left(x\right)\right)\le s_1-s_2-1$, then $q_3\left(x\right)\ast (h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right))\in Span\left({\mathit{F}}_4\right)$. Otherwise, using the division algorithm we obtain

$$q_3\left(x\right)={\displaystyle \frac{x^{\beta }-1}{h_2\left(x\right){\rho }_2\left(x\right)}}{q}_4\left(x\right)+r_4\left(x\right),$$

where $q_4\left(x\right),r_4\left(x\right)\in \mathfrak{R}\left[x\right]$ and $r_4\left(x\right)=0$ or $deg\left(r_4\left(x\right)\right)\le s_1-s_2-1$. Therefore,

$$\begin{array}{cc}\hfill q_3\left(x\right)\ast (h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right))=& \left({\displaystyle \frac{x^{\beta }-1}{h_2\left(x\right){\rho }_2\left(x\right)}}{q}_4\left(x\right)+r_4\left(x\right)\right)\ast \left(h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right)\right)\hfill \\ \hfill =& q_4\left(x\right)\ast \left({\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)|0\right)+r_4\left(x\right)\ast \left(h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right)\right).\hfill \end{array}$$

From Lemma 3, $q_4\left(x\right)\ast \left({\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right)|0\right)\in Span({\mathit{F}}_1\cup {\mathit{F}}_2)$ and $r_4\left(x\right)\ast \left(h_2\left(x\right)ß\left(x\right)|\varrho h_2\left(x\right){\rho }_2\left(x\right)\right)$$\in Span\left({\mathit{F}}_4\right).$ Therefore, $\mathit{F}$ is a spanning set for $\mathfrak{C}$. Since, no element in ${\mathit{F}}_1\cup {\mathit{F}}_2\cup {\mathit{F}}_3\cup {\mathit{F}}_4$ is linearly dependent with other elements. Therefore, it is a minimal spanning set for $\mathfrak{C}$. Clearly, $\mathfrak{C}$ has $4^{2\alpha +2\beta -t_1-s_1-t_2-s_2}$ codewords.    □

Example 1.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length (9,7) over $\mathfrak{R}$, where ${\xi }_1\left(x\right)={\rho }_1\left(x\right)=1+x^3+x^6$, $ß\left(x\right)=\varrho +\varrho x^3$, ${\xi }_2\left(x\right)=1+x+x^2+x^4$ and ${\rho }_2\left(x\right)=1+x^2+x^3$. We also have $x^9-1={\xi }_1\left(x\right)h_1\left(x\right)$$\Rightarrow h_1\left(x\right)=1+x^3$ and $x^7-1={\xi }_2\left(x\right)h_2\left(x\right)$$\Rightarrow h_2\left(x\right)=1+x+x^3$. Then, from Theorem 4, the spanning set gives the following generator matrix for $\mathfrak{C}$:

$$\left[\begin{array}{cccccccccccccccc}1+\varrho & 0& 0& 1+\varrho & 0& 0& 1+\varrho & 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1+\varrho & 0& 0& 1+\varrho & 0& 0& 1+\varrho & 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1+\varrho & 0& 0& 1+\varrho & 0& 0& 1+\varrho & 0& 0& 0& 0& 0& 0& 0\\ \varrho & 0& 0& \varrho & 0& 0& 0& 0& 0& 1+\varrho & 1& 1+\varrho & \varrho & 1& 0& 0\\ 0& \varrho & 0& 0& \varrho & 0& 0& 0& 0& 0& 1+\varrho & 1& 1+\varrho & \varrho & 1& 0\\ 0& 0& \varrho & 0& 0& \varrho & 0& 0& 0& 0& 0& 1+\varrho & 1& 1+\varrho & \varrho & 1\\ \varrho & \varrho & 0& 0& \varrho & 0& \varrho & 0& 0& \varrho & \varrho & \varrho & \varrho & \varrho & \varrho & \varrho \end{array}\right]$$

The Gray image $\mathsf{\Gamma }\left(\mathfrak{C}\right)$ of $\mathfrak{C}$ is a linear code over ${\mathbb{F}}_4$ with parameters $[32,7,3]$.

We construct some examples of $\mathfrak{R}$-double cyclic codes in

Table 1. Some $\mathfrak{R}$-double cyclic code with optimal or near optimal Gray images.

Generators of $\mathfrak{C}$	$(\mathit{\alpha },\mathit{\beta })$	Parameter of $\mathsf{\Gamma }\left(\mathfrak{C}\right)$
${\xi }_1\left(x\right)=1+x^3={\rho }_1\left(x\right)$,$ß\left(x\right)=\varrho (w+x)$, ${\xi }_2\left(x\right)=1+x^3$, ${\rho }_2\left(x\right)=w+x$	$(3,3)$	$[12,2,8]$
${\xi }_1\left(x\right)=1+x^5={\rho }_1\left(x\right)$, $ß\left(x\right)=\varrho (1+wx+w{x}^2+x^3)$,	$(5,5)$	$[20,2,16]$ †
${\xi }_2\left(x\right)=1+x^5$, ${\rho }_2\left(x\right)=1+wx+w{x}^2+x^3$
${\xi }_1\left(x\right)=1+x^3={\rho }_1\left(x\right)$, $ß\left(x\right)=\varrho (w+x)$, ${\xi }_2\left(x\right)=1+x^5$	$(3,5)$	$[16,2,12]$ †
${\rho }_2\left(x\right)=1+wx+w{x}^2+x^3$
${\xi }_1\left(x\right)=1+x^9={\rho }_1\left(x\right)$, $ß\left(x\right)=\varrho {\displaystyle \frac{x^9-1}{x-1}}$, ${\xi }_2\left(x\right)=1+x^9$, ${\rho }_2\left(x\right)={\displaystyle \frac{x^{9}1}{x-1}}$	$(9,9)$	$[36,1,36]$ †
${\xi }_1\left(x\right)=1+x^{15}={\rho }_1\left(x\right)$, $ß\left(x\right)=\varrho (w+w{x}^2+w^2{x}^3+w{x}^4+w^2{x}^6+w^2{x}^7$
$+w^2{x}^8+w{x}^9+w{x}^{10}+w^2{x}^{11}+x^{12})$, ${\xi }_2\left(x\right)=1+x^{15}$, ${\rho }_2\left(x\right)=w+w{x}^2$	$(15,15)$	$[60,3,44]$
$+w^2{x}^3+w{x}^4+w^2{x}^6+w^2{x}^7+w^2{x}^8+w{x}^9+w{x}^{10}+w^2{x}^{11}+x^{12}$
${\xi }_1\left(x\right)=1+x^3={\rho }_1\left(x\right)$, $ß\left(x\right)=\varrho (w+x)$, ${\xi }_2\left(x\right)=1+x^{25}$,
${\rho }_2\left(x\right)=1+wx+w{x}^2+x^3+x^5+w{x}^6+w{x}^7+x^8+x^{10}+w{x}^{11}+w{x}^{12}$	$(3,25)$	$[56,2,44]$ †
$+x^{13}+x^{15}+w{x}^{16}+w{x}^{17}+x^{18}+x^{20}+w{x}^{21}+w{x}^{22}+x^{23}$
${\xi }_1\left(x\right)=1+x^{25}={\rho }_1\left(x\right)={\xi }_2\left(x\right)$, $ß\left(x\right)=\varrho (1+wx+w{x}^2+x^3+x^5+w{x}^6$
$+w{x}^7+x^8+x^{10}+w{x}^{11}+w{x}^{12}+x^{13}+x^{15}+w{x}^{16}+w{x}^{17}+x^{18}$
$+x^{20}+w{x}^{21}+w{x}^{22}+x^{23})$, ${\rho }_2\left(x\right)=1+wx+w{x}^2+x^3+x^5+w{x}^6$
$+w{x}^7+x^8+x^{10}+w{x}^{11}+w{x}^{12}+x^{13}+x^{15}+w{x}^{16}+w{x}^{17}+x^{18}+x^{20}$	$(25,25)$	$[100,2,80]$ †
$+w{x}^{21}+w{x}^{22}+x^{23}$

, whose Gray images give optimal or near optimal parameters over ${\mathbb{F}}_4$ according to Grassl’s table. We compare their parameters $[n,k,d]$ with the best-known bounds listed in Grassl’s tables of linear codes. The parameters with † represent optimal codes according to the online Grassl database http://www.codetables.de/ (accessed on 17 December 2025).

4. $\mathfrak{R}$-Double Cyclic Codes for the Construction of DNA Codes

Deoxyribonucleic acid or DNA is an acid found in almost every living organism, mostly found in the nucleus of the cell (eukaryotic cell), which contains the genetic information of the living organism. It is a sequence of two long polymers called strands, which are composed of four nucleotide bases, namely Adenine (A), Guanine (G), Thymine (T) and Cytosine (C). Two strands are very twisted, forming a double helix, running in opposite directions from each other and joined together by hydrogen bonds between nucleotide bases. This attachment follows the Watson–Crick Complement rule. A pairs with T and G pairs with C, as per the Watson–Crick Complement rule. A and G are called the complements of T and C, respectively, and vice versa. The complement of a base X is denoted by $\overline{X}$. $\overline{G}=C$, for instance, is the complement of G. Thus, if $X=AGATT$ is a DNA strand, then $\overline{X}=TCTAA$ would be its complement. According to the Watson–Crick complement rule, a DNA strand $Y=y_1{y}_2\dots y_l$ will pair up with $Y^{rc}=\overline{y_l}\overline{y_{l-1}}\dots \overline{y_2}\overline{y_1}$, the reverse-complement of Y. For instance, a DNA strand $5^{\prime }-TCTAAGT-3^{\prime }$ will pair up with $3^{\prime }-ACTTAGA-5^{\prime }$. A DNA code $\mathsf{𝒞}$ with minimum distance d may satisfy some or all the following constraints:

(a)

The Hamming constraint:  $d_H(s_1,s_2)\ge d$, where $s_1,s_2\in \mathsf{𝒞}$ and $s_1\ne s_2$.

(b)

The reverse constraint: $d_H(s_1,s_2^r)\ge d$ including $s_1=s_2$, where $s_1,s_2\in \mathsf{𝒞}$ and $s_2^r$ is the reverse of $s_2$.

(c)

The reverse-complement constraint: $d_H(s_1^r,s_2^c)\ge d$ including $s_1=s_2$, where $s_1,s_2\in \mathsf{𝒞}$ and $s_2^c$ is the complement of $s_2$.

(d)

The $GC$-content constraint: Each codeword $s\in \mathsf{𝒞}$ has the same number of G or C.

First, three constraints ensure that the probability of non-specific hybridization is reduced. The fixed $GC$-content constraint ensures the similar melting point.

We divide this section into two subsections. In the first subsection, we study reversible double cyclic codes over $\mathfrak{R}$, and in the next subsection, we study reversible-complement double cyclic codes over $\mathfrak{R}$.

4.1. Reversible Double Cyclic Codes

In this section, we mainly focus on the reversibility of the double cyclic code over $\mathfrak{R}$. For any vector $c=\left(b\right|b^{\prime })=(b_0,b_1,\dots ,b_{\alpha -1}|b_0^{\prime },b_1^{\prime },\dots ,b_{\beta -1}^{\prime })$ in ${\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$, the reverse $c^r$ of c is defined as $c^r=(b_{\alpha -1},b_{\alpha -2},\dots ,b_0|b_{\beta -1}^{\prime },b_{\beta -2}^{\prime },\dots ,b_0^{\prime })=\left(b^r\right|b^{\prime r})$.

Definition 2.

“A double cyclic code $\mathfrak{C}$ of length $(\alpha ,\beta )$ over $\mathfrak{R}$ is said to be reversible double cyclic code if for all $c\in \mathfrak{C}$, $c^r\in \mathfrak{C}$”.

Lemma 4

([17]). “Let $f\left(x\right),g\left(x\right)$ be two polynomials in $\mathfrak{R}\left[x\right]$ with $deg\left(g\right(x\left)\right)\le deg\left(f\right(x\left)\right)$. Then,

(a)

${\left(f\left(x\right)g\left(x\right)\right)}^{\ast }=f^{\ast }\left(x\right)g^{\ast }\left(x\right)$,

(b)

${(f\left(x\right)+g\left(x\right))}^{\ast }=f^{\ast }\left(x\right)+x^{deg\left(f\right(x\left)\right)-deg\left(g\right(x\left)\right)}{g}^{\ast }\left(x\right)$”.

Theorem 5

([9]). “Let $\mathfrak{C}=\langle \xi \left(x\right),\varrho \rho \left(x\right)\rangle$ be an $\mathfrak{R}$-cyclic code with odd length n as in Theorem 1. Then $\mathfrak{C}$ is reversible if and only if $\xi \left(x\right)$ and $\rho \left(x\right)$ both are self-reciprocal polynomials”.

Let us now find the reversibility conditions for separable double cyclic codes over $\mathfrak{R}$.

Theorem 6.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),\left(0\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a separable double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$, where ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. Then, $\mathfrak{C}$ is a reversible $\mathfrak{R}$-double cyclic code if and only if ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials.

Proof. 

Let $\mathfrak{C}$ be reversible. Then, the coordinate projections ${\mathfrak{C}}_{\alpha }\mathrm{and}{\mathfrak{C}}_{\beta }$ of $\mathfrak{C}$ are reversible cyclic codes of length $\alpha \mathrm{and}\beta$, respectively, over $\mathfrak{R}$. Therefore, ${\varphi }_{\alpha }\left(\mathfrak{C}\right)=\langle {\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)\rangle$ and ${\varphi }_{\beta }\left(\mathfrak{C}\right)=\langle {\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)\rangle$ are reversible cyclic codes over $\mathfrak{R}$ of respective length $\alpha$ and $\beta$. Hence, by Theorem 5, ${\xi }_1\left(x\right),{\rho }_1\left(x\right)$ and ${\xi }_2\left(x\right),{\rho }_2\left(x\right)$ are self-reciprocal polynomials.

Conversely, suppose that ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials. This is because $\mathfrak{C}={\mathfrak{C}}_{\alpha }\times {\mathfrak{C}}_{\beta }$, where ${\mathfrak{C}}_{\alpha }$ is a cyclic code of length $\alpha$ over $\mathfrak{R}$ generated by ${\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)$. Then, by Theorem 5, ${\mathfrak{C}}_{\alpha }$ is a reversible cyclic code. A similar argument yields that ${\mathfrak{C}}_{\beta }$ is a reversible cyclic code. Now, take any $c=\left(b\right|b^{\prime })\in \mathfrak{C}={\mathfrak{C}}_{\alpha }\times {\mathfrak{C}}_{\beta }$. Then, $b\in {\mathfrak{C}}_{\alpha }$ and $b^{\prime }\in {\mathfrak{C}}_{\beta }$; therefore, $b^r\in {\mathfrak{C}}_{\alpha }$ and $b^{\prime r}\in {\mathfrak{C}}_{\beta }$. Hence, $c^r=\left(b^r\right|b^{\prime r})\in {\mathfrak{C}}_{\alpha }\times {\mathfrak{C}}_{\beta }=\mathfrak{C}$.    □

The following two lemmas are true in our ring setting.

Lemma 5

([26]). “Let $v\left(x\right),w\left(x\right),\left(b\left(x\right)\right|d\left(x\right))\in {\displaystyle \frac{\mathfrak{R}\left[\mathfrak{x}\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[\mathfrak{x}\right]}{\langle x^{\beta }-1\rangle }}$ and $i\in \mathbb{N}$. Then

1.

${[v\left(x\right)+w\left(x\right)]}^r={\left[v\left(x\right)\right]}^r+{\left[w\left(x\right)\right]}^r$,

2.

${\left(x^{i}b\left(x\right)\right|x^{i}d\left(x\right))}^r=x^{(m+1)\alpha -1-deg\left(x^{i}b\left(x\right)\right)}\ast \left(b^{\ast }\left(x\right)\right|0)+x^{(n+1)\beta -1-deg\left(x^{i}d\left(x\right)\right)}\ast \left(0\right|d^{\ast }\left(x\right))$.

where $m,n$ are 0 or the smallest positive integers such that

$$\begin{array}{cc}\hfill & m\alpha -deg\left(x^{i}b\left(x\right)\right)+deg\left(\left[x^{i}b\left(x\right)\right]\mathrm{mod}(x^{\alpha }-1)\right)\ge 0\mathrm{and}\hfill \\ \hfill & n\beta -deg\left(x^{i}d\left(x\right)\right)+deg\left(\left[x^{i}d\left(x\right)\right]\mathrm{mod}(x^{\beta }-1)\right)\ge 0”.\hfill \end{array}$$

Lemma 6

([26]). “Let $\left(b\left(x\right)\right|d\left(x\right))\in {\displaystyle \frac{\mathfrak{R}\left[\mathfrak{x}\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[\mathfrak{x}\right]}{\langle x^{\beta }-1\rangle }}$ and $i\in \mathbb{N}$. Suppose $\beta =(2k+1)\alpha$ and $deg\left(d\right(x\left)\right)=2k\alpha +deg\left(b\right(x\left)\right)$, where $k\in \mathbb{N}\cup \left\{0\right\}$. Then,

$${\left(x^{i}b\left(x\right)\right|x^{i}d\left(x\right))}^r=x^{(M(2k+1)+1)\alpha -1-deg\left(x^{i}b\left(x\right)\right)}\ast \left(b^{\ast }\left(x\right)\right|d^{\ast }\left(x\right)).$$

where M is 0 or the smallest positive integer such that

$$\begin{array}{cc}\hfill & M\alpha -deg\left(x^{i}b\left(x\right)\right)+deg\left(\left[x^{i}b\left(x\right)\right]\mathrm{mod}(x^{\alpha }-1)\right)\ge 0\mathrm{and}\hfill \\ \hfill & M\beta -deg\left(x^{i}d\left(x\right)\right)+deg\left(\left[x^{i}d\left(x\right)\right]\mathrm{mod}(x^{\beta }-1)\right)\ge 0.”\hfill \end{array}$$

Theorem 7.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. If $\mathfrak{C}$ is reversible, then ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials.

Proof. 

Let $\mathfrak{C}$ be reversible. Then, ${\varphi }_{\beta }\left(\mathfrak{C}\right)=\langle {\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)\rangle$ is a reversible cyclic code of length $\beta$ over $\mathfrak{R}$. Hence, by Theorem 5, ${\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal. Since $J=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\rangle =ker\left({\varphi }_{\beta }\right)$ is an $\mathfrak{R}\left[x\right]$-submodule of ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}\times {\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\beta }-1\rangle }}$, there is a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$. If J is not reversible, then there exist some $w\left(x\right)\in J$ such that $w{\left(x\right)}^r\in \mathfrak{C}\setminus J$. Then, $w\left(x\right)=\lambda \left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)$ and $w{\left(x\right)}^r=\left(x^{\alpha -1-deg\left(B\right(x\left)\right)}{B}^{\ast }\left(x\right)\right|0)$, where $B\left(x\right)=\lambda \left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\mathrm{mod}(x^{\alpha }-1)$. Thus,

$$\begin{array}{cc}\hfill w{\left(x\right)}^r& =\left(x^{\alpha -1-deg\left(B\right(x\left)\right)}{B}^{\ast }\left(x\right)\right|0)\hfill \\ \hfill & ={\lambda }_1\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+{\lambda }_2\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)),\hfill \end{array}$$

where ${\lambda }_1\left(x\right)\mathrm{and}0\ne {\lambda }_2\left(x\right)\in \mathfrak{R}\left[x\right]$. Then, we have

$$x^{\alpha -1-deg\left(B\right(x\left)\right)}{B}^{\ast }\left(x\right)={\lambda }_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+{\lambda }_2\left(x\right)ß\left(x\right)\mathrm{mod}(x^{\alpha }-1),$$

(4)

$$0={\lambda }_2\left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1).$$

(5)

Since ${\lambda }_2\left(x\right)\ne 0$, ${\lambda }_2\left(x\right)=\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}{\mu }_1\left(x\right)$ or ${\lambda }_2\left(x\right)={\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{\mu }_2\left(x\right)$ for some ${\mu }_1\left(x\right),{\mu }_2\left(x\right)\in \mathfrak{R}\left[x\right]$. If ${\lambda }_2\left(x\right)=\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}{\mu }_1\left(x\right)$, then from (4) and Lemma 3, we have $x^{\alpha -1-deg\left(B\right(x\left)\right)}{B}^{\ast }\left(x\right)=q_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\mathrm{mod}(x^{\alpha }-1)$, for some $q_1\left(x\right)\in \mathfrak{R}\left[x\right]$; hence, $w{\left(x\right)}^r\in J$. Similarly, if ${\lambda }_2\left(x\right)={\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{\mu }_2\left(x\right)$, we obtain $w{\left(x\right)}^r\in J$. Hence, J is a reversible double cyclic code; therefore, ${\varphi }_{\alpha }\left(J\right)=\langle {\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)\rangle$ is a reversible cyclic code of length $\alpha$ over $\mathfrak{R}$. Thus, by Theorem 5, ${\xi }_1\left(x\right)\mathrm{and}{\rho }_1\left(x\right)$ are self-reciprocal.    □

Theorem 8.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta =(2k+1\left)\alpha \right)$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$, ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$, and $k\in \mathbb{N}\cup \left\{0\right\}$. Suppose that $deg({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=2k\alpha +deg(ß\left(x\right))$. If $\mathfrak{C}$ is reversible, then $({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})$ in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}$, where $f\left(x\right),p\left(x\right)\in {\mathbb{F}}_4\left[x\right]$ and $f\left(x\right){\xi }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}=1$, and $j=deg\left({\xi }_2\left(x\right)\right)-deg\left({\rho }_2\left(x\right)\right)$.

Proof. 

Let $\mathfrak{C}$ be reversible; then, ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right),{\rho }_2\left(x\right)$ are self-reciprocal polynomials. This is because $(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\in \mathfrak{C}$. Then,

$${(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^r=\left(x^{\alpha -1-deg(ß(x\left)\right)}{ß}^{\ast }\left(x\right)\right|x^{\beta -1-deg({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast })\in \mathfrak{C}.$$

This means

$$\begin{array}{cc}\hfill x^{2k\alpha +1+deg(ß(x\left)\right)}\ast & \left(x^{\alpha -1-deg(ß(x\left)\right)}{ß}^{\ast }\left(x\right)\right|x^{\beta -1-deg({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast })\hfill \\ \hfill & =\left({ß}^{\ast }\left(x\right)\right|{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast })\in \mathfrak{C}.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \left({ß}^{\ast }\left(x\right)\right|{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast })& =\left({ß}^{\ast }\left(x\right)\right|{\xi }_2\left(x\right)+\varrho x^j{\rho }_2\left(x\right))\hfill \\ \hfill & =q_1\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+q_2\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)),\hfill \end{array}$$

where $q_1\left(x\right),q_2\left(x\right)\in \mathfrak{R}\left[x\right]$ and $j=deg\left({\xi }_2\left(x\right)\right)-deg\left({\rho }_2\left(x\right)\right)$. Then, we have

$${ß}^{\ast }\left(x\right)=q_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+q_2\left(x\right)ß\left(x\right)\mathrm{mod}(x^{\alpha }-1),$$

(6)

$${\xi }_2\left(x\right)+\varrho x^j{\rho }_2\left(x\right)=q_2\left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1).$$

(7)

Notice that

$$\begin{array}{cc}\hfill \varrho {\xi }_2\left(x\right)& =\varrho ({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1)\mathrm{and}\hfill \\ \hfill {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}\varrho {\rho }_2\left(x\right)& ={\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1)\hfill \end{array}$$

and $gcd({\xi }_2\left(x\right),{\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})=1$. Therefore, $f\left(x\right){\xi }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}=1$ for some $f\left(x\right),p\left(x\right)\in {\mathbb{F}}_2\left[x\right]$. Also,

$$\begin{array}{cc}\hfill \varrho {\rho }_2\left(x\right)& =\varrho {\rho }_2\left(x\right)(f\left(x\right){\xi }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})\hfill \\ \hfill & =\varrho {\rho }_2\left(x\right)f\left(x\right){\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}.\hfill \end{array}$$

Hence, we have

$$\varrho {\rho }_2\left(x\right)=(\varrho f\left(x\right){\rho }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1).$$

Thus,

$$\varrho x^j{\rho }_2\left(x\right)=x^j(\varrho f\left(x\right){\rho }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1)\mathrm{and}$$

$$\begin{array}{cc}\hfill {\xi }_2\left(x\right)& ={\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)+\varrho {\rho }_2\left(x\right)\hfill \\ \hfill & =({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))+[\varrho f\left(x\right){\rho }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}]({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1)\hfill \\ \hfill & =({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))[1+\varrho f\left(x\right){\rho }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}]\mathrm{mod}(x^{\beta }-1).\hfill \end{array}$$

This implies that

$${\xi }_2\left(x\right)+\varrho x^j{\rho }_2\left(x\right)=({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))[1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}]\mathrm{mod}(x^{\beta }-1).$$

Then, from (7), we have

$$[q_2\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})]({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=0\mathrm{mod}(x^{\beta }-1).$$

This means that $[q_2\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})]=0$ or $\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}{\mu }_1\left(x\right)$ or ${\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{\mu }_2\left(x\right)$ for some ${\mu }_1\left(x\right),{\mu }_2\left(x\right)\in \mathfrak{R}\left[x\right]$.

If $q_2\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})=0$, then

$$q_2\left(x\right)=(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}).$$

Then, from (6), we obtain

$$\begin{array}{cc}\hfill {ß}^{\ast }\left(x\right)=& q_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})ß\left(x\right)\mathrm{mod}(x^{\alpha }-1)\hfill \\ \hfill {ß}^{\ast }\left(x\right)+ß\left(x\right)& (1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})=q_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\mathrm{mod}(x^{\alpha }-1).\hfill \end{array}$$

This means that

$$({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})\mathrm{in}{\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}.$$

If $q_2\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})=\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}{\mu }_1\left(x\right)$, then

$$q_2\left(x\right)ß\left(x\right)=\varrho {\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}ß\left(x\right){\mu }_1\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}).$$

Then, from (6) and part $\left(ii\right)$ of Lemma 3, we have

$$({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})\mathrm{in}{\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}.$$

Similarly, if $q_2\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})={\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}{\mu }_2\left(x\right)$, then

$$q_2\left(x\right)ß\left(x\right)={\displaystyle \frac{x^{\beta }-1}{{\rho }_2\left(x\right)}}ß\left(x\right){\mu }_2\left(x\right)+(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}).$$

Then, from (6) and part $\left(i\right)$ of Lemma 3, we have

$$({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})\mathrm{in}{\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}.$$

□

Theorem 9.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta =(2k+1\left)\alpha \right)$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$, ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$, and $k\in \mathbb{N}\cup \left\{0\right\}$. Suppose that $deg({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=2k\alpha +deg(ß\left(x\right))$. Then, $\mathfrak{C}$ is reversible if and only if ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials and $({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})$ in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}$, where $f\left(x\right),p\left(x\right)\in {\mathbb{F}}_4\left[x\right]$ and $f\left(x\right){\xi }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}=1$ and $j=deg\left({\xi }_2\left(x\right)\right)-deg\left({\rho }_2\left(x\right)\right)$.

Proof. 

Let $\mathfrak{C}$ be reversible; then, the result follows from Theorems 7 and 8.

Conversely, suppose that ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials, and $({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$ divides ${ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})$ in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}$, $f\left(x\right){\xi }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}=1$ and $j=deg\left({\xi }_2\left(x\right)\right)-deg\left({\rho }_2\left(x\right)\right)$. Then,

$${\xi }_2\left(x\right)+\varrho x^j{\rho }_2\left(x\right)=[1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}]({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\mathrm{mod}(x^{\beta }-1)$$

and

$${ß}^{\ast }\left(x\right)=h\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})\mathrm{mod}(x^{\alpha }-1),$$

for some $h\left(x\right)\in \mathfrak{R}\left[x\right]$. Then, we have

$$\begin{array}{cc}\hfill \left({ß}^{\ast }\left(x\right)\right|{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast })& =\left({ß}^{\ast }\left(x\right)\right|{\xi }_2\left(x\right)+\varrho x^j{\rho }_2\left(x\right))\hfill \\ \hfill & =(h\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+\hfill \\ & (1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}\left)\right|[1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}]\hfill \\ & ({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)))\hfill \\ \hfill & =h\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))+[(1+x^j)\hfill \\ & \varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}]\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)).\hfill \end{array}$$

This means that $\left({ß}^{\ast }\left(x\right)\right|{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast })\in \mathfrak{C}.$ Also, in Theorem 7 we proved that $ker\left({\varphi }_{\beta }\right)=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)\rangle$ is reversible; hence, we conclude that $\left({({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))}^{\ast }\right|0)\in \mathfrak{C}.$ Now, take any $w\left(x\right)\in \mathfrak{C}$; then, $w\left(x\right)=m_1\left(x\right)\ast ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0)+m_2\left(x\right)\ast (ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))$ for some $m_1\left(x\right),m_2\left(x\right)\in \mathfrak{R}\left[x\right]$. Then,

$$\begin{array}{cc}\hfill w{\left(x\right)}^r& ={(m_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))+m_2\left(x\right)ß\left(x\right)|m_2\left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)))}^r\hfill \\ \hfill & ={\left(m_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right|0)}^r+{(m_2\left(x\right)ß\left(x\right)|m_2\left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)))}^r.\hfill \end{array}$$

Now we have

$$\begin{array}{cc}\hfill {\left(m_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right|0)}^r& ={\left(\sum _{i=0}^e{c}_i{x}^i({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|0\right)}^r,\mathrm{where}{m}_1\left(x\right)=\sum _{i=0}^e{c}_i{x}^i\hfill \\ \hfill & =\sum _{i=0}^e{c}_i{\left(x^i({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|0\right)}^r.\hfill \end{array}$$

From Lemma 5, there exists $t_i\in \mathbb{N}\cup \left\{0\right\}$ such that

$${\left(x^i({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right|0)}^r=x^{(t_i+1)\alpha -1-deg\left(x^i({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right)}\ast \left({({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))}^{\ast }\right|0).$$

This means that ${\left(m_1\left(x\right)({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))\right|0)}^r\in \mathfrak{C}$, and

$$\begin{array}{cc}\hfill {(m_2\left(x\right)ß\left(x\right)|m_2\left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)))}^r& ={(\sum _{i=0}^o{s}_i{x}^iß\left(x\right)|\sum _{i=0}^o{s}_i{x}^i(g_2\left(x\right)+\varrho {\rho }_2\left(x\right)))}^r\hfill \\ \hfill & =\sum _{i=0}^o{s}_i{\left(x^{i}t\left(x\right)|x^i(g_2\left(x\right)+\varrho {\rho }_2\left(x\right))\right)}^r,\hfill \end{array}$$

where, $m_2\left(x\right)={\sum }_{i=0}^o{s}_i{x}^i$. By Lemma 6, there exists $M_i\in \mathbb{N}\cup \left\{0\right\}$ such that

$${\left(x^iß\left(x\right)|x^i(g_2\left(x\right)+\varrho {\rho }_2\left(x\right))\right)}^r=x^{(M_i(2k+1)+1)\alpha -1-deg(x^iß\left(x\right)}\ast \left({ß}^{\ast }\left(x\right)\right|{({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))}^{\ast }).$$

This implies that ${(m_2\left(x\right)ß\left(x\right)|m_2\left(x\right)({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right)))}^r\in \mathfrak{C}$; this means $w{\left(x\right)}^r\in \mathfrak{C}$. Hence, $\mathfrak{C}$ is reversible.    □

Notice that if we take $deg\left({\xi }_2\left(x\right)\right)=deg\left({\rho }_2\left(x\right)\right)$, then $j=0$ in the above Theorem 9. Then, $({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)$, since $deg\left({ß}^{\ast }\left(x\right)\right)\le deg(ß\left(x\right))\le deg({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))$. This means that ${ß}^{\ast }\left(x\right)+ß\left(x\right)=0\Rightarrow {ß}^{\ast }\left(x\right)=ß\left(x\right)$, i.e., $ß\left(x\right)$ is a self-reciprocal polynomial. Hence, we have the following corollary.

Corollary 3.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta =(2k+1\left)\alpha \right)$ over $\mathfrak{R}$ and $k\in \mathbb{N}\cup \left\{0\right\}$. Suppose that $deg\left({\xi }_2\left(x\right)\right)=deg\left({\rho }_2\left(x\right)\right)=2k\alpha +deg(ß\left(x\right))$. Then, $\mathfrak{C}$ is reversible if and only if ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right),{\rho }_2\left(x\right)\mathrm{and}ß\left(x\right)$ are self-reciprocal polynomials.

4.2. Reversible-Complement Double Cyclic Codes

In this section, we study reversible-complement double cyclic codes over $\mathfrak{R}$ and construct some DNA codes derived from double cyclic DNA codes. For any vector $w=\left(b\right|b^{\prime })=(b_0,b_1,\dots ,b_{\alpha -1}|b_0^{\prime },b_1^{\prime },\dots ,d_{\beta -1})\in {\mathfrak{R}}^{\alpha }\times {\mathfrak{R}}^{\beta }$, the complement of w is defined as $w^c=(\overline{b_0},\overline{b_1},\dots ,\overline{b_{\alpha -1}}|\overline{b_0^{\prime }},\overline{b_1^{\prime }},\dots ,\overline{b_{\beta -1}^{\prime }})$, i.e., $w^c=\left(b^c\right|b^{\prime c})$, and the reverse-complement is defined as $w^{rc}=(\overline{b_{\alpha -1}},\overline{b_{\alpha -2}},\dots ,\overline{b_0}|\overline{b_{\beta -1}^{\prime }},\overline{b_{\beta -2}^{\prime }},\dots ,\overline{b_0^{\prime }})$, i.e., $w^{rc}=\left(b^{rc}\right|b^{\prime rc})$. An $\mathfrak{R}$-double cyclic code $\mathfrak{C}$ of length ($\alpha$,$\beta$) is said to be a reversible-complement double cyclic code if $w^{rc}\in \mathfrak{C}$, whenever $w\in \mathfrak{C}$. In [27], the authors provided a one-to-one correspondence $\zeta$ between the elements of the ring $\mathfrak{R}$ and the set of all DNA double base pairs $S_{D_{16}}$, which we provided in

Table 2. Correspondence ($\zeta$) of DNA base pair with elements of the ring $\mathfrak{R}$.

Elements of $\mathfrak{R}$ $\left(\mathit{a}\right)$	DNA Pairs $\left(\zeta \right(\mathit{a}\left)\right)$	Elements of $\mathfrak{R}$ $\left(\mathit{a}\right)$	DNA Pairs $\left(\zeta \right(\mathit{a}\left)\right)$
0	$AA$	$w\varrho$	$CC$
1	$AT$	$1+w\varrho$	$CG$
w	$AC$	$w+w\varrho$	$CA$
w	$AG$	$w^2+w\varrho$	$CT$
$\varrho$	$TT$	$w^2\varrho$	$GG$
$1+\varrho$	$TA$	$1+w^2\varrho$	$GC$
$w+\varrho$	$TG$	$w+w^2\varrho$	$GT$
$w^2+\varrho$	$TC$	$w^2+w^2\varrho$	$GA$

.

Notice that in Table 2, the mapping $\zeta$ preserves the complementary property by adding $\varrho$, and when an element of $\mathfrak{R}$ is multiplied by $(1+\varrho )$, the corresponding DNA pair is reversed. Therefore, if $c_1{c}_2\dots c_n$ is a DNA sequence of n-tuple c in ${\mathfrak{R}}^n$, then the DNA sequence of $(1+\varrho )c$ is $c_n{c}_{n-1}\dots w_1$; hence, we conclude the following.

Lemma 7.

Let $c=(b_0,\dots ,b_{\alpha -1}|d_0,\dots ,d_{\beta -1})$ be any codeword of a double cyclic code $\mathfrak{C}$ of length $(\alpha ,\beta )$ over $\mathfrak{R}$ and $X=x_1{x}_2\dots x_{2\alpha }{y}_1{y}_2\dots y_{2\beta }$ is the DNA sequence corresponding to c. Then, the DNA sequence corresponding to the codeword $(1+\varrho )c^r$ is $x_{2\alpha }{x}_{2\alpha -1}\dots x_1{y}_{2\beta }{y}_{2\beta -1}\dots y_1$.

Definition 3.

A double cyclic code $\mathfrak{C}$ of length $(\alpha ,\beta )$ over $\mathfrak{R}$ is said to be a double cyclic DNA code of length $(\alpha ,\beta )$ over $\mathfrak{R}$ if for any $w\in \mathfrak{C},w\ne w^{rc}$, $w^{rc}\in \mathfrak{C}$.

For an $\mathfrak{R}$-linear code $\mathfrak{C}$ of length $n=\alpha +\beta$, we have $0=(0,0,\dots ,0)\in \mathfrak{C}$. If $\mathfrak{C}$ is a reversible-complement $\mathfrak{R}$-double cyclic code of length ($\alpha$,$\beta$), then $0^{rc}={(0,\dots ,0|0,\dots ,0)}^{rc}\in \mathfrak{C}$, i.e., ${(0,\dots ,0|0,\dots ,0)}^{rc}=(\varrho ,\dots ,\varrho |\varrho ,\dots ,\varrho )\in \mathfrak{C}$. This fact gives us the following necessary and sufficient condition for an $\mathfrak{R}$-double cyclic code of length ($\alpha$,$\beta$) to be a reversible-complement $\mathfrak{R}$-double cyclic code.

Theorem 10.

Let $\mathfrak{C}$ be a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$. Then, $\mathfrak{C}$ is a reversible-complement double cyclic code if and only if

(a)

$\mathfrak{C}$ is a reversible double cyclic code and

(b)

$\varrho \mathbf{I}=(\stackrel{\alpha }{\overbrace{\varrho ,\dots ,\varrho }}|\stackrel{\beta }{\overbrace{\varrho ,\dots ,\varrho }})\in \mathfrak{C}$, where $\mathbf{I}$ is all one vector, i.e., $\mathbf{I}=(\stackrel{\gamma }{\overbrace{1,\dots ,1}}|\stackrel{\delta }{\overbrace{1,\dots ,1}})$.

Proof. 

Let $\mathfrak{C}$ be a reversible-complement code. Since $(\stackrel{\alpha }{\overbrace{0,\dots ,0}}|\stackrel{\beta }{\overbrace{0,\dots ,0}})\in \mathfrak{C}$, then we have ${(\stackrel{\alpha }{\overbrace{0,\dots ,0}}|\stackrel{\beta }{\overbrace{0,\dots ,0}})}^{rc}\in \mathfrak{C}$. Then,

$$\begin{array}{cc}\hfill {(\stackrel{\alpha }{\overbrace{0,\dots ,0}}|\stackrel{\beta }{\overbrace{0,\dots ,0}})}^{rc}& =(\stackrel{\alpha }{\overbrace{\overline{0},\dots ,\overline{0}}}|\stackrel{\beta }{\overbrace{\overline{0},\dots ,\overline{0}}})\hfill \\ \hfill & =(\stackrel{\alpha }{\overbrace{0+\varrho ,\dots ,0+\varrho }}|\stackrel{\beta }{\overbrace{0+\varrho ,\dots ,0+\varrho }})\hfill \\ \hfill & =(\stackrel{\alpha }{\overbrace{0,\dots ,0}}|\stackrel{\beta }{\overbrace{0,\dots ,0}})+(\stackrel{\alpha }{\overbrace{\varrho ,\dots ,\varrho }}|\stackrel{\beta }{\overbrace{\varrho ,\dots ,\varrho }}).\hfill \end{array}$$

Since ${(\stackrel{\alpha }{\overbrace{0,\dots ,0}}|\stackrel{\beta }{\overbrace{0,\dots ,0}})}^{rc}$$\in \mathfrak{C}$ and $(\stackrel{\alpha }{\overbrace{0,\dots ,0}}|\stackrel{\beta }{\overbrace{0,\dots ,0}})\in \mathfrak{C}$, this means that $(\stackrel{\alpha }{\overbrace{\varrho ,\dots ,\varrho }}|\stackrel{\beta }{\overbrace{\varrho ,\dots ,\varrho }})=\varrho \mathbf{I}\in \mathfrak{C}$. Let any $w=(b_0,b_1,\dots ,b_{\alpha -1}|b_0^{\prime },b_1^{\prime }\dots ,b_{\beta -1}^{\prime })\in \mathfrak{C}$. Then, $w^{rc}\in \mathfrak{C}$. Then,

$$\begin{array}{cc}\hfill w^{rc}& =(\overline{b_{\alpha -1}},\overline{b_{\alpha -2}},\dots ,\overline{b_0}|\overline{b_{\beta -1}^{\prime }},\overline{b_{\beta -2}^{\prime }},\dots ,\overline{b_0^{\prime }})\hfill \\ \hfill & =(b_{\alpha -1}+\varrho ,b_{\alpha -2}+\varrho ,\dots ,b_0+\varrho |b_{\beta -1}^{\prime }+\varrho ,b_{\beta -2}^{\prime }+\varrho ,\dots ,b_0^{\prime }+\varrho )\hfill \\ \hfill & =(b_{\alpha -1},b_{\alpha -2},\dots ,b_0|b_{\beta -1}^{\prime },b_{\beta -2}^{\prime },\dots ,b_0^{\prime })+(\stackrel{\alpha }{\overbrace{\varrho ,\dots ,\varrho }}|\stackrel{\beta }{\overbrace{\varrho ,\dots ,\varrho }})\hfill \\ \hfill & =w^r+\varrho \mathbf{I}.\hfill \end{array}$$

Since $w^{rc},(1+u)\mathbf{I}\in \mathfrak{C}$, this means $w^r\in \mathfrak{C}$; hence, $\mathfrak{C}$ is reversible.

Conversely, suppose that $\mathfrak{C}$ is reversible and $\varrho \mathbf{I}\in \mathfrak{C}$; then, any $w\in \mathfrak{C}$ implies $w^r\in \mathfrak{C}$ and we see that $w^{rc}=w^r+\varrho \mathbf{I}$. This means for any $w\in \mathfrak{C}$, $w^{rc}\in \mathfrak{C}$; hence, $\mathfrak{C}$ is reversible-complement.    □

Theorem 11.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),\left(0\right|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a separable double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$, where ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$ and ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$. Then, $\mathfrak{C}$ is a reversible-complement double cyclic code if and only if ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials and $\varrho \mathbf{I}\in \mathfrak{C}$.

Proof. 

From Theorems 6 and 10, we obtain the result.    □

For non-separable codes, we have the following results:

Theorem 12.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta =(2k+1\left)\alpha \right)$ over $\mathfrak{R}$, where $ß\left(x\right)\in \mathfrak{R}\left[x\right]$, ${\rho }_i\left(x\right)\mathrm{and}{\xi }_i\left(x\right)$ are polynomials in ${\mathbb{F}}_4\left[x\right]$, for $i=1,2$, such that ${\rho }_1\left(x\right)\left|{\xi }_1\left(x\right)\right|(x^{\alpha }-1)$, ${\rho }_2\left(x\right)\left|{\xi }_2\left(x\right)\right|(x^{\beta }-1)$ and $k\in \mathbb{N}\cup \left\{0\right\}$. Suppose that $deg({\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))=2k\alpha +deg(ß\left(x\right))$. Then, $\mathfrak{C}$ is reversible-complement if and only if

(i)

${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right)\mathrm{and}{\rho }_2\left(x\right)$ are self-reciprocal polynomials,

(ii)

$({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right))|{ß}^{\ast }\left(x\right)+ß\left(x\right)(1+(1+x^j)\varrho f\left(x\right){\rho }_2\left(x\right)+(1+x^j)p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}})$ in ${\displaystyle \frac{\mathfrak{R}\left[x\right]}{\langle x^{\alpha }-1\rangle }}$, where $f\left(x\right),p\left(x\right)\in {\mathbb{F}}_2\left[x\right]$ and $f\left(x\right){\xi }_2\left(x\right)+p\left(x\right){\displaystyle \frac{x^{\beta }-1}{{\xi }_2\left(x\right)}}=1$ and $j=deg\left({\xi }_2\left(x\right)\right)-deg\left({\rho }_2\left(x\right)\right)$ and

(iii)

$\varrho \mathbf{I}\in \mathfrak{C}$.

Proof. 

Theorems 9 and 10 directly establish the result.    □

Corollary 4.

Let $\mathfrak{C}=\langle ({\xi }_1\left(x\right)+\varrho {\rho }_1\left(x\right)|0),(ß\left(x\right)|{\xi }_2\left(x\right)+\varrho {\rho }_2\left(x\right))\rangle$ be a double cyclic code of length $(\alpha ,\beta =(2k+1\left)\alpha \right)$ over $\mathfrak{R}$ and $k\in \mathbb{N}\cup \left\{0\right\}$. Suppose that $deg\left({\xi }_2\left(x\right)\right)=deg\left({\rho }_2\left(x\right)\right)=2k\alpha +deg(ß\left(x\right))$. Then, $\mathfrak{C}$ is reversible-complement if and only if ${\xi }_1\left(x\right),{\rho }_1\left(x\right),{\xi }_2\left(x\right),{\rho }_2\left(x\right)\mathrm{and}ß\left(x\right)$ are self-reciprocal polynomials and $\varrho \mathbf{I}\in \mathfrak{C}$.

Let $\mathfrak{C}$ be a double cyclic code of length $(\alpha ,\beta )$ over $\mathfrak{R}$ and $S_{D_4}$ be the set of all DNA bases. Recall that $\theta$ shows the correspondence between $\mathfrak{R}$ and $S_{D_4}$. Now, consider a correspondence

$$\begin{array}{cc}\hfill \mathsf{\Theta }:\mathfrak{C}\to & S_{D_{16}}^{\alpha +\beta },\mathrm{defined}\mathrm{by}\hfill \\ \hfill \mathsf{\Theta }(b_0,b_1,\dots ,b_{\alpha -1}|b_0^{\prime },b_1^{\prime },\dots ,b_{\beta -1}^{\prime })& =(\theta \left(b_0\right),\theta \left(b_1\right),\dots ,\theta \left(b_{\alpha -1}\right),\theta \left(b_0^{\prime }\right),\theta \left(b_1^{\prime }\right),\dots ,\theta \left(b_{\beta -1}^{\prime }\right)),\hfill \end{array}$$

$(b_0,b_1,\dots ,b_{\alpha -1}|b_0^{\prime },b_1^{\prime },\dots ,b_{\beta -1}^{\prime })\in \mathfrak{C}$. For any double cyclic code $\mathfrak{C}$ of length $(\alpha ,\beta )$ over $\mathfrak{R}$, consider a set

$$\mathfrak{B}=\{\left(b^{\prime }\right|b)\in {\mathfrak{R}}^{\beta }\times {\mathfrak{R}}^{\alpha }|\left(b\right|b^{\prime })\in \mathfrak{C}\}\subseteq {\mathfrak{R}}^{\beta }\times {\mathfrak{R}}^{\alpha }.$$

Then, a DNA code $\mathfrak{D}$ is generated from $\mathfrak{C}\cup \mathfrak{B}$.

Example 2.

Let $\mathfrak{C}=\langle \left(\varrho (x^2+x+1)\right|0),\left(0\right|\varrho (x^2+wx+1)(x^2+w^{2}x+1))\rangle$ be a double cyclic code of length (3,5) over $\mathfrak{R}$. Since $\varrho \mathbf{I}=\left(\varrho (x^2+x+1)\right|0)+\left(0\right|\varrho (x^2+wx+1)(x^2+w^{2}x+1))\in \mathfrak{C}$, by Theorem 11, $\mathfrak{C}$ is a reversible-complement double cyclic code of length $(3,5)$. The DNA code $\mathfrak{D}$ of length 16 with minimum distance 4 obtained from $\mathfrak{C}\cup \mathfrak{B}$ has 14 codewords, given in

Table 3. DNA code $\mathfrak{D}$ of length 16 obtained from $\mathfrak{C}\cup \mathfrak{B}$.

$\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}$	$\mathit{T}\mathit{T}\mathit{T}\mathit{T}\mathit{T}\mathit{T}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{A}$
$AAAAAAGGGGGGGGGG$	$GGGGGGGGGGAAAAAA$
$AAAAAATTTTTTTTTT$	$TTTTTTTTTTAAAAAA$
$AAAAAACCCCCCCCCC$	$CCCCCCCCCCAAAAAA$
$TTTTTTGGGGGGGGGG$	$AAAAAAAAAATTTTTT$
$TTTTTTTTTTTTTTTT$	$GGGGGGGGGGTTTTTT$
$TTTTTTCCCCCCCCCC$	$CCCCCCCCCCTTTTTT$

.

Example 3.

Let $\mathfrak{C}=\langle \left(\varrho (x^2+x+1)\right|0),(x^2+x+1|x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)\rangle$ be an $\mathfrak{R}$-double cyclic code of length $(3,9)$. Then, $\varrho \mathbf{I}=\varrho (x^2+x+1|x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)\in \mathfrak{C}$. Hence, by Theorem 11, $\mathfrak{C}$ is a reversible-complement double cyclic code of length $(3,9)$. The DNA code $\mathfrak{D}$ of length 24 with minimum distance 6 obtained from $\mathsf{𝒞}\cup \mathcal{B}$ has 48 codewords, given in

Table 4. DNA code $\mathfrak{D}$ obtained from $\mathfrak{C}\cup \mathfrak{B}$.

$\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}$	$\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}\mathit{C}\mathit{G}$
$GAGAGAGAGAGAGAGAGAGAGAGA$	$AGAGAGAGAGAGAGAGAGAGAGAG$
$GTGTGTGTGTGTGTGTGTGTGTGT$	$TGTGTGTGTGTGTGTGTGTGTGTG$
$GGGGGGGGGGGGGGGGGGGGGGGG$	$AAAAAAAAAAAAAAAAAAAAAAAA$
$TATATATATATATATATATATATA$	$ATATATATATATATATATATATAT$
$TCTCTCTCTCTCTCTCTCTCTCTC$	$CTCTCTCTCTCTCTCTCTCTCTCT$
$CACACACACACACACACACACACA$	$ACACACACACACACACACACACAC$
$TTTTTTTTTTTTTTTTTTTTTTTT$	$CCCCCCCCCCCCCCCCCCCCCCCC$
$CTCTCTGAGAGAGAGAGAGAGAGA$	$AGAGAGAGAGAGAGAGAGTCTCTC$
$CGCGCGGCGCGCGCGCGCGCGCGC$	$CGCGCGCGCGCGCGCGCGGCGCGC$
$CACACAGTGTGTGTGTGTGTGTGT$	$TGTGTGTGTGTGTGTGTGACACAC$
$CCCCCCGGGGGGGGGGGGGGGGGG$	$GGGGGGGGGGGGGGGGGGCCCCCC$
$ATATATTATATATATATATATATA$	$ATATATATATATATATATTATATA$
$AGAGAGTCTCTCTCTCTCTCTCTC$	$CTCTCTCTCTCTCTCTCTGAGAGA$
$ACACACTGTGTGTGTGTGTGTGTG$	$GTGTGTGTGTGTGTGTGTCACACA$
$TCTCTCAGAGAGAGAGAGAGAGAG$	$GAGAGAGAGAGAGAGAGACTCTCT$
$GCGCGCCGCGCGCGCGCGCGCGCG$	$GCGCGCGCGCGCGCGCGCCGCGCG$
$GAGAGACTCTCTCTCTCTCTCTCT$	$TCTCTCTCTCTCTCTCTCAGAGAG$
$GTGTGTCACACACACACACACACA$	$ACACACACACACACACACTGTGTG$
$TTTTTTAAAAAAAAAAAAAAAAAA$	$AAAAAAAAAAAAAAAAAATTTTTT$
$AAAAAATTTTTTTTTTTTTTTTTT$	$TTTTTTTTTTTTTTTTTTAAAAAA$
$GGGGGGCCCCCCCCCCCCCCCCCC$	$CCCCCCCCCCCCCCCCCCGGGGGG$
$TGTGTGACACACACACACACACAC$	$CACACACACACACACACAGTGTGT$
$TATATAATATATATATATATATAT$	$TATATATATATATATATAATATAT$

.

Example 4.

Let $\mathfrak{C}=\langle \left(\varrho (x^2+x+1)\right|0),\left(\varrho \right|\varrho (x^6+x^3+1))\rangle$ be a double cyclic code of length (3,9) over $\mathcal{R}$. Then, $\varrho \mathbf{I}=\varrho (x^2+x+1|x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)\in \mathfrak{C}$. Hence, by Theorem 12, $\mathfrak{C}$ is a reversible-complement double cyclic code of length $(3,9)$. The DNA code $\mathfrak{D}$ of length 24 with minimum distance 6 obtained from $\mathfrak{C}\cup \mathfrak{B}$ has 190 words, given in

Table 5. DNA code $\mathfrak{D}$ of length 24 obtained from $\mathfrak{C}\cup \mathfrak{B}$.

$\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}$	$\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}\mathit{A}\mathit{A}\mathit{A}\mathit{A}\mathit{G}\mathit{G}$
$AAGGAAAAGGAAAAGGAAAAGGAA$	$GGAAGGGGAAGGGGAAGGGGAAGG$
$GGGGAAGGGGAAGGGGAAGGGGAA$	$AAGGGGAAGGGGAAGGGGAAGGGG$
$TTGGAATTGGAATTGGAATTGGAA$	$AAGGTTAAGGTTAAGGTTAAGGTT$
$CCGGAACCGGAACCGGAACCGGAA$	$AAGGCCAAGGCCAAGGCCAAGGCC$
$GGGGGGGGGGGGGGGGGGGGGGGG$	$GGTTGGGGTTGGGGTTGGGGTTGG$
$TTGGGGTTGGGGTTGGGGTTGGGG$	$GGGGTTGGGGTTGGGGTTGGGGTT$
$CCGGGGCCGGGGCCGGGGCCGGGG$	$GGGGCCGGGGCCGGGGCCGGGGCC$
$TTAAGGTTAAGGTTAAGGTTAAGG$	$GGAATTGGAATTGGAATTGGAATT$
$AATTGGAATTGGAATTGGAATTGG$	$GGTTAAGGTTAAGGTTAAGGTTAA$
$TTTTGGTTTTGGTTTTGGTTTTGG$	$GGTTTTGGTTTTGGTTTTGGTTTT$
$CCTTGGCCTTGGCCTTGGCCTTGG$	$GGTTCCGGTTCCGGTTCCGGTTCC$
$CCAAGGCCAAGGCCAAGGCCAAGG$	$GGAACCGGAACCGGAACCGGAACC$
$AACCGGAACCGGAACCGGAACCGG$	$GGCCAAGGCCAAGGCCAAGGCCAA$
$GGCCGGGGCCGGGGCCGGGGCCGG$	$AAAAAAAAAAAAAAAAAAAAAAAA$
$TTCCGGTTCCGGTTCCGGTTCCGG$	$GGCCTTGGCCTTGGCCTTGGCCTT$
$CCCCGGCCCCGGCCCCGGCCCCGG$	$GGCCCCGGCCCCGGCCCCGGCCCC$
$TTAAAATTAAAATTAAAATTAAAA$	$AAAATTAAAATTAAAATTAAAATT$
$AATTAAAATTAAAATTAAAATTAA$	$TTGGTTTTGGTTTTGGTTTTGGTT$
$TTTTAATTTTAATTTTAATTTTAA$	$AATTTTAATTTTAATTTTAATTTT$
$CCTTAACCTTAACCTTAACCTTAA$	$AATTCCAATTCCAATTCCAATTCC$
$CCGGTTCCGGTTCCGGTTCCGGTT$	$TTGGCCTTGGCCTTGGCCTTGGCC$
$TTAATTTTAATTTTAATTTTAATT$	$TTTTTTTTTTTTTTTTTTTTTTTT$
$CCTTTTCCTTTTCCTTTTCCTTTT$	$TTTTCCTTTTCCTTTTCCTTTTCC$
$CCAATTCCAATTCCAATTCCAATT$	$TTAACCTTAACCTTAACCTTAACC$
$AACCTTAACCTTAACCTTAACCTT$	$TTCCAATTCCAATTCCAATTCCAA$
$TTCCTTTTCCTTTTCCTTTTCCTT$	$AACCAAAACCAAAACCAAAACCAA$
$CCCCTTCCCCTTCCCCTTCCCCTT$	$TTCCCCTTCCCCTTCCCCTTCCCC$
$CCAAAACCAAAACCAAAACCAAAA$	$AAAACCAAAACCAAAACCAAAACC$
$CCCCAACCCCAACCCCAACCCCAA$	$AACCCCAACCCCAACCCCAACCCC$
$CCGGCCCCGGCCCCGGCCCCGGCC$	$CCTTCCCCTTCCCCTTCCCCTTCC$
$CCAACCCCAACCCCAACCCCAACC$	$CCCCCCCCCCCCCCCCCCCCCCCC$
$CCTTTTGGAAAAGGAAAAGGAAAA$	$AAAAGGAAAAGGAAAAGGTTTTCC$
$TTCCTTAAGGAAAAGGAAAAGGAA$	$AAGGAAAAGGAAAAGGAATTCCTT$
$CCCCTTGGGGAAGGGGAAGGGGAA$	$AAGGGGAAGGGGAAGGGGTTCCCC$
$AACCTTTTGGAATTGGAATTGGAA$	$AAGGTTAAGGTTAAGGTTTTCCAA$
$GGCCTTCCGGAACCGGAACCGGAA$	$AAGGCCAAGGCCAAGGCCTTCCGG$
$TTTTCCAAAAGGAAAAGGAAAAGG$	$GGAAAAGGAAAAGGAAAACCTTTT$
$CCTTCCGGAAGGGGAAGGGGAAGG$	$GGAAGGGGAAGGGGAAGGCCTTCC$
$TTCCCCAAGGGGAAGGGGAAGGGG$	$GGGGAAGGGGAAGGGGAACCCCTT$
$CCCCCCGGGGGGGGGGGGGGGGGG$	$GGGGGGGGGGGGGGGGGGCCCCCC$
$AACCCCTTGGGGTTGGGGTTGGGG$	$GGGGTTGGGGTTGGGGTTCCCCAA$
$GGCCCCCCGGGGCCGGGGCCGGGG$	$GGGGCCGGGGCCGGGGCCCCCCGG$
$AATTCCTTAAGGTTAAGGTTAAGG$	$GGAATTGGAATTGGAATTCCTTAA$
$TTAACCAATTGGAATTGGAATTGG$	$GGTTAAGGTTAAGGTTAACCAATT$
$CCAACCGGTTGGGGTTGGGGTTGG$	$GGTTGGGGTTGGGGTTGGCCAACC$
$AAAACCTTTTGGTTTTGGTTTTGG$	$GGTTTTGGTTTTGGTTTTCCAAAA$
$GGAACCCCTTGGCCTTGGCCTTGG$	$GGTTCCGGTTCCGGTTCCCCAAGG$
$GGTTCCCCAAGGCCAAGGCCAAGG$	$GGAACCGGAACCGGAACCCCTTGG$
$TTGGCCAACCGGAACCGGAACCGG$	$GGCCAAGGCCAAGGCCAACCGGTT$
$CCGGCCGGCCGGGGCCGGGGCCGG$	$GGCCGGGGCCGGGGCCGGCCGGCC$
$AAGGCCTTCCGGTTCCGGTTCCGG$	$GGCCTTGGCCTTGGCCTTCCGGAA$
$GGGGCCCCCCGGCCCCGGCCCCGG$	$GGCCCCGGCCCCGGCCCCCCGGGG$
$TTTTTTAAAAAAAAAAAAAAAAAA$	$AAAAAAAAAAAAAAAAAATTTTTT$
$AATTTTTTAAAATTAAAATTAAAA$	$AAAATTAAAATTAAAATTTTTTAA$
$CCAATTGGTTAAGGTTAAGGTTAA$	$AATTGGAATTGGAATTGGTTAACC$
$TTAATTAATTAAAATTAAAATTAA$	$AATTAAAATTAAAATTAATTAATT$
$AAAATTTTTTAATTTTAATTTTAA$	$AATTTTAATTTTAATTTTTTAAAA$
$GGAATTCCTTAACCTTAACCTTAA$	$AATTCCAATTCCAATTCCTTAAGG$
$CCTTAAGGAATTGGAATTGGAATT$	$TTAAGGTTAAGGTTAAGGAATTCC$
$CCCCAAGGGGTTGGGGTTGGGGTT$	$TTGGGGTTGGGGTTGGGGAACCCC$
$TTCCAAAAGGTTAAGGTTAAGGTT$	$TTGGAATTGGAATTGGAAAACCTT$
$AACCAATTGGTTTTGGTTTTGGTT$	$TTGGTTTTGGTTTTGGTTAACCAA$
$GGCCAACCGGTTCCGGTTCCGGTT$	$TTGGCCTTGGCCTTGGCCAACCGG$
$TTTTAAAAAATTAAAATTAAAATT$	$TTAAAATTAAAATTAAAAAATTTT$
$AATTAATTAATTTTAATTTTAATT$	$TTAATTTTAATTTTAATTAATTAA$
$CCAAAAGGTTTTGGTTTTGGTTTT$	$TTTTGGTTTTGGTTTTGGAAAACC$
$TTAAAAAATTTTAATTTTAATTTT$	$TTTTAATTTTAATTTTAAAAAATT$
$GGAAAACCTTTTCCTTTTCCTTTT$	$TTTTCCTTTTCCTTTTCCAAAAGG$
$GGTTAACCAATTCCAATTCCAATT$	$TTAACCTTAACCTTAACCAATTGG$
$CCGGAAGGCCTTGGCCTTGGCCTT$	$TTCCGGTTCCGGTTCCGGAAGGCC$
$TTGGAAAACCTTAACCTTAACCTT$	$TTCCAATTCCAATTCCAAAAGGTT$
$AAGGAATTCCTTTTCCTTTTCCTT$	$TTCCTTTTCCTTTTCCTTAAGGAA$
$GGGGAACCCCTTCCCCTTCCCCTT$	$TTCCCCTTCCCCTTCCCCAAGGGG$
$GGTTTTCCAAAACCAAAACCAAAA$	$AAAACCAAAACCAAAACCTTTTGG$
$CCGGTTGGCCAAGGCCAAGGCCAA$	$AACCGGAACCGGAACCGGTTGGCC$
$AAGGTTTTCCAATTCCAATTCCAA$	$AACCTTAACCTTAACCTTTTGGAA$
$TTGGTTAACCAAAACCAAAACCAA$	$AACCAAAACCAAAACCAATTGGTT$
$GGGGTTCCCCAACCCCAACCCCAA$	$AACCCCAACCCCAACCCCTTGGGG$
$CCTTGGGGAACCGGAACCGGAACC$	$CCAAGGCCAAGGCCAAGGGGTTCC$
$CCCCGGGGGGCCGGGGCCGGGGCC$	$CCGGGGCCGGGGCCGGGGGGCCCC$
$AACCGGTTGGCCTTGGCCTTGGCC$	$CCGGTTCCGGTTCCGGTTGGCCAA$
$TTCCGGAAGGCCAAGGCCAAGGCC$	$CCGGAACCGGAACCGGAAGGCCTT$
$GGCCGGCCGGCCCCGGCCCCGGCC$	$CCGGCCCCGGCCCCGGCCGGCCGG$
$AATTGGTTAACCTTAACCTTAACC$	$CCAATTCCAATTCCAATTGGTTAA$
$CCAAGGGGTTCCGGTTCCGGTTCC$	$CCTTGGCCTTGGCCTTGGGGAACC$
$AAAAGGTTTTCCTTTTCCTTTTCC$	$CCTTTTCCTTTTCCTTTTGGAAAA$
$TTAAGGAATTCCAATTCCAATTCC$	$CCTTAACCTTAACCTTAAGGAATT$
$GGAAGGCCTTCCCCTTCCCCTTCC$	$CCTTCCCCTTCCCCTTCCGGAAGG$
$TTTTGGAAAACCAAAACCAAAACC$	$CCAAAACCAAAACCAAAAGGTTTT$
$GGTTGGCCAACCCCAACCCCAACC$	$CCAACCCCAACCCCAACCGGTTGG$
$CCGGGGGGCCCCGGCCCCGGCCCC$	$CCCCGGCCCCGGCCCCGGGGGGCC$
$AAGGGGTTCCCCTTCCCCTTCCCC$	$CCCCTTCCCCTTCCCCTTGGGGAA$
$TTGGGGAACCCCAACCCCAACCCC$	$CCCCAACCCCAACCCCAAGGGGTT$
$GGGGGGCCCCCCCCCCCCCCCCCC$	$CCCCCCCCCCCCCCCCCCGGGGGG$

.

5. Conclusions

In this paper, we have studied the structure of double cyclic codes of length $(\alpha ,\beta )$ over the ring $\mathfrak{R}={\mathbb{F}}_4+\varrho {\mathbb{F}}_4,{\varrho }^2=0$ when both $\alpha$ and $\beta$ are odd integers. We have also studied the reversibility of double cyclic codes of length $(\alpha ,\beta =(2k+1\left)\alpha \right)$, where both $\alpha$ and $\beta$ are odd integers. Moreover, we have studied the reversible-complement codes over the ring $\mathfrak{R}$ that are suitable for DNA code construction. Further, we have provided some examples of these codes. For future studies, it will be interesting to study double cyclic codes over different rings with different lengths and extend these studies to DNA code construction and DNA computing.