On reversibility problem in DNA bases over a class of rings

Abstract

Let ${\mathcal{R}}_k={\mathbb{Z}}_4[u_1,u_2,\dots ,u_k]/⟨u_i^2-u_i,u_i{u}_j-u_j{u}_i⟩$ be a non-chain ring of characteristic 4, where $1\le i,j\le k$ and $k\ge 1$. In this article, we discuss reversible cyclic codes of odd lengths over the ring ${\mathcal{R}}_k$. We construct bijections between the elements of the ring ${\mathcal{R}}_k$ and DNA-$2^k$ bases for k = 1, 2 in such a manner that the reversibility problem is solved. Employing these bijections, reversible complement cyclic codes of odd lengths are generated. Furthermore, we construct a Gray map $\mathrm{\Phi }:{\mathcal{R}}_k^n\to {\mathbb{Z}}_4^{2^{k}n}$ and as an application of the Gray map Φ, we obtain the GC-content of cyclic codes of arbitrary odd length over the ring ${\mathcal{R}}_k$. Meanwhile, we provide some examples of reversible cyclic codes of odd lengths over the ring ${\mathcal{R}}_k$ for different values of k, and also obtain the Lee distances of these codes.

KEYWORDS:

• Cyclic codes
• reversible codes
• DNA codes
• GC-content

MATHEMATICS SUBJECT CLASSIFICATIONS:

• 94B05
• 94B15
• 94B60
• 92D20

1. Introduction

Identifying elements of an algebraic structure (fields, rings, etc.) having cardinality $4^k$, where $k\ge 1$ with DNA-k bases is a fascinating problem. Suppose an algebraic structure $X=\{x_1,x_2,x_3,x_4\}$ has only four elements and if we identify these elements with the four nucleotides $\{A,G,T,C\}$, then any reversible code over the algebraic structure X will always corresponds to a reversible DNA code. On the other hand, this is not true, if $|X|=4^k$, where k>1 then a reversibility problem occur. In order to clarify the reversibility problem we present an example. Suppose $\beta =({\beta }_1,{\beta }_2,{\beta }_3,{\beta }_4)$ is a codeword of length 4 over X. We identify code alphabets of the codeword β with DNA double pairs as follows: ${\beta }_1\to \mathrm{AA}$, ${\beta }_2\to \mathrm{GA}$, ${\beta }_3\to \mathrm{CT}$, and ${\beta }_4\to \mathrm{AT}$. Therefore, corresponding DNA sequence to the codeword β is AAGACTAT. The reverse codeword of β is given by ${\beta }^r=({\beta }_4,{\beta }_3,{\beta }_2,{\beta }_1)$ and corresponding DNA sequence to ${\beta }^r$ is ATCTGAAA. One can verify that the DNA sequence ATCTGAAA is not the reverse of AAGACTAT. In fact the reverse of the DNA sequence AAGACTAT is TATCAGAA. This reversibility problem was invented by Oztas and Siap [1], where the authors first made use of lifted polynomials over ${\mathbb{F}}_{16}$ to solve this problem.

Many researchers solved this reversibility problem by using different techniques. For example, Gursoy et al. [2] make use of skew cyclic codes over a non-chain ring to solve the reversibility problem. Oztas et al. [3] solved the reversibility problem that occur in DNA k-bases using the coterm polynomials. Further, Oztas and Siap [4] obtained the generalization of lifted polynomials and solved this reversibility problem. Moreover, Ashraf et al. [5] employed reversible cyclic codes over a non-chain ring to solve this reversibility problem.

The interest on DNA computing was initiated by Adleman [6]. In Adleman [6] performed an experiment involving the use of DNA molecules based on the Watson–Crick complement (WCC) model of a DNA strand in solving the famous directed salesman problem. Later on, a molecular program is provided by Adlemen et al. [7] for breaking the symmetric cryptographic algorithm (DES). Additionally, it was demonstrated in [8] that DNA molecules can be employed as a storage media. Multiple theories are necessary for developing DNA sequences that satisfy certain constraints. We call a given sequence $\alpha ={\alpha }_0{\alpha }_1\dots {\alpha }_n$, a quaternary n-sequence or a DNA sequence if ${\alpha }_i\in \mathbf{Y}$, where $\mathbf{Y}=\{A,G,T,C\}$. According to the Watson-Crick complement rule, a DNA sequence combined with its complement (reversible) forms a helix. The Complement of T is A (and vice versa), and the complement of G is C (and vice versa). We denote by $C^c=G$, $G^c=C$, $A^c=T$, $T^c=A$, where $C^c$, $G^c$, $A^c$, and $T^c$ are the complements of C, G, A and T respectively. For example, the Watson-Crick complement of $(\mathrm{ATTGACC}{)}^c=A^c{T}^c{T}^c{G}^c{A}^c{C}^c{C}^c=\mathrm{TAACTGG}$. A DNA code $\mathcal{C}$ of length n is the collection of DNA sequences of length n where each sequence occurs with its reverse complement sequence i.e. ${\alpha }^{\mathrm{rc}}\in \mathcal{C}$, where ${\alpha }^{\mathrm{rc}}={\alpha }_{n-1}^c{\alpha }_{n-2}^c\dots {\alpha }_0^c$ is the reverse complement of $\alpha ={\alpha }_0{\alpha }_1\dots {\alpha }_{n-1}$.

The ability to construct DNA codes that fulfill particular constraints is of utmost importance in various domains, including biotechnology and security. These applications encompass a range of areas, such as DNA computing, DNA cryptography, and DNA steganography (see [7, 9] and references therein). A cyclic DNA code $\mathcal{C}$ always satisfies the Hamming distance constraint and it may satisfy the other three constraints listed below

1. The Hamming constraint: If $H(c_1,c_2)\ge d$, where $c_1,c_2\in \mathcal{C}$ and $c_1\ne c_2$ for some Hamming distance d.

2. The Reverse constraint: If $H(c_1,c_2^r)\ge d$, where $c_1,c_2\in \mathcal{C}$ for some Hamming distance d, and $c_2^r=({\alpha }_{n-1},{\alpha }_{n-2},\dots ,{\alpha }_0)$ is the reverse of $c_2=({\alpha }_0,{\alpha }_1,\dots ,{\alpha }_{n-1})$.

3. The Reverse-complement constraint: If $H(c_1,c_2^{\mathrm{rc}})\ge d$, where $c_1,c_2\in \mathcal{C}$ for some Hamming distance d, and $c_2^{\mathrm{rc}}=({\alpha }_{n-1}^c,{\alpha }_{n-2}^c,\dots ,{\alpha }_0^c)$ is the reverse complement of $c_2=({\alpha }_0,{\alpha }_1,\dots ,{\alpha }_{n-1})$.

4. The Fixed GC-content constraint: If any codeword contains the same number of G and C.

The objective of the first three constraints is to reduce the probability of non-specific hybridization. The fixed GC-content constraint is used to obtain similar melting temperatures.

Reversible codes have versatile applications in the design of DNA codes as well as data storage and retrieval systems. In [10], the author started the study of reversible codes and provided conditions for a given cyclic code over a finite field to be reversible. In [11], a relation between the minimum distance of certain reversible cyclic codes and BCH bounds is obtained. In addition to the above studies, the authors in [12] constructed reversible cyclic codes over $\mathrm{GF}(q)$. Later on, Abualrab [13] obtained quaternary reversible cyclic codes. In [14–16] the authors obtained criteria for negacyclic, constacyclic and quasi-cyclic codes to be reversible respectively. Moreover, in [17] the authors provided a survey of DNA codes over finite rings. For the intensive study of reversible cyclic codes over different algebraic structures, we refer to the readers (see [18–27] and references therein). Meanwhile, Siap et al. [28] obtained cyclic DNA codes over the ring ${\mathbb{F}}_2[u]/⟨u^2-1⟩$ based on the deletion distance. Further, Yildiz and Siap [29] constructed cyclic DNA codes of odd length satisfying the Hamming distance constraint over the ring ${\mathbb{F}}_2[u]/⟨u^4-1⟩$. Later on, Dinh et al. [30] obtained cyclic DNA codes over the ring ${\mathbb{F}}_2+u{\mathbb{F}}_2+v{\mathbb{F}}_2+\mathrm{uv}{\mathbb{F}}_2+v^2{\mathbb{F}}_2+u{v}^2{\mathbb{F}}_2$, where $u^2=0,v^3=v$. Moreover, Guenda and Gulliver [31] constructed DNA codes over the ring ${\mathbb{F}}_2+u{\mathbb{F}}_2$, $u^2=0$ satisfying the reverse constraint, the reverse-complement constraint and the GC-content constraint.

Inspired by these studies, in the present article we construct reversible cyclic codes of any odd length over the ring ${\mathcal{R}}_k$. We construct bijections namely (${\mathrm{\Gamma }}_k$) between the elements of the ring ${\mathcal{R}}_k$ and DNA-$2^k$ bases in such a manner that the reversibility problem has been solved for every k. Furthermore, as an application of these bijections reversible complement codes of any odd length are generated over the ring ${\mathcal{R}}_k$. Moreover, we study the GC-content of DNA codes over the ring ${\mathcal{R}}_k$. In addition, some examples of reversible code are constructed.

The organization of the present article is as follows: In Section 2, we discuss some basic definitions. Section 3 comprises structure of cyclic codes of any odd length over the ring ${\mathcal{R}}_k$. In Section 4, we provide a necessary and sufficient condition for a given cyclic code $\mathcal{C}$ over the ring ${\mathcal{R}}_k$ to be reversible. Also, we construct bijections ${\mathrm{\Gamma }}_k$ for k = 1, 2 (see Tables 1 and 2) in such a manner that the reversibility problem has been solved. Similarly, one can construct ${\mathrm{\Gamma }}_k$ for rest of the values of k, with the aforementioned properties. Moreover, by utilizing these bijections reversible-complement cyclic codes of odd lengths are generated. In Section 5, we provide a Gray map $\mathrm{\Phi }:{\mathcal{R}}_k^n\to {\mathbb{Z}}_4^{2^{k}n}$ and discuss the GC-content in terms of the Gray images. Finally, in Section 6 we provide some examples of reversible cyclic codes of different odd lengths over the ring ${\mathcal{R}}_k$.

Table 1. ${\mathrm{\Gamma }}_1$ between ${\mathcal{R}}_1$ and $S_{D_{16}}$.

DNA 2-bases	Elements of ${\mathcal{R}}_1$	DNA 2-bases	Elements of ${\mathcal{R}}_1$
AA	0	TT	2
AT	1	TA	3
GT	$2+3u_1$	CA	$3u_1$
CT	$1+3u_1$	GA	$3+3u_1$
GG	$2+2u_1$	CC	$2u_1$
GC	$1+2u_1$	CG	$3+2u_1$
AC	$u_1$	TG	$2+u_1$
AG	$1+u_1$	TC	$3+u_1$

Table 2. ${\mathrm{\Gamma }}_2$ correspondence between ${\mathcal{R}}_2$ and $S_{D_{256}}$.

S.no.	Elements of ${\mathcal{R}}_2$	DNA 4-bases
1	0	GAAG
2	$u_1{u}_2$	AAAG
3	$2u_1{u}_2$	GTTG
4	$3u_1{u}_2$	GAAA
5	$u_2$	AAAT
6	$u_2+u_1{u}_2$	AAAC
7	$u_2+2u_1{u}_2$	AAGA
8	$u_2+3u_1{u}_2$	AAGG
9	$2u_1$	ACCT
10	$2u_1+u_1{u}_2$	AAGT
11	$2u_1+2u_1{u}_2$	AAGT
12	$2u_1+3u_1{u}_2$	TGAA
13	$3u_2$	TAAA
14	$3u_2+u_1{u}_2$	GGAA
15	$3u_2+2u_1{u}_2$	AGAA
16	$3u_2+3u_1{u}_2$	CAAA
17	$u_1$	AAGC
18	$u_1+u_1{u}_2$	AATA
19	$u_1+2u_1{u}_2$	AATG
20	$u+3u_1{u}_2$	AATC
21	$u_1+u_2$	AACA
22	$u_1+u_2+u_1{u}_2$	AACG
23	$u_1+u_2+2u_1{u}_2$	AACT
24	$u_1+u_2+3u_1{u}_2$	AACC
25	$u_1+2u_2$	AGAT
26	$u_1+2u_2+u_1{u}_2$	AGAG
27	$u_1+2u_2+2u_1{u}_2$	AGAC
28	$u_1+2u_2+3u_1{u}_2$	AGGG
29	$u_1+3u_2$	AGGT
30	$u_1+3u_2+u_1{u}_2$	AGGC
31	$u_1+3u_2+2u_1{u}_2$	AGTA
32	$u_1+3u_2+3u_1{u}_2$	AGTG
33	$2u_1$	ATTA
34	$2u_1+u_1{u}_2$	AGTC
35	$2u_1+2u_1{u}_2$	GGGG
36	$2u_1+3u_1{u}_2$	CTGA
37	$2u_1+u_2$	AGCA
38	$2u_1+u_2+u_1{u}_2$	AGCG
39	$2u_1+u_2+2u_1{u}_2$	AGCC
40	$2u_1+u_2+3u_1{u}_2$	ATAG
41	$2u_1+2u_2$	AAAA
42	$2u_1+2u_2+u_1{u}_2$	ATAC
43	$2u_1+2u_2+2u_1{u}_2$	GCCG
44	$2u_1+2u_2+3u_1{u}_2$	CATA
45	$2u_1+3u_2$	ACGA
46	$2u_1+3u_2+u_1{u}_2$	GATA
47	$2u_1+3u_2+2u_1{u}_2$	CCGA
48	$2u_1+3u_2+3u_1{u}_2$	GCGA
49	$3u_1$	CGAA
50	$3u_1+u_1{u}_2$	CTAA
51	$3u_1+2u_1{u}_2$	GTAA
52	$3u_1+3u_1{u}_2$	ATAA
53	$3u_1+u_2$	TGGA
54	$3u_1+u_2+u_1{u}_2$	GTGA
55	$3u_1+u_2+2u_1{u}_2$	ATGA
56	$3u_1+u_2+3u_1{u}_2$	CGGA
57	$3u_1+2u_2$	TAGA
58	$3u_1+2u_2+u_1{u}_2$	GGGA
59	$3u_1+2u_2+2u_1{u}_2$	CAGA
60	$3u_1+2u_2+3u_1{u}_2$	GAGA
61	$3u_1+3u_2$	ACAA
62	$3u_1+3u_2+u_1{u}_2$	CCAA
63	$3u_1+3u_2+2u_1{u}_2$	TCAA
64	$3u_1+3u_2+3u_1{u}_2$	GCAA
65	1	GATC
66	$1+u_1{u}_2$	ATGT
67	$1+2u_1{u}_2$	AATT
68	$1+3u_1{u}_2$	ACAT
69	$1+u_1$	ATGG
70	$1+u_1+u_1{u}_2$	ATGC
71	$1+u_1+2u_1{u}_2$	ATTG
72	$1+u_1+3u_1{u}_2$	ATTC
73	$1+2u_2$	ATAT
74	$1+2u_2+2u_1{u}_2$	AGCT
75	$1+2u_2+u_1{u}_2$	ATCA
76	$1+2u_2+3u_1{u}_2$	TGAT
77	$1+3u_2$	CCAT
78	$1+3u_2+u_1{u}_2$	GAAT
79	$1+3u_2+2u_1{u}_2$	CAAT
80	$1+3u_2+3u_1{u}_2$	GCAT
81	$1+u_1$	ATCG
82	$1+u_1+u_1{u}_2$	ATCC
83	$1+u_1+2u_1{u}_2$	ACAG
85	$1+u_1+u_2$	ACGG
84	$1+u_1+3u_1{u}_2$	ACAC
86	$1+u_1+u_2+u_1{u}_2$	ACGC
87	$1+u_1+u_2+2u_1{u}_2$	ACTG
88	$1+u_1+u_2+3u_1{u}_2$	ACTC
89	$1+u_1+2u_2$	ACCG
90	$1+u_1+2u_2+u_1{u}_2$	ACCC
91	$1+u_1+2u_2+2u_1{u}_2$	GCAG
92	$1+u_1+2u_2+3u_1{u}_2$	GCAC
93	$1+u_1+3u_2$	GCGG
94	$1+u_1+3u_2+u_1{u}_2$	GCTG
95	$1+u_1+3u_2+2u_1{u}_2$	GCTC
96	$1+u_1+3u_2+3u_1{u}_2$	GCCC
97	$1+2u_1$	ACGT
98	$1+2u_1+u_1{u}_2$	GAAC
99	$1+2u_1+2u_1{u}_2$	GGCC
100	$1+2u_1+3u_1{u}_2$	GTTC
101	$1+2u_1+u_2$	GAGG
102	$1+2u_1+u_2+u_1{u}_2$	GATG
103	$1+2u_1+u_2+2u_1{u}_2$	GACC
104	$1+2u_1+u_2+3u_1{u}_2$	GTGG
105	$1+2u_1+2u_2$	GCGC
106	$1+2u_1+2u_2+u_1{u}_2$	GTCC
107	$1+2u_1+2u_2+2u_1{u}_2$	GTAC
108	$1+2u_1+2u_2+3u_1{u}_2$	GGAC
109	$1+2u_1+3u_2$	CCTC
110	$1+2u_1+3u_2+u_1{u}_2$	CCAC
111	$1+2u_1+3u_2+2u_1{u}_2$	GGTC
112	$1+2u_1+3u_2+3u_1{u}_2$	CATC
113	$1+3u_1$	CGAT
114	$1+3u_1+u_1{u}_2$	GTGT
115	$1+3u_1+2u_1{u}_2$	CTGT
116	$1+3u_1+3u_1{u}_2$	GGAT
117	$1+3u_1+u_2$	CCGC
118	$1+3u_1+u_2+u_1{u}_2$	GGGC
119	$1+3u_1+u_2+2u_1{u}_2$	GAGC
120	$1+3u_1+u_2+3u_1{u}_2$	CAGC
121	$1+3u_1+2u_2$	CGGT
122	$1+3u_1+2u_2+u_1{u}_2$	GTGC
123	$1+3u_1+2u_2+2u_1{u}_2$	CTGC
124	$1+3u_1+2u_2+3u_1{u}_2$	GGGT
125	$1+3u_1+3u_2$	CCGT
126	$1+3u_1+3u_2+u_1{u}_2$	GATA
127	$1+3u_1+3u_2+2u_1{u}_2$	CAGT
128	$1+3u_1+3u_2+3u_1{u}_2$	GCGT
129	2	CTTC
130	$2+u_1{u}_2$	TTTC
131	$2+2u_1{u}_2$	CAAC
132	$2+3u_1{u}_2$	CTTT
133	$2+u_2$	TTTA
134	$2+u_2+u_1{u}_2$	TTTG
135	$2+u_2+2u_1{u}_2$	TTCT
136	$2+u_2+3u_1{u}_2$	TTCC
137	$2+2u_2$	TGGT
138	$2+2u_2+u_1{u}_2$	TTCA
139	$2+2u_2+2u_1{u}_2$	TCCT
140	$2+2u_2+3u_1{u}_2$	ACTT
141	$2+3u_2$	ATTT
142	$2+3u_2+u_1{u}_2$	CCTT
143	$2+3u_2+2u_1{u}_2$	TCTT
144	$2+3u_2+3u_1{u}_2$	GTTT
145	2 + u	TTCG
146	$2+u_1+u_1{u}_2$	TTAT
147	$2+u_1+2u_1{u}_2$	TTAC
148	$2+u_1+3u_1{u}_2$	TTAG
149	$2+u_1+u_2$	TTGT
150	$2+u_1+u_2+u_1{u}_2$	TTGC
151	$2+u_1+u_2+2u_1{u}_2$	TTGA
152	$2+u_1+u_2+3u_1{u}_2$	TTGG
153	$2+u_1+2u_2$	TCTA
154	$2+u_1+2u_2+u_1{u}_2$	TCTC
155	$2+u_1+2u_2+2u_1{u}_2$	GTCT
156	$2+u_1+2u_2+3u_1{u}_2$	TCCC
157	$2+u_1+3u_2$	TCCA
158	$2+u_1+3u_2+u_1{u}_2$	TCCG
159	$2+u_1+3u_2+u_1{u}_2$	TCAT
160	$2+u_1+3u_2+u_1{u}_2$	TCAC
161	$2+2u_1$	TAAT
162	$2+2u_1+u_1{u}_2$	TCAG
163	$2+2u_1+2u_1{u}_2$	CCCC
164	$2+2u_1+3u_1{u}_2$	GACT
165	$2+2u_1+u_2$	TCGT
166	$2+2u_1+u_2+u_1{u}_2$	TCGC
167	$2+2u_1+u_2+2u_1{u}_2$	TCGG
168	$2+2u_1+u_2+3u_1{u}_2$	TATC
173	$2+2u_1+3u_2$	TGCT
174	$2+2u_1+3u_2+u_1{u}_2$	CTAT
175	$2+2u_1+3u_2+2u_1{u}_2$	GGCT
176	$2+2u_1+3u_2+3u_1{u}_2$	CGCT
177	$2+3u_1$	GCTT
178	$2+3u_1+u_1{u}_2$	GATT
179	$2+3u_1+2u_1{u}_2$	CATT
180	$2+3u_1+3u_1{u}_2$	TATT
181	$2+3u_1+u_2$	ACCT
182	$2+3u_1+u_2+u_1{u}_2$	CACT
183	$2+3u_1+u_2+2u_1{u}_2$	TACT
184	$2+3u_1+u_2+3u_1{u}_2$	GCCT
185	$2+3u_1+2u_2$	ATCT
186	$2+3u_1+2u_2+u_1{u}_2$	CCCT
187	$2+3u_1+2u_2+2u_1{u}_2$	TCTG
188	$2+3u_1+2u_2+3u_1{u}_2$	CTCT
189	$2+3u_1+3u_2$	TGTT
190	$2+3u_1+3u_2+u_1{u}_2$	GGTT
191	$2+3u_1+3u_2+2u_1{u}_2$	AGTT
192	$2+3u_1+3u_2+3u_1{u}_2$	CGTT
193	3	CTAG
194	$3+u_1{u}_2$	TACA
195	$3+2u_1{u}_2$	TTAA
196	$3+3u_1{u}_2$	TGTA
197	$3+u_2$	TACC
198	$3+u_2+u_1{u}_2$	TACG
199	$3+u_2+2u_1{u}_2$	TAAC
200	$3+u_2+3u_1{u}_2$	TAAG
201	$3+2u_2$	TATA
202	$3+2u_2+u_1{u}_2$	TAGT
203	$3+2u_2+2u_1{u}_2$	TCGA
204	$3+2u_2+3u_1{u}_2$	ACTA
205	$3+3u_2$	GGTA
206	$3+3u_2+u_1{u}_2$	CTTA
207	$3+3u_2+u_1{u}_2$	GTTA
208	$3+3u_2+3u_1{u}_2$	CGTA
209	$3+u_1$	TAGC
210	$3+u_1+u_1{u}_2$	TAGG
211	$3+u_1+2u_1{u}_2$	TGTC
212	$3+u_1+3u_1{u}_2$	TGTG
213	$3+u_1+u_2$	TGCC
214	$3+u_1+u_2+u_1{u}_2$	TGCG
215	$3+u_1+u_2+2u_1{u}_2$	TGAC
216	$3+u_1+u_2+3u_1{u}_2$	TGAG
217	$3+u_1+2u_2$	TGGC
218	$3+u_1+u_2+u_1{u}_2$	TGGG
219	$3+u_1+u_2+2u_1{u}_2$	CGTC
220	$3+u_1+u_2+3u_1{u}_2$	CGTG
221	$3+u_1+3u_2$	CGCC
222	$3+u_1+3u_2+u_1{u}_2$	CGAC
223	$3+u_1+3u_2+2u_1{u}_2$	CGAG
224	$3+u_1+3u_2+3u_1{u}_2$	CGGG
225	$3+2u_1$	TGCA
226	$3+2u_1+u_1{u}_2$	CTTG
227	$3+2u_1+2u_1{u}_2$	CCGG
228	$3+2u_1+3u_1{u}_2$	CAAG
229	$3+2u_1+u_2$	CTCC
230	$3+2u_1+u_2+u_1{u}_2$	CTAC
231	$3+2u_1+u_2+2u_1{u}_2$	CTGG
232	$3+2u_1+u_2+3u_1{u}_2$	CACC
233	$3+2u_1+2u_2$	CGCG
234	$3+2u_1+2u_2+u_1{u}_2$	CAGG
235	$3+2u_1+2u_2+2u_1{u}_2$	CATG
236	$3+2u_1+2u_2+3u_1{u}_2$	CCTG
237	$3+2u_1+3u_2$	GGAG
238	$3+2u_1+3u_2+u_1{u}_2$	GGTG
239	$3+2u_1+3u_2+2u_1{u}_2$	CCAG
240	$3+2u_1+3u_2+3u_1{u}_2$	GTAG
241	$3+3u_1$	GCTA
242	$3+3u_1+u_1{u}_2$	CACA
243	$3+3u_1+2u_1{u}_2$	GACA
244	$3+3u_1+3u_1{u}_2$	CCTA
245	$3+3u_1+u_2$	GGCG
246	$3+3u_1+u_2+u_1{u}_2$	CCCG
247	$3+3u_1+u_2+u_1{u}_2$	CTCG
248	$3+3u_1+u_2+3u_1{u}_2$	GTCG
249	$3+3u_1+2u_2$	GCCA
250	$3+3u_1+2u_2+u_1{u}_2$	CACG
251	$3+3u_1+2u_2+2u_1{u}_2$	GACG
252	$3+3u_1+2u_2+3u_1{u}_2$	CCCA
253	$3+3u_1+3u_2$	GGCA
254	$3+3u_1+3u_2+u_1{u}_2$	CTCA
255	$3+3u_1+3u_2+2u_1{u}_2$	GTCA
256	$3+3u_1+3u_2+3u_1{u}_2$	CGCA

2. Preliminaries

We begin this section by characterizing the ring ${\mathcal{R}}_k$. Let ${\mathcal{R}}_k={\mathbb{Z}}_4[u_1,u_2,\dots ,u_k]/⟨u_i^2-u_i,u_i{u}_j-u_j{u}_i⟩$ be a finite ring. Here, we make use of results presented in (see [32–35]), where the authors obtained linear codes over the ring ${\mathcal{R}}_k$, for k = 1, 2. Suppose $u_B=\prod _{i\in B}{u}_i$, where B is a subset of the set $\{1,2,\dots ,k\}$. It is worth to notice that every element of the ring ${\mathcal{R}}_k$ can be written uniquely as $\sum _{B\in {\mathcal{P}}_k}{\alpha }_B{u}_B$, where ${\alpha }_B\in {\mathbb{Z}}_4$ and ${\mathcal{P}}_k$ is any subset of the set $\{1,2,\dots ,k\}$. In particular we assume that $u_{\varphi }=1$. Also, observe that for any subsets $M,N\subseteq \{1,2,\dots ,k\}$ we have $u_M{u}_N=u_{M\cup N}$ and therefore we obtain $\sum _{M\in {\mathcal{P}}_k}{\alpha }_M{u}_M\cdot \sum _{N\in {\mathcal{P}}_k}{\alpha }_N{u}_N=\sum _{R\in {\mathcal{P}}_k}\left(\sum _{M\cup N=R}{\alpha }_M{u}_M\right)u_R.$

Lemma 2.1

The ring ${\mathcal{R}}_k$ has characteristic 4 and cardinality $4^{2^k}$.

Proof.

Since the coefficients of $u_B$ are the members of ${\mathbb{Z}}_4$, it follows directly that characteristic of the ring ${\mathcal{R}}_k$ is 4. Also, observe that for any given element $\sum _{B\in {\mathcal{P}}_k}{\alpha }_B{u}_B$ there are four choices for each ${\alpha }_B$ and $2^k$ subsets of the set $\{1,2,\dots ,k\}$. This shows that cardinality of the ring ${\mathcal{R}}_k$ is $4^{2^k}$.

An element e is called an idempotent element if $e^2=e$. Suppose $e_1,e_2\in {\mathcal{R}}_k$ are two idempotent elements if $e_1{e}_2=0$, then $e_1$ and $e_2$ are said to be orthogonal idempotent. Now we list all orthogonal idempotent elements of the ring ${\mathcal{R}}_k$ in Table 3. Kumar and Singh [26], Li et al. [35] and Bustomi [36] obtained all orthogonal idempotent elements over the ring ${\mathcal{R}}_k$ for k = 1, k = 2 and k = 3 respectively. Here, we use the same technique which is presented in [26, 35, 36] to count all orthogonal idempotents in the ring ${\mathcal{R}}_k$.

Table 3. Orthogonal idempotent elements of the ring ${\mathcal{R}}_k$.

$\prod _{i=1}^k(1-u_i)$	$u_1\prod _{i\ne 1}^{}(1-u_i)$,	$u_1{u}_2\prod _{i\ne 1,2}^{}(1-u_i)$,	···	$\prod _{i\ne 1}{u}_i(1-u_1)$,	$\prod _{i=1}^k{u}_i$
	$u_2\prod _{i\ne 2}^{}(1-u_i)$,	$u_1{u}_3\prod _{i\ne 1,3}^{}(1-u_i)$,	···	$\prod _{i\ne 2}{u}_i(1-u_2)$,
	$u_3\prod _{i\ne 3}^{}(1-u_i)$,	⋮	···	$\prod _{i\ne 3}{u}_i(1-u_3)$,
	⋮	$u_1{u}_k\prod _{i\ne 1,k}^{}(1-u_i)$,	···	⋮
	$u_k\prod _{i\ne k}^{}(1-u_i)$	$u_2{u}_3\prod _{i\ne 2,3}^{}(1-u_i)$,	···	$\prod _{i\ne k}{u}_i(1-u_k)$
		$u_2{u}_4\prod _{i\ne 2,4}^{}(1-u_i)$,	···
		⋮
		$u_2{u}_k\prod _{i\ne 2,k}^{}(1-u_i)$,	···
		⋮
		$u_{k-1}{u}_k\prod _{i\ne k-1,k}^{}(1-u_i)$	···

By observing Table 3, the total number of orthogonal idempotent elements in the ring ${\mathcal{R}}_k$ are $(\genfrac{}{}{0ex}{}{k}{0})+(\genfrac{}{}{0ex}{}{k}{1})+(\genfrac{}{}{0ex}{}{k}{2})+\cdots +(\genfrac{}{}{0ex}{}{k}{k})=2^k$. We denote all idempotent elements by $e_1,e_2,\dots ,e_{2^k}$. All idempotent elements are also pairwise orthogonal, since $e_i^2=e_i$, for $1\le i\le 2^k$ and $e_i\cdot e_j=0$ for $i\ne j$ and also we have $\sum _{i=1}^{2^k}{e}_i=1$. Hence, by Chinese remainder theorem we obtain ${\mathcal{R}}_k={\mathbb{Z}}_4{e}_1⨁{\mathbb{Z}}_4{e}_2⨁\cdots ⨁{\mathbb{Z}}_4{e}_{2^k}.$Recall that a linear code $\mathcal{C}$ of length n over ${\mathcal{R}}_k$ is an ${\mathcal{R}}_k$-submodule of ${\mathcal{R}}_k^n$ and members of $\mathcal{C}$ are known as codewords. A cyclic shift on ${\mathcal{R}}_k^n$ is a permutation σ given by $\sigma (c_0,c_1,\dots ,c_{n-1})=(c_{n-1},c_0,\dots ,c_{n-2}).$A code $\mathcal{C}$ is said to be a cyclic code if it is invariant under σ. The inner-product of two given n-tuples $a=(a_0,a_1,\dots ,a_{n-1})$, $b=(b_0,b_1,\dots ,b_{n-1})$ is defined as $a\cdot b=\sum _{i=0}^{n-1}{a}_i{b}_i$, a and b are said to be orthogonal if $a\cdot b=0$. For a given linear code $\mathcal{C}$, we define the dual code ${\mathcal{C}}^{\perp }$ over the ring ${\mathcal{R}}_k$ in the following manner ${\mathcal{C}}^{\perp }=\{a\in {\mathcal{R}}_k^n|a\cdot b=0\text{for all}b\in \mathcal{C}\}.$One can verify that ${\mathcal{C}}^{\perp }$ is a linear code over the ring ${\mathcal{R}}_k$ of same length as $\mathcal{C}$. In addition, If $\mathcal{C}\subseteq {\mathcal{C}}^{\perp }$ then a linear code $\mathcal{C}$ is said to be self-orthogonal and if $\mathcal{C}={\mathcal{C}}^{\perp }$ then we call a linear code $\mathcal{C}$ is self-dual. The reciprocal polynomial $p^{\ast }(x)$ of a given polynomial $p(x)=p_0+p_{1}x+\cdots +p_{n-1}{x}^{n-1}\in {\mathcal{R}}_k[x]$ is defined as $p^{\ast }(x)=x^{\mathrm{deg}(p(x))}p(\frac{1}{x})=p_{n-1}+\cdots +p_0{x}^{n-1}$. It is worth to mention that $\mathrm{deg}(p(x))\ge \mathrm{deg}(p^{\ast }(x))$ and if $p_0\ne 0$, then $\mathrm{deg}(p(x))=\mathrm{deg}(p^{\ast }(x))$. A given polynomial $p(x)$ is said to be self-reciprocal if $p^{\ast }(x)=p(x)$.

We need the following result which provides the criteria for obtaining the reciprocal of sum and product of polynomials.

Lemma 2.2

[13]

Suppose $f_1(x)$ and $f_2(x)$ are two polynomials over the ring ${\mathcal{R}}_k$, where $\mathrm{deg}(f_2(x))\le \mathrm{deg}(f_1(x))$. Then the following statements hold:

(1)	$(f_1(x)+f_2(x){)}^{\ast }=f_1^{\ast }(x)+x^{\alpha }{f}_2^{\ast }(x)$,
(2)	$(f_1(x)\cdot f_2(x){)}^{\ast }=f_1^{\ast }(x)\cdot f_2^{\ast }(x)$, provided $\alpha =\mathrm{deg}(f_1(x))-\mathrm{deg}(f_2(x))$.

3. Structure of cyclic codes over ${\mathcal{R}}_k$

In this section, the structure of cyclic codes over the ring ${\mathcal{R}}_k$ is conferred. For obtaining the generating set of a cyclic code over the ring ${\mathcal{R}}_k$, we make use of cyclic codes over the ring ${\mathbb{Z}}_4$. In [37], the authors obtained non-linear codes over the ring ${\mathbb{Z}}_4$ as Gray images of binary codes. Many researchers have obtained the structure of cyclic codes over the ring ${\mathbb{Z}}_4$. For example, in [13] the author provided generating set of cyclic codes over the ring ${\mathbb{Z}}_4$ of arbitrary length. Similarly, Pless and Qian [38] obtained a different generating set for cyclic codes of odd lengths over the ring ${\mathbb{Z}}_4$. In the present article, we use generating set given in [38]. Therefore, we have following result.

Theorem 3.1

[38]

Suppose $\mathcal{C}$ is a cyclic code of odd length n over the ring ${\mathbb{Z}}_4$. Then $\mathcal{C}$ is given by $\mathcal{C}=⟨f(x)g(x),2f(x)h(x)⟩$, where $f(x)$, $g(x)$ and $h(x)$ are monic polynomials such that $x^n-1=f(x)g(x)h(x)$. Moreover, $\mathcal{C}$ is of the type $4^{n-\mathrm{deg}(f(x))}{2}^{\mathrm{deg}f(x)-\mathrm{deg}g(x)}$.

We obtain the decomposition of a linear code $\mathcal{C}$ over the ring ${\mathcal{R}}_k$ in terms of the linear codes ${\mathcal{C}}_i$, where each ${\mathcal{C}}_i$ is given as follows: $\begin{array}{rl}& {\mathcal{C}}_1=\{a_1\in {\mathbb{Z}}_4^n:e_1{a}_1+e_2{a}_2+\cdots +e_{2^k}{a}_{2^k}\in \mathcal{C}\mathrm{forsome}{a}_i\in {\mathbb{Z}}_4^n\},\\ & {\mathcal{C}}_2=\{a_2\in {\mathbb{Z}}_4^n:e_1{a}_1+e_2{a}_2+\cdots +e_{2^k}{a}_{2^k}\in \mathcal{C}\mathrm{forsome}{a}_i\in {\mathbb{Z}}_4^n\},\\ & {\mathcal{C}}_3=\{a_3\in {\mathbb{Z}}_4^n:e_1{a}_1+e_2{a}_2+\cdots +e_{2^k}{a}_{2^k}\in \mathcal{C}\mathrm{forsome}{a}_i\in {\mathbb{Z}}_4^n\},\\ & ⋮\\ & {\mathcal{C}}_{2^k}=\{a_{2^k}\in {\mathbb{Z}}_4^n:e_1{a}_1+e_2{a}_2+\cdots +e_{2^k}{a}_{2^k}\in \mathcal{C}\mathrm{forsome}{a}_i\in {\mathbb{Z}}_4^n\}.\end{array}$It is easy to observe that each ${\mathcal{C}}_i$ is a linear code over the ring ${\mathbb{Z}}_4$ and therefore $\mathcal{C}$ can be uniquely decomposed as (1) $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}.$(1) Now, we present the following results which are helpful in describing the structure of linear codes over the ring ${\mathcal{R}}_k$.

Theorem 3.2

Let $\mathcal{C}$ be a linear code of length n over the ring ${\mathcal{R}}_k$. Then $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}$, where each ${\mathcal{C}}_i$ is a linear code over ${\mathbb{Z}}_4$, and the direct sum decomposition is unique. Moreover, the dual code of $\mathcal{C}$ is given by ${\mathcal{C}}^{\perp }=e_1{\mathcal{C}}_1^{\perp }+e_2{\mathcal{C}}_2^{\perp }+\cdots +e_{2^k}{\mathcal{C}}_{2^k}^{\perp }$, where ${\mathcal{C}}_i^{\perp }$ is the dual code of the linear code ${\mathcal{C}}_i$.

Proof.

Proof of the first part follows directly from the Decomposition (1(1) $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}.$(1) ). For the dual code, suppose that ${\mathfrak{D}}^{\perp }=e_1{\mathcal{C}}_1^{\perp }+e_2{\mathcal{C}}_2^{\perp }+\cdots +e_{2^k}{\mathcal{C}}_{2^k}^{\perp }$. Take any $x\in \mathfrak{D}$ and $y\in \mathcal{C}$, where $x=\sum _{i=1}^{2^k}{e}_i{r}_i$, $r_i\in {\mathcal{C}}_i^{\perp }$ and $y=\sum _{i=1}^{2^k}{e}_i{t}_i$, $t_i\in {\mathcal{C}}_i$. Now, consider $x\cdot y=\sum _{i=1}^{2^k}{e}_i(r_i{t}_i)=0$, we must have $x\in {\mathcal{C}}^{\perp }$. Hence, we obtain $\mathfrak{D}\subseteq {\mathcal{C}}^{\perp }$. For the converse part, notice that ${\mathcal{R}}_k$ is a Frobenius ring. Thus $|\mathfrak{D}|=|{\mathcal{C}}_1^{\perp }||{\mathcal{C}}_2^{\perp }|\cdots |{\mathcal{C}}_{2^k}^{\perp }|=|{\mathcal{C}}^{\perp }|$. Therefore, we obtain $\mathfrak{D}={\mathcal{C}}^{\perp }$.

Corollary 3.3

Let $\mathcal{C}$ be a linear code over the ring ${\mathcal{R}}_k$ given by $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}$. Then $\mathcal{C}$ is a self-orthogonal if and only if each ${\mathcal{C}}_i$ is self-orthogonal over ${\mathbb{Z}}_4$. Furthermore, $\mathcal{C}$ is a self-dual if and only if each ${\mathcal{C}}_i$ is self-dual over ${\mathbb{Z}}_4$.

Proof.

The result directly follows from Theorem 3.2.

Theorem 3.4

Let $\mathcal{C}$ be a linear code of length n given by $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}$. Then $\mathcal{C}$ is a cyclic code over the ring ${\mathcal{R}}_k$ if and only if each ${\mathcal{C}}_i$ is a cyclic code over the ring ${\mathbb{Z}}_4$.

The following theorem provides the generating set of a given cyclic code of any odd length.

Theorem 3.5

Let $\mathcal{C}$ be a cyclic code of odd length n over the ring ${\mathcal{R}}_k$. Then $\mathcal{C}$ is given by $\mathcal{C}=⟨e_1{F}_1(x),2e_1{H}_1(x),e_2{F}_2(x),2e_2{H}_2(x),\dots ,e_{2^k}{F}_{2^k}(x),2e_{2^k}{H}_{2^k}(x)⟩,$where $F_i(x)=f_i(x)g_i(x)$; $H_i(x)=f_i(x)h_i(x)$ and $x^n-1=f_i(x)g_i(x)h_i(x);$ $1\le i\le 2^k$.

Proof.

The above result is a direct consequence of Theorem 3.1.

4. Reversible-complement DNA codes

In this section, we discuss reversibility of cyclic codes of any odd length over the ring ${\mathcal{R}}_k$. We provide a necessary and sufficient condition for a given cyclic code $\mathcal{C}$ of any odd length over the ring ${\mathcal{R}}_k$ to be reversible. Moreover, in this section we provide lemmas which are useful in solving the reversibility problem. Further, we discuss reversible-complement codes over the ring ${\mathcal{R}}_k$.

Definition 4.1

A block code will be called reversible if the block of digits formed by reversing the order of the digits in a codeword is always another codeword in the same code.

In [10], the author obtained a condition for a given cyclic code over a finite field ${\mathbb{F}}_q$ to be reversible.

Theorem 4.2

Suppose $\mathcal{C}$ is a cyclic code over ${\mathbb{F}}_q$ given by $\mathcal{C}=⟨g(x)⟩$. Then $\mathcal{C}$ is reversible if and only if $g(x)=g^{\ast }(x)$.

The following result provides a necessary and sufficient condition for the reversibility of cyclic codes of odd lengths over the ring ${\mathcal{R}}_k$ in terms of their generating polynomials.

Theorem 4.3

Let $\mathcal{C}$ be a cyclic code of odd length n over the ring ${\mathcal{R}}_k$ given by $\mathcal{C}=⟨e_1{F}_1(x),2e_1{H}_1(x),e_2{F}_2(x),2e_2{H}_2(x),\dots ,e_{2^k}{F}_{2^k}(x),2e_{2^k}{H}_{2^k}(x)⟩,$where $F_i(x)=f_i(x)g_i(x)$; $H_i(x)=f_i(x)h_i(x)$ and $x^n-1=f_i(x)g_i(x)h_i(x);$ $1\le i\le 2^k$. Then $\mathcal{C}$ is a reversible cyclic code over the ring ${\mathcal{R}}_k$ if and only if each $F_i(x)$ and $H_i(x)$ are self-reciprocal polynomials.

Proof.

Suppose $\mathcal{C}$ is a reversible cyclic code over the ring ${\mathcal{R}}_k$.Suppose on contrary, the polynomial $F_i(x)$ is not self-reciprocal i.e. $F_i(x)\ne F_i^{\ast }(x)$. Suppose $F(x)=gcd(F_i(x),F_i^{\ast }(x))$. Then there exist polynomials $\lambda (x)$, $\mu (x)\in {\mathcal{R}}_k[x]$ such that $F(x)=\lambda (x)F_i(x)+\mu (x)F_i^{\ast }(x)$ and $\mathrm{deg}F(x)<\mathrm{deg}{F}_i(x)$. Let $M=\{e_1,e_2,\dots ,e_{2^k}\}$ be the set of all orthogonal idempotents. Since $\mathcal{C}$ is a reversible cyclic code over the ring ${\mathcal{R}}_k$, $\mathrm{\alpha F}(x)\in \mathcal{C}$ for all $\alpha \in M$. This contradicts the fact that $F_i(x)$ is a minimal degree generating polynomial of $\mathcal{C}$. Therefore, we can conclude that $F_i(x)=F_i^{\ast }(x)$. Thus, each $F_i(x)$ is a self-reciprocal polynomial. Similarly, one can prove for $H_i(x)$.

Conversely, assume that $F_i(x)$ and $H_i(x)$ are self-reciprocal polynomials i.e. $F_i(x)=F_i^{\ast }(x)$ and $H_i(x)=H_i^{\ast }(x)$, where $1\le i\le 2^k$. Let $c(x)\in \mathcal{C}$ be any element. Then there exist polynomials ${\lambda }_1(x)$, ${\lambda }_2(x),\cdots ,{\lambda }_{2^{k+1}}(x)$, over the ring ${\mathcal{R}}_k$ such that $c(x)=e_1{\lambda }_1(x)F_1(x)+2e_1{\lambda }_2(x)H_1(x)+\cdots +e_{2^k}{\lambda }_{2^{k+1}-1}(x)F_{2^k}(x)+e_{2^k}{\lambda }_{2^{k+1}}(x)H_{2^k}(x).$The reciprocal polynomial $c^{\ast }(x)$ of the polynomial $c(x)$ can be obtained using Lemma 2.2. $\begin{array}{rl}{c}^{\ast }(x)& =e_1{\lambda }_1^{\ast }(x)F_1^{\ast }(x)+2e_1{\lambda }_2^{\ast }(x)H_1^{\ast }(x)+\cdots +e_{2^k}{\lambda }_{2^{k+1}-1}^{\ast }(x)F_{2^k}^{\ast }(x)\\ & +e_{2^k}{\lambda }_{2^{k+1}}^{\ast }(x)H_{2^k}^{\ast }(x).\end{array}$Since $F_i(x)=F_i^{\ast }(x)$ and $H_i(x)=H_i^{\ast }(x)$, where $1\le i\le 2^k$, we get $\begin{array}{rl}{c}^{\ast }(x)& =e_1{\lambda }_1^{\ast }(x)F_1(x)+2e_1{\lambda }_2^{\ast }(x)H_1(x)+\cdots +e_{2^k}{\lambda }_{2^{k+1}-1}^{\ast }(x)F_{2^k}(x)\\ & +e_{2^k}{\lambda }_{2^{k+1}}^{\ast }(x)H_{2^k}(x).\end{array}$Therefore, we obtain $c^{\ast }(x)\in \mathcal{C}$. Hence, $\mathcal{C}$ is a reversible cyclic code over the ring ${\mathcal{R}}_k$.

Example 4.4

Suppose k = 1, consider the ring ${\mathcal{R}}_1={\mathbb{Z}}_4+u_1{\mathbb{Z}}_4$, where $u_1^2=u_1$. We factorize the polynomial $x^{17}-1$ over the ring ${\mathbb{Z}}_4$ as follows: $x^{17}-1=(x-1)(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)(x^8+x^7+3x^6+3x^4+3x^2+x+1)=f_1{f}_2{f}_3$. We can verify that $f_2$ and $f_3$ both are self-reciprocal polynomials over the ring ${\mathbb{Z}}_4$. Several reversible cyclic codes with their Gray images are listed in Table 4.

Table 4. Reversible cyclic codes of length 17 over the ring ${\mathcal{R}}_1$.

Generating polynomial(s) of ${\mathcal{C}}_i$	${\mathbb{Z}}_4$-Gray images
$(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$	$[34,4^{18}{2}^0,7]$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^{18}{2}^0,7]$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^2{2}^0,17]$
$2(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$	$[34,4^0{2}^{18},10]$
$2(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^0{2}^{18},10]$
$2(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^0{2}^2,34]$
$(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$,
$2(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$	$[34,4^{18}{2}^{18},7]$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$,
$2(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^{18}{2}^{18},7]$
$(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$,
$2(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^{18}{2}^2,7]$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$,
$2(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)$
$(x^8+x^7+3x^6+3x^4+3x^2+x+1)$	$[34,4^{18}{2}^2,7]$

The following result provides the reversibility of a given cyclic code $\mathcal{C}$ in terms of ${\mathcal{C}}_i$.

Theorem 4.5

Suppose $\mathcal{C}$ is a cyclic code of odd length n over the ring ${\mathcal{R}}_k$ given by $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}$. Then $\mathcal{C}$ is a reversible cyclic code over the ring ${\mathcal{R}}_k$ if and only if each ${\mathcal{C}}_i$ is a reversible cyclic code over ${\mathbb{Z}}_4$.

Proof.

Assume that $\mathcal{C}$ is a reversible cyclic code of length n over the ring ${\mathcal{R}}_k$. Let $x\in \mathcal{C}$ be any element such that $x=e_1{x}_1+e_2{x}_2+\cdots +e_{2^k}{x}_{2^k}$, where $x_i\in {\mathcal{C}}_i$. Now, the reverse of x is given by $x^r=e_1{x}_1^r+e_2{x}_2^r+\cdots +e_{2^k}{x}_{2^k}^r\in \mathcal{C}$. Suppose $x^r=e_1{x}_1^r+e_2{x}_2^r+\cdots +e_{2^k}{x}_{2^k}^r=e_1{y}_1+e_2{y}_2\cdots +e_{2^k}{y}_{2^k}$, where $y_i\in {\mathcal{C}}_i$. Then, $e_1(x_1^r-y_1)+e_2(x_2^r-y_2)+\cdots +e_{2^k}(x_{2^k}^r-y_{2^k})=0$. Therefore, $x_1^r=y_1\in {\mathcal{C}}_1$, $x_2^r=y_2\in {\mathcal{C}}_2,\dots ,x_{2^k}^r=y_{2^k}\in {\mathcal{C}}_{2^k}$. Thus each ${\mathcal{C}}_i$ is a reversible cyclic code.

Conversely, assume that each ${\mathcal{C}}_i$ is a reversible cyclic code over the ring ${\mathbb{Z}}_4$. Let $x\in \mathcal{C}$ be any element such that $x=e_1{x}_1+e_2{x}_2+\cdots +e_{2^k}{x}_{2^k}$, where $x_i\in {\mathcal{C}}_i$. The reverse of x is given by $x^r=e_1{x}_1^r+e_2{x}_2^r+\cdots +e_{2^k}{x}_{2^k}^r$. Since $x_i^r\in {\mathcal{C}}_i$, $x^r\in \mathcal{C}$. Therefore, we can conclude that $\mathcal{C}$ is a reversible cyclic code.

Our current focus is on the reversible-complement cyclic codes characterized by odd lengths. We will establish a condition that is both necessary and sufficient for a cyclic code to be categorized as a reversible-complement cyclic code over the ring ${\mathcal{R}}_k$. To accomplish this, we discuss the foundational concepts of complement over the ring ${\mathcal{R}}_k$. For every k, we denote the complement of an element $z\in {\mathcal{R}}_k$ by $\overline{z}$ and define as: $\overline{z}=z+2$.  Let $p(x)\in {\mathcal{R}}_k[x]$, where $p(x)=p_0+p_{1}x+p_2{x}^2+\cdots +p_{n-1}{x}^{n-1}$. Then the complement of $p(x$) is defined as follows: $p^c(x)={\overline{p}}_0+{\overline{p}}_{1}x+{\overline{p}}_2{x}^2+\cdots +{\overline{p}}_{n-1}{x}^{n-1}$, where ${\overline{p}}_i$ denotes the complement of $p_i$ for $p_i\in {\mathcal{R}}_k$, $0\le i\le n-1$. The reverse-complement of $p(x)\in {\mathcal{R}}_k[x]$ is denoted by $p^{\mathrm{rc}}(x)$ and is defined as: $p^{\mathrm{rc}}(x)={\overline{p}}_{n-1}+{\overline{p}}_{n-2}x+\cdots +{\overline{p}}_0{x}^{n-1}$.

Recall that the Watson-Crick complement for a given DNA sequence $D=d_1{d}_2\cdots d_n$ is given by $D^c=d_1^c{d}_2^c\cdots d_n^c$.

From Table 1, we notice that if $d_1{d}_2$ is a DNA sequence corresponding to element $z\in {\mathcal{R}}_1$, then $d_1^c{d}_2^c$ corresponds to z + 2. Similarly, from Table 2, we notice that if $d_1{d}_2{d}_3{d}_4$ is a DNA sequence corresponding to element $z\in {\mathcal{R}}_2$, then $d_1^c{d}_2^c{d}_3^c{d}_4^c$ corresponds to z + 2. Next we discuss the construction of two bijections ${\mathrm{\Gamma }}_1$ and ${\mathrm{\Gamma }}_2$ separately. Based on these constructions, one can obtain a bijection ${\mathrm{\Gamma }}_k$ between the elements of the ring ${\mathcal{R}}_k$ and DNA-$2^k$ bases, where k is any positive integer.

For k = 1

We define the complement of an element $a\in {\mathcal{R}}_1$ as $\overline{a}=a+2$. In general, a given DNA double pair $d_1{d}_2$ is associated with an element $a\in {\mathcal{R}}_1$ such that the Watson-Crick complement $d_1^c{d}_2^c$ corresponds to $\overline{a}$. Further, the reverse $d_2{d}_1$ is associated with $3a$. Notice that there are four DNA double pairs namely $\{\mathrm{AA},\mathrm{GG},\mathrm{TT},\mathrm{CC}\}$ such that the reverse of each codeword coincides with itself. These four DNA double pairs are associated with the four elements of the set $M=\{0,2,2+2u_1,2u_1\}$, where 3b = b for all $b\in M$. Moreover, there are four DNA double pairs namely $\{\mathrm{AT},\mathrm{GC},\mathrm{TA},\mathrm{CG}\}$ whose reverse coincides with their Watson-Crick complement. We associate these pairs with the four elements of the set $M^{\prime }=\{1,3,1+2u_1,3+2u_1\}$, where 3b = b + 2 for all $b\in M^{\prime }$.

For k = 2

In general, any given DNA-4 base $d_1{d}_2{d}_3{d}_4$ can be identified with an element $a\in {\mathcal{R}}_2$ such that the Watson-Crick complement $d_1^c{d}_2^c{d}_3^c{d}_4^c$ is associated with $\overline{a}$. Moreover, the reverse $d_4{d}_3{d}_2{d}_1$ is associated with $3a$. There are some special DNA sequences which are associated as follows: Consider a set $P=\{d_1{d}_2{d}_3{d}_4\in S_{D_{256}}:d_1{d}_2{d}_3{d}_4=d_4{d}_3{d}_2{d}_1\}$. Total number of DNA-4 bases in the set P is 16. On the other hand, cardinality of the set $P^{\ast }=\{a\in {\mathcal{R}}_2:3a=a\}$ is also 16. We identify all DNA-4 bases of the set P with the elements of the set $P^{\ast }$. Furthermore, consider the following set $Q=\{d_1{d}_2{d}_3{d}_4\in S_{D_{256}}:d_4{d}_3{d}_2{d}_1=d_1^c{d}_3^c{d}_2^c{d}_4^c\}$ and $|Q|=16$. There exists a subset $Q^{\ast }=\{a\in {\mathcal{R}}_2:3a=a+2\}$ of the ring ${\mathcal{R}}_2$ with cardilaity 16. We associate DNA-4 bases of the set Q, with elements of the set $Q^{\ast }$. Rest of the DNA-4 bases are associated with the above general rule. Similarly, we can construct bijections for other values of k, where k is any positive integer.

Next we describe how these bijections are helpful in solving the reversibility problem. In this manuscript, we construct bijections (${\mathrm{\Gamma }}_1$ & ${\mathrm{\Gamma }}_2$) between the elements of the ring ${\mathcal{R}}_k$ and DNA-$2^k$ bases (where k = 1, 2) in Tables 1 and 2 such that the reversibility problem is solved. Therefore, the following lemma plays a key role in solving the reversibility problem that occurred due to the DNA double pair.

Lemma 4.6

Suppose $a=(a_0,a_1,\dots ,a_{n-1})$ is a codeword of length n over the ring ${\mathcal{R}}_1$. Let $D=d_1{d}_2\cdots d_{2n-1}{d}_{2n}$ be the corresponding DNA sequence to the codeword a. Then, DNA sequence corresponding to the codeword $3a^r$ is the reverse of D i.e. $d_{2n}{d}_{2n-1}\cdots d_2{d}_1$.

Let $\theta =(1+u_1,2,3u_1,2+u_1)$ be a codeword over the ring ${\mathcal{R}}_1$. According to Table 1, the DNA sequence corresponding to θ is AGTTCATG. Now consider the codeword $3{\theta }^r=(3u_1+2,u_1,2,3+3u_1)$. By using Table 1, corresponding DNA sequence to the codeword $(3u_1+2,u_1,2,3+3u_1)$ is GTACTTGA. We can verify that GTACTTGA is the reverse of AGTTCATG. Similarly for k = 2, we obtain the following result.

Lemma 4.7

Suppose $a=(a_0,a_1,\dots ,a_{n-1})$ is a codeword of length n over the ring ${\mathcal{R}}_2$. Let $D=d_1{d}_2\cdots d_{4n-1}{d}_{4n}$ be the DNA sequence corresponding to the codeword a. Then, DNA sequence corresponding to the codeword $3a^r$ is the reverse of D i.e. $d_{4n}{d}_{4n-1}\cdots d_2{d}_1$.

Theorem 4.8

Let $\mathcal{C}$ be a cyclic code of odd length n over the ring ${\mathcal{R}}_k$ given by $\mathcal{C}=⟨e_1{F}_1(x),2e_1{H}_1(x),e_2{F}_2(x),2e_2{H}_2(x),\dots ,e_{2^k}{F}_{2^k}(x),2e_{2^k}{H}_{2^k}(x)⟩,$where $F_i(x)=f_i(x)g_i(x)$; $H_i(x)=f_i(x)h_i(x)$ and $x^n-1=f_i(x)g_i(x)h_i(x);$ $1\le i\le 2^k$. Then $\mathcal{C}$ is a reversible complement cyclic code over the ring ${\mathcal{R}}_k$ if and only if

(i)	the element $\frac{2(x^n-1)}{x-1}\in \mathcal{C}$,
(ii)	$f_i(x),g_i(x)$ and $h_i(x)$ are self-reciprocal polynomials.

Proof.

Suppose $\mathcal{C}$ is a reversible-complement cyclic code over ${\mathcal{R}}_k$. Assume $\mathbf{0}=(0,0,\dots ,0)\in \mathcal{C}$. Since $\mathcal{C}$ is reversible-complement, ${\mathbf{0}}^{\mathrm{rc}}\in \mathcal{C}$. Now, $\begin{array}{rl}(\overline{0},\overline{0},\dots ,\overline{0})& =(2,2,\dots ,2)\in \mathcal{C}\\ & =2(1,1,\dots ,1)\in \mathcal{C}\\ & =\frac{2(x^n-1)}{x-1}\in \mathcal{C},\end{array}$where $(1,1,\dots ,1)\in \mathcal{C}$ can be identified by $\frac{x^n-1}{x-1}$ in $\mathcal{C}$.

Now, for the first condition take any $c(x)\in \mathcal{C}$. Let $c(x)=[F_1(x){\ell }_1(x)+2e_1{H}_1(x){\ell }_2(x)+\cdots +e_{2^k}{F}_{2^k}(x){\ell }_{2^{k+1}-1}(x)+2e_{2^k}{H}_{2^k}(x){\ell }_{2^{k+1}}(x)],$where ${\ell }_i(x)\in {\mathcal{R}}_k[x]$, $1\le i\le k-1$. Then the reciprocal polynomial $c^{\ast }(x)$ is given by (2) $\begin{array}{rl}{c}^{\ast }(x)& =[F_1(x){\ell }_1(x)+2e_1{H}_1(x){\ell }_2(x)+\cdots +e_{2^k}{F}_{2^k}(x){\ell }_{2^{k+1}-1}(x)\\ & +2e_{2^k}{H}_{2^k}(x){\ell }_{2^{k+1}}(x){]}^{\ast },\\ c^{\ast }(x)& =F_1^{\ast }(x){\ell }_1^{\ast }(x)+2e_1{x}^{{\alpha }_1}{H}_1^{\ast }(x){\ell }_2^{\ast }(x)+\cdots +e_{2^k}{x}^{{\alpha }_{2^{k+1}-2}}{F}_{2^k}^{\ast }(x){\ell }_{2^{k+1}-1}^{\ast }(x)\\ & +2e_{2^k}{x}^{{\alpha }_{2^{k+1}-1}}{H}_{2^k}^{\ast }(x){\ell }_{2^{k+1}}^{\ast }(x).\end{array}$(2) Let $F_1(x)={\beta }_0^{\mathrm{\prime }}+{\beta }_1^{\mathrm{\prime }}x+\cdots +{\beta }_{r-1}^{\mathrm{\prime }}{x}^{r-1}+{\beta }_r^{\mathrm{\prime }}{x}^r,$be a polynomial over ${\mathcal{R}}_k$. Then the reverse of $F_1(x)$ is given by $F_1^r(x)={\beta }_r^{\mathrm{\prime }}+{\beta }_{r-1}^{\mathrm{\prime }}x+\cdots +{\beta }_1^{\mathrm{\prime }}{x}^{r-1}+{\beta }_0^{\mathrm{\prime }}{x}^r,$and the reverse-complement is given by $F_1^{\mathrm{rc}}(x)={\overline{\beta }}_r^{\mathrm{\prime }}+{\overline{\beta }}_{r-1}^{\mathrm{\prime }}x+\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{r-1}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^r.$Now multiplying both sides of the above expression by $x^{n-r-1}$ we obtain (3) $\begin{array}{rl}(x^{n-r-1})F_1^{\mathrm{rc}}(x)& ={\overline{\beta }}_r^{\mathrm{\prime }}{x}^{n-r-1}+{\overline{\beta }}_{r-1}^{\mathrm{\prime }}{x}^{n-r}+\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{n-2}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^{n-1}\\ & =\overline{0}+\overline{0}x+\cdots +\overline{0}{x}^{n-r-2}+{\overline{\beta }}_r^{\mathrm{\prime }}{x}^{n-r-1}\\ & +{\overline{\beta }}_{r-1}^{\mathrm{\prime }}{x}^{n-r}+\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{n-2}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^{n-1}\\ (x^{n-r-1})F_1^{\mathrm{rc}}(x)& =2+2x+\cdots +2x^{n-r-2}+{\overline{\beta }}_r^{\mathrm{\prime }}{x}^{n-r-1}+{\overline{\beta }}_{r-1}^{\mathrm{\prime }}{x}^{n-r}\\ & +\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{n-2}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^{n-1}.\end{array}$(3) On adding $\frac{2(x^n-1)}{x-1}$ to both sides of (3(3) $\begin{array}{rl}(x^{n-r-1})F_1^{\mathrm{rc}}(x)& ={\overline{\beta }}_r^{\mathrm{\prime }}{x}^{n-r-1}+{\overline{\beta }}_{r-1}^{\mathrm{\prime }}{x}^{n-r}+\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{n-2}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^{n-1}\\ & =\overline{0}+\overline{0}x+\cdots +\overline{0}{x}^{n-r-2}+{\overline{\beta }}_r^{\mathrm{\prime }}{x}^{n-r-1}\\ & +{\overline{\beta }}_{r-1}^{\mathrm{\prime }}{x}^{n-r}+\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{n-2}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^{n-1}\\ (x^{n-r-1})F_1^{\mathrm{rc}}(x)& =2+2x+\cdots +2x^{n-r-2}+{\overline{\beta }}_r^{\mathrm{\prime }}{x}^{n-r-1}+{\overline{\beta }}_{r-1}^{\mathrm{\prime }}{x}^{n-r}\\ & +\cdots +{\overline{\beta }}_1^{\mathrm{\prime }}{x}^{n-2}+{\overline{\beta }}_0^{\mathrm{\prime }}{x}^{n-1}.\end{array}$(3) ), we obtain (4) $\begin{array}{rl}(x^{n-r-1})F_1^{\mathrm{rc}}(x)+\frac{2(x^n-1)}{x-1}& =0+0x+0x^2+\cdots +0x^{n-r-2}+{\beta }_r^{\mathrm{\prime }}{x}^{n-r-1}\\ & +{\beta }_{r-1}^{\mathrm{\prime }}{x}^{n-r}+\cdots +{\beta }_1^{\mathrm{\prime }}{x}^{n-2}+{\beta }_0^{\mathrm{\prime }}{x}^{n-1}\\ & =x^{n-r-1}[{\beta }_r^{\mathrm{\prime }}+{\beta }_{r-1}^{\mathrm{\prime }}x+\cdots +{\beta }_0^{\mathrm{\prime }}{x}^r]\\ (x^{n-r-1})F_1^{\mathrm{rc}}(x)+\frac{2(x^n-1)}{x-1}& =x^{n-r-1}{F}_1^{\ast }(x).\end{array}$(4) Notice that $\mathcal{C}$ is a reversible-complement cyclic code and condition (ii) holds, using (4(4) $\begin{array}{rl}(x^{n-r-1})F_1^{\mathrm{rc}}(x)+\frac{2(x^n-1)}{x-1}& =0+0x+0x^2+\cdots +0x^{n-r-2}+{\beta }_r^{\mathrm{\prime }}{x}^{n-r-1}\\ & +{\beta }_{r-1}^{\mathrm{\prime }}{x}^{n-r}+\cdots +{\beta }_1^{\mathrm{\prime }}{x}^{n-2}+{\beta }_0^{\mathrm{\prime }}{x}^{n-1}\\ & =x^{n-r-1}[{\beta }_r^{\mathrm{\prime }}+{\beta }_{r-1}^{\mathrm{\prime }}x+\cdots +{\beta }_0^{\mathrm{\prime }}{x}^r]\\ (x^{n-r-1})F_1^{\mathrm{rc}}(x)+\frac{2(x^n-1)}{x-1}& =x^{n-r-1}{F}_1^{\ast }(x).\end{array}$(4) ) we obtain $F_1^{\ast }(x)\in \mathcal{C}$. Using the same argument, we must have $F_2^{\ast }(x),F_3^{\ast }(x),\dots ,F_{2^k}^{\ast }(x)\in \mathcal{C};$ $H_1^{\ast }(x),H_2^{\ast }(x),\dots ,H_{2^k}^{\ast }(x)\in \mathcal{C}$, and hence from (2(2) $\begin{array}{rl}{c}^{\ast }(x)& =[F_1(x){\ell }_1(x)+2e_1{H}_1(x){\ell }_2(x)+\cdots +e_{2^k}{F}_{2^k}(x){\ell }_{2^{k+1}-1}(x)\\ & +2e_{2^k}{H}_{2^k}(x){\ell }_{2^{k+1}}(x){]}^{\ast },\\ c^{\ast }(x)& =F_1^{\ast }(x){\ell }_1^{\ast }(x)+2e_1{x}^{{\alpha }_1}{H}_1^{\ast }(x){\ell }_2^{\ast }(x)+\cdots +e_{2^k}{x}^{{\alpha }_{2^{k+1}-2}}{F}_{2^k}^{\ast }(x){\ell }_{2^{k+1}-1}^{\ast }(x)\\ & +2e_{2^k}{x}^{{\alpha }_{2^{k+1}-1}}{H}_{2^k}^{\ast }(x){\ell }_{2^{k+1}}^{\ast }(x).\end{array}$(2) ), we get $c^{\ast }(x)\in \mathcal{C}$ for ${\ell }_i^{\ast }(x)\in {\mathcal{R}}_k[x]$. Therefore, $\mathcal{C}$ is a reversible cyclic code.

Conversely, assume that conditions (i) & (ii) are satisfied. Then, we have to show that $\mathcal{C}$ is a reversible-complement cyclic code. Since $\mathcal{C}$ is a reversible cyclic code, $P^{\ast }(x)\in \mathcal{C}$ for all $P(x)\in \mathcal{C}$. Let $P(x)=p_0+p_{1}x+\cdots +p_m{x}^m\in \mathcal{C}$be an arbitrary codeword. Then the reverse-complement of $P(x)$ is given by $P^{\mathrm{rc}}(x)={\overline{p}}_m+{\overline{p}}_{m-1}x+\cdots +{\overline{p}}_1{x}^{m-1}+{\overline{p}}_0{x}^m,$multiplying by $x^{n-m-1}$ on both sides in the above polynomial, we obtain $(x^{n-m-1})P^{\mathrm{rc}}(x)={\overline{p}}_m{x}^{n-m-1}+{\overline{p}}_{m-1}{x}^{n-m}+\cdots +{\overline{p}}_1{x}^{n-2}+{\overline{p}}_0{x}^{n-1}.$On adding $\frac{2(x^n-1)}{x-1}$ to both sides of the above expression, we obtain $\begin{array}{rl}(x^{n-m-1})P^{\mathrm{rc}}+\frac{2(x^n-1)}{x-1}& =x^{n-m-1}[p_m+p_{m-1}x+\cdots +p_0{x}^m]\\ & =x^{n-m-1}{P}^{\ast }(x).\end{array}$Now, on simplifying the above expression we find that (5) $(x^{n-m-1})P^{\mathrm{rc}}(x)=x^{n-m-1}{P}^{\ast }(x)-\frac{2(x^n-1)}{x-1}.$(5) Utilizing conditions (i) and (ii) in theorem with Equation (5(5) $(x^{n-m-1})P^{\mathrm{rc}}(x)=x^{n-m-1}{P}^{\ast }(x)-\frac{2(x^n-1)}{x-1}.$(5) ) leads to the conclusion that $P^{\mathrm{rc}}(x)$ belongs to $\mathcal{C}$. Consequently, we must have $\mathcal{C}$ is a reversible-complement cyclic code over the ring ${\mathcal{R}}_k$.

5. The GC–content

The thermodynamic stability of the bond formed between the complementary nucleotide bases C and G exceeds that of the bond formed between the complementary nucleotide bases A and T. Consequently, DNA sequences tend to exhibit a preference for being rich in GC content. Moreover, in DNA codes the same GC-content in every codeword ensures that the codewords have similar hybridization energy and melting temperature. This motivates to consider such DNA codes in which all codewords have the same GC-content. In this section, we construct a Gray map Φ which can be utilize to obtain the GC-content. Suppose $S_{D_4}=\{A,G,T,C\}$ are the nucleotides. We identify these four nucleotides with the elements of the ring ${\mathbb{Z}}_4$ as follows $0\to A$, $1\to G$, $3\to C$ and $2\to T$. Note that every element $x\in {\mathcal{R}}_k$ can be uniquely written as $\begin{array}{rl}x& ={\alpha }_1+{\alpha }_2{u}_1+\cdots +{\alpha }_{k+1}{u}_k+u_1{u}_2{\alpha }_{k+1}+u_1{u}_3{\alpha }_{k+2}+\cdots +u_{k-1}{u}_k{\alpha }_{\frac{k^2+k+2}{2}}\\ & +\cdots +u_1{u}_2\cdots u_k{\alpha }_{2^k},\end{array}$where ${\alpha }_i\in {\mathbb{Z}}_4;$ $1\le i\le 2^k$. We construct a map $\mathrm{\Phi }:{\mathcal{R}}_k\to {\mathbb{Z}}_4^{2^k}$ as follows: $\mathrm{\Phi }(x)=({\alpha }_1,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_2+{\alpha }_3,\dots ,{\alpha }_1+{\alpha }_2+\cdots +{\alpha }_{2^k}).$It is worth to notice that $d_L(x,y)=d_L(\mathrm{\Phi }(x),\mathrm{\Phi }(y))=w_L(\mathrm{\Phi }(x)-\mathrm{\Phi }(y))$. Therefore, we can conclude that Φ is a Lee distance preserving map. Moreover, Φ is a ${\mathbb{Z}}_4$–linear. Similarly, we can extend the map $\mathrm{\Phi }:{\mathcal{R}}_k^n\to {\mathbb{Z}}_4^{2^{k}n}$.

Theorem 5.1

Suppose $\mathcal{C}$ is a linear code of length n over the ring ${\mathcal{R}}_k$ with minimum Lee distance $d_L$. Then, $\mathrm{\Phi }(\mathcal{C})$ is a ${\mathbb{Z}}_4-$linear code of length $2^{k}n$ and minimum Lee distance $d_L$.

Moreover, since each element of ${\mathbb{Z}}_4$ is identified by the nucleotides $S_{D_4}$, Φ can also be viewed as $\mathrm{\Phi }:{\mathcal{R}}_k\to S_{D_4}^{2^k}$.

The next theorem provides the minimal generating set of a given cyclic code of odd length over the ring ${\mathbb{Z}}_4$.

Theorem 5.2

[38]

Let $\mathcal{C}$ be a cyclic code of odd length n over the ring ${\mathbb{Z}}_4$ given by $\mathcal{C}=⟨f(x)g(x),2f(x)h(x)⟩$, where $f(x)$, $g(x)$ and $h(x)$ are monic polynomials such that $x^n-1=f(x)g(x)h(x)$. Then minimal generating set of $\mathcal{C}$ is given by $\{f(x)g(x),\mathrm{xf}(x)g(x),\dots ,x^{\mathrm{deg}(h(x))-1}f(x)g(x),2f(x)h(x),2\mathrm{xf}(x)h(x),\dots ,2x^{\mathrm{deg}(g(x))-1}f(x)h(x)\}$.

We obtain the minimal generating set for a cyclic code of odd lengths over the ring ${\mathcal{R}}_k$ as follows:

Theorem 5.3

Let $\mathcal{C}$ be a cyclic code of odd length n over the ring ${\mathcal{R}}_k$ given by $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}.$Then minimal generating set Γ of $\mathcal{C}$ is given by $\mathrm{\Gamma }=e_1{\mathrm{\Gamma }}_1+e_2{\mathrm{\Gamma }}_2+\cdots +e_{2^k}{\mathrm{\Gamma }}_{2^k},$where ${\mathrm{\Gamma }}_i$ is the minimal generating set of ${\mathcal{C}}_i$.

To obtain the GC-content over the ring ${\mathcal{R}}_k$ we use a method presented in [26], where the authors obtained the GC-content over the ring ${\mathcal{R}}_k$ for k = 1 in terms of Gray images of minimal generating set of a code over the ring ${\mathcal{R}}_k$. The Gray image of minimal generating set Γ is given by $\mathrm{\Phi }(\mathrm{\Gamma })=x^n{\mathrm{\Gamma }}_1+{\mathrm{\Gamma }}_2+\cdots +{\mathrm{\Gamma }}_{2^k}$. The GC-content is given by the Hamming weight of 2$\mathrm{\Phi }(\mathrm{\Gamma })$. Therefore, the next theorem provides the GC-content.

Theorem 5.4

Suppose $\mathcal{C}$ is a cyclic code of odd length n over the ring ${\mathcal{R}}_k$ given by $\mathcal{C}=e_1{\mathcal{C}}_1+e_2{\mathcal{C}}_2+\cdots +e_{2^k}{\mathcal{C}}_{2^k}$, where ${\mathcal{C}}_i=⟨f_i(x)g_i(x),2f_i(x)h_i(x)⟩$ and $f_i(x)g_i(x)h_i(x)=x^n-1$. Then the GC-content over the ring ${\mathcal{R}}_k$ is the Hamming weight enumerator of $\begin{array}{rl}\sum & =x^n\{f_1(x)g_1(x),x{f}_1(x)g_1(x),\dots ,x^{\mathrm{deg}(h_1(x))-1}{f}_1(x)g_1(x)\}\\ & +\{f_2(x)g_2(x),x{f}_2(x)g_2(x),\dots ,x^{\mathrm{deg}(h_2(x))-1}{f}_2(x)g_2(x)\}\\ & +\cdots +\{f_{2^k}(x)g_{2^k}(x),x{f}_{2^k}(x)g_{2^k}(x),\dots ,x^{\mathrm{deg}(h_{2^k}(x))-1}{f}_{2^k}(x)g_{2^k}(x)\}.\end{array}$

Proof.

Since GC-content can be obtained by multiplying Gray images of minimal generating sets of cyclic code $\mathcal{C}$ with 2, using Theorem 5.3 we obtain $2\mathrm{\Phi }(\mathrm{\Gamma })=2x^n{\mathrm{\Gamma }}_1+2{\mathrm{\Gamma }}_2+\cdots +2{\mathrm{\Gamma }}_{2^k}.$Therefore, the GC-content is given by the Hamming weight enumerator of $\begin{array}{rl}\sum & =x^n\{f_1(x)g_1(x),x{f}_1(x)g_1(x),\dots ,x^{\mathrm{deg}(h_1(x))-1}{f}_1(x)g_1(x)\}\\ & +\{f_2(x)g_2(x),x{f}_2(x)g_2(x),\dots ,x^{\mathrm{deg}(h_2(x))-1}{f}_2(x)g_2(x)\}\\ & +\cdots +\{f_{2^k}(x)g_{2^k}(x),x{f}_{2^k}(x)g_{2^k}(x),\dots ,x^{\mathrm{deg}(h_{2^k}(x))-1}{f}_{2^k}(x)g_{2^k}(x)\}.\end{array}$

Example 5.5

Let k = 1, ${\mathcal{R}}_1={\mathbb{Z}}_4+u_1{\mathbb{Z}}_4$ be the ring, where $u_1^2=u_1$. Factorize the polynomial $x^9-1$ over the ring ${\mathbb{Z}}_4$ as follows: $x^9-1=(x-1)(x^2+x+1)(x^6+x^3+1)$. Suppose $\mathcal{C}$ is a cyclic code over the ring ${\mathcal{R}}_1$ of length 9 given by $\mathcal{C}=⟨e_1{f}_1{g}_1,2e_1{f}_1{h}_1⟩$, where $f_1=(x-1)$, $g_1=(x^2+x+1)$ and $h_1=(x^6+x^3+1)$. Therefore, by using Theorem 5.4, the GC-content of the cyclic code $\mathcal{C}$ is given by $w{t}_H(x^9\{x^3+3,x(x^3+3),x^2(x^3+3),x^3(x^3+3),x^4(x^3+3),x^5(x^3+3)\})=2.$

6. Examples

Here, we provide some examples of reversible cyclic codes of different odd lengths over the ring ${\mathcal{R}}_k$, for k = 1, 2. After comparing the Gray images of these reversible cyclic codes with online data available in [39], we find that some reversible codes have good parameters. Moreover, we construct few reversible complement codes of different length over the ring ${\mathcal{R}}_k$.

Example 6.1

For k = 1 consider the ring ${\mathcal{R}}_1={\mathbb{Z}}_4+u_1{\mathbb{Z}}_4$, $u_1^2=u_1$. We factorize the polynomial $x^9-1$ over ${\mathbb{Z}}_4$ as follows: $x^9-1=(x-1)(x^2+x+1)(x^6+x^3+1).$Then various reversible cyclic codes of length 9 and their ${\mathbb{Z}}_4$ Gray images are listed in Table 5. Suppose $\mathcal{C}=⟨(x^2+x+1)(x^6+x^3+1)⟩$ be cyclic code of length 9 over the ring ${\mathcal{R}}_1$. It is worth to notice that $\frac{2(x^9-1)}{x-1}=2(x^2+x+1)(x^6+x^3+1)\in \mathcal{C}$. Therefore, by using Theorem 4.8, $\mathcal{C}$ is a reversible complement code over the ring ${\mathcal{R}}_1$. Moreover, ${\mathrm{\Gamma }}_1(\mathcal{C})$ is a DNA code of length 18, given in Table 6.

Table 5. Reversible cyclic codes of length 9 over the ring ${\mathcal{R}}_1$.

Generating polynomial(s) of ${\mathcal{C}}_i$	${\mathbb{Z}}_4$-Gray images
$(x^2+x+1)(x^6+x^3+1)$	$[18,4^2{2}^0,9]$
$(x^6+x^3+1)$	$[18,4^6{2}^0,3]$
$(x^2+x+1)(x^6+x^3+1),2(x^2+x+1)$	$[18,4^2{2}^{14},4]$
$(x^2+x+1)(x^6+x^3+1),2(x^6+x^3+1)$	$[18,4^2{2}^6,4]$
$2(x^6+x^3+1)$	$[18,4^0{2}^6,6]$
$2(x^2+x+1)$	$[18,4^0{2}^{14},4]$
$2(x^2+x+1)(x^6+x^3+1)$	$[18,4^0{2}^2,18]$

Table 6. DNA code of length 18.

TGTGTGTGTGTGTGTGTG

CGCGCGCGCGCGCGCGCG

ATATATATATATATATAT

AGAGAGAGAGAGAGAGAG

CTCTCTCTCTCTCTCTCT

CCCCCCCCCCCCCCCCCC

AAAAAAAAAAAAAAAAAA

TTTTTTTTTTTTTTTTTT

GAGAGAGAGAGAGAGAGA

GGGGGGGGGGGGGGGGGG

GTGTGTGTGTGTGTGTGT

GCGCGCGCGCGCGCGCGC

CACACACACACACACACA

TCTCTCTCTCTCTCTCTC

ACACACACACACACACAC

TATATATATATATATATA

Example 6.2

Consider the factorization $\begin{array}{rl}{x}^{17}-1& =(x-1)(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)\\ & \times (x^8+x^7+3x^6+3x^4+3x^2+x+1)\end{array}$over the ring ${\mathbb{Z}}_4$. Let k = 1 and ${\mathcal{R}}_1={\mathbb{Z}}_4+u_1{\mathbb{Z}}_4$, where $u_1^2=u_1$ be the ring. Then we provide some reversible cyclic codes over the ring ${\mathcal{R}}_1$ and their ${\mathbb{Z}}_4$ Gray images which are given in Table 4. Let $\mathcal{C}=⟨(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)(x^8+x^7+3x^6+3x^4+3x^2+x+1)⟩$ be a cyclic code of length 17 over the ring ${\mathcal{R}}_1$. Then $\frac{2(x^{17}-1)}{x-1}=2(x^8+2x^6+3x^5+x^4+3x^3+2x^2+1)(x^8+x^7+3x^6+3x^4+3x^2+x+1)\in \mathcal{C}$. Therefore, by using Theorem 4.8, $\mathcal{C}$ will be a reversible complement code. Moreover, ${\mathrm{\Gamma }}_1(\mathcal{C})$ is a DNA code of length 34, given in Table 7.

Table 7. DNA code of length 34.

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

GCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGC

TCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTC

ACACACACACACACACACACACACACACACACAC

CGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCG

AGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAG

TGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTG

ATATATATATATATATATATATATATATATATAT

CTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCT

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

GAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGA

GTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGT

CACACACACACACACACACACACACACACACACA

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

TATATATATATATATATATATATATATATATATA

Example 6.3

Suppose that k = 2, and consider the ring ${\mathcal{R}}_2={\mathbb{Z}}_4+u_1{\mathbb{Z}}_4+u_2{\mathbb{Z}}_4+u_1{u}_2{\mathbb{Z}}_4$ where $u_1^2=u_1$, $u_2^2=u_2$, and $u_1{u}_2=u_2{u}_1$. We obtain the factors of the polynomial $x^{25}-1$ over the ring ${\mathbb{Z}}_4$ as follows: $x^{25}-1=(x-1)(x^4+x^3+x^2+x+1)(x^{20}+x^{15}+x^{10}+x^5+1).$Then different reversible cyclic codes of length 25 are listed in Table 8.

Table 8. Reversible cyclic codes of length 25 over the ring ${\mathcal{R}}_2$.

Generating polynomial(s) of ${\mathcal{C}}_i$	${\mathbb{Z}}_4$-Gray images
$x^4+x^3+x^2+x+1$	$[100,4^{84}{2}^0,2]$
$x^{20}+x^{15}+x^{10}+x^5+1$	$[100,4^{20}{2}^0,5]$
$(x^4+x^3+x^2+x+1)(x^{20}+x^{15}+x^{10}+x^5+1)$	$[100,4^4{2}^0,25]$
$2(x^4+x^3+x^2+x+1)$	$[100,4^0{2}^{84},4]$
$2(x^{20}+x^{15}+x^{10}+x^5+1)$	$[100,4^0{2}^{20},10]$
$2(x^4+x^3+x^2+x+1)(x^{20}+x^{15}+x^{10}+x^5+1)$	$[100,4^0{2}^4,50]$

7. Conclusion

In this article, we have identified elements of the ring ${\mathcal{R}}_k$ with DNA-$2^k$ bases for k = 1, 2 in such a manner that the reversibility problem occur in DNA-$2^k$ bases is solved. Using this idea, one can construct bijections between the elements of the ring ${\mathcal{R}}_k$ and DNA-$2^k$ bases for other choices of k. We have constructed DNA codes which satisfy the Hamming constraint, the reverse constraint and the reverse-complement constraints. We have provided a necessary and sufficient for a given cyclic code of odd length over the ring ${\mathcal{R}}_k$ to be reversible and as an application, we have constructed some examples of reversible codes and obtained their ${\mathbb{Z}}_4$-Gray images. Furthermore, some reversible complement codes of any odd length over the ring ${\mathcal{R}}_k$ have been generated. We obtained the GC-content of a given cyclic code over the ring ${\mathcal{R}}_k$ as the Gray images of the Hamming weight enumerator. For future study, it would be fascinating to construct reversible cyclic codes of even lengths over the ring ${\mathcal{R}}_k$.

Remark

All codes in the above examples are calculated using Magma software [40].

Acknowledgments

The authors extend their gratitude to the referees for their meticulous examination of the manuscript, constructive comments, and valuable suggestions, all of which have greatly enhanced the paper's quality. Moreover, the authors extend their appreciation to Princess Nourah Bint Abdulrahman University (PNU), Riyadh, Saudi Arabia for funding this research under researchers supporting number (PNURSP2024R231).

Data availability statement

This manuscript has no associated data set.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

The authors extend their appreciation to Princess Nourah bint Abdulrahman University for funding this research under Research Supporting Project number (PNURSP2024R231).