ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.1000959 J. Algebra Comb. Discrete Appl. 8(3) • 219–231 Received: 07 December 2020 Accepted: 29 April 2021 Journal of Algebra Combinatorics Discrete Structures and Applications Cyclic DNA codes over the ring Z4 + uZ4 + u2Z4 Research Article Karthick Gowdhaman, Somi Gupta, Cruz Mohan, Kenza Guenda, Durairajan Chinnapillai Abstract: In this work, we have investigated the one generator cyclic DNA codes with reverse and reverse complement constraints over the ring R = Z4+uZ4+u2Z4 with u3 = 0. Skew cyclic codes with reverse complement constraint are constructed over R. We have also determined a one-to-one correspondence between the elements of the ring R and DNA codons satisfying the Watson-Crick complement. Finally, we have established some examples which satisfy the given constraints. 2010 MSC: 94B05, 94B15 Keywords: DNA codes, Skew codes, Reversible codes 1. Introduction DNA is shortened word of Deoxyribonucleic acid. It is a molecule made of two chains that curl around one another to frame a two-fold helix conveying the hereditary guidelines utilized in the development of all creatures. This two-fold helix is assembled by blending the four fundamental structure units A- (ADENINE), C-(CYTOSINE), G-(GUANINE) and T-(THYMINE) which are called the nucleotides held by Hydrogen ties. The DNA strand is held by an important feature called complementary base pairing which connects the Watson-Crick complementary bases with each other denoted by A = T,G = C,C = G,T = A. In [4] Adleman performed a successful experiment for DNA computing. The basic idea of his work was to use DNA which is an ideal source of computing due to its dense and self-replicating property to solve a mathematical problem. After this successful experiment, the area of DNA computing was flooded with different approaches such as DNA tile assembly, the building of DNA nanostructures, DNA-based data storage system and study of error-correcting properties of DNA. Karthick Gowdhaman, Cruz Mohan, Durairajan Chinnapillai; Department of Mathematics, Bharathidasan University, Tiruchirapalli, Tamil Nadu, India (email: karthimath123@bdu.ac.in, cruzmohan@gmail.com, cdu- rai66@rediffmail.com). Somi Gupta; Department of Mathematics and Applications "Renato Caccioppoli", University of Napoli Federico II, Naples, Italy (email: gupta7somi200@gmail.com). Kenza Guenda; Laboratory of Algebra and Number Theory, USTHB, Algiers (email: ken.guenda@gmail.com). 219 https://orcid.org/0000-0003-0104-2993 https://orcid.org/0000-0002-1482-7565 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Cyclic codes over rings have been studied by many authors (see for example [6] and [2]). Later DNA cyclic codes have gained interest of many researchers for their applications (see [16], [11], [18], [12], [14], [19], [5]). Abhay et al. [10] have studied DNA cyclic codes over the ring Z4 + uZ4 with u2 = 0. Cyclic codes over the ring R = Z4 + uZ4 + u2Z4 with u3 = 0 have been studied by Ozen et al. in [15]. Cyclic code over skew polynomial ring have been constructed by [7]. Boucher and Ulmer [8] have found a link between arithmetic structure of skew polynomials and existence of such codes. In [17] Siap et al. studied skew cyclic code of arbitrary length and established a strong connection with well known codes. Most recently, the DNA codes over ring of order 256 has been studied in [9]. They have obtained DNA skew cyclic codes over the ring F2 + uF2 + vF2 + wF2 + uvF2 + uwF2 + vwF2 + uvwF2, where, u2 = 0,v2 = v,w2 = w,uv = vu,uw = wu,vw = wv, addressing reversibility problem. In this work, we have investigated the one generator cyclic DNA codes with reverse and reverse complement constraint over the ring R = Z4 + uZ4 + u2Z4 with u3 = 0. Skew cyclic codes with reverse complement constraint are constructed over R. We have also determined a one-to-one correspondence between elements of the ring R and DNA codons satisfying Watson-Crick complement. Finally, we have established some examples which satisfy the given constraints. Beside the theoretical results concerning the reverse and the reverse complement codes over the ring R and the one-to-one correspondence with codes, our other motivation in choosing this ring is the fact that it contains an additive subgroup of order 16. Then the idea of the additive stem distance characterizing the hybridization energy given in [12] can be extended to this subgroup and then to the ring. This paper is structured in the following way: Section 2 contains basic definitions of cyclic DNA codes. We have established a one-to-one correspondence between elements of the ring R = Z4+uZ4+u2Z4 with u3 = 0, elements of Z34 and 64 codons. In Section 3, we have studied the reversibility condition of one generator cyclic codes over the ring R and have determined reverse complement codes. Binary image of the ring and skew cyclic codes are studied in Sections 4 and 5 respectively. In Section 6, we have obtained some examples. Finally, Section 7 concludes the paper with some establishments of future work that can be done using this work. 2. Preliminaries Ozen et al. in [15] have determined cyclic codes over the ring R = Z4 + uZ4 + u2Z4, where u3 = 0. They obtained that the total number of cyclic codes over R is 13r. They also found the general form of generator of cyclic codes over R and one generator cyclic codes (cyclic codes generated by one element) over the ring R. Let R = Z4 +uZ4 +u2Z4 = {a+ub+u2c | a,b,c ∈ Z4}, where u3 = 0. The ring R has 11 nontrivial ideals given by A2u2 = {0,2u2} Au2 = {0,u2,2u2,3u2} A2u = {0,2u,2u2,2u + 2u2} A2u+u2 = {0,2u2,2u + u2,2u + 3u2} A2u,u2 = {0,u2,2u2,3u2,2u,2u + u2,2u + 2u2,2u + 3u2} A2 = {0,2,2u,2u2,2 + 2u,2 + 2u2,2u + 2u2,2 + 2u + 2u2} A2+u2 = {0,2u,2u2,2u + 2u2,2 + u2,2 + 3u2,2 + 2u + u2,2 + 2u + 3u2} A2,u2 = {0,2,2u,u2,2u2,3u2,2 + 2u,2 + u2, 2 + 2u2,2 + 3u2,2u + u2,2u + 2u2, 2u + 3u2,2 + 2u + u2,2 + 2u + 2u2,2 + 2u + 3u2} 220 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Au = {0,u,2u,3u,u2,2u2,3u2,u + u2,u + 2u2,u + 3u2,2u + u2,2u + 2u2, 2u + 3u2,3u + u2,3u + 2u2,3u + 3u2} A2+u = {0,2u,u2,2u2,3u2,2 + u,2 + 3u,2u + u2,2u + 2u2,2u + 3u2,2 + u + u2, 2 + u + 2u2,2 + u + 3u2,2 + 3u + u2,2 + 3u + 2u2,2 + 3u + 3u2} A2,u = {0,2,u,2u,3u,u2,2u2,3u2,2 + u,2 + 2u,2 + 3u,2 + u2,2 + 2u2,2 + 3u2, u + u2,u + 2u2,u + 3u2,2u + u2,2u + 2u2,2u + 3u2,3u + u2,3u + 2u2, 3u + 3u2,2 + u + u2,2 + u + 2u2,2 + u + 3u,2 + 2u + u2,2 + 2u + 2u2, 2 + 2u + 3u2,2 + 3u + u2,2 + 3u + 2u2,2 + 3u + 3u2}. The ring R is a finite local ring with A2,u as its unique maximal ideal (as it contains all the non-zero divisors of R). The residue field of R is K = R/A2,u = {0 + A2, u,1 + A2, u}∼= Z2. Let p(x) be a monic basic irreducible polynomial of degree m in R[x], then the Galois ring extension over R is defined by the residue class ring Qm = R[x]/(p(x)) having 64m elements. The triplet of nucleotides called codons is the basic coding unit for amino acids during the protein synthesis in living organisms. Since the ring R has 64 elements, then it is convenient to describe a one-to-one correspondence between the elements of R and the codons. In the following table, we give a one-to-one correspondence between the elements of R, the elements of Z34 and the codons. The one- to-one correspondence, called Gray map between the rings R and Z34 is defined as ψ : R → Z34 with (a + ub + u2c) 7→ (a,b,c), where a,b,c ∈ Z4. Another important one-to-one correspondence between the elements of Z34 and D 3 is denoted by φ : Z34 → D3, where D denotes the set of all nucleotides. The full names of codons have been taken from http://www.hgmd.cf.ac.uk/docs/cd_amino.html. Table 1. Correspondence between elements of R, Z34 and Codons a ∈ R ψ(a) ∈ Z34 φ(a) ∈ D3 Full Name of Codons 0 (0, 0, 0) GTG Valine u (0, 1, 0) CCG Alanine u+u2 (0, 1, 1) CGA Arginine u+2u2 (0, 1, 2) CAA Glutamine u+3u2 (0, 1 ,3) TAG Termination (amber) 2u+u2 (0,2 , 1) CAT Histidine 2u+2u2 (0, 2, 2) CTC Serine 2u+3u2 (0, 2, 3) ATA Isoleucine 3u+u2 (0, 3, 1) GCG Alanine 3u+2u2 (0, 3, 2) TCG Serine 3u+3u2 (0, 3, 3) TAA Termination (ochre) 1 (1, 0, 0) GGC Arginine 1+u (1, 1, 0) TGA Termination (opal or umber) 1+u+u2 (1, 1, 1) GGG Glycine 1+ u+2u2 (1, 1, 2) GCA Alanine 1+u+3u2 (1, 1, 3) GAA Glutamate 1+2u (1, 2, 0) CCA Proline 1+ 2u+u2 (1, 2, 1) TTT Phenylalanine 1+2u+2u2 (1, 2, 2) CCT Proline 1+2u+3u2 (1, 2, 3) AGG Arginine 1+3u (1, 3, 0) TGG Trytophan 1+3u+u2 (1, 3, 1) TTA Leucine 1+3u+2u2 (1, 3, 2) TAC Tyrosine 1+3u+3u2 (1, 3, 3) TGC Cysteine 2 (2, 0, 0) GAG Arginine 2u (0, 2,0) ACA Threonine 2+u (2, 1, 0) AGC Serine 2+2u (2, 2, 0) AGA Glutamate 221 http://www.hgmd.cf.ac.uk/docs/cd_amino.html K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 2+ u+u2 (2, 1, 1) ATT Isoleucine 2+ u+2u2 ( 2, 1, 2) GAT Aspartate 2+u+3u2 (2, 1, 3) CGC Arginine 2+2u+u2 (2, 2, 1) CAG Glutamine 2+2u+2u2 (2, 2, 2) CAC Histidine 2+2 u+3u2 (2, 2, 3) GAC Aspartate 2+3u (2, 3, 0) GTT Valine 2+3 u+u2 (2, 3, 1) ATC Isoleucine 2+3u+2u2 (2, 3, 2) GGC Glycine 2+3 u+3u2 (2, 3, 3) GCT Valine 3 (3, 0, 0) GGA Glycine 3u (0, 3, 0) CTA Leucine 3+u (3, 1, 0) ATG Methionine 3+u+u2 (3, 1, 1) ACG Threonine 3+u+2u2 (3, 1, 2) ACC Threonine 3+u+3u2 (3, 1, 3) AAT Asparagine 3+2u (3, 2, 0) TCA Serine 3+2u+u2 (3, 2, 1) AAG Lysine 3+2u+2u2 (3, 2, 2) CCG Alanine 3+2u+3u2 (3, 2, 3) TTG Leucine 3+3u (3, 3, 0) CGT Arginine 3+3u+u2 (3, 3, 1) CTT Leucine 3+3u+2u2 (3, 3, 2) ACT Threonine 3+3u+3u2 (3 ,3 ,3) CCC Proline 1+u2 (1, 0, 1) AAC Asparagine 1+2u2 (1, 0, 2) AGT Serine 1+3u2 (1, 0, 3) TTC Phenylalanine 2+u2 (2, 0 , 1) TAT Tyrosine 2+2u2 (2, 0, 2) TGT Cysteine 2+3u2 (2, 0, 3) GTA Valine 3+u2 (3, 0, 1) TCC Serine 3+2u2 (3, 0, 2) GGT Glycine 3+3u2 (3, 0,3) AAA Lysine u2 (0, 0, 1) CTG Leucine 2u2 (0, 0, 2) TCT Leucine 3u2 (0, 0, 3) GTC Valine Definition 2.1. Let a,b ∈ Z4, then we define a distance called Gray distance by dG(a,b) = dH(φ(a),φ(b)), where φ(a),φ(b) ∈ D3. Note that it can be extended upto length n. Hence the map φ is a distance preserving map from (Rn,dG) to (D3n,dH). Example 2.2. Let a = 1 + u + u2 and b = 2 + 2u + 2u2, then dG(a,b) = dH(GGG,CAC) = 3. Definition 2.3. If C is invariant under the cyclic shift operator δ : Rn → Rn given by δ(c1,c2, · · · ,cn) = (cn,c1, · · · ,cn−1), then the code C is called a cyclic code of length n. Definition 2.4. Let C be a code of arbitrary length n over any finite set A. Then C is called reversible code if it remains invariant under the reversal of each codeword, i.e., if c = (c1,c2, . . . ,cn) ∈ C, then cR = (cn,cn−1, . . . ,c1) ∈C. While working with cyclic and reversible codes we need to deal with a polynomial called self- reciprocal polynomial defined in the following definition. Definition 2.5. Let p(x) = a0 + a1x + · · · + anxn be a polynomial of degree n. The polynomial g(x) = an + an−1x + · · · + a0xn is called reciprocal polynomial of p(x). A polynomial p(x) over R is said to be self-reciprocal polynomial if p(x) = p∗(x), where p∗(x) = xnp ( 1 x ) . 222 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 We follow the definition of a DNA code from [12]. A code C is called a DNA code if it satisfies some of the following properties: (i) The Hamming distance constraint is defined as dH(a, b) ≥ d for all a, b ∈ C and a 6= b for some predefined distance d. (ii) The reverse constraint is defined as dH(aR, b) ≥ d for all a,b ∈C and a 6= b for some predefined distance d and where aR denotes the reverse sequence of alphabets in a. (iii) The reverse complement constraint is defined as dH(aR,b) ≥ d for all a,b ∈ C,a 6= b for some predefined distance d and where b denotes a word in which each alphabet of b is replaced by its Watson-Crick complement. DNA code satisfying reverse complement constraint is called the reverse complement DNA code. (iv) The fixed GC-content constraint specifies that each codeword must have fixed number of G’s and C’s. Generally this fixed number is bn 2 c where n denotes the length of codewords. Definition 2.6. Let C be a code of arbitrary length n over D. Then C is called reverse complement code if it contains both the reverse and the complement of every codeword in C, that is, if (c1,c2, . . . ,cn) ∈ C, then (cn,cn−1, . . . ,c1) ∈C. It is interesting to observe that one-to-one correspondence we define is a customized gray map between the DNA codons and elements of the ring R. The elements of ideal A2 correspond only to self reversible DNA codons. This map take care of GC content for the ideals A2u2 and A2u. The GC content is 30 to 50 percentage in these ideals. Further, the map satisfies the following property. Lemma 2.7. Let a ∈ R, then a = a + 2(1 + u + u2). 3. Cyclic DNA codes over R In this section, we study the reverse and the reverse complement cyclic codes over R. For this, we will use the following Lemmas. Lemma 3.1. [1] Let p(x) and q(x) be polynomials over Z4 with deg p(x) ≥ deg q(x). Then (i) [p(x)q(x)]∗ = p∗(x)q∗(x) (ii) [p(x) + q(x)]∗ = p∗(x) + x(deg p(x)−deg q(x))q∗(x). Lemma 3.2. [3] Let C = 〈p(x)〉 be a cyclic code of odd length n over Z4; where p(x) is a monic polynomial of degree r over Z4. Then C is reversible if and only if p(x) is self reciprocal polynomial. Now, we are able to prove the following result which will be useful thereafter. Lemma 3.3. Let p1(x),p2(x), . . . ,pr(x) be polynomials over Z4 with deg pi(x) = ki and kr ≤ kr−1 ≤ ···≤ k1, then (i) [p1(x)p2(x) · · ·pr(x)]∗ = p∗1(x)p∗2(x) · · ·p∗r(x) and (ii) [p1(x) + p2(x) + · · ·+ pr(x)]∗ = p∗1(x) + xk1−k2p∗2(x) + · · ·+ xk1−kip∗i (x) + · · ·+ x k1−krp∗r(x). Proof. We prove this by induction on r. By Lemma 3.1, this is true for r = 2. Let us assume that this is true for less than or equal to l, that is, [p1(x)p2(x) · · ·pl(x)]∗ = p∗1(x)p ∗ 2(x) · · ·p ∗ l (x), [p1(x) + p2(x) + · · ·+ pl(x)]∗ = p∗1(x) + x k1−k2p∗2(x) + · · ·+ x k1−klp∗l (x). 223 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Now, we will check whether the hypothesis is true for r = l + 1. By using hypothesis for r = 2 and r = l consecutively, [p1(x)p2(x) · · ·pl(x)pl+1(x)]∗ = [p1(x)p2(x) · · ·pl(x)]∗p∗l+1(x) = p∗1(x)p ∗ 2(x) · · ·p ∗ l+1(x). Repeat the same process to prove the second identity. [p1(x) + p2(x) + · · ·+ pl+1(x)]∗ = [p1(x) + p2(x) + · · ·+ pl(x)]∗ + x(k1−kl+1)p∗l+1(x) = p∗1(x) + x k1−k2p∗2(x) + · · ·+ x k1−klp∗l (x) + x (k1−kl+1)p∗l+1(x). Hence the induction hypothesis is true for r = l+1. Therefore, by mathematical induction, the statement is true for all positive integer r. Proposition 3.4. Let pi(x) and qi(x) be in Z4[x] for i = 1,2 and 3. If the following equality holds p1(x) + up2(x) + u 2p3(x) = q1(x) + uq2(x) + u 2q3(x), (1) then pi(x) = qi(x) for i = 1,2 and 3. Proof. Assume that equation (1) is true. Multiplying it by u2 and using u3 = 0, we have u2p1(x) = u2q1(x). Therefore p1(x) = q1(x), because p1(x),q1(x) ∈ Z4[x]. Again multiplying (1) by u and substituting p1(x) = q1(x), we get u2p2(x) = u2q2(x) and hence p2(x) = q2(x). Similarly, we get p3(x) = q3(x). Now, we study the reverse constraint of one generator cyclic code of odd length n over R. For this, first we need the following result which states one generator cyclic codes of odd length n over the ring R. This result with its proof can be found in [15]. Theorem 3.5. [15] Let C be a cyclic code of odd length n over R. If C = 〈h1(x) + ug1(x) + u2b1(x),uh2(x) + u 2b2(x),u 2h3(x)〉 and h1(x) and h3(x) are equal, then C = 〈h1(x) + ug1(x) + u2b1(x)〉. Definition 3.6. Let R be a ring and p(x) ∈ R[x] is called regular polynomial if it is not a zero divisor in R[x]. Theorem 3.7. Let C be a cyclic code of odd length n over the ring R. Let p(x) be a regular polynomial in Z4[x] and C = 〈 p(x) + uq(x) + u2h(x) 〉 where p(x),q(x) and h(x) are in Z4[x]. If deg p(x) = r, deg q(x) = s and deg h(x) = t with r ≥ s ≥ t, then C is reversible if and only if (a) p(x) is a self reciprocal polynomial. (b) (i) xr−sq∗(x) = q(x) and xr−th∗(x) = h(x), or (ii) xr−sq∗(x) = p(x) + q(x) and xr−th∗(x) = q(x) + h(x), or (iii) xr−sq∗(x) = p(x) + q(x) and xr−th∗(x) = p(x) + q(x) + h(x), or (iv) xr−sq∗(x) = q(x) and xr−th∗(x) = p(x) + h(x). Proof. Let us assume that C is a reversible code over R. Then C mod u = 〈p(x)〉 is a reversible code over Z4. Therefore, by Lemma 3.2 we have that p(x) is a self reciprocal polynomial in Z4[x]. Now, by Lemma 3.3 we have [p(x) + uq(x) + u2h(x)]∗ =p∗(x) + uxr−sq∗(x) + u2xr−th∗(x) =p(x) + uxr−sq∗(x) + u2xr−th∗(x) =(p(x) + uq(x) + u2h(x))k(x) ∈C. 224 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Since the degrees of both sides are the same, we have k(x) is a constant in R, i.e., k = k1 + uk2 + u2k3 and hence p(x) + uxr−sq∗(x) + u2xr−th∗(x) =(p(x) + uq(x) + u2h(x))k =(p(x) + uq(x) + u2h(x))(k1 + uk2 + u 2k3) =p(x)k1 + u(p(x)k2 + q(x)k1) + u 2(p(x)k3 + q(x)k2 + h(x)k1) (2) Now by Proposition 3.4, we get k1 = 1 and each k2 and k3 have 4 possibilities. Therefore, we have 16 possible cases, i.e., conditions for k = 1,1 + u,1 + 2u,1 + 3u,1 + u + u2,1 + u + 2u2,1 + u + 3u2,1 + 2u + u2,1 + 2u + 2u2,1 + 2u + 3u2,1 + 3u + u2,1 + 3u + 2u2,1 + 3u + 3u2,1 + u2,1 + 2u2,1 + 3u2. We have investigated all these cases by substituting values of k in (2). Then we get the following summarized results: Constant k Conditions Obtained k = 1 xr−sq∗(x) = q(x) and xr−th∗(x) = h(x) k = 1 + u xr−sq∗(x) = p(x) + q(x) and xr−th∗(x) = q(x) + h(x) k = 1 + 2u 2xr−sq∗(x) = 2q(x) and 2xr−th∗(x) = 2h(x) k = 1 + 3u 2xr−sq∗(x) = 2p(x) + 2q(x) and 2xr−th∗(x) = 2q(x) + 2h(x) k = 1 + u + u2 xr−sq∗(x) = p(x) + q(x) and xr−th∗(x) = p(x) + q(x) + h(x) k = 1 + u + 2u2 2xr−sq∗(x) = 2p(x) + 2q(x) and 2xr−th∗(x) = 2q(x) + 2h(x) k = 1 + u + 3u2 xr−sq∗(x) = p(x) + q(x) and 2xr−th∗(x) = 2p(x) + 2q(x) + 2h(x) k = 1 + 2u + u2 2xr−sq∗(x) = 2q(x) and 2xr−th∗(x) = 2p(x) + 2h(x) k = 1 + 2u + 2u2 2xr−sq∗(x) = 2q(x) and 2xr−th∗(x) = 2h(x) k = 1 + 2u + 3u2 2xr−sq∗(x) = 2q(x) and 2xr−th∗(x) = 2p(x) + 2h(x) k = 1 + 3u + u2 2xr−sq∗(x) = 2p(x) + 2q(x) and 2xr−th∗(x) = 2p(x) + 2q(x) + 2h(x) k = 1 + 3u + 2u2 2xr−sq∗(x) = 2p(x) + 2q(x) and 2xr−th∗(x) = 2q(x) + 2h(x) k = 1 + 3u + 3u2 2xr−sq∗(x) = 2p(x) + 2q(x) and 2xr−th∗(x) = 2p(x) + 2q(x) + 2h(x) k = 1 + u2 xr−sq∗(x) = q(x) and xr−th∗(x) = p(x) + h(x) k = 1 + 2u2 xr−sq∗(x) = q(x) and 2xr−th∗(x) = 2h(x) k = 1 + 3u2 xr−sq∗(x) = q(x) and 2xr−th∗(x) = 2p(x) + 2h(x) From the above table, we can observe that there are 12 similar cases. Therefore, we can reduce above conditions into following four cases. (i) xr−sq∗(x) = q(x) and xr−th∗(x) = h(x), or (ii) xr−sq∗(x) = p(x) + q(x) and xr−th∗(x) = q(x) + h(x), or (iii) xr−sq∗(x) = p(x) + q(x) and xr−th∗(x) = p(x) + q(x) + h(x), or (iv) xr−sq∗(x) = q(x) and xr−th∗(x) = p(x) + h(x). For the converse part, let us assume that the hypothesis is true. Since C is a cyclic code over R, it is enough to show that [p(x) + uq(x) + u2h(x)]∗ belong to C. [p(x) + uq(x) + u2h(x)]∗ =p∗(x) + uxr−sq∗(x) + u2xr−th∗(x) =p(x) + uxr−sq∗(x) + u2xr−th∗(x). Now using conditions in the hypothesis, we have [p(x) + uq(x) + u2h(x)]∗ = (p(x) + uq(x) + u2h(x))k ∈C where the constant k ∈{1,1 + u,1 + 2u,1 + 3u,1 + u + u2,1 + u + 2u2,1 + u + 3u2,1 + 2u + u2,1 + 2u + 2u2,1 + 2u + 3u2,1 + 3u + u2,1 + 3u + 2u2,1 + 3u + 3u2,1 + u2,1 + 2u2,1 + 3u2}⊆ R. Therefore, C is a reversible cyclic code in R[x]. 225 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 By using Table 1 and Lemma 2.7, we have the following result. Theorem 3.8. Let C = 〈 p(x) + uq(x) + u2h(x) 〉 be a cyclic code over R where p(x), q(x),h(x) ∈ Z4[x]. If deg p(x) = r, deg q(x) = s and deg h(x) = t with r ≥ s ≥ t, then C is a reverse complement code if and only if 1. the element 2(1+u+u 2)(xn−1) (x−1) ∈C and 2. the cyclic code C is reversible. Proof. Let C be the code satisfying the hypothesis. Let c(x) = c0 + c1x + · · · + ckxk ∈ C. Since C is reversible, c∗(x) = ck + ck−1x + · · · + c0xk ∈ C. As C is an ideal of R[x] 〈xn−1〉, so x n−k−1c∗(x) = ckx n−k−1 + ck−1x n−k + · · · + c0xn−1 belongs to C. By hypothesis 2(1 + u + u2) (xn−1) (x−1) ∈ C, therefore 2(1 + u + u2) (xn−1) (x−1) + x n−k−1c∗(x) ∈C. Then 2(1 + u + u2) (xn −1) (x−1) + xn−k−1c∗(x) =2(1 + u + u2)(1 + x + x2 + · · ·+ xn−k−2) + (ck + 2(1 + u + u 2))xn−k−1 + · · · + (c0 + 2(1 + u + u 2))xn−1 =2(1 + u + u2)(1 + x + x2 + · · ·+ xn−k−2) + ckx n−k−1 + ck−1x n−k + · · ·+ c0xn−1 2(1 + u + u2) (xn −1) (x−1) + xn−k−1c∗(x) =c∗(x). (using Lemma(2.7)) Hence, we conclude that C is a reverse complement code. Conversely, we assume that C is a reverse complement code, i.e., if c(x) ∈ C, then c∗(x) ∈ C. First we observe that since C is linear this implies that the element a(x) = 0 ∈C and therefore a(x) = 2(1 + u + u2)(1 + x + · · ·+ xn−1) = 2(1 + u + u2) (xn −1) (x−1) ∈C. Now, let c(x) = c0 + c1x + · · ·+ ckxk ∈C, then c∗(x) =2(1 + u + u2)(1 + x + · · ·+ xn−k−2) + xn−k−1ck + · · ·+ c0 =2(1 + u + u2)(1 + x + · · ·+ xn−k−2) + xn−k−1(ck + 2(1 + u + u2))+ · · ·+ (c0 + 2(1 + u + u2))xn−1 Adding 2(1 + u + u2)(x n−1) (x−1) to the above equation, we get c∗(x) + 2(1 + u + u2) (xn −1) (x−1) =ckx n−k−1 + ck−1x n−k + · · ·+ c0xn−1 =xn−k−1(ck + ck−1x + · · ·+ c0xk) Multiplying both side by xk+1, we get xk+1(c∗(x) + 2(1 + u + u2) (xn −1) (x−1) ) = ck + ck−1x + · · ·+ c0xk = c∗(x). Thus, we have C is reversible code. 226 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Theorem 3.9. Let C be a cyclic code of odd length n which satisfies the conditions of Theorem 3.8, then C is a DNA code. Proof. Combining the proof of Theorem 3.7 and Theorem 3.8, we get the result. Example 3.10. Let x = (2u+ 2u2,2 + 2u2,2 + 2u) and y = (2 + 2u,2 + 2u2,2u+ 2u2). We define a code C consisting of all cyclic shifts and linear combinations of the vectors x and y over R. Thus we obtain corresponding DNA code of parameters (9,32,4) as follows, CTCTGTAGA GAGCACGAG CACGAGGAG CTCCTCGTG GTGGTGGTG AGATGTCTC TCTACAGAG CACCACCAC GTGCTCCTC GAGGAGCAC CTCGTGCTC GAGACATCT ACAGAGTCT CTCAGATGT AGAGTGAGA CACTCTTCT TCTTCTCAC TGTGTGTGT TCTGAGACA GTGTGTTGT TGTAGACTC ACAACACAC TGTTGTGAG ACATCTGAG CACACAACA AGACTCTGT ACACACACA AGAAGAGTG GTGAGAAGA TCTCACTCT GAGTCTACA TGTCTCAGA Example 3.11. Let x = (2,2 + 2u,2u,2 + 2u + 2u2), then we define a code generated by a generator matrix consisting of the cyclic shifts of the vector x over R. Thus, we have a DNA code of parameters (12,16,6) as follows, TGTGTGACAGTG TGTTGTACAACA GTGGTGGTGGTG GTGTGTGTGACA TGTACAACATGT TGTCACACACAC GTGACAGTGTGT GTGCACGTGCAC CACGTGCACGTG CACTGTCACACA ACAGTGTGTGTG ACATGTTGTACA CACACACACTGT CACCACCACCAC ACAACATGTTGT ACACACTGTCAC 4. Binary image of elements in R In this section, we define binary images of elements of the ring R which will be useful for DNA computing. An element in the ring R is of the form a + ub + u2c; where a,b,c ∈ Z4. Now, we can define a map between R and Z2. A one-to-one correspondence between R and Z34 is defined as ψ : R → Z34 with (a + ub + u2c) 7→ (a,b,c) where a,b,c ∈ Z4. We define a Gray map ϕ : Z4 → Z22 using 2-adic expansion of elements in Z4 which are as follows: c π(c) ρ(c) υ(c) 0 0 0 0 1 1 0 1 2 0 1 1 3 1 1 0 We have ϕ(c) = (ρ(c),υ(c)) for all c ∈ Z4 (as in [13]). Therefore, 0 → 00,1 → 01,2 → 11,3 → 10. For any v ∈ Z4, the Lee weight wL(v) is defined as min(v,4 − v). This Lee weight can be extended to the ring R as follow: for x = a + ub + u2c Lee weight of x is defined as wL(x) = wL(a,b,c). The Hamming distance dH(c1,c2) between two codewords c1 and c2 is the Hamming weight of the codeword wH(c1 −c2) that is the number of non zero element in (c1 −c2). Define χ : R → Z62 by χ(a+ub+u2c) = (ρ(a),υ(a),ρ(b),υ(b),ρ(c),υ(c)). The map χ is clearly a linear map. Lemma 4.1. The Gray map from Rn to Z6n2 is a distance preserving map. 227 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Proof. Let x1,x2 be two elements of Rn. Then dL(x1,x2) = wL(x1 −x2). Since the map is bijective, we have wL(x1 − x2) = wH(χ(x1 − x2)) = wH(χ(x1) − χ(x2)). Therefore, the Gray map χ is distance preserving. Given an element c of length n in a cyclic code C. If the cyclic shift δ(c) ∈ C then ψ(δ(C)) will be quasi cyclic code of length 3n with index 3 over Z3n4 . Then it can be easily seen that ϕ(ψ(δ(c))) satisfies quasi cyclic shift of length 6n with index 6 over Z6n2 . Thus we have the following theorem. Theorem 4.2. If C is a cyclic code of length n, then its image is a quasi cyclic code of length 6n and with index 6 over Z2. 5. Skew cyclic codes over R Let θ be a non-trivial automorphism defined by θ : R → R such that (a + ub + u2c) 7→ a−ub + u2c. The order of θ is 2 that is θ(θ(a + ub + u2c)) = a + ub + u2c. The ring R[x,θ] = {a0 + a1x + a2x2 + · · ·+ an−1x n−1 : ai ∈ R,n ∈ N}, a non commutative ring with usual addition and multiplication defined as axi · bxj = aθi(b)xi+j is called skew polynomial ring. Definition 5.1. A set C of codewords over Rn is skew cyclic code if it satisfies the following (i) C is a submodule over R. (ii) Whenever (c0,c1, . . . ,cn−1) ∈C then (θ(cn−1),θ(c0), . . . ,θ(cn−2)) ∈C. Let p(x) + 〈xn − 1〉 ∈ Rθ = R[x,θ]/〈xn − 1〉 and r(x) ∈ R[x,θ], then define multiplication as r(x)(p(x) + 〈xn −1〉) = r(x)p(x) + 〈xn −1〉, for any r(x) ∈ R[x,θ]. Clearly Rθ is left R[x,θ]−module. Theorem 5.2. A code C over R is a skew cyclic code of length n if and only if C is left ideal of R[x,θ]- module Rnθ . Proof. The proof is same as the proof of [7, Theorem 1]. Theorem 5.3. Let C be a skew cyclic code of length n over R. If C contains a monic polynomial of minimal degree p(x), then C = 〈p(x)〉, where p(x) is a right divisor of xn −1. Proof. The proof is same as the proof of [7, Lemma 1]]. Now, we will introduce reverse complement skew cyclic codes over R. For this we have to notice that the multiplication over R[x,θ] is not commutative therefore we need to see things differently. Let c = (c0,c1, . . . ,cn−1) be in R[x,θ] then reversal of c denoted by cR is given by cR = (cn−1,cn−2, · · · ,c0). c(x−1) ·xn−1 = (c0 + c1x−1 + · · ·+ cn−1x−n+1) ·xn−1 = c0 ·xn−1 + c1x−1 ·xn−1 + · · ·+ cn−1x−n+1 ·xn−1 = c0θ 0(1)xn−1 + c1θ −1xn−1−1 + · · ·+ cn−1θ−n+1x−n+1+n−1 = c0x n−1 + c1x n−2 + · · ·+ cn−1 = cR As θ is an automorphism we have θ(1) = 1 and θr(1) = 1 for all r ∈ Z. Note that the reciprocal polynomial c∗(x) and cR(x) are different due to the operations on them. Also see (p(x) · q(x))R 6= pR(x) · qR(x) as in Lemma 3.3. If pR(x) coincide with (rpR(x)), then p(x) is called self reciprocal polynomial where r is a constant in R. Theorem 5.4. Let C = 〈p(x)〉 be a skew cyclic code of length n, where p(x) is a monic polynomial. Then if C is reverse complement cyclic code then p(x) is a self reciprocal polynomial and 2(1 + u + u2)(xn − 1)/(x−1) ∈C. 228 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 Proof. Let C = 〈p(x)〉 be a reverse complement cyclic code. Since C is linear (0, · · · ,0) ∈C, therefore 2(1 + u + u2)(xn −1)/(x−1) ∈C. Let p(x) = 1 + p1x + p2x2 + · · ·+ xk. Then, p(x) R = (p(x−1) ·xk−1) = 2(1 + u + u2)(1 + x + · · ·+ xn−k−2) + 1xn−k−1 + pk−1xn−k + · · ·+ 1xn−1 = 2(1 + u + u2)(1 + x + · · ·+ xn−k−2) + (3 + 2(1 + u + u2))xn−k−1+ (3pk−1 + 1 + u + u 2)xn−k + · · ·+ (3 + 2(1 + u + u2))xn−1 Now as the C is a reverse complement code therefore p(x) R ∈C. Using linearity of C we have 2(1 + u + u2)(xn −1)/(x−1) + p(x) R = 3xn−k−1 + 3pk−1x n−k + · · ·+ 3xn−1 ∈C. Multiplying xk+1−n both side and making use of C is a left ideal of R[x,θ]-module Rnθ we get, (2(1 + u + u2)(xn −1)/(x−1) + p(x) R ) ·xk+1−n = 3(θn−k−1(1) + pk−1θn−k(1)x + · · ·+ θn−1(1)xk) = 3(1 + pk−1x + · · · + xk) = 3pR(x) ∈ C. Since C = 〈p(x)〉 implies 3pR(x) = h(x)p(x) for some h(x) ∈ R[x,θ] degree of p(x) and pR(x) is same implies h(x) is constant. Hence p(x) is self reciprocal. Example 5.5. Let p(x) = a+bx+cx2+dx3, where a = 1+u+u2,b = 2+2u+2u2 = c and d = 1+3u+u2. We define a code C obtained by using G =   p(x) xp(x) x2p(x) x3p(x)   as a generator matrix. Then C correspond to DNA code which is reverse complement skew cyclic code of length 21 and |C| = 48 ·22. 6. Examples Example 6.1. For n = 9, x9−1 = (x+3)(x2 +x+1)(x6 +x3 +1), let C = 〈g(x)+up(x)+u2h(x)〉, where g(x) = p(x) = h(x) = (x2 + x + 1)(x6 + x3 + 1). Clearly, this code satisfies g(x) = g∗(x), p(x) = xip∗(x), h(x) = xjh∗(x), where i,j = 0, gives that C is a reverse complement cyclic DNA code of length n = 27 with minimum distance d = 9 and cardinality | C |= 64. The codewords in C are as follows. ATCATCATCATCATCATCATCATCATC GCGGCGGCGGCGGCGGCGGCGGCGGCG GTAGTAGTAGTAGTAGTAGTAGTAGTA GGAGGAGGAGGAGGAGGAGGAGGAGGA TCATCATCATCATCATCATCATCATCA TCGTCGTCGTCGTCGTCGTCGTCGTCG AAAAAAAAAAAAAAAAAAAAAAAAAAA GCAGCAGCAGCAGCAGCAGCAGCAGCA CTACTACTACTACTACTACTACTACTA ACGACGACGACGACGACGACGACGACG TTTTTTTTTTTTTTTTTTTTTTTTTTT GATGATGATGATGATGATGATGATGAT ATAATAATAATAATAATAATAATAATA AGAAGAAGAAGAAGAAGAAGAAGAAGA CCCCCCCCCCCCCCCCCCCCCCCCCCC GAGGAGGAGGAGGAGGAGGAGGAGGAG ACAACAACAACAACAACAACAACAACA AATAATAATAATAATAATAATAATAAT AAGAAGAAGAAGAAGAAGAAGAAGAAG GACGACGACGACGACGACGACGACGAC AACAACAACAACAACAACAACAACAAC TTATTATTATTATTATTATTATTATTA TGATGATGATGATGATGATGATGATGA GTTGTTGTTGTTGTTGTTGTTGTTGTT TATTATTATTATTATTATTATTATTAT TAGTAGTAGTAGTAGTAGTAGTAGTAG TACTACTACTACTACTACTACTACTAC GTGGTGGTGGTGGTGGTGGTGGTGGTG TTGTTGTTGTTGTTGTTGTTGTTGTTG TTCTTCTTCTTCTTCTTCTTCTTCTTC TGTTGTTGTTGTTGTTGTTGTTGTTGT GTCGTCGTCGTCGTCGTCGTCGTCGTC TGGTGGTGGTGGTGGTGGTGGTGGTGG TGCTGCTGCTGCTGCTGCTGCTGCTGC TCTTCTTCTTCTTCTTCTTCTTCTTCT GGTGGTGGTGGTGGTGGTGGTGGTGGT TCCTCCTCCTCCTCCTCCTCCTCCTCC TAATAATAATAATAATAATAATAATAA ATTATTATTATTATTATTATTATTATT GGCGGCGGCGGCGGCGGCGGCGGCGGC ATGATGATGATGATGATGATGATGATG AGTAGTAGTAGTAGTAGTAGTAGTAGT AGGAGGAGGAGGAGGAGGAGGAGGAGG GCTGCTGCTGCTGCTGCTGCTGCTGCT 229 K. Gowdhaman et. al. / J. Algebra Comb. Discrete Appl. 8(3) (2021) 219–231 AGCAGCAGCAGCAGCAGCAGCAGCAGC ACTACTACTACTACTACTACTACTACT ACCACCACCACCACCACCACCACCACC GCCGCCGCCGCCGCCGCCGCCGCCGCC GAAGAAGAAGAAGAAGAAGAAGAAGAA CAACAACAACAACAACAACAACAACAA CGACGACGACGACGACGACGACGACGA CCACCACCACCACCACCACCACCACCA CATCATCATCATCATCATCATCATCAT CAGCAGCAGCAGCAGCAGCAGCAGCAG CACCACCACCACCACCACCACCACCAC CTTCTTCTTCTTCTTCTTCTTCTTCTT CTGCTGCTGCTGCTGCTGCTGCTGCTG CTCCTCCTCCTCCTCCTCCTCCTCCTC CGTCGTCGTCGTCGTCGTCGTCGTCGT CGTCGTCGTCGTCGTCGTCGTCGTCGT CGCCGCCGCCGCCGCCGCCGCCGCCGC CCTCCTCCTCCTCCTCCTCCTCCTCCT CCGCCGCCGCCGCCGCCGCCGCCGCCG GGGGGGGGGGGGGGGGGGGGGGGGGGG Example 6.2. For n = 5, x5 −1 = (x+ 3)(x4 +x3 +x2 +x+ 1), let C = 〈g(x) +up(x) +u2h(x)〉, where g(x) = p(x) = h(x) = (x2 + x + 1)(x6 + x3 + 1). Clearly, this code satisfies g(x) = g∗(x), p(x) = xip∗(x), h(x) = xjh∗(x) and hence this is a reverse complement DNA cyclic code of length n = 15 with minimum distance d = 5 and cardinality | C |= 64. The code C contains the following codewords. ATCATCATCATCATC GCGGCGGCGGCGGCG GTAGTAGTAGTAGTA GGAGGAGGAGGAGGA TCATCATCATCATCA TCGTCGTCGTCGTCG AAAAAAAAAAAAAAA GCAGCAGCAGCAGCA CTACTACTACTACTA ACGACGACGACGACG TTTTTTTTTTTTTTT GATGATGATGATGAT ATAATAATAATAATA AGAAGAAGAAGAAGA CCCCCCCCCCCCCCC GAGGAGGAGGAGGAG ACAACAACAACAACA AATAATAATAATAAT AAGAAGAAGAAGAAG GACGACGACGACGAC AACAACAACAACAAC TTATTATTATTATTA TGATGATGATGATGA GTTGTTGTTGTTGTT TATTATTATTATTAT TAGTAGTAGTAGTAG TACTACTACTACTAC GTGGTGGTGGTGGTG TTGTTGTTGTTGTTG TTCTTCTTCTTCTTC TGTTGTTGTTGTTGT GTCGTCGTCGTCGTC TGGTGGTGGTGGTGG TGCTGCTGCTGCTGC TCTTCTTCTTCTTCT GGTGGTGGTGGTGGT TCCTCCTCCTCCTCC TAATAATAATAATAA ATTATTATTATTATT GGCGGCGGCGGCGGC ATGATGATGATGATG AGTAGTAGTAGTAGT AGGAGGAGGAGGAGG GCTGCTGCTGCTGCT AGCAGCAGCAGCAGC ACTACTACTACTACT ACCACCACCACCACC GCCGCCGCCGCCGCC GAAGAAGAAGAAGAA CAACAACAACAACAA CGACGACGACGACGA CCACCACCACCACCA CATCATCATCATCAT CAGCAGCAGCAGCAG CACCACCACCACCAC CTTCTTCTTCTTCTT CTGCTGCTGCTGCTG CTCCTCCTCCTCCTC CGTCGTCGTCGTCGT CGTCGTCGTCGTCGT CGCCGCCGCCGCCGC CCTCCTCCTCCTCCT CCGCCGCCGCCGCCG GGGGGGGGGGGGGGG 7. 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