ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.864902 J. Algebra Comb. Discrete Appl. 8(1) • 1–8 Received: 28 December 2019 Accepted: 24 August 2020 Journal of Algebra Combinatorics Discrete Structures and Applications Reversible DNA codes from skew cyclic codes over a ring of order 256 Research Article Yasemin Cengellenmis, Nuh Aydin, Abdullah Dertli Abstract: We introduce skew cyclic codes over the finite ring F2+uF2+vF2+wF2+uvF2+uwF2+vwF2+uvwF2, where u2 = 0, v2 = v, w2 = w, uv = vu, uw = wu, vw = wv and use them to construct reversible DNA codes. The 4-mers are matched with the elements of this ring. The reversibility problem for DNA 4-bases is solved and some examples are provided. 2010 MSC: 94B05, 94B60 Keywords: DNA codes, Skew codes, Reversibility 1. Introduction It is well known that DNA contains genetic program for the biological development of life and has two strands which are linked by Watson-Crick pairing so that every A is linked with a T and every C with a G, and vice versa, where A,T,C,G are the four bases of a DNA sequence. DNA computing started in 1994 when Adleman showed how to solve a computationally difficult problem (traveling salesman problem, a well-known NP-complete problem) by manipulations of DNA molecules in [2]. Later, more applications of DNA codes were discovered such as using DNA codes to break a cryptosystem known as DES [3, 5], and using DNA codewords as high density storage media [16]. Devising methods to design DNA codes for DNA computing has been a major topic of research since the beginning of the century. A block code is called a DNA code if it satisfies the following constraints [8, 22]. 1. the Hamming constraint for minimum distance, Yasemin Cengellenmis; Department of Mathematics, Trakya University, Edirne, Turkey (email: ycengellen- mis@gmail.com). Nuh Aydin; Department of Mathematics and Statistic, Kenyon College, Gambier,OH, United States (email: aydinn@kenyon.edu). Abdullah Dertli (Corresponding Author); Department of Mathematics, Ondokuz Mayıs University, Samsun, Turkey (email: abdullah.dertli@gmail.com). 1 https://orcid.org/0000-0002-8133-9836 https://orcid.org/0000-0002-5618-2427 https://orcid.org/0000-0001-8687-032X Y. Cengellenmis, N. Aydin, A. Dertli / J. Algebra Comb. Discrete Appl. 8(1) (2021) 1–8 2. the reverse-complement constraint, 3. the reverse constraint, and 4. the fixed GC content. In [8], all of these constraints are translated in terms of coding theory. This enabled researchers to use the results from classical coding theory to design codes for DNA computation. The last decade witnessed an increased activity in this direction. Since DNA uses a four-letter alphabet, {A,G,T,C}, it is most natural to use a ring of size 4 in employing classical coding theory techniques for the design of DNA codes. Later, alphabets of size 4k for k ≥ 1 have been considered (e.g. [1, 4, 6–9, 13, 21? , 22]). It was also observed that having a cyclic structure in DNA codes has certain advantages in terms of complexities of algorithms [17, 21]. One of the challenging problems in the field is the reversibility problem [18]. This problem arises from the fact that the pairing of nucleotides in two different strands of a DNA sequence is done in opposite direction and reverse order. For example, let us consider the codeword (DNA string) GTTAGGCA which corresponds to a codeword (a1,a2). The reverse of (a1,a2) is (a2,a1). However, the vector (a2,a1) corresponds to GGCAGTTA which is not the reverse of GTTAGGCA. The reverse of GTTAGGCA is ACGGATTG. Some authors solved this problem by considering skew cyclic codes. The reversibility problem for DNA 8-bases and DNA 2s+1k-bases is solved in [10] and [11] respectively by using skew cyclic codes over the finite rings F16 + uF16 + vF16 + uvF16, where u2 = u,v2 = v,uv = vu and F42k [u1, ...,us]/〈u12 − u1, ...,us 2 −us〉 where k,s > 1,uiuj = ujui. Motivated by the previous work [10, 11], we study the reversibility problem for DNA 4-bases using skew cyclic codes over the finite ring R := F2 + uF2 + vF2 + wF2 + uvF2 + uwF2 + vwF2 + uvwF2 of order 256 where u2 = 0,v2 = v,w2 = w,uv = vu,uw = wu,vw = wv. In [6], cyclic DNA codes over R are studied. A map from R to R21 is given where R1 = F2 + uF2 + vF2 + uvF2 and u2 = 0,v2 = v,uv = vu. Moreover, cyclic codes of arbitrary length over R satisfying the reverse constraint and reverse complement constraint are studied and a one to one correspondence between the elements of the ring R and SD256 is established where SD256 = {AAAA,...,GGGG}. The binary image of a cyclic code over R is also determined. In this paper, by defining a non-trivial automorphism on R, skew cyclic codes over R are introduced. Thanks to these type of codes, reversible DNA codes are obtained and some examples are provided. 2. Preliminaries In [6], the finite ring R = F2 + uF2 + vF2 + wF2 + uvF2 + uwF2 + vwF2 + uvwF2 = {a1 + ua2 + va3 + wa4 + uva5 + uwa6 + vwa7 + uvwa8 : ai ∈ F2, i = 1, 2, ..., 8} with u2 = 0,v2 = v,w2 = w,uv = vu,uw = wu,vw = wv is introduced. The ring R is commutative with characteristic 2 and 256 elements. It can be viewed as R = (F2 + uF2 + vF2 + uvF2) + w(F2 + uF2 + vF2 + uvF2) = R1 + wR1, w2 = w where R1 is the ring F2 + uF2 + vF2 + uvF2, u2 = 0,v2 = v,uv = vu, introduced in [23]. By using the DNA alphabet SD4 = {A,T,C,G}, the authors define a correspondence τ between the elements of the finite ring R1 and DNA double pairs as in the following table, by means of a Gray map from R1 to (F2 + uF2) 2 with u2 = 0. 2 Y. Cengellenmis, N. Aydin, A. Dertli / J. Algebra Comb. Discrete Appl. 8(1) (2021) 1–8 Ring elements α DNA double pairs τ(α) 0 AA 1 GG u TT v AG uv AT 1 + u CC 1 + v GA u + v TC u + uv TA v + uv AC 1 + uv GC 1 + u + uv CG 1 + u + v CT 1 + v + uv GT u + v + uv TG 1 + u + v + uv CA We define a Gray map as follows φ : R −→ R21 x + yw 7−→ (x,x + y) where x,y ∈ R1. By using the matching and the Gray map above, we get a matching Ψ between the elements of R and a set of DNA 4-bases SD256 = {AAAA,TTTT,....} as follows. Ψ : R −→ SD256 x + wy 7−→ Ψ(x + wy) = γ(x,x + y) = (τ(x),τ(x + y)) where γ : R21 −→ SD256 (s,t) 7−→ (τ(s),τ(t)) for s,t ∈ R1. That is, Ψ = γ ◦φ. 3. Skew cyclic codes over R Definition 3.1. Let B be a finite ring and θ be a non-trivial automorphism on B. A subset C of Bn is called a skew cyclic code of length n if C satisfies the following conditions, 1. C is a submodule of Bn 2. If c = (c0,c1, ...,cn−1) ∈ C, then σθ(c) = (θ(cn−1),θ(c0), ...,θ(cn−2)) ∈ C, where σθ is the skew cyclic shift operator. By defining a non-trivial automorphism θ on R as follows, we can define skew cyclic codes over R. Let θ : R −→ R x + yw 7−→ θ(x + yw) = θ ′ (x + y) + wθ ′ (y) 3 Y. Cengellenmis, N. Aydin, A. Dertli / J. Algebra Comb. Discrete Appl. 8(1) (2021) 1–8 where x,y ∈ R1 and θ ′ is a non-trivial authomorphism on R1 defined by θ ′ : R1 −→ R1 a + bu + v(c + du) 7−→ (a + c + (b + d)u) + v(c + du) The order of each θ and θ ′ is 2. The set of polynomials R[x,θ] = {a0 + a1x + ... + an−1xn−1 : ai ∈ R,n ∈ N} is the skew polynomial ring over R with the usual addition of polynomials and the non-commutative multiplication given by (axi)(bxj) = aθi(b)xi+j. In polynomial representation, a skew cyclic code of length n over R is defined as a left ideal of the quotient ring Rθ,n = R[x,θ]/〈xn − 1〉, if the order of θ divides n, that is, if n is even. If the order of θ does not divide n, a skew cyclic code of length n over R is defined as a left R[x,θ]-submodule of Rθ,n, since the set Rθ,n = R[x,θ]/〈xn − 1〉 = {f(x) + 〈xn − 1〉 : f(x) ∈ R[x,θ]} is a left R[x,θ]-module with the multiplication from left defined by r(x)(f(x) + 〈xn − 1〉) = r(x)f(x) + 〈xn − 1〉 for any r(x) ∈ R[x,θ]. In either case, the following holds. Theorem 3.2. Let C be a skew cyclic code over R and let f(x) be a polynomial in C of minimal degree. If the leading coefficient of f(x) is a unit in R, then C = 〈f(x)〉, where f(x) is a right divisor of xn − 1. Proof. It can be proven similarly to the proof of Theorem 4 in [20]. 4. Reversible DNA codes from skew cyclic codes over R Definition 4.1. For x = (x0,x1, ...,xn−1) ∈ Rn, the vector (xn−1,xn−2, ...,x1,x0) is called the reverse of x and is denoted by xr. A linear code C of length n over R is said to be reversible if xr ∈ C for every x ∈ C. Each element α of R1 and θ ′ (α) are mapped to DNA pairs, which are reverses of each other. For example, τ(v) = AG, while τ ( θ ′ (v) ) = GA. This map can be extended to a map γ from R21 to 4-mers as follows, γ(a,b) = (τ(a),τ(b)) where a,b ∈ R1. By means of the map Ψ = γ ◦φ, we can find a relationship between skew cyclic codes over R and DNA codes. We note that Ψ(r) and Ψ (θ(r)) are DNA reverses of each other. Indeed, for r = x+yw ∈ R, we have Ψ(r) = γ (φ(x + yw)) = γ (x,x + y) = (τ(x),τ(x + y)) On the other hand, Ψ (θ(r)) = Ψ ( θ ′ (x + y) + wθ ′ (y) ) = γ ( φ ( θ ′ (x + y) + wθ ′ (y) )) = γ ( θ ′ (x + y),θ ′ (x) ) = ( τ ( θ ′ (x + y) ) ,τ ( θ ′ (x) )) 4 Y. Cengellenmis, N. Aydin, A. Dertli / J. Algebra Comb. Discrete Appl. 8(1) (2021) 1–8 This map can be extended as follows. For any r = (r0, ...,rn−1) ∈ Rn, (Ψ(r0), Ψ(r1), . . . , Ψ(rn−1)) r = (Ψ(θ(rn−1)), . . . , Ψ(θ(r1)), Ψ(θ(r0))) Example 4.2. If r = u + uv + w(1 + uv) ∈ R, then we get Ψ(r) = γ (φ(r)) = γ (u + uv, 1 + u) = (τ (u + uv) ,τ (1 + u)) = (TA,CC) On the other hand, Ψ (θ(r)) = Ψ ( θ ′ (1 + u) + wθ ′ (1 + uv) ) = γ ◦φ ( θ ′ (1 + u) + wθ ′ (1 + uv) ) = γ ( θ ′ (1 + u),θ ′ (u + uv) ) = ( τ ( θ ′ (1 + u) ) ,τ ( θ ′ (u + uv) )) = (CC,AT) Definition 4.3. Let C be a code of length n over R. If Ψ(c)r ∈ Ψ(C) for all c ∈ C, then C or equivalently Ψ(C) is called a reversible DNA code. Definition 4.4. Let g(x) = a0 + a1x + a2x2 + ... + asxs be a polynomial of degree s over R. g(x) is called a palindromic polynomial if ai = as−i for all i ∈{0, 1, ...,s}. g(x) is called a θ-palindromic polynomial if ai = θ(as−i) for all i ∈{0, 1, ...,s}. As the order of θ is 2, a skew cyclic code of odd length n over R with respect to θ is an ordinary cyclic code. So we will take the length n to be even. The next two theorems show that palindromic and θ-palindromic polynomials generate reversible DNA codes. They are analogous to Theorem 1 and Theorem 2 in [12] stated for codes over a field. Theorem 4.5. Let C = 〈f(x)〉 be a skew cyclic code of length n over R, where f(x) is a right divisor of xn−1 and deg(f(x)) is odd. If f(x) is a θ-palindromic polynomial, then Ψ(C) is a reversible DNA code. Proof. Let f(x) be a θ-palindromic polynomial and f(x) = a0 + a1x + ... + a2s−1x2s−1. So ai = θ(a2s−1−i), for all i = 0, 1, ...,s− 1. Let h(x) = h0 + h1x + · · · + h2k−1x2k−1. Let bl be the coefficient of xl in h(x)f(x), where l = 0, 1, . . . ,n− 1. For any t < n/2, the coefficient of xt in h(x)f(x) is bt = t∑ j=0 hjθ j(at−j) and the coefficient of xn−t is bn−t = ∑t j=0 h2k−1−jθ 2k−1−j(a2s−1−(t−j)). The polynomial h(x)f(x) = ∑2k−1 d=0 hdx df(x) corresponds to a vector b = (b0,b1, ..., bn−1) ∈ C. The vector Ψ(b)r = ((Ψ(b0), ..., Ψ(bn−1)))r is equal to the vector Ψ(z), where the vector z corresponds to the polynomial ∑2k−1 d=0 θ(hd)x 2k−1−df(x). So, Ψ(C) is a reversible DNA code. Theorem 4.6. Let C = 〈f(x)〉 be a skew cyclic code of length n over R, where f(x) is a right divisor of xn − 1 and deg(f(x)) is even. If f(x) is a palindromic polynomial, then Ψ(C) is a reversible DNA code. 5 Y. Cengellenmis, N. Aydin, A. Dertli / J. Algebra Comb. Discrete Appl. 8(1) (2021) 1–8 Proof. Let f(x) be a palindromic polynomial with even degree so that f(x) = a0 + a1x + ... + a2sx2s and ai = a2s−i, for all i = 0, 1, ...,s. Let h(x) = h0 + h1x + ... + h2kx2k. Let bl be the coefficient of xl in h(x)f(x), where l = 0, 1, ..,n− 1. For any t < n/2, the coefficient of xt in h(x)f(x) is bt = t∑ j=0 hjθ j(at−j) and the coefficient of xn−t is bn−t = ∑t j=0 h(2k)−jθ (2k)−j(a2s−(t−j)). The polynomial h(x)f(x) = ∑2k d=0 hdx df(x) corresponds to a vector b = (b0,b1, . . . ,bn−1) ∈ C. The vector Ψ(b)r = ((Ψ(b0), ..., Ψ(bn−1)))r is equal to the vector Ψ(z), where the vector z corresponds to the polynomial ∑2k d=0 θ(hd)x 2k−df(x). So, Ψ(C) is a reversible DNA code. The next two theorems show that palindromic and θ-palindromic polynomials come in pairs. They are analogous to Theorem 4 and Theorem 3 in [12] stated for skew polynomials over a field. Theorem 4.7. Let xn−1 = h(x)f(x) ∈ R[x,θ], where the degree of f(x) is odd. If f(x) is a θ-palindromic polynomial, then h(x) is a palindromic polynomial. Proof. Let f(x) = a0 + a1x + ... + a2s−1x2s−1. As the length n is even, then h(x) = h0 + h1x + ... + h2k−1x 2k−1. Since f(x) is a θ-palindromic polynomial, then ai = θ(a2s−1−i) for all i = 0, 1, ...,s− 1. Let bl be the coefficient of xl in h(x)f(x), where l = 0, 1, ..,n− 1. For any t < n/2, the coefficient of xt in h(x)f(x) is bt = t∑ j=0 hjθ j(at−j) and the coefficient of xn−t is bn−t = ∑t j=0 h2k−1−jθ 2k−1−j(a2s−1−(t−j)). By using the fact that b0 = bn = 0 and bi = 0 for all i = 1, 2, ...,n− 1, it can be shown that hi = h2k−1−i for all i = 0, 1, ..,k − 1 as in the proof of Theorem 4 in [12], by induction. A polynomial that is in the center Z(R[x,θ]) of R[x,θ] is called a central polynomial. A central polynomial commutes with every element of R[x,θ]. As in the field case, we can prove that xn − 1 is central when n is even. Proposition 4.8. The polynomial xn − 1 is central if and only if n is even. Proof. Let n be even. Let s(x) = a0 + a1x + ... + amxm ∈ R[x,θ]. Since n is even, θn(a) = a for any a ∈ R. So (xn−1)s(x) = xna0+xna1x+...+xnamxm−s(x) = xna0+θn(a1)xnx+...+θn(am)xnxm−s(x) = (a0 + a1x + ... + amx m)xn − s(x) = s(x)(xn − 1). Hence (xn − 1) ∈ Z(R[x,θ]). Conversely, assume (xn − 1) ∈ Z(R[x,θ]). Then (xn − 1) commutes with every element in R[x,θ]. In particular, we have (xn − 1)amxm = amxm(xn − 1) for any m and any am ∈ R. As (xn − 1)amxm = θn(am)xn+m −amxm and amxm(xn − 1) = (am)xn+m −amxm, we have θn(am) = am. This implies that n is even. Lemma 4.9. Let xn − 1 = hf ∈ Z(R[x,θ]). Then hf = fh in R[x,θ]. Proof. Since hf is a central element, we have (hf)h = h(hf). So h(fh−hf) = 0. Since the leading coefficient of h is a unit, h is not a zero divisor. Hence, fh = hf in R[x,θ]. Theorem 4.10. Let xn−1 = h(x)f(x) ∈ R[x,θ], where the degree of f(x) is even. If h(x) is a palindromic polynomial then so is f(x). 6 Y. Cengellenmis, N. Aydin, A. Dertli / J. Algebra Comb. Discrete Appl. 8(1) (2021) 1–8 Proof. This can be proven similarly to Theorem 3 in [12]. Since n is even, xn − 1 = hf ∈ Z(R[x,θ]) . So we get that any right divisor of xn − 1 is also a left divisor of xn − 1. Corollary 4.11. Let xn − 1 = h(x)f(x) ∈ R[x,θ], where the degree of f(x) is even. If f(x) is a palindromic polynomial, then h(x) is a palindromic polynomial as well. Finally, we give a few examples of palindromic and theta-palindromic polynomials over R. Hence, these polynomials generate reversible DNA codes. We found these polynomials using Magma software [15]. Example 4.12. There are at least 576 different factorizations of x4 − 1 in the form x4 − 1 = f(x)h(x) where deg(f(x)) = deg(h(x)) = 2, in the skew polynomial ring over R. Of these, 64 of the factorizations involve palindromic polynomials. One of these 64 pairs is f(x) = h(x) = x2 + ux + 1. Example 4.13. There are at least 990 different factorizations of x8 − 1 in the form x8 − 1 = h(x)f(x) over R where deg(h(x)) = 3 and deg(f(x)) = 5. Of these, 1 factorization involves theta-palindromic polynomials, namely h(x) = x3 + (uw + (uv + (u + 1)))x2 + (uw + (uv + (u + 1)))x + 1 and f(x) = x5 + (uw + (uv + (u + 1)))x4 + (uw + (uv + u))x3 + (uw + (uv + u))x2 + (uw + (uv + (u + 1)))x + 1. Example 4.14. The polynomial f(x) = x4 + (v + 1)x3 + (uvw + u)x2 + (v + 1)x + 1 divides x16 − 1 in the skew polynomial ring over R and it is palindromic. Hence it generates a reversible DNA code. Additionally, h(x) = x 16−1 f(x) (division in the skew polynomial ring over R) is also palindromic. Example 4.15. 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