www.biomathforum.org/biomath/index.php/biomath ORIGINAL ARTICLE On the cyclic DNA codes over the finite rings Z4 + wZ4 and Z4 + wZ4 + vZ4 + wvZ4 Abdullah Dertli1, Yasemin Cengellenmis2 1Ondokuz Mayıs University, Faculty of Arts and Sciences Mathematics Department, Samsun, Turkey abdullah.dertli@gmail.com 2Trakya University, Faculty of Sciences Mathematics Department, Edirne, Turkey ycengellenmis@gmail.com Received: 8 May 2017, accepted: 16 December 2017, published: 22 December 2017 Abstract—The structures of the cyclic DNA codes of odd length over the finite rings R = Z4 + wZ4, w2 = 2 and S = Z4 + wZ4 + vZ4 + wvZ4,w2 = 2,v2 = v,wv = vw are studied. The links between the elements of the rings R, S and 16 and 256 codons are established, respectively. The cyclic codes of odd length over the finite ring R satisfy reverse complement constraint and the cyclic codes of odd length over the finite ring S satisfy reverse constraint and reverse complement constraint are studied. The binary images of the cyclic DNA codes over the finite rings R and S are determined. Moreover, a family of DNA skew cyclic codes over R is constructed, its property of being reverse complement is studied. Keywords-DNA codes; cyclic codes; skew cyclic codes. I. INTRODUCTION DNA is formed by the strands and each strand is sequence consists of four nucleotides ; Adenine (A), Guanine (G), Thymine (T) and Cytosine (C). Two strands of DNA are linked with Watson-Crick Complement. This is as A = T , T = A, G = C, C = G. For example if c = (ATCCG) then its complement is c = (TAGGC). A code is called a DNA code if it satisfies some or all of the following conditions: i) The Hamming contraint, for any two different codewords c1,c2 ∈ C, H(c1,c2) ≥ d ii) The reverse constraint, for any two different codewords c1,c2 ∈ C, H(c1,cr2) ≥ d iii) The reverse complement constraint, for any two different codewords c1,c2 ∈ C, H(c1,c rc 2 ) ≥ d iv) The fixed GC content constraint, for any codeword c ∈ C contains the some number of G and C element. The purpose of the i)-iii) constraints is to avoid undesirable hybridization between different strands. DNA computing were started by Leonhard Adleman in 1994, in [3]. The special error correct- Copyright: c© 2017 Dertli et al. This article is distributed under the terms of the Creative Commons Attribution License (CC BY 4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. Citation: Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings Z4 + wZ4 and Z4 + wZ4 + vZ4 + wvZ4, Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 1 of 11 http://www.biomathforum.org/biomath/index.php/biomath https://creativecommons.org/licenses/by/4.0/ https://creativecommons.org/licenses/by/4.0/ http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... ing codes over some finite fields and finite rings with 4n elements where n ∈ N were used for DNA computing applications. In [12], the reversible codes over finite fields were studied, firstly. It was shown that C = 〈f(x)〉 is reversible if and only if f(x) is a self reciprocal polynomial. In [1], they developed the theory for constructing linear and additive cyclic codes of odd length over GF(4). In [13], they introduced a new family of polynomials which generates reversible codes over a finite field GF(16). In [2], the reversible cyclic codes of any length n over the ring Z4 were studied. A set of genera- tors for cyclic codes over Z4 with no restrictions on the length n was found. In [17], the cyclic DNA codes over the ring R = {0, 1,u, 1 + u} where u2 = 1 based on a similarity measure were constructed. In [9], the codes over the ring F2 + uF2,u 2 = 0 were constructed for using in DNA computing applications. I. Siap et al. considered the cyclic DNA codes over the finite ring F2[u]/ 〈 u2 − 1 〉 in [18]. In [10], Liang and Wang considered the cyclic DNA codes over F2 +uF2,u2 = 0. Yıldız and Siap stud- ied the cyclic DNA codes over F2[u]/ 〈 u4 − 1 〉 in [20]. Bayram et al. considered codes over the finite ring F4 + vF4,v2 = v in [3]. Zhu and Chan studied the cyclic DNA codes over the non-chain ring F2[u,v]/ 〈 u2,v2 −v,uv −vu 〉 in [21]. In [6], Bennenni at al. studied the cyclic DNA codes over F2[u]/ 〈 u6 〉 . Pattanayak et al. considered the cyclic DNA codes over the ring F2[u,v]/ < u 2 − 1,v3 − v,uv − vu > in [15]. Pattanayak and Singh studied the cyclic DNA codes over the ring Z4 + uZ4,u2 = 0 in [14]. J. Gao et al. studied the construction of the cyclic DNA codes by cyclic codes over the finite ring F4[u]/ 〈 u2 + 1 〉 , in [11]. Also, the construc- tion of DNA the cyclic codes has been discussed by several authors in [7,8,16]. We study families of DNA cyclic codes of the finite rings Z4 + wZ4, w2 = 2 and Z4 + wZ4 + vZ4 + wvZ4,w2 = 2,v2 = v,wv = vw. The rest of the paper is organized as follows. In section 2, details about algebraic structure of the finite ring Z4 + wZ4, w2 = 2 are given. We define a Gray map from R to Z4. In section 3, the cyclic codes of odd length over R satisfy the reverse complement constraint are determined. In section 4, the cyclic codes of odd length over S satisfy the reverse complement constraint and the reverse contraint are examined. A linear code over S is represented by means of two linear codes over R. In section 5, the binary image of cyclic DNA code over R is determined. In section 6, the binary image of cyclic DNA code over S is determined. In section 7, by using a non trivial automorphism, the DNA skew cyclic codes are introduced. In section 8, the design of linear DNA code is presented. II. PRELIMINARIES The algebraic structure of the finite ring R = Z4 + wZ4, w2 = 2 is given in [4]. R is the commutative, characteristic 4 ring Z4 + wZ4 = {a + wb : a,b ∈ Z4} with w2 = 2. R can also be thought of as the quotient ring Z4[w]/ 〈 w2 − 2 〉 . R is a principal ideal ring with 16 elements and finite chain ring. The units of the ring are 1, 3, 1 + w, 3 + w, 1 + 2w, 1 + 3w, 3 + 3w, 3 + 2w, and the non-units are 0, 2,w, 2w, 3w, 2 + w, 2 + 2w, 2 + 3w. R has 4 ideals: 〈0〉 = {0}, 〈1〉 = 〈3〉 = 〈1 + 3w〉 = ... = R, 〈w〉 = {0, 2,w, 2w, 3w, 2+w, 2+2w, 2+3w}, = 〈3w〉 = 〈2 + w〉 = 〈2 + 3w〉 , 〈2w〉 = {0, 2w}, 〈2〉 = 〈2 + 2w〉 = {0, 2, 2w, 2 + 2w}. We have 〈0〉⊂ 〈2w〉⊂ 〈2〉⊂ 〈w〉⊂ R. Moreover R is a Frobenious ring. We define φ : R −→ Z24 as φ (a + wb) = (a,b) . Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 2 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... The Gray map is extended component wise to φ : Rn −→ Z2n4 (α1,α2, ...,αn) , = (a1, ...,an,b1, ...,bn), where αi = ai + biw with i = 1, 2, ...,n. φ is a Z4 module isomorphism. A linear code C of length n over R is an R- submodule of Rn. An element of C is called a codeword. A code of length n is cyclic if the code is invariant under the automorphism σ which is σ (c0,c1, ...,cn−1) = (cn−1,c0, ...,cn−2) A cyclic code of length n over R can be identified with an ideal in the quotient ring R[x]/〈xn − 1〉 via the R–modul isomorphism Rn −→ R[x]/〈xn − 1〉 (c0,c1, ...,cn−1) 7−→ c0 +c1x+...+cn−1xn−1 +〈xn − 1〉 Theorem 1: Let C be a cyclic code in R[x]/〈xn − 1〉 .Then there exists polynomials g(x),a(x) such that a(x)|g(x)|xn − 1 and C = 〈g(x),wa(x)〉 . The ring R[x]/〈xn − 1〉 is a principal ideal ring when n is odd. So, if n is odd, then there exists s(x) ∈ R[x]/〈xn − 1〉 such that C = 〈s(x)〉, in [4,19]. III. THE REVERSIBLE COMPLEMENT CODES OVER R In this section, we study the cyclic code of odd length over R satisfies the reverse complement constraint. Let {A,T,G,C} represent the DNA al- phabet. DNA occurs in sequences with represented by sequences of the DNA alphabet. DNA code of length n is defined as a set of the codewords (x0,x1, ...,xn−1) where xi ∈{A,T,G,C}. These codewords must satisfy the four constraints which are mentioned in [21]. Since the ring R is of cardinality 16, we define the map φ which gives a one to one correspon- dence between the elements of R and the 16 codons over the alphabet {A,T,G,C}2 by using the Gray map as follows Elements Gray images DNA double pairs 0 (0, 0) AA 1 (1, 0) CA 2 (2, 0) GA 3 (3, 0) TA w (0, 1) AC 2w (0, 2) AG 3w (0, 3) AT 1 + w (1, 1) CC 1 + 2w (1, 2) CG 1 + 3w (1, 3) CT 2 + w (2, 1) GC 2 + 2w (2, 2) GG 2 + 3w (2, 3) GT 3 + w (3, 1) TC 3 + 2w (3, 2) TG 3 + 3w (3, 3) TT The codons satisfy the Watson-Crick Comple- ment. Definition 2: 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. For x = (x0,x1, ...,xn−1) ∈ Rn, the vector (x0,x1, ...,xn−1) is called the complement of x and is denoted by xc. A linear code C of length n over R is said to be complement if xc ∈ C for every x ∈ C. For x = (x0,x1, ...,xn−1) ∈ Rn, the vec- tor (xn−1,xn−2, ...,x1,x0) is called the reversible complement of x and is denoted by xrc. A linear code C of length n over R is said to be reversible complement if xrc ∈ C for every x ∈ C. Definition 3: Let f(x) = a0 +a1x+...+atxt ∈ R[x] ( S[x] ) with at 6= 0 be polynomial. The reciprocal of f(x) is defined as f∗(x) = xtf( 1 x ). It is easy to see that deg f∗(x) ≤ deg f(x) and if a0 6= 0, then deg f∗(x) = deg f(x). f(x) is called a self reciprocal polynomial if there is a constant m such that f∗(x) = mf(x). Lemma 4: Let f(x),g(x) be polynomials in R[x]. Suppose deg f(x) − deg g(x) = m then, Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 3 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... i) (f(x)g(x))∗ = f∗(x)g∗(x) ii) (f(x) + g(x))∗ = f∗(x) + xmg∗(x) Lemma 5: For any a ∈ R, we have a + a = 3 + 3w. Lemma 6: If a ∈{0, 1, 2, 3}, then we have (3+ 3w) −wa = wa. Theorem 7: Let C = 〈g(x),wa(x)〉 be a cyclic code of odd length n over R. If f(x)rc ∈ C for any f(x) ∈ C, then (1+w)(1+x+x2 +...+xn−1) ∈ C and there are two constants e,d ∈ Z∗4 such that g∗(x) = eg(x) and a∗(x) = da(x). Proof: Suppose that C = 〈g(x),wa(x)〉 , where a(x)|g(x)|xn − 1 ∈ Z4[x]. Since (0, 0, ..., 0) ∈ C, then its reversible complement is also in C. (0, 0, ..., 0)rc = (3 + 3w, 3 + 3w,..., 3 + 3w) = 3(1 + w)(1, 1, ..., 1) ∈ C. This vector corresponds of the polynomial (3 + 3w) + (3 + 3w)x + ... + (3 + 3w)xn−1 = (3 + 3w) xn − 1 x− 1 ∈ C. Since 3 ∈ Z∗4, then (1+w)(1+x+...+x n−1) ∈ C. Let g(x) = g0 + g1x + ... + gs−1xs−1 + gsxs. Note that g(x)rc= (3+3w)+(3+3w)x+...+(3+3w)xn−s−2 +gsx n−s−1 +...+g1x n−2 +g0x n−1 ∈ C. Since C is a linear code, then 3(1 + w)(1 + x + x2 + ... + xn−1) −g(x)rc ∈ C which implies that ((3 + 3w)−gs)xn−s−1 + ((3 + 3w)−gs−1)xn−s−2+...+((3+3w)−g0)xn−1 ∈ C. By using (3 + 3w) −a = a, this implies that xn−s−1(gs+gs−1x+...+g0x s) = xn−s−1g∗(x) ∈ C Since g∗(x) ∈ C, this implies that g∗(x) = g(x)u(x) + wa(x)v(x) where u(x),v(x) ∈ Z4[x]. Since gi ∈ Z4, for i = 0, 1, ...,s, we have that v(x) = 0. As deg g∗(x) = deg g(x), we have u(x) ∈ Z∗4. Therefore there is a constant e ∈ Z∗4 such that g ∗(x) = eg(x). So, g(x) is a self reciprocal polynomial. Let a(x) = a0 + a1x + ... + atxt. Suppose that wa(x) = wa0 + wa1x + ... + watx t. Then (wa(x))rc = (3 + 3w) + (3 + 3w)x + ... +watx n−t−1 + ... + wa1x n−2 +wa0x n−1 ∈ C As (3 + 3w)x n−1 x−1 ∈ C and C is a linear code, then −(wa(x))rc + (3 + 3w) xn − 1 x− 1 ∈ C Hence, xn−t−1[(−(wat)+(3+3w))+(−(wat−1)+ (3 + 3w))x+...+ (−(wa0) + (3 + 3w))xt]. By the Lemma 6, we get xn−t−1(wat + wat−1x + ... + wa0x t) xn−t−1wa∗(x) ∈ C. Since wa∗(x) ∈ C, we have wa∗(x) = g(x)h(x) + wa(x)s(x) Since w doesn’t appear in g(x), it follows that h(x) = 0 and a∗(x) = a(x)s(x). As deg a∗(x) = deg a(x), then s(x) ∈ Z∗4. So, a(x) is a self reciprocal polynomial. Theorem 8: Let C = 〈g(x),wa(x)〉 be a cyclic code of odd length n over R. If (1+w)(1+x+x2+ ... + xn−1) ∈ C and g(x),a(x) are self reciprocal polynomials, then c(x)rc ∈ C for any c(x) ∈ C. Proof: Since C = 〈g(x),wa(x)〉 , for any c(x) ∈ C, there exist m(x) and n(x) in R[x] such that c(x) = g(x)m(x) + wa(x)n(x). By using the Lemma 4, we have c∗(x) = (g(x)m(x) + wa(x)n(x)) = (g(x)m(x))∗ + xs(wa(x)n(x)) = g∗(x)m∗(x) + wa∗(x)(xsn∗(x)) Since g∗(x) = eg(x),a∗(x) = da(x), we have c∗(x) = eg(x)m∗(x) + dwa(x)(xsn∗(x)) ∈ C. So, c∗(x) ∈ C. Let c(x) = c0 + c1x + ... + ctxt ∈ C. Since C is a cyclic code, we get xn−t−1c(x) = c0x n−t−1+c1x n−t+...+ctx n−1 ∈ C Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 4 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... Since (1 + w) + (1 + w)x + ... + (1 + w)xn−1 ∈ C and C is a linear code we have −(1 + w) xn − 1 x− 1 −xn−t−1c(x) = −(1 + w) − (1 + w)x + ... + (−c0 − (1 + w))xn−t−1 +... + (−ct − (1 + w))xn−1 ∈ C. By using a + (1 + w) = −a, this implies that −(1 + w) − ... + c0xn−t−1 + ... + ctxn−1 ∈ C This shows that (c∗(x))rc ∈ C. ((c∗(x))rc)∗ = ct + ct−1x + ... + (3 + 3w)x n−1 This corresponds this vector (ct,ct−1, ...,c0, ..., 0). Since (c∗(x)rc)∗ = (xn−t−1c(x))rc, so c(x)rc ∈ C. Example 9: Let x3−1 = (x+ 3)(x2 +x+ 1) ∈ Z4[x]. Let C = 〈 x2 + x + 1 + w(x2 + x + 1) 〉 . C is a cyclic DNA code of length 3 over R. The Gray image of C under the Gray map φ is a DNA code of length 6, Hamming distance 3. These codewords are as follows All 16 codewords of C CCCCCC TGTGTG GGGGGG GTGTGT TTTTTT GCGCGC AAAAAA CGCGCG GAGAGA CTCTCT AGAGAG TCTCTC TATATA ACACAC ATATAT CACACA Example 10: Let x7 − 1 = (x + 3)(x3 − 2x2 + x − 1)(x3 − x2 + 2x − 1) ∈ Z4[x]. Let C =< x6−3x5 +x4−3x3 +x2−3x+ 1 +w(x6−3x5 + x4 − 3x3 + x2 − 3x + 1) >. C is a cyclic DNA code of length 7 over R. The Gray image of C under the Gray map φ is a DNA code of length 14, Hamming distance 7. These codewords are as follows All 16 codewords of C CCCCCCCCCCCCCC GGGGGGGGGGGGGG TTTTTTTTTTTTTT AAAAAAAAAAAAAA GAGAGAGAGAGAGA AGAGAGAGAGAGAG TATATATATATATA ATATATATATATAT TGTGTGTGTGTGTG GTGTGTGTGTGTGT GCGCGCGCGCGCGC CGCGCGCGCGCGCG CTCTCTCTCTCTCT TCTCTCTCTCTCTC ACACACACACACAC CACACACACACACA IV. THE REVERSIBLE AND REVERSIBLE COMPLEMENT CODES OVER S Throughout this paper, S denotes the commu- tative ring Z4 + wZ4 + vZ4 + wvZ4 = {b1 + wb2 + vb3 + wvb4 : bj ∈ Z4, 1 ≤ j ≤ 4} with w2 = 2,v2 = v,wv = vw, with characteristic 4. S can also be thought of as the quotient ring Z4[w,v]/ < w2 − 2,v2 −v,wv −vw > . Let S = Z4 + wZ4 + vZ4 + wvZ4 = (Z4 + wZ4) + v(Z4 + wZ4) = R + vR We define the Gray map φ1 from S to R as follows φ1 : S −→ R2 a + vb 7−→ (a,b) where a,b ∈ R. This Gray map is extended compenentwise to φ1 : S n −→ R2n x = (x1, ...,xn) 7−→ (a1, ...,an,b1, ...,bn) where xi = ai + vbi,ai,bi ∈ R for i = 1, 2, ...,n. In this section, we study the cyclic codes of odd length n over S satisfy reverse and reverse Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 5 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... complement constraint. Since the ring S is of the cardinality 44, then we define the map φ1 which gives a one to one correspondence between the element of S and the 256 codons over the alphabet {A,T,G,C}4 by using the Gray map. For example: 0 = 0 + v0 7−→ φ1(0) = (0, 0) −→ AAAA 2wv = 0 +v(2w) 7−→φ1(2wv) = (0,2w)−→AAAG 1+3v+3wv =1+v(3+3w) 7−→φ1(1+v(3+3w)) = (1, 3 + 3w) −→ CATT Definition 11: Let A1,A2 be linear codes. A1 ⊗A2 = {(a1,a2) : a1 ∈ A1,a2 ∈ A2} and A1 ⊕A2 = {a1 + a2 : a1 ∈ A1,a2 ∈ A2} Let C be a linear code of length n over S. Define C1 = {a : ∃ b ∈ Rn,a + vb ∈ C} C2 = {b : ∃ a ∈ Rn,a + vb ∈ C} where C1 and C2 are linear codes over R of length n. Theorem 12: Let C be a linear code of length n over S. Then φ1(C) = C1 ⊗ C2 and |C| = |C1| |C2| . Corollary 13: If φ1(C) = C1 ⊗ C2, then C = vC1 ⊕ (1 −v)C2. Theorem 14: Let C = vC1 ⊕ (1 − v)C2 be a linear code of odd length n over S. Then C is a cyclic code over S if and only if C1,C2 are cyclic codes over R. Proof: Let (a10,a 1 1, ...,a 1 n−1) ∈ C1, (a 2 0,a 2 1, ...,a 2 n−1) ∈ C2. Assume that mi = va 1 i + (1 − v)a 2 i for i = 0, 1, 2, ...,n − 1. Then (m0,m1, ...,mn−1) ∈ C. Since C is a cyclic code, it follows that (mn−1,m0,m1, ...,mn−2) ∈ C. Note that (mn−1,m0, ...,mn−2) = v(a 1 n−1,a 1 0, ...,a 1 n−2) + (1 − v)(a2n−1,a 2 0, ...,a 2 n−2). Hence (a1n−1,a 1 0, ...,a 1 n−2) ∈ C1, (a 2 n−1,a 2 0, ...,a 2 n−2) ∈ C2. Therefore C1,C2 are cyclic codes over R. Conversely, suppose that C1,C2 are cyclic codes over R. Let (m0,m1, ...,mn−1) ∈ C, where mi = va1i + (1 − v)a 2 i for i = 0, 1, 2, ...,n − 1. Then (a1n−1,a 1 0, ...,a 1 n−2) ∈ C1, (a 2 n−1,a 2 0, ...,a 2 n−2) ∈ C2. Note that (mn−1,m0, ...,mn−2) = v(a 1 n−1,a 1 0, ...,a 1 n−2) + (1 −v)(a2n−1,a 2 0, ...,a 2 n−2) ∈ C. So, C is a cyclic code over S. Theorem 15: Let C = vC1 ⊕ (1 − v)C2 be a linear code of odd length n over S. Then C is reversible over S iff C1,C2 are reversible over R. Proof: Let C1,C2 be reversible codes. For any b ∈ C,b = vb1 + (1 − v)b2, where b1 ∈ C1,b2 ∈ C2. Since C1 and C2 are reversible, br1 ∈ C1,b r 2 ∈ C2. So, b r = vbr1 + (1 −v)b r 2 ∈ C. Hence C is reversible. On the other hand, Let C be a reversible code over S. So for any b = vb1 + (1−v)b2 ∈ C, where b1 ∈ C1,b2 ∈ C2, we get br = vbr1 +(1−v)b r 2 ∈ C. Let br = vbr1 + (1−v)b r 2 = vs1 + (1−v)s2, where s1 ∈ C1,s2 ∈ C2. So C1 and C2 are reversible codes over R. Lemma 16: For any c ∈ S, we have c + c = (3 + 3w) + v(3 + 3w). Lemma 17: For any a ∈ S, a + 30 = 3a. Theorem 18: Let C = vC1 ⊕ (1 − v)C2 be a cyclic code of odd length n over S. Then C is reversible complement over S iff C is reversible over S and (0, 0, ..., 0) ∈ C. Proof: Since C is reversible complement, for any c = (c0,c1, ...,cn−1) ∈ C,crc = (cn−1,cn−2, ...,c0) ∈ C. Since C is a linear code, so (0, 0, ..., 0) ∈ C. Since C is reversible complement, so (0, 0, ..., 0) ∈ C. By using the Lemma 17, we have 3cr = 3(cn−1,cn−2, ...,c0) = (cn−1,cn−2, ...,c0) + 3(0, 0, ..., 0) ∈ C. So, for any c ∈ C, we have cr ∈ C. On the other hand, let C be reversible. So, for any c = (c0,c1, ...,cn−1) ∈ C, cr = (cn−1,cn−2, ...,c0) ∈ C. To show that C is re- versible complement, for any c ∈ C, crc = (cn−1,cn−2, ...,c0) = 3(cn−1,cn−2, ...,c0) + (0, 0, ..., 0) ∈ C. Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 6 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... So, C is reversible complement. Lemma 19: For any a,b ∈ S, a + b = a + b− 3(1 + w)(1 + v). Theorem 20: Let D1 and D2 be two reversible complement cyclic codes of length n over S. Then D1 + D2 and D1 ∩D2 are reversible complement cyclic codes. Proof: Let d1 = (c0,c1, ...,cn−1) ∈ D1,d2 = (c10,c 1 1, ...,c 1 n−1) ∈ D2. Then, (d1 +d2) rc= ( (cn−1 +c 1 n−1), ..., (c1 +c 1 1), (c0 +c 1 0) ) = ( cn−1 +c 1 n−1−3(1+w)(1+v), ..., c0 + c 1 0 − 3(1 + w)(1 + v) ) =(cn−1 − 3(1 + w)(1 + v), ...,c0 −3(1+w)(1+v))+ ( c1n−1, ...,c 1 0 ) = ( drc1 − 3(1 + w)(1 + v) xn − 1 x− 1 ) +drc2 ∈ D1 + D2. This shows that D1 +D2 is reversible complement cyclic code. It is clear that D1 ∩D2 is reversible complement cyclic code. V. BINARY IMAGES OF CYCLIC DNA CODES OVER R The 2-adic expansion of c ∈ Z4 is c = α(c) + 2β(c) such that α(c) + β(c) + γ(c) = 0 for all c ∈ Z4 c α(c) β(c) γ(c) 0 0 0 0 1 1 0 1 2 0 1 1 3 1 1 0 The Gray map is given by Ψ : Z4 −→ Z22 c 7−→ Ψ(c) = (β(c),γ(c)) for all c ∈ Z4 in [14]. Define Ŏ : R −→ Z42 a + bw 7−→ Ŏ(a + wb) = Ψ (φ (a + wb)) = Ψ(a,b) = (β(a),γ(a),β(b),γ(b)) Let a + wb be any element of the ring R. The Lee weight wL of the element of the ring R is defined as follows wL(a + wb) = wL(a,b) where wL(a,b) described the usual Lee weight on Z24. For any c1,c2 ∈ R the Lee distance dL is given by dL(c1,c2) = wL(c1 − c2). The Hamming distance d(c1,c2) between two codewords c1 and c2 is the Hamming weight of the codewords c1 − c2. AA −→ 0000 CA −→ 0100 GA −→ 1100 TA −→ 1000 AC −→ 0001 AG −→ 0011 AT −→ 0010 CC −→ 0101 CG −→ 0111 CT −→ 0110 GC −→ 1101 GG −→ 1111 GT −→ 1110 TC −→ 1001 TG −→ 1011 TT −→ 1010 Lemma 21: The Gray map Ŏ is a distance preserving map from (Rn, Lee distance) to (Z4n2 , Hamming distance). It is also Z2-linear. Proof: For c1,c2 ∈ Rn, we have Ŏ(c1 − c2) = Ŏ(c1) − Ŏ(c2). So, dL(c1,c2) = wL(c1 − c2) = wH(Ŏ(c1 − c2)) = wH(Ŏ(c1) − Ŏ(c2)) = dH(Ŏ(c1), Ŏ(c2)). So, the Gray map Ŏ is distance preserving map. For any c1,c2 ∈ Rn,k1,k2 ∈ Z2,we have Ŏ(k1c1 +k2c2) = k1Ŏ(c1) +k2Ŏ(c2). Thus, Ŏ is Z2-linear. Proposition 22: Let σ be the cyclic shift of Rn and υ be the 4-quasi-cyclic shift of Z4n2 . Let Ŏ be the Gray map from Rn to Z4n2 . Then Ŏσ = υŎ. Proof: Let c = (c0,c1, ...,cn−1) ∈ Rn, we have ci = a1i + wb2i with a1i,b2i ∈ Z4, 0 ≤ i ≤ n− 1. By applying the Gray map, we have Ŏ(c)=  β(a10),γ(a10),β(b20),γ(b20),β(a11),γ(a11),β(b21),γ(b21), ...,β(a1n−1), γ(a1n−1),β(b2n−1),γ(b2n−1)   . Hence υ(Ŏ(c)) =  β(a1n−1),γ(a1n−1),β(b2n−1),γ(b2n−1),β(a10),γ(a10),β(b20),γ(b20), ...,β(a1n−2), γ(a1n−2),β(b2n−2),γ(b2n−2)  . Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 7 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... On the other hand, σ(c) = (cn−1,c0,c1, ...,cn−2). We have Ŏ(σ(c)) =  β(a1n−1),γ(a1n−1),β(b2n−1),γ(b2n−1),β(a10),γ(a10),β(b20),γ(b20), ..., β(a1n−2),γ(a1n−2),β(b2n−2),γ(b2n−2)  . Therefore, Ŏσ = υŎ. Theorem 23: If C is a cyclic DNA code of length n over R then Ŏ(C) is a binary quasi-cyclic DNA code of length 4n with index 4. VI. BINARY IMAGE OF CYCLIC DNA CODES OVER S We define Ψ̃ : S −→ Z44 a0 + wa1 + va2 + wva3 7−→ (a0,a1,a2,a3) where ai ∈ Z4, for i = 0, 1, 2, 3. Now, we define Θ : S −→ Z82 as a0 + wa1 + va2 + wva3 7−→ Θ(a0 + wa1 + va2 + wva3) = Ψ(Ψ̃(a0 + wa1 + va2 + wva3)) = (β(a0),γ(a0),β(a1),γ(a1),β(a2),γ(a2),β(a3),γ(a3)), where Ψ is the Gray map Z4 to Z22. Let a0 + wa1 + va2 + wva3 be any element of the ring S. The Lee weight wL of the element of the ring S is defined as wL(a0 +wa1 +va2 +wva3) = wL((a0,a1,a2,a3)) where wL((a0,a1,a2,a3)) described the usual Lee weight on Z44. For any c1,c2 ∈ S, the Lee distance dL is given by dL(c1,c2) = wL(c1 − c2). The Hamming distance d(c1,c2) between two codewords c1 and c2 is the Hamming weight of the codewords c1 − c2. The binary images of cyclic DNA codes; AAAA −→ 00000000 AACA −→ 00000100 AAGA −→ 00001100 AATA −→ 00001000 ... ... ... Lemma 24: The Gray map Θ is a distance preserving map from (Sn, Lee distance) to (Z8n2 , Hamming distance). It is also Z2-linear. Proof: It is proved as in the proof of the Lemma 21. Proposition 25: Let σ be the cyclic shift of Sn and ′ υ be the 8-quasi-cyclic shift of Z8n2 . Let Θ be the Gray map from Sn to Z8n2 . Then Θσ = ′ υΘ. Proof: It is proved as in the proof of the Proposition 22. Theorem 26: If C is a cyclic DNA code of length n over S then Θ(C) is a binary quasi-cyclic DNA code of length 8n with index 8. Proof: Let C be a cyclic DNA code of length n over S. So, σ(C) = C. By using the Proposition 25, we have Θ(σ(C)) = ′ υ(Θ(C)) = Θ(C). Hence Θ(C) is a set of length 8n over the alphabet Z2 which is a quasi-cyclic code of index 8. VII. SKEW CYCLIC DNA CODES OVER R We will use a non trivial automorphism, for all a + wb ∈ R, it is defined by θ : R −→ R a + wb 7−→ a−wb The ring R[x,θ] = {a0 +a1x+...+an−1xn−1 : ai ∈ R,n ∈ N} is called skew polynomial ring. It is non commutative ring. The addition in the ring R[x,θ] is the usual polynomial and multiplication is defined as (axi)(bxj) = aθi(b)xi+j. The order of the automorphism θ is 2. Definition 27: A subset C of Rn is called a skew cyclic code of length n if C satisfies the following conditions, i) C is a submodule of Rn, ii) If c = (c0,c1, ...,cn−1) ∈ C, then σθ (c) = (θ(cn−1),θ(c0), ...,θ(cn−2)) ∈ C Let f(x) + 〈xn − 1〉 be an element in the set Řn = R [x,θ] /〈xn − 1〉 and let r(x) ∈ R [x,θ]. Define multiplication from left as follows, r(x)(f(x) + 〈xn − 1〉) = r(x)f(x) + 〈xn − 1〉 for any r(x) ∈ R [x,θ]. Theorem 28: Řn is a left R [x,θ]-module where multiplication defined as in above. Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 8 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... Theorem 29: A code C over R of length n is a skew cyclic code if and only if C is a left R [x,θ]- submodule of the left R [x,θ]-module Řn. Theorem 30: Let C be a skew cyclic code over R of length n and let f(x) be a polynomial in C of minimal degree. If f(x) is monic polynomial, then C = 〈f(x)〉 , where f(x) is a right divisor of xn − 1. For all x ∈ R, we have θ(x) + θ(x) = 3 − 3w. Theorem 31: Let C = 〈f(x)〉 be a skew cyclic code over R, where f(x) is a monic polynomial in C of minimal degree. If C is reversible comple- ment, the polynomial f(x) is self reciprocal and (3 + 3w) xn − 1 x− 1 ∈ C. Proof: Let C = 〈f(x)〉 be a skew cyclic code over R, where f(x) is a monic polynomial in C. Since (0, 0, ..., 0) ∈ C and C is reversible complement, we have ( 0, 0, ..., 0 ) = (3 + 3w, 3 + 3w,..., 3 + 3w) ∈ C. Let f(x) = 1 + a1x + ... + at−1xt−1 + xt. Since C is reversible complement, we have frc(x) ∈ C. That is frc(x) = (3+3w)+(3+3w)x+...+(3+3w)xn−t−2 +(2+3w)xn−t−1 +at−1x n−t+ ... +a1x n−2 +(2+3w)xn−1. Since C is a linear code, we have frc(x) − (3 + 3w) xn − 1 x− 1 ∈ C. This implies that −xn−t−1 + (at−1 − (3 + 3w))xn−t + ... + (a1 − (3 + 3w))xn−2 −xn−1 ∈ C. Multiplying on the right by xt+1−n, we have −1 + (at−1 − (3 + 3w))θ(1)x + ... + (a1 − (3 + 3w))θt−1(1)xt−1 −θt(1)xt ∈ C. By using a + a = 3 + 3w, we have −1 −at−1x−at−2x2 − ...−a1xt−1 −xt = 3f∗(x) ∈ C. Since C = 〈f(x)〉, there exist q(x) ∈ R [x,θ] such that 3f∗(x) = q(x)f(x). Since deg f(x) = deg f∗(x), we have q(x) = 1. Since 3f∗(x) = f(x), we have f∗(x) = 3f(x). So, f(x) is self reciprocal. Theorem 32: Let C = 〈f(x)〉 be a skew cyclic code over R, where f(x) is a monic polynomial in C of minimal degree. If (3 + 3w)x n−1 x−1 ∈ C and f(x) is self reciprocal, then C is reversible complement. Proof: Let f(x) = 1+a1x+...+at−1xt−1+xt be a monic polynomial of the minimal degree. Let c(x) ∈ C. So, c(x) = q(x)f(x), where q(x) ∈ R[x,θ]. By using Lemma 4, we have c∗(x) = (q(x)f(x))∗ = q∗(x)f∗(x). Since f(x) is self reciprocal, so c∗(x) = q∗(x)ef(x), where e ∈ Z4\{0}. Therefore c∗(x) ∈ C = 〈f(x)〉. Let c(x) = c0 + c1x + ... + ctx t ∈ C. Since C is a cyclic code, we get c(x)xn−t−1 =c0x n−t−1 +c1x n−t+...+ctx n−1∈C. The vector corresponding to this polynomial is (0, 0, ..., 0,c0,c1, ...,ct) ∈ C. Since (3 + 3w, 3 + 3w,..., 3 + 3w) ∈ C and C linear, we have (3+3w, 3+3w,..., 3+3w)−(0, 0, ..., 0,c0,c1, ...,ct) = (3+3w,..., 3+3w, (3+3w)−c0, ..., (3+3w)−ct)∈C. By using a + a = 3 + 3w, we get (3 + 3w, 3 + 3w,..., 3 + 3w,c0, ...,ct) ∈ C, which is equal to (c(x)∗)rc. This shows that ((c(x)∗) rc ) ∗ = c(x)rc ∈ C. VIII. DNA CODES OVER S Definition 33: Let f1 and f2 be polynomials with deg f1 = t1, deg f2 = t2 and both dividing xn − 1 ∈ R[x]. Let m = min{n − t1,n − t2} and f(x) = vf1(x) + (1 − v)f2(x) over S. The set L(f) is called a Γ-set, where the automorphism Γ : S −→ S is defined as follows: a+wb+vc+wvd 7−→a+b+w(b+d)−vc−wvdc. Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 9 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... L(f) =   a0 a1 a2 · · · at 0 · · · · · · · · · 0 0 Γ(a0) Γ(a1) · · · · · · Γ(at) 0 · · · · · · 0 0 0 a0 a1 · · · · · · at 0 · · · 0 0 0 0 Γ(a0) Γ(a1) · · · · · · Γ(at) · · · 0 ... · · · · · · · · · ... · · · · · · · · · · · · ...   (1) The set L(f) is defined as L(f) = {E0,E1, ...,Em−1}, where Ei = { xif if i is even xiΓ(f) if i is odd L(f) generates a linear code C over S denoted by C = 〈f〉Γ. Let f(x) = a0 + a1x + ... + atx t be over S and S-submodule generated by L(f) is generated by the matrix in Eq. (1). Theorem 34: Let f1 and f2 be self reciprocal polynomials dividing xn − 1 over R with degree t1 and t2, respectively. If f1 = f2, then f = vf1 + (1 −v)f2 and |〈L(f)〉| = 256m. C = 〈L(f)〉 is a linear code over S and Θ(C) is a reversible DNA code. Proof: It is proved as in the proof of the Theorem 5 in [5]. Corollary 35: Let f1 and f2 be self reciprocal polynomials dividing xn − 1 over R and C = 〈L(f)〉 be a cyclic code over S. If x n−1 x−1 ∈ C, then Θ(C) is a reversible complement DNA code. Example 36: Let f1(x) = f2(x) = x − 1 dividing x7 − 1 over R. Hence, C = 〈vf1(x) + (1 −v)f2(x)〉Γ = 〈x− 1〉Γ is a Γ-linear code over S and Θ(C) is a reversible complement DNA code, because of x7 − 1 x− 1 ∈ C. Acknowledgement 37: We wish to express sin- cere thanks to Steven Dougherty who gave helpful comments. REFERENCES [1] Abualrub T., Ghrayeb A., Zeng X., Construction of cyclic codes over GF(4) for DNA computing, J. Franklin Institute, 343, 448-457, 2006. [2] Abualrub T., Siap I., Reversible quaternary cyclic codes, Proc. of the 9th WSEAS Int. Conference on Appl. Math., Istanbul, 441-446, 2006. [3] Adleman L., Molecular computation of the solution to combinatorial problems, Science, 266, 1021-1024, 1994. [4] Aydın N., Dertli A., Cengellenmis Y., Cyclic and Con- stacyclic codes over Z4 + wZ4, preprint. [5] Bayram A., Oztas E., Siap I., Codes over F4 + vF4 and some DNA applications, Designs, Codes and Cryptogra- phy, DOI: 10.107/s10623-015-0100-8, 2015. [6] Bennenni N., Guenda K., Mesnager S., New DNA cyclic codes over rings, arXiv: 1505.06263v1, 2015. [7] Gaborit P., King O. D., Linear construction for DNA codes, Theor. Computer Science, 334, 99-113, 2005. [8] Guenda K., Gulliver T. A., Sole P,. On cyclic DNA codes, Proc., IEEE Int. Symp. Inform. Theory, Istanbul, 121- 125, 2013. [9] Guenda K., Gulliver T. A., Construction of cyclic codes over F2 + uF2 for DNA computing, AAECC, 24, 445- 459, 2013. [10] Liang J., Wang L., On cyclic DNA codes over F2 +uF2, J. Appl. Math. Comput., DOI: 10.1007/s12190-015-0892- 8, 2015. [11] Ma F., Yonglin C., Jian G., On cyclic DNA codes over F4[u]/ 〈 u2 + 1 〉 . [12] Massey J. L., Reversible codes, Inf. Control, 7, 369-380, 1964. [13] Oztas E. S., Siap I., Lifted polynomials over F16 and their applications to DNA codes, Filomat, 27, 459-466, 2013. [14] Pattanayak S., Singh A. K., On cyclic DNA codes over the ring Z4 + uZ4, arXiv: 1508.02015, 2015. [15] Pattanayak S., Singh A. K., Kumar P., DNA cyclic codes over the ring F2[u, v]/ 〈 u2 − 1, v3 − v, uv − vu 〉 , arXiv:1511.03937, 2015. [16] Pattanayak S., Singh A. K., Construction of Cyclic DNA codes over the ring Z4[u]/ 〈 u2 − 1 〉 based on deletion distance, arXiv: 1603.04055v1, 2016. [17] Siap I., Abualrub T., Ghrayeb A., Cyclic DNA codes over the ring F2[u]/ ( u2 − 1 ) based on the delition distance, J. Franklin Institute, 346, 731-740, 2009. Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 10 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Abdullah Dertli, Yasemin Cengellenmis, On the cyclic DNA codes over the finite rings ... [18] Siap I., Abualrub T., Ghrayeb A., Similarity cyclic DNA codes over rings, IEEE, 978-1-4244-1748-3, 2008. [19] Wan Z. X., Quaternary codes, vol.8., World Scientific, 1997. [20] Yıldız B., Siap I., Cyclic DNA codes over the ring F2[u]/ ( u4 − 1 ) and applications to DNA codes, Com- put. Math. Appl., 63, 1169-1176, 2012. [21] Zhu S., Chen X., Cyclic DNA codes over F2 + uF2 + vF2 + uvF2, arXiv: 1508.07113v1, 2015. Biomath 6 (2017), 1712167, http://dx.doi.org/10.11145/j.biomath.2017.12.167 Page 11 of 11 http://dx.doi.org/10.11145/j.biomath.2017.12.167 Introduction Preliminaries The reversible complement codes over R The reversible and reversible complement codes over S Binary images of cyclic DNA codes over R Binary image of cyclic DNA codes over S Skew cyclic DNA codes over R DNA codes over S References