Cyclic codes ‹#› Cyclic codes 1 CHAPTER 3: Cyclic and convolution codes Cyclic codes are special linear cods of interest and importance because • They posses a rich algebraic structure that can be utilized in a variety of ways. • They have extremely concise specifications. • They can be efficiently implemented using simple shift registers. • Many practically important codes are cyclic. Convolution codes allow to encode streams od data (bits). IV054 ‹#› Cyclic codes 2 Cyclic codes IMPORTANT NOTE In order to specify a binary code with 2k codewords of length n one may need to write down 2k codewords of length n. In order to specify a linear binary code with 2k codewords of length n it is sufficient to write down k codewords of length n. In order to specify a binary cyclic code with 2k codewords of length n it is sufficient to write down 1 codeword of length n. > ‹#› Cyclic codes 3 Cyclic codes BASIC DEFINITION AND EXAMPLES Definition A code C is cyclic if (i) C is a linear code; (ii) any cyclic shift of a codeword is also a codeword, i.e. whenever a0,… an -1 Î C, then also an -1 a0 … an –2 Î C. IV054 Example (i) Code C = {000, 101, 011, 110} is cyclic. (ii) Hamming code Ham(3, 2): with the generator matrix is equivalent to a cyclic code. (iii) The binary linear code {0000, 1001, 0110, 1111} is not a cyclic, but it is equivalent to a cyclic code. (iv) Is Hamming code Ham(2, 3) with the generator matrix (a) cyclic? (b) equivalent to a cyclic code? ‹#› Cyclic codes 4 Cyclic codes FREQUENCY of CYCLIC CODES Comparing with linear codes, the cyclic codes are quite scarce. For, example there are 11 811 linear (7,3) linear binary codes, but only two of them are cyclic. Trivial cyclic codes. For any field F and any integer n >= 3 there are always the following cyclic codes of length n over F: • No-information code - code consisting of just one all-zero codeword. • • Repetition code - code consisting of codewords (a, a, …,a) for a Î F. • • Single-parity-check code - code consisting of all codewords with parity 0. • • No-parity code - code consisting of all codewords of length n For some cases, for example for n = 19 and F = GF(2), the above four trivial cyclic codes are the only cyclic codes. IV054 > ‹#› Cyclic codes 5 Cyclic codes EXAMPLE of a CYCLIC CODE The code with the generator matrix has codewords c1 = 1011100 c2 = 0101110 c3 =0010111 c1 + c2 = 1110010 c1 + c3 = 1001011 c2 + c3 = 0111001 c1 + c2 + c3 = 1100101 and it is cyclic because the right shifts have the following impacts c1 ® c2, c2 ® c3, c3 ® c1 + c3 c1 + c2 ® c2 + c3, c1 + c3 ® c1 + c2 + c3, c2 + c3 ® c1 c1 + c2 + c3 ® c1 + c2 IV054 ‹#› Cyclic codes 6 Cyclic codes POLYNOMIALS over GF(q) A codeword of a cyclic code is usually denoted a0 a1…an -1 and to each such a codeword the polynomial a0 + a1 x + a2 x2 + … + an -1 xn -1 will be associated. NOTATION : Fq[x] denotes the set of all polynomials over GF(q ). deg (f(x )) = the largest m such that xm has a non-zero coefficient in f(x). IV054 Multiplication of polynomials If f(x), g(x) Î Fq[x], then deg (f(x) g(x)) = deg (f(x)) + deg (g(x)). Division of polynomials For every pair of polynomials a(x), b(x) ¹ 0 in Fq[x] there exists a unique pair of polynomials q(x), r(x) in Fq[x] such that a(x) = q(x)b(x) + r(x), deg (r(x)) < deg (b(x)). Example Divide x3 + x + 1 by x2 + x + 1 in F2[x]. Definition Let f(x) be a fixed polynomial in Fq[x]. Two polynomials g(x), h(x) are said to be congruent modulo f(x), notation g(x) º h(x) (mod f(x)), if g(x) - h(x) is divisible by f(x). ‹#› Cyclic codes 7 Cyclic codes RING of POLYNOMIALS The set of polynomials in Fq[x] of degree less than deg (f(x)), with addition and multiplication modulo f(x) forms a ring denoted Fq[x]/f(x). Example Calculate (x + 1)2 in F2[x] / (x2 + x + 1). It holds (x + 1)2 = x2 + 2x + 1 º x2 + 1 º x (mod x2 + x + 1). How many elements has Fq[x] / f(x)? Result | Fq[x] / f(x) | = q deg (f(x)). Example Addition and multiplication in F2[x] / (x2 + x + 1) IV054 Definition A polynomial f(x) in Fq[x] is said to be reducible if f(x) = a(x)b(x), where a(x), b(x) Î Fq[x] and deg (a(x)) < deg (f(x)), deg (b(x)) < deg (f(x)). If f(x) is not reducible, it is irreducible in Fq[x]. Theorem The ring Fq[x] / f(x) is a field if f(x) is irreducible in Fq[x]. + 0 1 x 1 + x 0 0 1 x 1 + x 1 1 0 1 + x x x x 1 + x 0 1 1 + x 1 + x x 1 0 · 0 1 x 1 + x 0 0 0 0 0 1 0 1 X 1 + x x 0 x 1 + x 1 1 + x 0 1 + x 1 x ‹#› Cyclic codes 8 Cyclic codes FIELD Rn, Rn = Fq[x] / (xn - 1) Computation modulo xn – 1 Since xn º 1 (mod (xn -1)) we can compute f(x) mod (xn -1) as follows: In f(x) replace xn by 1, xn +1 by x, xn +2 by x2, xn +3 by x3, … Identification of words with polynomials a0 a1… an -1 « a0 + a1 x + a2 x2 + … + an -1 xn -1 Multiplication by x in Rn corresponds to a single cyclic shift x (a0 + a1 x + … an -1 xn -1) = an -1 + a0 x + a1 x2 + … + an -2 xn -1 IV054 > ‹#› Cyclic codes 9 Cyclic codes Algebraic characterization of cyclic codes Theorem A code C is cyclic if C satisfies two conditions (i) a(x), b(x) Î C Þ a(x) + b(x) Î C (ii) a(x) Î C, r(x) Î Rn Þ r(x)a(x) Î C Proof (1) Let C be a cyclic code. C is linear Þ (i) holds. (ii) Let a(x) Î C, r(x) = r0 + r1x + … + rn -1xn -1 r(x)a(x) = r0a(x) + r1xa(x) + … + rn -1xn -1a(x) is in C by (i) because summands are cyclic shifts of a(x). (2) Let (i) and (ii) hold · Taking r(x) to be a scalar the conditions imply linearity of C. · Taking r(x) = x the conditions imply cyclicity of C. IV054 > ‹#› Cyclic codes 10 Cyclic codes CONSTRUCTION of CYCLIC CODES Notation If f(x) Î Rn, then áf(x)ñ = {r(x)f(x) | r(x) Î Rn} (multiplication is modulo xn -1). Theorem For any f(x) Î Rn, the set áf(x)ñ is a cyclic code (generated by f). Proof We check conditions (i) and (ii) of the previous theorem. (i) If a(x)f(x) Î áf(x)ñ and also b(x)f(x) Î áf(x)ñ, then a(x)f(x) + b(x)f(x) = (a(x) + b(x)) f(x) Î áf(x)ñ (ii) If a(x)f(x) Î áf(x)ñ, r(x) Î Rn, then r(x) (a(x)f(x)) = (r(x)a(x)) f(x) Î áf(x)ñ. IV054 Example C = á1 + x2 ñ, n = 3, q = 2. We have to compute r(x)(1 + x2) for all r(x) Î R3. R3 = {0, 1, x, 1 + x, x2, 1 + x2, x + x2, 1 + x + x2}. Result C = {0, 1 + x, 1 + x2, x + x2} C = {000, 011, 101, 110} ‹#› Cyclic codes 11 Cyclic codes Characterization theorem for cyclic codes We show that all cyclic codes C have the form C = áf(x)ñ for some f(x) Î Rn. Theorem Let C be a non-zero cyclic code in Rn. Then • there exists unique monic polynomial g(x) of the smallest degree such that • C = ág(x)ñ • g(x) is a factor of xn -1. IV054 Proof (i) Suppose g(x) and h(x) are two monic polynomials in C of the smallest degree. Then the polynomial g(x) - h(x) Î C and it has a smaller degree and a multiplication by a scalar makes out of it a monic polynomial. If g(x) ¹ h(x) we get a contradiction. (ii) Suppose a(x) Î C. Then a(x) = q(x)g(x) + r(x) (deg r(x) < deg g(x)) and r(x) = a(x) - q(x)g(x) Î C. By minimality r(x) = 0 and therefore a(x) Î ág(x)ñ. ‹#› Cyclic codes 12 Cyclic codes Characterization theorem for cyclic codes IV054 (iii) Clearly, xn –1 = q(x)g(x) + r(x) with deg r(x) < deg g(x) and therefore r(x) º -q(x)g(x) (mod xn -1) and r(x) Î C Þ r(x) = 0 Þ g(x) is a factor of xn -1. GENERATOR POLYNOMIALS Definition If for a cyclic code C it holds C = ág(x)ñ, then g is called the generator polynomial for the code C. ‹#› Cyclic codes 13 Cyclic codes HOW TO DESIGN CYCLIC CODES? The last claim of the previous theorem gives a recipe how to get all cyclic codes of the given length n. Indeed, all we need to do is to find all factors of xn -1. Problem: Find all binary cyclic codes of length 3. Solution: Since x3 – 1 = (x + 1)(x2 + x + 1) both factors are irreducible in GF(2) we have the following generator polynomials and codes. Generator polynomials Code in R3 Code in V(3,2) 1 R3 V(3,2) x + 1 {0, 1 + x, x + x2, 1 + x2} {000, 110, 011, 101} x2 + x + 1 {0, 1 + x + x2} {000, 111} x3 – 1 ( = 0) {0} {000} IV054 ‹#› Cyclic codes 14 Cyclic codes Design of generator matrices for cyclic codes Theorem Suppose C is a cyclic code of codewords of length n with the generator polynomial g(x) = g0 + g1x + … + grxr. Then dim (C) = n - r and a generator matrix G1 for C is IV054 Proof (i) All rows of G1 are linearly independent. (ii) The n - r rows of G represent codewords g(x), xg(x), x2g(x),…, xn -r -1g(x) (*) (iii) It remains to show that every codeword in C can be expressed as a linear combination of vectors from (*). Inded, if a(x) Î C, then a(x) = q(x)g(x). Since deg a(x) < n we have deg q(x) < n - r. Hence q(x)g(x) = (q0 + q1x + … + qn -r -1xn -r -1)g(x) = q0g(x) + q1xg(x) + … + qn -r -1xn -r -1g(x). ‹#› Cyclic codes 15 Cyclic codes EXAMPLE The task is to determine all ternary codes of length 4 and generators for them. Factorization of x4 - 1 over GF(3) has the form x4 - 1 = (x - 1)(x3 + x2 + x + 1) = (x - 1)(x + 1)(x2 + 1) Therefore there are 23 = 8 divisors of x4 - 1 and each generates a cyclic code. Generator polynomial Generator matrix 1 I4 x-1 x + 1 x2 + 1 (x - 1)(x + 1) = x2 - 1 (x - 1)(x2 + 1) = x3 - x2 + x - 1 [ -1 1 -1 1 ] (x + 1)(x2 + 1) [ 1 1 1 1 ] x4 - 1 = 0 [ 0 0 0 0 ] IV054 ‹#› Cyclic codes 16 Cyclic codes Check polynomials and parity check matrices for cyclic codes Let C be a cyclic [n,k]-code with the generator polynomial g(x) (of degree n - k). By the last theorem g(x) is a factor of xn - 1. Hence xn - 1 = g(x)h(x) for some h(x) of degree k (where h(x) is called the check polynomial of C). Theorem Let C be a cyclic code in Rn with a generator polynomial g(x) and a check polynomial h(x). Then an c(x) Î Rn is a codeword of C if c(x)h(x) º 0 – (this and next congruences are all modulo xn – 1). IV054 Proof Note, that g(x)h(x) = xn - 1 º 0 (i) c(x) Î C Þ c(x) = a(x)g(x) for some a(x) Î Rn Þ c(x)h(x) = a(x) g(x)h(x) º 0. º 0 (ii) c(x)h(x) º 0 c(x) = q(x)g(x) + r(x), deg r(x) < n – k = deg g(x) c(x)h(x) º 0 Þ r(x)h(x) º 0 (mod xn - 1) Since deg (r(x)h(x)) < n – k + k = n, we have r(x)h(x) = 0 in F[x] and therefore r(x) = 0 Þ c(x) = q(x)g(x) Î C. ‹#› Cyclic codes 17 Cyclic codes POLYNOMIAL REPRESENTATION of DUAL CODES Since dim (áh(x)ñ) = n - k = dim (C^) we might easily be fooled to think that the check polynomial h(x) of the code C generates the dual code C^. Reality is “slightly different'': Theorem Suppose C is a cyclic [n,k]-code with the check polynomial h(x) = h0 + h1x + … + hkxk, then (i) a parity-check matrix for C is (ii) C^ is the cyclic code generated by the polynomial i.e. the reciprocal polynomial of h(x). IV054 ‹#› Cyclic codes 18 Cyclic codes POLYNOMIAL REPRESENTATION of DUAL CODES Proof A polynomial c(x) = c0 + c1x + … + cn -1xn –1 represents a code from C if c(x)h(x) = 0. For c(x)h(x) to be 0 the coefficients at xk,…, xn -1 must be zero, i.e. Therefore, any codeword c0 c1… cn -1 Î C is orthogonal to the word hk hk -1…h000…0 and to its cyclic shifts. Rows of the matrix H are therefore in C^. Moreover, since hk = 1, these row-vectors are linearly independent. Their number is n - k = dim (C^). Hence H is a generator matrix for C^, i.e. a parity-check matrix for C. In order to show that C^ is a cyclic code generated by the polynomial it is sufficient to show that is a factor of xn -1. Observe that and since h(x -1)g(x -1) = (x -1)n -1 we have that xkh(x -1)xn -kg(x -1) = xn(x –n -1) = 1 – xn and therefore is indeed a factor of xn -1. IV054 ‹#› Cyclic codes 19 Cyclic codes ENCODING with CYCLIC CODES I Encoding using a cyclic code can be done by a multiplication of two polynomials - a message polynomial and the generating polynomial for the cyclic code. Let C be an [n,k]-code over an field F with the generator polynomial g(x) = g0 + g1 x + … + gr –1 x r -1 of degree r = n - k. If a message vector m is represented by a polynomial m(x) of degree k and m is encoded by m Þ c = mG, then the following relation between m(x) and c(x) holds c(x) = m(x)g(x). Such an encoding can be realized by the shift register shown in Figure below, where input is the k-bit message to be encoded followed by n - k 0' and the output will be the encoded message. Shift-register encodings of cyclic codes. Small circles represent multiplication by the corresponding constant, Å nodes represent modular addition, squares are delay elements IV054 01 ‹#› Cyclic codes 20 Cyclic codes Hamming codes as cyclic codes Definition (Again!) Let r be a positive integer and let H be an r * (2r -1) matrix whose columns are distinct non-zero vectors of V(r,2). Then the code having H as its parity-check matrix is called binary Hamming code denoted by Ham (r,2). It can be shown that binary Hamming codes are equivalent to cyclic codes. IV054 Theorem The binary Hamming code Ham (r,2) is equivalent to a cyclic code. Definition If p(x) is an irreducible polynomial of degree r such that x is a primitive element of the field F[x] / p(x), then p(x) is called a primitive polynomial. Theorem If p(x) is a primitive polynomial over GF(2) of degree r, then the cyclic code áp(x)ñ is the code Ham (r,2). ‹#› Cyclic codes 21 Cyclic codes Hamming codes as cyclic codes Example Polynomial x3 + x + 1 is irreducible over GF(2) and x is primitive element of the field F2[x] / (x3 + x + 1). F2[x] / (x3 + x + 1) = {0, x, x2, x3 = x + 1, x4 = x2 + x, x5 = x2 + x + 1, x6 = x2 + 1} The parity-check matrix for a cyclic version of Ham (3,2) IV054 ‹#› Cyclic codes 22 Cyclic codes PROOF of THEOREM The binary Hamming code Ham (r,2) is equivalent to a cyclic code. It is known from algebra that if p(x) is an irreducible polynomial of degree r, then the ring F2[x] / p(x) is a field of order 2r. In addition, every finite field has a primitive element. Therefore, there exists an element a of F2[x] / p(x) such that F2[x] / p(x) = {0, 1, a, a2,…, a2r –2}. Let us identify an element a0 + a1 + … ar -1xr -1 of F2[x] / p(x) with the column vector (a0, a1,…, ar -1)T and consider the binary r * (2r -1) matrix H = [ 1 a a2 … a2^r –2 ]. Let now C be the binary linear code having H as a parity check matrix. Since the columns of H are all distinct non-zero vectors of V(r,2), C = Ham (r,2). Putting n = 2r -1 we get C = {f0 f1 … fn -1 Î V(n, 2) | f0 + f1 a + … + fn -1 an –1 = 0 (2) = {f(x) Î Rn | f(a) = 0 in F2[x] / p(x)} (3) If f(x) Î C and r(x) Î Rn, then r(x)f(x) Î C because r(a)f(a) = r(a) · 0 = 0 and therefore, by one of the previous theorems, this version of Ham (r,2) is cyclic. IV054 > ‹#› Cyclic codes 23 Cyclic codes BCH codes and Reed-Solomon codes To the most important cyclic codes for applications belong BCH codes and Reed-Solomon codes. Definition A polynomial p is said to be minimal for a complex number x in Zq if p(x) = 0 and p is irreducible over Zq. IV054 Definition A cyclic code of codewords of length n over Zq, q = pr, p is a prime, is called BCH code1 of distance d if its generator g(x) is the least common multiple of the minimal polynomials for w l, w l +1,…, w l +d –2 for some l, where w is the primitive n-th root of unity. w If n = qm - 1 for some m, then the BCH code is called primitive. 1BHC stands for Bose and Ray-Chaudhuri and Hocquenghem who discovered these codes. Definition A Reed-Solomon code is a primitive BCH code with n = q - 1. Properties: • Reed-Solomon codes are self-dual. ‹#› Cyclic codes 24 Cyclic codes CONVOLUTION CODES Very often it is important to encode an infinite stream or several streams of data – say of bits. Convolution codes, with simple encoding and decoding, are quite a simple generalization of linear codes and have encodings as cyclic codes. An (n,k) convolution code (CC) is defined by an k x n generator matrix, entries of which are polynomials over F2. For example, is the generator matrix for a (2,1) convolution code CC1 and is the generator matrix for a (3,2) convolution code CC2 IV054 > ‹#› Cyclic codes 25 Cyclic codes ENCODING of FINITE POLYNOMIALS An (n,k) convolution code with a k x n generator matrix G can be used to encode a k-tuple of plain-polynomials (polynomial input information) I=(I0(x), I1(x),…,Ik-1(x)) to get an n-tuple of crypto-polynomials C=(C0(x), C1(x),…,Cn-1(x)) As follows C= I . G IV054 > ‹#› Cyclic codes 26 Cyclic codes EXAMPLES EXAMPLE 1 (x3 + x + 1).G1 = (x3 + x + 1) . (x2 + 1, x2 + x + 1] = (x5 + x2 + x + 1, x5 + x4 + 1) EXAMPLE 2 > ‹#› Cyclic codes 27 Cyclic codes ENCODING of INFINITE INPUT STREAMS The way infinite streams are encoded using convolution codes will be Illustrated on the code CC1. An input stream I = (I0, I1, I2,…) is mapped into the output stream C= (C00, C10, C01, C11…) defined by C0(x) = C00 + C01x + … = (x2 + 1) I(x) and C1(x) = C10 + C11x + … = (x2 + x + 1) I(x). The first multiplication can be done by the first shift register from the next figure; second multiplication can be performed by the second shift register on the next slide and it holds C0i = Ii + Ii+2, C1i = Ii + Ii-1 + Ii-2. That is the output streams C0 and C1 are obtained by convolving the input stream with polynomials of G1’ IV054 > ‹#› Cyclic codes 28 Cyclic codes ENCODING The first shift register Å 1 x x2 input output will multiply the input stream by x2+1 and the second shift register Å 1 x x2 input output will multiply the input stream by x2+x+1. IV054 > ‹#› Cyclic codes 29 Cyclic codes ENCODING and DECODING Å 1 x x2 I C00,C01,C02 Å C10,C11,C12 Output streams The following shift-register will therefore be an encoder for the code CC1 For encoding of the convolution codes so called Viterbi algorithm Is used. IV054 >