Algorithms in the Real World Error Correcting Codes II – Cyclic Codes – Reed-Solomon Codes Page 1 Viewing Messages as Polynomials A (n, k, n-k+1) code: Consider the polynomial of degree k-1 p(x) = ak-1 xk-1 + L + a1 x + a0 Message: (ak-1, …, a1, a0) Codeword: (p(y0),p(y1), …, p(yn-1)) for distinct y0,…,yn-1 To keep the p(yi) fixed size, we use yi, ai GF(pr) To make the yi distinct, n < pr Unisolvence Theorem: Any subset of size k of (p(y0), p(y2), …, p(yn-1)) is enough to (uniquely) reconstruct p(x) using polynomial interpolation, e.g., LaGrange’s Formula. Page 2 Polynomial-Based Code A (n, k, 2s +1) code: k 2s n Can detect 2s errors Can correct s errors Can correct 2s erasures Generally can correct a erasures and b errors if a + 2b 2s 2s = n-k, so this is an (n,k,n-k+1) code Page 3 Correcting Errors Correcting s errors: 1. Find k + s symbols that agree on a polynomial p(x). These must exist since originally k + 2s symbols agreed and only s are in error 2. There are no k + s symbols that agree on the wrong polynomial p’(x) - Any subset of k symbols will define p’(x) - Since at most s out of the k+s symbols are in error, p’(x) = p(x) Page 4 A Systematic Code Message: (m0, m1, …, mk-1) Find polynomial p(x) = ak-1 xk-1 + L + a1 x + a0 such that p(y0)= m0, p(y2)= m2, …, p(yk-1) = m0 Codeword: (m0, m1, …, mk-1, p(yk), p(yk+1), …, p(yn-1)) This has the advantage that if we know there are no errors (e.g., all points lie on the same degree-(k-1) polynomial), decoding is trivial. The version of RS used in practice uses something slightly different. This will allow us to use the “Parity Check” ideas from linear codes (i.e., HcT = 0?) to quickly test for errors. Page 5 Reed-Solomon Codes in the Real World (204,188,17)256 : ITU J.83(A)2 (128,122,7)256 : ITU J.83(B) (255,223,33)256 : Common in Practice – Note that they are all byte based (i.e., symbols are from GF(28)). Decoding rate on 1.8GHz Pentium 4: – (255,251,5) = 89Mbps – (255,223,33) = 18Mbps Dozens of companies sell hardware cores that operate 10x faster (or more) – (204,188,17) = 320Mbps (Altera decoder) 296.3 Page 6 Applications of Reed-Solomon Codes • • • • • Storage: CDs, DVDs, “hard drives”, Wireless: Cell phones, wireless links Sateline and Space: TV, Mars rover, … Digital Television: DVD, MPEG2 layover High Speed Modems: ADSL, DSL, .. Good at handling burst errors. Other codes are better for random errors. – e.g., Gallager codes, Turbo codes Page 7 RS and “burst” errors Let’s compare to Hamming Codes (which are “optimal”). code bits check bits RS (255, 253, 3)256 2040 16 Hamming (211-1, 211-11-1, 3)2 2047 11 They can both correct 1 error, but not 2 random errors. – The Hamming code does this with fewer check bits However, RS can fix 8 contiguous bit errors in one byte – Much better than lower bound for 8 arbitrary errors n n log1 L 8 log( n 7) 88 check bits 8 1 Page 8 Discrete Fourier Transform (DFT) Evaluating polynomial at n points via matrix multiply: a is a primitive nth root of unity (an = 1) – a generator 1 1 1 a2 1 a 4 T 1 a 2 a n 1 2 ( n 1) 1 a a L L L L n 1 a a 2( n 1) a ( n 1)(n 1) 1 c0 m0 c m k 1 T k 1 ck 0 0 c n 1 Evaluate polynomial mk-1xk-1 + + m1x + m0 at n distinct roots of unity, 1, a, a2, a3, , an-1 DFT: c Tm 1 Inverse DFT: m T c FFT is fast way to compute DFT Page 9 Galois Field GF(23) with irreducible polynomial: x3 + x + 1 a = x is a generator 0 0 000 y0 a x 010 y1 a2 x2 100 y2 a3 x+1 011 y3 a4 x2 + x 110 y4 a5 x2 + x +1 111 y5 a6 x2 + 1 101 y6 a7 1 001 y7 Will use this as an example. Page 10 DFT Example a = x is 7th root of unity in GF(23)/x3 + x + 1 (i.e., multiplicative group, which excludes additive inverse) Recall a = y1, a2 = y2, … 1 1 1 T 1 1 1 1 1 1 1 1 1 a a2 a3 a4 a5 a2 a4 a6 a3 a6 a4 a5 a6 1 1 6 a 1 1 1 1 1 1 1 y1 y2 y3 y4 y6 y7 1 y12 y 22 y 32 1 y13 y 32 1 y14 1 y15 1 6 y1 y 67 Should be clear that c = T (m0,m1,…,mk-1,0,…)T is the same as evaluating p(x) = m0 + m1x + … + mk-1xk-1 at n points y0, y1, …, yn-1. Page 11 Decoding Why is it hard? k 2s possibilities and solve for Brute Force: try ks each. Page 12 Cyclic Codes A linear code is cyclic if: (c0, c1, …, cn-1) C (cn-1, c0, …, cn-2) C Both Hamming and Reed-Solomon codes are cyclic. Note: we might have to reorder the columns to make the code “cyclic”. Motivation: Cyclic codes are easier to decode than general codes. Page 13 Linear Code Generator and Parity Check Matrices View message, codeword as vectors (m0, m1, …,mk-1) and (c0, c1, …,cn-1) Generator Matrix: A k x n matrix G such that: C = {m G | m k} Made from stacking the basis vectors Parity Check Matrix: A (n – k) x n matrix H such that: C = {v n | H vT = 0} Codewords are the nullspace of H These always exist for linear codes H GT = 0 Page 14 RS Generator and Parity Check Polynomials View message (m0, m1, …,mk-1) as polynomial m0 + m1x + … + mk-1xk-1, codeword (c0, c1, …,cn-1) as polynomial c0 + c1x + … + cn-1xn-1 Generator Polynomial: A degree (n-k) polynomial g(x) = g0 + g1x + … + gn-kxn-k such that g | xn – 1 C = {m g | m m0 + m1x + … + mk-1xk-1} Parity Check Polynomial: A degree k polynomial h(x) = h0 + h1x + … + hkxk such that h | xn - 1 C = {v n [x] | h v = 0 (mod xn –1)} These always exist for linear cyclic codes h g = xn - 1 Page 15 Poly multiplication via matrix multiplication If g(x) = g0 + g1x + … + gn-kxn-k We can put this generator in matrix form (k x n): k-1 n-k+1 g0 0 G 0 g1 L g k 1 L g0 L L L 0 L g0 g nk g n k 1 0 g nk L L g n 2 k 1 L 0 0 L g n k 0 k-1 Write m = m0 + m1x +…+ mk-1xk-1 as (m0, m1, …, mk-1) Then c = mG Page 16 g generates cyclic codes 0 0 0 g g 0 g1 L g k 1 L g n k g nk 0 xg 0 g 0 L L L g nk G L k 1 0 0 L g x g L g L L g 0 n 2 k 1 nk Codes consist of all linear combinations of the rows: c = (c0,c1,…,cn-1) = m0g + m1xg + m2x2g+ … + mk-1xk-1g Claim: c’ = (cn-1,c0,c1,…,cn-2) is also a codeword. Right shift of every row but last is next row. Right shift of last row is xkg mod (xn –1) = gn-k,0,…,g0,g1,…,gn-k-1 Will show xkg mod (xn –1) is a linear combination of other rows. Consider h = h0 + h1x + … + hkxk (gh = xn –1) h0g + (h1x)g + … + (hk-1xk-1)g + (hkxk)g = xn – 1 (hkxk)g = (xn – 1) – (h0g + (h1x)g + … + (hk-1xk-1)g ) (hkxk)g mod (xn –1) = -(h0g + h1(xg) + … + hk-1(xk-1g)) xkg mod (xn –1) = -hk-1(h0g + h1(xg) + … + hk-1(xk-1g)) Therefore right cyclic shift of every row is a linear Page 17 combination of other rows. Viewing h as a matrix If h = h0 + h1x + … + hkxk we can put this parity check poly. in matrix form ((n-k) x n) : k+1 n-k-1 0 0 H h k h0 0 0 L 0 L 0 L hk hk L h1 hk 1 L h0 L h1 h0 n-k-1 HcT = 0 (syndrome gives coefficients of xn-1 through xk in h c, which are the same as in h c mod xn-1) Page 18 Hamming Codes Revisited The Hamming (7,4,3)2 code. g = 1 + x + x3 1 0 G 0 0 1 0 1 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 1 0 1 h = x4 + x2 + x + 1 0 0 1 0 1 1 1 H 0 1 0 1 1 1 0 1 0 1 1 1 0 0 gh = x7 – 1, GHT = 0 The columns are not identical to the previous example Hamming code. (And note that coefficients are from GF(2), but n > 2.) Page 19 Factors of xn -1 Intentionally left blank Page 20 Another way to write g Let a be a generator of GF(pr). Let n = pr - 1 (the size of the multiplicative group) Then we can write a generator polynomial as g(x) = (x-a)(x-a2) … (x - an-k), h = (x- an-k+1)…(x-an) Lemma: g | xn – 1, h | xn – 1, gh = xn – 1 (a | b means a divides b) Proof: – an = 1 (because of the size of the group) an – 1 = 0 a root of xn – 1 (x - a) | xn -1 – similarly for a2, a3, …, an – therefore xn - 1 is divisible by (x - a)(x - a2) … Page 21 Back to Reed-Solomon Consider a generator polynomial g GF(pr)[x], s.t. g | (xn – 1) Recall that n – k = 2s (the degree of g is n-k, n-k+1 coefficients) Encode (trick to make code systematic): – m’ = m x2s (basically shift by 2s) – b = m’ (mod g), m’ = qg + b for some q – c = m’ – b = (mk-1, …, m0, -b2s-1, …, -b0) – Note that c is a cyclic code based on g - c = m’ – b = qg (i.e., given m we found another message q such that qg is “systematic” for m) Parity check: - hc=0? Page 22 Example Lets consider the (7,3,5)8 Reed-Solomon code. We use GF(23)/x3 + x + 1 a x 010 a2 x2 100 a3 x+1 011 a4 x2 + x 110 a5 x2 + x + 1 111 a6 x2 + 1 101 a7 1 001 Page 23 Example RS (7,3,5)8 n = 7, k = 3, n-k = 2s = 4, d = 2s+1 = 5 g = (x - a)(x - a2)(x - a3)(x - a4) a 010 4 3 3 2 3 = x + a x + x + ax + a a2 100 h = (x - a5)(x - a6)(x - a7) a3 011 = x3 + a3x2 + a2x + a4 a4 110 gh = x7 - 1 a5 111 Consider the message: 110 000 110 a6 101 m = (a4, 0, a4) = a4x2 + a4 a7 001 m’ = x4m = a4x6 + a4x4 = (a4 x2 + x + a3)g + (a3x3 + a6x + a6) c = (a4, 0, a4, a3, 0, a6, a6) ch = 0 (mod x7 –1) = 110 000 110 011 000 101 101 Page 24 A useful theorem Theorem: For any b, if g(b) = 0 then b2sm(b) = b(b) Proof: x2sm(x) = m’(x) = g(x)q(x) + b(x) b2sm(b) = g(b)q(b) + b(b) = b(b) Corollary: b2sm(b) = b(b) for b {a, a2, a3,…, a2s=n-k} Proof: {a, a2, …, a2s} are the roots of g by definition. Page 25 Fixing erasures Theorem: Any k symbols from c can reconstruct c and hence m Proof: We can write 2s equations involving m (cn-1, …, c2s) and b (c2s-1, …, c0). These are a2s m(a) = b(a) a4s m(a2) = b(a2) … a2s(2s) m(a2s) = b(a2s) We have at most 2s unknowns (erasures), so we can solve for them. (I’m skipping showing that the equations are linearly independent). Page 26 Efficient Decoding I don’t plan to go into the Reed-Solomon decoding algorithm, other than to mention the steps. c Syndrome Calculator Error Polynomial Error Locations Error Magnitudes Berlekamp Massy Chien Search Forney Algorithm Error Corrector This is the hard part. CD players use this algorithm. (Can also use Euclid’s algorithm.) Page 27 m