Download presentation

Presentation is loading. Please wait.

Published byErika Pipkin Modified over 2 years ago

1
296.3Page1 296.3:Algorithms in the Real World Error Correcting Codes II – Cyclic Codes – Reed-Solomon Codes

2
296.3Page2 Viewing Messages as Polynomials A (n, k, n-k+1) code: Consider the polynomial of degree k-1 p(x) = a k-1 x k-1 + + a 1 x + a 0 Message: (a k-1, …, a 1, a 0 ) Codeword: (p(y 0 ),p(y 1 ), …, p(y n-1 )) for distinct y 0,…,y n-1 To keep the p(y i ) fixed size, we use y i, a i GF(p r ) To make the y i distinct, n < p r Unisolvence Theorem: Any subset of size k of (p(y 1 ), p(y 2 ), …, p(y n )) is enough to (uniquely) reconstruct p(x) using polynomial interpolation, e.g., LaGrange’s Formula.

3
296.3Page3 Polynomial-Based Code A (n, k, 2s +1) code: k2s Can detect 2s errors Can correct s errors Generally can correct erasures and errors if + 2 2s n

4
296.3Page4 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)

5
296.3Page5 A Systematic Code Message: (m 0, m 1, …, m k-1 ) Find polynomial p(x) = a k-1 x k-1 + + a 1 x + a 0 such that p(y 0 )= m 0, p(y 2 )= m 2, …, p(y k-1 ) = m 0 Codeword: (m 0, m 1, …, m k-1, p(y k ), p(y k+1 ), …, p(y n-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), it is trivial to decode. 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., Hc T = 0?) to quickly test for errors.

6
296.3Page6 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(2 8 )). Decoding rate on 1.8GHz Pentium 4: –(255,251) = 89Mbps –(255,223) = 18Mbps Dozens of companies sell hardware cores that operate 10x faster (or more) – (204,188,17) = 320Mbps (Altera decoder)

7
296.3Page7 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

8
296.3Page8 RS and “burst” errors 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 code bitscheck bits RS (255, 253, 3) 256 204016 Hamming (2 11 -1, 2 11 -11-1, 3) 2 204711 Let’s compare to Hamming Codes (which are “optimal”).

9
296.3Page9 Galois Field GF(2 3 ) with irreducible polynomial: x 3 + x + 1 = x is a generator x0102 22 x2x2 1003 33 x + 10114 44 x 2 + x1105 55 x 2 + x + 11116 66 x 2 + 11017 77 10011 Will use this as an example.

10
296.3Page10 Discrete Fourier Transform (DFT) Evaluating polynomial at n points via matrix multiply: is a primitive n th root of unity ( n = 1) – a generator Inverse DFT: Evaluate polynomial m k-1 x k-1 + + m 1 x + m 0 at n distinct roots of unity, 1, , 2, 3, , n-1

11
296.3Page11 DFT Example = x is 7 th root of unity in GF(2 3 )/x 3 + x + 1 (i.e., multiplicative group, which excludes additive inverse) Recall = “2”, 2 = “3”, …, 7 = 1 = “1” Should be clear that c = T (m 0,m 1,…,m k-1,0,…) T is the same as evaluating p(x) = m 0 + m 1 x + … + m k-1 x k-1 at n points.

12
296.3Page12 Decoding Why is it hard? Brute Force: try k+2s choose k + s possibilities and solve for each.

13
296.3Page13 Cyclic Codes A linear code is cyclic if: (c 0, c 1, …, c n-1 ) C (c n-1, c 0, …, c n-2 ) C Both Hamming and Reed-Solomon codes are cyclic. Note: we might have to reorder the columns to make the code “cyclic”. Motivation: They are more efficient to decode than general codes.

14
296.3Page14 Linear Code Generator and Parity Check Matrices View message, codeword as vectors (m 0, m 1, …,m k-1 ) and (c 0, c 1, …,c n-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 v T = 0} Codewords are the nullspace of H These always exist for linear codes H G T = 0

15
296.3Page15 RS Generator and Parity Check Polynomials View message (m 0, m 1, …,m k-1 ) as polynomial m 0 + m 1 x + … + m k-1 x k-1, codeword (c 0, c 1, …,c n-1 ) as polynomial c 0 + c 1 x + … + c n-1 x n-1 Generator Polynomial: A degree (n-k) polynomial g(x) = g 0 + g 1 x + … + g n-k x n-k such that: C = {m g | m m 0 + m 1 x + … + m k-1 x k-1 } such that g | x n – 1 Parity Check Polynomial: A degree k polynomial h(x) = h 0 + h 1 x + … + h k x k such that: C = {v n [x] | h v = 0 (mod x n –1)} such that h | x n - 1 These always exist for linear cyclic codes h g = x n - 1

16
296.3Page16 Poly multiplication via matrix multiplication If g(x) = g 0 + g 1 x + … + g n-k x n-k We can put this generator in matrix form (k x n): Write m = m 0 + m 1 x +…+ m k-1 x k-1 as (m 0, m 1, …, m k-1 ) Then c = mG k-1 n-k+1

17
296.3Page17 g generates cyclic codes Codes are linear combinations of the rows. All but last row is clearly cyclic (left shift of next row) Right shift of last row is x k g mod (x n –1) = g n-k,0,…,g 0,g 1,…,g n-k-1 Will show x k g mod (x n –1) is a linear combination of other rows. Consider h = h 0 + h 1 x + … + h k x k (gh = x n –1) h 0 g + (h 1 x)g + … + (h k-1 x k-1 )g + (h k x k )g = x n – 1 (h k x k )g = (x n – 1) – (h 0 g + (h 1 x)g + … + (h k-1 x k-1 )g ) (h k x k )g mod (x n –1) = -(h 0 g + h 1 (xg) + … + h k-1 (x k-1 g)) x k g mod (x n –1) = -h k -1 (h 0 g + h 1 (xg) + … + h k-1 (x k-1 g))

18
296.3Page18 Viewing h as a matrix If h = h 0 + h 1 x + … + h k x k we can put this parity check poly. in matrix form ((n-k) x n) : Hc T = 0 (syndrome gives coefficients of x n-1 through x k in h c, which are the same as in h c mod x n -1) k+1 n-k-1

19
296.3Page19 Hamming Codes Revisited The Hamming (7,4,3) 2 code. g = 1 + x + x 3 h = x 4 + x 2 + x + 1 The columns are not identical to the previous example Hamming code. gh = x 7 – 1, GH T = 0

20
296.3Page20 Factors of x n -1 Intentionally left blank

21
296.3Page21 Another way to write g Let be a generator of GF(p r ). Let n = p r - 1 (the size of the multiplicative group) Then we can write a generator polynomial as g(x) = (x- )(x- 2 ) … (x - n-k ), h = (x- n-k+1 )…(x- n ) Lemma: g | x n – 1, h | x n – 1, gh | x n – 1 (a | b means a divides b) Proof: – n = 1 (because of the size of the group) n – 1 = 0 root of x n – 1 (x - ) | x n -1 –similarly for 2, 3, …, n –therefore x n - 1 is divisible by (x - )(x - 2 ) …

22
296.3Page22 Back to Reed-Solomon Consider a generator polynomial g GF(p r )[x], s.t. g | (x n – 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 x 2s (basically shift by 2s) –b = m’ (mod g), m’ = qg + b for some q –c = m’ – b = (m k-1, …, m 0, -b 2s-1, …, -b 0 ) –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: -h c = 0 ?

23
296.3Page23 Example Lets consider the (7,3,5) 8 Reed-Solomon code. We use GF(2 3 )/x 3 + x + 1 x0102 22 x2x2 1003 33 x + 10114 44 x 2 + x1105 55 x 2 + x + 11116 66 x 2 + 11017 77 10011

24
296.3Page24 Example RS (7,3,5) 8 g = (x - )(x - 2 )(x - 3 )(x - 4 ) = x 4 + 3 x 3 + x 2 + x + 3 h = (x - 5 )(x - 6 )(x - 7 ) = x 3 + 3 x 3 + 2 x + 4 gh = x 7 - 1 Consider the message: 110 000 110 m = ( 4, 0, 4 ) = 4 x 2 + 4 m’ = x 4 m = 4 x 6 + 4 x 4 = ( 4 x 2 + x + 3 )g + ( 3 x 3 + 6 x + 6 ) c = ( 4, 0, 4, 3, 0, 6, 6 ) = 110 000 110 011 000 101 101 010 22 100 33 011 44 110 55 111 66 101 77 001 ch = 0 (mod x 7 –1) n = 7, k = 3, n-k = 2s = 4, d = 2s+1 = 5

25
296.3Page25 A useful theorem Theorem: For any , if g( ) = 0 then 2s m( ) = b( ) Proof: x 2s m(x) = m’(x) = g(x)q(x) + b(x) 2s m( ) = g( )q( ) + b( ) = b( ) Corollary: 2s m( ) = b( ) for { , 2, 3,…, 2s=n-k } Proof: { , 2, …, 2s } are the roots of g by definition.

26
296.3Page26 Fixing errors Theorem: Any k symbols from c can reconstruct c and hence m Proof: We can write 2s equations involving m (c n-1, …, c 2s ) and b (c 2s-1, …, c 0 ). These are 2s m( ) = b( ) 4s m( 2 ) = b( 2 ) … 2s(2s) m( 2s ) = b( 2s ) We have at most 2s unknowns, so we can solve for them. (I’m skipping showing that the equations are linearly independent).

27
296.3Page27 Efficient Decoding I don’t plan to go into the Reed-Solomon decoding algorithm, other than to mention the steps. Syndrome Calculator Error Polynomial Berlekamp Massy Error Locations Chien Search Error Magnitudes Forney Algorithm Error Corrector cm This is the hard part. CD players use this algorithm. (Can also use Euclid’s algorithm.)

Similar presentations

Presentation is loading. Please wait....

OK

Error detection and correction

Error detection and correction

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Best ppt on online education Ppt on online banking system Ppt on operating system structure Ppt on bresenham's line drawing algorithm Ppt on web designing Ppt on ozone layer depletion and its effects Ppt on earth hour video Ppt on different solid figures chart Ppt on history of atomic structure Download ppt on teamviewer 9