Section 3 Error Correcting Codes (ECC): Fundamentals

Transcription

1 Section 3 Error Correcting Codes (ECC): Fundamentals Communication systems and channel models Definition and examples of ECCs Distance For the contents relevant to distance, Lin & Xing s book, Chapter 2, should be helpful. W. Dai (IC) EE4.07 Coding Theory ECC 2017 page 3-1

2 Communication Systems W. Dai (IC) EE4.07 Coding Theory ECC Systems & Models 2017 page 3-2

3 Abstract Channel Model: Binary Symmetric Channel (BSC) Binary Symmetric Channel: a memoryless channel such that Pr (0 received 1 sent) = Pr (0 received 1 sent) = p, Pr (1 received 1 sent) = Pr (0 received 0 sent) = 1 p. p is called the transition (cross-over) probability. Memoryless channel: A channel that satisfies Pr (y received x sent) = n i=1 Pr (y i received x i sent). W. Dai (IC) EE4.07 Coding Theory ECC Systems & Models 2017 page 3-3

4 The Memoryless q-ary Symmetric Channel Define an alphabet set F q. Both channel input x i and channel output y i are from F q. Crossover probability p: Pr (y i x i ) = { 1 p if y i = x i p/ (q 1) if y i x i W. Dai (IC) EE4.07 Coding Theory ECC Systems & Models 2017 page 3-4

5 The Memoryless Binary Erasure Channel (BEC) Binary Erasure Channel: Internet traffic: a package got lost. Cloud storage: one copy of file messed up. W. Dai (IC) EE4.07 Coding Theory ECC Systems & Models 2017 page 3-5

6 What is a Code? Definition 3.1 (Code) A code is a set C containing (row) vectors of elements from F q. An (n, M) block code: C F n q and C = M. A codeword: a vector in C. Codeword length: n Dimension: k = log q M. Code size: M Rate: r = k/n. Example 1: F 2 = {0, 1}. C = {0000, 1100, 1111}. n = 4. M = 3. k = log 2 3 = r = Example 2: F 3 = {0, 1, 2}. C = {00000, 12121, 20202}. n = 5. M = 3. k = log 3 3 = 1. r = 0.2. W. Dai (IC) EE4.07 Coding Theory ECC Definitions & Examples 2017 page 3-6

7 Triple Repetition Code Encoding Decoding: majority voting 111, 110, 101, , 001, 010, Error probability computation: P (ŝ = 1 s = 0) = P (111 0) + P (110 0) + P (101 0) + P (011 0) = p 3 + 3p 2 (1 p) = , when p = Much better than an uncoded system. The price to pay: data rate 1/3. Coding theory: tradeoff between error correction and data rate. W. Dai (IC) EE4.07 Coding Theory ECC Definitions & Examples 2017 page 3-7

8 Performance Comparison Uncoded case (f=0.1) ¹ ½ µ ¼ ¼ Ê ¹ ½ ½ ½ µ Triple repetition code ÒÓ Ö Ø ÒÒ Ð ¹ ½¼± ¹ Ö Ó Ö ¹ From David J.C. MacKay, Information Theory, Inference, and Learning Algorithms, Cambridge University Press, W. Dai (IC) EE4.07 Coding Theory ECC Definitions & Examples 2017 page 3-8

9 The 2nd example: (7, 4) Hamming code Encoding: encode every 4 bit information into 7 bits. (Details are omitted.) Code rate: r = 4/ Much higher rate but slightly larger P e. From David J.C. MacKay, Information Theory, Inference, and Learning Algorithms, Cambridge University Press, W. Dai (IC) EE4.07 Coding Theory ECC Definitions & Examples 2017 page 3-9

10 Another example - low-density parity-check code Details are omitted here. Only simulation is presented BSC with p = 7.5%. LDPC (20 000, ) r = 0.5 From David J.C. MacKay, Information Theory, Inference, and Learning Algorithms, Cambridge University Press, W. Dai (IC) EE4.07 Coding Theory ECC Definitions & Examples 2017 page 3-10

11 Distance: Definition Definition 3.2 (Distance) A distance d on a set X is a function d : X X R such that for all x, y, z X, the following conditions hold: Positive definite: Symmetry: Triangle inequality: d (x, y) 0 where = holds iff x = y. d (x, y) = d (y, x). d (x, z) d (x, y) + d (y, z). In this course, d is also translation invariant, that is, d (x, y) = d (x + z, y + z). W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-11

12 Examples of Commonly Used Distances Let x, y R n be two vectors of length n, for example, x = [9, 1, 0], y = [6, 1, 4] R 3 l 2 -norm distance: Euclidean distance d 2 n d 2 (x, y) = i=1 (x i y i ) 2 = = 5. l 1 -norm distance: d 1 d 1 (x, y) = n i=1 x i y i = = 7. Hamming distance: d H d H (x, y) = n i=1 δ x i y i = = 2, where δ xi y i = 1 if x i y i and δ xi y i = 0 if x i = y i. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-12

13 Hamming Distance Definition 3.3 (Hamming Distance) For x, y F n, the Hamming distance is given by d H (x, y) = n i=1 δ x i y i = {i : x i y i }. Fact 3.4 Hamming distance is a well defined distance. To prove this fact, the only non-trivial part is the triangle inequality. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-13

14 Proof of the Triangle Inequality for Hamming Distance 1. scalar case: d H (x i, z i ) d H (x i, y i ) + d H (y i, z i ): If x i = z i, then the equality holds obviously. If x i z i, LHS= 1. We have three cases: y i = x i y i z i y i = z i y i x i RHS 1. y i x i and y i z i 2. vector case: d H (x, z) = n i=1 d H (x i, z i ) n i=1 (d H (x i, y i ) + d H (y i, z i )) = d H (x, y) + d H (y, z). W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-14

15 Hamming Distance: Properties Fact 3.5 Hamming distance is translation invariant: d H (x 1, x 2 ) = d H (x 1 + y, x 2 + y). Definition 3.6 (Weight) A weight of a vector x F n q is defined as its Hamming distance from the zero vector: wt (x) = d H (x, 0). Example: x = [9, 1, 4], y = [0, 1, 4] d H (x, y) = 1. x = [1, 2, 1, 2, 1], y = [2, 0, 2, 0, 2] d H (x, y) = 5. x = [0, 1, 0, 1] wt (x) = 2. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-15

16 Decoding Techniques Suppose that a codeword c C F n q is transmitted and a word y is received. The decoding function is defined as the mapping D : F n q C y ĉ C. Popular decoding strategies include Maximum likelihood decoding: ĉ ML = D ML (y) = arg max c C Pr (y received c sent). Minimum distance decoding: ĉ MD = D MD (y) = arg min c C d H (y, c). They are equivalent for many channels. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-16

17 Equivalence Between ML and MD decoding Theorem 3.7 Consider a memoryless binary symmetric channel (BSC) with cross-over probability p < 1/2. Then ĉ ML = ĉ MD. Proof: Pr (y received c sent) = n Pr (y i received c i sent) i=1 = p dh(y,c) (1 p) n d H(y,c) ( ) = (1 p) n p dh (y,c). 1 p That p < 1/2 implies that p/ (1 p) < 1. Hence, Pr (y received c sent) is a monotonically decreasing function of d H (y, c). The maximum Pr (y c) is achieved when d H (y, c) is minimized. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-17

18 Distance of a Code Definition 3.8 The distance of a code C is defined as d H (C) = min d H (x 1, x 2 ). x 1,x 2 C, x 1 x 2 Notation: An (n, M, d)-code: a code of codeword length n, size M, and distance d. Example: Consider the binary code C = {00000, 00111, 11111}. It is a binary (5, 3, 2)-code. Example: Consider the ternary code C = {000000, , }. It is a ternary (6, 3, 3)-code. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-18

19 Error Detection Error detector: if the received word y C, let ĉ = y and claim no error; if y / C, claim that errors happened. Theorem 3.9 Let C F n q be an (n, M, d) code. The above error detector detects every pattern of up to d 1 many errors. Proof: 1. Every pattern of d 1 many errors are detectable. Since at most d 1 many errors happened, 0 < d H (c, y) < d := d (C) and y / C. The receiver will claim that errors happened. 2. Exists a pattern of d many errors that is not detectable. By the definition of the code distance, there exist c 1, c 2 C s.t. d H (c 1, c 2 ) = d. Suppose that c 1 is the transmitted codeword and the channel errors happen to be e = c 2 c 1 (d errors happened). Then y = c 2 is received. As c 2 C, the detector claims that no error happened and set ĉ = c 2. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-19

20 Error Correction Theorem 3.10 Let C F n q be an (n, M, d) code. The minimum distance decoder can correct every pattern of up to t := (d 1) /2 many errors. Side mark: and d 1 t = = 2 { d if d is odd, d 2 1 if d is even. 2t + 1 d (C) 2t + 2 Examples: The previous ternary (6, 3, 3) code is exactly 1-error-detecting. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-20

21 Error Correction: Proof Proof: Let D be the minimum distance decoder. Let c and y be the transmitted codeword and received word respectively. Let ĉ = D MD (y). 1. If d H (y, c) t = (d 1) /2, then ĉ = c. Suppose that ĉ c. By the way the decoder D MD is defined, d H (y, ĉ) d H (y, c) t. On the other hand, by the definition of the code distance, d d H (c, ĉ) d H (c, y) + d H (y, ĉ) 2t d 1, which is a contradiction. 2. a pair (c, y) C F n q s.t. d H (y, c) = t + 1 and it may happen that D MD (y) c. By the definition of the code distance, c, c C s.t. d H (c, c ) = d. W.l.o.g., assume the first d positions of c, c are different. Define y such that y i = c i, i = 1, 2,, t + 1 and y i = c i, i = t + 2,, n. It is clear that d H (y, c) = t + 1 and d H (y, c ) = d (t + 1) t + 1 = d H (y, c). Hence, it may happen that ĉ = D MD (y) c. W. Dai (IC) EE4.07 Coding Theory ECC Code Distance 2017 page 3-21

22 Section 4 Linear Codes Definition. Decoding. Generator matrices. Parity-check matrices. Remark: Why linear codes? Low complexity in encoding, decoding, and distance computation. For the contents relevant to distance, Lin & Xing s book, Chapter 2, should be helpful. W. Dai (IC) EE4.07 Coding Theory Linear Codes 2017 page 4-1

23 Linear Codes: Definition Block codes: all codewords are of the same length C F n q. Definition 4.1 (Linear Codes) A linear code is a code for which any linear combination of codewords is also a codeword. That is, let u, v C F n q. Then λu + µv C, λ, µ F q. Example of linear codes: C = {0000, 0011, 1100, 1111} F 4 2. C = { v F 4 2 : wt (v) is even.}. Example of nonlinear codes: C = {0000, 1100, 1111}. C = { v F 4 3 : wt (v) is even.}. W. Dai (IC) EE4.07 Coding Theory Linear Codes Definition 2017 page 4-2

24 Basis Definition 4.2 (Basis) Let B = {v 1,, v k } F n. It is a basis of a set C F n if it satisfies the following conditions: Linear independence property: For all λ 1,, λ k F, if λ i v i = 0, then necessarily λ 1 = = λ k = 0. The spanning property: For every c C, there exist λ 1,, λ k F s.t. c = i λ iv i. dim (C) = k: the # of vectors in a basis. The basis B is not unique in general, but the dimension is. Example: Let C = {0000, 0011, 1100, 1111}. B 1 = {0011, 1100} is a basis for C. B 2 = {0011, 1111} is also a basis for C. dim (C) = 2. W. Dai (IC) EE4.07 Coding Theory Linear Codes Generator Matrix 2017 page 4-3

25 Construct a Basis Definition 4.3 (Linear Span) For any subset V F n, define V as the linear span of V: { } V = λi v i : λ i F, v i V. Construct a basis for a linear code C F n : 1. From C, take a nonzero vector, say v Take a nonzero vector, say v 2, from C {v 1 }. 3. Take a nonzero vector, say v 3, from C {v 1, v 2 }. 4. Continue this process, until C {v 1, v 2,, v k } = φ. 5. Set B = {v 1, v 2,, v k }. W. Dai (IC) EE4.07 Coding Theory Linear Codes Generator Matrix 2017 page 4-4

26 The Size of a Linear Code Theorem 4.4 Let C F n q be a linear code and dim (C) = k, then C = q k. Proof: 1. dim (C) = k a basis B = {v 1,, v k } for C. 2. C q k : { k } Definition of the basis suggests C = B = i=1 λ iv i : λ i F q. There are q k many possible linear combinations. Hence, C q k (repetition may exist). 3. C = q k : It suffices to show that there is no repetition. Let λ (1) λ (2). Let x (1) = k v 1 and x (2) = k Then x (1) x (2) = k i=1 ( i=1 λ(1) i λ (1) i λ (2) i ) i=1 λ(2) i v 1. v i 0 by linear independence of v i s and the fact that λ (1) λ { (2). k } There is no repetition in the set i=1 λ iv i : λ i F q. W. Dai (IC) EE4.07 Coding Theory Linear Codes Generator Matrix 2017 page 4-5

27 Generator Matrix Definition 4.5 (Generator Matrix) A generator matrix G for a linear code C F n is a matrix whose rows form a basis for C. For a given (n, k) linear code C F n, it can be defined using its generator matrix G F k n. The encoding function that maps information symbols to a codeword is given by E : F k C F n s c = sg C. Remark: Encoding of a linear code is efficient: vector-matrix product. Encoding of a nonlinear code is via a look-up table and hence computationally less efficient. W. Dai (IC) EE4.07 Coding Theory Linear Codes Generator Matrix 2017 page 4-6

28 Examples Example 1: the (3, 1) repetition code: G = [ ]. Example 2: the (7, 4) Hamming code G = Example [ 3: the generator ] matrix[ is not unique. ] G = and G = generate the same code C = {0000, 1010, 0101, 1111} F 4 2. W. Dai (IC) EE4.07 Coding Theory Linear Codes Generator Matrix 2017 page 4-7

29 Dual Code Definition 4.6 (Dual Code) Let C F n q be a non-empty code. Its dual code C is defined as { C = v F n q : vc T = } v i c i = 0 for all c C. i Lemma 4.7 For any non-empty code C F n q (linear or nonlinear), its dual code C is always linear. Proof: Take arbitrary v 1, v 2 C. Then for all λ 1, λ 2 F q and for all c C, (λ 1 v 1 + λ 2 v 2 ) c T = λ 1 v 1 c T + λ 2 v 2 c T = 0, which implies λ 1 v 1 + λ 2 v 2 C. W. Dai (IC) EE4.07 Coding Theory Linear Codes Parity Check Matrix 2017 page 4-8

30 Parity Check Matrix Definition 4.8 (Parity-Check Matrix) A parity-check matrix H for a linear code C F n q is a generator matrix for the dual code C. For a code C [n, k], it holds that G F k n q H G T = 0. and H F (n k) n q. Define a linear code via its parity-check matrix: C = { c F n q : ch T = 0 }. W. Dai (IC) EE4.07 Coding Theory Linear Codes Parity Check Matrix 2017 page 4-9

31 Examples The (3, 1) repetition code: G = [ ] and H = The (7, 4) Hamming code: G = and H = [ A self-dual code is a code s.t. C = C, Example: C = {0000, 1010, 0101, 1111}, where [ ] G = = H Self-dual codes do not exist for vector space R n or C n. ] W. Dai (IC) EE4.07 Coding Theory Linear Codes Parity Check Matrix 2017 page 4-10.

32 Relation Between G and H Consider C [n, k] F n q. Write G and H in systematic forms: Let G = [I k A] Fq k n, where A Fq k (n k). Let H = [B I n k ] F (n k) n q, where B F q (n k) k. Lemma 4.9 It holds that B = A T. Proof: [ HG T Ik = [B I n k ] A T ] = B I k + I n k A T = A T + A T = 0 F (n k) k q. Systematic form: Why? Easy to compute H from G, and vice versa. How? Gaussian-Jordan elimination. W. Dai (IC) EE4.07 Coding Theory Linear Codes Parity Check Matrix 2017 page 4-11

33 Hamming Weight Hamming Weight of x: wt (x) = {i : x i 0} = d (x, 0). Theorem 4.10 For a linear code C, d H (C) = min wt (x). x C\{0} Proof: d H (c 1, c 2 ) = wt (c 1 c 2 ) = wt (c ) for some c C (by the definition of linear codes). Notation: C [n, k, d]: n: codeword length; k: dimension; d: distance. W. Dai (IC) EE4.07 Coding Theory Linear Codes Distance 2017 page 4-12

34 Distance and Parity Check Matrix Theorem 4.11 Let C be a linear code defined by the parity-check matrix H. Then that d (C) = d is equivalent to that 1. Every d 1 columns of H are linearly independent. 2. There exist d linearly dependent columns. W. Dai (IC) EE4.07 Coding Theory Linear Codes Distance 2017 page 4-13

35 Two Confusing Concepts Given a matrix H, spark: minimum number of linearly dependent columns column rank: maximum number of linearly independent columns. Theorem 4.11 suggests that spark (H) = d (C). Example 4.12 [ ] H = : spark (H) = 3 and column rank (H) = H =..... is an n (n + 1) matrix: spark (H) = 2 and column rank (H) = n + 1. W. Dai (IC) EE4.07 Coding Theory Linear Codes Distance 2017 page 4-14

36 Application of Theorem 4.11: Binary Hamming Codes Definition 4.13 (Binary Hamming Codes ) The parity-check matrix of the binary Hamming code H [2 r 1, 2 r 1 r, 3] consists of all the nonzero binary vectors (columns) of length r. (Also denoted by H r.) Example 4.14 The H 2 [3, 1, 3] is given by H = [ ], and the H 3 [7, 4, 3] is given by H = W. Dai (IC) EE4.07 Coding Theory Linear Codes Distance 2017 page 4-15

37 The Distance of Binary Hamming Codes Corollary 4.15 The distance of a binary Hamming code is 3, i.e., d (H r ) = 3. Proof: We apply Theorem That there is no zero column implies that the minimum number of linearly dependent columns is at least 2, i.e., d (C) = spark (H) 2. In the binary field, that every two columns are distinct implies that every two columns are linearly independent. Hence, d (C) = spark (H) 3. It is easy to see that there exist three columns that are linearly dependent (for example the first three columns). Therefore d (C) = 3. Corollary 4.16 Binary Hamming codes correct up to one error. W. Dai (IC) EE4.07 Coding Theory Linear Codes Distance 2017 page 4-16

38 Theorem 4.11: Proof Proof: Let h i be the i th column of H. c C, let i 1,, i K be the locations where c i 0. By the definition of parity-check matrix, 0 = n K c i h i = c ik h ik. i=1 k=1 d (C) = d Claim 2: d (C) = d implies that c C s.t. wt (c) = d. That is, d k=1 c i k h ik = 0, or, h i1,, h ik are linearly dependent. d (C) = d Claim 1: Suppose not. h i1, h id 1 are linear dependent, i.e., d 1 k=1 x i k h ik = 0. Let x = [ 0 x i1 x ik x id 1 0 ]. Then wt (x) d 1 and x C. Hence d (C) d 1. A contradiction with d (C) = d. Claims 1&2 d (C) = d: That every d 1 columns are linearly independent implies no nonzero codeword of weight d 1. That there exists d columns that are linearly dependent means the existence of a codeword of weight d. Hence d (C) = min x C\{0} wt (x) = d. W. Dai (IC) EE4.07 Coding Theory Linear Codes Distance 2017 page 4-17

39 Syndrome Vector Let H F (n k) n q be a parity-check matrix of a linear code C [n, k] F n q. Suppose that the received word is given by y F n q. Define the syndrome vector s := yh T. It depends only on the error vector not the transmitted codeword. In particular, let y = x + e where x C is the transmitted codeword and e F n q is the error vector introduced by the channel. It holds that s = yh T = (x + e) H T = eh T. W. Dai (IC) EE4.07 Coding Theory Linear Codes Syndrome Decoding 2017 page 4-18

40 Syndrome Decoding MD decoding: Find ĉ = arg min c C Syndrome decoding: d H (c, y). 1. For the received word y, compute the syndrome vector: s := yh T. 2. Find the error vector e with the minimum weight: (MD decoding) 3. Decode y as ĉ = y ê. ê = arg min e wt (e) s.t. s = eh T. (1) Comments: In general, it is computationally challenging to solve (1). However, all practical codes have particular structures in the parity-check matrix so that the decoding problem can be solved efficiently. W. Dai (IC) EE4.07 Coding Theory Linear Codes Syndrome Decoding 2017 page 4-19

41 Decoding of Binary Hamming Codes Take H 3 (Definition 4.13) as an example. Assume that y = [ ]. Find the MD decoded codeword ĉ C. Since d (H 3 ) = 3, it corrects up to 1 error. For any e s.t. wt (e) = 1, let e i 0 for some i [n]. Then s = eh T = e i h T i = h T i. In the example, s = [001], e = [ ] and ĉ = [ ]. W. Dai (IC) EE4.07 Coding Theory Linear Codes Syndrome Decoding 2017 page 4-20

42 Section 5 Coding Bounds Sphere packing (Hamming) bound Sphere covering (Gilbert-Varshamov) bound Singleton bound and MDS codes The lectures will only cover sphere packing, sphere covering, singleton bounds and relevant contents. Reference: Lin & Xing s book, Chapter 5. W. Dai (IC) EE4.07 Coding Theory Coding Bounds 2017 page 5-1

43 Coding Bounds: Motivation Consider the Hamming code H r : r = 2: [3, 1, 3] r = 3: [7, 4, 3] r = 4: [15, 11, 3] Questions: Can we do better? What is the best that we can do? It is possible to construct linear codes with parameters [7, 4, 4] over F 8. [15, 11, 5] over F 16. W. Dai (IC) EE4.07 Coding Theory Coding Bounds 2017 page 5-2

44 Hamming Bound Theorem 5.1 (Hamming bound, sphere-packing bound) For a code of length n and distance d, the number of codewords is upper bounded by ( t ( ) ) n M q n / (q 1) i, i i=0 where t := d 1 2. W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Packing 2017 page 5-3

45 Examples Definition 5.2 (Perfect Codes) A perfect code is a code that attains the Hamming bound. Binary Hamming code H r [2 r 1, 2 r 1 r, 3] is a perfect code. d = 3 t = d 1 2 = 1. Ball Volume: t ( n i=0 i) (q 1) i = 1 + (2 r 1) = 2 r. Hamming bound: q n / t ( n i=0 i) (q 1) i = 2 2r 1 /2 r = 2 2r r 1 = 2 k. Perfect codes are rare (binary Hamming codes & Golay codes). W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Packing 2017 page 5-4

46 Hamming Bound: Proof (1) Define a ball in F n q centered at x F n q with radius t by B (x, t) = { y F n q : d (x, y) t }. Step one: the balls B (c, t), c C, are disjoint. For all c c C, it holds that B (c, t) B (c, t) = φ. For a y B (c, t), then y / B (c, t) for all c c. By triangle inequality: d d H (c, c ) d H (c, y) + d H (y, c ). Then ( d H y, c ) d 1 d d H (c, y) d t = d 2 d 1 > = t, 2 which implies y / B (c, t). W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Packing 2017 page 5-5

47 Hamming Bound: Proof (2) Step two: Consider the union of these balls. Clearly c C B (c, t) Fn q. Hence, ( ) Vol B (c, t) = Vol (B (c, t)) Vol ( F n ) q = q n, c C c C where the first equality holds because the balls do not overlap. The volume of each ball is Therefore Vol (B (c, t)) = t i=0 MVol (B (c, t)) q n M q n / ( ) n (q 1) i. i t i=0 ( ) n (q 1) i. i W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Packing 2017 page 5-6

48 Gilbert-Varshamov Bound Theorem 5.3 (Gilbert-Varshamov bound, sphere covering bound) For given code length n and distance d, there exists a code such that q n /Vol (d 1) M, where Vol (d 1) := d 1 ( n ) i=0 i (q 1) i. Comment: Different from the sphere packing bound, which holds for all codes, the sphere covering bound claims the existence of a code. That means, some badly designed codes may not satisfy this bound. W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Covering 2017 page 5-7

49 Gilbert-Varshamov Bound: Proof It s proved by construction. Let M 0 = q n /Vol (d 1) > 1. It suffices to show that exists a code with M 0 codewords. Take an arbitrary word c 1 F n q. Since M 0 > 1, or q n > Vol (d 1), it holds F n q \B (c 1, d 1) φ. Take an arbitrary word c 2 F n q \B (c 1, d 1). It is clear that d (c 1, c 2 ) d (c 2 / B (c 1, d 1)). Continue this process inductively. Suppose to ( obtain codewords c 1,, c M0 1 in this way. M0 ) 1 Since Vol i=1 B (c i, d 1) (M 0 1) Vol (d 1) < q n, it holds that F n q \ M 0 1 i=1 B (c i, d 1) φ. Take an arbitrary c M0 F n q \ M 0 1 i=1 B (c i, d 1) φ. Let C = {c 1,, c M0 }. By construction, d (c, c ) > d 1 for all c c C. Hence d (C) d. W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Covering 2017 page 5-8

50 Illustration for Sphere Packing and Covering Sphere Packing Sphere Covering W. Dai (IC) EE4.07 Coding Theory Coding Bounds Sphere Covering 2017 page 5-9

51 Singleton Bound and MDS Theorem 5.4 (Singleton Bound) The distance of any code C F n q with M codewords satisfies M q n d+1. In particular, if the code is linear and M = q k, then d n k + 1. Definition 5.5 (MDS) Codes that attain the singleton bound are maximum distance separable (MDS). Binary Hamming codes H r [2 r 1, 2 r 1 r, 3] are not MDS in general. n k + 1 = r + 1 > 3 for all r 2. W. Dai (IC) EE4.07 Coding Theory Coding Bounds MDS 2017 page 5-10

52 Singleton Bound: Proof Proof of the general case: Let C be of length n and distance d. c C, let c 1:n d+1 F n d+1 be the vector containing the first n d + 1 entries of c, and c n d+2:n F d 1 be the vector composed of the last d 1 elements of c. c c C, d d H (c, c ) = d H ( c1:n d+1, c 1:n d+1) + dh ( cn d+2:n, c n d+2:n). But d H ( cn d+2:n, c n d+2:n) d 1. Hence, d H ( c1:n d+1, c 1:n d+1) d (d 1) = 1. The truncated codewords are all distinct. Hence, M q n d+1. Proof for linear codes: Note that the parity-check matrix H F (n k) n contains n k rows. Every n k + 1 columns must be linearly dependent. By Theorem 4.11, d n k + 1. W. Dai (IC) EE4.07 Coding Theory Coding Bounds MDS 2017 page 5-11

53 Dual of MDS Codes Theorem 5.6 If a linear code C is MDS, then its dual code C is also MDS. Let the linear code C [n, k] be MDS. According to Theorem 5.6, one has Parity-check matrix Generator Matrix Parameters C H F (n k) n G F k n (n, k, n k + 1) C G F k n H F (n k) n (n, n k, k + 1) Key for the proof: Theorem If C [n, k] is MDS, then every set of n k columns of H is linear independent. W. Dai (IC) EE4.07 Coding Theory Coding Bounds MDS 2017 page 5-12

54 Dual of MDS Codes (Theorem 5.6): Proof Suppose d ( C ) < k + 1. Then there exists a nonzero codeword c C with at most k nonzero entries and at least n k zeros. Since permuting the coordinates reserves the codeword weights (i.e., the distance), w.l.o.g., assume that the last n k coordinates of c are zeros. Write the generator matrix of C (the parity-check matrix of C) as H = [A, H ], where A F (n k) k and H F (n k) (n k). By definition of the generator matrix, there exists s F n k such that c = sh. As C is MDS, by Theorem 4.11 the columns of H are linearly independent. That is, H is invertible. That the last n k coordinates of c are zeros implies that s = c k+1:n (H ) 1 = 0. But s = 0 implies c = sh = 0 which contradicts the assumption that c 0. Hence, d ( C ) k + 1. By the Singleton bound, d ( C ) = k + 1. W. Dai (IC) EE4.07 Coding Theory Coding Bounds MDS 2017 page 5-13

55 Section 6 RS & BCH Codes Reed-Solomon Codes Definition and properties. Decoding Cyclic and BCH codes The contents in this section are significant re-organization and condensation of the materials of many sources, including Lin & Xing s book, Chapters 7 and 8, and Roth s book, Chapters 5, 6 and 8. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes 2017 page 6-1

56 Reed-Solomon Codes Our Heroes: Irving S. Reed and Gustave Solomon Used in Magnetic recording (all computer hard disks use RS codes) Digital versatile disks (CDs, DVDs, etc.) Optical fiber networks (ITU-TG.795) ADSL transceivers (ITU-TG.992.1) Wireless telephony (3G systems, 4G systems) Digital satellite broadcast (ETS S, ETS ) Deep space exploration (all NASA probes) W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-2

57 RS Codes: Evaluation Mapping Definition 6.1 (Evaluation Mapping) Let F q be a finite field. Let n q 1 (typically n = q 1). Let A = {α 1,, α n } F q. For any polynomial f (x) F q [x], define the evaluation mapping eval (f (x)) that maps f to a vector c (F q ) n Example 6.2 f c = [c 1,, c n ] where c i = f (α i ). F 7 = {0, 1, 2, 3, 4, 5, 6}. Choose the primitive element α = 3. Let A = { 1, α,, α 5} = {1, 3, 2, 6, 4, 5}. f (x) = 2x + 1, c = eval (f) = [3, 0, 5, 6, 2, 4]. f (x) = 3x 2 + x + 2, c = eval (f) = [6, 4, 2, 4, 5, 5]. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-3

58 RS Codes: Definition Definition 6.3 (Reed-Solomon Codes) Given A = {α 1,, α n } F q, an [n, k] q-ary RS code C = {eval (f), 0 deg f k 1}. The set A is called a defining set of points of C. A common choice of defining set of points of C is A = { 1, α,, α q 2} where α is a primitive element in F q. In this case, n = q 1. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-4

59 RS Codes: Properties Theorem RS codes are linear codes. 2. RS codes are MDS, i.e., The distance of the RS code is d = n k + 1. Proof: 1. Let c 1 = eval (f 1 ) and c 2 = eval (f 2 ) where deg f 1 k 1 and deg f 2 k 1. Then αc 1 + βc 2 = eval (g) with g = αf 1 + βf 2. Since deg g k 1, eval (g) C. 2. A polynomial of degree k 1 can have at most k 1 zeros. Hence, c C s.t. c 0, c = eval (f) has weight at least n k + 1. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-5

60 RS Codes: Conventional Definition Theorem 6.5 Let the defining set of points is { 1, α,, α n 1} with order (α) = n (typically n = q 1). The generated RS code has generator matrix and parity-check matrix given by α α n 1 G = α k 1 α (k 1)(n 1) and H = 1 α α n 1 1 α 2 α 2(n 1) α n k α (n k)(n 1) W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-6

61 Generator Matrix: Justification For any c C, there exists a polynomial f (x) = a 0 + a 1 x + + a k 1 x k 1 s.t. c = eval (f) = [ f (1), f (α),, f ( α n 1)] C. Note that i [n], c i = f ( α i) = k 1 l=0 a ( l α i 1 ) l = [a0,, a k 1 ] G i where G i is the i-th column of the G matrix. One has C = { sg : s F k q} and G is a generator matrix of C. Remark: In the definition of the generator matrix (Def. 4.5), the rows of G are required to be linearly independent. We shall prove it later. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-7

62 Parity-Check Matrix: Justification Lemma 6.6 The G and H matrices defined in Theorem 6.5 satisfy GH T = 0. Proof: Let A := GH T F k (n k) q. i [k] and j [n k], it holds that A i,j = n α (l 1)(i 1) α (l 1)j = l=1 (a) = α(i+j 1)n 1 (b) α i+j 1 = 0, 1 n l=1 α (i+j 1)(l 1) where (a) comes from that i + j 1 < n and α i+j 1 1, and (b) holds because α n = 1. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-8

63 Row Rank of the G/H Matrix Theorem 6.7 The rows of the G/H matrix in Theorem 6.5 are linearly independent. Proof: It is sufficient to show that any k-column sub-matrix of G ((n k)-column sub-matrix of H) has full rank. Note that a k-column sub-matrix of G is of the form G α i 1 α i 2 α i k = α (k 1)i 1 α (k 1)i 2 α (k 1)i k, which is a Vandermonde matrix (defined and analysed later). A Vandermonde matrix has full rank. Hence the rows of G are linearly independent. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-9

64 Vandermonde Matrix Definition 6.8 (Vandermonde Matrix) A Vandermonde matrix V F n n is of the form α 1 α 2 α n V = α n 1 1 α n 1 2 αn n 1 Theorem 6.9 The determinant of a Vandermonde matrix V F n n is V = i<j (α j α i ). As a result, if α i α j, 1 i j n, then V 0 and V is of full rank. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-10

65 Determinant: A Recap Definition 6.10 (Determinant) A F n n, its determinant A is computed via A = n j=1 ( 1)i+j a i,j M i,j, where M i,j is the minor matrix obtained by deleting row i and column j from A. Lemma AB = A B. 2. If B results from A by adding a multiple of one row/column to another row/column, then B = A. 3. A 0 A is of full rank. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-11

66 Theorem 6.9: Proof (1) We prove Theorem 6.9 by using induction. Recall that V n = α 1 α 2 α n 1 α n α n 2 1 α n 2 2 α n 2 n 1 α n 1 1 α n 1 2 α n 1 n 1 α n 2 n α n 1 n Let ( V n ) :,2 = (Vn) :,2 (Vn) :,1,, ( V n ) :,n = (Vn) :,n (Vn) :,1. We obtain V n = α 1 α 2 α 1 α n 1 α 1 α n α α n 2 1 α n 2 2 α n 2 1 α n 2 n 1 αn 2 α n 1 1 α n 1 2 α n 1 1 α n 1 n 1 αn 1 1 α n 2 n 1 α n 1 n α n 2 1 α n 1 1 W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-12

67 Theorem 6.9: Proof (2) Let ( V n ) n,: = ( V n ) n,: α ( 1 V n )n 1,:,, ( V n ) 2,: = ( V n ) 2,: α ( 1 V ) n. We obtain 1,: V n = α 2 α 1 α n 1 α 1 α n α α n 2 2 α 1 α n 3 2 α n 2 n 1 α 1α n 3 n 1 0 α n 1 2 α 1 α n 2 2 α n 1 n 1 α 1α n 2 n 1 α n 2 n α 1 α n 3 n α n 1 n α 1 α n 2 n α 2 α 1 α n 1 α 1 α n α 1... = (α 2 α 1 ) α n 3 ( ) 2 αn 1 α 1 α n 3 n 1 0 (α 2 α 1 ) α n 2 ( ) 2 αn 1 α 1 α n 2 n 1 (α n α 1 ) α n 3 n (α n α 1 ) α n 2 n [ 1 = Vn 1 (α2,, αn) ] 1 α 2 α 1... αn 1 α1 α n α 1 Hence V n = V n = V n = Vn 1 j>1 ( αj α 1). W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Codes 2017 page 6-13

68 Decoding with Known Error Locations Let e be the error vector. Let I = {i : e i 0} be the set of error locations. e I, H I : sub-vector and sub-matrix of e and H respectively. If we knew error locations I: Solve H I e T I = st. (e T I = H I st ) Complexity of pseudo-inverse H I : O ( d 3). W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-14

69 Erasure Correction Recall the erasure channel model. Suppose that c C was transmitted. Receive r = [c 1 c i 1? c i+1 c n ] (at most d 1 symbols erased). Decoding: Set the missing symbols to zero, i.e., r I = 0. Then r = c + e, where e I c = 0. s T = Hr T = He T = H I e T I. α i 1 1 α 2(i 1 1). α s(i 1 1) α i 2 1 α 2(i 2 1). α s(i 2 1) α is 1 α 2(is 1) α s(is 1) e 1 e 2. e s = s 1 s 2. s s. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-15

70 A Specific Example Consider a [7, 4, 4] RS code over F 8 (F 2 [x] /x 3 + x + 1). Let α be a primitive element (a root of f (x) = x 3 + x + 1). F 8 : α 3 = α α α 2 α 3 α 4 α 5 α 6 Encoded message m (x) = αx 3 + αx 2 + x. c = eval (m) = [ 1 α 5 α 1 α 5 α 6 α 5 ]. r = [ 1 α 5 α 1?? α 5 ]. H = ch T =0 1 α α 2 α 3 α 4 α 5 α 6 1 α 2 α 4 α 6 α α 3 α 5 1 α 3 α 6 α 2 α 5 α α 4 α 4 α 5 α α 3 α 5 α [ c5 c 6 ] = α 1 α matrix inverse [ c5 c 6 ] [ α 5 = α 6 ]. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-16

71 Error Correction In previous example: Error (erasure) locations are known. The error values are found via matrix inverse. For error correction: Find error locations ( Exhaustive search: complexity n t) = O (n t ). In practice, methods to find error locations efficiently. Correct errors with given error locations Methods to avoid matrix inverse. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-17

72 Efficient Error Correction: Definitions Definitions 6.12 Syndrome polynomial: S (z) = n k 1 j=0 s j z j, where s = rh T = eh T. Error locator polynomial: (information about error locations) L (z) = ( i I 1 α i z ). Error evaluator polynomial: (information about errors) E (z) = L (z) i I e i α i (1 α i z) = i I e iα i j I\{i} ( 1 α j z ). Remark: The receiver can compute the syndrome vector and the syndrome polynomial easily. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-18

73 Information Encoded in L (z) and E (z) If we know L (z), we can find error locations. { L ( α k) = 0 if k I L ( α k) 0 if k / I. Error locations can be found by exhaustively computing L ( α k), 0 k n 2. E (z) helps find errors e i, i I, without matrix inverse. ( ) k I, E α k = e k α k ( j k 1 α j α k) 0. e k = E ( α k) ( / α k ( j k 1 α j α k)). or ek = E ( α k) / d dz L ( α k), where d dz L (z) is the derivative of L (z). Complexity is reduced from O ( n 3) to O ( n 2). Decoding strategy: from S (z) to find L (z) and E (z). W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-19

74 An Example of L (z) and E (z) I = {1, 2, 5}. L (z) = (1 αz) ( 1 α 2 z ) ( 1 α 5 z ) L (z) = 0 if z = α 1, α 2, or α 5. L (z) 0 otherwise. E (z) = e 1 α 1 ( 1 α 2 z ) ( 1 α 5 z ) T1 + e 2 α 2 (1 αz) ( 1 α 5 z ) T2 + e 5 α 5 (1 αz) ( 1 α 2 z ) T3 T1 T2 T3 E (z) z = α 1 0 = 0 = 0 0 z = α 2 = 0 0 = 0 0 z = α 5 = 0 = W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-20

75 Properties of L (z) and E (z) Let t = d 1 2. Theorem gcd (L (z), E (z)) = The key equation: E (z) = L (z) S (Z) mod z d (Uniqueness) Let a (z), b (z) F q [z] be such that deg (a (z)) t 1, deg (b (z)) t, gcd (a (z), b (z)) = 1 and ( a (z) S (z) b (z) mod z d 1). Then a (z) and b (z) are unique up to a constant. That is, we can treat a (z) = ce (z), b (z) = cl (z), and E (z) and L (z) are generated from an error vector e s.t. wt (e) t. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-21

76 Decoding Process 1. Compute the syndrome vector and polynomial s and S (z) respectively. 2. Apply Euclidean algorithm to z d 1 and S (z), i.e., z d 1 = q 1 (z) S (z) + r 1 (z) S (z) = q 2 (z) r 1 (z) + r 2 (z). r l 2 (z) = q l (z) r l 1 (z) + r l (z), where deg (r l (z)) t By Bézout s Identity (Lem. 1.5), one has r l (z) = a (z) S (z) + b (z) z d 1 a (z) S (z) mod z d Let c be the leading coefficient of the polynomial a (z), i.e., c 1 a (z) is a monic polynomial. By Theorem 6.13, set L (z) = c 1 a (z), and E (z) = c 1 r l (z). 5. Find the error locations i I from L (z) and the errors e i from E (z). ĉ = y e. The complexity is highly reduced! W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-22

77 Theorem 6.13, Part 1: Proof Proof: L (z) has roots α i, i I. None of them is a root of E (z). L (z) and E (z) does not share any roots. gcd (L (z), E (z)) = 1. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-23

78 Theorem 6.13, Part 2: Proof Theorem 6.13 part 2 is a direct consequence of the lemma below. Lemma 6.14 S (z) i I e iα i 1 α i z mod zd 1 Proof: As s = rh T = eh T, it follows that Hence, s j = n 1 i=0 e iα i(j+1) = i I e iα i(j+1), 0 j d 2. By the definition of S (z), it holds that S (z) = d 2 j=0 s jz j = d 2 j=0 i I e iα i(j+1) z j = ( i I e d 2 iα i ( j=0 α i z ) ) j = i I e iα i ( j=0 ( α i z ) j ) mod z d 1 = i I e iα i 1 1 α i z. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-24

79 Theorem 6.13, Part 3: Proof Proof: To prove the uniqueness, we assume that there exist (E (z), L (z)) (E (z), L (z)) s.t. E (z) = S (z) L (z) mod z d 1 and E (z) = S (z) L (z) mod z d 1. It follows that E (z) L (z) = S (z) L (z) L (z) mod z d 1 = E (z) L (z) mod z d 1. (2) By assumption, deg (E (z)) t 1 and deg (L (z)) t. It is clear that deg (E (z) L (z)) 2t 1 d 2. The same is true for E (z) L (z). As a result, (2) becomes E (z) L (z) = E (z) L (z) Note gcd (E (z), L (z)) = 1. By Lemma 1.12, E (z) E (z) and L (z) L (z). Similarly from gcd (E (z), L (z)) = 1, E (z) E (z) and L (z) L (z). Hence, E (z) = ce (z) and L (z) = cl (z) for some nonzero c F q. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-25

80 An Example Example: Consider the [7, 3] RS code over F 8 (F 8 is given as follows). 0 1 α α 2 α 3 α 4 α 5 α Let the received signal be y = [ α 3, α, 1, α 2, 0, α 3, 1 ]. Find ĉ. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-26

81 Solutions to the Example 1. Parameters: n k = 4, d = 5 (t = 2), and H F 4 7 H = 2. Syndromes: s = yh T = [ α 3, α 4, α 4, 0 ]. S (z) = α 4 z 2 + α 4 z + α α α 2 α 3 α 4 α 5 α 6 1 α 2 α 4 α 6 α α 3 α 5 1 α 3 α 6 α 2 α 5 α α 4 1 α 4 α α 5 α 2 α 6 α 3 3. Key polynomials: apply Euclidean algorithm to z 4 and S (z). 3.1 z 4 = ( α 3 z 2 + α 3 z + α 5) S (z) + (z + α). 3.2 L (z) = α 3 z 2 + α 3 z + α 5. E (z) = z + α. 3.3 L (z) = α 5 z 2 + α 5 z + 1. E (z) = α 2 z + α Find ĉ: 4.1 Plug 1, α 1, into L (z). L ( α 2) = L ( α 3) = According E (z), we have e 2 = α 3 and e 3 = α ĉ = y e = y + e = [ α 3, α, α, 1, 0, α 3, 1 ].. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes RS Code - Decoding 2017 page 6-27

82 Towards Cyclic and BCH Codes Have seen Binary Hamming codes: d = 3. Reed-Solomon codes: MDS (d = n k + 1) and requires large fields (typically q = n + 1). Will introduce cyclic codes Reed-Solomon codes are a special case of cyclic codes. BCH codes as another special case. Systematic way to construct binary codes with large distance. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-28

83 Cyclic Codes Definition 6.15 An [n, k] linear code is cyclic if for every codeword c = c 0 c 1 c n 2 c n 1, the right cyclic shift of c, c n 1 c 0 c 1 c n 2, is also a codeword. Example: The H [7, 3] has the parity-check matrix H = It s dual code H3 (view H as the generator matrix) is cyclic. (The codewords are , , , , , , , ) W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-29

84 Generating Function Definition 6.16 The generating function of a codeword c = [c 0 c n 1 ] is c (x) = c 0 + c 1 x + + c n 1 x n 1. It will be convenient to use c (x) to represent a codeword c. The right cyclic shift c = (c 0,, c n 1 ) c = (c n 1, c 0,, c n 2 ) can be obtained by c (x) = x c (x) mod x n 1 as x c (x) = c 0 x + c 1 x c n 2 x n 1 + c n 1 x n = c n 1 + c 0 x + c 1 x c n 2 x n 1 mod x n 1. Lemma 6.17 Let c (x) C. For an arbitrary u (x), u (x) c (x) mod x n 1 is in C. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-30

85 Generator Polynomial Theorem 6.18 For a cyclic code C, a unique monic polynomial g (x) s.t. for all c (x) C, c (x) = u (x) g (x) for some u (x). Proof: Let g (x) C be the nonzero polynomial of least degree. Since C is linear, w.l.o.g., assume that g (x) is monic. Then c (x) C, write c (x) = u (x) g (x) + r (x). By definition of cyclic codes, u (x) g (x) C. Hence, r (x) C by linearity of C. But deg (r (x)) < deg (g (x)), which implies r (x) = 0. The uniqueness of g (x) can be proved by contradiction. Suppose that there are two monic polynomials g 1 (x) g 2 (x) of the same degree that both generate C. Then g 1 (x) g 2 (x) C and deg (g 1 g 2 ) < deg (g 1 ), which forces g 1 (x) g 2 (x) = 0. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-31

86 Properties of the Generator Polynomial Corollary 6.19 g (x) x n 1. Proof: Write x n 1 = q (x) g (x) + r (x). Take mod x n 1 on both sides. 0 = x n 1 mod x n 1 C. q (x) g (x) mod x n 1 C (x). Hence r (x) mod x n 1 C r (x) C r (x) = 0. Remark: Let n = q m 1. We know how to factor x n 1 in terms of minimal polynomials. g (x) must be a product of minimal polynomials. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-32

87 Generator Matrices of Cyclic Codes Theorem 6.20 The generator matrix of a cyclic code C [n, k]: g (x) xg (x) G =. x k 1 g (x) = g 0 g 1 g n k g 0 g 1 g n k g 0 g 1 g n k. Observations: deg (g (x)) = n k. Easy for implementation: can be implemented by using flip-flops. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-33

88 Parity-Check Matrices of Cyclic Codes Recall g (x) x n 1. Define h (x) such that g (x) h (x) = x n 1. Then h (x) is a monic polynomial with degree k. Write h (x) = k i=0 a ix i. Definition 6.21 The reciprocal polynomial h R (x) of h (x) is given by h R (x) = a k + a k 1 x + + a 0 x k = x k h (1/x). Example: h (x) = 1 + x 2 + x 3 h R (x) = 1 + x + x 3. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-34

89 Parity-Check Matrix Theorem 6.22 The parity-check matrix of the cyclic code C [n, k] is h R (x) xh R (x) H =. x n k 1 h R (x) = h k h k 1 h 0 h k h k 1 h h k h k 1 h 0. Corollary 6.23 The dual of a cyclic code, C, is also cyclic. h 1 0 h R (x) is the generator polynomial of C. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-35

90 Theorem 6.22: Proof By assumption, x n 1 = ( g (x) h (x). Note that n k ) ( g (x) h (x) = i=0 g k ) ix i i=0 h ix i = n i=0 ( i l=0 g lh i l ) x i = n i=0 a ix i, where a 0 = g 0 h 0 = 1, a n = g n k h k = 1 1 = 1, and a i = i l=0 h lg i l = 0, 1 i n 1. Let A = GH T with A i,j = [0,, 0, g }{{} 0,, g n k, 0, 0] [0,, 0, h }{{} k,, h 0, 0,, 0] T. i 1 j 1 It can be verified that A 1,1 = a k, A 1,2 = a k+1,, and a k a k+1 a n 1 A = GH T a k 1 a k a n 2 = = 0 Fk (n k) a 1 a 2 a n k W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes Cyclic Codes 2017 page 6-36

91 Cyclic Codes: An Example To construct a cyclic code on F q, we realize that M (i) (x) F q [x] M (i) (x) x qm 1 1. Definition 6.24 A BCH code over F q of length n = q m 1 is the cyclic code generated by g (x) = lcm ( M (a) (x),, M (a+δ 2) (x) ) for some integer a. (The code is called narrow-sense if a = 1.) Lemma 6.25 A BCH code defined in Definition 6.24 has d δ. δ is referred to the designed distance. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes BCH Codes 2017 page 6-37

92 Distance of BCH Codes: Proof of Lemma 6.25 Let α be the primitive element in F q m. By construction, α a,, α a+δ 2 are roots of the generator polynomial g (x). That is, c C, the generating function c (x) satisfies c ( α i) = 0, a i a + δ 2. In matrix format, 1 α a α a(n 1) c 0 1 α a+1 α (a+1)(n 1) c = 0 (3) 1 α a+δ 2 α (a+δ 1)(n 1) c n 1 Any (δ 1)-column submatrix is a Vandermonde matrix and hence of full rank. This implies d δ. Remark: The matrix in (3) is in F (δ 1) n q m while the vector c C Fn q. Hence the matrix is not a parity-check matrix when m > 1. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes BCH Codes 2017 page 6-38

93 Example: Reed-Solomon Codes Recall that a RS-code C [n, k, n k + 1] is built on F q with typically n = q 1. Compare the parity-check matrix of a RS-code (Theorem 6.5) and Equation (3). It is clear that a RS-code is a special case of a BCH code with m = 1. In particular, suppose that we are asked to build a BCH code over F q with n = q 1 and d δ = n k + 1. We find M (i) (x) F q [x], 1 i 1 + δ 2 = n k. Since M (i) (x) F q [x], it follows that M (i) (x) = x α i. Hence g (x) = n k ( i=1 x α i ) and the generator matrix can be constructed (in a different form of that in Theorem 6.5) and good for implementation). Its parity-check matrix is given by the matrix in Equation (3). RS decoder can be directly applied for decoding. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes BCH Codes 2017 page 6-39

94 Example: Binary BCH Codes We have learned binary Hamming codes. The distance is always 3. The question is how to construct a binary code with large distance. For example, how to construct a binary code of length 15 and d 5? 1. For binary codes, use F 2. n = 15 = hence m = δ = 5 implies g (x) = lcm ( M (1) (x), M (2) (x), M (3) (x), M (4) (x) ). 3. The relevant cyclotomic cosets of 2 modulo 15 include C 1 = {1, 2, 4, 8} and C 3 = {3, 6, 9, 12}. Hence M (1) (x) = i C 1 ( x α i ) = M (2) (x) = M (4) (x) and M (3) (x) = i C 3 ( x α i ). Furthermore, g (x) = M (1) (x) M (3) (x). 4. Find the generator matrix and parity-check matrix according to Theorems 6.20 and 6.22 respectively. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes BCH Codes 2017 page 6-40

95 From Hamming to BCH Example: A binary code of length 15 and d 5? g (x) = lcm ( M (1) (x), M (2) (x), M (3) (x), M (4) (x) ) = lcm ( M (1) (x), M (3) (x) ) = M (1) (x) M (3) (x). RS codes are special cases of BCH codes (m = 1). Have learned [7, 4, 3] Hamming code H = Another view: Let α be a primitive element of F 8 that satisfies α 3 = α + 1. The parity check matrix can be written as [ 1 α α 2 α 3 α 4 α 5 α 6 ]. (Column vectors in F 3 2 Elements in F 8) W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes BCH Codes 2017 page 6-41

96 From Hamming to BCH: Larger Distance Binary BCH codes with d 5: g (x) = lcm ( M (1) (x), M (2) (x), M (3) (x), M (4) (x) ) It holds that = lcm ( M (1) (x), M (3) (x) ). 1 α α 2 α 3 α 4 α 5 α 6 1 α 2 α 4 α 6 α α 3 α 5 1 α 3 α 6 α 2 α 5 α α 4 1 α 4 α α 5 α 2 α 6 α 3 { c ( α 2) = c (α) 2 = 0 c = 0 But c (α) = 0 c ( α 4) = c (α) 4 = 0 [ 1 α α 2 α H = 3 α 4 α 5 α 6 ] 1 α 3 α 6 α 2 α 5 α α 4 Eventually, we get a [7, 1, 7] code C = { , }. W. Dai (IC) EE4.07 Coding Theory RS & BCH Codes BCH Codes 2017 page 6-42