Non-Standard Coding Theory

Transcription

1 Non-Standard Coding Theory Steven T. Dougherty July 3, 2013

2 Rosenbloom-Tsfasman Metric Codes with the Rosenbloom-Tsfasman Metric

3 Rosenbloom-Tsfasman Metric Mat n,s (F q ) denotes the linear space of all matrices with n rows and s columns with entries from a finite field F q of q elements.

4 Rosenbloom-Tsfasman Metric Mat n,s (F q ) denotes the linear space of all matrices with n rows and s columns with entries from a finite field F q of q elements. A linear code is a subspace of Mat n,s (F q ).

5 Rosenbloom-Tsfasman Metric Define ρ on Mat n,s (F q )

6 Rosenbloom-Tsfasman Metric Define ρ on Mat n,s (F q ) Let n = 1 and ω = (ξ 1, ξ 2,..., ξ s ) Mat 1,s (F q ). Then, we put ρ(0) = 0 and ρ(ω) = max{i ξ i 0} (1) for ω 0.

7 Rosenbloom-Tsfasman Metric Define ρ on Mat n,s (F q ) Let n = 1 and ω = (ξ 1, ξ 2,..., ξ s ) Mat 1,s (F q ). Then, we put ρ(0) = 0 and ρ(ω) = max{i ξ i 0} (1) for ω 0. Ex: ρ(1, 0, 0, 1, 0) = 4.

8 Rosenbloom-Tsfasman Metric Now let Ω = (ω 1,..., ω n ) T Mat n,s (F q ), ω j Mat 1,s (F q ), 1 j n, and ( ) T denotes the transpose of a matrix. Then, we put n ρ(ω) = ρ(ω j ) (2) j=1

9 Rosenbloom-Tsfasman Metric Now let Ω = (ω 1,..., ω n ) T Mat n,s (F q ), ω j Mat 1,s (F q ), 1 j n, and ( ) T denotes the transpose of a matrix. Then, we put n ρ(ω) = ρ(ω j ) (2) j=1 Ex: ρ = = 12.

10 Weight Distribution For a given linear code C Mat n,s (F q ) the following set of nonnegative integers w r (C) = {Ω C ρ(ω) = r}, 0 r ns (3) is called the ρ weight spectrum of the code C.

11 Weight Enumerator Define the ρ weight enumerator by W (C z) = ns r=0 w r (C)z r = Ω C z ρ(ω) (4)

12 Weight Enumerator Define the ρ weight enumerator by W (C z) = ns r=0 w r (C)z r = Ω C z ρ(ω) (4) Note that if s = 1, it reduces to the Hamming weight enumerator.

13 Inner-Product Introduce the following innerproduct on Mat n,s (F q ). At first, let n = 1 and ω 1 = (ξ 1,..., ξ s), ω 2 = (ξ 1,..., ξ s ) Mat 1,s (F q ). Then we put ω 1, ω 2 = ω 2, ω 1 = s i=1 ξ iξ s+1 i (5)

14 Inner-Product Introduce the following innerproduct on Mat n,s (F q ). At first, let n = 1 and ω 1 = (ξ 1,..., ξ s), ω 2 = (ξ 1,..., ξ s ) Mat 1,s (F q ). Then we put ω 1, ω 2 = ω 2, ω 1 = s i=1 ξ iξ s+1 i (5) Ex: q = 5, (1, 2, 1, 3, 4), (2, 1, 4, 3, 4) = 1(4) + 2(3) + 1(4) + 3(1) + 4(2) = 3.

15 Inner-Product Introduce the following innerproduct on Mat n,s (F q ). At first, let n = 1 and ω 1 = (ξ 1,..., ξ s), ω 2 = (ξ 1,..., ξ s ) Mat 1,s (F q ). Then we put ω 1, ω 2 = ω 2, ω 1 = s i=1 ξ iξ s+1 i (5) Ex: q = 5, (1, 2, 1, 3, 4), (2, 1, 4, 3, 4) = 1(4) + 2(3) + 1(4) + 3(1) + 4(2) = 3. Note that this is a non-standard inner-product on rows.

16 Inner-Product Now, let Ω i = (ω (1) i,..., ω (n) i ) T Mat n,s (F q ), i = 1, 2, ω (j) i Mat 1,s (F q ), 1 j n. Then we put Ω 1, Ω 2 = Ω 2, Ω 1 = n j=1 ω (j) 1, ω(j) 2 (6)

17 Orthogonal Let C Mat n,s (F q ). C Mat n,s (F q ) is defined by C = {Ω 2 Mat n,s (F q ) Ω 2, Ω 1 = 0 for all Ω 1 C}. (7)

18 Orthogonal Let C Mat n,s (F q ). C Mat n,s (F q ) is defined by C = {Ω 2 Mat n,s (F q ) Ω 2, Ω 1 = 0 for all Ω 1 C}. (7) C is a linear code, and (C ) = C.

19 Orthogonal Let C Mat n,s (F q ). C Mat n,s (F q ) is defined by C = {Ω 2 Mat n,s (F q ) Ω 2, Ω 1 = 0 for all Ω 1 C}. (7) C is a linear code, and (C ) = C. We have d + d = ns, C C = q ns, C = q d, C = q ns d, (8) where d is the dimension of C and d is the dimension of C.

20 Examples q = 2, n = s = 2

21 Examples q = 2, n = s = 2 ( 0 0 C 1 = { 0 0 ) ( 1 0, 1 0 ) ( 0 0 }, C 2 = { 0 0 ) ( 0 0, 0 1 ) } (9)

22 Examples q = 2, n = s = 2 ( 0 0 C 1 = { 0 0 ) ( 1 0, 1 0 Both codes have ρ weight enumerator ) ( 0 0 }, C 2 = { 0 0 ) ( 0 0, 0 1 ) } (9) 1 + z 2 (10)

23 Duals ( C1 0 0 = { 0 0 ( ) ( 0 1, 0 1 ), ( ) ( 1 1, 1 1 ), ( ) ( 1 1, 0 1 ), ( ), ) }

24 Duals ( C1 0 0 = { 0 0 ( ) ( 0 1, 0 1 ), ( ) ( 1 1, 1 1 ), ( ) ( 1 1, 0 1 ), ( ), ) } ( C2 0 0 = { 0 0 ( ) ( 1 0, 0 0 ), ( ) ( 0 1, 0 0 ), ( ) ( 0 0, 0 1 ), ( ), ) }

25 Weight Enumerators The ρ weight enumerator for C 1 and C 2 turns out to be different: W (C 1 z) = 1 + 4z 4 + 2z + z 2

26 Weight Enumerators The ρ weight enumerator for C 1 and C 2 turns out to be different: W (C 1 z) = 1 + 4z 4 + 2z + z 2 W (C 2 z) = 1 + 2z 4 + z 3 + 3z 2 + z

27 Weight Enumerators The ρ weight enumerator for C 1 and C 2 turns out to be different: W (C 1 z) = 1 + 4z 4 + 2z + z 2 W (C 2 z) = 1 + 2z 4 + z 3 + 3z 2 + z Therefore, the ρ weight enumerators cannot be related by a MacWilliams type relation.

28 Is it a problem with the inner-product We shall compare the first innerproduct with the common one: [ω 1, ω 2 ] = s i=1 ξ iξ i. (11)

29 Is it a problem with the inner-product We shall compare the first innerproduct with the common one: [ω 1, ω 2 ] = s i=1 ξ iξ i. (11) Consider two linear codes C 1 and C 2 Mat 1,4 (F 2 ), C 1 = {0000, 1100, 1001, 0101}, C 2 = {0000, 0100, 0001, 0101}.

30 Is it a problem with the inner-product We shall compare the first innerproduct with the common one: [ω 1, ω 2 ] = s i=1 ξ iξ i. (11) Consider two linear codes C 1 and C 2 Mat 1,4 (F 2 ), C 1 = {0000, 1100, 1001, 0101}, C 2 = {0000, 0100, 0001, 0101}. Notice that these codes have the same ρ weight enumerators: W (C i z) = W (C i z) = 1 + z 2 + 2z 4, i = 1, 2. (12)

31 Is it a problem with the inner-product Denote by C 1 and C 2 codes dual to C 1 and C 2 with respect to the common inner product. We have C 1 = {0000, 0010, 1111, 1101}

32 Is it a problem with the inner-product Denote by C 1 and C 2 codes dual to C 1 and C 2 with respect to the common inner product. We have C 1 = {0000, 0010, 1111, 1101} C 2 = {0000, 0010, 1000, 1010}

33 Is it a problem with the inner-product Denote by C 1 and C 2 codes dual to C 1 and C 2 with respect to the common inner product. We have C 1 = {0000, 0010, 1111, 1101} C 2 = {0000, 0010, 1000, 1010} The ρ weight enumerators are different: W (C 1 Z) = 1 + z 3 + 2z 4, W (C 2 z) = 1 + z + 2z 3. (13)

34 Is it a problem with the inner-product Therefore, the ρ weight enumerators W (C z) and W (C z) cannot be related by a MacWilliams-type identity with the common inner-product.

35 Is it a problem with the inner-product Therefore, the ρ weight enumerators W (C z) and W (C z) cannot be related by a MacWilliams-type identity with the common inner-product. It is not a problem with the inner-product but rather with the weights.

36 T -Weight Enumerator T (C Z 1,..., Z n ) = Ω C Υ(Ω Z 1,..., Z n ) (14) where Υ(Ω) = z (1) a 1 z a (2) 2... z (n) a n and ρ(ω i ) = a i, 1 i n.

37 T -Weight Enumerator T (C Z 1,..., Z n ) = Ω C Υ(Ω Z 1,..., Z n ) (14) where Υ(Ω) = z (1) a 1 z a (2) 2... z (n) a n and ρ(ω i ) = a i, 1 i n. The Z i are n complex vectors with s + 1 components, Z j = (z (j) 0,..., z(j) s ).

38 T -Weight Enumerator Example: Υ = z1 3 z 2 4 z 3 1 z 4 2

39 H-Weight Enumerator H(C Z) = T (C Z, Z,..., Z).

40 H-Weight Enumerator H(C Z) = T (C Z, Z,..., Z). In the previous example the monomial becomes z 3 z 4 z 1 z 2.

41 H-Weight Enumerator H(C Z) = T (C Z, Z,..., Z). In the previous example the monomial becomes z 3 z 4 z 1 z 2. Notice that the first enumerator is a polynomial of degree at most one in each of n(s + 1) variables z (j) i, 0 i s 1 j n, while the second enumerator has degree at most n in each of s + 1 variables z i, 0 i s.

42 Linear Transformation Introduce a linear transformation Θ s : C s+1 C s+1 by setting Z = Θ s Z, where z 0 = z 0 + (q 1)z 1 + q(q 1) + q 2 (q 1)z q s 2 (q 1)z s 1 + q s 1 (q 1)z s

43 Linear Transformation z 1 = z 0 + (q 1)z 1 + q(q 1) + q 2 (q 1)z q s 2 (q 1)z s 1 + q s 1 z s... z s 2 = z 0 + (q 1)z 1 + q(q 1) q 2 z 3 z s 1 = z 0 + (q 1)z 1 qz 2 z s = z 0 z 1

44 Linear Transformation We assume that Z = (z 0, z 1, z 2,... ) is an infinite sequence with z i = 0 for i > s. Thus the s + 1 by s + 1 matrix Θ s = θ lk, 0 l, k s, has the following entries

45 Linear Transformation 1 if l = 0, q θ lk = l 1 (q 1) if 0 < l s k, q l 1 if l + k = s + 1, 0 if l + k > s + 1.

46 Linear Transformation 1 if l = 0, q θ lk = l 1 (q 1) if 0 < l s k, q l 1 if l + k = s + 1, 0 if l + k > s + 1. Θ 1 = ( 1 q )

47 Linear Transformation 1 if l = 0, q θ lk = l 1 (q 1) if 0 < l s k, q l 1 if l + k = s + 1, 0 if l + k > s + 1. Θ 1 = ( 1 q ) Θ 2 = 1 q 1 q(q 1) 1 q 1 q 1 1 0

48 Linear Transformation 1 if l = 0, q θ lk = l 1 (q 1) if 0 < l s k, q l 1 if l + k = s + 1, 0 if l + k > s + 1. Θ 1 = ( 1 q ) Θ 3 = Θ 2 = 1 q 1 q(q 1) 1 q 1 q q 1 q(q 1) q 2 (q 1) 1 q 1 q(q 1) q 2 1 q 1 q

49 MacWilliams Relations Theorem The T -enumerators of mutually dual linear codes C, C Mat n,s (F q ) are related by T (C Z 1,..., Z n ) = 1 C T (C Θ sz 1,..., Θ s Z n ).

50 MacWilliams Relations Theorem The H-enumerator of mutually dual linear codes C, C Mat n,s (F q ) are related by H(C Z) = 1 C H(C Θ sz)

51 MacWilliams Relations Hence by expanding the amount of information in the weight enumerator MacWilliams relations can be found!

52 Singleton Bound The minimum weight of a code C is given by ρ(c) = min{ρ(ω, Ω ) Ω, Ω C, Ω Ω }.

53 Singleton Bound The minimum weight of a code C is given by ρ(c) = min{ρ(ω, Ω ) Ω, Ω C, Ω Ω }. If the code is linear (i.e. A is a finite ring and the code is a submodule) then ρ(c) = min{ρ(ω) Ω C, } where ρ(ω) = ρ(ω, 0).

54 Singleton Bound Theorem Let A be any finite alphabet with q elements and let C Mat n,s (A), be an arbitrary code, then C q n d+1.

55 Singleton Bound Theorem Let A be any finite alphabet with q elements and let C Mat n,s (A), be an arbitrary code, then C q n d+1. Proof. Mark the first d 1 positions lexicographically. Two elements of C never coincide in all other positions since otherwise the distance between them would be less than d. Hence C q n d+1.

56 Singleton Bound Corollary Let C Mat n,s (A), where A = q, be an arbitrary code consisting of q k, 0 k ns, points. Then ρ(c) ns k + 1.

57 Singleton Bound Corollary Let C Mat n,s (A), where A = q, be an arbitrary code consisting of q k, 0 k ns, points. Then ρ(c) ns k + 1. Naturally, we define a code meeting this bound as a Maximum Distance Separable Code with respect to the ρ metric.

58 MDS Codes Theorem (Skriganov) If C is a linear MDS code in Mat n,s (F q ), then C is also an MDS code.

59 MDR Bound Theorem If C is a linear code in Mat n,s (Z k ) of rank h, then ρ(c) ns h + 1.

60 MDR Bound Theorem If C is a linear code in Mat n,s (Z k ) of rank h, then ρ(c) ns h + 1. Codes meeting this bound are called MDR codes.

61 MDR Codes Theorem Let C 1, C 2,..., C r be linear codes in Mat n,s (Z k1 ),..., Mat n,s (Z kr ), respectively, where k 1,..., k r are positive integers with gcd(k i, k j ) = 1 for i j. If C i is an MDR code for all i, then C = CRT(C 1, C 2,..., C r ) is an MDR code.

62 Uniform Distributions Let U denote the interval [0, 1) and M A = [ m 1 k a, m k a )... [ m n 1 k, m n + 1 ) U n an k an an elementary box, where M = (m 1,..., m n ) and A = (a 1,..., a n ).

63 Uniform Distributions Let U denote the interval [0, 1) and M A = [ m 1 k a, m k a )... [ m n 1 k, m n + 1 ) U n an k an an elementary box, where M = (m 1,..., m n ) and A = (a 1,..., a n ). Definition Given an integer 0 h n, a subset D U n consisting of k h points is called an optimum [ns, h] s distribution in base k if each elementary box M A of volume k h contains exactly one point of D.

64 Uniform Distributions For a point X in Q n (k s ) define the following matrix which is an element of Mat n,s (Z k ): where and x = s i=1 ξ i(x)k i s 1. Ω X = (ω(x 1 ), ω(x 2 ),..., ω(x n )) T ω x = (ξ 1 (x), ξ 2 (x),..., ξ s (x))

65 Uniform Distributions Theorem Let C be an optimum distribution in Q n (k s ) for any k and C its corresponding code then the following are equivalent: D is an optimum [ns, λ] s distribution in base k C is an MDS code in the ρ metric in Mat n,s (Z k ).

66 Z 2 Z 4 Codes Codes over Z 2 Z 4 and their Gray Map

67 Delsarte Delsarte defines additive codes as subgroups of the underlying abelian group in a translation association scheme.

68 Delsarte Delsarte defines additive codes as subgroups of the underlying abelian group in a translation association scheme. For the binary Hamming scheme, the only structures for the abelian group are those of the form Z α 2 Zβ 4, with α + 2β = n.

69 Delsarte Delsarte defines additive codes as subgroups of the underlying abelian group in a translation association scheme. For the binary Hamming scheme, the only structures for the abelian group are those of the form Z α 2 Zβ 4, with α + 2β = n. Thus, the subgroups C of Z α 2 Zβ 4 binary Hamming scheme. are the only additive codes in a

70 Gray Map Φ : Z α 2 Z β 4 Zn 2 where n = α + 2β.

71 Gray Map Φ : Z α 2 Z β 4 Zn 2 where n = α + 2β. Φ(x, y) = (x, φ(y 1 ),..., φ(y β )) for any x Z α 2 and any y = (y 1,..., y β ) Z β 4, where φ : Z 4 Z 2 2 is the usual Gray map.

72 Gray Map Φ : Z α 2 Z β 4 Zn 2 where n = α + 2β. Φ(x, y) = (x, φ(y 1 ),..., φ(y β )) for any x Z α 2 and any y = (y 1,..., y β ) Z β 4, where φ : Z 4 Z 2 2 is the usual Gray map. The map Φ is an isometry which transforms Lee distances in Z α 2 Zβ 4 to Hamming distances in Zα+2β 2.

73 Weights Denote by wt H (v 1 ) the Hamming weight of v 1 Z α 2 wt L (v 2 ) the Lee weight of v 2 Z β 4. and by

74 Weights Denote by wt H (v 1 ) the Hamming weight of v 1 Z α 2 wt L (v 2 ) the Lee weight of v 2 Z β 4. and by For a vector v = (v 1, v 2 ) Z α 2 Zβ 4, define the weight of v, denoted by wt(v), as wt H (v 1 ) + wt L (v 2 ), or equivalently, the Hamming weight of Φ(v).

75 Generator Matrix The generator matrix for a Z 2 Z 4 -additive code C of type (α, β; γ, δ; κ): G S = I κ T 2T T 1 2I γ κ 0 0 S S R I δ where T, T 1, T 2, R, S are matrices over Z 2 and S is a matrix over Z 4.,

76 Inner-Product The following inner product is defined for any two vectors u, v Z α 2 Zβ 4 : u, v = 2( α u i v i ) + i=1 α+β j=α+1 u j v j Z 4.

77 Inner-Product The following inner product is defined for any two vectors u, v Z α 2 Zβ 4 : u, v = 2( α u i v i ) + i=1 α+β j=α+1 u j v j Z 4. The additive dual code of C, denoted by C, is defined in the standard way C = {v Z α 2 Z β 4 u, v = 0 for all u C}.

78 MacWilliams Relations Define WL(x, y) = c C x n wt L(c) y wt L(c).

79 MacWilliams Relations Define WL(x, y) = c C x n wt L(c) y wt L(c). Theorem Let C be a Z 2 Z 4 code, then WL C (x, y) = 1 C WL C (x + y, x y).

80 Bounds Theorem Let C be a Z 2 Z 4 -additive code of type (α, β; γ, δ; κ), then d(c) 1 α β γ δ; (15) 2 d (C) 1 α + β γ δ. (16) 2

81 Separable Let C be a Z 2 Z 4 -additive code. If C = C X C Y, then C is called separable.

82 Separable Let C be a Z 2 Z 4 -additive code. If C = C X C Y, then C is called separable. Theorem If C is a Z 2 Z 4 -additive code which is separable, then the minimum distance is given by d (C) = min {d (C X ), d (C Y )}.

83 MDS We say that a Z 2 Z 4 -additive code C is maximum distance separable (MDS) if d (C) meets the bound given in The usual Singleton bound for a code C of length n over an alphabet of size q is given by d(c) n log q C + 1..

84 MDS We say that a Z 2 Z 4 -additive code C is maximum distance separable (MDS) if d (C) meets the bound given in The usual Singleton bound for a code C of length n over an alphabet of size q is given by d(c) n log q C In the first case, we say that C is MDS with respect to the Singleton bound, briefly MDSS. If it meets the second bound, C is MDS with respect to the rank bound, briefly MDSR.

85 MDSS Theorem Let C be an MDSS Z 2 Z 4 -additive code of type (α, β; γ, δ; κ) such that 1 < C < 2 α+2β. Then C is either (i) (ii) the repetition code of type (α, β; 1, 0; κ) and minimum distance d(c) = α + 2β, where κ = 1 if α > 0 and κ = 0 otherwise; or the even code with minimum distance d(c) = 2 and type (α, β; α 1, β; α 1) if α > 0, or type (0, β; 1, β 1; 0) otherwise.

86 MDSR Theorem Let C be an MDSR Z 2 Z 4 -additive code of type (α, β; γ, δ; κ) such that 1 < C < 2 α+2β. Then, either (i) (ii) (iii) C is the repetition code as in (i) of Theorem 3 with α 1; or C is of type (α, β; γ, α + β γ 1; α), where α 1 and d(c) = 4 α {3, 4}; or C is of type (α, β; γ, α + β γ; α), where α 1 and d(c) 2 α {1, 2}.