Coding Theory: Equivalent Codes Fundamental Problem Equivalent codes Given n and k, when are two (n; k) codes \essentially the same"? Yes they are, since permuting the rows of G2 (putting G2 in a systematic form) gives G1. Equivalent Codes Example Given the following two generator matrices G1 and G2 over F2 1 0 1 0 1 1 G = ;G = ; 1 0 1 1 2 1 0 1 are the corresponding codes C1, C2 \the same"? Equivalent Codes Example Given the following two generator matrices G1 and G2 over F2 1 0 1 0 1 1 G = ;G = ; 1 0 1 1 2 1 0 1 are the corresponding codes C1, C2 \the same"? Yes they are, since permuting the rows of G2 (putting G2 in a systematic form) gives G1. In G2, add the second row to the first one to get: 1 0 1 1 0 1 2 1 Equivalent Codes Example Given the following two generator matrices G1 and G2 over F3 1 0 2 2 1 2 2 0 G = ;G = ; 1 0 1 1 2 2 0 1 2 1 are the corresponding codes C1, C2 \the same"? Equivalent Codes Example Given the following two generator matrices G1 and G2 over F3 1 0 2 2 1 2 2 0 G = ;G = ; 1 0 1 1 2 2 0 1 2 1 are the corresponding codes C1, C2 \the same"? In G2, add the second row to the first one to get: 1 0 1 1 0 1 2 1 Yes they are essentially the same codes, since we have the same codewords up to permuting coordinates. Equivalent Codes Example Given the following two generator matrices G1 and G2 over F3 1 0 2 2 (x ; x )G = (x ; x ) = (x ; x ; 2x + x ; 2x + 2x ); 1 2 1 1 2 0 1 1 2 1 2 1 2 1 2 1 2 2 0 (x ; x )G = (x ; x ) = (x ; 2x + x ; 2x + 2x ; x ); 1 2 2 1 2 0 1 2 1 1 1 2 1 2 2 Equivalent Codes Example Given the following two generator matrices G1 and G2 over F3 1 0 2 2 (x ; x )G = (x ; x ) = (x ; x ; 2x + x ; 2x + 2x ); 1 2 1 1 2 0 1 1 2 1 2 1 2 1 2 1 2 2 0 (x ; x )G = (x ; x ) = (x ; 2x + x ; 2x + 2x ; x ); 1 2 2 1 2 0 1 2 1 1 1 2 1 2 2 Yes they are essentially the same codes, since we have the same codewords up to permuting coordinates. Equivalent Codes Example (x1; x2)G1 = (x1; x2; 2x1 + x2; 2x1 + 2x2); (x1; x2)G2 = (x1; 2x1 + x2; 2x1 + 2x2; x2) 21 0 0 03 60 0 1 07 (x1; 2x1+x2; 2x1+2x2; x2) 6 7 = (x1; x2; 2x1+x2; 2x1+2x2) 40 0 0 15 0 1 0 0 Equivalent Codes Example Given the following two generator matrices G1 and G2 over F3 1 0 2 2 (x ; x )G = (x ; x ) = (x ; x ; 2x + x ; 2x + 2x ); 1 2 1 1 2 0 1 1 2 1 2 1 2 1 2 21 0 0 03 21 0 0 03 60 0 1 07 1 2 2 0 60 0 1 07 (x1; x2) G2 6 7 = (x1; x2) 6 7 40 0 0 15 0 1 2 1 40 0 0 15 0 1 0 0 0 1 0 0 | {z } G1 21 0 0 03 60 0 1 07 = (x1; 2x1 + x2; 2x1 + 2x2; x2) 6 7 = (x1; x2)G1: 40 0 0 15 0 1 0 0 Permutation equivalent Applying P to a generator matrix rearrange the Two linear codes C and 1 columns of the generator C are permutation 2 matrix. equivalent provided there is a permutation matrix P such that G1 is a generator matrix of C1 if and only if G1P is a permutation matrix of C2. A linear (n; k) code C is Each row contains a pivot, a 1st permutation equivalent nonzero number from the left, to a code which has always strictly to the right of the generator matrix in pivot of the row above. systematic form. Multiply every row so that the Apply elementary row pivot is 1, repeat row operations operations to any k × n so that zeroes are introduced generator matrix of C to above pivots. Because of the pivot transform it into row positions, all the columns of Ik are echelon form: every row present, permute then. will be nonzero (it has rank k). A linear (n; k) code C is Row echelon form (pivots are 1): permutation equivalent 21 0 1 0 03 to a code which has ! 40 1 1 0 15 generator matrix in 0 0 0 1 1 standard form. All the columns of Ik are present, permute then. Consider the following 21 0 0 1 03 generator matrix (k = 3 ! 0 1 0 1 1 over ): 4 5 F2 0 0 1 0 1 21 0 1 0 03 (here best to permute rather than 0 1 1 0 1 4 5 removing 0 above pivots.) 1 0 1 1 1 Monomial matrix Over F2, every monomial matrix is a permutation A square matrix with matrix: exactly one nonzero 20 1 03 entry in each row and 0 0 1 column. 4 5 1 0 0 20 a 03 M = 40 0 b5 c 0 0 Every monomial matrix can be written either as 2a 0 03 20 1 03 0 0 DP or PD , D; D are DP = 40 b 05 40 0 15 diagonal matrices, P is 0 0 c 1 0 0 a permutation matrix. 20 1 03 2c 0 03 0 2 3 PD = 40 0 15 40 a 05 0 a 0 1 0 0 0 0 b M = 40 0 b5 c 0 0 Monomial matrix: convention 2a 0 03 20 1 03 (x1; x2; x3) 40 b 05 40 0 15 We usually choose 0 0 c 1 0 0 M = DP and apply it | {z } scaling on the right of row vectors. 20 1 03 xM = = (ax1; bx2; cx3) 40 0 15 1 0 0 20 a 03 | {z } permutation (x1; x2; x3) 40 0 b5 c 0 0 Monomially equivalent We can also say that C2 = C1M. Monomial equivalence and Two linear codes C1 (with generator matrix permutation equivalence are the same for binary codes. G1) and C2 are monomially equivalent provided there is a monomial matrix M such that G1M is a generator matrix of C2. Monomially Equivalent Codes Example Given the following two generator matrices G1 and G2 over F3 1 0 2 2 G = ; 1 0 1 1 2 21 0 0 03 1 0 1 1 1 0 2 2 60 1 0 07 G2 = = 6 7 ; 0 1 2 1 0 1 1 2 40 0 2 05 0 0 0 2 there is a monomial matrix M such that G1M = G2, so C2 = C1M. Code Equivalences One more equivalence Suppose there exists an element ! which is a zero of X2 + X + 1 (mod 2). Then ! 6= 0; 1, !2 = ! + 1 (mod 2);!3 = !(! + 1) = !2 + ! = 1 (mod 2): F4 + 0 1 !!2 · 0 1 !!2 0 0 1 !!2 0 0 0 0 0 1 1 0 !2 ! 1 0 1 !!2 ! !!2 0 1 ! 0 !!2 1 !2 !2 ! 1 0 !2 0 !2 1 ! 2 Consider F4 = f0; 1; w; w = w + 1g. • σ(0) = 0; σ(1) = 1. • σ(w2) = σ(w)σ(w) = σ(w) + σ(1) = σ(w) + 1. So x = σ(w) satisfies x2 = x + 1. So σ(w) can be either w or w2. • Thus σ : a + bw 7! a + bw or 2 σ : a + bw 7! a + bw , a; b 2 F2. Automorphisms of Fq A map σ from Fq to itself, such that σ(x + y) = σ(x) + σ(y), σ(xy) = σ(x)σ(y), σ(0) = 0 and σ(1) = 1. • σ(0) = 0; σ(1) = 1. • σ(w2) = σ(w)σ(w) = σ(w) + σ(1) = σ(w) + 1. So x = σ(w) satisfies x2 = x + 1. So σ(w) can be either w or w2. • Thus σ : a + bw 7! a + bw or 2 σ : a + bw 7! a + bw , a; b 2 F2. 2 Automorphisms of Fq Consider F4 = f0; 1; w; w = w + 1g. A map σ from Fq to itself, such that σ(x + y) = σ(x) + σ(y), σ(xy) = σ(x)σ(y), σ(0) = 0 and σ(1) = 1. • σ(w2) = σ(w)σ(w) = σ(w) + σ(1) = σ(w) + 1. So x = σ(w) satisfies x2 = x + 1. So σ(w) can be either w or w2. • Thus σ : a + bw 7! a + bw or 2 σ : a + bw 7! a + bw , a; b 2 F2. 2 Automorphisms of Fq Consider F4 = f0; 1; w; w = w + 1g. A map σ from Fq to • σ(0) = 0; σ(1) = 1. itself, such that σ(x + y) = σ(x) + σ(y), σ(xy) = σ(x)σ(y), σ(0) = 0 and σ(1) = 1. • Thus σ : a + bw 7! a + bw or 2 σ : a + bw 7! a + bw , a; b 2 F2. 2 Automorphisms of Fq Consider F4 = f0; 1; w; w = w + 1g. A map σ from Fq to • σ(0) = 0; σ(1) = 1. itself, such that • σ(w2) = σ(w)σ(w) σ(x + y) = σ(x) + σ(y), = σ(w) + σ(1) = σ(w) + 1.