Question

Let C be the code generated by the matrix [1 0 0 11 G= 0 1 0 2 over Fz. Lo 0 1 1] (i) How many codewords will have, and why? (ii) Give three distinct codewords of C and find their Hamming weights. (iii) List all the steps required for finding the minimum distance of any code. 7

Answer 1

1 TOO! A @ . G= lo 10zl over IF Since be has lank 3 Ge is a generator matrix of a linear code C. Then e is [4,3) - code (i) Since C is a subspace of F of dimension 3, e is canonically isomorphic to F as a vectos space oves Ę. Since I t 1 = 3 = 27 , 1c1=27. Thus c has 27 codewords (ii) Since epoh sow vectors of G form a basis of e one may take 3 distinct codewords as 6, = (1,0,0,1), C = (0,1,0,2), (3 = (0,0,1,1). Hamming weights of of a codeword c = Number of nonzero which we denote by Wt (C). coo coordinates. *ht Ci) = 2, Ht (C) = 2, W+CC3) = 2. (1) By definition minimum distance of a code c is Id(e) = min {wt (x) : XEC, #0}, There is a characterization of minimum distace of a [n, kJ linear code e interms of its parity check matrix .. Def of parity check matrix of C: Let ċ be an [n, k) = code. A (n-k x n) matrix H is said to be arity check matrix if c= {xEE: HIC=0}, Then an [n, KJ-code c has minimum distance d iff every (d-xcolumns of H are linearly independent and there exists d columns of H which are linearly dependent. There is no such particular way to determine minimum distance of a code. For example: To determine minimum distance of Grassmann code Nogin (1993) used decomposable elements characterization in Hei Exterior product space. 0