Abstract Algebra Theory and Applications - Computer Science ...

EXERCISES 131A potential problem with this method of decoding is that we might haveto examine every coset for the received codeword. The following propositiongives a method of implementing coset decoding. It states that we canassociate a syndrome with each coset; hence, we can make a table that designatesa coset leader corresponding to each syndrome. Such a list is calleda decoding table.Proposition 7.16 Let C be an (n, k)-linear code given by the matrix H andsuppose that x and y are in Z n 2 . Then x and y are in the same coset of Cif and only if Hx = Hy. That is, two n-tuples are in the same coset if andonly if their syndromes are the same.Proof. Two n-tuples x and y are in the same coset of C exactly whenx − y ∈ C; however, this is equivalent to H(x − y) = 0 or Hx = Hy. □Example 20. Table 7.6 is a decoding table for the code C given in Example18. If x = (01111) is received, then its syndrome can be computed tobe⎛ ⎞0Hx = ⎝ 1 ⎠ .1Examining the decoding table, we determine that the coset leader is (00010).It is now easy to decode the received codeword.Given an (n, k)-block code, the question arises of whether or not cosetdecoding is a manageable scheme. A decoding table requires a list of cosetsand syndromes, one for each of the 2 n−k cosets of C. Suppose that we havea (32, 24)-block code. We have a huge number of codewords, 2 24 , yet thereare only 2 32−24 = 2 8 = 256 cosets.Table 7.6. Syndromes for each cosetSyndrome Coset Leader(000) (00000)(001) (00001)(010) (00010)(011) (10000)(100) (00100)(101) (01000)(110) (00110)(111) (10100)