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I have to solve a problem on Linear Code, I've studied the theory but since I've never saw nor done an exercise on it I'm pretty stuck with lot of different definitions and method and I'm not sure how to solve this problem.

Is given a linear code C, binary, generated from the set of codeword {1011,1101}.Question:

  • Write the control matrix
  • Write all the lateral class of $F_4^2/C$
  • Write all the leaders, in case they're multiple choose one
  • Write the table of leaders and all their sydromes
  • Decode the word 1110 1111 0011
    I wrote the generatrix matrix just using the 2 codeword at the beginning, then I modified the matrix with linear operations to get a standard form.And using that matrix I got the Control Matrix which is at the end

    \begin{matrix} 1 & 1 & 1 & 0\\ 1 & 0 & 0 & 1\\ \end{matrix}

But now I don't know how to continue. Hope someone can help me.

Update1 I have $F_4^2$:
0000 1000 0100 0010 0001 1100 0110 0011 1010 0101 1001 0111 1110 1101 1011 1111
and C:{1011,0000,1101,0110} so $F_4^2/C$: {0000,1000,0100,0010,0001,0110,1011,1101} And I have to find the coset of each of these elements is it correct?

Update 2: $[0000]=(0000; 1011; 1101; 0110) \\ [1000] = (1000; 0011; 0101; 1110) \\ [0100] = (0100; 1111; 1001; 0010) \\ [0001] = (0001; 1010; 1100; 0111)$
The representant of the coset is the leader.

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The proper English terms are: (parity) check matrix (not control matrix) and coset (not lateral class). Your check matrix is fine. Your syndrome is two bits. Your check matrix has all possible non-zero combos of two bits as columns, so it is easy to find coset leaders: place a single one to the position of a column with the desired syndrome (of course, when the syndrome is zero, you have a valid codeword, so the leader has all zeros). The message to be decoded seems to have three words. –  Jyrki Lahtonen Oct 25 '12 at 21:37
    
Sorry, I was translating it from Spanish and Spanish has always a particular translation for everything.I'm trying to finish the exercise and post the next steps to check if I'll work properly –  Laura Oct 28 '12 at 23:37
    
Your Update 2 seems to be correct. In Update 1 (as well as earlier) it should (IMHO) read $F_2^4/C$, because it is more common that $F_4^2$ denotes the space of vectors of length 2 over the field of four elements as opposed to the space of vectors of length 4 over the field of two elements, which is clearly intended here. Sorry about not noticing that earlier. Anyway, the elements of $F_2^4/C$ are exactly the cosets: $[0000]=0000+C=C$, $[1000]=1000+C$, $[0100]$ and $[0001]$. –  Jyrki Lahtonen Oct 30 '12 at 8:42
    
@JyrkiLahtonen: I've actually heard the term control matrix used to refer to a matrix whose rows span the dual code, but are not necessarily linearly independent. (So a check matrix is a control matrix whose rows form a basis of the dual code.) –  Snowball Oct 30 '12 at 19:26
    
@Snowball: can't say I'm overly surprised to hear that. I would freely use the term "check matrix" irrespective of whether there are redundant rows or not (may happen easily e.g. in the case of a BCH code). I guess it is useful to make the distinction in some cases. Thanks for pointing it out! –  Jyrki Lahtonen Oct 31 '12 at 19:54

1 Answer 1

up vote 2 down vote accepted

At the end I think I can answer my own question summing up all the things I've done. The check matrix is written over there and it's correct. The cosets are: $[0000]=(0000; 1011; 1101; 0110) \\ [1000] = (1000; 0011; 0101; 1110) \\ [0100] = (0100; 1111; 1001; 0010) \\ [0001] = (0001; 1010; 1100; 0111)$
aswering the point b)

The representant of the coset is the leader.In the third coset I had 2 members with the same weights so I chose one. answering the point c)
The sydromes are respectively:
$[0 0];[1 1];[1 0];[0 1];$ answering the point d)

Decoding the word $[1110]$: $H*w^t=[11]$ it is in the coset of leader [1000] so $c=(1110)+(1000)=(0110)$
In the same way we can decode the other words which are respectively: $(1011); (1011) $
answering the point e)

share|improve this answer
    
+1 Well done.${}$ –  Jyrki Lahtonen Oct 31 '12 at 9:02

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