1
$\begingroup$

I have the parity matrix for the Extended Hamming Code [16, 11, 4] : \begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix}

The received code I have is:

$r=10?1$ $1001$ $?001$ $1001$

How would I go about finding the most likely code vector that was sent? Normally, if it wasn't the Binary Erasure Channel, I'd just multiply the received code by the parity check matrix transposed, and then find the error position, thus allowing me to figure out the original code-vector. But I'm having a hard time finding information for this related to the BEC.

$\endgroup$

1 Answer 1

2
$\begingroup$

The erased positions must have been $0$ or $1$. So, you can construct $4$ possible received vectors without any erasures in them from the one given to you. Compute the syndrome for all four of them. The one with zero syndrome is the transmitted codeword.

Faster way: if the syndrome of your first hypothesized received vector is zero, you are done. If the syndrome is non-zero, attempt to complete the decoding process. It will either give you the transmitted codeword (and you are done) or it will tell you that a detectable but uncorrectable error has occurred. In the latter case, just flip the two bits that you put into the erased positions and you are done.

$\endgroup$
1
  • $\begingroup$ Thank you! The second way you listed is nice. $\endgroup$
    – pfinferno
    Commented Nov 4, 2015 at 0:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .