1
$\begingroup$

I have a problem in error-correcting codes.

Say we have a generator matrix of a linear binary code

$$g=\begin{pmatrix} 10011 \\ 01101 \end{pmatrix} $$

Q1: How many different codeword do we have? How many code bits can the code correct?

Say the generator matrix of a $(6,2)$ binary code is $$g=\begin{pmatrix} 101001\\ 011111 \end{pmatrix}$$

Q2: How many errors can it guarantee to detect?

My try:

For Q1:

The way I find the number of codeword is count the number of rows in $g$ and raise it as a power to 2. So the answer is $2^2=4$.

For detecting and correcting I’m not sure, should I find the hamming distance and subtract 1?

I hope someone can explain to me how I can find the number bits I can detect and correct.

$\endgroup$

1 Answer 1

0
$\begingroup$

If the minimal distance is $d$, then

  • Up to $d-1$ errors can be detected. This means that if a sent codeword is modified by flipping at least one and up to $d-1$ bits, then the receiver can tell that some bits were flipped.

Indeed, suppose that $w$ is the codeword that was sent, and that $w'$ is obtained from $w$ by flipping at least one and up to $d-1$ bits. Since $1 \leq d_H(w,w') < d$ (where $d_H$ is Hamming distance), we see that $w'$ is not a codeword (since the distance between any two different codewords is always at least $d$), and so we can detect that errors have occurred.

  • Up to $\lfloor \frac{d-1}{2} \rfloor$ can be corrected. This means that if a sent codeword is modified by flipping at most $\lfloor \frac{d-1}{2} \rfloor$ bits, then the original codeword can be recovered.

Indeed, suppose that $w$ is the codeword that was sent, and that $w'$ is obtained from $w$ by flipping up to $\lfloor \frac{d-1}{2} \rfloor$ bits. I claim that $w$ is the unique codeword at distance at most $\lfloor \frac{d-1}{2} \rfloor$ from $w'$. Indeed, if $z$ were another such codeword then $$ d_H(w,z) \leq d_H(w,w') + d_H(w',z) \leq 2 \lfloor \tfrac{d-1}{2} \rfloor < d, $$ contrary to the definition of minimal distance.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.