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Let $C$ be a $[n, k]$ linear code over $\mathbb{F}_q$.

I want to calculate the covering radius of the Hamming codes.

I have thought the following:

Since the Hamming distance is $3$, the coverig radius will always be $3$.

Am I right?

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    $\begingroup$ We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. $\endgroup$ – D.W. Mar 23 '16 at 22:26
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    $\begingroup$ What have you tried? If you want to know whether you are right, one starting point would be to try some examples and see if it seems to hold true for all values of $n,k,q$ you've tried. Have you tried that? If so, you should put that in the question. If not, in the future you should try that before asking here. $\endgroup$ – D.W. Mar 23 '16 at 22:27
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The Hamming codes are perfect codes. This means that balls of radius $(d-1)/2$ (where $d$ is the minimal distance) centered around the codewords partition the space. In particular, the covering radius is $(d-1)/2$.

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  • $\begingroup$ With space you mean the set of the Hamming codes, right? @YuvalFilmus $\endgroup$ – Evinda Mar 23 '16 at 23:55
  • $\begingroup$ Look up the definition of perfect codes and see what I mean. You don't need me for that. Whenever you have a question, try to locate the information yourself. Think about where it could be found. These skills are more important than the actual material. $\endgroup$ – Yuval Filmus Mar 23 '16 at 23:58

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