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Math Help - The [5,2] error correcting code

  1. #1
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    The [5,2] error correcting code

    First some theory:

    A Hamming code is defined for numbers [n,k] where n=2^r-1 and k=n-r. Every k bits are encoded into n-bit codewords. Furthermore, every codeword in a Hamming code has a distance of minimum 3 to another codeword, which gives the ability of correcting 1 corrupt bit in each codeword (corrects 1 error per n-bits)

    Having that said, please consider the [5,2] code (a non Hamming code, as per definition above). Every 2 bits are encoded into a 5 bit codeword, i.e.
    <br />
00 \to 00000
    01 \to 01011
    10 \to 10101
    11 \to 11110

    This code have the exact characteristics, as far as I can see, as the Hamming code - the distance between codewords is minimum 3, and it corrects 1 error in 5 bits.

    How come, then, that it is not classified as a Hamming code? Is there any difference that I have missed?
    Last edited by mewoshh; July 22nd 2010 at 07:34 AM. Reason: add info
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  2. #2
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    Red Flag

    Knowing nothing about the subject, here is something that I don't quite get about the definition as it appears here... You gave an example of a [5,2] hamming code, where each 2-bit word is encoded as a 5-bit word. This makes perfect sense, but what is the purpose of r? If k=n-r, then r is 3 in this example, but it also says that n=2^r-1 which is not true (2^3-1=7). Perhaps we need a more detailed definition.

    Also, again this is just a curiosity, but there is nothing magical about these codes, all it does is repeat the 2-bit word twice, so that if any one bit is missing the other occurrence of the word is intact. The fifth bit is arbitrary.

    00 = 00 00 0
    01 = 01 01 1
    10 = 10 10 1
    11 = 11 11 0
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  3. #3
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    Quote Originally Posted by Media_Man View Post
    You gave an example of a [5,2] hamming code, where each 2-bit word is encoded as a 5-bit word. This makes perfect sense, but what is the purpose of r? If k=n-r, then r is 3 in this example, but it also says that n=2^r-1 which is not true (2^3-1=7). Perhaps we need a more detailed definition.
    Quote Originally Posted by mewoshh
    Having that said, please consider the [5,2] code (a non Hamming code, as per definition above).
    I hope this clears things out.
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  4. #4
    MHF Contributor chisigma's Avatar
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    In fact the code proposed by mewoshh is a standard Hamming [7,4] code 'shortened' , i.e. the three redundancy bits are computed for a group of four information bits, two of them 'fixed' to 0...

    Kind regards

    \chi \sigma
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