for an information word M with m bits that is coded as following:

  • M is coded into a word A using an unknown code that allows detection of not more than one error.
  • the code word is the word obtained by a concatenation of A to itself. so A becomes AA and etc..

1)is it possible to know the overhead in the given code?

2)is it possible to know the distance of the code?

EDIT: I don't know how to solve it and i've tried several times and apparently i am going to the wrong directions. please solve it if you can so i can learn the correct way of approaching and solving this kind of questions.

thank you very much

my attempt:

after thinking on it again and rereading the text, i began to realize(hopefully i am right now) that the question is about parity bit and not hamming. so if it can detect not more than one error, there's a parity bit, i think. based on that thought i tried to solve it:

1)I began to realize it's not about hamming code. if we can only detect not more than one error, then it's about the parity bit. so i think that if we add a bit to each byte of message, then the overhead will be 9 bytes total. so it can be calculated, i think.

2)since this sub problem is about the concatenation of A to itself, then like @YuvalFilmus said, the minimal distance should be twice now.

is it correct now? would really appreciate your comments and help.

thank you very much

  • $\begingroup$ Does "identification of not more than one error" mean "identification of one error, but not two", or is it possible that the code doesn't allow identification of any errors? Am I correct to interpret identification as detection, not correction? $\endgroup$ Commented Apr 1, 2019 at 9:06
  • $\begingroup$ it means, according to the text, that it can detect not more than one error. it doesn't say anything about correction(in my attempt i assumed that because i thought it means hamming code). so the correct phrasing should be "detection of one error, but not more"? @PeterTaylor $\endgroup$
    – hps13
    Commented Apr 1, 2019 at 9:50
  • $\begingroup$ could anyone please check what i've done and correct me for my mistakes? $\endgroup$
    – hps13
    Commented Apr 2, 2019 at 19:30
  • 1
    $\begingroup$ The minimal distance of the new code is twice that of the original code. $\endgroup$ Commented Apr 6, 2019 at 13:54
  • $\begingroup$ @YuvalFilmus, i am really stuck on the first part and would appreciate help with it. according to the details, if we can detect not more than one error, then there's a parity bit from what i understand. now if i put the parity bit on each byte it's like adding one bit to eight, so this is the overhead? or it cannot be calculated? $\endgroup$
    – hps13
    Commented Apr 8, 2019 at 10:13


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