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I am struggling to find how can I know the number of errors that can be corrected and detected using (n=10) bit code with a (k=550) codewords. As far as I know, to calculate the number of errors to be corrected and/or detected someone must know the Hamming distance between 2 codewords. However, I do not have any relevant information to find out a solution to this question. I tried to calculate the efficiency of this code using the formula: (E = Number of original message bits / length of the codeword). The efficiency in my case is 1/5. But, I do not know how can I use this information to calculate the number of errors that can be corrected and detected.

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  • $\begingroup$ Welcome to COMPUTER SCIENCE @SE. Can you please spell out how to arrive at E = 1/5? The numbers don't seem to fit. $\endgroup$
    – greybeard
    Dec 2 '20 at 8:49
  • $\begingroup$ E= 2/10 =1/5; since the number of original message bits is 2(binary can be 0 or 1). $\endgroup$ Dec 2 '20 at 10:18
  • $\begingroup$ You misunderstood what is meant by original message bits. This means the main terms that can be used to construct a set of codewords can be either 0 or 1. For example, 0010110010, 1100101100, 0101001110 are considered to be 3 codewords of 10 bits with a 2 as an original message bits. $\endgroup$ Dec 2 '20 at 10:33
  • $\begingroup$ the number of original message bits is 2 that sounds extremely unlikely - there would be exactly four messages possible. Can you quote or at least hyperlink the original problem? $\endgroup$
    – greybeard
    Dec 2 '20 at 20:57
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There are $2^{10}$ binary strings of length $10$. We can partition them into pairs $x_1\ldots x_90,x_1\ldots x_91$. Since your code contains more than $2^9$ codewords, it will contain at least one such pair. Hence the minimum distance of your code is only $1$.

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  • $\begingroup$ If the minimum distance is 1. Therefore, no errors can be corrected and/or detected. Am I right? $\endgroup$ Dec 2 '20 at 13:37
  • $\begingroup$ Right, the code is useless. $\endgroup$ Dec 2 '20 at 14:15

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