Investigating the generalised Hamming distance of some code and information transmission of a binary code.

Kale Sampson

Kale Sampson

Answered question

2022-11-17

Investigating the generalised Hamming distance of some code and information transmission of a binary code.
Letting A be an alphabet of size a and S = A n . Meaning the set of all worlds of length n with bits chosen from the alphabet A. We now take w S.
Assume n = 14. How many words in S are of Hamming distance less than or equal to 4 from w?
Secondly, assume n = 14 and a = 2 and that w S is transmitted through a Binary Symmetric Channel with probability p of correct transmission for each individual bit. How would you generalise a formula for the probability that 5 or fewer errors will occur during transmission of w?

Answer & Explanation

Aliya Moore

Aliya Moore

Beginner2022-11-18Added 17 answers

Step 1
To have Hamming distance 4, we pick those positions that are intended to be different from w and purposely choose them to be different from w at that position, of which we have a 1 choices, hence the desired answer is
i = 0 4 ( 14 i ) ( a 1 ) i .
Step 2
Similarly, for the case where n = 14, a = 2, and p being the probability of success, we choose the position of mistakes. At those positions, we multiply by probability 1 p and at the correct positions, we multiply by p. Hence the expression:
i = 0 5 ( 14 i ) p 14 i ( 1 p ) i .
That is your idea was right, just write down 6 terms.

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