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1 votes
We have to transfer m message bits,  and employ error correction. For that purpose let no of parity  bits added be p. Now error can occur in m message bits or p parity bits or no error therefore total

m+p+1 possibilities , one bit error at any m+p places because parity bit can also get corrupted and  + 1 if there is no error. so

2^p >=m+p+1

coz after receiving parity bits using those p bits we must be able to identify all these possibilities.

eg

0    0    0     ( no change  in p4,p2,p1 after calculating at receiver end)    it means  no error in data and parity

If its

1   0    1   means error in pos 5 (  p1 covering 1,3,5,7 and p4 covering 4,5,6,7

Both doesnt match means error at pos 5)

Thus with 3 bits u can cover 8 possibilities
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4 votes
Actually location of parity bits are fixed at the power of $2'$s

as $2^0,2^1,2^2,2^3.....$

so if I say there are $3$ parity bits at location $1,2,4$ then we can have maximum of hamming code till $7$th or $(2^3-1)$ location, as soon as we get 8th bit then it should be for parity bit , then we get 4 parity bit.

so hamming code bits , $m+r \leq 2^r-1$

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