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23:  We need a dataword of at least 11 bits. Find the values of k and n in the Hamming
code C(n, k) with dmin :3.

Soln: We need to find k = 2m −1 − m ≥ 11. We use trial and error to find the right
a. Let m = 1 k = 2m −1 − m = 21 −1 − 1 = 0 (not acceptable)
b. Let m = 2 k = 2m −1 − m = 22 −1 − 2 = 1 (not acceptable)
c. Let m = 3 k = 2m −1 − m = 23 −1 − 3 = 4 (not acceptable)
d. Let m = 4 k = 2m −1 − m = 24 −1 − 4 = 11 (acceptable)
Comment: The code is C(15, 11) with dmin = 3.

How this n=15 came?? please explain

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$n=k+m=11+4=15$. where $n$ is the length of the codeword. $k$ is the length of the message and $m$ is the length of the parity bits. For hamming code, $2^{m} = k+m+1$ , here $k\geq 11$ , So, $2^{m} -m-1 \geq 11$ , So, $m=4$. Since, $n$ is defined as $k+m$. So, it will be $11+4=15$