For a hamming code,

n = (2^m)-1

k = (2^m)-1-m

n-k = m, where m>=3.

If this condition is satisfied then the code is called hamming code.

Hamming distance is actually defined as the difference in number of elements in the respective locations.

Consider an example:

1101 1111

Each code vector will have a hamming weight. Hamming weight is the number of non zero elements in the code vector.

T = the number of bits up to which the linear block code can correct.

In other words, it can correct the number of bits with in the circle.

If Ci and Cj are together taken, then the correct bits will be:

D (Ci, Cj) >= (2*t)+1

(t+t+1)

t – Ci , t – Cj .

i.e. d (Ci, Cj) >= (2*t)+1

That is d min >= (2*t)+1

d min-1 >= (2*t)

(2*t) <= d min-1

t <= ½*(d min– 1)

d min

The minimum distance (d min) between two codes is the minimum code vector of the linear block codes.

n = (2^m)-1

k = (2^m)-1-m

n-k = m, where m>=3.

If this condition is satisfied then the code is called hamming code.

**Hamming Distance:**Hamming distance is actually defined as the difference in number of elements in the respective locations.

Consider an example:

1101 1111

**Hamming Weight:**Each code vector will have a hamming weight. Hamming weight is the number of non zero elements in the code vector.

Ci, t Cj, t

T = the number of bits up to which the linear block code can correct.

In other words, it can correct the number of bits with in the circle.

If Ci and Cj are together taken, then the correct bits will be:

D (Ci, Cj) >= (2*t)+1

(t+t+1)

t – Ci , t – Cj .

i.e. d (Ci, Cj) >= (2*t)+1

That is d min >= (2*t)+1

d min-1 >= (2*t)

(2*t) <= d min-1

t <= ½*(d min– 1)

d min

The minimum distance (d min) between two codes is the minimum code vector of the linear block codes.

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