The matrix G I4 A where G 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
Solution
The minimum distance of Hamming’s (7,4) code is three.
Hence the 1-spheres centered on the codewords are disjoint.
Now the Hamming 1 sphere of any codeword must contain
7 + 7
There are 16 codewords.
Hence the 16 Hamming 1-spheres with center at each of the 16
Code words must contain a total of 16X8=128.
0 1 1 1 1 0 0
H=10 1 1 0 1 0
1 10 1 0 0 1
Linear equations of Hamming’s (7,4) code:
C2+c3+c4+c5=0
C1+c3+c4+c6=0
C1+c2+c4+c7=0
No column of H can be all zeros,or else an error in the corresponding code vector position would not affect the syndrome and would be undetectable.
All columns of H must be unique.If two columns are Identical errors corresponding to these code word
Locations will be indistinguishable.
A matrix H is called a parity-check matrix for a linear code C if the columns of H form a basis for the dual code Ci.
![The matrix G = [I_4 | A], where G = [1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | 0 1 1 1 0 1 1 1 0 1 1 1] is a generator matrix in standard form for a [7, 4] that we den The matrix G = [I_4 | A], where G = [1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | 0 1 1 1 0 1 1 1 0 1 1 1] is a generator matrix in standard form for a [7, 4] that we den](/WebImages/36/the-matrix-g-i4-a-where-g-1-0-0-0-0-1-0-0-0-0-1-0-0-0-0-1-1108392-1761587124-0.webp)