Construct the hamming code for the bit sequence10011101

Bit sequence 10011101

For finding Redundant bits ,we check-

        2^r >= r+m+1

      m is no of bits.

We check from 1,m=8

2>10 which is wrong.

We check from 2,m=8

4>11 which is wrong.

We check from 3,m=8

8>12 which is wrong.

We check from 4,m=8

16>13 which is right

So r=4, there are 4 redundant bits;

r1,r2,r4,r8        

Inserting redundant bits in bit sequence-

1  0  0  1  r8  1  1  0 r4  1 r2  r1

r1 is the first parity bit

r2 is the second parity bit

r4 is the third parity bit

r8 is the forth parity pit

We will use even parity bit-we have to make no of 1's = even , if already no of ones is even, then simply r = 0 else r=1.

for r1, we will check at the positions which consist one's at the first position(1,3,5,7,9,11) .

for r2, we will check at the positions which consist one's at the second position (2,3,6,7,10,11) .

for r4, we will check at the positions which consist one's at the forth position(4,5,6,7,12) .

for r8, we will check at the positions which consist one's at the eighth  position(8,9,10,11,12) .

Calculating r1

as bit no. 3 = 1

     bit no. 5 = 0

     bit no. 7 = 1

     bit no. 9 = 1

     bit no. 11 = 0

to make no. of 1's even we have r1 = 1

For r2

count no of 1's at bit (2,3,6,7,10,11)

No. of 1's are  = 3 (on bit 3,6,7)  

As 3 is odd

for even parity,we have to make it 1

Similarly ,

r4 =1

r8 =0

Hence Following are the redundancy bits.

r1 = 1  

r2 = 1  

r4 = 1  

r8 = 0

So the Hamming code is 100101101111