HAMMING CODE (ERROR DETECTION AND CORRECTION)MORRIS MANO 5ED ,PAGE NO 313

Question

P1 = XOR of bits (3, 5, 7, 9, 11) = 1 ⊕  1 ⊕  0 ⊕  0 ⊕ 0 = 0
P2 = XOR of bits (3, 5, 7, 10, 11) = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕  0 = 0
P4 = XOR of bits (5, 6, 7, 12) = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
P8 = XOR of bits (9, 10, 11, 12) = 0 ⊕  1 ⊕ 0 ⊕ 0 = 1

my doubt is how to take min-terms of xor bits shown in parenthesis ??

Answer 1

P8  P4  P2  P1     m
0   0   0   0   |   0   
0   0   0   1   |   1   
0   0   1   0   |   2   
0   0   1   1   |   3   
0   1   0   0   |   4   
0   1   0   1   |   5  
0   1   1   0   |   6   
0   1   1   1   |   7   
1   0   0   0   |   8   
1   0   0   1   |   9   
1   0   1   0   |   10  
1   0   1   1   |   11  
1   1   0   0   |   12  
1   1   0   1   |   13  
1   1   1   0   |   14 
1   1   1   1   |   15  

Start counting from 0 to 15  skip (0,1,4,8)as 1 is P1 and 2 is P2  4 is P4 and 8 is P8 in hamming  code generation

P1 = XOR of all the minterm whose P1 is 1   (3, 5, 7, 9, 11)

P2 = XOR of all the minterm whose P2 is 1   (  3, 5, 7, 10, 11)

P2 = XOR of all the minterm whose P4 is 1   ( 5, 6, 7, 12)

P8 = XOR of all the minterm whose P8 is 1   (9, 10, 11, 12)

hence  for  sequence 11000100

P1 = XOR of bits (3, 5, 7, 9, 11) = 1 ⊕  1 ⊕  0 ⊕  0 ⊕ 0 = 0
P2 = XOR of bits (3, 5, 7, 10, 11) = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕  0 = 0
P4 = XOR of bits (5, 6, 7, 12) = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
P8 = XOR of bits (9, 10, 11, 12) = 0 ⊕  1 ⊕ 0 ⊕ 0 = 1