Problem I:
1101011 , 1010110 , 0000011 , 0001100 , MHD is the number smallest Hamming distance between all possible pairs.
HD (a,b) = 5 , HD (a,c) =3 , HD (a,d) = 5 , HD (b,c) = 4 , HD (b,d) = 4 , HD (c,d) = 4 .
Therefor , MHD = 3.
To increase the MHD by 1.
1101011 should become 1101111 or 1101010. So, now
HD (a,b) = 4 , HD (a,c) =4 , HD (a,d) = 4 , HD (b,c) = 4 , HD (b,d) = 4 , HD (c,d) = 4 .
Thus, MHD = 4.
Problem II:
1. Golden Arches Food
2. Children's TV Show
Problem IV :
a. 11 = 0001 0001 and 26 = 0010 0110 , 11 + 26 = 0011 0111
b. 541 = 0101 0100 0001 and 216 = 0010 0001 0110 , 541 - 216 =0011 0010 0101
c. 27 = 0010 0111 and 36 = 0011 0110 , 27 + 36 = 0110 0011
explanation - in BCD 10,11,12,13,14,15 in binary representation are forbidden so we add 6 whenever the sum goes beyond 9.
Problem V|| :
Problem I (Hamming Distance) a) Compute the Minimum Hamming Distance (MHD) for the coding (1101011, 1010110,...