Question

Q3. In a (7,4) Hamming Code, three parity bits p1, p2, p3 are added to four data bits dl, d2, d3, and d4, and the coverage of each parity bit is as shown in the table below: Bit position 2 3 4 5 6 7 Encoded data bits p1 p2 di p3 d2 d3 d4 da X p1 X X X x Parity bit coverage p2 х X X p3 X X X х 1) (3 pts) Assume even parity is used. If a code is received such that d1 -0,d2=1, d3=1, 04-0, and p2 = 1. Is p2 in parity violation? Briefly explain why. (ii) (3 pts) Assume another code is received such that pl and p2 are found to be in parity violation, while p3 is not. Assume there is only a single-bit error, which bit should be corrected?

Answer 1

Even Parity means that the no. of '1's in the given set of bits should be even.

i.) Now as given d1=0, d2=1, d3=1, d4=0 and p2=1 . To check whether in a voilation of Hamming Code or not we need to check a specific set of bits to show even parity for p2 bit and that specific set of bits to check are at position 2, 3, 6 and 7 i.e we need to check bits p2, d1, d3 and d4 to show even parity as it is the Rule of Hamming Code that the parity bit p2 value is determined by the bits that have '1' at 2nd position in their binary representation like 2: 0010 and similarly 3, 6 and 7. Now d1 = 1, d3=1 and d4=0 so no. of '1's is 2 i.e. even and hence value of p2 should be '0' whereas it is '1' in the question thats why it is voilation of Hamming Rule .

ii.) Now to check which bit is wrong or needs to be corrected we need to know how the parity bits values are choosen:

• p1 value is determined by positions whose binary representation contains '1' as least significant bit in their binary representation like 1, 3, 5, 7 contains 1 at position 1 in their binary form such as 1: 0001.
• p2 value is determined by positions whose binary representation contains '1' as second least significant bit in their binary representation like 2, 3, 6, 7 contains 1 at position 2 in their binary form such as 2: 0010.
• p3 value is determined by positions whose binary representation contains '1' as third least significant bit in their binary representation like 4, 5, 6, 7 contains 1 at position 3 in their binary form such as 4: 0100.

Now as p3 value is right so positions 4,5,6 and 7 position bits are correct. Now both values of p1 and p2 are in parity violation means that the same bit is affecting both parity bits and that is bit at position 3 needs to be corrected.

For any further doubt write it in the comments section.....