ITPEC FE Morning October 2022 Question 5

Hamming Distance — count positions where two bit strings differ.

Compare 10101 and 11110 bit by bit:

• •Position 1: 1 vs 1 → same
• •Position 2: 0 vs 1 → different
• •Position 3: 1 vs 1 → same
• •Position 4: 0 vs 1 → different
• •Position 5: 1 vs 0 → different

Hamming distance = 3

Why not others:
- (a) 0 — would mean identical strings
- (b) 2 — undercounts by one position
- (d) 5 — would require all bits to differ

Key rule: XOR the strings and count the 1s: 10101 XOR 11110 = 01011 → three 1s → distance is 3.

AI-generated — may contain errors