ITPEC FE Morning October 2019 Question 2

Source exam: ITPEC FE Morning October 2019Topic: Basic Theory & Math

ITPEC FE Morning October 2019 — Question 2 of 80

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

Hamming distance = 3

Quick method: XOR the strings → count 1s:
10101 ⊕ 11110 = 01011 → three 1s → distance = 3

Why not others:
- (a) 0 — would mean identical strings

- (b) 2 — undercounts; misses one differing position

- (d) 5 — would require all bits to differ

Key rule: Hamming distance = number of 1s in the XOR of two bit strings.

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