Odd-parity validation: Which data/parity pair correctly implements an ODD-parity data transmission, ensuring the total number of 1s (data + parity) is odd?
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Adata = 1101 1011 parity = 1
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Bdata = 1101 0010 parity = 0
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Cdata = 0001 0101 parity = 1
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Ddata = 1010 1111 parity = 0
Answer
Correct Answer: data = 1101 1011 parity = 1
Explanation
Introduction / Context:Parity bits provide a simple integrity check by constraining the total count of 1s in a codeword. For odd parity, the parity bit is selected so that data plus parity contains an odd number of 1s. This question checks your ability to count 1s correctly and apply the rule.
Given Data / Assumptions:
- Odd parity is used.
- Each option lists 8 data bits and a single parity bit.
- Total ones across data and parity must be odd.
Concept / Approach:Count the number of 1s in the data field. If the count is even, parity must be 1 to make the total odd. If the count is already odd, parity must be 0 to keep the total odd. Verify each option against this rule.
Step-by-Step Solution:
Option A: 1101 1011 → ones = 6; parity = 1 → total = 7 (odd) ✓Option B: 1101 0010 → ones = 4; parity = 0 → total = 4 (even) ✗Option C: 0001 0101 → ones = 3; parity = 1 → total = 4 (even) ✗Option D: 1010 1111 → ones = 6; parity = 0 → total = 6 (even) ✗Verification / Alternative check:A quick check: for odd parity, parity = NOT(even(data_ones)). Only option A uses parity = 1 when data has an even count, satisfying the requirement.
Why Other Options Are Wrong:
B: Even data ones with parity 0 leaves total even.C: Odd data ones with parity 1 yields even total.D: Even data ones with parity 0 yields even total.Common Pitfalls:Miscounting ones; forgetting spaces in visual groups do not affect counting; mixing odd/even rules; assuming parity “fixes” any error (it only detects many single-bit flips).
Final Answer:data = 1101 1011 parity = 1