ITPEC FE Morning April 2022 Question 22
ITPEC FE Morning April 2022 — Question 22 of 80
Boolean Logic — Tape Pattern Validation — identify valid patterns using a negated OR expression.
A valid pattern has at least one hole but not all five. Invalid = all zeros or all ones.
Expression (d): ¬[ (A·B·C·D·E) + (Ā·B̄·C̄·D̄·Ē) ]
- •
A·B·C·D·E=1only when all five = 1 (all holes) - •
Ā·B̄·C̄·D̄·Ē=1only when all five = 0 (no holes) - •OR combines both invalid cases
- •NOT inverts: valid →
1, invalid →0
Verification:
- All holes (11111): inside = 1 → result = 0 ✓
- No holes (00000): inside = 1 → result = 0 ✓
- Mixed (e.g. 10010): inside = 0 → result = 1 ✓
Why not others:
- (a) ¬(Ā·B·C·D·Ē) + (A·B·C·D·E) — negation is only over the first term; gives 1 for the all-ones invalid case
- (b) (A+B+C+D+E)·(Ā+B̄+C̄+D̄+Ē) — close idea but depends on exact negation placement per variable
- (c) (Ā·B̄·C̄·D̄·Ē) + ¬(A+B+C+D+E) — second term (De Morgan) equals Ā·B̄·C̄·D̄·Ē, so this is just the no-holes case doubled
Key rule: To exclude exactly two patterns, OR them together and negate the whole expression.
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