What is the minimum number of 2-input NOR gates required to implement a…
2019
What is the minimum number of 2-input NOR gates required to implement a 4-variable function expressed in sum-of-minterms form as f = (0, 2, 5, 7, 8, 10, 13, 15)? Assume that all the inputs and their complements are available.
Answer: _______________.
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Correct answer: 3
Simplify the function: f = Σ(0, 2, 5, 7, 8, 10, 13, 15).
Using a Karnaugh map the ones form two 2×2 blocks giving the product terms BD and B' D'. Thus the simplified expression is BD + B' D', which is the logical equivalence (B XNOR D).
Minimum 2-input NOR gate implementation (inputs and their complements are available):
Compute p = B NOR D' = ¬(B + D') = B'·D (one 2-input NOR gate).
Compute q = B' NOR D = ¬(B' + D) = B·D' (one 2-input NOR gate).
Compute f = p NOR q = ¬(p + q) = ¬(B'·D + B·D') = B XNOR D (one 2-input NOR gate).
Total number of 2-input NOR gates used: 3
Reason this is minimal:
The simplified function is the sum of two distinct product terms that depend on both B and D in complementary polarities; at least one gate is needed to form each of these product-like signals from the available inputs, requiring at least two gates.
A final gate is required to combine those two intermediate signals into the output. Therefore fewer than three 2-input NOR gates cannot realize the function, so 3 is minimal.
Answer: 3