Question 60

We are given the 4-variable function:

Grouping the minterms in a 4-variable K-map:

• Minterms $0, 2, 8, 10$ form a 4-corner group corresponding to $B'D'$.
• Minterms $5, 7, 13, 15$ form a 4-cell group corresponding to $BD$.

Thus, the simplified Boolean expression is:

Notice that variables $A$ and $C$ are redundant and do not affect the output.

We are given that all inputs and their complements ($B, B', D, D'$) are available. A 2-input NOR gate computes $\text{NOR}(X, Y) = (X + Y)' = X'Y'$.

We can synthesize $f$ using three 2-input NOR gates as follows:

• Gate 1: $\text{NOR}(B, D') = (B + D')' = B'D$
• Gate 2: $\text{NOR}(B', D) = (B' + D)' = BD'$
• Gate 3: $\text{NOR}(\text{Gate 1}, \text{Gate 2}) = (B'D + BD')' = (B \oplus D)' = B'D' + BD = f$

All 3 gates are 2-input NOR gates, and every gate input is an available literal or an output of a preceding gate. Hence, the minimum number of 2-input NOR gates required is 3.