GATE CS 2019 Q50 — Boolean Algebra

First, let's simplify the given 4-variable function $f(A, B, C, D) = \Sigma(0, 2, 5, 7, 8, 10, 13, 15)$ using a Karnaugh map:

• Group 1: $\{0, 2, 8, 10\} \implies B'D'$
• Group 2: $\{5, 7, 13, 15\} \implies BD$

Thus, the simplified Boolean expression is:
$$f = B'D' + BD$$

Since all inputs and their complements ($B, B', D, D'$) are freely available, we can express $f$ in Product-of-Sums (POS) form:
$$f = (B' + D)(B + D')$$

Applying De Morgan's Law to express this in terms of NOR operations:
$$f = \left( (B' + D)' + (B + D')' \right)'$$

This can be implemented using exactly 3 two-input NOR gates as follows:

1. $\text{Gate 1} = (B' + D)'$
2. $\text{Gate 2} = (B + D')'$
3. $\text{Gate 3} = (\text{Gate 1} + \text{Gate 2})' = f$