Gate-count before simplification For the Boolean expression XY + X(X + Z) + Y(X + Z), determine the minimum number of logic gates required to implement it as written (before algebraic simplification), assuming you may reuse the subexpression (X + Z) and that multi-input OR is available.

Introduction / Context:
Estimating gate count prior to simplification helps designers appreciate the impact of algebraic reduction on area, power, and delay. Even before minimization, smart sharing of common subexpressions can reduce hardware compared to a naive implementation.

Given Data / Assumptions:

• Expression: XY + X(X + Z) + Y(X + Z).
• You may share (X + Z) between terms.
• Assume one 3-input OR gate is available for the final sum; AND/OR gates are otherwise 2-input.

Concept / Approach:
Implement inner subexpressions first, then feed them into outer operators. Reusing (X + Z) avoids duplicating hardware. Finally, sum the three product terms with a single 3-input OR gate.

Step-by-Step Solution:

Verification / Alternative check:
If only 2-input ORs are allowed at the output, you would need two 2-input ORs to sum the three products, making 6 gates. With a shared 3-input OR, 5 is minimal before algebraic reduction.

Why Other Options Are Wrong:

• 1 or 2: Far too few to realize three product terms and a sum.
• 4: Would require eliminating a necessary stage or duplicating an impossible sharing assumption.

Common Pitfalls:
Double-counting (X + Z) by building it twice; forgetting the final OR gate; or assuming simplification beforehand. Always clarify sharing assumptions and gate arity when counting.

Final Answer:
5