GATE CS 2011 Q.14 (CS Q.14): Boolean Algebra and Logic Minimization

The simplified SOP (Sum of Product) form of the Boolean expression $(P + \overline{Q} + \overline{R}) \cdot (P + \overline{Q} + R) \cdot (P + Q + \overline{R})$ is

$(\overline{P} \cdot Q + \overline{R})$

$(P + \overline{Q} \cdot \overline{R})$

$(\overline{P} \cdot Q + R)$

$(P \cdot Q + R)$

Official answer and solution

Official answer: B
From: Official GATE 2011 answer key

Solution

Apply Boolean algebra minimization and distributive laws on the POS expression.

The given expression is $E = (P + \overline{Q} + \overline{R}) \cdot (P + \overline{Q} + R) \cdot (P + Q + \overline{R})$ .

Combine the first two maxterms using the distributive law $(A + \overline{B})(A + B) = A$ where $A = (P + \overline{Q})$ and $B = R$ :

(P + \overline{Q} + \overline{R})(P + \overline{Q} + R) = P + \overline{Q}

Multiply this result with the third term:

E = (P + \overline{Q})(P + Q + \overline{R})

Apply distributive law $(A + B)(A + C) = A + B C$ with $A = P$ :

E = P + \overline{Q}(Q + \overline{R}) = P + \overline{Q} Q + \overline{Q}\,\overline{R} = P + 0 + \overline{Q}\,\overline{R} = P + \overline{Q}\,\overline{R}

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2011 Computer Science and Information Technology (CS)
Question: 14 of 65, printed as CS Q.14
Section: Computer Science and Information Technology

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