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Official PYQ · 2010

GATE CS 2010 Q6 — Boolean Algebra

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The minterm expansion of f(P,Q,R)=PQ+QRˉ+PRˉf(P,Q,R) = PQ +Q \bar{R}+P\bar{R} is

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Answer: A

To find the minterm expansion of the Boolean function f(P,Q,R)=PQ+QRˉ+PRˉf(P, Q, R) = PQ + Q\bar{R} + P\bar{R}, we can expand each term to include all three variables PP, QQ, and RR (where PP is the MSB and RR is the LSB):

  1. Term PQPQ:
    PQ=PQ(R+Rˉ)=PQR+PQRˉPQ = PQ(R + \bar{R}) = PQR + PQ\bar{R}Converting to binary representation:
    • PQR  ⟹  1112=m7PQR \implies 111_2 = m_7
    • PQRˉ  ⟹  1102=m6PQ\bar{R} \implies 110_2 = m_6

  2. Term QRˉQ\bar{R}:
    QRˉ=(P+Pˉ)QRˉ=PQRˉ+PˉQRˉQ\bar{R} = (P + \bar{P})Q\bar{R} = PQ\bar{R} + \bar{P}Q\bar{R}Converting to binary representation:
    • PQRˉ  ⟹  1102=m6PQ\bar{R} \implies 110_2 = m_6
    • PˉQRˉ  ⟹  0102=m2\bar{P}Q\bar{R} \implies 010_2 = m_2

  3. Term PRˉP\bar{R}:
    PRˉ=P(Q+Qˉ)Rˉ=PQRˉ+PQˉRˉP\bar{R} = P(Q + \bar{Q})\bar{R} = PQ\bar{R} + P\bar{Q}\bar{R}Converting to binary representation:
    • PQRˉ  ⟹  1102=m6PQ\bar{R} \implies 110_2 = m_6
    • PQˉRˉ  ⟹  1002=m4P\bar{Q}\bar{R} \implies 100_2 = m_4

Combining all the unique minterms obtained, we get:

f(P,Q,R)=m2+m4+m6+m7=∑m(2,4,6,7)\begin{aligned}f(P, Q, R) &= m_2 + m_4 + m_6 + m_7 \\&= \sum m(2, 4, 6, 7)\end{aligned}

Therefore, the correct option is A.

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