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Reduce the Boolean function by using 4-variable Karnaugh map, showing the various groups (i.e…
(a)[4.0]
Reduce the Boolean function $F(P, Q, R, S) = (P + Q + R + S) \cdot (P + Q + R + S') \cdot (P + Q + R' + S) \cdot (P + Q' + R + S) \cdot (P + Q' + R + S') \cdot (P + Q' + R' + S) \cdot (P + Q' + R' + S') \cdot (P' + Q + R + S) \cdot (P' + Q + R + S')$ by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
(b)[1.0]
Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.
Answer
Answer (a)
Official answer keyThe given POS expression is $F(P,Q,R,S) = \pi(0,1,2,4,5,6,7,8,9)$, i.e. these 9 maxterms (0's) are plotted on a 4-variable K-map (rows in order PQ, PQ', P'Q', P'Q; columns in order RS, RS', R'S', R'S).
Groups formed (all quads of 0's):
Quad 1: cells (0,1,8,9) -> $Q + R$
Quad 2: cells (4,5,6,7) -> $P + Q'$
Quad 3: cells (0,2,4,6) -> $P + S$
Reduced POS expression:
$F(P,Q,R,S) = (Q+R) \cdot (P+Q') \cdot (P+S)$
Answer (b)
Official answer keyLogic gate diagram for $F = (Q+R) \cdot (P+Q') \cdot (P+S)$, drawn from the three OR-gate outputs feeding a 3-input AND gate.
Three 2-input OR gates: OR gate 1 has inputs Q and R, giving output (Q+R). OR gate 2 has inputs P and S, giving output (P+S). OR gate 3 has inputs P and Q', giving output (P+Q'). The three OR-gate outputs are wired into the three inputs of a 3-input AND gate, whose output is labelled F(P,Q,R,S) = (Q+R).(P+Q').(P+S).
From ISC 2025 Specimen Computer Science Paper 1, question 4(i).