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Reduce the following using a Karnaugh map.
Reduce the Boolean function by using 4-variable Karnaugh map, showing the various groups (i.e…
$F(P, Q, R, S) = (P + Q + R + S) \cdot (P + Q + R + S') \cdot (P + Q + R' + S) \cdot (P + Q' + R + S) \cdot (P + Q' + R + S') \cdot (P + Q' + R' + S) \cdot (P + Q' + R' + S') \cdot (P' + Q + R + S) \cdot (P' + Q + R + S')$
(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.
Draw: logic gate diagram for the reduced expression
Answer
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).
Check your working with the Boolean algebra solver: the steps law by law, the K-map and the logic circuit for any expression.