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Computer Science20255 marksCase basedBoolean AlgebraTool: Boolean algebra solver

Reduce the following using a Karnaugh map.

F(P, Q, R, S) = (P + Q + R + S) · (P + Q + R + S') · (P + Q + R' + S) · (P + Q' + R + S) · (P + Q'…

$F(P, Q, R, S) = (P + Q + R + S) \cdot (P + Q + R + S') \cdot (P + Q + R' + S) \cdot (P + Q' + R + S) \cdot (P + Q' + R + S') \cdot (P + Q' + R' + S) \cdot (P + Q' + R' + S') \cdot (P' + Q + R + S) \cdot (P' + Q + R + S')$
(a)[4]
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]
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

Open in the Boolean algebra solver

Answer

Answer (a)

Official answer key

Groups: (0, 1, 8, 9) → (Q + R); (4, 5, 6, 7) → (P + Q'); (0, 2, 4, 6) → (P + S)

Reduced expression: F = (Q + R)(P + Q')(P + S)

Three quads of 0s: (0,1,8,9) gives Q + R, (4,5,6,7) gives P + Q', (0,2,4,6) gives P + S.

Answer (b)

Official answer key
Logic 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).