← 2016 Paper 2
UPSC 2016 Maths Optional Paper 2 Q8c — Step-by-Step Solution
15 marks · Section B
Boolean algebra · Numerical Analysis · asked 15× in 13 yrs · Read the full method →
Question
Let A,B,C be Boolean variables, Aˉ denote complement of A, A+B is an expression for A OR B and A⋅B is an expression for A AND B. Then simplify the following expression and draw a block diagram of the simplified expression, using AND and OR gates.
A⋅(A+B+C)⋅(Aˉ+B+C)⋅(A+Bˉ+C)⋅(A+B+Cˉ).
Technique
Absorption A⋅(A+X)=A kills the three maxterms that contain a bare A; the remaining A⋅(Aˉ+B+C)=A(B+C) by AAˉ=0 and distribution; confirmed by an 8-row truth table.
Solution
Setup. Call the expression
F=A⋅(A+B+C)⋅(Aˉ+B+C)⋅(A+Bˉ+C)⋅(A+B+Cˉ).
We simplify by absorption, then confirm against a truth table.
Step 1 — Absorb the maxterms containing A
The leading factor is A. Any other factor of the form (A+⋯) is absorbed by A in a product, because A⋅(A+X)=A:
A⋅(A+B+C)=A,A⋅(A+Bˉ+C)=A,A⋅(A+B+Cˉ)=A.
So three of the four bracketed factors vanish, leaving only the one without a bare A:
F=A⋅(Aˉ+B+C).
Step 2 — Eliminate the complement
A⋅(Aˉ+B+C)=AAˉ+A(B+C)=0+A(B+C).
Answer
F=A⋅(B+C)=AB+AC.