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Consider the circuit in figure. $f$ implements

1. $\overline{A} \overline{B}C + \overline{A}B \overline{C} + ABC$

2. $A + B + C$

3. $A \oplus B \oplus C$

4. $AB + BC + CA$

### 1 comment

edited
Since $1$ and $2$ both point to $\bar{C}$  it doesn't matter whether $S_{1}$ points to $A$ or $B$ and similarly $S_{0}$ points to $B$ or $A$.

• $0 - C$ will be selected for $A = 0, B=0$.
• $1 - \bar C$ will be selected for $A = 0, B = 1.$
• $2 -\bar C$ will be selected for $A = 1, B = 0.$
• $3 -C$ will be selected for $A = 1, B = 1.$

So, $f = \bar A \bar B C + \bar A B \bar C +A \bar B \bar C + ABC$

$\qquad = \bar A (\bar BC + B \bar C) + A (\bar B \bar C + BC)$

$\qquad = \bar A (B \oplus C) + A (B \odot C)$

$\qquad = \bar A (B \oplus C) + A (\overline{B \oplus C})$

$\qquad = A \oplus B \oplus C$

Correct Answer: $C$

by

See the last part...

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