GATE CS 2024 Set 2 — Question 50

The circuit consists of an AND gate, an OR gate, and an XOR gate. Let the output of the AND gate be $A$.
$A = f1 \cdot f2$
$f1 = \sum(0,2,3,5,7,8,11,13)$
$f2 = \sum(1,3,5,7,11,13, 15)$
The intersection of minterms is:
$A = \sum(3, 5, 7, 11, 13)$ Let the output of the OR gate be $B$.
$B = f3 + f4$
$f3 = \sum(0,1,4,11)$
$f4 = \sum(0,2,6,13)$
The union of minterms is:
$B = \sum(0, 1, 2, 4, 6, 11, 13)$ $Y = A \oplus B$
The XOR operation includes minterms present in $A$ or $B$ but not both (Symmetric Difference).

Minterms in $A$: $\{3, 5, 7, 11, 13\}$
Minterms in $B$: $\{0, 1, 2, 4, 6, 11, 13\}$

Common minterms ($A \cap B$): $\{11, 13\}$
Union ($A \cup B$): $\{0, 1, 2, 3, 4, 5, 6, 7, 11, 13\}$

Removing common minterms from the union:
$Y = \{0, 1, 2, 3, 4, 5, 6, 7\}$

So, in sum-of-minterms form:
$Y = \sum(0, 1, 2, 3, 4, 5, 6, 7)$
This matches Option (C). The maxterms correspond to the minterms not present in the function.
The missing minterms from the set $\{0, ..., 15\}$ are $\{8, 9, 10, 11, 12, 13, 14, 15\}$.
Therefore, $Y = \Pi(8, 9, 10, 11, 12, 13, 14, 15)$.
This matches Option (D).Thus, both (C) and (D) are correct representations of the function $Y$.