GATE CS 2025 Set 2 Q.43: Boolean Algebra and Logic Minimization

Given the following Karnaugh Map for a Boolean function $F(w, x, y, z)$ :

4 by 4 Karnaugh Map for F(w, x, y, z) with row variables wx and column variables yz; rows 1 and 4 are 1, 0, 0, 1 and rows 2 and 3 are 0, 1, 1, 0.

Which one or more of the following Boolean expression(s) represent(s) $F$ ?

$\bar{w}\bar{x}\bar{y}\bar{z} + w\bar{x}\bar{y}\bar{z} + \bar{w}\bar{x}y\bar{z} + w\bar{x}y\bar{z} + xz$

$\bar{w}\bar{x}\bar{y}\bar{z} + \bar{w}\bar{x}y\bar{z} + w\bar{x}yz + xz$

$\bar{w}\bar{x}\bar{y}\bar{z} + w\bar{x}\bar{y}\bar{z} + w\bar{x}\bar{y}z + xz$

$\bar{x}\bar{z} + xz$

Official answer and solution

Official answer: A and D
From: Official GATE 2025 answer key

Solution

Group the four central 1s ( $xz$ ) and the four corner 1s ( $\bar{x}\bar{z}$ ) using standard Gray-code K-map labeling $(00, 01, 11, 10)$ .

With standard Gray-code row/column ordering $00, 01, 11, 10$ for both $wx$ and $yz$ :

Central $2 \times 2$ block of 1s: Lies in rows $wx \in \{01, 11\}$ (where $x = 1$ ) and columns $yz \in \{01, 11\}$ (where $z = 1$ ), which simplifies to $xz$ .
Four corner 1s: Lie in rows $wx \in \{00, 10\}$ (where $x = 0$ ) and columns $yz \in \{00, 10\}$ (where $z = 0$ ). Their individual minterms are: $\bar{w}\bar{x}\bar{y}\bar{z} + w\bar{x}\bar{y}\bar{z} + \bar{w}\bar{x}y\bar{z} + w\bar{x}y\bar{z} = \bar{x}\bar{z}(\bar{w} + w)(\bar{y} + y) = \bar{x}\bar{z}$

Therefore, $F = \bar{x}\bar{z} + xz$ (Option D), which is also identically equal to the uncombined corner-minterm form $\bar{w}\bar{x}\bar{y}\bar{z} + w\bar{x}\bar{y}\bar{z} + \bar{w}\bar{x}y\bar{z} + w\bar{x}y\bar{z} + xz$ (Option A).

Final answer

Options (A), (D)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
Question: 43 of 65
Section: Computer Science and Information Technology

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