GATE CS 2025 Set 1 Q.42: Boolean Algebra and Logic Minimization

Consider the following four variable Boolean function in sum-of-product form

F(b_3, b_2, b_1, b_0) = \sum(0, 2, 4, 8, 10, 11, 12).

where the value of the function is computed by considering $b_3 b_2 b_1 b_0$ as a 4-bit binary number, where $b_3$ denotes the most significant bit and $b_0$ denotes the least significant bit. Note that there are no don't care terms. Which ONE of the following options is the CORRECT minimized Boolean expression for $F$ ?

$\overline{b_1}\,\overline{b_0} + \overline{b_2}\,\overline{b_0} + b_1 \overline{b_2} b_3$

$\overline{b_1}\,\overline{b_0} + \overline{b_2}\,\overline{b_0}$

$\overline{b_2}\,\overline{b_0} + b_1 b_2 b_3$

$\overline{b_0}\,\overline{b_2} + \overline{b_3}$

Official answer and solution

Official answer: A
From: Official GATE 2025 answer key

Solution

Group the minterms on a 4-variable Karnaugh map to find all prime implicants.

Listing the 4-bit binary representations $(b_3 b_2 b_1 b_0)$ of the minterms $\{0, 2, 4, 8, 10, 11, 12\}$ :

Quad 1 $\{0, 4, 8, 12\} = \{0000, 0100, 1000, 1100\}$ : $b_1 = 0, b_0 = 0 \implies \overline{b_1}\,\overline{b_0}$ (essential because of minterms $4$ and $12$ ).
Quad 2 $\{0, 2, 8, 10\} = \{0000, 0010, 1000, 1010\}$ : $b_2 = 0, b_0 = 0 \implies \overline{b_2}\,\overline{b_0}$ (essential because of minterm $2$ ).
Pair 3 $\{10, 11\} = \{1010, 1011\}$ : $b_3 = 1, b_2 = 0, b_1 = 1 \implies b_1 \overline{b_2} b_3$ (essential because of minterm $11$ ).

Combining these three essential prime implicants gives:

F = \overline{b_1}\,\overline{b_0} + \overline{b_2}\,\overline{b_0} + b_1 \overline{b_2} b_3

Final answer

Option (A)

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 1 (CS1)
Question: 42 of 65
Section: Computer Science and Information Technology

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