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Official PYQ · 2025Set 1 · Q32Minimization (K-Maps etc.)MCQ2 marks

GATE CS 2025 Set 1 Q32 — Minimization (K-Maps etc.)

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

F(b3,b2,b1,b0)=∑(0,2,4,8,10,11,12)F\left(b_{3}, b_{2}, b_{1}, b_{0}\right)=\sum(0,2,4,8,10,11,12)

where the value of the function is computed by considering b3b2b1b0b_{3} b_{2} b_{1} b_{0} as a 44-bit binary number, where b3b_{3} denotes the most significant bit and b0b_{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 FF ?

Choose one option
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Answer: A

To find the minimized Boolean expression for F(b3,b2,b1,b0)=∑(0,2,4,8,10,11,12)F(b_3, b_2, b_1, b_0) = \sum(0, 2, 4, 8, 10, 11, 12), we can plot the minterms on a 4-variable Karnaugh Map (K-map):

b3b2\b1b0b_3b_2 \backslash b_1b_0 00 01 11 10
00 1 (0) 0 (1) 0 (3) 1 (2)
01 1 (4) 0 (5) 0 (7) 0 (6)
11 1 (12) 0 (13) 0 (15) 0 (14)
10 1 (8) 0 (9) 1 (11) 1 (10)

We can group the 1s to find the essential prime implicants:

  • Quad 1: Combining the four corners: minterms (0,2,8,10)(0, 2, 8, 10).
    This yields the term: bˉ2bˉ0\bar{b}_2 \bar{b}_0.
  • Quad 2: Combining the entire first column: minterms (0,4,12,8)(0, 4, 12, 8).
    This yields the term: bˉ1bˉ0\bar{b}_1 \bar{b}_0.
  • Pair 1: Combining minterms (10,11)(10, 11) to cover the remaining 11 at minterm 1111.
    This yields the term: b3bˉ2b1b_3 \bar{b}_2 b_1.

Summing these essential prime implicants gives the minimized expression:

F=bˉ1bˉ0+bˉ2bˉ0+b1bˉ2b3F = \bar{b}_1 \bar{b}_0 + \bar{b}_2 \bar{b}_0 + b_1 \bar{b}_2 b_3

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