Official PYQ · 2017Set 1 · Q21Minimization (K-Maps etc.)NAT1 mark

GATE CS 2017 Set 1 Q21 — Minimization (K-Maps etc.)

Consider the Karnaugh map given below, where $X$ represents "don't care" and blank represents $0$ .

Assume for all inputs $\left ( a,b,c,d \right )$ , the respective complements $\left ( \bar{a}, \bar{b}, \bar{c}, \bar{d} \right )$ are also available. The above logic is implemented using $2$ -input $\text{NOR}$ gates only. The minimum number of gates required is ____________ .

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Answer: 1

From the given Karnaugh map, we can identify the minterms and don't-care conditions (with $ab$ representing the rows and $cd$ representing the columns):

Minterms (1s): $m_0 (0000)$ , $m_2 (0010)$
Don't cares (X): $m_1 (0001)$ , $m_3 (0011)$ , $m_4 (0100)$ , $m_5 (0101)$ , $m_6 (0110)$ , $m_7 (0111)$
Maxterms (0s): All other cells

To find the minimized expression, we group the 1s and don't-cares into a quad of size 4 using the cells $\{m_0, m_2, m_4, m_6\}$ :

f(a, b, c, d) = \bar{a}\bar{d}

Using De Morgan's Law, this expression can be rewritten as:

f = \overline{a + d}

Since the complements of all inputs are freely available, we can implement this function using a single $2$ -input $\text{NOR}$ gate with inputs $a$ and $d$ .

Thus, the minimum number of gates required is 1.

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Published 10 Oct 2026

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