GATE 2020 IN Q.8: solution and answer

Official answer: BSource: Official answer key, GATE 2020 IN (IIT Delhi)

1. Complements are written with $\overline{}$. Draw the 4-variable K-map with rows $AB=00,01,11,10$ and columns $CD=00,01,11,10$. Put 1s at $0,1,2,6,8,9,10,11$ and X (don't-care) at $3,7,14,15$. The remaining cells, $4,5,12,13$, are 0s.

2. Octet 1: rows $AB=00$ and $AB=10$ are adjacent because the map wraps around. Together they hold cells $0,1,3,2,8,9,11,10$. Every one of these is a 1, except cell 3, which is an X and is taken as 1. The only variable that stays constant over the octet is $B=0$, so the term is $\overline{B}$.

3. Octet 2: columns $CD=11$ and $CD=10$ hold cells $3,2,7,6,15,14,11,10$, and each of these is a 1 or an X. The only constant variable is $C=1$, so the term is $C$.

4. Coverage check. $\overline{B}$ covers the minterms $0,1,2,8,9,10,11$, and $C$ covers minterm $6$, so every minterm is covered. Neither group contains a 0-cell, because each of $4,5,12,13$ has $B=1$ and $C=0$.

5. Minimality. No single literal covers every 1: $\overline{B}$ misses minterm 6, and $C$ misses minterm 0. So the minimal SOP is two one-literal prime implicants: $$f=\overline{B}+C$$

6. Check the other options against the map. (A) $A\overline{B}+BC$ misses minterms $0,1,2$, where $A=0$ and $B=0$. (C) $\overline{D}+A$ is 1 at cells 4 and 12, which are 0s. (D) $\overline{A}B+\overline{C}D$ misses minterm 0, and its $\overline{C}D$ term is 1 at cells 5 and 13, which are 0s. Only (B) is correct.