Q.Simplify the Boolean expression Y = $\Sigma m (0, 4, 5, 7)$ using K-map. Realize the simplified expression by using both AND-OR logic and NAND – NAND gates.

The three-variable minterms 0, 4, 5, 7 are plotted on a K-map, grouped, and the result is drawn first as AND-OR logic and then as an equivalent NAND-NAND network. …

[!TLDR]

K-map grouping gives $Y=\overline{B}\,\overline{C}+AC$, realised with two AND gates + one OR gate, or equivalently with three NAND gates.

Step 1 — plot the map (three variables A, B, C; weights 4 2 1). Ones at minterms 0 (000), 4 (100), 5 (101) and 7 (111).

Step 2 — loop the groups:

• Pair (0, 4): both have $B=0, C=0$ (A varies) → term $\overline{B}\,\overline{C}$.
• Pair (5, 7): both have $A=1, C=1$ (B varies) → term $AC$.

Each pair eliminates one variable, and together they cover all four minterms.

Step 3 — simplified expression: $Y = \overline{B}\,\overline{C} + AC$.

AND-OR realization: an AND gate forms $\overline{B}\,\overline{C}$ (inputs $\overline{B}, \overline{C}$) and a second AND gate forms $AC$ (inputs $A, C$); their outputs feed a two-input OR gate giving $Y$. …