Q.Simplify the Boolean expression Y= $\Sigma m(0,2,4,8,10)+\Sigma d(12,14)$ using K-map. Draw the NAND gate equivalent circuit to realize the simplified equation.

The four-variable minterms are plotted on a K-map with 12 and 14 as don't-cares, grouped into quads, and the simplified SOP is drawn as a NAND-NAND circuit. …

[!TLDR]

K-map grouping gives $Y=\overline{B}\,\overline{D}+\overline{C}\,\overline{D}$, realised with three NAND gates.

Step 1 — plot the map (variables A, B, C, D; weights 8 4 2 1). Ones at 0, 2, 4, 8, 10 and don't-cares (X) at 12, 14.

Step 2 — loop the largest groups (quads):

• Quad (0, 2, 8, 10): all have $B=0$ and $D=0$ → term $\overline{B}\,\overline{D}$ (eliminates A and C).
• Quad (0, 4, 8, 12): all have $C=0$ and $D=0$ (using don't-care 12 as 1) → term $\overline{C}\,\overline{D}$ (eliminates A and B).

Together these cover every required 1 (0, 2, 4, 8, 10); the don't-care 14 is left unused.

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

NAND equivalent circuit: by double-inversion, $Y=\overline{\overline{\overline{B}\,\overline{D}}\cdot\overline{\overline{C}\,\overline{D}}}$. So:

• NAND gate 1: inputs $\overline{B}, \overline{D}$ → output $\overline{\overline{B}\,\overline{D}}$.
• NAND gate 2: inputs $\overline{C}, \overline{D}$ → output $\overline{\overline{C}\,\overline{D}}$. …