Q.Simplify the Boolean expression Y= $\Sigma m(4,5,7,9,11,12,13,15)+\Sigma d(1,3,8)$ using K-map. Draw the NAND gate equivalent circuit to realize the simplified equation.

The four-variable minterms are plotted with 1, 3 and 8 as don't-cares; an octet and a quad are looped and the result is drawn as a NAND-NAND circuit. …

[!TLDR]

K-map grouping gives $Y=D+B\overline{C}$.

Step 1 — plot the map (variables A, B, C, D; weights 8 4 2 1). Ones at 4,5,7,9,11,12,13,15 and don't-cares (X) at 1, 3, 8.

Step 2 — loop the largest groups:

• Octet (group of 8): every cell with $D=1$ — minterms 1, 3, 5, 7, 9, 11, 13, 15 — is 1 or don't-care, so this octet gives the single-variable term $D$ (eliminates A, B, C).
• Quad (4, 5, 12, 13): all have $B=1$ and $C=0$ → term $B\overline{C}$ (eliminates A and D). This covers the remaining ones 4 and 12 (5 and 13 are already covered by the octet).

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

NAND equivalent circuit: $Y = D + B\overline{C} = \overline{\overline{D}\cdot\overline{B\overline{C}}}$. So:

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