K-map Σ(1,3,7,11,15)+d, NAND

Given $F(A,B,C,D) = \Sigma m(1,3,7,11,15) + d(0,2,5,8,14)$.

K-map simplification

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | X | 1 | 1 | X |
      +---+---+---+---+
 01   | 0 | X | 1 | 0 |
      +---+---+---+---+
 11   | 0 | 0 | 1 | X |
      +---+---+---+---+
 10   | X | 0 | 1 | 0 |
      +---+---+---+---+

• Quad m(3,7,11,15) → $CD$
• Quad m(1,3,5,7) → $\overline{A}D$

$$F = CD + \overline{A}D = D(C + \overline{A})$$

Check: 3, 7, 11, 15 by $CD$; 1 (with 3, 5, 7) by $\overline{A}D$. Don't-care 5 helps form the $\overline{A}D$ quad. ($\overline{A}\,\overline{B}$ in place of $\overline{A}D$ is an equally small alternative.)

Realization using NAND gates

$$\begin{aligned} F &= \overline{\overline{CD + \overline{A}D}} \\ &= \overline{\overline{CD}\cdot\overline{\overline{A}D}} \end{aligned}$$

A ──[NAND]── A'   (inputs tied)

C, D  ──[NAND]── P1  ─┐
                      ├──[NAND]── F
A', D ──[NAND]── P2  ─┘

Gates: one NAND inverter for $\overline{A}$, two 2-input NANDs and one 2-input output NAND: 4 NAND gates.