K-map Σm(0,2,4,6,8,10,15)+d, NOR

Given $F(A,B,C,D) = \Sigma m(0,2,4,6,8,10,15) + d(3,11,14)$.

K-map simplification

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

• Quad m(0,2,4,6) → $\overline{A}\,\overline{D}$
• Quad m(0,2,8,10) → $\overline{B}\,\overline{D}$
• Quad m(10,11,14,15) → $AC$

$$F = \overline{A}\,\overline{D} + \overline{B}\,\overline{D} + AC$$

Check: 0, 2, 4, 6 by $\overline{A}\,\overline{D}$; 8, 10 by $\overline{B}\,\overline{D}$; 15 by $AC$ (using don't-cares 11 and 14). Don't-care 3 is not needed.

Reduced circuit using NOR gates

For NOR-only realization of an SOP expression, write each product term as a NOR of complemented literals (De Morgan: $\overline{X}\,\overline{Y} = \overline{X+Y}$), feed these to a NOR gate (which gives $\overline{F}$), and invert with one more NOR whose inputs are tied.

$$\begin{aligned} \overline{A}\,\overline{D} &= \overline{A+D} \\ \overline{B}\,\overline{D} &= \overline{B+D} \\ AC &= \overline{\overline{A}+\overline{C}} \\ F &= \overline{\overline{P_1+P_2+P_3}} \end{aligned}$$

A ──[NOR]── A'   (inputs tied)
C ──[NOR]── C'   (inputs tied)

A, D   ──[NOR]── P1  ─┐
B, D   ──[NOR]── P2  ─┤──[NOR]── F'
A', C' ──[NOR]── P3  ─┘

F' ──[NOR]── F   (inputs tied)

Gates: 2 NOR inverters, 3 two-input NORs, one 3-input NOR and one NOR inverter: 7 NOR gates.