K-map Σ(0,2,4,5,11)+d, NOR-only

Given $F(A,B,C,D) = \Sigma m(0,2,4,5,11) + d(3,7,12,15)$.

K-map simplification

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

Groups (don't-cares 3, 7, 15 are used for the quad; 12 is not needed):

• Pair m(0,2) → $\overline{A}\,\overline{B}\,\overline{D}$
• Pair m(4,5) → $\overline{A}B\overline{C}$
• Quad m(3,7,11,15) → $CD$

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

Check: 0, 2 by $\overline{A}\,\overline{B}\,\overline{D}$; 4, 5 by $\overline{A}B\overline{C}$; 11 by $CD$. (The minimal POS needs four sum terms, so SOP is simpler here.)

Realization using NOR gates only

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.

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

where $P_1, P_2, P_3$ are the three first-level NOR outputs.

B ──[NOR]── B'   (inputs tied)
C ──[NOR]── C'   (inputs tied)
D ──[NOR]── D'   (inputs tied)

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

F' ──[NOR]── F   (inputs tied)

Gates used: 3 NOR inverters (for $\overline{B}, \overline{C}, \overline{D}$), three first-level NORs, one 3-input NOR and one NOR inverter: 8 NOR gates in all.