K-map Σ(0,1,4,8,10,11,12)+d, NOR

Given $F(A,B,C,D) = \Sigma m(0,1,4,8,10,11,12) + d(2,3,6,9,15)$.

K-map simplification

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

Groups (don't-cares 2, 3, 9 complete the octet; 6 and 15 are not used):

• Octet m(0,1,2,3,8,9,10,11) → $\overline{B}$
• Quad m(0,4,8,12) → $\overline{C}\,\overline{D}$

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

Check: the octet covers 0, 1, 8, 10, 11 and the quad covers 0, 4, 8, 12.

Realization using NOR gates

Write the product term as a NOR: $\overline{C}\,\overline{D} = \overline{C+D}$. Then

$$\begin{aligned} F &= \overline{B} + \overline{C+D} \\ &= \overline{\overline{\overline{B} + \overline{C+D}}} \end{aligned}$$

So: $P =$ NOR($C$, $D$); $\overline{F}$ = NOR($\overline{B}$, $P$); $F$ = NOR($\overline{F}$, $\overline{F}$).

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

C ──┐
    [NOR]── P ──┐
D ──┘           [NOR]── F' ──[NOR]── F
B' ─────────────┘          (inputs tied)

Total: 4 two-input NOR gates.

(Alternatively, the POS form $F = (\overline{B}+\overline{C})(\overline{B}+\overline{D})$ gives a NOR-NOR circuit, but it needs 6 NOR gates including inverters.)