K-map ΠM(0,2,3,8,10,11,12,15)+d, NOR

Given $y(A,B,C,D) = \Pi M(0,2,3,8,10,11,12,15)$ with don't-cares $d = 7, 13, 14$. The listed maxterms are the 0-cells; the remaining cells 1, 4, 5, 6, 9 are 1s.

K-map

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

Simplified POS (grouping the 0s)

• Quad of 0s M(12,13,14,15) → $(\overline{A}+\overline{B})$
• Quad of 0s M(0,2,8,10) → $(B+D)$
• Quad of 0s M(3,7,11,15) → $(\overline{C}+\overline{D})$

$$y = (\overline{A}+\overline{B})(B+D)(\overline{C}+\overline{D})$$

Simplified SOP (grouping the 1s)

• Quad m(4,5,6,7) → $\overline{A}B$
• Quad m(1,5,9,13) → $\overline{C}D$

$$y = \overline{A}B + \overline{C}D$$

Both forms are correct. The SOP form has only 2 terms and 4 literals, so it gives the smaller NOR circuit.

Reduced circuit using NOR gates

Each product term becomes a NOR of complemented literals:

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

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

A, B' ──[NOR]── P1  ─┐
                     ├──[NOR]── y'
C, D' ──[NOR]── P2  ─┘

y' ──[NOR]── y   (inputs tied)

Gates: 2 NOR inverters, 2 first-level NORs, 1 NOR giving $\overline{y}$ and 1 NOR inverter: 6 two-input NOR gates.

(The POS form could be built as a NOR-NOR circuit, $y = \overline{\overline{\overline{A}+\overline{B}} + \overline{B+D} + \overline{\overline{C}+\overline{D}}}$, but it needs 8 NOR gates including inverters.)