K-map Σm(0,1,2,4,7,8,9,10,12,15)+d, NOR-only

Given $F(A,B,C,D) = \Sigma m(0,1,2,4,7,8,9,10,12,15) + d(5,11,13)$.

K-map simplification (grouping 1s)

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

• Octet m(0,1,4,5,8,9,12,13) → $\overline{C}$
• Quad m(0,2,8,10) → $\overline{B}\,\overline{D}$
• Quad m(5,7,13,15) → $BD$

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

Variable A drops out completely.

POS form for NOR realization

NOR gates give a POS expression directly (NOR-NOR = OR-AND). The 0s are at cells 3, 6 and 14; grouping them (with don't-care 11):

• Pair of 0s M(3,11) → $(B+\overline{C}+\overline{D})$
• Pair of 0s M(6,14) → $(\overline{B}+\overline{C}+D)$

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

Check: the SOP and POS forms agree on every cell; e.g. cell 6 (B=1, C=1, D=0) makes the second sum term 0, so F = 0.

NOR-only circuit

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

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

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

Gates: 3 NOR inverters, 2 three-input NORs and 1 two-input NOR: 6 NOR gates. (Implementing the SOP form instead would need more gates.)