K-map Σ(1,2,3,8,9,10,11,13,14)+d, POS

Given $F(A,B,C,D) = \Sigma m(1,2,3,8,9,10,11,13,14) + d(0,4,7,12)$. The 0-cells are 5, 6, 15.

K-map

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

Simplified SOP (grouping 1s)

• Octet m(0,1,2,3,8,9,10,11) → $\overline{B}$
• Quad m(8,9,12,13) → $A\overline{C}$
• Quad m(8,10,12,14) → $A\overline{D}$

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

Simplified POS (grouping 0s)

Treat don't-cares 4 and 7 as 0 so the 0s form large groups:

• Quad of 0s M(4,5,6,7) → $(A+\overline{B})$
• Pair of 0s M(7,15) → $(\overline{B}+\overline{C}+\overline{D})$

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

Check: cell 5 (A=0, B=1) makes $A+\overline{B} = 0$; cell 15 (B=C=D=1) makes the second factor 0. All 1-cells give both factors = 1.

Standard POS expression

The standard (canonical) POS is the product of the maxterms where F = 0. For the simplified POS, the 0s are cells 4, 5, 6, 7 (from $A+\overline{B}$) and 7, 15 (from $\overline{B}+\overline{C}+\overline{D}$):

$$F = \Pi M(4,5,6,7,15)$$ $$\begin{aligned} F = {} & (A+\overline{B}+C+D)(A+\overline{B}+C+\overline{D})(A+\overline{B}+\overline{C}+D) \\ & (A+\overline{B}+\overline{C}+\overline{D})(\overline{A}+\overline{B}+\overline{C}+\overline{D}) \end{aligned}$$

In terms of the given data only, $F = \Pi M(5,6,15)\cdot d(0,4,7,12)$; the simplification chose don't-cares 4 and 7 as 0.