K-map Π(0,4,5,8,9,11,15), standard SOP

Given $F(A,B,C,D) = \Pi M(0,4,5,8,9,11,15)$. These cells are 0; the remaining cells are 1:

$$F = \Sigma m(1,2,3,6,7,10,12,13,14)$$

K-map

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

Grouping the 1s:

• Quad m(2,3,6,7) → $\overline{A}C$
• Quad m(2,6,10,14) → $C\overline{D}$
• Pair m(12,13) → $AB\overline{C}$
• Pair m(1,3) → $\overline{A}\,\overline{B}D$

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

Check: 2, 3, 6, 7 by $\overline{A}C$; 10, 14 (with 2, 6) by $C\overline{D}$; 12, 13 by $AB\overline{C}$; 1 (with 3) by $\overline{A}\,\overline{B}D$.

(Grouping the 0s gives the POS form $F = (A+\overline{B}+C)(A+C+D)(\overline{A}+B+C)(\overline{A}+\overline{C}+\overline{D})$, which has more literals, so the SOP form is the simplified answer.)

Standard SOP expression

The standard (canonical) SOP has one 4-literal minterm for each 1-cell:

$$\begin{aligned} F = {} & \overline{A}\,\overline{B}\,\overline{C}D + \overline{A}\,\overline{B}C\overline{D} + \overline{A}\,\overline{B}CD + \overline{A}BC\overline{D} + \overline{A}BCD \\ & + A\overline{B}C\overline{D} + AB\overline{C}\,\overline{D} + AB\overline{C}D + ABC\overline{D} \end{aligned}$$ $$= \Sigma m(1,2,3,6,7,10,12,13,14)$$