K-map Σ(1,2,3,8,10,13)+d, standard SOP

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

K-map simplification

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

The many don't-cares allow three large groups (two octets and one quad):

• Octet m(0,1,2,3,4,5,6,7) → $\overline{A}$
• Octet m(0,1,4,5,8,9,12,13) → $\overline{C}$
• Quad m(0,2,8,10) → $\overline{B}\,\overline{D}$

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

Check: 1, 2, 3 by $\overline{A}$; 8, 13 (and 1) by $\overline{C}$; 10 (and 2, 8) by $\overline{B}\,\overline{D}$. All six minterms are covered; don't-cares 0, 4, 5, 6, 7, 9, 12 are taken as 1, so the 0s are only 11, 14, 15.

Standard SOP expression

The standard (canonical) SOP lists every minterm where the simplified F is 1. Expanding:

• $\overline{A}$ → $m_0$ to $m_7$
• $\overline{C}$ → $m_0, m_1, m_4, m_5, m_8, m_9, m_{12}, m_{13}$
• $\overline{B}\,\overline{D}$ → $m_0, m_2, m_8, m_{10}$

$$F = \Sigma m(0,1,2,3,4,5,6,7,8,9,10,12,13)$$ $$\begin{aligned} F = {} & \overline{A}\,\overline{B}\,\overline{C}\,\overline{D} + \overline{A}\,\overline{B}\,\overline{C}D + \overline{A}\,\overline{B}C\overline{D} + \overline{A}\,\overline{B}CD + \overline{A}B\overline{C}\,\overline{D} + \overline{A}B\overline{C}D + \overline{A}BC\overline{D} \\ & + \overline{A}BCD + A\overline{B}\,\overline{C}\,\overline{D} + A\overline{B}\,\overline{C}D + A\overline{B}C\overline{D} + AB\overline{C}\,\overline{D} + AB\overline{C}D \end{aligned}$$

Without the don't-cares, the required minterms alone are $\Sigma m(1,2,3,8,10,13)$.