Karnaugh

$$ \renewcommand{\T}[1]{\mbox{#1}} \renewcommand{\Tb}[1]{{\bf \mbox{#1}}} \renewcommand{\F}[1]{\overline{\mbox{#1}}} \renewcommand{\Fb}[1]{\overline{{\bf \mbox{#1}}}} \newcommand{\AND}{\!\;\!\cdot\!\;\!} $$

MPOS

The following example assumes you were asked to construct the product-of-sums form for a three-input function $$\Tb{h}$$: $$\begin{align*} \Tb{h}(\T{A},\T{B},\T{C})= \F{A}\AND\F{B}\AND\F{C}&+ \F{A}\AND\F{B}\AND\T{C}+ \F{A}\AND\T{B}\AND\F{C}+ \F{A}\AND\T{B}\AND\T{C}+ \T{A}\AND\F{B}\AND\T{C}=\sum m(0, 1, 2, 3, 5) \end{align*}$$

you would proceed as follows:

• Determine the simplified function for the zeros - let's call it $$\Tb{g}$$. In this case, it turns out

$$\begin{align*} \Tb{g}=\Fb{h}=\F{A}\AND\T{C}+\T{B}\AND\T{C} \end{align*}$$

• Recognize that the function we are looking for is the opposite of $$\Tb{g}$$:

$$\begin{align*} \Tb{h}&=\Fb{g}\\ \Tb{h}&=\F{ $\F{A}\AND\T{C}+\T{B}\AND\T{C}$ } \end{align*}$$

• Use DeMorgan's Theorem $$(\F{$\T{x}+\T{y}$}=\F{x}\AND\F{y})$$ to rewrite the overall sum as a product:

$$\begin{align*} \Tb{h}=\F{$\F{A}\AND\T{C}+\T{B}\AND\T{C}$}&= \F{($\F{A}\AND\T{C}$)}\AND\F{$(\T{B}\AND\T{C})$} \end{align*}$$

• Again use DeMorgan's Theorem $$\left(\F{$\T{x}\AND\T{y}$}=\F{x}+\F{y}\right)$$ to rewrite the component products as sums:

$$ \begin{align*} \Tb{h}= \F{$(\F{A}\AND\T{C})$}\AND\F{$(\T{B}\AND\T{C})$}&= (\T{$\T{A}+\F{C}$})\AND (\T{$\F{B}+\F{C}$}) \end{align*} $$ You should see a pattern - the product-of-sums ends up being products of the sums of the negated variables that make up the sum-of-products of the function defining the zeroes. The product-of-sums form is not obtained from the MSOP!