From my understanding, to use karnaugh maps to simplify an expression, you would have an SOP expression that does not have to be canonical and then group the 1s in the table.

So that got me wondering. Does this mean I can then have a POS expression and then group the 0s instead of the 1s?

  • \$\begingroup\$ apply de morgan as needed \$\endgroup\$ Nov 7, 2018 at 16:06
  • \$\begingroup\$ but without using morgans, is there a way to use K maps with POS \$\endgroup\$
    – fred
    Nov 7, 2018 at 16:08
  • 2
    \$\begingroup\$ Yes. Highlight 0's. Write OR'ed equations with inverted inputs. Easiest way is to take an existing example and try it. \$\endgroup\$ Nov 7, 2018 at 16:22
  • \$\begingroup\$ thank you. What exactly do you mean by inverted inputs? \$\endgroup\$
    – fred
    Nov 7, 2018 at 16:36

1 Answer 1


Given PoS: \$ f=\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}\$

Becomes SoP: \$ f= (\overline B + \overline{D})\cdot (A + \overline C)\cdot (A + \overline D ) \$

Circle 0's. Write OR'ed equations with inverted inputs.

Intersection of B & D have 4 zeros. This becomes \$ \overline B + \overline D\$

wide demo pic


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