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\$ – ratchet freak Nov 7 '18 at 16:06
  • \$\begingroup\$ but without using morgans, is there a way to use K maps with POS \$\endgroup\$ – fred Nov 7 '18 at 16:08
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    \$\begingroup\$ Yes. Highlight 0's. Write OR'ed equations with inverted inputs. Easiest way is to take an existing example and try it. \$\endgroup\$ – StainlessSteelRat Nov 7 '18 at 16:22
  • \$\begingroup\$ thank you. What exactly do you mean by inverted inputs? \$\endgroup\$ – fred Nov 7 '18 at 16:36

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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