Reduce Truth table with Karnaugh Map

Question

I am trying to reduce a boolean expression with a K-map. But when I make the map I end up with three true values which is not possible to make a map with.

Is there a way to simplify this map?

The original expression was:

Y1 = A1 B1 A2' B2' + A1 B1 A2 B2' + A1 B1 A2' B2

Thanks a lot for the help. The picture shows the truth tables and the k-map

score 1 · Accepted Answer · answered May 26 '16 at 18:24

1

Simplifying the map does not make much sense, I suppose you mean simplify the expression, in which case yes, you can. You would have to make two groups of two ones (trues), getting Y1 = A1B1A2' + A1B1B2'.

answered May 26 '16 at 18:24

Nevado

162
7

Thanks a lot that's exactly what I was looking for. – May 26 '16 at 18:32

score 1 · Answer 2 · answered May 26 '16 at 18:34

1

Thanks to the response for Nevado above I was able to come up with a solution by grouping two adjacent values in two groups.

answered May 26 '16 at 18:34

score 1 · Answer 3 · answered Dec 11 '16 at 14:51

If you are using it in a logical circuit it might be sometimes a surprisingly better approach to use the CNF(POS) instead of DNF(SOP).

If you take the simplified DNF: y1 = a1·b1·¬a2 + a1·b1·¬b2 as is (without any optimization using NANDs or NORs), you'll end up needing two 3-input ANDs, one 2-input OR and two invertors.

It might be better to use the CNF: y1 = a1·b1·(¬a2 + ¬b2), where only one of each logical function is used along with the two invertors.

Circling the larger bunches of 0 results in that:

Reduce Truth table with Karnaugh Map

3 Answers3