# Do Karnaugh maps yield the simplest solution possible?

I'm learning to use a Karnaugh map, but I'm not sure if I obtained the simplest expression possible. Have a look at this example:

Truth table (inputs are A B C; output is F):

A   B   C   |   F
0   0   0   |   1
0   0   1   |   0
0   1   0   |   1
1   0   0   |   0
0   1   1   |   1
1   0   1   |   1
1   1   0   |   0
1   1   1   |   1


The Karnaugh map looks like this:

AB
--
C|

00  01  11  10
--------------
0|  (1  1)  0   (1)
1|  0   (1  1)  (1)


And this yields $\bar{C}\bar{A}+CB+A\bar{B}$. Is there any simpler way of choosing the 1s from the map and getting a simpler result? Are Karnaugh maps guaranteed to always yield the simplest result possible?

The only other way to group the 1's into pairs would be to group the ones in indices 0,2/1,5/and 6,7. All of these are groups of two, like your original solution, and no 1 is left on it's own. This solution would give you a result of equal simplicity. No, this is the minimum representation.

If it were not the minimum representation, you would see terms that differ by only one term in the disjunctive normal form.

Karnaugh maps do not always give the simplest expression possible, but they do always give the simplest "Sum of Products" expression possible (https://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Logic/Logic3.html).