# Is a Karnaugh map always a good way to simplify a Boolean expression?

I want to simply the following expression:

(NOT A AND NOT B AND NOT C) OR (NOT A AND B) OR (A AND B AND NOT C) OR (A AND C)

I created a truth table first, then a K map from the truth table. (See below.)

As far as I can tell, the simplest expression I can derive purely from the K-map is:

(A AND NOT C) OR B OR (A AND C)

But I can reduce this further with Boolean algebra to:

A OR B

Is it fair to say that a K-map will not always give the best possible solution?

• Something is wrong here, your truth table doesn't match what you're saying: In the first row A=B=C=0 and F=1, but with A OR B it should be F=0. (Similar problem in the fifth row.) Commented Aug 13, 2016 at 10:51
• Your derived expression is incorrect. It's (!A*!C)+B+(A*C)
– W5VO
Commented Aug 13, 2016 at 10:56
• @Michael The truth table and k map look right to me.
– W5VO
Commented Aug 13, 2016 at 11:05
• When I look at the K-map again, I can see the top left group of two ones represents (NOT A AND NOT C). So the final expression is indeed (NOT A AND NOT C) OR B OR (A AND C) and this agrees with Woolfram Alpha's calculator. I think my truth table does represent the original expression and the K-map reflects the truth table. I interpreted the K-map wrongly. Sloppy of me. I feel silly now. Commented Aug 13, 2016 at 11:57