Simpler sum of products from boolean algebra than from karnaugh map

I was given a question, simplify the expression represented by the sum of minterms 0,1,3 and 7 for the 3 parameter function f. Writing this out I got $$f = A'B'C' + A'B'C + A'BC + ABC = A'B' + BC$$. However, using a karnaugh map  BC 00 10 11 01 A 0 1 0 1 1 1 0 0 1 0  I found 3 groups of 2 ones, which gave me the expression $$A'B' + A'C + BC$$. I know the two expressions are equivalent, but I was under the impression that a karnaugh map would produce the simplest possible sum of products, which it demonstrably did not do here. Did I make a mistake somewhere or are my assumptions wrong?

• You have a mistake in the Karnaugh map. The term A'B'C is misplaced.
– Jens
Commented Mar 17, 2019 at 22:04
• Whoops, thank you. I had B'C and BC' reversed from my scratch paper. Fixed now.
– Ben
Commented Mar 17, 2019 at 22:13
• The Karnaugh now agrees with the first simplification.
– Jens
Commented Mar 17, 2019 at 22:18

I looked up k-maps and saw that you edited your to reflect the proper Karnaugh map. $$\begin{array}{|c|c|c|c|c|} & B'C' & B'C & BC & BC' \\ \hline A' & 1 & 1 & 1 & 0 \\ A & 0 & 0 & 1 & 0 \end{array}$$
I noticed that whenever neither $$A$$ nor $$B$$ the result is $$1$$. Ergo: $$A'B'$$
Next, I also noticed that whenever $$B$$ and $$C$$ are both true, the result is $$1$$. Ergo: $$A'B' + BC$$