Simpler sum of products from boolean algebra than from karnaugh map

Question

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?

Note: the second answer was also the answer given.

Answer 1

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: $$A'B' + BC$$