I came across one question in which the Karnaugh map for some function is given and using it, i have to find a simple function which gets mapped onto that map.
My Attempt:
Corresponding to every shaded square there will be a boolean expression of the form, $0000,0010,1010,..$ etc.
What I did was that I added the minterms corresponding to those expressions which had value $1$ i.e. I added the minterms corresponding to $0000,0010,1010,1000,1011,1111,1101,1110$.
My questions:
First, i am not sure about, what, is actually meant by "$simple function$" here, does it simply mean that i have to add the minterms which have value $1$ and find a simplified expression or do i have to make pairs of ones in the karnaugh diagram and find the minimal form for the function ?
Second,will there be a unique simple function ? or would it depend on the way in which sequence $0's$ and$1's$ are chosen?
In your example (i) on the left:
The four top-row squares are covered by $b_2\land b_4'$. Similarly, the second column from the right is covered by $b_1' \land b_3$. Now, there is only one square left to be covered: $b_1' \land b_2' \land b_3' \land b_4$. But this can be simplified to $b_1' \land b_2' \land b_4$ by omitting $b_3'$ and combining it with the already covered column.
The resulting simple function is a disjunction of conjunctions:
$$(b_2\land b_4') \lor (b_1' \land b_3) \lor (b_1' \land b_2' \land b_4)$$
A similar reasoning for example (ii) on the right
also leads to a sum of three expressions:
$$(b_1 \land b_2) \lor (b_1' \land b_2' \land b_4) \lor (b_1' \land b_2' \land b_3')$$
"Simple" usually means "minimum number of terms and/or literals". The selection of terms is not always unique, but in your examples, there are no other choices which exhibit the same number of terms.