Simplify boolean expression using Karnaugh map

Question

I have the following problem:

Express the following boolean expressions as sums of products and simplify as much as possible using a Karnaugh map

I drew the Karnaugh map and then placed my values in the table as true (First one, B non D meaning 10 and non B and D meaning 01) We then have the following values: 0100,0110,1100,1110 (as A and C can be either 0 or 1). So we get:

We notice we only have one group (which is circled in blue) and then we have:

0100
0110
1100
1110

We see that the only variables that don't modify their values are B and D and therefore we get the following simplified version:

B non D

But this is the answer only for the expression in brackets, without the minus. Any ideas how I can solve it in case I have a minus in front of the expression? How does it change my expression?

My second question is how I should solve it when I have a double negation like this one

When mapping does the first one mean 1111 and the rest 0101, 1101, 0101 and then I solve it the same way? Any ideas? Thank you!

Answer 1

For the first question, the expression without the negation can be called (B XOR D) so XOR with a negation is basically XNOR. it can be represented in sums of products as (BD + B'D')

Answer 2

(A.C.!(!B.!D)) + (!A.C.!(!B.!D)) + (A!C!(!B.!D))+(!A!C!(!B.!D)) Steps

1. = C.!(!B.!D).(A+!A) + !C.!(!B.!D).(A+!A)
2. = C.!(!B.!D).1 + !C.!(!B.!D).1
3. = C.!(!B.!D) + !C.!(!B.!D)
4. = !(!B.!D).(C+!C)
5. = !(!B.!D)
6. = !B+!D