I have an exam for a university course shortly, and upon reviewing one of my assignments I have come to realize that I don't understand why I have lost marks/how to do a couple of questions. Hopefully someone can shed some light on the subject for me! The questions were as follows:
Use K-Maps to simplify the following boolean functions (Note that d() represents a don't care minterm):
1.) F(w, x, y, z) = ∑(1,3,5,7,11,12,13,15)
My answer:
Prime Implicants: yz, w'z, xz, wxy'
Essential Prime Implicants: yz, w'z, wxy'
Possible Minimal Expression(s): yz + w'z + wxy'
Answer sheet (professor's answer):
Prime Implicants: yz, w'z, xz, wxy'
Essential Prime Implicants: Same as prime implicants
Possible Minimal Expression(s): yz + w'z + xz + wxy'
2.) F(w, x, y, z) = ∑(1,2,5,7,12) + d(0,9,13)
My answer:
Prime Implicants: w'x'z', y'z, w'xz, wxy', w'x'y'
Essential Prime Implicants: w'x'z', w'xz, wxy'
Possible Minimal Expression(s): w'x'z' + w'xz + wxy'
Answer sheet (professor's answer):
Prime Implicants: w'x'z', y'z, w'xz, wxy', w'x'y'
Essential Prime Implicants: w'x'z', w'xz, wxy'
Possible Minimal Expression(s): w'x'z' + w'xz + wxy' + y'z
I suppose I should add that I asked my professor after he returned my assignment to me if he had made a mistake and explained my point of view. He seemed pretty certain that he was correct, but couldn't really explain why because he speaks poor English (well, that's university for you..).
Thanks in advance to anyone who can help! This has been quite a task to try to figure out on my own!
1.) You are correct: XY is no essential prime implicant. It does not cover any minterm which is not covered by other prime implicants. Thus, is can be removed from the solution.
The Karnaugh map might help to see this more clearly:
wx
00 01 11 10
+---+---+---+---+
00 | 0 | 0 | 1 | 0 |
+---+---+---+---+
01 | 1 | 1 | 1 | 0 |
yz +---+---+---+---+
11 | 1 | 1 | 1 | 1 |
+---+---+---+---+
10 | 0 | 0 | 0 | 0 |
+---+---+---+---+
I am not sure what is meant by "possible minimal expressions". If you enumerate all potential encircled blocks in the map, XY would also be one.
2.) Your solution and the official solution are the same. Again - like in 1.) - the solution sheet also includes the non-essential terms as "possible minimal expressions".
F = w x y' + w' x z + w' x' z' + w' x' y'