F(A,B,C) = Σm(0,1,2,5,6,7) — Karnaugh map
introcyclic chart, no essential prime implicants; 2 minimal covers
Answer
A′ · B′ + B · C′ + A · C | A′ · C′ + B′ · C + A · B
2 equally minimal forms exist, all of the same cost: A′ · B′ + B · C′ + A · C | A′ · C′ + B′ · C + A · B.
Why this example is worth doing
The best three-variable map in the syllabus, because it has no essential prime implicants at all. Every one of its six 1s can be covered two different ways, and the two minimal answers cost exactly the same: three terms of two literals each. A student who groups it correctly and gets a different expression from the back of the book has not made a mistake, and this page says so in as many words. It is also the standard demonstration case for Petrick's method on the Quine–McCluskey page.
Try your own input in the Karnaugh map solver. Group a 2- to 6-variable map yourself and have every group marked right or wrong.
How the answer is reached
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A′ · B′ |
| 2 | A′ · C′ |
| 3 | B′ · C |
| 4 | B · C′ |
| 5 | A · C |
| 6 | A · B |
| # | Cover |
|---|---|
| 1 | A′ · B′ + B · C′ + A · C |
| 2 | A′ · C′ + B′ · C + A · B |
3 term(s), 6 literal(s); 0 essential prime implicant(s).
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 0 |
| 5 | 1 | 0 | 1 | 1 |
| 6 | 1 | 1 | 0 | 1 |
| 7 | 1 | 1 | 1 | 1 |