F(A,B,C) = Σm(0,1,2,5,6,7) — Quine–McCluskey
intro6 prime implicants, 0 essential, 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 canonical cyclic chart. Every prime implicant covers exactly two minterms and every minterm is covered by exactly two prime implicants, so no row is essential and the usual essential-then-dominance reduction stalls immediately with the chart untouched. That is precisely the situation Petrick's method exists for: write a product of sums over the rows that cover each column, multiply it out, absorb, and read off the cheapest product. This example is where the page introduces the algorithm, because here it is not optional.
Try your own input in the Quine–McCluskey solver. Minimise past the K-map limit with the full tabular method and Petrick’s step.
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 |