F(A,B,C,D) = Σm(2,3,7,9,11,13) — Quine–McCluskey
core5 prime implicants, 3 essential, 2 minimal covers
Answer
A′ · B′ · C + A′ · C · D + B′ · C · D + A · C′ · D | A′ · B′ · C + A′ · C · D + A · B′ · D + A · C′ · D
2 equally minimal forms exist, all of the same cost: A′ · B′ · C + A′ · C · D + B′ · C · D + A · C′ · D | A′ · B′ · C + A′ · C · D + A · B′ · D + A · C′ · D.
Why this example is worth doing
After the essential prime implicants are pulled out, this chart still has rows left, and it reduces by dominance: a row that covers a superset of another row's remaining columns at no greater cost makes the dominated row unnecessary. The page is careful about which kind of dominance is applied where. Row dominance is always safe; column dominance is safe for finding one minimum but provably discards alternative optima, so it is used in the narrative walkthrough and excluded from the enumeration that reports how many minimal covers exist.
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′ · C |
| 2 | A′ · C · D |
| 3 | B′ · C · D |
| 4 | A · B′ · D |
| 5 | A · C′ · D |
| # | Cover |
|---|---|
| 1 | A′ · B′ · C + A′ · C · D + B′ · C · D + A · C′ · D |
| 2 | A′ · B′ · C + A′ · C · D + A · B′ · D + A · C′ · D |
4 term(s), 12 literal(s); 3 essential prime implicant(s).
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 2 | 0 | 0 | 1 | 0 | 1 |
| 3 | 0 | 0 | 1 | 1 | 1 |
| 4 | 0 | 1 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 0 |
| 6 | 0 | 1 | 1 | 0 | 0 |
| 7 | 0 | 1 | 1 | 1 | 1 |
| 8 | 1 | 0 | 0 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 | 1 |
| 10 | 1 | 0 | 1 | 0 | 0 |
| 11 | 1 | 0 | 1 | 1 | 1 |
| 12 | 1 | 1 | 0 | 0 | 0 |
| 13 | 1 | 1 | 0 | 1 | 1 |
| 14 | 1 | 1 | 1 | 0 | 0 |
| 15 | 1 | 1 | 1 | 1 | 0 |