F(A,B,C,D) = Σm(4,8,10,11,12,15) + d(9,14) — Quine–McCluskey
core4 prime implicants, 2 essential, 2 minimal covers
Answer
B · C′ · D′ + A · B′ + A · C | B · C′ · D′ + A · D′ + A · C
2 equally minimal forms exist, all of the same cost: B · C′ · D′ + A · B′ + A · C | B · C′ · D′ + A · D′ + A · C.
Why this example is worth doing
Don't-cares behave asymmetrically in Quine–McCluskey and this is the example that shows it. They take part in the combining rounds, so they can enlarge implicants exactly as they enlarge K-map groups — but they get no column in the prime-implicant chart, because nothing obliges you to cover them. The consequence is a rule with teeth: a prime implicant covering only don't-cares must be dropped before the chart is built. Leaving it in yields covers that are not minimal, and that is the known defect in several popular library implementations.
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 | B · C′ · D′ |
| 2 | A · B′ |
| 3 | A · D′ |
| 4 | A · C |
| # | Cover |
|---|---|
| 1 | B · C′ · D′ + A · B′ + A · C |
| 2 | B · C′ · D′ + A · D′ + A · C |
3 term(s), 7 literal(s); 2 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 | 0 |
| 3 | 0 | 0 | 1 | 1 | 0 |
| 4 | 0 | 1 | 0 | 0 | 1 |
| 5 | 0 | 1 | 0 | 1 | 0 |
| 6 | 0 | 1 | 1 | 0 | 0 |
| 7 | 0 | 1 | 1 | 1 | 0 |
| 8 | 1 | 0 | 0 | 0 | 1 |
| 9 | 1 | 0 | 0 | 1 | X |
| 10 | 1 | 0 | 1 | 0 | 1 |
| 11 | 1 | 0 | 1 | 1 | 1 |
| 12 | 1 | 1 | 0 | 0 | 1 |
| 13 | 1 | 1 | 0 | 1 | 0 |
| 14 | 1 | 1 | 1 | 0 | X |
| 15 | 1 | 1 | 1 | 1 | 1 |