F(A,B,C,D) = Σm(0,5,7,8,9,10,11,14,15) — Quine–McCluskey
core5 prime implicants, 4 essential
Answer
B′ · C′ · D′ + A′ · B · D + A · B′ + A · C
Why this example is worth doing
Minterm 0 has no adjacent 1 anywhere on the map, so it never combines with anything and survives the tabulation as a full four-literal prime implicant. It is trivially essential, being the only thing that covers its own column. The page uses it to make the point that the tabulation is exhaustive rather than greedy: an isolated minterm is not a sign of an error in the working, and a minimal expression can legitimately contain a term with every variable in it.
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 · D |
| 3 | B · C · D |
| 4 | A · B′ |
| 5 | A · C |
| # | Cover |
|---|---|
| 1 | B′ · C′ · D′ + A′ · B · D + A · B′ + A · C |
Warning:
4 term(s), 10 literal(s); 4 essential prime implicant(s).
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 2 | 0 | 0 | 1 | 0 | 0 |
| 3 | 0 | 0 | 1 | 1 | 0 |
| 4 | 0 | 1 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 1 |
| 6 | 0 | 1 | 1 | 0 | 0 |
| 7 | 0 | 1 | 1 | 1 | 1 |
| 8 | 1 | 0 | 0 | 0 | 1 |
| 9 | 1 | 0 | 0 | 1 | 1 |
| 10 | 1 | 0 | 1 | 0 | 1 |
| 11 | 1 | 0 | 1 | 1 | 1 |
| 12 | 1 | 1 | 0 | 0 | 0 |
| 13 | 1 | 1 | 0 | 1 | 0 |
| 14 | 1 | 1 | 1 | 0 | 1 |
| 15 | 1 | 1 | 1 | 1 | 1 |