F(A,B,C,D) = Σm(0,1,3,7,8,9,11,15) — Quine–McCluskey
exam standard3 prime implicants, 2 essential
Answer
B′ · C′ + C · D
Why this example is worth doing
Eight minterms that pair off cleanly into two quads, and the resulting expression contains no B at all. Seeing a variable vanish is unsettling the first time and worth an explicit note: it means the function's value never depends on that input, so the synthesised circuit leaves the pin unconnected. The page shows the tabulation reaching a dash in B's position in the very first round for every pair, which is the mechanical signature of an irrelevant input.
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′ |
| 2 | B′ · D |
| 3 | C · D |
| # | Cover |
|---|---|
| 1 | B′ · C′ + C · D |
Warning:
2 term(s), 4 literal(s); 2 essential prime implicant(s).
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 2 | 0 | 0 | 1 | 0 | 0 |
| 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 | 1 |
| 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 | 0 |
| 14 | 1 | 1 | 1 | 0 | 0 |
| 15 | 1 | 1 | 1 | 1 | 1 |