F(A,B,C) = ΠM(0,1,4,6) — Karnaugh map
exam standardentered as maxterms; minimal POS derived
Answer
A′ · B + A · C
Why this example is worth doing
Entered as a list of maxterms rather than minterms, which is how roughly a third of problem sets phrase the question. The tool converts to the ON-set, maps that, and reports both forms. The subtlety the page insists on: minimal POS is obtained by minimising the OFF-set as a sum of products and applying De Morgan to each resulting term, not by dualising the minimal SOP. Dualising computes f(x̄)′, which is a different function, and it is a common and silent bug in competing tools.
Try your own input in the Karnaugh map solver. Group a 2- to 6-variable map yourself and have every group marked right or wrong.
How the answer is reached
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A′ · B |
| 2 | B · C |
| 3 | A · C |
| # | Cover |
|---|---|
| 1 | A′ · B + A · C |
Warning:
2 term(s), 4 literal(s); 2 essential prime implicant(s).
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 1 |
| 4 | 1 | 0 | 0 | 0 |
| 5 | 1 | 0 | 1 | 1 |
| 6 | 1 | 1 | 0 | 0 |
| 7 | 1 | 1 | 1 | 1 |