F(A,B,C) = Σm(0,2,5,7) — Karnaugh map
corecheckerboard; only 2-cell groups exist
Answer
A′ · C′ + A · C
Why this example is worth doing
The 1s sit in a checkerboard, so no four of them are ever adjacent and the largest legal group is a pair. Two terms of two literals is the floor. The point of showing it is that map area is not the same as grouping potential: four 1s that happen to be spread diagonally are worth less than four in a row, and eyeballing the count tells you nothing. The function is A ⊙ C in disguise — B is irrelevant — which the page notes, and which explains why XOR-shaped functions map so badly.
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′ · C′ |
| 2 | A · C |
| # | Cover |
|---|---|
| 1 | A′ · C′ + A · C |
Warning:
2 term(s), 4 literal(s); 2 essential prime implicant(s).
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 0 |
| 5 | 1 | 0 | 1 | 1 |
| 6 | 1 | 1 | 0 | 0 |
| 7 | 1 | 1 | 1 | 1 |