F(A,B,C) = Σm(0,1,2,3,4,5,6,7) — Karnaugh map
introconstant 1; the whole map is one group
Answer
1
Why this example is worth doing
Every cell is a 1, so the single legal grouping is the entire map and the answer is the constant 1 — no variables at all. This is the degenerate case that breaks naive implementations, which try to emit a product term with zero literals and print an empty string. It also makes the size rule concrete: groups must be rectangles of 1, 2, 4 or 8 cells, and 8 is allowed on a 3-variable map. The tool draws the eight-cell group rather than refusing to draw anything.
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 | 1 |
| # | Cover |
|---|---|
| 1 | 1 |
Warning:
1 term(s), 0 literal(s); 1 essential prime implicant(s).
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 1 |
| 4 | 1 | 0 | 0 | 1 |
| 5 | 1 | 0 | 1 | 1 |
| 6 | 1 | 1 | 0 | 1 |
| 7 | 1 | 1 | 1 | 1 |