F(A,B,C,D) = Σm(0,2,8,10) — Karnaugh map
corefour-corner wrap group
Answer
B′ · D′
Why this example is worth doing
The four corners of the map, which are all adjacent to each other because the map is a torus: the left edge wraps to the right and the top to the bottom. They form one legal quad giving B̄·D̄. This is the grouping students miss most often, because on paper the corners look maximally far apart. The tool draws the group as four segments of one rectangle rather than four separate boxes, so the wrap is visible, and the grading rule accepts the selection as a single group rather than four singletons.
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 | B′ · D′ |
| # | Cover |
|---|---|
| 1 | B′ · D′ |
Warning:
1 term(s), 2 literal(s); 1 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 | 1 |
| 3 | 0 | 0 | 1 | 1 | 0 |
| 4 | 0 | 1 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 0 |
| 6 | 0 | 1 | 1 | 0 | 0 |
| 7 | 0 | 1 | 1 | 1 | 0 |
| 8 | 1 | 0 | 0 | 0 | 1 |
| 9 | 1 | 0 | 0 | 1 | 0 |
| 10 | 1 | 0 | 1 | 0 | 1 |
| 11 | 1 | 0 | 1 | 1 | 0 |
| 12 | 1 | 1 | 0 | 0 | 0 |
| 13 | 1 | 1 | 0 | 1 | 0 |
| 14 | 1 | 1 | 1 | 0 | 0 |
| 15 | 1 | 1 | 1 | 1 | 0 |