F(A,B,C,D) = Σm(1,3,5,7,9,11,13,15) — Karnaugh map
edge caseone 8-cell group; answer is D
Answer
D
Why this example is worth doing
Every odd minterm, which is exactly the set of rows where D is 1, so the whole function collapses to a single literal and the circuit is a wire. It is the cleanest illustration of what minimisation is for: the canonical form has eight terms of four literals, the minimal form has one literal, and the two are the same function. It also gives the tool's group-grading a clean test, because the eight cells form one octet and any attempt to cover them as four pairs is valid, complete and four times too expensive.
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 | D |
| # | Cover |
|---|---|
| 1 | D |
Warning:
1 term(s), 1 literal(s); 1 essential prime implicant(s).
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 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 | 1 |
| 6 | 0 | 1 | 1 | 0 | 0 |
| 7 | 0 | 1 | 1 | 1 | 1 |
| 8 | 1 | 0 | 0 | 0 | 0 |
| 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 | 1 |
| 14 | 1 | 1 | 1 | 0 | 0 |
| 15 | 1 | 1 | 1 | 1 | 1 |