F(A,B,C,D) = Σm(0,1,2,4,5,6,8,9,12,13,14) — Karnaugh map
coreeleven 1s; minimise the OFF-set instead
Answer
C′ + A′ · D′ + B · D′
Why this example is worth doing
Eleven 1s and five 0s. Grouping the 1s directly is slow and error-prone; grouping the five 0s gives the complement in a couple of terms, and De Morgan turns that straight into a product of sums. The rule worth carrying away is that whichever of the ON-set and the OFF-set is smaller is the one to map. The page shows both routes and confirms the minimal SOP obtained each way is the same, which is not obvious when the second route never draws a group around a 1.
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 | C′ |
| 2 | A′ · D′ |
| 3 | B · D′ |
| # | Cover |
|---|---|
| 1 | C′ + A′ · D′ + B · D′ |
Warning:
3 term(s), 5 literal(s); 3 essential prime implicant(s).
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 2 | 0 | 0 | 1 | 0 | 1 |
| 3 | 0 | 0 | 1 | 1 | 0 |
| 4 | 0 | 1 | 0 | 0 | 1 |
| 5 | 0 | 1 | 0 | 1 | 1 |
| 6 | 0 | 1 | 1 | 0 | 1 |
| 7 | 0 | 1 | 1 | 1 | 0 |
| 8 | 1 | 0 | 0 | 0 | 1 |
| 9 | 1 | 0 | 0 | 1 | 1 |
| 10 | 1 | 0 | 1 | 0 | 0 |
| 11 | 1 | 0 | 1 | 1 | 0 |
| 12 | 1 | 1 | 0 | 0 | 1 |
| 13 | 1 | 1 | 0 | 1 | 1 |
| 14 | 1 | 1 | 1 | 0 | 1 |
| 15 | 1 | 1 | 1 | 1 | 0 |