F = A·B·C·D + Ā·B̄·C̄·D̄ — SOP and POS
exam standard2 minterms, 14 maxterms
Answer
Σm(0,15) = ΠM(1,2,3,4,5,6,7,8,9,10,11,12,13,14)
Why this example is worth doing
Two minterms over four variables, so the canonical POS has fourteen factors — a lopsidedness that is itself the lesson. The canonical form's size depends on how many rows the function is true on, not on how complicated it looks, and a function with a two-term SOP can have a fourteen-term POS. The page uses it to justify the practical rule of thumb: expand toward whichever canonical form has fewer terms, and get the other by complementing the index set.
Try your own input in the SOP & POS canonical forms. Expand to canonical minterms and maxterms, and convert between the two.
How the answer is reached
Canonical sum of products
A′ · B′ · C′ · D′ + A · B · C · D
Canonical product of sums
(A + B + C + D′) · (A + B + C′ + D) · (A + B + C′ + D′) · (A + B′ + C + D) · (A + B′ + C + D′) · (A + B′ + C′ + D) · (A + B′ + C′ + D′) · (A′ + B + C + D) · (A′ + B + C + D′) · (A′ + B + C′ + D) · (A′ + B + C′ + D′) · (A′ + B′ + C + D) · (A′ + B′ + C + D′) · (A′ + B′ + C′ + D)
Truth table
| # | A | B | C | D | F |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 2 | 0 | 0 | 1 | 0 | 0 |
| 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 | 0 |
| 9 | 1 | 0 | 0 | 1 | 0 |
| 10 | 1 | 0 | 1 | 0 | 0 |
| 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 | 1 |