F(A,B,C,D) = Σm(0,3,5,6,9,10,12,15) — SOP and POS
edge case8 minterms; XNOR chain
Answer
Σm(0,3,5,6,9,10,12,15) = ΠM(1,2,4,7,8,11,13,14)
Why this example is worth doing
Even parity over four variables: exactly the eight rows with an even number of 1s. In canonical SOP it is eight four-literal terms, in canonical POS eight four-literal factors, and no two-level minimisation reduces either — the function is maximally hostile to SOP and POS alike. As an XNOR chain it is three gates. The page uses it as the closing argument that canonical forms are a normal form for comparison and not a target for implementation, and links to the parity tool.
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 + A′ · B · C′ · D + A′ · B · C · D′ + A · B′ · C′ · D + A · B′ · C · D′ + 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)
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 | 1 |
| 4 | 0 | 1 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 1 |
| 6 | 0 | 1 | 1 | 0 | 1 |
| 7 | 0 | 1 | 1 | 1 | 0 |
| 8 | 1 | 0 | 0 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 | 1 |
| 10 | 1 | 0 | 1 | 0 | 1 |
| 11 | 1 | 0 | 1 | 1 | 0 |
| 12 | 1 | 1 | 0 | 0 | 1 |
| 13 | 1 | 1 | 0 | 1 | 0 |
| 14 | 1 | 1 | 1 | 0 | 0 |
| 15 | 1 | 1 | 1 | 1 | 1 |