NOR from NAND — universal gates
corecrossing between the families
Answer
4 NAND gates realising A′ · B′
Why this example is worth doing
Building one universal gate out of the other takes four gates, which is the price of crossing between the NAND and NOR families. The page notes that no real design does this — you pick a family and stay in it — but that the construction matters as a proof: it shows the two families have exactly the same expressive power, so the choice between them is an engineering one and never a question of what can be built.
Try your own input in the Universal gates. Rebuild any of the other gates using only NAND, or only NOR, with the gate count.
How the answer is reached
Gate list
| Node | Gate | Inputs |
|---|---|---|
| g1 | nand | A, A |
| g2 | nand | B, B |
| g3 | nand | g1, g2 |
| g4 | nand | g3, g3 |
NAND-only realisation
A ↑ A ↑ (B ↑ B) ↑ (A ↑ A ↑ (B ↑ B))A′ · B′— 4 NAND gates, verified by reading the network back out.
Truth table
| # | A | B | A ↑ A ↑ (B ↑ B) ↑ (A ↑ A ↑ (B ↑ B)) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 0 |