# How many parties are needed to compute multiplication with BGW?

Let's say we have 3 parties and each one of them has a different secret number. Every party wants to learn the product of all the 3 numbers without learning about the other inputs.

With the BGW protocol: Can we do this with 3 parties, or do we need 5 parties to do this? I am confused because BGW states that we need $$2t+1$$ parties to reconstruct the solution, which would be 5.

• t is the degree of the polynomial that they select. Thus for 3 parties t = 2. Thus 2t+1=5 – macco Nov 11 '19 at 17:15

The BGW protocol has a semi-honest and a malicious version. For semi-honest, a simple honest majority is enough. In that case, 3 parties can run the protocol, with security against one (semi-honest) corrupted party. For malicious adversaries, BGW requires $$t < n/3$$ meaning that with one corrupted party, you need at least 4 parties; with up to two corrupted parties, you need at least 7 parties.