4

My research is to propose highly secure MPC protocol with some conditions.

Especially, I want to consider that

  1. security against malicious (active) adversary
  2. dishonest majority setting
  3. information-theoretic security

I know SPDZ family that achieve 1 and 2 above and some protocols that achieve 1 and 3 above.

Could you tell me the MPC that achieves 1, 2 and 3 above? I want to know this type of MPC as an example, however, I cannot find it.

I think this type of MPC cannot exist if it doesn't have very strong conditions.

The answer "I don't know this type of MPC" is also welcome.

shoy700
  • 43
  • 3

1 Answers1

7

What you are asking for is not possible, not even if you ask for passive security. Here is a sketch of a proof.

Suppose for sake of contradiction you have an $n$-party MPC protocol $\Pi$ for $f(x_1, \ldots, x_n) = x_1 \land x_n$, where the $x_i$'s are bits. This protocol is secure against a computationally unbounded adversary who can passively corrupt $\ge n/2$ parties.

You can take this $n$-party protocol $\Pi$ and construct a related 2-party protocol $\Pi^*$. Player 1 in $\Pi^*$ plays the role of parties 1 through $n/2$ in $\Pi$, and Player 2 in $\Pi^*$ plays the role of parties $n/2+1$ through $n$ in $\Pi$. So $\Pi^*$ is a 2 party protocol that takes input $x_1$ for Party 1 and $x_n$ for Party 2 and computes $x_1 \land x_n$.

The 2-party protocol $\Pi^*$ is secure against 1 corrupt party. Corrupting 1 party in $\Pi^*$ is like simultaneously corrupting $n/2$ parties in $\Pi$, and by our assumption $\Pi$ is secure in that scenario.

So now we have a 2-party protocol $\Pi^*$ for securely computing the AND of two bits, which is secure against a passive, computationally unbounded adversary (who corrupts 1 party). But this is known to be impossible. If you're asking about perfect security, it was proven in:

The proof was later generalized to statistical security in:

Mikero
  • 14,908
  • 2
  • 35
  • 58