When Arthur Has Neither Random Coins Nor Time to Spare: Superfast Derandomization of Proof Systems

What is the actual cost of derandomization? And can we get it for free? These questions were recently raised by Doron et. al (FOCS 2020) and have been attracting considerable interest. In this work we extend the study of these questions to the setting of derandomizing interactive proofs systems.

First, we show conditional derandomization of MA and of AM with optimal runtime overhead, where optimality is under the #NSETH assumption. Specifically, denote by AMTIME^{[⇌ c]}[T] a protocol with c turns of interaction in which the verifier runs in polynomial time T. We prove that, for every constant є>0,

MATIME[T] ⊆ NTIME[T^2+є] ,  

AMTIME^[⇌ c][T] ⊆ NTIME[n· T^{⌈ c/2 ⌉ + є}] ;  

assuming the existence of properties of Boolean functions that can be recognized quickly from the function’s truth-table such that functions with the property are hard for proof systems that receive near-maximal amount of non-uniform advice.

To obtain faster derandomization, we introduce the notion of a deterministic effective argument system. This is an NP-type proof system in which the verifier is deterministic, and the soundness is relaxed to be computational, as follows: For every probabilistic polynomial-time adversary P, the probability that P finds an input x∉ L and misleading proof π such that V(x,π)=1 is negligible.

Under strong hardness assumptions, we prove that any constant-round doubly efficient proof system can be compiled into a deterministic effective argument system, with essentially no time overhead. As one corollary, under strong hardness assumptions, for every є>0 there is a deterministic verifier V that gets an n-bit formula Φ of size 2^o(n), runs in time 2^{є · n}, and satisfies the following: An honest prover running in time 2^O(n) can print, for every Φ, a proof π such that V(Φ,π) outputs the number of satisfying assignments for Φ; and for every adversary P running in time 2^O(n), the probability that P finds Φ and π such that V(Φ,π) outputs an incorrect count is 2^−ω(n).