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I came across interactive proofs and randomized computation, in particular, i read about the complexity classes $\text{IP}, \text{BPP}, \text{RP}$, etc.

Since the above classes are well-known, I will give a hand-wavy definition of $\text{IP}$. Then, I will state my question. Roughly speaking, a language $L \in \text{IP}$ iff there exist a pair of interactive algorithms Prover $P$ and Verifier $V$, with $V$ running in probabilistic polynomial time (in the length of the input $x$: $x$ is common to both $P$ and $V$), such that:

  • If $x\in L$, and the Prover $P$ follows the protocol, then the probability that the Verifier $V$ accepts is 1, and
  • if $x\notin L$, then for any Prover $P$ (even a cheating one), the probability that the Verifier $V$ accepts is at most $\frac{1}{2}$.

I note that $P$ is allowed to be computationally unbounded.

Now to understand the power of interaction, I started to see what happens when restrict or give up the interaction between the prover and the verifier. I read the following in this lecture here, given by Jonathan Katz:

1- Assume that the verifier has input $x$. If the protocol is such that the prover $P$ sends a proof $\pi$ (to the fact that $x\in L$) to the verifier $V$, and then the interaction ends. The verifier has then to verify that $x \in L$, given $\pi$. If the verifier is deterministic and runs in polynomial time in $|x|$. Then, it is not hard to see that such interactive protocol is equivalent to the complexity class $\text{NP}$.

2- Now if we consider the interaction protocol in item 1, and additionally allow the verifier to be probabilistic, then we get more than $\text{NP}$. Specifically, it is known that in this case, any language in $L \in \text{BPP}$ can be handled by the protocol. Note that this actually suggest that we can eliminate the error when $x\in L$.

The protocol in item 2, can be thought of as a randomized version of $\text{NP}$.

Here comes my question: can the protocol given in item 2, handle any language in $\text{IP}$? That is, given a language $L\in \text{IP}$, can we give up the interaction entirely and still be able to verify that $x \in L$ in probabilistic polynomial time with completeness $1$, and soundness at most $\frac{1}{2}$? Intuitively, I think the answer is unknown because randomized complexity classes are conjectured to be contained in $\text{P}$ and we don't even know whether $\text{P}$ is a strict subset of $\text{PSPACE} = \text{IP}$.

Bader Abu Radi
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1 Answers1

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The class you described is MA (interactive proofs consisting of one round where the prover, Merlin, sends one meassage to the verifier, Arthur, which then has to decide in probabilistic polynomial time whether to accept or reject).

Since $BPP\subseteq \Sigma_2\cap \Pi_2$ it immediately follows that $MA\subseteq \Sigma_2\cap \Pi_2$, which is believed to be a strict subset of PSPACE which contains the entire polynomial hierarchy.

Ariel
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