patternMinor
Interactive Proofs for coNP
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conpinteractiveforproofs
Problem
I am trying to understand interactive proof systems and tried the following problem as an exercise. We know that $PH \subseteq PSPACE$ and $IP=PSPACE$, so come up with (easy to understand) interactive proof systems for $PH$?
An interactive proof system for $NP$ is trivial, but I failed to get an interactive proof system even for $coNP$. Do you know of an explicit interactive proof system (by explicit I mean without going through the $IP=PSPACE$ route) for $coNP$?
An interactive proof system for $NP$ is trivial, but I failed to get an interactive proof system even for $coNP$. Do you know of an explicit interactive proof system (by explicit I mean without going through the $IP=PSPACE$ route) for $coNP$?
Solution
Wikipedia outlines such an example. Consider the coNP-complete problem UNSAT: given a CNF $\varphi$ on $n$ variables, we want to convince the verifier that $\varphi$ is not satisfiable. We arithmetize $\varphi$ to a polynomial $p$ and choose some large prime $q$. Let
$$p(x_1,\ldots,x_k) = \sum_{x_{k+1}=0}^1 \cdots \sum_{x_n=0}^1 p(x_1,\ldots,x_n).$$
The protocol proceeds as follows:
Because the degree of $p$ is small compared to $q$, if the prover is cheating then the verifier will probably catch her (see Wikipedia for the proof, or work it out yourself using the Schwartz-Zippel lemma).
$$p(x_1,\ldots,x_k) = \sum_{x_{k+1}=0}^1 \cdots \sum_{x_n=0}^1 p(x_1,\ldots,x_n).$$
The protocol proceeds as follows:
- Prover sends verifier a prime $q \in (2^n,2^{n+1})$, and the latter verifies that $q$ is prime.
- Prover sends verifier $p(z) \in \mathbb{Z}_q[z]$. Verifier verifies that $p(0) + p(1) = 0$, and sends prover a random $r_1$.
- Prover sends verifier $p(r_1,z) \in \mathbb{Z}_q[z]$. Verifier verifies that $p(r_1,0) + p(r_1,1) = p(r_1)$, and sends prover a random $r_2$.
- Eventually, verifier gets $p(r_1,\ldots,r_n) \in \mathbb{Z}_q$, and verifies that it has the correct value by evaluating $p$ directly.
Because the degree of $p$ is small compared to $q$, if the prover is cheating then the verifier will probably catch her (see Wikipedia for the proof, or work it out yourself using the Schwartz-Zippel lemma).
Context
StackExchange Computer Science Q#13286, answer score: 9
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