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Given code to just an NP-verifier, where the certificate/witness is required to be of size polynomial in the instance, for a language L, can you, from that data alone, construct code for a solver, or generate / get back the language L itself?

At a glance, the answer seems to be yes--you could just try every word, certificate pair; however, there is a part I'm not sure about with this process: sure there are only a finite number of words of each size, but there are an infinte number of words that are poly|w| for any given size |w| that are potential certificates. So, without the certificates, you could be trying different strings as the second input for the same word as first input forever to no avail.

Thanks!

DeeDee
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  • Please define what kind of verifier you are talking about. A NP-verifier, where the certificate/witness is required to be of size polynomial in the instance? An arbitrary verifier? Why do you think there are an infinite number of words that are of size $|w|$? It seems there are obviously exactly $2^{|w|}$, assuming a binary alphabet. – D.W. Nov 25 '20 at 20:23
  • @D.W. there are a finite number of words of size |w| but there are an infinte number of words of size poly|w| (infinite number of potential certificates for any word of any given size) – DeeDee Nov 25 '20 at 20:25
  • No, there aren't: there are only $2^{\text{poly}(|w|)}$, which is finite. – D.W. Nov 25 '20 at 20:26
  • @D.W. oh really? So you can brute force check all possible certificates. made the approriate update to question – DeeDee Nov 25 '20 at 20:26
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    Perhaps you can write an answer to your own question now? – D.W. Nov 25 '20 at 20:27

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Having a verifier for a language in general is known as semidecidability, which is actually weaker than decidability. So in general, the answer is no, we can't build a decider for $L$.

But if the verifier is efficient (i.e., $L \in $ NP), then indeed $L$ is decidable, and your argument is pretty close for why that is.

Let's say we have an efficient verifier $V(x,c)$.

Given an $x \in \Sigma^*$, the "certificate space" is $C=\{c \in \Sigma^* \mid |c| \leq p_L(|x|)\}$, where $p_L$ is a polynomial. That means the certificate space is actually finite. $x \in L$ if and only if there exists a $c \in C$ such that $V(x,c)$ accepts. So indeed we can just check every possible certificate for a given input $x$ to construct a decider for $L$.

In fact with a bit more work you can show that $L \in$ EXP, meaning $L$ is decidable in exponential time.

CSSTUDENT
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