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In every definition of zero-knowledge (ZK) proof systems that I have seen (see, e.g., this one on Wikipedia, or this primer by Goldreich), a proof system $\langle p, v \rangle$ for a language $L$ is defined as being ZK if for every $v'$ there is some $z$ (of the same computational power as $v$ and $v'$) such that for every $x \in L$, the distributions $\langle p, v' \rangle(x)$ and $z(x)$ are indistinguishable. (I am ignoring all the nuances regarding the different ways of defining "indistinguishable".)

If the distributions $\langle p, v' \rangle(x)$ and $z(x)$ were indistinguishable for every $x$, then I suppose that this would mean there is no point in having a prover. However, if the motivation in defining ZK via simulation (of $\langle p, v' \rangle$ by $z$) is that the verifier doesn't gain any additional computational power by interacting with $p$, then it's not clear to me why this definition is appropriate.

For example, if there were some $v'$ and $x \notin L$ such that $\Pr \left[ \langle p, v' \rangle(x) \text{ rejects} \right] \gg \Pr \left[ z(x) \text{ rejects} \right]$ for any machine $z$ of the same computational power as $v$ and $v'$, would this not imply that the verifier was learning something more by interacting with $p$? Is there something in the standard definition of ZK that means this can't be the case? Or should we simply not care if it is the case? I would very much appreciate any clarification on this point!

(I have seen some related discussion on the question of when knowledge soundness implies soundness for ZK proofs of knowledge, e.g. here, but the answers do not go into why there is sometimes an 'asymmetry' in such definitions.)

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  • $\begingroup$ While you have a great technical answer below, an intuitive way to look as this is as follows. The simulator must produce transcripts indistinguishable from the real interaction. If $x\in L$, then it should be an accepting transcript, whereas if $x\notin L$ then the transcript should be rejecting. A simulator that works for every $x$ would need to be able to decide if $x\in L$ to know if it should produce an accepting or a rejecting transcript. $\endgroup$
    – lamontap
    Commented Apr 15 at 16:53
  • $\begingroup$ Have you tried looking at examples of simulators for classic zero knowledge problems (proving quadratic residuosity of some number, for example)? I think it could help with intuition for what this definition is supposed to capture $\endgroup$ Commented Apr 15 at 17:08

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So let me try and rephrase your question (please correct me if I'm wrong) -- you're asking why ZK is defined for only the YES instances and not for YES and NO instances both?

EDIT: There's a more intuitive answer (which I should have mentioned in the first place).

In a proof system, an honest verifier is trying to not be cheated i.e. soundness implies that that no prover can force the verifier to output $\mathtt{Accept}$ when the instance $x \in \mathtt{No}$ with high probability.Soundness protects the honest verifier.

When $x \in \mathtt{Yes}$, some prover that is not the honest prover could try to get the verifier to output $\mathtt{Reject}$, but we do not care about this (neither completeness or soundness handles this case). One may think of the verifier outputting $\mathtt{Accept}$ to be a very expensive operation, and outputting $\mathtt{Reject}$ to be a cheap operation. Thus, outputting $\mathtt{Accept}$ when the answer should be $\mathtt{Reject}$ is very bad. We do not care about the other case.

Now given the above description of a proof system, we say the proof system is ZK, if in helping the verifier output the correct decision, the prover does not reveal anymore information other than the fact $x \in \mathtt{YES}$. So ZK protects the honest prover. Here the game is reversed. Now the verifier is trying to get the prover to reveal extra information.

If $x \in \mathtt{No}$, the honest prover does not need protection from ZK. It does not need to help honest verifier come up with the right answer at all. It could send nothing, or simply the statement that $x \in \mathtt{No}$. The cheating verifier has no incentive to think differently. So it does not make any sense to define ZK for the $\mathtt{No}$ instances.


Now you could say, what if we didn't make this restriction that outputting $\mathtt{Reject}$ was cheap. Then one argument is that it is impossible to define ZK for both yes and no instances. The argument as to why follows from Chapter III of Salil Vadhan's thesis (https://people.seas.harvard.edu/~salil/research/phdthesis.pdf) pages 44-46. I'll summarise the idea below:

The definition of ZK roughly states that for every instance $x$ in the language, and for every verifier strategy $V^*$, there exists a PPT algorithm called the simulator which has blackbox access to $V^*$, that can produce a transcript called simulator output that is statistically indistinguishable from the transcript in the real protocol (the messages that $V^*$ receives during the protocol $\Pi(P, V^*)$ from an honest prover $P$ which we are claiming to be ZK)

The really short answer as to why we cannot define ZK for YES and NO instances is that it is impossible to define ZK to apply to both YES and NO instances. For89 show that for statistical zero knowledge proofs, the simulators output distribution can be used to almost perfectly distinguish YES and NO instances.

In more detail, the act of simulation can be viewed as an interaction between a virtual verifier and a virtual prover (remember, the simulator does not actually talk to the prover, it pretends to be the prover (virtual prover), when interacting with $V^*$ which can be thought of as the virtual verifier).

If $x \in L$ (YES) instance, then both of the following must be true:

  • The simulators messages would be accepted by the virtual verifier with high probability (otherwise the two transcripts would not be indistinguishable).
  • The virtual verifier behaves like the real verifier (this is saying if the virtual prover interacted with the real verifier instead of the virtual verifier, nothing would really change).

Now if $x \notin L$ (NO), both of the above cannot hold. More simply, one if not not both of the above statements must be false. Assume that this was not the case -- then the virtual prover can convince the real verifier with high probability (this is what both of the above statements being true mean). Then the real prover could just use the virtual prover as the proving strategy and fool the real verifier (this would break soundness).

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  • $\begingroup$ Thanks Ari! So, returning to the original question, the answer seems to be (and please correct me if I'm wrong) that if $\langle p, v \rangle$ is ZK then in fact there will be always be a $z$ such that $\Pr \left[ \langle p, v' \rangle(x) \text{ rejects} \right]$ and $\Pr \left[ z(x) \text{ rejects} \right]$ are indistinguishable for any $v'$ and $x \notin L$. I.e., having a simulator for positive instances implies that we can get a simulator for distinguishing positive and negative instances, and so there will also be a simulator for negative instances (for SZK, at least). $\endgroup$ Commented Apr 14 at 14:01
  • $\begingroup$ The quantifiers say for any $v'$ there is an $z$. So there is not one unique $z$ but we just to find one for the given $v'$. $\endgroup$
    – Ari
    Commented Apr 14 at 16:14
  • $\begingroup$ Once you have this simulator $z$, you can use it to distinguish between Yes and No instances (based on the For89). $\endgroup$
    – Ari
    Commented Apr 14 at 16:15
  • $\begingroup$ Oops, yes, I switched the order of the quantifiers in my comment, sorry! $\endgroup$ Commented Apr 15 at 17:15

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