1
$\begingroup$

I am reading this explanation of zkSnark written by Maksym Petkus - http://www.petkus.info/papers/WhyAndHowZkSnarkWorks.pdf

From Section 3.5

Because verifier can extract knowledge about the unknown polynomial $p(x)$ only from the data sent by the prover, let us consider those provided values (the proof): $g^p$, $g^{p'}$, $g^h$. They participate in the following checks:

$g^p = (g^{h})^{t(s)}$ (polynomial $p(x)$ has roots of $t(x)$)

$(g^p)^\alpha = g^{p'}$ (polynomial of a correct form is used)

The question is how do we alter the proof such that the checks still hold, but no knowledge can be extracted? One answer can be derived from the previous section: we can "shift" those values by some random number $\delta$ (delta), e.g., $(g^p)^{\delta}$. Now, in order to extract the knowledge, one first needs to find $\delta$ which is considered infeasible. Moreover, such randomization is statistically indistinguishable from random.

We already have Strong Homomorphic Encryption (as stated in Section 3.3.3),

$E(v) = g^v \pmod n$

The $g^p$, $g^{p'}$, $g^h$ are created as above. I mean how do you extract info about p, p' & h from $g^p$, $g^{p'}$, $g^h$ that the shift by $\delta$ is required?

So then why is the $\delta$ shift required for Zero Knowledge?

$\endgroup$

1 Answer 1

1
$\begingroup$

Zero-knowledge means zero knowledge, that we learn literally nothing except the validity of the proof.

Even if we don't learn $p, p'$ and $h$ themselves, we still learn $g^p$ which we could not have known before the protocol took place. So we have learned something, even if it isn't $p$.

Shifting by $\delta$ removes this leakage. Then we learn absolutely nothing, because the $g^{p\delta}$ completely hides $g^p$. As they say, "such randomization is statistically indistinguishable from random." If I gave you a random $X = g^x$, where $x$ is generated as $p\delta$ for a random $\delta$, you have no way of knowing anything about $\delta$ or $p$ (including $g^p$) from it.

$\endgroup$
2
  • $\begingroup$ Ok, got it. I guess it's like a using an IV or nonce in symmetric encryption to randomize so that the same input twice gives different output. $\endgroup$
    – user93353
    Jan 28 at 1:34
  • $\begingroup$ Yep, exactly :) $\endgroup$ Jan 28 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.