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In the Shnorr protocol where the prover wants to prove he has a witness $w$ for $g^w$ the following interactions happen:

  • the prover chooses a random $r$, calculates $y=g^r$ and sends $y$

  • the verifier sends a challenge $x$

  • the prover calculates $t=xw+r$ and sends it to the verifier

  • the verifier tests if $g^t=(g^w)^x.y$

My question is why the verifier sends a challenge. Would he be convinced if the prover just sends $t=r+x$ and the verifier tests if $g^t=g^w.y$? Plus $t$ won't reveal the witness.

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my question is why the verifier sends a challenge, would he be convinced if the prover just sends $t=r+x$ and the verifier tests if $g^t=g^w \cdot y$ ?

That is, why doesn't the prover just send $t$ and $y$? Well, anyone can pick a random $t$ and compute $y = g^t \cdot (g^w)^{-1}$. Because $g^w$ is public, this can be computed by anyone, and so wouldn't serve as a proof of knowledge.

And, it is easy to find a solution to $g^t=(g^w)^x \cdot y$ (without knowing $w$), if you know the $x$ value before selecting the $y$ (and you suggested a constant $x=1$, hence the prover knows it up-front). However, if you can find a solution for $g^t=(g^w)^x \cdot y$ for two different $x$'s, that's different; it's easy to show that with solutions to two different $x$'s (and the same $y$), we can recover $w$ (and hence someone who can do that must know $w$). On the other hand, we can't just give out two solutions (as that means the verifier would then be able to deduce $x$).

So, what we do is get the prover to give a solution for an $x$ he cannot predict in advance; either he got extremely lucky (and he guessed the correct $x$ value when he generated $y$), or he does in fact know multiple solutions (and hence knows $w$).

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