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Assume the zero knowledge protocol where the prover knows a $x$ such that: $g^x = h \pmod{p}$.

  • The prover chooses a random $t \in \mathbb{Z}^*_m$ and sends $y = g^t \pmod{p}$
  • The verifier sends random $c \in \mathbb{Z}^*_m$ and sends it
  • The prover calculates $s = t + c + x$ and sends it
  • The verifier accepts if and only if $g^s = yg^ch \pmod{p}$

Is this for honest verifiers?

Attempt:

The real transcript is $(t, g^t \pmod{p}, c, t +c+x)$, but I cannot find a simulated transcript $(t, \cdot, c, \cdot)$ that has the the same distribution as the real one. Can you give me some hints?

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Can you give me some hints?

It's not a proof of knowledge, as someone without knowledge of $x$ can complete this protocol successfully with an honest verifier.

Hint: what happens if the ignorant prover sends $y = h^{-1}$?

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  • $\begingroup$ Then the verifier will choose a $c$ and the prover will send back $s=c$ so that $g^s = h^{-1} g^c h = yg^ch$ and the prover will pass the challenge without knowing $x$ as you said. Can you elaborate on what does that show for the protocol? Does it mean that it's not a zero knowledge protocol or that it's not a zero knowledge protocol for honest verifiers? $\endgroup$ – Paris Jan 23 at 17:03
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    $\begingroup$ @Paris: it's not a zero knowledge protocol for proof of knowledge of $x$, as you can 'prove' it without knowing $x$ $\endgroup$ – poncho Jan 23 at 17:08

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