The prover has $n$ group elements $g_1, ..., g_n$ and wishes to demonstrate the knowledge of the discrete logarithm to base $g$ for each of them, i.e, for each $i \in [1,n]$ she knows some $e_i$ s.t. $g_i = g^{e_i}$. We know that by applying Fiat-Shamir to the Schnorr's protocol, we can get $n$ non-interactive proofs in the form of $(R_i, c_i, s_i)$ where $c_i$ is the hash of $(g_i, R_i)$.

The question is: can we use a same challenge $c$ which is defined as: $c = hash((R_1,..., R_n), (g_1, ..., g_n))$? Each proof will look like $(R_i, c, s_i)$ sharing the same $c$. Would that change the soundness of the scheme? I vaguely feel that this may work through correlation intractability?


1 Answer 1


One way to answer your question is to check if the following proof system is sound:

  • Prover sends $R_1,\dots,R_n$
  • Verifier sends challenge $c$
  • Prover responds with $s_1,\dots,s_n$
  • Verifier checks for each $i$ that $(R_i,c,s_i)$ is an accepting transcript of the Schnorr protocol

Note that this proves a different language, namely $L=\{(g_1,\dots,g_n)\,|\, \forall i\exists w_i:\,g_i=g^{w_i}\}$. Now, if the original protocol was special sound, which is the case for Schnorr's protocol, then this is also special sound. Given two accepting transcripts $(R_{1..n},c,s_{1..n})$ and $(R_{1..n},c',s'_{1..n})$ with $c\neq c'$, you can use the witness extractor of Schnorr's protocol for each $i$ on input $(R_i,c,c',s_i,s'_i)$ to recover every $w_i$.

So soundness is preserved, what about zero-knowledge? You can just invoke the special HVZK simulator for Schnorr's protocol $n$ times on input $c$ to get $n$ accepting transcripts of the form $(R_i,c,s_i)$. This should have the same distribution as a real transcript.

In short, the new protocol is also a $\Sigma$-protocol, i.e. it satisfies special soundness (it's a proof of knowledge), special HVZK and public randomness for the verifier. So applying the Fiat-Shamir transform to it should get you a non-interactive zero-knowledge proof of knowledge with provable security in the random oracle model.

  • $\begingroup$ Many thanks for the insights. $\endgroup$
    – Sean
    Commented Jul 7, 2023 at 19:41

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