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Unbounded ZK vs perfect ZK?

ng flag

In the "On the Non-malleability of the Fiat-Shamir Transform" article by Faust et.al., there is a definition for unbounded NIZK. What unbounded means in this context? It does not look like perfect ZK because the distinguisher is PPT and not unbounded as required here.

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Lev avatar
jp flag
Lev
The only thing this could be, if not a typo, is that the prover is computationally unbounded.
pintor avatar
ng flag
@Lev, I've contacted the authors and the answer is "unbounded means that the distinguisher can obtain an unbounded (polynomial) number of proofs". I'll delete a question
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