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Isn't this just a wasteful Schnorr? You don't have to send both X0 and X1 since given X0, X1 is uniquely determined. So if you don't send X1 you are back to binary challenge Schnorr.
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twitter.com/CouteauGeoffro Well, I don't know about "just", but you're right in that it certainly leads down that road
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Replying to @cronokirby and @Ivan__Visconti
actually you don't even need to do that, just pick a random x0 and send only g^x0, the g^x1 part can be homomorphically computed from the public information anyway. Then it's back to being plain ol' Schnorr, but with bit challenges
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