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Mar 3, 2022 at 9:45 comment added meshcollider No, I'm not aware off the top of my head. It sounds like it would be a weird algorithm.
Mar 3, 2022 at 8:27 comment added Napoleon I agree that all of those error are theoretically possible, but practically -- are you aware for example about a distinguisher with advantage $n^{-\log \log n}$?
Mar 3, 2022 at 7:58 comment added meshcollider Theoretically, any of those things could exist, the security assumption says nothing about it. We just draw the line at "negligible" and leave it there.
Mar 3, 2022 at 7:55 comment added Napoleon I agree that with the specific machine $M$ you mentioned, that simply guesses elements, has $1/2^{O(n)}$ advantage. Nevertheless, there might be other PPTs that do something different, and able to gain better advantage, say, $1/n^{\log n}$, which is still negligible. My question concerns more how much negligible the advantage can be: $n^{-\log \log n}, n^{-\log n}, 2^{-O(n)}$?
Mar 3, 2022 at 7:31 history answered meshcollider CC BY-SA 4.0