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by Alex Appel 26 days ago | link | parent

Update: This isn’t really an issue, you just need to impose an assumption that there is some function \(f\) such that \(f(n)>n\), and \(f(n)\) is computable in time polynomial in \(f(n)\), and you always find out whether exploration happened on turn \(f(n)\) after \(\mathcal{O}(f(n+1))\) days.

This is just the condition that there’s a subsequence where good feedback is possible, and is discussed significantly in section 4.3 of the logical induction paper.

If there’s a subsequence B (of your subsequence of interest, A) where you can get good feedback, then there’s infinite exploration steps on subsequence B (and also on A because it contains B)

This post is hereby deprecated. Still right, just not that relevant.





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