by Alex Appel 144 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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