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 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.
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 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.