Is there a non-computable set $X\subset\omega$ such that, for some $Y\subset\omega$, any infinite subset or cosubset (=subset of the complement) of $Y$ computes $X$?
More generally, call a set $X$ $n$-hintable if there is some $f\in n^\omega$ such that, whenever I am told infinitely many bits of $f$, I can compute all of $X$. Then: What are the $n$-hintable sets?
(I ran into this question while thinking about stable ramsey's theorem: given a stable $k$-coloring of pairs - that is, a $k$-coloring of pairs $c$ such that $\lim_yc(x<y)$ always exists - the map $f: x\mapsto \lim_yc(x<y)$ is an element of $k^\omega$ such that, given infinitely many bits of $f$, we can compute an infinite $c$-homogeneous set. So we could say that the mass problem of solutions to $c$ is $k$-hintable; and now the question arises whether there are individual reals which are hintable.)
I suspect the answer is "every $n$-hintable set is computable," but I can't see how to prove it.
By way of background, a set $X$ is bi-introreducible if any infinite subset or cosubset of $X$ computes $X$. So 2-hintability is a much weaker property than bi-introreducibility. Seetapun (http://projecteuclid.org/download/pdfview_1/euclid.ndjfl/1040136917, theorem 2.19) showed that the bi-introreducible sets are all computable, but his argument doesn't answer my question, as far as I can tell. EDIT: As Liang Yu's answer below shows, Seetapun's arguments do answer my question, at least for $n=2$.