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 $n=2$.



Sets computable from enough hints
There is no 2-hintable set by the same argument from Seetpun.


Suppose that there is such a pair $X$ and $Y$. By Freidberg's Theorem, there is a set $A$ such that $A$ is not above either $X$ or $Y$, but $A'$ computes $X\oplus Y$. By Theorem 2.18 in the paper, there is a function $f\leq\_T A:[\omega]^2\to 2$ such that any infinite $f$-homogeneous set is either subset of $Y$ or of the complement of $Y$. Now by Theorem 2.1 in the paper, there is an $f$-homogeneous set $H$ not computing $X$.


I think that, by a simple modification of this argument, one may show that there is no $n$-hintable set for any $n\geq 2$.


