Consider $(X, \leq)$ a finite poset over $n$ items, and $P$ an unknown monotonic predicate over $X$ (i.e., for any $x$, $y \in X$, if $P(x)$ and $x \leq y$ then $P(y)$). I can evaluate $P$ by providing one node $x \in X$ and finding out if $P(x)$ holds or not. My goal is to determine exactly the set of nodes $x \in X$ such that $P(x)$ holds, using as few evaluations of $P$ as possible. (I can choose my queries depending on the answer of all previous queries, I am not required to plan all queries in advance.)

A strategy $S$ over $(X, \leq)$ is a function which tells me, as a function of the queries that I ran so far and their answers, which node to query, and which ensures that on any predicate $P$, by following the strategy, I will reach a state in which I know the value of $P$ on all nodes. The running time $r(S, P)$ of $S$ on a predicate $P$ is the number of queries required to know the value of $P$ on all nodes. The worst running time of $S$ is $wr(S) = \max\_P r(S, P)$. An optimal strategy $S'$ is such that $wr(S') = \min\_S wr(S)$.

My question is the following: given as input the poset $(X, \leq)$, how can I determine the worst running time of the optimal strategies?

[It is clear that for an empty poset $n$ queries will be needed (we need to ask about each single node), and that for a total order around $\lceil \log\_2 n \rceil$ queries will be needed (doing a binary search to find the frontier). A more general result is the following information-theoretic lower bound: the number of possible choices for the predicate $P$ is the number $N\_X$ of antichains of $(X, \leq)$ (because there is a one-to-one mapping between monotonic predicates and antichains interpreted as the maximal elements of $P$), so, since each query gives us one bit of information, we will need at least $\lceil \log\_2 N\_X \rceil$ queries, subsuming the two previous cases. Is this bound tight, or are them some posets whose structure is such that learning can require asymptotically more queries than the number of antichains?]