Given 2 finite sets S and M, with card(S) >= card(M), and an item z not in M. There is an unknown function f: S --> M u {z}, which is known to be one-to-one for all s in S for which f(s) in M (i.e. for which f(s) != z). The goal is to find f. To this end, I can query an oracle by sending it a question Q subset of S, and getting back from it answer A = f(Q) subset of M u {z}. Obviously, I could use the trivial strategy and sequentially ask the questions Q = {s} over all s in S, but querying the oracle is very costly. Is there a questioning strategy that is less costly than the trivial one? Or can one prove that there is no strategy less costly than the trivial one?

Given 2 finite sets $S$ and $M$, with $\operatorname{card}(S) \geq \operatorname{card}(M)$, and an item $z \not\in M$. There is an unknown function $f: S \to M \cup \{z\}$, which is known to be one-to-one for all $s \in S$ for which $f(s) \in M$ (i.e. for which $f(s) \neq z$). The goal is to find $f$. To this end, I can query an oracle by sending it a question $Q \subseteq S$, and getting back from it answer $A = f(Q) \subseteq M \cup \{z\}$. Obviously, I could use the trivial strategy and sequentially ask the questions $Q = \{s\}$ over all $s \in S$, but querying the oracle is very costly. Is there a questioning strategy that is less costly than the trivial one? Or can one prove that there is no strategy less costly than the trivial one?