Least number of guesses needed to determine all unknown subsets of a set

Say I have a set $$\mathbb{S}=\{1,2,...,n\}$$. I have an adversary who breaks up $$\mathbb{S}$$ into $$k$$ unknown and disjoint subsets. Denote this new set $$\mathbb{A}$$. I can guess any combination $$s$$ and my adversary has to tell me if $$s$$ is itself a subset of any element in $$\mathbb{A}$$.

For example, if $$n=4$$, a valid $$\mathbb{A}$$ might be $$\{ \{2\}, \{1,3,4\}\}$$. All "true" guesses are $$\{1\}$$, $$\{2\}$$, $$\{3\}$$, $$\{4\}$$, $$\{1,3\}$$, $$\{3,4\}$$, $$\{1,4\}$$, $$\{1,3,4\}$$.

What's the fastest algorithm and the upper bound on number of guesses I need to make to fully reverse engineer $$\mathbb{A}$$?

Obviously the brute force algorithm takes $$^nC_1 + ^nC_2 + ... + ^nC_n$$ guesses for every possible subset.

But I think I only need to test all possible subsets of size 2 i.e. exactly $$^nC_2$$ guesses?

Then, I can represent each of the correct guesses of size 2 as an undirected edge in a graph. Every member of $$\mathbb{A}$$ of size 3 or greater should form a simple cycle in this graph. So I just need to DFS on every node in this graph, removing the longest cycle detected from the graph, and repeat for all remaining nodes until only lone edges (subsets of size 2) or unconnected vertices (subsets of size 1) remain.

This graph should have $$n$$ vertices and at most $$^nC_2$$ edges. So the fastest algorithm has a time complexity of $$\mathbb{o}(n^{2}\cdot{}^{n}C_{2})$$. Is there an approach with fewer guesses or a faster graph algorithm?

P.S.: In case anyone's interested about the context: One example use case is if $$n$$ known services are hosted on unknown $$k$$ servers with unknown grouping, and you have some way to overload any subset of services to mount a denial attack on a server. The problem becomes if you can profile the servers with the least number of queries. I came across a similar scenario and thought this was a pretty interesting problem with no literature on.

• "Every member of $\mathbb{A}$ of size 3 or greater should form a simple cycle in this graph" should be "Every member of $\mathbb{A}$ of size 3 or greater forms a connected component that is a complete graph. – John L. Dec 20 '18 at 6:24
• There is a simple algorithm using $O(kn)$ queries. Find the part of 1 using $n-1$ queries. Then pick an element not in the part of 1, find its part, and so on. – Yuval Filmus Dec 20 '18 at 6:32
• There’s a simple $\Omega(n\log n)$ information-theoretic lower bound when $k$ is not fixed (or when $k=n/2$), by counting the number of possible partitions. – Yuval Filmus Dec 20 '18 at 10:15
• Here's an easy proof of $\Theta(n^2)$ worst-case time when $k$ is not fixed, or is $\Theta(n)$: Suppose $k=n-1$. Then there is exactly one edge in the graph. If you test fewer than $n(n-1)/2-1$ unordered vertex pairs, then there are at least two, say $ab$ and $cd$, that you haven't tested. The input in which $ab$ is the only subset, and the input in which $cd$ is the only subset, therefore cannot be distinguished by any such algorithm. – j_random_hacker Dec 20 '18 at 10:50
• @j_random_hacker In your example they still can't do better because testing anything bigger than a pair would result in false. – orlp Dec 20 '18 at 11:18