It is not uncommon to have interactive tree problems where you are allowed to query some connected component of the tree and use the return value to determine whether the answer is in the connected component or outside the connected component (Link to example problem). The general approach for these kinds of problems is to always choose a connected component of size $$$\frac{N}{2}$$$. However, there are also problems where the allowed queries are more restricted, preventing $$$\frac{N}{2}$$$ halving from being possible. This blog covers one of those types of problems.

Definitions

• Subtree: $$$S(r, u)$$$ contains the set of vertices in the subtree of vertex $$$u$$$ if the tree was rooted at vertex $$$r$$$.
• Neighbour: $$$N(u)$$$ contains the set of vertices that are directly adjacent to vertex $$$u$$$.
• Extended subtree: $$$ES(r, V) = \bigcup_{v\in V} S(r, v)\text{ if } v\in N(r)$$$. In other words, an extended subtree is a combination of the subtrees of a chosen set of vertices that are directly adjacent to the root.

Problem Structure

There is a hidden special vertex in a tree with $$$n$$$ vertices. Find the special vertex using at most $$$\left\lceil\log_{1.5}n\right\rceil$$$ of the following query:

• Choose an extended subtree of the tree. The grader will return whether the special vertex is in the chosen extended subtree. More formally, choose any vertex $$$r$$$ and a subset of neighbours $$$V \subseteq N(r)$$$, then the grader will return whether the special vertex $$$x \in ES(r, V)$$$.
  

Solution

Let us denote the size of the chosen extended subtree as $$$x$$$. If the grader returns that the special vertex is in the extended subtree, we can combine the root together with the extended subtree to form a new tree with $$$x + 1$$$ vertices. Otherwise, the vertices outside of the extended subtree form a tree with $$$n - x$$$ vertices. This means that with each query, we can reduce the size of the tree by at least $$$\max(x + 1, n - x)$$$.

Since we are allowed to use $$$\left\lceil\log_{1.5}n\right\rceil$$$ queries, we will be able to solve this problem if the size of the tree reduces to at least $$$\frac{2n}{3}$$$ after each query. Solving the inequality $$$\max(x + 1, n - x) \le \frac{2n}{3}$$$, we get $$$\frac{n}{3} \le x \le \frac{2n}{3} - 1$$$.