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In the following, we consider binary trees where only the leaves have weights. Let $T$ be a binary tree and $W(T)$ be the sum of the weights of its leaves. Let $T.l$ and $T.r$ be the left child and right child respectively.

We define the imbalance $I(T)$ of a binary tree $T$ as follows:

$$I(T) = \begin{cases} 0 & size(T) = 1\\ |W(T.l) - W(T.r)| + I(T.l) + I(T.r) & size(T)>0 \end{cases}$$

Examples

  a.)  *        b.)  *
      / \          /   \
     *   5        *     *
    / \          / \   / \
   *   3        1   5 3   3
  / \
 1   3

Tree a. has imbalance 5 (2 + 1 + 2) and tree b. has imbalance 4 (4 + 0 + 0).

Consider the following problem:

Input: a list of $n$ positive integers $W = (a_1, \ldots, a_n)$, and an integer $k$.

Query: is there a binary tree $T$ with leaf weights $W$ such that $I(T) \leq k$?

Is this problem $\mathsf{NP}$-hard?

The case $k = 0$ is easy. There is a simple algorithm which goes as follows ($Q$ is a priority queue):

  1. for $w \in W$: $Q.push(w)$
  2. while $Q.size > 1$:
    2.1. pop the two smallest elements $w_1$ and $w_2$ form $Q$.
    2.2. if $w_1 = w_2$ then $Q.push(2*w_1)$
    2.3. else return "No"
  3. return "Yes"
Kaveh
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rex123
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1 Answers1

0

Not a complete answer, but an intuition to see that the problem is highly likely to be NP-complete.

This problem has optimal substructure property, i.e., in an optimal tree $T$, trees $T.l$ and $T.r$ will also be optimal for their corresponding subset of leaves.

Then, lets say you have two polynomial time methods $w, s: \mathcal{P}(W) \setminus \{\emptyset, W\} \rightarrow \mathbb{N}_0$ which denote the $W(T)$ and $I(T)$ of the optimal tree for any proper subset $H\subset W$.

Then our task remains to find the subset $H_0 \subset W$ where $s(W) = |w(H_0) - w(W\setminus H_0) | + s(H_0) + s(W\setminus H_0)$ is $\le k$. This is very similar to the partition problem to be solvable in polynomial time. Idea is that there can be an adversary which fine tunes on the list $W$ and $k$ such that the above problem becomes ewuivalent to the partition problem.

EnEm
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