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arXiv:2604.24237 · v3

Computational Complexity of the Interval Ordering Problem

首次提交 2026-04-27 17:46
版本更新 2026-11-11 16:21
雷达观测 2026-11-13 14:16
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摘要

We study an interval ordering problem introduced by Dürr et al. [Discrete Appl. Math. 2012] which is motivated by applications in bioinformatics. The task is to order a given set of n intervals with the goal of minimizing a certain objective which is defined via a given cost function $f$ which assigns a cost to the exposed part of each interval (that is, the pieces not covered by previous intervals). We develop a dynamic programming approach which solves the problem with $O(2^n\text{poly}(n))$ oracle calls to $f$ and arithmetic operations. Moreover, our approach yields polynomial-time algorithms for all cost functions $f$ such that $f-f(0)$ is subadditive or superadditive. This answers an open question for the function $f(x)=2^x$. We contrast these results by proving a running time lower bound of $2^{n-1}$ for any algorithm that solves the problem for every function $f$ (with oracle access only) and further proving NP-hardness for some classes of simple functions. Thus, we significantly narrow the gap regarding the computational complexity of the problem.

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v3 · 2026-11-11 16:21 · 最新观测版本

Computational Complexity of the Interval Ordering Problem

We study an interval ordering problem introduced by Dürr et al. [Discrete Appl. Math. 2012] which is motivated by applications in bioinformatics. The task is to order a given set of n intervals with the goal of minimizing a certain objective which is defined via a given cost function $f$ which assigns a cost to the exposed part of each interval (that is, the pieces not covered by previous intervals). We develop a dynamic programming approach which solves the problem with $O(2^n\text{poly}(n))$ oracle calls to $f$ and arithmetic operations. Moreover, our approach yields polynomial-time algorithms for all cost functions $f$ such that $f-f(0)$ is subadditive or superadditive. This answers an open question for the function $f(x)=2^x$. We contrast these results by proving a running time lower bound of $2^{n-1}$ for any algorithm that solves the problem for every function $f$ (with oracle access only) and further proving NP-hardness for some classes of simple functions. Thus, we significantly narrow the gap regarding the computational complexity of the problem.

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