基于反例的排序

中文摘要 AI 辅助

考虑如下问题：学习n个未知项的线性顺序，每一轮中学习者猜测项的完整排序，会收到猜测正确的确认，或一个反例——一对顺序错误的项。目标是用尽可能少的查询识别未知顺序。我们研究返回的反例中最多k个可能不真实的情况，其中k事先未知。我们确定了最优查询复杂度（忽略常数因子）：Θ(n log n + nk)。因此，无噪声情况下的复杂度与经典排序复杂度匹配，而每个不真实的反例会产生n阶的额外成本。上界基于排列的几何表示和Grünbaum定理，下界结合排序论证与Condorcet型构造。我们还研究目标排序具有低维几何表示的情况：每个项由R^d中的一个点表示，排序通过将点投影到未知方向得到。对于这些类别，我们给出上界O(d² log n + dk)和下界Ω(d log n + dk)，在无噪声项中留下d倍的差距。

英文摘要

Consider the following problem of learning an unknown linear order on $n$ items. In each round, the learner guesses a complete ordering of the items and receives either confirmation that the guess is correct or a counterexample: a pair of items in the wrong order. The goal is to identify the unknown order using as few queries as possible. We study this problem when up to $k$ of the returned counterexamples may be untruthful, where $k$ is not known in advance. We determine the optimal query complexity up to constant factors: \[ Θ(n\log n + nk). \] Thus, while the noiseless complexity matches the classical complexity of sorting, each untruthful counterexample incurs an additional cost of order $n$. The upper bound is based on a geometric representation of permutations and Grünbaum's theorem, while the lower bound combines sorting arguments with a Condorcet-type construction. We also study the case where the target ranking has a low-dimensional geometric representation: each item is represented by a point in $\mathbb{R}^d$, and the ranking is obtained by projecting the points onto an unknown direction. For these classes we give an upper bound of $O(d^2\log n+dk)$ and a lower bound of $Ω(d\log n+dk)$, leaving a factor of $d$ gap in the noiseless term.

发表机构

• Princeton University（普林斯顿大学）
• Tel Aviv University（特拉维夫大学）
• Technion – Israel Institute of Technology（以色列理工学院）
• Google Research（谷歌研究院）

机构由 AI 辅助整理，请以论文原文为准。