← 研踪 PaperTrailarXiv:2611.08397 · v1
A Near-Optimal Generalized Sorting Algorithm
首次提交 2026-11-06 02:08
版本更新 2026-11-06 02:08
雷达观测 2026-11-06 12:09
摘要
We continue the study of the classic \emph{generalized sorting problem}, a basic variant of ordinary sorting. There are $n$ elements and an (initially hidden) total ordering $π$ over them. Unlike in ordinary sorting, however, only \emph{some} comparisons are allowed: those given by the edges of an undirected graph $G = (V, E)$ on the elements. How many comparisons are needed to find the total ordering? In ordinary sorting, $Θ(n \log n)$ comparisons are necessary and sufficient. For generalized sorting, however, the best known upper bound has been $\tilde{O}(n^{3/2})$ since the work of Huang, Kannan, and Khanna (FOCS 2011), who even asked whether the problem might in fact require $n^{1+Ω(1)}$ comparisons. We show that it does not: we give a randomized algorithm that uses $O(n \log^4 n)$ comparisons in expectation. Thus, up to logarithmic factors, generalized sorting is no harder than ordinary sorting.
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v1 · 2026-11-06 02:08 · 最新观测版本
A Near-Optimal Generalized Sorting Algorithm
We continue the study of the classic \emph{generalized sorting problem}, a basic variant of ordinary sorting. There are $n$ elements and an (initially hidden) total ordering $π$ over them. Unlike in ordinary sorting, however, only \emph{some} comparisons are allowed: those given by the edges of an undirected graph $G = (V, E)$ on the elements. How many comparisons are needed to find the total ordering? In ordinary sorting, $Θ(n \log n)$ comparisons are necessary and sufficient. For generalized sorting, however, the best known upper bound has been $\tilde{O}(n^{3/2})$ since the work of Huang, Kannan, and Khanna (FOCS 2011), who even asked whether the problem might in fact require $n^{1+Ω(1)}$ comparisons. We show that it does not: we give a randomized algorithm that uses $O(n \log^4 n)$ comparisons in expectation. Thus, up to logarithmic factors, generalized sorting is no harder than ordinary sorting.
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