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arXiv 2203.09496math.COcs.ITmath.IT
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具有整数反馈的组合查询问题的最优方案

Optimal schemes for combinatorial query problems with integer feedback

Anders Martinsson

AI总结:

本文提出查询游戏短解的一般框架,并据此给出硬币称重变体的新证明及 Mastermind 在两种反馈信息下的确定性查询复杂度最优界。

摘要:

一个查询游戏是由一个查询集合 $Q$ 和一个函数集合 $\mathcal{F}$(或称为码字)$f:Q\rightarrow \mathbb{Z}$ 组成的一对。我们将其视为一个双人游戏。一个玩家,编码者,选择一个隐藏的码字 $f\in \mathcal{F}$。另一个玩家,解码者,然后尝试通过询问一系列查询 $q\in Q$ 来确定 $f$,每次查询之后编码者必须回答值 $f(q)$。解码者的目标是使用尽可能少的查询来唯一确定 $f$。这类游戏的两个经典例子是弹簧秤硬币称重和 Mastermind,它们既作为娱乐游戏,也因其与信息论的联系而受到关注。在本文中,我们将提出一个用于寻找查询游戏短解的一般框架。作为应用,我们给出了硬币称重问题变体的查询复杂度的新的自包含证明,并证明了新的结果:如果只提供黑色桩信息,具有 $n$ 个位置和 $k$ 种颜色的 Mastermind 的确定性查询复杂度为 $Θ(n \log k/ \log n + k)$;如果同时提供黑色和白色桩信息,则为 $Θ(n \log k / \log n + k/n)$。在确定性设置下,对于任何 $k\geq n^{1-o(1)}$,这些是已知的 Mastermind 的首个达到常数因子最优的解。

英文摘要:

A query game is a pair of a set $Q$ of queries and a set $\mathcal{F}$ of functions, or codewords $f:Q\rightarrow \mathbb{Z}.$ We think of this as a two-player game. One player, Codemaker, picks a hidden codeword $f\in \mathcal{F}$. The other player, Codebreaker, then tries to determine $f$ by asking a sequence of queries $q\in Q$, after each of which Codemaker must respond with the value $f(q)$. The goal of Codebreaker is to uniquely determine $f$ using as few queries as possible. Two classical examples of such games are coin-weighing with a spring scale, and Mastermind, which are of interest both as recreational games and for their connection to information theory. In this paper, we will present a general framework for finding short solutions to query games. As applications, we give new self-contained proofs of the query complexity of variations of the coin-weighing problems, and prove new results that the deterministic query complexity of Mastermind with $n$ positions and $k$ colors is $Θ(n \log k/ \log n + k)$ if only black-peg information is provided, and $Θ(n \log k / \log n + k/n)$ if both black- and white-peg information is provided. In the deterministic setting, these are the first up to constant factor optimal solutions to Mastermind known for any $k\geq n^{1-o(1)}$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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