静态二、三钉黑钉AB博弈的最优策略

IF 0.6 Q4 MATHEMATICS, APPLIED Discrete Mathematics Algorithms and Applications Pub Date : 2022-10-10 DOI:10.1142/s1793830923500490
Gerold Jager, F. Drewes
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引用次数: 0

摘要

AB游戏是一款类似于流行游戏《Mastermind》的游戏。我们研究了这个游戏的一个版本,叫做Static Black-Peg AB game。它由两个玩家,编码者和密码破译者来玩。编码者通过将一组$c$颜色中的一种颜色放在每个$p \le c$标记上来创建一个所谓的秘密,但每种颜色最多只能使用一次。密码破译者试图通过提问来确定秘密,所有的问题都是一次给出的,每个问题都是一个可能的秘密。作为答案,编码人员会显示每个问题正确放置颜色的数量。在那之后,密码破译者只需要再尝试一次来确定密码,从而赢得游戏。对于给定的$p$和$c$,我们的目标是找到密码破译者需要赢得的最小问题数$k$,而不管秘密是什么,以及相应的问题列表,称为$(k+1)$ -策略。我们提出了针对所有$c \ge 2$的$p=2$的$\lceil 4c/3 \rceil-1)$ -策略,以及针对所有$c \ge 4$的$p=3$的$\lfloor (3c-1)/2 \rfloor$ -策略,并展示了这两种策略的最优性,即,我们证明了对于较小的$k$不存在$(k+1)$ -策略。
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Optimal Strategies for Static Black-Peg AB Game with Two and Three Pegs
The AB~Game is a game similar to the popular game Mastermind. We study a version of this game called Static Black-Peg AB~Game. It is played by two players, the codemaker and the codebreaker. The codemaker creates a so-called secret by placing a color from a set of $c$ colors on each of $p \le c$ pegs, subject to the condition that every color is used at most once. The codebreaker tries to determine the secret by asking questions, where all questions are given at once and each question is a possible secret. As an answer the codemaker reveals the number of correctly placed colors for each of the questions. After that, the codebreaker only has one more try to determine the secret and thus to win the game. For given $p$ and $c$, our goal is to find the smallest number $k$ of questions the codebreaker needs to win, regardless of the secret, and the corresponding list of questions, called a $(k+1)$-strategy. We present a $\lceil 4c/3 \rceil-1)$-strategy for $p=2$ for all $c \ge 2$, and a $\lfloor (3c-1)/2 \rfloor$-strategy for $p=3$ for all $c \ge 4$ and show the optimality of both strategies, i.e., we prove that no $(k+1)$-strategy for a smaller $k$ exists.
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来源期刊
CiteScore
1.50
自引率
41.70%
发文量
129
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