arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 2312.02477math.CO
收藏

约瑟夫斯尼姆

Josephus Nim

Shoei Takahashi, Hikaru Manabe, Aoi Murakami, Ryohei Miyadera

AI总结:

本文研究一种两堆尼姆变体,揭示其先手必败局面与约瑟夫斯问题最后剩余编号之间的对应关系,并推广到格伦迪数与删除编号的关联。

摘要:

本文提出一种两堆尼姆游戏的变体。第一堆中放置重量为1的石子,第二堆中放置重量为-2的石子。两名玩家轮流从其中一堆取石子,且每次取出的石子总重量必须小于或等于该堆石子总重量的一半。取走最后一颗或最后一组石子的玩家获胜。作者发现,当(n,m)是先手必败局面时,2m+1恰好是约瑟夫斯问题中最后剩下的编号,其中共有n个数,且每隔一个数删除一个。类似地,格伦迪数为s的局面与第(n-s)个被删除的编号之间也存在对应关系。

英文摘要:

Here, we present a variant of Nim with two piles. In the first pile, we have stones with a weight of 1, and in the second pile, we have stones with a weight of -2. Two Players take turns to take stones from one of the piles, and the total weight of stones to be removed should be equal to or less than half of the total weight of the stones in the pile. The player who removed the last stone or stones is the winner of the game. The authors discovered that when (n,m) is a previous player's winning position, 2m+1 is the last remaining number of the Josephus problem, where there are n -numbers, and every second number is to be removed. There are similar relations between the position of which the Grundy number is s and the (n-s)-th removed number.

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。

↑