Josephus Nim

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 tha...

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Bibliographic Details
Main Authors Takahashi, Shoei, Manabe, Hikaru, Murakami, Aoi, Miyadera, Ryohei
Format Journal Article
LanguageEnglish
Published 04.12.2023
Subjects
Mathematics - Combinatorics
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Summary: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.
DOI:10.48550/arxiv.2312.02477