arXiv subjects

Shoei Takahashi

Publications and source records attributed to Shoei Takahashi.

7 recordsLinked to original sources

arXiv2024

Maximum Nim and Josephus Problem algorithm

In this study, we study a Josephus problem algorithm. Let $n,k$ be positive integers and $g_k(n) = \left\lfloor \frac{n}{k-1} \right\rfloor +1$, where $ \left\lfloor \ \ \right\rfloor$ is a floor function. Suppose that there exists $p$ such that $g_{k}^{p-1}(0) < n(k-1) \leq g_{k}^{p}(0)$, where $g_{k}^p$ is the $p$-th functional power of $g_k$. Then, the last number that remains is $nk-h2_{k}^{p}(0)$ in the Josephus problem of $n$ numbers, where every $k$-th numbers are removed. This algorithm is based on Maximum Nim with the rule function $f_k(n)=\left\lfloor \frac{n}{k} \right\rfloor$. Using the present article's result, we can build a new algorithm for Josephus problem.

math.CO↗

arXiv2024

A New 0(klog n) Algorithm for Josephus Problem

We present a new O(k log n) algorithm of the Josephus problem. The time complexity of our algorithm is O(k log n), and this time complexity is on a par with the existing O(k log n) algorithm. We do not have any recursion overhead or stack overflow because we do not use any recursion. Therefore, the space complexity of our algorithm is O(1), and ours is better than the existing O(k log n) algorithm in this respect. When k is small and n is large, our algorithm is better than the existing O(k log n) algorithm. This new algorithm is based on a relation between the Josephus problem and a maximum Nim of combinatorial game theory.

cs.DS↗

arXiv2024

Maximum Nim and Josephus Problem

In this study, we study the relation between Grundy numbers of a Maximum Nim and Josephus problem. Let f(x) = floor(x/k), where floor( ) is the floor function and k is a positive integer. We prove that there is a simple relation with a Maximum Nim with the rule function f and the Josephus problem in which every k-th numbers are to be removed.

math.CO↗

arXiv2023

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 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.

math.CO↗

arXiv2023

Games of Nim with Dynamic Restrictions

The authors present formulas for the previous player's winning positions of two variants of restricted Nim. In both of these two games, there is one pile of stones, and in the first variant, we investigate the case that in k-th turn, you can remove f(k) stones at most, where f is a function whose values are natural numbers. In the second variant, there are two kinds of stones. The Type 1 group consists of stones with the weight of one, and the Type 2 group consists of stones with the weight of two. When the total weight of stones is a, you can remove stones whose total weight is equal to or less than half of a.

math.CO↗

arXiv2023

Curious Properties of Iterative Sequences

In this study, several interesting iterative sequences were investigated. First, we define the iterative sequences. We fix function f(n). An iterative sequence starts with a natural number n, and calculates the sequence f(n),f(f(n)), ...f(f(f(f(n)))),... We then search for interesting features in this sequence. We study Kaprekar's routine, the digit factorial process, and the digit power process. The authors presented new variants of Kaprekar's routine.

math.GM↗

arXiv2023

Impartial and Partizan Restricted Chocolate Bar Games

In this paper, we consider impartial and partizan restricted chocolate bar games. In impartial restricted chocolate bar games, players cut a chocolate bar into two pieces along any horizontal or vertical line and eat whichever piece is smaller. If the two pieces are the same size, a player can eat either one. In constrast, partizan restricted chocolate bar games include players designated as Left and Right and chocolate bars with black and white stripes. Left cuts the chocolate bar in two as above and eats the part with fewer black blocks. Similarly, Right cuts the bar and eats the part with fewer white blocks. A player loses when they cannot eat the remaining chocolate bar. We provide formulas that describe the winning positions of the previous player, Right, and Left players. We also present an interesting similarity in the graphs of previous players' winning positions for impartial and partizan chocolate bar games.

math.CO↗