arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 2009.09510math.NT
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Zeckendorf对策的界

Bounds on Zeckendorf Games

Anna Cusenza, Aiden Dunkelberg, Kate Huffman, Dianhui Ke, Micah McClatchey, Steven J. Miller, Clayton Mizgerd, Vashisth Tiwari, Jingkai Ye, Xiaoyan Zheng

摘要:

Zeckendorf证明了每一个正整数$n$都可以唯一地写成不相邻的Fibonacci数之和。我们使用这种分解来构造一个两人博弈。给定一个固定的整数$n$和$n=n F_1$的初始分解,两个参与者通过使用与递归关系$F_{n+1}=F_n+F_{n-1}$相关的移动来交替,最后移动的人获胜。博弈总是以Zeckendorf分解结束;根据移动的选择,博弈的长度和赢家可以变化,尽管对于$n\ge 2$,有一个非构造性的证据表明参与人2有一个获胜的策略。最初,游戏长度的下限是$n$阶(并且已知是尖锐的),而上限是大小$n \log n$。最近的工作减少了上限的大小$n$,但有一个更大的常数比所示。我们改进了上界,得到了$n$阶的$\frac{\sqrt{5}+3}{2}\ n - IZ(n) - \frac{1+\sqrt{5}}{2}Z(n)$的精确上界,$Z(n)$是$n$的Zeckendorf分解中的项数,$IZ(n)$是$n$的Zeckendorf分解中的指数之和(它们的大小最多分别为$\log n$和$\log^2 n$).我们还介绍了一个贪婪的算法,实现了上限,并表明,最长的游戏上的任何$n$是通过应用分裂移动尽可能实现的。

英文摘要:

Zeckendorf proved that every positive integer $n$ can be written uniquely as the sum of non-adjacent Fibonacci numbers. We use this decomposition to construct a two-player game. Given a fixed integer $n$ and an initial decomposition of $n=n F_1$, the two players alternate by using moves related to the recurrence relation $F_{n+1}=F_n+F_{n-1}$, and whoever moves last wins. The game always terminates in the Zeckendorf decomposition; depending on the choice of moves the length of the game and the winner can vary, though for $n\ge 2$ there is a non-constructive proof that Player 2 has a winning strategy. Initially the lower bound of the length of a game was order $n$ (and known to be sharp) while the upper bound was of size $n \log n$. Recent work decreased the upper bound to of size $n$, but with a larger constant than was conjectured. We improve the upper bound and obtain the sharp bound of $\frac{\sqrt{5}+3}{2}\ n - IZ(n) - \frac{1+\sqrt{5}}{2}Z(n)$, which is of order $n$ as $Z(n)$ is the number of terms in the Zeckendorf decomposition of $n$ and $IZ(n)$ is the sum of indices in the Zeckendorf decomposition of $n$ (which are at most of sizes $\log n$ and $\log^2 n$ respectively). We also introduce a greedy algorithm that realizes the upper bound, and show that the longest game on any $n$ is achieved by applying splitting moves whenever possible.

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