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## 数学代写|组合学代写Combinatorics代考|Conway’s Game of Reaching a Level

A finite number of coins are arranged in some points with integer coordinates $(x, y)$ on the Cartesian plane. Only one coin can be placed at one point. Coins may be moved and removed according to the following rules. Let us consider three points $A, B$, and $C$ with integer coordinates such that the following conditions are satisfied:
(C1) The points $A, B$, and $C$ belong to the same horizontal line or to the same vertical line; the point $B$ is the midpoint of the segment $A C$; the distance between the points $A$ and $C$ is equal to 2 .
(C2) There is a coin on points $A$ and $B$, and there is no coin on point $C$, see Figures 11.4.1 and 11.4.2.

Then, it is allowed to move the coin from point $A$ to point $C$, and remove the coin from point $B$ at the same time.

If a coin is placed on the point whose coordinates are $(x, y)$, where $y=k$, we say that this coin is at the level $k$. Suppose that a finite number of coins are arranged on some points with integer coordinates, and that all these points are on the $x$-axis or below the $x$-axis. The player chooses how many coins will be used, and the points where these coins will be placed at the beginning of the game. The goal of the game is to put a coin at the highest possible level.

## 数学代写|组合学代写Combinatorics代考|Two More Games

Example 11.5.1. Suppose that $n$ points are given on a circle, where $n \geqslant 4$. Assume that the points are labeled $1,2, \ldots, n$, one after the other in a chosen direction. Two players, $A$ and $B$, play the game in which they alternately draw a chord which endpoints belong to the set ${1,2, \ldots, n}$. The endpoints of any chord should be of the same parity. It is not allowed for the chords to have points of intersection. Player $A$ starts the game. At the end of the game the winner is the player who has drawn the last chord. Which player has a winning strategy?

We shall prove that player $A$ has a winning strategy if $n=4 k, n=4 k+1$, or $n=4 k+3$, where $k \in \mathbb{N}$. Player $B$ has a winning strategy if $n=4 k+2$, where $k \in \mathbb{N}$. Without loss of generality we can assume that the given points are the vertices of a regular $n$-gon.

Case $n=4 k, k \in \mathbb{N}$. Player $A$ draws a chord with endpoints 1 and $2 k+1$ as the first move of the game. If player $B$ draws chord $i j$, then, in the next move, player $A$ draws chord $i^{\prime} j^{\prime}$ that is symmetric to $i j$ around the center of the circle.

Case $n=4 k+2, k \in \mathbb{N}$. In this case there is no a diameter $i j$ of the circle, such that $i, j \in{1,2, \ldots, n}$, where $i$ and $j$ are of the same parity. The winning strategy for player $B$ is to draw chord $i^{\prime} j^{\prime}$ that is symmetric to $i j$ around the center of the circle, where $i j$ is the chord drawn by $A$ in the previous move of the game.

Case $n=4 k+1, k \in \mathbb{N}$. As the first move, player $A$ draws a chord with endpoints 1 and 3 . In the sequel, the points $4,5, \ldots, 4 k+1$ can be chosen as the endpoints of the new chords. There are $4 k-2=4(k-1)+2$ such points, and, using the strategy of player $A$ in the previous case, now player $B$ will always win the game.

Case $n=4 k+3, k \in \mathbb{N}$. As the first move player $A$ draws a chord with endpoints $2 k+1$ and $2 k+3$. The points $1,2, \ldots, 2 k$ and $2 k+4,2 k+5$, $\ldots, 4 k+3$ can be chosen in the next moves. Without loss of generality we can suppose that the $4 k$ points in this order are vertices of a regular $4 k$-gon. The diameters $(1,2 k+4),(2,2 k+5), \ldots,(2 k, 4 k+3)$ are not allowed to be drawn, because the endpoints are not of the same parity. Using a strategy based on symmetry around the center of the circle, player $A$ can win the game. $\triangle$

## 数学代写|组合学代写Combinatorics代考|康威达到一个水平的游戏

(C1 $A, B$，以及 $C$ 同一的:属于同一水平线或同一垂直线的;重点是 $B$ 是线段的中点吗 $A C$;两点之间的距离 $A$ 和 $C$ = 2
(C2)点上有硬币吗 $A$ 和 $B$在这一点上没有硬币 $C$，见图11.4.1和11.4.2

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。