# 数学代写|组合数学代写Combinatorial mathematics代考|MATH418

#### Doug I. Jones

Lorem ipsum dolor sit amet, cons the all tetur adiscing elit

couryes-lab™ 为您的留学生涯保驾护航 在代写组合数学Combinatorial mathematics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写组合数学Combinatorial mathematics代写方面经验极为丰富，各种代写组合数学Combinatorial mathematics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
couryes™为您提供可以保分的包课服务

## 数学代写|组合数学代写Combinatorial mathematics代考|Solutions to Exercises

9.1. (a) We can prove by induction on $n$ that there is a party of $n$ people such that every integer $0,1, \ldots, n-2$ appears as the number of acquaintances of a party member.

Two non-acquainted people deliver an example of such a party for $n=2$.

Assume that there is a party $P$ of $n$ people such that every integer $0,1, \ldots, n-2$ appears as the number of acquaintances. In fact, by the Pigeonhole Principle, one of these numbers, say $k(0 \leq k \leq n-2)$, must appear twice.

Now we have to construct a party $P^{\prime}$ of $n+1$ people such that every integer $0,1, \ldots, n-1$ appears as the number of acquaintances. We start with the party $P$ of $n$ people and add one more person, who is acquainted with exactly one of two people with $k$ acquaintances, and with everyone having more than $k$ acquaintances. You can easily verify that $P^{\prime}$ satisfies the required condition.
(b) There is such a party. The construction is similar to the one in Exercise 9.1(a).
9.2. Due to Theorem 9.1, the total number $T$ of acquaintances is even, on the other hand, if we assume that there exists a party of 21 such that everyone has exactly seven acquaintances, we have $T=21 \times 7$, an odd number. This contradiction proves that such a party does not exist.
We would like to mention here that similar reasoning proves the following result: there is no polyhedron with an odd number of oddsided faces.

Graphs appear in our discussion as diagrams of acquaintances. Thus, the only thing that matters when we represent a graph in the plane is the set of vertices (but not their positions on the plane) and which vertices are adjacent (but not the shapes of the edges, which we presume have no points in common except the vertices of the graph). In fact, think of a graph as a set of pins, some of which are connected by rubber bands. A graph remains the same if we reposition the pins and stretch the rubber bands.

Two graphs are called isomorphic if “pins” of one of them can be repositioned and its “rubber bands” stretched so that the two become identical.

More formally, two graphs $G$ and $G_1$ are said to be isomorphic if there is a one-to-one correspondence $f: V \rightarrow V_1$ of their vertex sets that preserves adjacency, i.e., vertices $v_1$ and $v_2$ of $G$ are adjacent if and only if $f\left(v_1\right)$ and $f\left(v_2\right)$ of $G_1$ are adjacent.
We denote the isomorphism of graphs $G$ and $G_1$ by $G \cong G_1$.

Solution. Let us manipulate “pins” and “rubber bands” of $G$ : first we flip $v_2 v_5$, then stretch it (Figure 10.2).

Lo and behold, we end up with a graph identical to $G_1$. $G$ and $G_1$ are isomorphic.

Needless to say, two isomorphic graphs must have equal numbers of vertices and edges; therefore, the numbers of vertices and edges are what we call invariants of graphs (invariants are characteristics shared by all isomorphic graphs). The equality of these two invariants, however, is not sufficient to prove the isomorphism of two graphs.
Example 10.2. Are the graphs in Figure $10.3$ isomorphic?
Solution. Even though $G$ and $G_1$ have equal numbers of vertices and edges, they are not isomorphic. Under no one-to-one correspondence of vertices can adjacency be preserved, because $G_1$ contains a vertex $v$ of degree one. This vertex must correspond under isomorphic correspondence to a vertex of degree one in $G$. But $G$ has no such vertex!

While solving Example 10.2, we discovered a new invariant of graphs: the degrees of its vertices. Do we have enough invariants to guarantee isomorphism of two graphs? In other words, given $G$ and $G_1$ with equal numbers of vertices and edges and equal sequences of vertex degrees, do $G$ and $G_1$ have to be isomorphic?

## 数学代写|组合数学代写Combinatorial mathematics代考|Solutions to Exercises

9.1. (a) 我们可以通过对n的归纳来证明n，有一个n人的聚会，使得每个整数0,1,…,n−2都表现为一个聚会成员的熟人数目。

GG1G≅G1

GG1G1v一级的。该顶点必须在同构对应下对应于中的一阶顶点。但是没有这样的顶点！GG

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

Days
Hours
Minutes
Seconds

# 15% OFF

## On All Tickets

Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)