### 数学代写|图论作业代写Graph Theory代考|COMPLETE GRAPHS

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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

## 数学代写|图论作业代写Graph Theory代考|BIGRAPHS

In a complete graph, every vertex connects to each of the other vertices. The notation $K_{V}$ represents a complete graph with $\mathrm{V}$ vertices. For example, the complete graph shown below, which has $\mathrm{V}=8$ vertices, is represented by $\mathrm{K}_{8}$.A complete graph can be used to represent the following classic handshaking problem. If $\mathrm{V}$ people are in a room and each person shakes hands once with each of the $V-1$ other people, how many handshakes will there be all together? The answer is given by the handshaking lemma: $\mathrm{V}(\mathrm{V}-1) / 2$ equals the number of handshakes. Why? Multiply $\mathrm{V}$ by $(\mathrm{V}-1)$ and then divide by 2 to correct for double counting. To better understand the formula associated with the handshaking lemma, let’s consider the $\mathrm{K}_{6}$ graph shown below.

Let’s list the handshakes:

• A’s handshakes include $\mathrm{AB}, \mathrm{AC}, \mathrm{AD}, \mathrm{AE}$, and $\mathrm{AF}$. That’s five.
• B’s handshakes include BA, BC, BD, BE, and BF. That’s also five, but note that $\mathrm{BA}$ is the same as $\mathrm{AB}$.
• C’s handshakes include CA, CB, CD, CE, and CF. That’s also five, but CA and CB are the same as AC and BC.
• D’s handshakes include DA, DB, DC, DE, and DF. That’s another five, but DA, DB, and DC are the same as AD, BD, and CD.
• E’s handshakes include EA, EB, EC, ED, and EF. That’s five more, but $\mathrm{EA}, \mathrm{EB}, \mathrm{EC}$, and ED are the same as AE, BE, CE, and DE.
• F’s handshakes include FA, FB, FC, FD, and FE. That’s five again, but all of these are repeats. They are the same as AF, BF, CF, DF, and EF.
The previous graph, $\mathrm{K}_{6}$, has $\mathrm{V}=6$ vertices. According to the handshaking lemma, there are $\mathrm{V}(\mathrm{V}-1) / 2=6(5) / 2=15$ handshakes. Look at our list. For A thru F, there are 5 handshakes each, giving us $6(5)=30$ handshakes, but 15 of these are repeated so that there really are only $30 / 2=15$ handshakes. That’s what we meant about dividing by 2 to correct for double counting. For example, BC and CB are the same handshake because they involve the same two people. A complete graph is really asking: given $V$ vertices, how many ways are there to make pairs of them. The answer is $\mathrm{V}(\mathrm{V}-1) / 2$.

Complete graphs with 1 to 10 vertices are illustrated on the following page. A couple of noteworthy complete graphs concerning the four-color theorem include:

• $\mathrm{K}{4}$ is the largest complete graph that is a MPG. $\mathrm{K}{4}$ is also the largest complete graph that can be colored such that the coloring satisfies the fourcolor theorem. As we will explore in later chapters, MPG’s with $\mathrm{K}{4}$ subgraphs tend to have more restrictive coloring (meaning that there tend to be fewer ways to color the graphs compared to graphs that lack $K{4}$ ‘s), vertices with degree three tend to take part in $\mathrm{K}{4}$ subgraphs, and even $\mathrm{K}{4}$ subgraphs without any vertices with degree three have separating triangles. $\mathrm{K}_{4}$ is sometimes referred to as the tetrahedral graph.
• $\mathrm{K}{5}$ is the smallest complete graph that isn’t a MPG. $\mathrm{K}{5}$ is also the smallest complete graph that isn’t four-colorable. As we will explore in Chapter 6, $\mathrm{K}{5}$ plays an important role in determining whether or not a graph is a PG or if it is nonplanar. $\mathrm{K}{5}$ is sometimes referred to as the pentatope graph.

## 数学代写|图论作业代写Graph Theory代考|A complete bipartite graph

The number of edges on a complete graph with $\mathrm{V}$ vertices is given by the formula from the handshaking lemma: $E=V(V-1) / 2$. For example, you can verify that the $\mathrm{K}_{6}$ graph shown above with $\mathrm{V}=6$ vertices has $\mathrm{E}=6(6-1) / 2=$ $6(5) / 2=30 / 2=15$ edges.

A complete bipartite graph has two sets of vertices where every vertex of one set connects to every vertex of the other set. The notation $\mathrm{K}{\mathrm{X}, \mathrm{Y}}$ denotes a complete bipartite graph where set $P$ has $X$ vertices and set $Q$ has $Y$ vertices. A complete bipartite graph has $\mathrm{E}=\mathrm{XY}$ edges because each of the $\mathrm{X}$ vertices in set $P$ connects to each of the $Y$ vertices in set Q. For example, the complete bipartite graph $\mathrm{K}{3,4}$ shown on the following page has $\mathrm{X}=3$ vertices in set $\mathrm{P}, \mathrm{Y}=4$ vertices in set $\mathrm{Q}$, and $\mathrm{E}=\mathrm{XY}=3(4)=12$ edges.

A bipartite graph is also called a bigraph. A bipartite graph (without the word “complete”) is defined to have two sets of vertices where no two vertices of a single set connect to one another. A complete bipartite graph has every vertex of one set connected to every vertex of the other set and viceversa.

The $\mathrm{K}{3,3}$ graph shown above is noteworthy as it concerns $\mathrm{PG}$ ‘s. In Chapter 6 , we will see that $\mathrm{K}{3,3}$ is nonplanar and that $\mathrm{K}{3,3}$ and $\mathrm{K}{5}$ play an important role in determining whether or not a graph is a $P G$ or if it is nonplanar. The $K_{3,3}$ graph is also referred to as the utility graph, based on the following puzzle [Ref. 5]. Can you connect three cottages $(A, B$, and $C$ ) to three utilities (D, E, and F) without crossing the lines? Since the utility graph isn’t planar (as we will show in Chapter 6), the answer is no. Recall that we are defining PG to include any graph that can be drawn in the plane without crossings (even if the graph happens to be drawn in a form that has avoidable crossings).

Observe that complete bipartite graphs are two-colorable. Since the vertices of set P don’t connect to one another, every vertex in set P can be one color $\mathrm{~ ( s u c h h ~ a ̂ s ~ r e ̣ d ) . ~ S i m i l a ̆ r l y , ~ s i n n c e ̀ ~ t h e e ~ v e r r t i c e ́ s}$ another, every vertex in set $\mathrm{Q}$ can be another color (such as blue). For example, in the $\mathrm{K}_{3,4}$ graph shown above, $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ can each be red and $\mathrm{D}$, $\mathrm{E}, \mathrm{F}$, and $\mathrm{G}$ can each be blue.

## 数学代写|图论作业代写Graph Theory代考|EXERCISES

1. A total of 12 people attend a conference. If every person at the conference shakes hands once with every other person, how many handshakes will occur in total?
2. A research project is started with 9 female mathematicians and 7 male mathematicians. If each female shakes hands once with all of the males, each male shakes hands once with all of the females, no female shakes hands with another female, and no male shakes hands with another male, how many handshakes will occur in total?
3. Draw $\mathrm{K}{4}$ as it was drawn in this chapter. Now redraw $\mathrm{K}{4}$ to show that it is a PG. Is $K_{4}$ a MPG? How can you tell?
4. Draw $\mathrm{K}_{5}$. Now redraw the graph with one of its edges removed. Is this new graph (with one edge removed) planar? Is it a MPG? How can you tell?
5. Draw $\mathrm{K}{6}$. How many edges must be removed from $\mathrm{K}{6}$ in order to for the new graph (with edges removed) to be a MPG? Draw the new graph (with edges removed), showing that it can be a MPG. Draw a second graph with the same number of edges removed, which is also a MPG, but where the vertices have different degrees from the first MPG. Now draw a third graph with the same number of edges removed, but which isn’t a MPG.
6. How many edges must be removed from a complete graph with 9 vertices in order to make a MPG with 9 vertices? Will the resulting graph necessarily be a MPG?
7. Show that if you remove $(V-3)(V-4) / 2$ edges from a complete graph with $\mathrm{V}$ vertices that the resulting graph will have the right number of edges to be a MPG with $\mathrm{V}$ vertices. Will the resulting graph necessarily be a MPG?
8. Draw $\mathrm{K}{5}$. How many edges must be removed from $\mathrm{K}{5}$ in order to for the new graph (with edges removed) to be $\mathrm{K}{2,3}$ ? Draw $\mathrm{K}{2,3}$ by removing this

number of edges from $K_{5}$.

1. Draw $\mathrm{K}{6}$. How many edges must be removed from $\mathrm{K}{6}$ in order to for the new graph (with edges removed) to be $\mathrm{K}{3,3}$ ? Draw $\mathrm{K}{3,3}$ by removing this number of edges from $\mathrm{K}{6}$. Can you find more than one way to remove edges from $\mathrm{K}{6}$ to draw a graph that is isomorphic to $\mathrm{K}_{3,3}$ ?
2. If $\mathrm{V}$ is an even number, show that $\mathrm{V}(\mathrm{V}-2) / 4$ edges need to be removed from $K_{V}$ in order for the new graph (with edges removed) to be $K_{N, N}$ with $N=$ $\mathrm{V} / 2$. Will the resulting graph necessarily be $\mathrm{K}_{\mathrm{N}, \mathrm{N}}$ ?

## 数学代写|图论作业代写Graph Theory代考|BIGRAPHS

• A的握手包括一种乙,一种C,一种D,一种和， 和一种F. 那是五个。
• B的握手包括BA、BC、BD、BE和BF。这也是五个，但请注意乙一种是相同的一种乙.
• C 的握手包括 CA、CB、CD、CE 和 CF。那也是五个，但 CA 和 CB 与 AC 和 BC 相同。
• D 的握手包括 DA、DB、DC、DE 和 DF。那是另外五个，但 DA、DB 和 DC 与 AD、BD 和 CD 相同。
• E的握手包括EA、EB、EC、ED和EF。还有五个，但是和一种,和乙,和C, 和 ED 与 AE、BE、CE 和 DE 相同。
• F的握手包括FA、FB、FC、FD和FE。又是五次，但所有这些都是重复的。它们与 AF、BF、CF、DF 和 EF 相同。
上一张图，ķ6， 拥有在=6顶点。根据握手引理，有在(在−1)/2=6(5)/2=15握手。看看我们的清单。对于 A 到 F，每个有 5 次握手，给我们6(5)=30握手，但其中 15 次是重复的，所以真的只有30/2=15握手。这就是我们要除以 2 来纠正重复计算的意思。例如，BC 和 CB 是相同的握手，因为它们涉及相同的两个人。一个完整的图表真的在问：给定在顶点，有多少种方法可以使它们成对。答案是在(在−1)/2.

• ķ4是最大的完整图，它是 MPG。ķ4也是可以着色的最大完全图，使得着色满足四色定理。正如我们将在后面的章节中探讨的那样，MPG 与ķ4子图往往具有更多限制性的着色（这意味着与缺少的图相比，为图着色的方法往往更少ķ4’s)，度数为 3 的顶点倾向于参与ķ4子图，甚至ķ4没有任何三度顶点的子图具有分隔三角形。ķ4有时称为四面体图。
• ķ5是不是 MPG 的最小完整图。ķ5也是最小的非四色完整图。正如我们将在第 6 章中探讨的那样，ķ5在确定图是 PG 还是非平面图方面起着重要作用。ķ5有时称为五边形图。

## 数学代写|图论作业代写Graph Theory代考|EXERCISES

1. 共有12人参加会议。如果会议上每个人与其他人握手一次，总共会发生多少次握手？
2. 一个研究项目由 9 名女性数学家和 7 名男性数学家开始。如果每个女性与所有男性握手一次，每个男性与所有女性握手一次，没有女性与另一个女性握手，也没有男性与另一个男性握手，总共会发生多少次握手？
3. 画ķ4正如本章所描绘的那样。现在重绘ķ4表明它是一个PG。是ķ4MPG？你怎么知道？
4. 画ķ5. 现在重新绘制图形，删除其中一条边。这个新图（去掉一条边）是平面的吗？是MPG吗？你怎么知道？
5. 画ķ6. 必须删除多少条边ķ6为了使新图（已删除边缘）成为 MPG？绘制新图（去掉边），表明它可以是 MPG。绘制第二个图，删除相同数量的边，这也是一个 MPG，但顶点与第一个 MPG 的度数不同。现在绘制第三张图，删除了相同数量的边，但它不是 MPG。
6. 必须从具有 9 个顶点的完整图中删除多少条边才能生成具有 9 个顶点的 MPG？结果图一定是 MPG 吗？
7. 显示，如果你删除(在−3)(在−4)/2来自完整图的边在结果图将具有正确数量的边以成为 MPG 的顶点在顶点。结果图一定是 MPG 吗？
8. 画ķ5. 必须删除多少条边ķ5为了使新图（已删除边缘）成为ķ2,3? 画ķ2,3通过删除这个

1. 画 $\mathrm{K} {6 }.H这在米一种n是和dG和s米在s吨b和r和米这在和dFr这米\数学{K{6一世n这rd和r吨这F这r吨H和n和在Gr一种pH(在一世吨H和dG和sr和米这在和d)吨这b和\数学{K {3,3?Dr一种在\数学{K {3,3b是r和米这在一世nG吨H一世sn在米b和r这F和dG和sFr这米\数学{K{6.C一种n是这在F一世nd米这r和吨H一种n这n和在一种是吨这r和米这在和和dG和sFr这米\数学{K{6吨这dr一种在一种Gr一种pH吨H一种吨一世s一世s这米这rpH一世C吨这\ mathrm {K} _ {3,3}$?
2. 如果在是偶数，证明在(在−2)/4边缘需要从ķ在为了使新图（已删除边缘）成为ķñ,ñ和ñ= 在/2. 结果图一定是ķñ,ñ ?

## 有限元方法代写

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## 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。