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

statistics-lab™ 为您的留学生涯保驾护航 在代写图论Graph Theory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。我们的专家在代写图论Graph Theory代写方面经验极为丰富，各种代写图论Graph Theory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

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

Coloring a map or graph in such a way that it satisfies the four-color theorem is like solving an ill-conditioned logic problem. Why “ill-conditioned”? A typical logic problem gives you exactly the right amount of information such that if you solve the problem correctly, there is exactly one answer to the problem. When coloring an empty map or graph, the information given in the problem is insufficient, which results in multiple solutions to the problem.
A graph that is constrained such that it is four-colorable only 24 ways can be solved like a well-conditioned logic problem once you set the colors of the vertices of one triangular face. Once these three colors are set, for the most restrictive graphs there is only one way to color the remaining vertices. Logic may be applied to work out the colors of the remaining regions.

If the graph it isn’t highly constrained, meaning that that there are more than 24 ways to color it (or equivalently, after setting the colors of the vertices of one triangular face, there are multiple ways to color the remaining vertices), then a logic table will be ill-conditioned. There won’t be a single solution for coloring the remaining vertices.

Let’s first look at a highly restrictive graph (the corresponding map is shown on the left). If we set B to be color 1, C to be color 2, and D to be color 3 , there is only one solution.

For the map and corresponding graph shown on the previous page:

• A connects to B, C, D, F, and G (but not E).
• B connects to $\mathrm{A}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}$, and $\mathrm{G}$.
• C connects to $A, B, D, E$, and $F$ (but not $G$ ).
• $\mathrm{D}$ connects to $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ (but not $\mathrm{E}, \mathrm{F}$, or $\mathrm{G}$ ).
• E connects to $\mathrm{B}, \mathrm{C}$, and $\mathrm{F}$ (but not $\mathrm{A}, \mathrm{D}$, or $\mathrm{G}$ ).
• F connects to $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{E}$, and $\mathrm{G}$ (but not $\mathrm{D}$ ).
• $\mathrm{G}$ connects to $\mathrm{A}, \mathrm{B}$, and $\mathrm{F}$ (but not $\mathrm{C}, \mathrm{D}$, or $\mathrm{E}$ ).
We can set up an equivalent logic problem based on the sharing b regions. Since $B, C$, and D are the vertices of a triangular face, thes regions must be three different colors. We may assign colors 1,2 , ar these regions in our logic table.

## 数学代写|图论作业代写Graph Theory代考|three-coloring

In this example, even after setting the first three colors, the problem remains ill-conditioned since the problem still has multiple solutions. For example, B can’t be the same as $\Lambda$ or $C$, but be could be the same as E. Therefore, B can be color 3 or 4 . Similarly, D can be color 1 or 4 and $F$ can be color 2 or 4 . This allows four possible solutions. One is $B=3, D=1$, and $F=2$. For the other three solutions, change one of these colors to 4 . Note that two regions can’t both be color 4 since regions $B, D$, and $F$ are all connected to one another.

What would a logic table look like if a solution didn’t exist? One example is to work out a logic table for three-coloring (not four-coloring) of the $\mathrm{K}_{4}$ graph. In the following logic table, region $D$ can’t be colors 1,2 , or 3 because it can’t be the same color as regions $A, B$, or $C$, which results in no solution (unless you introduce color 4). The only way to solve the logic table above is to allow region $\mathrm{D}$ to be a fourth color. This proves that $\mathrm{K}_{4}$ is four=colorable, but not three-colorable.

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

Proceed to complete the logic table for the partially colored graph below. If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means. Proceed to complete the logic table for the partially colored graph below. If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means.、

Proceed to complete the logic table for the partially colored graph below. If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means.Proceed to complete the logic table for the partially colored graph below. If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means. Note: This logic table only has three colors. The goal for this logic table is to determine if this graph is or isn’t three-colorable (not whether it is four-colorable).

Challenge problem 1: A $\mathrm{K}{4}$ subgraph is four-colorable, whereas a $\mathrm{K}{5}$ subgraph isn’t four-colorable. Is it possible for a graph to contain a subgraph that is a subdivision of $\mathrm{K}{5}$ (which would make the graph nonplanar according to Kuratowski’s theorem) to be four-colorable? Either provide an example, prove that it isn’t possible, or argue why it would be very difficult to determine. Similarly, is it possible for a graph to contain a subgraph that is a subdivision of $\mathrm{K}{4}$ to be three-colorable? Either provide an example, prove that it isn’t possible, or argue why it would be very difficult to determine.
Could the answers to these questions help to prove the four-color theorem? Explain. Are the answers to the above questions relevant to the challenge problem from Chapter 6? Explain.

Note: The answer key doesn’t include answers to the challenge problems. These problems are intended to encourage you to think about the ideas. However, you may wish to consider how this problem relates to Chapter $27 .$

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

• A 连接到 B、C、D、F 和 G（但不是 E）。
• B 连接到一种,C,D,和,F， 和G.
• C 连接到一种,乙,D,和， 和F（但不是G ).
• D连接到一种,乙， 和C（但不是和,F， 或者G).
• E 连接到乙,C， 和F（但不是一种,D， 或者G).
• F 连接到一种,乙,C,和， 和G（但不是D).
• G连接到一种,乙， 和F（但不是C,D， 或者和）。
我们可以建立一个基于共享 b 区域的等价逻辑问题。自从乙,C, D 是三角形面的顶点，这些区域必须是三种不同的颜色。我们可以在逻辑表中分配颜色 1,2 ，这些区域。

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## 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。

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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。

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The graphs above are incomplete. These figures only show a vertex with degree four (vertex E), its nearest neighbors (A, B, C, and D), and segments of A-C Kempe chains. The entire graphs would also contain several other vertices (especially, more colored the same as B or D) and enough edges to be MPG’s. The left figure has A connected to $C$ in a single section of an A-C Kempe chain (meaning that the vertices of this chain are colored the same as A and C). The left figure shows that this A-C Kempe chain prevents B from connecting to $\mathrm{D}$ with a single section of a B-D Kempe chain. The middle figure has A and C in separate sections of A-C Kempe chains. In this case, B could connect to D with a single section of a B-D Kempe chain. However, since the A and C of the vertex with degree four lie on separate sections, the color of C’s chain can be reversed so that in the vertex with degree four, C is effectively recolored to match A’s color, as shown in the right figure. Similarly, D’s section could be reversed in the left figure so that D is effectively recolored to match B’s color.

Kempe also attempted to demonstrate that vertices with degree five are fourcolorable in his attempt to prove the four-color theorem [Ref. 2], but his argument for vertices with degree five was shown by Heawood in 1890 to be insufficient [Ref. 3]. Let’s explore what happens if we attempt to apply our reasoning for vertices with degree four to a vertex with degree five.

## 数学代写|图论作业代写Graph Theory代考|The previous diagrams

The previous diagrams show that when the two color reversals are performed one at a time in the crossed-chain graph, the first color reversal may break the other chain, allowing the second color reversal to affect the colors of one of F’s neighbors. When we performed the $2-4$ reversal to change B from 2 to 4 , this broke the 1-4 chain. When we then performed the 2-3 reversal to change E from 3, this caused C to change from 3 to 2 . As a result, F remains connected to four different colors; this wasn’t reversed to three as expected.
Unfortunately, you can’t perform both reversals “at the same time” for the following reason. Let’s attempt to perform both reversals “at the same time.” In this crossed-chain diagram, when we swap 2 and 4 on B’s side of the 1-3 chain, one of the 4’s in the 1-4 chain may change into a 2, and when we swap 2 and 3 on E’s side of the 1-4 chain, one of the 3’s in the 1-3 chain may change into a 2 . This is shown in the following figure: one 2 in each chain is shaded gray. Recall that these figures are incomplete; they focus on one vertex (F), its neighbors (A thru E), and Kempe chains. Other vertices and edges are not shown.

Note how one of the 3’s changed into 2 on the left. This can happen when we reverse $\mathrm{C}$ and $\mathrm{E}$ (which were originally 3 and 2 ) on E’s side of the 1-4 chain. Note also how one of the 4’s changed into 2 on the right. This can happen when we reverse B and D (which were originally 2 and 4) outside of the 1-3 chain. Now we see where a problem can occur when attempting to swap the colors of two chains at the same time. If these two 2’s happen to be connected by an edge like the dashed edge shown above, if we perform the double reversal at the same time, this causes two vertices of the same color to share an edge, which isn’t allowed. We’ll revisit Kempe’s strategy for coloring a vertex with degree five in Chapter $25 .$

## 数学代写|图论作业代写Graph Theory代考|The shading of one section of the B-R

• MPG 是三角测量的。它由具有三个边和三个顶点的面组成。
• 每个面的三个顶点必须是三种不同的颜色。
• 每条边由两个相邻的三角形共享，形成一个四边形。
• 每个四边形将有 3 或 4 种不同的颜色。如果与共享边相对的两个顶点恰好是相同的颜色，则它有 3 种颜色。
• 对于每个四边形，四个顶点中的至少 1 个顶点和最多 3 个顶点具有任何颜色对的颜色。例如，具有 R、G、B 和G有 1 个顶点R−是和3个顶点乙−G，或者您可以将其视为 1 个顶点乙−是和3个顶点G−R，或者您可以将其视为 BR 的 2 个顶点和 GY 的 2 个顶点。在后一种情况下，2G’ 不是同一链的连续颜色。
• 当您将更多三角形组合在一起（四边形仅组合两个）并考虑可能的颜色时，您将看到 Kempe 的部分

• 画一张R顶点和一个是由边连接的顶点。
• 如果一个新顶点连接到这些顶点中的每一个，它必须是乙或者G.
• 如果一个新顶点连接到 R 而不是是，可能是是,乙， 或者G.
• 如果一个新的顶点连接到是但不是R，可能是R,乙， 或者G.
• RY 链要么继续增长，要么被 B 包围，G.
• 如果你关注 B 和 G，你会为它的链条得出类似的结论。
• 如果一条链条完全被其对应物包围，则链条的新部分可能会出现在其对应物的另一侧。
Kempe 证明了所有具有四阶的顶点（那些恰好连接到其他四个顶点的顶点）都是四色的 [Ref. 2]。例如，考虑下面的中心顶点。

## 数学代写|图论作业代写Graph Theory代考|In the previous figure

• A 和 C 或者是 AC Kempe 链的同一部分的一部分，或者它们各自位于 AC Kempe 链的不同部分。（如果一种和C例如，是红色和黄色的，则 AC 链是红黄色链。） – 如果一种和C每个位于 AC Kempe 链的不同部分，其中一个部分的颜色可以反转，这有效地重新着色 C 以匹配 A 的颜色。如果 A 和 C 是 AC Kempe 链的同一部分的一部分，则 B 和 D每个都必须位于 BD Kempe 链的不同部分，因为 AC Kempe 链将阻止任何 BD Kempe 链从 B 到达 D。（如果乙和D是蓝色和绿色，例如，那么一种BD Kempe 链是蓝绿色链。）在这种情况下，由于 B 和 D 分别位于 BD Kempe 链的不同部分，因此 BD Kempe 链的其中一个部分的颜色可以反转，这有效地重新着色 D 以匹配 B颜色。– 因此，可以使 C 与 A 具有相同的颜色或使 D 具有与 A 相同的颜色乙通过反转 Kempe 链的分离部分。

Kempe 还试图证明五阶顶点是可四色的，以证明四色定理 [Ref. 2]，但 Heawood 在 1890 年证明他关于五次顶点的论点是不充分的 [Ref. 3]。让我们探讨一下如果我们尝试将我们对度数为四的顶点的推理应用于度数为五的顶点会发生什么。

## 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。