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

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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|GRAPHS

Recall from Chapter 3 that a MPG (which stands for “maximal planar graph”) is a PG that is triangulated, meaning that every face is a triangle (keeping in mind that one of its edges may be curved), including the infinite area outside of the graph. A MPG is “maximal” in the sense that if another edge were added to anywhere to the MPG, then the graph would no longer be a PG (there would be an unavoidable crossing).

This chapter focuses on the important question, “How can you tell whether or not a graph is a MPG?” That is, if a graph is drawn with a crossing, is the crossing avoidable or not?

The simplest test is an exclusion test. What is an exclusion test? If a graph fails the exclusion test, then it isn’t a MPG. However, if the graph passes the exclusion test, we will need more information before we can determine whether or not it is a MPG. This exclusion test is useful because it can rule many graphs out very quickly, but the exclusion test has limited use because when a graph passes the exclusion test, another test is still needed. To perform the exclusion test, count the number of vertices and edges and see if these values satisfy Euler’s formula (or related formulas from Chapter 4) for a MPG. If a graph with at least 3 vertices is a MPG, the number of edges must equal $E=3 V-6$. If $E$ is greater than $3 V-6$, the graph isn’t MPG (and it isn’t even a PG).

For example, consider the graphs on the next page which have $\mathrm{V}=6$ vertices. A MPG with 6 vertices should have $E=3(6)-6=18-6=12$ edges.

Counting edges can get tricky when a graph has numerous vertices, but fortunately there is a simple trick to make this easy. The number of edges equals the sum of the degrees of the vertices divided by two.

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

The diagrams above show how the exclusion test can determine that the right graph above isn’t a MPG. As mentioned previously, a MPG with $V=6$ vertices should have $\mathrm{E}=12$ edges.

• The sum of the degrees for the left graph is $4+3+4+3+4+4=22$, so it has $22 \div 2=11$ edges. Since 11 is less than 12 , the left graph isn’t a MPG. Euler’s formula doesn’t prove that it’s a PG; this we were able to determine by redrawing the graph without crossings.
• The sum of the degrees for the middle graph is $4+4+4+4+4+4=24$, so it has $24 \div 2$ edges. Since this equals 12 , the middle graph could be a MPG. Euler’s formula doesn’t prove that it’s a MPG, but it doesn’t exclude this graph from being one. We were able to determine that it was a PC by redrawing the graph without crossings. The redrawn graph plus Euler’s formula then tells us that it’s a MPG.
• The sum of the degrees for the right graph is $4+5+4+4+5+4=26$, so it has $26 \div 2=13$ edges. Since 13 is greater than 12 , the right graph isn’t planar (it has too many edges to be a MPG, and a MPG has the maximum number of edges for a PG). We don’t need to try to redraw this graph to see whether or not it has unavoidable crossings. Since Euler’s formula excludes the right graph from being a MPG (or a PG), we know that it can’t be redrawn without at least one crossing.
Now let’s look at an example where Euler’s formula doesn’t help. Both graphs below have $\mathrm{V}=6$ vertices. Both graphs agree with Euler’s formula for a MPG: $E=3 V-6=3(6)-6=18-6=12$ edges. Yet the left graph is a MPG, whereas the right graph isn’t planar. In this example, the exclusion test doesn’t help.
• The sum of the degrees for the left graph is $4+5+3+4+5+3=24$, so it has $24 \div 2$ edges. Since this equals 12 , the left graph could be a MPG. Euler’s formula doesn’t prove that it’s a MPG, but it doesn’t exclude this graph from being one. We were able to determine that it was a PG by redrawing the graph without crossings. The redrawn graph plus Euler’s formula then tells us that it’s a MPG.
• The sum of the degrees for the right graph is $4+4+4+4+5+3=24$, so it has $24 \div 2$ edges. Since this equals 12 , the right graph could be a MPG. Euler’s formula doesn’t prove whether it’s a MPG, but it doesn’t exclude this graph from being one. If you try to redraw the graph with the same edge-sharing, it will have at least one crossing. Why is the left graph planar, but not the right graph? In the left graph, we were able to separate the inside edges into a group of three inside edges and another group of three outside edges without crossings. In the right graph, if you attempt to do this, it won’t work because AD, BE, and CF triple cross (whereas AD, AE, and BF do not), such that at least one pair of these edges will cross inside or outside the polygon.

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

Fortunately, there are other tests besides using Euler’s formula as an exclusion test.

One of these tests we have already been using: the redrawing test. If it is possible to redraw a graph without any crossings (meaning any crossings previously shown were avoidable), then the graph is planar. If a graph is planar and is also fully triangulated (which is the case if it satisfies Euler’s formula for a MPG), then the graph is a MPG. On the other hand, if it isn’t

possible to redraw a graph without at least one crossing, the graph isn’t planar.

The redrawing test is inconvenient, especially for a graph with a large number of vertices (and thus a large number of edges, too). Sometimes, there is a way to redraw a graph without crossings that isn’t easy to think of.

The redrawing test is simpler when all of the vertices lie at the corners of a closed polygon. It’s important that the polygon be closed; if it’s missing an edge, the following rule won’t apply. As we’ll explore in Chapter 13 (regarding Hamiltonian cycles), any MPG can be drawn with all of its vertices on the corners of a closed polygon unless it has separating triangles (which we’ll define in Chapter 12), so the polygon version of the redrawing test will actually apply to the most important examples concerning the fourcolor theorem. For a graph where all of the vertices lie at the corners of a closed polygon, it is a MPG if all of these apply:

• It has $\mathrm{E}=3 \mathrm{~V}-6$ edges, as required by Euler’s formula for a MPG.
• $V$ edges form the outline of a closed polygon.
• $V-3$ edges can be drawn inside the polygon without crossing.
• $\mathrm{V}-3$ different edges can be drawn outside the polygon without crossing.
It’s interesting to note that the two sets of $\mathrm{V}-3$ edges are interchangeable; you can put the inside edges outside and vice-versa. We will explore this more fully in Chapter 14. For now, we will focus on how this helps us determine whether or not a graph is a MPG. (Note that our polygon redrawing test is focused on possible MPG’s, not more general PG’s.)

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

• 左图的度数之和为4+3+4+3+4+4=22，所以它有22÷2=11边缘。由于 11 小于 12 ，左图不是 MPG。欧拉公式并不能证明它是 PG；我们可以通过重绘没有交叉的图形来确定这一点。
• 中间图的度数总和是4+4+4+4+4+4=24，所以它有24÷2边缘。因为这等于 12 ，所以中间的图可以是 MPG。欧拉公式并不能证明它是 MPG，但它并不排除这张图是一张图。我们能够通过重新绘制没有交叉的图形来确定它是一台 PC。重绘的图形加上欧拉公式告诉我们它是一个 MPG。
• 右图的度数总和是4+5+4+4+5+4=26，所以它有26÷2=13边缘。由于 13 大于 12 ，因此右图不是平面图（它的边太多而不能成为 MPG，而 MPG 具有 PG 的最大边数）。我们不需要尝试重新绘制此图来查看它是否有不可避免的交叉点。由于欧拉公式将右图排除在 MPG（或 PG）之外，因此我们知道它不能在没有至少一个交叉点的情况下重新绘制。
现在让我们看一个欧拉公式不起作用的例子。下面两张图都有在=6顶点。两张图都符合欧拉的 MPG 公式：和=3在−6=3(6)−6=18−6=12边缘。然而，左图是 MPG，而右图不是平面图。在这个例子中，排除测试没有帮助。
• 左图的度数之和为4+5+3+4+5+3=24，所以它有24÷2边缘。因为这等于 12 ，所以左图可以是 MPG。欧拉公式并不能证明它是 MPG，但它并不排除这张图是一张图。我们能够通过重新绘制没有交叉的图形来确定它是一个 PG。重绘的图形加上欧拉公式告诉我们它是一个 MPG。
• 右图的度数总和是4+4+4+4+5+3=24，所以它有24÷2边缘。因为这等于 12 ，所以右图可能是 MPG。欧拉公式并不能证明它是否是一个 MPG，但它并不排除这个图是一个。如果您尝试使用相同的边共享重新绘制图形，它将至少有一个交叉点。为什么左图是平面的，而右图不是？在左图中，我们能够将内边分成一组三个内边和另一组三个没有交叉的外边。在右图中，如果您尝试这样做，它将无法正常工作，因为 AD、BE 和 CF 三重交叉（而 AD、AE 和 BF 没有），这样至少一对这些边将在内部交叉或多边形之外。

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

• 它有和=3 在−6根据欧拉公式对 MPG 的要求。
• 在边形成封闭多边形的轮廓。
• 在−3边缘可以在多边形内部绘制而不会交叉。
• 在−3可以在多边形之外绘制不同的边而不会交叉。
有趣的是，这两组在−3边缘是可互换的；您可以将内边缘放在外面，反之亦然。我们将在第 14 章更全面地探讨这一点。现在，我们将关注它如何帮助我们确定一个图是否是 MPG。（请注意，我们的多边形重绘测试侧重于可能的 MPG，而不是更一般的 PG。）

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