数学代写|图论作业代写Graph Theory代考|EULER’S FORMULA

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数学代写|图论作业代写Graph Theory代考|EULER’S FORMULA

The number of vertices, edges, and faces of a map or a graph are related by Euler’s formula, provided that we consider the unbounded surrounding area as one of the faces. We define the symbols $\mathrm{V}, \mathrm{E}$, and $\mathrm{F}$ as follows:

• $\mathrm{V}$ is the number of vertices.
• $E$ is the number of edges.
• $F$ is the number of faces. Remember to count the unbounded surrounding area.

According to Euler’s formula [Ref. 4], for a map or a graph (or even a polyhedron):
$$\mathrm{V}+\mathrm{F}=\mathrm{E}+2$$
The number of vertices plus the number of faces is two more than the number of edges.

For the map shown above:

• $\mathrm{V}=20$. On a map, vertices are where the edges intersect. We marked the vertices with small dots $(\bullet)$ on the diagram above to help you count them. There are 5 for the inner pentagon, 10 for the decagon, and another 5 for the outer pentagon.
• $E=30$. On a map, an edge is any line segment (or part of one) or curve that separates two regions (or which separates a region from the unbounded surrounding area). The inner pentagon has 5 , there are 5 connecting the inner pentagon to the decagon, there are 10 along the decagon, there are 5

connecting the decagon to the outer pentagon, and there are 5 along the outer pentagon.

• $\mathrm{F}=12$. On a map, the faces are regions. The twelve regions are $\mathrm{A}, \mathrm{B}, \mathrm{C}$, $\mathrm{D}, \mathrm{E}, \mathrm{F}, \mathrm{G}, \mathrm{H}, \mathrm{I}, \mathrm{J}, \mathrm{K}$, and L. This includes the unbounded surrounding area as a face.

Check the formula for the map on the left: $V+F=20+12=32$ and $E+2=$ $30+2=32$. Since $\mathrm{V}+\mathrm{F}$ and $\mathrm{E}+2$ both equal 32 for this map, we see that Euler’s formula agrees with it.

数学代写|图论作业代写Graph Theory代考|For the graph shown above

For the graph shown above:

• $\mathrm{V}=12$. On a graph, the vertices are regions where lines intersect. These are regions $A, B, C, D, E, F, G, H, I, J, K$, and L.
• $E=30$. On a graph, each edge is a line or curve that connects a pair of regions. The edges are $A B, A E, A F, A G, A K, B C, B F, B G, B H$ (but don’t count BA since that’s the same as AB, which was already counted), CD, CF, CH, CI, DE, DF, DI, DJ, EF, EJ, EK, GH, GK, GL, HI, HL, IJ, IL, JK, $\mathrm{JL}$, and KL.
• $F=20$. On a graph, the faces are areas formed between edges. The faces are $A B F, A B G, A E F, A E K, A G K, B C F, B C H, B G H$ (but don’t count BAF or $B A G$ since they are the same as $A B F$ and $A B G$, which were already counted), CDF, CDI, CHI, DEF, DEJ, DIJ, EJK, GKL, HIL, IJL, JKL, and the infinite area lying outside of the graph.

Note that the graph shown above corresponds to the map on the previous page, where we included region $L$ explicitly on the graph and still counted the infinite area lying outside of the graph as its own face. On the map, L represented the unbounded surrounding area and was counted only as a face. If you are wondering if this may be inconsistent, there is a reason for drawing

the map and its corresponding graph as we have done here. The values of $\mathrm{V}$ and $\mathrm{F}$ are swapped for the map compared to the graph when we draw them this way. Doing so emphasizes the fact that the graph is considered to be a dual representation of the map. (We’ll elaborate on what we mean by a dual representation in the solution to Problem 1.)

For the graph, V + F $=12+20=32$ and $\mathrm{E}+2=30+2=32$, agreeing with Euler’s formula. Note that the map and graph give the same values of V, F, and $\mathrm{E}$ as a dodecahedron (a 12 -sided polyhedron) and its dual polyhedron, which is an icosahedron (a 20-sided polyhedron). If you imagine cutting the map along the five outer radial edges and then folding the map up, you might be able to visualize how it can fold into the shape of a dodecahedron.

Note that if you remove region L from the previous graph, Euler’s formula will still work. In that case, you would have one less vertex (region) so $\mathrm{V}=$ 11, five fewer edges (since L connects to $\mathrm{G}, \mathrm{H}$, I, J, and $\mathrm{K}$ ) so $\mathrm{E}=25$, and four fewer faces (you wouldn’t have GKL, HIL, IJL, or JKL) so $\mathrm{F}=16$ (note that we still include the infinite area outside the graph as a face). In this case $\mathrm{V}+\mathrm{F}=11+16=27$ and $\mathrm{E}+2=25+2=27$, still satisfying Euler’s formula.

数学代写|图论作业代写Graph Theory代考|in the graph

When we include region $\mathrm{L}$ in the graph, note that $2 \mathrm{E}=3 \mathrm{~F}$. That is, two times the number of edges is equal to three times the number of faces. When we include reggion $L, 2 E=2(30)=60$ and $3 F=3(20)=60$. The formula $2 E=3 F$ applies to any MPG. (Recall that MPG stands for “maximal planar graph” and that for a MPG every face in the graph has three “sides;” we’ll call each edge a side whether it appears straight or curved.) To get the number of faces $\mathrm{F}$ from the number of edges for a MPG, divide $\mathrm{E}$ by 3 because each face has 3 edges and then multiply this by 2 because each edge is shared by 2 faces: $F$ $=2 E / 3$, from which it follows that $F / E=2 / 3$. In the previous graph, for example, note that edge $\mathrm{AE}$ is part of both triangles AEF and AEK. The ratio of $F$ to $E$ is a maximum for a MPG. In this case, the ratio is $2 / 3$, which is approximately $0.67$ when rounded to two decimal places. (Note that if region $\mathrm{L}$ isn’t included in the graph, it would be a PG, not a MPG. In that case, the ratio is slightly smaller: $16 / 25=0.64$.) In contrast, if every face were a pentagon, the ratio would be $2 / 5=0.40$.

Now we will do a little arithmetic to obtain a classic result related to the fourcolor theorem. This is generally done in a more formal, technical manner, but

we’ll try to keep it simple. Note that the formulas from this chapter apply if $\mathrm{V}$ is at least $3 .$

Consider the special case where every vertex of the graph has the same degree, meaning that the same number of edges intersect at every vertex. If every vertex has the same degree and if we define $\mathrm{D}$ to be the degree of each vertex (equal to the number of edges intersecting at each vertex), then $\mathrm{DV} / 2$ $=\mathrm{E}$, which is equivalent to $\mathrm{DV}=2 \mathrm{E}$. This well-known result is referred to as the handshaking lemma. If $\mathrm{V}$ people each shake hands with $\mathrm{D}$ other people, DV/2 equals the number of handshakes that will occur. We divide by two in order to avoid double counting. For example, if there are 8 people at a gathering and each person shakes hands with 7 other people, there are 8(7)/2 $=56 / 2=28$ handshakes. If their names are Mr. A, Mr. B, Mr. C, etc., thru Mr. H, then we divide by 2 because 56 counts Mr. A shaking hands with Mr. B and Mr. B shaking hands with Mr. A separately, and similarly for all other pairs (like Mr. A and Mr. C which is the same as Mr. C with Mr. A).

What if the vertices don’t all have the same degree? This will be the case with most graphs. We can still use the formula DV $=2 E$ provided that we interpret D as the average degree of the vertices.

If we multiply both sides of Euler’s formula by 2, we get $2 \mathrm{~V}+2 \mathrm{~F}=2 \mathrm{E}+4$. Substitute the equation $\mathrm{DV}=2 \mathrm{E}$ into this equation to obtain $2 \mathrm{~V}+2 \mathrm{~F}=\mathrm{DV}+$ 4. Multiply both sides by 3 to get $6 \mathrm{~V}+6 \mathrm{~F}=3 \mathrm{DV}+12$.

For a MPG, we noted that $2 \mathrm{E}=3 \mathrm{~F}$. Combine $\mathrm{DV}=2 \mathrm{E}$ with $2 \mathrm{E}=3 \mathrm{~F}$ to see that $\mathrm{DV}=3 \mathrm{~F}$. Multiply both sides by 2 to get $2 \mathrm{DV}=6 \mathrm{~F}$. Substitute this into the last equation from the previous paragraph to obtain $6 \mathrm{~V}+2 \mathrm{DV}=3 \mathrm{DV}+$ 12. Subtract 3DV from both sides to get $6 \mathrm{~V}-\mathrm{DV}=12$. Factor out the $\mathrm{V}$ to obtain $\mathrm{V}(6-\mathrm{D})=12$. We finally obtain the formula $\mathrm{V}=12 /(6-\mathrm{D})$ for a MPG (recall that every face of a MPG has three sides). Recall that $V$ is the number of vertices (which are regions) and that $\mathrm{D}$ is the average number of edges intersecting at each vertex on the graph.

When each vertex has degree $\mathrm{D}=2$, we get $\mathrm{V}=12 /(6-2)=12 / 4=3$. Every vertex on the MPG will connect to 2 edges if there are exactly 3 vertices.

数学代写|图论作业代写Graph Theory代考|EULER’S FORMULA

• 在是顶点的数量。
• 和是边数。
• F是面数。记得计算无界的周边区域。

• 在=20. 在地图上，顶点是边相交的地方。我们用小点标记了顶点(∙)在上图中帮助您计算它们。内五边形有 5 个，十边形有 10 个，外五边形有 5 个。
• 和=30. 在地图上，边是分隔两个区域（或将一个区域与无界周边区域分隔开）的任何线段（或其中的一部分）或曲线。内五边形有 5 个，连接内五边形和十边形的有 5 个，沿十边形有 10 个，有 5 个

• F=12. 在地图上，面是区域。这十二个地区是一种,乙,C, D,和,F,G,H,一世,Ĵ,ķ, 和 L. 这包括作为面的无界周围区域。

数学代写|图论作业代写Graph Theory代考|For the graph shown above

• 在=12. 在图上，顶点是线相交的区域。这些是地区一种,乙,C,D,和,F,G,H,一世,Ĵ,ķ， 和我。
• 和=30. 在图上，每条边都是连接一对区域的直线或曲线。边缘是一种乙,一种和,一种F,一种G,一种ķ,乙C,乙F,乙G,乙H（但不要计算 BA，因为它与 AB 相同，已计算在内），CD，CF，CH，CI，DE，DF，DI，DJ，EF，EJ，EK，GH，GK，GL，HI， HL，IJ，伊利诺伊州，JK，Ĵ大号, 和 KL。
• F=20. 在图上，面是在边之间形成的区域。面孔是一种乙F,一种乙G,一种和F,一种和ķ,一种Gķ,乙CF,乙CH,乙GH（但不计 BAF 或乙一种G因为它们与一种乙F和一种乙G，已计算在内）、CDF、CDI、CHI、DEF、DEJ、DIJ、EJK、GKL、HIL、IJL、JKL 以及位于图外的无限区域。

有限元方法代写

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MATLAB代写

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