如果你也在 怎样代写图论Graph Theory 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。图论Graph Theory有趣的部分原因在于，图可以用来对某些问题中的情况进行建模。这些问题可以在图表的帮助下进行研究(并可能得到解决)。因此，图形模型在本书中经常出现。然而，图论是数学的一个领域，因此涉及数学思想的研究-概念和它们之间的联系。我们选择包含的主题和结果是因为我们认为它们有趣、重要和/或代表主题。

图论Graph Theory通过熟悉许多过去和现在对图论的发展负责的人，可以增强对图论的欣赏。因此，我们收录了一些关于“图论人士”的有趣评论。因为我们相信这些人是图论故事的一部分，所以我们在文中讨论了他们，而不仅仅是作为脚注。我们常常没有认识到数学是一门有生命的学科。图论是人类创造的，是一门仍在不断发展的学科。

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。

数学代写|图论作业代写Graph Theory代考|Union and Intersection of Graphs

Let $G_1=\left(V_1, E_1\right)$ and $G_2=\left(V_2, E_2\right)$ be two graphs. The union of $G_1$ and $G_2$, denoted by $G_1 \cup G_2$, is another graph $G_3=\left(V_3, E_3\right)$, whose vertex set $V_3=V_1 \cup V_2$ and edge set $E_3=E_1 \cup E_2$.

Similarly, the intersection of $G_1$ and $G_2$, denoted by $G_1 \cap G_2$, is another graph $G_4=\left(V_4, E_4\right)$, whose vertex set $V_4=V_1 \cap V_2$ and edge set $E_4=E_1 \cap E_2$.

Figures 2.10(a) and (b) show two graphs $G_1$ and $G_2$, and Figs. 2.10(c) and (d) illustrate their union and intersection, respectively.

Clearly, we can define the union and intersection of more than two graphs in a similar way. These operations on graphs can be used to solve many problems very easily. We now present such an application of these operations on two graphs [2].
Suppose there are $h+g$ people in a party; $h$ of them are hosts and $g$ of them are guests. Each person shakes hands with each other except that no host shakes hands with any other host. The problem is to find the total number of handshakes. As usual, we transform the scenario into a graph problem as follows. We form a graph with $h+g$ vertices; $h$ of them are black vertices, representing the hosts and the other $g$ vertices are white, representing the guests. The edges of the graph represent the handshakes. Thus, there is an edge between every pair of vertices except for that there is no edge between any pair of black vertices. Thus, the problem now is to count the number of edges in the graph thus formed. The graph is illustrated for $h=3$ and $g=4$ in Fig. 2.11(a).

To solve the problem, we note that the graph can be thought of as a union of two graphs: a complete graph $K_g$ and a complete bipartite graph $K_{h, g}$ as illustrated in Fig. 2.11(b). Since there is no common edge between the two graphs, their intersection contains no edges. Thus, the total number of edges in the graph (i.e., the total number of handshakes in the party) is $n(n-1) / 2+m \times n$.

数学代写|图论作业代写Graph Theory代考|Complement of a Graph

The complement of a graph $G=(V, E)$ is another graph $\bar{G}=(V, \bar{E})$ with the same vertex set such that for any pair of distinct vertices $u, v \in V,(u, v) \in \bar{E}$ if and only if $(u, v) \notin E$. We often denote the complement of a graph $G$ by $\bar{G}$. Figure 2.12(b) illustrates the complement of the graph in Fig. 2.12(a). A null graph is the complement of the complete graph with the same number of vertices and vice versa. The following lemma is an interesting observation in terms of the complement of a graph.
Lemma 2.5.1 For any graph of six vertices, $G$ or $\bar{G}$ contains a triangle.
Proof Let $G$ be a graph of six vertices, and let $v$ be a vertex of $G$. Since the total number of neighbors of $v$ in $G$ and $\bar{G}$ is five, $v$ has at least three neighbors either in $G$ or in $\bar{G}$ by the pigeonhole principle. Without loss of generality we can assume that $v$ has three neighbors $x, y$ and $z$ in $G$. If any two of $x, y$, and $z$ are adjacent to each other, then $G$ contains a triangle. If no two of $x, y$, and $z$ are adjacent, then $x$, $y$, and $z$ will form a triangle in $\bar{G}$.

图论代考

数学代写|图论作业代写Graph Theory代考|Union and Intersection of Graphs

设$G_1=\left(V_1, E_1\right)$和$G_2=\left(V_2, E_2\right)$为两个图。$G_1$和$G_2$的并集，用$G_1 \cup G_2$表示，是另一个图$G_3=\left(V_3, E_3\right)$，其顶点集$V_3=V_1 \cup V_2$和边集$E_3=E_1 \cup E_2$。

同样，$G_1$和$G_2$的交集，用$G_1 \cap G_2$表示，是另一个图$G_4=\left(V_4, E_4\right)$，其顶点集$V_4=V_1 \cap V_2$和边集$E_4=E_1 \cap E_2$。

图2.10(a)和(b)显示了两个图$G_1$和$G_2$，图2.10(c)和(d)分别表示了它们的并集和交集。

显然，我们可以用类似的方法定义两个以上图的并和交。图上的这些操作可以很容易地用于解决许多问题。现在我们在两个图[2]上给出了这些运算的应用。
假设聚会中有$h+g$人;其中$h$是主人，$g$是客人。除了主人不与其他主人握手外，每个人都互相握手。问题是要找出握手的总数。像往常一样，我们将场景转换为图问题，如下所示。我们形成一个有$h+g$个顶点的图;其中$h$为黑色顶点，代表主人，$g$为白色顶点，代表客人。图的边缘表示握手。因此，除了任何一对黑色顶点之间没有边之外，每对顶点之间都有一条边。因此，现在的问题是计算这样形成的图中的边的数量。$h=3$和$g=4$的曲线图如图2.11(a)所示。

为了解决这个问题，我们注意到图可以被认为是两个图的并:如图2.11(b)所示的完全图$K_g$和完全二部图$K_{h, g}$。由于两个图之间没有共同的边，因此它们的交点不包含边。因此，图中边的总数(即一方中握手的总数)为$n(n-1) / 2+m \times n$。

数学代写|图论作业代写Graph Theory代考|Complement of a Graph

图$G=(V, E)$的补是另一个图$\bar{G}=(V, \bar{E})$，它具有相同的顶点集，使得对于任意一对不同的顶点$u, v \in V,(u, v) \in \bar{E}$当且仅当$(u, v) \notin E$。我们经常用$\bar{G}$表示图的补$G$。图2.12(b)为图2.12(a)图的补图。空图是具有相同顶点数的完全图的补图，反之亦然。下面的引理是关于图的补的一个有趣的观察。
引理2.5.1对于任何有六个顶点的图，$G$或$\bar{G}$包含一个三角形。
证明设$G$是一个有六个顶点的图，设$v$是$G$的一个顶点。由于$v$在$G$和$\bar{G}$中的邻居总数为5，因此根据鸽子洞原理，$v$在$G$或$\bar{G}$中至少有3个邻居。在不丧失一般性的情况下，我们可以假设$v$在$G$中有三个邻居$x, y$和$z$。如果$x, y$和$z$中的任意两个相邻，则$G$包含一个三角形。如果$x, y$和$z$中没有相邻的两个，那么$x$、$y$和$z$将在$\bar{G}$中形成一个三角形。

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