Posted on Categories:Graph Theory, 图论, 数学代写

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|图论代写GRAPH THEORY代考|Properties of a Tree

There are many different sets of properties of trees; any of these sets of properties can be taken as a definition of a tree. In this section, we discuss these properties of a tree. We first observe the following two trivial properties of a tree which play crucial roles while dealing with trees.
Lemma 4.2.1 Every tree with two or more vertices has at least two leaves.
Proof Let $T$ be a tree of two or more vertices and let $P$ be a maximal path in $T$. Then the end vertices $u$ and $v$ of $P$ have degree 1 (see Fig.4.1(f)), otherwise $P$ would not be a maximal path in $T$.

Lemma 4.2.2 Every edge in a tree is a cut edge.
Proof Immediate from Lemma 3.1.4.
The following lemma gives some characterization of trees.
Lemma 4.2.3 Let $G$ be a graph with $n$ vertices. Then, any two of the following three statements imply the third (and characterize a tree of $n$ vertices).
(a) $G$ is connected.
(b) $G$ contains no cycle.
(c) $G$ has $n-1$ edges.
Proof (a) \& (b) $\Rightarrow$ (c). We first prove that a connected and acyclic graph $G$ with $n$ vertices has $n-1$ edges. The claim is obvious for $n=1$ since a graph with a single vertex and no cycle has no edges. We thus assume that $n>1$ and the claim is true for any connected and acyclic graph with less than $n$ vertices. We now show that the graph $G$ with $n$ vertices has $n-1$ edges. Since $G$ contains no cycle, every edge $e$ of $G$ is a cut edge. Let $H_1$ and $H_2$ be the two connected components of $G-e$ with $n_1$ and $n_2$ vertices, respectively, where $n_1+n_2=n$. Since both $H_1$ and $H_2$ are acyclic and connected, they contain $n_1-1$ and $n_2-1$ edges respectively. Then the total number of edges in $G$ is $n_1-1+n_2-1+1=n-1$.

## 数学代写|图论代写GRAPH THEORY代考|Rooted Trees

A rooted tree is a tree in which one of the vertices is distinguished from the others. The distinguished vertex is called the root of the tree. The root of a tree is usually drawn at the top. In Fig. 4.3, the root is $v_1$. If a rooted tree is regarded as a directed graph in which each edge is directed from top to bottom, then every vertex $u$ other than the root is connected by an edge from some other vertex $p$, called the parent of $u$. We also call $u$ a child of vertex $p$. We draw the parent of a vertex above that vertex. For example, in Fig. 4.3, $v_1$ is the parent of $v_2, v_3$, and $v_4$, while $v_2$ is the parent of $v_5$ and $v_6 ; v_2, v_3$, and $v_4$ are the children of $v_1$, while $v_5$ and $v_6$ are the children of $v_2$. A leaf is a vertex of a tree that has no children. An internal vertex is a vertex that has one or more children. Thus every vertex of a tree is either a leaf or an internal vertex. In Fig. 4.3, the leaves are $v_4, v_5, v_6$, and $v_7$, and the vertices $v_1, v_2$, and $v_3$ are internal vertices.

The parent-child relationship can be extended naturally to ancestors and descendants. Suppose that $u_1, u_2, \ldots, u_l$ is a sequence of vertices in a tree such that $u_1$ is the parent of $u_2$, which is a parent of $u_3$, and so on. Then vertex $u_1$ is called an ancestor of $u_l$ and vertex $u_l$ a descendant of $u_1$. The root is ancestor of every other vertex in a tree and every other vertex is a descendant of the root. In Fig. 4.3, $v_1$ is an ancestor of all other vertices, and all other vertices are descendants of the root $v_1$. Note that the definition of ancestor (descendant) does not allow a vertex to be an ancestor (descendant) of itself. However, there are some definitions of ancestor (descendant) which allow a vertex to be an ancestor (descendant) of itself.

A rooted tree is called a binary tree if each vertex has at most two children. In general, a rooted tree is called a $k$-ary tree if each vertex has at most $k$ children. That is, the maximum number of children of a vertex in a $k$-ary tree is $k$.

An ordered rooted tree is a rooted tree in which the children of a vertex is somehow ordered. For example, in a ordered rooted binary tree the children of a vertex are ordered as the left child and the right child. The children of a vertex in a ordered rooted tree may also be ordered in a clockwise or in a counterclockwise order.

## 数学代写|图论代写GRAPH THEORY代考|Properties of a Tree

(a)连接$G$。
(b) $G$不包含循环。
(c) $G$有$n-1$条边。

Posted on Categories:Graph Theory, 图论, 数学代写

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|图论代写GRAPH THEORY代考|Connected Separable Graphs

A connected graph is separable if $G$ has at least one cut vertex, otherwise $G$ is nonseparable. Thus a nonseparable graph is 2-connected. However, $K_1$ and $K_2$ are considered nonseparable. A maximal nonseparable connected subgraph of $G$ is called a block of $G$. Thus a block of a connected graph $G$ of $n \geq 2$ vertices is either a biconnected component or a bridge of $G$. We now have the following lemma.

The blocks and cut vertices in $G$ can be represented by a tree $T$, called the $B C$-tree of $G$. In $T$ each block is represented by a $B$-node and each cut vertex of $G$ is represented by a $C$-node. The graph in Fig. 3.11(a) has the blocks $B_1, B_2, \ldots, B_9$ depicted in Fig.3.11(b). The $B C$-tree $T$ of the plane graph $G$ in Fig.3.11(a) is depicted in Fig.3.11(c), where each $B$-node is represented by a rectangle and each $C$-node is represented by a circle.

2-Connected Graphs
The reliability of a computer network can be increased by providing alternative paths between workstations. In fact the connectivity of a graph is a measure of number of alternative paths. If the graph is 1-connected then there is a path between any two workstations. We will show that there are two alternative paths between any two workstations in a 2-connected network, as in Theorem 3.4.5 below due to Whitney [3]. Two paths $P_1$ and $P_2$ with the same end vertices are internally disjoint if $P_1$ and $P_2$ do not share any internal vertex.

Theorem 3.4.5 A graph $G$ of three or more vertices is 2-connected if and only if there are two internally disjoint paths between every pair of vertices in $G$.

Proof Sufficiency Assume that there are two internally disjoint paths between every pair $u, v \in V(G)$. Then deletion of one vertex cannot separate $u$ from $v$. Since this condition is valid for every pair of vertices in $G, G$ is 2-connected.

Necessity Assume that $G$ is 2 -connected. We show that $G$ has two internally disjoint paths between every pair $u, v \in V(G)$ by induction on the length $l$ of a shortest path between $u$ and $v$.

## 数学代写|图论代写GRAPH THEORY代考Ear Decomposition

An ear of a graph $G$ with $\delta(G)=2$ is a maximal path with distinct end vertices whose internal vertices have degree 2 in $G$. Note that an edge $(u, v), u \neq v$ in $G$ with $d_G(u) \geq 3$ and $d_G(v) \geq 3$ is also an ear of $G$. An ear decomposition of $G$ is a decomposition $P_0, \ldots, P_k$ of edges such that $P_0$ is a cycle and $P_i$ for $i \geq 1$ is an ear of the graph induced by $P_0 \cup \cdots P_i$. An ear decomposition $P_0, P_1, \ldots, P_7$ of a graph is illustrated in Fig. 3.14 where the ear $P_6$ is an edge. Whitney [3] in 1932 gave a characterization of a 2-connected graph in terms of ear decomposition as in the following theorem.

Theorem 3.4.8 A graph $G$ has an ear decomposition if and only if $G$ is 2-connected.
Proof Necessity Assume that $G$ has an ear decomposition $P_0, P_1, \ldots, P_k$. By definition of ear decomposition $P_0$ is a cycle, which is 2-connected. Assume that $G_i=P_0 \cup \cdots \cup P_i$ is biconnected. We now show that $G_{i+1}=G_i \cup P_{i+1}$ is biconnected. $G_i$ is biconnected and the two end vertices of $P_{i+1}$ are on $G_i$. Then adding $P_{i+1}$ to $G_i$ will not introduce any cut vertex in $G_{i+1}$, and hence $G_{i+1}$ is biconnected.
Sufficiency We give a constructive proof. Assume that $G$ is 2 -connected. Then $G$ has a cycle. Let $C$ be a cycle in $G$. We choose $C$ as $P_0=G_0$. If $G_0 \neq G$, we can choose an edge $(u, v)$ of $G-E\left(P_0\right)$ and an edge $(x, y) \in E\left(P_0\right)$. Since $G$ is 2-connected, $(u, v)$ and $(x, y)$ lie on a cycle $C^{\prime}$ by Lemma 3.4.7. Let $P$ be a path in $C^{\prime}$ that contains $(u, v)$ and exactly two vertices of $G_0$. We choose $P$ as $P_1$. Clearly $G_1=P_0 \cup P_1$ is biconnected and $P_1$ is an ear of $G_1$. Let $G_i$ be the graph obtained by successively adding ears $P_1, P_2, \ldots, P_i$. We can find an ear $P_{i+1}$ similarly as we found $P_1$. The process ends only by absorbing all edges of $G$.

## 数学代写|图论代写GRAPH THEORY代考|Connected Separable Graphs

$G$中的块和切割顶点可以用树$T$表示，称为$G$的$B C$ -tree。在$T$中，每个块由一个$B$ -节点表示，$G$的每个切割顶点由一个$C$ -节点表示。图3.11(a)中的图中有图3.11(b)中所示的块$B_1, B_2, \ldots, B_9$。图3.11(a)中平面图$G$的$B C$ -tree $T$如图3.11(c)所示，其中每个$B$ -节点用矩形表示，每个$C$ -节点用圆形表示。

## 数学代写|图论代写GRAPH THEORY代考Ear Decomposition

Posted on Categories:Graph Theory, 图论, 数学代写

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|图论代写GRAPH THEORY代考|Supply Gas to a Locality

A gas company wants to supply gas to a locality from a single gas source. They are allowed to pass the underground gas lines along the road network only, because no one allows to pass gas lines through the bottom of one’s building. The road network divides the locality into many regions as illustrated in Fig. 1.4(a), where each road is represented by a line segment and a point at which two or more roads meet is represented by a small black circle. A point at which two or more roads meet is called an intersection point. Each region is bounded by some line segments and intersection points. These regions need to be supplied gas. If a gas line reaches an intersection point on the boundary of a region, then the region may receive gas from the line at that intersection point. Thus, the gas lines should reach the boundaries of all the regions of the locality. Gas will be supplied from a gas field which is located outside of the locality, and a single pipe line will be used to supply gas from the gas field to an intersection point on the outer boundary of the locality.

The gas company wants to minimize the establishment cost of gas lines by selecting the roads for laying gas lines such that the total length of the selected roads is minimal. Since gas will be supplied from the gas field using a single line to the locality, the selected road network should be connected and contain an intersection point on the outer boundary of the locality. Thus, the gas company needs to find a set of roads that induces a connected road network, supply gas in all the regions of the locality and the length of the induced road network is minimum. Such a set of roads is illustrated by thick lines in Fig. 1.4(b).The problem mentioned above can be modeled using a “plane graph.” A graph is planar if it can be embedded in the plane without edge crossings. A plane graph is a planar graph with a fixed planar embedding in the plane. A plane graph divides the plane into connected regions called faces. Let $G=(V, E)$ be an edge-weighted connected plane graph, where $V$ and $E$ are the sets of vertices and edges, respectively. Let $F$ be the set of faces of graph $G$. For each edge $e \in E, w(e) \geq 0$ is the weight of the edge $e$ of $G$. A face-spanning subgraph of $G$ is a connected subgraph $H$ induced by a set of edges $S \subseteq E$ such that the vertex set of $H$ contains at least one vertex from the boundary of each face $f \in F$ of $G$ [6]. Figure 1.5 shows two face-spanning subgraphs drawn by thick lines where the cost of the face-spanning subgraph in Fig. 1.5(a) is 22 and the cost of the face-spanning subgraph in Fig. 1.5(b) is 16. Thus, a plane graph may have many face-spanning subgraphs whose costs are different. A minimum face-spanning subgraph $H$ of $G$ is a face-spanning subgraph of $G$, where $\sum_{e \in S} w(e)$ is minimum, and a minimum face-spanning subgraph problem asks to find a minimum face-spanning subgraph of a plane graph. If we represent each road of the road network by an edge of $G$, each intersection point by a vertex of $G$, each region by a face of $G$, and assign the length of a road to the weight of the corresponding edge, then the problem of finding a minimum face-spanning subgraph of $G$ is the same as the problem of the gas company mentioned above [6]. A minimum face-spanning subgraph problem often arises in applications like establishing power transmission lines in a city, power wires layout in a complex circuit, planning irrigation canal networks for irrigation systems, etc.

## 数学代写|图论代写GRAPH THEORY代考|Floorplanning

Graph modeling has applications in VLSI floorplanning as well as architectural floorplaning [7]. In a VLSI floorplanning problem, an input is a plane graph $F$ as illustrated in Fig. 1.6(a); $F$ represents the functional entities of a chip, called modules, and interconnections among the modules; each vertex of $F$ represents a module, and an edge between two vertices of $F$ represents the interconnections between the two corresponding modules. An output of the problem for the input graph $F$ is a partition of a rectangular chip area into smaller rectangles as illustrated in Fig. 1.6(d); each module is assigned to a smaller rectangle, and furthermore, if two modules have interconnections, then their corresponding rectangles must be adjacent, i.e., they must have a common boundary. A similar problem may arise in architectural floorplanning also. When building a house, the owner may have some preference; for example, a bedroom should be adjacent to a reading room. The owner’s choice of room adjacencies can be easily modeled by a plane graph $F$, as illustrated in Fig. 1.6(a); each vertex represents a room and an edge between two vertices represents the desired adjacency between the corresponding rooms.

A “rectangular drawing” of a plane graph may provide a suitable solution to the floorplanning problem described above. (In a rectangular drawing of a plane graph each vertex is drawn as a point, each edge is drawn as either a horizontal line segment or a vertical line segment and each face including the outer face is drawn as a rectangle.) First, obtain a plane graph $F^{\prime}$ by triangulating all inner faces of $F$ as illustrated in Fig. 1.6(b), where dotted lines indicate new edges added to $F$. Then obtain a “dual-like” graph $G$ of $F^{\prime}$ as illustrated in Fig. 1.6(c), where the four vertices of degree 2 drawn by white circles correspond to the four corners of the rectangular area. Finally, by finding a rectangular drawing of the plane graph $G$, obtain a possible floorplan for $F$ as illustrated in Fig. 1.6(d).

## 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。