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

# 数学代写|图论代考GRAPH THEORY代写|MA57500

avatest™

## 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。