# 数学代写|图论作业代写Graph Theory代考|MATH360

## 数学代写|图论作业代写Graph Theory代考|Connectivity and Paths

Now that we have some familiarity with connectivity, we turn to its relationship to paths within a graph. Note that for the remainder of this section, we will assume the graphs are connected, as otherwise the results are trivial. We begin by relating cut-vertices and bridges to paths. You should notice that almost every result for vertices has an edge analog. We begin with the most simple results relating a cut-vertex or a bridge to its presence on a path.
Theorem 4.6 A vertex $v$ is a cut-vertex of a graph $G$ if and only if there exist vertices $x$ and $y$ such that $v$ is on every $x-y$ path.
Proof: First suppose $v$ is a cut-vertex in a graph $G$. Then $G-v$ must have at least two components. Let $x$ and $y$ be vertices in different components of $G-v$. Since $G$ is connected, we know there must exist an $x-y$ path in $G$ that does not exist in $G-v$. Thus $v$ must lie on this path. Conversely, let $v$ be a vertex and suppose there exist vertices $x$ and $y$ such that $v$ is on every $x-y$ path. Then none of these paths exist in $G-v$, and so $x$ and $y$ cannot be in the same component of $G-v$. Thus $G$ must have at least two components and so $v$ is a cut-vertex.
Below is the edge version of the result above, the proof of which appears in Exercise 4.17.

Theorem 4.7 An edge $e$ is a bridge of $G$ if and only if there exist vertices $x$ and $y$ such that $e$ is on every $x-y$ path.

The theorem to follow has a similar feel to that about leaves and trees (see Exercise 3.18) and allows us to know we can pick some vertex of a graph not to be a cut-vertex. The second theorem listed below relates bridges to cycles. It should be obvious that any edge along a cycle cannot be a bridge since its removal will only break the cycle, not disconnect the graph; perhaps more surprising is that all edges not on a cycle are in fact bridges. The proof of these results appear in Exercise $4.18$ and $4.19$.

## 数学代写|图论作业代写Graph Theory代考|2-Connected Graphs

As we have already seen, 1-connected graphs are simply those graphs that we more commonly call connected and $k$-connected graphs can be described in terms of $k$ number of paths between two vertices. So why then do we single out 2-connected graphs? This is in part because they hold a special area in the study of connectivity – they are known to be connected and as we will see cannot contain any cut-vertices. The class of 2-connected graphs provide both some easy results and some more technical and complex areas of study. We begin with a review of our graphs $G_2$ and $G_3$ from page 169, reproduced below.

Recall that we showed $\kappa\left(G_2\right)-2-\kappa^{\prime}\left(G_2\right)$ but that $\kappa\left(G_3\right)-1$ and $\kappa^{\prime}\left(G_3\right)=2$. So what is the structural difference between $G_2$ and $G_3$ that provides the difference in the connectivity measures? Obviously, we can describe it in terms of internally disjoint paths, but there must be some more basic property that separates them. In particular, notice how vertex $c$ seems to be the connecting point between the two halves of $G_3$; that it, $c$ is a cut-vertex.
Theorem 4.17 A graph $G$ with at least 3 vertices is 2-connected if and only if $G$ is connected and does not have any cut-vertices.

Proof: Assume $G$ is 2-connected. Then any cut-set of $G$ must be of size at least 2. Therefore $G$ must be connected and cannot have a cut-vertex, as in either of these situations we would have a cut-set of size less than 2 .
Convèrsëly, suppóse $G$ is not 2-connèctèd. Thẻn èithèr $G$ is discoñnected or by Menger’s Theorem there exist two non-adjacent vertices $x$ and $y$ for which there is exactly one path between them. Removing any vertex along this path will disconnect $x$ and $y$, and so that vertex serves
In Exercise $3.19$ we proved that every non-leaf of a tree is a cut-vertex. Therefore no tree is 2-connected and so every 2-connected graph must contain a cycle, or more specifically every vertex lies on a cycle. The proof of the fóllowing corollary appeảars in Exercisése $4.21$.

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## 数学代写|图论作业代写图论代考|2-连通图

Convèrsëly, suppóse $G$ 不是2-connèctèd。Thẻn èithèr $G$ 是discoñnected还是根据门格尔定理存在两个不相邻顶点 $x$ 和 $y$ 它们之间只有一条路。沿着这条路径移除任何顶点都会断开连接 $x$ 和 $y$，因此在练习中，顶点服务
$3.19$ 我们证明了树的每一个非叶都是一个切顶点。因此，没有树是二连通的，所以每个二连通图必须包含一个循环，或者更确切地说，每个顶点都位于一个循环上。fóllowing推论的证明出现在Exercisése中 $4.21$.

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