## 数学代写|图论作业代写Graph Theory代考|k-Edge-Connected

A similar notion with regards to edges exists, where we now look at how many edges need to be removed before the graph is disconnected. Recall that when we remove an edge $e=x y$ from a graph, we are not removing the endpoints $x$ and $y$.

Definition 4.3 A bridge in a graph $G=(V, E)$ is an edge $e$ whose removal disconnects the graph, that is, $G$ is connected but $G-e$ is not. An edge-cut is a set $F \subseteq E$ so that $G-F$ is disconnected.

Clearly every connected graph has an edge-cut since removing all the edges from a graph will result in just a collection of isolated vertices. As with the vertex version, we are more concerned with the smallest size of an edge-cut.
Definition 4.4 We say $G$ is $\boldsymbol{k}$-edge-connected if the smallest edge-cut is of size at least $k$.

Define $\kappa^{\prime}(G)=k$ to be the maximum $k$ such that $G$ is $k$-edgeconnected, that is there exists a edge-cut $F$ of size $k$, yet no edge-cut exists of size $k-1$.

## 数学代写|图论作业代写Graph Theory代考|Whitney’s Theorem

Can you discern any relationship between the vertex and edge connectivity measures? The examples above should demonstrate that these measures need not. be equal, though they can be. How does the minimum degree of a graph play a role in these? Notice how in both $G_2$ and $G_3$ above we found an edge-cut by removing both edges incident to a specific vertex.

Theorem 4.5 (Whitney’s Theorem) For any graph $G, \kappa(G) \leq \kappa^{\prime}(G) \leq$ $\delta(G)$

Proof: Let $G$ be a graph with $n$ vertices and $\delta(G)=k$ and suppose $x$ is a vertex with $\operatorname{deg}(x)=k$. Let $F$ be the set of all edges incident to $x$. Then $G-F$ is disconnected, since $x$ is now isolated, and so $F$ is an edge-cut.
Thus $\kappa^{\prime}(G) \leq k$.
It remains to show that $\kappa(G) \leq \kappa^{\prime}(G)$. If $G=K_n$, then $\delta(G)=n-1$ and $\kappa(G)=n-1$ by definition, and so $\kappa(G)=\kappa^{\prime}(G)$. Otherwise, let $F$ be a minimal edge-cut of $G$ and define $G_1$ and $G_2$ to be the two components of $G-F$. We will consider how these components are related, and in both cases find a cut-set $S$ of size less than that of $F$.

Case 1: Every vertex of $G_1$ is adjacent to every vertex of $G_2$. Then $|F|=\left|G_1\right| \cdot\left|G_2\right| \geq n-1$ since each component has at least one vertex. Since $G \neq K_n$, at least one of $G_1$ and $G_2$, say $G_1$, has a pair of vertices $x$ and $y$ that are not adjacent. Let $S$ be all vertices of $G$ except $x$ and $y$, that is $S=V(G)-{x, y}$. Then $S$ is a cut-set of size $n-2$ and so $\kappa(G) \leq n-2<n-1 \leq \kappa^{\prime}(G)$

## 数学代写|图论作业代写Graph Theory代考|Whitney’s Theorem

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