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# 计算机代写|算法代写Algorithm代考|CSE431/531 Minimum-Capacity Cuts

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## 计算机代写|算法代写Algorithm代考|Minimum-Capacity Cuts

We may naturally think of the Maximum Flow problem as an attempt to maximize the number of resources that reach a collection of destinations. In this vein, there is a natural dual problem: what are the minimum number of disruptions needed to prevent any resources from reaching any of the destinations? The motivation for these problems arose during the Cold War. Here, the United States was interested in the Soviet Union Railway System that connected Eastern Europe- particularly, East Germany- and the western region of the Soviet Union. In particualr, the United States wanted to identify the minimum number of points to bomb in order to disrupt the flow of resources along this system [Ano, Sch02]. We will see later that the Maximum Flow problem is equivalent to finding the minimum number of disruptions. This is the celebrated Max-Flow Min-Cut Theorem.
We now turn to formalizing the Minimum Cut problem.
Definition 93. Let $\mathcal{N}(G, c, S, T)$ be a flow network. A cut of $\mathcal{N}$ is a partition of the vertices $(X, Y)$, where $S \subseteq X$ (that is, $X$ contains the source vertices), and $T \subseteq Y$ (that is, $Y$ contains the sink vertices). Note that as $(X, Y)$ is a partition, we have that $X$ and $Y$ are disjoint.

The capacity of the partition $(X, Y)$ is the sum of the edge capacities with the initial endpoint in $X$ and the destination vertex in $Y$. That is:
$$c(X, Y):=\sum_{x \in X} \sum_{y \in Y} c(x, y) .$$
Recall that if $(x, y)$ is not an edge of the flow network, then $c(x, y)=0$.
Definition 94. The Minimum Cut problem is defined as follows.

Instance: Let $\mathcal{N}(G, c, S, T)$ be a flow network.

Solution: A cut $(X, Y)$ such that $c(X, Y)$ is minimized.

## 计算机代写|算法代写Algorithm代考|Max-Flow Min-Cut Theorem

In this section, we prove the Max-Flow Min-Cut Theorem. Our proof is based on [Mou16b].
Theorem 97 (Max-Flow Min-Cut Theorem (Ford-Fulkerson, 1956).). Let $\mathcal{N}(G, c, S, T)$ be a flow network. Let $f^$ be a maximum-valued flow, and let $(X, Y)$ be a minimum-capacity cut. We have that val $\left(f^\right)=c(X, Y)$.
We prove Theorem 97. We first show that the value of a maximum flow is no bigger than the capacity of a minimum cut. We then show that there exists a cut whose capacity is no bigger than the value of a maximum flow. It follows from this second claim that the capacity of a minimum cut is no bigger than the value of a maximum flow.

We begin by showing that the value of a maximum flow is no bigger than the capacity of a minimum cut. To this end, we introduce the following lemma, which intuitively states that the amount of flow that we can push from the source nodes to the sink nodes cannot exceed the capacity of a cut.

Lemma 98. Let $\mathcal{N}(G, c, S, T)$ be a flow network, and let $f$ be a flow. Let $(X, Y)$ be a cut. We have that $\operatorname{val}(f) \leq c(X, Y)$

Proof. As $(X, Y)$ is a cut, we have by the conservation of flow the total flow that makes it from the source vertices to the sink vertices is the amount of flow leaving $X$, minus the amount of flow returning to $X$ from $Y$. This is precisely:
$$\operatorname{val}(f)=\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v) .$$
By ignoring the flow coming back into $X$, we have that:
\begin{aligned} \operatorname{val}(f) & =\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v) \ & =\sum_{u \in X} \sum_{v \in Y} f(u, v) \ & \leq \sum_{u \in X} \sum_{v \in Y} c(u, v) \ & =c(X, Y) \end{aligned}

## 计算机代写|算法代寻Algorithm代考|Minimum-Capacity Cuts

$$c(X, Y):=\sum_{x \in X} \sum_{y \in Y} c(x, y)$$

## 计算机代写|算法代写Algorithm代考|Max-Flow Min-Cut Theorem

\operatorname{val}(f)=\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v)
$$通过芴略返回的流量 X ，我们有:$$
\operatorname{val}(f)=\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v) \quad=\sum_{u \in X} \sum_{v \in Y} f(u, v) \leq \sum_{u \in X} \sum_{v \in Y} c(u, v) \quad=c(X, Y)


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