# 数学代写|组合优化代写Combinatorial optimization代考|APM6664

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## 数学代写|组合优化代写Combinatorial optimization代考|Dinitz Algorithm

In this and the next sections, we present more algorithms for the maximum flow problem. They have running time better than Edmonds-Karp algorithm.

First, we note that the idea in Hopcroft-Karp algorithm can be extended from matching to flow. This extension gives a variation of Edmonds-Karp algorithm, called Dinitz algorithm.

Consider a flow network $G=(V, E)$. The algorithm starts with a zero flow $f(u, v)=0$ for every arc $(u, v)$. In each substantial iteration, consider residual network $G_f$ for flow $f$. Start from source node $s$ to do the breadth-first search until node $t$ is reached. If $t$ cannot be researched, then algorithm stops, and the maximum flow is already obtained. If $t$ is reached with distance $\ell$ from node $s$, then the breadth-first-search tree contains $\ell$ level, and its nodes are divided into $\ell$ classes $V_0, V_1, \ldots, V_{\ell}$ where $V_i$ is the set of all nodes each with distance $i$ from $s$ and $\ell \leq|V|$. Collect all arcs from $V_i$ to $V_{i+1}$ for $i=0,1, \ldots, \ell-1$. Let $L(s)$ be the obtained levelable subnetwork. Above computation can be done in $O(|E|)$ time.
Next, the algorithm finds augmenting paths to do augmentations in the following way.

Step 1. Iteratively, for $v \neq t$ and $u \neq s$, remove, from $L(s)$, every arc $(u, v)$ with no coming arc at $u$ or no outgoing arc at $v$. Denote by $\hat{L}(s)$ the obtained levelable network.
Step 2. If $\hat{L}(s)$ is empty, then this iteration is completed, and go to the next iteration. If $\hat{L}(s)$ is not empty, then it contains a path of length $\ell$, from $s$ to $t$. Find such a path $P$ by using the depth-first search. Do augmentation along the path $P$. Update $L(s)$ by using $\hat{L}(s)$ and deleting all critical arcs on $P$. Go to Step 1.
This algorithm has the following property.
Lemma 5.5.1 Let $\delta_f(s, t)$ denote the distance from $s$ to $t$ in residual graph $G_f$ of flow $f$. Suppose flow $f^{\prime}$ is obtained from flow $f$ through an iteration of Dinitz algorithm. Then $\delta_{f^{\prime}}(s, t) \geq \delta_f(s, t)+2$.
Proof The proof is similar to the proof of Lemma 5.2.1.
The correctness of Dinitz algorithm is stated in the following theorem.

## 数学代写|组合优化代写Combinatorial optimization代考|Goldberg-Tarjan Algorithm

In this section, we study a different type of incremental method for maximum network flow. In this method, a valid label will play an important role. This valid label will be on each arc to guide the incremental direction.

Consider a flow network $G=(V, E)$ with capacity $c(u, v)$ for each arc $(u, v) \in$ $E ; s$ and $t$ are source and sink, respectively. As usual, for simplicity of description, we extend capacity $c(u, v)$ to every pair of nodes $u$ and $v$ by defining $c(u, v)=0$ if $(u, v) \notin E$.
A function $f: V \times V \rightarrow R$ is called a preflow if

1. (Capacity constraint) $f(u, v) \leq c(u, v)$ for every $u, v \in V$.
2. (Skew symmetry) $f(u, v)=-f(v, u)$ for all $u, v \in V$.
3. For every $v \in V \backslash{s, t}, \sum_{v \in V \backslash{u}} f(u, v) \geq 0$, i.e., $\sum_{(u, v) \in E} f(u, v) \geq$ $\sum_{(v, w) \in E} f(v, w)$
Compared with those three conditions in the definition of flow, the first two are the same, and the third one is different. The flow conservation condition is relaxed to allow more flow coming than going out at any node other than $s$ and $t$. This difference is called the excess at node $v$ and denotes
$$e(v)=\sum_{(u, v) \in E} f(u, v) \geq \sum_{(v, w) \in E} f(v, w) .$$
A node $v$ is said to be active if $e(v)>0, v \neq s$, and $v \neq t$. In preflow-relabel algorithm, the excess will be pushed from an active node toward the sink, relying on the valid distance label $d(v)$ for $v \in V$, satisfying the following conditions.
• $d(t)=0$.
• $d(u) \leq d(v)+1$ for $(u, v) \in E$.
An $\operatorname{arc}(u, v)$ is said to be admissible if $d(u)=d(v)+1$ and $c(u, v)>0$. Note that if we consider a residual graph, then $c(u, v)$ should be considered as updated capacity.

Lemma 5.6.1 Let dist $(u, v)$ denote the minimum number of arcs on the path from $u$ to $v$. Then $d(u) \leq \operatorname{dist}(u, t)$.

Proof It can be proved by induction on $\operatorname{dist}(u, t)$. For $\operatorname{dist}(u, t)=0, u$ must be $t$, and hence $d(t) \leq \operatorname{dist}(t, t)$. For $\operatorname{dist}(u, t)=k>0$, suppose $\left(u, u_1, \ldots, u_k=t\right)$ is the shortest path from $u$ to $t$. Then $\operatorname{dist}\left(u_1, t\right)=k-1$.

# 组合优化代考

## 数学代写|组合优化代写Combinatorial optimization代考|Goldberg-Tarjan Algorithm

1. (容量限制) $f(u, v) \leq c(u, v)$ 每一个 $u, v \in V$.
2. (斜对称) $f(u, v)=-f(v, u)$ 对全部 $u, v \in V$.
3. 对于每一个v$\in V \backslash s, t, \sum_{v \in V \backslash u} f(u, v) \geq 0$ ，那是， $\sum_{(u, v) \in E} f(u, v) \geq \sum_{(v, w) \in E} f(v, w)$
与流定义中的那三个条件相比，前两个相同，第三个不同。放 宽流量守恒条件，允许在除 $s$ 和 $t$. 这种差异称为节点处的多余 部分 $v$ 并表示
$$e(v)=\sum_{(u, v) \in E} f(u, v) \geq \sum_{(v, w) \in E} f(v, w) .$$
一个节点 $v$ 据说是活跃的，如果 $e(v)>0, v \neq s$ ，和 $v \neq t$. 在 preflow-relabel 算法中，多余部分将根据有效距离标签从 活动节点推向汇点 $d(v)$ 为了 $v \in V$ ，满足以下条件。$d(t)=0$
• $d(u) \leq d(v)+1$ 为了 $(u, v) \in E$.
一个 $\operatorname{arc}(u, v)$ 据说是可以接受的，如果 $d(u)=d(v)+1$ 和 $c(u, v)>0$. 请注意，如果我们考虑残差图，则 $c(u, v)$ 应被 视为更新容量。
引理 5.6.1 让 $\operatorname{dist}(u, v)$ 表示路径上的最小弧数 $u$ 到 $v$. 然后 $d(u) \leq \operatorname{dist}(u, t)$
证明可以用归纳法证明 $\operatorname{dist}(u, t)$. 为了 $\operatorname{dist}(u, t)=0, u$ 一定是 $t$ ， 因此 $d(t) \leq \operatorname{dist}(t, t)$. 为了dist $(u, t)=k>0$ ， 认为 $\left(u, u_1, \ldots, u_k=t\right)$ 是从 $u$ 到 $t$. 然后 dist $\left(u_1, t\right)=k-1$.

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## MATLAB代写

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