# 计算机代写|组合优化代写Combinatorial optimization代考|COMP567

## 计算机代写|组合优化代写Combinatorial optimization代考|Matching

In this section, we study matching in a directed way. First, we define the augmenting path as follows.

Consider a matching $M$ in a bipartite graph $G=(U, V, E)$. Let us call every edge in $M$ as matched edge and every edge not in $M$ as unmatched edge. A node $v$ is called a free node if $v$ is not an ending point of a matched edge.

Definition 5.4.1 (Augmenting Path) The augmenting path is now defined to be a path satisfying the following:

• It is an alternating path, that is, edges on the path are alternatively unmatched and matched.
• The path is between two free nodes.
There are totally odd number of edges in an augmenting path. The number of unmatched edges is one more than the number of matched edges. Therefore, on an augmenting path, turn all matched edges to unmatched and turn all unmatched edges to matched. Then considered matching will become a matching with one more edge. Therefore, if a matching $M$ has an augmenting path, then $M$ cannot be maximum. The following theorem indicates that the inverse holds.

Theorem 5.4.2 A matching $M$ is maximum if and only if $M$ does not have an augmenting path.

Proof Let $M$ be a matching without augmenting path. For contradiction, suppose $M$ is not maximum. Let $M^$ be a maximum matching. Then $|M|<\left|M^\right|$. Consider $M \oplus M^=\left(M \backslash M^\right) \cup\left(M^* \backslash M\right)$, in which every node has degree at most two (Fig. 5.11).

Hence, it is disjoint union of paths and cycles. Since each node with degree two must be incident to two edges belonging to $M$ and $M^{\prime}$, respectively. Those paths and cycles must be alternative. They can be classified into four types as shown in Fig. 5.12.

Note that in each of the first three types of connected components, the number of edges in $M$ is not less than the number of edges in $M^$. Since $|M|<\left|M^\right|$, we have $\left|M \backslash M^\right|<\left|M^ \backslash M\right|$. Therefore, the connected component of the fourth type must exist, that is, $M$ has an augmenting path, a contradiction.

## 计算机代写|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 $\ell \leq|V|$. Cólléct âll ārcs from $V_l$ tō $V_{l+1}$ för $i=0,1, \ldots, \ell-1$. Lêt $L(s)$ bẻ 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.

# 组合优化代考

## 计算机代写|组合优化代写Combinatorial optimization代考|Matching

• 它是一条交替路径，即路径上的边交替不匹配和匹配。
• 该路径位于两个空闲节点之间。
在增广路径中有完全奇数个边。不匹配边的数量比匹配边的数量多一。因此，在增广路径上，将所有匹配的边变为不匹配的，并将所有不匹配的边变为匹配的。然后考虑的匹配将成为与多一条边的匹配。因此，如果匹配米有一条增广路径，那么米不能是最大值。以下定理表明逆成立。

## 计算机代写|组合优化代写Combinatorial optimization代考|Dinitz Algorithm

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