# 数学代写|图论作业代写Graph Theory代考|Matching in bipartite graphs

#### Doug I. Jones

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## 数学代写|图论作业代写Graph Theory代考|Matching in bipartite graphs

For this whole section, we let $G=(V, E)$ be a fixed bipartite graph with bipartition ${A, B}$. Vertices denoted as $a, a^{\prime}$ etc. will be assumed to lie in $A$, vertices denoted as $b$ etc. will lie in $B$.

How can we find a matching in $G$ with as many edges as possible? Let us start by considering an arbitrary matching $M$ in $G$. A path in $G$ which starts in $A$ at an unmatched vertex and then contains, alternately, edges from $E \backslash M$ and from $M$, is an alternating path with respect to $M$. An alternating path $P$ that ends in an unmatched vertex of $B$ is called an augmenting path (Fig. 2.1.1), because we can use it to turn $M$ into a larger matching: the symmetric difference of $M$ with $E(P)$ is again a matching (consider the edges at a given vertex), and the set of matched vertices is increased by two, the ends of $P$.

Alternating paths play an important role in the practical search for large matchings. In fact, if we start with any matching and keep applying augmenting paths until no further such improvement is possible, the matching obtained will always be an optimal one, a matching with the largest possible number of edges (Exercise 1). The algorithmic problem of finding such matchings thus reduces to that of finding augmenting paths – which is an interesting and accessible algorithmic problem.

Our first theorem characterizes the maximal cardinality of a matching in $G$ by a kind of duality condition. Let us call a set $U \subseteq V$ a (vertex) cover of $E$ if every edge of $G$ is incident with a vertex in $U$.

## 数学代写|图论作业代写Graph Theory代考|Second proof

Second proof. We apply induction on $|A|$. For $|A|=1$ the assertion is true. Now let $|A| \geqslant 2$, and assume that the marriage condition is sufficient for the existence of a matching of $A$ when $|A|$ is smaller.
If $|N(S)| \geqslant|S|+1$ for every non-empty set $S \varsubsetneqq A$, we pick an edge $a b \in G$ and consider the graph $G^{\prime}:=G-{a, b}$. Then every non-empty set $S \subseteq A \backslash{a}$ satisfies
$$\left|N_{G^{\prime}}(S)\right| \geqslant\left|N_G(S)\right|-1 \geqslant|S|,$$
so by the induction hypothesis $G^{\prime}$ contains a matching of $A \backslash{a}$. Together with the edge $a b$, this yields a matching of $A$ in $G$.

Suppose now that $A$ has a non-empty proper subset $A^{\prime}$ with $\left|B^{\prime}\right|=$ $\left|A^{\prime}\right|$ for $B^{\prime}:=N\left(A^{\prime}\right)$. By the induction hypothesis, $G^{\prime}:=G\left[A^{\prime} \cup B^{\prime}\right]$ contains a matching of $A^{\prime}$. But $G-G^{\prime}$ satisfies the marriage condition too: for any set $S \subseteq A \backslash A^{\prime}$ with $\left|N_{G-G^{\prime}}(S)\right|<|S|$ we would have $\left|N_G\left(S \cup A^{\prime}\right)\right|<\left|S \cup A^{\prime}\right|$, contrary to our assumption. Again by induction, $G-G^{\prime}$ contains a matching of $A \backslash A^{\prime}$. Putting the two matchings together, we obtain a matching of $A$ in $G$.

For our last proof, let $H$ be a spanning subgraph of $G$ that satisfies the marriage condition and is edge-minimal with this property. Note that $d_H(a) \geqslant 1$ for every $a \in A$, by the marriage condition with $S={a}$.

# 图论代考

## 数学代写|图论作业代写Graph Theory代考|Second proof

$$\left|N_{G^{\prime}}(S)\right| \geqslant\left|N_G(S)\right|-1 \geqslant|S|,$$

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