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\lecture{17} {Matching in Bipartite Graphs}{Ramesh Nidadavolu}
A set of disjoint edges in $G$ is called a {\it matching}. {\it Maximum matching} in a graph $G$ is a {\it Matching} of maximum size.

First we discuss a few details. If $M$ is {\it matching} and $M_0$ is a {\it maximum matching} then $M \Delta M_0$\footnote{${A \Delta B} =(A-B)\cup(B-A)$ } has connected components which are either even cycles or simple paths. Also in a odd path the first edge in the path cannot be an edge from the matching $M$, else we could interchange the $M_0$ edges and the $M$ edges in this path and increase the size of the matching $M_0$ by 1. 

Another interesting thing to note is that if ${A \Delta B} = C$ then ${A \Delta C} = B$. (Proof is simple. Draw a venn diagram to see that it should hold)

We also state that $M \Delta M_0$ has atleast one component which is a path of odd length.(Proof is easy. Solve the exercises below to figure out the proof)
\begin{Ex}
If $M$ is a matching of size $|M_0| - 1$ then there exists an odd length path in $M \Delta M_0$. 
\end{Ex}

\begin{Ex}
For any matching $M$ there is atleast one odd length path in $M \Delta M_0$. (Hint: If all components would be even then $|M| = |M_0|$ ? )
\end{Ex}

Also as we pointed out earlier any odd length path has to start with an edge from $M_0$


Now given a graph $G$ and a matching $M$ we define a augumenting path as a path in $G$ whose start and end edges are not in $M$ and alternate edges are in $M$ along the path.(so an augumenting path would have odd number of edges with more unmatched edges than matched edges.)

Now we prove that if a matching $M$ is not a maximum matching there must exist an augumenting path wrt to $M$. (Actually a stronger result is proven below)

\section{Main result}
We prove the following result:
{\it A matching $M$ is maximum iff there does not exist any augumenting path in $G$.}

If the matching is maximum and there exists an augumenting path $P$ then $M \Delta P$ would be a matching of greater size.Contradicton.
Also If the matching is not maximum then there must exist an odd length path given by $M \Delta M_0$ (as proven earlier).  
 
Thus proven.

\section{Why this result?}
The idea is that in searching for a matching that is maximum we can start of with some arbitary matching $M$ and then look for augementing paths and then construct a new matching $M \Delta P$ which would be of a size greater than the matching we started of with. Thus repeatedly looking for augumenting paths and incrementing our matching size we hope to end up with a maximum matching. More on this in the next lecture.

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