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\lecture{20}{Algorithm for Non-bipartite Matching}{Lakulish Antani}

\section{Recap}

We have seen an algorithm for finding a maximal matching in a bipartite graph. The algorithm
is based on a modified BFS, which tries to find an augmenting path with respect to the 
current matching, if one exists; if no such augmenting path exists, then we conclude that
the current matching is maximal. However, the algorithm cannot be used (without
modifications) for general non-bipartite graphs.

We have seen that if we encounter an even-even edge during the search, we have an odd-length
cycle in the graph. We handle blossoms by ``shrinking" them to a single vertex; the only
modification required to the previous algorithm is basically to detect and shrink blossoms.
This works because of the following theorem (which we have proved):

\begin{Thm}
Let $G = (V,E)$ be a graph, and $M$ be a matching in $G$. Suppose $B$ is a
blossom such that its base is unmatched. Consider the graph $G' = (V',E')$ obtained by
shrinking $B$, and the matching $M'$ composed of the matched edges of $M$ which are still
present in $E'$. Then there is an augmenting path in $G$ w.r.t. $M$ if and only if there is
an augmenting path in $G'$ w.r.t. $M'$.
\end{Thm}

\section{Finding an Augmenting Path}

To find an augmenting path for general graphs, we start the (modified) BFS from an unmatched
vertex, as before. However, we may encounter an even-even edge between two even vertices
$v_1$ and $v_2$ (say). These are part of a blossom in the graph, and we first find the
base of the blossom. This is easily seen to be the common ancestor $v_0$ of $v_1$ and $v_2$.

Now we need to make sure that $v_0$ is unmatched for the above theorem to apply. This is
clearly true if $v_0$ is the starting vertex ($u$). If it is not the starting vertex, then since it has been reached from $u$, it is matched to its parent in the search graph. 
Note that there will be an even-length path from $v_0$ to $u$. If, on the path $p$ from $u$ to $v_0$, we change all matched edges to unmatched and vice-versa, we are still left with a
valid matching $M''$. Now it can be shown that there is an augmenting path in $G$ w.r.t $M$ if and only if there is an augmenting path in $G$ w.r.t. $M''$.

Hence the algorithm is roughly as follows:

\begin{program}
\> perform the modified BFS on $G$ as before\\
\> $u$ := unmatched vertex from which search begins\\
\> \IF when expanding an even vertex $v_1$, we find an edge to another even vertex $v_2$\\
\>\> $v_0$ := common ancestor of $v_1$ and $v_2$ in the search graph\\
\>\> \IF\ $v_0$ = $u$\\
\>\>\> shrink $v_0$\\
\>\> \ELSE \\
\>\>\> $p$ := path from $u$ to $v_0$\\
\>\>\> change all matched edges in $p$ to unmatched, and vice-versa\\
\>\>\> shrink $v_0$\\
\end{program}

In the next lecture, we will see a detailed proof of why the \textbf{else} clause above
works, and a proof of correctness for the algorithm.

\section{Time Complexity}

We shall show a generous bound on the time complexity of the algorithm. Suppose the graph
is $G = (V,E)$, and that $|V| = n$ and $|E| = m$. Observe that:

\begin{itemize}
\item The number of shrinkages performed can clearly be no more than the number of vertices
in the graph, i.e. $n$.
\item In the worst case, BFS is performed $n$ times.
\item Each BFS takes $O(m)$ time.
\end{itemize}

So we get a generous bound of $O(n^{2}m)$ on the complexity.

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