Improve language
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@ -542,16 +542,16 @@ and there are some edges between the sets:
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\end{center}
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\end{center}
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The size of the cut is the sum of the edges
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The size of the cut is the sum of the edges
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that go from the set $A$ to the set $B$.
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that go from $A$ to $B$.
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This is an upper bound for the flow
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This is an upper bound for the flow
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in the graph, because the flow has to proceed
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in the graph, because the flow has to proceed
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from the set $A$ to the set $B$.
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from $A$ to $B$.
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Thus, a maximum flow is smaller than or equal to
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Thus, the size of a maximum flow is smaller than or equal to
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any cut in the graph.
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the size of any cut in the graph.
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On the other hand, the Ford–Fulkerson algorithm
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On the other hand, the Ford–Fulkerson algorithm
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produces a flow that is \emph{exactly} as large
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produces a flow whose size is \emph{exactly} as large
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as a cut in the graph.
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as the size of a cut in the graph.
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Thus, the flow has to be a maximum flow
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Thus, the flow has to be a maximum flow
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and the cut has to be a minimum cut.
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and the cut has to be a minimum cut.
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@ -773,14 +773,14 @@ from the source to the sink is 1.
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\index{maximum matching}
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\index{maximum matching}
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The \key{maximum matching} problem asks to find
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The \key{maximum matching} problem asks to find
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a maximum-size set of node pairs of a graph
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a maximum-size set of node pairs in an undirected graph
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such that each pair is connected with an edge and
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such that each pair is connected with an edge and
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each node belongs to at most one pair.
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each node belongs to at most one pair.
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There are polynomial algorithms for finding
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There are polynomial algorithms for finding
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maximum matchings in general graphs \cite{edm65},
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maximum matchings in general graphs \cite{edm65},
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but such algorithms are complex and do
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but such algorithms are complex and
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not appear in programming contests.
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rarely seen in programming contests.
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However, in bipartite graphs,
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However, in bipartite graphs,
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the maximum matching problem is much easier
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the maximum matching problem is much easier
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to solve, because we can reduce it to the
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to solve, because we can reduce it to the
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@ -813,7 +813,7 @@ the groups are $\{1,2,3,4\}$ and $\{5,6,7,8\}$.
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\path[draw,thick,-] (4) -- (7);
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\path[draw,thick,-] (4) -- (7);
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\end{tikzpicture}
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\end{tikzpicture}
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\end{center}
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\end{center}
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The size of a maximum matching of the graph is 3:
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The size of a maximum matching of this graph is 3:
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\begin{center}
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\begin{center}
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\begin{tikzpicture}[scale=0.60]
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\begin{tikzpicture}[scale=0.60]
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\node[draw, circle] (1) at (2,4.5) {1};
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\node[draw, circle] (1) at (2,4.5) {1};
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@ -843,8 +843,8 @@ to the maximum flow problem by adding two new nodes
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to the graph: a source and a sink.
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to the graph: a source and a sink.
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We also add edges from the source
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We also add edges from the source
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to each left node and from each right node to the sink.
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to each left node and from each right node to the sink.
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After this, the maximum flow of the graph
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After this, the size of a maximum flow in the graph
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equals the maximum matching of the original graph.
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equals the size of a maximum matching in the original graph.
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For example, the reduction for the above
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For example, the reduction for the above
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graph is as follows:
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graph is as follows:
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@ -895,11 +895,8 @@ The maximum flow of this graph is as follows:
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\node[draw, circle] (a) at (-2,2.25) {\phantom{0}};
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\node[draw, circle] (a) at (-2,2.25) {\phantom{0}};
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\node[draw, circle] (b) at (12,2.25) {\phantom{0}};
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\node[draw, circle] (b) at (12,2.25) {\phantom{0}};
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%\path[draw,thick,->] (1) -- (5);
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%\path[draw,thick,->] (2) -- (7);
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\path[draw,thick,->] (3) -- (5);
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\path[draw,thick,->] (3) -- (5);
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\path[draw,thick,->] (3) -- (6);
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\path[draw,thick,->] (3) -- (6);
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%\path[draw,thick,->] (3) -- (8);
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\path[draw,thick,->] (4) -- (7);
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\path[draw,thick,->] (4) -- (7);
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\path[draw,thick,->] (a) -- (1);
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\path[draw,thick,->] (a) -- (1);
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@ -1007,7 +1004,7 @@ we cannot form such a matching.
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Since $X$ contains more nodes than $f(X)$,
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Since $X$ contains more nodes than $f(X)$,
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there are no pairs for all nodes in $X$.
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there are no pairs for all nodes in $X$.
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For example, in the above graph, both nodes 2 and 4
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For example, in the above graph, both nodes 2 and 4
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should be connected with node 7 which is not allowed.
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should be connected with node 7 which is not possible.
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\subsubsection{Kőnig's theorem}
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\subsubsection{Kőnig's theorem}
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@ -1052,9 +1049,9 @@ with a maximum matching of size 3:
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\path[draw=red,thick,-,line width=2pt] (3) -- (6);
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\path[draw=red,thick,-,line width=2pt] (3) -- (6);
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\end{tikzpicture}
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\end{tikzpicture}
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\end{center}
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\end{center}
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Kőnig's theorem tells us that the size
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Now Kőnig's theorem tells us that the size
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of a minimum node cover is also 3.
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of a minimum node cover is also 3.
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It can be constructed as follows:
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Such a cover can be constructed as follows:
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\begin{center}
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\begin{center}
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\begin{tikzpicture}[scale=0.60]
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\begin{tikzpicture}[scale=0.60]
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@ -1116,7 +1113,7 @@ set is as follows:
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\index{path cover}
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\index{path cover}
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A \key{path cover} is a set of paths of a graph
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A \key{path cover} is a set of paths in a graph
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such that each node of the graph belongs to at least one path.
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such that each node of the graph belongs to at least one path.
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It turns out that in directed, acyclic graphs,
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It turns out that in directed, acyclic graphs,
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we can reduce the problem of finding a minimum
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we can reduce the problem of finding a minimum
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@ -1177,12 +1174,12 @@ by constructing a \emph{matching graph} where each node
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of the original graph is represented by
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of the original graph is represented by
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two nodes: a left node and a right node.
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two nodes: a left node and a right node.
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There is an edge from a left node to a right node
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There is an edge from a left node to a right node
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if there is a such an edge in the original graph.
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if there is such an edge in the original graph.
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In addition, the matching graph contains a source and a sink,
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In addition, the matching graph contains a source and a sink,
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and there are edges from the source to all
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and there are edges from the source to all
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left nodes and from all right nodes to the sink.
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left nodes and from all right nodes to the sink.
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A maximum matching of the resulting graph corresponds
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A maximum matching in the resulting graph corresponds
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to a minimum node-disjoint path cover in
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to a minimum node-disjoint path cover in
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the original graph.
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the original graph.
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For example, the following matching graph
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For example, the following matching graph
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