Corrections

luku20.tex

@@ -556,17 +556,25 @@ as a cut in the graph.
 Thus, the flow has to be a maximum flow,
 and the cut has to be a minimum cut.
 
-\section{Parallel paths}
+\section{Disjoint paths}
 
-As a first application for flows,
-we consider a problem where the task is to
-form as many parallel paths as possible
-from the starting node of the graph
-to the ending node.
-It is required that no edge appears
-in more than one path.
+Many graph problems can be solved by reducing
+them to the maximum flow problem.
+Our first example of such a problem is
+as follows: we are given a directed graph
+with a source and a sink,
+and our task is to find the maximum number
+of disjoint paths from the source to the sink.
 
-For example, in the graph
+\subsubsection{Edge-disjoint paths}
+
+We will first focus on the problem of
+finding the maximum number of
+\key{edge-disjoint paths} from the source to the sink.
+This means that we should construct a set of paths
+such that each edge appears in at most one path.
+
+For example, consider the following graph:
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
 \node[draw, circle] (1) at (1,2) {$1$};
@@ -586,10 +594,12 @@ For example, in the graph
 \path[draw,thick,->] (5) -- (6);
 \end{tikzpicture}
 \end{center}
-we can form two parallel paths from node 1 to node 6.
-This can be done by choosing paths
+
+In this graph, the maximum number of edge-disjoint
+paths is 2.
+We can choose the paths
 $1 \rightarrow 2 \rightarrow 4 \rightarrow 3 \rightarrow 6$
-and $1 \rightarrow 4 \rightarrow 5 \rightarrow 6$:
+and $1 \rightarrow 4 \rightarrow 5 \rightarrow 6$ as follows:
 
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
@@ -620,19 +630,29 @@ and $1 \rightarrow 4 \rightarrow 5 \rightarrow 6$:
 \end{tikzpicture}
 \end{center}
 
-It turns out that the maximum number of parallel paths
-equals the maximum flow in the graph when the weight
-of each edge is 1.
+It turns out that the maximum number of
+edge-disjoint paths
+equals the maximum flow of the graph,
+assuming that the capacity of each edge is one.
 After the maximum flow has been constructed,
-the parallel paths can be found greedily by finding
-paths from the starting node to the ending node.
+the edge-disjoint paths can be found greedily
+by following paths from the source to the sink.
 
-Let's then consider a variation for the problem
-where each node (except for the starting and ending nodes)
-can appear in at most one path.
-After this restriction, we can construct only one path
-in the above graph, because node 4 can't appear
-in more than one path:
+\subsubsection{Node-disjoint paths}
+
+Let us now consider another problem:
+finding the maximum number of
+\key{node-disjoint paths} from the source
+to the sink.
+In this problem, every node,
+except for the source and sink,
+may appear in at most one path.
+The number of node-disjoint paths is
+often smaller than the number of
+edge-disjoint paths.
+
+For example, in the previous graph,
+the maximum number of node-disjoint paths is 1:
 
 \begin{center}
 \begin{tikzpicture}
@@ -659,14 +679,17 @@ in more than one path:
 \end{tikzpicture}
 \end{center}
 
-A standard way to restrict the flow through a node
-is to divide the node into two parts.
-All incoming edges are connected to the first part,
-and all outgoing edges are connected to the second part.
-In addition, there is an edge from the first part
-to the second part.
+We can reduce also this problem to the maximum flow problem.
+Since each node can appear in at most one path,
+we have to limit the flow that goes through the nodes.
+A standard method for this is to divide each node into
+two nodes such that the first node has the incoming edges
+of the original node and the second node has the outgoing
+edges of the original node.
+In addition, there is a new edge from the first node
+to the second node.
 
-In the current example, the graph becomes as follows:
+In our example, the graph becomes as follows:
 \begin{center}
 \begin{tikzpicture}
 \node[draw, circle] (1) at (1,2) {$1$};
@@ -742,35 +765,34 @@ The maximum flow for the graph is as follows:
 \end{tikzpicture}
 \end{center}
 
-This means that it is possible to form exactly
-one path from the starting node to the ending node
-when a node can't appear in more than one path.
+Thus, the maximum number of node-disjoint paths
+from the source to the sink is 1.
 
-\section{Maximum matching}
+\section{Maximum matchings}
 
 \index{matching}
 \index{maximum matching}
 
-A \key{maximum matching} is the largest possible
-set of pairs of nodes in a graph
-such that there is an edge between each pair of nodes,
-and each node belongs to at most one pair.
+The \key{maximum matching} problem asks to find
+a maximum-size set of node pairs in a graph
+such that each pair is connected with an edge and
+each node belongs to at most one pair.
 
-There is a polynomial algorithm for finding
-a maximum matching in a general graph,
-but it is very complex.
-For this reason, we will restrict ourselves to the
-case where the graph is bipartite.
-In this case we can easily find the maximum matching
-using a maximum flow algorithm.
+There are polynomial algorithms for finding
+maximum matchings in general graphs,
+but such algorithms are complex and do
+not appear in programming contests.
+However, in bipartite graphs,
+the maximum matching problem is much easier
+to solve, because we can reduce it to the
+maximum flow problem.
 
-\subsubsection{Finding a maximum matching}
+\subsubsection{Finding maximum matchings}
 
-A bipartite graph can be always presented so
-that it consists of left-side and right-side nodes,
-and all edges in the graph go between
-left and right sides.
-As an example, consider the following graph:
+The nodes in a bipartite graph can be always
+divided into two groups such that all edges
+of the graph go from the left group to the right group.
+For example, consider the following bipartite graph:
 
 \begin{center}
 \begin{tikzpicture}[scale=0.60]
@@ -813,20 +835,20 @@ In this graph, the size of a maximum matching is 3:
 
 \path[draw=red,thick,-,line width=2pt] (1) -- (5);
 \path[draw=red,thick,-,line width=2pt] (2) -- (7);
-\path[draw=red,thick,-,line width=2pt] (3) -- (6);
+\path[draw=red,thick,-,line width=2pt] (3) -- (8);
 \end{tikzpicture}
 \end{center}
 
-A maximum matching in a bipartite graph
-corresponds to a maximum flow in an extended graph
-that contains a starting node,
-an ending node and all the nodes of the original graph.
-There is an edge from the starting node to
-each left-side node, and an edge from 
-each right-side node to the ending node.
-The capacity of each edge is 1.
+We can reduce the bipartite maximum matching problem
+to the maximum flow problem by adding two new nodes
+to the graph: a source and a sink.
+In addition, we add edges from the source
+to each left node and from each right node to the sink.
+After this, the maximum flow of the graph
+equals the maximum matching of the original graph.
 
-In the example graph, the result is as follows:
+For example, the reduction for the above
+graph is as follows:
 \begin{center}
 \begin{tikzpicture}[scale=0.60]
 \node[draw, circle] (1) at (2,4.5) {1};
@@ -859,35 +881,74 @@ In the example graph, the result is as follows:
 \end{tikzpicture}
 \end{center}
 
-The size of a maximum flow in this graph
-equals the size of a maximum matching
-in the original graph,
-because each path from the starting node
-to the ending node adds one edge to the matching.
-In this graph, the maximum flow is 3,
-so the maximum matching is also 3.
+The maximum flow of this graph is as follows:
+\begin{center}
+\begin{tikzpicture}[scale=0.60]
+\node[draw, circle] (1) at (2,4.5) {1};
+\node[draw, circle] (2) at (2,3) {2};
+\node[draw, circle] (3) at (2,1.5) {3};
+\node[draw, circle] (4) at (2,0) {4};
+\node[draw, circle] (5) at (8,4.5) {5};
+\node[draw, circle] (6) at (8,3) {6};
+\node[draw, circle] (7) at (8,1.5) {7};
+\node[draw, circle] (8) at (8,0) {8};
+
+\node[draw, circle] (a) at (-2,2.25) {\phantom{0}};
+\node[draw, circle] (b) at (12,2.25) {\phantom{0}};
+
+%\path[draw,thick,->] (1) -- (5);
+%\path[draw,thick,->] (2) -- (7);
+\path[draw,thick,->] (3) -- (5);
+\path[draw,thick,->] (3) -- (6);
+%\path[draw,thick,->] (3) -- (8);
+\path[draw,thick,->] (4) -- (7);
+
+\path[draw,thick,->] (a) -- (1);
+\path[draw,thick,->] (a) -- (2);
+\path[draw,thick,->] (a) -- (3);
+\path[draw,thick,->] (a) -- (4);
+\path[draw,thick,->] (5) -- (b);
+\path[draw,thick,->] (6) -- (b);
+\path[draw,thick,->] (7) -- (b);
+\path[draw,thick,->] (8) -- (b);
+
+\path[draw=red,thick,->,line width=2pt] (1) -- (5);
+\path[draw=red,thick,->,line width=2pt] (2) -- (7);
+\path[draw=red,thick,->,line width=2pt] (3) -- (8);
+
+\path[draw=red,thick,->,line width=2pt] (a) -- (1);
+\path[draw=red,thick,->,line width=2pt] (a) -- (2);
+\path[draw=red,thick,->,line width=2pt] (a) -- (3);
+
+\path[draw=red,thick,->,line width=2pt] (5) -- (b);
+\path[draw=red,thick,->,line width=2pt] (7) -- (b);
+\path[draw=red,thick,->,line width=2pt] (8) -- (b);
+
+\end{tikzpicture}
+\end{center}
 
 \subsubsection{Hall's theorem}
 
 \index{Hall's theorem}
 \index{perfect matching}
 
-\key{Hall's theorem} describes when a bipartite graph
-has a matching that contains all nodes
-in one side of the graph.
-If both sides contain the same number of nodes,
-Hall's theorem tells us if it's possible to
-construct a \key{perfect matching} where
-all nodes are paired with each other.
+\key{Hall's theorem} can be used to find out
+whether a bipartite graph has a matching
+that contains all left or right nodes.
+If the number of left and right nodes is the same,
+Hall's theorem tells us if it is possible to
+construct a \key{perfect matching} that
+contains all nodes of the graph.
 
-Assume that we want to construct a matching
-that contains all left-side nodes.
-Let $X$ be a set of left-side nodes,
+Assume that we want to find a matching
+that contains all left nodes.
+Let $X$ be any set of left nodes
 and let $f(X)$ be the set of their neighbors.
-According to Hall's theorem, a such matching exists
+According to Hall's theorem, a matching
+that contains all left nodes exists
 exactly when for each $X$, the condition $|X| \le |f(X)|$ holds.
 
-Let's study Hall's theorem in the example graph.
+Let us study Hall's theorem in the example graph.
 First, let $X=\{1,3\}$ and $f(X)=\{5,6,8\}$:
 
 \begin{center}
@@ -935,18 +996,19 @@ Next, let $X=\{2,4\}$ and $f(X)=\{7\}$:
 \end{center}
 
 In this case, $|X|=2$ and $|f(X)|=1$,
-so the condition of Hall's theorem doesn't hold.
-This means that it's not possible to form
+so the condition of Hall's theorem does not hold.
+This means that it is not possible to form
 a perfect matching in the graph.
 This result is not surprising, because we already
-knew that the maximum matching in the graph is 3 and not 4.
+know that the maximum matching of the graph is 3 and not 4.
 
-If the condition of Hall's theorem doesn't hold,
-the set $X$ provides an explanation why we can't form a matching.
+If the condition of Hall's theorem does not hold,
+the set $X$ provides an explanation \emph{why}
+we cannot form such a matching.
 Since $X$ contains more nodes than $f(X)$,
 there is no pair for all nodes in $X$.
 For example, in the above graph, both nodes 2 and 4
-should be connected to node 7 which is not possible.
+should be connected with node 7 which is not allowed.
 
 \subsubsection{Kőnig's theorem}
 
@@ -954,23 +1016,19 @@ should be connected to node 7 which is not possible.
 \index{node cover}
 \index{minimum node cover}
 
-\key{Kőnig's theorem} provides an efficient way
-to construct a \key{minimum node cover} for a
-bipartite graph.
-This is a minimum set of nodes such that
-each edge in the graph is connected to at least
-one node in the set.
-
+A \key{minimum node cover} of a graph
+is a set of nodes such that each edge of the graph
+has at least one node in the set.
 In a general graph, finding a minimum node cover
 is a NP-hard problem.
-However, in a bipartite graph,
-the size of
-a maximum matching and a minimum node cover
-is always the same, according to Kőnig's theorem.
-Thus, we can efficiently find a minimum node cover
+However, according to \key{Kőnig's theorem},
+the size of a minimum node cover
+and the size of a maximum matching is always the same
+if the graph is bipartite.
+Thus, we can calculate the size of a minimum node cover
 using a maximum flow algorithm.
 
-Let's consider the following graph
+Let us consider the following graph
 with a maximum matching of size 3:
 \begin{center}
 \begin{tikzpicture}[scale=0.60]
@@ -995,7 +1053,7 @@ with a maximum matching of size 3:
 \path[draw=red,thick,-,line width=2pt] (3) -- (6);
 \end{tikzpicture}
 \end{center}
-Using Kőnig's theorem, we know that the size
+Kőnig's theorem tells us that the size
 of a minimum node cover is also 3.
 It can be constructed as follows:
 
@@ -1018,19 +1076,16 @@ It can be constructed as follows:
 \path[draw,thick,-] (4) -- (7);
 \end{tikzpicture}
 \end{center}
-For each edge in the maximum matching,
-exactly one of its end nodes belongs to
-the minimum node cover.
 
 \index{independent set}
 \index{maximum independent set}
 
-The set of all nodes that do \emph{not}
+The nodes that do \emph{not}
 belong to a minimum node cover
-forms a \key{maximum independent set}.
+form a \key{maximum independent set}.
 This is the largest possible set of nodes
-where there is no edge between any two nodes
-in the graph.
+such that no two nodes in the set
+are connected with an edge.
 Once again, finding a maximum independent
 set in a general graph is a NP-hard problem,
 but in a bipartite graph we can use
@@ -1063,22 +1118,16 @@ set is as follows:
 \index{path cover}
 
 A \key{path cover} is a set of paths in a graph
-that is chosen so that each node in the graph
-belongs to at least one path.
-It turns out that we can reduce the problem
-of finding a minimum path cover in a
-directed, acyclic graph into a maximum flow problem.
+such that each node of the graph belongs to at least one path.
+It turns out that in a directed, acyclic graph,
+we can reduce the problem of finding a minimum
+path cover to the problem of finding a maximum
+flow in another graph.
 
-There are two variations for the problem:
-In a \key{node-disjoint cover},
-every node appears in exactly one path,
-and in a \key{general cover},
-a node may appear in more than one path.
-In both cases, the minimum path cover can be
-found using a similar idea.
-
-\subsubsection{Node-disjoint cover}
+\subsubsection{Node-disjoint path cover}
 
+In a \key{node-disjoint path cover},
+each node belongs to exactly one path.
 As an example, consider the following graph:
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
@@ -1099,7 +1148,8 @@ As an example, consider the following graph:
 \end{tikzpicture}
 \end{center}
 
-In this case, the minimum node-disjoint path cover
+A minimum node-disjoint path cover
+of this graph
 consists of three paths.
 For example, we can choose the following paths:
 
@@ -1121,18 +1171,23 @@ For example, we can choose the following paths:
 \end{center}
 
 Note that one of the paths only contains node 2,
-so it is possible that a path doesn't contain any edges.
+so it is possible that a path does not contain any edges.
 
-Finding a path cover can be interpreted as finding
-a maximum matching in a graph where each node
-in the original graph is represented by two nodes:
-a left node and a right node.
-There is an edge from a left node to a right node,
-if there is such an edge in the original graph.
-The idea is that the matching determines which
-edges belong to paths in the original graph.
+We can find a minimum node-disjoint path cover
+by constructing a matching graph so that each node
+in the original graph corresponds to two
+nodes in the matching graph: a left and right node.
+There is an edge from a left node to a right node
+if there is a such an edge in the original graph.
+In addition, the matching graph contains a source and a sink
+such that there are edges from the source to all
+left nodes and from all right nodes to the sink.
 
-The matching in the example case is as follows:
+A maximum matching in the resulting graph corresponds
+to a minimum node-disjoint path cover in
+the original graph.
+For example, the following graph contains
+a maximum matching of size 4:
 
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
@@ -1186,26 +1241,21 @@ The matching in the example case is as follows:
 \end{tikzpicture}
 \end{center}
 
-In this case, the maximum matching consists of four edges
-that corresponds to edges
-$1 \rightarrow 5$, $3 \rightarrow 4$,
-$5 \rightarrow 6$ and $6 \rightarrow 7$ in the original graph.
-Thus, a minimum node-disjoint path cover consists of paths
-that contain these edges.
+Each edge in the maximum matching of the matching graph corresponds
+to an edge in the minimum node-disjoint path cover
+of the original graph.
+Thus, the size of the minimum node-disjoint path cover is $n-c$,
+where $n$ is the number of nodes in the original graph
+and $c$ is the size of the maximum matching.
 
-The size of a minimum path cover is $n-c$ where
-$n$ is the number of nodes in the graph,
-and $c$ is the number of edges in the maximum matching.
-For example, in the above graph the size of the
-minimum path cover is $7-4=3$.
-
-\subsubsection{General cover}
-
-In a general path cover, a node can belong to more than one path
-which may decrease the number of paths needed.
-In the example graph, the minimum general path cover
-consists of two paths as follows:
+\subsubsection{General path cover}
 
+A \key{general path cover} is a path cover
+where a node can belong to more than one path.
+A minimum general path cover may be smaller
+than a minimum node-disjoint path cover,
+because a node can used multiple times in paths.
+Consider again the following graph:
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
 \node[draw, circle] (1) at (0,0) {1};
@@ -1216,28 +1266,58 @@ consists of two paths as follows:
 \node[draw, circle] (6) at (2,-2) {6};
 \node[draw, circle] (7) at (4,-2) {7};
 
-\path[draw=blue,thick,->,line width=2pt] (1) -- (5);
-\path[draw=blue,thick,->,line width=2pt] (5) -- (6);
-\path[draw=blue,thick,->,line width=2pt] (6) -- (3);
-\path[draw=blue,thick,->,line width=2pt] (3) -- (4);
-\path[draw=green,thick,->,line width=2pt] (2) -- (6);
-\path[draw=green,thick,->,line width=2pt] (6) -- (7);
+\path[draw,thick,->,>=latex] (1) -- (5);
+\path[draw,thick,->,>=latex] (2) -- (6);
+\path[draw,thick,->,>=latex] (3) -- (4);
+\path[draw,thick,->,>=latex] (5) -- (6);
+\path[draw,thick,->,>=latex] (6) -- (3);
+\path[draw,thick,->,>=latex] (6) -- (7);
 \end{tikzpicture}
 \end{center}
 
-In this graph, a minimum general path cover contains 2 paths,
-while a minimum node-disjoint path cover contains 3 paths.
-The difference is that in the general path cover,
-node 6 appears in two paths.
+The minimum general path cover in this graph
+consists of two paths.
+For example, the first path may be as follows:
+\begin{center}
+\begin{tikzpicture}[scale=0.9]
+\node[draw, circle] (1) at (0,0) {1};
+\node[draw, circle] (2) at (2,0) {2};
+\node[draw, circle] (3) at (4,0) {3};
+\node[draw, circle] (4) at (6,0) {4};
+\node[draw, circle] (5) at (0,-2) {5};
+\node[draw, circle] (6) at (2,-2) {6};
+\node[draw, circle] (7) at (4,-2) {7};
+
+\path[draw=red,thick,->,line width=2pt] (1) -- (5);
+\path[draw=red,thick,->,line width=2pt] (5) -- (6);
+\path[draw=red,thick,->,line width=2pt] (6) -- (3);
+\path[draw=red,thick,->,line width=2pt] (3) -- (4);
+\end{tikzpicture}
+\end{center}
+And the second path may be as follows:
+\begin{center}
+\begin{tikzpicture}[scale=0.9]
+\node[draw, circle] (1) at (0,0) {1};
+\node[draw, circle] (2) at (2,0) {2};
+\node[draw, circle] (3) at (4,0) {3};
+\node[draw, circle] (4) at (6,0) {4};
+\node[draw, circle] (5) at (0,-2) {5};
+\node[draw, circle] (6) at (2,-2) {6};
+\node[draw, circle] (7) at (4,-2) {7};
+
+\path[draw=red,thick,->,line width=2pt] (2) -- (6);
+\path[draw=red,thick,->,line width=2pt] (6) -- (7);
+\end{tikzpicture}
+\end{center}
 
 A minimum general path cover can be found
 almost like a minimum node-disjoint path cover.
-It suffices to extend the matching graph
+It suffices to add some new edges to the matching graph
 so that there is an edge $a \rightarrow b$
-always when there is a path from node $a$ to node $b$
+always when there is a path from $a$ to $b$
 in the original graph (possibly through several edges).
 
-The matching graph for the example case looks as follows:
+The matching graph for the above graph is as follows:
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
 \node[draw, circle] (1a) at (0,6) {1};
@@ -1327,17 +1407,17 @@ The matching graph for the example case looks as follows:
 \index{Dilworth's theorem}
 \index{antichain}
 
-\key{Dilworth's theorem} states that the size of
-a minimum general path cover in a directed, acyclic graph
-equals the maximum size of an \key{antichain}, i.e.,
-a set of nodes such that there is no path
-from any node to another node.
+An \key{antichain} is a set of nodes of a graph
+such that there is no path
+from any node to another node
+using the edges of the graph.
+\key{Dilworth's theorem} states that
+in a directed acyclic graph, the size of
+a minimum general path cover
+equals the size of a maximum antichain.
 
-For example, in the example graph, the minimum
-general path cover contains two paths,
-so the largest antichain contains two nodes.
-We can construct such an antichain
-by choosing nodes 3 and 7:
+For example, nodes 3 and 7 form an antichain
+in the following graph:
 
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
@@ -1358,9 +1438,7 @@ by choosing nodes 3 and 7:
 \end{tikzpicture}
 \end{center}
 
-There is no path from node 3 to node 7,
-and no path from node 7 to node 3,
-so nodes 3 and 7 form an antichain.
-On the other hand, if we choose any three
-nodes in the graph, there is certainly a
-path from one node to another node.
+This is a maximum antichain, because it is not possible
+to construct any antichain that would contain three nodes.
+We have seen before that the size of a minimum
+general path cover of this graph consists of two paths.