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