552 lines
17 KiB
TeX
552 lines
17 KiB
TeX
\chapter{Strongly connectivity}
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\index{strongly connected graph}
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In a directed graph, the directions of
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the edges restrict possible paths in the graph,
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so even if the graph is connected,
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this doesn't guarantee that there would be
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a path between any two nodes.
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Thus, it is meaningful to define a new concept
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for directed graphs that requires more than connectivity.
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A graph is \key{strongly connected}
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if there is a path from any node to all
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other nodes in the graph.
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For example, in the following picture,
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the left graph is strongly connected,
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while the right graph is not.
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (1,1) {$1$};
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\node[draw, circle] (2) at (3,1) {$2$};
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\node[draw, circle] (3) at (1,-1) {$3$};
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\node[draw, circle] (4) at (3,-1) {$4$};
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\path[draw,thick,->] (1) -- (2);
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\path[draw,thick,->] (2) -- (4);
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\path[draw,thick,->] (4) -- (3);
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\path[draw,thick,->] (3) -- (1);
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\node[draw, circle] (1b) at (6,1) {$1$};
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\node[draw, circle] (2b) at (8,1) {$2$};
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\node[draw, circle] (3b) at (6,-1) {$3$};
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\node[draw, circle] (4b) at (8,-1) {$4$};
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\path[draw,thick,->] (1b) -- (2b);
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\path[draw,thick,->] (2b) -- (4b);
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\path[draw,thick,->] (4b) -- (3b);
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\path[draw,thick,->] (1b) -- (3b);
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\end{tikzpicture}
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\end{center}
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The right graph is not strongly connected
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because, for example, there is no path
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from node 2 to node 1.
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\index{strongly connected component}
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\index{component graph}
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The \key{strongly connected components}
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of a graph divide the graph into strongly connected
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subgraphs that are as large as possible.
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The strongly connected components form an
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acyclic \key{component graph} that represents
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the deep structure of the original graph.
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For example, for the graph
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\begin{center}
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\begin{tikzpicture}[scale=0.9,label distance=-2mm]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\path[draw,thick,->] (2) -- (1);
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\path[draw,thick,->] (1) -- (3);
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\path[draw,thick,->] (3) -- (2);
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\path[draw,thick,->] (2) -- (4);
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\path[draw,thick,->] (3) -- (5);
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\path[draw,thick,->] (4) edge [bend left] (6);
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\path[draw,thick,->] (6) edge [bend left] (4);
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\path[draw,thick,->] (4) -- (5);
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\path[draw,thick,->] (5) -- (7);
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\path[draw,thick,->] (6) -- (7);
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\end{tikzpicture}
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\end{center}
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the strongly connected components are as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\path[draw,thick,->] (2) -- (1);
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\path[draw,thick,->] (1) -- (3);
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\path[draw,thick,->] (3) -- (2);
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\path[draw,thick,->] (2) -- (4);
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\path[draw,thick,->] (3) -- (5);
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\path[draw,thick,->] (4) edge [bend left] (6);
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\path[draw,thick,->] (6) edge [bend left] (4);
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\path[draw,thick,->] (4) -- (5);
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\path[draw,thick,->] (5) -- (7);
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\path[draw,thick,->] (6) -- (7);
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\draw [red,thick,dashed,line width=2pt] (-0.5,2.5) rectangle (-3.5,-0.5);
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\draw [red,thick,dashed,line width=2pt] (-4.5,2.5) rectangle (-7.5,1.5);
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\draw [red,thick,dashed,line width=2pt] (-4.5,0.5) rectangle (-5.5,-0.5);
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\draw [red,thick,dashed,line width=2pt] (-6.5,0.5) rectangle (-7.5,-0.5);
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\end{tikzpicture}
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\end{center}
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The corresponding component graph is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (-3,1) {$B$};
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\node[draw, circle] (2) at (-6,2) {$A$};
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\node[draw, circle] (3) at (-5,0) {$D$};
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\node[draw, circle] (4) at (-7,0) {$C$};
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\path[draw,thick,->] (1) -- (2);
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\path[draw,thick,->] (1) -- (3);
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\path[draw,thick,->] (2) -- (3);
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\path[draw,thick,->] (2) -- (4);
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\path[draw,thick,->] (3) -- (4);
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\end{tikzpicture}
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\end{center}
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The components are $A=\{1,2\}$,
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$B=\{3,6,7\}$, $C=\{4\}$ and $D=\{5\}$.
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A component graph is an acyclic, directed graph,
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so it is easier to process than the original
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graph because it doesn't contain cycles.
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Thus, as in Chapter 16, it is possible to
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construct a topological sort for a component
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graph and also use dynamic programming algorithms.
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\section{Kosaraju's algorithm}
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\index{Kosaraju's algorithm}
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\key{Kosaraju's algorithm} is an efficient
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method for finding the strongly connected components
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of a directed graph.
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It performs two depth-first searches:
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the first search constructs a list of nodes
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according to the structure of the graph,
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and the second search forms the strongly connected components.
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\subsubsection{Search 1}
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The first phase of the algorithm constructs
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a list of nodes in the order in which a
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depth-first search processes them.
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The algorithm goes through the nodes,
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and begins a depth-first search at each
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unprocessed node.
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Each node will be added to the list
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after it has been processed.
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In the example graph the nodes are processed
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in the following order:
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\begin{center}
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\begin{tikzpicture}[scale=0.9,label distance=-2mm]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\node at (-7,2.75) {$1/8$};
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\node at (-5,2.75) {$2/7$};
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\node at (-3,2.75) {$9/14$};
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\node at (-7,-0.75) {$4/5$};
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\node at (-5,-0.75) {$3/6$};
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\node at (-3,-0.75) {$11/12$};
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\node at (-1,1.75) {$10/13$};
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\path[draw,thick,->] (2) -- (1);
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\path[draw,thick,->] (1) -- (3);
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\path[draw,thick,->] (3) -- (2);
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\path[draw,thick,->] (2) -- (4);
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\path[draw,thick,->] (3) -- (5);
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\path[draw,thick,->] (4) edge [bend left] (6);
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\path[draw,thick,->] (6) edge [bend left] (4);
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\path[draw,thick,->] (4) -- (5);
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\path[draw,thick,->] (5) -- (7);
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\path[draw,thick,->] (6) -- (7);
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\end{tikzpicture}
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\end{center}
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The notation $x/y$ means that
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processing the node started at moment $x$
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and ended at moment $y$.
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When the nodes are sorted according to
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ending times, the result is the following list:
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\begin{tabular}{ll}
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\\
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node & ending time \\
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\hline
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4 & 5 \\
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5 & 6 \\
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2 & 7 \\
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1 & 8 \\
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6 & 12 \\
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7 & 13 \\
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3 & 14 \\
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\\
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\end{tabular}
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In the second phase of the algorithm,
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the nodes will be processed
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in reverse order: $[3,7,6,1,2,5,4]$.
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\subsubsection{Search 2}
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The second phase of the algorithm
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forms the strongly connected components
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of the graph.
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First, the algorithm reverses every
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edge in the graph.
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This ensures that during the second search,
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we will always find a strongly connected
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component without extra nodes.
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The example graph becomes as follows
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after reversing the edges:
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\begin{center}
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\begin{tikzpicture}[scale=0.9,label distance=-2mm]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\path[draw,thick,<-] (2) -- (1);
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\path[draw,thick,<-] (1) -- (3);
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\path[draw,thick,<-] (3) -- (2);
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\path[draw,thick,<-] (2) -- (4);
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\path[draw,thick,<-] (3) -- (5);
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\path[draw,thick,<-] (4) edge [bend left] (6);
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\path[draw,thick,<-] (6) edge [bend left] (4);
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\path[draw,thick,<-] (4) -- (5);
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\path[draw,thick,<-] (5) -- (7);
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\path[draw,thick,<-] (6) -- (7);
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\end{tikzpicture}
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\end{center}
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After this, the algorithm goes through the nodes
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in the order defined by the first search.
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If a node doesn't belong to a component,
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the algorithm creates a new component
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and begins a depth-first search
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where all new nodes found during the search
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are added to the new component.
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In the example graph, the first component
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begins at node 3:
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\begin{center}
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\begin{tikzpicture}[scale=0.9,label distance=-2mm]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\path[draw,thick,<-] (2) -- (1);
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\path[draw,thick,<-] (1) -- (3);
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\path[draw,thick,<-] (3) -- (2);
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\path[draw,thick,<-] (2) -- (4);
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\path[draw,thick,<-] (3) -- (5);
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\path[draw,thick,<-] (4) edge [bend left] (6);
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\path[draw,thick,<-] (6) edge [bend left] (4);
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\path[draw,thick,<-] (4) -- (5);
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\path[draw,thick,<-] (5) -- (7);
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\path[draw,thick,<-] (6) -- (7);
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\draw [red,thick,dashed,line width=2pt] (-0.5,2.5) rectangle (-3.5,-0.5);
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%\draw [red,thick,dashed,line width=2pt] (-4.5,2.5) rectangle (-7.5,1.5);
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%\draw [red,thick,dashed,line width=2pt] (-4.5,0.5) rectangle (-5.5,-0.5);
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%\draw [red,thick,dashed,line width=2pt] (-6.5,0.5) rectangle (-7.5,-0.5);
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\end{tikzpicture}
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\end{center}
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Note that since we reversed all edges in the graph,
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the component doesn't ''leak'' to other parts in the graph.
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\begin{samepage}
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The next nodes in the list are nodes 7 and 6,
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but they already belong to a component.
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The next new component begins at node 1:
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\begin{center}
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\begin{tikzpicture}[scale=0.9,label distance=-2mm]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\path[draw,thick,<-] (2) -- (1);
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\path[draw,thick,<-] (1) -- (3);
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\path[draw,thick,<-] (3) -- (2);
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\path[draw,thick,<-] (2) -- (4);
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\path[draw,thick,<-] (3) -- (5);
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\path[draw,thick,<-] (4) edge [bend left] (6);
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\path[draw,thick,<-] (6) edge [bend left] (4);
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\path[draw,thick,<-] (4) -- (5);
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\path[draw,thick,<-] (5) -- (7);
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\path[draw,thick,<-] (6) -- (7);
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\draw [red,thick,dashed,line width=2pt] (-0.5,2.5) rectangle (-3.5,-0.5);
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\draw [red,thick,dashed,line width=2pt] (-4.5,2.5) rectangle (-7.5,1.5);
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%\draw [red,thick,dashed,line width=2pt] (-4.5,0.5) rectangle (-5.5,-0.5);
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%\draw [red,thick,dashed,line width=2pt] (-6.5,0.5) rectangle (-7.5,-0.5);
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\end{tikzpicture}
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\end{center}
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\end{samepage}
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Finally, the algorithm processes nodes 5 and 5
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that create the remaining strongy connected components:
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\begin{center}
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\begin{tikzpicture}[scale=0.9,label distance=-2mm]
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\node[draw, circle] (1) at (-1,1) {$7$};
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\node[draw, circle] (2) at (-3,2) {$3$};
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\node[draw, circle] (4) at (-5,2) {$2$};
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\node[draw, circle] (6) at (-7,2) {$1$};
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\node[draw, circle] (3) at (-3,0) {$6$};
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\node[draw, circle] (5) at (-5,0) {$5$};
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\node[draw, circle] (7) at (-7,0) {$4$};
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\path[draw,thick,<-] (2) -- (1);
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\path[draw,thick,<-] (1) -- (3);
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\path[draw,thick,<-] (3) -- (2);
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\path[draw,thick,<-] (2) -- (4);
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\path[draw,thick,<-] (3) -- (5);
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\path[draw,thick,<-] (4) edge [bend left] (6);
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\path[draw,thick,<-] (6) edge [bend left] (4);
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\path[draw,thick,<-] (4) -- (5);
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\path[draw,thick,<-] (5) -- (7);
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\path[draw,thick,<-] (6) -- (7);
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\draw [red,thick,dashed,line width=2pt] (-0.5,2.5) rectangle (-3.5,-0.5);
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\draw [red,thick,dashed,line width=2pt] (-4.5,2.5) rectangle (-7.5,1.5);
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\draw [red,thick,dashed,line width=2pt] (-4.5,0.5) rectangle (-5.5,-0.5);
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\draw [red,thick,dashed,line width=2pt] (-6.5,0.5) rectangle (-7.5,-0.5);
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\end{tikzpicture}
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\end{center}
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The time complexity of the algorithm is $O(n+m)$
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where $n$ is the number of nodes and $m$
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is the number of edges.
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The reason for this is that the algorithm
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performs two depth-first searches and
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each search takes $O(n+m)$ time.
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\section{2SAT problem}
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\index{2SAT problem}
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Strongly connectivity is also linked with the
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\key{2SAT problem}.
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In this problem, we are given a logical formula
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\[
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(a_1 \lor b_1) \land (a_2 \lor b_2) \land \cdots \land (a_m \lor b_m),
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\]
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where each $a_i$ and $b_i$ is either a logical variable
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($x_1,x_2,\ldots,x_n$)
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or a negation of a logical variable
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($\lnot x_1, \lnot x_2, \ldots, \lnot x_n$).
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The symbols ''$\land$'' and ''$\lor$'' denote
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logical operators ''and'' and ''or''.
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Our task is to assign each variable a value
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so that the formula is true or state
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that it is not possible.
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For example, the formula
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\[
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L_1 = (x_2 \lor \lnot x_1) \land
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(\lnot x_1 \lor \lnot x_2) \land
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(x_1 \lor x_3) \land
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(\lnot x_2 \lor \lnot x_3) \land
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(x_1 \lor x_4)
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\]
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is true when $x_1$ and $x_2$ are false
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and $x_3$ and $x_4$ are true.
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However, the formula
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\[
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L_2 = (x_1 \lor x_2) \land
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(x_1 \lor \lnot x_2) \land
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(\lnot x_1 \lor x_3) \land
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(\lnot x_1 \lor \lnot x_3)
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\]
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is always false.
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The reason for this is that we can't
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choose a value for variable $x_1$
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without creating a contradiction.
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If $x_1$ is false, both $x_2$ and $\lnot x_2$
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should hold which is impossible,
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and if $x_1$ is true, both $x_3$ and $\lnot x_3$
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should hold which is impossible as well.
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The 2SAT problem can be represented as a graph
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where the nodes correspond to
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variables $x_i$ and negations $\lnot x_i$,
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and the edges determine the connections
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between the variables.
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Each pair $(a_i \lor b_i)$ generates two edges:
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$\lnot a_i \to b_i$ and $\lnot b_i \to a_i$.
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This means that if $a_i$ doesn't hold,
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$b_i$ must hold, and vice versa.
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The graph for formula $L_1$ is:
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\\
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\begin{center}
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\begin{tikzpicture}[scale=1.0,minimum size=2pt]
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\node[draw, circle, inner sep=1.3pt] (1) at (1,2) {$\lnot x_3$};
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\node[draw, circle] (2) at (3,2) {$x_2$};
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\node[draw, circle, inner sep=1.3pt] (3) at (1,0) {$\lnot x_4$};
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\node[draw, circle] (4) at (3,0) {$x_1$};
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\node[draw, circle, inner sep=1.3pt] (5) at (5,2) {$\lnot x_1$};
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\node[draw, circle] (6) at (7,2) {$x_4$};
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\node[draw, circle, inner sep=1.3pt] (7) at (5,0) {$\lnot x_2$};
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\node[draw, circle] (8) at (7,0) {$x_3$};
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\path[draw,thick,->] (1) -- (4);
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\path[draw,thick,->] (4) -- (2);
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\path[draw,thick,->] (2) -- (1);
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\path[draw,thick,->] (3) -- (4);
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\path[draw,thick,->] (2) -- (5);
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\path[draw,thick,->] (4) -- (7);
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\path[draw,thick,->] (5) -- (6);
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\path[draw,thick,->] (5) -- (8);
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\path[draw,thick,->] (8) -- (7);
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\path[draw,thick,->] (7) -- (5);
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\end{tikzpicture}
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\end{center}
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And the graph for formula $L_2$ is:
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\\
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\begin{center}
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\begin{tikzpicture}[scale=1.0,minimum size=2pt]
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\node[draw, circle] (1) at (1,2) {$x_3$};
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\node[draw, circle] (2) at (3,2) {$x_2$};
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\node[draw, circle, inner sep=1.3pt] (3) at (5,2) {$\lnot x_2$};
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\node[draw, circle, inner sep=1.3pt] (4) at (7,2) {$\lnot x_3$};
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\node[draw, circle, inner sep=1.3pt] (5) at (4,3.5) {$\lnot x_1$};
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\node[draw, circle] (6) at (4,0.5) {$x_1$};
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\path[draw,thick,->] (1) -- (5);
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\path[draw,thick,->] (4) -- (5);
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\path[draw,thick,->] (6) -- (1);
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\path[draw,thick,->] (6) -- (4);
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\path[draw,thick,->] (5) -- (2);
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\path[draw,thick,->] (5) -- (3);
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\path[draw,thick,->] (2) -- (6);
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\path[draw,thick,->] (3) -- (6);
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\end{tikzpicture}
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\end{center}
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The structure of the graph indicates whether
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the corresponding 2SAT problem can be solved.
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If there is a variable $x_i$ such that
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both $x_i$ and $\lnot x_i$ belong to the
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same strongly connected component,
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then there are no solutions.
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In this case, the graph contains
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a path from $x_i$ to $\lnot x_i$,
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and also a path from $\lnot x_i$ to $x_i$,
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so both $x_i$ and $\lnot x_i$ should hold
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which is not possible.
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However, if the graph doesn't contain
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such a variable, then there is always a solution.
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In the graph of formula $L_1$
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no nodes $x_i$ and $\lnot x_i$
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belong to the same strongly connected component,
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so there is a solution.
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In the graph of formula $L_2$
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all nodes belong to the same strongly connected component,
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|
so there are no solutions.
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|
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If a solution exists, the values for the variables
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|
can be found by processing the nodes of the
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component graph in a reverse topological sort order.
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At each step, we process and remove a component
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|
that doesn't contain edges that lead to the
|
|
remaining components.
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If the variables in the component don't have values,
|
|
their values will be determined
|
|
according to the component,
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|
and if they already have values,
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they are not changed.
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The process continues until all variables
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|
have been assigned a value.
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|
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The component graph for formula $L_1$ is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=1.0]
|
|
\node[draw, circle] (1) at (0,0) {$A$};
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|
\node[draw, circle] (2) at (2,0) {$B$};
|
|
\node[draw, circle] (3) at (4,0) {$C$};
|
|
\node[draw, circle] (4) at (6,0) {$D$};
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|
|
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\path[draw,thick,->] (1) -- (2);
|
|
\path[draw,thick,->] (2) -- (3);
|
|
\path[draw,thick,->] (3) -- (4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The components are
|
|
$A = \{\lnot x_4\}$,
|
|
$B = \{x_1, x_2, \lnot x_3\}$,
|
|
$C = \{\lnot x_1, \lnot x_2, x_3\}$ and
|
|
$D = \{x_4\}$.
|
|
When constructing the solution,
|
|
we first process component $D$
|
|
where $x_4$ becomes true.
|
|
After this, we process component $C$
|
|
where $x_1$ and $x_2$ become false
|
|
and $x_3$ becomes true.
|
|
All variables have been assigned a value,
|
|
so the remaining components $A$ and $B$
|
|
don't change the variables anymore.
|
|
|
|
Note that this method works because the
|
|
structure of the graph is special.
|
|
If there are paths from node $x_i$ to node $x_j$
|
|
and from node $x_j$ to node $\lnot x_j$,
|
|
then node $x_i$ never becomes true.
|
|
The reason for this is that there is also
|
|
a path from node $\lnot x_j$ to node $\lnot x_i$,
|
|
and both $x_i$ and $x_j$ become false.
|
|
|
|
\index{3SAT problem}
|
|
|
|
A more difficult problem is the \key{3SAT problem}
|
|
where each part of the formula is of the form
|
|
$(a_i \lor b_i \lor c_i)$.
|
|
No efficient algorithm for solving this problem
|
|
is known, but it is a NP-hard problem.
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