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Antti H S Laaksonen 2017-05-29 20:39:30 +03:00
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chapter17.tex
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@ -250,7 +250,7 @@ the example graph is as follows:
\end{center} \end{center}
After this, the algorithm goes through After this, the algorithm goes through
the list of nodes created by the first search the list of nodes created by the first search,
in \emph{reverse} order. in \emph{reverse} order.
If a node does not belong to a component, If a node does not belong to a component,
the algorithm creates a new component the algorithm creates a new component
@ -283,9 +283,6 @@ begins at node 3:
\path[draw,thick,<-] (6) -- (7); \path[draw,thick,<-] (6) -- (7);
\draw [red,thick,dashed,line width=2pt] (-0.5,2.5) rectangle (-3.5,-0.5); \draw [red,thick,dashed,line width=2pt] (-0.5,2.5) rectangle (-3.5,-0.5);
%\draw [red,thick,dashed,line width=2pt] (-4.5,2.5) rectangle (-7.5,1.5);
%\draw [red,thick,dashed,line width=2pt] (-4.5,0.5) rectangle (-5.5,-0.5);
%\draw [red,thick,dashed,line width=2pt] (-6.5,0.5) rectangle (-7.5,-0.5);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -294,8 +291,8 @@ the component does not ''leak'' to other parts in the graph.
\begin{samepage} \begin{samepage}
The next nodes in the list are nodes 7 and 6, The next nodes in the list are nodes 7 and 6,
but they already belong to a component. but they already belong to a component,
The next new component begins at node 1: so the next new component begins at node 1:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9,label distance=-2mm] \begin{tikzpicture}[scale=0.9,label distance=-2mm]
@ -499,10 +496,10 @@ In the graph of the formula $L_1$
there are no nodes $x_i$ and $\lnot x_i$ there are no nodes $x_i$ and $\lnot x_i$
such that both nodes such that both nodes
belong to the same strongly connected component, belong to the same strongly connected component,
so there is a solution. so a solution exists.
In the graph of the formula $L_2$ In the graph of the formula $L_2$
all nodes belong to the same strongly connected component, all nodes belong to the same strongly connected component,
so there are no solutions. so a solution does not exist.
If a solution exists, the values for the variables If a solution exists, the values for the variables
can be found by going through the nodes of the can be found by going through the nodes of the
@ -544,13 +541,13 @@ where $x_4$ becomes true.
After this, we process the component $C$ After this, we process the component $C$
where $x_1$ and $x_2$ become false where $x_1$ and $x_2$ become false
and $x_3$ becomes true. and $x_3$ becomes true.
All variables have been assigned a value, All variables have been assigned values,
so the remaining components $A$ and $B$ so the remaining components $A$ and $B$
do not change the variables. do not change the variables.
Note that this method works, because the Note that this method works, because the
graph has a special structure. graph has a special structure:
If there are paths from node $x_i$ to node $x_j$ if there are paths from node $x_i$ to node $x_j$
and from node $x_j$ to node $\lnot x_j$, and from node $x_j$ to node $\lnot x_j$,
then node $x_i$ never becomes true. then node $x_i$ never becomes true.
The reason for this is that there is also The reason for this is that there is also
@ -559,7 +556,7 @@ and both $x_i$ and $x_j$ become false.
\index{3SAT problem} \index{3SAT problem}
A more difficult problem is the \key{3SAT problem} A more difficult problem is the \key{3SAT problem},
where each part of the formula is of the form where each part of the formula is of the form
$(a_i \lor b_i \lor c_i)$. $(a_i \lor b_i \lor c_i)$.
This problem is NP-hard, so no efficient algorithm This problem is NP-hard, so no efficient algorithm