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Improve language

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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}
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.
If a node does not belong to a component,
the algorithm creates a new component
@ -283,9 +283,6 @@ begins at node 3:
\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] (-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{center}
@ -294,8 +291,8 @@ the component does not ''leak'' to other parts in the graph.
\begin{samepage}
The next nodes in the list are nodes 7 and 6,
but they already belong to a component.
The next new component begins at node 1:
but they already belong to a component,
so the next new component begins at node 1:
\begin{center}
\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 $\lnot x_i$
such that both nodes
belong to the same strongly connected component,
so there is a solution.
so a solution exists.
In the graph of the formula $L_2$
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
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 $C$
where $x_1$ and $x_2$ become false
and $x_3$ becomes true.
All variables have been assigned a value,
All variables have been assigned values,
so the remaining components $A$ and $B$
do not change the variables.
Note that this method works, because the
graph has a special structure.
If there are paths from node $x_i$ to node $x_j$
graph has a special structure:
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
@ -559,7 +556,7 @@ and both $x_i$ and $x_j$ become false.
\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
$(a_i \lor b_i \lor c_i)$.
This problem is NP-hard, so no efficient algorithm