1.1 --- a/demo/graph_orientation.cc	Thu Jul 20 06:20:27 2006 +0000
     1.2 +++ b/demo/graph_orientation.cc	Thu Jul 20 14:12:01 2006 +0000
     1.3 @@ -74,7 +74,7 @@
     1.4    IterableBoolMap<ListGraph, Node> active(g,false);
     1.5    for(NodeIt n(g);n!=INVALID;++n) active[n]=(def[n]>0);
     1.6    
     1.7 -  ListGraph::EdgeMap<bool> rev(g,false); // rev[e]==true <=> e is be 
     1.8 +  ListGraph::EdgeMap<bool> rev(g,false); // rev[e]==true <=> e is to be 
     1.9                                           //                  reversed
    1.10  
    1.11    int nodeNum=countNodes(g);

     2.1 --- a/doc/graph_orientation.dox	Thu Jul 20 06:20:27 2006 +0000
     2.2 +++ b/doc/graph_orientation.dox	Thu Jul 20 14:12:01 2006 +0000
     2.3 @@ -11,7 +11,7 @@
     2.4  
     2.5  The input of the problem is a(n undirected) graph and an integer value
     2.6  <i>f(n)</i> assigned to each node \e n. The task is to find an orientation
     2.7 -of the edges for which the number of edge arriving to each node \e n is at
     2.8 +of the edges for which the number of edge arriving at each node \e n is at
     2.9  least least <i>f(n)</i>.
    2.10  
    2.11  In fact, the algorithm reads a directed graph and computes a set of edges to
    2.12 @@ -113,14 +113,15 @@
    2.13  The variable \c nodeNum will refer to the number of nodes.
    2.14  \skipline nodeNum
    2.15  
    2.16 -Here comes the algorithms itself. 
    2.17 +Here comes the algorithm itself. 
    2.18  In each iteration we choose an active node (\c act will do it for us).
    2.19  If there is
    2.20  no such a node, then the orientation is feasible so we are done.
    2.21  \skip act
    2.22  \until while
    2.23  
    2.24 -Then we check if there exists an edge leaving this node that steps down exactly
    2.25 +Then we check if there exists an edge leaving this node and
    2.26 +stepping down exactly
    2.27  one level.
    2.28  \skip OutEdge
    2.29  \until while