/*assign lowalso check for articulation points *void graph::assignlowvertex low num/rule for each vertex adjacent to ifw num num /forward edge assignloww )ifw low > num cout < <is an articulation point<endlv low minv loww low )/rule else ifv parent ! /back edge low minv loww num )/rule figure pseudocode to compute low and to test for articulation points (test for the root is omittedvthen the recursive call to will find adjacent to this is not back edgeonly an edge that has already been considered and needs to be ignored otherwisethe procedure computes the minimum of the various low and num entriesas specified by the algorithm there is no rule that traversal must be either preorder or postorder it is possible to do processing both before and after the recursive calls the procedure in figure combines the two routines assignnum and assignlow in straightforward manner to produce the procedure findart euler circuits consider the three figures in figure popular puzzle is to reconstruct these figures using pendrawing each line exactly once the pen may not be lifted from the paper while the drawing is being performed as an extra challengemake the pen finish at the same point at which it started this puzzle has surprisingly simple solution stop reading if you would like to try to solve it the first figure can be drawn only if the starting point is the lower leftor right-hand cornerand it is not possible to finish at the starting point the second figure is easily drawn with the finishing point the same as the starting pointbut the third figure cannot be drawn at all within the parameters of the puzzle we can convert this problem to graph theory problem by assigning vertex to each intersection then the edges can be assigned in the natural manneras in figure