CGAL users discussion list

[houes <>]

Dear Efi,

Thank you very much for providing the solution!

I kind of understood your first solution: the halfplane intersection is the
face that is on the positive direction of all lines in 2D arrangment. So in
each overlay, if the new face is on the negative side of any line, then it
is not in the intersection, otherwise the face is the intersection. Time
complexity is O(n*n) for arrangement, n is the number of lines.

I am not very clear with the second solution. But I guess the intersection
face should be the one with the highest count? Since when you are at the
intersection, you go to the highest place. The max count difference among
all faces is n (supposing we have n lines). 

Both solution seem good to me. The only thing is that I am kind of new to
CGAL, and am not very familiar about how to define and use extend data
structure like the face in DCEL or the triats class. So I wonder if the
above solution has better performance than the Nef_polygon_2 data structure?
Since it looks like, Nef_2 can just do the intersection and save me some
time to write my own arraignment code.

Thank you.




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