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[Mael <>]

Hello,

Assuming you are using a basic Delaunay_triangulation_2: for the maximal alpha value, the alpha shape is the Delaunay triangulation of the point set, thus the status of a vertex cannot be exterior since the Delaunay triangulation is a triangulation of the convex hull of the point set.

If you are using a weighted triangulation (Regular_triangulation_2), things are bit trickier: if the point has a weight such that it does not appear in the (weighted) Delaunay triangulation, then its status should be exterior, even at the greatest alpha value of the family. However, there is a known bug in Alpha_shape_2 when using the General mode that makes it so a vertex is never marked as exterior. I am not sure if its also present in regular mode.

Best,
Mael

On 03/08/2018 09:46, nsoonhui wrote:

I have the following (Pseudo) code:
 
      
       Alpha_shape_2 *A =new Alpha_shape_2(first,last, 100000,Mode.Regular);
      auto maxLevel =  A->number_of_alphas()-1;
      auto alphaLevel = A->get_nth_alpha(maxLevel);
	for(fit=A->finite_faces_begin(); fit!=A->finite_faces_end(); ++fit)
	{


		     A->set_alpha(alphaLevel);
	           auto classType2 = A->classify(fit);   //can this be EXTERIOR?
			
	}


I know that if alphaLevel is not maximum, then    A->classify(fit) can be
EXTERIOR.

But, my question is that, is it even possible for the maximum alpha level to
classify one of the face as EXTERIOR?





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