CGAL users discussion list

A point to polyhedron test is O(n) for every test. That will be quite
cumbersome, especially for a large number of points needed for mesh
generations. You might try to obtain some code to compute the medial surface 
of
the polyhedron, in order to efficiently compute the distance function, but I'm
not sure how that works for self-intersecting meshes. Besides, the complexity
is at least O(n^2).

Why don't you clip the intersecting parts? See what triangle intersect badly,
cut off the inside-intersecting parts and then glue the intersections
boundaries together. You can do this in O(n) quite easily.

Amir.

-----Original Message-----
From: Qianqian Fang 
[mailto:]
 
Sent: Sunday, February 22, 2009 10:59 PM
To: 

Subject: [cgal-discuss] use Surface mesher to re-mesh self-intersecting
surfaces

hi

In the past, I asked in this list about removing self-intersecting
elements from triangular surface. I was recommended to use
the surface mesher avoid repairing a bad mesh as it guarantees to
have no self-intersecting elements.

Now, I need to process a surface generated from other software,
unfortunately, this mesh contains self-intersecting elements. So, I
still need a program to fix this surface (and resample it to my
desired density).

Reading the documents of the Surface Mesher, I have the impression
that it can be applied for this situation. The only difference
is that the surface is not defined by gray-scale image, or an implicit
function, but by a triangular surface mesh.

Indeed, I can even define the surface by an implicit function, which
returns the distance from a Point_3 to a Polyhedron_3, similar to
the sphere case in the manual. Does this sound like a viable
approach to you?

If it may work, what I need is to find a subroutine to calculate the
distance. I only found a squared_distance() for Point_3
to Plane_3 in include/CGAL/squared_distance_3_2.h. I am wondering
if a similar function to calculate Point3 to a polyhedron exists?
if it does, can anyone show me an example how to use it?

thank you

Qianqian
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