CGAL users discussion list

[Mariette Yvinec <>]

There are two ways to triangulate a cube,
using 5 or 6 tetrahedra.
Both  ways yield Delaunay triangulations.
When using a 6 tetrhadra triangulation, you get
the diagonal of the cube in the triangulation.

Juliano Costa wrote:

Hello,

 

I'm researching CGAL for 3D Triangulations.

To do so I'm providing to the CGAL::Delaunay_triangulation_3, 8 points of a 3D Cube.

 

Then I ran an Edge iterator and I got a result which is one of the diagonals of my cube.

I wasn't expecting to get this Edge.

 

If I run the finite_cells_iterator the result is even more strange to me because, If I try to draw the cells

in the order they are given my cube doesn't look right.

 

What is the right way to draw the triangles? Using edges iterators?

 

It seems that I didn't understand Delaunay triangulations 3 right.

 

Of course I'm aware that CGAL result is correct, or maybe my source code is wrong but, can someone please help me to understand it?

 

Is there any way to use CGAL library to get only the 12 triangles from the cube surface (2 triangles per side of the cube)?

 

The code I've been compiling on VS2005 32bit is right below.

 

Thanks anyone in advance!

 

#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Delaunay_triangulation_3.h>
#include <CGAL/Triangulation_vertex_base_with_info_3.h>
#include <conio.h>

struct K : CGAL::Exact_predicates_inexact_constructions_kernel {};
typedef CGAL::Triangulation_vertex_base_with_info_3<unsigned long, K> Vb;
typedef CGAL::Triangulation_data_structure_3<Vb> Tds;
typedef CGAL::Delaunay_triangulation_3<K, Tds> Delaunay;
typedef Delaunay::Point Point;
typedef Delaunay::Vertex_handle Vertex_handle;

int main()
{
  Delaunay T;

  Vertex_handle vh;

 

  vh = T.insert(Point(-1,0,0));
  vh->info() = 1; //attaching an index for each point from 1 to 8.

  vh = T.insert(Point(1,0,0));
  vh->info() = 2;
  vh = T.insert(Point(-1,0,1));
  vh->info() = 3;
  vh = T.insert(Point(1,0,1));
  vh->info() = 4;
  vh = T.insert(Point(-1,1,0));
  vh->info() = 5;
  vh = T.insert(Point(1,1,0));
  vh->info() = 6;
  vh = T.insert(Point(-1,1,1));
  vh->info() = 7;
  vh = T.insert(Point(1,1,1));

  vh->info() = 8;

   int ct=0;

   std::cout<<std::endl<<"Finite_edges_iterator:"<<std::endl;

   for(Delaunay::Finite_edges_iterator eit=T.finite_edges_begin();eit!=T.finite_edges_end();eit++)

   {

       unsigned long i1 =eit->first->vertex(eit->second)->info();

       unsigned long i2=eit->first->vertex(eit->third)->info();

       std::cout<<" Edge: ("<<i1<<","<<i2<<")."<<std::endl;

   }

   ct=0;

   std::cout<<std::endl<<"Finite_cells_iterator:"<<std::endl;

   for(Delaunay::Finite_cells_iterator fcit=T.finite_cells_begin();fcit!=T.finite_cells_end();fcit++)
   {
     std::cout<<"Cell "<<ct++<<": "<<(fcit->vertex(0))->info()<<","<<(fcit->vertex(1))->info()<<","<<(fcit->vertex(2))->info()<<","<<(fcit->vertex(3))->info()<<std::endl;
   }

   std::cout<<"Type anything to close!";

   _getch();

   return 0;

}

 

The output /*(in reduced number of lines)*/ is:

Finite_edges_iterator:
 Edge: (1,3).  Edge: (1,5).  Edge: (3,5).  Edge: (2,1).  Edge: (2,3).  Edge: (2,5).  Edge: (6,8).  Edge: (6,7).
 Edge: (8,7).  Edge: (8,4).  Edge: (4,7).  Edge: (2,4).  Edge: (4,3).  Edge: (2,6).  Edge: (4,6).  Edge: (6,5).
 Edge: (3,7).  Edge: (2,7).  Edge: (7,5).

 

//Edge 2,7 is a diagonal of my cube!!!!! Is that right for a delaunay triangulation????

 

Finite_cells_iterator:
Cell 0: 2,1,3,5 Cell 1: 6,4,8,7  Cell 2: 3,4,2,7 Cell 3: 2,4,6,7 Cell 4: 2,7,6,5 Cell 5: 3,7,2,5

 

 

-- 
Mariette Yvinec
Geometrica project team
INRIA  Sophia-Antipolis