CGAL users discussion list

[Manuel Caroli <>]

Hi Juliano,

the Triangulation_3 package computes a triangulation of R^3 with the 
input points as vertices. If you want to compute a triangular mesh 
approximating a surface you should use a surface mesher. Maybe you could 
be more specific about your problem.

best

Manuel

Juliano Costa wrote:
> Thanks a lot for your attention Mariette,
>  
> Ok, so the source code I used gives me the 6 tetrhadra triangulation, is 
> that right? 
> But can I achieve this 5 tetrhadra triangulation using the CGAL library?
> Can you give me a clue of the source code I need?
> Do you think I should use the mesh algorithm instead of triangulation to 
> achieve 
> what I was expecting? 
>  
> Best regards,
>  
> Juliano Costa
> Brazil.
>
>  
> 2009/8/3 Mariette Yvinec < 
> <mailto:>>
>
>     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(0,0,0));
> >       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(0,0,1));
> >       vh->info() = 3;
> >       vh = T.insert(Point(1,0,1));
> >       vh->info() = 4;
> >       vh = T.insert(Point(0,1,0));
> >       vh->info() = 5;
> >       vh = T.insert(Point(1,1,0));
> >       vh->info() = 6;
> >       vh = T.insert(Point(0,1,1));
> >       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