CGAL users discussion list

["Sebastien Loriot (GeometryFactory)" <>]

To see how to build a Delaunay triangulation, have a look here
http://www.cgal.org/Manual/latest/doc_html/cgal_manual/Triangulation_3/Chapter_main.html#Subsection_37.5.1

To get facets that are on the convex hull of your point set,
use this function

template <class OutputIterator>
OutputIterator  t.incident_cells ( Vertex_handle v, OutputIterator cells)

using a handle to the infinite vertex ( t.infinite_vertex() ).

For each cell c found, you can get the facet as
std::make_pair(c,c.index(t.infinite_vertex())).

Remember that as documneted here:
http://www.cgal.org/Manual/latest/doc_html/cgal_manual/TriangulationDS_3_ref/Concept_TriangulationDataStructure_3.html
Facet (c,i) is the facet of c opposite to the vertex of index i.


To get vertices on the convex hull use this function:
template <class OutputIterator>
OutputIterator 	t.adjacent_vertices ( Vertex_handle v, OutputIterator 
vertices)

using a handle to the infinite vertex.


These functions are documented here:
http://www.cgal.org/Manual/latest/doc_html/cgal_manual/Triangulation_3_ref/Class_Triangulation_3.html
The cell type is documented here:
http://www.cgal.org/Manual/latest/doc_html/cgal_manual/TriangulationDS_3_ref/Concept_TriangulationDataStructure_3--Cell.html


S.


piroshiki wrote:
> Hello,
>  
> I'm researching CGAL for 3D Triangulations and surface mesh.
>
> I want to get surface facets and its vertices of the convex hull after
> execute 3D Delaunay Triangulation from point sets. How can I do this with
> CGAL ?
>
> Please teach me a concrete example code like this:
>
> 1. Give the coordinates (x, y, z) of the points to insert. For example, cube
> consists of eight points，
>     (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (0, 1, 1), (1, 0,
> 1), (1, 1, 1)
>
> 2. Execut 3D Delaunay Triangulation from these points. 
>
> 3. Access all facets, and determine which are surface facets of the cube
> model's convex hull. 
>    If the facet is surface, output its vertices' cooridnates like p1(x1, y1,
> z1), p2(x2, y2, z2), p3(x3, y3,  z3)
>
>
> In this case, I think I can get 12 facets from cube.  What I think is
> correct ?
>
> I am sorry if my English is hard to understand. Please ask if you would like
> me to explain something again.