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[Sebastien Loriot <>]

I'm not sure to understand what is your problem. A minimal self 
contained example that we can run and that shows your problem would
be perfect.

What I can tell from the code I see:
You are computing a 3 triangulations that consists of tetrahedra.
`finite_facets()` provides the range of face that are not connected
to the infinite vertex. A facet is given a std::pair of a Cell_handle
(tetrahedron) and the id of the vertex in the cell opposite to the
facet. About the access to the vertices of a facet, the %4 is here so
that the index stays in [0,3] while ignoring pair.second (that might or
might not be infinite).

HTH,

Sebastien.

On 7/5/22 14:46, KL ( via cgal-discuss Mailing List) 
wrote:
> Hi,
> The issue: while the code below works, it returns too many triangles for 
> a simple shape.
> So, for a square, instead of two triangles made by one of the diagonal 
> of the square, I get four tris, because the two diagonals are connecting.
> I get my vertices using:
> using K = CGAL::Exact_predicates_inexact_constructions_kernel;
> using Triangulation = CGAL::Triangulation_3 <K>;
> using Cell_handle = Triangulation::Cell_handle;
> using Vertex_handle = Triangulation::Vertex_handle;
> using Locate_type = Triangulation::Locate_type;
> using Point_CGAL = Triangulation::Point;
> ...
> Triangulation tri;
> ...
> // load data
> ...
> //loop
> for ( auto &e : tri.finite_facets() ){
>          auto &vert = *( e.first );
>          auto starter_id = e.second;
>          for ( u32 x = 1; x < 4; x += 1 ){
>                  auto &pt = *( vert.vertex( ( starter_id + x ) % 4 ) 
> );                 // saw that line on the stackoverflow -> 3 finite 
> vertices and 1 infinite vertice, hence the 4.
>                  // store my result
>          }
> }
> I need the triangulation to be the minimum number of tris for any given 
> shape. A square should have 2 tris.
> I read the doc, but it is not among the easiest to wrap one's head around.
> What should I modify, please?
> Thanks
> --
> MKL
>
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