CGAL users discussion list

[Adam Hopkins <>]

Hi,

thank you for your answers. When I was talking about triangulations, I was thinking on non-convex regions, but the triangulations obtained with CGAL always are convex regions. For this reason, all vertexes in the convex hull are the external vertexes.

On the other hand, I think that the solution of Sebastien is better because dont need to compute the convex hull. But,

If I have the infinite Cells of a triangulation, and I take the infinite vertexes of each of these cells, Are always the three adjacent vertexes of each infinite vertex, the most external ones?

Thank agian
A.

On Mon, Jan 30, 2012 at 4:30 PM, Sebastien Loriot (GeometryFactory) <> wrote:

Use adjacent_vertices with the infinite vertex.

http://www.cgal.org/Manual/latest/doc_html/cgal_manual/Triangulation_3_ref/Class_Triangulation_3.html#Function_OutputIterator_adjacent_vertices6Vertex_handle_v+_OutputIterator_vertices9_const;

Sebastien.

On 01/30/2012 04:18 PM, Daniel Duque wrote:

On Monday 30 January 2012 16:07:18 Adam Hopkins wrote:

Hi Daniel and Sebastien:

Again, thank you for your answers. When I say external vertexes, I am
refering to all vertexes in external faces, is to say: using the example
code of my first email, I can compute the external cells of a
triangulation, and each of them has an external face defined for three
vertexes, which are the external vertexes.

Hi,

That's right: unless I am very wrong, one vertex is the infinite one and the
other two belong to the convex hull...

Best,

Daniel

For example assume that I have a triangulation of a sphere, I want to know
which are the vertexes in the external surface. Please, see the code in my
first email for more references.

thank in advance and waiting suggestions
A.

On Mon, Jan 30, 2012 at 3:58 PM, Sebastien Loriot (GeometryFactory)<

>  wrote:

What is your definition of external vertices?

Sebastien.

On 01/30/2012 03:41 PM, Adam Hopkins wrote:

Hi Daniel:

Thank you for your help, but I think that it is not the correct way to
solve my problem. When the convex hull is computed not all the external
vertexes are included. For this reason, I need other form to know which
are the real external vertexes of a triangulation.

Regards and waiting more suggestions
A.

On Mon, Jan 30, 2012 at 3:29 PM, Daniel Duque<

<mailto:>>  wrote:
   On Monday 30 January 2012 15:20:52 Adam Hopkins wrote:
    >  Hi all,
    >
    >  Thank you for your previous answers, but I have a last issue

   related with

    >  triangulations.
    >
    >  Using the example code below, I can know which are the external

   cells of a

    >  triangulations, but I also need to know which are the external

   vertex of

    >  the same triangulation.

   I guess you mean the vertices that form the convex hull. Have a look
   here:

   http://www.cgal.org/Manual/**latest/doc_html/cgal_manual/**

Triangulation_2/Chapter_main.**html#Subsection_36.4.2<http://www.cgal.or
g/Manual/latest/doc_html/cgal_manual/Triangulation_2/Chapter_main.html#S
ubsection_36.4.2>

   Best,

   Daniel

    >  Is there any form to know which are the external
    >  facets of each of these cells in order to know which are the most

   external

    >  vertexes of this triangulation?
    >
    >  Concretely, How Can I know which are the external vertices of a
    >  triangulation?
    >
    >  std::vector<Cell_handle>  infinite_cells;
    >  T.incident_cells(T.infinite_**vertex(),

   std::back_inserter(infinite_**cells));

    >  for(int ii = 0; ii<  (int)infinite_cells.size(); ii++)
    >  {
    >  Cell_handle c = infinite_cells[ii];
    >  Cell_handle e = c->neighbor(c->index(T.**infinite_vertex()));
    >  }
    >
    >  Regards and thank you for all
    >  A.

   --
   http://dcain.etsin.upm.es/~**daniel<http://dcain.etsin.upm.es/~daniel
   >

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