CGAL users discussion list

[Manuel Caroli <>]

Hi Ariel,

sorry for the late answer, I somehow missed your question.

Unfortunately I don't see any better solution than the function you 
posted below. The problem is that your point p1 is not part of the 
triangulation, so there is no implicit way to find out how it has to be 
translated in order to be located close to the given cell.

best

Manuel


On 13/05/10 08:18, Ariel Keselman wrote:
> Dear all,
>
> I need to calculate the distance from a point to the
> vertices of the cell that contains it, in a periodic
> Delaunay triangulation. This is done to interpolate
> a function with the inverse distance method.
>
> The most obvious way to do this, and the one I
> implemented, is to take the periodic_point from the
> cell, and then calcule dx,dy,dz between the
> periodic_point and the point. In those directions
> where the difference is too large, (say,>box_len/2)
> the point is translated by box_len. Here is the
> function I use:
>
> inline const tvc
> periodic_r(const PDT&  T, const tch&  ch,
>      const unsigned int i, const tpt&  p1) {
>
>      const double blen2=T.domain().xmax()/2.0;
>      const tpt p2=T.point(ch,i);
>      const tvc rv(p1,p2);
>      bool needs_trans=false;
>      int ox=0,oy=0,oz=0;
>      if (rv.x()>=blen2) {
>          needs_trans=true;
>          ox=+1;
>      }
>      if (rv.y()>=blen2) {
>          needs_trans=true;
>          oy=+1;
>      }
>      if (rv.z()>=blen2) {
>          needs_trans=true;
>          oz=+1;
>      }
>      if (!needs_trans) return rv;
>      const tpp tpp1(p1,PDT::Offset(ox,oy,oz));
>      const tpt tp1=T.point(tpp1);
>      const tvc trv(tp1,p2);
>      return trv;
> }
>
> My question is, is there a way to do this more efficient?
> This function increases the time of  the main loop by
> 0.1 secs, which is about 5%. It just feels this should
> be a one-liner, but being new to CGAL, I can't see how
> to improve it...
>
> Thanks in advance,
> Ariel.