Re: [Libmesh-users] non-active elements

[Michael P. <pov...@in...>]

if I want to refine, than I can put REFINE flag, of course.
The thing is that I usually don't want to refine anything, but I have to 
remove some elements from the simulation.
I understand that I "abuse" libmesh features, but I didn't find a better 
solution.
Michael.


John Peterson wrote:
> What about setting the REFINE flag for elements which you'd like
> to refine, and then calling the mesh refinement function?  As Roy
> pointed out, INACTIVE has a special meaning for the library...
>
> Your patch seems OK to me, but I wonder if it should really be
>
> assert (this->has_children());
>
> to enforce the idea that "inactive" elements have children.
> Your code would fail this assert of course, but would it work
> if you set the REFINE flag instead?
>
> -John
>
>
> Michael Povolotskyi writes:
>  > Dear developers,
>  > I'm currently working on a problem, where I have to solve two different 
>  > problems on different regions of the same mesh.
>  > In order to do so, I programmed a code in which I make elements to be 
>  > active or not, depending  on which region they belong to.
>  > I set a flag INACTIVE or DO_NOTHING to the elements and call the 
>  > refinement function.
>  > 
>  > 
>  > Then I found a problem. As far as I can see, the element activity status 
>  > is introduced in order to control the refinement only.
>  > So, it is assumed that if an element is not active, it must have children.
>  > Due to this, the function
>  > void Elem::active_family_tree_by_neighbor (std::vector<const Elem*>& 
>  > family,
>  >                                            const Elem* neighbor,
>  >                                            const bool reset) const
>  > 
>  > crashed.
>  > So, I modified it a bit by adding an if(this->has_children()) statement:
>  > 
>  > void Elem::active_family_tree_by_neighbor (std::vector<const Elem*>& 
>  > family,
>  >                                            const Elem* neighbor,
>  >                                            const bool reset) const
>  > {
>  >   // Clear the vector if the flag reset tells us to.
>  >   if (reset)
>  >     family.clear();
>  > 
>  >   // This only makes sense if we're already a neighbor
>  >   assert (this->is_neighbor(neighbor));
>  > 
>  >   // Add an active element to the family tree.
>  >   if (this->active())
>  >     family.push_back(this);
>  > 
>  >   // Or recurse into an ancestor element's children.
>  >   // Do not clear the vector any more.
>  >   else
>  >   {
>  >     if (this->has_children())  //added by me
>  >       for (unsigned int c=0; c<this->n_children(); c++)
>  >         if (this->child(c)->is_neighbor(neighbor))
>  >           this->child(c)->active_family_tree_by_neighbor (family, 
>  > neighbor, false);
>  >   }
>  > }
>  > 
>  > Now it works.
>  > What do you think about it? May I go this way?
>  > Thank you,
>  > Michael.
>
>   


-- 
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Michael Povolotskyi, Ph.D.
University of Rome "Tor Vergata"
Department of Electronic Engineering
Viale Politecnico, 1 - 00133 Rome - Italy 

Phone + 39 06 72597781
Fax   + 39 06 2020519
http://www.optolab.uniroma2.it/pages/moshe/moshe.html
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