CGAL users discussion list

[Everton Constantino <>]

Hi Efi,
thanks for the tips, the overlay gives the opposite of what I want,
sorry I should have made myself clearer I meant union and intersection
as set theoretic words. Anyway you sorta gave me the path, I'm iterating
edges, getting the associated curve and using Kernel::Intersect_2. Now
everything is working as intended.

Best regards,
E.

On 23:50, Efi Fogel wrote:
> Hi Everton,
> 
> 1. The equivalent of the '==' operator is the 'Equal_2' functor; see
> http://doc.cgal.org/latest/Arrangement_on_surface_2/classArrTraits_1_1Equal__2.html
> 2. The overlay gives you the overlay. What do you mean by 'union of
> arrangements' or 'intersection of arrangements'?
> 
> Best,
> Efi
> 
>    ____  _        ____             _
>   /_____/_) o    /__________  __  //
>  (____ (   (    (    (_/ (_/-(-'_(/
>                          _/
> 
> 
> 
> On Tue, May 17, 2016 at 9:40 PM, Everton Constantino <
> >
>  wrote:
> 
> > Hi Efi,
> > reading CGAL's documentation I failed to find the definition of the ==
> > operator for edges so I ended up using brute force. The overlay free
> > function gives me the union of the arrangements (I dunno if there's an
> > option for the intersection).
> >
> > Thanks,
> > E.
> >
> > On 21:14, Efi Fogel wrote:
> > > I'm not sure why you compare points but say that you intend to intersect
> > > edges, but it seems to me that you should use the overlay() free
> > function;
> > > see, e.g.,
> > >
> > http://doc.cgal.org/latest/Arrangement_on_surface_2/group__PkgArrangement2Funcs.html#gaddb9e44b14b27e4cf0c7bb26b27d8518
> > >
> > > Also, notice that the traits contains a predicate that compares two
> > points.
> > > You should use it instead of the '==' operator.
> > >
> > >
> > >    ____  _        ____             _
> > >   /_____/_) o    /__________  __  //
> > >  (____ (   (    (    (_/ (_/-(-'_(/
> > >                          _/
> > >
> > >
> > >
> > > On Tue, May 17, 2016 at 8:45 PM, Everton Constantino <
> > > >
> > >  wrote:
> > >
> > > > So, changing the algorithm as you mentioned I realised something. One
> > > > edge on a Arrangement might represent the same line as several edges 
> > > > on
> > > > the other or, even worse, one edge of an Arrangement might contain
> > > > several edges of the other and even more so my way of checking the
> > > > intersection fails.
> > > >
> > > > What I think I should be doing is to check which segments of one
> > > > arrangement are contained in the other not edges in particular. I know
> > a
> > > > naive, brute force, way to solve this. Do I have something in CGAL 
> > > > that
> > > > could help me?
> > > >
> > > > E.
> > > > On 18:14, Sebastien Loriot (GeometryFactory) wrote:
> > > > > One way to speed things up is to put arrangement edges in a 
> > > > > std::set<
> > > > > std::pair<vertex_handle,vertex_handle> > with a custom less functor
> > > > > comparing the points of the vertices (if you don't care about
> > > > > orientation then also sort the pair before insertion).
> > > > > You can then use the set for queries.
> > > > >
> > > > > About the missing edges, it would be interesting to see those
> > missing.
> > > > > Did you try another visibility algorithm provided in CGAL, just so
> > see
> > > > > if the results are the same?
> > > > >
> > > > > Sebastien.
> > > > >
> > > > >
> > > > > On 05/17/2016 05:50 PM, Everton Constantino wrote:
> > > > > > The source is on github https://github.com/everton1984/Malex the
> > file
> > > > is
> > > > > > src/get_chunks.cpp. Thanks for the == tip, I already started using
> > it
> > > > > > btw.
> > > > > >
> > > > > > The data, unfortunately, is private and I can't share it.
> > > > > >
> > > > > > E.
> > > > > >
> > > > > > On 17:17, Sebastien Loriot (GeometryFactory) wrote:
> > > > > > > We're looking for a pb here. If we can't see a full example
> > > > > > > (in particular the typedefs) and try compiling it to reproduce
> > the
> > > > > > > issue it is harder to help because we can only guess.
> > > > > > >
> > > > > > > I know it is some extra work but it is needed if you want to
> > maximize
> > > > > > > the chances to get an answer.
> > > > > > >
> > > > > > > Sebastien.
> > > > > > >
> > > > > > > PS: Note that operator==()(Point_2, Point_2) is implemented in
> > CGAL.
> > > > > > >
> > > > > > > On 05/17/2016 04:59 PM, Everton Constantino wrote:
> > > > > > > > Oh ok, here is the part where I do the intersection:
> > > > > > > >
> > > > > > > > Arrangement_2 res;
> > > > > > > > vector<Kernel::Segment_2> bounding_roads;
> > > > > > > > for( Edge_const_iterator eit = output_arr.edges_begin(); eit 
> > > > > > > > !=
> > > > output_arr.edges_end(); ++eit){
> > > > > > > >       bool found = false;
> > > > > > > >       for( Edge_const_iterator envit = env.edges_begin();
> > envit !=
> > > > env.edges_end() && !found; ++envit ){
> > > > > > > >           if( envit->source()->point().x() ==
> > > > eit->source()->point().x() &&
> > > > > > > >                   envit->source()->point().y() ==
> > > > eit->source()->point().y() &&
> > > > > > > >                   envit->target()->point().x() ==
> > > > eit->target()->point().x() &&
> > > > > > > >                   envit->target()->point().y() ==
> > > > eit->target()->point().y() ){
> > > > > > > >               found = true;
> > > > > > > >           }else if(envit->source()->point().x() ==
> > > > eit->target()->point().x() &&
> > > > > > > >                   envit->source()->point().y() ==
> > > > eit->target()->point().y() &&
> > > > > > > >                   envit->target()->point().x() ==
> > > > eit->source()->point().x() &&
> > > > > > > >                   envit->target()->point().y() ==
> > > > eit->source()->point().y() ){
> > > > > > > >               found = true;
> > > > > > > >           }
> > > > > > > >       }
> > > > > > > >       if( found ) {
> > > > > > > >           Kernel::Segment_2 seg( eit->source()->point(),
> > > > eit->target()->point() );
> > > > > > > >           bounding_roads.push_back(seg);
> > > > > > > >       }
> > > > > > > > }
> > > > > > > > CGAL::insert(res, bounding_roads.begin(),
> > bounding_roads.end());
> > > > > > > >
> > > > > > > >
> > > > > > > > On 16:26, Sebastien Loriot (GeometryFactory) wrote:
> > > > > > > > > Sorry I meant a minimal code example to see where the pb
> > could
> > > > be.
> > > > > > > > > The problem was clear.
> > > > > > > > >
> > > > > > > > > Sebastien.
> > > > > > > > >
> > > > > > > > > On 05/17/2016 04:22 PM, Everton Constantino wrote:
> > > > > > > > > > Hi Sebastien,
> > > > > > > > > > you can find a file at http://braindump.com.br/arr.pdf
> > with
> > > > sets of 3
> > > > > > > > > > images, one is a map of roads, the 'raw' shows the 
> > > > > > > > > > computed
> > > > visibility
> > > > > > > > > > Arrangement (for the point on the image, there's also a 
> > > > > > > > > > not
> > > > shown square
> > > > > > > > > > drawn around the point) and the other shows the
> > intersection
> > > > of edges
> > > > > > > > > > form by the roads and the visibility region.
> > > > > > > > > >
> > > > > > > > > > What I want to do is to calculate the roads that belong to
> > the
> > > > > > > > > > visibility of a certain point. You will notice that some
> > roads
> > > > that are
> > > > > > > > > > present both on the 'raw' image and on the full map are 
> > > > > > > > > > not
> > > > present on
> > > > > > > > > > the intersection.
> > > > > > > > > >
> > > > > > > > > > The way I'm calculating right now is iterating through
> > edges
> > > > of both
> > > > > > > > > > arrangements and comparing source() and target() points
> > (since
> > > > they are
> > > > > > > > > > directed segments I compare both directions because I 
> > > > > > > > > > don't
> > > > care).
> > > > > > > > > >
> > > > > > > > > > Hope that clarifies something.
> > > > > > > > > >
> > > > > > > > > > Thanks,
> > > > > > > > > > E.
> > > > > > > > > >
> > > > > > > > > > On 16:02, Sebastien Loriot (GeometryFactory) wrote:
> > > > > > > > > > > Please share a minimal example showing what is not
> > working
> > > > correctly.
> > > > > > > > > > >
> > > > > > > > > > > Thanks,
> > > > > > > > > > >
> > > > > > > > > > > Sebastien.
> > > > > > > > > > >
> > > > > > > > > > > On 05/17/2016 03:52 PM, Everton Constantino wrote:
> > > > > > > > > > > > Hi,
> > > > > > > > > > > > I'm trying to intersect two Arrangement_2 edge's but
> > > > iterating through
> > > > > > > > > > > > both sets of edges and checking the source(), target()
> > is
> > > > not only slow
> > > > > > > > > > > > but wrong (some edges are missing). Is there a
> > canonical
> > > > way to do this?
> > > > > > > > > > > >
> > > > > > > > > > > > Thanks,
> > > > > > > > > > > >
> > > > > > > > > > >
> > > > > > > > > > >
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> > > > > > > > >
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> > > >
> > > > --
> > > > Everton R. Constantino 
> > > > <>
> > > > "Success is overrated"
> > > >
> > >
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> >
> > --
> > Everton R. Constantino 
> > <>
> > "Success is overrated"
> >
> 
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-- 
Everton R. Constantino 
<>
"Success is overrated"

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