t On Sat, Oct 19, 2019 at 1:59 PM 'Jonathan Kliem' via sage-devel <sage-devel@googlegroups.com> wrote: > > Let me try again. This is the current behaviour: > > sage: P = Polyhedron(vertices=[[-1,0],[1,0]],lines=[[0,1]]); P > A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 2 vertices > and 1 line > sage: P.vertex_graph() > Graph on 2 vertices > sage: P.faces(0) > () > > Let P be a polyhedron defined by vertices/rays/lines. The vertex graph of P > returns basically the vertex graph for Q, where Q is defined by the same > vertices and rays, but not lines. > > So > sage: P = Polyhedron(vertices=my_vertices, rays=my_rays, lines=my_lines) > sage: Q = Polyhedron(vertices=my_vertices, rays=my_rays) > > have vertex graphs that are canonically isomorphic.
V-representation is only unique for compact polyhedra, anyway, so one can talk about the graph of a V-representation, not about the graph of a polyhedron, in general. To me, a polyhedron with a linearity space should have no vertices, full stop. Compute a partial decomposition into the linearity space and something poined, then do something with it... > > I don't think this is the way it should be and the vertex graph in the first > case should be empty. If no one disagrees I will apply this change in #28626. > > However, the current behaviour is somewhat fixed by a doctest. > > As I understand it, the mentioned doctest tries to make a point about > > sage: P.combinatorial_automorphism_group(vertex_graph_only=True) > > In this case it is stated that the automorphism group can only interchange > vertices with vertices, rays with rays and lines with lines. Strictly > speaking, this is correct. But an automorphism group obtained from this will > never interchange rays with rays or lines with lines. > > Even worse, it will completely ignore the rays and lines, so that > sage: P = > Polyhedron(vertices=[[0,0,0],[0,1,1],[1,0,1],[-1,-1,1]],rays=[[0,0,1]]); P > A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 4 vertices > and 1 ray > sage: len(P.combinatorial_automorphism_group(vertex_graph_only=True)) > 24 > > This is why I consider the doctest strange. > > To me it does not make sense to use the vertex graph to obtain the > automorphism group. > > Btw (and really not the point of this post), I'm not sure that the > automorphism group of the vertex-facet graph is what one should consider for > the automorphism group of an unbounded polyhedron. At least in dimension > higher than 4 this might behave differently than at least I would define it. > I would instead consider the vertex-facet-graph with an extra marked vertex > at infinity. > > Am Samstag, 19. Oktober 2019 11:52:24 UTC+2 schrieb jplab: >> >> Hi Jonathan, >> >> Could you be more precise with what you mean with "projection"? >> >> What is the status quo? (Please provide a minimal reproducible working >> example that shows the contrast along with the proposed wished change that >> you have in mind, otherwise we are left to guess what you mean and that is >> leaving the door open to a lot of misunderstanding) >> >> What is the proposed changed? >> >> What is the assumption in the mentioned doctest? >> >> For computational aid, when no vertices exists (i.e. there is a lineality >> space) the current implementation creates an appropriate vertex. I guess we >> have to deal with this eventhough the object does not have a 0-dimensional >> face in reality. This is to have consistency between H- and >> V-representations. In order to have a consistent V-representation, you need >> an anchor point, otherwise just having rays and lines does not determine >> uniquely the object. >> >> That said, if you are interested in making a vertex-graph for unbounded >> polyhedra, then, edges should have two vertices. >> >> One way is to ask to keep only bounded edges in the definition, and the >> default behavior gives the vertex graph on bounded edges. If one wants the >> full graph with a point at infinity that compactifies the full vertex graph, >> then one could do so by adding an optional parameter "unbounded=False/True". >> Where the default would be False and return only compact edges and True >> would return one more vertex "the vertex at infinity". >> >> To me this seems to be the most reasonable behavior. I don't know if this >> answer your question. >> >> Le vendredi 18 octobre 2019 12:09:53 UTC+2, Jonathan Kliem a écrit : >>> >>> In #28626 I want to use `CombinatorialPolyhedron` to obtain the vertex >>> graph of polyhedra. >>> >>> However, I seem to have a different opinion of a vertex graph than is >>> currently implemented in sage. >>> >>> Should the vertex graph of an unpointed polyhedron return the vertex graph >>> of the projection or should it be empty? >>> >>> As of now the method `vertices` returns the defining vertices, which are >>> the proper vertices except for unpointed polyhedra. >>> >>> I don't think this ambiguous meaning of vertex should continue for the >>> vertex graph. >>> >>> However the current behavior is at the moment assumed in a strange doctest >>> of `combinatorial_automorphism_group`. >>> (E.g. the order of the combinatorial automorphism group does not change >>> when you add lines.) > > -- > You received this message because you are subscribed to the Google Groups > "sage-devel" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to sage-devel+unsubscr...@googlegroups.com. > To view this discussion on the web visit > https://groups.google.com/d/msgid/sage-devel/91149799-da9b-49a4-b9c0-bd9629a386e3%40googlegroups.com. -- You received this message because you are subscribed to the Google Groups "sage-devel" group. 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