combinatorial equivalence for Polyhedra / isomorphism for lattices

asked 2014-07-03 10:20:12 +0200

mf
161 ●5 ●11 ●20

updated 2019-04-05 20:35:06 +0200

5125 ●3 ●43 ●111

Is there an easy way to check whether two polyhedra are combinatorial equivalent, i.e. have have isomorphic face lattices.

this does not work:

Poly1=Polyhedron(vertices=[[0,1],[1,0],[1,1]], base_ring=QQ)
Poly2=Polyhedron(vertices=[[2,0],[2,2],[0,2]], base_ring=QQ)
Poly1.face_lattice()==Poly2.face_lattice()

In this case this would work:

str(Poly1.faces(1))==str(Poly2.faces(1))

but what would be a good way to check this in general?

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answered 2014-07-03 10:47:43 +0200

vdelecroix
7397 ●19 ●81 ●164 http://www.labri.fr/pe...

Hi,

sage:  Poly1.face_lattice().is_isomorphic(Poly2.face_lattice())
True

You can use X.is_isomorphic(Y) on graphs as well.

Vincent

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perfect, thanks! I don't know how i missed this..

mf ( 2014-07-03 11:04:16 +0200 )edit

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answered 2017-10-09 15:59:14 +0200

jipilab
533 ●17 ●18

Hi,

Since version 7.6 of Sagemath, there is exactly the required method for Polyhedron objects:

sage: Poly1=Polyhedron(vertices=[[0,1],[1,0],[1,1]], base_ring=QQ)
sage: Poly2=Polyhedron(vertices=[[2,0],[2,2],[0,2]], base_ring=QQ)
sage: Poly1.is_combinatorially_isomorphic(Poly2)
True

It does not requires the computation of the full face lattice (checks vertex/facet adjacency graph isomorphism), hence should be faster than asking for isomorphic face lattices. The algorithm option can still be set to face_lattice to get another certificate option.

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