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combinatorial equivalence for Polyhedra / isomorphism for lattices

asked 2014-07-03 03:20:12 -0500

mf gravatar image

updated 2015-01-13 13:34:55 -0500

FrédéricC gravatar image

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 03:47:43 -0500

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 gravatar imagemf ( 2014-07-03 04:04:16 -0500 )edit

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Asked: 2014-07-03 03:20:12 -0500

Seen: 299 times

Last updated: Jul 03 '14