Le 25/01/2016 07:15, Pranav a écrit :

> I also would like to add my original problem to the description:

> /Given 2 linear cell complexes(say lcc1 and lcc2), determine if all

> 2-cells(facets) of lcc1 are present in lcc2. /

>

> For this problem, is it meaningful to convert lcc1 and lcc2 to corresponding

> Nef polyhedrons and perform boolean operations to extract facets which are

> present in lcc1 but not in lcc2? Actually, I suspect validity of this

> approach because from documentation of Linear cell complex and Nef

> polyhedron it seems that Linear cell complex are more general in terms of

> their capability to model domains than Nef polyhedron.

I am not sure converting the two lcc into nef polyhedrons is the best

way to do that.

It is easy to modify the are_facets_same_geometry function to test if

two faces have the same geometry and the same orientation (replace 0 by 1).

Then you can use the same principle than the one used in the function

sew3_same_facets to retreive the faces having the same geometry.

Guillaume

>

>

> --

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http://cgal-discuss.949826.n4.nabble.com/Linear-cell-complex-facet-membership-query-tp4661486p4661514.html> Sent from the cgal-discuss mailing list archive at Nabble.com.

>

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