Thanks for the reponse: in the above solution, the plane intersects or cut to the cube generating two polyhedra, so how could get these polihedros to manipulate?
I don't understand your question. The intersection of two convex sets is always a single convex set (perhaps empty).
Define the cube as a polyhedron (its a canned example):
sage: cube = polytopes.n_cube(3)
sage: cube.Hrepresentation()
(An inequality (0, 0, -1) x + 1 >= 0, An inequality (0, -1, 0) x + 1 >= 0, An inequality (-1, 0, 0) x + 1 >= 0, An inequality (1, 0, 0) x + 1 >= 0, An inequality (0, 0, 1) x + 1 >= 0, An inequality (0, 1, 0) x + 1 >= 0)
Define the plane as a polyhedron:
sage: plane = Polyhedron(eqns=[(0,1,0,0)])
sage: plane.Hrepresentation()
(An equation (1, 0, 0) x + 0 == 0,)
Compute the intersection:
sage: cube.intersection(plane)
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 4 vertices
Asked: 2012-07-31 16:21:52 +0100
Seen: 533 times
Last updated: Jul 31 '12
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.