# CUBE+PLANES = POLYHEDRONS I have an cube and an collention of planes ( normal and one point, Ax = b ). how can I get the polyhedrons formed by the intersection of the cube and the planes?

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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

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Thanks for the response: the plane intersects or cut to the cube generating two polyhedra, so how could get these polihedros to manipulate?

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?

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I don't understand your question. The intersection of two convex sets is always a single convex set (perhaps empty).

may not explain well . I have a polyhedron and I need to cut it with several planes and I want to count the number of generated polihedros and I also need to draw