Thread: Ployhedron and Rectangle View Single Post

## Ployhedron and Rectangle

 All polyhedra/polytopes that you make by defining constraints of the form Ax <= b are convex. That's not the only way to define a polyhedron.. I think all you need for a polyhedron is for it to have faces that are polygons. You can glue together all sorts of shapes with polygonal faces cut out of cardboard.
Ok, I see. Now it makes sense.

 It just my preliminary guess at a systematic way of representing the vertices. Its a form of mental filing and sorting. Is it clear that all vertices can be represented that way? For example, in 2D, L = (1,2) , U = (5,7) , U-L = (4,5) vertex (1,2) = (1+ (0)4, 2 + (0)5) vertex (1,7) = (1 +(0)4, 2 + (1)5) vertex (5,2) = (1 + (1)4, 2 + (0)5) vetex (5,7) = (1 + (1)4, 2 + (1) 5)
Ok, I agree on that formulating.

 I say that a polytope is formed by "intersecting" hyperplanes (if we mean only the boundary of the volume, not the interior). The edges are intersections of hyperplanes but we should say more since the intersection of of two hyperplanes is usually a line, not a line segment. An edge is a line segment. We need some way to describe that.
I have a book in geometry defines a convex polyhedron as following (which means that there are non convex polyhedra as you said):

 A convex polyhedron is a system of finitely many convex polygons, with the property that each side of every polygon is also aside of a second polygon and for every polygon in the system the vertices of the polyhedron that do not lie on the given polygon are all in the same halfspace determined by the plane of the polygon
(I do not understand the underlined part). I think, we can say that each hyperplane is cut by two other hyperplanes to form a polytope. This process makes a polytope intersected polygons, and a polygon is bounded, and its edges are line segments. Am I right on this?