Thursday, May 7, 2015
In practice, we cannot list every boundary in the same table. Granted, the number of regions grows no faster than 2^m, where m is the number of boundaries, but f(n,m), the number of regions in a space of n dimensions and m boundaries, grows no slower than m. Thus, it is practical to consider only subspaces of the universe. To claim subspaces are in the same universe, they must share boundaries, directly or indirectly, and shared subspaces must be equivalent. To crawl from one known subspace to another, there must be a way to find subspaces that share boundaries. To determine whether a polytope embedded in one subspace overlaps a polytope embedded in another subspace, we must be able to choose a superspace that is in the universe of possible subspaces. If B1, B2, ...Ba are boundaries of a new subspace of the universe, find S1, S2, ...Sb, all known subspaces with boundaries including at least one of B1, B2, ...Ba. Find subspaces T1, T2, ...Tb of S1, S2, ...Sb, restricted to B1, B2, ...Ba. Now find a superspace of T1, T2, ...Tb. First consider the case of several 1-dimensional subspaces of a 1-dimensional universe. 1-dimensional subspaces impose orderings on their boundaries. Finding a superspace of 1-dimensional subspaces involves choosing the first unchosen boundary from one of the subspaces. Note, we can only choose a boundary if it is the first unchosen from all subspaces it is in. Use recursion and the antisection theorem to construct higher dimensional superspaces.