Date: 07/04/2003 at 19:55:13 From: Paul Lynch Subject: Higher-Dimensional Cubes I've written a program that draws cubes in any number of dimensions between 0 and 15. It's really a simple one; it only shows perspective in our three natural dimensions. I've noticed, however, that the resulting cubes actually do have the properties that a 3-dimensional cube does compared to a square. I've been expanding on this program to have it count the number of edges each cube has, which is as simple as counting the number of lines the computer draws for each cube. What I want to do now though is have it display how many faces and cells of each dimension each cube has; right now it only shows how many cubes of each dimension it *draws*, not how many actually exist. I've noticed, for example, that a hypercube actually has eight cubic cells - I think. There's almost no way of telling how many 2-dimensional faces it has. Finally, I would like to know if there is any way I can make my program accurately display how many cells of each dimension actually exist in each higher-dimensional cube. possibly in a progressive way so it can count them (or seem as though it is). Are there any formulas I can use?
Date: 07/05/2003 at 10:32:12 From: Doctor Jacques Subject: Re: Higher-Dimensional Cubes Hi Paul, If you have an n-dimensional cube ("n-cube"), you can put it into a coordinate system: * Pick one vertex as the origin. * Select axes in each of the n dimensions, along the edges adjacent to the vertex in question. * Choose the unit of length equal to the length of the side of the cube. In this way, each vertex of the n-cube is represented by a set of n coordinates, and each of those coordinates is equal to either 0 or 1. For example, for a 2-cube (a square), we have: (0,1) (1,1) +---------+ | | | | | | | | | | +---------+ (0,0) (1,0) This shows that the number of vertices is equal to the number of ordered sets of n numbers, each equal to 0 or 1. As you have two possibilities for each coordinate, and there are n coordinates, this gives a total of 2^n vertices. Note that we have to extend the definition for n = 0 and 1: * A 0-cube is a single vertex * A 1-cube is a line segment Let us now consider the number of k-cells in an n-cube. A k-cell is a k-cube, so it will consist of 2^k vertices. A k-cell is defined by: * A set of k coordinates, that define the "orientation" of the cell. * A set of values for the remaining coordinates, that define the position of the cell compared to other "parallel" cells. For example, consider the 1-cells (edges) in a 3-cube. As an edge is one-dimensional, you define its orientation by selecting one of the three dimensions, for example, x (if the three directions are x,y,z). The edge will contain two vertices, corresponding to x = 0 and x = 1. To select a particular edge in the x direction, you need to specify y and z; each of these can be either 0 and 1, for a total of 4 possibilities - you have 4 edges in the x direction. As there are three possible directions, you find a total of 12 edges. Let now consider the 2-cells (faces) in a 3-cube. A 2-cell has 2 dimensions, so you must select two of the axes to define its orientation - let us choose x and y. The face will contain vertices where x and y each takes the values 0 and 1, for a total of 4 vertices. For a given orientation, the position of the face is given by the z coordinate, which can be 0 or 1: there are two faces in any given orientation. You can select the two axes of the cell in 3 ways, giving a total of 6 faces. In general, to define a k-cell in an n-cube: * You select an orientation, by choosing k coordinates from the n available - you can do this in C(n,k) ways. * You assign values to the remaining (n-k) coordinates - you can do this in 2^(n-k) ways. So the total number of k-cells is given by: T(k,n) = C(n,k)*2^(n-k) Notice that this is the term in 2^(n-k) in the binomial expansion of: (1 + 2)^n = 3^n and this gives you a nice identity: the total number of cells (of all dimensions from 0 to n) in an n-cube is 3^n. For example, with n = 3 (a cube): * There are 8 0-cells (vertices) * There are 12 1-cells (edges) * There are 6 2-cells (faces) * There is 1 3-cell (the cube itself) for a total of 3^3 = 27. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum