On 23/07/2013 09:06 PM, Nekar Xenos wrote:
> If we take a vertical 2D slice of a tin can we get a rectangle.
> If we take a horizontal 2D slice of the same tin, we get a rectangle.

Try this:

A line is a 1D object. A line has two ends. Ends are points, which are 
0D objects.

A square is a 2D object. A square has four edges. Edges are 1D objects. 
By extruding a line into a square, you're turning one line into two, and 
you're turning the two end-points into additional lines as well.

A cube is a 3D object. A cube has six surfaces. Surfaces are 2D objects. 
Again, when you extrude a square into a cube, that one square becomes a 
pair of squares, but in addition all four edges become new squares. So 
you have 2 + 4 = 6 new surfaces.

Now, come with me on this journey. A 4-cube is a 4D object. If it allows 
the pattern above, that means it should have 3D *volumes* as its "faces" 
(weird as that sounds). Presumably each one is cube-shaped. It also 
means that by extruding a cube into a 4-cube, the original cube becomes 
two cubes, and each surface on the original cube becomes an additional, 
new cube. That gives us 2 + 6 = 8 cuboid volumes.

A simpler way to work this out: A cube is given by
   -1 <= x <= +1
   -1 <= y <= +1
   -1 <= z <= +1
This defines a 3D region of space. By setting (say) x = -1, we now have 
only two degrees of freedom left (y and z). If we choose to fix one 
variable, there are three variables to choose, and two values to fix it 
to (-1 or +1), giving a total of six possibilities - which is why a cube 
has six sides.

We can extend this idea; what if we fix *two* variables? Then only one 
degree of freedom remains. Taking this approach, we see that we can pick 
two variables out of three - which is equivalent to NOT picking one out 
of the three. We can then set each variable to one of two values, for a 
total of four combinations. 4 * 3 = 12, and hence a cube has 12 
boundaries of the 1D type - i.e., 12 edges.

Fixing all three variables, we have 2^3 = 8 possible combinations, and 
sure enough a cube has eight 1D corners.

Taking all this, we see that a hypercube ought to have

   (4 choose 1) * 2^1 = 4 *  2 =  8 volumes
   (4 choose 2) * 2^2 = 6 *  4 = 24 surfaces
   (4 choose 3) * 2^3 = 4 *  8 = 32 edges
   (4 choose 4) * 2^4 = 1 * 16 = 16 corners

Another way of saying all this is that each time you extrude an N-cube 
into an (N+1)-cube, the number of K-dimensional things is doubled, and 
the number of (K-1)-dimensional things is then added to that.

          |   Points |         Lines |      Squares |       Cubes |
   -------+----------+---------------+--------------+-------------+
   Point  |        1 |               |              |             |
   Line   | 1*2 =  2 |             1 |              |             |
   Square | 2*2 =  4 |  1*2 + 2 =  4 |            1 |             |
   Cube   | 4*2 =  8 |  4*2 + 4 = 12 | 1*2 + 4 =  6 |           1 |
   4cube  | 8*2 = 16 | 12*2 + 8 = 32 | 8*2 + 6 = 24 | 2*1 + 6 = 8 |

Reassuringly, this agrees with the previous calculation.

Post a reply to this message