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## Hypercubes

A hypercube in n dimensions, or an n cube, is the n dimensional analog of a cube.  It has two hyperfaces on each axis; the hyperfaces are n-1 dimensional hypercubes.

In general, we call the volume enclosed by a hypercube an n-volume.  Ordinary "volume" (measured in things like quarts and liters) is 3-volume.  Area (measured in things like acres and square meters) is 2-volume.  The magnitude of the volume of an n-cube with edge length r is rn.  We can form such an n-cube by "sweeping" an (n-1)-cube with edge length r in a direction perpendicular to itself, through a distance of r units.  The rest of this page just expands on, explains, and attempts to justify those statements.

Some examples of "low-dimension" n-cubes:
• A 0-cube is a point, and has no defined volume.
• A 1-cube is a line segment; its 1-volume is  its length.
• A 2-cube is a square; its 2-volume is its area.
• A 3-cube is an ordinary cube.
• A 4-cube is a tesseract, and is occasionally just called a "hypercube".
 Figure 1:  Sweeping out a cube
To construct an n cube, we start with an n-1 cube, and sweep it through space perpendicular to the hyperplane in which it lies.   In figure 1 we show this operation for the first three dimensions.  We start with a point, at the origin, which sweeps out a line segment; the line segment then moves perpendicularly to its length, and sweeps out a square; the square then moves perpendicularly to its surface, and sweeps out a cube.  We have a hard time drawing anything higher than 3 dimensional figures so we stopped there.

By paying attention to what happens as we do this, we can learn what we need to know about how an n cube looks. For clarity, we'll usually call 2-dimensional faces "2-faces", and n-1 dimensional faces, "hyperfaces".  We want to know how many vertices, edges, and hyperfaces an n-cube has; each of those is described by a "recurrence relation", based on the number of vertices, edges, and hyperfaces an (n-1)-cube has, and by paying attention to the details as we build a cube we can see what those relations must be.

• We started with a point (0 dimensions).  A point has 0 edges, 0 "2-faces", and 1 vertex.  It also has zero hyperfaces.  It has no volume.

•  Figure 2: "1-cube"
We sweep the point through space r units to form a line segment, or 1-cube, with length r.  It has 1 edge, zero "2-faces", and 2 vertices:
• The single vertex of the point was replaced with two new vertices when it swept along:  one (old) vertex at its starting position, and one (new) vertex where it ended.  We observe that the single vertex was doubled as a result.  In general, we will see that an n-dimensional hypercube will have 2n vertices.
• In general, just as the single point formed two images of itself -- one where it started sweeping, one where it finished sweeping -- we would expect any hypercube to form two images of itself if it were swept through space perpendicular to the plane in which it lies.  Thus, each edge will form two "image edges", each face will form two "image faces", and so forth.
• The edge was created as the vertex swept along.  It seems clear that every vertex will form an edge when it's swept through space.
• The 1-volume of the line segment is r, the distance which was "swept out" by the point.
• The line segment has two "hyperfaces", which are its vertices; they lie on the axis along which the point swept.

•  Figure 3: "2-cube"
We sweep the line segment through space to form a square, or 2-cube.  The square has four vertices, four edges, and four "hyperfaces" (the edges).
• The two vertices on the line segment each "doubled", giving us four vertices.  The number of vertices is, again, 2n.
• The line segment "doubles", leaving an edge where it started, and forming another where it ended.
• Each of the two vertices of the original line segment has swept out a new edge.  Thus, we have four edges altogether:  two for each edge in the 1-cube, plus one for each vertex in the 1-cube.
• The 2-volume of the square is the distance swept times the "1-volume" of the 1-cube, or r2.
• The square has four "hyperfaces", which are its edges.  The line segment moved along the y axis, and left two images of itself along that axis. The two vertices of the segment each swept out new edges, and they lie on the x axis.
We see that in general, (n-1)-cube forms two images of itself, which become two hyperfaces of the n-cube.  In other words, the n-cube is the volume swept out by one of its hyperfaces.
In addition, each hyperface of the (n-1)-cube sweeps out a new (n-1)-cube, which in turns becomes a face of the n-cube.
So, we acquired one hyperface for each hyperface of the (n-1)-cube, plus two new hyperfaces:

Since we're "sweeping" along each axis in turn, this means we have one pair of hyperfaces along each axis; the total number of hyperfaces is

•  Figure 4: "3-cube"
We sweep the square through space to form a cube, or 3-cube.  The 3-cube has eight vertices, 10 edges, and six "hyperfaces" (in the case of a 3-cube, they're the ordinary 2-faces).
• The four vertices on the square each "doubled", giving us eight vertices.  Again, we have 2n vertices.
• Again, each edge of the square "doubled", giving us two new edges in place of each old one.
• In addition, each vertex on the square swept out a new edge as it moved.  Thus, the number of edges on the 3-cube is equal to the number of vertices on the 2-cube plus twice the number of edges on the 2-cube:

• The 3-volume of the 3-cube is the area of the square, times the distance the square moved, or r3.
• The square formed two faces for our cube: one where it started, and one where it ended.  In addition, each edge (hyperface) of the square formed a new square.  So the number of hyperfaces on the 3-cube is 6, which is, as we expected, 2n.

### Reflective Symmetry Along the Axes

There is an additional item here worth emphasizing, which is that it does not make any difference what axis we start with.  We can sweep a point along any axis, then sweep the resulting line segment along any of the remaining axes, and so forth.

In other words, the hypercube which results is symmetric in swaps of the axes.

Furthermore -- and related -- it is symmetric in reflections through a hyperplane which passes through the center of the cube, and which is perpendicular to one of the axes.  Less obscurely, the cube is r units on an edge and we formed it by sweeping along each axis in turn, then if we reflect the cube across the plane xi = r/2, for any coordinate xi, it will be unchanged.

To see this, consider first the last axis, along which we swept an (n-1)-cube to form the n-cube.  We have two images of the (n-1)-cube "lying on" that axis:  One at 0, and one at r.  But think about what happens on the other axes during the final sweep:  Each of the faces of the (n-1)-cube, which is, itself, an (n-2)-cube, will sweep out a new (n-1)-cube, which will become a face of the newly formed n-cube.   But because of the way we performed the earlier sweeps, there will be a pair of these along each axis, and we'll end up with a pair of faces along each axis.

This is, of course, directly related to the fact that there are 2n faces on an n-cube.

This argument could be formalized into an induction proof fairly easily, but we won't be depending on it in any formal proofs so I won't go through that here.

### Coordinates of the Vertices

Let's consider the process of "sweeping" which we used to construct the hypercubes.

We start with a single point in a 0 dimensional space, which has no coordinates.

We then introduce a single coordinate (the x coordinate), in which our point is at the origin; when we "sweep" it along the newly introduced axis, we create a second coordinate r units from the origin.  So, we have the vertices:

(0), (r)

Next we introduce a second coordinate (the y coordinate), and place each existing vertex at 0 on the y axis:

(0, 0), (r, 0)

That's still a line segment lying on the x axis.  When we sweep this segment r units along the y axis, we introduce additional vertices which have y values of r.  We now have four vertices:

(0,0), (0, r), (r,0), (rr)

And so it goes.  With each "sweep" operation, we add a new "0" coordinate to each existing vertex and we add new vertices which have the same old coordinate values as the old vertices, but have "r" for the new coordinate value instead of 0.  For the cube, we thus obtain the set:

(0,0,0), (0,r,0), (r,0,0),  (r,r,0),
(0,0,r), (0,r,r), (r,0,r),  (r,r,r)

If we do this again, the vertices of the 4 cube will be:

(0,0,0,0), (0,r,0,0), (r,0,0,0),  (r,r,0,0),
(0,0,r,0), (0,r,r,0), (r,0,r,0),  (r,r,r,0),
(0,0,0,r), (0,r,0,r), (r,0,0,r),  (r,r,0,r),
(0,0,r,r), (0,r,r,r), (r,0,r,r),  (r,r,r,r)

In general, we see (and we can prove by induction) that every possible combination of "r" and "0" is a vertex.

There are two possible values for each coordinate, and n coordinates, so the number of vertices must be 2n, which matches the value we already determined, above.

### The Set Enclosed Within an n-Cube

Each time we "sweep" along an axis as we construct the cube, we start at 0 and move r units.  Thus, the length spanned by the cube on each axis is r units, and it includes all points in that region.

Since we sweep along each axis, all the points along each axis which fall between 0 and r must be included in the cube.

So, the set of points inside the cube must be all those points for which all coordinates are less than or equal to r. (This, by the way, shows that the faces must be where one or more coordinates take on the values r, and from that, we can in turn deduce that the coordinates of the vertices must all be 0 or r, as we already pointed out in the last section.)

Formally, we can say:

Or, if we wish to include the "skin", we could say the volume occupied by the cube is the set:

Page first posted on 11/04/2007