Interesting Hypercube Problem?

[Note - I have removed all previous posts and am archiving at a blog page, to which i hope to add some diagrams later to make the whole thing more palatable. Everything relevant to this discussion from previous and current posts is pointed at by the link in the 'source' section: http://www.thoughts.com/index.php?_action=blog_vie...

ADDED 1:14 am GMT 24/12

Bad news on the cube with side length sqrt(5)/2 front - I have determined it is impossible to draw a cube in the hypercube with a side length of sqrt(5)/2 in the manner that I describe. You just can't find three mutually perpendicular lines from any given point to form the corner of the cube.

It also occurred to me that it might be trivial to try to fit a 4-cube into a 5-cube; analogising down to a lower case, the biggest square that can fit into a cube with edges that touch the boundaries of the cube is just the same as one of the face of the cube. And anyway, in the 3D case, we resorted to a hexagon, not a square.

This, coupled with the fact that fitting a tetrahedron inside didn't seem to work either (though I'm having conceptual problems with the pentachoron at the moment), means that I have changed my tune and think a not-entirely regular - but still uniform - convex polychoron of 10 cells would be the best idea. Given that, I'm trying a couple of others (truncated and rectified 5-cells) which have 10 faces each.

@Torquestomp - very good thinking on the orthonormal basis vectors thing; If your assumption could be proven,and especially if you could find some way of getting that set of orthonormal vectors to establish the side length of the inscribed polytope, that could be a powerful result. However - note that an octahedron only has 6 vertices, so it can't be made up of those 8 points you say earlier. Unless that wasn't what you meant?

ADDED 14:22 pm GMT 23/12:

Okay, I was up till 6am today and I think I've solved the n = 5 case, and through that discovered a direction for induction to go in. I'll explain the n = 5 case first.

Continuing on from the same analogy as before (see below), what we have here is a bounded 5-cube intersected by a 4-region. A 5-cube has 10 4-faces (which I'll call 'facets' now as the higher dimensional analogy to 'faces' (2D) and 'cells' (3D)) which are all 4-cubes. So bounding our inscribed polychoron we expect a maximum of 10 3D cells, one at most inscribed inside each 4-facet of the 5-cube. Now there is no regular polychoron which has 10 facets; the biggest one we can find has 8 facets, and is the 4-cube. This means the biggest regular polychoron we can inscribe in the 5-cube is the 4-cube, which is bounded by 3-cubes. So the question becomes, how do we draw a 'maximal' 3-cube inside a unit 4-cube?

[By 'maximal' here we mean 'of greatest volume - just as we attempted to find the biggest area triangle that would fit in the cube for the n = 4 case.]

The way I found was as follows. I began by drawing a hypercube in stereographic projection (the usual one you see, with the small cube embedded inside the larger cube.) I am going to put axes on this for convenience, so imagine the unit 4-cube is centred at the origin. Imagine the system to be right-handed so that the x-axis is pointing at the 'bottom left' of the page, the y-axis at the bottom right and the z-axis at the top. (Obviously, each axis is perpendicular to 6 faces and 4 cells of the hypercube.) By construction, the x, y and z co-ordinates will run between [-1/2, 1/2]^3, so we want the w co-ordinate to do the same. So assign the 'outer cube' to w = 1/2, and the 'inner cube' to w = -1/2. I am now going to use 8 points drawn on the vertices and edge of this hypercube draw a cube inside this hypercube that seems to have the maximum 3-volume.

Now I am going to describe 8 points, in co-ordinate form, and join them up into a cube. i strongly urge you to try and copy my diagram if you can see how I described it, because it really is enlightening. Basically, getting from any vertex of this cube to any other adjacent vertex involves travelling from a corner along the hypotenuse of a triangle for which one of the shorter sides is a whole edge of the hypercube and the other shorter side is half of an adjacent edge - or starting at the mid-point of an edge, going along to a corner and then tracing out a further whole edge (again, the diagram would be most helpful if you can draw it - i don;'t have a scanner but I'll try and post this drawing somewhere.)

Start with the points marked on the 'outer' cube:

A: (-1/2, -1/2, 1/2, 1/2), joined to

B: (-1/2, 1/2, 0, 1/2), joined to

C: (1/2, 1/2, -1/2, 1/2), joined to

D: (1/2, -1/2, 0, 1/2), which when joined back to A completes one square.

[if you really want to check this is a cube at the end, you can use the vectors AB, BC etc and use the scalar product and the magnitude formula to prove all edges are equal and mutually orthogonal.]

[[Aaargh! I've just noticed a flaw - the scalar products of BA.BC, BC.DC, DC.DA and DA.BA are 1/4, 1/4, -1/4, -1/4 - so is ABCD a trapezium? This needs further work... I think maybe the w-co-ordinates need to be adjusted, since I've just noticed that the x, ys and zs are variously -1/2, 0 and 1/2, but there are no 0 w co-ordinates, which to me says a symmetry violation.]]

[[[Actually, I just noticed that if I move B to (-1/2, 0, -1/2, 1/2) then BA.DA= 0! So it's just a matter of how I arrange my points - I shall update them soon. The rest of the proof should work now.]]]

Now, connect to the points on the inner square:

A is joined to E (-1/2, 0, 1/2, -1/2);

B is joined to F (-1/2, 1/2, -1/2, -1/2);

C is joined to G (1/2, 0, -1/2, -1/2);

D is joined to H (1/2, -1/2, 1/2, -1/2).

Now obviously EFGH is joined up into a square.

Insert usual caveat about this not being rigorous but me needing help making it so :)

So if this is indeed the 'maximal' cube you can inscribe in a hypercube, not what its edge length is: Each of AB, BC etc were made by a triangle which goes all the way along one edge of a unit 4-cube, then halfway along a perpendicular edge. this means that the hypercube we have drawn has side length sqrt(1^2 + (1/2)^2) = sqrt(5/4).

We already know that the largest circle we can draw in a unit hypercube has radius sqrt(1/2), so therefore the biggest circle we can draw in a hypercube of side length sqrt(5/4) is

sqrt(5/4)*sqrt(1/2) = sqrt(5/8)!

Now I can see a direction for future work. Doing some research, we find that for n >= 5, there are only 3 regular n-polytopes: http://en.wikipedia.org/wiki/List_of_regular_polyt...

the n-simplex (analogue of the tetrahedron);

the n-cube (analogue of the cube), and

the n-orthoplex (analogue of the octahedron).

So above n = 5, the problem might become much easier. Considering the 'n + 1' case: We know an (n + 1)-cube has 2(n + 1) n-dimensional hyperfaces. Therefore we are looking for a regular n-polytope with at maximum 2(n + 1) hyperfaces.

So I imagine we can prove that the n-orthoplex has more hyperfaces than the (n + 1)-cube (so it won't fit) but the n-cube has less hyperfaces - so it will fit. And if we can prove that for n>= 5, the maximal shape is always a hypercube, then all that is left is to prove that the side length ratio 'increases' by sqrt((n + 1)/n) each time, e.g. for n = 6, we want the side length of the maximal 5-cube to be sqrt(6/5), so that the circle will have radius sqrt(6/5)*sqrt(5/8) = sqrt(6/8).

Hence the proof I am going to look at will use an inductive argument, and will consist of proving the following:

1) Prove that for (n + 1)-dimensions, the (n + 1)-cube has less hyperfaces than the n-orthoplex but more than the n-cube. this should be fairly trivial and algebraic but shows that the maximal inscribed polytope we are using at any stage is the n-cube.

2) Prove the ratio of the side length of the (n + 1)-cube to the maximal n-cube inscribing it is equal to sqrt((n + 1)/n). This will be the tricky part; and will involve some argument about the maximal n-cube you can draw in an (n + 1)-cube.

3) The case is proven for n = 2, 3 and 4; Use induction to show that since the radius of the circle for n = 5 dimensions is sqrt(5/8); and since the inductive hypothesis says the radius for (k>=5)-dimensions is sqrt(k/8), then for (k + 1) dimensions it is

sqrt((k + 1)/k)*sqrt(k/8) = sqrt((k + 1)/8).

I feel like I'm finally getting somewhere :)

ADDED 1:32 am GMT 22/12:

Jjust a note - by 'simplex' i mean the simplest shape you can make in n dimensions. So the 1-simplex is the line, the 2-simplex is the triangle, the 3-simplex is the tetrahedron, the 4-simplex is the pentachoron, and so on. To make each 'next' shape, we take the previous shape, add a point in the added dimension, and join it to all previous points (so e.g. adding a 4th point 'off the plane' to a triangle makes it into a tetrahedron.) This 'regular simplex' of yours will either usually not be regular, or not have the 2(n +1) relation you assume. As an example, for the n = 4 case, we'd need a regular polychoron (4D shape) of 10 hyperfaces (3D shapes), and such a thing doesn't exist; we'll have to assume that the 4D hyperplane doesn't quite go through all hyperfaces, just as a plane only gets to intersect 4 faces of an octahedron. Just to clarify.

interesting hypercube problem

It strikes me that this problem is dual to the following problem:

What is the edge length of the smallest n dimensional hypercube which encloses a unit circle?

Once having solved this problem and found the edge length L, it seems easy to transform the solution to the solution of the original problem by scaling everything by 1/L.

The dual may be easier to solve because it seems easier to define the smallest enclosing hypercube about a given circle than it does to define the largest circle enclosed by a given hypercube.

back when i took calculus,

there was those triple integrals that where for a hyper sphere.

Ii think the toughest part is finding where this sphere is tangent to the h-cube.

once we find that we can find the h spheres radius

but how?

**and i think there might be a relation to the biggest circle inside a 3 dcube

and perhaps we can use the same ratios to get the radius of the h sphere within a h cube?

let me know