Distance Across Hypercubes

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Distance Across Hypercubes

Postby anderscolingustafson » Sat Mar 08, 2014 10:34 pm

In 1d the longest distance across a line is just the distance from the top to the bottom.

1d line.png
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In 1d the longest distance across a line is the same as the diameter of the line.

In 2d the longest distance across a square is from the bottom left to the top right.

2d Square.png
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In 2d the longest distance across a square is sqrt((x1-x2)2+(y1-y2)2) which for a cube is the same as sqrt(l2+l2) which is the same as l(sqrt(2)). So the longest length across a square is the sqrt(2) times the shortest length.

In 3d the longest distance across a cube is from the front bottom left to the back top right.

3d cube.png
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In 3d the longest distance across a cube is sqrt(l2+l2+l2) or sqrt(3) times longer than the longest length.

In 4d the longest distance across a Hypercube is from the ana front bottom left to the kata back top right.

4d Hypercube.png
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In 4d the longest distance across a hypercube is 2 times the shortest distance.

In any number of dimensions the ratio between the longest distance across a hypercube and the shortest distance is the square root of the number of dimensions. So the more dimensions there are the greater the longest distance across a hypercube. The more dimensions there are the harder it is to fit a hypercube of a given diameter into a hypersphere of a given diameter.
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Re: Distance Across Hypercubes

Postby quickfur » Sun Mar 09, 2014 1:01 am

Yes, this is a well-known fact, and also one of the counterintuitive things that happen when you go up to higher dimensions. The circumradius of the n-dimensional hypercube with coordinates (±1, ±1, ±1, ...) is √n, and as we know, as n grows without bound, √n also grows without bound. However, the distance from the center of the n-cube to the center of one of its hyperfaces is still just 1. So we have the counterintuitive fact that when n is very large, you need a very large n-dimensional sphere to wrap around the n-cube, but only an n-sphere of unit radius can fit inside the cube!

A related consequence is that the volume of an n-sphere of unit radius begins to shrink after a certain dimension (IIRC somewhere between n=6 and n=7), so that when n is very large, the n-sphere occupies almost zero volume, whereas the unit n-cube occupies a volume of 1.

Now, these two things lead to the very counterintuitive fact that in n-dimensional space (for large n), almost all of the volume of the n-cube is concentrated near its corners, rather than its center (because the volume around its center is just the volume of the unit n-sphere, which is almost zero when n is very large). So the cube takes on strange properties, as though it was shaped like a pointed star, rather than a squarish shape. Well, it's actually still a squarish shape, but it behaves like a pointed star because it has almost no volume around its center, and most of its volume is concentrated towards its corners, which are very far away from its center even though the middle of its facets are only 1 unit away.
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Re: Distance Across Hypercubes

Postby anderscolingustafson » Sun Mar 09, 2014 4:43 am

I also found that for triangle like shapes in which all the sides and angles are the same it seems like the more dimensions it has the smaller the height is in comparison to the edges. I wonder if this property applies to all polytopes.
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Re: Distance Across Hypercubes

Postby wendy » Sun Mar 09, 2014 7:17 am

The densest packing of spheres in 9 dimensions, you can rattle the thing apart. But things pick up in 10 dimensions, and this does not happen again until 25 dimensions.

In 19 dimensions, the volume of the sphere of radius 1, is less than the cube of edge 1. So the radian^19 is more than the sphere.
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Re: Distance Across Hypercubes

Postby quickfur » Mon Mar 10, 2014 3:05 am

anderscolingustafson wrote:I also found that for triangle like shapes in which all the sides and angles are the same it seems like the more dimensions it has the smaller the height is in comparison to the edges. I wonder if this property applies to all polytopes.

You're talking about the n-simplex? (I.e. higher dimensional equivalent of tetrahedron). One interesting thing about the n-simplex is that as n increases, the difacet angle (angle between two (n-1)-simplex facets) approaches 90°, yet the height of the n-simplex decreases. :) So you get the paradoxical situation where the "walls" of an n-simplex shaped pyramid get more and more vertical, but the height of the pyramid gets shorter and shorter!

As for whether it applies to all polytopes, obviously it doesn't apply to the n-cube. :) The height of the n-cube stays constant as n increases.
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Re: Distance Across Hypercubes

Postby anderscolingustafson » Mon Mar 10, 2014 9:59 pm

As for whether it applies to all polytopes, obviously it doesn't apply to the n-cube. :) The height of the n-cube stays constant as n increases.


Actually what I was asking is does the property in which the ratio of the largest distance across to the smallest distance across increases with the number of dimensions apply to all polytopes? I mean does the lonest distance across a polytope get longer in comparison to the shortest distance as the number of dimensions increases no matter what the polytope?
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Re: Distance Across Hypercubes

Postby quickfur » Mon Mar 10, 2014 11:22 pm

anderscolingustafson wrote:
As for whether it applies to all polytopes, obviously it doesn't apply to the n-cube. :) The height of the n-cube stays constant as n increases.


Actually what I was asking is does the property in which the ratio of the largest distance across to the smallest distance across increases with the number of dimensions apply to all polytopes? I mean does the lonest distance across a polytope get longer in comparison to the shortest distance as the number of dimensions increases no matter what the polytope?

So you're talking about circumradius (i.e., the radius of an n-sphere that encloses the polytope) vs. edge length? Well, it depends on the polytope.

The n-cube of edge length 2, for example, has a circumradius of √n (so distance between opposite corners = 2√n), so the circumradius will increase with increasing n. The ratio of the circumradius to the edge length will grow without bound as n increases.

The n-simplex (n-dimensional tetrahedron) of edge length 2, however, has circumradius √(2n/(n+1)), which is always < 2. So the circumradius is always smaller than the edge length, no matter how big n is. As n approaches infinity, the circumradius approaches the edge length. So in this case, the longer distance is the edge length, and it does not get longer compared to the shorter distance (the circumradius); in fact, the two get closer the bigger n is.

So, the answer depends on which polytope you're talking about. There is no universal trend here.
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Re: Distance Across Hypercubes

Postby wendy » Tue Mar 11, 2014 7:40 am

The ratios of the in-diameter to the out-diameter of a regular simplex is n. For a cube or cross polytope, it's sqrt(n). But there are figures which can make this very small.
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Re: Distance Across Hypercubes

Postby quickfur » Thu Oct 23, 2014 4:55 pm

It's a 4D apartment, I take it? :P
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Re: Distance Across Hypercubes

Postby ICN5D » Mon Oct 27, 2014 10:52 pm

raima55 is a bot, I fear. We've become inundated with them lately. All have the same generic username/sig combo, and comments.
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Re: Distance Across Hypercubes

Postby quickfur » Mon Oct 27, 2014 11:05 pm

How disappointing. I was about to ask where I could rent a 4D apartment myself...
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Re: Distance Across Hypercubes

Postby ICN5D » Tue Oct 28, 2014 2:56 am

Gosh, I know. That's what I was thinking, too. I mean, how awesome would that be?
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Re: Distance Across Hypercubes

Postby quickfur » Tue Oct 28, 2014 3:08 am

It would be really awesome 'cos we could decorate the apartment with 4D furniture, like 24-cells and swirlprisms, laced with Klein "bottles"... well, until we collapse in a crumpled 3D heap on the floor and are unable to get up again. :P
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Re: Distance Across Hypercubes

Postby Prashantkrishnan » Wed Dec 24, 2014 6:39 pm

So the tesseract happens to be the only hypercube for which the circumradius is equal to the side length...

I wonder what Quickfur meant about the n-volume of the n-sphere reducing after n=6 or n=7
Comparing the ratios of the n-volumes of inscribed n-sphere to n-hypercube for the first few dimensions
n=1
inradius = 1 if length = 2
Ratio of lengths is 1:1

n=2
Ratio of areas is pi:4

n=3
Ratio of volumes is (4pi/3):8

n=4
Ratio of bulks is (pi^2/2):16

n=5
Ratio of 5-volumes is (8pi^2/15):32

n=6
(pi^3/6):64

It does indeed show that the inscribed n-sphere contains less and less part of the n-volume of the n-hypercube.
But what does it mean to say that the hypercube becomes more star like with most of its volume concentrated near its corners?
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Re: Distance Across Hypercubes

Postby Prashantkrishnan » Wed Dec 24, 2014 6:53 pm

This seems out of the topic, but I'm curious to know if there is a way to find out the formula for the n-volume of the n-sphere without actually deriving it by integration. The general format being some constant multiplied by pi[n/2]rn, is there a pattern which that constant follows?(The square brackets denote greatest integer function)
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Re: Distance Across Hypercubes

Postby Klitzing » Thu Dec 25, 2014 9:09 am

You either can calculate the volume of the unit hyperball recursively:
Code: Select all
V_0 = 1
V_1 = 2
V_n = (2 pi / n) . V_(n-2)

or explicitely, but separate for even and odd numbers, by means of
Code: Select all
V_(2k)   = pi^k / k!
V_(2k+1) = 2 . k! . (4 pi)^k / (2k+1)!

or by means of the Gamma function for integral and half-integral values in a single term:
Code: Select all
V_n = pi^(n/2) / Gamma(n/2 + 1)


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Re: Distance Across Hypercubes

Postby Prashantkrishnan » Thu Dec 25, 2014 1:10 pm

Klitzing wrote:You either can calculate the volume of the unit hyperball recursively
or explicitely, but separate for even and odd numbers, by means of
Code: Select all
V_(2k)   = pi^k / k!
V_(2k+1) = 2 . k! . (4 pi)^k / (2k+1)!



This is such a simple way of putting the whole formula :D
I suppose we could derive the units for solid angles too the same way
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Re: Distance Across Hypercubes

Postby wendy » Fri Dec 26, 2014 10:48 am

There are different approaches to solid angles.

The one i use is to just measure the fraction of space occupied by the solid at the point. Radians etc suppose that you have an integration formula at hand.

I found the solid angle of the {3,3,5} to be :33.15.60 (base 120), long before i determined the radian at :06.09.60. You don't need to reference things to area of surface.

The solid angle of the simplex is s(n)/(n-1)! where 1<s(n-1)<s(n)<sqrt(n/4). The value is known exactly to me for only n=1, 2, 3, 4. For s(8)>1.333, and s(24)>1.6 are both known loosely.
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