Roundness of Polytopes

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

Roundness of Polytopes

Postby Klitzing » Tue Nov 13, 2018 11:59 am

In a Facebook discussion Tadeusz Dorozinski asks about any kind of measure of roundness of polytopes.

I remember that we already had a similar type of research when dealing about the stariness of the hypercubes, i.e. that most of their body volume would be maintained close to their corners, because the ratio of hypervolume of hypercube and inscribed hyperball in its dimensional limit would diverge.

The main arguments in the FB discussion so stick to just one and the same dimension. In fact his question itself was concerned with 3D merely. Thus answers like ratio of polyhedral volume to volume of inball, or even to maximal inball (i.e. with radius upto the outermost faces) seem likely.

One clearly might want to dualize that very setup and consider the ratio of the volume of the circumball to the volume of the polyhedron. This ratio clearly generalizes to every dimension, and moreover would easily encompass all orbiform polytopes.

But then there is our interest in CRFs (in 4D and even beyond), which is the Johnson solids within 3D, not at all to mention the toroids, which all would fall short here.

So I brought in a further idea for a potential roundness measure, the ratio of volume to surface content. That one clearly maximises for the ball and similarily can be applied to 2D setups, then maximising for the disk. Quite generally we would consider here in a first run vol(P)/vol(∂P), where ∂P denotes the surface or boundary of P and vol simply is the according dimensional volume function. But this measure lacks the disadvantage that it obviously would be size depending. For the D-dimensional volume of the body content scales differently with the absolute size of P than the D-1-dimensional surface content.

This disadventage can be overcome quite simply. Thus consider vol(P)dim(∂P)/vol(∂P)dim(P) instead. This value now will be size Independent, and still is being maximized for the according dimensional hyperball. In fact this measure now serves very well for comparision of any polytopes within one and the same dimensionality. Sure, it is a global measure. Thus the seeming counter argument of an icosahedron with an attached needle, which still has nearly the same roundness amount as the icosahedron itself, does not count here. That very isolated excess is kind a Lesbeque null set, which thus does not change the global behaviours of both, the hypersurface and the hypervolume.

Still there is a further issue to be overcome. When we consider shapes of various dimensionalities, their values differ widely. In fact even the according values vol(O)dim(∂O)/vol(∂O)dim(O) for the hyperballs O themselves would decrease quite fast with the increase of the dimensions. This is why it would be much better to take refuge to a normalized setup instead. This is why I opted within that FB discussion for
roundness(P) = (vol(P)dim(∂P)/vol(∂P)dim(P)) / (vol(O)dim(∂O)/vol(∂O)dim(O))
where O again is the according dimensional hyperball, i.e. dim(O)=dim(P).

Sure, the roundness values according to this definition are not too easy to calculate. You not only have to evaluate the body volume and the surface content of your own to be considered polytope only, moreover you will have to get the accoding values for the hyperball as well. And those body volume and the surface content values of the hyperball either are known recursively only, are provided explicitely in 2 separate formula for even and odd dimensions, or would relate to the Gamma function.

Nonetheless I managed since to evaluate the according values for all three, the dimensional series of regular simplices, of hypercubes, and of cross-polytopes (aka orthoplexes). Here they come:
roundness( 2D-simplex ) = (2D)! πD / (D! sqrt[(2D)2D(2D+1)2D+1])
roundness( 2D+1-simplex ) = D! (2π)D / ((D+1)D+1 sqrt[(2D+1)2D+1])

roundness( line ) = 1 = 100 %
roundness( {3} ) = π sqrt(3) / 32 = 60.459979 %
roundness( tet ) = π sqrt(3) / (2 32) = 30.229989 %
roundness( pen ) = 3 π2 sqrt(5) / (22 53) = 13.241464 %
roundness( hix ) = 23 π2 sqrt(5) / (33 53) = 5.231196 %
roundness( hop ) = 5 π3 sqrt(7) / (32 74) = 1.898165 %
roundness( oca ) = 3 π3 sqrt(7) / (24 74) = 0.640631 %
roundness( ene ) = 5·7 π4 / (28 38) = 0.202982 %
...
roundness( 2D-hypercube ) = (π/4)D / D!
roundness( 2D+1-hypercube ) = πD D! / (2D+1)!

roundness( line ) = 1 = 100 %
roundness( {4} ) = π / 22 = 78.539816 %
roundness( cube ) = π / (2·3) = 52.359878 %
roundness( tes ) = π2 / 25 = 30.842514 %
roundness( pent ) = π2 / (22 3·5) = 16.449341 %
roundness( ax ) = π3 / (27 3) = 8.074551 %
roundness( hept ) = π3 / (23 3·5·7) = 3.691223 %
roundness( octo ) = π4 / (211 3) = 1.585434 %
...
roundness( 2D-orthoplex ) = πD (2D)! / (23D DD D!)
roundness( 2D+1-orthoplex ) = πD D! / sqrt[(2D+1)2D+1]

roundness( line ) = 1 = 100 %
roundness( {4} ) = π / 22 = 78.539816 %
roundness( oct ) = π sqrt(3) / 32 = 60.459979 %
roundness( hex ) = 3 π2 / 26 = 46.263771 %
roundness( tac ) = 2 π2 sqrt(5) / 53 = 35.310570 %
roundness( gee ) = 5 π3 / (26 32) = 26.915171 %
roundness( zee ) = 2·3 π3 sqrt(7) / 74 = 20.500183 %
roundness( ek ) = 3·5·7 π4 / 216 = 15.606620 %
...


Now, what are your thoughts?
    - To Tadeusz' original question?
    - and to all those reported initial answers?
    - To my setup of that roundness measure definition?
    - To the provided list of those calculated values?
    - and esp. their dimensional behavior?

--- rk
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Re: Roundness of Polytopes

Postby wendy » Wed Nov 14, 2018 8:38 am

The ultimate irony of roundness, is that polytopes with the sharper corners tend to be rounder.

The {3,3,5} is rounder than {5,3,3} although its vertex-angle is :33.15 against :38.24, and the cross polytope at 0:05 is rounder then the tesseract at 0:15. It has more to to with a measure of faces, the greater number of faces leads to a rounder polytope.

Something like a oblate ellipsoid, still has a²/c², is less round than the prolate ellipsoid a/c. A torus is less round again.

It seems a good idea, but the more round objects will in the main, have pointier things. A measure against the circumsphere might show polytopes that spend more time there might be rounder. The omnitruncate simplex (Hamilton-tiling) is the tiling that provides the easiest cover of spheres to cover all of the space, but it is by no means the best tiling for packing spheres.
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Re: Roundness of Polytopes

Postby wendy » Fri Nov 16, 2018 4:28 am

I had another thought about this last night.

Would largest insphere :: smallest out-sphere work. In essence, relating volume to smallest insphere allows figures to have very un-round features like a long rod sticking out of it, like an all-day-sucker (a kind of sweet ball on a stick).
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Re: Roundness of Polytopes

Postby mr_e_man » Wed Dec 25, 2019 8:03 pm

While working with multivariable integrals, I encountered the quantity (1/2)*diam(P)*vol(∂P)/vol(P), where diam(P) is the maximum distance between any two points in P. Have you considered this as a roundness measure? (Actually it's the inverse, sharpness.) This is also independent of size, and minimized by the n-dimensional ball, with value n.

The icosahedron would have its diameter increased by having a needle attached (pointing outward); and its diameter would remain unchanged by having a needle excavated (or attached pointing inward). But the surface has the same intrinsic geometry whether the needle points out or in.

That makes me wonder about an intrinsic sharpness measure. It would involve the Gaussian curvature, or equivalently the angle defects at vertices. For smooth surfaces, one possibility is the integral of the square of Gaussian curvature:

(int K2 dA) * (int dA) / (int K dA)2

= (int K2 dA) * A / (2π χ)2

Of course A is area, and χ is the Euler characteristic. I think the Cauchy-Schwarz inequality applies here, in the form (int f*g)2 <= (int f2)*(int g2), so that the sharpness is always at least 1. But it's infinite for the torus (χ=0), even though the torus feels round. We could drop the denominator, to get (int K2 dA)*A, which is finite for the torus, and still independent of size. In any case, int K2 dA is infinite for polyhedra.

The analogous quantity for polyhedra involves the sum over vertices of the square of the angle defect:

(sum δ2) * (sum 1) / (sum δ)2

= (sum δ2) * NV / (2π χ)2

Again this is always at least 1, and is infinite for toroidal polyhedra. For ordinary (spherical, χ=2) polyhedra, if a face is divided into several faces, then the number of vertices increases, and the new vertices have angle defect δ=0, so the sharpness increases, even though the intrinsic geometry remains unchanged. So that doesn't work. I suppose sum δ2, by itself, could work.
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Re: Roundness of Polytopes

Postby mr_e_man » Tue Mar 23, 2021 11:24 pm

wendy wrote:I had another thought about this last night.

Would largest insphere :: smallest out-sphere work. In essence, relating volume to smallest insphere allows figures to have very un-round features like a long rod sticking out of it, like an all-day-sucker (a kind of sweet ball on a stick).


Yes, I think that's a good idea, squeezing the object's boundary between two spheres.

Another possibility is to use calliper measure: the distance between two parallel (hyper)planes enclosing and touching the object. We can consider the ratio of the smallest calliper measure to the largest calliper measure, as a roundness measure. (In fact the maximum calliper measure is the same as the diameter, that is, the maximum distance between any two points in the object.)

For example, an n-cube with edge length 1 has minimum calliper measure 1, and maximum calliper measure √n, so its roundness would be 1/√n.

An n-simplex with edge length 1 has distance √((n+1)/(2n)) between a facet and the opposite vertex, which I guess is the minimum calliper measure, and thus also the roundness (since the diameter is 1).

An n-orthoplex with diameter 1 (and edge length 1/√2) has distance 1/√n between opposite facets, so its roundness is 1/√n.

Of course a unit n-sphere always has calliper measure 2, so its roundness is 1.

...According to this measure, an n-simplex is rounder than an n-cube. To understand this, let's consider the 2D case: For an even-sided polygon, the two calliper lines are either both touching vertices (far from the polygon's centre) or both touching edges (closer to the polygon's centre) so there's a large variation in distance between the lines. For an odd-sided polygon, if one calliper line is touching an edge then the other line is touching a vertex, so there's less variation in distance between the lines, hence more roundness.

See also curves of constant width and linked topics.
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