Coordinates of all uniform hypercubic n-polytopes

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

Coordinates of all uniform hypercubic n-polytopes

Postby quickfur » Fri Nov 21, 2008 7:00 pm

Dear all,

Last night I had an epiphany regarding the construction of the various uniform truncates/rectates/rhombates/etc. of the n-dimensional cubes/crosses. While attempting to compute the Cartesian coordinates of a few interesting 4D polychora, I stumbled upon the following pattern that completely enumerates all uniform n-polytopes derived from the n-cube/n-cross, and gives the full Cartesian coordinates for their vertices.

Notation

For conciseness' sake, I shall write [x1,x2,...xn] to denote the convex hull of all possible permutations of coordinates and signs of the point <x1,x2,...xn>. Thus, [1,1,1] is the cube, and [1,1,1,1] is the tetracube. [0,1,1] is the cuboctahedron.

If P is a point of the form <x1,...xn>, then I shall write [P] to mean [x1,...xn].

Definition

I shall call the class of enumerated the n-truncates (is there a better term for this?), and denote it by T.

Firstly, the point [0] and the line segment [1] are in T. (We may regard the point as a "degenerate truncate", represent in all dimensions, as we shall see.)

Next, given any (n-1)-dimensional truncate of the form [P], the n-dimensional truncates are defined to be:

1) Any polytope [P+], where P+ is P concatenated with its largest coordinate. For example, since [1] is a 1-truncate, one possible 2-truncate is [1,1], formed by repeating the largest coordinate 1.

2) Any polytope [P#], where P# is P concatenated with x+√2, where x is the largest coordinate of P. For example, from [1] we may form [1,1+√2], which is the octagon.

Note that all polytopes generated by this definition have an edge length of 2. (This is vacuously true for the point as well.)

Intuition

The basic intuition behind this definition is that we take any (n-1)-truncate and lay them in the hyperplanes of an n-cube's facets, with two possible distances along the coordinate axes: one where their axial facets coincide (thus joining to each other via their axial facets), and one where their axial faces are 2 units apart, the gap to be filled in by an appropriate prism. By construction, all edge lengths will be equal. Furthermore, all polytopes thus generated are uniform (proof left as an exercise for the reader).

Enumeration

1D:
[0] is a point.
[1] is a line segment.

2D:
From [0], we may repeat the largest coordinate to get [0,0], which is again the point in 2D, or we may add √2 to its largest coordinate to get [0,√2], which is the square diamond with edge length 2.

From [1], repeating the largest coordinate gives [1,1], the square, and adding √2 gives [1,1+√2], the octagon.

These are all the possibilities.

3D:
From [0,0], we get [0,0,0], the point again, and [0,0,√2], the octahedron of edge length 2.

From [0,√2], we get [0,√2,√2], the cuboctahedron, and [0,√2,2√2], the truncated octahedron.

From [1,1], we get [1,1,1], the cube, and [1,1,1+√2], the (small) rhombicuboctahedron.

From [1, 1+√2], we get [1, 1+√2, 1+√2], the truncated cube, and [1, 1+√2, 1+2√2], the great rhombicuboctahedron ("truncated" cuboctahedron).

These account for all the "cube-like" or "octahedron-like" Archimedean polyhedra.

4D:
[0,0,0,0] is the point in 4D.
[0,0,0,√2] is the 16-cell.
[0,0,√2,√2] is the 24-cell (rectified 16-cell).
[0,0,√2,2√2] is the truncated 16-cell.
[0,√2,√2,√2] is the rectified 8-cell.
[0,√2,√2,2√2] is the cantellated 16-cell (rectified 24-cell).
[0,√2,2√2,2√2] is the bitruncated 16-cell (bitruncated 8-cell).
[0,√2,2√2,3√2] is the cantitruncated 16-cell (truncated 24-cell).
[1,1,1,1] is the tesseract (8-cell).
[1,1,1,1+√2] is the runcinated tesseract (runcinated 16-cell).
[1,1,1+√2,1+√2] is the cantellated 8-cell.
[1,1,1+√2,1+2√2] is the runcitruncated 16-cell.
[1,1+√2,1+√2,1+√2] is the truncated tesseract.
[1,1+√2,1+√2,1+2√2] is the runcitruncated 8-cell.
[1,1+√2,1+2√2,1+2√2] is the cantitruncated 8-cell.
[1,1+√2,1+2√2,1+3√2] is the omnitruncated 8-cell.

These account for all the uniform polychora in the tesseractic/16-cell family.

I won't try to enumerate them in 5D, but it is clear that there will be precisely 32 cubic truncates in 5D, or 31 if we exclude the degenerate point. Excluding the 5-cube and 5-cross themselves, we see there are 29 uniform 5-polytopes in the pentarectic family.

In general, in n-dimensions, there will be 2^n-1 non-degenerate n-cubic uniform polytopes (including the n-cube and n-cross). Their Cartesian coordinates can be derived simply by enumerating in the manner exemplified above.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby wendy » Sat Nov 22, 2008 7:57 am

The standard coordinates for any figure, derived from the symmetry of the cube/cross series, can be derived from the dynkin symbol, and these vectors.

The edge is 2, and all permutations, all change of signs. q = sqrt(2).

0. (q,0,0,....)
1. (q,q,0,....)
2. (q,q,q,0,...)

the last is
n. (1,1,1,....1).

For eg o3x3x4o one has (q,q,0,0) + (q,q,q,0) = (2q,2q,q,0), APACS.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby quickfur » Sat Nov 22, 2008 5:08 pm

wendy wrote:The standard coordinates for any figure, derived from the symmetry of the cube/cross series, can be derived from the dynkin symbol, and these vectors.

The edge is 2, and all permutations, all change of signs. q = sqrt(2).

0. (q,0,0,....)
1. (q,q,0,....)
2. (q,q,q,0,...)

the last is
n. (1,1,1,....1).

For eg o3x3x4o one has (q,q,0,0) + (q,q,q,0) = (2q,2q,q,0), APACS.

Neat. So I've just rediscovered the dynkin symbol without knowing it. :P

Now, I've found this interesting property that the corner facet of the hypercubic truncates (the facet corresponding with an n-cross's facets) run through all possible truncates of the (n-1)-simplex as well. In fact, given the 2^n truncates in n dimensions, the corner facets of the first half and the corner facets of the second half are identical up to a translation. So that gives 2^(n-1) truncations of the (n-1)-simplex, some of which are identical because simplices are self-dual. This gives us Cartesian coordinates for all the simplicial truncates as well (they are not origin-centered, but this is easily rectified... heh, pun intended).
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby begvend » Sun Nov 23, 2008 9:53 pm

Hi all,

What is APACS ?
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby wendy » Sun Nov 23, 2008 11:33 pm

APACS = all permutations, all change of sign. You can replace either of the A's with E or O to halve the vertices, eg EPECS
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Another system of shape notation

Postby quickfur » Tue Nov 25, 2008 12:26 am

(The title is to catch Hayate's attention. :lol:)

I've refined by ideas in the previous topic about computing coordinates for n-dimensional hypercubic truncates, and now I have an elegant system for naming these things which I think is worthwhile to present as a whole.

Notation

In terms of syntax, the notation is dead simple. An n-dimensional polytope is written as an n-digit binary string, enclosed in either square brackets or angle brackets.

Interpretation

When a binary string is enclosed in square brackets, it denotes a hypercubic truncate. The coordinates of the truncate's vertices are obtained in the following manner: we first translate the binary string into an n-dimensional point, and then take all permutations of coordinates and sign. The translation is performed as follows:

- The first coordinate of the point is equal to the first digit of the binary string.
- For each subsequent digit, if the digit is 0, the corresponding coordinate is set to the largest coordinate seen so far. If the digit is 1, the corresponding coordinate is set to the largest coordinate seen so far plus √2.

When a binary string is enclosed in angle brackets, it denotes a simplicial truncate. The coordinates of the simplicial truncate is obtained as follows: first, prefix the binary string with 0, then translate it into an (n+1)-dimensional point as above. Then, take all permutation of coordinates (but not sign---all coordinates are non-negative). The convex hull of these points is the simplicial truncate, embedded in (n+1) dimensions. To obtain coordinates in n-space, let A be the rotation matrix that transforms <1,1,1,...,1> to <0,0,0,... √n>. Apply this rotation to all points. This will produce a set of points with the last coordinate equal to √n. Dropping this last coordinate yields the coordinates of the simplicial truncate in n dimensions.

Note that due to the self-duality of the n-simplex, there will be some duplicates among the simplicial truncates. This duplication follows a regular pattern, which can be applied to remove duplicates in a full enumeration of simplicial truncates in n dimensions.

Also note that when all digits of the binary string is 0, the result is just a single point for both square brackets and angle brackets.

Examples

Since I've already listed most of the lower dimensional hypercubic truncates in the previous topic, I won't bother to list them again. I will list the simplicial truncates, though, to illustrate how the notation works:

<0> - point.
<1> - line segment.
<00> - point
<01> - triangle: first, prepend 0 to get [001] which is the octahedron; taking permutations of non-negative coordinates give us the triangle <1,0,0>, <0,1,0>, <0,0,1>. Applying the rotation and translation will map this into 2D coordinates (exercise for the reader).
<10> - dual triangle: first, prepend 0 to get [010], which is the cuboctahedron; permutations of non-negative coordinates give us the dual triangle <1,1,0>, <1,0,1>, <0,1,1>.
<11> - hexagon: first, prepend 0 to get [011], which is the truncated octahedron; the non-negative coordinates give us a hexagon.
<000> - point
<001> - the tetrahedron: prepending 0 gives [0001], which is the 16-cell; non-negative coordinates define a single tetrahedral cell.
<010> - octahedron (rectified tetrahedron): prepending 0 gives [0010], which is the 24-cell; taking non-negative coordinates yield a single octahedral cell.
<011> - truncated tetrahedron: prepending 0 gives [0011], which is the truncated 16-cell; non-negative coordinates yield a truncated tetrahedron.
<100> - dual tetrahedron: prepending 0 gives [0100], the rectified tesseract; non-negative coordinates give a tetrahedron in dual configuration to <001>.
<101> - cuboctahedron (cantellated tetrahedron): prepending 0 gives [0101], the cantellated 16-cell, whose non-negative coordinates yield a cantellated tetrahedron.
<110> - truncated dual tetrahedron (bitruncated tetrahedron): prepending 0 gives [0110], the bitruncated 16-cell, whose non-negative coordinates yield a truncated tetrahedron in dual orientation to <011>.
<111> - truncated octahedron (omnitruncated tetrahedron): prepending 0 gives [0111] which is the cantitruncated 16-cell; non-negative coordinates give a truncated octahedron.

Notice that reversing the binary string in an angle bracket yields a simplicial truncate in dual orientation; so only half the non-palindromic binary strings yield a unique simplicial truncate. The palindromic binary strings yield unique simplicial truncates.
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Re: Another system of shape notation

Postby wendy » Tue Nov 25, 2008 8:26 am

The notations here offer little that can not be determined from the schläfli symbol. The former amounts to reading the dodecahedral presentation of the octahedral groups where x before Q stands for 1, and after the 4, stand for q. The second is also a known coordinate system, in the general (sum=0) axis for the simplex (eg tetra = 3,-1,-1,-1).

It really does not offer much over the schläfli symbol, and does not even take advantage of the rules allocated under the name of schläfli (each symbol has a value, for which one can find the diameter2 by way of 2(vertex figure)/(figure), where the prism-product is the product of the prisms.

for simplex, S=(n+1); for the cubic (4, 34, 334, ...), S=2; for the halfcubics (33A, 333A, ...) S=4; for Gosset's figures, S=9-n; for the pentagonals, S=(n+1)-(n-1); for the 343, S=1.

Something like 3;3,5 gives 2 * 2 * (3-f) / (5-3f), for 333;334 we have 2 S(33) . S(34) / S(333334) = 2 . 4 . 2 / 2 = 8.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby Keiji » Tue Nov 25, 2008 12:22 pm

You didn't have to start a new topic - I was reading the old one (as I always read all new posts on the board apart from Religious Debates). I just have nothing to say here as, as wendy rightly says, you're just reinventing the wheel (Schlaefli symbol in this case).
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby quickfur » Wed Nov 26, 2008 5:36 pm

Hayate wrote:You didn't have to start a new topic - I was reading the old one (as I always read all new posts on the board apart from Religious Debates). I just have nothing to say here as, as wendy rightly says, you're just reinventing the wheel (Schlaefli symbol in this case).

Ah, but it's fun to reinvent the square wheel! :P

OK, I created this notation mainly for ease of translation into Cartesian coordinates. It doesn't actually express anything that can't already be expressed in existing notation, as you all said. Right now I'm more concerned about computing coordinates so that I can render these things with my polytope viewer.

When I get time, I'll revisit ways of generating new shapes, although I doubt I can beat Wendy here; her lace prisms and geometric products cover just about every conceivable shape that might be interesting.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby wendy » Thu Nov 27, 2008 7:41 am

The current notations cover only a small bit of what's out there. It's more or less what Coxeter did in 1935, just a little further, really.

Many of the things that i toss around don't really fit in the notations. I am still trying to figure out whether the real hulls of o3o3o4s%h and o3x3o4%h, figures which exist in eight dimensions (as the %h suggests a dimension doubling), since both of these give a real figure with 196 (twe) vertices and d2=8.

Another figure i tangle with is the pd 3,5,3, ; a chiral hyperbolic tiling of dodecahedra and pentagonal antiprisms. It has been derived down to a convex flag, transimtted by rotations and reflections, and ought exist.

Figures in the MME FE and FC group are not expressed by the notation, are believed to produce a number of uniform figures, of which six are known. Lots of work needed here.

Generally speaking, the notation is about a year behind the bleeding edge.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby quickfur » Thu Nov 27, 2008 4:42 pm

wendy wrote:The current notations cover only a small bit of what's out there. It's more or less what Coxeter did in 1935, just a little further, really.[...]

What about non-polytopic shapes, though, such as the roundish shapes and crinds we've been playing with? Did Coxeter do much work in that direction? I remember compiling a list of distinct 4D quadratic 3-manifolds some time ago, but I don't recall much off the top of my head.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby wendy » Fri Nov 28, 2008 8:24 am

Coxeter does not discuss any of the sphere examples, except to give general formulae for the sphere volume, this by way of saying one ought avoid generalisations. The works are very 'vertex-centric' in as much that polytopes are things for holding vertices that in turn behave under certian ways under symmetry operations. Very little interest exist in finding things like the catalans, which are more important in crystalography.

One must recall that Coxeter seems more interested in polytopes as a framework for group theory rather than the way they are presented in "Regular Polytopes". Books like "Generators and Relations in Discrete Groups" (1962), "Twelve Essays" (1968), "Regular Complex Polytopes" (1971), and even "The Coxeter Legacy" (2008) are more into presenting polytopes as frameworks for presenting group theory, and then reading the group theory.

The symbol i designate Dynkin symbol, aka Coxeter-Dynkin symbol, makes no sense, unless you first realise that the structure comes from the group theory, and the means to decorate it come from reading Wythoff's paper dealing with mirror-edge constructions that replicate Mrs Stott's edge-construction. Wythoff had nothing to do with 'wythoff symbols', which are really decorated schwarz triangles.

The dynkin symbol was invented three times, by Coxeter (1935), Dynkin, and by de Witt (both in the 1940s). They arise quite naturally in Lie groups, where a single operator [mirror] is written as a point or node, while the relation between two points [mirror-angle] is written as a branch. Coxeter discovered the manner of placing dots in the thing to represent mirrors.

Rotopes, Bracketopes, etc are new here, but the thing had been experimented by me very early in the piece (radiant forms). In any particular form, you can construct polytopes by combining lines in products of various bracketope operators (prism, tegum, crind, ...) [yes there are more!] and look at the resulting figure. Once ye figure out that P, T, C are coherent in the metrological sense (ie one can do physics where volumes are defined by T, and that m³ is a tegum of unit diameter (1/6 cubic metre = 1 octahedric metre). [it's actually quite natural in form: the constant of rationalisation is 8 pi, not 4 pi!

Escher spent an awful amount of time dinking through Coxeter's projection of the group [4,6] and [3,7], by way of understanding the Poincare projection, before embarking on his "circle limits" prints.

Likewise, I spent many years wrangling with different forms of the dynkin notation, polytopes as quantum objects, and trying to extend the notation like 2_21 into a general notation for all things. This is done by defining the concepts, and then writing the notations as a separate device. You can do a quick transform from notation to notation by bouncing off the underlying notation. 2_21 becomes /4B. The system was from the outset made to be very flexable, so the dual of any figure with / markers is the WMM figure with \. eg CO = 1/Q, rhombic dodeca = 1\Q.

The object was to render the calculations to work with minimal resources. Most of the initial work was done with a four-function calculator, pen and paper, far from any book on the subject. The numbers selected to represent polygons were chosen, and quickly learnt to the width of the calculator, eg 1.93185165259 is a relatively important number in the scheme of things = (r6+r2)/2.

The order of groups, count of surtopes, and some radii are given in twelfty. This greatly simplifies calculation: these were always hand-calculated. The choice and form of calculating in base 120 was some lengthy decision after considering much input from different experimental forms. Lingering problems were overcome in the next decade.

All in all, its much profound insights.
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Re: Coordinates of all uniform hypercubic n-polytopes

Postby quickfur » Fri Nov 28, 2008 6:00 pm

wendy wrote:Coxeter does not discuss any of the sphere examples, except to give general formulae for the sphere volume, this by way of saying one ought avoid generalisations. The works are very 'vertex-centric' in as much that polytopes are things for holding vertices that in turn behave under certian ways under symmetry operations. Very little interest exist in finding things like the catalans, which are more important in crystalography.

I'm quite interested in whether it is possible to have a facet-transitive polytope that is not a dual of a uniform polytope. This is equivalent to whether there exists vertex-transitive polytopes whose facets are not uniform surtopes. Is there anything that might exclude this possibility?

One must recall that Coxeter seems more interested in polytopes as a framework for group theory rather than the way they are presented in "Regular Polytopes". Books like "Generators and Relations in Discrete Groups" (1962), "Twelve Essays" (1968), "Regular Complex Polytopes" (1971), and even "The Coxeter Legacy" (2008) are more into presenting polytopes as frameworks for presenting group theory, and then reading the group theory.

Whereas I am interested in the polytopes themselves as objects of interest.

The symbol i designate Dynkin symbol, aka Coxeter-Dynkin symbol, makes no sense, unless you first realise that the structure comes from the group theory, and the means to decorate it come from reading Wythoff's paper dealing with mirror-edge constructions that replicate Mrs Stott's edge-construction. Wythoff had nothing to do with 'wythoff symbols', which are really decorated schwarz triangles.

FWIW, I find group theory much more accessible after learning about higher-dimensional polytopes.

[...]Rotopes, Bracketopes, etc are new here, but the thing had been experimented by me very early in the piece (radiant forms). In any particular form, you can construct polytopes by combining lines in products of various bracketope operators (prism, tegum, crind, ...) [yes there are more!] and look at the resulting figure. Once ye figure out that P, T, C are coherent in the metrological sense (ie one can do physics where volumes are defined by T, and that m³ is a tegum of unit diameter (1/6 cubic metre = 1 octahedric metre). [it's actually quite natural in form: the constant of rationalisation is 8 pi, not 4 pi!

I'm personally quite interested in objects, not necessarily reducible to polytopes, that have circular shapes, such as crinds, intersections of hyperspherical cylinders, and other such things.

[...]The object was to render the calculations to work with minimal resources.[...]

I'm more interested in the geometric aspect of it, especially in the way of visualizing higher-dimensional spaces. I rediscovered duoprisms by noticing two cycles of cubic cells in the projection of the tesseract, and arriving at the duocylinder by taking the limit of duoprisms (not knowing it was the duocylinder at the time--I had known of the duocylinder but did not know its appearance). I also independently derived many of the tesseractic truncates purely by visualization (and verifying my results with George Olshevsky's webpage afterwards)--not even by way of truncating the tesseract, but by assembling projections of cells into a closed form in the mind's eye.
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