Polynomials can be used find the elements of a hypercube.

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

Polynomials can be used find the elements of a hypercube.

Postby W axis » Tue Nov 11, 2008 9:00 pm

I found that using (x+2)^n will give you a polynomial with the n-dimensional elements that make up the n-dimensional hypercube.

Example a hendekeract is:



X^11+22x^10+220x^9+1320x^8+5280x^7+14784x^6+29568x^5+42240x^4+42240x^3+28160x^2+11266x+2048

That factored is (x+2)^11
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Re: Polynomials can be used find the elements of a hypercube.

Postby wendy » Wed Nov 12, 2008 8:03 am

It works with all of the regular polytopes, and all of the regular products (prism, tegum, pyramid, comb). With these, one supposes that any given polytope has a '1' at each end (eg cube = 1, 6, 12, 8, 1).

The four products then equate to including (*) or excluding (#) the first and last digit, eg prism product = *#. so

line = 1,2,1 -> 1,2,#. The prism-power of a line is (1,2)^n, #
= 1,2,1 => #,2,1. The tegum power of a line is #,(2,1)^n

point = 1,1 -> 1,1 The pyramid power of a point is eg (1,1)^[n+1]

comb product of a polygon 1,p,p,1 -> #,p,p,# -> (p,p)^[n-1].

You can multiply any polygon to get the appropriate product, eg

triangle-prism = 1,3,3,# * 1,3,3,# = 1,6,15,18,9,#
triangle-dodecahedron tegum = 1,3,3,# * 1,12,30,20,# = 1,15,69,146,150,60,#

Even the 4d torus (comb) of dodechedron * pentagon

#,12,30,20,# * #,5,5,# = #,60,210,250,100,#.
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Re: Polynomials can be used find the elements of a hypercube

Postby Keiji » Sat May 25, 2013 9:20 am

Hi wendy,

Although this is an old topic, I just found it while moving the forums around. Can you give a little more explanation on this, in particular, how a pyramid has two arguments?

Am I right in saying

pyramid = **
prism = *#
tegum = #*
comb = ##

Can you give some examples of these with say, all three dimensional arguments?
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Re: Polynomials can be used find the elements of a hypercube

Postby Klitzing » Sat May 25, 2013 10:37 am

Coming back to the original polynomials:

  • Simplex product "polynomial": (1/x)*[(x+1)^(D+1)],
    e.g. for D=2 (triangle): x^2+3x+3+1x^(-1)
    i.e. 1 body term (x^2) and 1 nulloid term (x^(-1))
  • Prism product polynomial: (x+2)^D,
    e.g. for D=3 (cube): x^3+6x^2+12x+8
    i.e. 1 body term (x^3) but no nulloid term (x^(-1))
  • Tegum product "polynomial": (1/x)*[(2x+1)^D],
    e.g. for D=3 (octahedron): 8x^2+12x+6+1x^(-1)
    i.e. no body term (x^3) but 1 nulloid term (x^(-1))
  • Honeycomb product polynomial: [N*(x+1)]^D, where N-> infinity
    e.g. for D=3 (cubical honeycomb): N^3*(x^3+3x^2+3x+1)
    i.e. no body term (as a horochoron, when used in hyperbolic tesselations: x^4), no nulloid (x^(-1))

--- rk
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Re: Polynomials can be used find the elements of a hypercube

Postby wendy » Sun May 26, 2013 8:56 am

Hi Kenji

Yes, the idea is that you add one to the various ends of the product. For example, if the polytopes are a pentagon and a line, their surtope consists are

v = vertex, e = edge, h = hedron (2d), c = choron (3d), t = teron (4d)

pentagon = 5e + 5v line = 2v

The prism product is *#, means that you add a 1 in front, ie pentagon = 1h+5e+5v, line = 1e+2v, and 1.5.5 * 1.2 = 1.7.15.10. This is the *# form of the pentagon prism, which has 7h+15e+10v Here "vertex * vertex = vertex"

The tegum product is #*, which means, that you add the 1 to the tail, ie 5e+5v+1n * 2v+n = answer + n. Now it's n (-1 d), not v(0) that stands in the units column, and "vertex * vertex = line". So 5.5.1 * 2.1 gives 10.15.7.1, or 10h, 15e, 7v (pentagonal tegum or "bipyramid").

The comb product ## just multiplies the numbers as they stand: a polygon * polygon gives a polyhedron, say as a kind of torus.

The pyramid product ** adds 1 to both ends, so 1.5.5.1 * 1.2.1 = 1.7.16.16.7.1, a polychoron, with 7c, 16h, 16v, 7e.

The products ending in # are repetition products (ie there's a point ab for each point a in one base, and b in the second base)

The products ending in * are draught-products: points are connected by a line. There's a line for each point a in A and b in B. This increases the dimension.

The products beginning in * are content products: the whole content of the bases are considered. eg In the triangle, there's a ray radiating from a vertex to every point of the base, and in the prism, there's a copy of the base for every point in the height.

The products beginning # are surface products: In the pentagonal tegum, only the surface is drawn to the apex.
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Re: Polynomials can be used find the elements of a hypercube

Postby wendy » Sun May 26, 2013 9:16 am

A pyramid can have two or more elements.

In three dimensions, consider the tetrahedron. Instead of thinking of it as a triangle-point pyramid, think of it as a line*line pyramid.

You can see that if the lines are held in parallel planes, then the intermediate parallel planes, are prism products of one line at size (a), and the other at size (1-a).

In four dimensions, you can have a double-pyramid, ie a pyramid whose base is a pyramid, ie point ** point ** polygon. Since the order is not dependent, one can write point ** point = line, and write line ** polygon. When this is held in parallel planes (3-spaces), then sections between the top and bottom become a series of prisms, of bases a line of length a and a polygon of size (1-a).

One does not get a general polygon ** polygon pyramid until five dimensions, where intermediate sections give a polygon-polygon pyramid.

One notes that if said pyramids have zero height, the total projection is a tegum-product of the bases.
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Re: Polynomials can be used find the elements of a hypercube

Postby Keiji » Sun May 26, 2013 10:29 am

Thanks both! I understand this a lot better now.

I had ruled out ** before, thinking it would produce something with n+m dimensions which led to silly results, but to consider it as producing something with n+m+1 dimensions leads to useful results.

I tend to include the 1s on either end, so # removes them and re-adds after the operation, but that's equivalent to excluding the 1s and having * add them and remove after the operation.
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Re: Polynomials can be used find the elements of a hypercube

Postby wendy » Mon May 27, 2013 7:00 am

One should note that a 'product' supposes that there is a property that multipies algebraically. Likewise 'sum' or 'progression' supposes that there is some additive term to hand, that is, something that resloves down to an algebraic addition.

There are five products, corresponding to the five regular solids. Four of these are surtope products, which means the elements of the product, is the product of the elements. This is implemented as f(AB) = f(A) f(B), by removal of 1's from various ends.

Three of the radiant constructions give rise to products: the prism or *#, the tegum or #*, and the crind. These can be implemented as functions over solids, when the solid is constructed as a radiant field, with 0 at the centre, and 1 at the surface. The octahedron, cube, and sphere in their canonical forms, might be constructed as a radiant product of the three axies from -1 to +1. The radiant value of such becomes abs(x), and the various products are

cube = max( abs(x), abs(y), abs(z))
sphere = rss( abs(x), abs(y), abs(z)) rss = root-sum-square, r² = x²+y²+z²
octahedron = sum(abs(x), abs(y), abs(z)

The three radiant products define coherent units, when multiplication is mapped onto these.

A progression is a sum-operator, specifically P(A, B, t) = A -tA + tB by coordinates. Such are the mainstay of interpreting the intermediates between sections of polytopes, of lace prisms, etc.
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Re: Polynomials can be used find the elements of a hypercube

Postby Keiji » Mon May 27, 2013 8:07 am

wendy wrote:There are five products, corresponding to the five regular solids.


Cube and octahedron I already knew about, tetrahedron you explained above... but what would the products for the dodecahedron and icosahedron be?
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Re: Polynomials can be used find the elements of a hypercube

Postby wendy » Mon May 27, 2013 10:09 am

It's actually for the sphere and the cubic lattice. The sphere is counted as regular solid, but it's not a polytope, the cubic lattice is a regular polytope in hyperbolic space. The dodecahedron belongs to the five irregular regulars (along with the icosa, the 343, the 335 and the 533.)
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Re: Polynomials can be used find the elements of a hypercube

Postby wendy » Wed May 29, 2013 10:50 am

There is the matter of 'progressions'. This is a kind of additive process that changes one polytope into another. It is used in figuring out what sort of things make intermediate sections.

For example, the progression between the icosahedron and dodecahedron, the first and second vertex-first sections of {3,3,5}, runs like this.

o5o3x becomes o3o5x by way of xa3o5xb (a+b=1). This polytope is a rhombo-icosadodecahedron. The pentagons expand, while the triangles shrink. This is what one expects as one descends a pentagonal pyramid, and ascends a triangle pyramid. What also happens, is there is a series of rectangles, of constant perimeter 2(a+b), which correspond to the sections of the line-line pyramid. These are the 30 tetrahedra whose bases are in fact edges of x5o3o and of x3o5o.

There is also an apiculation, or pyramid-raising, on the pentagons. This is caused by edges that run from o5o3x to o5o3f (girthing decagons). This edge is one of the bases of a tetrahedron, the other base is the edge of the dodecahedron x5o3o. This gives sections that raise pyramids on the twelve faces of the dodecahedron (and the intermediate xa3o5xb, as well).

Progressions can be used in some form of generalised lace-prism, which i origionally implemented as an extended pyramid product.

Code: Select all
         A2    B1    B2     B3         (A)    B1    B2   B3      (A)   X   Y   Z

         a     x3o   o3o   o3o           a    x3o  o3x  o3o       a   x   f   o
         b     o3o   x3o   o3o           b    o3o  x3o  o3x       b   o   x   f
         c     o3o   o3o   x3o           c    o3x  o3o  x3o       c   f   o   x

           simplex in 8 dim             The 2_21, or /4B         The icosahedron



We see here that the altitude A can be multi-dimensional. In the first, A is a smallish equalateral triangle, perpendicular to the spaces B1, B2, and B3. The combination of A2 + B1 + B2 + B3 gives eight dimensions.

In the second and third examples, the altitude is 0, so one gets a figure inscribed in another (like the hexagonal cupola ox6xx, which is flat in the plane). The second are a set of coordinates for a complex polytope 3{3}3{3}3 in eisenstein complex integers. It maps neatly without distortion to a tripple {3,6} coordinate system for the 2_21, which is what is given here. One can see that in B1, the figure given is a oxo3oox&x, a segmentotope corresponding to an octahedral pyramid. In practice, the two tetrahedra are the opposite faces of an octahedron, which appears in flattened pyramid (ie tegum) product, with the nine squares that form in the bi-triangular prism in x3o o3x. The squares are actually 2d tegums, which join up with the 3d tegum to make 9 of the 27 2_11 faces of this polytope.

The third is a crossing of three golden triangles to make an icosahedron.

The finding of the relevant faces happens in the same way in all cases. One draws faces from one section onto faces of an other, giving rise to prism and pyramid products, as the elements align or fall against each other.

Most of this i figured out long before i understood the dynkin symbol.
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