New Uniform tilings in H3

Higher-dimensional geometry (previously "Polyshapes").

New Uniform tilings in H3

Postby wendy » Sun May 25, 2014 9:37 am

There are a lot of new tilings in H3, all having their vertices at the horizon (aka ideal points).

In essence, any uniform polyhedron, with three or four faces at a vertex, can tile space where its vertices at infinity. This leaves {3,5}, the snub cube, and the snub dodecahedron, none of which do.

But, it does contain for instance, all the uniforms, including the prisms and antiprisms. So for example, there is a tiling of heptagonal antiprisms.

One can do a series of 'cut and pastes', of the laminate style. So for example, one can have a tiling of heptagonal prisms and antiprisms, or a tiling of truncated cubes and octahedral prisms (or antiprisms), or say alternating bands of x4x3o and x4x3x, or even x3o4x and x3x4x.

Apart from that, Roice Nelson managed to do a shot of the pd{3,5,3}. It is more beautiful than i had envisaged.
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Re: New Uniform tilings in H3

Postby wendy » Mon May 26, 2014 11:11 am

This has now been revised.

Every polygon of three or four sides tiles the plane. Using this as an ideal vertex-figure, we note that all but three of the twenty uniform polyhedra have vertex figures that tile the plane. The three that have pentagons for vertex-figures (I, sC, sD) do not tile space in this way.

The way that the tiling happens is that there are parallel lines, which means that they can be laminated. So most of the new tilings are laminates. We classify such tilings by their laminate structures, when they are not one of the seventeen given above.

The layers consist of one of the sides of the triangle, or the two unequal sides of a trapezium, or the two opposite sides of a rectangle. The middle one gives rise to an etching of alternating triangles and P-gons (ie o3xPoW4z ), while the remainder give rise to a regular tiling ie xPoW4o.

Although some of the seventeen are earlier-known, eg x3o3o6o, x3o4o4o, and x3o6o3o, they take part in the laminate processes, rather like the oct-tet tiling is both the wythoffian x3x4oAo and the laminate LA2.

So there are something like fourteen new tilings, and three further players elsewise known.

Laminate terms

A laminate tiling is one where there are non-crossing planes bounding a region like a layer. Such a thing is called a "laminatope", and is thought to be full of whatever. The exposed faces form an 'etching' on the plane. One simply joins layers of the same etching to form a tiling.

For example, there is a plane through the tiling of triangular prisms that make triangles. The etching is {3,6}. Another {3,6} comes from the layers in the oct-tet tiling. Relative to the etching, we see that the triangle prisms is symmetric, while the oct-tet tiling is not. This means that the face of the laminatope can be used as a mirror in one case, but not the other.

We designate the octtet tiling as LA2, and the prism as LPP. Both of these are wythoffian, but the LPA2 tiling (alternating oct-tet layers and prisms), is not wythoffian, and is counted as a laminate tiling.

Because of this, we can see that we *could* use the etching as a mirror, giving rise to a reversal. This is different to LA2, and does not have a wythoffian construction. We call this LB2. We can insert prismatic layers between slices of LB2, to get LPB2.

The remaining two laminates come from a {4,4} etching, which occurs in the cubic lattice, and again in the triangular lattice. The effect of a triangle prism on its 'back' is to push the squares half a unit in the x or y direction. But in order to be a complete laminate tiling, it must push it in both the x and y axis, so alternate bands of triangle prisms are rotated 90 degrees to each other, giving a tiling LC2. The prismatic form is LPC2.

In all, there is one tiling in 2d that is not wythoffian (triangle + square bands = LPC1), and five in 3d (LB2, LC2, LPA2, LPA3, and LPA4). There are eight in four dimensions (LB3, LC3, LPA3, LPB3, LPC3, LC1A2, LC1B2, and LC1C2) &c.

Here the vertex-figure is laminate, and therefore the tiling becomes laminate.

Qn - o3oPoW4z

The uniform figures that have a vertex figure of ( P Q 3 Q ) are the antiprisms, where Q=3, and the runcinates x3oPx (where Q=4). Since the Q=3 and Q=4 become as struts inside the parallel lines, we can variously alternate these. But the thing is symmetric so reversals are not allowed.

Runcinates occur for P= 4, 5, 6.

Q4, Q5, Q6 p-Antiprism x3oPx + laminate tiling of alternating layers (3 all together)
Q7+ p-Antiprism

Pn - oPoW4z

The etching is a straight layer of polygons which appear at the vertex-figure as p-gon shortchords. Like Q, they appear for all values of n, but we need only consider those values that occur outside of the prisms for the laminates.

Pn p-prism. Complete for 7,9,11,13+

P3 3p x3o3o x3o4o o3x4o o3x5o, x3o6o, o3x6o + 21 laminates
P4 x2xPo x4o3o o4x3o + 3 laminates (2 infinite families)
P5 5p o3o5x o3x5o + 3 laminates
P6 6p. o3x3x, x3x4o x3x4x x3x5o x3x5x x6o3o + 57 laminates.
P8 8p, o3x4x x3x4x x4x4o + 12
P10 10p, o3x5x x3x5x + 6 laminates
P12 12p , o3x6x x3x6x + 6 laminates

The above list shows of the fourteen representatives in this group, which of the laminates they may be part of. Some are crossed by two or three different laminates: the tiling of ID for example, consists of horizontal rows of triangles and vertical rows of pentagons. We can cut these in either way, and use the pentagon layers or triangle layers.

The vertex-figures for all of the P3 and P5 layers are symmetric, in that there is a mirror that bisects the vertex-figure orthogonal to the 3 or 5 edge. So there are no reversals. So there are 7*6/2 = 21 laminates in P3, and 3*2/2=3 laminates in the P5 group.

Reversals in P6 to P12

An example of a reversal in ordinary euclidean space, is the oct-tet tiling. This has co-planar faces, which form a {3,6} etching. If instead of following the oct-tet truss, one can use the layer-levels as mirrors, and get a different tiling (LB2), whose vertices correspond to the hexagonal close-pack of spheres.

In the present case, we have layers which are based on a series of {10} or {12}, the interior of the layers consist of triangles, whose sides correspond to one of the three uniform tilings crossed by {10, W4}.

10p \ 4 /4 \4 /4 result = decagonal prisms
o3x5x \3 /10 \3 /10 result = trunc dodecahedra
x3x5x \4 /6 \4 /6 result = trunc ID

The primary layers preserve the orientation from layer to layer, so that the 3-edges form through-lines too. However, if we use the etching as a mirror, the 3-edges form a zigzag. Therefore we count o3x5x as 'reversable'. On the other hand, a similar treatment to the decagonal-prism layer, gives exactly the same result, and so is 'symmetric'.

Placing a 'reversable' layer against a non-reversable layer, gives rise to two different tilings, only when both sides of the etching are reversable, so that ++ = -- and +- = -+.

So for this group we count 3 as with the pentagon series, and a furtger 3 from reversals of BB, BC, CC. Together 6 for each of P10 and P12

For P8, we have P8 " 8p, o3x4x x3x4x x4x4o " of which the last three are reversal, and the first symmetric.

So we have 4*3/2 = 6 of the ordinary count, and BB, BC, BD, CC, CD, DD = 6, by the reversal count, giving together 12.

For P6, we have P6 6p. o3x3x, x3x4o x3x4x x3x5o x3x5x x6o3o o3x6o x3x6x, +

Of these 6p, x6o3o, and o6x3o are symmetric, and x3x3o, x3x4o, x3x4x, x3x5o, x3x5x x3x6x, are reversable. This gives 36 by the first layer, and 21 by the second count, all together, some 57 examples.

All together, g1 brings together some 42+69=111 laminates.
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Re: New Uniform tilings in H3

Postby wendy » Tue May 27, 2014 10:02 am

I fiddled around getting the calculations right. The method was right, but the actual consists are rather hard to do in the head. I used pen and paper, several calculations, to the same result.

Any triangle or quadralateral, can, by rotation on its edges, tile the euclidean plane. The symmetry is 2 2 2 2 in both cases: that is, the tiling is of the nature of snubs and wrap-snubs. The edges of of the triangle or quadralateral can be of any measure that forms such a figure in the euclidean plane.

Most of the uniform polyhedra have vertex-figures of this nature, and so there is a tiling of these uniform polyhedra, with ideal vertices (horoteelons) that have the 2 2 2 2 symmetry. The ones with five or six polygons need to be addressed separately.

The following table shows the 20 spherous (compact), and 11 horous (paracompact) uniform polytopes.

Code: Select all
  1   C     4,3,6    9   ID   17   tCO             24    SH 
  2   Pp     (u)    10  rCO   18   tID             25   rSH
  3   Ap     (u)    11  rID   19    sC   (*)       26    tQ
  4   T     3,3,6   12   tT   20    sD   (*)       27    tH
  5   O     3,4,4   13   tO   ---------------      28   tSH
  6   I      (*)    14   tC   21     Q  4,4,4      29    sQ  (*)
  7   D     6,3,5   15   tI   22     S  3,6,3 #    30    sH  (#)
  8   CO            16   tD   23     H  6,3,6      31  LPC1  (*)
  p,q,r    This construction has a wythoff construction, as given.
   (u)     An infinite class, for p>5, and P3, A4. 

   (*)     The vertex figure does not tile the euclidean plane.
    S      The vertex figure tiles, but leads to no apeirohedra
   sH      The vertex-figure tiles with vertices of two kinds.   

Most of these are laminate: they have one or more classes of apeirohedra (planes made entirely of cell-faces), suitable for making laminatopes. All of the examples have reflexes in the ordinary uniform tilings, so we should use these to explain etc.

An apeirohedron is made of hedra laid without a perimeter: that is, an unbounded cover of polygons. While one might consider these in isolation, they occur in higher structures too. For example, all of the faces of a cubic fall into apeirohedra {4,4}. The oct-tet tiling has apeirotopes {3,6}. There is a tiling of triangular prisms, which has squares falling in {4,4} and the triangles in {3,6}. The tiling of hexagonal prisms has apeirohedra of the form {6,3}, but the squares form no such apeirohedron.

S = {3,6,3} forms something like the tiling of tO, which lets no apeirohedra, even though the tiling exists.

sH = s3s6s, forms something like x3x3o3o3z, a tiling of tetrahedra and truncated tetrahedra. All faces fall on apeirohedra, but you can not select a set of parallels that contain all vertices. It does not form a uniform laminate.

In this count, we ignore the wythoffs, the non-functional ones, and count the rest.

(u) 2 ( ) 16 Together, 2u+16.

Vertex Figures

All triangles tile the plane under the orbifold 2 2 2 2, such that there are through-lines that contain lines of one type (ie a or b or c).

The quadralaterals to form through-line tilings under 2 2 2 2 are parallelograms and trapezia. In the first case, it is as if were formed from two triangles, the second case alternates the two bases.

All of the quadralateral vertex-figures are of this type.

Code: Select all

       P    3     P    3                           P   P    P
    o------o--o------o--o--                     o----o----o----
     \    /    \    /    \  str = 3 or 4      Q |    |   xPoQx has
    --o--o------o--o------o--                   |    |   this form
     /    \    /    \    /                      o----o
    o------o--o------o--o----                     P

For triangles, the triangle can be placed on the apeirohedron so that they are assymetric. The example is the oct-tet tiling, whose continuation over the {3,6} is not a mirror image. This leads to the A and B laminates, ie LA2 and LB2.

Code: Select all

  \ //   \ //   \ //       \//    \//    \//
--o------o -----o----    --o------o------o--
  // \   // \   //\        /\\    /\\    /\\
       Normal (A form)      Reversed (B form)
     LA2 = {3,4,A}              LB2.

The polytopes mentioned above are all 'A' form, there are crossing apeirogons formed by the three kinds of edge. The B form is like LB2, the tiling with the close-pack vertices.

Laminate Polytopes

We now turn to laminates. One can cut along a series of apeirotopes, such that every vertex is exposed. The apeirotopes would need to be identical. Given such a set, one can create alternating layers as long as the etching is identical. That is, one can make a tiling of alternating triangular-prisms and oct-tet layers, etc.

Code: Select all

  \ //   \ //   \ //   A   \//    \//    \//    A
--o------o -----o----    --o------o------o--   
   |      |      |     P    |      |      |     P
--o------o -----o----    --o------o------o--
  // \   // \   //\    A   /\\    /\\    /\\    Ar
      Normal (A form)      Reversed (B form)
           LPA2                  LPB2.

Here there are alternating layers of a symmetric form P, with a reversable set A, Ar. Note here that the A's must be equal, and the P's must be equal.

The number of Laminates

The following table shows all possible apeirohedra, along with the symmetric (f) and assymetric (h) tilings that support this kind of layer. There are five kinds of laminates, which have these numbers.

C LC2 f(f+1)/2 symmetric, excluding XX
A LA2 h(h-1)/2 normal assymetric, excluding XX
B LB2 h(h+1)/2 reversed assymetric, including XX'
PA LPA2 fh normal extended
PB LPB2 fh reversed extended.

The sum of the last four is h(h+2f).

We then list the various apeirohedra, which are p = xPoUo and P3 = o3xPoUo. These are lines variously of alternating 3,P and of simple P.

Code: Select all

   3.    P3   T  O   CO  ID
         tT  tC  tD  SH  tQ    45
   4.    C   CO  tO  Q   tQ    10
   4:    (Pp)   tCO tID tSH          39
   43.   A4  rCO                1
   5.    P5  D   ID  tI         6
   53.   A5  rID                1
   6.    P6  H   SH             3
   6:    tT  tO  tI tCO tID tSH      72
   63.   A6  rSH                1
   8.    P8                     0
   8:    tC tCO  tD                  15
  10.    P10                    0
  10:    tD  tID                      8
  12.    P12                    0
  12:    tH  tSH                      8
                              67    142  = 209

The above table gives the 'sundries'. Specifically, 39 is found with f=5, h=3, excludes the infinite families. When we put f=5,h=4, one gets 56, meaning that there are 17 classes of infinite laminates. This includes the reversed layering of (Pp) / (Pp), which we shall remove from here, to get 16 infinite families ie 16u.

The form Pp can alternate with Pq, both normal and reversed, which leads to two bi-infinite classes. So the total of laminates is

Primitives 2u + 16
Laminates 2u^2 + 16u + 209
Total 2u^2 + 18u + 225 or 245 separate entries.
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Re: New Uniform tilings in H3

Postby Marek14 » Tue Apr 28, 2015 12:29 am

Hm, an idea -- not uniform, as it contains nonuniform cells solids, but scalliform:

Imagine we start with a square tiling x4o4o. We know that this is isomorphic to rectified form o4x4o and with small rhombated form x4o4x. In hyperbolic space, this tiling exists on a horosphere.

How would a lace tower look?

x4o4o || o4x4o: contains square pyramids and square antiprisms. And since we can have infinite sequence of concentric horospheres with constant separation, we can make a laminate out of this -- every vertex is surrounded with two square antiprisms and two square pyramids on the "inner" side of the horosphere, and with four square antiprisms and one square pyramid on the "outer" side.

x4o4o || x4o4x: contains cubes, triangular prisms and square pyramids. The laminate would have each vertex surrounded by one cube, two triangular prisms and one square pyramid on the inner side, and by four cubes, four triangular prisms and one square pyramid on the outer side.

o4x4o || x4o4x: just another form of x4o4o || o4x4o

We can find two other laminate tilings that work this way:

x4x4o || x4x4x: contains octagonal prisms, triangular prisms and square cupolas. There might be some blending between square cupolas and octagonal prisms.

o3o6x || x3x6o: contains hexagonal pyramids, tetrahedra and hexagonal antiprisms.
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Re: New Uniform tilings in H3

Postby Klitzing » Tue Apr 28, 2015 2:39 pm

Marek14 wrote:Hm, an idea -- not uniform, as it contains nonuniform cells solids, but scalliform:

Hmm, x4o4o || o4x4o cannot be scaliform. Most probably it could be compared somehow to J37 (esquigybcu): even so having all identical vertex figures, these vertices still are not globally equivalent. In fact, as you said, at the top-side of each infinite segment (between 2 neighbouring square tilings) you'll get 4 antiprisms and the tip of a pyramid per vertex, while on the bottom side of each segment you'll get 2 antiprisms and 2 base corners of the pyramid per vertex. Thus in total you have a ratio of 1:2 of tip to base corner of the square pyramids, while a scaliform figure rather should use a ratio of 1:4, ain't it?

The problem here won't be so much the pyramids. Because at each base corner there will be a further pyramid incident by its tip. But consider the antiprisms. When starting with a top pyramid, its base square has to connect to an antiprism. But from the opposite side it has to adjoin a further antiprism. And this then continues for ever on. Thus we always have one sided yeakles, starting with a pyramid atop an one sided infinite tower of antiprisms.

Thus we have to divide the squares into red (pyr-ap) and green ones (ap-ap). Accordingly we likewise have to color the antiprisms themselves into A ones (red-green) and B ones (green-green). Or rather we even could apply an infinite-coloring according to pyramids = 0, and each antiprism will be labeled according to its number of to be crossed squares up to the final pyramid, i.e. having yeakle towers as 0-1-2-3-4-...

And as we thus have infinitely many to be distinguished antiprisms, but at every vertex there are just 6 (2 from above, 4 from below), these vertices too are to be diversed at least according to such order 6 subsets of integers.

Here we are: the vertices are disproved to be globally equivalent in contrast to the required vertex transity constraint

Your x4o4o || x4o4x even has a pyramid vertex ratio of 1:1 of tip to base corner per vertex. Also not the required 1:4. Thus it most probably lacks a similar problem as outlined above for the other laminate.

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Re: New Uniform tilings in H3

Postby Marek14 » Tue Apr 28, 2015 3:16 pm

Well, yeah, it's not scalliform. Instead it has a curious structure where you have square pyramids at the bottom of infinite towers of antiprisms. Each tower of antiprisms definitely has a bottom. Maybe there can be one bottomless tower (which leads to a question -- are there two versions of this structure, one with a bottomless tower and one without?).

Now, you say that the tip to base corner is 1:2, while it should be 1:4. I am not actually sure of that. The problem is that while it surely HAS to be 1:4 for finite structures, this might not be the case in the infinity.

You see, the 1:4 ratio is not limited to scalliform figures -- EVERY polychoron, scalliform or not, that contains square pyramids, MUST contain four times as many (3,3,4) corners as (4,4,4,4) corners.

And for shapes like J37 (is there a name for shapes constrained only by identical vertex figures?), this must hold as well.

The reason why my tiling has different ratio that 1:4 has nothing to do with scalliformity -- not being scalliform, by itself, wouldn't allow it to change the ratio. The ratio is different because the figure is infinite.

Consider: Each square pyramid has apex in a "lower" layer and four vertices of the base in the "upper" layer. But what happens if you overlay o4x4o over x4o4o? The vertices of o4x4o lie on edges of x4o4o. Since each square has 4 edges and each edge is shared by 2 squares, any finite region of this tiling will have twice as many vertices in the upper layer as in the lower layer. THIS is why the ratio is different -- although both layers have the same (infinite) amount of vertices, their local approximations have ratio of 2:1.
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