infinte repetitions of some finite layer sequence

[Klitzing]

 
Euclidean honeycombs and higher dimensional space filling like tetracombs etc. often can be seen as an infinte repetition of some finite layer sequence.
This deserves an own notational concept.


Well we have already lace towers as a notational aid to denote stacked vertex layers in finite polytopes. Thus e.g. ofx-&#xt denotes a regular pentagon, xux-&#xt denotes a regular hexagon, xfoo-5-oofx-&#xt denotes a regular dodecahedron, xxxx-4-oxxo-&#xt denotes a small rhombicuboctahedron, etc. (Dashes here only were inserted for the ease of reading, those are not a part of the notation itself.) Thus we have already some notational aid wrt. the multilayer concept, esp. when cosidering true facets (at the antipodal ends of that axial direction) and about pseudo facets as the other vertex layer sections, as well as wrt. to the lacing elements.

But when it comes to some kind of repetinioning, we are still without a clou on how to denote that. - For that reason, I'd like to take some rescue to the notation in musics. There too sometimes some passages are to be repeated. Notationally those usually are denoted that way:

Code: Select all

||: .... | .... | .... | .... :||

So what about enclosing repeating layers in our context as well within ':' characters? And, in order to provide a further additional clue on how often the repetition would have to take place, we could add at the end of that axial notation part ('&#xt') the corresponding repetition number, or, in case of our infinite structures, simply an 'i' (for 'infinitely often').

Thus chon = x4o3o4o would become chon = :x:4:o:4:o:&#xti or alternatively (in a different orientation) as chon = :qoo:3:oqo:3:ooq:3*a&#xti. And rich = o4x3o4o could be denoted as rich = :oq:4:xo:4:oo:&#xti = :xxo:3:xox:3:oxx:3*a&#xti. I think, that these few examples already make the concept quite clear, and that this concept does allow for lots of applications.

(For sure, whenever we have just a single repetition layer, as in the first provided example, this is just a different description of a cross-product with aze, i.e. :x:4:o:4:o:&#xti = x-infin-o x4o4o.)

Wendy, as I suppose, would make lots of use from this new notational concept, as she already had considered lots of high-dimensional lattice geometries, esp. those A_n, D_n and E_n geometries, and has deduced from such repeating layerwise considerations most of her layerwise understandings of the corresponding cell polytopes. So it would be quite easy to her, to put her already derived applications into that new notational representation.

Esp. wrt. her 'laminates', which she so far only denotes by some L (laminate), some letter (representing type) and optionally some number (representing subtype) - e.g. LA1. That L-notation might serve for her purposes quite well, as she knows what these character-number-suffixes are chosen for, but for all the other willing readers that notation is a bluddy horror. I suppose, that this - much more Dynkin style - notation finally could help out of that mess too.


The most important thing here, for sure, is that this new notation then allows for a corresponding concise description of all the respective subelements as well, which no longer have to take refuge to totally different notational concepts any more.

--- rk

[wendy]

The laminate tilings in euclidian space is quite thorny, and using the CD diagram globally simply does not work.

Something like LC4C5 repeats over forty layers, even though it's a ten-dimensional figure. It has the same vertex figure and surtope consist as LC3C6, (12 layers), LC2C7 (28 layers), LC1C8 (16 layers) and LC9 (nine layers). You can't tell what you are looking at until you get a repetition in the same parity.

The uniform laminates in euclidean space, are based on layers A, B, C, and P. A is 'advance' and B is 'backwards', both apply to layers of the A_n tiling. In the A mode, the next layer is placed above the 'forward' motion of the tiling, in the hexagonal, the 'upwards-pointing simplex'. LA2 is simply the oct-tet tiling. The B is backwards, so if one layer is on an up triangle, the next is on a down one, giving the hexagonal close-pack LB2, for which the verf is xxx3o3o&#xt.

When one introduces a prism layer, these become LPA2 and LPB2. The vertex figure here is x3o || x3x |q| o3o in both cases, but they do not correspond.

The C layer is a shunt of cubes in one of the available axies of the layer. These are formed by a layer of triangle-prisms, in the form ox2xx, or xo2xx. These would push it half an edge in the x or y axis. This supports primatic layers.

The alternate bands of squares and triangles, is LPC1. LC1, like LA1 and LB1, makes bands of triangles, is no different to the triangular tiling.

In 3d, you have LC2 (layers of digonal cupolae = triangular prisms) that push the etching 1/2 X or 1/2 Y-wards. To these, one adds the prismatic layer of cubes, to get LPC2.

In 4d and higher, it is possible to have bands that operate in different etchings. When two different bands are chosen, one can not have prismatic layers, though. So LA2C1, would project onto 3d as LPA2, with short prisms. What is happening is that between each layer of the oct-tet truss, there is a shuttle pushing it in and out relative to the W axis. LA2C1 and LB2C1 have the same vertex figure, and it is only after four layers that one gets a difference.

In the hyperbolic cases, the etchings (markings on the layer surfaces), are pretty restricted and not scalable. None the same, one must entertain the possibility that the etchings can belong to several different figures, and in the case of the octagon ball, 4 different layers are represented at each vertex.

[Klitzing]

Thus we'd have LA1 = LB1 = LC1 = trat =

Code: Select all

:xo:∞:ox:&#xti   (N → ∞)   → heights = sqrt(3)/2 = 0.866025

 o. ∞ o.     | N * | 2  2 0  2 | 2 1 2 1
 .o ∞ .o     | * N | 0  2 2  2 | 1 2 1 2
-------------+-----+-----------+--------
 x.   ..     | 2 0 | N  * *  * | 1 0 1 0
 oo ∞ oo &#x | 1 1 | * 2N *  * | 1 1 0 0
 ..   .x     | 0 2 | *  * N  * | 0 1 0 1
:oo:∞:oo:&#x | 1 1 | *  * * 2N | 0 0 1 1
-------------+-----+-----------+--------
 xo   .. &#x | 2 1 | 1  2 0  0 | N * * *
 ..   ox &#x | 1 2 | 0  2 1  0 | * N * *
:xo: :..:&#x | 2 1 | 1  0 0  2 | * * N *
:..: :ox:&#x | 1 2 | 0  0 1  2 | * * * N

--- rk

[Klitzing]

And what then is squat in your laminate notation?
(Would it perhaps LP1?)

In this repetitive layers notation, it clearly could be described as:

Code: Select all

:x:∞:o:&#xti   (N → ∞)   → heights = 1

 o ∞ o       | N | 2 2 | 4
-------------+---+-----+--
 x   .       | 2 | N * | 2
:o:∞:o:&#x   | 2 | * N | 2
-------------+---+-----+--
:x:  . &#x   | 4 | 2 2 | N


or, in different orientation, it well could also described as a 2 layer repetitive pattern:

Code: Select all

:qo:∞:oq:&#xti   (N → ∞)   → heights = 1/sqrt(2) = 0.707107

 o. ∞ o.      | N * |  2  2 | 2 2
 .o ∞ .o      | * N |  2  2 | 2 2
--------------+-----+-------+----
 oo ∞ oo &#x  | 1 1 | 2N  * | 1 1
:oo:∞:oo:&#x  | 1 1 |  * 2N | 1 1
--------------+-----+-------+----
:qo: :..:&#xt | 2 2 |  2  2 | N *
:..: :oq:&#xt | 2 2 |  2  2 | * N

--- rk

[Klitzing]

And this then would be etrat = LPC1:

Code: Select all

:xxoo:∞:ooxx:&#xti   (N → ∞)   → heights alternatingly = sqrt(3)/2 = 0.866025 resp. = 1

 o... ∞ o...     | N * * * | 2 1 0  0 0 0 0  2 | 2 0 0 0 2 1
 .o.. ∞ .o..     | * N * * | 0 1 2  2 0 0 0  0 | 2 2 1 0 0 0
 ..o. ∞ ..o.     | * * N * | 0 0 0  2 2 1 0  0 | 0 1 2 2 0 0
 ...o ∞ ...o     | * * * N | 0 0 0  0 0 1 2  2 | 0 0 0 2 1 2
-----------------+---------+-------------------+------------
 x...   ....     | 2 0 0 0 | N * *  * * * *  * | 1 0 0 0 1 0
 oo.. ∞ oo.. &#x | 1 1 0 0 | * N *  * * * *  * | 2 0 0 0 0 0
 .x..   ....     | 0 2 0 0 | * * N  * * * *  * | 1 1 0 0 0 0
 .oo. ∞ .oo. &#x | 0 1 1 0 | * * * 2N * * *  * | 0 1 1 0 0 0
 ....   ..x.     | 0 0 2 0 | * * *  * N * *  * | 0 0 1 1 0 0
 ..oo ∞ ..oo &#x | 0 0 1 1 | * * *  * * N *  * | 0 0 0 2 0 0
 ....   ...x     | 0 0 0 2 | * * *  * * * N  * | 0 0 0 1 0 1
:o..o:∞:o..o:&#x | 1 0 0 1 | * * *  * * * * 2N | 0 0 0 0 1 1
-----------------+---------+-------------------+------------
 xx..   .... &#x | 2 2 0 0 | 1 2 1  0 0 0 0  0 | N * * * * *
 .xo.   .... &#x | 0 2 1 0 | 0 0 1  2 0 0 0  0 | * N * * * *
 ....   .ox. &#x | 0 1 2 0 | 0 0 0  2 1 0 0  0 | * * N * * *
 ....   ..xx &#x | 0 0 2 2 | 0 0 0  0 1 2 1  0 | * * * N * *
:x..o: :....:&#x | 2 0 0 1 | 1 0 0  0 0 0 0  2 | * * * * N *
:....: :o..x:&#x | 1 0 0 2 | 0 0 0  0 0 0 1  2 | * * * * * N


--- rk

[Klitzing]

Stepping up to the next dimension, LA2 = octet =

Code: Select all

:xoo:3:oxo:3:oox:3*a&#xti   (N → ∞)   → all heights = sqrt(2/3) = 0.816497

 o.. 3 o.. 3 o.. 3*a    | N * * |  6  3  3  0  0  0 | 3 3  6  3  6  3 0 0  0  0 0 0 | 3 3 1 3 3 1 0 0 0
 .o. 3 .o. 3 .o. 3*a    | * N * |  0  3  0  6  3  0 | 0 0  3  6  0  0 3 3  6  3 0 0 | 3 1 3 0 0 0 3 1 3
 ..o 3 ..o 3 ..o 3*a    | * * N |  0  0  3  0  3  6 | 0 0  0  0  3  6 0 0  3  6 3 3 | 0 0 0 1 3 3 1 3 3
------------------------+-------+-------------------+-------------------------------+------------------
 x..   ...   ...        | 2 0 0 | 3N  *  *  *  *  * | 1 1  1  0  1  0 0 0  0  0 0 0 | 1 1 0 1 1 0 0 0 0
 oo. 3 oo. 3 oo. 3*a&#x | 1 1 0 |  * 3N  *  *  *  * | 0 0  2  2  0  0 0 0  0  0 0 0 | 2 1 1 0 0 0 0 0 0
:o.o:3:o.o:3:o.o:3*a&#x | 1 0 1 |  *  * 3N  *  *  * | 0 0  0  0  2  2 0 0  0  0 0 0 | 0 0 0 1 2 1 0 0 0
 ...   .x.   ...        | 0 2 0 |  *  *  * 3N  *  * | 0 0  0  1  0  0 1 1  1  0 0 0 | 1 0 1 0 0 0 1 0 1
 .oo 3 .oo 3 .oo 3*a&#x | 0 1 1 |  *  *  *  * 3N  * | 0 0  0  0  0  0 0 0  2  2 0 0 | 0 0 0 0 0 0 1 1 2
 ...   ...   ..x        | 0 0 2 |  *  *  *  *  * 3N | 0 0  0  0  0  1 0 0  0  1 1 1 | 0 0 0 0 1 1 0 1 1
------------------------+-------+-------------------+-------------------------------+------------------
 x.. 3 o..   ...        | 3 0 0 |  3  0  0  0  0  0 | N *  *  *  *  * * *  *  * * * | 1 0 0 1 0 0 0 0 0
 x..   ...   o.. 3*a    | 3 0 0 |  3  0  0  0  0  0 | * N  *  *  *  * * *  *  * * * | 0 1 0 0 1 0 0 0 0
 xo.   ...   ...    &#x | 2 1 0 |  1  2  0  0  0  0 | * * 3N  *  *  * * *  *  * * * | 1 1 0 0 0 0 0 0 0
 ...   ox.   ...    &#x | 1 2 0 |  0  2  0  1  0  0 | * *  * 3N  *  * * *  *  * * * | 1 0 1 0 0 0 0 0 0
:x.o: :...: :...:   &#x | 2 0 1 |  1  0  2  0  0  0 | * *  *  * 3N  * * *  *  * * * | 0 0 0 1 1 0 0 0 0
:...: :...: :o.x:   &#x | 1 0 2 |  0  0  2  0  0  1 | * *  *  *  * 3N * *  *  * * * | 0 0 0 0 1 1 0 0 0
 .o. 3 .x.   ...        | 0 3 0 |  0  0  0  3  0  0 | * *  *  *  *  * N *  *  * * * | 1 0 0 0 0 0 1 0 0
 ...   .x. 3 .o.        | 0 3 0 |  0  0  0  3  0  0 | * *  *  *  *  * * N  *  * * * | 0 0 1 0 0 0 0 0 1
 ...   .xo   ...    &#x | 0 2 1 |  0  0  0  1  2  0 | * *  *  *  *  * * * 3N  * * * | 0 0 0 0 0 0 1 0 1
 ...   ...   .ox    &#x | 0 1 2 |  0  0  0  0  2  1 | * *  *  *  *  * * *  * 3N * * | 0 0 0 0 0 0 0 1 1
 ..o   ...   ..x 3*a    | 0 0 3 |  0  0  0  0  0  3 | * *  *  *  *  * * *  *  * N * | 0 0 0 0 1 0 0 1 0
 ...   ..o 3 ..x 3*a    | 0 0 3 |  0  0  0  0  0  3 | * *  *  *  *  * * *  *  * * N | 0 0 0 0 0 1 0 0 1
------------------------+-------+-------------------+-------------------------------+------------------
 xo. 3 ox.   ...    &#x | 3 3 0 |  3  6  0  3  0  0 | 1 0  3  3  0  0 1 0  0  0 0 0 | N * * * * * * * *
 xo.   ...   oo. 3*a&#x | 3 1 0 |  3  3  0  0  0  0 | 0 1  3  0  0  0 0 0  0  0 0 0 | * N * * * * * * *
 ...   ox.   oo.    &#x | 1 3 0 |  0  3  0  3  0  0 | 0 0  0  3  0  0 0 1  0  0 0 0 | * * N * * * * * *
:x.o:3:o.o: :...:   &#x | 3 0 1 |  3  0  3  0  0  0 | 1 0  0  0  3  0 0 0  0  0 0 0 | * * * N * * * * *
:x.o: :...: :o.x:3*a&#x | 3 0 3 |  3  0  6  0  0  3 | 0 1  0  0  3  3 0 0  0  0 1 0 | * * * * N * * * *
:...: :o.o: :o.x:   &#x | 1 0 3 |  0  0  3  0  0  3 | 0 0  0  0  0  3 0 0  0  0 0 1 | * * * * * N * * *
 .oo 3 .xo   ...    &#x | 0 3 1 |  0  0  0  3  3  0 | 0 0  0  0  0  0 1 0  3  0 0 0 | * * * * * * N * *
 .oo   ...   .ox 3*a&#x | 0 1 3 |  0  0  0  0  3  3 | 0 0  0  0  0  0 0 0  0  3 1 0 | * * * * * * * N *
 ...   .xo 3 .ox    &#x | 0 3 3 |  0  0  0  3  6  3 | 0 0  0  0  0  0 0 1  3  3 0 1 | * * * * * * * * N

--- rk

[wendy]

The current lace tower notation handles it quite well, except for the coda. The coda is here ##, means that a cycle of the lace-tower is given.

LPC1 xxooUooxx&##

LA2 xoo3oxo3oox3z&## = x3o3o3o3z
LB2 xo3ox3oo3z&##
LPA2 xxoooo3ooxxoo3ooooxx3z&##
LPB2 xxoo3ooxx3oooo3z&##
LC2 xxooUooxx2oxxoUxoox&##
LPC2 xxxxooooUooooxxxx2ooxxxxooUxxooooxx&##

LA3 xooo3oxoo3ooxo3ooox3z&## = x3o3o3o3o3z
LB3 xo3ox3oo3oo3z&##

[Klitzing]

I well could adopt that new coda of yours ...

But then, is that second hash a leading or a trailing one?
That is, when also providing the specific length of to be used lacing edges (x, q, f, ...), is it more some &#x#, or rather a &##x ?

--- rk

[wendy]

Suppose it could be trailing, eg ##x I neglected the lacing-edge.

[Klitzing]

One more question.
Or, rather, quest for some further helping advice.

For more or less spherical polytopes the instances, where corealmic cells would have to be encountered, within all are negelectable. Here, in purely euclidean geometry, this clearly is rather the opposite. Thus all these multy-layer-notations - in general, but esp. here - have the problem, that those do not tell anything about usage or rejection of such joining faces.

E.g. etoh, the elongated tetrahedron-octahedron honeycomb, clearly uses tetrahedra, octahedra, and triangular prisms. At any prism on one triangular side attaches a tetrahedron, at the other an octahedron. - But we well could combine those pairings of prism and tetrahedron into a single cell each, i.e. consider etripies (J7) and octahedra only. The corresponding honeycomb then was coined to be gyeditoh. - But both honeycombs would be described by  :xxoooo:3:ooxxoo:3:ooooxx:3*a&##x.

Or, when considering gyetoh, the gyroelongated tetrahedron-octahedron honeycomb, then the same cells are to be used as for etoh, but now at half of the prisms tetrahedra attach on both its triangular sides. - Again those local configurations each could well be melted into a single cell then, into etridpies (J14). Thus there is a different honeycomb here too, having the same edge skeletton, but with cells being etridpies, octs, and trips (the remaining ones). - Both these honeycombs then ought be described by  :xxoo:3:ooxx:3:oooo:3*a&##x.

Thus, we clearly need a further discrimination, I suppose. Else those multy-layer-notations would not describe unique cellstructures, only unique vertex sets ...

--- rk

[wendy]

Technically the lace notation can only go as far as cells bound in the layers. Even something like xfo3oox&#x really describes xf3oo&#x + fo3ox&#x. It's the magic of what happens afterwards.

1. In a lace tower, we are supposing it's one polytope, and we imagine that the f3o triangle can be removed. This supposes that xfo&#x is actually coplanar, which from this instance we know is. We also look for missed edges, such as in xof3ooo5oxo&#xt, which has an extra set of lacing from layers a to c. You could show this in Klitzing-style ac **ac i suppose.

2. Secondly, we imagine that it is possible to merge cells on the surface, so xfo&#x is not a trapezium and triangle but a pentagon.

Of course, in the present instance, we are still supposing a polytope, but this is simply the surface, and nothing is happening inside.

The angles are all 1/2 circle, but the test for uniformity here is that each cell has a vertex-tangent sphere. This means that it freely deforms correctly in hyperbolic space.

None the same, it is possible to remove internal layers, such as xxoo4xxxx4ooxx&##x, which leads directy to a tiling of square cupolae, octagonal prisms, tetrahedra and cubes, but the first two merge into x4o3x, and the tiling is x4o3xAo. In your case, you want to merge various layers into CRF polytopes, such as oxxo4oooo&#xt, or even oxx2xxo&#xt. Both would work, but the notation is not really up to differentiating these by surtope removal.

It should be noted that past nine or so dimensions, the actual cell consist of the dense packings really are not known. It was many years before the 22 deep holes of the Leech lattice was classified, but i know an easy proof that there is no copy of it that is no closer than 1/q to the it (like the body-centred points of the square lattice). The resulting lattice, did it exist, would exceed the maximum packing density in 120 dimensions when sixteen copies of the fifth-moment are applied.

Still, I have not heard much of the Coxeter-Todd lattice, it is slightly more efficient than my best effort (it is 64/27 q-units, where i can only get 54/27 q-units).

[Klitzing]

Yep, you are totally right. It is kind a true magic, what applies when considering multi layer descriptions of polytopes - or, in our case, of honeycombs etc.
For, those descriptions mainly describe vertex sets only, at most, by means of lacings, some edges are described additionally. But the rest of its cell filling remains more or less undefined.


It is more our aim of finding CRFs e.g., which then reminds us for example that all f spaced vertex pairs would be considered false edges. Moreover, when considering convex polytopes, cells usually would not become corealmic very often. And, in the rare cases, where those do none the less, these just are the exceptional cases to be considered of special interest. E.g. when implementing some Johnsonian solids instead of Wythoffians only. Therefore, in those spherical geometries, we usually are silently assuming cells to be maximal.

We well could drag this mindset over to the Euclidean geometry as well. But then we would trail away most of such descriptions of the uniform tesselations. Simply for the evident reason that in Euclidean geometry the corealmic adjoin of cells is no longer exceptional, rather it is standard.


Thus, I think we shall have to take refuge to some additional adjectivic attribution, describing which cell filling is to be understood in every single case.
I.e. specifying whether we are aiming for maximized cells, for minimized ones (provided this would be uniquely possible, and this solution moreover is conform to the describing symmetry), for uniform cells only, or any further ones the like.

Thus we would have the need for some well chosen addressing of either of these possibilities. That is, according adjectives ought be well chosen and then settled in an unambiguous way.
Any input from either side on how we could come up with such?


By doing so we thus kind of have to concede that such multi layer Dynkin symbols finally by no means are that strong than the single layer ones had been. Infact, the main strength of the single layer Dynkin symbols just was, that within this tiny symbol the whole structure was coded completely. For those multi layer symbols OTOH we now observe that those give room for some arbitrariness.

This by no means is restricted to that new &##x coda. That disambiguity, I suppose, would be all the same for &#x, &#xt, &#zx, etc.

--- rk

[wendy]

You can remove cell walls on a simple node-layer basis. This is because the walls that one removes to join the layers, always surround exactly one node, and one layer. You simply indicate, with a non-spacing rune (here, r), which has the effect of removing the wall between the cells above and below it.

xxo3xox3oxx3z&##x is a tiling of triangular cupolae and octahedra, the etching is full, is o6x3o.

xxor3xorx3orxx3z&##x removes the walls that form at the nodes marked r. These here are hexagons, and the resulting tiling is one of cuboctahedra and octahedra. It is in fact, the same as o4x3o4o.

xx3xor3orx&##x is a tiling of xxx3oxo&#x (it's a double-cupola), and octahedra. The hexagons at the base of the cupolae have been removed.

[Klitzing]

Wow Wendy, once more a brilliant idea on how to solve that problem!

But sadly there are some minor issues in there, which ought to be still optimized.
Thus I'll take it more as a first shot.

First of all that 'r' clashes with your zero (+ epsilon) edge size, also being called 'r'. So we should introduce an different unambiguous letter / operant here.

Then, as I recognize, that 'r' is not an own edge symbol itself, making up an own further layer, instead it is meant as a postfixing operant onto that true edge mark - 'o' in those cases you provided. Moreover, it then does not apply to that node 'o' itself, instead it applies to the hexagon x3x, which occurs here as the full layer, except that thus marked 'o': x3x3or3*a.

Thus, this is deciphered and understood. But sadly that complementary marking only works for the &##x coda. It does not work e.g. for the &#xt coda:
How could such an 'r' be applied to xxo3oxx&#xt in order to result in the cuboctahedron instead of a mere stack of 2 cupolae?

Moreover, my feeling would prefer more the opposite approach. So far usually maximized cells are understood (that co is already "known" to be xxo3oxx&#xt, teddi is "known" to be ofx3xoo&#xt, and similarly for all that &#zx stuff too). So it would be somehow better to have a notational aid in order to explicitly insert separating membranes... (or ought we have both?)

--- rk

[wendy]

'r' does not so much clash with the other uses it has. In one case, you simply don't use double-point lines in a lace tower, in the other case, using 'r' to reduce p/d does not make sense in the lower case.

It works in both &# and &##, but the thing to realise here is that the first we are usually talking of polytopes, and the second, we are talking of tilings. In essence, the r-mark can only remove walls between cells, and is perfectly workable in xfo3oorx&#tx, where the r removes the 'f' etching that would fall here.

Tilings are in effect, only surfaces of polytopes, and the CD diagram only ever describes a tiling / polytope-surface. But then 'polyhedra' is only a description of a surface, not a solid.

When we create the solid, we suppose that it's permissible to join cells whose angles remain 180deg together, etc, so it's not needed to show the r in xfo3oorx&#xt. If on the other hand you describe the teddy as a stack of segmentohedra, then it is perfectly possible that the pentagons = trapezium + triangle, as xf&#x + fo&#x. The way that i resolve many of these figures is simply to write the columns out in a table, and see if there are removable walls, ie hunt down the possible 'r' locations.

But calling it a 'polytope', you effectively have removed all but one content (upper incidence). You don't need to do anything to the symbol: the name already does it.

The idea of making it a non-edge, is because the edges themselves define the return to the next row, and if 'r' or any other symbol was also doing it, you could not read the table. It is not an edge modifier, in the way that i does (i is in effect a post-mark negative sign, so 'fi' creates an edge of -1.618&c. It's the magic behind the spreadsheet thing). So r is a kind of notation to say that the wall that normally forms at this point is removed. The actual shape of the wall, is set by all of the remaining nodes.

[Klitzing]

hmmm, it kind of converges then.
You're right, in ofx3xoo&#xt I was fixed too much to the f3o, but it is already the f edge itself, which has to be suppressed.

- But still, I'm not fully convinced.

That inclusion or removal could occur in several different dimensional elements at the same time. Consider some .4.3....3.3.3.4. symmetry, where ...3.3.3.4. elements become subelements of further such elements. (Could that add some problems here, or would that one just be the same thing as for the teddi example above?)
But then we might have also different inter layer cells to be united or separated (or even one yes, one not) while crossing the same layer level.

Wrt. the 'r' symbol I'd then suggest a '\' character instead. - It both is used in informatics (in programming languages, RegEx, or even in LaTeX too) to mask some other elements, and in mathematical set theory as a negation or exclusion sign. Usually that then is placed before the to be excluded thingy, whereas you promote postfixing. That in turn could be supported by the Reversed Polish Notation, so would be admittable, I think.
And additionally, this '\', in the postfixed version, might also have the connotation of an ASCII shorthand for a downward, i.e. dropping arrow.
(Hope that this '\' would have less clashes, then.)

--- rk

[wendy]

xrxrx3xrxxr3xxrxr3z&##x, is a tiling of hexagonal prisms, the height three times the base edge. The prisms have been symmetically displaced according to the colouring of the hexagonal board. In the first column, we see xr xr x which is three x3x2x in a stack with the intermediate heaxgons removed.

One notes that r here does not 'add', that is, one treats each r separately, and suppose that no other r exists in the line, so eg

xr3xr3x3z is the first layer-etching. The first r means we pop out the ,3x3x3z hexagon. The second r means we pop out the x3,3x3z hexagon. This leaves just one solid hexagon, connected to the others only by the edges.

Likewise, we use bevels, for example b3b4o is a cube, with its edges beveled (planed at an angle, a woodwork craft) by the rhombic dodecahedron. In practice, b's represent an intersection of the duals of single-marked figures, the face-planes of the wythoff figures.

[student91]

Wrt. the 'r' symbol I'd then suggest a '\' character instead. - It both is used in informatics (in programming languages, RegEx, or even in LaTeX too) to mask some other elements, and in mathematical set theory as a negation or exclusion sign. Usually that then is placed before the to be excluded thingy, whereas you promote postfixing. That in turn could be supported by the Reversed Polish Notation, so would be admittable, I think.
And additionally, this '\', in the postfixed version, might also have the connotation of an ASCII shorthand for a downward, i.e. dropping arrow.
(Hope that this '\' would have less clashes, then.)

--- rk

I myself already used that symbol for the de-activation of a given mirror node. E.g. x4o3o means a cube, and x4o3\o then means the square of the cube. Likewise, ofox3xofo&#xt means an icosahedron, and ofox3\xofo&#xt means two laced triangles. This is better than saying e.g. ofox3....&#xt, because you can still read the other coordinates. From a fundamental viewpoint, this '\' can be used to axiomatize the construction device implied by the dynkin notation. I myself will be using this decoration.

[Klitzing]

Ain't that structure not a mere topologically slightly squashed xrxru3xruxr3uxrxr3z&##x = batch (bitruncated cubic honeycomb, o4x3x4o) in hexagon first axial orientation?!

(By using these different edge sizes, the r symbol kind of becomes obsolet, as "usually" only x edges are "understood" to be effectively used.
As in xr3xr3u3*a just the x3x3.3*a hexagon survives.)
In fact the lacing faces indeed here are tetragons and hexagons, because the vertices and edges still all are used (independent of the topological deformation to be applied, or not).

--- rk

[Klitzing]

I myself already used that symbol for the de-activation of a given mirror node. E.g. x4o3o means a cube, and x4o3\o then means the square of the cube. Likewise, ofox3xofo&#xt means an icosahedron, and ofox3\xofo&#xt means two laced triangles. This is better than saying e.g. ofox3....&#xt, because you can still read the other coordinates. From a fundamental viewpoint, this '\' can be used to axiomatize the construction device implied by the dynkin notation. I myself will be using this decoration.

So, to conclude,

• as '\' thus kind of clashes with your already working usage, and
• as Wendy seems not to be willing to abandon her 'r' in that respect,

I fear we will have to live with 'r' meaning both,

• the reduction operant (e.g. as in the last few posts in this thread) AND
• a zero sized edge (as in r4o3o being a zero-sized true cube, e.g. as used when considering Stott expansions / contractions).

Makes it kind of impossible to decide, when using both readings within the same symbol ...
Or we could decide then for a different (still unused) character in the second reading of Stott operations ...

--- rk

[wendy]

r already had two meanings before this, This just adds an extra meaning to it. The other meaning is as in P6D2R2 ie (6÷2)/(2÷2) But this is well past the scope of the CRF project.

I use \ in some contexts to mean dual, the opposite of /. But the present form uses m for this. The dual of /4B (2_21) is \4B.

r as zero-edge has meaning only when one is discussing the micro-polytope at the beginning of Stott's expansions. It has the meaning of an edge of length zero, rather than a point, like between edge and digon.

As long as we can agree on the name of the operation, that's a start.

[quickfur]

How do you account for Penrose tilings? Or is that not considered here?

[Klitzing]

Repetition is meant to be periodic.
Penrose tiling OTOH is non-periodic.

In fact, it just is quasiperiodic. Quasiperiodicity allows for any radius r to provide a distance R such that any patch of size r will be found again at most at distance R. The truth behind that statement is just, that quasiperiodic tilings are nothing but an irrational section through some higher dimensional periodic structure (cut and project description). Quasicrystals in the late 80th were discovered by means of their diffraction spectrums. Such spectra of usual (periodic) crystals allow for 3 Miller indices - in conformance to the 3D space the crystal resides in. Quasicrystalls OTOH provide such spectra, which ask for more than just 3 Miller indices.

For further details cf. eg. here: http://bendwavy.org/klitzing/explain/quasi.htm.

[quickfur]

Well, the currently known Penrose tilings are quasiperiodic... I've read that the aperiodic tiling problem is equivalent to the Halting Problem, which means that there ought to be other kinds of aperiodic tilings that may actually be completely "random" (i.e. have no discernible pattern, or the pattern is extremely complex and cannot be reduced to a simple repetition of some kind). In fact, such tilings ought to more numerous than the set of periodic tilings (correspondence to the Halting Problem means some kind of homomorphism onto the set of Turing machines, and there are far more "aperiodic" Turing machines than "periodic", or terminating, ones). Unfortunately, we don't know of any examples beyond the current quasiperiodic Penrose tilings!

[Klitzing]

Well, the terminus "quasiperiodic" dates back onto Harald Bohr (mathematician), the brother of Nils Bohr (physician). He moreover coined "almost periodic" for the case, when the periodic embeding dimension becomes countably infinite. And yes, beyond then finally comes true aperiodicity or randomness.

But the main issue for quaiscrystals was the existance of highly symmetrical diffraction patterns. Just the like as for the known periodic crystals. Only that those now showed up 5fold, 8fold, 10fold, 12fold axial symmetries as well as icosahedral symmetries. All these are incompatible with 3D periodicity. But when going into (still finite) higher dimensional setups, such symmetries clearly become possible via projection in a well-chosen direction.

Sure, you are right. Pure randomness is quite different. But not too much. E.g. the random square-triangle pattern at best allows for 12fold symmetry. Accordingly it well could be lifted up into a corresponding 4D lattice (either D4 or A2xA2 could be used for that purpose). But such random tilings then do not show up nice polygonal acceptance domains within perp space, rather would provide some fractal over there.

--- rk

[quickfur]

Well, between regular penrose and completely random, I'm sure there ought to be other interesting patterns. For example, maybe there is some kind of spiralling tiling that doesn't evenly divide the circle but fits into an ever-expanding spiral tiling? Or some other kind of interesting pattern of this sort. This isn't really related to your original topic but it's definitely a very interesting idea to me. I think there's a lot of room to explore here.

[Klitzing]

indeed there is. There also is a slightly different Ansatz than projection, the iterated substitutions. Definitely well-ordered structures, but conceptually quite different. There are some patterns which can be described by both, but generally there is no inclusion in either direction. The university of Bielefeld holds an online encyclopedia of substitutional tilings, cf. http://tilings.math.uni-bielefeld.de/ if you'd be interested.
--- rk

[quickfur]

Thanks for the link, it's very, very interesting!

[wendy]

The Penrose tiling is iso-bounded, and therefore must be aperiodic at every scale. The proof is relatively straight forward, and left as an exercise for the reader. In effect, the Penrose tiles are never form an infinite tiling because their isomorph would also be infinite, and thus not bounded.

There are an infinitude of such tilings, one might consult Dr Klitzing's paper on this, although he takes only thin slices through the forrest here. What is actually happening is truly interesting, though few people might grasp the concept.

The quasi-periodic groupings i have seen in nature are less iso-bounded then the Penrose tilings, the defraction layers come from distances between atom layers. Because these are in turn derived by imagining the atoms have definite bond positions, the particular layer formations are more likely than others, which is why we see these things.

In four dimensions the corresponding tilings are based on 5,3,3,5/2 and its kith, and on the octagonny o3x4x3o3o4/3o3z the dual here is o3o4o3o4x4/3x3z.

[Klitzing]

Wendy, I ought be supposed to know most of what you are talking about.
Sadly I'm rather in a complete loss. Could you please extend your points, so that others have a chance to grasp what you are saying?

Aperiodicity as such simply is the complement to periodicity.
Quasiperiodicity and almost peridicity OTOH are 2 further qualifiers within that complement.

--- rk