(nsi-)CvRF

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

(nsi-)CvRF

Postby Klitzing » Fri Aug 26, 2016 9:54 pm

 
  1. We already have a high-level thread about CRFs (convex regular faced polytopes - mostly polychora only).
  2. So I wonder whether it might be desirable to consider the concave counterpart as well (CvRF, concave regular faced polytopes)?
  3. At least those special concave ones, which still are non-self-intersecting (nsiCvRF), i.e. which are kind of potato-shaped, might be of higher interest.
    And only if those would have appeared as failing to pass the CRF test, and thus elsewise would no-where be mentioned any more, even so already having stumbled upon.

Two such nsiCvRF examples we already have come accross the last week.

One is pautpen (the penta-augmented truncated pentachoron). There the augmenting external blends are octatuts (ox3xx3oo&#x). The full shape is describable as xo3xx3oo3oA&#zx, where A evaluates as A:x=3:2. The total cell count here is:  5 oct + 20 trip + 20 tricu + 5 tet.  It already occured here (even so, at that time, erroniously assuming that it would be convex). It has just a single concave dihedral angle type. That one is at the hexagons between the 2 incident tricues. That one measures 360 deg.-arccos(-11/16) = 226.567 deg. (All other ones are convex.)

The other one is spysp (small pyramidic swirlprism). That one was found already in 2000 by George Olshevsky, but recently, meanwhile having forgotten about it, I stumbled upon it again, cf. here and then esp. here. That one is special in that it has just a single vertex type and just a single cell type. Thus it qualifies as noble polychoron. It has  240 peppies  for cells. It has just a single concave dihedral angle type. That one is at the pentagons between the pair of incident pyramids. That one happens to be 216 deg. exactly. (The single other one at the triangles is convex.)


Any additions here?

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Re: (nsi-)CvRF

Postby Klitzing » Mon Sep 19, 2016 6:16 pm

In the transition between convexity and concavity there are boarder-convex ones, where some dihedral angles would be exactly 180 degrees, but still none is beyond this value. This assumption then already extends the Johnson solids within 3D.

Sure, when doing so, we well could dissect any square into N² smaller squares and likewise dissecting triangles. Therefore any convex polyhedron with only triangles and squares gives rise for an infinite series of such boarder-convex figures. Therefore one then usually aims for some restriction of this explosion of numbers.

One way to do so would be by disallowing for collinear edges extending in opposite directions from a single vertex. A similar restricting definition was chosen in the research for convex regular-faced polyhedra with conditional edges, which should be included here. On that page all the 78 convex regular-faced polyhedra with conditional edges and all proper vertices are listed and displayed.

A research really worth noticing!

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Re: (nsi-)CvRF

Postby quickfur » Tue Oct 11, 2016 6:20 pm

I'm pretty sure the 600-cell would yield a very large number of concave but non-self-intersecting polytopes. Just take any floret of 5 tetrahedra: this represents a 4D wedge in the shape of pentagon || line, embedded in the body of the 600-cell with its 5 tetrahedra exposed on the surface. Cutting this wedge out yields a modified 600-cell with 5 less tetrahedra and two new pentagonal pyramids, which are concave. Any non-overlapping set of florets on the 600-cell can be substituted this way, and AFAICT the result should meet your definition. So there should be a very large number of these cuttings.
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Re: (nsi-)CvRF

Postby Klitzing » Tue Oct 11, 2016 7:25 pm

Yes, quickfur. :P
But then one usually would look for a maximized "diminishing" (here: excavating).
And that then is the mentioned spysp - with just 240 peppies for cells.
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Re: (nsi-)CvRF

Postby quickfur » Tue Oct 11, 2016 8:10 pm

Hmm. I've yet to check whether this construction remains non-self-intersecting, but I think (almost?) all of the 600-cell family uniform polytopes can probably have the element corresponding with the x5.3.3. position excavated with an appropriate segmentochoron of the form x5.3. || o5.3. (or some such derivation). Basically, replace each of the 120 cells in that position with shallow pits. If adjacent such excavations remain non-self-intersecting, this immediately gives us a large number of polytopes that fall under your definition, all with high symmetry.
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Re: (nsi-)CvRF

Postby Klitzing » Tue Oct 11, 2016 8:31 pm

Ah, I get, what you are after. You kind of are looking at doe and dimple each pentagon in by excavating it by a peppy each. That thing then is a 3D representant of what our aim is in 4D.

But the pentagon has a circumradius smaller than unity. Therefore it works here. The dodecahedron OTOH has a circumradius larger than unity. Therefore a corresponding pyramid does not exist in 4D. And thus it cannot be used for excavation either.

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Re: (nsi-)CvRF

Postby quickfur » Tue Oct 11, 2016 8:34 pm

The dodecahedron pyramid is well-known to be non-CRF, but what about excavating, say, dodecahedron || icosahedron from the 120-cell? AFAIK, the icosahedron's circumradius is smaller than the dodecahedron's inradius, so a single excavation ought to remain non-self-intersecting. But what I'm not sure about, is whether two adjacent excavations would intersect.
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Re: (nsi-)CvRF

Postby Klitzing » Tue Oct 11, 2016 9:26 pm

Okay, I accept that challange.
But it kind is too easy. :XP:

The dihedral angle between x5o3o (doe) and the lacing xo5oo&#x (peppy) within xo5oo3ox&#x (ikadoe = ike || doe) is 72 degrees.
OTOH the dihedral angle between 2 neighbouring x5o3o (doe) within x5o3o3o (hi) equals 144 degrees.

In fact, xo5oo3ox3oo&#x (roxahi = rox || hi) is a degenerate segmentopeton of zero height, and therefore rather classifies as decomposition of hi into a centrral rox (o5o3x3o) and other chunks. For sure, there are included those 120 ikadoes. :P

That is, according excavation is possible after all!
It just does not support the outermost pentagons. Those likewise are undermined, as the doe dimples by construction were. The remainder accoringly to that decomposition then is nothing but a central rox with 600 attached octpies (oo3ox3oo&#x) - supporting the old vertices. And possibly, if you'd like, also some further 1200 pens (xo ox3oo&#x) - then supporting additionally the old edges. :nod:

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Re: (nsi-)CvRF

Postby quickfur » Tue Oct 11, 2016 11:16 pm

Nice, very nice. :)
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Re: (nsi-)CvRF

Postby Klitzing » Wed Oct 12, 2016 7:41 pm

:)

... and here comes the incidence matrix of that "dimpled in" hecatonicosachoron:
Code: Select all
720   * |   10    5    0 |   10   20    5   0 |   2   10   10  (rox kernel vertices)
  * 600 |    0    6    4 |    0   12   12   6 |   0    4   12  (hi hull vertices)
--------+----------------+--------------------+--------------
  2   0 | 3600    *    * |    2    2    0   0 |   1    2    1
  1   1 |    * 3600    * |    0    4    2   0 |   0    2    4
  0   2 |    *    * 1200 |    0    0    3   3 |   0    0    3
--------+----------------+--------------------+--------------
  3   0 |    3    0    0 | 2400    *    *   * |   1    1    0  (type A)
  2   1 |    1    2    0 |    * 7200    *   * |   0    1    1  (type B)
  1   2 |    0    2    1 |    *    * 3600   * |   0    0    2  (type C)
--------+----------------+--------------------+--------------
 12   0 |   30    0    0 |   20    0    0   0 | 120    *    *  (ike)
  3   1 |    3    3    0 |    1    3    0   0 |   * 2400    *  (tet, type a)
  2   2 |    1    4    1 |    0    2    2   0 |   *    * 3600  (tet, type b)


The according dihedral angles then are:
  • at {3}type A between ike and tettype a:   360°-arccos[-sqrt(5/8)] = 217.761244°
  • at {3}type B between tettype a and tettype b:   360°-arccos[-(1+3 sqrt(5))/8] = 195.522488°
  • at {3}type C between tettype b and tettype b:   arccos(1/4) = 75.522488°
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Re: (nsi-)CvRF

Postby Klitzing » Sun Jan 01, 2017 9:16 am

Happy New Year

Robert Austin mentioned yesterday in FB a new post on his lovesPi-Pages. So I went there and then went on to an older post of his.
Thought it might serve well as a belated Christmas star or rather already an Happy New Year star too: the 41st stellation of the truncated icosahedron. - Thus we are back to 3D here.

Image
© https://robertlovespi.net/2014/08/06/four-stellations-of-the-truncated-icosahedron/

Thought a bit about it. Then I found that the pentagons not only are regular, but that the rhombs here are pairs of coplanar triangles. I.e. those have angles of 60 resp. 120 degrees. - In Wendy’s tegum sum notation that one also could be given as xfoa-3-fooo-5-oxfo-&#zx, where |a| shall represent an edge length of sqrt(5).

After staring a bit onto that non-convex (but still non-intersecting) and not-all-regular-faced polyhedron it occured to me, that this figure allows for some expansion too, which then returns to an all-regular-faced (non-convex) polyhedron. Just expand those yellow rhombs into complete hexagons. Then the dimpled vertices at their tips would become further pentagons. The old triples of pentagons remain as before. But the outermost edges also would have to expand into squares then.

I myself have no graphical device at hand. But perhaps someone might want to come in and provide an according picture of that expansion?

If - to this end - someone would ask for the respective coordinates, these then would be readily accessible from the according expanded tegum sum notation here: xfoa-3-fooo-5-xuFx-&#zx. (Again: |o|=0, |x|=1, |f|=1.618, |u|=2, |a|=2.236, |F|=2.618.)

Well to be fair, "tegum sum" here is a bit a streched term, as it isn't a convex figure and so no hull operation applies. But still the vertex set of that figure is represented by the union of the vertex sets of x3f5o, of f3o5u, of o3o5F, and of a3o5x.

The starting figure displayed above by itself does not belong into this thread because the existance of rhombs, which clearly are no regular faces. But the aimed for result then does. It just consiste from squares, pentagons, and hexagons.

Best regards,
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Re: (nsi-)CvRF

Postby wendy » Sun Jan 01, 2017 12:31 pm

The rhombs lie by threes at the vertices of the triangle. But then the pentagons have their ends clipped off, so they are x5/2xx (where xx>x).

But happy new year just þe same.
The dream you dream alone is only a dream
the dream we dream together is reality.
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Re: (nsi-)CvRF

Postby Klitzing » Sun Jan 01, 2017 1:15 pm

??? - Don't get what you want to tell here, Wendy.

But the outermost edges in the displayed stellation are just the wedge parts of the metabidiminished icosahedra. Accordingly the dihedral angle between the pentagons there (i.e. the convex bi-pentagonal one!) is just the same as for the great dodecahedron.

In fact, the whole displayed stellation is nothing but a great dodecahedron, where the vertices have been bitten out, to produce those 5-rhomb-dimples.

And within the described expansion then these outermost edges become squares with alternatingly attached pentagons (the same red ones, just shifted apart) respectively hexagons. And those 5-rhomb-dimples then will become the pentagon neighbourhoods of truncated icosahedra.

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Re: (nsi-)CvRF

Postby quickfur » Sun Jan 01, 2017 9:31 pm

Interesting. If you expand it, then where the pentagons currently are would become gaps that can be filled in with x5o3o's, and where the rhombs currently are would become hexagons surrounding pentagons, so they could be filled with o5x3x's. Can this cell complex serve as the seed for a 3D tessellation? Not sure what other pieces would be needed to fill in all the gaps.
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Re: (nsi-)CvRF

Postby Klitzing » Sun Jan 01, 2017 9:51 pm

quickfur wrote:Interesting. If you expand it, then where the pentagons currently are would become gaps that can be filled in with x5o3o's, and where the rhombs currently are would become hexagons surrounding pentagons, so they could be filled with o5x3x's. Can this cell complex serve as the seed for a 3D tessellation? Not sure what other pieces would be needed to fill in all the gaps.


Nice idea to use that fellow as a seed of an honeycomb. But I suppose I've to disappoint you with that.
The truncated icosahedra already would be matched each with one pentagon and all its 5 adjoining hexagons onto the seed. And the 5 next pentagons furthermore would attach directly to the dodecahedra. This leaves just a second opposing such patch to the truncated icosahedron - and the alternating girdle of 2*5 equatorial hexagons. These then ought to be matched somehow...
But I suppose that there is no larger periodic structure for an overlap-free continuation...

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Re: (nsi-)CvRF

Postby Klitzing » Thu Jan 05, 2017 5:44 pm

Klitzing wrote:... After staring a bit onto that non-convex (but still non-intersecting) and not-all-regular-faced polyhedron it occured to me, that this figure allows for some expansion too, which then returns to an all-regular-faced (non-convex) polyhedron. Just expand those yellow rhombs into complete hexagons. Then the dimpled vertices at their tips would become further pentagons. The old triples of pentagons remain as before. But the outermost edges also would have to expand into squares then.

I myself have no graphical device at hand. But perhaps someone might want to come in and provide an according picture of that expansion?

If - to this end - someone would ask for the respective coordinates, these then would be readily accessible from the according expanded tegum sum notation here: xfoa-3-fooo-5-xuFx-&#zx. (Again: |o|=0, |x|=1, |f|=1.618, |u|=2, |a|=2.236, |F|=2.618.)

Well to be fair, "tegum sum" here is a bit a streched term, as it isn't a convex figure and so no hull operation applies. But still the vertex set of that figure is represented by the union of the vertex sets of x3f5o, of f3o5u, of o3o5F, and of a3o5x. ...


Since no one grasped the opportunity, I myself had to dive in, calculate all the coordinates, and then write up an according VRML file by hand. :cry:
But at least the outcome is worth its work.

Below is a screenshot of that announced expanded polyhedron:
Image

Its incidence matrix could be calculated too:

Code: Select all
xfoa3fooo5xuFx&#zx  → height = 0
                      a = sqrt(5) = 2.236068
(tegum sum of (x,f,x)-grid, (f,u)-srid, F-doe, and (a,x)-srid)

o...3o...5o...     | 120  *  *  * |  1  1   1  0  0  0 |  1  1  1  0  [4,5_a,6]
.o..3.o..5.o..     |   * 60  *  * |  0  0   2  1  1  0 |  0  2  2  0  [(5_a)^2,6^2]
..o.3..o.5..o.     |   *  * 20  * |  0  0   0  3  0  0 |  0  3  0  0  [(5_a)^3]
...o3...o5...o     |   *  *  * 60 |  0  0   0  0  1  2 |  0  0  2  1  [5_b,6^2]
-------------------+--------------+--------------------+------------
x... .... ....     |   2  0  0  0 | 60  *   *  *  *  * |  1  1  0  0
.... .... x...     |   2  0  0  0 |  * 60   *  *  *  * |  1  0  1  0
oo..3oo..5oo..&#x  |   1  1  0  0 |  *  * 120  *  *  * |  0  1  1  0
.oo.3.oo.5.oo.&#x  |   0  1  1  0 |  *  *   * 60  *  * |  0  2  0  0
.o.o3.o.o5.o.o&#x  |   0  1  0  1 |  *  *   *  * 60  * |  0  0  2  0
.... .... ...x     |   0  0  0  2 |  *  *   *  *  * 60 |  0  0  1  1
-------------------+--------------+--------------------+------------
x... .... x...     |   4  0  0  0 |  2  2   0  0  0  0 | 30  *  *  *  {4}
xfo. .... ....&#xt |   2  2  1  0 |  1  0   2  2  0  0 |  * 60  *  *  {5} (type a)
.... .... xu.x&#xt |   2  2  0  2 |  0  1   2  0  2  1 |  *  * 60  *  {6}
.... ...o5...x     |   0  0  0  5 |  0  0   0  0  0  5 |  *  *  * 12  {5} (type b)


It should be noted furthermore, that this highly symmetrical figure also matches that old research of Jim McNeill on acrohedra (cf. http://www.orchidpalms.com/polyhedra/acrohedra/acrohedra.html) by virtue of that outermost vertex figure [4,5,6].

For those, who still like VRML files, the respective file can be found attached in zipped form. (Hope it gets through.) :D

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Re: (nsi-)CvRF

Postby Klitzing » Thu Jan 05, 2017 11:46 pm

Klitzing wrote:... Robert Austin mentioned yesterday in FB a new post on his lovesPi-Pages. So I went there and then went on to an older post of his.
Thought it might serve well as a belated Christmas star or rather already an Happy New Year star too: the 41st stellation of the truncated icosahedron. - Thus we are back to 3D here.

Image
© https://robertlovespi.net/2014/08/06/four-stellations-of-the-truncated-icosahedron/

Thought a bit about it. Then I found that the pentagons not only are regular, but that the rhombs here are pairs of coplanar triangles. I.e. those have angles of 60 resp. 120 degrees. - In Wendy’s tegum sum notation that one also could be given as xfoa-3-fooo-5-oxfo-&#zx, where |a| shall represent an edge length of sqrt(5).

After staring a bit onto that non-convex (but still non-intersecting) and not-all-regular-faced polyhedron it occured to me, that this figure allows for some expansion too, which then returns to an all-regular-faced (non-convex) polyhedron. ...


Just occured to me, that there is an even simpler way to obtain a non-intersecting non-convex all-regular-faced polyhedron from that depicted polyhedron: Just erase the deepest vertex type, providing then xfo-3-foo-5-oxf-&#zx only, by filling in these 12 yellow dimples half way. Then the rhombs will decompose into just their triangular tips, and you'll get spanned in pentagons there. - As has been described above, or could be read from the tegum notation, all edges then would have the same size, i.e. the triangles would be regular.

Will see who could provide an according picture first... :lol:

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Re: (nsi-)CvRF

Postby Klitzing » Sat Jan 07, 2017 9:51 pm

Klitzing wrote:Just occured to me, that there is an even simpler way to obtain a non-intersecting non-convex all-regular-faced polyhedron from that depicted polyhedron: Just erase the deepest vertex type, providing then xfo-3-foo-5-oxf-&#zx only, by filling in these 12 yellow dimples half way. Then the rhombs will decompose into just their triangular tips, and you'll get spanned in pentagons there. - As has been described above, or could be read from the tegum notation, all edges then would have the same size, i.e. the triangles would be regular.

Will see who could provide an according picture first... :lol:

--- rk


Haha,
once again won the prize myself:
here it comes ...
Image
obviously being obtained from 20 concave corners of the dodecahedron plus
12 concave pentagon surroundings of the icosidodecahedron.

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Re: (nsi-)CvRF

Postby Klitzing » Sun Jan 08, 2017 11:14 pm

Klitzing wrote:
quickfur wrote:Interesting. If you expand it, then where the pentagons currently are would become gaps that can be filled in with x5o3o's, and where the rhombs currently are would become hexagons surrounding pentagons, so they could be filled with o5x3x's. Can this cell complex serve as the seed for a 3D tessellation? Not sure what other pieces would be needed to fill in all the gaps.


Nice idea to use that fellow as a seed of an honeycomb. But I suppose I've to disappoint you with that.
The truncated icosahedra already would be matched each with one pentagon and all its 5 adjoining hexagons onto the seed. And the 5 next pentagons furthermore would attach directly to the dodecahedra. This leaves just a second opposing such patch to the truncated icosahedron - and the alternating girdle of 2*5 equatorial hexagons. These then ought to be matched somehow...
But I suppose that there is no larger periodic structure for an overlap-free continuation...

--- rk


In a private message, Roger Kaufman provided to me several VRMLs. One of these is exactly what you were after here, quickfur, cf. the according screenshot, depicting the cage of 20 dodecahedra and 12 truncated icosahedra around the (not shown) xfoa3fooo5xuFx&#zx:
Image
Here the then furtheron needed 4-5-6-acrohedra can be spotted clearly. And those cannot be closed without self-intersection or other weird stuff.

An other VRML of his provides a similar cage of 20 dodecahedra and 30 metabidiminished icosahedra (= J62 = mibdi) around the (not shown) xfo3foo5oxf&#zx (of my later post):
Image
It then clearly could be continnued by 12 icosidodecahedra (or at least by pentagonal rotundae = J6 = pero) at the decagonal holes.

And as mere addition he provided a similar cage of 20 triangular hebesphenorotunda (= J92 = thawro) around a (not shown) truncated icosahedron:
Image

--- rk
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