Bid 3.3.3.5

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

Bid 3.3.3.5

Postby wendy » Fri Dec 28, 2012 12:11 pm

This ought work. This is the latest entry in the non-wythoff tilings.

Code: Select all
Andrew Weimholt discovered in 2004, the bi-diminished 335. 

   The diminished 335 gives the s3s4o3o.  This can be further
   diminished to give a polychoron bid.ex with faces of
   xfo3oox, the vf of s3s4o3o.

   This makes a tiling of s3s4o3o, in x3o3o3o5o,
   whose faces are all snub 24chora.     

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Re: Bid 3.3.3.5

Postby Klitzing » Fri Dec 28, 2012 6:13 pm

would you mind to elaborate a bit, Wendy?
How does the sadi (1-idex) and bidex (2-idex) relate here?
Sure sadi shall be the only cell. But you do relate it somehow to bidex.
I can see that ex is the verf of your hyperbolic pentachoric tesselation.
But beyond...

Btw. what is that list of non-Wythoffian tilings you refer to?
A private one, I suppose?
Somewhere accessible?
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Re: Bid 3.3.3.5

Postby wendy » Sat Dec 29, 2012 7:07 am

I started with the Bid.ike, the vertex-figure of Weimholt's figure.

The first was to calculate, what solid angle it would subtend where it falls in the first vertex-ring of the {3,3,5}, the unit-edge icosahedron. You can cut it up into lots of little peices and make a bid.ike. The ike can divides into 20 faces, 30 edge bits, 12 point-edge and the core dodecahedron. Each class makes a total of 1s 0f from the centre of the 3,3,5.

The bid.ike then is made of the central ike, six points, and five edges. These add to 1 + 6/12 + 5/30 s, or 1 + 0:60 + 0:20 = 1:80 s. This means that 72 of these make a sphere.

I decided then to make an bid.ike out of zome.tools. It's truly interesting. It looks like a kind of litroform octahedron. Apart from this, it is quite apparent that it's self dual. Self-dual.

The vertices of bid.ex fall on 72 of the 1.00 vertices of the 3,3,5, and the 48 faces make up the rest. The dual of course, has faces which are dual of the vertex-figures, which are the same as the vertex-figure itself. The dual of the bid.ex has 72 bid.ikes, and is a diminished bid.ex, ie a tred.ex.

In H5, we see that the bid.ex is a vertex-figure of a tiling of tilings of the snub 24ch ( i suppose you're calling this sadi).

The list of non-wythoff uniforms is the reference list i use to keep tabs on this. It's private at this moment.

Code: Select all
Non-wythoff bollochora of type FE

  1.  tC, vf = octa tegum, e=a(8), octagon = a(3)
  2.  vf = octagon ball  = oxqxo8ooooo    edge= 4,8
  3.  3,5,3  vf = tetrahedrally dim dodecahedron =
  4.  3,5,3  vf = diametriclly dim dodecahedron = fxfo3ofxf
  5.  x3o:s3s4s   vf = snub cube, + o3x: + x3x
  6.  x3o:s3s5s   vf = snub dodeca + rect, trunc
  7.  borrochemochora - for 3, 5 and larger - an infinite set
  8.  extended borroch - 3 /a * /b 2 /c    a,b,c = 2,3,P. 

Horris wythoff-snubs groups of type FC
 
  1.  s3s4o3o    = snub 24choron.
  2.  s4s4o3o
      s3s4o4o    gives cells of sq prisms, not uniform   
  3.  o4s4o3o    vf = rectified q2q3o  (Anton Sherwood)  e
                 sn cell = triangle prism.


Non-wythoff bollotera of FE

  1.  tiling of octagonny o3x4x3o, vf = o8m2m8o
  2.  tiling of bi-octagon o8x2x8o, vf = o3m4m3o, dual of (1)
  3.  Andrew Weimholt tiling of s3s4o3o, 48 at a vertex.


Andrew Weimholt discovered in 2004, the bi-diminished 335. 

   The diminished 335 gives the s3s4o3o.  This can be further
   diminished to give a polychoron bid.ex with faces of
   xfo3oox, the vf of s3s4o3o.

   This makes a tiling of s3s4o3o, in x3o3o3o5o,
   whose faces are all snub 24chora.     


Other vertex figures known to work

    o3o3m3o3o.   
       vf = dual simplex, edge = h, lacing = q
       cells = 6,3 ## 6,3 combs. 

    there should be other gosset figures, since the dual is
    made of triangles h:q:q -> hexagonal prisms.

   s4s4s4s4z = o4s4s4o 
      o4x4x4o  vf = qo2oq, lacing as {8}, so cells are x4x4o.
      snub appears to give cells as vf above, and make the
      cells into snub s4s4o. 


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Re: Bid 3.3.3.5

Postby Klitzing » Sat Dec 29, 2012 8:31 am

Good morning dear Wendy,

here are some of Jonathans acronyms:
ex = hexacosachoron (x3o3o5o)
ico = icositetrachoron (x3o4o3o) - in contextes sometimes even reduced to -i
ike = icosahedron (x3o5o) - in contextes sometimes also reduced to -i
teddi = tridiminished icosahedron (xfo3oox&#xt)
bidex = bi-icositetra-diminished hexacosachoron
sadi = snub dis-icositetrachoron

Recently I wrote to Jonathan, that sadi could be seen as an idex as well, and he replied: yes, the same way as ike could be seen as a snit (snub tetrahedron). - Well here do some naming differences of Kepler and Norman interfere: sometimes the -dis- is used, sometimes not. Anyway.

In my view we are considering here the sequence
0-idex = ex
1-idex = sadi
2-idex = bidex

Further it occured to me that the respective vertex figures are corresponding diminishings of ike:
vf(0-idex) = 3*0-di = 0-di = ike
vf(1-idex) = 3*1-di = 3-di = teddi
vf(2-idex) = 3*2-di = 6-di

You are right in that decomposition of ike into stella chunks (as dinogeorge called them): we have the sequence ike - (inscribed) gad (= x5o5/2o) - (inscribed) sissid (= o5o5/2x) - (kernel) doe (= o3o5x).

It even is clear how those diminishings do apply. Consider some triangle of ike. Take the tips of all 3 adjacent ones as positions for the first diminishing triple. Likewise for the other triples. How are those centre-triangles to be chosen? Here comes snit in! In fact, those 4 tetrahedrally aligned faces play the right role. Then the 3*4 tip positions give a complete overlap-free decomposition of the vertices of ike.

Playing that game we even could consider:
3*0-di = ike (= dual of doe)
3*1-di = teddi
3*2-di = Andrews verf, which is selfdual
3*3-di = the dual of teddi
3*4-di = the kernel doe

Those clearly relate to the 4d shapes as vf(n-idex) = 3*n-di. In fact, it is chi (chirohecatonicosachoron = chiral compound of 5 icoes), which can be vertex inscribed into ex (at least a tau scaled version of chi). Any individual of those inscribed icoes gives rise for a further diminishing. This runs thru till 4 icoes, still having some of the ex vertices left, giving rise to the 3*4-di as vertex figure. Only for the 5-idex there are no ex vertices left. Accordingly we likewise cannot have an ike-diminishing anymore.

Okay. so far up to the sadi / bidex -topics.

Now to your bid 3.3.3.5 . Still too sparse, your infos about that one.

A list is quite good for keeping track. But if its private not of much use for others... - Ah, just saw: you included that as code above. Well, have to admit, all those infos are a bit sparse too...

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Re: Bid 3.3.3.5

Postby wendy » Sat Dec 29, 2012 10:00 am

I will try to explain the set.

The polyhedra are: teddi (tri-diminished icosa), and a bid-35. This corresponds to removing one or two of the tetrahedral faces of the inscribed {3,5/2}

bid.ike (or 3,5), corresponds to half of the vertices of the ike, basically, the North pole, and a line that includes one other northern hemisphere, and then four southern hemisphere points. The face centres are the same shape on the southern face.

The polychora are: the monid.ex or 'sadi' or s3s4o3o., the bid.ex, which has 48 teddi, and the trid.ex, with 72 bid.ikes. and 48 faces. These correspond to two and three of the inscribed 3,4,3 in 3,3,5. Sadi corresponds to 4 rows of the same table. Sadi's verf is the teddi, the teddi's verf is half of the six faces of bid.ike.

The bollotera are supposed at a tiling of sadi, with a verf of bid.ex, a tiling of bid.ex, with a vertex-figure of trid.ex.

The varuous non-wythoff groups are

FE = finite extent
FC = finite content (points go to infinity)

Apart from 7,8 in that list, we see most of these are new. vf = verf. a(x) = shortchord of x.

Horris = euclidean, ie horizon-centred.
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Re: Bid 3.3.3.5

Postby Klitzing » Sat Dec 29, 2012 7:37 pm

wendy wrote:I will try to explain the set.
...
The bollotera are supposed at a tiling of sadi, with a verf of bid.ex, a tiling of bid.ex, with a vertex-figure of trid.ex.
...


So you now even have got 2 tilings.
Hmmm.

Like 1-idex was derived from ex by diminishing at the vertex positions of a (scaled) vertex inscribed ico, and 2-idex was derived from ex by diminishing at the vertex positions of 2 (scaled) vertex inscribed icoes, like furthermore teddi = 3*1-di was derived from ike by diminishing at the vertex positions of a (scaled) vertex inscribed (affine) triangle, and Andrew's bidex-verf = 3*2-di was derived from ike by diminishing at the vertex positions of 2 (scaled) vertex inscribed (affine) triangles, I suppose your hyperbolic tilings too should be derivable from x3o3o3o5o by "diminishing" at the vertex positions of some (scaled) vertex inscribed (to be determined) structures. - Do you've any idea what those are?

Further the relative interrelation of those structures for the higher diminishing is of interest each. For bidex and its verf this is solved too. But for your hyperbolic tiling its most probably still open.

Having answered these questions, this should also give some better names for your tilings, than just adapt "bid" for that analogue here. (As "bid" there stood for bi-icositetra-diminished).

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Re: Bid 3.3.3.5

Postby wendy » Sun Dec 30, 2012 7:21 am

There are two tilings, but the bi.dex is not uniform, so it isn't listed as such.
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Re: Bid 3.3.3.5

Postby wendy » Sun Dec 30, 2012 8:59 am

It's interesting to see that richard is complaining about the sort of things i have been complaining for: hijacking terms.
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Re: Bid 3.3.3.5

Postby Klitzing » Sun Dec 30, 2012 11:23 am

wendy wrote:It's interesting to see that richard is complaining about the sort of things i have been complaining for: hijacking terms.

:lol:
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Re: Bid 3.3.3.5

Postby wendy » Tue Jan 01, 2013 10:48 am

I found the volume of the bid.ike to be 0:v0 phi, that is 1/phi of the icosahedron. Bid.ex makes a dandy litroform. (one is a rabid metric-hater, since this is what common sense dictates).

One plan is to use zometools to make the composite dual thingie.

Of course, you could use the -i- in bid to be a dimensional context prefix. i in bid.ex is a specifically 4d thing, one could use it for three and five with different meanings to preserve the -i- here. The relation to the 24choron is not lost by me.
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Re: Bid 3.3.3.5

Postby Klitzing » Wed Jan 02, 2013 12:48 pm

Still am not convinced in "bid" being a good chosen name for that operant you're applying onto 3.5, 3.3.5, and 3.3.3.5; neither for the -i- within ("i" were stemming from icositetra=24 within that acronym "bidex"), nor for the b- part ("b" were stemming from bi=2). You more likely are generalizing what I was refering as n-idex (n-fold icositetra-diminished hexacosachoron). The vertex figures of n-idex I was refering to as n*3-di (n-fold tri-diminished icosahedron). Esp. -i- is not applicable here, as a triangle rather would associated with -t- than -i-. You now are extrapolating to hyperbolic 5d figures, which do have n-idex for verf.
(On the other hand it might well include the -d- of bidex, as all clearly are some diminishing.))

So I'd rather suggest the following:
  • First figure out what vertex inscribed structure describes the positions of diminishings here.
  • Then put that in a sequence with tip-centered triangle pyramid and icositetrachoron.
  • Find some common features of those and coin some thereon based naming.

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Re: Bid 3.3.3.5

Postby Klitzing » Fri Jan 18, 2013 12:11 pm

Klitzing wrote:
wendy wrote:I will try to explain the set.
...
The bollotera are supposed at a tiling of sadi, with a verf of bid.ex, a tiling of bid.ex, with a vertex-figure of trid.ex.
...


So you now even have got 2 tilings.
Hmmm.

Like 1-idex was derived from ex by diminishing at the vertex positions of a (scaled) vertex inscribed ico, and 2-idex was derived from ex by diminishing at the vertex positions of 2 (scaled) vertex inscribed icoes, like furthermore teddi = 3*1-di was derived from ike by diminishing at the vertex positions of a (scaled) vertex inscribed (affine) triangle, and Andrew's bidex-verf = 3*2-di was derived from ike by diminishing at the vertex positions of 2 (scaled) vertex inscribed (affine) triangles, I suppose your hyperbolic tilings too should be derivable from x3o3o3o5o by "diminishing" at the vertex positions of some (scaled) vertex inscribed (to be determined) structures. - Do you've any idea what those are?
[...]
--- rk


If I get your post from 2005 correctly, you there were already mentioning candidates to look for? I.e. bollotera vertex inscribable to x3o3o3o5o ?

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Re: Bid 3.3.3.5

Postby wendy » Sat Jan 19, 2013 7:25 am

The 2005 post mentioned is not a search for anything.

What i did there was to evaluate what polytopes would exist at those distances, where the expression base phi2 does not go past the radix. I enumerated every possible ring up to that radius, and then picked out the ten or so that correspond to hyperbolic radii. Since the shell is euclidean, there is no further thinning needed.

101. 6000 5000 0.1.f.0 f0.0.0.1 {3,5,3,3}

There are three different bases here! 101. is written in base 2.61803398875, with digits 0, 1, f=1.61803398875. It's the square of the nominated radius, the indicated radius is f.sqrt(3) as 101 = 3*f*f. You calculate the diameter by squaring this, and dividing by "f.1". The second number is decimal, being a translation of the third number (twelfty). In turn, the twelfty translates 0.1.f.0, and f0.0.0.1. The former has 30.00 vertices, the latter has 20.00, which adds to 50.00, which gives 6000 decimal. The value in curlies are polytopes of the same number of edges, ie {3,5,3,3}

The edges of the various bids correspond to 1 and 10 in this table. That is, the edges are 1 and f of the {3,3,5}. However, the vertices of this polytope form a primary quarterion integer set of class 2, (one of two!), so it means that nearly everything that 1.00 divides, then it is a possible collection of that many {3,3,5}. Likewise, bid-2 and bid-3 of 48 and 72 vertices, are sums of 2 and 3 copies of {3,4,3}, as s{3,4,3} and {3,3,5} are 4 and 5 of the same.

Most of the indicated bollotera indicated are actually inscribed, but you can't always take this so. The {3,3,3,5} shares many, but not all, of its vertices with {5,3,3,3}. In practice, E2 (112) of the 100 (120) vertices in ring 1 are found, while the remaining 16 are found at the cell centres of (5,3,3,3).

If one supposes that {3,3,3,5} has a substrate with a vertex-figure of s{3,4,3}, then the pentachora are not arranged with a central one and five others, but each pentachoron belongs to 4 others and one {3,3,5}. In turn, a {3,3,5} has 24 pentachora it owns, which belong to one of the 25 special colouring groups. This preserves the general overall symmetry of {3,4,3+} indicated. The pentachora are not then simple clusters but yickloid strings.

It was done in a project which enumerated the rings of 5_21, which confirmed a known thing directly in coordinates for each of the first fifty rings. The octagonny was not enumerated, because it uses a more subtle technique.
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Re: Bid 3.3.3.5

Postby Klitzing » Wed Jan 23, 2013 12:02 pm

Coming back to the subsymmetrical diminishings mentioned lately in the other thread (Johnsonian Polytopes):
  • ex, diminished at ico vertex directions, results in sadi (the former vertex figure thereby becomes tridiminished)
  • ex, diminished at the vertex directions of 2 icoes, results in bidex (the former vertex figure becomes chirally hexadiminished)
  • ico, diminished at hex vertex directions, results in tes (the former vertex figure thereby becomes tetradiminished)
  • ico, diminished at the vertex directions of 2 hexes, results in a third hex (the former vertex figure becomes octadiminished)
By re-reading my own webpage I just stumbled across similar cases, which Wendy mentioned already way back in 2011. (I'll post that - for record - in this thread, because it did not apply to objects of spherical geometry, but rather on an object of hyperbolic space - like the leading figure of this very thread. :) )

She reduced the vertex figure of the hyperbolic tiling {3,5,3} (i.e. a doe) at opposite vertices. The resulting hyperbolic tiling is what she had called spt(3,5,3), semi partially truncated {3,5,3}, having for cells does, paps, and ikes.

Moreover, the vertex figure doe likewise could be diminished/truncated at a tetrahedral subset of vertices. This results in what she called pt(3,5,3). That honeycomb wouldn't have any former ikes left, cells being does and paps only.

PS:
The 2 figures discribed so far in that topic, thus likewise might be called (better?) spt(3,3,3,5) resp. pt(3,3,3,5). (At least this does not take wrong refuge onto that -icositetra- part of the recently used "bid" operand.)

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Re: Bid 3.3.3.5

Postby wendy » Thu Jan 24, 2013 6:57 am

The various figures Richard mentions as 'truncated' are actually 'diminished'. There is a difference. It ought be spd(3,5,3) and pd(3,5,3), rather than a 't' there.

In any case, i come to the conclusion that a polyteron based on a verf of s{3,4,3} is unlikely.

The margins (walls between cells), are always simplexes, and stand in the same ratio as its edge figure (teddi), ie 1:3:1. This means that there's one between the pentagons, three between the pentagon and triangles, and one between the triangles.

ALL NUMBERS IN TWELFTY

In the case of teddi, we see these numbers are 3:9:3. In the case of s(3,4,3), the numbers are 96 248 96, giving the 400 triangles this figure demands. In the case of <3,3,3,5>, for each {3,3,5}, the ratio is 100:300:100. The 100 between the two {3,3,5} "ex" would indicate that each ex has 200 tetrahedra that interface to another ex. It's quite possible to make such an arrangement. The special colouring of the faces of {5,3,3} is into 25, being a letter-number each over 5, eg A5. The inscribed {3,3,5} corresponds the five sets with the same letter or number, eg A1, A2, A3, A4, A5. Picking two letters like A and B, would bring the two sets of 100 too close to each other for what is demanded. However, the set A1 A2 B2 B3 C3 C4 D4 D5 E5 A5, places them at the required distance from each other. So it's possible to pick 200 faces of {3,3,5} that do not share a common triangle.

Where we run into problems, is the simplex. The indication here is that the number of simplex-simplex margins equals the number of simplexes themselves. This supposes at best, branched chains or branched loops. However, we see that the clusters of four tetrahedra at a vertex figure, suggest that there are points where there ought be four links. This suggests that there ought be simplexes with no adjacent ends (ie the vertex opposite the end of a chain), but no such things exist: therefore the requirements of the figure are not met here.
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Re: Bid 3.3.3.5

Postby Klitzing » Thu Jan 24, 2013 9:33 am

wendy wrote:The various figures Richard mentions as 'truncated' are actually 'diminished'. There is a difference. It ought be spd(3,5,3) and pd(3,5,3), rather than a 't' there.[...]

:P Admitted.
So we have now
  • spd{3,5,3} with verf = paradim. doe = fxfo3ofxf&#xt, tiles being ike, pap, doe
  • pd{3,5,3} with verf = tetradim. doe, tiles being pap, doe
  • spd{3,3,3,5} with verf = bidex, tiles being sadi
  • pd{3,3,3,5} with verf = tridex, tiles being bidex
(The first ones are done as IncMats already, cf. the link provided above, to my website. The second ones still are based only on your statements, so far I still did not manage to wrap my mind fully around those. :\ )

[...]In any case, i come to the conclusion that a polyteron based on a verf of s{3,4,3} is unlikely. [...]

Are you saying here that there will be no such thing like a sspd{3,3,3,5}, i.e. being so super-sparsely diminished that the verf would be sadi, while cells would be pen and ex?

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Re: Bid 3.3.3.5

Postby wendy » Thu Jan 24, 2013 11:54 am

I did incidence diagrams on 'spd{3,3,3,5}'. Spent all day doing them. That's where i picked the fault. vf is my form of verf.

Of the remainder, we have still ones i call IIt, IIIt, and IVt.

It, having ex and pent as cells, and sadi as a verf, does not seem to have wessian (existance), though i need to go through the figures again.

IIt has sadi as cells, and Weimholt's polytope as the vf. It's dual is the IVt

IIIt has the bid.ex as a cell and the tri.dex as a vertex figure. It's self dual.

IVt has tri.dex as cells, and the dual of sadi as a vertex figure.

I don't know if IIt to IVt actually exist.
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Re: Bid 3.3.3.5

Postby Klitzing » Thu Jan 24, 2013 12:29 pm

wendy wrote:I did incidence diagrams on 'spd{3,3,3,5}'. Spent all day doing them. That's where i picked the fault. vf is my form of verf.

Of the remainder, we have still ones i call IIt, IIIt, and IVt.

It, having ex and pent as cells, and sadi as a verf, does not seem to have wessian (existance), though i need to go through the figures again.

IIt has sadi as cells, and Weimholt's polytope as the vf. It's dual is the IVt
[...]

Don't you are contradicting yourself, now? :(
IIt was the one you originally called bid(3,3,3,5) and we'd thus went to name spd(3,3,3,5) recently.
At any hand, you've thus answered my question: It is what doesn't exist (by your reasoning of that former mail, and that one would have dealt with sadi for verf, not for cells. I.e. those would have been truely different).
[...]
IIIt has the bid.ex as a cell and the tri.dex as a vertex figure. It's self dual.
[...]

Oh, self dual: interesting!
[...]
IVt has tri.dex as cells, and the dual of sadi as a vertex figure.
[...]

That one is once more a new one...
[...]
I don't know if IIt to IVt actually exist.

Ah, this is essential input! Today I struggled with the IncMat of your IIt. But so far did not succeed. :( (So, would not dare to take any implications therefrom, so far...)

Annotation: consider the special structure of the yellow edges in quickfurs pic at http://teamikaria.com/hddb/forum/viewtopic.php?f=25&t=1719#p17835.

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Re: Bid 3.3.3.5

Postby Klitzing » Thu Jan 24, 2013 8:42 pm

Wrapped my mind around that dual thing.
It is that
  • 1-idex (sadi) is dual to 4-idex, and
  • 2-idex (bidex) is dual to 3-idex.
Just as already
  • 1*3-di (teddi) is dual 3*3-di, while
  • 2*3-di (Weimholt-hexahedron) is self-dual.

Here:
n-idex = n-times icositetra-diminished hexacosachoron;
n*3-di = 3n-times diminished icosahedron = vertex figure of n-idex;

1-idex had 3 types of cells: 2 different tet and ike. Accordingly 4-idex has 3 types of vertices. The 4*3-di clearly is the kernel doe, i.e. the dual of those ikes only.
2-idex is noble (isogonal and isochoral). Therefore that diversity does not occure for 2-idex and 3-idex.

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Re: Bid 3.3.3.5

Postby Klitzing » Thu Jan 24, 2013 9:27 pm

Have not considered that It so far. But you said that it would be a wessian.

But I considered now your IIt. And it seems to be one too!

Just assume that IIt = spd3335, i.e. that noble hyperbolic tiling with sadi for tiles and bidex for verf, would exist. Then we would have to consider the dual of the vertex figure as being the fundamental domain of that tiling, i.e. its Voronoi domain. That dual is tridex.

So first consider that one closer. What would be its IncMat? Being the dual of something known (bidex), the IncMat can be derived therefrom just be rotation about 180 degrees (and reconsidering), i.e.
Code: Select all
tridex:

48 |  3  3  2 |  6   9 |  9
---+----------+--------+---
 2 | 72  *  * |  3   2 |  5  : x-edges
 2 |  * 72  * |  1   2 |  3  : f-edges: trigon-to-trapezium + trigon-to-anitialigned trigon
 2 |  *  * 48 |  0   3 |  3  : f-edges: trigon-to-para trigon
---+----------+--------+---
 4 |  3  1  0 | 72   * |  2  : x-x-x-f trapeziums
 3 |  1  1  1 |  * 144 |  2  : f-f-x triangles
---+----------+--------+---
 6 |  5  3  2 |  2   4 | 72  : Weimholt-hexahedron


Thus each tridex of the Voronoi complex then corresponds to a vertex of the spd3335 tiling.
Each tridex-cell corresponds to an edge of that tiling.
But there is just a single such cell, so all edges seem to be alike.
We have 2 types of 2-boundaries of tridex: trapeziums and triangles.
Accordingly we will have 2 types of 2-faces in that tiling too.
(As sadi has triangles for 2-faces only, those should fall in 2 classes then.)
Further tridex has 3 types of 1-boundaries, so we will get 3 types of 3-boundaries in that tiling.
(Sadi uses ikes and tets, these are already 2, i.e. one of those classes has to be further divided.
Probably the tets, as those form 4-tet-complexes, the central one thus distinguished from the other ones.)
And finally, the fundamental region is isogonal, this should result in a single 4-face of the tiling (i.e. the supposed sadi itself).

Thus we deduce for the spd3335 IncMat
Code: Select all
2N |  72 | 144  72 |  48  72  72 | 48
---+-----+---------+-------------+---
 2 | 72N |   4   2 |   2   3   5 |  6
---+-----+---------+-------------+---
 3 |   3 | 96N   * |   1   1   1 |  3
 3 |   3 |   * 48N |   0   1   3 |  4
---+-----+---------+-------------+---
 4 |   6 |   4   0 | 24N   *   * |  2
 4 |   6 |   #   # |   * 36N   * |  2 -> 8/3 4/3 ???
12 |  30 |   8  12 |   *   * 12N |  2
---+-----+---------+-------------+---
96 | 432 | 288 192 |  48  72  24 |  N


It might well occur from outer geometry (outside a single domain) that some classes, considered different in the above descriptive listing, would have to be combined. But clearly tets cannot be combined with ikes. Those triangles further cannot be combined either, as one will have 3 incident cells while the other one needs 4 (*). From that then follows that the tets cannot be combined either.

(*) Note that the numbers of the top line are the numbers of the vertex figure (if read from left to right, resp. of the fundamental domain, if read from right to left). That further the superdiagonal part of the second row is not only the edge figure but in fact the vertex figure of that vertex figure. Accordingly the entries of the third block of rows are nothing but the different vertex figures of the row before. - This is why there is a 3fold and a 4fold.

So in total, that places marked by # cannot be filled consistently. (The numbers induced by the corresponding IncMats rules are displayed above too, becoming impossible fractions.)
This usually is a sure sign that somewhere we made an error. But I cannot see where in my reasoning here might be one. Thus it needs to be the assumption itself, i.e. the existance of that thing.

What would you think?

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Re: Bid 3.3.3.5

Postby wendy » Fri Jan 25, 2013 7:03 am

The reason i went for using bid etc across several dimensions, is because they are all kinds of partial diminishings repeatedly used.

In 3d, the vertex group is formed by the four faces of a tetrahedron, inscribed in a great dodecahedron (5/2,3). Removing one face and its vertices gives the tridiminished icosahedron, ie teddi. Two faces gives Weimholt's cube. weimholt's cube is chiral and self-dual, but the dual is of the same hand as the figure itself. Three faces obviously leaves a triangle.

ie an id.ike = teddi, a bid.ike = weimholt's cube.

In 4d, the vertex-group is formed by five 343's 'ico' in a 335 'ex'. One can remove one, two, three, or four of these 343's to get various figures

1: Ic = id.ex = sardi = s3s4o3o
2: IIc = bid.ex = weimholt's polychoron: 48 faces = id.ike = teddi, 72 verf = bid.ike = weimholt's cube.
3. IIIc = trid.ex. faces = 72 bid.ike = weimholt's cube, 48 verf = dual of id.ike
4. ivc = 343 itself.

IIc and IIIc are duals. ic has an irregular dual, consisting of a partially apiculated 533.

The group [3,4,3+] is a subgroup of [3,3,5], the quotent being an LR group formed by five letters and five numbers, eg A1, C5 etc. Each of the five inscribed 343 in 335 represent a common letter or number, eg A1, A2, A3, A4, A5. The dual of these 343 form the vertices of a 335, the group of all of the same letters is a 335 in 533.

Taken in its lesser form (5*24=100), we get one row (eg A), gives iv c, two rows gives iiic, three rows = iic, four rows = i c, and 5 rows = 335.

The groups in 3335.

One might suppose that this extends into higher dimensions, specifically the tiling of 3335. The various subgroups would then be

i t = id.3335, being a tiling of 335 and 333, verf = id.ex
ii t = tiling of id.ex, verf = bid.ex , being dual of iv, and uniform
iii t = tiling of bid.ex, verf = trid.ex, being self dual
iv t = tiling of trid.ex, irregular verf. being dual of ii.

This would suppose that the vertices of 3335 has a subgroup of six, which permits free removal of the elements. From calculations, it appears that i t, does not have existance. I have not done the large scale grouping on the group of six as yet.

It should be seen that id.ike, id.ex, and id.335 have faces from the original surface, which means that its dual has vertices from 53, 533, and 5333. The remainder have the vertices and faces split between the vertices of 35, 335, and 3335.
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Re: Bid 3.3.3.5

Postby Klitzing » Fri Jan 25, 2013 9:42 am

wendy wrote:[...]ie an id.ike = teddi, a bid.ike = weimholt's cube.[...]

Still, -i- represents 24 (icositetra). We already introduced -p- instead (for partial). Or you could use the appropriate count as operator (if the arrangement becomes clear from context). Thus id.ike is nominal nonsense. You better uses pd.ike or 3-dimin.ike a.k.a. 1*3-di.
[...]
1: Ic = id.ex = sardi = s3s4o3o
2: IIc = bid.ex = weimholt's polychoron: 48 faces = id.ike = teddi, 72 verf = bid.ike = weimholt's cube.
3. IIIc = trid.ex. faces = 72 bid.ike = weimholt's cube, 48 verf = dual of id.ike
[...]

Would not like to speak of a cube here. Yes, it is a hexahedron (as the cube is too). And indeed it has 2 tetrahedral faces. But the majotity is trigonal. And it more resembles a trigonal prism, one of its squares being diametrically halved! In fact, this is, what its topology truely is.

(Btw. here the use -i- is clearly correct: id.ex = 24-dimin.ex.)
[...]
4. ivc = 343 itself.

IIc and IIIc are duals. ic has an irregular dual, consisting of a partially apiculated 533.
[...]

Thus you proclaim that Ic and IVc are not duals? This seems wrong, as any such diminishing reduces the count of vertices by 24. Starting with ex (v=120) we get v(Ic=sadi)=96, v(IIc=bidex)=72, v(IIIc=tridex=dual of bidex)=48, v(IVc)=24 - so far this would match with ico, as you said. But likewise we get 24 deep cells (dc) for each applied diminishing, all those being at the same depth. So we have dc(Ic)=24 ikes, dc(IIc)=48 teddies, dc(IIIc)=72 Weimholts hexahedra, dc(IVc) then shall have a count of 96. But ico only has 24 octs. Whereas the dual of sadi clearly would have the required 96 cells.
[...]
The group [3,4,3+] is a subgroup of [3,3,5], the quotent being an LR group formed by five letters and five numbers, eg A1, C5 etc. Each of the five inscribed 343 in 335 represent a common letter or number, eg A1, A2, A3, A4, A5. The dual of these 343 form the vertices of a 335, the group of all of the same letters is a 335 in 533.

Taken in its lesser form (5*24=100), [...]

hihi, 100 (twelfty) = 120 (decimal)
[...] we get one row (eg A), gives iv c, two rows gives iiic, three rows = iic, four rows = i c, and 5 rows = 335.

The groups in 3335.
[...]
(finally, we get to the actual point :) )
[...]
One might suppose that this extends into higher dimensions, specifically the tiling of 3335. The various subgroups would then be

i t = id.3335, being a tiling of 335 and 333, verf = id.ex
ii t = tiling of id.ex, verf = bid.ex , being dual of iv, and uniform
iii t = tiling of bid.ex, verf = trid.ex, being self dual
iv t = tiling of trid.ex, irregular verf. being dual of ii.

This would suppose that the vertices of 3335 has a subgroup of six, which permits free removal of the elements. From calculations, it appears that i t, does not have existance. I have not done the large scale grouping on the group of six as yet.

It should be seen that id.ike, id.ex, and id.335 [...]

  • "id" still missused
  • 335 here probably should read: 3335
[...] have faces from the original surface, which means that its dual has vertices from 53, 533, and 5333. The remainder have the vertices and faces split between the vertices of 35, 335, and 3335.

Well, above I stressed your words "might suppose".
I too stumbled upon the true geometry of bidex when doing its IncMat for the first time. This is because the inner symmetry of teddi is not transfered to bidex. Teddi has axial symmetry. So we have 3 times 3 equivalent vertices. (This is reflected in its description as ofx3xoo&#xt.) Knowing that the teddies would pile up as circular columns, we might even admit to identify the top and bottom edges. This was my first intend. But the outside symmetry (surroundings) force one chiral set of the lacing edges oo.3oo.&#x to be of the same type as those top- and bottom-base edges, while the other chiral set has to be united with the lacing edges .oo3.oo&#x! This is, what makes up the swirl-symmetry of bidex. - Cf. the pic of quickfur I mentioned already in a recent post. There he just colored the edges of one of the teddies of bidex according to the adjacent cell counts:
Image

Somehow I've the feeling that this very swirl-symmetry is what might not can be transported into 3335. (I'm not so good in group theory to estimate thereon. - At least this might be a reason for your dis-proof of I t and my dis-proof of IIt to take place.)

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Re: Bid 3.3.3.5

Postby wendy » Sat Jan 26, 2013 6:59 am

Here, 'id' is being used as an operator in 3, 4, 5 dimensions, and really does not conflict with its 4d usage.

I had a bit of a rethink on id.ike/id.ex/ figures. and their duals. The first cut of 'id' does not remove all of the original faces of ike/ex, so the dual has vertices which are not part of the original ike/ex. The figures i qouted at the dual positions correspond to the convex hull of one set of points.

On the other hand, if one supposes that id is a kind of organised diminishing, which can be applied over several dimensions, then the dual of i is the next on the series, so we get

IIIh trid.ike = dual of id.ike (teddi)
IVc quid.ex = dual of id.ex (sadi)

trid.ike is a tricorn, or thrice-apiculated doe. It's what you get when you mount pyramids on three faces of a doe. The triangle formerly described has its vertices in this triangle, the remaining five vertices are from the core dodecahedron.

quid.ex has 96 faces, all being trid.ikes. This amounts to a 533, with one of the inscribed 343. The dodecahedral faces are apiculated to meet these 24 vertices (which form the set of 24 originally described as IVc

The cuts are still made against the missing vertices of 35, 335, but now the cuts are hitting the core of the dual 53, 533. The additional vertices of this series correspond to the faces of 35, 335, left in the first cut. The next set of cuts would leave the inner core 53 'doe' and 533 'twech' themselves.

This brings us to the next, interesting point.

The whole series of id's are made on the same eutactic stars. This means that all of the edges of these figures are perpendicular to mirrors of the underlying [3,5] and [3,3,5], although not necesssarily bisected by them. This is different to the catalans, which have other kinds of edge.

The same would hold for IVt, were it to exist.
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Re: Bid 3.3.3.5

Postby Klitzing » Sat Jan 26, 2013 10:08 am

wendy wrote:Here, 'id' is being used as an operator in 3, 4, 5 dimensions, and really does not conflict with its 4d usage. [...]

Yes, Wendy, I do know.
It is just that "id" was taken over from the 4D case, and there its origin was "icositetra-diminished".
Esp. as you are quite keen in re-organizing wordings according to their original stems, I'm stressing that point all the time in return.
I clearly understand your intend when applying "id" to other dimensions too. It is just that "id" there has no ethymological support...
[...]
I had a bit of a rethink on id.ike/id.ex/ figures. and their duals. The first cut of 'id' does not remove all of the original faces of ike/ex, so the dual has vertices which are not part of the original ike/ex. The figures i qouted at the dual positions correspond to the convex hull of one set of points.

On the other hand, if one supposes that id is a kind of organised diminishing, which can be applied over several dimensions, then the dual of i is the next on the series, so we get

IIIh trid.ike = dual of id.ike (teddi)
IVc quid.ex = dual of id.ex (sadi)

trid.ike is a tricorn, or thrice-apiculated doe. It's what you get when you mount pyramids on three faces of a doe. The triangle formerly described has its vertices in this triangle, the remaining five vertices are from the core dodecahedron.
[...]

Yep. that thrice apiculated doe or teddi dual has edges of 3 lengths. It has 6 triangles of the form x-x-f and 3 tetragons of the form x-x-v-v.
[...]
quid.ex has 96 faces, all being trid.ikes. This amounts to a 533, with one of the inscribed 343. [...]

Inscribed? I'd suppose circumscribed, as those 343 vertices would correspond to the apiculated doe vertices. - Cf. the dualization effect too: id.ex (sadi) has the 24 diminishings (ikes) deeper than the remaining boundary. Thus quid.ex, being the dual of id.ex, then would have 24 vertices farer out than the other ones.
[...]The dodecahedral faces are apiculated to meet these 24 vertices (which form the set of 24 originally described as IVc

The cuts are still made against the missing vertices of 35, 335, but now the cuts are hitting the core of the dual 53, 533. The additional vertices of this series correspond to the faces of 35, 335, left in the first cut. The next set of cuts would leave the inner core 53 'doe' and 533 'twech' themselves.

This brings us to the next, interesting point.

The whole series of id's are made on the same eutactic stars. This means that all of the edges of these figures are perpendicular to mirrors of the underlying [3,5] and [3,3,5], although not necesssarily bisected by them. This is different to the catalans, which have other kinds of edge.

The same would hold for IVt, were it to exist.


:idea: Finally just rethought your "id" once more, while writing this up.
Meanwhile we were reusing your "pt" of pt.3.5.3 for that reason; in fact we corrected that "t" ("t"runcated) towards "d" ("d"iminished). Thus "pd" ("p"artially "d"iminished) would be appropriate instead.
On the other hand, when applying that in an increasing sequence, we no longer could use your then used qualifier "s" ("s"emi/"s"parse), at least as soon as we have more than just 2 cases. For ike we have the need of at least 3, for ex we have the need of at least 4 subsequent further diminishings. So some number count should apply, rather than mere adjectives. Number counts could be given in greekish or latinised prefixes too. But it is then that these get an nicer ring to them, if those can be joined to the stem operant by a vowel rather than a consonant. So "-id" would be clearly more desirable than just that clumsy "-pd", admitted.
Therefore I rethought to re-define that "i" in such a way, as to have just that meaning of that "p" (partial), and thus being fully supported for its extended use through all dimensions (i.e. getting released from its mere use of "i"cositetra).
And, what shall I say, I was successful! We could translate that "i" as being the abbreviation of "i"ncomplete (= partial, subsymmetrical, ...).

Thus your usage of "id", "bid", "trid", "quid", ... gets finally cross-dimensional ethymological support, after all. Hurray! 8)

Still, we have to fiddle out, whether the application of those operants onto 3.3.3.5 are admittable or not...

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Re: Bid 3.3.3.5

Postby wendy » Sun Jan 27, 2013 7:17 am

You can always make 'i' mean different things, though. Still, i get ugly feelings that the division of 2/5 of 533's vertices is going to produce some instances of unit edges of 533, rather than the phi demanded by theory. It was interesting while it lasted. though.
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Re: Bid 3.3.3.5

Postby Klitzing » Thu Jan 31, 2013 11:52 am

Just to mention, that I recently started discussing related aspects of this topic with Marek14 in the more related section "other geometry", cf. that specific post, rather than within "polytopes", esp. as this specific point of interest clearly is hyperbolic and further meanwhile turned out to be more group theoretical.

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Re: Bid 3.3.3.5

Postby Klitzing » Thu Jan 31, 2013 11:42 pm

Getting a bit closer to our problem with incidence matrices for k-id.3335 :

  • Consider a 6-coloring of the vertices of x3o3o3o5o to exist.
  • Then, by restriction to the neighbouring vertices of any single vertex thereof, there will have to be a 5-coloring of the vertex figure of that tiling, i.e. of x3o3o5o or ex. (That one we already know to exist. This is just the tau-scaled inscribed chiral compound of 5 icoes.)
  • Similarily, by restriction to the neighbouring vertices of any single vertex of ex, there will be a 4-coloring of its vertex figure, i.e. of x3o5o or ike. (That one too is known to exist. Just consider a tet being face-center inscribed. Any tet-vertex then induces exactly 3 (mutually disjoined!) tips of the adjacent triangles to that centered one of consideration.)
  • Again likewise, by restriction onto the neighbouring vertices of any single vertex of ike, there will be a 3-coloring of its vertex figure, i.e. of the pentagon x5o.

At least at this point this seems to become freaky: what? a 3-coloring of a 5-gon?
But sure, yes, this is possible. And the sole possibility, which uses no adjacent identical colors, clearly is  a-b-c-a-b- . This obviously breaks down any symmetry of the pentagon: That single c clearly fixes that point. And the remaining vertices (locally) would allow a rotation only - contradicting to that fixing. I.e. the only symmetry here is the identity!

Considering next the ike. Consider that already described 4-coloring. It becomes clear that any symmetry, which would interchange any 3 vertices of the same color, would be a symmetry of the corresponding tip of the inscribed tetrahedron. But any such action, applied onto the total tetrahedron, thereby would interchange the remaining vertices of that tetrahedron. Thus the other colors of the 4-colored ike thus would interchange as well. - This shows that here too the only symmetry of the so colored ike would be the identity.

Now coming onto incidence matrices. Incidence matrices always are based on symmetry. In fact, those are essentially based on 0-1-matrices, which describe the incidences of any vertex, any edge, any face, etc. If there is a symmetry of the described thingy, then the elements of the same track, can be grouped. This is how numbers larger than one occure. Well, as this coloring induces the identity to be the only symmetry at all, the according matrices would be 0-1-matrices only. And those, for sure, do exist - at least in the finite cases.

OTOH, we are not considering multicoloring of those vertex sets, we are effectively considering bi-colorings only: those vertices, which have to be chopped off, and those, which aren't. But chopping here is related to that so far considered multi-color scheme. We select the vertices of k (arbitrary) colors for those k-id-operations. Thus both, these k and the remaining n-k colors too will be considered alike (in these 2 groups) then first. Here some non-identity transformation of the bi-coloring might take place again. And by the action of this symmetry that single class of non-diminished vertices gets re-divided again. - This is why we still have numbers larger than one in the matrices of teddi, sadi, bidex, et al.

E.g. teddi is derived from that 4-coloring of ike by chopping off the vertices of one color, while considering the other 3 colors alike. Just as the inscribed tet mentioned in the coloring-description, this bi-coloring will show up axial 3fold symmetry. Accordingly those remaining vertices will fall into different orbits according to this 3fold axial symmetry. In fact those are 3 cycles of 3 vertices.

For x3o3o3o5o we then seem to have 6 colors. Those define Wendy's proposed hyperbolic structures:
We could divide into 1 : 5, defining 1-id.3335. Cells ought to be pen + ex (= 0-id.335), vertex figure ought to be sadi (= 1-id.335).
We could use 2 : 4, defining 2-id.3335. Cells ought to be sadi (= 1-id.335), vertex figure ought to be bidex (= 2-id.335).
We could use 3 : 3, defining 3-id.3335. Cells ought to be bidex (= 2-id.335), vertex figure ought to be tridex (= 3-id.335).
We could use 4 : 2, defining 4-id.3335. Cells ought to be tridex (= 3-id.335), vertex figure ought to be quidex (= 4-id.335).
And we could use 5 : 1, defining 5-id.3335. Cells ought to be quidex (= 4-id.335), while here 2 types of vertex figures should occure.

Btw., even so this seems to be a big number, we already found some relations:
1-id.35 (teddi) is dual to 3-id.35
2-id.35 (Weimholt's hexahedron) is self-dual
1-id.335 (sadi) is dual to 4-id.335
2-id.335 (bidex) is dual to 3-id.335
So seemingly
1-id.3335 ought to be dual to 5-id.3335,
2-id.3335 ought to be dual to 4-id.3335,
3-id.3335 ought to be self-dual.

Now coming back onto our incidence matrix problems. Main difficulty here is: not to know that acting symmetry of the diminished structure. As stated above this most probably would result in dealing with every vertex on its own. Dealing with a (hyperbolic) tiling, we have an infinite amount of (remaining) vertices.
One could start with the counts of the provided by the asked for vertex figure. But it is not clear (and, from our so-far troubles, seemingly not true), whether the full symmetry of that would be supported.

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