CRFs with duoprismic symmetry

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

CRFs with duoprismic symmetry

Postby Klitzing » Wed Mar 12, 2014 10:39 pm

quickfur wrote:
Klitzing wrote:And still I have to consider my own ideas: "Idea 3" was disproven, "idea 2" was elaborated. One of those 2 was definitely non-convex, as including stips for cells. So far I suppose that the other one so would be convex. - Any support for that assumption? - And "idea 1" still remains to be evaluated by my side. (quickfur here already gave some Input with respect to some dihedral angle restrictions. But so far I had no sparetime to look more closely into these...)

Which one would be convex again? I only remember two convex findings from your ideas, one was the same as an n,12-duoprism, and the other was the same as x4o3o3x. Did I miss another possible convex figure in there?


Yes, idea 3 was that n,12-douprism, and thus disproven to be of interest here.
But within idea 2 I had that thingy:
Code: Select all
    x4x   x4x   
                
x4x x4o   x4o x4x
                
                
x4x x4o   x4o x4x
                
    x4x   x4x   

which had the inverse usage of octagons / squares within the lace city as sidpith has!

Here is the corresponding incidence matrix as well:
Code: Select all
64  * |  1  1  1  1  1  0  0 | 1  1  1  1  1  1  1  1  1  0 0 | 1 1  1  1  1  1 0
 * 16 |  0  0  0  0  4  2  2 | 0  0  0  0  0  4  4  2  2  4 1 | 0 0  4  2  2  1 2
------+----------------------+--------------------------------+------------------
 2  0 | 32  *  *  *  *  *  * | 1  1  1  0  0  1  0  0  0  0 0 | 1 1  1  1  0  0 0 {8}-sides parallel to pseudo {4}
 2  0 |  * 32  *  *  *  *  * | 1  0  0  1  1  0  0  1  0  0 0 | 1 1  0  0  1  1 0 {8}-sides rel. dual to pseudo {4}
 2  0 |  *  * 32  *  *  *  * | 0  1  0  1  0  0  1  0  0  0 0 | 1 0  1  0  1  0 0 extension-para. op lacings
 2  0 |  *  *  * 32  *  *  * | 0  0  1  0  1  0  0  0  1  0 0 | 0 1  0  1  0  1 0 former op lacings
 1  1 |  *  *  *  * 64  *  * | 0  0  0  0  0  1  1  1  1  0 0 | 0 0  1  1  1  1 0
 0  2 |  *  *  *  *  * 16  * | 0  0  0  0  0  2  0  0  0  2 0 | 0 0  2  1  0  0 1 pseudo {4}-sides
 0  2 |  *  *  *  *  *  * 16 | 0  0  0  0  0  0  2  0  0  2 1 | 0 0  2  0  1  0 2 extension edges
------+----------------------+--------------------------------+------------------
 8  0 |  4  4  0  0  0  0  0 | 8  *  *  *  *  *  *  *  *  * * | 1 1  0  0  0  0 0
 4  0 |  2  0  2  0  0  0  0 | * 16  *  *  *  *  *  *  *  * * | 1 0  1  0  0  0 0 adj. cubes
 4  0 |  2  0  0  2  0  0  0 | *  * 16  *  *  *  *  *  *  * * | 0 1  0  1  0  0 0 between former op and trips
 4  0 |  0  2  2  0  0  0  0 | *  *  * 16  *  *  *  *  *  * * | 1 0  0  0  1  0 0 between ext. op and trip
 4  0 |  0  2  0  2  0  0  0 | *  *  *  * 16  *  *  *  *  * * | 0 1  0  0  0  1 0 adj. squippies
 2  2 |  1  0  0  0  2  1  0 | *  *  *  *  * 32  *  *  *  * * | 0 0  1  1  0  0 0
 2  2 |  0  0  1  0  2  0  1 | *  *  *  *  *  * 32  *  *  * * | 0 0  1  0  1  0 0
 2  1 |  0  1  0  0  2  0  0 | *  *  *  *  *  *  * 32  *  * * | 0 0  0  0  1  1 0
 2  1 |  0  0  0  1  2  0  0 | *  *  *  *  *  *  *  * 32  * * | 0 0  0  1  0  1 0
 0  4 |  0  0  0  0  0  2  2 | *  *  *  *  *  *  *  *  * 16 * | 0 0  1  0  0  0 1
 0  4 |  0  0  0  0  0  0  4 | *  *  *  *  *  *  *  *  *  * 4 | 0 0  0  0  0  0 2
------+----------------------+--------------------------------+------------------
16  0 |  8  8  8  0  0  0  0 | 2  4  0  4  0  0  0  0  0  0 0 | 4 *  *  *  *  * * extension-para. op
16  0 |  8  8  0  8  0  0  0 | 2  0  4  0  4  0  0  0  0  0 0 | * 4  *  *  *  * * former op
 4  4 |  2  0  2  0  4  2  2 | 0  1  0  0  0  2  2  0  0  1 0 | * * 16  *  *  * * lacing cube
 4  2 |  2  0  0  2  4  1  0 | 0  0  1  0  0  2  0  0  2  0 0 | * *  * 16  *  * * former op adj. trip
 4  2 |  0  2  2  0  4  0  1 | 0  0  0  1  0  0  2  2  0  0 0 | * *  *  * 16  * * ext. op adj. trip
 4  1 |  0  2  0  2  4  0  0 | 0  0  0  0  1  0  0  2  2  0 0 | * *  *  *  * 16 * squippies
 0  8 |  0  0  0  0  0  4  8 | 0  0  0  0  0  0  0  0  0  4 2 | * *  *  *  *  * 4 cube


So any support on the convexity of that fellow?

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Mar 12, 2014 11:41 pm

Klitzing wrote:[...]
Code: Select all
    x4x   x4x   
                
x4x x4o   x4o x4x
                
                
x4x x4o   x4o x4x
                
    x4x   x4x   

which had the inverse usage of octagons / squares within the lace city as sidpith has!

I'm struggling to understand that lace city... doesn't it indicate that it would be possible to take a cross-section of it that would produce x4x||x4o||x4o||x4x as a section? Wouldn't that make it non-convex?

Here is the corresponding incidence matrix as well:
Code: Select all
64  * |  1  1  1  1  1  0  0 | 1  1  1  1  1  1  1  1  1  0 0 | 1 1  1  1  1  1 0
 * 16 |  0  0  0  0  4  2  2 | 0  0  0  0  0  4  4  2  2  4 1 | 0 0  4  2  2  1 2
------+----------------------+--------------------------------+------------------
 2  0 | 32  *  *  *  *  *  * | 1  1  1  0  0  1  0  0  0  0 0 | 1 1  1  1  0  0 0 {8}-sides parallel to pseudo {4}
 2  0 |  * 32  *  *  *  *  * | 1  0  0  1  1  0  0  1  0  0 0 | 1 1  0  0  1  1 0 {8}-sides rel. dual to pseudo {4}
 2  0 |  *  * 32  *  *  *  * | 0  1  0  1  0  0  1  0  0  0 0 | 1 0  1  0  1  0 0 extension-para. op lacings
 2  0 |  *  *  * 32  *  *  * | 0  0  1  0  1  0  0  0  1  0 0 | 0 1  0  1  0  1 0 former op lacings
 1  1 |  *  *  *  * 64  *  * | 0  0  0  0  0  1  1  1  1  0 0 | 0 0  1  1  1  1 0
 0  2 |  *  *  *  *  * 16  * | 0  0  0  0  0  2  0  0  0  2 0 | 0 0  2  1  0  0 1 pseudo {4}-sides
 0  2 |  *  *  *  *  *  * 16 | 0  0  0  0  0  0  2  0  0  2 1 | 0 0  2  0  1  0 2 extension edges
------+----------------------+--------------------------------+------------------
 8  0 |  4  4  0  0  0  0  0 | 8  *  *  *  *  *  *  *  *  * * | 1 1  0  0  0  0 0
 4  0 |  2  0  2  0  0  0  0 | * 16  *  *  *  *  *  *  *  * * | 1 0  1  0  0  0 0 adj. cubes
 4  0 |  2  0  0  2  0  0  0 | *  * 16  *  *  *  *  *  *  * * | 0 1  0  1  0  0 0 between former op and trips
 4  0 |  0  2  2  0  0  0  0 | *  *  * 16  *  *  *  *  *  * * | 1 0  0  0  1  0 0 between ext. op and trip
 4  0 |  0  2  0  2  0  0  0 | *  *  *  * 16  *  *  *  *  * * | 0 1  0  0  0  1 0 adj. squippies
 2  2 |  1  0  0  0  2  1  0 | *  *  *  *  * 32  *  *  *  * * | 0 0  1  1  0  0 0
 2  2 |  0  0  1  0  2  0  1 | *  *  *  *  *  * 32  *  *  * * | 0 0  1  0  1  0 0
 2  1 |  0  1  0  0  2  0  0 | *  *  *  *  *  *  * 32  *  * * | 0 0  0  0  1  1 0
 2  1 |  0  0  0  1  2  0  0 | *  *  *  *  *  *  *  * 32  * * | 0 0  0  1  0  1 0
 0  4 |  0  0  0  0  0  2  2 | *  *  *  *  *  *  *  *  * 16 * | 0 0  1  0  0  0 1
 0  4 |  0  0  0  0  0  0  4 | *  *  *  *  *  *  *  *  *  * 4 | 0 0  0  0  0  0 2
------+----------------------+--------------------------------+------------------
16  0 |  8  8  8  0  0  0  0 | 2  4  0  4  0  0  0  0  0  0 0 | 4 *  *  *  *  * * extension-para. op
16  0 |  8  8  0  8  0  0  0 | 2  0  4  0  4  0  0  0  0  0 0 | * 4  *  *  *  * * former op
 4  4 |  2  0  2  0  4  2  2 | 0  1  0  0  0  2  2  0  0  1 0 | * * 16  *  *  * * lacing cube
 4  2 |  2  0  0  2  4  1  0 | 0  0  1  0  0  2  0  0  2  0 0 | * *  * 16  *  * * former op adj. trip
 4  2 |  0  2  2  0  4  0  1 | 0  0  0  1  0  0  2  2  0  0 0 | * *  *  * 16  * * ext. op adj. trip
 4  1 |  0  2  0  2  4  0  0 | 0  0  0  0  1  0  0  2  2  0 0 | * *  *  *  * 16 * squippies
 0  8 |  0  0  0  0  0  4  8 | 0  0  0  0  0  0  0  0  0  4 2 | * *  *  *  *  * 4 cube


So any support on the convexity of that fellow?

--- rk

I'm having a hard time seeing how it could be convex, seeing that the x4x's will overhang the x4o's in the interior of the lace city. Am I missing something obvious?
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Mar 13, 2014 7:36 am

Haha, this is exactly, what I was doubting as well.

But I cannot see that (so far) from the mere construction itself: considering the 8 octagonal prism ring as an alternating one of A and B ops. Then consider the outer squares of either of those ops as alternating as well, i.e. Aa, Ab, Ba, Bb. Attach to Aa cubes, to Ab trips (in the obvious orientation), to Bb squippies, and to Ba also trips (in the then again obvious orientation). Bend that adjoined cells into 4D such that the lacing faces connect. (The geometry was chosen such that those will do!) Then the still open faces would be the opposite squares of the 4x4 cubes. Here you can attach now an orthogonal ring of 4 cubes, just as being used within the 4,4-duoprism (= tes).

Thus, more like an augmented 8,8-duoprism (= odip). And odip clearly is convex.

Any idea where non-convexity would slip in? (if indeed non-convex.)

--- rk
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Mar 13, 2014 7:56 am

How exactly would the coordinates look? I tried to simply combine coordinates for octagonal duoprism (the outer ring) and tesseract (the inner ring), but the result was just octagonal duoprism (the "inner" vertices got swallowed), so there should be some trick in it...

But the x4x x4o x4o x4x reminds me of drilled truncated cube...
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Re: Johnsonian Polytopes

Postby wendy » Thu Mar 13, 2014 8:07 am

I'm pretty sure that the internal x4o sections in Richard klitzing's lace cities are actually w4o, where w=q+1. That would give the middle section as a rCO. This would be consistant with what the surrounding text is saying.

I browse the list, but i think a good number of examples given here are just a tad hard to do by hand. One of the reasons i have the transport of numbers and other weird things to tell me if something is worth the effort.

But i look at the posts for obvious errors and requests.
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Mar 13, 2014 8:16 am

With that edge length, it lokos like it's vice versa -- the tesseract swallows the octagonal duoprism.
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Mar 13, 2014 1:23 pm

wendy wrote:I'm pretty sure that the internal x4o sections in Richard klitzing's lace cities are actually w4o, where w=q+1. That would give the middle section as a rCO. This would be consistant with what the surrounding text is saying.
...


(Suppose you meant tC instead ...).

Sad to disappoint you on that. Just as quickfur already pointed out, it would be in fact an exterior blend of 4x
Code: Select all
    x4x
        
x4x x4o
i.e. op||{4}, and 4x
Code: Select all
x4x   x4x
        
x4o   x4o
i.e. op||cube, and 1x tes: The formers build the outer ring, pairwise blending out the squacues. Then the tes comes in the sense of a 4,4-duoprism, blending out the top cubes of the (op||cube)s, replacing that op-ring parallel ring of cubes by the orthogonal ring of cubes of that duoprism. Thus the central region of the lace city should get filled then.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Mar 13, 2014 1:41 pm

Marek14 wrote:With that edge length, it lokos like it's vice versa -- the tesseract swallows the octagonal duoprism.

No Marek, Wendy was not intending to replace x4o3o3o = x4o x4o by w4o3o3o = w4o w4o, but rather by the mixed duoprism x4o w4o. (The lace city positions would remain, just the perp space assignments will increase.)

And yes, the result then should be nothing but a combinatorical variant of my figure, which then clearly shall be disigned to have the right dihedral angle between the (deformed) squacues. - But that figure then, even so being definitely convex, is clearly not unit edged. You will need some edges x, some edges w, and moreover some further ones, depending on the outcome of the lacings of those "squacues"...

But my figure was designed with all unit edges so.

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temp

Postby Klitzing » Sat Mar 15, 2014 7:37 am

Klitzing wrote:...
Code: Select all
    x4x   x4x   
                
x4x x4o   x4o x4x
                
                
x4x x4o   x4o x4x
                
    x4x   x4x   

which had the inverse usage of octagons / squares within the lace city as sidpith has!
...
So any support on the convexity of that fellow?


Haha, finally fiddled it out myself: one should "read" what I painted just correctly.

Then it will be obvious, that this one is non-convex after all:
  • The dihedral angle between the op and the adjoining lacing cube surely is 45 degrees.
  • The dihedral angle between the lacing cube and the cube of the orthogonal ring is 135+90=225 degrees, i.e. reflex.
So it is kind a 4D analogue of a blood cell: thick rim and shallow central region... - OMG. :oops:

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Re: Johnsonian Polytopes

Postby quickfur » Sat Mar 15, 2014 2:24 pm

That's what I thought, the rows and columns across a lace city, as I understand it, represent the cross-sections of the polytope in the corresponding directions in 4D. In your given lace city, these cross-sections are all non-convex: you have x4x x4o x4o x4x in each case, which is a stack of octagon-square-square-octagon. The 3D shape would be therefore a cube in the middle with two square cupola attached to two opposite faces, a kind of square pillar like shape, and obviously non-convex. Since convex polytopes always have convex cross-sections, whereas here all rows and columns give the same non-convex shape, it seemed pretty clear to me that the overall polytope cannot be convex either.

However, in spite of being non-convex, the shape itself is rather interesting: it seems to be a kind of octagonal duoprism ring that wraps around a tesseract, leaving only a ring of 4 cubes exposed to the exterior. So you have a duoprism-like structure of a "fat" outer ring of octagonal prisms, and an orthogonal inner ring of 4 cubes, laced by some manner of cupolaeic cells in a concave manner. It's sorta like the ring of cubes is a "tight belt" around the orthogonal space ("waist") of the fat ring, "squeezing" it to become narrow (concave) around that space. :lol:
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Mar 15, 2014 2:54 pm

Could it be some 4D analogy of a Stewart toroid?
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Re: Johnsonian Polytopes

Postby quickfur » Sat Mar 15, 2014 3:01 pm

It has no holes, though. Only a concave ring. ;)
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Mar 15, 2014 3:15 pm

But maybe if you removed some faces, but kept the edges...
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Re: Johnsonian Polytopes

Postby quickfur » Sat Mar 15, 2014 6:33 pm

That's easy, just take out the ring of cubes. Or rather, cut out the tesseract core of it, then you'll have a 4D Stewart-style toroid. :) It will have a ring of octagonal prisms, like an 8,8-duoprism, but the orthogonal ring would be missing, and there would be a toroidal hole in it. (The lacing cells would still be there, too.)
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Re: CRFs with duoprismic symmetry

Postby Marek14 » Tue Apr 28, 2015 2:05 am

Got a strange idea tonight, not sure if it works.

Take two S1 x S1 surfaces tiled by squares. Let's say that the first one is 10 x 20 and the other is 20 x 10. Then, we should be able to join them by a combination of cubes, tetrahedra and triangular prisms.

On the 10 x 20, the 20 columns would have cubes on odd columns and triangular prisms on the even ones. When joined together, the even columns will vanish in the next layer, leaving only 10 columns.
But, another triangular prism is inserted between each pair of cubes, so the 10 rows double to 20 in the next layer. Tetrahedra are inserted between adjacent triangular prisms, transforming vertical edges in one layer into horizontal edges in the other.

Would this kind of layer work? I drew a rudimentary vertex figure -- it makes it clear that the prisms outside of the layer could be theoretically in two different directions.

Coordinates of vertex figure with prisms lined in one direction are (edges are 2-4, 2-5 and 4-5 with length 1, 0-1, 0-3, 0-4, 0-5, 1-2, 1-4, 2-3 and 3-5 with length sqrt(2) and 1-3 with length corresponding to n-gon) shows real solutions for n=4 and n=6:
n = 4
0: 1.00000 0.00000 0.00000 0.00000
1: 0.00000 1.00000 0.00000 0.00000
2: 0.70711 0.00000 0.70711 0.00000
3: 0.00000 0.00000 0.00000 1.00000
4: 0.00000 0.00000 0.70711 0.70711
5: 0.00000 0.70711 0.70711 0.00000

n = 6
0: 1.00000 0.00000 0.00000 0.00000
1: 0.00000 1.00000 0.00000 0.00000
2: 0.79057 0.00000 0.61237 0.00000
3: 0.00000 -0.50000 0.00000 0.86603
4: 0.00000 0.00000 0.81650 0.57735
5: 0.00000 -0.50000 0.81650 -0.28868

In the other direction, the results show that this configuration should be possible for cubes or hexagonal, octagonal and decagonal prisms - not for dodecagonal and 14-gonal prisms give hyperbolic solution.

The coordinates for vertex figure are (edges are 2-4, 2-5 and 4-5 with length 1, 0-1, 0-3, 0-4, 0-5, 1-2, 1-4, 2-3 and 3-5 with length sqrt(2) and 0-2 with length corresponding to n-gon):
n = 4
0: 1.00000 0.00000 0.00000 0.00000
1: 0.00000 1.00000 0.00000 0.00000
2: 0.00000 0.00000 1.00000 0.00000
3: 0.00000 0.33333 0.00000 0.94281
4: 0.00000 0.00000 0.50000 0.86603
5: 0.00000 -0.81650 0.50000 0.28868

n = 6
0: 1.00000 0.00000 0.00000 0.00000
1: 0.00000 1.00000 0.00000 0.00000
2: -0.50000 0.00000 0.86603 0.00000
3: 0.00000 0.25000 0.00000 0.96825
4: 0.00000 0.00000 0.57735 0.81650
5: 0.00000 -0.79057 0.57735 0.20412

n = 8
0: 1.00000 0.00000 0.00000 0.00000
1: 0.00000 1.00000 0.00000 0.00000
2: -0.70711 0.00000 0.70711 0.00000
3: 0.00000 0.00000 0.00000 1.00000
4: 0.00000 0.00000 0.70711 0.70711
5: 0.00000 0.70711 0.70711 0.00000

n = 10
0: 1.00000 0.00000 0.00000 0.00000
1: 0.00000 1.00000 0.00000 0.00000
2: -0.80902 0.00000 0.58779 0.00000
3: 0.00000 0.80902 0.00000 0.58779
4: 0.00000 0.00000 0.85065 0.52573
5: 0.00000 0.30902 0.85065 -0.42533

I would guess that n=6 in first mode and n=10 in second mode has nonconvex vertex figure because of the w-coordinate of vertex dropping to negative. n=4 in first mode and n=8 in second is also weird since the vertex 5 is in the same plane as vertices 0, 1 and 2. But the options n=4 and n=6 in second mode at least seem locally convex.

n=4 would be two rings of 8 cubes joined by 16 cubes, 32 triangular prisms and 16 tetrahedra. Hm, I've seen those numbers before -- in fact, this is simply small prismated tesseract x4o3o3x, just seen from an unusual symmetry. n=6 is more interesting. It would contain two rings of 12 hexagonal prisms joined by 36 cubes, 72 triangular prisms and 36 tetrahedra. I suspect there's a mistake somewhere, since with this construction, it would have to be uniform.

EDIT: Realized where's the problem: each hexagonal prism would be surrounded by a ring of cubes and triangular prisms -- but the dihedral angles would add up to 360, so it would be flat. And increasing it to dodecagonal prism doesn't work since the cubes are common to both, so you'd get a structure made of intersecting dodecagonal prisms and tetrahedra. It might still look interesting, though... and it might make a great 4D puzzle if you can rotate the hexagonal prisms in both rings and move the cubes :)
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Re: CRFs with duoprismic symmetry

Postby Klitzing » Tue Apr 28, 2015 3:00 pm

"small prismated tesseract x4o3o3x" - small dis-prismated tesseracti-hexadecachoron (by symmetry of diagram), sidpith.

"it might make a great 4D puzzle if you can rotate the [...]" - hehe a 4D Rubik's cube, that sidpith!

--- rk
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Re: CRFs with duoprismic symmetry

Postby Marek14 » Tue Apr 28, 2015 3:18 pm

Klitzing wrote:"small prismated tesseract x4o3o3x" - small dis-prismated tesseracti-hexadecachoron (by symmetry of diagram), sidpith.

"it might make a great 4D puzzle if you can rotate the [...]" - hehe a 4D Rubik's cube, that sidpith!

--- rk


The thing is that a 4D Rubik's cube, as the name itself shows, is just an extension of a 3D idea. This, however, is a puzzle that can't be easily imagined in 3D, as it's based on two partially intersecting rings of dodecagonal prisms.
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Re: CRFs with duoprismic symmetry

Postby Klitzing » Fri May 01, 2015 7:43 am

Marek14 wrote:Got a strange idea tonight, not sure if it works.

Take two S1 x S1 surfaces tiled by squares. Let's say that the first one is 10 x 20 and the other is 20 x 10. Then, we should be able to join them by a combination of cubes, tetrahedra and triangular prisms.

On the 10 x 20, the 20 columns would have cubes on odd columns and triangular prisms on the even ones. When joined together, the even columns will vanish in the next layer, leaving only 10 columns.
But, another triangular prism is inserted between each pair of cubes, so the 10 rows double to 20 in the next layer. Tetrahedra are inserted between adjacent triangular prisms, transforming vertical edges in one layer into horizontal edges in the other.

Would this kind of layer work? I drew a rudimentary vertex figure -- it makes it clear that the prisms outside of the layer could be theoretically in two different directions.

[...]

n=4 would be two rings of 8 cubes joined by 16 cubes, 32 triangular prisms and 16 tetrahedra. Hm, I've seen those numbers before -- in fact, this is simply small prismated tesseract x4o3o3x, just seen from an unusual symmetry.

[...]


Unusual symmetry? just duoprismatic symmetry, or Clifford symmetry if you'd like.

Here comes the incidence matrix of sidpith under this special symmetry:
Code: Select all
xo4xx ox4xx&#zx   → height = 0
(tegum sum of 2 interchanged sodips)

o.4o. o.4o.     & | 64 |  1  1  2  2 |  2  2  1  3  4 | 1 1  1  3  2
------------------+----+-------------+----------------+-------------
x. .. .. ..     & |  2 | 32  *  *  * |  2  0  0  2  0 | 1 0  1  2  0
.. x. .. ..     & |  2 |  * 32  *  * |  0  2  0  0  2 | 0 1  0  1  2
.. .. .. x.     & |  2 |  *  * 64  * |  1  1  1  0  1 | 1 1  0  1  1
oo4oo oo4oo&#x    |  2 |  *  *  * 64 |  0  0  0  2  2 | 0 0  1  2  1
------------------+----+-------------+----------------+-------------
x. .. .. x.     & |  4 |  2  0  2  0 | 32  *  *  *  * | 1 0  0  1  0
.. x. .. x.     & |  4 |  0  2  2  0 |  * 32  *  *  * | 0 1  0  0  1
.. .. o.4x.     & |  4 |  0  0  4  0 |  *  * 16  *  * | 1 1  0  0  0
xo .. .. ..&#x  & |  3 |  1  0  0  2 |  *  *  * 64  * | 0 0  1  1  0
.. xx .. ..&#x  & |  4 |  0  1  1  2 |  *  *  *  * 64 | 0 0  0  1  1
------------------+----+-------------+----------------+-------------
x. .. o.4x.     & |  8 |  4  0  8  0 |  4  0  2  0  0 | 8 *  *  *  *
.. x. o.4x.     & |  8 |  0  4  8  0 |  0  4  2  0  0 | * 8  *  *  *
xo .. ox ..&#x    |  4 |  2  0  0  4 |  0  0  0  4  0 | * * 16  *  *
xo .. .. xx&#x  & |  6 |  2  1  2  4 |  1  0  0  2  2 | * *  * 32  *
.. xx .. xx&#x    |  8 |  0  4  4  4 |  0  2  0  0  4 | * *  *  * 16

--- rk
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Re: CRFs with duoprismic symmetry

Postby wendy » Fri May 01, 2015 8:56 am

You could use the lacing-height prism to work these out. In essence, they are oxPxx2xxPxo&#xz. I doubt if you would find many successes, because by the time you hit 6, xxPxo is already flat. The case for x=5 is ox5xf2fx5xo&#xz gives the grand antiprism,
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Re: CRFs with duoprismic symmetry

Postby Klitzing » Fri May 01, 2015 9:58 am

Hmm, Marek, you mentioned in your listings of cases a lot of conditions in x and q short chords. And then added the one with the {n} short chord. But you did not mention the other one with the {2n} short chord each.

Btw., you purposedly skipped the case n=2?

In the first case this would result in your vertices 1 and 3 would have to coincide. While in the latter it results in coincidence of 0 and 2.
The latter obviously brings up a contradiction, as then 0=2 ought be connected to 4 resp. 5 on the one hand by x short chords, on the other by q short chords. Thus this rules out.
But for the first case you'd get
n = 2
0: 1.00 0.00 0.00 0.00
1: 0.00 1.00 0.00 0.00
2: 0.00 0.00 1.00 0.00
4: -0.50 -0.50 0.50 0.50
5: -0.50 -0.50 0.50 -0.50
That one then accordingly ought just have 4 cubes, 8 trips, and 4 tets. If existent. ...
--- rk
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Re: CRFs with duoprismic symmetry

Postby Klitzing » Fri May 01, 2015 10:14 am

Hehe, and Wendy's comment, added meanwhile, would recognize this n=2 case being just some ox2xx2xx2xo&#yz with the lacing edges y to be derived according to the z at the end, i.e. to have a height of zero. This then results due to Wendy's spreadsheet in y = 0.5*q which thus is not the desired x. Therefore that case would not exist either.

In fact as estimated, for I did not know of some polychoron with that cell numbers.

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Re: CRFs with duoprismic symmetry

Postby Klitzing » Fri May 01, 2015 10:19 am

But when solving oxPxx2xxPxo&#xz for P, we just get P=4 or P=4/3. And those then are nothing but the known ones: sidpith and the conjugate quidpith.
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