Bilbirothawroids (D4.3 to D4.9)

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

Re: Johnsonian Polytopes

Postby Keiji » Tue Feb 04, 2014 11:34 pm

Castellated_rhodoperihedral_prism

I'll just leave this here, it's now way past my bedtime ;)
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Feb 04, 2014 11:38 pm

student91 wrote:A further investigation of xfox3oxFx3oooo&#xt gave the following:
the first part works out well.
the second part is, according to xfox3oxFx3oooo&#xt, build up by placing a trigonal prism at a triangle of the ....3ox..3oo..&#xt-tetahedron, and then filled up with square pyramids.
The problem here is that [dichoral angle of trigonal prism]+[dichoral angle of square pyramid]=144.7356103<159.0948426=[dichoral angle of J92 at 4.3]. This means it is not possible with a trigonal prism and a square pyramid. you might want to insert a polytope in beween somewhere, it's dihedral angle should then be between 14.3592323 and 56.169471. Of course you could also use different polytopes. fact is that a lot of what is already build is rigid.


Hm, and it should have triangles in that dihedral angle to fit with the two? Then I don't think you can get that sort of dihedral angle between two triangles. Anything less than 70.5288 will be too narrow to fit anything else to the vertices.

Minimum dihedral angle between two squares seems to be, based on the same argument, 60 and for triangle and square 54.7356.
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Re: Johnsonian Polytopes

Postby student91 » Tue Feb 04, 2014 11:49 pm

maybe xfox3oxFx3ooox&#xt is possible, the triangular cupola (....3..x.x3..o.x&#xt) have a bigger dihedral angle. furthermore, the previous attempt was invalid anyway, as it implied the ...o3...o-vertices to be coplanar with all the o...3o..., .o..3.o.. and ..o.3..o.-vertices.
anyway, xfox3oxFx3oooA&#xt with A being a specific number is possible and convex. if A=x, we have a new CRF-polytope. I suspect A not to be 1 though.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 05, 2014 12:19 am

some cumbersome calculations gave me A=1.288678. That's a pitty

EDIT:Those calculations were wrong, A=f should be the solution, see D4.8 etc.
Last edited by student91 on Thu Apr 03, 2014 4:42 pm, edited 2 times in total.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 05, 2014 12:21 am

student91 wrote:
Marek14 wrote:
quickfur wrote:
Surrounding a tetrahedron is a shard of o5x3o3o. The only other uniform polychoron that uses icosidodecahedra is o5x3o3x.

This would lead into cuboctahedron surrounded by 4 J92's, 4 other cuboctahedra (or possibly triangular cupolas/gyrobicupolas, if those would fit better) and 6 pentagonal prisms. This is a second form that would have to be checked if it can be completed.

What's wrong with putting 4 J92's around a tetrahedron? It would appear that we should be able to at least insert 4 more tetrahedra into the result (i.e. following the pattern of o5x3o3o), which then leaves only the outer surface of hexagons, squares, and triangles, to be closed up. Seems like it should be possible? Or am I missing something obvious?


Well, that was something which was already suggested. My suggestion was meant to be "in addition to", not "in place of" the previous one.

it seems that your polytope is an expansion of mine, so xfox3oxFx3oooo&#xt => xfox3oxFx3xxxx&#xt.

Hmm. Your xfox3oxFx3oooo&#xt looks very promising, actually.

I've worked out, topologically, that attaching 4 J92's around a tetrahedron and adding nothing else, just taking the convex hull, will give a (topological) truncated octahedron cell opposite the tetrahedron. If this truncated octahedron is uniform (very big if!) then the result will be CRF, and will have 1+4+12=17 tetrahedra, 6 square pyramids, 6 triangular cupola, 4 J92's, and 1 truncated octahedron. But all this rests on whether all of the hexagons in the truncated octahedron cell are regular. If not, the square pyramids will not fit, and the shape will not be CRF, and some other way will need to be found to close it up in a CRF way.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 05, 2014 12:26 am

Heh, it seems you've already worked out the same conclusion. :D So the question now is whether the last node is x, or some other quantity. If it's x, then we certainly have a new CRF. But if not, then we'll have to find larger closures for this J92 configuration. Now I'm all tense and expectant, I'm hoping it will work out but I'm not sure. :sweatdrop:

Edit: gah, missed your second post. Are you sure A≠1? That's a pity indeed. :( Maybe I'll try to recalculate and see if it's actually 1 after all. The reason I'm somewhat hopeful is because the dihedral angles of the icosidodecahedron are related to the dihedral angles of the tridiminished icosahedron, and the latter has been proven to have CRF results with tetrahedral symmetry. Of course, the non-icosidodecahedral faces of J92 does throw a monkey wrench into the mix, so maybe my hope is ill-founded after all. Hmm.
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Feb 05, 2014 12:34 am

student91 wrote:That thing is just super awesomely cool. :o_o:
I mean, ursachora are pretty cool, and this one is like 20 times cooler.

Speaking of which, it has some similarity with ursachora. Ursachora are based on the xfo3oox-buildup of a trid.ike, and this one is based on the xFoFx(2)xofox-buildup of the bilunabirotunda. Therefore, as the ursachora are possible with other symmetries as well, this one might be too, so e.g. xFoFx3ooooo3xofox&#xt might be possible as well, and maybe expanded ones, so xFoFx3xxxxx5xofox&#xt too. I'm not sure about this, and just extrapolated the ursachora to this one. Of course I couldn't've done this extrapolation without your awesome discovery. It's real awesome :) :D


student91


Others symmetries? - hmm I have to think that through...

Stott expansion? - surely is!
That is, the pentagons, so far connecting 2 bilbiroes (bilunabirotundae), now would spread apart, giving place for pips (5-prisms) inbetween. The tets (tetrahedra) likewise get extended to tricues (3-cupulae), the peppies (5-pyramids) become pecues (5-cupolae). Further those bilbiro edges, which connect 2 pentagons, formerly were incident to 5 bilbiroes. But now they would spread apart, giving room for further pips. And for sure, the srids (small rhombicosidodecahedra) become grids (great ones).

That is, the cell count of xFoFx3xxxxx5xofox&#xt thus is: 2x grid + 40x tricu + 24x pecu + 30x bilbiro + (60+12)x pip.

If you called the former (contracted) version cool, the expanded one is even cooler! As here there are even more Johnson solids being used, both by types and by total count.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 05, 2014 12:39 am

Klitzing wrote:
student91 wrote:That thing is just super awesomely cool. :o_o:
I mean, ursachora are pretty cool, and this one is like 20 times cooler.

Speaking of which, it has some similarity with ursachora. Ursachora are based on the xfo3oox-buildup of a trid.ike, and this one is based on the xFoFx(2)xofox-buildup of the bilunabirotunda. Therefore, as the ursachora are possible with other symmetries as well, this one might be too, so e.g. xFoFx3ooooo3xofox&#xt might be possible as well, and maybe expanded ones, so xFoFx3xxxxx5xofox&#xt too. I'm not sure about this, and just extrapolated the ursachora to this one. Of course I couldn't've done this extrapolation without your awesome discovery. It's real awesome :) :D


student91


Others symmetries? - hmm I have to think that through...

I'm pretty sure that's not possible, unfortunately.

Stott expansion? - surely is!
That is, the pentagons, so far connecting 2 bilbiroes (bilunabirotundae), now would spread apart, giving place for pips (5-prisms) inbetween. The tets (tetrahedra) likewise get extended to tricues (3-cupulae), the peppies (5-pyramids) become pecues (5-cupolae). Further those bilbiro edges, which connect 2 pentagons, formerly were incident to 5 bilbiroes. But now they would spread apart, giving room for further pips.

Ah, so I miscounted again. :oops: I already gave the cell counts for this Stott expansion, but I missed the pentagonal prisms introduced by the spreading apart of the degree-5 edges. Thanks for giving the right counts! ;)

And for sure, the srids (small rhombicosidodecahedra) become grids (great ones).

That is, the cell count of xFoFx3xxxxx5xofox&#xt thus is: 2x grid + 40x tricu + 24x pecu + 30x bilbiro + (60+12)x pip.

If you called the former (contracted) version cool, the expanded one is even cooler! As here there are even more Johnson solids being used, both by types and by total count.

Yes, and now I have to go and compute the coordinates for the expanded version so I can render it. :mrgreen: Unless you're kind enough to do that for me. :XP:
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Feb 05, 2014 1:02 am

quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.


I still have not fully wrangled my mind around those ones. But surely yours is no valide argument!
Note that the pentagon-pentagon edge of bilbro has 3-fold vertices and as such is fixed. Moreover it connects to a triangle. I.e. we have a 3-5-5 acron. This very acron likewise occurs in teddi!
Note also that the dihedral angle at that mentioned edge between the pentagons of bilbiro is given as arccos(1/sqrt(5)) = 63.434949°. That is, neither 3, nor 4, nor 5 of them fill up a full circuit. But it is this very defekt, which is needed to bend into 4D.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 05, 2014 1:34 am

Klitzing wrote:
quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.


I still have not fully wrangled my mind around those ones. But surely yours is no valide argument!
Note that the pentagon-pentagon edge of bilbro has 3-fold vertices and as such is fixed. Moreover it connects to a triangle. I.e. we have a 3-5-5 acron. This very acron likewise occurs in teddi!
Note also that the dihedral angle at that mentioned edge between the pentagons of bilbiro is given as arccos(1/sqrt(5)) = 63.434949°. That is, neither 3, nor 4, nor 5 of them fill up a full circuit. But it is this very defekt, which is needed to bend into 4D.

--- rk

Sorry, I was unclear in what I wrote. In a later post I clarified what I meant: the 5.3.5.3 vertex of each J91 cell has exactly 3 J91's surrounding it; this forces the dichoral angles of the J91's to be a specific value. So if this pattern of attaching 3 J91's to each other around the 5.3.5.3 vertex is continued, it will force the pattern to close up into an icosahedral symmetry (because of the dihedral angle between two pentagons around that vertex, since it will be analogous to the dodecahedra in the 120-cell). Therefore, this in turn forces the top/bottom edges of the J91's to have degree 5, unless somewhere we break the pattern of attaching 3 J91's around the 5.3.5.3 vertex. The latter is possible, but it will also mean that the result will not look like the current polychoron, since it will require a different arrangement of J91's.
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Feb 05, 2014 1:37 am

quickfur wrote:
student91 wrote:
quickfur wrote:But yeah, I think the trigonal/square versions are not CRF, because the 3 J91's around the vertex are rigid, which in turn fixes the dihedral angle of the triangular faces of the pentagonal pyramids, so they have to be pentagonal pyramids and not tetrahedra or square pyramids, otherwise things won't close up.

Note that the ursachora have similar edges (3 tridiminished icosahedra placed face-to-face, and a pyramid filling the gap). This one also works out.

The difference there is that the number of tridiminished icosahedra around an edge is flexible. Here, the number of J91's around the 5.3.5.3 vertex can only be 3, because otherwise it becomes hyperbolic (the angle of the two pentagons is too wide to fit more than 3 J91's around that vertex). So the dichoral angles between the J91's is fixed, which means that if you continue the pattern of 3 J91's around the 5.3.5.3 vertices, it will eventually form an icosahedral framework of J91's, which means the top/bottom edges will have 5 J91's surrounding it.

Now, it may be possible to have other numbers of J91's around that edge, but it means that you cannot also have the repeating pattern of 3 J91's around the 5.3.5.3 vertices anymore, but you'll need some other arrangement of cells to close the shape up.

Quickfur, no-one is arguing about an other count at that 3.5.3.5 vertex. That one is indeed fixed. Thus you'll have a shard of 3 bilbroes and 2 tets as larger bilding block. But that one can be used freely I think, just as similar patches of triangles and pentagons not only occur within the icosidodecahedron, but also in bilbro (J91) and thawro (triangular hebesphenorotunda, J92), extending it quite differently.

I also think that this shard has kind a triangular shape, when seen from above. So it might well be suitable to build an icosahedron, an octahedron, or an tetrahedron therefrom. And this is exactly what is under discussion. That is, whether those pentagon-pentagon edges of bilbro might become incident by 3, 4, and/or 5.


So, let's try the other way round. Asume their existance, and derive how those then would look like.

xAoAx-3-ooooo-P-xoBox-&#xt, with some A=A(P) and B=B(P). And xAoAx-3-xxxxx-P-xoBox-&#xt for sure.

  • The contracted version for P=5 was Quickfur's first shot. Cells are: 2x srid + 40x tet + 24x peppy + 30x bilbro.
  • The expanded version with P=5 then has cells: 2x grid + 40x tricu + 24x pecu + 30x bilbro + (60+12)x pip.
  • The contracted version for P=4 should have for cells: 2x sirco + 16x tet + 12x squippy + 12x bilbro.
  • The expanded version with P=4 then should have: 2x girco + 16x tricu + 12x squacu + 12x bilbro + 24x pip + 6x cube.
  • The contracted version for P=3 should have for cells: 2x co + (8+8)x tet + 6x bilbro.
  • The expanded version with P=3 then should have: 2x toe + (8+8)x tricu + 6x bilbro + 12x pip + 4x trip.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 05, 2014 2:42 am

Klitzing wrote:[...]
Quickfur, no-one is arguing about an other count at that 3.5.3.5 vertex. That one is indeed fixed. Thus you'll have a shard of 3 bilbroes and 2 tets as larger bilding block. But that one can be used freely I think, just as similar patches of triangles and pentagons not only occur within the icosidodecahedron, but also in bilbro (J91) and thawro (triangular hebesphenorotunda, J92), extending it quite differently.

I also think that this shard has kind a triangular shape, when seen from above. So it might well be suitable to build an icosahedron, an octahedron, or an tetrahedron therefrom. And this is exactly what is under discussion. That is, whether those pentagon-pentagon edges of bilbro might become incident by 3, 4, and/or 5.

If you can prove me wrong, that will be great, since that means more CRFs. :D But the problem I see is this: suppose you try to make an octahedron with J91's as the "edges". This will retain the 3 J91's surrounding each 3.5.3.5 vertex, but have degree 4 at the pentagon-pentagon edge. So let's construct the shape piece-by-piece. Say we make the first octahedron face ABC, where A, B, C are J91's. So far so good. Now let's say we want to make the second octahedron face. Topologically, we can do this by adding a 4th J91, D, to ABC, say to the free pentagonal face of A. The problem is, once you attach D to A, it now shares the 3.5.3.5 vertex with A, and so it makes a fixed dichoral angle with A (since otherwise, it will be impossible to add another J91 to the other pentagon of A on the other side of the 3.5.3.5, since that vertex is rigid.) Since A itself is rigid, you cannot just bend the ABC part and the AD part around the pentagon-pentagon edge to make a different angle; the dichoral angles of AB and AD are fixed. This in turn causes each additional J91 you add to the complex to always make the same dichoral angles with the preexisting cells, so eventually it will fold into icosahedral symmetry. There's no degree of freedom here to allow for alternatives, unfortunately.

The situation is analogous to the rhombic triacontahedron. You can't make a closed shape out of configurations of 1:phi rhombuses without causing the shape not to close up. You especially can't make it into octahedral symmetry; the only way that can happen is if you modify the rhombus to 1:√2 instead, then it will close up into octahedral symmetry. The J91's are like 1:phi rhombuses, if you try to put their acute vertices in a non-pentagonal configuration, you won't be able to close up the shape.

You can, of course, insert other stuff in between the J91's, and you will be able to modify the global closure properties, but if you only use J91's to fold around each other, it will always end up with icosahedral symmetry.

If you can prove me wrong, please do; I'd like to have more interesting CRFs than not. :D But it seems clear to me that it's not possible.



So, let's try the other way round. Asume their existance, and derive how those then would look like.

xAoAx-3-ooooo-P-xoBox-&#xt, with some A=A(P) and B=B(P). And xAoAx-3-xxxxx-P-xoBox-&#xt for sure.

  • The contracted version for P=5 was Quickfur's first shot. Cells are: 2x srid + 40x tet + 24x peppy + 30x bilbro.
  • The expanded version with P=5 then has cells: 2x grid + 40x tricu + 24x pecu + 30x bilbro + (60+12)x pip.
  • The contracted version for P=4 should have for cells: 2x sirco + 16x tet + 12x squippy + 12x bilbro.
  • The expanded version with P=4 then should have: 2x girco + 16x tricu + 12x squacu + 12x bilbro + 24x pip + 6x cube.
  • The contracted version for P=3 should have for cells: 2x co + (8+8)x tet + 6x bilbro.
  • The expanded version with P=3 then should have: 2x toe + (8+8)x tricu + 6x bilbro + 12x pip + 4x trip.

--- rk

If you have coordinates for these, I'll run them through my polytope viewer to prove myself wrong. If such coordinates exist, that is. ;)
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 05, 2014 4:40 am

Klitzing wrote:[...]
  • The contracted version for P=5 was Quickfur's first shot. Cells are: 2x srid + 40x tet + 24x peppy + 30x bilbro.
  • The expanded version with P=5 then has cells: 2x grid + 40x tricu + 24x pecu + 30x bilbro + (60+12)x pip.
[...]

OK, I managed to modify my polytope dissection program that I used to make the uniform truncates, to allow Stott expansion with user-specified resultant edge lengths, so with that, I managed to compute the coordinates for the expanded castellated x5o3x prism (using the fact that it's equal to x5x3x || o5x3F || f5x3o || o5x3F || x5x3x). So here's a first look at this beauty:

Image

For ease of comparison, I use the same coloring scheme as for the non-expanded version, except that here, since the J91's are well-separated, I can show all of them simultaneously without ambiguity with the cell colorings (and without having to solve the graph coloring problem for the rhombic triacontahedron :P). Again, the J91's are seen from a 90° angle so their shape is a bit hard to discern, even though this projection does give you a good idea of the global structure of the polychoron. Next, we do a side-view, as before:

Image

Again, I tweaked the 3D viewpoint a bit so that the J91's are clearly visible. The shape of the projection is actually a flat cylinder, in case it's not obvious. Here you can see the triangular cupolae, pentagonal prisms, and pentagonal cupola laid out in a nice symmetric pattern. Pretty neat, huh?

Anyway, here are the coordinates I used:

Code: Select all
# The two x5x3x's:
apacs<1, 1, 4*phi+1> ~ <±phi>
epacs<1, phi^3, 3+2*phi> ~ <±phi>
epacs<2, phi^2, phi^4> ~ <±phi>
epacs<phi^2, 3*phi, 2*phi^2> ~ <±phi>
epacs<2*phi, 1+3*phi, 2+phi> ~ <±phi>

# The two o5x3F's:
epacs<0, 3+2*sqrt(5), (3+sqrt(5))/2> ~ <±1>
epacs<1, (5+3*sqrt(5))/2, 3+sqrt(5)> ~ <±1>
epacs<(1+sqrt(5))/2, (7+3*sqrt(5))/2, (5+sqrt(5))/2> ~ <±1>

# The central f5x3o:
epacs<0, (1+sqrt(5))/2, 3+2*sqrt(5)> ~ <0>
epacs<1, 1+sqrt(5), (7+3*sqrt(5))/2> ~ <0>
epacs<(5+sqrt(5))/2, 2+sqrt(5), 3+sqrt(5)> ~ <0>


This is the actual input to my new polytope program (which I'm in the process of rewriting); the '~' operator just means concatenation, and 'phi' is a built-in constant equal to the Golden Ratio (1+√5)/2. The '~' operator is a convenient way to factor coordinates in terms of Wendy's epacs/apacs permutation generators for only a subset of the coordinates, without having to list them all out just because the last coordinate isn't permuted.

P.S. I haven't tried computing the P=4 or P=3 variants that Klitzing proposed yet, since I don't think they exist, but I'll be happy to be proven wrong if Klitzing can calculate concrete coordinates for them.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 05, 2014 7:40 am

Klitzing wrote:[...]
So, let's try the other way round. Asume their existance, and derive how those then would look like.

xAoAx-3-ooooo-P-xoBox-&#xt, with some A=A(P) and B=B(P). And xAoAx-3-xxxxx-P-xoBox-&#xt for sure.

  • The contracted version for P=5 was Quickfur's first shot. Cells are: 2x srid + 40x tet + 24x peppy + 30x bilbro.
  • The expanded version with P=5 then has cells: 2x grid + 40x tricu + 24x pecu + 30x bilbro + (60+12)x pip.
  • The contracted version for P=4 should have for cells: 2x sirco + 16x tet + 12x squippy + 12x bilbro.
  • The expanded version with P=4 then should have: 2x girco + 16x tricu + 12x squacu + 12x bilbro + 24x pip + 6x cube.
  • The contracted version for P=3 should have for cells: 2x co + (8+8)x tet + 6x bilbro.
  • The expanded version with P=3 then should have: 2x toe + (8+8)x tricu + 6x bilbro + 12x pip + 4x trip.

--- rk

The problem here is that xAoAx(2)xoBox&#xt isn't always a bilbro. this is only the case with Quickfurs thing. It took a while before I was conviced of their nonexistence too. The vertex quickfur is talking about fixes the dihedraldichoral angle for the pentagons of the bilbro's. this means that the angle of the to-be-inserted pyramid is also fixed, because the bilbro-bilbro-pyramid edge is rigid, and thus fixed with one imput. It might be possible, however, that something with prism-symmetry is possible. e.g. xFoFx3ooooo2xofox&#xt (not very promising) or xFoFx5ooooo2xofox&#xt. they don't close up very nicely though.
Last edited by student91 on Wed Feb 05, 2014 4:56 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 05, 2014 10:39 am

I've been thinking about polystratic polytopes, and got something which i'll post
most polystratic polytopes are based on the buildup of a Johnson solid (ursachora are based on xfo3oox&#xt, the new thing is based on xFoFx2xofox&#xt).
What we do is extend the diagram, giving xfo3oox3ooo&#xt etc.
When I tried J92, I had to insert an A in the diaram, and got xfox3oxFx3oooA&#xt.
In general, I think if you extend something, so add (n)?????(n), you have to choose the new things(????) such that it works out. this way xfo3oox&3xt can be seen as xfo&#xt extended with a 3 to xfo3???&#xt, and then ??? has to be oox to make it close.
Now say we want to try the snub disphenoid. it has D2h-symmetry, so we can use xoAo2oAox&#xt with A being something. we now have to add another line, and hope we get a CRF. Let's try xoAo(n)????(n)oAox&#xt. here ???? are some numbers, and (n) are two integers.
Let's go to xoAo3BooB3oAox&#xt as an example. B is yet another number. if B=1, we have another CRF. probably B≠1, so thats nothing.
We could also use two different things in the middle, so eg xoAo3ABCD3oAox&#xt. This would make the J84's point a little bit outwards, and thus the "base"...o3...D3...x&#xt will be very big. maybe it works if A=0, it might be that D=1 then, but I don't think it will be.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 05, 2014 11:43 am

furthermore, the "rotunda"(in the diminished uniforms section ion the wiki) are also based on some buildup of a polytope:

The icosidodecahedral rotunda can be seen as based on the xox5ofx&#xt-build of the pentagonal rotunda, giving ooo3xox5ofx&#xt and xxx3xox5ofx&#xt

The icosahedral rotunda and the octahedral rotunda are based on the xux3oox&#xt-build of the truncated tetrahedron, giving ooo5xoo3xux&#xt and xxx5xoo3xux&#xt. And ooo4xoo3xux&#xt and xxx4xoo3xux&#xt.

I think those build-ups are what prevents some uniforms from being diminished.

What I'm trying to say is that we could try to make new polytopes by taking the buildup of a polytope, and then add a line to it. If all the points lay in the same plane, the new line can be xxx or ooo.(e.g. xoo3xux&#xt, the xux&#xt forms a hexagon, and thus the vertices of those layers lay in the same plane. At the J92, xfo.3oFx.&3xt makes a pentagon, and thus lays in the same plane, but the last layer doesn't lay in the same plane anymore, and thus the added line should be BBBA). the planes I'm talking about are the planes intersect all paralell hyperplanes.
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Feb 05, 2014 2:16 pm

quickfur wrote:If you can prove me wrong, that will be great, since that means more CRFs. :D But the problem I see is this: suppose you try to make an octahedron with J91's as the "edges". This will retain the 3 J91's surrounding each 3.5.3.5 vertex, but have degree 4 at the pentagon-pentagon edge. So let's construct the shape piece-by-piece. Say we make the first octahedron face ABC, where A, B, C are J91's. So far so good. Now let's say we want to make the second octahedron face. Topologically, we can do this by adding a 4th J91, D, to ABC, say to the free pentagonal face of A. The problem is, once you attach D to A, it now shares the 3.5.3.5 vertex with A, and so it makes a fixed dichoral angle with A (since otherwise, it will be impossible to add another J91 to the other pentagon of A on the other side of the 3.5.3.5, since that vertex is rigid.) Since A itself is rigid, you cannot just bend the ABC part and the AD part around the pentagon-pentagon edge to make a different angle; the dichoral angles of AB and AD are fixed. This in turn causes each additional J91 you add to the complex to always make the same dichoral angles with the preexisting cells, so eventually it will fold into icosahedral symmetry. There's no degree of freedom here to allow for alternatives, unfortunately.

The situation is analogous to the rhombic triacontahedron. You can't make a closed shape out of configurations of 1:phi rhombuses without causing the shape not to close up. You especially can't make it into octahedral symmetry; the only way that can happen is if you modify the rhombus to 1:√2 instead, then it will close up into octahedral symmetry. The J91's are like 1:phi rhombuses, if you try to put their acute vertices in a non-pentagonal configuration, you won't be able to close up the shape.

You can, of course, insert other stuff in between the J91's, and you will be able to modify the global closure properties, but if you only use J91's to fold around each other, it will always end up with icosahedral symmetry.

If you can prove me wrong, please do; I'd like to have more interesting CRFs than not. :D But it seems clear to me that it's not possible.

So, let's try the other way round. Asume their existance, and derive how those then would look like.

xAoAx-3-ooooo-P-xoBox-&#xt, with some A=A(P) and B=B(P). And xAoAx-3-xxxxx-P-xoBox-&#xt for sure.

  • The contracted version for P=5 was Quickfur's first shot. Cells are: 2x srid + 40x tet + 24x peppy + 30x bilbro.
  • The expanded version with P=5 then has cells: 2x grid + 40x tricu + 24x pecu + 30x bilbro + (60+12)x pip.
  • The contracted version for P=4 should have for cells: 2x sirco + 16x tet + 12x squippy + 12x bilbro.
  • The expanded version with P=4 then should have: 2x girco + 16x tricu + 12x squacu + 12x bilbro + 24x pip + 6x cube.
  • The contracted version for P=3 should have for cells: 2x co + (8+8)x tet + 6x bilbro.
  • The expanded version with P=3 then should have: 2x toe + (8+8)x tricu + 6x bilbro + 12x pip + 4x trip.

--- rk

If you have coordinates for these, I'll run them through my polytope viewer to prove myself wrong. If such coordinates exist, that is. ;)


Hehe, even so not coordinates based, your arguments sound rather convincing.
Thus, most probably the last 4 of the above zoo won't exist (as already was somehow doubted).

And, to contrast the ursachora, where not only such P <> 5 did exist, but also the same dihedral angle of used cells (bilbiro resp. teddi) occurs, it seems to be indeed the existance of those rigid (equatorial) vertices, which at first govern the slope of the incident bilbiroes, and thus, by means of directly or parallelly attached similar sharps, the global behaviour...

This special lordship of medial vertices in the consideration of multistratic axial polytopes then seems new. At least I cannot remember having come across such a surprising feature within the monostratics so far...

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Feb 05, 2014 3:59 pm

student91 wrote:The problem here is that xAoAx(2)xoBox&#xt isn't always a bilbro. this is only the case with Quickfurs thing. It took a while before I was conviced of their nonexistence too. The vertex quickfur is talking about fixes the dihedral angle for the pentagons of the bilbro's. this means that the angle of the to-be-inserted pyramid is also fixed, because the bilbro-bilbro-pyramid edge is rigid, and thus fixed with one imput. It might be possible, however, that something with prism-symmetry is possible. e.g. xFoFx3ooooo2xofox&#xt (not very promising) or xFoFx5ooooo2xofox&#xt. they don't close up very nicely though.

Still wrong, student91! Quickfiur was not dealing with the 3-5-5 vertex of the bilbiroes, which surely makes that cell rigid within 3D, but tells nothing about its surroundings. He rather was dealing with the 3-5-3-5 vertex of the bilbiroes, which makes the surrounding cells become rigid within 4D, as that one allows in here for 3 bilbiroes and 2 tets only! (Or its Stott expansion ...)

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 06, 2014 5:33 am

Earlier today, I discovered a most fascinating polychoron involving J92 cells, but unfortunately it is only a near-miss crown jewel, not an actual one, because it cannot be made completely CRF. :( Nevertheless, it has an extremely interesting structure, and the last bits that aren't CRF have edge lengths that almost admit a CRF modification, so I thought I'd post it here.

I found this polychoron almost "by mistake", because I was experimenting with attaching two J92's together at their hexagonal faces in gyro orientation at various dichoral angles, and I randomly tried 60° (after trying a few others with no interesting results). I was instantly confronted with the most amazing series of coincidences involving certain edge lengths and coordinates that somehow just work out right: first, almost all lacing edges between the J92's became unit length, resulting in CRF square pyramids connecting the two J92's, and then I noticed that just by adding a single vertex antipodal to the hexagonal face (at a distance of exactly the Golden Ratio φ away!), I was able to make almost the entire polytope CRF. To give you an idea of how amazing this was, take a look at this:

Image

This is a view of the polychoron from a viewpoint opposite of the hexagonal face between the two J92's. The vertex in the middle where the two tetrahedra meet is the vertex exactly φ units from the hexagonal face on the far side of the polytope. Surrounding these two tetrahedra is a most amazing configuration of alternating tridiminished icosahedra that somehow just exactly fit into the gaps between the two J92's!

Now let's look at the two J92's:

Image

As you can see, the two J92's are laced by a series of square pyramids, of all things, that somehow just work out to be CRF at this dichoral angle!

The only remaining edges that are not unit length is exactly φ in length (you can see them between the tips of the respective pentagons of the J92's). These edges are surrounded by tetrahedra that are unit edge length except for one edge that is length φ. Given the tridiminished icosahedra and coincidences involving φ popping up everywhere in the coordinates, my immediate thought was that adding a few more vertices should allow us to replace these scalene tetrahedra with CRF florets of tetrahedra, like in the 600-cell, since the φ edge length suggests that we could break up the edge by "completing the pentagon". But alas, when I calculated the coordinates of the prospective vertices, I found that the points would define a phi-scaled hexagon that lies in the same 2-plane as the hexagon between the two J92's and in the same 3-plane as the J92's, which means that the hexagon will become internal to the polychoron, the square pyramids will disappear, and the J92's merge into non-CRF cells. So close, yet so far! :cry:
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 06, 2014 7:48 am

quickfur wrote:[...]
The only remaining edges that are not unit length is exactly φ in length (you can see them between the tips of the respective pentagons of the J92's). These edges are surrounded by tetrahedra that are unit edge length except for one edge that is length φ. Given the tridiminished icosahedra and coincidences involving φ popping up everywhere in the coordinates, my immediate thought was that adding a few more vertices should allow us to replace these scalene tetrahedra with CRF florets of tetrahedra, like in the 600-cell, since the φ edge length suggests that we could break up the edge by "completing the pentagon". But alas, when I calculated the coordinates of the prospective vertices, I found that the points would define a phi-scaled hexagon that lies in the same 2-plane as the hexagon between the two J92's and in the same 3-plane as the J92's, which means that the hexagon will become internal to the polychoron, the square pyramids will disappear, and the J92's merge into non-CRF cells. So close, yet so far! :cry:

cool thing!!! I think your calculations went wrong somewhere, or you're trying to make an other pentagon than I think you want. That thing has a lace-city that looks like two towers xFxo3xofx&#xt and xofx3xFxo&#xt attached at the x3x with an 60degree angle. The edges connectiong F3o with f3x have length 1. The edges connection F3o with o3F have length phi. the edges connection x3x with F3o have length 1. I thought you would make a pentagon containing the x3x-F3o edges and the F3o-o3F phi-edge. This pentagon would clearly not be coplanar with the hexagon, as the x3x-F3o edges are not coplanar with it as well. So or you're trying to make a different pentagon, or you are doing something wrong, or making a pentagon won't work out. Very cool thing though :D
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 06, 2014 12:05 pm

as far as I see it, with the vertices inserted, the lace-city would look somewhat like this:
Code: Select all
                       o3x

                x3f

       F3o
            A3B
x3x                        o3o
            B3A
       o3F

                f3x

                       x3o

I get stuck at calculating A and B.
But In this picture, you can see that when A and B are calculated, the vertices won't lay in the same plane, and also not in the same 3-space as the J92's. I hope we can get a CRF out of this
EDIT: it appears B=sqrt(3)/(2phi)
And A=sqrt((3-phi)/5) +sqrt(4/5) -1/(2phi)
I'm not sure about those calculations, I tend to miscalculate often.

the far part is especially interestion, it has ooxfx3oxfxo&#xr, another ring!!!. this is even more interesting than K4.109, as that one was a 4-ring
Last edited by student91 on Thu Feb 06, 2014 1:15 pm, edited 2 times in total.
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 06, 2014 12:52 pm

Klitzing wrote:Still wrong, student91! Quickfiur was not dealing with the 3-5-5 vertex of the bilbiroes, which surely makes that cell rigid within 3D, but tells nothing about its surroundings. He rather was dealing with the 3-5-3-5 vertex of the bilbiroes, which makes the surrounding cells become rigid within 4D, as that one allows in here for 3 bilbiroes and 2 tets only! (Or its Stott expansion ...)

--- rk

I'm not dealing with that vertex either. I'm dealing with quickfurs vertex, esp. with the bilbro-bilbro-tet edge, which fixes the dichoral angle for the pentagon. This is the same pentagon at the bilbro-bilbro-pyramid edge, so the dihedral angle for the pyramid is fixed. I know I mistyped the "dichoral" in my previous post, and that's probably what caused the misconception.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 06, 2014 3:28 pm

student91 wrote:as far as I see it, with the vertices inserted, the lace-city would look somewhat like this:
Code: Select all
                       o3x

                x3f

       F3o
            A3B
x3x                        o3o
            B3A
       o3F

                f3x

                       x3o

I get stuck at calculating A and B.
But In this picture, you can see that when A and B are calculated, the vertices won't lay in the same plane, and also not in the same 3-space as the J92's. I hope we can get a CRF out of this
[...]

Hmm. I think you're right!!! There may be a way to get a CRF out of this yet! I'll see if I can figure out the coordinates for those new points...
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Feb 06, 2014 4:19 pm

Haha,
consider the hexacosachoron (ex = x3o3o5o) being viewed along a 3-fold axis. The corresponding lace city is
Code: Select all
                o3o               
                                  
        x3o   o3f f3o   o3x       
                                  
                                  
 o3o o3f   f3x       x3f   f3o o3o
                                  
   f3o       o3F   F3o       o3f   
                                  
                                  
o3x   x3f F3o   f3f   o3F f3x   x3o
                                  
                                  
   f3o       o3F   F3o       o3f   
                                  
 o3o o3f   f3x       x3f   f3o o3o
                                  
                                  
        x3o   o3f f3o   o3x       
                                  
                o3o               

You recognize that pattern? In fact it is nothing but the figure shown by "student91" (with the missing numbers A and B already being calculated), multiplied by 6 around the central hexagon. - Really? - Not quite! The central hexagon should have been x3x, not f3f. (But reserve that for later.)

You can also "see" the 6 diametral lines (one of which is horizontally displayed) through the center. Those represent the equatorial icosidodecahedra (id, o3x5o). If such an id itself would be cut into 2 (by a plane parallel to a triangular face), it would result in the tower, displayed above as a radius, i.e. the lace towers oxFf3xfof&#xt. That plane moreover cuts all 6 tropal pentagons into golden x-x-f triangles respectively golden x-x-x-f trapeziae.

Now consider that half of id. Reject for a moment the bottom tau scaled hexagon and all those golden polygons (parts of the former pentagons). Then you'll get 6 triangles, each with 2 open sides, the tip of which exactly connecting to the bottom plane. Consider further that those triangles could be spun around the remaining connected side. Then there would be a different angle too, again having the tip connected to the bottom plane. This time they won't point out, rather they would point in. - We could now ask about the pairwise distances of those 6 tips, when bowing to this new constraint. - Well, the answer here is surprisingly simple. They would be spaced at unit distance! In fact, this is how thawro (J92, triangular hebesphenorotunda) is related to id! And yes, those former golden triangles resp. tetragons then become regular ones. And sure, the bottom hexagon thus becomes unit edged too. (Esp. the height of thawros, when measured between the hexagon and the opposite triangle, is exaclty the inradius of id, whith resp. to its triangles.)

That is, the dissected pentagons of that half of id, represent equatorial sections of rosettes of 5 tets (tetrahedra) of ex. the regular triangles then are bottom faces of other tets. That is, when bending in the triangles around the still connected edge, this represents a bending in of the corresponding tets around the then still connected face. The trapezium deforms into a square. Accordingly that rosette of 5 tets gets replaced by a squippy. And the golden triangle becomes a regular one. That is the other partial rosette of 5 tets becomes replaced by a single tet. That's all. - The remaining structure can be kept. I.e., when finally taking one 6th, you'd re-introduce those so far blended out thawros. You'd get exactly the desired structure!

Thus, finally, this new find of quickfur indeed seems valide. (In fact, that doubted f-edge can be seen from his own renders as being nothing but a false edge, the diagonal of 2 bottom-attached tetrahedra.)

But we not only get this figure now, but a second one for free as well! This is, as quickfur's find in fact is a lune polychoron. As he already pointed out himself, a 60 degree lune. - What about the corresponding 120 degree lune? (Perhaps we might have to fiddle a bit next to the x3x, so. Similar to the n/10-lunes of ex.)

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 06, 2014 5:11 pm

Klitzing wrote:Haha,
consider the hexacosachoron (ex = x3o3o5o) being viewed along a 3-fold axis. The corresponding lace city is
Code: Select all
                o3o               
                                  
        x3o   o3f f3o   o3x       
                                  
                                  
 o3o o3f   f3x       x3f   f3o o3o
                                  
   f3o       o3F   F3o       o3f   
                                  
                                  
o3x   x3f F3o   f3f   o3F f3x   x3o
                                  
                                  
   f3o       o3F   F3o       o3f   
                                  
 o3o o3f   f3x       x3f   f3o o3o
                                  
                                  
        x3o   o3f f3o   o3x       
                                  
                o3o               

You recognize that pattern? In fact it is nothing but the figure shown by "student91" (with the missing numbers A and B already being calculated), multiplied by 6 around the central hexagon. - Really? - Not quite! The central hexagon should have been x3x, not f3f. (But reserve that for later.)

After φ popped up everywhere in my coordinates and the tridiminished icosahedra appeared, I pretty much realized that we're dealing with some sort of diminished 600-cell, albeit with some non-obvious cuts that introduces J92's and square pyramids. In fact, if you add tridiminished icosahedra pyramids to the far side of my polychoron, you'll get a patch of the 600-cell's surface.

[...]
Now consider that half of id. Reject for a moment the bottom tau scaled hexagon and all those golden polygons (parts of the former pentagons). Then you'll get 6 triangles, each with 2 open sides, the tip of which exactly connecting to the bottom plane. Consider further that those triangles could be spun around the remaining connected side. Then there would be a different angle too, again having the tip connected to the bottom plane. This time they won't point out, rather they would point in. - We could now ask about the pairwise distances of those 6 tips, when bowing to this new constraint. - Well, the answer here is surprisingly simple. They would be spaced at unit distance! In fact, this is how thawro (J92, triangular hebesphenorotunda) is related to id! And yes, those former golden triangles resp. tetragons then become regular ones. And sure, the bottom hexagon thus becomes unit edged too. (Esp. the height of thawros, when measured between the hexagon and the opposite triangle, is exaclty the inradius of id, whith resp. to its triangles.)

I had already realized (thanks to Wikipedia's triangular hebesphenorotunda article :P) that J92 is nothing but a bisected o5x3o, with the 6 equatorial vertices scaled down by phi (conversely, if you scale the hexagonal face of J92 by phi, you'll end up with a bisected o5x3o). So when those tridiminished icosahedra showed up, I immediately recognized it as some kind of diminished 600-cell. But what I wasn't sure about, was whether scaling those equatorial vertices down by phi would maintain CRF-ness in the result, and my initial calculations suggest no. But after student91 kindly provided the lace city for the polychoron, I realized that I've been unconsciously lapsing into 3D-centric thinking again, and so my calculations went in the wrong direction -- in fact, the coplanar coordinates I found were exactly those that turned the J92's back into bisected o5x3o's, which is not what is desired in this case. Nevertheless, on the way to work this morning, I realized that it must be possible to insert two points where the phi-scaled edges are, and get a CRF patch of the 600-cell again.

That is, the dissected pentagons of that half of id, represent equatorial sections of rosettes of 5 tets (tetrahedra) of ex. the regular triangles then are bottom faces of other tets. That is, when bending in the triangles around the still connected edge, this represents a bending in of the corresponding tets around the then still connected face. The trapezium deforms into a square. Accordingly that rosette of 5 tets gets replaced by a squippy. And the golden triangle becomes a regular one. That is the other partial rosette of 5 tets becomes replaced by a single tet. That's all. - The remaining structure can be kept. I.e., when finally taking one 6th, you'd re-introduce those so far blended out thawros. You'd get exactly the desired structure!

Thus, finally, this new find of quickfur indeed seems valide. (In fact, that doubted f-edge can be seen from his own renders as being nothing but a false edge, the diagonal of 2 bottom-attached tetrahedra.)

Wow. Two crown jewels in two days??! I think I just died and went to heaven...

But we not only get this figure now, but a second one for free as well! This is, as quickfur's find in fact is a lune polychoron. As he already pointed out himself, a 60 degree lune. - What about the corresponding 120 degree lune? (Perhaps we might have to fiddle a bit next to the x3x, so. Similar to the n/10-lunes of ex.)

--- rk

Not only so, I think it might be possible to get another 600-cell diminishing that has just a single J92 cell, maybe with a triangular cupola and some square pyramids to interface the hexagon with the rest of the 600-cell's surface.

This is pretty crazy. We knew the 600-cell had a lot of diminishings, but after the discovery of 600-cell lunae and then these J92 diminishings, I think it has far more CRFs hiding in it than we ever guessed. For example, I have some ideas that may work out to be 600-cell diminishings with J91 cells, although I haven't actually worked out any actual calculations yet.
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Feb 06, 2014 5:32 pm

Maybe you should publish somewhere about these crown jewels...
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 06, 2014 6:24 pm

Marek14 wrote:Maybe you should publish somewhere about these crown jewels...

You mean in a math journal, or just online? 'cos they are already listed on the wiki that goes with this forum: http://teamikaria.com/hddb/wiki/Crown_jewel
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Feb 06, 2014 6:53 pm

No, I meant in a journal. After all, we have quite a lot of unique results on the forum, don't we?
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 06, 2014 7:15 pm

Cool, it exists, and it's extremely awesome.
I still have a little doubt, and I hope you can prove me wrong easilly.
doesn't scaling down the hexagon place the hexagon inside of everything, making it non-convex? I mean, quickfurs program wouldn't've drawn that edge for nothing, or quickfur just told the program to.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 06, 2014 7:27 pm

student91 wrote:Cool, it exists, and it's extremely awesome.
I still have a little doubt, and I hope you can prove me wrong easilly.
doesn't scaling down the hexagon place the hexagon inside of everything, making it non-convex? I mean, quickfurs program wouldn't've drawn that edge for nothing, or quickfur just told the program to.

Wait, you mean Klitzing is claiming that those vertices will give a CRF result just by themselves? Because that is false, the images my program outputs is based on a convex hull calculation from the vertices; I did not insert the edges by hand!

If that's the case, then this is not a true CRF after all. :(

OTOH, I think it's possible to insert two new vertices per phi-scaled edge, as you suggested; that should complete the pentagon in the correct way and should yield a CRF, or something that's easily CRF-able (I think it might introduce more phi-scaled edges with the vertices on the far side, but that's easily fixed by adding more vertices from the 600-cell to turn them into tetrahedral florets).
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