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 » Thu Feb 20, 2014 8:20 am

Marek14 wrote:Why is face 39 not closed...?


I don't know, ask quickfur... all I'm doing is passing your input to his program after all. ;)
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 20, 2014 3:11 pm

This error sometimes happens when you have redundant vertices on the same hyperplane (e.g., if you have a hexagon with an extra coplanar point in the middle, or a cell with extra points inside it). The cause is that the polytope building code tries to construct a polygon from the vertices, and gets confused because it can't find a unique circuit of edges that closes the polygon (extra edges are produced because the redundant vertex is incident on that face, and the lattice enumeration algorithm doesn't realize it's redundant).
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 20, 2014 6:00 pm

quickfur wrote:[...]
I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...

I just discovered that if you delete all vertices of which abs( x-coordinate ) <phi/2 (or just phi if the edge-length is two ;) ), and then add the vector (-phi/2,0,0,0) to the vertices with x-coordinate>=phi/2 and add the vector (phi/2,0,0,0) to the vertices with x-coordinate <=-phi/2, you should get the bilbiro'd o5x3o3x. you only have to insert the pentagonal prisms to make it CRF then.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 20, 2014 6:30 pm

student91 wrote:
quickfur wrote:[...]
I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...

I just discovered that if you delete all vertices of which abs( x-coordinate ) <phi/2 (or just phi if the edge-length is two ;) ), and then add the vector (-phi/2,0,0,0) to the vertices with x-coordinate>=phi/2 and add the vector (phi/2,0,0,0) to the vertices with x-coordinate <=-phi/2, you should get the bilbiro'd o5x3o3x. you only have to insert the pentagonal prisms to make it CRF then.

Yes, that's what I did last night, but it turns out that 12 of the o5x3o's have their pentagons removed, resulting in non-CRF cells with phi-scaled pentagons. These pentagons moreover are too close to each other in the northern/southern hemispheres, resulting in sub-unit edges, and also introduce a whole bunch of non-CRF pyramids of the triangular cupola fragments. However, deleting these pentagons from these non-CRF diminished o5x3o's turns them into CRF pentagonal rotunda, and introduces some double-length edges. Probably adding a few more vertices (your pentagonal prisms?) will make it CRF. We shall see. I'm currently trying to fix an error I made in the model somewhere else, but the lattice enumeration algorithm is taking a long time (too many cells in o5x3o3x!).

Edit: the problematic pentagons have vertices at x=±1, so subtracting phi/2 causes them to introduce sub-unit edges.
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 20, 2014 8:24 pm

quickfur wrote:
student91 wrote:
quickfur wrote:[...]
I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...

I just discovered that if you delete all vertices of which abs( x-coordinate ) <phi/2 (or just phi if the edge-length is two ;) ), and then add the vector (-phi/2,0,0,0) to the vertices with x-coordinate>=phi/2 and add the vector (phi/2,0,0,0) to the vertices with x-coordinate <=-phi/2, you should get the bilbiro'd o5x3o3x. you only have to insert the pentagonal prisms to make it CRF then.

Yes, that's what I did last night, but it turns out that 12 of the o5x3o's have their pentagons removed, resulting in non-CRF cells with phi-scaled pentagons. These pentagons moreover are too close to each other in the northern/southern hemispheres, resulting in sub-unit edges, and also introduce a whole bunch of non-CRF pyramids of the triangular cupola fragments. However, deleting these pentagons from these non-CRF diminished o5x3o's turns them into CRF pentagonal rotunda, and introduces some double-length edges. Probably adding a few more vertices (your pentagonal prisms?) will make it CRF. We shall see. I'm currently trying to fix an error I made in the model somewhere else, but the lattice enumeration algorithm is taking a long time (too many cells in o5x3o3x!).

Edit: the problematic pentagons have vertices at x=±1, so subtracting phi/2 causes them to introduce sub-unit edges.

Those f-pentagon vertices should be removed. To make it CRF again, you should insert one pentagonal prism per "gap", as shown in my lace city's. After that's done, everything should be CRF. :) I don't know a fast way to calculate the coordinates of the pentagonal prisms, but they should be located at two x5o3A's, with A some number. I know it's not the easiest CRF to calculate coordinates for :\ .
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 20, 2014 9:27 pm

student91 wrote:[...]
Those f-pentagon vertices should be removed. To make it CRF again, you should insert one pentagonal prism per "gap", as shown in my lace city's. After that's done, everything should be CRF. :) I don't know a fast way to calculate the coordinates of the pentagonal prisms, but they should be located at two x5o3A's, with A some number. I know it's not the easiest CRF to calculate coordinates for :\ .

Good, so we're on the same page. :D Here's where I got so far:

Image

This is a view of (half of) the equatorial cells, with the southern hemisphere rendered in solid yellow (so that other northern hemisphere cells don't obscure the image), and the northern hemisphere cells completely omitted. The green cells are the J91's obtained after squashing the vertices together. (In order to save a bit of time, I also cut off the tristratic caps from the north/south poles to reduce the number of vertices, that's why the top/bottom of the image appears slightly flattened from the usual spherical shape. But that shouldn't matter since the J91's are far away from that region.)

You can see long decagonal prisms between the J91's obtained after deleting the f-pentagons; this is where the double-length edges are. So if you look carefully at the non-CRF hexahedra to the left and right of the center green J91, for example, you can see that it's actually a bisected hexagonal prism. To either side of this bisected prism, are non-CRF cells with 2 hexagons sharing an edge, with triangles at either end of the edge. Basically, this looks like some kind of bisected truncated tetrahedron.

So, it appears that if we "complete the hexagons" by extending those bisected hexagonal prisms into full CRF hexagonal prisms, then the bisected truncated tetrahedra should become CRF truncated tetrahedra surrounding the tall decagonal prisms, which would be broken up into 2 pentagonal cupola touching at a pentagonal prism each. Basically, exactly as student91 says: we have to add those pentagonal prisms (and hope they don't protrude too far out in 4D to cause adjacent cells to become internal!). The good news is that their first coordinates are known to be ±1, and their configuration is known to be x5o3A, so the trick here is just to find A such that we get unit edges from the top/bottom edges of those truncated hexagonal prisms. I'm going to try to calculate A, but anyone else who'd like to volunteer for that would be greatly appreciated. ;)
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 12:25 am

Hmm. Using the trick of "completing the hexagon", that is, by adding the edge vectors of the incomplete hexagons in the bisected hexagonal prisms to find the missing points, I managed to obtain two sample points of the pentagonal prisms on either side of a J91, and measuring the distance between them yielded A=6*phi. I'm going to try to compute x5o3A now and see if that works...
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Feb 21, 2014 6:26 am

quickfur wrote:Hmm. Using the trick of "completing the hexagon", that is, by adding the edge vectors of the incomplete hexagons in the bisected hexagonal prisms to find the missing points, I managed to obtain two sample points of the pentagonal prisms on either side of a J91, and measuring the distance between them yielded A=6*phi. I'm going to try to compute x5o3A now and see if that works...

If phi = 1.618, then it cannot be true, at least as long you use unit edges.

Consider the following:
those pentagon vertices will have a distance which allows for a (non-straight) path, being a shortchord of a hexagon (at hip), then a shortchord of a pentagon (at bilbiro), and then again of a hexagon (at hip). That is, according to the triangle inequality (or rather: tetragon inequality, hehe) you'd have A ≤ h + f + h = 1.732 + 1.618 + 1.732 = 5.082.

Whereas your value would be 9.708.

Still, your idea by completing the hexagons should work - if done correctly...

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Re: Johnsonian Polytopes

Postby Marek14 » Fri Feb 21, 2014 6:35 am

BTW, does anyone have off files for all the ursachora and 600-cell lunes?
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 6:39 am

Klitzing wrote:
quickfur wrote:Hmm. Using the trick of "completing the hexagon", that is, by adding the edge vectors of the incomplete hexagons in the bisected hexagonal prisms to find the missing points, I managed to obtain two sample points of the pentagonal prisms on either side of a J91, and measuring the distance between them yielded A=6*phi. I'm going to try to compute x5o3A now and see if that works...

If phi = 1.618, then it cannot be true, at least as long you use unit edges.

Consider the following:
those pentagon vertices will have a distance which allows for a (non-straight) path, being a shortchord of a hexagon (at hip), then a shortchord of a pentagon (at bilbiro), and then again of a hexagon (at hip). That is, according to the triangle inequality (or rather: tetragon inequality, hehe) you'd have A ≤ h + f + h = 1.732 + 1.618 + 1.732 = 5.082.

Whereas your value would be 9.708.

Still, your idea by completing the hexagons should work - if done correctly...

--- rk

Sorry, I forgot to mention that I customarily work with edge length 2, so the value of A for me is 6*phi, but if you're using unit edge it should be 3*phi. And I just built the model, and all edge lengths are equal, so looks like it's CRF after all! I'll post images later -- I just tweaked something and running the convex hull / lattice enumeration again, so by tomorrow I'll be able to post images.

Good job, student91, looks like your idea worked! :D
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 6:41 am

Marek14 wrote:BTW, does anyone have off files for all the ursachora and 600-cell lunes?

I haven't built all of the ursachora, but everything that I made renders of, I have the .def files, so I can convert them to .off easily. So whatever has images posted here or on the wiki, I can give you the .off files. I'll see if I can post some of them tomorrow, but I might be busy, so maybe Saturday.
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Re: Johnsonian Polytopes

Postby Marek14 » Fri Feb 21, 2014 6:42 am

quickfur wrote:
Marek14 wrote:BTW, does anyone have off files for all the ursachora and 600-cell lunes?

I haven't built all of the ursachora, but everything that I made renders of, I have the .def files, so I can convert them to .off easily. So whatever has images posted here or on the wiki, I can give you the .off files. I'll see if I can post some of them tomorrow, but I might be busy, so maybe Saturday.


No problem :)
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 3:28 pm

Marek14 wrote:BTW, does anyone have off files for all the ursachora and 600-cell lunes?

Here are the .off files for the 600-cell lunae:

Hemi-600-cell (pseudo-bisected 600-cell, 5/10 luna)
Wedge #1 (4/10 luna)
Wedge #2 (3/10 luna)
Wedge #3 (2/10 luna)
Wedge #4 (1/10 luna, aka augmented pentagonal gyrobirotundular ring)
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Re: Johnsonian Polytopes

Postby student91 » Fri Feb 21, 2014 3:45 pm

quickfur wrote:Sorry, I forgot to mention that I customarily work with edge length 2, so the value of A for me is 6*phi, but if you're using unit edge it should be 3*phi. And I just built the model, and all edge lengths are equal, so looks like it's CRF after all! I'll post images later -- I just tweaked something and running the convex hull / lattice enumeration again, so by tomorrow I'll be able to post images.

Good job, student91, looks like your idea worked! :D

Hurray!! :D I think we've now found most of the bilbiro'd polytopes. :) It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed. :)
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 3:49 pm

quickfur wrote:[...] And I just built the model, and all edge lengths are equal, so looks like it's CRF after all! I'll post images later -- I just tweaked something and running the convex hull / lattice enumeration again, so by tomorrow I'll be able to post images.

Good job, student91, looks like your idea worked! :D

Alright, here is a quick render from the same viewpoint as before:

Image

As you can see, the long decagonal prisms have now been replaced by two pentagonal cupolae and a pentagonal prism each, and the previous non-CRF bisected cells are now CRF hexagonal prisms and truncated tetrahedra. My polytope viewer confirms that all edge lengths are equal.

Here's a projection from the usual viewpoint, with the J91's lying at the limb of the projection image:

Image

The green cells are the J91's; the hexagons between them are the bases of triangular cupolae. The circles of 5 alternating pentagons (hexagon+triangle) and rectangles (2 squares) are the images of the new truncated tetrahedra and hexagonal prisms,and the pentagons in the middle are the new pentagonal prisms. You can see the outlines of the pentagonal cupolae on either end of these prisms. All in all, a rather interesting configuration of cells.

I have to run now, will post more renders later.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 5:51 pm

student91 wrote:[...]
quickfur wrote:Good job, student91, looks like your idea worked! :D

Hurray!! :D I think we've now found most of the bilbiro'd polytopes. :)

Oh? How do you know that? Did you discover something we don't know, that limits the number of bilbiro-able polytopes?

It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed. :)

I'd love to hear about that! :)
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Re: Johnsonian Polytopes

Postby Marek14 » Fri Feb 21, 2014 6:17 pm

Here's a question:

I had a look at the hemi-600cell now. Of course, the dihedral angles at its icosidodecahedron won't allow gluing two together (which is also clear from its construction).
But -- what if the bilbiro process was used on this icosidodecahedron? Could that allow for duplicating the shape? Basically, it would be bilbiroing the middle layer of x3o3o5o, despite the fact that it usually doesn't contain any icosidodecahedra.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 8:09 pm

Marek14 wrote:Here's a question:

I had a look at the hemi-600cell now. Of course, the dihedral angles at its icosidodecahedron won't allow gluing two together (which is also clear from its construction).
But -- what if the bilbiro process was used on this icosidodecahedron? Could that allow for duplicating the shape? Basically, it would be bilbiroing the middle layer of x3o3o5o, despite the fact that it usually doesn't contain any icosidodecahedra.

Not sure what you mean, the hemi-600cell basically has a bunch of bisected tetrahedra deleted and replaced with inward-slanting pentagonal pyramids (in the sense that the bisected tetrahedra are perpendicular to the bisecting hyperplane, but the pentagonal pyramids are curved inwards towards the center of the original 600-cell, rather than being perpendicular to the bisecting hyperplane). Gluing two hemi-600cells together would yield the 600-cell with concave holes precisely where these pentagonal pyramids are. Of course, they happen to be closeable with florets of 5 tetrahedra (or equivalently, plugging the holes with line||pentagon pieces), then you get the full 600-cell back.

I'm not sure where bilbiroing comes in here. Are you suggesting to glue two hemi-600cells together with a belt of bilbiros in the middle? I'm not sure how that would work. What orientation should the bilbiroes be in? Should they touch the two halves with their top/bottom edges, or their square faces, or their pentagon-triangle-pentagon-triangle faces?

Or are you suggesting to contract the icosidodecahedron into a bilbiro, and somehow modifying the tetrahedra above it to fit?

One alternative CRF route I have been thinking of, is to make 600-cell wedges that are not aligned to cut the icosidodecahedron into two pentagonal rotundae, but instead aligned to cut them parallel to two opposite triangles. Then the non-CRF bisected icosidodecahedra in the result can have their phi-scaled hexagons contracted to unit edge, to transform them into J92's, with some bridging cells to link them to the corona of tetrahedra above. I suspect we might run into some close relatives of the J92 rhombochoron here.
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Re: Johnsonian Polytopes

Postby student91 » Fri Feb 21, 2014 8:19 pm

quickfur wrote:
student91 wrote:[...]
quickfur wrote:Good job, student91, looks like your idea worked! :D

Hurray!! :D I think we've now found most of the bilbiro'd polytopes. :)

Oh? How do you know that? Did you discover something we don't know, that limits the number of bilbiro-able polytopes?
well, I thought bilbiro's are only derivable from id's, and also only from the .2.-view of the id. Now we've bilbiro'd all things with o5x3o3. So that means any similar bilbiro-ing (we might call it "obvious" bilbiro-ing) doesn't exist. Any less-obvious bilbiro-ing of course is not excluded, but I suspect it to be not very productive.
It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed. :)

I'd love to hear about that! :)

I don't have anything yet, I just was saying what I was going to innvestigate next, and when I'm done investigating (or stuck) I'll of course share my results with this forum.
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Re: Johnsonian Polytopes

Postby Marek14 » Fri Feb 21, 2014 8:37 pm

quickfur: Right, I guess it wouldn't work...
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 21, 2014 9:11 pm

Marek14 wrote:quickfur: Right, I guess it wouldn't work...

I didn't say it wouldn't work, I just wasn't clear what you meant. Perhaps there is a way to insert bilbiro's between the two hemi-600cells that would produce a CRF. Might be worth looking into?
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Re: Johnsonian Polytopes

Postby Marek14 » Fri Feb 21, 2014 9:18 pm

quickfur wrote:
Marek14 wrote:quickfur: Right, I guess it wouldn't work...

I didn't say it wouldn't work, I just wasn't clear what you meant. Perhaps there is a way to insert bilbiro's between the two hemi-600cells that would produce a CRF. Might be worth looking into?


Well, my original idea was to replace the icosidodecahedron, but in the previous replacements it was side element, which is not the case here.
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Feb 21, 2014 11:21 pm

quickfur wrote:
Marek14 wrote:Here's a question:

I had a look at the hemi-600cell now. Of course, the dihedral angles at its icosidodecahedron won't allow gluing two together (which is also clear from its construction).
But -- what if the bilbiro process was used on this icosidodecahedron? Could that allow for duplicating the shape? Basically, it would be bilbiroing the middle layer of x3o3o5o, despite the fact that it usually doesn't contain any icosidodecahedra.

Not sure what you mean, ...


To me it's clear.
Look at this lace city of ex:
Image
Top red Point is just a point. Next layer is an ike, next a doe, next an f-ike, and equatorial is id.
Note that this projection is done with respect to digonal symmetry. Thus red points represent a single point, orange ones represent a stack of 2 points, yellow ones a stack of 4 points.

Now consider just the top half. This introduces a bottom id, some peppies to connect its pentagons, and still a lot of tets.

Next cut out the central vertical 3 layers, i.e. identifying the left halved doe (vertical tower) with the right one. This would again reduce some former rosettes into peppies. In fact those correspond to the remaining complete 4 pentagons of those halved does. The bottom id then gets reduced to that bilbiro. Suppose moreover that the lateral squares of that one would get incident to squippies.

But cannot see at the moment what else is needed near the bottom of that halved doe section, i.e. near to the pentagon-pentagon edges of the bilbiro to make that figure become convex...

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Re: Johnsonian Polytopes

Postby student91 » Sat Feb 22, 2014 12:11 pm

student91 wrote:
quickfur wrote:
It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed. :)

I'd love to hear about that! :)

I don't have anything yet, I just was saying what I was going to innvestigate next, and when I'm done investigating (or stuck) I'll of course share my results with this forum.

I got something! When you take the o5o3x3o, cut it at the x5x3o's, delete the equatorial vertices, glue the x5x3o's together, there should occur some pentagonal prisms (the two pentagonal cupola's should get automatically relpaced with the other half of xxx5xoo&#x). Now if you delete a 10,10 edge of the x5x3o, a bilbiro should occr. Don't have much time to be clearer though.
Last edited by student91 on Mon Feb 24, 2014 11:22 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Feb 22, 2014 4:19 pm

I just updated the crown jewel page and added Keiji's proposed CJx.y.z numbering system for them. These include all of the confirmed CRF crown jewels we found so far.

There are still 4 as-yet unnamed crown jewels, CJ4.5.1, CJ4.5.2, CJ4.5.3, and CJ4.6. Any suggestions? :)

I gotta run now, I'll create stub wiki pages for these crown jewels later, and link the images and .off files to them.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Feb 22, 2014 11:45 pm

quickfur wrote:I just updated the crown jewel page and added Keiji's proposed CJx.y.z numbering system for them. These include all of the confirmed CRF crown jewels we found so far.

There are still 4 as-yet unnamed crown jewels, CJ4.5.1, CJ4.5.2, CJ4.5.3, and CJ4.6. Any suggestions? :)

I gotta run now, I'll create stub wiki pages for these crown jewels later, and link the images and .off files to them.

Alright, I've added stub pages for the latest crown jewels, including Stella4D .off files. Please let me know if there are any missing .off files.

Also, it would be nice if somebody could help reorganize the CRF polychora discovery project page; it's currently rather unwieldy, and should probably be split off into sub-pages, maybe one for all the monostratics, one for uniform polychora augmentations / diminishings (or maybe one page per family: 5-cell, tesseract/16-cell, 24-cell, and 120-/600-cell), and the existing crown jewels page. I'm not sure what to do with the counts, I think it's a bit premature to count all CRF polychora at the moment, maybe the monostratics can be counted, and the duoprism augmentation counts can be mentioned on the respective subpage.

Ideally, every CRF that has an image should go in its own page (because it means I have the .def file that I can convert to .off). But that's a lot of clerical work... hopefully more of us can help with it rather than just Keiji & myself. (wink wink nudge nudge) 8)
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Re: Johnsonian Polytopes

Postby Keiji » Sat Feb 22, 2014 11:58 pm

Thanks for the hard work quickfur!

I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.
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Re: Johnsonian Polytopes

Postby quickfur » Sun Feb 23, 2014 12:12 am

Keiji wrote:Thanks for the hard work quickfur!

I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.

I didn't build it because basically it's CJ4.5.1 with the top part of the rectified 120-cell glued on both ends. It's just extra vertices (=slower convex hull times, slower rendering, etc.) away from the exciting action (i.e. where the J92's are). You could still create a page for it, and mention the cell counts, etc., which are all already known. It's just that it's not that much more interesting than CJ4.5.1 itself. :)
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Re: Johnsonian Polytopes

Postby quickfur » Sun Feb 23, 2014 12:53 am

Hmm. Today I was reading up on the snub disphenoid on Wolfram, and it claims that the analytic solution for its coordinates requires solving a cubic equation?? How did Marek get a solution in terms of square roots??
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Re: Johnsonian Polytopes

Postby Marek14 » Sun Feb 23, 2014 6:03 am

quickfur wrote:
Keiji wrote:Thanks for the hard work quickfur!

I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.

I didn't build it because basically it's CJ4.5.1 with the top part of the rectified 120-cell glued on both ends. It's just extra vertices (=slower convex hull times, slower rendering, etc.) away from the exciting action (i.e. where the J92's are). You could still create a page for it, and mention the cell counts, etc., which are all already known. It's just that it's not that much more interesting than CJ4.5.1 itself. :)


If the top part of rectified 120-cell is glued on "both ends", does it mean you can get anohter CRF that would count as crown jewel by gluing it on only one end?

quickfur wrote:Hmm. Today I was reading up on the snub disphenoid on Wolfram, and it claims that the analytic solution for its coordinates requires solving a cubic equation?? How did Marek get a solution in terms of square roots??


Hm. Let's have a look.

Snub disphenoid has 8 vertices which can be separated into 4 layers. Let's use edge length 2 here:

1st layer: (1,0,0) and (-1,0,0)
2nd layer: (0,x,y) and (0,-x,y)
3rd layer: (x,0,z) and (-x,0,z)
4th layer: (0,1,w) and (0,-1,w)

Each layer has two vertices and the line of these two is always perpendicular between layers. Further, snub disphenoid has dihedral symmetry, meaning that distance between layers 1 and 2 is the same as distance between 3 and 4, so we can replace w with sum y+z.

Now then, snub disphenoid has four kinds of edges:

1. Edge joining two vertices in layers 1 and 4 -- this is 2 by definition.
2. Edge joining layers 1-2 or 3-4 -- this means distance (1,0,0) and (0,x,y) is 2.
3. Edge joining layers 1-3 or 2-4 -- this means distance (1,0,0) and (x,0,z) is 2.
4. Edge joining layers 2-3 -- this means distance (0,x,y) and (x,0,z) is 2.

So all the conditions mean that it's sufficient to ensure the triangle (1,0,0), (0,x,y), (x,0,z) is equilateral with side 2.

We have equations:

x^2+y^2+1 = 4
(x-1)^2+z^2 = 4
2x^2+(z-y)^2 = 4

By solving these three equations in Mathematica, I now get cubic solutions as well (there are quadratic solutions, but these have z = 0). The positive solution has x = 1.28917, y = 1.15674 and z = 1.97898. Not sure what went wrong before...
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