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 student91 » Wed Feb 26, 2014 11:12 am

Wait, we've thought of something like this before!! it was about xfox3oxFx3oooA&#xt:

quickfur wrote: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.

You seemed to be right after all :oops: :oops:
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 26, 2014 11:56 am

I found the Thawro-ing of the rox (o5o3x3o). it is oxFx3xfox3oox&#xt. (of course) it isn't closed yet, but maybe you can make something of it.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 26, 2014 3:59 pm

student91 wrote:
quickfur wrote:Last night I discovered another CRF, this time with tetrahedral symmetry, and with both J91 and J92 cells! Here is the lace tower:

Code: Select all
x3o3o
f3x3o
o3F3o
x3x3f
F3o3x
f3o3x
o3x3o


Cool CRF, really cool CRF!! :D :o_o: . I think you made a typo, shouldn't the f3o3x be x3o3f?

Good catch! You're right, it's x3o3f, while constructing the model I flipped the sign of one coordinate in order to make things line up, but forgot to adjust the CD symbol in the comments. :oops: Thanks, I'll update my post and the wiki page (and my model source file). :)

This is a really cool polytope, it has both J91 and J92, and it isn't directly related to an 120-uniform, so It's even cooler than the J92-rhombochoron!! :D .

It's actually a piece of the rectified 120-cell's surface, centered on a tetrahedron, with additional bits added to close it up. :) Just as you pointed out below. :D Originally, I had an even grander vision: to have 8 thawroes surrounding the tetrahedron in 2 alternating layers (like the first two layers of o5x3o's in the tetrahedron-first projection of o5x3o3o), but it turns out that the modification required for the first layer of 4 id's to become thawroes destroyed the possibility of the second layer of thawroes in that position.

However, this isn't the end of the story by any means! When I got to the x3x3f layer, there were at least two possible ways to proceed, both by "completing the pentagon" at a phi-edge. After staring at it for a while, I noticed the possibility of obtaining J91's in one way, so I chose that route. However, the other route remains to be explored!

[...]
This seems to be a tetrahedron-oriented thawroing of o5x3o3o. The thawroings we've seen before were all icosahedron-oriented. Doing a tetraheron-oriented one is really revolutionary :)

It's not that revolutionary, actually. I had already thought of this possibility quite a while back, and had been meaning to try it out, but just didn't get around to it until now.

Now we come to the far side, where things begin to get really interesting. The pentagonal faces from the near side are touching 6 J91s which surround an antipodal octahedron in alternated orientation:

Image

For clarity, I didn't color the J91 nearest to the 3D viewpoint, but its outline should be obvious. The red outlined triangular faces show where 4 tridiminished icosahedra (J63) touch the octahedra on the near side of the polytope. They alternate with 4 other octahedra that surround the antipodal octahedron in a formation alternating with the J63's. The remaining gaps are filled by tetrahedra and triangular cupolae.
[...]

this side looks like a tetrahedron-oriented bilbiroing of the o5o3x3o. Apparantely, a tetrahedron-oriented bilbiroing isn't happening at the equator, but at the second set of ids. The tetrahedron-oriented bilbiro-ing resp. thawro-ing certainly opens new possibilities :) , great find :D :D
It's especially cool that you integrated a thawro-ing and a bilbiro-ing of two different 120-uniforms, this one definately classifies as exceptional crown jewel

To be honest, I wasn't intentionally trying to combine two different 120-cell uniform patches; I just started with the tetrahedron + 4*J92 patch from the rectified 120-cell, and built the rest of the vertex layers piecemeal by looking for ways to make the surface close up in a CRF way. The octahedron + 6 J91 structure was basically "accidentally" discovered, based on a whim I had that the initial tetrahedron + 4*J92 patch looks like it could fit in some J91 cells. You have no idea how excited I was when the shape finally closed up and turned out to be CRF. :D :D :D I almost couldn't sleep all night.

Now that you noted the connection of the octahedron + 6*J91 patch with the rectified 600-cell o5o3x3o, it strengthens a hypothesis I have, that there should be many possibilities for combining various surface patches of the 120-cell/600-cell family uniforms into CRFs. While working with the coordinates, I noticed that the presence of the golden ratio causes many coordinates to keep simplifying into nicely-closed forms, with the consequence that you can just keep building upon the structure by borrowing various elements from the pentagonal polytopes, and each time the golden ratio causes the coordinates to simplify into nice closed forms, so the process repeats itself until things close up in a CRF way. This happened not just for this CRF, but also for many of the previous CRFs I constructed (both the ones I discovered myself, and the ones the rest of you guys described).
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 26, 2014 5:51 pm

quickfur wrote:[...]
It's actually a piece of the rectified 120-cell's surface, centered on a tetrahedron, with additional bits added to close it up. :) Just as you pointed out below. :D Originally, I had an even grander vision: to have 8 thawroes surrounding the tetrahedron in 2 alternating layers (like the first two layers of o5x3o's in the tetrahedron-first projection of o5x3o3o), but it turns out that the modification required for the first layer of 4 id's to become thawroes destroyed the possibility of the second layer of thawroes in that position.

A symmetric thawroing can only be done on ids that are oriented in a .3.-way. The second layer of ids you're talking about are .2.-oriented, so they will never be thawro-able. A bilbiroing remains possible for these id's. Furthermore, thawroings most of the time have equatorial counterparts. Maybe an equitorial thawroing is possible then (although it might get hard to close it up.)

However, this isn't the end of the story by any means! When I got to the x3x3f layer, there were at least two possible ways to proceed, both by "completing the pentagon" at a phi-edge. After staring at it for a while, I noticed the possibility of obtaining J91's in one way, so I chose that route. However, the other route remains to be explored!

I'm interested to hear your outcome (or the part you got stuck if that happens).

[...]
Now that you noted the connection of the octahedron + 6*J91 patch with the rectified 600-cell o5o3x3o, it strengthens a hypothesis I have, that there should be many possibilities for combining various surface patches of the 120-cell/600-cell family uniforms into CRFs. While working with the coordinates, I noticed that the presence of the golden ratio causes many coordinates to keep simplifying into nicely-closed forms, with the consequence that you can just keep building upon the structure by borrowing various elements from the pentagonal polytopes, and each time the golden ratio causes the coordinates to simplify into nice closed forms, so the process repeats itself until things close up in a CRF way. This happened not just for this CRF, but also for many of the previous CRFs I constructed (both the ones I discovered myself, and the ones the rest of you guys described).

This repetitive simplification probably has something to to with phi being a CVP2-value. basically all formulas you use to calculate coordinates are based on pythagoras. pythagoras squares thing. this means a CVP2-value can get rearranged, whereas a CVP3-value won't have a cube root simplified by this. I think this means that CVP3-things should be build "all-at-once", making the cube-roots very case-specific. In 4D, thingsdon't like to be build all-at-once, so I hope we might get to constrain the cases in which a CVP3-thing will be able to occur.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 26, 2014 7:01 pm

student91 wrote:
quickfur wrote:[...]
It's actually a piece of the rectified 120-cell's surface, centered on a tetrahedron, with additional bits added to close it up. :) Just as you pointed out below. :D Originally, I had an even grander vision: to have 8 thawroes surrounding the tetrahedron in 2 alternating layers (like the first two layers of o5x3o's in the tetrahedron-first projection of o5x3o3o), but it turns out that the modification required for the first layer of 4 id's to become thawroes destroyed the possibility of the second layer of thawroes in that position.

A symmetric thawroing can only be done on ids that are oriented in a .3.-way. The second layer of ids you're talking about are .2.-oriented, so they will never be thawro-able.

Don't be too sure about that. ;) Take a look, for example, at this fragment of the bicyclopentadiminished o5x3o3o (from this post):

Image

This is a chiral arrangement of pentagonal cupolae around a tetrahedron, with tetrahedral symmetry. Nothing says we can't produce a chiral thawro-ing of the .2.-oriented id's. (In fact, this is one of the "alternative orientations" of J91/J92 that I alluded to in an earlier post.) Since the above arrangement actually closes up (it forms part of a CRF, after all), it suggests that a chiral arrangement of J92's might have the possibility of closing up too. Of course, it would be more difficult to construct, since I'm not sure how a chiral cutting would be represented in a lace tower.

A bilbiroing remains possible for these id's. Furthermore, thawroings most of the time have equatorial counterparts. Maybe an equitorial thawroing is possible then (although it might get hard to close it up.)

These are all exciting possibilities to explore. :)

However, this isn't the end of the story by any means! When I got to the x3x3f layer, there were at least two possible ways to proceed, both by "completing the pentagon" at a phi-edge. After staring at it for a while, I noticed the possibility of obtaining J91's in one way, so I chose that route. However, the other route remains to be explored!

I'm interested to hear your outcome (or the part you got stuck if that happens).

Well, I was hoping to construct your bilbro'ed o5o3x3o first...

[...]
Now that you noted the connection of the octahedron + 6*J91 patch with the rectified 600-cell o5o3x3o, it strengthens a hypothesis I have, that there should be many possibilities for combining various surface patches of the 120-cell/600-cell family uniforms into CRFs. While working with the coordinates, I noticed that the presence of the golden ratio causes many coordinates to keep simplifying into nicely-closed forms, with the consequence that you can just keep building upon the structure by borrowing various elements from the pentagonal polytopes, and each time the golden ratio causes the coordinates to simplify into nice closed forms, so the process repeats itself until things close up in a CRF way. This happened not just for this CRF, but also for many of the previous CRFs I constructed (both the ones I discovered myself, and the ones the rest of you guys described).

This repetitive simplification probably has something to to with phi being a CVP2-value. basically all formulas you use to calculate coordinates are based on pythagoras. pythagoras squares thing. this means a CVP2-value can get rearranged, whereas a CVP3-value won't have a cube root simplified by this.

It's not just about CVP2 values being rearrangeable, though. The reason is that solving via pythagoras (which is implicitly what's happening in all of these constructions) requires a computation of the form X = √(edge_length2 - Y2), where X is the coordinate being solved for, and Y is usually a difference between two vectors. The thing is, usually the value of Y will itself involve square roots (e.g., the golden ratio, or √2, or √3, etc.), and in theory the difference between the edge length squared and Y is, generally speaking, not a perfect square, so usually X will involve nested square roots.

However, what I found was that when the golden ratio is involved in constructions derived from the 120-cell family uniforms, somehow (edge_length2 - Y2) keeps working out to be some kind of perfect square, so X becomes again something with only a single square root that can be rewritten in terms of combinations of the golden ratio. So that means something special is going on with constructions involving the golden ratio, probably due to its connection with the 120-cell family polychora, that causes these values to repeatedly work out to nicely-closed forms.

I think this means that CVP3-things should be build "all-at-once", making the cube-roots very case-specific. In 4D, thingsdon't like to be build all-at-once, so I hope we might get to constrain the cases in which a CVP3-thing will be able to occur.

Yep. So that means the best approach to find a CVP3 4D CRF may be to work topologically, rather than by direct construction as we have been doing so far. Once we have a topologically-closed cell complex that is likely to be CRF, we can then try to solve for the coordinates to see if it is CRF-able.

I've actually tried topological constructions before, but they are pretty hard because you have to be able to visualize 4D and perform topological assembly operations in your head. Drawing diagrams on paper quickly becomes a nasty tangle of incomprehensible lines that lead nowhere. It can be alleviated somewhat if you're a careful artist and use colored pens/pencils to indicate edges belonging to different cells, but that distracts you with the mechanics of 2D diagramming rather than letting you focus on the 4D construction at hand. Perhaps what is needed is some kind of interactive 4D-doodle program that lets you manipulate topological polyhedra with 3D rendering that helps prevent the clutter of edges that will quickly become unmanageable.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 26, 2014 9:44 pm

quickfur wrote:[...]
Don't be too sure about that. ;) Take a look, for example, at this fragment of the bicyclopentadiminished o5x3o3o (from this post):

Image

This is a chiral arrangement of pentagonal cupolae around a tetrahedron, with tetrahedral symmetry. Nothing says we can't produce a chiral thawro-ing of the .2.-oriented id's. (In fact, this is one of the "alternative orientations" of J91/J92 that I alluded to in an earlier post.) Since the above arrangement actually closes up (it forms part of a CRF, after all), it suggests that a chiral arrangement of J92's might have the possibility of closing up too. Of course, it would be more difficult to construct, since I'm not sure how a chiral cutting would be represented in a lace tower.

Hmm, i'm indeed good at assuming things to be impossible too easilly :D . It's nice you guys are always pointing out if I'm doing so :)
[...]
Well, I was hoping to construct your bilbro'ed o5o3x3o first...

Yes, I guess I'm rushing a bit, letting you do the rendering and doing the fun myself. You have to know I really appreciate your renderings, and am very reliable on them. For all discoveries I've done so far have I have heavilly used your renders (and klitzings site for the non-visual info I needed). Your sites are indispensable.
[...]
It's not just about CVP2 values being rearrangeable, though. The reason is that solving via pythagoras (which is implicitly what's happening in all of these constructions) requires a computation of the form X = √(edge_length2 - Y2), where X is the coordinate being solved for, and Y is usually a difference between two vectors. The thing is, usually the value of Y will itself involve square roots (e.g., the golden ratio, or √2, or √3, etc.), and in theory the difference between the edge length squared and Y is, generally speaking, not a perfect square, so usually X will involve nested square roots.

However, what I found was that when the golden ratio is involved in constructions derived from the 120-cell family uniforms, somehow (edge_length2 - Y2) keeps working out to be some kind of perfect square, so X becomes again something with only a single square root that can be rewritten in terms of combinations of the golden ratio. So that means something special is going on with constructions involving the golden ratio, probably due to its connection with the 120-cell family polychora, that causes these values to repeatedly work out to nicely-closed forms.

I think it is more related to phi itself than the 120-family. The fact that phi^2=phi+1 probably highly contributes to this. Furthermore, A*sqrt(5) + B might just always be a perfect square, I'll try to work this out tomorrow.

.[...]
Yep. So that means the best approach to find a CVP3 4D CRF may be to work topologically, rather than by direct construction as we have been doing so far. Once we have a topologically-closed cell complex that is likely to be CRF, we can then try to solve for the coordinates to see if it is CRF-able.

I've actually tried topological constructions before, but they are pretty hard because you have to be able to visualize 4D and perform topological assembly operations in your head. Drawing diagrams on paper quickly becomes a nasty tangle of incomprehensible lines that lead nowhere. It can be alleviated somewhat if you're a careful artist and use colored pens/pencils to indicate edges belonging to different cells, but that distracts you with the mechanics of 2D diagramming rather than letting you focus on the 4D construction at hand. Perhaps what is needed is some kind of interactive 4D-doodle program that lets you manipulate topological polyhedra with 3D rendering that helps prevent the clutter of edges that will quickly become unmanageable.

The doodle program sounds fun to play with. Maybe even people that are not interested in polytopes would like to play with it.
The topological constructions might just be more difficult that we would think. I can think of a lot of polytopes consisting of tetrahedra that work out topologically, yet they should be ,impossible according to mr and mrs Blind. I don't know how this would be extednded to CRF's, but I think not everything that works topologically will work "in real" as well.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 26, 2014 9:58 pm

student91 wrote:[...]
I already tried to explain it in my previous posts, but I think everyone missed that:
student91 wrote:
student91 wrote:[...]
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 occur. Don't have much time to be clearer though.

Basically all info needed to build a bilbiro'd is included in this post. Any attempt I'l make to explain this is just saying the same in different words.

Bilbiroing means taking ids, and deleting some vertices such that you're left with two lunes. Then you glue these two lunes together and Tadaa!. :D
Now o5o3x3o has .2.-oriented ids at the equator, so some equatorial vertices should be deleted. These vertices are the vertices between the x5x3o-cuts of the rox. (Basically any bilbiroing deletes every vertex with |x|<phi/2, because the ids are perpendicular to a specific plane). This basically gives us the lace tower (something)||x5x3o||(the same something). The somethings are what is on the "pole" side of x5x3o in a normal rox . Now we can cut off an J91-pseudopyramid, (this is done by deleting an 10,10 edge of the x5x3o) resulting in a bilbiro. The maximal diminishing here is 10 cuts corresponding to an pentagonal antiprism inscribed in an id, but maybe a cut based on an inscribed octahedron is more interesting, and looks better if combined with the thawroing.

Alright, so I confirmed that the initial step (delete equatorial vertices between two x5x3o's and compressing the two remaining halves of o5o3x3o together) is CRF. So I proceeded to create J91's by deleting 10,10 edges from the merged x5x3o. Turns out that the specific coordinates I was using lend themselves most readily to the octahedral diminishing; this results in something with pyritohedral symmetry:

Image

In order to reduce the number of vertices unrelated to the J91's, I cut off two bistratic caps from the initial o5o3x3o, resulting in two x5o3x's (the one on the near side is seen in yellow above). The red patches are the J91's, lying on the equator of the polytope.

Now, this suggests an interesting further diminishing: if you delete the 8 icosahedra located at the corners of an imaginary cube inscribed in this projection image, it should produce 8 icosidodecahedral cells that connect the J91's together. (Of course, I'd have to add parts of the bistratic caps back in, in order for the icosidodecahedra to be complete.) They will all touch each other at a triangular face. This probably cannot be done on both sides of the polytope, since the icosahedra from either hemisphere share a triangular face, and deleting both icosahedra will probably make it non-CRF. However, it may be possible to do this in alternated tesseract symmetry, in which case we'll have 8 icosidodecahedra in skewed tesseractic orientation. :lol:

Now since these icosidodecahedra are in .2.-orientation, it looks like it should be possible to thawro it as well. We'd then have 8 J92's in alternated tesseractic symmetry coupled with 6 J91's in pyritohedral symmetry, capped at the remaining two ends with either icosahedra or icosidodecahedra (I think the latter is possible). :lol: Lots of fun possible here.

The other way to diminish the initial compressed o5o3x3o, as suggested by student91, is to delete 10 edges, creating 10 J91's in pentagonal antiprimic symmetry. In that case, it's possible to further delete pentagonal prism pyramids between the J91's, to create a CRF where the 10 J91's are linked to each other by pentagonal prisms. Lots of possibilities here.

All these variations probably deserve a crown jewel category, perhaps CJ4.9.x? But there are a lot of members in this category, so I'm not sure how to assign a consistent numbering for them.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 26, 2014 10:28 pm

quickfur wrote:Alright, so I confirmed that the initial step (delete equatorial vertices between two x5x3o's and compressing the two remaining halves of o5o3x3o together) is CRF. So I proceeded to create J91's by deleting 10,10 edges from the merged x5x3o. Turns out that the specific coordinates I was using lend themselves most readily to the octahedral diminishing; this results in something with pyritohedral symmetry:

Image

In order to reduce the number of vertices unrelated to the J91's, I cut off two bistratic caps from the initial o5o3x3o, resulting in two x5o3x's (the one on the near side is seen in yellow above). The red patches are the J91's, lying on the equator of the polytope.

Nice renders!! :D
Now, this suggests an interesting further diminishing: if you delete the 8 icosahedra located at the corners of an imaginary cube inscribed in this projection image, it should produce 8 icosidodecahedral cells that connect the J91's together. (Of course, I'd have to add parts of the bistratic caps back in, in order for the icosidodecahedra to be complete.) They will all touch each other at a triangular face. This probably cannot be done on both sides of the polytope, since the icosahedra from either hemisphere share a triangular face, and deleting both icosahedra will probably make it non-CRF. However, it may be possible to do this in alternated tesseract symmetry, in which case we'll have 8 icosidodecahedra in skewed tesseractic orientation. :lol:
I'm sorry to dissapont you here, but the rearrangements of the equator causes the ike-gap be irregular. (It goes up to F3o3o, and after that it becomes non-CRF.it migh4 be posible to insert a patch from another CRF, just as I did with the bilbiro'd o5x3o3x.

Now since these icosidodecahedra are in .2.-orientation, it looks like it should be possible to thawro it as well. We'd then have 8 J92's in alternated tesseractic symmetry coupled with 6 J91's in pyritohedral symmetry, capped at the remaining two ends with either icosahedra or icosidodecahedra (I think the latter is possible). :lol: Lots of fun possible here.

The other way to diminish the initial compressed o5o3x3o, as suggested by student91, is to delete 10 edges, creating 10 J91's in pentagonal antiprimic symmetry. In that case, it's possible to further delete pentagonal prism pyramids between the J91's, to create a CRF where the 10 J91's are linked to each other by pentagonal prisms. Lots of possibilities here.

All these variations probably deserve a crown jewel category, perhaps CJ4.9.x? But there are a lot of members in this category, so I'm not sure how to assign a consistent numbering for them.
I'd suggest calling them as such:
The undiminished stack will be called 4.9.0 or 4.9.1 (4.9.0 because it's not really special)
Now every diminishing will get a new number, the octagon-diminishing 4.9.1 and the pent. ap. 4.9.2 and so on.
The further derivatives will get another number, e.g. 4.9.1.1 and 4.9.1.2 etc.
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Feb 26, 2014 10:35 pm

Basically, this "patching" is analogical to the process which creates bilbiro from parts of icosidodecahedron and rhombicosidodecahedron, right?
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 26, 2014 10:36 pm

student91 wrote:[...]
Yes, I guess I'm rushing a bit, letting you do the rendering and doing the fun myself. You have to know I really appreciate your renderings, and am very reliable on them. For all discoveries I've done so far have I have heavilly used your renders (and klitzings site for the non-visual info I needed). Your sites are indispensable.

Thanks! I have to say, I also found Klitzing's incmats site extremely valuable for the lace towers it provides for the uniform polychora. It greatly simplified many of my recent CRF constructions, because having the CD symbols readily available means I only need to compute the height of the layer (and sometimes not even, if I'm copying a section of a 4D uniform intact -- the height is already encoded in the coordinates, just have to sort by the first coordinate (or 4th, if you align your construction that way)), and the rest can be automatically generated with wendy's apacs/epacs/apecs/etc. operators from known coordinates of 3D uniforms.

quickfur wrote:[...]
It's not just about CVP2 values being rearrangeable, though. The reason is that solving via pythagoras (which is implicitly what's happening in all of these constructions) requires a computation of the form X = √(edge_length2 - Y2), where X is the coordinate being solved for, and Y is usually a difference between two vectors. The thing is, usually the value of Y will itself involve square roots (e.g., the golden ratio, or √2, or √3, etc.), and in theory the difference between the edge length squared and Y is, generally speaking, not a perfect square, so usually X will involve nested square roots.

However, what I found was that when the golden ratio is involved in constructions derived from the 120-cell family uniforms, somehow (edge_length2 - Y2) keeps working out to be some kind of perfect square, so X becomes again something with only a single square root that can be rewritten in terms of combinations of the golden ratio. So that means something special is going on with constructions involving the golden ratio, probably due to its connection with the 120-cell family polychora, that causes these values to repeatedly work out to nicely-closed forms.

I think it is more related to phi itself than the 120-family. The fact that phi^2=phi+1 probably highly contributes to this. Furthermore, A*sqrt(5) + B might just always be a perfect square, I'll try to work this out tomorrow.

No, that's not true. The pentagon on the plane, for example, has nested square roots resulting from a combination of phi that isn't a perfect square.

Also, not just phi but any quadratic rational (i.e., numbers of the form x+y√z, for fixed z) will exhibit similar properties to phi, because numbers of this form are actually closed under field operations (+,-, and * should be obvious; division also applies because you just multiply a denominator x+y√z by its conjugate x-y√z to get rid of the reciprocal square root term and turn the denominator into a non-square root number, then it can be rewritten once again in the form x+y√z). Given any quadratic polynomial Ax^2 + Bx + C, a root R of this polynomial will (obviously) satisfy the relation AR^2 + BR + C = 0, or, in other words, AR^2 = -BR-C. Dividing through by R produces the relation AR = -B-C/R, or, in more familiar form, C/R = -B-AR. In the case of phi, A=1, B=-1, C=-1, so we have phi^2 = phi+1 and 1/phi = phi-1. But as you can see, other values of A, B, and C will produce numbers that also exhibit special relationships with their square and reciprocal. And, as I mentioned, since these quadratic rationals are closed under field operations, adding, subtracting, or multiplying them will all, again, produce numbers of the same form.

So you see, phi isn't that special after all. Well, it still is, but not just because of its relations with its square and reciprocal. :) The reason phi crops up more often than not is only because it is one of the simplest quadratic rationals, and because of its connection with the pentagon.

The special behaviour of phi that I've been observing, though, appears to be intimately tied to the 120-cell family polychora, in the sense that certain kinds of combinations of phi seem to somehow gravitate towards perfect squares, even though there are many situations where this doesn't happen (e.g., in the pentagon on the 2D plane). This makes me suspect that there might be some kind of underlying phi-based tessellation, or something like that, in 4D, that causes these numbers to "magically" work out in nice forms in many cases where it would normally be unexpected. IOW, the CRFs we've been finding so far are all just CRF fragments of this underlying tiling or grid or whatever it is, and that many more such fragments exist.


.[...]
Yep. So that means the best approach to find a CVP3 4D CRF may be to work topologically, rather than by direct construction as we have been doing so far. Once we have a topologically-closed cell complex that is likely to be CRF, we can then try to solve for the coordinates to see if it is CRF-able.

I've actually tried topological constructions before, but they are pretty hard because you have to be able to visualize 4D and perform topological assembly operations in your head. Drawing diagrams on paper quickly becomes a nasty tangle of incomprehensible lines that lead nowhere. It can be alleviated somewhat if you're a careful artist and use colored pens/pencils to indicate edges belonging to different cells, but that distracts you with the mechanics of 2D diagramming rather than letting you focus on the 4D construction at hand. Perhaps what is needed is some kind of interactive 4D-doodle program that lets you manipulate topological polyhedra with 3D rendering that helps prevent the clutter of edges that will quickly become unmanageable.

The doodle program sounds fun to play with. Maybe even people that are not interested in polytopes would like to play with it.
The topological constructions might just be more difficult that we would think. I can think of a lot of polytopes consisting of tetrahedra that work out topologically, yet they should be ,impossible according to mr and mrs Blind. I don't know how this would be extednded to CRF's, but I think not everything that works topologically will work "in real" as well.

Of course, there will be many more topological constructions that are possible to realize in a CRF way. So you can't just blindly enumerate topological constructions and hope to hit lots of CRFs, because chances are that the vast majority of them are not CRF-able. :) The idea is to explore topological constructions that are likely to be CRF-able, but without needing to actually check whether they are actually CRF at every step, and only verify that at the end.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 26, 2014 10:44 pm

Marek14 wrote:Basically, this "patching" is analogical to the process which creates bilbiro from parts of icosidodecahedron and rhombicosidodecahedron, right?

Yep! The direct analogues, even. So in that sense the CRFs we've been constructing are all analogues of J91 and J92, even the ones that don't immediately contain J91 and J92 themselves as cells (e.g., the modified polystratic stacks created from the 120-cell uniforms, that student91 uses as the initial step before introducing the J91's and J92's).

I suppose you could say that our recent string of successes is mainly because these things are so easy to construct, and there are so many of them, that randomly groping in the dark will land on one with high probability.

The CVP3 polychora, though, if they exist, remain ever so elusive. :\

student91 wrote:[...]Nice renders!! :D

Thanks!

Now, this suggests an interesting further diminishing: if you delete the 8 icosahedra located at the corners of an imaginary cube inscribed in this projection image, it should produce 8 icosidodecahedral cells that connect the J91's together. (Of course, I'd have to add parts of the bistratic caps back in, in order for the icosidodecahedra to be complete.) They will all touch each other at a triangular face. This probably cannot be done on both sides of the polytope, since the icosahedra from either hemisphere share a triangular face, and deleting both icosahedra will probably make it non-CRF. However, it may be possible to do this in alternated tesseract symmetry, in which case we'll have 8 icosidodecahedra in skewed tesseractic orientation. :lol:
I'm sorry to dissapont you here, but the rearrangements of the equator causes the ike-gap be irregular. (It goes up to F3o3o, and after that it becomes non-CRF.it migh4 be posible to insert a patch from another CRF, just as I did with the bilbiro'd o5x3o3x.

You're right. :oops: :( So it is not possible after all, at least not as I described. But might still be worth a look to see if CRF pieces can be inserted to close it up.

[...] All these variations probably deserve a crown jewel category, perhaps CJ4.9.x? But there are a lot of members in this category, so I'm not sure how to assign a consistent numbering for them.
I'd suggest calling them as such:
The undiminished stack will be called 4.9.0 or 4.9.1 (4.9.0 because it's not really special)
Now every diminishing will get a new number, the octagon-diminishing 4.9.1 and the pent. ap. 4.9.2 and so on.
The further derivatives will get another number, e.g. 4.9.1.1 and 4.9.1.2 etc.

Please update the crown jewel page accordingly. ;)
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Re: Johnsonian Polytopes

Postby student5 » Thu Feb 27, 2014 1:06 am

hello everybody :D

I have been reading the posts, and might have found a CRF, thanks to Quickfur's latest find :mrgreen:
the outer layer (the weird shape you encountered, which looked like o3x5o) might be constructed using 4 thawro's, some triangular cupola, and an x3x3o, and I'll try to explain how;

take the picture here
Image
if you insert thawro's at the red triangles, and rotate the cupolae by 60 degrees, the prisms could become octahedra and the hole in the center would, if I have figured well, be an x3x3o, so it could be another CRF :D :P but I have no idea if the edges are indeed 1 or if it is still convex, I just lego'd the shape from my earlier idea :roll:

I'll make an attempt for the lace tower (my first :) ) the first part has to be correct anyway, I am not very good at these, it's probably incorrecct :sweatdrop:
Code: Select all
x3o3o
f3x3o
o3F3o
x3x3f
F3o3x
o3x3x

well.... I'm stuck, so I just added o3x3x :\ , because that seemed like the only reasonable thing to do :roll: maybe you could help me out here?
I do hope it is CRF :nod:
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Re: Johnsonian Polytopes

Postby student5 » Thu Feb 27, 2014 1:22 am

But, just wondering, would the hemisphere of the shape you encountered be another 3D CRF polytope? because it is both convex and built from regular faces...
Image
that'd mean there is still more to discover in three dimensions
but you must've figured that out already, since you were baffled by it aswell, it could maybe be added it to wikipedia or something?
and if you'd find mutiple different ways of "filling" it, you'd find a lot of new polytopes, since any combination comes with any other combination (the original coul be built either north||north south||south or north||south
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 2:23 am

student5 wrote:But, just wondering, would the hemisphere of the shape you encountered be another 3D CRF polytope? because it is both convex and built from regular faces...
Image
that'd mean there is still more to discover in three dimensions
[...]

Unfortunately, interesting as this polyhedron may be, it is not CRF. Its edges are only equal length in 4D, where it is a skew polyhedron, meaning that it does not lie within a single hyperplane. Its vertices are not coplanar. If you were to squish it onto a hyperplane (basically what projection does :lol: ), then its edges would no longer be equal length. So, fascinating as it may be, it's not another 3D CRF, sad to say.

Besides, it has already been proven rigorously that Johnson's 92 solids are the only ones possible in 3D (besides the 3D regulars and uniforms). There are no others. :(

Now arguably, this polyhedron could qualify as a "near-miss" Johnson, but I'm skeptical about how "near" it is to being Johnson, since the curvature of its skew decagon great circles implies a significant amount of distortion from a flat 3D figure. It's not hard to figure out its 3D coordinates; it's just the union of the vertex layers o3F3o, x3x3f and F3o3x in 3D (i.e., ignore the 4th coordinate).
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 3:48 am

Most fascinating!! I just realized that this skew polyhedron, or near-miss Johnson if we squash it into 3D, is basically the rectified version of near miss Johnson solid #22 on orchidpalms.com! It is actually quite close to being Johnson; the hexagons are regular, though the pentagons are not (the triangles are probably not as well). The edge lengths vary only about 6.6% from their average value, and the circumradii of the vertices vary about 2% from their average value. Perhaps it's a good near-miss candidate after all! :D
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Re: Johnsonian Polytopes

Postby wendy » Thu Feb 27, 2014 9:42 am

This is an impressive project. I read it every night, but i don't see how i can really contribute anything to it.

But quickfur's renders are really interesting. I went through the wiki and added lace towers to the 3d crf's, as far as i could read the names.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 3:39 pm

wendy wrote:This is an impressive project. I read it every night, but i don't see how i can really contribute anything to it.

You already have. :) Without your lace tower / lace city notation, it would have been far more difficult to compute the coordinates / communicate the structure of these CRFs. And besides, you did discover the construction that led to the ursatopes in all dimensions! :nod:

I did see your post about the criteria for the existence / convex hemisphericity of the ursatopes... I've been meaning to get around to studying that in more depth, but things have just been very busy on the J91/J92 front. :D
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 27, 2014 6:26 pm

And I've derived the lace-tower for the equitorially thawro'd o5o3x3o: xoffxffox3xxxoooxxx5oxoxfxoxo&#xt. In order to get a Thawro, you have to delete a triangle of the o3x5x. cool thing about this one is that the "north" and the "south" thawro will be toutching each other at their triangles.


The way I derived this thawroing was very interesting, it ables me to do more magic with the CD-diagrams. :) . I already encountered such a way of thinking when I derived the polar thawroing, but back then I didn't know why it exactly worked:
student91 wrote:The way I derived the thawro'd o5o3x3o was by looking at the f3o5x as a f3f5A, with A=-1. I'm not sure thats the way negative nodes are used, but when I used it this way it worked. I ment with the negative number that the edge had a length 1 the other way around. Now you have a f3f.. that can be shrinked to x3x. The A has to become 0, so that gives me x3x5o. I think I cheated a bit with the negative numbers, but at least it worked. :)

It's negative numbers on CD-diagrams!! (I guess Klitzing and wendy already know this, according to their spreadsheeds, but I will just post it for others that are interested.)
I discovered this by trying to thawro rox tetrahedral-oriented. it's lace-tower looks like o3x3o || x3o3f || x3f3o || o3f3f || ... The o3f3f-thing should be shrunken to A3x3x, but I got stuck calculating A, so I went on doing something else. I wanted to get the thawro (o3x || f3x || o3F || (x3x) ) be visible in the lace-tower.
The o3x is clearly present in the o3x3o. The f3x should be present in the x3o3f, and it is!! Just think of x3o3f as build up of four x3f-hexagons (just as x3o3x is build up of four x3x's). Now the F3o. it should be present in the x3f3o. After making a nice drawing I concluded it was below the x3f-hexagon. When I drew only these F-triangles and the x-edges, I got a non-convex thing, and saw it could be written as (-x)3F3o !!. This figure has an F3o, just as we needed, and a (-x)3F, which I could see in my drawing. (It resembles the white line in this picture:)Imagethe (-x)2o isn't anything.
So basically we can say that, when only looking at vertices, x3f3o "=" (-x)3F3o. Why is that?
I didn't have an answer to this in an instand, but let's first look at x3(-x). what does it look like?. It looks like a triangle, with a double turning number, so you ca say o3x "=" x3(-x).
After I understood this, I went on to see why x3f "=" (-x)3F. When you draw x3f, the longest diagonals are the diagonals that become the A's in (-x)3A. These diagonals have length f+(shortchord of a triangle)=f+1=F. :o_o: (that was my face when I discovered that).
Now we've fouund the F3o in the x3f3o. now can we write f3o3x in a way that shows us the hidden x3f? it took a while till I got that, but the answer is yes!! it's f3x3(-x).
What about something with a 5? what does o5x "equal"? the answer is just the same as with the triangle: when x becomes negative, o becomes o+(shortchord of a pentagon)=f. This means o5x "=" f5(-x).
In general it is true that A5B "=" (A+fB)5(-B)

This, not the general note, is all I needed to see the hidden thawro in Klitzings lace-tower of rox: ... || o3f5o || f3o5x || o3x5x || f3x5o || f3o5x || (x3o5f + V3o5o) || (the same backwards). This becomes (... || o3f5o ) || f3f5(-x) || o3F5(-x) || f3x5o || x3o5f || (the same backwards). The .3.'s now give f3f || o3F || f3x || x3o, i.e. an unajusted thawro :D . Now I only have to shrink the f3f5(-x) into x3x5o, and the thawroing is done. Of course such a thawroing is also possible without everything described above, but this just gave me a much better understanding of it all.

btw. I'm going on a vacation, so I won't be active for a week.
Last edited by student91 on Thu Feb 27, 2014 8:43 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 27, 2014 7:59 pm

student5 wrote:hello everybody :D

I have been reading the posts, and might have found a CRF, thanks to Quickfur's latest find :mrgreen:
the outer layer (the weird shape you encountered, which looked like o3x5o) might be constructed using 4 thawro's, some triangular cupola, and an x3x3o, and I'll try to explain how;

take the picture here
Image
if you insert thawro's at the red triangles, and rotate the cupolae by 60 degrees, the prisms could become octahedra and the hole in the center would, if I have figured well, be an x3x3o, so it could be another CRF :D :P but I have no idea if the edges are indeed 1 or if it is still convex, I just lego'd the shape from my earlier idea :roll:

I'll make an attempt for the lace tower (my first :) ) the first part has to be correct anyway, I am not very good at these, it's probably incorrecct :sweatdrop:
Code: Select all
x3o3o
f3x3o
o3F3o
x3x3f
F3o3x
o3x3x

well.... I'm stuck, so I just added o3x3x :\ , because that seemed like the only reasonable thing to do :roll: maybe you could help me out here?
I do hope it is CRF :nod:

Well, you got a point here!! your lace-tower isn't anything, but what you're describing certainly makes sense!! What you're saying is to fill it up with J92's, in a way that indeed works (although I had a hard time checking that it did). so finally whatyou get is o3x3o || x3o3f || F3o3x || x3x3f || o3F3o || o3x3x. It works!! (I think).
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 8:04 pm

student91 wrote:And I've derived the lace-tower for the equitorially thawro'd o5o3x3o: xoffxffox3xxxoooxxx5oxoxfxoxo&#xt. In order to get a Thawro, you have to delete a triangle of the o3x5x. cool thing about this one is that the "north" and the "south" thawro will be toutching each other at their triangles.
[...]

It's interesting that so far, we've had repeated success "bilbiro-ing" and "thawro-ing" the 120-cell family uniform polychora. It seems that J91 and J92 in 3D are not so unusual after all: they are just compressed icosidodecahedra and pseudo-bisected icosidodecahedra. This compression process can be applied to 4D uniforms too, yielding J91 cells in the process, and similarly this pseudo-bisection can be carried out in 4D by substituting the f-hexagon-containing cell introduced by the bisection with another cell that has a unit-edged hexagon.

These two operations are not restricted to just CRFs that contain J91's and J92's in the final product; in a general sense they can be applied anywhere. The strata of the 600-cell come to mind: pretty much every stratum can be cut off as an independent polychora, and made into CRF segmentochora by substituting the bisected tetrahedra with various pieces (pentagonal pyramids, and such). The hemi-600-cell and the associated lunae are examples of polychora produced by various applications of this operation. You could also cut off the middle two strata from the 600-cell between two dodecahedra, and glue the two halves together at the dodecahedral interface, and perhaps there's something that can be inserted around the concave bits to make it CRF. But CRFability is secondary; the point is that this is a general "compaction" operation. The fact that it sometimes produces J91 cells is just an added bonus.

Similarly, pseudo-bisection is generally applicable, also illustrated by the hemi-600-cell: the bisected 600-cell is non-CRF, but deleting the bisected tetrahedra, which equals deleting the non-CRF pentagonal pyramids from the bottom cell introduced by the cut, makes it CRF again. So we're in essence substituting the pentakis icosidodecahedron with a uniform dodecahedron. This is not so much different from substituting, say, x5o3x (introduced by the cut) with o5x3o like we have been doing when "thawro-ing" a 120-cell family uniform. And as student91's crown jewels have been showing us, pseudo-bisection is not limited to cutting the polytope exactly in half, but it can be applied to just a cap, or even in oblique orientation (e.g. tetrahedral symmetry centered instead of icosahedral symmetry centered).

We may even say, in summary, that the various compacted (bilbiro'd) polychora are just generalizations of J91 itself (which is basically the compaction of the icosidodecahedron in 3D), and that the various pseudo-bisected (thawro'd) polychora are just generalizations of J92 itself (basically the pseudo-bisection of the icosidodecahedron by a plane parallel to two opposite triangular faces, followed by a substitution of the bottom f-hexagon with a unit hexagon).

So this suggests a consistent naming scheme for the crown jewels constructed in this manner: "compacted" refers to the deletion of some layers of vertices and joining the two halves together (not necessarily with J91's resulting, though the cases we've considered so far mostly do so), and "pseudo-bisected" or "pseudo-diminished" refers to a diminishing of the polytope followed by a substitution of the bottom facet introduced by the cut with something else (usually chosen to make the result CRF). So CJ4.5.1, for example, would be the parabidiminished compacted n-stratic rectified 120-cell cap (where n=number of strata in the rectified 120-cell cap), and CJ4.5.2 simply the compacted n-stratic rectified 120-cell cap. CJ4.5.3 would be the pseudo-parabidiminished rectified 120-cell. CJ4.5.4 then is just the compacted version of CJ4.5.3.

CJ4.6 can be the compacted cantellated 600-cell, and CJ4.7 the pseudo-octadiminished diminished tristratic rectified 600-cell cap. Then CJ4.9.x will just be various diminishings of the compacted rectified 600-cell.

I know this nomenclature is still a bit rough around the edges, so suggestions for improvements are welcome, but the basic idea is that we have two operations, compaction and pseudo-bisection/pseudo-diminishing, that can be applied in various combinations, orientations, and depths to produce new CRFs from existing ones.

Polymorphic (in the sense of being built from surface patches of different things) crown jewels like CJ4.8 can probably adopt a naming scheme similar to the Johnson solids with compound names like sphenocorona (spheno patch + corona patch).

In this light, the J92 rhombochoron can probably be reanalysed as a Siamese bi-J92 wedge (it's vaguely reminiscient of the snub disphenoid, also known as the siamese dodecahedron, which can be understood as two copies of 3 quadrants of an octahedron fused together -- the J92 rhombochoron is formed by fusing together two J92 wedges).
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 8:42 pm

student91 wrote:
student5 wrote:hello everybody :D

I have been reading the posts, and might have found a CRF, thanks to Quickfur's latest find :mrgreen:
the outer layer (the weird shape you encountered, which looked like o3x5o) might be constructed using 4 thawro's, some triangular cupola, and an x3x3o, and I'll try to explain how;

take the picture here
Image
if you insert thawro's at the red triangles, and rotate the cupolae by 60 degrees, the prisms could become octahedra and the hole in the center would, if I have figured well, be an x3x3o, so it could be another CRF :D :P but I have no idea if the edges are indeed 1 or if it is still convex, I just lego'd the shape from my earlier idea :roll:

I'll make an attempt for the lace tower (my first :) ) the first part has to be correct anyway, I am not very good at these, it's probably incorrecct :sweatdrop:
Code: Select all
x3o3o
f3x3o
o3F3o
x3x3f
F3o3x
o3x3x

well.... I'm stuck, so I just added o3x3x :\ , because that seemed like the only reasonable thing to do :roll: maybe you could help me out here?
I do hope it is CRF :nod:

Well, you got a point here!! your lace-tower isn't anything, but what you're describing certainly makes sense!! What you're saying is to fill it up with J92's, in a way that indeed works (although I had a hard time checking that it did). so finally whatyou get is o3x3o || x3o3f || F3o3x || x3x3f || o3F3o || o3x3x. It works!! (I think).

Hold on, how come the first layer is o3x3o? Is that a typo, or is this a completely new construction? (I'm interested to find out if this works!!)
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 27, 2014 8:49 pm

quickfur wrote:[...]
Hold on, how come the first layer is o3x3o? Is that a typo, or is this a completely new construction? (I'm interested to find out if this works!!)

look at the first part of my lace-tower: o3x3o || x3o3f || F3o3x || x3x3f || o3F3o. that's just the bilbiro-part of your CJ4.8. student5 was describing a construction as if it were on the "north", but it should be applied to the "south" in order to work. What's really weird is that I cant see what the last part comes from, it doesn't seem to be a thawro-ing of any known 120-uniform. That's why I would like this one to be triple-checked, I'm still not sure if it will work.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 9:35 pm

student91 wrote:
quickfur wrote:[...]
Hold on, how come the first layer is o3x3o? Is that a typo, or is this a completely new construction? (I'm interested to find out if this works!!)

look at the first part of my lace-tower: o3x3o || x3o3f || F3o3x || x3x3f || o3F3o. that's just the bilbiro-part of your CJ4.8. student5 was describing a construction as if it were on the "north", but it should be applied to the "south" in order to work. What's really weird is that I cant see what the last part comes from, it doesn't seem to be a thawro-ing of any known 120-uniform. That's why I would like this one to be triple-checked, I'm still not sure if it will work.

I'm not sure if I made a mistake, but are you sure o3F3o || x3x3f is possible to have unit edges? From what I can tell, the circumradius of x3x3f and the circumradius of o3F3o differs by more than an edge length. :( (But I may have made a mistake somewhere...)
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 27, 2014 9:46 pm

quickfur wrote:[...]I'm not sure if I made a mistake, but are you sure o3F3o || x3x3f is possible to have unit edges? From what I can tell, the circumradius of x3x3f and the circumradius of o3F3o differs by more than an edge length. :( (But I may have made a mistake somewhere...)

yeah, pretty sure. in fact, it's just part of your J4.8:
quickfur wrote:[...]
Code: Select all
x3o3o
f3x3o
o3F3o
x3x3f
F3o3x
x3o3f
o3x3o


Here's a projection of the near side (seen from the top of the lace tower):

Image
you can see the vertex of o3F3o being attached to the hexagons of x3x3f. furthermore, Klitzings Circumradius-spreadsheet gives for -x3f3-f a lacing height of 0,218508.., which is real.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 27, 2014 11:29 pm

student91 wrote:
quickfur wrote:[...]I'm not sure if I made a mistake, but are you sure o3F3o || x3x3f is possible to have unit edges? From what I can tell, the circumradius of x3x3f and the circumradius of o3F3o differs by more than an edge length. :( (But I may have made a mistake somewhere...)

yeah, pretty sure. in fact, it's just part of your J4.8:
[...]

Argh, I knew it was a silly careless calculation mistake on my part. :oops: :oops: :oops: :oops:

I seriously need to write a program that automates the computation of lace towers...
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Re: Johnsonian Polytopes

Postby quickfur » Fri Feb 28, 2014 12:22 am

student91 wrote:
student5 wrote:hello everybody :D

I have been reading the posts, and might have found a CRF, thanks to Quickfur's latest find :mrgreen:
the outer layer (the weird shape you encountered, which looked like o3x5o) might be constructed using 4 thawro's, some triangular cupola, and an x3x3o, and I'll try to explain how;
[...]
Well, you got a point here!! your lace-tower isn't anything, but what you're describing certainly makes sense!! What you're saying is to fill it up with J92's, in a way that indeed works (although I had a hard time checking that it did). so finally whatyou get is o3x3o || x3o3f || F3o3x || x3x3f || o3F3o || o3x3x. It works!! (I think).

Woohoo, it does work!!! The polychoron is CRF!!! :D :D :D

Congratz, student5, you've found your first CRF crown jewel. :P

Well here's a render centered on the o3x3o:

Image

This is basically identical to CJ4.8. But the other side, centered on o3x3x, is different:

Image

This is weird, so the same octahedron + 4*J91 surface patch is compatible with two different patches, one with a tetrahedron + 4*J92, and the other with a truncated tetrahedron + 4*inverted_J92. Amazing!

I think I'll assign this CRF as CJ4.8.2, since it appears to be related to CJ4.8 (which maybe should be renamed CJ4.8.1?). I wonder if there are any other CRFs that can be joined together in this way via that unusual skew polyhedron?

EDIT: Created a wiki page for it: CJ4.8.2 (you can find the Stella4D model on that page)
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Re: Johnsonian Polytopes

Postby student91 » Fri Feb 28, 2014 12:47 am

quickfur wrote:[...]I think I'll assign this CRF as CJ4.8.2, since it appears to be related to CJ4.8 (which maybe should be renamed CJ4.8.1?). I wonder if there are any other CRFs that can be joined together in this way via that unusual skew polyhedron?

EDIT: Created a wiki page for it: CJ4.8.2 (you can find the Stella4D model on that page)

Again, nice renders.
I think there are other ways to close it up, just found a new one (I hope): (whatever)|| o3F3o || x3x3f || F3o3x || x3f3x || o3x3x. let's hope it works :) :\ :)
(for whatever you can either use o3x3x or f3x3o || x3o3o.)
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Re: Johnsonian Polytopes

Postby student5 » Fri Feb 28, 2014 12:54 am

quickfur wrote:[...]
Woohoo, it does work!!! The polychoron is CRF!!! :D :D :D

Congratz, student5, you've found your first CRF crown jewel. :P


:lol: :lol: :lol: wooohooooo :D :D :D this thing looks so simple, it's so coool, you could consider it art :nod: :roll: :P

Well here's a render centered on the o3x3o:

Image

This is basically identical to CJ4.8. But the other side, centered on o3x3x, is different:

Image

This is weird, so the same octahedron + 4*J91 surface patch is compatible with two different patches, one with a tetrahedron + 4*J92, and the other with a truncated tetrahedron + 4*inverted_J92. Amazing!

well, I think o3x3o || x3o3f || F3o3x || x3x3f || o3F3o || x3x3x could be possible aswell. if you fit the thawro's the same way, you'd get some 4 hexagonal prisms, 4 thawro's and 12 ??square pyramids??
I think I'll assign this CRF as CJ4.8.2, since it appears to be related to CJ4.8 (which maybe should be renamed CJ4.8.1?). I wonder if there are any other CRFs that can be joined together in this way via that unusual skew polyhedron?

well, actually I think there might ba a range of polychodra which could be derived from that polyhedron, I just do not know exactly how, but there might be quite some...
take the queer polyhedron Image it could be seen, in a way, as a tetrahedral id-thingy or something. but, if a tetrahedral id-thingy is possible, what about an oct/cub id-thingy or an ike/doe id-thingy ( :lol: ike/doe id)
the lace towers of those shapes would be Fxo3oxF4xfo&#xt, Fxo4oxF3xfo&#xt, Fxo3oxF5xfo&#xt or Fxo5oxF3xfo&#xt I think quite a lot may be possible with these shapes, I'll just pass most of the possibilities during my holiday
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Re: Johnsonian Polytopes

Postby Keiji » Fri Feb 28, 2014 7:16 am

I have a slight suggestion. As quickfur has said, all the polytopes mentioned lately have been analogs of J91/J92. Could we perhaps just call these set of figures the bilbiro/thawro polychora, and give them BT numbers instead of CJ numbers?

With how many we are finding, they shouldn't really be called crown jewels. It would certainly be nice to enumerate all possible BT polychora, though! :)
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Feb 28, 2014 8:45 am

Great finds lately!

Just an idea: are those hexagons of that "equatorial" skew neerly-icosidodecahedron kind of flat, i.e. are either all their vertices at the same circumradius, or otherwise are those those faces "parallel" in some sense, i.e. do they all have a symmetrical slope? Then it could be possible to join either one of these J92 containing halfs as well. Thus providing a third relative. Or am I wrong in this?

--- rk
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