Trixylodiminished hydrochoron

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

Trixylodiminished hydrochoron

Postby Keiji » Wed Mar 12, 2014 6:38 pm

(Mono)xylodiminished hydrochoron is the snub demitesseract. Bixylodiminished hydrochoron is that awesome cell-transitive CRF we know and love. Tetraxylodiminished hydrochoron would obviously be the xylochoron itself and pentaxylodiminished hydrochoron would be the empty set. But what is trixylodiminished hydrochoron? Has anyone attempted to construct this? Presumably it is not CRF (if it was it would have been mentioned already), but I am curious. :)
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Re: Johnsonian Polytopes

Postby quickfur » Wed Mar 12, 2014 7:02 pm

Hmm. This is probably not hard to construct, because I already listed the vertices of the BXD on my website, so all you have to do, is to go through the 600-cell's vertices and delete all of the BXD vertices. Since the BXD contains 3 copies of the 24-cell, this should leave the vertices of 2 24-cells, which is what you want. Then you just send the result to makepoly, and off you go. :)
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Trixylodiminished hydrochoron

Postby Keiji » Wed Mar 12, 2014 10:48 pm

Okay, I worked out the vertices, and here's the links to the .def and .off files.

Sadly it's not* CRF, but guess what it is?

Image

That cell right there is the cell that shows up in the dual of BXD. The element counts are BXD's element counts reversed, so I'm quite inclined to suggest this polytope is indeed the dual of BXD!

edit: *unless it actually is CRF and the cells are skew polyhedra, somehow? How do you verify all edge lengths equal with polyview?
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Re: Johnsonian Polytopes

Postby quickfur » Wed Mar 12, 2014 11:29 pm

Keiji wrote:
quickfur wrote:Hmm. This is probably not hard to construct, because I already listed the vertices of the BXD on my website, so all you have to do, is to go through the 600-cell's vertices and delete all of the BXD vertices. Since the BXD contains 3 copies of the 24-cell, this should leave the vertices of 2 24-cells, which is what you want. Then you just send the result to makepoly, and off you go. :)


Okay, I worked out the vertices, and here's the links to the .def and .off files.

Sadly it's not* CRF, but guess what it is?

Image

That cell right there is the cell that shows up in the dual of BXD. The element counts are BXD's element counts reversed, so I'm quite inclined to suggest this polytope is indeed the dual of BXD!

Whoa. That's insane, if it's actually the dual of BXD!! It would be the first example of a polytope (that I know of, anyway), where the dual is derived by deleting 1/3 the vertices of the original!

edit: *unless it actually is CRF and the cells are skew polyhedra, somehow? How do you verify all edge lengths equal with polyview?

Code: Select all
calc edgelen
# or "calc elen" for short

will tell you the minimum, maximum, and average edge lengths. If all three are equal, then you know it has equal edge lengths. (Sadly, it doesn't currently tell you all occurring edge lengths -- that's something I sorely miss recently while constructing CRFs, so I might add that in a future version.)

Also, since makepoly uses a convex hull algorithm, it's not possible for cells to be skew polyhedra; by definition, the cells output by a convex hull algorithm must lie flat on their respective hyperplanes. (It is possible to feed polyview a non-convex polytope, if you hand-doctor the .def files, but if you got the definition from makepoly, then it is definitely convex, and facets are definitely non-skew. Note that feeding polyview doctored .def files may produce strange output, since many places in the code assume convexity.)
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Re: Johnsonian Polytopes

Postby quickfur » Thu Mar 13, 2014 12:07 am

Keiji wrote:[...]Image

That cell right there is the cell that shows up in the dual of BXD. The element counts are BXD's element counts reversed, so I'm quite inclined to suggest this polytope is indeed the dual of BXD!
[...]

Hmm. Here's a nicer ;) render with two cells sharing only a vertex highlighted:

Image
It does look indeed like it's cell-transitive (though that's currently not proven). Polyview reports ("count c with v 0", "count c with v 1", ..., etc.) that 9 cells surround each vertex (OK, I didn't prove this one either, but all randomly picked vertices have 9 cells around them), so it appears to be vertex-transitive as well (well OK, cell counts don't prove transitivity, but the circumstantial evidence is rather strong). Does the dual of BXD also have 9 cells around each vertex? If so, I'd say the evidence is pretty compelling! Better yet, it appears to consistently have 8 edges around every vertex ("count e with v 0", "count e with v 13", ...). Since there are 8 faces in each J63, this certainly looks like the dual of BXD indeed!! :o_o: :o_o:

My mind is totally blown as to how deleting 1/3 of the vertices of BXD produces its dual... that's just ... unreal. :sweatdrop:
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Mar 13, 2014 5:31 am

Removed full last post quote. ~Keiji

I took a look at it in Stella -- it IS cell-transitive and vertex-transitive, but has two different edge-lengths. And yes, taking a dual of it results in the polychoron made of 48 tridiminished icosahedra :)
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Re: Johnsonian Polytopes

Postby Keiji » Thu Mar 13, 2014 6:57 am

quickfur wrote:Does the dual of BXD also have 9 cells around each vertex?


Surely that should be easy to answer as BXD consists of TDI cells which each have 9 vertices? ;)

Looks like Marek has confirmed it anyway. My first thought was to load the .off file in Stella for myself, but you need to buy the full version to load .off files :(

Well, in any case - I'll be naming the BXD's dual as the trixylodiminished hydrochoron, or TrXD for short. (I'm thinking of pronouncing them bexy and trexy. :D )

Any suggestions for the name of that cell, which so far I've been calling the "triangulated triangular prism"? Perhaps take some inspiration from the naming of sphenocorona, etc.? It's shown up in two CRF duals so far, and I'm inclined to expect it to appear in more.

Also, is there a way to "extract" a cell in polyview to its own .def file, so that it can be rendered independently as a 3D figure?
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Mar 13, 2014 7:11 am

quickfur wrote:
Keiji wrote:
quickfur wrote:Hmm. This is probably not hard to construct, because I already listed the vertices of the BXD on my website, so all you have to do, is to go through the 600-cell's vertices and delete all of the BXD vertices. Since the BXD contains 3 copies of the 24-cell, this should leave the vertices of 2 24-cells, which is what you want. Then you just send the result to makepoly, and off you go. :)


Okay, I worked out the vertices, and here's the links to the .def and .off files.

Sadly it's not* CRF, but guess what it is?

Image

That cell right there is the cell that shows up in the dual of BXD. The element counts are BXD's element counts reversed, so I'm quite inclined to suggest this polytope is indeed the dual of BXD!

Whoa. That's insane, if it's actually the dual of BXD!! It would be the first example of a polytope (that I know of, anyway), where the dual is derived by deleting 1/3 the vertices of the original!


Well, that was already well-known.
And moreover, this is far more general, valide for the whole family:

ike = x3o5o --> ike - 3gon = teddi --> ike - 2x 3gon = some self-dual figure (in fact the cell highlighted above) --> ike - 3x 3gon = teddi dual --> ike - 4x 3gon = ike dual = doe
ex = x3o3o5o --> ex - ico = sadi --> ike - 2x ico = bidex --> ex - 3x ico = bidex dual --> ex - 4x ico = sadi dual --> ex - 5x ico = ex dual = hi

cf. to that topic eg. this separate thread.
(We even where discussing the hyperbolic space extension of x3o3o3o5o. But that one somehow eluded so far the accurate imagination...)

--- rk

Edit: just corrected the cursive parts above...
Last edited by Klitzing on Thu Mar 13, 2014 7:49 am, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby Keiji » Thu Mar 13, 2014 7:22 am

Klitzing wrote:Well, that was already well-known.
And moreover, this is far more general, valide for the whole family:

ike = x3o5o --> ike - 3gon = teddi --> ike - 2x 3gon = teddi dual --> ike - 3x 3gon = ike dual = doe
ex = x3o3o5o --> ex - ico = sadi --> ike - 2x ico = bidex --> ex - 3x ico = bidex dual --> ex - 4x ico = sadi dual --> ex - 5x ico = ex dual = hi


Hold up, something doesn't seem right here. If I'm reading your shorthand correctly, ex = hydrochoron (120 = 5x24 vertices), sadi = snub demitesseract (96 = 4x24 vertices), bidex = BXD (72 = 3x24 vertices), bidex dual = TrXD (48 = 2x24 vertices), that looks good so far.

But the snub demitesseract has 144 cells, so its dual has 144 vertices. Tetraxylodiminished hydrochoron, or your ex - 4x ico, would surely have 24 = 1x24 vertices, so cannot be the snub demitesseract dual. Moreover, there is only one xylochoron left to remove - so the tetraxylodiminished hydrochoron would surely be the xylochoron itself.

And pentaxylodiminished hydrochoron, or your ex - 5x ico, would surely have 0 = 0x24 vertices, the empty set, so cannot be the cosmochoron, or hydrochoron dual.

Furthermore, in your 3D "analog" with icosahedron, you are removing triangles, a 2D figure? In the 4D case, we are removing vertices of a xylochoron (4D figure) from a hydrochoron (another 4D figure) - not removing (copies of) a 3D figure from a 4D figure...
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Mar 13, 2014 7:54 am

The ike part just has been corrected. There indeed slipped in a mistake. (Cf. the original mail.)

With respect to the higher withdrawels, you should consider that those are diminishings at former vertex positions, not hulls of remainder vertices! That is, even if you omitt all former vertices, you still will keep up with some kernel figure.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Thu Mar 13, 2014 3:14 pm

Keiji wrote:[...]
Looks like Marek has confirmed it anyway. My first thought was to load the .off file in Stella for myself, but you need to buy the full version to load .off files :(

I'm so stupid... there is a very obvious method to check: run it through the `dissect` utility to compute its dual:
Code: Select all
dissect polytope.def dual | makepoly dual_polytope - dual.def

That would have cleared all doubts immediately. :sweatdrop:

[...]
Also, is there a way to "extract" a cell in polyview to its own .def file, so that it can be rendered independently as a 3D figure?

Currently, this isn't directly supported, but you can do it by creating the following script:
Code: Select all
load polytope.def
select ... [some cell specification here]
list vertices in selection

Let's say you name this script "getcell.pvs". Then you can do this:
Code: Select all
polyview -qes getcell.pvs | sed -e's/.*: //' | makepoly -fperm single_cell - single_cell.def

That should give you a single-cell (subdimensional) polytope containing only that cell. Unfortunately, you won't be able to treat it as a 3D model, because the coordinates will still be 4D (or higher).

To get a 3D model out of it, one thing you can do is to use parallel projection and dig out the coordinates from the .pov file:
Code: Select all
load single_cell.def
parallel
translate to center
povfile single_cell.pov
render

You may have to do a "lookfrom <....>", where <...> is the vector for that single cell in the .def file (i.e., the normal to the cell's hyperplane), in order to avoid projective foreshortening.

Now edit single_cell.pov and look for the vertex_vectors block in the mesh2 object. These coordinates will have been projected into 3D, so you should be able to feed them back into makepoly to make a 3D model of it.

(Yeah I know, this is very involved... I really should add this as a feature sometime. :sweatdrop: )
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Re: Johnsonian Polytopes

Postby quickfur » Thu Mar 13, 2014 3:28 pm

quickfur wrote:
Keiji wrote:[...]
Also, is there a way to "extract" a cell in polyview to its own .def file, so that it can be rendered independently as a 3D figure?

[...]

Another, cleaner way to do this, is if you're lucky enough that the cell you want lies on a hyperplane perpendicular to one of the coordinate axes (i.e., all its vertices has one coordinate with the same value), then you can just delete that coordinate to get a 3D figure out of it. Many of the .def files I have has this property, due to the way they are constructed (e.g., 120cell.def has a cell perpendicular to <1,0,0,0>, so deleting the first coordinate from the vertices of that cell gives you a dodecahedron in 3D). But unfortunately, this isn't always true (600cell.def doesn't, for example).

I wonder if I should add a subcommand to the dissect utility to do this, since you aren't the first one who has wanted to do this -- I've wanted to do this myself in the past too.
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