Johnsonian Polytopes

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 quickfur » Sat Jan 07, 2012 7:19 pm

OK, I've enumerated the n,6-duoprism augmentations:

Code: Select all
3,6-duoprism:
- First ring can be augmented with triangle||6-prism. Adjacent augments
  allowed.
   - 2 augments can be ortho (o) or gyro (g)
   - 3 augments can be ortho-ortho (oo) or ortho-gyro (og)
- Second ring augmentable with 3-prism pyramid.
- Enumeration:
   - 1 augment
   - 2o augments
   - 2g augments
   - 3oo augments
   - 3og augments
- Total: 5+4=9 augmentations (4 augmentations from 3-prism pyramids in the
  second ring)

4,6-duoprism:
- First ring augmentable with triangle||6-prism. Adjacent augments NOT allowed.
   - So only up to 2 augments, ortho (o) or gyro (g).
- Second ring augmentable with 4-prism pyramid.
- Cross-ring angle sum ~= 102.69°, so only one ring can be augmented at a time.
- Enumeration:
   - 1 augment
   - 2o augments
   - 2g augments
- Total: 3+4=7 augmentations (4 augmentations with 4-prism pyramid).

5,6-duoprism:
- First ring augmentable with triangle||6-prism. Adjacent augments NOT allowed.
   - Only up to 2 augments, meta (m) or para (p). Both cases can be either
     ortho (o) or gyro (g).
- Second ring augmentable with 5-prism pyramid.
- Cross-ring angle sums:
   - Vertex-aligned: 70.97°
   - Edge-aligned: 79.19°
   - So both v-aligned and e-aligned cross-ring augments are allowed.
- Cross-ring constraints:
   - When the first ring has no augments, the second ring has the set of 12
     augmentations by 5-prism pyramids.
   - When the first ring has 1 augment, the second ring becomes oriented, and
     so will have > 12 augmentations.
   - When the first ring has 2 ortho augments, the second ring also becomes
     oriented, and has the same number of 5-prism pyramid augmentations as
     the 1-augment case.
   - When the first ring has 2 gyro augments, the split orientations become
     equivalent again, so it has the set of 12 augmentations as in the
     no-augment case.
- Enumeration:
   - No augment in 1st ring:
      - .....*
      - ....**
      - ...*.*
      - ...***
      - ..*..*
      - ..*.**
      - ..****
      - .*.*.*
      - .*.***
      - .**.**
      - .*****
      - ******
      - Subtotal: 12 augmentations
   - 1 augment in 1st ring (imposes alternating configuration on 2nd ring):
     (* = vertex-aligned, # = edge-aligned)
      - .....*
      - ....#.
      - ....#*
      - ...*#.
      - ...*.*
      - ..#.#.
      - ...*#*
      - ..#*#.
      - ..#..*
      - ..#.#*
      - .*.*#.
      - ..#*#*
      - .*.*.*
      - #.#.#.
      - .*.*#*
      - #.#*#.
      - .*#.#*
      - .*#*#*
      - #*#*#.
      - #*#*#*
      - Subtotal: 20 augmentations
   - 2 ortho augments in 1st ring:
      - Same as 1 augment case
      - Subtotal: 20 augmentations
   - 2 gyro augments in 1st ring:
      - Same as 0 augment case
      - Subtotal: 12 augmentations
Total: 12+20+20+12 = 64 augmentations

6,6-duoprism:
- Only non-adjacent augments allowed, can be meta (m) or para (p). No
  cross-ring augments allowed.
- Enumeration:
   - 1 augment
   - 2m augments
   - 2p augments
   - 3 augments (can only be meta)
- Total: 4 augmentations.

As you can see, the number of augmentations is quite small (<10) except for the 5,6-duoprism where the count jumps to 64 suddenly, and then drops back down quickly.
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Re: Johnsonian Polytopes

Postby quickfur » Sun Jan 08, 2012 12:31 am

It's been a while since I did any CRF renders. So here's a pentagon||decagonal_prism, or pentagonal magnabicupolic ring, according to Keiji's terminology.

Image Image

This is what we'll be using to augment n,10-duoprisms.

Its height (from decagonal prism to pentagon) is sqrt((5-2*sqrt(5))/5).
Its intra-ring angle (angle between pentagonal cupola and decagonal prism) is exactly 18° (just like the pentagonal prism pyramid);
Its vertex-aligned cross-ring angle (angle between square pyramid and decagonal prism) is ~13.28°;
Its edge-aligned cross-ring angle (angle between triangular prism and decagonal prism) is ~10.81°.

These shallow angles allow us to go up to 10,20-duoprisms (!!!) in terms of augmentations. The 10,10-duoprism can take all combinations of augmentations in both rings simultaneously. AND each augment can be freely oriented in one of two orientations. That is a lot of combinations! And from 10,11 to 10,20 only non-adjacent augments are permitted.
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Re: Johnsonian Polytopes

Postby Marek14 » Sun Jan 08, 2012 8:35 am

Do we have a complete list of "towers", i.e. several segmentochora put on top of each other? For example, could you have a pentagonal prism pyramid put on top of pentagonal prism||decagonal prism?
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Re: Johnsonian Polytopes

Postby quickfur » Sun Jan 08, 2012 3:29 pm

Marek14 wrote:Do we have a complete list of "towers", i.e. several segmentochora put on top of each other? For example, could you have a pentagonal prism pyramid put on top of pentagonal prism||decagonal prism?

I don't think anyone made a list. I would imagine segmentochora towers would have quite a large number of combinations as well. If you have the list of all the segmentochora in digital form, you might be able to write a program to enumerate all towers.
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Re: Johnsonian Polytopes

Postby quickfur » Sun Jan 08, 2012 4:23 pm

I have just discovered a new kind of duoprism augment that is possible with the 6,6-, 6,8-, and 6,10-duoprisms.

You can augment the prisms in the 6-membered ring with the segmentochora 3-prism||6-prism, cube||8-prism, and 5-prism||10-prism, respectively. This makes the duoprism non-convex, because the dichoral angle between the cupola cells and the 6-, 8-, 10-prism base is 90°. However, the angle between the cupola cells of two adjacent augments is exactly 60°, which means that you can fill in the gap between two adjacent augments by the segmentochora hexagon||3-prism, octagon||4-prism, and decagon||8-prism, respectively. In Keiji's terminology, these are ortho-bicupolic rings. So if you augment all the prisms, the result can be made convex by inserting ortho-bicupolic rings.

If I didn't make any mistake in the construction, the result will be a kind of "expanded" duoprism consisting of a 12-membered ring of 3-, 4-, 5-prisms, respectively, with 6 of them surrounded alternately by cubes and triangular prisms, and the other 6 by triangular prisms and tetrahedra. As far as I can tell, these are valid CRF polychora. I'll try to derive some coordinates and double-check them with my polytope viewer soon.

Note that this augmentation only works for 6-membered ring; for other kinds of rings, the angles are wrong and the result will have mismatching edge lengths, so they won't be CRF. I haven't checked this yet, but the 6,6-duoprism may be able to be augmented in both rings.
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Re: Johnsonian Polytopes

Postby wintersolstice » Sun Jan 08, 2012 7:58 pm

Marek14 wrote:Hmm, if I remember correctly, the vertices of a 24-cell can be expressed as (2,0,0,0) and all permutations and sign changes of it (8 vertices) plus (1,1,1,1) and all sign changes (16 vertices) with edge length 2, edges being of type (2,0,0,0)-(1,1,1,1) or (1,1,1,1)-(-1,1,1,1).

Since 24-cell is self-dual, we can use the same structure for cells/faces.

We should also notice that there are three tesseractic-symmetry groups of cells embedded in the structure. One of them are orthogonal faces ((2,0,0,0) and its ilk), second is (+-1,+-1,+-1,+-1) with even sign parity, third is the same, but with odd sign parity.

Let's start by augmenting the "top" face, (2,0,0,0). This eliminates all faces (+1,+-1,+-1,+-1), so next face to augment must be (0,<+-2,0,0>) [<+-2,0,0> indicates permutation], (-1,+-1,+-1,+-1) or (0,0,0,-2).

Let's make a handy table:
Fixed: (2,0,0,0)
Possible 1: (0,2,0,0),(0,-2,0,0),(0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (first "ring", the same group as first cell)
Possible 2: (-1,1,1,1),(-1,-1,1,1),(-1,1,-1,1),(-1,-1,-1,1),(-1,1,1,-1),(-1,-1,1,-1),(-1,1,-1,-1),(-1,-1,-1,-1) (second "ring", different groups than first cell)
Possible 3: (-2,0,0,0) (antipode of first cell)

This leads to three biaugmented 24-cells:

1: orthobiaugmented 24-cell
Fixed: (2,0,0,0),(0,2,0,0)
Possible 1: (0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to both fixed cells)
Possible 2: (-1,-1,1,1),(-1,-1,-1,1),(-1,-1,1,-1),(-1,-1,-1,-1) (all remaining cells in other groups, in "meta" position to both fixed cells)
Possible 3: (-2,0,0,0),(0,-2,0,0) (antipodes to two fixed cells, "ortho" position to one and "para" to other)

2: metabiaugmented 24-cell:
Fixed: (2,0,0,0),(-1,1,1,1)
Possible 1: (0,-2,0,0),(0,0,-2,0),(0,0,0,-2),(-1,1,-1,-1),(-1,-1,1,-1),(-1,-1,-1,1) (cells in "ortho" position to one fixed cell and "meta" to the other)
Possible 2: (-1,-1,-1,-1) (sort of "antipode" to the particular "meta" combination, it has "meta" position to both fixed cell)

3: parabiaugmented 24-cell:
Fixed: (2,0,0,0),(-2,0,0,0)
Possible: (0,2,0,0),(0,-2,0,0),(0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (the remaining group of the fixed cells, all in "ortho" position to both)

This leads to 4 possible triaugments:

1: tesedge-triaugmented 24-cell (ooo)
Name explanation: it augments three cells in one tesseractic group around the "edge" of this fictious tesseract.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to all three fixed cells)
Possible 2: (-1,-1,-1,1),(-1,-1,-1,-1) (all remaining cells in other groups, in "meta" position to all three fixed cells)
Possible 3: (-2,0,0,0),(0,-2,0,0),(0,0,-2,0) (antipodes to the three fixed cells, "ortho" position to two of them and "para" to the third)

2: assymetrical triaugmented 24-cell (omm)
Name explanation: it's the least symmetrical of the four.
Fixed: (2,0,0,0),(0,2,0,0),(-1,-1,1,1)
Possible 1: (0,0,-2,0),(0,0,0,-2) ("ortho" position to two fixed cells, "meta" to one)
Possible 2: (-1,-1,-1,-1) ("ortho" position to one cell, "meta" to two)

3: arc-triaugmented 24-cell (oop)
Name explanation: The three augmented cells lie on the same tesseract and form an arc of two antipodes and one lateral cell
Fixed: (2,0,0,0),(0,2,0,0),(-2,0,0,0)
Possible 1: (0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to all three fixed cells)
Possible 3: (0,-2,0,0) (antipode to the unpaired cell, "ortho" to two fixed cell and "para" to the third)

4: symmetrical triaugmented 24-cell: (mmm)
Name explanation: the most symmetrical triaugmentation using one augment from each tesseractic group
Fixed: (2,0,0,0),(-1,1,1,1),(-1,-1,-1,-1)
This augment cannot be augmented any further.

Now for tetraaugments. There is 5 of those:

1: tesvertex-tetraaugmented 24-cell
Name explanation: it augments four cells in one tesseractic group around the "vertex" of this fictious tesseract.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2)
Possible 1:(-2,0,0,0),(0,-2,0,0),(0,0,-2,0),(0,0,0,-2) (antipodes to the four fixed cells, "ortho" position to three of them and "para" to the fourth)
Possible 2:(-1,-1,-1,-1) (the only cell in other group, in "meta" position to all four fixed cells)

2: edge+1-tetraaugmented 24-cell
Name explanation: it augments three cells in one tesseractic group around the "edge" of this fictious tesseract plus one cell from another group.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-1,-1,-1,1)
Possible:(0,0,0,-2) (the only possible augmentation cell, "ortho" position to three fixed cells and "meta" to the fourth)

3: tesassymetrical-tetraaugmented 24-cell
Name explanation: it augments four cells in one tesseractic group in "assymetrical" configuration as opposed to "vertex" or "ring" configuration
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-2,0,0,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (two cells in "ortho" position to all four fixed cells)
Possible 2: (0,-2,0,0),(0,0,-2,0) (antipodes to two of the fixed cells, "ortho" position to three fixed cells and "para" to the fourth)

4: skewed tetraaugmented 24-cell
Name explanation: augments two and two cells in two different tesseractic groups
Fixed: (2,0,0,0),(0,2,0,0),(-1,-1,1,1),(-1,-1,-1,-1)
Cannot be augmented any further.

5: ring-triaugmented 24-cell
Name explanation: The four augmented cells lie on the same tesseract and form a ring aroung it.
Fixed: (2,0,0,0),(0,2,0,0),(-2,0,0,0),(0,-2,0,0)
Possible: (0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to all four fixed cells)

Now for pentaaugmentations. There are 3 of them.

1: tesantiedge-pentaaugmented 24-cell
Name explanation: it augments five cells in one tesseractic group such that three cells around one "edge" remains unaugmented.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0)
Possible:(0,-2,0,0),(0,0,-2,0),(0,0,0,-2) (antipodes to three unpaired fixed cells, "ortho" position to four fixed cells and "para" to the fifth)

2: vertex+1-pentaaugmented 24-cell
Name explanation: it augments four cells in one tesseractic group around the "vertex" of this fictious tesseract plus one cell in another group.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-1,-1,-1,-1)
Cannot be augmented any further.

3: tesantiarc-pentaaugmented 24-cell
Name explanation: it augments five cells in one tesseractic group such that the unaugmented cells form an "arc".
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-2,0,0,0),(0,-2,0,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (two cells in "ortho" position to all five fixed cells)
Possible 2: (0,0,-2,0) (antipode to the only unpaired cell, "ortho" position to four fixed cells and "para" to the fifth)

Now for hexaaaugmentations. There are 2 of them.

1: tesantiface-hexaaugmented 24-cell
Name explanation: it augments six cells in one tesseractic group such that two cells around one "face" remains unaugmented.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0),(0,-2,0,0)
Possible: (0,0,-2,0),(0,0,0,-2) (antipodes to two unpaired fixed cells, "ortho" position to five fixed cells and "para" to the sixth)

2: tesantipode-hexaaugmented 24-cell
Name explanation: it augments six cells in one tesseractic group such that the unaugmented cells are antipodal.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-2,0,0,0),(0,-2,0,0),(0,0,-2,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (two cells in "ortho" position to all six fixed cells)

Now for the only possible heptaaugmentation:

1: heptaaugmented 24-cell
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0),(0,-2,0,0),(0,0,-2,0)
Possible: (0,0,0,-2) (antipodes to the only unpaired fixed cell, "ortho" position to six fixed cells and "para" to the seventh)

And the only octaaugmentation:
1: octaaugmented 24-cell
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0),(0,-2,0,0),(0,0,-2,0),(0,0,0,-2)
Cannot be augmented any further.

So, for number of augments:
1 - 1
2 - 3
3 - 4
4 - 5
5 - 3
6 - 2
7 - 1
8 - 1

total: 20

are these augmentations or diminishes because I did a caculation (well I didn't work it out myself)a while ago that shows you can't augment the 24 cell because you ends up blending tetrahedra to octahedra which is a co-planar polyhedron and therefore not valid

second of all I posted the diminishes (which are also augemented tesseracts and no-one has aknoweledged them :( )

the dihedral angle of the 24-cell must be 120 degrees becausde it tesselates 4D space with 3 round a face: regular apeiroteron {3,4,3,3}
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Re: Johnsonian Polytopes

Postby wintersolstice » Sun Jan 08, 2012 8:06 pm

also I saw there was mention of towers of segementachora (there could be other shapes thatcan form part of these :D ) I call these "towertopes" you could also create "planartopes" and possibly "realmictopes" (I might explain these later)
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Re: Johnsonian Polytopes

Postby Marek14 » Sun Jan 08, 2012 8:09 pm

Somebody claimed that non-adjacent augments are possible -- I just went with the idea, didn't check it.
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Re: Johnsonian Polytopes

Postby wintersolstice » Sun Jan 08, 2012 8:15 pm

Marek14 wrote:Somebody claimed that non-adjacent augments are possible -- I just went with the idea, didn't check it.

I wasn't having a go at you :D I was just mentioning that there not valid

here is the info

24-cell honeycomb

In fact because the 24 is the rectate of the 16 cell it means if you augemnt 8 cells (in the form of the tesseract) each octahedron will become a larger tetrahedron, an octahdron being a rectate of the tetrahedron :D so it sort of makes sense

Has anyone included my list of the augemented tesseracts yet?
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jan 09, 2012 12:16 am

wintersolstice wrote:[...]are these augmentations or diminishes because I did a caculation (well I didn't work it out myself)a while ago that shows you can't augment the 24 cell because you ends up blending tetrahedra to octahedra which is a co-planar polyhedron and therefore not valid

OK, I think we need to post all our calculations so that we can check each other's work. Otherwise we're just blindly deriving results that are possibly invalid, or just misunderstanding each other.

I just tried to calculate the 24-cell's dichoral angles again, and realized I made a mistake in my previous post. Here's my recalculation of the 24-cell's dichoral angle, if it's wrong please point out which step is incorrect:

- Starting with with a 24-cell with coordinates apacs<2,0,0,0> and apacs<1,1,1,1>, I picked two adjacent cells, whose centers are <1,1,0,0> and <1,0,1,0>, respectively.
- The triangular ridge they meet at has vertices <1,1,1,1>, <1,1,1,-1>, and <2,0,0,0>. Which means the center of the triangular ridge is <4/3, 2/3, 2/3, 0>.
- Now let A be the vector from <4/3, 2/3, 2/3, 0> to <1,1,0,0>, and let B be the vector from <4/3, 2/3, 2/3, 0> to <1,0,1,0>. Then the angle between A and B must be the 24-cell's dichoral angle, correct?
- So A = <-1/3, 1/3, -2/3, 0>, and B = <-1/3, -2/3, 1/3, 0>. Then the dot product A.B = |A|*|B|*cos(angle), so cos(angle) = (A.B) / (|A|*|B|).
- The value of A.B = 1/9 - 2/9 - 2/9 + 0 = -3/9 = -1/3.
- And |A| = |B| = sqrt(2/3).
- Therefore, cos(angle) = (-1/3) / (sqrt(2/3) * sqrt(2/3)) = -1/2.
- So the angle is acos(-1/2) = 60°.
- CORRECTION: I see where I went wrong here. The dot product is negative, so the angle is obtuse. Therefore it's 120°, not 60°. :oops: :oops:

My bad. So the 24-cell is not augmentable. :(

second of all I posted the diminishes (which are also augemented tesseracts and no-one has aknoweledged them :( )

Actually, I think Keiji did acknowledge them.

And actually when I enumerated the tesseract augments I also knew you had enumerated them -- but i did it anyway because it's always good to check each other's work in case there's an overlooked mistake. I didn't intend my enumeration to be snubbing your work or anything like that. I just wanted to re-derive them to see if the two lists matched up.

But anyway, why don't you post your stuff on the wiki page? There's a wiki page dedicated to the CRF discovery project. This topic has too many posts, and it's getting very confusing to figure out what results are still valid, and what has been invalidated by a later post. We should be consolidating the most up-to-date info on the wiki page instead, so that we're not wasting time working with disproved results or repeating work that has already been done (other than double-checking, of course).

the dihedral angle of the 24-cell must be 120 degrees becausde it tesselates 4D space with 3 round a face: regular apeiroteron {3,4,3,3}

I'm an idiot, I should be checking my own work more carefully. :( And I need to stop doing 4D calculations while working on something else at the same time... all sorts of stupid errors creep into my calculations when I do that, as has been happening very frequently lately. :( I'm very sorry about my wrong posts... it's too easy to be caught up in the excitement at the moment of discovery and fail to verify that my results are actually correct.
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Re: Johnsonian Polytopes

Postby Keiji » Mon Jan 09, 2012 8:49 am

quickfur wrote:There's a wiki page dedicated to the CRF discovery project.


CRF polychora discovery project :)
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Jan 09, 2012 10:25 am

OK, so my augmentation list for 24-cell doesn't work. But -- it still showcases possible symmetries and maybe it could be used for augmenting something else with the same symmetry.

Options are:

Truncated 24-cell: could get cubic pyramids?
Rectified 24-cell: possibilities are cubic pyramids or perhaps something on the cuboctahedral cells? The lowest segmentochora would be topped with squares, octahedra or square pyramids.
Small rhombated 24-cell: triangular prisms, cuboctahedra and rhombicuboctahedra. Octahedron || Rhombicuboctahedron has height only 1/2.
Octagonny with its truncated cubes: cuboctahedron || truncated cube is once again only 1/2 high.
Great rhombated 24-cell: includes triangular prisms, truncated cubes and truncated cuboctahedra. Truncated octahedron || truncated cuboctahedron is once again 1/2 high.
Great prismated 24-cell: hexagonal prisms and truncated cuboctahedra. Could the prisms be augmented with triangle||hexagonal prism?
Prismatorhombated 24-cell: includes rhombicuboctahedra, truncated octahedra and triangular and hexagonal prisms.
Small prismated 24-cell: Octahedra + triangular prisms.

Some of them might be augmentable... Basically, maybe we need a document showing the dichoral angles for all uniform polychora...

BTW, do we have this group of CRFs?

Start with rectified 600-cell. We know you can diminish icosahedral faces by cutting icosahedron|| icosidodecahedron or the towertope icosahedron || icosidodecahedron || rhombicosidodecahedron. But, once you cut these off, what else can you glue in their place?

The icosidodecahedron can be topped not only by icosahedron, but by every diminishing of it apart from pentagonal pyramid. Rhombicosidodecahedron can be topped by dodecahedron, which has smaller height than icosidodecahedron || rhombicosidodecahedron and therefore should definitely fit.
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Re: Johnsonian Polytopes

Postby wintersolstice » Mon Jan 09, 2012 1:00 pm

quickfur wrote:I have just discovered a new kind of duoprism augment that is possible with the 6,6-, 6,8-, and 6,10-duoprisms.

You can augment the prisms in the 6-membered ring with the segmentochora 3-prism||6-prism, cube||8-prism, and 5-prism||10-prism, respectively. This makes the duoprism non-convex, because the dichoral angle between the cupola cells and the 6-, 8-, 10-prism base is 90°. However, the angle between the cupola cells of two adjacent augments is exactly 60°, which means that you can fill in the gap between two adjacent augments by the segmentochora hexagon||3-prism, octagon||4-prism, and decagon||8-prism, respectively. In Keiji's terminology, these are ortho-bicupolic rings. So if you augment all the prisms, the result can be made convex by inserting ortho-bicupolic rings.

If I didn't make any mistake in the construction, the result will be a kind of "expanded" duoprism consisting of a 12-membered ring of 3-, 4-, 5-prisms, respectively, with 6 of them surrounded alternately by cubes and triangular prisms, and the other 6 by triangular prisms and tetrahedra. As far as I can tell, these are valid CRF polychora. I'll try to derive some coordinates and double-check them with my polytope viewer soon.

Note that this augmentation only works for 6-membered ring; for other kinds of rings, the angles are wrong and the result will have mismatching edge lengths, so they won't be CRF. I haven't checked this yet, but the 6,6-duoprism may be able to be augmented in both rings.


I think you need to change "3-prism || 6-prism" (which I think is actually a triangular-cupola prism on its side) to the "triangle || 6-prism" etc, since the height of each prism is one edge length, the 3-prism || 6-prism would have a 90 angle between the base and cupola cells (these joined to the base of the prism) the height of the each cupola cells are the same and so are the widths of the bases so they form a rectangle which is what creates right angles.

Plus if you use these segmentachora they alternate between 3-prisms and cubes on the square faces so after you work out how many augmentations you can increase the list by rotating the segmentochora around the prisms (only when you have more than one of them of course) if the other girdle is augmented then one square of each prism is joined to a diferent configuartion of 3-prism/cubes on the other girdle, so you'd have to be more careful there aswell.

the latter part doesn't create problems just more possibities :D

dichoral angles might need checking though :D

also I don't have any software to filter out duplicates (rotations and reflections etc) and it seems that some of you do, I also don't know anything about programing so have no-idea where to start :(

If someone could help me with that I would appreciate it
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Jan 09, 2012 3:22 pm

The "software" I used was a very simple algorithm that depended on enumerating all possible combinations and then weeding through them. That algorithm would be impractical for more than, say, 30 active augmentation sites.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jan 09, 2012 4:45 pm

wintersolstice wrote:
quickfur wrote:I have just discovered a new kind of duoprism augment that is possible with the 6,6-, 6,8-, and 6,10-duoprisms.

You can augment the prisms in the 6-membered ring with the segmentochora 3-prism||6-prism, cube||8-prism, and 5-prism||10-prism, respectively. This makes the duoprism non-convex, because the dichoral angle between the cupola cells and the 6-, 8-, 10-prism base is 90°. However, the angle between the cupola cells of two adjacent augments is exactly 60°, which means that you can fill in the gap between two adjacent augments by the segmentochora hexagon||3-prism, octagon||4-prism, and decagon||8-prism, respectively. In Keiji's terminology, these are ortho-bicupolic rings. So if you augment all the prisms, the result can be made convex by inserting ortho-bicupolic rings.
[...]


I think you need to change "3-prism || 6-prism" (which I think is actually a triangular-cupola prism on its side) to the "triangle || 6-prism" etc, since the height of each prism is one edge length, the 3-prism || 6-prism would have a 90 angle between the base and cupola cells (these joined to the base of the prism) the height of the each cupola cells are the same and so are the widths of the bases so they form a rectangle which is what creates right angles.

What I had in mind was that the 3-prism||6-prism (which is, as you say, a cupola prism placed sideways) would augment at 90°, which makes the duoprism non-convex, but then a cupola wedge (hexagon||cube) is inserted between them to make it convex again. Partial augmentation would all be non-convex, because you need to fully augment the entire ring and insert wedges between every adjacent pair of augments in order to remain convex. Here's a diagram to illustrate what I mean:

Image

The P represents the duoprism, and A1 and A2 are the cupola prism augments. W is the wedge inserted to make the result convex.

However, I did make a mistake again. :( I wrongly assumed that the height of the augment, which is the height of the cupola, is equal to the edge length, but it is not. With the reduced height of the augment, the 60° angle in a 6-membered ring would cause the wedge to be non-CRF. It turns out that in order for the wedge to be CRF, the height of the cupola has to satisfy certain constraints. The dihedral angle of the duoprism P is (n-2)*180/n, and the sum of all the angles around the center must be 360, so the dihedral angle of W must be 360-(n-2)*180/n. But the length of the top of W must be equal to the edge length in order for the result to be CRF. If the height of A1 and A2 are equal to the edge length, then n=6. However, the height of the 3-, 4-, and 5-cupola are less than the edge length. In that case, n=6 would not yield a CRF.

So I tried to find what values of n would work for each of the cupola. It turns out that this constraint is impossible to satisfy with the 3-cupola and the 5-cupola, because if W's width is equal to the edge length, then the dihedral angle of W is an irrational fraction of 360° and so n would not be an integer, which means the duoprism is not uniform. The only case that works is the 4-cupola, which gives n=4.

Unfortunately, the 4,4-duoprism augmented in this manner is identical to the runcinated tesseract. :) So my method doesn't actually produce any new CRFs. :(

Now, if A1 and A2 are m,4-duoprisms instead, that is, their height is equal to edge length, then the constraints are solvable with n=6 (because the dihedral angle of W would be 60°). The wedge W would then be a diprism wedge, but then the dihedral angle with the other ring of the duoprism would be 180°, which means the triangular prism facets of W merge with the 6-prisms in the other ring, turning them into 12-prisms. So augmenting the m,6-duoprism in this way turns into an m,12-duoprism, which again doesn't produce anything new. :\

[...]also I don't have any software to filter out duplicates (rotations and reflections etc) and it seems that some of you do, I also don't know anything about programing so have no-idea where to start :( [...]

Like Marek said, this "software" is really just programs we wrote (separately) to check all possible reflections/rotations and weed out duplicates by brute force (i.e., exhaustive enumeration). Which is OK for relatively small and simple polychora like duoprisms, but which becomes unacceptably slow when dealing with more complicated polytopes (looking at ya, 600-cell family polytopes! :evil: ).
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Re: Johnsonian Polytopes

Postby quickfur » Mon Jan 09, 2012 6:19 pm

Alright, i've made a stub wiki page with some dichoral angles of segmentotopes that we can use for augmenting duoprisms, etc., here:

http://teamikaria.com/hddb/wiki/Segmentochoron#Properties

Please make corrections if you find any wrong info, or add more info if you have it so that we can make good use of it.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 10, 2012 1:47 am

Alright, I've finally discovered a real, valid, CRF augment of a duoprism. Well, it happens to be a 4,4-duoprism == tesseract, but the augmentation treats it as a duoprism. Basically, we augment each cube in a ring of 4 cubes with line||cube (i.e., the square pyramid prism), the height of which is exactly edgelen*sqrt(2)/2. These augments have 90° intra-ring angle, so the result is non-convex, but the distance between the peaks in two adjacent augments is exactly equal to the edge length, so the gaps can be filled in with line||square (basically, a wedge of two square pyramids with a line joining their apices -- it consists of two square pyramids plus a ring of 4 tetrahedra). The result is convex and CRF.

Another way to think about it, which is probably easier than assembling various pieces of 4D Lego, is that it is just the convex hull of a tesseract and an octagon. It has 4 cubes, 16 triangular prisms, and 16 tetrahedra.

I've verified that this construction is valid; I built a model of it and checked that the convex hull has equal edge lengths. Here's a projection of the thing:

Image

It retains the edge skeleton of the tesseract, which is quite visible here. I turned off visibility clipping so that all the cells can be seen.

Here's another view of it:

Image

In this render I used red for edges on the far side. At the center is a circle of 4 tetrahedra, with an (non-regular) octahedra outline. Above and below it are the circles of 4 triangular prisms, quite distorted because they are viewed at a 45° angle. If you look carefully at the top and bottom of the image, you can see how the edges on the far side form two more circles of 4 tetrahedra.

I don't know what's a good name for it, though. Currently I have it on file as tesseract+octagon, but I'm not particularly fond of the name.

EDIT: Just for completeness' sake, the coordinates of this augment are as follows, it's literally just the tesseract's vertices plus an octagon of equal edge length:

apacs<1,1,1,1>
<±1, ±(1+sqrt(2)), 0, 0>
<±(1+sqrt(2)), ±1, 0, 0>
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Re: Johnsonian Polytopes

Postby Keiji » Tue Jan 10, 2012 11:29 am

I'd say this was analogous to the square orthobicupola, which can be seen as the convex hull of a cuboid and octagon. The fact that the 8-cell is regular in the 4D version would be because of the √4 = 2 coincidence.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 10, 2012 5:20 pm

Keiji wrote:I'd say this was analogous to the square orthobicupola, which can be seen as the convex hull of a cuboid and octagon. The fact that the 8-cell is regular in the 4D version would be because of the √4 = 2 coincidence.

Yeah it does resemble the square orthobicupola in some respects. Although in this case, you can't really decompose it into two cupolae. (Or can you??)
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Re: Johnsonian Polytopes

Postby Keiji » Tue Jan 10, 2012 7:03 pm

No, but you should be able to decompose it into four copies of some object. I'm not sure what the component would be though; can you render a projection with the octagon reduced to a line, and the other three dimensions fully extended? If I'm correct, this should make the decomposition obvious.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 10, 2012 7:36 pm

Keiji wrote:No, but you should be able to decompose it into four copies of some object. I'm not sure what the component would be though; can you render a projection with the octagon reduced to a line, and the other three dimensions fully extended? If I'm correct, this should make the decomposition obvious.

Already did, it's the second projection i have above.

EDIT: P.S. you're probably right, I think the individual pieces are the segmentotope octagon||cube (K 4.73, aka square||square cupola).
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Re: Johnsonian Polytopes

Postby Keiji » Tue Jan 10, 2012 8:15 pm

Ah, I didn't recognise it. I was expecting to see a cube occupy the "middle" of the shape - it looks like that projection's been rotated 45 degrees from what I had in mind.
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Re: Johnsonian Polytopes

Postby wintersolstice » Tue Jan 10, 2012 8:33 pm

How can the "aerochoron" (16-cell) be joined to itself?

The 24-cell honeycomb {3,4,3,3} is the dual to the 16-cell honeycomb {3,3,4,3} (I mentioned the former recently)

bascialy the dichoral angle for the 16 cell must be 120 degrees to allow the tessalation (the 3 at the end of the Schaffli symbol means 3 round a face) this means the angle of the octahedron pyramid must be 60 degrees, subtract that from 180 degrees (when you rectate it) and you get another shape with dichoral angle 120 degrees (the 24-cell that tesselates with 3 round an edge)

so if the angle of the 16-cell is 120 degrees, then joining two of them will make it concave

PS I wasn't sure how I should edit it because someone might put it back if I remove it.

also where is the list of dichoral angles on the wiki, I could be a good contributor, :D I know a lot of strategies for ditopal angles.
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Jan 10, 2012 9:57 pm

Cupolas and rotundas derived from uniform polychora: attempt of definitive enumeration. I use the method of tracing vertex figures of polyhedra on the vertex figure of the polychoron.
Each of the results can be studied further.
Basically, diminishings are possible with polychora whose vertex figure is not a tetrahedron.

16-cell: can be cut into two octahedral pyramids. Octahedral pyramid itself can be cut in two square duopyramids.
24-cell: can be cut into two octahedron || cuboctahedron segmentochora. Since the dichoral angle of 24-cell is 120°, the angle between cuboctahedron and octahedron must be 60. The angle between cuboctahedron and square pyramid should be 90. Lots of cuboctahedron-based segmentochora should fit here.
24-cell can be also diminished by cutting cubic pyramids off. Both methods can be used at once to obtain segmentochora square pyramid || cuboctahedron and square || cuboctahedron.
600-cell: icosahedral pyramids can be cut off.

Truncated 16-cell: Can be cut in two rotunda-like objects looking like octahedron || truncated octahedron -- not a segmentochoron, lateral cells are 8 truncated tetrahedra and 6 square pyramids.
Truncated 600-cell: Similar rotunda can be cut off, looking like icosahedron || truncated icosahedron with lateral cells being 20 truncated tetrahedra and 12 pentagonal pyramids.

Rectified 5-cell: Triangular prism pyramid can be cut off, leaving a segmentochoron triangular prism || gyrated triangle.
Rectified tesseract: Can be cut in three layers, tetrahedron || truncated tetrahedron, truncated tetrahedron || dual truncated tetrahedron and truncated tetrahedron || tetrahedron
Rectified 24-cell: Can have cuboctahedron || truncated octahedron cut off. If you cut two at opposite sides, you get a layered truncated octahedron with 6 cuboctahedra, 8 triangular cupolas and 12 cubes as lateral cells. This could serve as one possible "elongation" mode for segmentochora that use truncated octahedron.
Rectified 120-cell: Segmentochoron dodecahedron || truncated dodecahedron can be cut off.
Rectified 600-cell: Several possible tracings. You can cut off pentagonal prism pyramid, segmentochoron icosahedron || icosidodecahedron, or towertope icosahedron || icosidodecahedron || rhombicosidodecahedron.

Small rhombated 5-cell: One possibility is to cut it in half, into octahedron || truncated tetrahedron and cuboctahedron || truncated tetrahedron. Another possibility is to cut triangle || hexagonal prism. The rest will be funny triangular prism || hexagonal prism with 1 intermediate layer of vertices.
Small rhombated tesseract: You can cut two caps looking like rhombicuboctahedron || truncated cube and leave truncated cube prism. Another diminishing is by square || octagonal prism.
Small rhombated 24-cell: Cuboctahedra can be cut either in single layer (cuboctahedron || truncated cube) or in two layers (cuboctahedron || truncated cube || truncated cuboctahedron).
Small rhombated 120-cell: You can cut off rhombicosidodecahedron || truncated dodecahedron or pentagon || decagonal prism.
Small rhombated 600-cell: You can cut off icosidodecahedron || truncated icosahedron. You can also cut two layers deeper to get icosidodecahedron || truncated icosahedron || truncated icosidodecahedron, where truncated icosahedron || truncated icosidodecahedron is a three-layer polytope whose lateral cells are 20 pentagonal rotundas, 40 triangular cupolas and 30 pentagonal prisms.

Prismatotruncated 5-cell: You can cut off truncated tetrahedron || truncated octahedron. The remaining shape is three-layered cuboctahedron || truncated octahedron.
Prismatotruncated tesseract: The central truncated cuboctahedral prism is capped by two three-layered rhombicuboctahedron || truncated cuboctahedron whose lateral cells are 6 cubes, 6 square cupolas, 12 hexagonal prisms and 8 truncated tetrahedra.
Prismatotruncated 16-cell: You can cut off truncated cube || truncated cuboctahedron.
Prismatotruncated 24-cell: You can cut off truncated octahedron || truncated cuboctahedron.
Prismatotruncated 120-cell: You can cut off three-layered rhombicosidodecahedron || truncated icosidodecahedron, with lateral faces being 20 truncated tetrahedra, 30 hexagonal prisms, 12 pentagonal prisms and 12 pentagonal cupolas.
Prismatotruncated 600-cell: You can cut off truncated dodecahedron || truncated icosidodecahedron.

Small prismated 10-cell: Can be cut in two tetrahedron || cuboctahedron. It might be possible to glue them together in different orientation.
Small prismated tesseract: Can be cut in two cube || rhombicuboctahedron and central rhombicuboctahedral prism.
Small prismated 120-cell: You can cut off dodecahedron || rhombicosidodecahedron
Small prismated 48-cell: You can cut off octahedron || rhombicuboctahedron

I posted a similar list before, but that was based on my trials with sections. This one uses a different method and should be more exact.
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Re: Johnsonian Polytopes

Postby Marek14 » Tue Jan 10, 2012 10:35 pm

The same method can also be used for constructing CRF tilings of space/hyperbolic space. Those can't be enumerated, obviously, but it shows the KINDS of tilings possible. For example, if you are building a fortress of evil in hyperbolic space, this could be useful in telling you where you can join several room into one bigger room without compromising the larger esthetics of your building.

Cubic tiling: Can be bordered with {4,4} planar tilings.
Rectified cubic tiling: Made of octahedra and cuboctahedra, it's possible to enlarge an octahedral cell to truncated octahedron (cutting six nearby octahedra to square pyramids and eight nearby cuboctahedra to triangular cupolae). It's also possible to border it with a {4,4} tiling (cutting octahedra in half) or with a rectified {3,6} tiling (cutting cuboctahedra in half).
Truncated cubic tiling: Made of octahedra and truncated cubes. It's possible to border it with truncated square tiling (cutting octahedra in half).
Small rhombated cubic tiling: Made of cubes, cuboctahedra and rhombicuboctahedra. A rhombicuboctahedral cell can be enlarged to truncated cuboctahedron by cutting eight nearby cuboctahedra into triangular cupolae, cutting six nearby rhombicuboctahedra into elongated square cupolas and swallowing twelve nearby cubes. It's also possible to border this tiling with a truncated square tiling. There are two possible kinds of border, then, either it's formed by cubes and rhombicuboctahedra cut into elongated square cupolas, or by cuboctahedra and rhombicuboctahedra cut into square cupolas.
Prismatotruncated cubic tiling: Made of cubes, truncated cubes, rhombicuboctahedra and octagonal prisms. It can be bordered by truncated square tiling, which will either create border of cubes, octagonal prisms and elongated square cupolas, or a border of octagonal prisms, truncated cubes and square cupolas.

Alternated cubical tiling: Made of tetrahedra and octahedra. One possibility includes enlarging a vertex by cutting six nearby octahedra in half and swallowing eight nearby tetrahedra, creating a cuboctahedral cell. You can border it by a square tiling made of square pyramids, or by a triangular tiling made of alternating tetrahedra and octahedra.
Truncated form of this tiling: Made of cuboctahedra, truncated tetrahedra and truncated octahedra. It can be bordered with hexagonal tiling made of truncated tetrahedra, truncated octahedra and triangular cupolas.
Small rhombated form: Made of tetrahedra, cubes and rhombicuboctahedra. A cubic cell can be enlarged to truncated cube by cutting square cupolas from six nearby rhombicuboctahedra and swallowing eight nearby tetrahedra. Or it can be bordered with truncated square tiling made either of cubes and elongated square cupolas or from rhombicuboctahedra and square cupolas.

The tiling of tetrahedra and truncated tetrahedra: Can be bordered by rectified triangular tiling that doesn't cut any cell.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 11, 2012 12:02 am

wintersolstice wrote:[...] also where is the list of dichoral angles on the wiki, I could be a good contributor, :D I know a lot of strategies for ditopal angles.

http://teamikaria.com/hddb/wiki/Segmentochoron
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 11, 2012 12:26 am

Marek14 wrote:[...]Small rhombated 5-cell: One possibility is to cut it in half, into octahedron || truncated tetrahedron and cuboctahedron || truncated tetrahedron. Another possibility is to cut triangle || hexagonal prism. The rest will be funny triangular prism || hexagonal prism with 1 intermediate layer of vertices.[...]

What is the CD diagram for these? I'm having trouble understanding what "rhombated" and "prismated" means in various combinations.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 11, 2012 12:33 am

wintersolstice wrote:How can the "aerochoron" (16-cell) be joined to itself?

By attaching two copies by its tetrahedral facets, of course. The result is non-convex and is bounded by 31 tetrahedra.

The 24-cell honeycomb {3,4,3,3} is the dual to the 16-cell honeycomb {3,3,4,3} (I mentioned the former recently)

bascialy the dichoral angle for the 16 cell must be 120 degrees to allow the tessalation (the 3 at the end of the Schaffli symbol means 3 round a face) this means the angle of the octahedron pyramid must be 60 degrees, subtract that from 180 degrees (when you rectate it) and you get another shape with dichoral angle 120 degrees (the 24-cell that tesselates with 3 round an edge)

so if the angle of the 16-cell is 120 degrees, then joining two of them will make it concave [...]

Concave is not a problem in a tessellation, as long as the gaps are also in the shape of 16-cells and can be filled by it.
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Jan 11, 2012 4:15 pm

quickfur wrote:
Marek14 wrote:[...]Small rhombated 5-cell: One possibility is to cut it in half, into octahedron || truncated tetrahedron and cuboctahedron || truncated tetrahedron. Another possibility is to cut triangle || hexagonal prism. The rest will be funny triangular prism || hexagonal prism with 1 intermediate layer of vertices.[...]

What is the CD diagram for these? I'm having trouble understanding what "rhombated" and "prismated" means in various combinations.


I use terms I learned on Bower's page years ago, I think... it's like this:

Regular - 1000
Rectified - 0100
Truncated - 1100
Small rhombated - 1010
Small prismated - 1001
Bitruncated - 0110
Great rhombated - 1110
Prismatotruncated - 1101
Great prismated - 1111
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Re: Johnsonian Polytopes

Postby wintersolstice » Wed Jan 11, 2012 6:33 pm

quickfur wrote:
wintersolstice wrote:How can the "aerochoron" (16-cell) be joined to itself?

By attaching two copies by its tetrahedral facets, of course. The result is non-convex and is bounded by 31 tetrahedra.

The 24-cell honeycomb {3,4,3,3} is the dual to the 16-cell honeycomb {3,3,4,3} (I mentioned the former recently)

bascialy the dichoral angle for the 16 cell must be 120 degrees to allow the tessalation (the 3 at the end of the Schaffli symbol means 3 round a face) this means the angle of the octahedron pyramid must be 60 degrees, subtract that from 180 degrees (when you rectate it) and you get another shape with dichoral angle 120 degrees (the 24-cell that tesselates with 3 round an edge)

so if the angle of the 16-cell is 120 degrees, then joining two of them will make it concave [...]

Concave is not a problem in a tessellation, as long as the gaps are also in the shape of 16-cells and can be filled by it.


Sorry I know how they can be joined I meant how can it be done in a convex way :D (you've just said it is non-convex, so why is it on the the list of CRF polychora?) that's what I was asking :Dbased on what yous aid it should be under a list of shapes that tesselate!!!

Basically i'm asking: "why is it it listed on Convex Regular Faced Polychora page (on the wiki) if it isn't convex?" and based on what you've said: "should it be on a list of shapes that tesselate instead?"

Sorry I should have been more specific!
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