## Possible self-intersecting polychoron?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### Possible self-intersecting polychoron?

I was trying to look for CRFs when I found this thing.

Start with an octahedron, and attach 4 thawros to 4 non-adjacent faces on it (by their "top" triangles, the ones with all 3.5.3.5 vertices). We'll also connect the other "icosidodecahedral" triangles of the thawros (at six places total), bringing their squares closer together. This sets up all 54 vertices of the polytope (6 from the octahedron, 12 (2x6) shared by 2 thawros each, and 36 ([3+6]*4) that are part of only 1 thawro each).
Attach 4 teddis to the remaining exposed faces of the octahedron (by the singular triangles, of course). Let their pentagonal faces connect with those of the thawros.
Fill in the six "trenches" between the thawros with a Trip between the thawros' squares and with two Squippys on each side of the trip. For each squippy, one of its triangles connected to the trip's, the two adjacent triangles connected to the two neighboring thawros, and the opposite triangle connected to the neighboring teddi.
Bring in a truncated octahedron. Connect its 6 square faces to the Trips' remaining faces, and 4 of its hexagonal faces to the "backs" of the thawros. Then on each of the other 4 hexagonal faces, place a tricu. Each tricu's "top" triangle connects to the last exposed triangle of a teddi, the squares connect to the "bases" of the three nearby squippys, and the 3 other triangles of the tricu finish the polytope off by connecting to the last remaining triangles on the thawros (the ones adjacent to the thawro's hexagon).

It has 54 vertices, 150 edges, 124 faces, and 32 cells. This makes its Euler characteristic -4 (https://people.math.osu.edu/fiedorowicz.1/math655/HyperEuler.html), although it's supposed to be 0. In my very limited experience, even a polychoron with angle sums greater than 360° has an Euler characteristic of 0. This indicates to me that it may be "self-intersecting," or whatever the 4D analog of that property is.
Also checked the Excel file with all the CRFs in it. Nothing matches its cell counts.

Not sure how relevant this is, but there are 10 different kinds of edges, all of which have angle sums of less than 360° (although some of them come very close).
Code: Select all
`Many thanks to Marek14 for the complete listing of cell dihedral angles!1. 4-6 tricu (54.7356°),      4-6 toe (125.264°),      3-4 trip (90°),   3-4 squippy (54.7356°)      (total 324.7352°) (x12)2. 4-6 toe (54.7356°),      4-6 thawro (110.905°),   4-4 trip (60°)                     (total 225.6406°) (x12)3. 6-6 toe (109.471°),      3-6 tricu (70.5288°),      3-6 thawro (138.19°)                     (total 318.1898°) (x12)4. 3-4 tricu (125.264°),      3-4 squippy (54.7356°),   3-3 thawro (138.19°)                     (total 318.1896°) (x24)5. 3-4 tricu (125.264°),      3-4 squippy (54.7356°),   3-3 teddi (138.19°)                     (total 318.1896°) (x12)6. 4-4 trip (60°),      3-4 thawro (110.905°) X2                                 (total 281.81°) (x6)7. 3-4 trip (90°),      3-4 thawro (159.095°),   3-3 squippy (109.471°)                  (total 358.566°) (x24)8. 3-5 teddi (100.812°),   3-3 squippy (109.471°),   3-5 thawro (100.812°)                  (total 311.095°) (x24)9. 5-5 teddi (63.4349°),   3-5 thawro (142.623°) X2                                 (total 348.6809°) (x12)10. 3-3 oct (109.471°),      3-5 thawro (142.623°),   3-5 teddi (100.812°)                     (total 352.906°) (x12)`

I'm not good enough with lace-tower notation to get these edge lengths right, but I know the vertices lie on o3x3o, x3x3o, x3o3x, x3x3x. One of the x's in x3x3o should be f, and the x3x3x term should have a u and another one I don't know the exact length of (but one of the x's is correct). o3x3o is the octahedron (with 6 vertices, of course), "x3x3o" has the 12 vertices shared by 2 thawros, "x3o3x" has 12 of the vertices exclusive to certain thawros, and "x3x3x" has the 24 vertices of the thawros' hexagons.

Anyway, I'm almost completely sure this polytope has some kind of self-intersection. Could one of you please use your advanced methods to determine that this thing isn't worth our time?
I don't know if this explanation is sufficient, so I can post some pictures of it if necessary. My methods are laughably primitive, though, so don't expect renders of Quickfur quality.
Thanks in advance, and sorry for wasting your time if this turns out to be an insignificant result.

New Kid on the 4D analog of a Block
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### Re: Possible self-intersecting polychoron?

Quite an interesting find of yours, New Kid!
Esp. as there was no new CRF find since quite a while.

Here are the relevant elements, as you described, given in the lace tower description:
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`1x octo3x3o. . .. . .. . .`
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`4x thawro. x3o. f3x. o3F. x3x`
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`4x teddio3x .o3f .x3o .. . .`
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`4x tricu. . .. . .x3o .x3x .`
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`6x trip. . .o . x. . .x . x`
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`12x squippy. . .o . .x . .x . .`
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`1x toe. . .. . .. . .x3x3x`

And here is the total lace tower:
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`totalo3x3oo3f3xx3o3Fx3x3x`

And further the full incidence matrix:
Code: Select all
`ooxx3xfox3oxFx&#xt - height(1,2) = 1/sqrt(8) = 0.353553                     height(2,3) = sqrt[3-sqrt(5)]/4 = 0.218508                     height(3,4) = sqrt[3+sqrt(5)]/4 = 0.572061o...3o...3o...     | 6  *  *  * |  4  2 0  0  0  0  0  0  0  0 | 2 2  4 1  0  0  0  0 0  0  0 0 0 0 | 1 2 2  0 0 0 0.o..3.o..3.o..     | * 12  *  * |  0  1 1  2  2  0  0  0  0  0 | 0 0  2 1  1  1  2  2 0  0  0 0 0 0 | 0 1 2  1 1 0 0..o.3..o.3..o.     | *  * 12  * |  0  0 0  2  0  2  2  0  0  0 | 0 0  1 0  2  0  0  2 1  2  1 0 0 0 | 0 1 1  2 0 1 0...o3...o3...o     | *  *  * 24 |  0  0 0  0  1  0  1  1  1  1 | 0 0  0 0  0  1  1  1 0  1  1 1 1 1 | 0 0 1  1 1 1 1-------------------+------------+------------------------------+------------------------------------+---------------.... x... ....     | 2  0  0  0 | 12  * *  *  *  *  *  *  *  * | 1 1  1 0  0  0  0  0 0  0  0 0 0 0 | 1 1 1  0 0 0 0oo..3oo..3oo..&#x  | 1  1  0  0 |  * 12 *  *  *  *  *  *  *  * | 0 0  2 1  0  0  0  0 0  0  0 0 0 0 | 0 1 2  0 0 0 0.... .... .x..     | 0  2  0  0 |  *  * 6  *  *  *  *  *  *  * | 0 0  0 1  0  0  2  0 0  0  0 0 0 0 | 0 0 2  0 1 0 0.oo.3.oo.3.oo.&#x  | 0  1  1  0 |  *  * * 24  *  *  *  *  *  * | 0 0  1 0  1  0  0  1 0  0  0 0 0 0 | 0 1 1  1 0 0 0.o.o3.o.o3.o.o&#x  | 0  1  0  1 |  *  * *  * 24  *  *  *  *  * | 0 0  0 0  0  1  1  1 0  0  0 0 0 0 | 0 0 1  1 1 0 0..x. .... ....     | 0  0  2  0 |  *  * *  *  * 12  *  *  *  * | 0 0  0 0  1  0  0  0 1  1  0 0 0 0 | 0 1 0  1 0 1 0..oo3..oo3..oo&#x  | 0  0  1  1 |  *  * *  *  *  * 24  *  *  * | 0 0  0 0  0  0  0  1 0  1  1 0 0 0 | 0 0 1  1 0 1 0...x .... ....     | 0  0  0  2 |  *  * *  *  *  *  * 12  *  * | 0 0  0 0  0  1  0  0 0  1  0 1 1 0 | 0 0 0  1 1 1 1.... ...x ....     | 0  0  0  2 |  *  * *  *  *  *  *  * 12  * | 0 0  0 0  0  0  0  0 0  0  1 1 0 1 | 0 0 1  0 0 1 1.... .... ...x     | 0  0  0  2 |  *  * *  *  *  *  *  *  * 12 | 0 0  0 0  0  0  1  0 0  0  0 0 1 1 | 0 0 1  0 1 0 1-------------------+------------+------------------------------+------------------------------------+---------------o...3x... ....     | 3  0  0  0 |  3  0 0  0  0  0  0  0  0  0 | 4 *  * *  *  *  *  * *  *  * * * * | 1 1 0  0 0 0 0.... x...3o...     | 3  0  0  0 |  3  0 0  0  0  0  0  0  0  0 | * 4  * *  *  *  *  * *  *  * * * * | 1 0 1  0 0 0 0.... xfo. ....&#xt | 2  2  1  0 |  1  2 0  2  0  0  0  0  0  0 | * * 12 *  *  *  *  * *  *  * * * * | 0 1 1  0 0 0 0.... .... ox..&#x  | 1  2  0  0 |  0  2 1  0  0  0  0  0  0  0 | * *  * 6  *  *  *  * *  *  * * * * | 0 0 2  0 0 0 0.ox. .... ....&#x  | 0  1  2  0 |  0  0 0  2  0  1  0  0  0  0 | * *  * * 12  *  *  * *  *  * * * * | 0 1 0  1 0 0 0.o.x .... ....&#x  | 0  1  0  2 |  0  0 0  0  2  0  0  1  0  0 | * *  * *  * 12  *  * *  *  * * * * | 0 0 0  1 1 0 0.... .... .x.x&#x  | 0  2  0  2 |  0  0 1  0  2  0  0  0  0  1 | * *  * *  *  * 12  * *  *  * * * * | 0 0 1  0 1 0 0.ooo3.ooo3.ooo&#x  | 0  1  1  1 |  0  0 0  1  1  0  1  0  0  0 | * *  * *  *  *  * 24 *  *  * * * * | 0 0 1  1 0 0 0..x.3..o. ....     | 0  0  3  0 |  0  0 0  0  0  3  0  0  0  0 | * *  * *  *  *  *  * 4  *  * * * * | 0 1 0  0 0 1 0..xx .... ....&#x  | 0  0  2  2 |  0  0 0  0  0  1  2  1  0  0 | * *  * *  *  *  *  * * 12  * * * * | 0 0 0  1 0 1 0.... ..ox ....&#x  | 0  0  1  2 |  0  0 0  0  0  0  2  0  1  0 | * *  * *  *  *  *  * *  * 12 * * * | 0 0 1  0 0 1 0...x3...x ....     | 0  0  0  6 |  0  0 0  0  0  0  0  3  3  0 | * *  * *  *  *  *  * *  *  * 4 * * | 0 0 0  0 0 1 1...x .... ...x     | 0  0  0  4 |  0  0 0  0  0  0  0  2  0  2 | * *  * *  *  *  *  * *  *  * * 6 * | 0 0 0  0 1 0 1.... ...x3...x     | 0  0  0  6 |  0  0 0  0  0  0  0  0  3  3 | * *  * *  *  *  *  * *  *  * * * 4 | 0 0 1  0 0 0 1-------------------+------------+------------------------------+------------------------------------+---------------o...3x...3o...     | 6  0  0  0 | 12  0 0  0  0  0  0  0  0  0 | 4 4  0 0  0  0  0  0 0  0  0 0 0 0 | 1 * *  * * * * octoox.3xfo. ....&#xt | 3  3  3  0 |  3  3 0  6  0  3  0  0  0  0 | 1 0  3 0  3  0  0  0 1  0  0 0 0 0 | * 4 *  * * * * teddi.... xfox3oxFx&#xt | 3  6  3  6 |  3  6 3  6  6  0  6  0  3  3 | 0 1  3 3  0  0  3  6 0  0  3 0 0 1 | * * 4  * * * * thawro.oxx .... ....&#x  | 0  1  2  2 |  0  0 0  2  2  1  2  1  0  0 | 0 0  0 0  1  1  0  2 0  1  0 0 0 0 | * * * 12 * * * squippy.o.x .... .x.x&#x  | 0  2  0  4 |  0  0 1  0  4  0  0  2  0  2 | 0 0  0 0  0  2  2  0 0  0  0 0 1 0 | * * *  * 6 * * trip..xx3..ox ....&#x  | 0  0  3  6 |  0  0 0  0  0  3  6  3  3  0 | 0 0  0 0  0  0  0  0 1  3  3 1 0 0 | * * *  * * 4 * tricu...x3...x3...x     | 0  0  0 24 |  0  0 0  0  0  0  0 12 12 12 | 0 0  0 0  0  0  0  0 0  0  0 4 6 4 | * * *  * * * 1 toe`

Summing up the respective diagonal elements of that matrix, you get your problem being solved:
• v =  54
• e = 150
• f = 128 (not 124!)
• c =  32
therefore v-e+f-c = 0, as desired.

@Quickfur: definitely worth a pic, ain't it?

--- rk
Klitzing
Pentonian

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### Re: Possible self-intersecting polychoron?

Just to bring up a further lace tower CRF, which I dreamed up tonight. It is a bistratic one only, with 64 cells:

x3o5x
o3f5o
o3x5x
Infact, it has 1 small rhombicosidodecahedron (srid) for top base, 20 lacing tridiminished icosahedra (J63, teddi), 12 lacing pentagonal rotundae (J6, pero), 30 lacing square pyramids (J1, squippy), dangling from the roof, and 1 truncated dodecahedron for base.

--- rk
Klitzing
Pentonian

Posts: 1621
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Possible self-intersecting polychoron?

Thanks for finding the error in the math; it turns out I forgot to put four of the toe's hexagons into the model. All better now.

I rendered my polychoron as best I could. Some of the toe's (and tricus') edges are less bold than the others. The projection is centered on the octahedron.
Screen Shot 2018-08-19 at 2.35.18 AM.png (42.35 KiB) Viewed 7050 times

I checked your polychoron's dihedrals, I'm pretty sure they're all less than 360°. So, does this mean we have D4.17 and D4.18 now?

New Kid on the 4D analog of a Block
Dionian

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Joined: Mon Jul 16, 2018 2:27 am
Location: Pacific Northwest

### Re: Possible self-intersecting polychoron?

New Kid on the 4D analog of a Block wrote:I rendered my polychoron as best I could.

Thanks for your well-done pic. - Sure, being a multistratic CRF, the cells with nearly flat projection are hard to detect. But from your original descriptions it is quite possible to search them all up.
So, does this mean we have D4.17 and D4.18 now?

Haha, we are already way beyond those numbers. Cf. https://bendwavy.org/klitzing/technical/CRF-list.xls.

--- rk
Klitzing
Pentonian

Posts: 1621
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Possible self-intersecting polychoron?

Klitzing wrote:Quite an interesting find of yours, New Kid!
Indeed!
Klitzing wrote:[...]
Code: Select all
`ooxx3xfox3oxFx&#xt - height(1,2) = 1/sqrt(8) = 0.353553                     height(2,3) = sqrt[3-sqrt(5)]/4 = 0.218508                     height(3,4) = sqrt[3+sqrt(5)]/4 = 0.572061o...3o...3o...     | 6  *  *  * |  4  2 0  0  0  0  0  0  0  0 | 2 2  4 1  0  0  0  0 0  0  0 0 0 0 | 1 2 2  0 0 0 0.o..3.o..3.o..     | * 12  *  * |  0  1 1  2  2  0  0  0  0  0 | 0 0  2 1  1  1  2  2 0  0  0 0 0 0 | 0 1 2  1 1 0 0..o.3..o.3..o.     | *  * 12  * |  0  0 0  2  0  2  2  0  0  0 | 0 0  1 0  2  0  0  2 1  2  1 0 0 0 | 0 1 1  2 0 1 0...o3...o3...o     | *  *  * 24 |  0  0 0  0  1  0  1  1  1  1 | 0 0  0 0  0  1  1  1 0  1  1 1 1 1 | 0 0 1  1 1 1 1-------------------+------------+------------------------------+------------------------------------+---------------.... x... ....     | 2  0  0  0 | 12  * *  *  *  *  *  *  *  * | 1 1  1 0  0  0  0  0 0  0  0 0 0 0 | 1 1 1  0 0 0 0oo..3oo..3oo..&#x  | 1  1  0  0 |  * 12 *  *  *  *  *  *  *  * | 0 0  2 1  0  0  0  0 0  0  0 0 0 0 | 0 1 2  0 0 0 0.... .... .x..     | 0  2  0  0 |  *  * 6  *  *  *  *  *  *  * | 0 0  0 1  0  0  2  0 0  0  0 0 0 0 | 0 0 2  0 1 0 0.oo.3.oo.3.oo.&#x  | 0  1  1  0 |  *  * * 24  *  *  *  *  *  * | 0 0  1 0  1  0  0  1 0  0  0 0 0 0 | 0 1 1  1 0 0 0.o.o3.o.o3.o.o&#x  | 0  1  0  1 |  *  * *  * 24  *  *  *  *  * | 0 0  0 0  0  1  1  1 0  0  0 0 0 0 | 0 0 1  1 1 0 0..x. .... ....     | 0  0  2  0 |  *  * *  *  * 12  *  *  *  * | 0 0  0 0  1  0  0  0 1  1  0 0 0 0 | 0 1 0  1 0 1 0..oo3..oo3..oo&#x  | 0  0  1  1 |  *  * *  *  *  * 24  *  *  * | 0 0  0 0  0  0  0  1 0  1  1 0 0 0 | 0 0 1  1 0 1 0...x .... ....     | 0  0  0  2 |  *  * *  *  *  *  * 12  *  * | 0 0  0 0  0  1  0  0 0  1  0 1 1 0 | 0 0 0  1 1 1 1.... ...x ....     | 0  0  0  2 |  *  * *  *  *  *  *  * 12  * | 0 0  0 0  0  0  0  0 0  0  1 1 0 1 | 0 0 1  0 0 1 1.... .... ...x     | 0  0  0  2 |  *  * *  *  *  *  *  *  * 12 | 0 0  0 0  0  0  1  0 0  0  0 0 1 1 | 0 0 1  0 1 0 1-------------------+------------+------------------------------+------------------------------------+---------------o...3x... ....     | 3  0  0  0 |  3  0 0  0  0  0  0  0  0  0 | 4 *  * *  *  *  *  * *  *  * * * * | 1 1 0  0 0 0 0.... x...3o...     | 3  0  0  0 |  3  0 0  0  0  0  0  0  0  0 | * 4  * *  *  *  *  * *  *  * * * * | 1 0 1  0 0 0 0.... xfo. ....&#xt | 2  2  1  0 |  1  2 0  2  0  0  0  0  0  0 | * * 12 *  *  *  *  * *  *  * * * * | 0 1 1  0 0 0 0.... .... ox..&#x  | 1  2  0  0 |  0  2 1  0  0  0  0  0  0  0 | * *  * 6  *  *  *  * *  *  * * * * | 0 0 2  0 0 0 0.ox. .... ....&#x  | 0  1  2  0 |  0  0 0  2  0  1  0  0  0  0 | * *  * * 12  *  *  * *  *  * * * * | 0 1 0  1 0 0 0.o.x .... ....&#x  | 0  1  0  2 |  0  0 0  0  2  0  0  1  0  0 | * *  * *  * 12  *  * *  *  * * * * | 0 0 0  1 1 0 0.... .... .x.x&#x  | 0  2  0  2 |  0  0 1  0  2  0  0  0  0  1 | * *  * *  *  * 12  * *  *  * * * * | 0 0 1  0 1 0 0.ooo3.ooo3.ooo&#x  | 0  1  1  1 |  0  0 0  1  1  0  1  0  0  0 | * *  * *  *  *  * 24 *  *  * * * * | 0 0 1  1 0 0 0..x.3..o. ....     | 0  0  3  0 |  0  0 0  0  0  3  0  0  0  0 | * *  * *  *  *  *  * 4  *  * * * * | 0 1 0  0 0 1 0..xx .... ....&#x  | 0  0  2  2 |  0  0 0  0  0  1  2  1  0  0 | * *  * *  *  *  *  * * 12  * * * * | 0 0 0  1 0 1 0.... ..ox ....&#x  | 0  0  1  2 |  0  0 0  0  0  0  2  0  1  0 | * *  * *  *  *  *  * *  * 12 * * * | 0 0 1  0 0 1 0...x3...x ....     | 0  0  0  6 |  0  0 0  0  0  0  0  3  3  0 | * *  * *  *  *  *  * *  *  * 4 * * | 0 0 0  0 0 1 1...x .... ...x     | 0  0  0  4 |  0  0 0  0  0  0  0  2  0  2 | * *  * *  *  *  *  * *  *  * * 6 * | 0 0 0  0 1 0 1.... ...x3...x     | 0  0  0  6 |  0  0 0  0  0  0  0  0  3  3 | * *  * *  *  *  *  * *  *  * * * 4 | 0 0 1  0 0 0 1-------------------+------------+------------------------------+------------------------------------+---------------o...3x...3o...     | 6  0  0  0 | 12  0 0  0  0  0  0  0  0  0 | 4 4  0 0  0  0  0  0 0  0  0 0 0 0 | 1 * *  * * * * octoox.3xfo. ....&#xt | 3  3  3  0 |  3  3 0  6  0  3  0  0  0  0 | 1 0  3 0  3  0  0  0 1  0  0 0 0 0 | * 4 *  * * * * teddi.... xfox3oxFx&#xt | 3  6  3  6 |  3  6 3  6  6  0  6  0  3  3 | 0 1  3 3  0  0  3  6 0  0  3 0 0 1 | * * 4  * * * * thawro.oxx .... ....&#x  | 0  1  2  2 |  0  0 0  2  2  1  2  1  0  0 | 0 0  0 0  1  1  0  2 0  1  0 0 0 0 | * * * 12 * * * squippy.o.x .... .x.x&#x  | 0  2  0  4 |  0  0 1  0  4  0  0  2  0  2 | 0 0  0 0  0  2  2  0 0  0  0 0 1 0 | * * *  * 6 * * trip..xx3..ox ....&#x  | 0  0  3  6 |  0  0 0  0  0  3  6  3  3  0 | 0 0  0 0  0  0  0  0 1  3  3 1 0 0 | * * *  * * 4 * tricu...x3...x3...x     | 0  0  0 24 |  0  0 0  0  0  0  0 12 12 12 | 0 0  0 0  0  0  0  0 0  0  0 4 6 4 | * * *  * * * 1 toe`

[...]
--- rk
Hmm, if the thawro x3o||f3x||o3F||x3x should close up in a CRF-way, the respective heights of the layers of your stack should be similar to the heights in the 3D-stack of thawro.
As the height of the stack of thawro are x3o |(1/sqrt(3))| f3x |(1/φsqrt(3))| o3F |(1/sqrt(3))| x3x, the heights you gave should be in the ratio (1:1/φ:1) as well.
As the first height is exactly φ times the second height, I guess the third height should be adjusted a bit. I guess the tower
Code: Select all
`o3x3o (octahedron) ||o3f3x (truncated tetrahedron with triangles of size φ, and the other edges of size 1) ||x3o3F (cuboctahedron with half of the triangles of size φ+1, the other half (the ones above the triangles of size  of the layer belowφ) of size 1)||f3x3o (truncated tetrahedron with triangles of size 1, and the other edges of size φ)`
does make the thawro's close up nicely. However, this one is not closed yet. I wonder how we can get it closed up...

New Kid on the 4D analog of a Block wrote:[...]I'm not good enough with lace-tower notation to get these edge lengths right, but I know the vertices lie on o3x3o, x3x3o, x3o3x, x3x3x.[...]
I myself make quite some use of the spreadsheet of Klitzings (or Wendy's, I believe it's some kind of collaboration), found here. It has been added on quite a bit lately, for a start you only use the part above row 13 and left of column K (and possibly cell H29), and then read of what you want to know in cell E30. I don't use the rest that much either, and it might look a bit scary, bit it really is very usefull!
student91
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### Re: Possible self-intersecting polychoron?

Sadly you got a typo in your correction, which indeed has a clever argument, student5. It then rather ought be f3x3x instead for the last layer.
But then the CRF of New Kid does not close. Nor does the squippy (J1) and trip part seem to be valid after all.
Thus we have to search for alternate closings...
--- rk
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### Re: Possible self-intersecting polychoron?

hmm, you could try to continue above those teddies with a further occurance of thawro (instead of the tricues), thereby replacing the gap of these erronious trip and squippies by a pair of squippies in an other orientation and a further pair of peppies. That one then would start accordingly by
Code: Select all
`o3x3oo3f3xx3o3Ff3x3xo3F3ox3x3A...`
(where |A|=2 sqrt(5)/3). The height sizes then indeed would be in the right ratio (1:phi alternatingly), as required by the teddies and thawroes. But for now I cannot figure out how to continue with that pseudo edge A.
Alternatively you might try to use full icosidodecahedra instead of the 2nd occurances of thawroes, but that would ask for even larger gaps to close...
Thus to me the lately erroniously enjoyed new find looks now quite hard to recover, if at all possible.

BTW, the correction according to student5, the required larger f sized edges of the priviously final layer f3x3x, are easily visible in the pic of New Kid.

--- rk
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### Re: Possible self-intersecting polychoron?

I'm not sure I understand what causes the polychoron to fail. Could you guide me through it so I don't get you guys excited with false alarms in the future?

Also, I found another one. Hopefully it doesn't succumb to whatever doomed the first one.
In a way, it resembles a bipyramid in 4D, but with two different types of pyramid (with the same base). Like a 4D analog of the pentagonal cupolarotunda, with tetrahedral symmetry? Anyway, I've projected it as two "pyramids" with the same "base."
The base is the polyhedron seen here, which isn't CRF but can be allowed here because of distortion from the projection. It's important that both pyramids share this base to complete the polychoron. Just treat it as an intermediate step on the pathway from oct to tut; it has no faces in the polychoron, just as the decagon has no faces in the cupolarotunda.

This one's "peak" is an oct. Like in my previous attempt, it's surrounded by four teddis (one at the top, with two right angles in each pentagon) and four thawros (one in the foreground). Again, the thawros are connected by the 3-5 triangles. But instead of a trip and two squippys, the six "edges" of the "tetrahedron" are each filled in by two squippys (connected to the thawros' squares, and sharing a triangular face like their thawro neighbors) and two peppys (connected to some triangular faces of the squippys, thawros, and teddis).
Screen Shot 2018-08-25 at 7.22.29 PM.png (45.97 KiB) Viewed 7012 times

Screen Shot 2018-08-25 at 7.20.27 PM.png
This one's "peak" is a tut (at center of image). Its hexagonal faces are each connected to a thawro (one is visible at top of image), and the thawros' 3-6 triangles are connected to those of the other thawros. The neighboring 3-3 triangles of the thawros are covered by 12 tets, and four tricus (with the central triangle connected to the tut, squares connected to the thawros, and other triangles connected to the tets) finish this "pyramid."

Total cell count: 54. 1 oct, 1 tut, 4 teddi, 12 peppy, 12 squippy, 4+4 thawro, 12 tet, 4 tricu. And 72 vertices, if I'm not mistaken. (42 on the "base," 18 on the oct side, 12 on the tut side.) My modeling software didn't do its best here, so at the moment I'm without face and edge counts.
Actually, upon reading your last post more thoroughly, it looks like you may have laid the groundwork to this figure. I apologize if it appears I've attempted to steal one of your ideas.

New Kid on the 4D analog of a Block
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### Re: Possible self-intersecting polychoron?

New Kid on the 4D analog of a Block wrote:I'm not sure I understand what causes the polychoron to fail. Could you guide me through it so I don't get you guys excited with false alarms in the future?
I think it is really good that you get us excited, even though it is with false alarms. Lately it has been very silent here (partly because I became quite inactive), so anything that is being said is really uplifting.

The main problem in your previous attempt is explained best using the Coxeter-Dynkin notation we use here all the time, but I can try to explain it without that notation.
Suppose you try to build a cube from squares. Then you start by taking 3 squares and putting them around a vertex. (If you have some squares lying around, DO this, it helps ) By doing this, the distance between the "far" vertices gets fixed at a value of sqrt(2) relative to the edge length.
Similarly, when you take a tetrahedron and two thawroes, and put them around an edge, the "far" edges of the thawroes end up at a distance of (1+sqrt(5))/2 relative to the edge-length. This is the main problem, all else I said is a means of showing that this is the case. This also means that the triangular prisms you thought to have found are not really triangular prisms; one edge of each triangle is actually of length (1+sqrt(5))/2 times the other edge length.
Now let's take the example of building a cube and do the same as I did with that one: A square can be written as (square)= (point) atop (edge of length sqrt(2)) atop (point) = o || q || o (where the latter is just a notation for the former).
Now when we want to build a cube, we put 3 squares around a vertex, so we write o3o || A3q || B3o = (point) atop (hexagon with 3 A-edges and 3 q-edge) atop (triangle with 3 B-edges) for some values A and B that have to be determined.
In fact, we directly know that A=0, as we want to glue the edges of the squares to each other (a non-zero value of A would insert some additional triangles between the squares).
The value of B is more difficult. As you have taken your squares and build a cube, you can see that it should be sqrt(2) = q. However, in 4D this is not so easy to see, so we do the following:
When we write (square) = o || q || o, we have that the height between the point and q-edge is 1/sqrt(2), and similarly the height between the q-edge and the other point is 1/sqrt(2). When putting the squares around the vertex, the squares get slanted a bit, so the height between the layers is smaller than if you put your squares upright (You can see this with the squares I told you you should take, do you have them yet?) However, the ratio between the first height and the second height stays the same, as your square is indeed a square. So that's how you know that you should insert a q-edge, as then all vertices of a given square are in the same plane.
We could stop now, and say that when you put 3 squares around a vertex and top it off with a triangle, you get a polyhedron with 3 squares and 4 triangles. However, in 3D you quickly see this is not true.
So maybe if you like to use dihedral angles to check those things, I'd say this means that if you see a square as two o||q-triangles glued along their q-edge, you should check the dihedral angle along the q-edge is 180 degrees.

I hope this helped to understand what went on in the 4D-case. I tried to type up something in the 4D-case, but it ended up quite messy. I leave the explanation of the spreadsheet and the standard edge lengths, and all other wasn't informative anyways.

You have the "standard" lengths x=1, f=(1+sqrt(5))/2, F=(3+sqrt(5))/2=f+x, v=(sqrt(5)-1)/2=f-x, o=0, q=sqrt(2), h=sqrt(3), etc.)
A thawro can be written as o3x||x3f||F3o||x3x, which then means (triangle with 3 x-edges) atop (hexagon with 3 x-edges and 3 f-edges) atop (triangle with 3 F-edges, oriented different from the x-triangle at the bottom) atop (hexagon with 6 x-edges).
Here the distance between the different layers can be calculated with the spreadsheed of Klitzings and Wendy's. For example, when you want to know the height between o3x and x3f, you fill in (x-o=1) at the first field and (f-x=0.618...) at the second field, and all other fields in this row should be 0. You have now put o(some number)x || x(some number)f in the spreadsheet. Then fill the triangle of fields with a 3 in the first field and a 2 in all other fields. This is how you put (some edge-length)3(some edge-length) in the spreadsheet. Then put the field next to "lacing edge" to 1 and the other field gives you the height with the given lacing edge

New Kid on the 4D analog of a Block wrote:Also, I found another one. Hopefully it doesn't succumb to whatever doomed the first one.
In a way, it resembles a bipyramid in 4D, but with two different types of pyramid (with the same base). Like a 4D analog of the pentagonal cupolarotunda, with tetrahedral symmetry? Anyway, I've projected it as two "pyramids" with the same "base."
The base is the polyhedron seen here, which isn't CRF but can be allowed here because of distortion from the projection. It's important that both pyramids share this base to complete the polychoron. Just treat it as an intermediate step on the pathway from oct to tut; it has no faces in the polychoron, just as the decagon has no faces in the cupolarotunda.

Screen Shot 2018-08-25 at 7.22.29 PM.png

Screen Shot 2018-08-25 at 7.20.27 PM.png

Total cell count: 54. 1 oct, 1 tut, 4 teddi, 12 peppy, 12 squippy, 4+4 thawro, 12 tet, 4 tricu. And 72 vertices, if I'm not mistaken. (42 on the "base," 18 on the oct side, 12 on the tut side.) My modeling software didn't do its best here, so at the moment I'm without face and edge counts.
Very nice find! actually, you rediscovered D4.8.4, which is a real accomplishment as when we discovered it we did have a hard time, well done!
New Kid on the 4D analog of a Block wrote:Actually, upon reading your last post more thoroughly, it looks like you may have laid the groundwork to this figure. I apologize if it appears I've attempted to steal one of your ideas.
Don't mind if you rediscover old things, it is a great way to learn, and over here it happens all the time. In fact, I started out with rediscovering something known as Lace Simplices (Those only get interesting at dimensions higher than 4, so you can look into that if that is something you like), and Klitzing just rediscovered part of an ex-EKF up a few posts (As x3o5x || o3f5o || o3x5x = x3o5(o+x) || o3f5(-x+x) || o3x5(o+x) is a partial Stott-expansion of x3o5o || o3f5(-x) || o3x5o "=" x3o5o || o3o5x || o3x5o, which is part of ex). It's these things that keep you going when nothing new seems to pop up, and thinking of these things is a greart way to learn! (By the way, you seem to have very nice rendering software )
student91
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### Re: Possible self-intersecting polychoron?

Thanks for helping me out! I hope I'm not simplifying your lesson too much, but something I'm getting from it is "make sure your squares really are planar." Now I think it's also important to give the cells degrees of freedom; i.e. if there are two cells that share a face (are connected), then you shouldn't add another cell with all of its vertices being spread over both of those two cells.
And thank you very much for explaining how to use that spreadsheet; I'm sure it'll prove to be a powerful tool.
Actually, I'm having trouble processing the F-containing parts of a J92. At first I thought it was because F wasn't listed in the A34:B38 area, but it looks like the cells below are supposed to process compounds of the "standard lengths." Any idea how to fix this? The spreadsheet gives me the correct height for the o3x-x3f example you provided, but it fails for x3f-F3o and F3o-x3x (because E30 tries to take the square root of a negative number).
For x3f-F3o, I entered -0.618 and 2.618, with 3 in the 2D symmetry cell. (0.618 and -2.618 would have yielded the same result.)

Upon further review, I realize that I forgot to check if the cell counts had already been listed in CRF-list.xls. Unsurprisingly, they were. I'll pay more attention to that next time.

The software is Sketchup, but I don't know if I can recommend it for polychoral use. It's 3D drawing software; the user needs to manually deal with the 4th dimension by drawing inwards as a stand-in for a fourth position coordinate. The model is nothing more than a 3D drawing of a 4D shape, analogous to a sketch on paper of a 3D shape. Not a true 4D model, unfortunately. And while its tools can be useful, they weren't exactly designed for the creation of polyhedra/cells.
Surely there's better software out there, but Stella isn't free...

New Kid on the 4D analog of a Block
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### Re: Possible self-intersecting polychoron?

Dear New Kid,

consider the beginning of a dodecahedron, i.e. a single pentagon and the 5 neighbouring ones. Thus you will have a stack
Code: Select all
`x5of5oo5A`

here the first layer clearly has unit size edges ("x"), the next layer has pseudo edges scaled up in the golden ratio ("f") as those are the diagonals of the lacing pentagons, and then there are further (pseudo) edges of some size "A", which are the distances of the tips of those lateral pentagons. You might consider to try A=x here, but obviously this would contradict to the continuation of a dodecahedron, as that one requires A=f. So, what goes wrong here with A=x? - It is that you would apply some folding onto the lacing pentagons, the former pseudo edges thus become true large edges, i.e. giving rise to trapezoids instead, and the remaining flat triangles would dimple inwards again. This then is what allows for the aimed for regular triangles to be inserted there as further lacing faces.

This illustrative 3D example shows what is going wrong with both your proposals. You would have to assert not only some vertex layers, which match nicely, but even the relative spacing would have to be such that this would not contradict to flat lacing cells. But furthermore, even when such a spacing itself would not contradict, you still will have to check, whether those lacing cells continue within the former gradient, or would fold into the other direction instead.

Code: Select all
`o3x3oo3f3xx3o3FA3x3x`

where you proposed A=o, but instead it would have been A=f, which would make the lacing thawroes (J92) to become flat. - The second proposal now is
Code: Select all
`o3x3oo3f3xx3o3Ff3x3xo3F3ox3x3A`

where you propose again A=o, but as I already mentioned above in my prior mail, in order to have a second flat set of further thawroes, it would require some pseudo edges A of some size 1.49071198 = 2 sqrt(5)/3. - Therefore at least the second half of your proposal does not work, sorry.

--- rk

Edit:
got trapped by my own advices! Both A=o and A=x*1,4907... would provide the correct height ratio. Then I had considered the layerwise circumradii and thought that the radius of layer x3x3A would match closer to the previous sequence when A is the larger value. So I subsumed that one to be the right value. - Now I cross-checked that by evaluating the lace heights not only from nearest neighbours, but also for farer vertex distances as well. And thus I found your refind of student5's early find is correct after all.
Last edited by Klitzing on Mon Aug 27, 2018 8:28 pm, edited 1 time in total.
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### Re: Possible self-intersecting polychoron?

New Kid on the 4D analog of a Block wrote:Actually, I'm having trouble processing the F-containing parts of a J92. At first I thought it was because F wasn't listed in the A34:B38 area, but it looks like the cells below are supposed to process compounds of the "standard lengths." Any idea how to fix this? The spreadsheet gives me the correct height for the o3x-x3f example you provided, but it fails for x3f-F3o and F3o-x3x (because E30 tries to take the square root of a negative number).
For x3f-F3o, I entered -0.618 and 2.618, with 3 in the 2D symmetry cell. (0.618 and -2.618 would have yielded the same result.)
I guess your input turned out wrong somehow; F means the value F+x=f*f=2.618... so the height between F3o and x3f you get by doing (F-x) 3 (o-f) = f3(-f) = 1.618... and -1.618... in the upper row, and the other one should be (F-x) 3 (o-x) = f3(-x)= 1.618... and -1 in the upper row. I guess that does work
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### Re: Possible self-intersecting polychoron?

student91 is correct after all.
That refound one indeed is one combination out of meanwhile 6 known ones, all of which use the same equatorial segments.
Cf. https://bendwavy.org/klitzing/explain/johnson.htm#tet-axial. - There those are being simply called
--- rk
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### Re: Possible self-intersecting polychoron?

Ah, I see where I went wrong with the spreadsheet! For x3f-F3o, I was entering (x-f) 3 (F-o). I ended up doing this because I tried o3x-x3f right before this, and whether inputted the right way or inputted in my wrong way, it'd be inputted as (o-x) 3 (x-f). The mistakes always look obvious in hindsight... thank you so much for your help!

On a mostly unrelated note, have we looked into the possibility of polychora similar to the D4.8.x series, but with a different near-miss polyhedron as a base? My untrained eye saw little promise in the near-miss Johnson solids with 2D rotational symmetry, so I'm about to try some with more "complete" symmetries: chamfered Platonics, rectified truncated deltahedra (Conway polyhedron notation atT, atO, atI), and what can be interpreted as octahedral and icosahedral equivalents of D4.8.x's "boundary." I hypothesize that hexagons will provide plenty of space for thawros and tricus, as they did for D4.8.x.
On the other hand, you guys probably looked into this sort of thing soon after discovering D4.8.x, and if you had found something, it'd be on the Discovery Index, probably under the name D4._.x and with a mention of its symmetry group. Can you offer any insights into this?

New Kid on the 4D analog of a Block
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### Re: Possible self-intersecting polychoron?

The mentioned "discovery index" of this website's wiki is just keiji's child and way too incomplete. He wasn't himself mainly involved into the according research, just referencing to some of the herein mentioned findings of the time of compilation. A much more complete list of references could be found here:
--- rk
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### Re: Possible self-intersecting polychoron?

I should have done this six months ago, but I made some modified versions of the reduced-ofx3/2oxx3ooo&#xt figure.
First one is eight sidtid-0-3-3-3s around the apex point instead of four. Lace might be a reduced ofx3/2oxx4ooo&#xt. I wasn't able to get very good pictures of it, but its other cells are 6 cubes and 6 squippies. The cubes fill in the "square" stratum of the sidtid-0-3-3-3s while the squippies fill in the "triangle" stratum (and their bases are connected to the cubes), and like in the original figure, the pentagons of the sidtid-0-3-3-3s are directly connected to each other.
8sidtid0333 sidtid cells.png
Left: four of the sidtid-0-3-3-3 cells in their true (projected) position. The edges and vertices of a square pyramid and cube are visible on its top. Right: seven of these cells in exploded view.
8sidtid0333 + other cells.png
Left: a single sidtid-0-3-3-3 cell for reference: Center: projection of the eight of those cells together. Right: one of three "patches" of two cubes and two squippies that completes the polychoron.

The next one is an expansion of the original. Its "base" is a pseudo-co and its "apex" a pseudo-tet, with four stacks of two thahs going in between them. (Each stack has the thahs join at a triangle in the middle layer.) Six tets cover the remaining three triangles of each lower thah, six trips cover those of the upper thahs. and six pips cover the squares of all the thahs. The remaining faces of these three cell types then cover the sidtid-0-3-3-3s (except the square faces of the prisms at the pseudo-co boundary, they're joined to each other). I couldn't figure out the lace (it should just be adding some x's to the "ooo" part, but I didn't see how that'd make the pseudo-co), but I did get better pictures than the previous idea.
exp-4sidtid0333 view4a.png
One sidtid-0-3-3-3 cell hidden. The nine faces it connects to are visible on the purple pips, blue trips, and red tets.
exp-4sidtid0333 view4b.png
One stack of two thahs hidden
exp-4sidtid0333 view6.png
One pip and one trip hidden. The faces that join the trip are visible on the white sidtid-0-3-3-3s and bright green thahs.

Of course, the next idea was to expand the octahedrally-symmetric figure. The result has a pseudo-sirco base and pseudo-cube apex. Six stacks of two tutrips go in between them, and like the previous "expanded" one, tets and trips connect to the "stacks' " triangles, pips connect to their squares (this time 12 of each of these cells), and sidtid-0-3-3-3s connect to those three cell types.
exp-8sidtid0333 view8.png
One sidtid-0-3-3-3 hidden. The pentagon and triangle faces it joins to are most visible, in the green pips and pink tets, with the purple trips and their squares hiding behind the pips.
exp-8sidtid0333 view6.png
One stack of two tutrips hidden
exp-8sidtid0333 view12.png
One pip and one trip hidden

I also faceted some segmentochora, but that's probably pretty trivial.

New Kid on the 4D analog of a Block
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### Re: Possible self-intersecting polychoron?

ain't you just describing, starting from
Code: Select all
`reduced( ofx3/2oxx3ooo&#xt by 2tet )`

the further related possibilities
Code: Select all
`ofx3/2oxx3xxx&#xtofx3/2oxx4ooo&#xtofx3/2oxx4xxx&#xt`

(respectively the according reductions of those again)?

--- rk
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### Re: Possible self-intersecting polychoron?

Yes. I'm not sure why I didn't do such simple modifications earlier.
I just put together the ofx3/2oxx5xxx&#xt, presumably the "ooo" one works too. Now it's clear to me that in between each pair of N-gonal pseudofaces on the top and bottom of ofx3/2oxxNxxx&#xt (besides the pseudo-triangles at the bases of the sidtid-0-3-3-3s), there's an "N/2" face.

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