Hi.gher. Space

by **mr_e_man** » Tue Jan 12, 2021 2:49 am

Suppose we want to make a 2D tiling with regular (convex) polygons. The sum of the angles around a vertex must be exactly 360°. There are only a few combinations of polygons that satisfy this condition:

60° + 128.5714° + 171.4286° (3.7.42)
60° + 135° + 165° (3.8.24)
60° + 140° + 160° (3.9.18)
60° + 144° + 156° (3.10.15)
60° + 150° + 150° (3.12.12)
90° + 108° + 162° (4.5.20)
90° + 120° + 150° (4.6.12)
90° + 135° + 135° (4.8.8)
108° + 108° + 144° (5.5.10)
120° + 120° + 120° (6.6.6)

60° + 60° + 90° + 150° (3.3.4.12, 3.4.3.12)
60° + 60° + 120° + 120° (3.3.6.6, 3.6.3.6)
60° + 90° + 90° + 120° (3.4.4.6, 3.4.6.4)
90° + 90° + 90° + 90° (4.4.4.4)

60° + 60° + 60° + 60° + 120° (3.3.3.3.6)
60° + 60° + 60° + 90° + 90° (3.3.3.4.4, 3.3.4.3.4)

60° + 60° + 60° + 60° + 60° + 60° (3.3.3.3.3.3)

Many of these configurations cannot be extended to a complete tiling: Going around a triangle or a pentagon, the polygons must alternate between the other two types (for example 4-gons and 20-gons); but this is impossible because 3 and 5 are odd. Thus, only 3, 4, 6, 8, and 12-gons can be used in a tiling. If there's an octagon anywhere, it must be the uniform 4.8.8 tiling. Anything else is essentially a tiling of triangles and squares; hexagons can be "augmented with hexagonal pyramids" (or cut into triangles), and dodecagons can be "augmented with hexagonal cupolas" (or cut into triangles, squares, and hexagons).

Now let's consider 3D honeycombs with CRF cells. This has been mentioned here and here; especially significant is the cube-doe-bilbiro honeycomb.

The first task is to find combinations of dihedral angles that sum to 360°.

I started with large prisms and antiprisms (in pairs sharing a large face), systematically adding up angles with other CRF polyhedra. One difficult case to consider was two (n-gon) antiprisms meeting a third (m-gon) antiprism at two of its triangle faces: The first two dihedral angles, between a triangle and the n-gon, have a sum slightly greater than 180°, and the third angle, between the two triangles, is slightly less than 180°, so the sum is arbitrarily close to 360° (both below it and above it) when n and m are large enough. It should be possible to rule out this case using algebraic number theory, but I just noted that the m-gon antiprism must be paired with another m-gon antiprism or prism at the same vertex, and the four large solid angles (each slightly less than 180°) don't leave enough space for anything else at the vertex.

Then I wrote a program to check the finitely many remaining angles. Here are my findings:

Nothing fits with an n-gon antiprism with n=4 or n≥6. Prisms are more compatible, but the large ones, including 7,9,10,11,≥13, can't be used in a honeycomb for the same reason as in 2D. A 12-gon prism can only appear in a 2D tiling stacked on top of itself.

As you probably expected, the crown jewels J84-90 and snic and snid don't fit with anything. Also the augmented pentagonal prism (angle 162.74°) doesn't fit.

From the polyhedra with 3,4,5,6,8,10-gons, there are 55 combinations of 3 angles adding to 360°, 128 combinations of 4 angles, 112 of 5, 56 of 6, 16 of 7, 3 of 8, and no combinations of 9 or more angles.

Some dihedral angles sum to 360° but there's no way to make the faces match. This rules out the augmented tridiminished icosahedron (angle 171.34°), the biaugmented triangular prism (angle 169.47°), and the gyrate rhombicosidodecahedron and relatives (angle 153.43°).

The edge between the hexagon and a square in thawro can be completed (for example, with another thawro in gyro orientation, and a pocuro), but the vertex cannot be completed (the vertical edge of the square has a remaining angle 41.81°). So, while J91 appears in a honeycomb, J92 does not.

Similarly a vertex of grid (10.6.4) cannot be completed.

Is there any CRF honeycomb involving 10-gons?

I did find a complete vertex configuration with 10-gons; it has an augmented truncated dodecahedron, a pentagonal pyramid, a diminished rhombicosidodecahedron, and two pentagonal rotundas. I don't know whether this can be extended.

As of now, this search for vertex configurations is being done manually. It would be harder to automate than the dihedral angle sums....

by **mr_e_man** » Tue Jan 12, 2021 4:26 am

Also I noticed that most Archimedean solids in the cubic family can be decomposed into smaller polyhedra while maintaining symmetry:

tut = tet + 4 oct + 6 tet
tricu = tet + 3 squippy + 3 tet
(These break each hexagon into 6 triangles. The following decompositions preserve the external faces.)
tic = cube + 6 squacu + 8 tet
toe = oct + 8 tricu + 6 squippy
co = 2 tricu = 6 squippy + 8 tet
sirco = op + 2 squacu
girco = sirco + 6 squacu + 8 tricu + 12 cube

Another way to describe this is that these segmentochora are degenerate:

tic || cube
toe || oct
co || point
girco || sirco

It seems that no such decompositions are possible with the dodecahedral family.

by **mr_e_man** » Wed Jan 13, 2021 12:26 am

I forgot to mention that the pentagonal prism's angle 108° only fits with a decagonal prism; that was part of my reasoning to prove that 10-gon prisms don't work. The only prisms that appear in honeycombs are 3,4,6,8,12.

Now I've also ruled out the truncated icosahedron, and the small rhombicosidodecahedron, and indeed anything with a 5.4.3.4 vertex; the two large angles 159.09°, at the triangle, conspire so they can't both be surrounded by polyhedra, though one can be surrounded.

A 5.3.3.3 vertex can only be completed in one way: The middle triangle attaches to a truncated dodecahedron (10.10.3), whose two decagons attach to pentagonal rotundas (10.5.3), and the three exposed pentagons are covered by a dodecahedron (5.5.5). (Of course the doe might be augmented on its far faces. The tid can't be augmented; that would involve 5.4.3.4.) So a pentagonal antiprism doesn't work; its top and bottom vertices would require different polyhedra (tid and pero) to attach to one triangle. This also rules out the icosahedron vertex 3.3.3.3.3 A = 3.3.5 + 5.3.3.3, and thus the pentagonal pyramid. The 5.3.3.3 vertex must be that of a tridiminished icosahedron.

That really cuts down the number of Johnson solids available! These are all we have left from the pentagonal/dodecahedral family for making honeycombs: doe, tid, id, pero, pobro, teddi, bilbiro.

In fact pobro can't be used either. The 5.5.3.3 vertex can only be completed with a pair of dodecahedra (attaching to the two pentagons) and a pair of tridiminished icosahedra (with their sharp ends attaching to the two triangles). But a teddi, with its three 5.3.3.3 vertices, needs to attach to three dodecahedra, not another teddi.

Two tids, sharing a decagon, don't fit with anything else. So one tid must be surrounded by 12 peros. Each pero has two possible orientations; we might call them ortho and gyro. If two triangles in two adjacent peros meet along an edge (with dihedral angles 79.19°+116.57°+79.19°), then nothing else fits at that edge; so we can't have two gyro peros next to each other. An ortho pero has a triangle meeting a triangle of tid (with dihedral angles 79.19°+142.62° = 360° - 138.19°); this only fits with the blunt end of a teddi. And we already know that this allows only one configuration; all peros must be ortho. The result is symmetrical:

tid + 12 pero + 20 teddi + 30 doe

This leaves 12 pero pentagons exposed, which can't be covered by id or pero or bilbiro (142.62°+142.62°) and must be covered by 12 doe's. Now we have a triangular depression between three doe's, which must be filled by a teddi. So any one of the 20 original teddis shares an edge with a new teddi, pointing in the opposite direction. But two of these new teddis, sharing an edge, point in the same direction; the required pattern cannot be continued. So tid and teddi are out.

The only thing left with decagons is pero. Thus, one pero must be paired with another, effectively forming an icosidodecahedron. As before, the pentagons can't be covered by id or pero or bilbiro and must be covered by doe's. Again we get these triangular depressions which can only be filled by teddis. So id and pero are out.

Finally, a doe or a bilbiro can only appear in the cube-doe-bilbiro honeycomb. (I won't show the steps I took here, but it's no more difficult than what I've already shown.)

That is all!

Any other honeycomb must be made of polyhedra like tut and squacu, which involve √2 and √3 and not φ.

...Did Weimholt know this?

by **mr_e_man** » Thu Jan 14, 2021 12:20 am

mr_e_man wrote:Two tids, sharing a decagon, don't fit with anything else. So one tid must be surrounded by 12 peros. Each pero has two possible orientations; we might call them ortho and gyro. If two triangles in two adjacent peros meet along an edge (with dihedral angles 79.19°+116.57°+79.19°), then nothing else fits at that edge; so we can't have two gyro peros next to each other. An ortho pero has a triangle meeting a triangle of tid (with dihedral angles 79.19°+142.62° = 360° - 138.19°); this only fits with the blunt end of a teddi. And we already know that this allows only one configuration; all peros must be ortho. The result is symmetrical:

tid + 12 pero + 20 teddi + 30 doe

This leaves 12 pero pentagons exposed, which can't be covered by id or pero or bilbiro (142.62°+142.62°) and must be covered by 12 doe's. Now we have a triangular depression between three doe's, which must be filled by a teddi. So any one of the 20 original teddis shares an edge with a new teddi, pointing in the opposite direction. But two of these new teddis, sharing an edge, point in the same direction; the required pattern cannot be continued. So tid and teddi are out.

: tidTeddi1.png (58.05 KiB) Viewed 178674 times

: tidTeddi2.png (119.22 KiB) Viewed 178674 times

by **Klitzing** » Thu Jan 14, 2021 9:49 pm

mr_e_man wrote:Also I noticed that most Archimedean solids in the cubic family can be decomposed into smaller polyhedra while maintaining symmetry:

tut = tet + 4 oct + 6 tet
tricu = tet + 3 squippy + 3 tet
(These break each hexagon into 6 triangles. The following decompositions preserve the external faces.)
tic = cube + 6 squacu + 8 tet
toe = oct + 8 tricu + 6 squippy
co = 2 tricu = 6 squippy + 8 tet
sirco = op + 2 squacu
girco = sirco + 6 squacu + 8 tricu + 12 cube

Another way to describe this is that these segmentochora are degenerate:

tic || cube
toe || oct
co || point
girco || sirco

It seems that no such decompositions are possible with the dodecahedral family.

This all is not new. Those well can be found on https://bendwavy.org/klitzing/explain/complexes.htm#degenerate-segmentochora.
Moreover it is shown there that dodecahedral ones do exist - provided the non-intersection property would get released.

--- rk

by **mr_e_man** » Wed Jan 20, 2021 5:16 am

Klitzing, do you know how Weimholt discovered the cube-doe-bilbiro honeycomb? Did he know that it's so unique? Did he try to find others like it, as we've tried here?

by **Klitzing** » Wed Jan 20, 2021 6:21 pm

Here is the relevant part from the private Polyhedron Mailing Archive:

On Wed 25 Aug 2004 (01:44:23), [andrew weimholt] wrote:
>Here's an interesting 3-honeycomb I came across while playing with
>Great Stella. It consists of Dodecahedra, Cubes, and Bilunabirotundas
>(J91s)

So he came across that find simply by hazard. But then it should be noted that this arrangement of 6 pyritohedrially arranged bilbiros around a central cube already are the main constituent of "E5 \ 6J91(P4)", i.e. a construction described in Steward's booklet Adventures Among the Toroids, eg. cf. to http://wwww.doskey.com/polyhedra/DecomposablePolyhedra.html or https://bendwavy.org/klitzing/explain/complexes.htm#toroids, and as such it is already much older.

--- rk

by **Klitzing** » Wed Jan 20, 2021 6:29 pm

But when speaking about scaliform honeycombs:
do you remember 5Y4-3T-3Q3-T3, which I've found in 2020:

Code: Select all: (N→∞) 12N | 2 4 1 2 | 3 6 4 3 1 2 | 1 2 5 3 1 ----+----------------+-----------------------+--------------- 2 | 12N * * * | 2 2 0 0 0 0 | 1 2 1 0 0 2 | * 24N * * | 0 2 1 1 0 0 | 0 1 2 1 0 2 | * * 6N * | 2 0 2 0 0 2 | 1 0 2 2 1 2 | * * * 12N | 0 0 1 1 1 1 | 0 0 1 2 1 ----+----------------+-----------------------+--------------- 3 | 2 0 1 0 | 12N * * * * * | 1 0 1 0 0 3 | 1 2 0 0 | * 24N * * * * | 0 1 1 0 0 4 | 0 2 1 1 | * * 12N * * * | 0 0 1 1 0 3 | 0 2 0 1 | * * * 12N * * | 0 0 1 1 0 3 | 0 0 0 3 | * * * * 4N * | 0 0 0 1 1 6 | 0 0 3 3 | * * * * * 4N | 0 0 0 1 1 ----+----------------+-----------------------+--------------- 4 | 4 0 2 0 | 4 0 0 0 0 0 | 3N * * * * tet (T) 1 4 | 2 4 0 0 | 0 4 0 0 0 0 | * 6N * * * tet (T) 2 5 | 2 4 1 1 | 1 2 1 1 0 0 | * * 12N * * squippy (Y4) 9 | 0 6 3 6 | 0 0 3 3 1 1 | * * * 4N * tricu (Q3) 12 | 0 0 6 12 | 0 0 0 0 4 4 | * * * * N tut (T3)

--- rk

by **mr_e_man** » Thu Jan 21, 2021 1:07 am

That falls into the category of honeycombs involving only √2 and √3. (I could make this more precise, but I think you know what I mean.)

It is interesting. If we ignore the triangular and square pyramids, it reminds me of the diamond crystal structure. Each tut is connected by 4 outgoing and 4 incoming tricus to 8 other tuts. I guess we could divide the tuts into 2 types, and consider only outgoing tricus on one type, and incoming tricus on the other type.

: 5Y4-3T-3Q3-T3.png (130.78 KiB) Viewed 178607 times

by **mr_e_man** » Thu Jan 21, 2021 1:13 am

Because of those decompositions I mentioned, any √2+√3 honeycomb is essentially made of just 6 polyhedra: trip, cube, op, tet, squippy, squacu.

by **Klitzing** » Thu Jan 21, 2021 5:36 pm

In fact that latter scaliform honeycomb 5Y4-3T-3Q3-T3 is closely related to octet.
In order to see this, you observe first that each tricu will be attached by 3 squippies at its squares.
Thus all squares now are covered.
The remainder then is just an according dissection of tut and of that tricu+squippy-combo into octs and tets.

--- rk

by **mr_e_man** » Sun Jun 13, 2021 10:55 pm

From viewtopic.php?p=27971#p27971

Klitzing wrote:When I read your starting article here I can get the followings

2D convex tilings with regular polygons only can use 3-, 4-, 6-, 8-, and 12-gons,
as any other being used regular convex polygon ultimatly results in a non-continuable configuration
morover there is only a single 2D convex tiling with regular polygons only using 8-gons: the uniform 4.8.8-tiling
occurances of 6-gons variously might be decomposed into 6 3-gons each (6-pyramid)
occurances of 12-gons variously might be decomposed into 6 3-gons, 6 4-gons, and 1 6-gon each (6-cupola),
in fact within 2 different orientations.
3D convex tilings with regular polygons should similarily be restricted to use 3-, 4-, 5-, 6-, 8-, 10-, and 12-gons only,
as any other being used regular convex polygon ultimatly results in a non-continuable configuration
morover usages of 12-gons only can occur within infinite stacks of according 2D tilings.
sevaral CRF cells can also be excluded for similar reasons:
4-ap, n-ap with n>5, 7-p, 9-p, 10-p, 11-p, n-p with n>12, J52-53, J84-90, J92, snic, snid, grid
About further restrictions on the usage of 10-gons - at least therein - you still seem unfixed.
Within a later post of that very thread you show up a configuration with a tid, but I don't get exactly why it isn't continuable.
Still, that one alone doesn't rule out any other 10-gon usage.

A bit later in that thread you further observe the (non-new) facts of degenerate segmentochora
tic || cube
toe || oct
co || point
girco || sirco
as well as the obvious decomposition of sirco into the lace tower squacu || op || squacu.

From that you now deduce that ANY 3D convex tiling with regular polygons should be decomposable into P3 (trip), P4 (cube), P8 (op), T (tet), Y4 (squippy), and Q4 (squacu)
- except for the single case of Weimholt's cube-doe-bilbiro honeycomb.

First of all I don't see the final 10-gon exclusion. But esp. I don't see why that mentioned exception should be the only possible 5-gon usage.

--- rk

by **mr_e_man** » Mon Jun 14, 2021 12:19 am

Klitzing wrote:A bit later in that thread you further observe the (non-new) facts of degenerate segmentochora
tic || cube
toe || oct
co || point
girco || sirco
as well as the obvious decomposition of sirco into the lace tower squacu || op || squacu.

From that you now deduce that ANY 3D convex tiling with regular polygons should be decomposable into P3 (trip), P4 (cube), P8 (op), T (tet), Y4 (squippy), and Q4 (squacu)
- except for the single case of Weimholt's cube-doe-bilbiro honeycomb.

The decomposition is not the main point; it's just a short way to describe all of the possible cells remaining. Another way to describe them (at least most of them) is that they're related to the square or cube or triangle somehow. Another way is that they can be given coordinates involving only √2 and √3. The longer way to describe them is:

tet, cube, oct, tut, tic, toe, co, sirco, girco, trip, hip, op, twip, J1, 3-4, 7-8, 12, 14-15, 18-19, 26-29, 35-37, 49-51, 54-57, 65-67.

All other CRF polyhedra have been tried and eliminated. (I suppose twip, the 12-gon prism, has also been eliminated, in that we know how to describe all of the honeycombs it can appear in. And J50-51 have been eliminated. But still they fit in this family.)

Klitzing wrote:About further restrictions on the usage of 10-gons - at least therein - you still seem unfixed.
Within a later post of that very thread you show up a configuration with a tid, but I don't get exactly why it isn't continuable.
Still, that one alone doesn't rule out any other 10-gon usage.

How far exactly do you follow my first and third posts here? Which polyhedra involving decagons do we have remaining? The original complete list is:

tid, grid, dip, dap, J5-6, 20-21, 24-25, 68-71, 76-83,

and it seems you've ruled out at least these:

tid, ~~grid, dip, dap~~, J5-6, ~~20-21, 24-25~~, 68-71, 76-83.

mr_e_man wrote:Now I've also ruled out [...] indeed anything with a 5.4.3.4 vertex; the two large angles 159.09°, at the triangle, conspire so they can't both be surrounded by polyhedra, though one can be surrounded.

I could explain this and fill in the details, perhaps in my next post. Anyway, this would give us

tid, ~~grid, dip, dap, J5~~, 6, ~~20-21, 24-25, 68-71, 76-83~~,

that is, just tid and J6...

by **mr_e_man** » Mon Jun 14, 2021 4:47 am

Let's consider the 5.4.3.4 vertex. From the output of my program, 159.0948° combines with other dihedral angles to make 360° in the following ways:

(1): 169.1877° + 159.0948° + 31.7175°
(2): 159.0948° + 121.7175° + 79.1877°
(3): 159.0948° + 110.9052° + 90°
(4): 159.0948° + 110.9052° + 45° + 45°
(5): 159.0948° + 90° + 79.1877° + 31.7175°
(6): 159.0948° + 79.1877° + 45° + 45° + 31.7175°

These angles appear in the following verfs (see my list), with bold indicating the relevant pairs of faces:

31.7175°: 10.4.3
45°: 8.4.3
79.1877°: 10.5.3
90°: 10.4.4, 8.8.3, 8.4.4, 6.4.4, 5.4.4, 4.4.4, 4.4.3, 6.4.3.3 A, 5.4.3.3 B, 4.4.3.3 B, 4.3.4.3 B, 4.3.3.3 B
110.9052°: 6.4.3.3 B, 5.3.4.3
121.7175°: 10.5.4, 4.4.4.3 C
169.1877°: 5.4.4.3 A

That last one, 5.4.4.3 A, only appears in polyhedra made from a decagonal prism (J21 and 40-43), which we've already ruled out. So we can't use sum (1).

Suppose we try to surround an edge of the triangle in 5.4.3.4 using sum (2). The 79.1877° angle is that of a rotunda, between the decagon and a triangle; neither of these faces are squares, so the rotunda can't be attached to a square in 5.4.3.4; it must be attached to the triangle in 5.4.3.4 . But then the other edge of the triangle gets 159.0948°+142.6226° from the other part of the rotunda; that partial sum doesn't appear in the list above, so the edge can't be completed. That illustrates what I meant by the two edges conspiring.

Suppose we try to surround an edge of the triangle in 5.4.3.4 using sum (3). The 110.9052° angle can't be part of the verf 6.4.3.3 B, because that's unique to thawro, which we've already ruled out; so it must be part of 5.3.4.3 . If the triangle in 5.4.3.4 is attached to a triangle in 5.3.4.3, then one edge gets 159.0948°+110.9052°, while the other edge gets 159.0948°+142.6226°; again, that partial sum is invalid. So the triangle in 5.4.3.4 must be attached to something involving the 90° angle. This would have to be 4.4.3 (trip) or 4.3.3.3 B (autrip) or 4.3.4.3 B (gybef). But now the 159.0948°+90° edge has two exposed squares and no exposed triangles; there is nothing for the triangle in 5.3.4.3 to attach to.

Suppose we try sum (4). 45° necessarily involves an octagon, while 159.0948° and 110.9052° don't involve octagons, so the two 45° angles must combine according to 3.4.8 + 8.4.3 = 4.4.3.3 B, forming a 90° angle. Then we get sum (3) again.

Suppose we try sum (5). As in the case of sum (2), the triangle in 5.4.3.4 can't attach to that in the rotunda (the source of the 79.1877° angle); and the triangle can't attach to the cupola, because 31.7175° is between a decagon and a square (no triangle); so it must attach to something involving the 90° angle. As in the case of sum (3), now the edge has two exposed squares and no exposed triangles. There is nothing at this edge (even considering the cupola we haven't placed yet) that the rotunda's triangle could attach to.

Sum (6) reduces to sum (5) in the same way that sum (4) reduces to sum (3).

Thus, 5.4.3.4 cannot be used.

by **mr_e_man** » Mon Jun 14, 2021 5:43 am

Klitzing wrote:Within a later post of that very thread you show up a configuration with a tid, but I don't get exactly why it isn't continuable.

In the third post here, I tried to show that a teddi must be attached to a tid, and conversely a tid must be attached to teddis in a certain way.

Starting from the tid, we end up with 20 yellow teddis pointing outward, and 60 orange teddis pointing inward.

Instead starting from one teddi, we get one tid, and then 19 other teddis pointing out from the tid, and 60 teddis pointing in to the tid. 3 of those 60 teddis are adjacent to the first teddi, and point in the opposite direction. Therefore, any one teddi must be surrounded by (in addition to doe's etc.) three other teddis pointing in the opposite direction.

But that configuration of 60 teddis already violates this condition.

Starting from an orange teddi, we get a different tid, which itself must be surrounded by 19+60 other teddis. But some of these 60 new teddis overlap with the 59 old orange teddis. The old ones point in the same direction, while the new ones point in the opposite direction from the first orange teddi.

by **Klitzing** » Mon Jun 14, 2021 4:51 pm

You still don't show, that tid-10-tid is impossible, that pero-10-pero is impossible - in either orientation each.
The argument about the orange teddis also misses mentioning of further reseach.
So eg. why can't you insert into those light green / dark green gaps ids? Do you really need further peroes?
Might be you can solve all those loop holes about decagons.

But then still what about pentagons generally?
What makes them be allowed within Weimholt's honeycomb, but nowhere else?

In fact, I didn't re-check all your statements of the first post.
At least those sound reasonable.

In fact you should compile a complete chain of arguments, which one by one reduces the list of allowed cells.

--- rk

by **mr_e_man** » Mon Jun 14, 2021 9:33 pm

I didn't show everything because it would be tedious. I hoped the readers would be able to verify my statements on their own. But I see that further explanation and guidance is necessary.

I suppose I should also show the program that gives these combinations of angles.

Should I restart from the beginning, and show the details of how I ruled out large prisms and antiprisms?

Klitzing wrote:You still don't show, that tid-10-tid is impossible

mr_e_man wrote:Two tids, sharing a decagon, don't fit with anything else.

Let's put two tids together and see what doesn't fit. From the output of my program, the partial sums 142.6226°+142.6226° and 116.5651°+116.5651° (for the ortho case) and 142.6226°+116.5651° (for the gyro case) appear in the following combinations:

142.6226° + 142.6226° + 74.7547°
126.8699° + 116.5651° + 116.5651°
142.6226° + 142.6226° + 37.3774° + 37.3774°
116.5651° + 116.5651° + 95.1524° + 31.7175°
116.5651° + 116.5651° + 63.4349° + 63.4349°
116.5651° + 116.5651° + 63.4349° + 31.7175° + 31.7175°
116.5651° + 116.5651° + 31.7175° + 31.7175° + 31.7175° + 31.7175°

142.6226° + 116.5651° + 100.8123°
142.6226° + 116.5651° + 69.0948° + 31.7175°
142.6226° + 116.5651° + 63.4349° + 37.3774°
142.6226° + 116.5651° + 37.3774° + 31.7175° + 31.7175°

In the ortho case, the 142.6226°+142.6226° edge has two triangles exposed, and the remaining space must be filled by either a polyhedron with a 74.7547° angle, or two polyhedra with 37.3774° angles. 74.7547° only appears in J13 and J30, and 37.3774° only appears in J2 and J5. But we've already ruled out J5 and J30 which have 5.4.3.4 vertices. And by this point in time we're supposed to have also ruled out J2 and J13 which have 3.3.3.3.3 A.

In the gyro case, the combination of angles must be 142.6226°+116.5651°+100.8123°, because 31.7175° only appears in J5, and 37.3774° only appears in J2 and J5, which have been ruled out. The angle 100.8123° appears in the verfs 5.5.3, 5.3.4.3, and 5.3.3.3 . But the 142.6226°+116.5651° edge has a triangle and a decagon exposed, and none of those three possible verfs involve decagons; the edge can't be completed.

Klitzing wrote:that pero-10-pero is impossible - in either orientation

The ortho case is described here

mr_e_man wrote:In fact pobro can't be used either. The 5.5.3.3 vertex can only be completed with a pair of dodecahedra (attaching to the two pentagons) and a pair of tridiminished icosahedra (with their sharp ends attaching to the two triangles). But a teddi, with its three 5.3.3.3 vertices, needs to attach to three dodecahedra, not another teddi.

and the gyro case is described later

mr_e_man wrote:The only thing left with decagons is pero. Thus, one pero must be paired with another, effectively forming an icosidodecahedron. As before, the pentagons can't be covered by id or pero or bilbiro and must be covered by doe's. Again we get these triangular depressions which can only be filled by teddis. So id and pero are out.

The statements depend on the previous ones; this is indeed a chain, where the list of possible cells is reduced successively.

by **mr_e_man** » Mon Jun 14, 2021 10:34 pm

Klitzing wrote:The argument about the orange teddis also misses mentioning of further reseach.
So eg. why can't you insert into those light green / dark green gaps ids? Do you really need further peroes?

The orange teddis have 5.3.3.3 vertices. The necessity of peros follows from this previous statement:

mr_e_man wrote:A 5.3.3.3 vertex can only be completed in one way: The middle triangle attaches to a truncated dodecahedron (10.10.3), whose two decagons attach to pentagonal rotundas (10.5.3), and the three exposed pentagons are covered by a dodecahedron (5.5.5).

So it's that statement which needs to be verified, before looking at the tid-teddi combination.

by **mr_e_man** » Wed Jun 16, 2021 4:49 pm

Here's the essence of my code, for sums of 3 decreasing angles:

Code: Select all: int N = 109; // total number of dihedral angles, plus one int a[N] = { // list of angles, in units of 0.0001 degree 0, 317175, 373774, 450000, 547356, 600000, 634349, 690948, 705288, 729730, 747547, 791877, 867268, 900000, 951524, 952466, 961983, 965945, 974555, 988994, 997356, 1001939, 1008123, 1025238, 1038362, 1080000, 1094712, 1095240, 1109052, 1117348, 1146452, 1147356, 1165651, 1170190, 1173556, 1188922, 1200000, 1217175, 1217432, 1247019, 1252644, 1268699, 1269641, 1273774, 1275516, 1284960, 1294446, 1314416, 1326240, 1335912, 1339728, 1350000, 1359915, 1363359, 1372401, 1381897, 1410576, 1413411, 1415945, 1426226, 1429834, 1434787, 1437383, 1440000, 1441436, 1447356, 1452219, 1454406, 1482825, 1484340, 1495648, 1500000, 1513301, 1521911, 1529299, 1529756, 1532346, 1534349, 1536350, 1539424, 1539624, 1544188, 1547223, 1571481, 1583754, 1585718, 1586816, 1590948, 1591865, 1598924, 1605288, 1614828, 1627356, 1641754, 1642068, 1642574, 1642596, 1664406, 1668114, 1691877, 1694282, 1694712, 1702644, 1713411, 1716457, 1717546, 1743401, 1744343, 1747356 }; float b[N]; // list of angles, in units of 1 degree for(int i = 0; i < N; i++) { b[i] = (float)a[i] * 1.0e-4; } int k = 3; // number of angles being added int S0 = 0; // initial value of the sum int S1, S2, S3, S4, S5, S6, S7, S8, S9, S10, S11; // partial sums; only the first k will be used int lo = 3600000 - ((k>>1)+1), hi = 3600000 + ((k>>1)+1); // error bounds for the sum for(int i1 = N-1; i1 > 0; i1--) { S1 = S0 + a[i1]; if(S1 + (k-1)*a[i1] < lo) break; // sum will be too small, for this and later values of i1; end the loop if(S1 + (k-1)*a[1] > hi) continue; // sum will be too large; skip to next value of i1 for(int i2 = i1; i2 > 0; i2--) { S2 = S1 + a[i2]; if(S2 + (k-2)*a[i2] < lo) break; // sum will be too small, for this and later values of i2; skip to next value of i1 if(S2 + (k-2)*a[1] > hi) continue; // sum will be too large; skip to next value of i2 for(int i3 = i2; i3 > 0; i3--) { S3 = S2 + a[i3]; if(S3 + (k-3)*a[i3] < lo) break; // sum is too small, for this and later values of i3; skip to next value of i2 if(S3 + (k-3)*a[1] > hi) continue; // sum is too large; skip to next value of i3 printf("%8.4f + %8.4f + %8.4f\n", b[i1], b[i2], b[i3]); } } } printf("Done");

and for sums of 3 increasing angles:

Code: Select all: int N = 109; // total number of dihedral angles, plus one int a[N] = { // list of angles, in units of 0.0001 degree 0, 317175, 373774, 450000, 547356, 600000, 634349, 690948, 705288, 729730, 747547, 791877, 867268, 900000, 951524, 952466, 961983, 965945, 974555, 988994, 997356, 1001939, 1008123, 1025238, 1038362, 1080000, 1094712, 1095240, 1109052, 1117348, 1146452, 1147356, 1165651, 1170190, 1173556, 1188922, 1200000, 1217175, 1217432, 1247019, 1252644, 1268699, 1269641, 1273774, 1275516, 1284960, 1294446, 1314416, 1326240, 1335912, 1339728, 1350000, 1359915, 1363359, 1372401, 1381897, 1410576, 1413411, 1415945, 1426226, 1429834, 1434787, 1437383, 1440000, 1441436, 1447356, 1452219, 1454406, 1482825, 1484340, 1495648, 1500000, 1513301, 1521911, 1529299, 1529756, 1532346, 1534349, 1536350, 1539424, 1539624, 1544188, 1547223, 1571481, 1583754, 1585718, 1586816, 1590948, 1591865, 1598924, 1605288, 1614828, 1627356, 1641754, 1642068, 1642574, 1642596, 1664406, 1668114, 1691877, 1694282, 1694712, 1702644, 1713411, 1716457, 1717546, 1743401, 1744343, 1747356 }; float b[N]; // list of angles, in units of 1 degree for(int i = 0; i < N; i++) { b[i] = (float)a[i] * 1.0e-4; } int k = 3; // number of angles being added int S0 = 0; // initial value of the sum int S1, S2, S3, S4, S5, S6, S7, S8, S9, S10, S11; // partial sums; only the first k will be used int lo = 3600000 - ((k>>1)+1), hi = 3600000 + ((k>>1)+1); // error bounds for the sum for(int i1 = 1; i1 < N; i1++) { S1 = S0 + a[i1]; if(S1 + (k-1)*a[i1] > hi) break; // sum will be too large, for this and later values of i1; end the loop if(S1 + (k-1)*a[N-1] < lo) continue; // sum will be too small; skip to next value of i1 for(int i2 = i1; i2 < N; i2++) { S2 = S1 + a[i2]; if(S2 + (k-2)*a[i2] > hi) break; // sum will be too large, for this and later values of i2; skip to next value of i1 if(S2 + (k-2)*a[N-1] < lo) continue; // sum will be too small; skip to next value of i2 for(int i3 = i2; i3 < N; i3++) { S3 = S2 + a[i3]; if(S3 + (k-3)*a[i3] > hi) break; // sum is too large, for this and later values of i3; skip to next value of i2 if(S3 + (k-3)*a[N-1] < lo) continue; // sum is too small; skip to next value of i3 printf("%8.4f + %8.4f + %8.4f\n", b[i1], b[i2], b[i3]); } } } printf("Done");

For a different number of angles, we need to change the value of k, and add more 'for' loop blocks, and more terms in 'printf'. We can't have more than 11 angles, because the smallest angle is 31.7175°, and 12 times that is already larger than 360°. We can lump together all possible numbers of angles in a single list, by making the indices go down to (or up from) 0 instead of 1, so that 0° may be included as an angle in the sum.

The initial value of the sum can be changed from 0, for example, to 159.0948°, to find combinations with this specific angle that sum to 360°.

I used integers for internal calculations, to eliminate unpredictable rounding errors. Each angle is accurate to within 0.00005°, so the sum of k angles is accurate to within k * 0.00005°. I used floating-point only for display at the end (because it's easier than telling 'printf' to put a decimal point in the middle of an integer).

So here's the output, for 3 angles:

Code: Select all: 174.7356 + 125.2644 + 60.0000 174.7356 + 114.7356 + 70.5288 174.3401 + 153.9424 + 31.7175 174.3401 + 148.2825 + 37.3774 174.3401 + 116.5651 + 69.0948 174.3401 + 110.9052 + 74.7547 171.3411 + 109.4712 + 79.1877 170.2644 + 144.7356 + 45.0000 170.2644 + 135.0000 + 54.7356 170.2644 + 99.7356 + 90.0000 169.4712 + 120.0000 + 70.5288 169.1877 + 159.0948 + 31.7175 169.1877 + 153.4349 + 37.3774 169.1877 + 127.3774 + 63.4349 169.1877 + 121.7175 + 69.0948 169.1877 + 100.8123 + 90.0000 164.2068 + 141.0576 + 54.7356 164.2068 + 125.2644 + 70.5288 160.5288 + 144.7356 + 54.7356 160.5288 + 109.4712 + 90.0000 160.5288 + 99.7356 + 99.7356 159.0948 + 121.7175 + 79.1877 159.0948 + 110.9052 + 90.0000 158.3754 + 138.1897 + 63.4349 158.3754 + 126.8699 + 74.7547 158.3754 + 100.8123 + 100.8123 153.9424 + 142.6226 + 63.4349 153.9424 + 126.8699 + 79.1877 153.9424 + 110.9052 + 95.1524 153.4349 + 127.3774 + 79.1877 153.4349 + 116.5651 + 90.0000 150.0000 + 150.0000 + 60.0000 150.0000 + 120.0000 + 90.0000 148.2825 + 148.2825 + 63.4349 148.2825 + 142.6226 + 69.0948 148.2825 + 121.7175 + 90.0000 148.2825 + 116.5651 + 95.1524 148.2825 + 110.9052 + 100.8123 144.7356 + 144.7356 + 70.5288 144.7356 + 125.2644 + 90.0000 144.0000 + 108.0000 + 108.0000 142.6226 + 142.6226 + 74.7547 142.6226 + 138.1897 + 79.1877 142.6226 + 127.3774 + 90.0000 142.6226 + 116.5651 + 100.8123 141.0576 + 109.4712 + 109.4712 138.1897 + 110.9052 + 110.9052 135.0000 + 135.0000 + 90.0000 135.0000 + 125.2644 + 99.7356 127.3774 + 121.7175 + 110.9052 126.8699 + 116.5651 + 116.5651 125.2644 + 125.2644 + 109.4712 125.2644 + 120.0000 + 114.7356 121.7175 + 121.7175 + 116.5651 120.0000 + 120.0000 + 120.0000

for 4 angles:

Code: Select all: 174.7356 + 70.5288 + 60.0000 + 54.7356 174.3401 + 116.5651 + 37.3774 + 31.7175 174.3401 + 110.9052 + 37.3774 + 37.3774 174.3401 + 79.1877 + 74.7547 + 31.7175 174.3401 + 79.1877 + 69.0948 + 37.3774 171.3411 + 79.1877 + 54.7356 + 54.7356 170.2644 + 99.7356 + 45.0000 + 45.0000 170.2644 + 90.0000 + 54.7356 + 45.0000 169.4712 + 70.5288 + 60.0000 + 60.0000 169.1877 + 127.3774 + 31.7175 + 31.7175 169.1877 + 121.7175 + 37.3774 + 31.7175 169.1877 + 100.8123 + 45.0000 + 45.0000 169.1877 + 90.0000 + 69.0948 + 31.7175 169.1877 + 90.0000 + 63.4349 + 37.3774 164.2068 + 70.5288 + 70.5288 + 54.7356 160.5288 + 109.4712 + 45.0000 + 45.0000 160.5288 + 99.7356 + 54.7356 + 45.0000 160.5288 + 90.0000 + 54.7356 + 54.7356 159.0948 + 110.9052 + 45.0000 + 45.0000 159.0948 + 90.0000 + 79.1877 + 31.7175 158.3754 + 138.1897 + 31.7175 + 31.7175 158.3754 + 126.8699 + 37.3774 + 37.3774 158.3754 + 100.8123 + 69.0948 + 31.7175 158.3754 + 100.8123 + 63.4349 + 37.3774 158.3754 + 95.1524 + 74.7547 + 31.7175 158.3754 + 95.1524 + 69.0948 + 37.3774 158.3754 + 74.7547 + 63.4349 + 63.4349 158.3754 + 69.0948 + 69.0948 + 63.4349 153.9424 + 142.6226 + 31.7175 + 31.7175 153.9424 + 110.9052 + 63.4349 + 31.7175 153.9424 + 95.1524 + 79.1877 + 31.7175 153.9424 + 79.1877 + 63.4349 + 63.4349 153.4349 + 116.5651 + 45.0000 + 45.0000 153.4349 + 90.0000 + 79.1877 + 37.3774 150.0000 + 120.0000 + 45.0000 + 45.0000 150.0000 + 90.0000 + 60.0000 + 60.0000 148.2825 + 148.2825 + 31.7175 + 31.7175 148.2825 + 142.6226 + 37.3774 + 31.7175 148.2825 + 135.0000 + 45.0000 + 31.7175 148.2825 + 125.2644 + 54.7356 + 31.7175 148.2825 + 121.7175 + 45.0000 + 45.0000 148.2825 + 120.0000 + 60.0000 + 31.7175 148.2825 + 116.5651 + 63.4349 + 31.7175 148.2825 + 110.9052 + 69.0948 + 31.7175 148.2825 + 110.9052 + 63.4349 + 37.3774 148.2825 + 109.4712 + 70.5288 + 31.7175 148.2825 + 100.8123 + 79.1877 + 31.7175 148.2825 + 95.1524 + 79.1877 + 37.3774 148.2825 + 90.0000 + 90.0000 + 31.7175 148.2825 + 79.1877 + 69.0948 + 63.4349 144.7356 + 125.2644 + 45.0000 + 45.0000 144.7356 + 99.7356 + 70.5288 + 45.0000 144.7356 + 90.0000 + 70.5288 + 54.7356 142.6226 + 142.6226 + 37.3774 + 37.3774 142.6226 + 135.0000 + 45.0000 + 37.3774 142.6226 + 127.3774 + 45.0000 + 45.0000 142.6226 + 125.2644 + 54.7356 + 37.3774 142.6226 + 120.0000 + 60.0000 + 37.3774 142.6226 + 116.5651 + 69.0948 + 31.7175 142.6226 + 116.5651 + 63.4349 + 37.3774 142.6226 + 110.9052 + 74.7547 + 31.7175 142.6226 + 110.9052 + 69.0948 + 37.3774 142.6226 + 109.4712 + 70.5288 + 37.3774 142.6226 + 100.8123 + 79.1877 + 37.3774 142.6226 + 90.0000 + 90.0000 + 37.3774 142.6226 + 79.1877 + 74.7547 + 63.4349 142.6226 + 79.1877 + 69.0948 + 69.0948 141.0576 + 109.4712 + 54.7356 + 54.7356 138.1897 + 110.9052 + 79.1877 + 31.7175 138.1897 + 79.1877 + 79.1877 + 63.4349 135.0000 + 135.0000 + 45.0000 + 45.0000 135.0000 + 125.2644 + 54.7356 + 45.0000 135.0000 + 120.0000 + 60.0000 + 45.0000 135.0000 + 116.5651 + 63.4349 + 45.0000 135.0000 + 110.9052 + 69.0948 + 45.0000 135.0000 + 109.4712 + 70.5288 + 45.0000 135.0000 + 100.8123 + 79.1877 + 45.0000 135.0000 + 99.7356 + 70.5288 + 54.7356 135.0000 + 90.0000 + 90.0000 + 45.0000 127.3774 + 121.7175 + 79.1877 + 31.7175 127.3774 + 110.9052 + 90.0000 + 31.7175 127.3774 + 90.0000 + 79.1877 + 63.4349 126.8699 + 116.5651 + 79.1877 + 37.3774 126.8699 + 79.1877 + 79.1877 + 74.7547 125.2644 + 125.2644 + 54.7356 + 54.7356 125.2644 + 120.0000 + 60.0000 + 54.7356 125.2644 + 116.5651 + 63.4349 + 54.7356 125.2644 + 114.7356 + 60.0000 + 60.0000 125.2644 + 110.9052 + 69.0948 + 54.7356 125.2644 + 109.4712 + 70.5288 + 54.7356 125.2644 + 100.8123 + 79.1877 + 54.7356 125.2644 + 99.7356 + 90.0000 + 45.0000 125.2644 + 90.0000 + 90.0000 + 54.7356 121.7175 + 121.7175 + 79.1877 + 37.3774 121.7175 + 116.5651 + 90.0000 + 31.7175 121.7175 + 110.9052 + 90.0000 + 37.3774 121.7175 + 90.0000 + 79.1877 + 69.0948 120.0000 + 120.0000 + 60.0000 + 60.0000 120.0000 + 116.5651 + 63.4349 + 60.0000 120.0000 + 114.7356 + 70.5288 + 54.7356 120.0000 + 110.9052 + 69.0948 + 60.0000 120.0000 + 109.4712 + 70.5288 + 60.0000 120.0000 + 100.8123 + 79.1877 + 60.0000 120.0000 + 90.0000 + 90.0000 + 60.0000 116.5651 + 116.5651 + 95.1524 + 31.7175 116.5651 + 116.5651 + 63.4349 + 63.4349 116.5651 + 110.9052 + 100.8123 + 31.7175 116.5651 + 110.9052 + 95.1524 + 37.3774 116.5651 + 110.9052 + 69.0948 + 63.4349 116.5651 + 109.4712 + 70.5288 + 63.4349 116.5651 + 100.8123 + 79.1877 + 63.4349 116.5651 + 95.1524 + 79.1877 + 69.0948 116.5651 + 90.0000 + 90.0000 + 63.4349 114.7356 + 114.7356 + 70.5288 + 60.0000 110.9052 + 110.9052 + 100.8123 + 37.3774 110.9052 + 110.9052 + 74.7547 + 63.4349 110.9052 + 110.9052 + 69.0948 + 69.0948 110.9052 + 109.4712 + 70.5288 + 69.0948 110.9052 + 100.8123 + 79.1877 + 69.0948 110.9052 + 95.1524 + 79.1877 + 74.7547 110.9052 + 90.0000 + 90.0000 + 69.0948 109.4712 + 109.4712 + 70.5288 + 70.5288 109.4712 + 100.8123 + 79.1877 + 70.5288 109.4712 + 90.0000 + 90.0000 + 70.5288 100.8123 + 100.8123 + 79.1877 + 79.1877 100.8123 + 90.0000 + 90.0000 + 79.1877 99.7356 + 99.7356 + 90.0000 + 70.5288 90.0000 + 90.0000 + 90.0000 + 90.0000

for 5 angles:

Code: Select all: 174.3401 + 79.1877 + 37.3774 + 37.3774 + 31.7175 170.2644 + 54.7356 + 45.0000 + 45.0000 + 45.0000 169.1877 + 90.0000 + 37.3774 + 31.7175 + 31.7175 169.1877 + 69.0948 + 45.0000 + 45.0000 + 31.7175 169.1877 + 63.4349 + 45.0000 + 45.0000 + 37.3774 160.5288 + 54.7356 + 54.7356 + 45.0000 + 45.0000 159.0948 + 79.1877 + 45.0000 + 45.0000 + 31.7175 158.3754 + 100.8123 + 37.3774 + 31.7175 + 31.7175 158.3754 + 95.1524 + 37.3774 + 37.3774 + 31.7175 158.3754 + 74.7547 + 63.4349 + 31.7175 + 31.7175 158.3754 + 69.0948 + 69.0948 + 31.7175 + 31.7175 158.3754 + 69.0948 + 63.4349 + 37.3774 + 31.7175 158.3754 + 63.4349 + 63.4349 + 37.3774 + 37.3774 153.9424 + 110.9052 + 31.7175 + 31.7175 + 31.7175 153.9424 + 79.1877 + 63.4349 + 31.7175 + 31.7175 153.4349 + 79.1877 + 45.0000 + 45.0000 + 37.3774 150.0000 + 60.0000 + 60.0000 + 45.0000 + 45.0000 148.2825 + 116.5651 + 31.7175 + 31.7175 + 31.7175 148.2825 + 110.9052 + 37.3774 + 31.7175 + 31.7175 148.2825 + 90.0000 + 45.0000 + 45.0000 + 31.7175 148.2825 + 79.1877 + 69.0948 + 31.7175 + 31.7175 148.2825 + 79.1877 + 63.4349 + 37.3774 + 31.7175 148.2825 + 70.5288 + 54.7356 + 54.7356 + 31.7175 148.2825 + 60.0000 + 60.0000 + 60.0000 + 31.7175 144.7356 + 70.5288 + 54.7356 + 45.0000 + 45.0000 142.6226 + 116.5651 + 37.3774 + 31.7175 + 31.7175 142.6226 + 110.9052 + 37.3774 + 37.3774 + 31.7175 142.6226 + 90.0000 + 45.0000 + 45.0000 + 37.3774 142.6226 + 79.1877 + 74.7547 + 31.7175 + 31.7175 142.6226 + 79.1877 + 69.0948 + 37.3774 + 31.7175 142.6226 + 79.1877 + 63.4349 + 37.3774 + 37.3774 142.6226 + 70.5288 + 54.7356 + 54.7356 + 37.3774 142.6226 + 60.0000 + 60.0000 + 60.0000 + 37.3774 141.0576 + 54.7356 + 54.7356 + 54.7356 + 54.7356 138.1897 + 79.1877 + 79.1877 + 31.7175 + 31.7175 135.0000 + 116.5651 + 45.0000 + 31.7175 + 31.7175 135.0000 + 110.9052 + 45.0000 + 37.3774 + 31.7175 135.0000 + 90.0000 + 45.0000 + 45.0000 + 45.0000 135.0000 + 79.1877 + 69.0948 + 45.0000 + 31.7175 135.0000 + 79.1877 + 63.4349 + 45.0000 + 37.3774 135.0000 + 70.5288 + 54.7356 + 54.7356 + 45.0000 135.0000 + 60.0000 + 60.0000 + 60.0000 + 45.0000 127.3774 + 110.9052 + 45.0000 + 45.0000 + 31.7175 127.3774 + 90.0000 + 79.1877 + 31.7175 + 31.7175 127.3774 + 79.1877 + 63.4349 + 45.0000 + 45.0000 126.8699 + 79.1877 + 79.1877 + 37.3774 + 37.3774 125.2644 + 116.5651 + 54.7356 + 31.7175 + 31.7175 125.2644 + 110.9052 + 54.7356 + 37.3774 + 31.7175 125.2644 + 99.7356 + 45.0000 + 45.0000 + 45.0000 125.2644 + 90.0000 + 54.7356 + 45.0000 + 45.0000 125.2644 + 79.1877 + 69.0948 + 54.7356 + 31.7175 125.2644 + 79.1877 + 63.4349 + 54.7356 + 37.3774 125.2644 + 70.5288 + 54.7356 + 54.7356 + 54.7356 125.2644 + 60.0000 + 60.0000 + 60.0000 + 54.7356 121.7175 + 116.5651 + 45.0000 + 45.0000 + 31.7175 121.7175 + 110.9052 + 45.0000 + 45.0000 + 37.3774 121.7175 + 90.0000 + 79.1877 + 37.3774 + 31.7175 121.7175 + 79.1877 + 69.0948 + 45.0000 + 45.0000 120.0000 + 116.5651 + 60.0000 + 31.7175 + 31.7175 120.0000 + 110.9052 + 60.0000 + 37.3774 + 31.7175 120.0000 + 90.0000 + 60.0000 + 45.0000 + 45.0000 120.0000 + 79.1877 + 69.0948 + 60.0000 + 31.7175 120.0000 + 79.1877 + 63.4349 + 60.0000 + 37.3774 120.0000 + 70.5288 + 60.0000 + 54.7356 + 54.7356 120.0000 + 60.0000 + 60.0000 + 60.0000 + 60.0000 116.5651 + 116.5651 + 63.4349 + 31.7175 + 31.7175 116.5651 + 110.9052 + 69.0948 + 31.7175 + 31.7175 116.5651 + 110.9052 + 63.4349 + 37.3774 + 31.7175 116.5651 + 109.4712 + 70.5288 + 31.7175 + 31.7175 116.5651 + 100.8123 + 79.1877 + 31.7175 + 31.7175 116.5651 + 95.1524 + 79.1877 + 37.3774 + 31.7175 116.5651 + 90.0000 + 90.0000 + 31.7175 + 31.7175 116.5651 + 90.0000 + 63.4349 + 45.0000 + 45.0000 116.5651 + 79.1877 + 69.0948 + 63.4349 + 31.7175 116.5651 + 79.1877 + 63.4349 + 63.4349 + 37.3774 116.5651 + 70.5288 + 63.4349 + 54.7356 + 54.7356 116.5651 + 63.4349 + 60.0000 + 60.0000 + 60.0000 114.7356 + 70.5288 + 60.0000 + 60.0000 + 54.7356 110.9052 + 110.9052 + 74.7547 + 31.7175 + 31.7175 110.9052 + 110.9052 + 69.0948 + 37.3774 + 31.7175 110.9052 + 110.9052 + 63.4349 + 37.3774 + 37.3774 110.9052 + 109.4712 + 70.5288 + 37.3774 + 31.7175 110.9052 + 100.8123 + 79.1877 + 37.3774 + 31.7175 110.9052 + 95.1524 + 79.1877 + 37.3774 + 37.3774 110.9052 + 90.0000 + 90.0000 + 37.3774 + 31.7175 110.9052 + 90.0000 + 69.0948 + 45.0000 + 45.0000 110.9052 + 79.1877 + 74.7547 + 63.4349 + 31.7175 110.9052 + 79.1877 + 69.0948 + 69.0948 + 31.7175 110.9052 + 79.1877 + 69.0948 + 63.4349 + 37.3774 110.9052 + 70.5288 + 69.0948 + 54.7356 + 54.7356 110.9052 + 69.0948 + 60.0000 + 60.0000 + 60.0000 109.4712 + 90.0000 + 70.5288 + 45.0000 + 45.0000 109.4712 + 79.1877 + 70.5288 + 69.0948 + 31.7175 109.4712 + 79.1877 + 70.5288 + 63.4349 + 37.3774 109.4712 + 70.5288 + 70.5288 + 54.7356 + 54.7356 109.4712 + 70.5288 + 60.0000 + 60.0000 + 60.0000 100.8123 + 90.0000 + 79.1877 + 45.0000 + 45.0000 100.8123 + 79.1877 + 79.1877 + 69.0948 + 31.7175 100.8123 + 79.1877 + 79.1877 + 63.4349 + 37.3774 100.8123 + 79.1877 + 70.5288 + 54.7356 + 54.7356 100.8123 + 79.1877 + 60.0000 + 60.0000 + 60.0000 99.7356 + 99.7356 + 70.5288 + 45.0000 + 45.0000 99.7356 + 90.0000 + 70.5288 + 54.7356 + 45.0000 95.1524 + 79.1877 + 79.1877 + 74.7547 + 31.7175 95.1524 + 79.1877 + 79.1877 + 69.0948 + 37.3774 90.0000 + 90.0000 + 90.0000 + 45.0000 + 45.0000 90.0000 + 90.0000 + 79.1877 + 69.0948 + 31.7175 90.0000 + 90.0000 + 79.1877 + 63.4349 + 37.3774 90.0000 + 90.0000 + 70.5288 + 54.7356 + 54.7356 90.0000 + 90.0000 + 60.0000 + 60.0000 + 60.0000 79.1877 + 79.1877 + 74.7547 + 63.4349 + 63.4349 79.1877 + 79.1877 + 69.0948 + 69.0948 + 63.4349

for 6 angles:

Code: Select all: 169.1877 + 45.0000 + 45.0000 + 37.3774 + 31.7175 + 31.7175 158.3754 + 74.7547 + 31.7175 + 31.7175 + 31.7175 + 31.7175 158.3754 + 69.0948 + 37.3774 + 31.7175 + 31.7175 + 31.7175 158.3754 + 63.4349 + 37.3774 + 37.3774 + 31.7175 + 31.7175 153.9424 + 79.1877 + 31.7175 + 31.7175 + 31.7175 + 31.7175 148.2825 + 79.1877 + 37.3774 + 31.7175 + 31.7175 + 31.7175 148.2825 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 31.7175 142.6226 + 79.1877 + 37.3774 + 37.3774 + 31.7175 + 31.7175 142.6226 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 135.0000 + 79.1877 + 45.0000 + 37.3774 + 31.7175 + 31.7175 135.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 127.3774 + 79.1877 + 45.0000 + 45.0000 + 31.7175 + 31.7175 125.2644 + 79.1877 + 54.7356 + 37.3774 + 31.7175 + 31.7175 125.2644 + 54.7356 + 45.0000 + 45.0000 + 45.0000 + 45.0000 121.7175 + 79.1877 + 45.0000 + 45.0000 + 37.3774 + 31.7175 120.0000 + 79.1877 + 60.0000 + 37.3774 + 31.7175 + 31.7175 120.0000 + 60.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 116.5651 + 116.5651 + 31.7175 + 31.7175 + 31.7175 + 31.7175 116.5651 + 110.9052 + 37.3774 + 31.7175 + 31.7175 + 31.7175 116.5651 + 90.0000 + 45.0000 + 45.0000 + 31.7175 + 31.7175 116.5651 + 79.1877 + 69.0948 + 31.7175 + 31.7175 + 31.7175 116.5651 + 79.1877 + 63.4349 + 37.3774 + 31.7175 + 31.7175 116.5651 + 70.5288 + 54.7356 + 54.7356 + 31.7175 + 31.7175 116.5651 + 63.4349 + 45.0000 + 45.0000 + 45.0000 + 45.0000 116.5651 + 60.0000 + 60.0000 + 60.0000 + 31.7175 + 31.7175 110.9052 + 110.9052 + 37.3774 + 37.3774 + 31.7175 + 31.7175 110.9052 + 90.0000 + 45.0000 + 45.0000 + 37.3774 + 31.7175 110.9052 + 79.1877 + 74.7547 + 31.7175 + 31.7175 + 31.7175 110.9052 + 79.1877 + 69.0948 + 37.3774 + 31.7175 + 31.7175 110.9052 + 79.1877 + 63.4349 + 37.3774 + 37.3774 + 31.7175 110.9052 + 70.5288 + 54.7356 + 54.7356 + 37.3774 + 31.7175 110.9052 + 69.0948 + 45.0000 + 45.0000 + 45.0000 + 45.0000 110.9052 + 60.0000 + 60.0000 + 60.0000 + 37.3774 + 31.7175 109.4712 + 79.1877 + 70.5288 + 37.3774 + 31.7175 + 31.7175 109.4712 + 70.5288 + 45.0000 + 45.0000 + 45.0000 + 45.0000 100.8123 + 79.1877 + 79.1877 + 37.3774 + 31.7175 + 31.7175 100.8123 + 79.1877 + 45.0000 + 45.0000 + 45.0000 + 45.0000 99.7356 + 70.5288 + 54.7356 + 45.0000 + 45.0000 + 45.0000 95.1524 + 79.1877 + 79.1877 + 37.3774 + 37.3774 + 31.7175 90.0000 + 90.0000 + 79.1877 + 37.3774 + 31.7175 + 31.7175 90.0000 + 90.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 90.0000 + 79.1877 + 69.0948 + 45.0000 + 45.0000 + 31.7175 90.0000 + 79.1877 + 63.4349 + 45.0000 + 45.0000 + 37.3774 90.0000 + 70.5288 + 54.7356 + 54.7356 + 45.0000 + 45.0000 90.0000 + 60.0000 + 60.0000 + 60.0000 + 45.0000 + 45.0000 79.1877 + 79.1877 + 74.7547 + 63.4349 + 31.7175 + 31.7175 79.1877 + 79.1877 + 69.0948 + 69.0948 + 31.7175 + 31.7175 79.1877 + 79.1877 + 69.0948 + 63.4349 + 37.3774 + 31.7175 79.1877 + 79.1877 + 63.4349 + 63.4349 + 37.3774 + 37.3774 79.1877 + 70.5288 + 69.0948 + 54.7356 + 54.7356 + 31.7175 79.1877 + 70.5288 + 63.4349 + 54.7356 + 54.7356 + 37.3774 79.1877 + 69.0948 + 60.0000 + 60.0000 + 60.0000 + 31.7175 79.1877 + 63.4349 + 60.0000 + 60.0000 + 60.0000 + 37.3774 70.5288 + 70.5288 + 54.7356 + 54.7356 + 54.7356 + 54.7356 70.5288 + 60.0000 + 60.0000 + 60.0000 + 54.7356 + 54.7356 60.0000 + 60.0000 + 60.0000 + 60.0000 + 60.0000 + 60.0000

for 7 angles:

Code: Select all: 158.3754 + 37.3774 + 37.3774 + 31.7175 + 31.7175 + 31.7175 + 31.7175 116.5651 + 79.1877 + 37.3774 + 31.7175 + 31.7175 + 31.7175 + 31.7175 116.5651 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 31.7175 + 31.7175 110.9052 + 79.1877 + 37.3774 + 37.3774 + 31.7175 + 31.7175 + 31.7175 110.9052 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 + 31.7175 90.0000 + 79.1877 + 45.0000 + 45.0000 + 37.3774 + 31.7175 + 31.7175 90.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 79.1877 + 79.1877 + 74.7547 + 31.7175 + 31.7175 + 31.7175 + 31.7175 79.1877 + 79.1877 + 69.0948 + 37.3774 + 31.7175 + 31.7175 + 31.7175 79.1877 + 79.1877 + 63.4349 + 37.3774 + 37.3774 + 31.7175 + 31.7175 79.1877 + 70.5288 + 54.7356 + 54.7356 + 37.3774 + 31.7175 + 31.7175 79.1877 + 69.0948 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 31.7175 79.1877 + 63.4349 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 79.1877 + 60.0000 + 60.0000 + 60.0000 + 37.3774 + 31.7175 + 31.7175 70.5288 + 54.7356 + 54.7356 + 45.0000 + 45.0000 + 45.0000 + 45.0000 60.0000 + 60.0000 + 60.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000

and for 8 angles:

Code: Select all: 79.1877 + 79.1877 + 37.3774 + 37.3774 + 31.7175 + 31.7175 + 31.7175 + 31.7175 79.1877 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 + 31.7175 + 31.7175 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000

mr_e_man wrote:As you probably expected, the crown jewels J84-90 and snic and snid don't fit with anything. Also the augmented pentagonal prism (angle 162.74°) doesn't fit.

From the polyhedra with 3,4,5,6,8,10-gons, there are 55 combinations of 3 angles adding to 360°, 128 combinations of 4 angles, 112 of 5, 56 of 6, 16 of 7, 3 of 8, and no combinations of 9 or more angles.

(Does anyone know why this website's 'code' feature gets rid of single spaces at the beginning of a line? It disrupts alignment; I had to add two spaces everywhere to fix it.)

by **mr_e_man** » Mon Aug 16, 2021 8:30 pm

Sorry, I was offline for a month or two....

Let's restart from the beginning. I'll do it a little differently this second time, to make it easier for everyone. I'll consider only elementary polyhedra: those CRF polyhedra which can't be cut by a single plane into two smaller CRF polyhedra. (We might consider cutting into more than two smaller polyhedra, as in tic = cube + 6 J4 + 8 tet, but that doesn't help much.) Here's the list thus shortened:

tet, cube, ~~oct~~, doe, ~~ike~~
tut, tic, toe, ~~co, sirco~~, girco, snic, tid, ti, ~~id, srid~~, grid, snid
prisms, antiprisms
J1-6, ~~7-62~~, 63, ~~64-79~~, 80, ~~81-82~~, 83-86, 87, 88-92

Obviously, any honeycomb involving non-elementary polyhedra can be made by combining cells from a honeycomb involving only elementary polyhedra.

And here's the list of dihedral angles, from elementary polyhedra with only 3,4,5,6,8,10-gon faces, to replace the longer list in the code above:

Code: Select all: int N = 76; // total number of dihedral angles, plus one int a[N] = { // list of angles, in units of 0.0001 degree 0, 317175, 373774, 450000, 547356, 600000, 634349, 705288, 729730, 791877, 867268, 900000, 952466, 961983, 965945, 974555, 988994, 1001939, 1008123, 1025238, 1038362, 1080000, 1094712, 1095240, 1109052, 1117348, 1146452, 1165651, 1170190, 1173556, 1188922, 1200000, 1217175, 1217432, 1247019, 1252644, 1275516, 1284960, 1294446, 1314416, 1335912, 1339728, 1350000, 1359915, 1363359, 1372401, 1381897, 1413411, 1426226, 1429834, 1434787, 1437383, 1440000, 1441436, 1447356, 1452219, 1454406, 1482825, 1484340, 1495648, 1529299, 1529756, 1532346, 1539624, 1544188, 1547223, 1571481, 1590948, 1591865, 1598924, 1614828, 1641754, 1642574, 1664406, 1668114, 1716457 };

by **mr_e_man** » Mon Aug 16, 2021 8:35 pm

tet, cube, doe
tut, tic, toe, girco, snic, tid, ti, grid, snid
prisms, antiprisms
J1-6, 63, 80, 83-86, 88-92

mr_e_man wrote:I started with large prisms and antiprisms (in pairs sharing a large face), systematically adding up angles with other CRF polyhedra.

I did that manually the first time. But now we can use the program to do some of the work.

There are three cases to consider here: a pair of prisms, a pair of antiprisms, or a prism and an antiprism. That is assuming n=7,9,11 or larger.

Two prisms

Code: Select all: | vertex | edges | squares | n-gon | n-prisms +--------+---------+---------+--------+---------- | * | 1 1 1 1 | 1 1 1 1 | 1 | 1 1 +--------+---------+---------+--------+---------- | 1 | * | 1 0 0 1 | 1 | 1 1 | 1 | * | 1 1 0 0 | 0 | 1 0 | 1 | * | 0 1 1 0 | 1 | 1 1 | 1 | * | 0 0 1 1 | 0 | 0 1 +--------+---------+---------+--------+---------- | 1 | 1 1 0 0 | * | | 1 0 | 1 | 0 1 1 0 | * | | 1 0 | 1 | 0 0 1 1 | * | | 0 1 | 1 | 1 0 0 1 | * | | 0 1 +--------+---------+---------+--------+---------- | 1 | 1 0 1 0 | | * | 1 1 +--------+---------+---------+--------+---------- | 1 | 1 1 1 0 | 1 1 0 0 | 1 | * | 1 | 1 0 1 1 | 0 0 1 1 | 1 | *

This configuration is so simple that we shouldn't need an incmat. But we'll build upon it, and we'll use the ordering to "name" different elements, which would otherwise be difficult to distinguish. (Also, I don't think I've made incmats before, so this gives me some experience.)

The 1st edge has dihedral angle sum 90° + 90°, with two exposed square faces and a hidden n-gon. To complete that edge, we need to find combinations of dihedral angles that sum to 180°. With 180° as the initial value of the sum, the program gives us

(1): 148.2825° + 31.7175°
(2): 142.6226° + 37.3774°
(3): 135° + 45°
(4): 125.2644° + 54.7356°
(5): 120° + 60°
(6): 116.5651° + 63.4349°
(7): 109.4712° + 70.5288°
(8): 100.8123° + 79.1877°
(9): 90° + 90°

(10): 116.5651° + 31.7175° + 31.7175°
(11): 110.9052° + 37.3774° + 31.7175°
(12): 90° + 45° + 45°
(13): 79.1877° + 63.4349° + 37.3774°
(14): 70.5288° + 54.7356° + 54.7356°
(15): 60° + 60° + 60°

(16): 79.1877° + 37.3774° + 31.7175° + 31.7175°
(17): 45° + 45° + 45° + 45°

(18): We also need to consider putting more prisms or antiprisms around the 1st edge, with 7,9,11-gons or larger.

In a honeycomb, each face is part of two cells; so, if the polyhedra are considered separately, then for any k, the total number of k-gons around an edge must be an even number (since each face is counted twice). Specifically, the 1st edge already has two squares not attaching to each other, so the number of squares from other polyhedra must be an even number, at least 2. This rules out cases (2), (6), (7), (8), (13). (Refer to the list of dihedral angles, and note that 4.4.3.3 E is from a non-elementary polyhedron, J30.)

For case (1), the configuration of faces is 4.10 + 10.4 (attaching to both sides of 4.n + n.4); there's one J5 and one 'grid'. Here's the incmat for more detail (omitting the single vertex):

Code: Select all: | edges | {4} |{n}|{10}|{3}|{6}| n-prisms | J5 | grid +---------+---------+---+----+---+---+----------+----+------ | * | 1 0 0 1 | 1 | 1 | 0 | 0 | 1 1 | 1 | 1 | * | 1 1 0 0 | 0 | 0 | 1 | 0 | 1 0 | 1 | 0 | * | 0 1 1 0 | 1 | 0 | 0 | 0 | 1 1 | 0 | 0 | * | 0 0 1 1 | 0 | 0 | 0 | 1 | 0 1 | 0 | 1 +---------+---------+---+----+---+---+----------+----+------ | 1 1 0 0 | * | | | | | 1 0 | 1 | 0 | 0 1 1 0 | * | | | | | 1 0 | 0 | 0 | 0 0 1 1 | * | | | | | 0 1 | 0 | 0 | 1 0 0 1 | * | | | | | 0 1 | 0 | 1 +---------+---------+---+----+---+---+----------+----+------ | 1 0 1 0 | | * | | | | 1 1 | 0 | 0 +---------+---------+---+----+---+---+----------+----+------ | 1 0 0 0 | | | * | | | 0 0 | 1 | 1 +---------+---------+---+----+---+---+----------+----+------ | 0 1 0 0 | | | | * | | 0 0 | 1 | 0 +---------+---------+---+----+---+---+----------+----+------ | 0 0 0 1 | | | | | * | 0 0 | 0 | 1 +---------+---------+---+----+---+---+----------+----+------ | 1 1 1 0 | 1 1 0 0 | 1 | 0 | 0 | 0 | * | | | 1 0 1 1 | 0 0 1 1 | 1 | 0 | 0 | 0 | * | | +---------+---------+---+----+---+---+----------+----+------ | 1 1 0 0 | 1 0 0 0 | 0 | 1 | 1 | 0 | | * | +---------+---------+---+----+---+---+----------+----+------ | 1 0 0 1 | 0 0 0 1 | 0 | 1 | 0 | 1 | | | *

The 2nd edge gets the prism's 4.4 dihedral angle and 159.0948° from J5. For a heptagonal prism n=7 we have 128.5714° + 159.0948° = 360° - 72.3338°; that remainder is clearly not a sum of CRF angles. For an enneagonal prism n=9 or larger, we get a remainder of 60.9052° or less, which is not a sum of 2 or more angles, but may be a single CRF angle, for some n. This angle would have to be one of 31.7175°, 37.3774°, 45°, 54.7356°, 60°. But adding each of these to 159.0948°, and subtracting the result from 360°, gives a remainder of 169.1877°, 163.5278°, 155.9052°, 146.1696°, 140.9052°, none of which is a prism angle.

The above paragraph shows that 159.0948° doesn't fit with the n-gon prism's 4.4 angle. In fact we haven't referred to the 'grid' at all, though it happens that grid's 4.6 angle is also 159.0948°. And we haven't referred to the second n-gon prism; it could instead be an antiprism.

Case (3) is analogous to case (1); there's one J4 and one girco. The 2nd edge gets the prism's 4.4 angle and 144.7356°. For n=7,9,11,12,13,..., (respectively) we get a remainder of

86.6930°, 75.2644°, 67.9917°, 65.2644°, 62.9567°, ... .

None of these can be a sum of 3 or more CRF angles, since they're less than the minimum 3*31.7175° = 95.1524°. None beyond the first four can be a sum of 2 angles, since 62.9567° < 2*31.7175° = 63.4349°. If one of the first four is a sum of 2 angles, those would have to be 31.7175°, 37.3774°, 45°, or 54.7356°, since anything bigger would be at least 31.7175° + 60° > 86.6930°. The sums of two of these first few CRF angles are as follows: (Let's include more angles for later reference, so the sum goes up to ~~106°~~ ~~117°~~ 122°:)

31.7175° + 31.7175° = 63.4349°
31.7175° + 37.3774° = 69.0948°
31.7175° + 45° = 76.7175°
31.7175° + 54.7356° = 86.4531°
(31.7175° + 60° = 91.7175°)
(31.7175° + 63.4349° = 95.1524°)
(31.7175° + 70.5288° = 102.2463°)
(31.7175° + 72.9730° = 104.6905°)
(31.7175° + 79.1877° = 110.9052°)
(31.7175° + 86.7268° = 118.4443°)
(31.7175° + 90° = 121.7175°)
(31.7175° + 95.2466° = 126.9641°)
37.3774° + 37.3774° = 74.7547°
37.3774° + 45° = 82.3774°
37.3774° + 54.7356° = 92.1130°
(37.3774° + 60° = 97.3774°)
(37.3774° + 63.4349° = 100.8123°)
(37.3774° + 70.5288° = 107.9062°)
(37.3774° + 72.9730° = 110.3504°)
(37.3774° + 79.1877° = 116.5651°)
(37.3774° + 86.7268° = 124.1042°)
45° + 45° = 90°
45° + 54.7356° = 99.7356°
(45° + 60° = 105°)
(45° + 63.4349° = 108.4349°)
(45° + 70.5288° = 115.5288°)
(45° + 72.9730° = 117.9730°)
(45° + 79.1877° = 124.1877°)
54.7356° + 54.7356° = 109.4712°
(54.7356° + 60° = 114.7356°)
(54.7356° + 63.4349° = 118.1705°)
(54.7356° + 70.5288° = 125.2644°)
(60° + 60° = 120°)
(60° + 63.4349° = 123.4349°)

These don't match any of the first four remainders above. So none of the remainders is a sum of 2 or more CRF angles; it might yet be a single CRF angle, which would have to be one of the five 31.7175°, ... , 60°. Adding each of these to 144.7356°, and subtracting the result from 360°, gives 183.5469°, 177.8870°, 170.2644°, 160.5288°, 155.2644°, none of which is a prism angle, though the 170-gon prism comes close: 177.8824° + 144.7356° + 37.3774° = 359.9954°.

This shows that 144.7356° doesn't fit with the n-gon prism's 4.4 angle.

Case (4) is also similar to case (1). The configuration of faces completing the 1st edge is either 4.6 + 6.4 (with J3 and 'toe') or 4.3 + 3.4 (with J1 and J3). In either case, there's a 125.2644° angle at the 2nd or 4th edge at this vertex, or at a corresponding edge at an adjacent vertex. For n=7 we have 128.5714° + 125.2644° = 360° - 106.1642°; for n=7,9,11,12,..., the remainder is

106.1642°, 94.7356°, 87.4629°, 84.7356°, 82.4279°, 80.4499°, 78.7356°, 77.2356°, 75.9121°, 74.7356°,
73.6830°, 72.7356°, 71.8785°, 71.0992°, 70.3878°, 69.7356°, 69.1356°, 68.5818°, 68.0689°, ... .

Clearly none of these can be a sum of 4 or more CRF angles. Only the first one might be a sum of 3, but it's easy to check that it's not: (Again let's include more angles for later reference:)

3*31.7175° = 95.1524°
2*31.7175° + 37.3774° = 100.8123°
2*31.7175° + 45° = 108.4349°
2*31.7175° + 54.7356° = 118.1705°
2*31.7175° + 60° = 123.4349°
31.7175° + 2*37.3774° = 106.4722°
31.7175° + 37.3774° + 45° = 114.0948°
31.7175° + 37.3774° + 54.7356° = 123.8304°
31.7175° + 2*45° = 121.7175°
31.7175° + 45° + 54.7356° = 131.4531°
3*37.3774° = 112.1321°
2*37.3774° + 45° = 119.7547°
2*37.3774° + 54.7356° = 129.4903°
37.3774° + 2*45° = 127.3774°

Any other sum of 3 will exceed 106.1642°. And you can check that none of the 19 remainders shown is a sum of 2 angles, using the list above (for case (3)). If one beyond those shown is a sum of 2, it would have to be 31.7175° + 31.7175°, since 31.7175° + 37.3774° > 68.0689°. But that sum of 2 is also a sum of 1, 63.4349°, and may be considered with other sums of 1. None of the 19 remainders shown is a single CRF angle; any beyond those shown would have to be one of the six 31.7175°, ... , 63.4349°. Adding each of these to 125.2644°, and subtracting the result from 360°, gives 203.0181°, 197.3582°, 189.7356°, 180°, 174.7356°, 171.3007°, none of which is a prism angle.

This shows that 125.2644° doesn't fit with the n-gon prism's 4.4 angle.

Case (5): There's a triangular and a hexagonal prism, with axes perpendicular to the n-gon prism's axis. We leave this as a possibility for now.

Case (9): The 1st edge is surrounded by four 90° angles. The only elementary polyhedron with a 90° dihedral angle, other than a prism, is 'tic'; but this has octagons, and can't attach to the squares from the n-gon prism. So the 90° angles are all from prisms. We leave this also as a possibility.

Case (10): The configuration of faces is 4.10 + 10.10 + 10.4, with two J5 and one 'tid'. But this puts a 159.0948° angle at the 2nd edge, and is thus ruled out as in case (1).

Case (11): The configuration of faces is 4.10 + 10.3 + 3.4, with two J5 and one J91 or 92. Again this puts 159.0948° at the 2nd edge.

Case (12): The configuration of faces is either 4.8 + 8.8 + 8.4, with two J4 and one 'tic'; or 4.8 + 8.4 + 4.4, with two J4 and one prism. In either case, there's a 144.7356° angle at the 2nd or 4th edge, so this is ruled out as in case (3).

Case (14): The configuration of faces is either 4.3 + 3.3 + 3.4, or 4.6 + 6.6 + 6.4, or 4.6 + 6.3 + 3.4; but 4.6 is from J3 which puts 125.2644° at the 2nd edge, so that's ruled out as in case (4). We're left with 4.3 + 3.3 + 3.4 (two J1 and one tet) as a possibility.

Case (15): There are three triangular prisms, with axes perpendicular to the n-gon prism's axis. This also remains a possibility.

Case (16): 79.1877° and 37.3774° don't involve squares, so we have 31.7175° where a square in J5 attaches to a square in the n-gon prism; this puts 159.0948° at the 2nd and 4th edges, and is thus ruled out as in case (1).

Case (17): 45° only appears in J4, so we get 144.7356° at the 2nd and 4th edges; this is ruled out as in case (3).

Case (18): The relevant prism angles, other than 90°, are

128.5714°, 140°, 147.2727°, 150°, 152.3077°, ... .

But, adding the smallest CRF angle, 150° + 31.7175° is already greater than 180°. So only those first three need to be considered. Take 128.5714°, 140°, or 147.2727°, and add to it 31.7175°, 37.3774°, 45°, or 54.7356°. Clearly it never equals 180°. Adding any more angles, or larger angles, will exceed 180°.

The angle in an antiprism, between two triangles, is at least 150.2223° and at most 180°; again, adding 31.7175° already exceeds 180°.

The angle in an antiprism, between a triangle and an m-gon, is at most 97.5723° and at least 90°. The m-gon must attach to another m-gon, from another prism or antiprism. But the prism 4.m angle is 90°, and the antiprism 3.m angle is greater than 90°, so the sum is greater than 180°.

Here are the cases we haven't ruled out yet, for configurations around the 1st edge:

(5): 4.4 + 4.4; 120° + 60°
(9): 4.m + m.4; 90° + 90° (here m is not restricted to 7,9,11+; it could be 3 or anything)
(14): 4.3 + 3.3 + 3.4; 54.7356° + 70.5288° + 54.7356°
(15): 4.4 + 4.4 + 4.4; 60° + 60° + 60°

We'll try putting various combinations of these at the 1st and 3rd edges. Case (5) puts 90° at the 2nd and 4th edges, as does case (15). Case (9) puts an m-gon prism's 4.4 angle at the 2nd and 4th edges. Case (14) puts 54.7356° at the 2nd and 4th edges. So we have 10 possible partial configurations of faces and angles around the 2nd edge:

(5,5): 6.4 + 4.4 + 4.6; 90° + (180°-360°/n) + 90°
(5,9): 6.4 + 4.4 + 4.4; 90° + (180°-360°/n) + (180°-360°/m)
(5,14): 6.4 + 4.4 + 4.3; 90° + (180°-360°/n) + 54.7356°
(5,15): 6.4 + 4.4 + 4.3; 90° + (180°-360°/n) + 90°
(9,9): 4.4 + 4.4 + 4.4; (180°-360°/m) + (180°-360°/n) + (180°-360°/k)
(9,14): 4.4 + 4.4 + 4.3; (180°-360°/m) + (180°-360°/n) + 54.7356°
(9,15): 4.4 + 4.4 + 4.3; (180°-360°/m) + (180°-360°/n) + 90°
(14,14): 3.4 + 4.4 + 4.3; 54.7356° + (180°-360°/n) + 54.7356°
(14,15): 3.4 + 4.4 + 4.3; 54.7356° + (180°-360°/n) + 90°
(15,15): 3.4 + 4.4 + 4.3; 90° + (180°-360°/n) + 90°

(Actually, since case (5) is asymmetric, (5,5) should be broken into two cases: The 2nd edge gets two hexagons while the 4th edge gets two triangles; or the 2nd and 4th edge each get one hexagon and one triangle. But that second case has the same effect as (5,15) if we just look at one edge.)

Cases (5,5), (5,15), and (15,15) are easily ruled out by the same reasoning as in the first part of case (18), because 90° + 90° = 180°. (We are looking at it from a different perspective. In case (18), 180° was the "base" to which we tried to add a prism and other things. Here, the prism is the "base" to which we try to add 180° and other things.)

Case (5,9), m=3: The sum of angles around the 2nd edge is at least 90° + 128.5714° + 60° = 278.5714° = 360° - 81.4286°. For n=7,9,11,12,..., the remainder is

81.4286°, 70°, 62.7273°, 60°, 57.6923°, 55.7143°, 54°, ... .

None of these is a sum of 2 CRF angles; check the list from case (3). Obviously 60° is a CRF angle, but since it only appears in the 3-gon prism, this would give a configuration 6.4 + 4.4 + 4.4 + 4.4 around the 2nd edge; the hexagon is trying to attach to a square. So 60° doesn't work here. One of the remainders might be 31.7175°, 37.3774°, or 45°; adding each of these to 90° + 60°, and subtracting the result from 360°, gives 178.2825°, 172.6226°, 165°. The first two are not prism angles, but the third one is, with n=24. The full sum of angles around the 2nd edge is then 90° + 165° + 60° + 45°, corresponding to a configuration 6.4 + 4.4 + 4.4 + 4.8; the hexagon is trying to attach to an octagon. So that doesn't work either.

Case (5,9), m≥4: m=4 is ruled out as in the first part of case (18), because 90° + 90° = 180°. Hence m≥5. The sum of angles around the 2nd edge, subtracted from 360°, is, for various combinations of n,m:

n=7: 33.4286°, 21.4286°, 12.8571°, 6.4286°, 1.4286°, -2.5714°
n=9: 22°, 10°, 1.4286°, -5°
n=11: 14.7273°, 2.7273°, -5.8442°
n=12: 12°, 0°, -8.5714°
n=13: 9.6923°, -2.3077°
...
n=19: 0.9474°, -11.0526°
n=20: 0°, -12°
n=21: -0.8571°
...

We see there are two special cases here, corresponding to the 2D vertex configurations 4.12.6 and 4.20.5 . But again the hexagon is trying to attach to a square.

Case (5,14): For n=7, the 2nd edge has 90° + 128.5714° + 54.7356° = 360° - 86.6930°... In fact we've seen this same remainder before. Since 90° + 54.7356° = 144.7356°, this is ruled out as in case (3).

Case (9,9): We leave this as a possibility for now. We know from 2D that some vertex configurations are valid locally, and can be ruled out only by going beyond just one vertex. At this moment, we are focusing on one vertex.

Case (9,14), m=3: Around the 2nd edge there's a triangular prism and a J1, and of course the n-gon prism between them. For n=7,9,11,..., the remaining angle is

116.6930°, 105.2644°, 97.9917°, 95.2644°, 92.9567°, 90.9787°, 89.2644°, 87.7644°, 86.4409°, 85.2644°, ... (limit is 65.2644°).

Check with the lists from cases (3) and (4) that none of the 10 remainders shown is a sum of 2 or 3 CRF angles. One not shown may yet be a sum of 2 CRF angles no larger than 45° (since 31.7175° + 54.7356° > 85.2644°). Adding each of 63.4349°, 69.0948°, 76.7175°, 74.7547°, 82.3774° (the relevant sums of 2) to 60° + 54.7356°, and subtracting the result from 360°, gives 181.8295°, 176.1696°, 168.5469°, 170.5097°, 162.8870°, none of which is a prism angle, though the 94-gon prism comes close: 176.1702° + 60° + 54.7356° + 31.7175° + 37.3774° = 360.0006°.

One of the remainders could be a single CRF angle. Indeed, one of the first six, being greater than 90°, could be an antiprism angle. But a peek at the list, along with a calculation of 90.9807° and 90.9626° for 53 and 54-gon antiprisms, shows no matches. (Note that 95.2644° ≠ 95.2466°; the 12-gon prism does not fit with the 10-gon antiprism.) Clearly none of the 10 remainders shown is a (non-antiprism) CRF angle; one not shown would have to be one of the three 70.5288°, 72.9730°, 79.1877°. Adding each of these to 60° + 54.7356°, and subtracting the result from 360°, gives 174.7356°, 172.2914°, 166.0767°, none of which is a prism angle.

Case (9,14), m≥4: m=4 is ruled out as in case (3), because 90° + 54.7356° = 144.7356°. Hence m≥5. There are two prisms and one J1 around the 2nd edge. The remaining angle is, for various combinations of n,m:

n=7: 68.6930°, 56.6930°, 48.1215°, 41.6930°, 36.6930°, 32.6930°, 29.4202°, ... , 0.0263°, -0.0043° (m=109)
n=9: 57.2644°, 45.2644°, 36.6930°, 30.2644°, ... , 0.2644°, -0.3356° (m=25)
n=11: 49.9917°, 37.9917°, 29.4202°, ... , 0.4917°, -0.8319° (m=17)
n=12: 47.2644°, 35.2644°, 26.6930°, ... , 0.9787°, -0.7356° (m=15)
n=13: 44.9567°, 32.9567°, 24.3853°, ... , 0.6490°, -1.3290° (m=14)
n=14: 42.9787°, 30.9787°, ... , 0.9787°, -1.3290° (m=13)

This will never be exactly 0°, because the two prisms give a rational multiple of 360° (whose doubled cosine is always an algebraic integer), while arccos(1/√3) = 54.7356° is an irrational multiple of 360° (because x=2/√3, with minimal polynomial 3x²-4, is not an algebraic integer). So the only possible values are 31.7175° and 37.3774°, for n>14 and m=5. Adding each of these two to 108° + 54.7356°, and subtracting the result from 360°, gives 165.5469°, 159.8870°, neither of which is a prism angle.

Case (9,15): See case (5,9), and replace "hexagon" with "triangle".

Case (14,14): This one is harder. For n=7 we have 54.7356° + 128.5714° + 54.7356° = 360° - 121.9574°. That remainder could be a sum of 3 CRF angles; but the list from case (4) reaches up to 122°, and shows no matches. It could be a sum of 2 CRF angles; but the list from case (3) also reaches up to 122°, and shows no matches. Yet, that list didn't account for 7,9,11+-gon prisms or antiprisms; 31.7175° plus an antiprism angle close to 90° could equal 121.9574°. Check that 121.9574° - 31.7175° = 90.2399° is not an antiprism angle (the closest are the 216 and 217-gon antiprisms).

For n=9,11,12,..., the remainder is

110.5288°, 103.2561°, 100.5288°, 98.2211°, 96.2431°, 94.5288°, 93.0288°, 91.7052°, 90.5288°, 89.4761°, ... (limit is 70.5288°).

Check that these greater than 90° are not sums of 2 or 3 CRF angles, nor single antiprism angles. One of those less than 90° could be a sum of 2; it could only be (see the list from case (3)) 63.4349°, 69.0948°, 76.7175°, 86.4531°, 74.7547°, 82.3774°. Adding each of these to 54.7356° + 54.7356°, and subtracting the result from 360°, gives 187.0939°, 181.4340°, 173.8113°, 164.0757°, 175.7741°, 168.1514°, none of which is a prism angle.

One of the remainders could be a single CRF angle, which would have to be one of the three 72.9730°, 79.1877°, 86.7268° (since the larger ones don't match the shown remainders). Adding each of these to 54.7356° + 54.7356°, and subtracting the result from 360°, gives 177.5558°, 171.3411°, 163.8020°, none of which is a prism angle.

This shows that 109.4712° = 2*54.7356° doesn't fit with the n-gon prism's 4.4 angle.

Case (14,15): This is ruled out as in case (3), because 54.7356° + 90° = 144.7356°.

Now all we have left is case (9,9): The n-gon prism must be surrounded by other prisms, with its axis parallel to their axes. (This is still assuming that the n-gon prism is stacked on another prism, not an antiprism.) We know from 2D that the only possible configurations with n=7,9,11+ are:

42.7.3
24.8.3
20.5.4
18.9.3
15.10.3
12.12.3
12.6.4
12.4.3.3
12.3.4.3

42.7.3 doesn't work because the 7-gon would have to be surrounded by alternating 42-gons and 3-gons, but 7 is an odd number. Similarly, 18.9.3 doesn't work because 9 is odd, and 15.10.3 doesn't work because 15 is odd.

24.8.3 might work; 3 is odd, but we haven't concluded that a triangular prism must be surrounded by other prisms, in general. Yet the 24-gon prism attached to the 3-gon prism does require two 8-gon prisms to attach to the other two squares of the 3-gon prism. So the far vertex of the triangle (in 2D) gets a partial configuration 8.3.8; the remaining angle is 30°, which cannot be filled by any CRF dihedral angle.

Similarly, 20.5.4 might work; we haven't concluded that a pentagonal prism must be surrounded by other prisms. The 20-gon prism does require two cubes on the near sides of the 5-gon prism: (converting to the 2D view)

Code: Select all: | vertices | edges | {20} | {5} | {4} +----------+-------+------+-----+----- | * | 1 1 0 | 1 | 1 | 1 0 | * | 1 0 1 | 1 | 1 | 0 1 | * | 0 1 0 | 0 | 1 | 1 0 | * | 0 0 1 | 0 | 1 | 0 1 +----------+-------+------+-----+----- | 1 1 0 0 | * | 1 | 1 | 0 0 | 1 0 1 0 | * | 0 | 1 | 1 0 | 0 1 0 1 | * | 0 | 1 | 0 1 +----------+-------+------+-----+----- | 1 1 0 0 | 1 0 0 | * | | +----------+-------+------+-----+----- | 1 1 1 1 | 1 1 1 | | * | +----------+-------+------+-----+----- | 1 0 1 0 | 0 1 0 | | | * | 0 1 0 1 | 0 0 1 | | | *

The 1st and 2nd vertices have the configuration 20.5.4; the 3rd and 4th vertices get the partial configuration 5.4, with 162° of empty space remaining.

(Reverting to 3D) 162° can't fit a large prism or antiprism (except the 20-gon prism itself): For the 7-gon prism, adding 128.5714° to 31.7175° or 37.3774° doesn't give 162°. For a 9-gon or larger prism, we have at least 140° + 31.7175° > 162°. For a 7-gon or larger antiprism, 150.2223° + 31.7175° > 162°. This rules out the 4.4 and 3.3 angles. If there's a 4.m or 3.m angle, it must be paired with another, so that the m-gon attaches to another m-gon; but these angles are at least 90°, so the sum is at least 180° > 162°.

With large prisms and antiprisms out of the way, we can use the program to check the 3,4,5,6,8,10-gon polyhedra. If we put 108° + 90° = 198.0000° as the initial value of the sum, we get no results (or, the result that there are no such 360° sums). Therefore, 162° can only be another 20-gon prism.

Code: Select all: | vertices | edges | {20} | {5} | {4} +-----------+-----------+-------+-----+----- | * | 1 1 0 0 0 | 1 0 0 | 1 | 1 0 | * | 1 0 1 0 0 | 1 0 0 | 1 | 0 1 | * | 0 1 0 1 0 | 0 1 0 | 1 | 1 0 | * | 0 0 1 0 1 | 0 0 1 | 1 | 0 1 | * | 0 0 0 1 1 | 0 1 1 | 1 | 0 0 +-----------+-----------+-------+-----+----- | 1 1 0 0 0 | * | 1 0 0 | 1 | 0 0 | 1 0 1 0 0 | * | 0 0 0 | 1 | 1 0 | 0 1 0 1 0 | * | 0 0 0 | 1 | 0 1 | 0 0 1 0 1 | * | 0 1 0 | 1 | 0 0 | 0 0 0 1 1 | * | 0 0 1 | 1 | 0 0 +-----------+-----------+-------+-----+----- | 1 1 0 0 0 | 1 0 0 0 0 | * | | | 0 0 1 0 1 | 0 0 0 1 0 | * | | | 0 0 0 1 1 | 0 0 0 0 1 | * | | +-----------+-----------+-------+-----+----- | 1 1 1 1 1 | 1 1 1 1 1 | | * | +-----------+-----------+-------+-----+----- | 1 0 1 0 0 | 0 1 0 0 0 | | | * | 0 1 0 1 0 | 0 0 1 0 0 | | | *

Now the 5th vertex gets a configuration 20.5.20, with angle sum 432° > 360°. Thus 20.5.4 is ruled out.

This concludes the case of two n-gon prisms, with n=7,9,11 or larger: The only possibility is n=12.

by **mr_e_man** » Mon Aug 16, 2021 8:48 pm

Prism and antiprism

(Omitting the single vertex)

Code: Select all: | edges | {4} | {3} | {n} | n-prism, antiprism +-----------+-----+-------+-----+-------------------- | * | 1 0 | 1 0 0 | 1 | 1 1 | * | 1 1 | 0 0 0 | 0 | 1 0 | * | 0 1 | 0 0 1 | 1 | 1 1 | * | 0 0 | 1 1 0 | 0 | 0 1 | * | 0 0 | 0 1 1 | 0 | 0 1 +-----------+-----+-------+-----+-------------------- | 1 1 0 0 0 | * | | | 1 0 | 0 1 1 0 0 | * | | | 1 0 +-----------+-----+-------+-----+-------------------- | 1 0 0 1 0 | | * | | 0 1 | 0 0 0 1 1 | | * | | 0 1 | 0 0 1 0 1 | | * | | 0 1 +-----------+-----+-------+-----+-------------------- | 1 0 1 0 0 | | | * | 1 1 +-----------+-----+-------+-----+-------------------- | 1 1 1 0 0 | 1 1 | 0 0 0 | 1 | * | 1 0 1 1 1 | 0 0 | 1 1 1 | 1 | *

The 1st edge has the 4.n angle (90°) and the n.3 angle (slightly greater than 90°), so the remainder is slightly less than 180°. For n=7,9,11,12,..., this is

172.4277°, 174.1570°, 175.2384°, 175.6408°, 175.9801°, 176.2702°, 176.5210°, 176.7402°, 176.9332°, 177.1047°,
177.2579°, 177.3957°, 177.5202°, 177.6334°, 177.7367°, 177.8313°, 177.9183°, 177.9986°, 178.0730°, ... .

Suppose we try to put another large prism or antiprism at the 1st edge. The configuration can't include 4.m + m.4, or 4.m + m.3 or 3.m + m.3, because those exceed 180°. It can't include 3.3, from a heptagonal or larger antiprism, because 150.2223° + 31.7175° > 180°, and 3.3 alone can't attach to the square from the n-gon prism. If it includes 4.4, from a heptagonal or larger prism, then the remaining angle is less than 52°, so it would have to be a single CRF angle. (4.4 alone can't attach to the triangle from the n-gon antiprism.) The only possibilities are 31.7175°, 37.3774°, 45°; but these would give complete configurations of

3.n + n.4 + 4.4 + 4.10,
3.n + n.4 + 4.4 + 10.3,
3.n + n.4 + 4.4 + 5.3,
3.n + n.4 + 4.4 + 4.8,

none of which is valid. (A k-gon must attach to another k-gon.)

So the 1st edge must be completed by polyhedra with only 3,4,5,6,8,10-gons. A single polyhedron won't work, because the largest relevant CRF angle is 171.6457° < 172.4277°. Six or more polyhedra won't work, because 6*31.7175° > 180°. The number of polyhedra here (in addition to the prism and antiprism) must be 2, 3, 4, or 5.

Let 'A' be the polyhedron attaching to the prism's square, and 'B' the polyhedron attaching to the antiprism's triangle. Let 'A1' and 'B1' be their dihedral angles at the 1st edge, and 'A2' and 'B4' their dihedral angles at the adjacent edges.

A1 = 31.7175°: This would imply A2 = 159.0948°, which was ruled out in 'Two prisms' case (1).

A1 = 37.3774°: That dihedral angle doesn't involve squares, but A does need a square.

A1 = 45°: This would imply A2 = 144.7356°, which was ruled out in 'Two prisms' case (3).

Hence A1 ≥ 54.7356°.

B1 = 31.7175°: That dihedral angle doesn't involve triangles, but B does need a triangle.

B1 = 37.3774°: This would imply either B4 = 159.0948° (from J5) or B4 = 138.1897° (from J2). Let's consider the larger one first. For n=7 we have at the 4th edge 150.2223° + 159.0948° = 360° - 50.6829°; for n=7,9,11,..., the remainder is

50.6829°, 44.0389°, 39.8219°, 38.2423°, 36.9064°, ... , 31.8486°, 31.3010° (n=20).

None of these is a CRF angle. Next, for n=7 we have 150.2223° + 138.1897° = 360° - 71.5881°; for n=7,9,11,..., the remainder is

71.5881°, 64.9441°, 60.7271°, 59.1475°, 57.8115°, 56.6669°, 55.6752°,
54.8077°, 54.0424°, ... , 45.0081°, 44.9596° (n=66), ... (limit is 41.8103° at n = ∞).

None of these is a CRF angle, nor a sum of 2; refer to the list from 'Two prisms' case (3). This shows that the n-gon antiprism's 3.3 angle doesn't fit with 159.0948° nor 138.1897°.

B1 = 45°: That dihedral angle doesn't involve triangles, but B does need a triangle.

B1 = 54.7356°: This would imply either B4 = 144.7356° (from J4) or B4 = 109.4712° (from J1). Let's consider the larger one first. For n=7 we have 150.2223° + 144.7356° = 360° - 65.0421°; for n=7,9,11,..., the remainder is

65.0421°, 58.3982°, 54.1811°, ... , 45.1649°, 44.7146°, ... , 37.3853°, 37.3639° (n=99) (limit is 35.2644°).

None of these is a CRF angle, nor a sum of 2. Next, for n=7 we have 150.2223° + 109.4712° = 360° - 100.3065°; for n=7,9,11,..., the remainder is

100.3065°, 93.6625°, 89.4455°, 87.8659°, 86.5300°, 85.3854°, 84.3937°, 83.5262°,
82.7608°, 82.0807°, ... , 79.1911°, 78.8445°, ... , 76.8279°, 76.6426°,
... , 74.7708°, 74.6859°, ... , 72.9741°, 72.9456° (n=86) (limit is 70.5288°).

None of these is a sum of 1 or 2 or 3 CRF angles; refer to the lists from 'Two prisms' cases (3) and (4). (Also compare 93.6625° to the 14,15-gon antiprisms: 93.7298°, 93.4790°.) This shows that the n-gon antiprism's 3.3 angle doesn't fit with 144.7356° nor 109.4712°.

B1 = 60°: That doesn't involve triangles.

B1 = 63.4349°: That doesn't involve triangles either.

Hence B1 ≥ 70.5288°.

This eliminates the possibility of four or more polyhedra completing the 1st edge, since 54.7356° + 70.5288° + 2*31.7175° > 180°.

There might be two or three polyhedra completing the 1st edge. Let's call the one in the middle (if it exists) 'C', and its dihedral angle 'C1'.

Code: Select all: | | | | | other | n-p, | | edges | {4} | {3} | {n} | faces | ap | A,B,C +-----------+-----+-------+-----+-------+------+------- | * | 1 0 | 1 0 0 | 1 | 1 1 | 1 1 | 1 1 1 | * | 1 1 | 0 0 0 | 0 | 0 0 | 1 0 | 1 0 0 | * | 0 1 | 0 0 1 | 1 | 0 0 | 1 1 | 0 0 0 | * | 0 0 | 1 1 0 | 0 | 0 0 | 0 1 | 0 1 0 | * | 0 0 | 0 1 1 | 0 | 0 0 | 0 1 | 0 0 0 +-----------+-----+-------+-----+-------+------+------- | 1 1 0 0 0 | * | | | | 1 0 | 1 0 0 | 0 1 1 0 0 | * | | | | 1 0 | 0 0 0 +-----------+-----+-------+-----+-------+------+------- | 1 0 0 1 0 | | * | | | 0 1 | 0 1 0 | 0 0 0 1 1 | | * | | | 0 1 | 0 0 0 | 0 0 1 0 1 | | * | | | 0 1 | 0 0 0 +-----------+-----+-------+-----+-------+------+------- | 1 0 1 0 0 | | | * | | 1 1 | 0 0 0 +-----------+-----+-------+-----+-------+------+------- | 1 0 0 0 0 | | | | * | 0 0 | 1 0 1 | 1 0 0 0 0 | | | | * | 0 0 | 0 1 1 +-----------+-----+-------+-----+-------+------+------- | 1 1 1 0 0 | 1 1 | 0 0 0 | 1 | 0 0 | * | | 1 0 1 1 1 | 0 0 | 1 1 1 | 1 | 0 0 | * | +-----------+-----+-------+-----+-------+------+------- | 1 1 0 0 0 | 1 0 | 0 0 0 | 0 | 1 0 | | * | 1 0 0 1 0 | 0 0 | 1 0 0 | 0 | 0 1 | | * | 1 0 0 0 0 | 0 0 | 0 0 0 | 0 | 1 1 | | *

C1 = 31.7175°: In this case, C = J5, and either A or B must have a decagon at the 1st edge.

If A has a decagon, then its edge configuration is 4.10, so A1 must be 90° (from 'dip'), 121.7175° (from J80,83), or 148.2825° (from 'grid'). But with A1 ≥ 90°, B1 ≥ 70.5288°, and C1 = 31.7175°, the angle sum is A1 + C1 + B1 ≥ 192.2463° > 180°.

If B has a decagon, then its edge configuration is 3.10, so B1 must be 79.1877° (from J6), 95.2466° (from 'dap'), or 142.6226° (from 'tid'). Let's consider the larger ones first. With B1 ≥ 95.2466°, A1 ≥ 54.7356°, and C1 = 31.7175°, the angle sum is B1 + C1 + A1 ≥ 181.6997° > 180°. Hence B1 = 79.1877°, and thus B4 = 142.6226°. For n=7 we have 150.2223° + 142.6226° = 360° - 67.1551°; for n=7,9,11,..., the remainder is

67.1551°, 60.5111°, 56.2941°, 54.7145°, ... , 45.0768°, 44.8017° (n=28) (limit is 37.3774°).

None of these is a CRF angle, nor a sum of 2. This shows that the n-gon antiprism's 3.3 angle doesn't fit with 142.6226°.

C1 = 37.3774°: In this case, C is either J5 or J2; but the former is ruled out by the same reasoning as with C1=31.7175°. So C = J2, and either A or B must have a pentagon at the 1st edge.

If A has a pentagon, then, since the smallest 4.5 angle is 90°, the angle sum is A1 + C1 + B1 ≥ 90° + 37.3774° + 70.5288° > 180°.

If B has a pentagon, then, since the smallest 3.5 angle larger than 70.5288° is 100.8123°, the angle sum is B1 + C1 + A1 ≥ 100.8123° + 37.3774° + 54.7356° > 180°.

C1 = 45°: In this case, C = J4, and either A or B must have an octagon at the 1st edge.

If A has an octagon, then, since the smallest 4.8 angle larger than 54.7356° is 90°, the angle sum is A1 + C1 + B1 ≥ 90° + 45° + 70.5288° > 180°.

If B has an octagon, then, since the smallest 3.8 angle larger than 70.5288° is 96.5945°, the angle sum is B1 + C1 + A1 ≥ 96.5945° + 45° + 54.7356° > 180°.

C1 ≥ 54.7356°: Now the angle sum is A1 + C1 + B1 ≥ 54.7356° + 54.7356° + 70.5288° = 180°.

Thus, there can be no C, only A and B.

If A1 ≥ 109.4712°, then A1 + B1 ≥ 109.4712° + 70.5288° = 180°. Hence A1 < 109.4712°.

A1 = 108°: In this case, A has edge configuration 4.4, so B has edge configuration 3.4; but the smallest 3.4 angle larger than 70.5288° is 90°, so the angle sum is A1 + B1 ≥ 108° + 90° > 180°.

A1 = 103.8362°: In this case, A has edge configuration 4.3, so B has edge configuration 3.3 . Now if B1 = 70.5288°, then A1 + B1 = 174.3650°, which doesn't match any of the remainders above. And the smallest 3.3 angle larger than 70.5288° is 86.7268°, which gives A1 + B1 ≥ 190.5630° > 180°.

A1 = 102.5238° or 100.1939°: These won't work for the same reason as A1 = 108°.

A1 = 97.4555°: In this case, A has edge configuration 4.3, so B has edge configuration 3.3 . Now if B1 = 70.5288°, then A1 + B1 = 167.9843° < 172°, which is smaller than any of the remainders above. And, again, the smallest 3.3 angle larger than 70.5288° is 86.7268°, which gives A1 + B1 ≥ 184.1823° > 180°.

Hence A1 ≤ 90°. This rules out B1 = 70.5288° or 79.1877°, as that would give A1 + B1 ≤ 169.1877° < 172°.

A1 = 90°: If B1 = 86.7268°, then we have A1 + B1 = 176.7268°, which doesn't match any of the remainders above. If B1 ≥ 90°, then A1 + B1 ≥ 180°.

A1 = 72.9730°: In this case, A = J88, with a large angle A2 = 154.7223°. For n=7,9,11,..., the remaining angle at the 2nd edge between the prism and J88 is

76.7063°, 65.2777°, 58.0050°, 55.2777°, 52.9700°, ... , 45.2777°, 44.2251°,
... , 37.6915°, 37.2777°, ... , 31.8232°, 31.7063° (n=56) (limit is 25.2777°).

None of these is a CRF angle, nor a sum of 2. So the 2nd edge can't be completed.

A1 = 60°: In this case, A = 'trip', and thus A2 = 90°. We leave this as a possibility for now.

A1 = 54.7356°: We can't have A = J3, as that would make A2 = 125.2644°, which was ruled out in 'Two prisms' case (4). So it must be J1, with A2 = 54.7356°. We leave this also as a possibility, for a moment.

We can apply the same reasoning to the 3rd edge as to the 1st edge. So two adjacent squares in the n-gon prism must attach to 'trip' (with its axis perpendicular to the prism's axis) or J1. But this was ruled out in 'Two prisms' cases (14,14), (14,15), and (15,15).

This concludes the case of an n-gon prism and antiprism, with n=7,9,11 or larger: Nothing fits.

by **mr_e_man** » Mon Aug 16, 2021 9:01 pm

Two antiprisms

Code: Select all: | edges | {3} | {n} | n-antiprisms +-------------+-------------+-----+-------------- | * | 1 0 0 0 0 1 | 1 | 1 1 | * | 1 1 0 0 0 0 | 0 | 1 0 | * | 0 1 1 0 0 0 | 0 | 1 0 | * | 0 0 1 1 0 0 | 1 | 1 1 | * | 0 0 0 1 1 0 | 0 | 0 1 | * | 0 0 0 0 1 1 | 0 | 0 1 +-------------+-------------+-----+-------------- | 1 1 0 0 0 0 | * | | 1 0 | 0 1 1 0 0 0 | * | | 1 0 | 0 0 1 1 0 0 | * | | 1 0 | 0 0 0 1 1 0 | * | | 0 1 | 0 0 0 0 1 1 | * | | 0 1 | 1 0 0 0 0 1 | * | | 0 1 +-------------+-------------+-----+-------------- | 1 0 0 1 0 0 | | * | 1 1 +-------------+-------------+-----+-------------- | 1 1 1 1 0 0 | 1 1 1 0 0 0 | 1 | * | 1 0 0 1 1 1 | 0 0 0 1 1 1 | 1 | *

The 1st edge has configuration 3.n + n.3, so the remaining angle is slightly less than 180°. For n=7,9,11,12,..., this is

164.8555°, 168.3141°, 170.4767°, 171.2815°, 171.9602°, 172.5404°, 173.0421°, 173.4803°, 173.8665°, 174.2093°,
174.5158°, 174.7913°, 175.0405°, 175.2668°, 175.4734°, 175.6626°, 175.8367°, 175.9973°, 176.1459°, ... .

Suppose we try to put another large prism or antiprism at the 1st edge. The configuration can't include 4.m + m.4, or 4.m + m.3 or 3.m + m.3, because those exceed 180°. If it includes 4.4, from a heptagonal or larger prism, it would have to be sandwiched between 2 other polyhedra (since 4 ≠ 3), but then the angle sum would be at least 128.5714° + 2*31.7175° > 180°. It might include 3.3, from a heptagonal or larger antiprism, though there would be no space for anything more at this edge, since 150.2223° + 31.7175° > 180°.

mr_e_man wrote:One difficult case to consider was two (n-gon) antiprisms meeting a third (m-gon) antiprism at two of its triangle faces: The first two dihedral angles, between a triangle and the n-gon, have a sum slightly greater than 180°, and the third angle, between the two triangles, is slightly less than 180°, so the sum is arbitrarily close to 360° (both below it and above it) when n and m are large enough. It should be possible to rule out this case using algebraic number theory, but I just noted that the m-gon antiprism must be paired with another m-gon antiprism or prism at the same vertex, and the four large solid angles (each slightly less than 180°) don't leave enough space for anything else at the vertex.

In fact there's an easier way. Either the 2nd or the 6th edge gets another 3.3 angle from the m-gon antiprism, while the other edge gets the 3.m angle. By symmetry, it doesn't really matter, so let's say the 2nd edge gets 3.3 :

Code: Select all: | edges | {3} | {n},{m} | antiprisms +-------------+---------------+---------+------------ | * | 1 0 0 0 0 1 0 | 1 0 | 1 1 1 | * | 1 1 0 0 0 0 1 | 0 0 | 1 0 1 | * | 0 1 1 0 0 0 0 | 0 0 | 1 0 0 | * | 0 0 1 1 0 0 0 | 1 0 | 1 1 0 | * | 0 0 0 1 1 0 0 | 0 0 | 0 1 0 | * | 0 0 0 0 1 1 0 | 0 1 | 0 1 1 +-------------+---------------+---------+------------ | 1 1 0 0 0 0 | * | | 1 0 1 | 0 1 1 0 0 0 | * | | 1 0 0 | 0 0 1 1 0 0 | * | | 1 0 0 | 0 0 0 1 1 0 | * | | 0 1 0 | 0 0 0 0 1 1 | * | | 0 1 0 | 1 0 0 0 0 1 | * | | 0 1 1 | 0 1 0 0 0 0 | * | | 0 0 1 +-------------+---------------+---------+------------ | 1 0 0 1 0 0 | | * | 1 1 0 | 0 0 0 0 0 1 | | * | 0 0 1 +-------------+---------------+---------+------------ | 1 1 1 1 0 0 | 1 1 1 0 0 0 0 | 1 0 | * | 1 0 0 1 1 1 | 0 0 0 1 1 1 0 | 1 0 | * | 1 1 0 0 0 1 | 1 0 0 0 0 1 1 | 0 1 | *

Now the 2nd edge has at least 2*150.2223° = 360° - 59.5554°; that remainder can't be a sum of 2 or more CRF angles (since 2*31.7175° > 59.5554°), and it can't be a single CRF angle either, because the smallest 3.3 angle is 70.5288°.

So the 1st edge must be completed by polyhedra with only 3,4,5,6,8,10-gons. A single polyhedron won't work, because the largest relevant CRF angles are 164.2574°, 166.4406°, 166.8114°, and 171.6457°; these don't match the first five remainders above (starting with 164.8555°), and they're smaller than any of the further remainders. Six or more polyhedra won't work, because 6*31.7175° > 180°. The number of polyhedra here (in addition to the two antiprisms) must be 2, 3, 4, or 5.

Let 'A' be the polyhedron attaching directly to the 1st antiprism (by the 1st triangle), and 'B' the polyhedron attaching to the 2nd antiprism (by the 6th triangle). Let 'A1' and 'B1' be their dihedral angles at the 1st edge, and 'A2' and 'B6' their dihedral angles at the adjacent edges.

By the same reasoning as in 'Prism and antiprism', we must have both A1 and B1 ≥ 70.5288°.

This eliminates the possibility of four or more polyhedra completing the 1st edge, since 2*70.5288° + 2*31.7175° > 180°.

There might be two or three polyhedra completing the 1st edge. Let's call the one in the middle (if it exists) 'C', and its dihedral angle 'C1'.

C1 = 31.7175°: In this case, C = J5, and either A or B must have a decagon at the 1st edge; But the smallest 3.10 angle larger than 70.5288° is 79.1877°, so the angle sum is A1 + C1 + B1 ≥ 79.1877° + 31.7175° + 70.5288° > 180°.

C1 = 37.3774°: In this case, C is either J5 or J2; but the former is ruled out by the same reasoning as with C1=31.7175°. For the latter, if C = J2, then either A or B must have a pentagon at the 1st edge; but the smallest 3.5 angle larger than 70.5288° is 100.8123°, so the angle sum is A1 + C1 + B1 ≥ 100.8123° + 37.3774° + 70.5288° > 180°.

C1 ≥ 45°: Now the angle sum is A1 + C1 + B1 ≥ 70.5288° + 45° + 70.5288° > 180°.

Thus, there can be no C, only A and B.

Let 'AB' be the face connecting A to B.

If A1 ≥ 109.4712°, then A1 + B1 ≥ 109.4712° + 70.5288° = 180°. Hence A1 ≤ 108°, and similarly B1 ≤ 108°.

If AB is a square, then, since the smallest 3.4 angle larger than 70.5288° is 90°, the angle sum is A1 + B1 ≥ 90° + 90° = 180°. Similarly, if AB is a pentagon or octagon, then, since the smallest 3.5 and 3.8 angles larger than 70.5288° are (respectively) 100.8123° and 96.5945°, the angle sum is A1 + B1 ≥ 2*96.5945° > 180°. Hence, AB is a triangle, hexagon, or decagon.

Suppose AB is a hexagon. The only 3.6 angles between 70.5288° and 108° (inclusive) are 70.5288° itself and 98.8994°. So the angle sum is either 2*70.5288° = 141.0576°, which is too small; or 70.5288° + 98.8994° = 169.4282°, which doesn't match any of the remainders above; or 2*98.8994° = 197.7988°, which is too large.

Suppose AB is a decagon. The only 3.10 angles between 70.5288° and 108° are 79.1877° and 95.2466°. So the angle sum is either 2*79.1877° < 164°; or 79.1877° + 95.2466° = 174.4343°, which doesn't match any of the remainders; or 2*95.2466° > 180°.

Suppose AB is a triangle. The only 3.3 angles between 70.5288° and 108° (inclusive) are 70.5288° itself, 86.7268°, and 96.1983°. So the possible angle sums are:

2*70.5288° < 164°
70.5288° + 86.7268° < 164°
70.5288° + 96.1983° = 166.7271°
2*86.7268° = 173.4536°
86.7268° + 96.1983° > 180°
2*96.1983° > 180°

And those two in the middle don't equal any of the remainders.

This concludes the case of two n-gon antiprisms, with n=7,9,11 or larger: Nothing fits.

(I think, the first time I did this, I merely added up dihedral angles at a single edge and compared to 360°, and didn't look at the face types or the adjacent edges. I would have had to consider sums of as many as 5 (plus the two antiprisms), from 31.7175° all the way up to 174.7356° (in the non-elementary polyhedra J54-57). It goes quickly, just using a calculator and not writing down the numbers or putting thoughts into words.)

by **mr_e_man** » Mon Aug 16, 2021 9:08 pm

There's still one loose end concerning large prisms and antiprisms:

mr_e_man wrote:A 12-gon prism can only appear in a 2D tiling stacked on top of itself.

Obviously, since we've ruled out the prism-antiprism combination, a single 'twip' (twelve-gon prism) must be part of an infinite stack of twips. But that doesn't immediately imply that the whole honeycomb is an infinite stack of 2D tilings.

We can prove that later. For now, let's just include twip in the list of possible cells, and add 150° to the program's list of dihedral angles.

by **mr_e_man** » Mon Aug 16, 2021 9:12 pm

tet, cube, doe
tut, tic, toe, girco, snic, tid, ti, grid, snid
trip, pip, hip, op, dip, twip, squap, pap, hap, oap, dap
J1-6, 63, 80, 83-86, 88-92

Running the program with the angles from these polyhedra gives the following results:

Code: Select all: 159.0948 + 121.7175 + 79.1877 159.0948 + 110.9052 + 90.0000 150.0000 + 150.0000 + 60.0000 150.0000 + 120.0000 + 90.0000 148.2825 + 148.2825 + 63.4349 148.2825 + 121.7175 + 90.0000 148.2825 + 110.9052 + 100.8123 144.7356 + 144.7356 + 70.5288 144.7356 + 125.2644 + 90.0000 144.0000 + 108.0000 + 108.0000 142.6226 + 138.1897 + 79.1877 142.6226 + 116.5651 + 100.8123 138.1897 + 110.9052 + 110.9052 135.0000 + 135.0000 + 90.0000 125.2644 + 125.2644 + 109.4712 121.7175 + 121.7175 + 116.5651 120.0000 + 120.0000 + 120.0000

Code: Select all: 159.0948 + 110.9052 + 45.0000 + 45.0000 159.0948 + 90.0000 + 79.1877 + 31.7175 150.0000 + 120.0000 + 45.0000 + 45.0000 150.0000 + 90.0000 + 60.0000 + 60.0000 148.2825 + 148.2825 + 31.7175 + 31.7175 148.2825 + 142.6226 + 37.3774 + 31.7175 148.2825 + 135.0000 + 45.0000 + 31.7175 148.2825 + 125.2644 + 54.7356 + 31.7175 148.2825 + 121.7175 + 45.0000 + 45.0000 148.2825 + 120.0000 + 60.0000 + 31.7175 148.2825 + 116.5651 + 63.4349 + 31.7175 148.2825 + 110.9052 + 63.4349 + 37.3774 148.2825 + 109.4712 + 70.5288 + 31.7175 148.2825 + 100.8123 + 79.1877 + 31.7175 148.2825 + 90.0000 + 90.0000 + 31.7175 144.7356 + 125.2644 + 45.0000 + 45.0000 144.7356 + 90.0000 + 70.5288 + 54.7356 142.6226 + 142.6226 + 37.3774 + 37.3774 142.6226 + 135.0000 + 45.0000 + 37.3774 142.6226 + 125.2644 + 54.7356 + 37.3774 142.6226 + 120.0000 + 60.0000 + 37.3774 142.6226 + 116.5651 + 63.4349 + 37.3774 142.6226 + 109.4712 + 70.5288 + 37.3774 142.6226 + 100.8123 + 79.1877 + 37.3774 142.6226 + 90.0000 + 90.0000 + 37.3774 138.1897 + 110.9052 + 79.1877 + 31.7175 138.1897 + 79.1877 + 79.1877 + 63.4349 135.0000 + 135.0000 + 45.0000 + 45.0000 135.0000 + 125.2644 + 54.7356 + 45.0000 135.0000 + 120.0000 + 60.0000 + 45.0000 135.0000 + 116.5651 + 63.4349 + 45.0000 135.0000 + 109.4712 + 70.5288 + 45.0000 135.0000 + 100.8123 + 79.1877 + 45.0000 135.0000 + 90.0000 + 90.0000 + 45.0000 125.2644 + 125.2644 + 54.7356 + 54.7356 125.2644 + 120.0000 + 60.0000 + 54.7356 125.2644 + 116.5651 + 63.4349 + 54.7356 125.2644 + 109.4712 + 70.5288 + 54.7356 125.2644 + 100.8123 + 79.1877 + 54.7356 125.2644 + 90.0000 + 90.0000 + 54.7356 121.7175 + 121.7175 + 79.1877 + 37.3774 121.7175 + 116.5651 + 90.0000 + 31.7175 121.7175 + 110.9052 + 90.0000 + 37.3774 120.0000 + 120.0000 + 60.0000 + 60.0000 120.0000 + 116.5651 + 63.4349 + 60.0000 120.0000 + 109.4712 + 70.5288 + 60.0000 120.0000 + 100.8123 + 79.1877 + 60.0000 120.0000 + 90.0000 + 90.0000 + 60.0000 116.5651 + 116.5651 + 63.4349 + 63.4349 116.5651 + 110.9052 + 100.8123 + 31.7175 116.5651 + 109.4712 + 70.5288 + 63.4349 116.5651 + 100.8123 + 79.1877 + 63.4349 116.5651 + 90.0000 + 90.0000 + 63.4349 110.9052 + 110.9052 + 100.8123 + 37.3774 109.4712 + 109.4712 + 70.5288 + 70.5288 109.4712 + 100.8123 + 79.1877 + 70.5288 109.4712 + 90.0000 + 90.0000 + 70.5288 100.8123 + 100.8123 + 79.1877 + 79.1877 100.8123 + 90.0000 + 90.0000 + 79.1877 90.0000 + 90.0000 + 90.0000 + 90.0000

Code: Select all: 159.0948 + 79.1877 + 45.0000 + 45.0000 + 31.7175 150.0000 + 60.0000 + 60.0000 + 45.0000 + 45.0000 148.2825 + 116.5651 + 31.7175 + 31.7175 + 31.7175 148.2825 + 110.9052 + 37.3774 + 31.7175 + 31.7175 148.2825 + 90.0000 + 45.0000 + 45.0000 + 31.7175 148.2825 + 79.1877 + 63.4349 + 37.3774 + 31.7175 148.2825 + 70.5288 + 54.7356 + 54.7356 + 31.7175 148.2825 + 60.0000 + 60.0000 + 60.0000 + 31.7175 144.7356 + 70.5288 + 54.7356 + 45.0000 + 45.0000 142.6226 + 116.5651 + 37.3774 + 31.7175 + 31.7175 142.6226 + 110.9052 + 37.3774 + 37.3774 + 31.7175 142.6226 + 90.0000 + 45.0000 + 45.0000 + 37.3774 142.6226 + 79.1877 + 63.4349 + 37.3774 + 37.3774 142.6226 + 70.5288 + 54.7356 + 54.7356 + 37.3774 142.6226 + 60.0000 + 60.0000 + 60.0000 + 37.3774 138.1897 + 79.1877 + 79.1877 + 31.7175 + 31.7175 135.0000 + 116.5651 + 45.0000 + 31.7175 + 31.7175 135.0000 + 110.9052 + 45.0000 + 37.3774 + 31.7175 135.0000 + 90.0000 + 45.0000 + 45.0000 + 45.0000 135.0000 + 79.1877 + 63.4349 + 45.0000 + 37.3774 135.0000 + 70.5288 + 54.7356 + 54.7356 + 45.0000 135.0000 + 60.0000 + 60.0000 + 60.0000 + 45.0000 125.2644 + 116.5651 + 54.7356 + 31.7175 + 31.7175 125.2644 + 110.9052 + 54.7356 + 37.3774 + 31.7175 125.2644 + 90.0000 + 54.7356 + 45.0000 + 45.0000 125.2644 + 79.1877 + 63.4349 + 54.7356 + 37.3774 125.2644 + 70.5288 + 54.7356 + 54.7356 + 54.7356 125.2644 + 60.0000 + 60.0000 + 60.0000 + 54.7356 121.7175 + 116.5651 + 45.0000 + 45.0000 + 31.7175 121.7175 + 110.9052 + 45.0000 + 45.0000 + 37.3774 121.7175 + 90.0000 + 79.1877 + 37.3774 + 31.7175 120.0000 + 116.5651 + 60.0000 + 31.7175 + 31.7175 120.0000 + 110.9052 + 60.0000 + 37.3774 + 31.7175 120.0000 + 90.0000 + 60.0000 + 45.0000 + 45.0000 120.0000 + 79.1877 + 63.4349 + 60.0000 + 37.3774 120.0000 + 70.5288 + 60.0000 + 54.7356 + 54.7356 120.0000 + 60.0000 + 60.0000 + 60.0000 + 60.0000 116.5651 + 116.5651 + 63.4349 + 31.7175 + 31.7175 116.5651 + 110.9052 + 63.4349 + 37.3774 + 31.7175 116.5651 + 109.4712 + 70.5288 + 31.7175 + 31.7175 116.5651 + 100.8123 + 79.1877 + 31.7175 + 31.7175 116.5651 + 90.0000 + 90.0000 + 31.7175 + 31.7175 116.5651 + 90.0000 + 63.4349 + 45.0000 + 45.0000 116.5651 + 79.1877 + 63.4349 + 63.4349 + 37.3774 116.5651 + 70.5288 + 63.4349 + 54.7356 + 54.7356 116.5651 + 63.4349 + 60.0000 + 60.0000 + 60.0000 110.9052 + 110.9052 + 63.4349 + 37.3774 + 37.3774 110.9052 + 109.4712 + 70.5288 + 37.3774 + 31.7175 110.9052 + 100.8123 + 79.1877 + 37.3774 + 31.7175 110.9052 + 90.0000 + 90.0000 + 37.3774 + 31.7175 109.4712 + 90.0000 + 70.5288 + 45.0000 + 45.0000 109.4712 + 79.1877 + 70.5288 + 63.4349 + 37.3774 109.4712 + 70.5288 + 70.5288 + 54.7356 + 54.7356 109.4712 + 70.5288 + 60.0000 + 60.0000 + 60.0000 100.8123 + 90.0000 + 79.1877 + 45.0000 + 45.0000 100.8123 + 79.1877 + 79.1877 + 63.4349 + 37.3774 100.8123 + 79.1877 + 70.5288 + 54.7356 + 54.7356 100.8123 + 79.1877 + 60.0000 + 60.0000 + 60.0000 90.0000 + 90.0000 + 90.0000 + 45.0000 + 45.0000 90.0000 + 90.0000 + 79.1877 + 63.4349 + 37.3774 90.0000 + 90.0000 + 70.5288 + 54.7356 + 54.7356 90.0000 + 90.0000 + 60.0000 + 60.0000 + 60.0000

Code: Select all: 148.2825 + 79.1877 + 37.3774 + 31.7175 + 31.7175 + 31.7175 148.2825 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 31.7175 142.6226 + 79.1877 + 37.3774 + 37.3774 + 31.7175 + 31.7175 142.6226 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 135.0000 + 79.1877 + 45.0000 + 37.3774 + 31.7175 + 31.7175 135.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 125.2644 + 79.1877 + 54.7356 + 37.3774 + 31.7175 + 31.7175 125.2644 + 54.7356 + 45.0000 + 45.0000 + 45.0000 + 45.0000 121.7175 + 79.1877 + 45.0000 + 45.0000 + 37.3774 + 31.7175 120.0000 + 79.1877 + 60.0000 + 37.3774 + 31.7175 + 31.7175 120.0000 + 60.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 116.5651 + 116.5651 + 31.7175 + 31.7175 + 31.7175 + 31.7175 116.5651 + 110.9052 + 37.3774 + 31.7175 + 31.7175 + 31.7175 116.5651 + 90.0000 + 45.0000 + 45.0000 + 31.7175 + 31.7175 116.5651 + 79.1877 + 63.4349 + 37.3774 + 31.7175 + 31.7175 116.5651 + 70.5288 + 54.7356 + 54.7356 + 31.7175 + 31.7175 116.5651 + 63.4349 + 45.0000 + 45.0000 + 45.0000 + 45.0000 116.5651 + 60.0000 + 60.0000 + 60.0000 + 31.7175 + 31.7175 110.9052 + 110.9052 + 37.3774 + 37.3774 + 31.7175 + 31.7175 110.9052 + 90.0000 + 45.0000 + 45.0000 + 37.3774 + 31.7175 110.9052 + 79.1877 + 63.4349 + 37.3774 + 37.3774 + 31.7175 110.9052 + 70.5288 + 54.7356 + 54.7356 + 37.3774 + 31.7175 110.9052 + 60.0000 + 60.0000 + 60.0000 + 37.3774 + 31.7175 109.4712 + 79.1877 + 70.5288 + 37.3774 + 31.7175 + 31.7175 109.4712 + 70.5288 + 45.0000 + 45.0000 + 45.0000 + 45.0000 100.8123 + 79.1877 + 79.1877 + 37.3774 + 31.7175 + 31.7175 100.8123 + 79.1877 + 45.0000 + 45.0000 + 45.0000 + 45.0000 90.0000 + 90.0000 + 79.1877 + 37.3774 + 31.7175 + 31.7175 90.0000 + 90.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 90.0000 + 79.1877 + 63.4349 + 45.0000 + 45.0000 + 37.3774 90.0000 + 70.5288 + 54.7356 + 54.7356 + 45.0000 + 45.0000 90.0000 + 60.0000 + 60.0000 + 60.0000 + 45.0000 + 45.0000 79.1877 + 79.1877 + 63.4349 + 63.4349 + 37.3774 + 37.3774 79.1877 + 70.5288 + 63.4349 + 54.7356 + 54.7356 + 37.3774 79.1877 + 63.4349 + 60.0000 + 60.0000 + 60.0000 + 37.3774 70.5288 + 70.5288 + 54.7356 + 54.7356 + 54.7356 + 54.7356 70.5288 + 60.0000 + 60.0000 + 60.0000 + 54.7356 + 54.7356 60.0000 + 60.0000 + 60.0000 + 60.0000 + 60.0000 + 60.0000

Code: Select all: 116.5651 + 79.1877 + 37.3774 + 31.7175 + 31.7175 + 31.7175 + 31.7175 116.5651 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 31.7175 + 31.7175 110.9052 + 79.1877 + 37.3774 + 37.3774 + 31.7175 + 31.7175 + 31.7175 110.9052 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 + 31.7175 90.0000 + 79.1877 + 45.0000 + 45.0000 + 37.3774 + 31.7175 + 31.7175 90.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 79.1877 + 79.1877 + 63.4349 + 37.3774 + 37.3774 + 31.7175 + 31.7175 79.1877 + 70.5288 + 54.7356 + 54.7356 + 37.3774 + 31.7175 + 31.7175 79.1877 + 63.4349 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 79.1877 + 60.0000 + 60.0000 + 60.0000 + 37.3774 + 31.7175 + 31.7175 70.5288 + 54.7356 + 54.7356 + 45.0000 + 45.0000 + 45.0000 + 45.0000 60.0000 + 60.0000 + 60.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000

Code: Select all: 79.1877 + 79.1877 + 37.3774 + 37.3774 + 31.7175 + 31.7175 + 31.7175 + 31.7175 79.1877 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 37.3774 + 31.7175 + 31.7175 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000 + 45.0000

Notice that the angles from most antiprisms, and J84-90 (except obviously 109.4712° from J87, but that's non-elementary anyway), and snic and snid, don't appear anywhere in this list. (If you don't want to search this long list looking for a single angle, you may run the program again with that angle as the initial value. You'll get no results.)

The pentagonal prism's angle 108° appears only in the sum 144° + 108° + 108°. Therefore, a pentagonal prism must be surrounded by alternating pentagonal and decagonal prisms; but this is impossible since 5 is odd. So 'pip' can't be used in a honeycomb.

And the decagonal prism's angle 144° also appears only in 144° + 108° + 108°, so 'dip' can't be used either.

tet, cube, doe
tut, tic, toe, girco, ~~snic~~, tid, ti, grid, ~~snid~~
trip, ~~pip~~, hip, op, ~~dip~~, twip, ~~squap~~, pap, ~~hap, oap, dap~~
J1-6, 63, 80, 83, ~~84-86, 88-90~~, 91-92

by **mr_e_man** » Mon Aug 16, 2021 9:14 pm

tet, cube, doe
tut, tic, toe, girco, tid, ti, grid
trip, hip, op, twip, pap
J1-6, 63, 80, 83, 91-92

Let's try to eliminate J92 now. The remaining polyhedra with hexagons, along with the relevant verfs, are:

tut: {6} 70.5288° {6} 109.4712° {3} 109.4712°
toe: {6} 109.4712° {6} 125.2644° {4} 125.2644°
girco: {8} 125.2644° {6} 144.7356° {4} 135°
ti: {6} 138.1897° {6} 142.6226° {5} 142.6226°
grid: {10} 142.6226° {6} 159.0948° {4} 148.2825°
hip: {6} 90° {4} 120° {4} 90°
J3: {6} 54.7356° {4} 125.2644° {3} 70.5288°
J92: {6} 110.9052° {4} 159.0948° {3} 138.1897° {3} 138.1897°

Let 'X' be the 6.4 edge, 'Y' the 4.3 edge, 'Z' the 3.3 edge, and 'W' the 6.3 edge, all at a single vertex in J92.

If the hexagon in J92 attaches to a hexagon in tut, toe, girco, hip, or J3, then the angle sum at W is one of the following:

138.1897° + 54.7356°
138.1897° + 70.5288°
138.1897° + 90°
138.1897° + 109.4712°
138.1897° + 125.2644°
138.1897° + 144.7356°

But none of these is valid as part of a 360° sum. (Again, if you don't want to search the list, you may run the program with that partial sum as the initial value.)

If the hexagon in J92 attaches to a hexagon in ti, then the angle sum at W is either 2*138.1897°, or 138.1897° + 142.6226°. But the former is invalid, and the latter combines only with 79.1877° (from J6), which would give an edge configuration 3.6 + 6.5 + 10.3; we can't have a pentagon attaching to a decagon.

If the hexagon in J92 attaches to a hexagon in grid, then the angle sum at W is either 138.1897° + 159.0948°, or 138.1897° + 142.6226°. The former is invalid. The latter combines only with 79.1877°, which gives an edge configuration 3.6 + 6.10 + 10.3 . Now, since J6 has a 142.6226° angle, the adjacent edge Z also gets 138.1897° + 142.6226°, and thus another 79.1877°, but this time the edge configuration is 3.3 + 3.5 + 10.3 which doesn't work.

Code: Select all: | edges | faces | | X,Y,Z,W | 6,4,3,3, 10,5 | J92,grid,J6 +---------+---------------+------------- | * | 1 1 0 0 0 0 | 1 1 0 | * | 0 1 1 0 0 0 | 1 0 0 | * | 0 0 1 1 0 1 | 1 0 1 | * | 1 0 0 1 1 0 | 1 1 1 +---------+---------------+------------- | 1 0 0 1 | * | 1 1 0 | 1 1 0 0 | * | 1 0 0 | 0 1 1 0 | * | 1 0 0 | 0 0 1 1 | * | 1 0 1 | 0 0 0 1 | * | 0 1 1 | 0 0 1 0 | * | 0 0 1 +---------+---------------+------------- | 1 1 1 1 | 1 1 1 1 0 0 | * | 1 0 0 1 | 1 0 0 0 1 0 | * | 0 0 1 1 | 0 0 0 1 1 1 | *

If the hexagon in J92 attaches to another J92, then the angle sum at W is either 2*138.1897° (in the ortho case), or 138.1897° + 110.9052° (in the gyro case). The former is invalid. The latter appears in the following combinations:

(1): 138.1897° + 110.9052° + 110.9052°
(2): 138.1897° + 110.9052° + 79.1877° + 31.7175°

In case (1) the edge configuration is 3.6 + 6.4 + 4.3, where the 4.3 is from the verf 5.3.4.3 . In case (2) the edge configuration is 3.6 + 6.4 + 4.10 + 10.3, where the 10.3 is from the verf 10.5.3 . In either case, the adjacent edge Z gets 138.1897° + 142.6226°, and thus 79.1877°; again the edge configuration at Z is 3.3 + 3.5 + 10.3 which doesn't work.

With J92 gone, we can also eliminate 5.4.3.4.

tet, cube, doe
tut, tic, toe, girco, tid, ti, grid
trip, hip, op, twip, pap
J1-4, 5, 6, 63, ~~80, 83~~, 91, 92

by **mr_e_man** » Mon Aug 16, 2021 9:17 pm

tet, cube, doe
tut, tic, toe, girco, tid, ti, grid
trip, hip, op, twip, pap
J1-4, 6, 63, 91

Next are grid and ti, now that we have hexagons in mind.

The 6.4 angle in grid is 159.0948°, which appears in 6 sums (from the larger list of angles), as we've seen in the case of 5.4.3.4:

(1): 159.0948° + 169.1877° + 31.7175°
(2): 159.0948° + 121.7175° + 79.1877°
(3): 159.0948° + 110.9052° + 90°
(4): 159.0948° + 110.9052° + 45° + 45°
(5): 159.0948° + 90° + 79.1877° + 31.7175°
(6): 159.0948° + 79.1877° + 45° + 45° + 31.7175°

But 169.1877°, 121.7175°, and 31.7175° only appear in non-elementary polyhedra and in J5,80,83 which have been ruled out. So we're left with cases (3) and (4). Note that 110.9052° must be in the verf 5.3.4.3, since J92 has also been ruled out.

Case (3): Since there are only three polyhedra around the edge, any two must share a face. 110.9052° is between a triangle and a square, so the partial configuration for 159.0948° + 110.9052° is 6.4 + 4.3; this would imply that the third polyhedron has faces 3.6 with 90° between them. But there is no such CRF polyhedron.

Case (4): The faces are 6.4, 4.3, 8.4, and 8.4; these don't fit together.

The 6.6 angle in ti is 138.1897°, which appears in 5 sums (from the smaller list), some of which are ruled out along with J5:

(1): 138.1897° + 142.6226° + 79.1877°
(2): 138.1897° + 110.9052° + 110.9052°
~~(3): 138.1897° + 110.9052° + 79.1877° + 31.7175°~~
(4): 138.1897° + 79.1877° + 79.1877° + 63.4349°
~~(5): 138.1897° + 79.1877° + 79.1877° + 31.7175° + 31.7175°~~

Case (1): Since there are only three polyhedra around the edge, any two must share a face. But 79.1877° doesn't involve hexagons.

Case (2): 110.9052° doesn't involve hexagons, since J92 has been eliminated.

Case (4): 63.4349° doesn't involve hexagons either.

tet, cube, doe
tut, tic, toe, girco, tid, ~~ti, grid~~
trip, hip, op, twip, pap
J1-4, 6, 63, 91

by **mr_e_man** » Mon Aug 16, 2021 9:29 pm

tet, cube, doe
tut, tic, toe, girco, tid
trip, hip, op, twip, pap
J1-4, 6, 63, 91

Now we'll consider the 5.3.3.3 vertex, and 3.3.3.3.3 A at the same time.

Let 'X' and 'Y' be two adjacent 3.3 edges (with dihedral angle 138.1897°). Let 'Z' be the edge adjacent to Y, and 'W' the edge adjacent to X.

We just saw that 138.1897° appears in these combinations:

(1): 138.1897° + 142.6226° + 79.1877°
(2): 138.1897° + 110.9052° + 110.9052°
(4): 138.1897° + 79.1877° + 79.1877° + 63.4349°

Case (4): The faces are 3.3, 10.3, 10.3, and either 10.5 or 5.5; so the pentagon has nothing to attach to.

Case (2): The configuration at X is 3.3 + 3.4 + 4.3, where both appearances of 4.3 are from 5.3.4.3 . So the adjacent edge Y gets 138.1897° + 142.6226°, and thus 79.1877°; but then the configuration at Y is 3.3 + 3.5 + 10.3 which doesn't work.

Code: Select all: | edges | faces | | W,X,Y,Z | 5,3,3,3, 4,5 | (5.3.3.3), (5.3.4.3) +---------+--------------+---------------------- | * | 1 1 0 0 0 0 | 1 0 | * | 0 1 1 0 1 0 | 1 1 | * | 0 0 1 1 0 1 | 1 1 | * | 1 0 0 1 0 0 | 1 0 +---------+--------------+---------------------- | 1 0 0 1 | * | 1 0 | 1 1 0 0 | * | 1 0 | 0 1 1 0 | * | 1 1 | 0 0 1 1 | * | 1 0 | 0 1 0 0 | * | 0 1 | 0 0 1 0 | * | 0 1 +---------+--------------+---------------------- | 1 1 1 1 | 1 1 1 1 0 0 | * | 0 1 1 0 | 0 0 1 0 1 1 | *

Case (1): Since 79.1877° involves a decagon, and 138.1897° does not, 142.6226° must also involve a decagon. The only possibility is from the verf 10.10.3 (note that 10.6.4 has been ruled out along with 'grid', and 10.3.4.3 is from non-elementary polyhedra). So the configuration at X is 3.3 + 3.10 + 10.3, and similarly at Y, since X and Y are congruent. This puts a 'tid' on the triangle between X and Y, and a J6 on the triangle between X and W.

Suppose the verf is 3.3.3.3.3 A, so all edges are congruent. Since X and W are two adjacent 3.3 edges, the above logic implies that there should be a tid on the triangle between X and W. But this contradicts J6 being there. Here's another way to look at it: There would have to be tid and J6 alternating around the vertex, which is impossible because the vertex is 5-valent, and 5 is odd.

That rules out J2. Now suppose the verf is 5.3.3.3 . The edge W gets the partial configuration 5.3 + 3.5 and angle sum 100.8123° + 142.6226°, which appears in these 360° sums:

100.8123° + 142.6226° + 116.5651°
100.8123° + 142.6226° + 79.1877° + 37.3774°

But 37.3774° only appeared in J2 and J5, so we can't use the second sum. The edge has two pentagons exposed, so the 116.5651° angle must be from the verf 5.5.5 .

Thus, there is only one possible honeycomb verf containing the polyhedron verf 5.3.3.3 :

Code: Select all: | edges | faces | | W,X,Y,Z,T | 5,3,3,3, 10, 5 | (5.3.3.3), (10.10.3), (10.5.3), (5.5.5) +-----------+------------------+----------------------------------------- | * | 1 1 0 0 0 0 1 0 | 1 0 1 0 1 | * | 0 1 1 0 1 0 0 0 | 1 1 1 0 0 | * | 0 0 1 1 0 1 0 0 | 1 1 0 1 0 | * | 1 0 0 1 0 0 0 1 | 1 0 0 1 1 | * | 0 0 0 0 1 1 1 1 | 0 1 1 1 1 +-----------+------------------+----------------------------------------- | 1 0 0 1 0 | * | 1 0 0 0 1 | 1 1 0 0 0 | * | 1 0 1 0 0 | 0 1 1 0 0 | * | 1 1 0 0 0 | 0 0 1 1 0 | * | 1 0 0 1 0 | 0 1 0 0 1 | * | 0 1 1 0 0 | 0 0 1 0 1 | * | 0 1 0 1 0 | 1 0 0 0 1 | * | 0 0 1 0 1 | 0 0 0 1 1 | * | 0 0 0 1 1 +-----------+------------------+----------------------------------------- | 1 1 1 1 0 | 1 1 1 1 0 0 0 0 | * | 0 1 1 0 1 | 0 0 1 0 1 1 0 0 | * | 1 1 0 0 1 | 0 1 0 0 1 0 1 0 | * | 0 0 1 1 1 | 0 0 0 1 0 1 0 1 | * | 1 0 0 1 1 | 1 0 0 0 0 0 1 1 | *

(Now I notice why Klitzing puts the labels on the side of the incmat. On the top, it takes too much space.)

This rules out the pentagonal antiprism: Obviously all vertices are 5.3.3.3 . Consider three vertices A,B,C, forming a triangular face ABC, with A,B on the bottom and C on the top of the antiprism. The vertex A requires ABC to attach to J6, while the vertex C requires ABC to attach to tid, a contradiction.

tet, cube, doe
tut, tic, toe, girco, tid
trip, hip, op, twip, ~~pap~~
J1, 2, 3-4, 6, 63, 91

by **mr_e_man** » Mon Aug 16, 2021 9:43 pm

tet, cube, doe
tut, tic, toe, girco, tid
trip, hip, op, twip
J1, 3-4, 6, 63, 91

And now we can return to the tid-teddi combination.

If we have a single J63, we see from the case of 5.3.3.3 that it must be attached to a 'tid', three J6, and three 'doe'. (I don't want to write "does" because that has another meaning, as in "do", "did", "done", "does". Similarly, "J6s" doesn't look right; it could be misinterpreted as "J6-snub" or "J(6 times s)" or something else.)

Suppose we have a single tid. The only remaining polyhedra with decagons are tid itself and J6.

If a decagon in tid attaches to another tid in ortho orientation, then one edge gets the partial configuration 3.10 + 10.3 and dihedral angle sum 2*142.6226°, which appears in only one 360° sum:

2*142.6226° + 2*37.3774°

But 37.3774° was from J2 and J5 which have been ruled out.

If a decagon in tid attaches to another tid in gyro orientation, then one edge gets the partial configuration 3.10 + 10.10 and dihedral angle sum 142.6226° + 116.5651°, which appears in these 360° sums, two of which are ruled out along with J2 and J5:

142.6226° + 116.5651° + 100.8123°
~~142.6226° + 116.5651° + 63.4349° + 37.3774°~~
~~142.6226° + 116.5651° + 37.3774° + 31.7175° + 31.7175°~~

And 100.8123° doesn't involve decagons. So the edge can't be completed.

If a decagon in tid attaches to J6 in gyro orientation, then one edge gets the partial configuration 3.10 + 10.5 . In this case the angle sum is 142.6226° + 63.4349°, which appears in

~~142.6226° + 63.4349° + 116.5651° + 37.3774°~~
~~142.6226° + 63.4349° + 79.1877° + 37.3774° + 37.3774°~~

So any decagon in tid must attach to J6 in ortho orientation: One type of edge gets 3.10 + 10.3, with angle sum 142.6226° + 79.1877°; and the other type of edge gets 10.10 + 10.5, with angle sum 116.5651° + 63.4349° = 180°. The former appears in the following 360° sums:

142.6226° + 79.1877° + 138.1897°
~~142.6226° + 79.1877° + 100.8123° + 37.3774°~~
~~142.6226° + 79.1877° + 63.4349° + 37.3774° + 37.3774°~~
~~142.6226° + 79.1877° + 37.3774° + 37.3774° + 31.7175° + 31.7175°~~

The only remaining polyhedron with 138.1897° is J63.

Thus, a tid in a honeycomb must be surrounded by 12 J6 (one for each decagon) and 20 J63 (one for each triangle). Since each J63 must attach to 3 doe, the tid must be surrounded by 30 doe, as shown in the first image here: viewtopic.php?p=27628#p27628

Now each J6 has one pentagon exposed. The remaining polyhedra with pentagons are doe, J6, 63, 91.

If one of those twelve J6 pentagons attaches to another J6, then the angle sum is 2*142.6226°; we saw in the case of 'ortho tid' that that doesn't work. If one of those pentagons attaches to J91, then at least one edge gets 2*142.6226° again. And the pentagon in J6 can't attach to J63, because J63 needs to attach to 3 doe. Therefore, each of those twelve J6 pentagons must attach to a doe. (This is light green in the second image.)

There's a J6 triangle surrounded by 3 doe (2 dark green and 1 light green). The dihedral angle sum, at any edge of the triangle, is 142.6226° + 116.5651°; we see from the case of 'gyro tid' that the remaining 100.8123° must be from a single polyhedron. Obviously it must involve pentagons. It can't be doe or J6, as they don't have that angle anywhere. It can't be J91; it fits at two of the triangle's edges, but the third edge gets 159.0948° ≠ 100.8123°. So it must be J63. (This is orange in the second image.) And its orientation is determined, as 138.1897° ≠ 100.8123°.

In summary, if there's a J63 or a tid anywhere in a honeycomb, the only possibility is that shown in the images:

tid + 12 J6 + 20 J63 + 30 doe + 12 doe + 12*5 J63

But this leads to a contradiction, as the former 20 and those latter 60 are not equivalent. I've described this before: viewtopic.php?p=28018#p28018

I'll try again. In the second image, the orange teddi in the middle requires its own tid, and 3 J6; the 3 doe are already there. One of those new J6 intrudes into the space of the orange teddi on the left, giving one of its edges a configuration 3.3 + 3.5 and angle sum 138.1897° + 142.6226°; the remaining angle 79.1877° must be from another J6, but then the completed edge configuration would be 3.3 + 3.5 + 10.3 which doesn't work.

: tidTeddi3.png (44.13 KiB) Viewed 177724 times

Thus tid and J63 are ruled out.

The only remaining polyhedron with decagons is J6. Here I can almost repeat two previous paragraphs: If one of the six pentagons attaches to another J6, then the angle sum is 2*142.6226°; we saw in the case of 'ortho tid' that that doesn't work. If one of the pentagons attaches to J91, then at least one edge gets 2*142.6226° again. And a pentagon in J6 can't attach to J63 which has been ruled out. Therefore, each of those six pentagons must attach to a doe.

There's a J6 triangle surrounded by 3 doe. The dihedral angle sum, at any edge of the triangle, is 142.6226° + 116.5651°; we see from the case of 'gyro tid' that the remaining 100.8123° must be from a single polyhedron. Obviously it must involve pentagons. It can't be doe or J6, as they don't have that angle anywhere. It can't be J91; it fits at two of the triangle's edges, but the third edge gets 159.0948° ≠ 100.8123°. And it can't be J63 which has been ruled out.

tet, cube, doe
tut, tic, toe, girco, ~~tid~~
trip, hip, op, twip
J1, 3-4, ~~6, 63~~, 91

by **mr_e_man** » Mon Aug 16, 2021 9:47 pm

tet, cube, doe
tut, tic, toe, girco
trip, hip, op, twip
J1, 3-4, 91

Only doe and J91 remain with pentagonal faces. Here are the relevant verfs:

{5} 116.5651° {5} 116.5651° {5} 116.5651°

{5} 63.4349° {5} 100.8123° {3} 100.8123°
{5} 142.6226° {3} 142.6226° {5} 142.6226° {3} 142.6226°
{5} 142.6226° {3} 110.9052° {4} 159.0948° {3} 100.8123°

Suppose doe attaches to another doe. 2*116.5651° appears in these 360° sums:

2*116.5651° + 2*63.4349°
~~2*116.5651° + 63.4349° + 2*31.7175°~~
~~2*116.5651° + 4*31.7175°~~

That 63.4349° angle appears in J91 (and none of the other remaining polyhedra) at a 5.5 "wedge". So we have 2 doe and 2 J91 around an edge; but this gives another edge a configuration 3.5 + 5.3 and angle sum 2*142.6226°, which appears only in 2*142.6226° + 2*37.3774°. Remember 37.3774° was from J2 and J5 which have been ruled out.

Suppose a pentagon in J91 attaches to another J91. There are 3 possible configurations, depending on the relative orientations of their wedge edges: These edges coincide, or they're 72° apart, or they're 144° apart. (The dihedral angles around the pentagon in J91 are 63.4349°, 100.8123°, 142.6226°, 142.6226°, 100.8123°; there are 3 ways to fit two of these together.) But in the first two cases, at least one other edge gets 2*142.6226° and thus can't be completed. In the third case, one edge gets 2*100.8123° which appears only in 2*100.8123° + 2*79.1877°; that angle was from J6 which has been ruled out.

Therefore, a pentagon in J91 must attach to a doe, and vice versa.

Starting with a single J91, we find that we're forced to build the cube-doe-bilbiro honeycomb. The 4 pentagons in J91 attach to 4 doe's. A pair of doe's meeting at a wedge edge of J91 leave a 63.4349° gap which must be filled by another J91 wedge. A pair of doe's meeting at a 5.3.5.3 vertex of J91, and thus meeting at an edge extending from that vertex, leave two 63.4349° gaps which must be filled by two J91 wedges. Continuing in this way, we get the entire structure of doe's and bilbiros; and obviously the cube-shaped gaps must be filled by cubes.

Hi.gher. Space

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