3D CRF honeycombs

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

Re: 3D CRF honeycombs

Postby mr_e_man » Mon Aug 16, 2021 9:58 pm

Finally, let's try to prove that any honeycomb involving twip must be an infinite stack of copies of a single layer which is the prism of a 2D tiling.

Since twap is ruled out, a single twip must be part of an infinite stack of twips. We saw in 'Two prisms' that a twip must be surrounded by other prisms (with parallel axes); so at least any prism directly touching the twip is also part of an infinite stack of prisms. A vertical edge of the twip has a configuration corresponding to one in 2D:

12.12.3
12.6.4
12.4.3.3
12.3.4.3

A 12-gon or 6-gon can be converted to 3-gons and 4-gons (as described in the first post here), without changing the validity of any 2D tiling it's a part of. Similarly, in 3D, an infinite stack of twips or hips can be converted to a stack of trips and cubes, without changing the validity of any honeycomb it's a part of. Let's do so, whenever possible.

twipHoneycombs1.png
twipHoneycombs1.png (26.6 KiB) Viewed 189 times

In 2D, the 12-gon produces 4.3.4.3.3 vertices inside of it (and one 3.3.3.3.3.3 vertex), and incomplete 4.3 vertices at the boundary.

Suppose there's an infinite stack of cubes attached to an infinite stack of trips (with both stacks extending in the same direction, of course). Notice that the analysis of 'Two prisms' was mostly local, considering only a neighbourhood of a vertex. It doesn't matter whether there are other trips and cubes forming a complete twip; all that matters is this configuration {4} 180° {4} 150° {4} 180° {4} 150°. It was shown in 'Two prisms' that these squares must attach to other prisms with parallel axes. Therefore, the two stacks of trips and cubes must attach to other stacks of prisms (except possibly on the far side of the cube, opposite the trip).

Suppose there are three infinite stacks of cubes, corresponding to the 2D configuration 4.4.4 . For the partial sum 3*90°, the program gives us 4*90°, and 3*90° + 2*45°. The latter sum can't be used, as J4 would give an adjacent edge 2*90° + 144.7356° which can't be completed. The former sum can only mean another cube (since the other 90° dihedral angles in elementary polyhedra are not 4.4 as required). So we get a fourth infinite stack of cubes, corresponding to the 2D configuration 4.4.4.4 .

Suppose there are four infinite stacks of trips, corresponding to the 2D configuration 3.3.3.3 . For the partial sum 4*60°, the program gives us 6*60°, and 4*60° + 120°. The dihedral angle 60° appears only in trip, and 120° appears only in hip. And a triangle in trip (in one layer of the honeycomb) can't attach to a hexagon in hip (in the next layer). So we get either two more infinite stacks of trips, corresponding to the 2D configuration 3.3.3.3.3.3, or an infinite stack of hips, corresponding to 3.3.3.3.6 (but then the hips decompose into trips anyway).

twipHoneycombs2.png
twipHoneycombs2.png (27.8 KiB) Viewed 189 times


Now we only need to prove the following: If we're fitting squares and triangles together in 2D, starting with a 4.3.4.3.3 vertex, completing any vertices including the partial configuration 4.3, or 4.4.4 or 3.3.3.3, and never completing a vertex with just 4.4 or 3.3 or 3.3.3, then the entire plane will be covered. And this should be true regardless of what choices are made in completing vertices (that is, whether any 4.3 becomes 4.3.4.3.3 or 4.3.3.4.3 or 4.3.3.3.4).

If we can prove that, then the statement about twip honeycombs will follow. Each time we complete a vertex in 2D, the corresponding honeycomb gets extended, necessarily with infinite stacks of prisms. When the triangles and squares cover the plane, the corresponding stacks of trips and cubes fill the entire space. When the triangles and squares overlap, invalidating the tiling, the corresponding honeycomb is also invalid. When there's a 30° gap in the tiling, there's also a 30° gap in the honeycomb, which can't be filled by any CRF dihedral angle. I make no claims about a twip honeycomb that's incomplete and cannot be completed. My claim is that any complete honeycomb containing twip must be made of infinite stacks of prisms with parallel axes.
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Re: 3D CRF honeycombs

Postby mr_e_man » Mon Aug 16, 2021 10:08 pm

If we're fitting squares and triangles together in 2D, starting with a 4.3.4.3.3 vertex, completing any vertices including the partial configuration 4.3, or 4.4.4 or 3.3.3.3, and never completing a vertex with just 4.4 or 3.3 or 3.3.3, then the entire plane will be covered.


Here are some examples to show the necessity of some of those conditions. If we start not with 4.3.4.3.3 but with 4.4.4.4, then the only incomplete vertices have 4.4 (or just 4), so we can't do anything, and the plane is not covered. And similarly if we start with 3.3.3.3.3.3 . If we start with 4.4.3.3.3 and complete each 4.3 into 4.3.3.3.4, then we get a strip of triangles and squares, with 3.3.3 vertices along one side of the strip, and 4.4 along the other side; so we can't do anything more, and the plane is not covered. If we start with 4.3.4.3.3 and complete each 4.3 into 4.3.3.3.4, but don't complete any 4.4.4 into 4.4.4.4, then we get an empty 90° sector of the plane, bounded by squares (4.4 and 4.4.4).

In fact we might want another condition, to guarantee that the result is a valid tiling, with no 30° gaps or overlapping tiles. But I don't know what that would be. Perhaps, require all 4.4.4 and 3.3.3.3 vertices (and 3.3.3.3.3 which includes 3.3.3.3) to be completed before those with 4.3; or require concave vertices to be completed before convex ones.


Here's my attempt at a proof by contradiction. Suppose that there are no more incomplete vertices including 4.3 or 4.4.4 or 3.3.3.3, and the plane is not covered. There must be some incomplete vertices (otherwise the plane is covered). Since their configurations don't include 4.3, any such vertex must have only triangles, or only squares.

An incomplete vertex with squares would have to be 4.4 or just 4 (a corner of a square with both edges not attaching to other tiles), and the adjacent incomplete vertices would also have squares. An incomplete vertex with triangles would have to be 3.3.3 or 3.3 or just 3, and the adjacent incomplete vertices would also have triangles.

If there's only 4.4, then we have (at least) an infinite strip of squares. But, on either side of the strip, the (incomplete) tiling can be extended only with another infinite strip of squares or triangles; it may contain 4.4.4.4, 3.3.3.3.3.3, and 4.4.3.3.3, but never 4.3.4.3.3 .

If there's (any number of) 4.4 and one 4, then we have two half-infinite (extending in only one direction, not the opposite direction) strips of squares, meeting at 90°, and the rest of the tiling is inside of this 90° sector. But the inside must be filled by squares (a triangle would leave a 30° gap somewhere), so it never contains 4.3.4.3.3 .

If there's 4.4 and two 4, then we have two half-infinite strips of squares, meeting a finite strip at 90°. (This could be described by, for example, x>0 and 0<y<5.) But again the inside must be filled by squares, so it never contains 4.3.4.3.3 .

4.4 and three 4 is not possible. If there's 4.4 and four 4, then we have a finite rectangle made of squares; this doesn't contain 4.3.4.3.3 .

If there's 4 and no 4.4, then we have just a single square; none of its vertices are 4.3.4.3.3 .

And similarly for triangles: If there's only 3.3.3, then we have an infinite strip of triangles, which can be extended only with more infinite strips, and thus never contains 4.3.4.3.3 . If there's 3.3.3 and some number of 3.3 and 3, then we have a region bounded by strips of triangles meeting at 60° or 120°; such a region must be filled by triangles, as a square would leave a 30° gap somewhere. So it doesn't contain 4.3.4.3.3 .

twipHoneycombs3.png
twipHoneycombs3.png (22.65 KiB) Viewed 188 times


And here's an idea for a more direct proof.

Starting with 4.3.4.3.3, we have an irregular heptagon, with angles 90°, 150°, 150°, 90°, 150°, 120°, 150°. The vertices with 150° angles have configuration 4.3, so they may be completed in any way, thus forming a concave polygon. But any concave vertex can be completed; the only excepted vertices have 180°, 120°, 90°, 60°. In fact, 60° will never appear; a triangle must attach to something on at least two of its sides, because it was made by completing a vertex.

We need to show that, as we build up the polygon by adding triangles and squares around concave vertices and 150° vertices, any point in the plane will be reached eventually. Equivalently, we need to show that any edge of the polygon will be expanded outward eventually. This will follow, if any edge (ignoring 180° vertices, thus combining collinear adjacent edges) always has at least one concave or 150° vertex.

The starting heptagon does have that property: Each edge has a 150° vertex. So, for induction, suppose we have a polygon, made of triangles and squares, with no 60° vertices, where each edge has a concave or 150° vertex. We need to show that a new polygon, formed by completing one of those vertices, also has that property.

There are many cases to consider here. The vertex being completed could have the configuration 4.4.4, 3.3.3.3, 3.3.3.3.3, 4.3, 4.4.3, 4.3.4, 4.3.3, 3.4.3, 4.4.3.3, 4.3.4.3, 4.3.3.4, 4.3.3.3, or 3.4.3.3 (summarized by the angles 150°, 210°, 240°, 270°, 300°); and the two adjacent vertices could also have various configurations. But they're all quite similar and easy (I've checked all of them), and it may be possible to reduce the number of cases somehow, or check several at once.

Below I've illustrated two cases, along with the starting heptagon. In the first case, 4.3 is being completed to 4.3.3.3.4 . The old edge on the left has one 120° vertex; by the induction hypothesis, its other vertex (not shown) must be concave or 150°; this implies that the new edge on the left (with length ≥2) has that same vertex which is concave or 150°. The edge on the right doesn't change, but its 150° vertex changes to 240° which is concave. The two edges in the middle disappear, or we might say they're expanded outward; and the three new edges in the middle each have a 150° or 240° vertex. The second case is simple: The edges just change length, and their vertex types don't change.

twipHoneycombs4.png
twipHoneycombs4.png (14.61 KiB) Viewed 188 times
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Re: 3D CRF honeycombs

Postby mr_e_man » Mon Sep 20, 2021 2:16 am

Klitzing wrote: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

Well, what do you think? Have I resolved all of this now?

Is there any part of the proof that I should clarify?
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Re: 3D CRF honeycombs

Postby Klitzing » Mon Sep 20, 2021 5:03 pm

At least your post http://hi.gher.space/forum/viewtopic.php?p=28084#p28084
finally shows the exceptionality of the cube-doe-bilbiro honeycomb.

--- rk
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