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....
3D CRF honeycombs
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3D CRF honeycombs
by mr_e_man » Tue Jan 12, 2021 2:49 am
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: 3D CRF honeycombs
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.
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.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: 3D CRF honeycombs
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?
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?
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: 3D CRF honeycombs
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 12 times
- tidTeddi2.png (119.22 KiB) Viewed 12 times
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: 3D CRF honeycombs
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
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