mr_e_man wrote:Interesting. After seeing this, I considered "almost-uniform" tilings in the Euclidean plane, and found that no such thing exists; any CRF Euclidean tiling with congruent vertices must be uniform. (You probably already knew this.) What about 3D Euclidean space?
Marek14 wrote:Yesterday, we spent some time considering (3,4,5,5) before concluding that this particular combination cannot be made into tiling, not even an aperiodic one.
mr_e_man wrote:Marek14 wrote:Yesterday, we spent some time considering (3,4,5,5) before concluding that this particular combination cannot be made into tiling, not even an aperiodic one.
I can see that easily: both vertices shared by the triangle and the square must be completed by pairs of pentagons; but then the triangle's third vertex has the form 5.3.5.x, which is not 3.4.5.5, regardless of what x is.
Marek14 wrote:As for what "almost uniform" means: there is more than one vertex type, but there is still a finite number of the types and the tiling is overall periodic.
mr_e_man wrote:Marek14 wrote:As for what "almost uniform" means: there is more than one vertex type, but there is still a finite number of the types and the tiling is overall periodic.
Do you allow several vertex configurations?
Do you allow several sets of faces?
Examples: 4.4.3.4.3 and 4.4.4.3.3 are different vertex configurations with the same set of faces. 3^{8} and 3^{2}.8^{3} are different vertex configurations with different sets of faces.
wendy wrote:John Conway described that while they are described by the vertex figure, the cycle of polygons is not enough to describe it.
What you further need is to describe the outbound and inbound edges, where they fall in the cycle, and if they change parity. This is essentially the orbifold.
The first in Mr_e's diagrams is [1,2] [3,5] [4], where [1,2] is the triangle, and [3,5] is the purple squares and [4] are between the purple triangles.
The second is [1,2] [3,5] (4), where the (4) edge is the centre of an order-2 rotation.
It looks like 3% 2% 2% 2% or (1,2) (3) (4) (5). This is an ordinary snub.
In any case, the first two have orbifold nodes of wanders or miracles, which you tell by the presence of non-consecutive numbers in the edges. The third one is a fairly ordinary snub, with squares (as does Miller's mosnster). A pair of consecutive edges in brackets is a cone, or rotation-polygon.
W
mr_e_man wrote:I thought that uniform tilings were uniquely determined by their vertex configurations. But it looks like there are three different uniform 3.4.4.4.4 tilings!
Marek14 wrote:Well, sure. See here: https://bendwavy.org/klitzing/explain/t ... .htm#aaaab
I've known this for a long time now
Marek14 wrote:I am aware that edge of {3,2n} is generally equal to {2n,n}, and also that edge of (3,4,n,4) is equal to (4,n,2n) (this is the principle behind Johnson solids like diminished rhombicosidodecahedra, I've been able to produce several "diminished" and "gyrated" periodic variants of (3,4,7,4) thanks to that).
mr_e_man wrote:Can anyone explain why, with the edge length of the regular {4,7} tiling, the angle formed by two squares at a vertex is the same as the angle formed by a triangle and a heptagon at a vertex?
2 * 51.428571° = 35.733235° + 67.123908°
I discovered this numerically, and verified it algebraically, but I want to understand it geometrically.
I proved that any CRF hyperbolic tiling with at least one vertex of the form 3.m.n or 4.m.n must be uniform, the only exception being 4.n.2n which can combine with 4.n.4.3 or 4.n.3.4 to make a modified uniform tiling:Marek14 wrote:I am aware that edge of {3,2n} is generally equal to {2n,n}, and also that edge of (3,4,n,4) is equal to (4,n,2n) (this is the principle behind Johnson solids like diminished rhombicosidodecahedra, I've been able to produce several "diminished" and "gyrated" periodic variants of (3,4,7,4) thanks to that).
In the process, looking at a 4.n.2n vertex next to a 4.4.3.n vertex, I needed to consider fitting other polygons instead of two squares on the other side of the 3.n edge. A single polygon's angle is always smaller than that of two squares. Three triangles, and thus any three or more polygons, is always larger than two squares. A square and anything larger, obviously, is larger than two squares. A triangle and an octagon is larger than two squares. A triangle and a hexagon is smaller than two squares. All that remains is a triangle and a heptagon; hence the above discovery.
(But the edge of {4,7} is too long, longer than that of any 4.n.2n, to be used in such a tiling.)
Marek14 wrote:The mystery family: And finally one thing I don't quite understand yet. It turns out that the edge of (n,n,2n,2n,2n,2n) is twice the edge of (4,4,n,n). My search has confirmed this for n=3 to 10, so it's unlikely to be a coincidence. I think this might be a general relationship, but I lack the knowledge to prove it. I should note that the case n=5 is anomalous because (4,4,10,10), as we already know, also shares the edge with (5,5,6,6) and (4,5,6,10), allowing those polygons to also mingle with doubly sized (5,5,10,10,10,10) pentagons and decagons.
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