## Quasiperiodic self-intersecting euclidean tiling

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

### Quasiperiodic self-intersecting euclidean tiling

Take a pentagon, now attach pentagons to all of the edges with only 1 neighboring pentagon, keep doing this until you've completely filled the plane with pentagons.
From what I can tell, the resulting tiling has regular pentagons as its faces and regular decagrams as its vertex figure, but the whole tiling itself is not regular.
gwa
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### Re: Quasiperiodic self-intersecting euclidean tiling

A tiling of pentagons, ten at a vertex, is {5,10/3} or x5o10/3o.

The vertices correspond to the cyclotomic numbers CZ5, which is of some interest to those of us who play around with the connection between bases and polytopes. I discuss a corresponding calculation of the CZ15, the 15-gon being the smallest polygon of odd sides that one can reach the centre of. (Any polygon, that is not prime or a prime power will do).

See eg https://www.tapatalk.com/groups/twelfty ... s-t69.html

These tilings are "piecewise discrete", which means that you can completely build any surtope and anything attached to it. The intersections are not 'seen' if there is no direct connection. It's taken to be on different floors or planes of existance. If you stood in a cell of these, you would see pentagons and bits of pentagons, but you can't "see around corners", that is, you can't see for example, the complete vertex-figure, just 360 of the 1080 degrees on offer.

The cyclotomic numbers CZ5 are C2D2 numbers, class-2 2-dimensional numbers. This equates to for example, folding a four-dimensional lattice onto two dimensions. The tilings {8,8/3} and {12,12/5} can actually be represented as for example {4,4}+i{4,4} and {3.6}+i{3,6}, that is, an ordinary square lattice, and one rotated 45 degrees. Every point of {4,3,3,4} appears as a different spot, and the petrie polygon of a tesseract, if in one projection is an octagon, in the other is an octagram.

The quasic-periodic nature comes when one is restricted to one floor (ie no intersections). You still get the various pentagonal defraction-lines, but it's not a pretty lattice like the tiling of squares. You can create rules as to what is allowable at a vertex, such as in Klitzing's thesis. For the class-2 systems, the general pattern is for closed holes of varying size. For class-3 and above, (such as {7,14/5} and {15,30/13}, the holes never close up generally.
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wendy
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### Re: Quasiperiodic self-intersecting euclidean tiling

Like what wendy said above, it can't be regular in euclidean space, but rather in hyperbolic space. This is because there is not enough space to fit a decagrammic position of pentagons in euclidean space.
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### Re: Quasiperiodic self-intersecting euclidean tiling

I think the idea here runs something like tihs: The vertex angle of a pentagon is 108º. 360/108 = 10/3, so {5,10/3} is a valid Euclidean tiling.

I've seen talk of this type before, but I don't think you can construct it actually.

By that logic, there should be a {5,3,3,5/2} tiling in 4d space because the dichoral angle of {5,3,3} is 144º, and 360/144 = 2.5. But such a thing clearly doesn't exist.
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### Re: Quasiperiodic self-intersecting euclidean tiling

The thing is that there is a huge difference between euclidean tilings and hyperbolic ones. Euclidean tilings increase at the rate of $$r^n$$, while the hyperbolic ones run to $$n^r$$. You can "unfold" a connected net of the hexagons of a 3d tiling like o5x3x5o onto the cells of x6o4o, without distortion. You can't do this in 3d to 2d euclidean, for example.

The unwritten extra rule is a requirement to be sparse. This means that a point is occupied by a finite number of objects. For example, {3,3,5/2} has at each point on its surface, 191 tetrahedra, because the surface is finitely covered, it is accepted. Something like {5/2,4} can start being constructed, but you can find out that it will never close up. You can enter p=5/2, q=4 into polytope formulae, and get things like the circumradius, the vertex-edge-face ratios etc, but the density does not resolve.

Something like {5,10/3} is a perfectly legitimate lattice in 2d space, except it keeps on winding on itself. But for any finite number of steps, there is a finite number of vertices (in the style $$r^n$$ ) that can be reached, and it essentially corresponds to a mapping of a 4d tiling onto 2d. The quasicrystaline nature comes from limiting choices in the selection of numbers. In essence, an atom occupies real space, and that space can not be double-booked.

The underlying number system is $$a + b\sqrt(5)$$. You can plot this onto a space $$x=a+b\sqrt(5), \ y = a-b\sqrt(5)$$, and exclude any overlap. The result is that an atom occupies not just a point on the x-axis, but a whole column of y, with the connections at one or two points near the middle. The next atom is a column, and makes its connections, and so forth. So you get sort of a straight line near y=0, as the quasicrystal expands.

If you viewed this from the y direction, the roughly straight line becomes a twisted back-and-forth near the origin, and you never really get far. The tighter the ball is near the y-space, the more pleasing the thing in the x-space is. Penrose's tiles are set up so that the y-space never gets out of an extremely tight area.
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### Re: Quasiperiodic self-intersecting euclidean tiling

I'm pretty sure this tiling actually exists since I've been able to create one vertex of it. It just happens that the pentagons are able to line up with themselves, so you don't get infinite pentagons per vertex.
gwa
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### Re: Quasiperiodic self-intersecting euclidean tiling

{5,10/3} does not have infinitely many pentagons per vertex, there are only 10 throughout all vertices. But vertices become infinitely dense, in fact they live on a module (with 4 Bragg indices within a 2D tiling space). So not the first statement is what hinders, it is the second one (at least from usual point of view).

--- rk
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### Re: Quasiperiodic self-intersecting euclidean tiling

Yeah, I think that's what causes it to be quasiperiodic.
gwa
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### Re: Quasiperiodic self-intersecting euclidean tiling

Not at all. Quasiperiodic tilings are, with respect to the 2nd point, very much like usual periodic tilings: they have fully separable vertices. In fact they usually not even have overlapping tiles. It is only that the vertices form a specific subset of some module. The points of that to be considered module generally can be obtained by oblique projection of some higher dimensional lattice. The vertices of the quasiperiodic tiling then are being selected therefrom as that subset which corresponds to some segment of that higher dimensional (embedding) space, which thereby is unbounded wrt. the tiling spaces parallel directions and quite bounded in the perpendicular directions. In fact, that perpendicular crosssection of that segment usually is either being called the atomic surface or alternatively the acceptance region. This perpendicular space restriction is what guarantees that the obtained tiling will not be dense any longer. This scenario is being known as "cut and project method".

Thus applying these ideas to your tiling of pentagons, you surely would have to tile the 2D space by pentagons without any overlap. But that then would ask for various gaps. You might then like to introduce several (few) further tiles of such gap shapes. That then might provide you with some aperiodic random tiling based on that module. But still it is not guaranteed to be truely quasiperiodic. Quasiperiodicity would require the existance of an according lifting into an accordingly restricted region of perp space!

Btw., quasiperiodicity here requires also that the module itself is being obtained from a finite dimensional embedding space. Else it is called limit periodic.

--- rk
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