pentagrid to generate penrose tiling

[sobriquet]

Hi.

Recently I've been exploring penrose tilings a bit and I was trying to implement a penrose tiling in desmos by means of a pentagrid as outlined here:

http://www.ams.org/publicoutreach/feature-column/fcarc-ribbons

However, when I try to line up the given pentagrid with the penrose tiling it supposedly generates, I can't get them to line up in a way that makes sense.
I superimpose the pentagrid on the penrose tiling, but it doesn't seem to work out however I try to scale it from the given intersection point.

https://i.imgur.com/ucloKMJ.png

Is there anyone who knows what's up with that? Is there any image online anywhere that actually shows how a pentagrid matches up exactly with the penrose tiling it generates?

Kind regards and thanks in advance for any feedback, Niek

[Mike Terry]

On 18/02/2021 02:27, sobriquet wrote:
> Hi.
>
> Recently I've been exploring penrose tilings a bit and I was trying to implement a penrose tiling in desmos by means of a pentagrid as outlined here:
>
> http://www.ams.org/publicoutreach/feature-column/fcarc-ribbons
>

Interesting article!

> However, when I try to line up the given pentagrid with the penrose tiling it supposedly generates, I can't get them to line up in a way that makes sense.
> I superimpose the pentagrid on the penrose tiling, but it doesn't seem to work out however I try to scale it from the given intersection point.
>

I think the problem is you're expecting the lines of the pentagrid to
fall neatly onto the lines of the tiling, actually coinciding with
tiling edges. That's not how the pentagrids work.

You could think of the pentagrids as a kind of template that gives the
rules for which types of rhombuses are encountered as you walk along one
of the ribbons in the tiling. Walking along a ribbon corresponds to
moving along one of the grid lines - as you walk along a grid line, it
is intersected by all the grid lines which are not parallel to it. The
orientation of the successive intersecting lines determines the type and
orientation of the rhombus next encountered in the tiling ribbon.

If you change the pentagrid, i.e. you change the defining offsets
gamma_i (explained in the article), then as you walk along a grid line,
the /order/ of the orientations of intersecting lines you encounter will
be different, and correspondingly the orientations and types of rhombus
that construct the tiling ribbon will be different - i.e. you end up
generating a different tiling.

I don't think it is easy to see in a simple way that the given pentagrid
in the article does indeed generate the given tiling, but I'm prepared
to accept this - it looks plausible at least. We can at least confirm
that the rhombuses comprising a particular ribbon are in the order
demanded by the pentagrid.

Hmmm, this might work - since each vertex in the tiling corresponds to a
particlular polygon in the pentagrid (and vice-versa), you could place
dots in each polygon in the pentagrid, and join the dots to make the
appropriate 4-gons which will correspond with the rhombuses of the
associated tiling. Of course, the 4-gons will be heavily distorted, but
will be correctly matched up with each other. (Put another way, the map
of the 4-gons will be topologically equivalent to the tiling of
rhombuses, so you could maybe visualise how it hangs together.)

> https://i.imgur.com/ucloKMJ.png
>
> Is there anyone who knows what's up with that? Is there any image online anywhere that actually shows how a pentagrid matches up exactly with the penrose tiling it generates?
>

Well, in the article there is a pentagrid alongside the tiling that it
generates, but as I said above, superimposing them will be tricky. To
generate the tiling from the pentagrid, you will need to follow the
coordinate rules in the article.

Is your aim to generate your own tiling from the pentagrid rules? If
so, you need to:

a) First decide on your offsets gamma_i that determine the pentagrid.

b) Next consider in turn the intersections of the various
pentagrid lines. (Only two such lines ever intersect at a
given point, never three.) Each such intersection determines
one of the rhombuses to be plotted in the tiling.

c) To work out /where exactly/ to plot the rhombus, follow the
rules in the article! :) The key here is working out the
integer "coordinates" which apply for each of the five unit vectors,
for the four corners of the rhombus. Two of the five coordinates
will be obvious (following from the choice of intersecting
grid lines and their orientations). The remaining three
"coordinates" you will have to /calculate/,
as they depend on the choice of the gamma_i you made in (a) above.
(This is where different pentagrids generate distinct tilings!)
To calculate the three "tricky" coordinates you will need to
work out which two corresponding parallel grid lines
your intersection point falls between. I.e. you need to
have the linear equations for all the grid lines, then it
should be straightforward enough.

d) Once you have all five coordinates (a0, a1, a2, a3, a4) for
a particular vertex, you can plot that vertex by
calculating the vector sum a0(e_0) + a1(e_1) .. +a4(e_4).

The article gives an example of doing this (visually) for one of the
rhombuses, where we can just see by inspection which of the strips the
considered intersection falls between.

HTH,
Mike.

[sobriquet]

Thanks for the response. After exploring some more websites/articles I
had also figured out that my assumption that the pentagrid somehow ought
to line up with the vertices in a penrose tiling was too simplistic. But
it seems that different articles tend to use different rules for relating
pentagrids to penrose tilings (like in some articles the rule imposed
seems to be that the offsets for the parallel lines should sum to one while
in other articles they are said to sum to 0).

https://www.mdpi.com/2073-4352/7/10/304/htm
http://www.math.utah.edu/~treiberg/PenroseSlides.pdf
https://www.math.brown.edu/reschwar/M272/pentagrid.pdf


I'm still at the stage of trying to figure out some basic premises, like whether
or not every part of the plane delimited by lines in a regular pentagrid is supposed
to contain a single vertex point and whether or not every point where two lines
of the pentagrid intersect corresponds to a thin or thick rombus in the associated
tiling.

For instance, in this image found online, it seems that only some regions enclosed by
gridlines end up with a point, but even more confusing is that in the bottom right area
there is a region that has two points.

https://i.imgur.com/qrTaZc0.png

Creating a pentagrid in desmos wasn't too hard, but I'm trying to find an article that
has a clear explanation/demonstration of how to relate the vertices of the penrose
tilings to the grid intersections and figuring out the logic behind the way potential
regular pentagrids (with at most two lines intersecting at any point) match up with
the complete spectrum of possible penrose tilings.

https://www.desmos.com/calculator/sfdkriwohe

[Mike Terry]

I don't see that that would matter - differences like this seem to be
just changing the origin point chosen for the pentagrid - that has no
substantial impact on the associated tiling. For example in the
original article you quoted, it is possible to construct the given
tiling from the given pentagrid (give or take a plane translation) with
no knowledge about where the origin is located. And given /any/ choice
of offsets, we can clearly choose a new origin that "balances" the
offsets so that they sum to zero. And adding 1 to any offset has no
effect on the pentagrid etc..

>
> https://www.mdpi.com/2073-4352/7/10/304/htm
> http://www.math.utah.edu/~treiberg/PenroseSlides.pdf
> https://www.math.brown.edu/reschwar/M272/pentagrid.pdf
>
>
> I'm still at the stage of trying to figure out some basic premises, like whether
> or not every part of the plane delimited by lines in a regular pentagrid is supposed
> to contain a single vertex point and whether or not every point where two lines
> of the pentagrid intersect corresponds to a thin or thick rombus in the associated
> tiling.

I believe, just from my understanding of your original linked article,
that there would be one such point in each such polygon. The number of
sides the polygon has is going to correspond to the number of rhombus
tiles that meet at that vertex. E.g. in the png file linked below, the
origin is surrounded by a decagon (10 sides), and in the generated
tiling we see 10 thin rhombuses meeting at the corresponding vertex.

>
> For instance, in this image found online, it seems that only some regions enclosed by
> gridlines end up with a point, but even more confusing is that in the bottom right area
> there is a region that has two points.
>
> https://i.imgur.com/qrTaZc0.png
>

Yeah, I don't really know what those images are showing with the dots in
diagram (b). Looking at the tiling in (c) it seems to me that there is
a vertex for /every/ polygon in (b), regardless of whether (b) shows a
dot - so I think the dots in (b) are for some different purpose. (Also
the original article seems to clearly imply (to me at least!) that this
is what is intended. Every polygon area is clearly the intersection of
exactly one pentagrid "stripe" of each of the five orientations, which
gives it a unique set of integer "coordinates", i.e. a unique vertex
location. And as the article covered, as you cross one of the bounding
lines to the polygon on the other side of the line, 4 of the coordinates
are unchanged, and the remaining increases or decreases by 1, which
corresponds to a tiling edge that is going to be one of the five unit
orientation vectors (or its negation, which amounts to the same).

> Creating a pentagrid in desmos wasn't too hard, but I'm trying to find an article that
> has a clear explanation/demonstration of how to relate the vertices of the penrose
> tilings to the grid intersections and figuring out the logic behind the way potential
> regular pentagrids (with at most two lines intersecting at any point) match up with
> the complete spectrum of possible penrose tilings.
>
> https://www.desmos.com/calculator/sfdkriwohe
>

I'd go with your original article, unless something I don't understand
stops that from working. The only vagueness there seems to be deciding
whether the stripe to the left or right of the "zero" line (for a given
family of parallel lines) should be called the "zero" strip. Again,
this hardly affects anything, at most introducing a translation to the
final tiling. I'd go with labelling the zero strips of each colour to
be those containing the graph origin - that seems intuitive.

So in your desmos pentagrid, looking at the graph origin I would say
that belongs to the 0 red strip, and similarly to the 0 blue, green,
orange and purple strips. So it's vertex coordinate (in the tiling
graph) would be
0(e_1) + 0(e_2) + 0(e_3) + 0(e_4) + 0(e_5) = 0
[The vertex is at the origin of the tiling graph, which seems
reasonable, but if we'd decreed the red strip containing the grid origin
was to be labelled (red) strip 1 it would just mean the every tiling
vertex will be translated by a fixed offset, no great shakes...]

And similarly the graph point (x,y)=(3,1) on your desmos page is part of
a grid octagon (just!) having strip coordinates (3,3,2,0,-2)
[corresponding to your red, orange, blue, purple, green strips
respectively], giving a vertex point of
3(e_1) + 3(e_2) + 2(e_3) + 0(e_4) - 2(e_5) = ???
[I'm just too lazy to work this out, but it's a clear and definite point
in the tiling plane...]

Ok, at this point I confess I had a bit of a panic/confusion! I
thought: this doesn't make sense because there will be multiple "strip
coordinates" that map to exactly the same tiling coordinate -
specifically the intersection of all the zero strips has "strip
coordinates" (0,0,0,0,0) which corresponds to a vertex at the (tiling)
origin. BUT ALSO (1,1,1,1,1) in strip coordinates corresponds to a
tiling vertex at
1(e_1) + 1(e_2) + 1(e_3) + 1(e_4) + 1(e_5) = 0 [by symmetry!]
i.e. also a duplicate vertex at the (tiling) origin. Aaaagh! WTF?

But... on reflection I'm worrying about something that never happens:
with the strip labelling I chose, the five "1-strips" (i.e. the strip 1
strips for each colour) have an empty intersection - there's no polygon
in the pentagrid with grid coordinates (1,1,1,1,1) or (2,2,2,2,2) etc..
Problem solved, sort of, I think. :)

Finally, you mention that /every/ tiling of the plane has a
corresponding pentagrid that generates it. The first article you linked
claims that, but it's not obvious to me at all how to prove it, so I
can't say about that! (It's a nice result, assuming that's been proved...)

Mike.

[Mike Terry]

Sorry I got confused with the e_i directions and consequent grid strip
labelling directions. So with more care I reckon it should be:

"strip coordinates" : (3,2,-2,-3,0)

so, vertex is at : 3(e_1) + 2(e_2) - 2(e_3) - 3(e_4) + 0(e_5)
= (7.66311.., 2.48989...)

(Your e_i might be labelled differently, but still the x-y graph coords
I calculated should come out the same for you.)

[sobriquet]

In this video it is claimed that there is an infinite number of penrose tilings
and there is no way to tell them apart, but that seems to contradict the claim
that suitable pentagrids determine penrose tilings and every penrose
tiling having an associated pentagrid.

https://www.youtube.com/watch?v=48sCx-wBs34

I guess the best thing to do would be to see if I can work out the details
from the original paper by De Bruijn, since he came up with the method in the
first place.
It's a fascinating topic for sure with an intriguing mix or ordered and chaotic
patterns.

[sobriquet]

> It's a fascinating topic for sure with an intriguing mix of ordered and chaotic
> patterns.

*mix of ordered and chaotic patterns that is.

And also aesthetically quite pleasing.

https://dogfeathers.com/quilt/images/PENROSE2.JPG

[Mike Terry]

The claim is not "that there is no way to tell them apart" exactly. He
says that if we were inside the tiling, wandering around between
adjacent cells and taking notes etc., then we would not be able to tell
which (infinite) tiling we were in. This is because the infinite tiling
is never characterised by any finite portion of the tiling - given a
finite section of the tiling, there will be multiple ways of extending
the tiling to the full plain - so with knowledge of just a finite
portion of the tiling [which is what the youtube videa is talking
about], we can't tell which full infinite tiling we are in. That's not
an automatic property of tilings, so it requires proof, but it has been
proved for the Penrose tilings, so I understand.

Mike.

[sobriquet]

But I assume you can use a pentagrid to determine a partial penrose
tiling and then the complete penrose tiling that has that partial section
embedded in it is still dictated by that particular pentagrid.

It also seems like claims that partial sequences of digits of arbitrary length
in the decimal expansion of irrational numbers occur in the decimal
expansion of other irrational numbers.

https://i.imgur.com/1OmvKdD.jpg

https://www.youtube.com/watch?v=4DNlEq0ZrTo

[Mike Terry]

If you have a partial Penrose tiling there will be /multiple/ ways of
completing it. One of those will correspond to the original pentagrid,
but the others will have different pentagrids.

Mike.

[sobriquet]

Yeah, I guess that seems right.
I've found some more pages that have useful clues about how to construct
the penrose tiling from the grid.

It'll take a while to work it out, but I hope I can make an interactive version.

Desmos is nice for interactive tilings.

https://www.desmos.com/calculator/g13zpznnjz

[sobriquet]

Finally got it working (took a bit of debugging).

https://www.desmos.com/calculator/j4inpqllzs

This page was useful and also gave a nice concrete example of a pentagrid with numeric specifics
and the associated penrose tiling:

https://www.vivat-geo.de/de_bruijn_1.html

It's always nice to get a visual proof by matching up your results with another source.

https://i.imgur.com/ZaSlS2w.png

[Chris M. Thomasson]

Oh yeah, you got it! Very nice.

[sobriquet]

Thx.
Next challenge would be to turn it into an infinite pattern like this one:

https://www.desmos.com/calculator/y8afui25zf

Should be possible I reckon, but I'll have to experiment a bit more to see if it works out ok.

Desmos is nice for math-oriented abstract art, especially now that it allows one to use
the full color palate.

https://www.desmos.com/calculator/ndk5zmig45

[sobriquet]

This version works a bit better, with improved colors.

https://www.desmos.com/calculator/givvd6odiw

[James Waldby]

sobriquet <dohd...@yahoo.com> wrote:
...

> This version works a bit better, with improved colors.
>
> https://www.desmos.com/calculator/givvd6odiw

That looks impressive! However, when I used the scroll wheel to zoom
out, the display has large gray areas out at the edges with separate
tiles scattered around. I don't know if tiles are missing vs. in the
wrong places. Would it be a desmos problem, a browser problem, or do
your formulas not work when there are a couple of hundred tiles on the
screen?

[sobriquet]

You can increase the value of the h0 variable to tile a larger area.
I'm still working on a version that will be tiling the entire plane, but that will be
tricky and I'm not sure it can be done in desmos.