[seqfan] Re: This is a Penrose tiling?
Brad Klee
bradklee at gmail.com
Wed Jan 8 20:30:22 CET 2020
For a relevant discussion, including definitions of "MLD" see,
E.O. Harriss, "On Canonical Substitution Tilings":
http://www.mathematicians.org.uk/eoh/files/Harriss_Thesis.pdf
If the tiling is MLD to Penrose, a substitution rule must exist and
must be equivalent to the canonical Penrose rule. Chapter 4 of
the above suggests simply scaling the window by 1/goldenratio:
Tiling: https://0x0.st/z5l0.png
Window: https://0x0.st/z5ln.png
( also: http://www.mathematicians.org.uk/eoh/files/Harriss_CSTABT.pdf )
This check does not apparently reveal a substitution rule. The
red subset doesn't scale down by golderatio to the black whole
set. This supports the hypothesis that such a rule does not exist,
i.e., that the set in question *is not* a Penrose Tiling by MLD.
However, it is still possible that a valid sub-window exists, but
has some complicated definition??
Another possibility is that the tiling is a Penrose Tiling by MLD,
but only up to a set of measure zero. This remains unproven as
far as I know.
It seems that the issue is insufficiently resolved. The comment
added to A302176 may need to be improved if we find out
otherwise.
I will ask Edmund about it later.
--Brad
On Tue, Jan 7, 2020 at 3:32 PM Brad Klee <bradklee at gmail.com> wrote:
>
> J-PA: To me this is not "the" Penrose tiling but "a" Penrose tiling.
> (of course depending of what the definition of a Penrose tiling is
> or should be)
>
> One definition would be: A tiling is *a* Penrose tiling if and only
> if it is "Mutually Locally Derivable" to *the* Penrose tiling.
>
> This was not proven but implicitly assumed(?) in the Shutov article,
> so I think it's fair to start with a skeptical attitude. If it is a Penrose
> tiling by MLD, what is the transformation to the canonical form?
>
> NJAS: Maybe you could generate 1000 terms and send them to me,
> if you don't know how to extend the sequence in the OEIS entry?
>
> Using optimized code, I can generate 100 terms in under 15 minutes,
> see follow-up post at:
>
> https://community.wolfram.com/groups/-/m/t/1854730
>
> Using the same code, 1000 terms should be doable in a day using
> one processor at 2.7Ghz, but I think it's a distraction. If the tiling is
> MLD then we should find how to obtain it from the Penrose tiling
> before doing anything else.
>
> Even then, it seems the R. Sigrist's sequences are a better place
> to start. We know that there are finitely many vertex types, so the
> sun and star sequences can each be projected to component
> sequences, one for each vertex type.
>
> Will the component sequences tell us anything interesting about
> the four separate trends on the scatter plots? I don't know.
>
> See also:
>
> https://oeis.org/A302841/graph
> https://oeis.org/A302842/graph
>
> --Brad
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