On 28/01/2021 12:28, Roger Bagula wrote:
Stewart Robert Hinsley
I'm hoping you are alive an well in England/ UK!
I was using Riddle rotations on N Apollonians and had been experimenting with quasiconformal rotations for several months when this theorem occurred to me.
Post comment in the Facebook group on my nApollonian Riddle Rotation movies: r.a->g.a.Inverse[g]:
Quasiconformal homomorphism to the Riddle rotation.
"r" rotation, "a" Kleinian group matrix, "g" quasiconformal matrix transform: as r->g^2.
So these n Apollonian movies are quasiconformal equivalents.
The theorem based on the twin dragon -Heighway dragon tiling pairing is based on me doing
a Riddle rotation on a tame dragon and finding an Heighway's equivalent tiling.
Theorem: All two transform tilings have a pair tiling by quasiconformal/ Riddle rotation.
If the theorem can be proved it disproves the Kenyon 17 tile number limit for two transform tilings.
There would have to be an always even number of such tiles.
Roger Lee Bagula
https://archive.org/details/apollonian-peanut-butter
It's not a theorem if you don't have a proof. If it's only based on one identified pairing it's hardly even a conjecture.
Your explanation is unclear; I don't understand how you relate the two
tiles.
All known order 2 self-similar tiles not involving reflections are
Perron tiles, i.e. the expansion factor of a tiling is a Perron number.
http://www.stewart.hinsley.me.uk/Fractals/IFS/Tiles/Perron.php
If we accept this as a conjecture I can apply my heuristics
(conjectures) relating to Perron tiles. I have two distinct heuristics
the involve rotations.
As described in the link above I have a heuristic for what rotation
angles are allowed in an IFS { t(i) }. This is equivalent to a rule for modifying an IFS by substituting r(i).t(i) for t(i).
In this case it is the IFSs, not tiles, that come in regular groups.
There are 4 IFSs for most combinations of dissection equation and Perron number, but 16 for the Perron number 1 + i. For order 2 tiles this can
be shown to give 40 IFSs. For order two tiles for the linear and cubic
Perron numbers it is empirically found that this process transforms an
IFS whose attractor is a tile to another IFS whose attractor is a tile,
but this does not generalise - an IFS for the terdragon is a member of a
group of 216 IFSs, but only 144 of these have tiles as attractors (the
number of distinct tiles is appreciably smaller).
These 40 IFS reduce to 17 tiles
* the 24 IFSs where the magnitude of the Perron number reduce to the 6 rep-2-tiles. (There is a formal proof that there are only 6.) This
reduction is a consequence of the two elements being not only similar
but also congruent. (The twindragon, Harter-Heighway dragon, right
isoceles triangle and Levy curve form a group of 4 tiles and 16 IFSs;
the sqrt(2):1 rectangle and tame twindragon are singletons.)
* the 8 IFSs for the 1st cubic Pisot (2 dissection numbers) reduce to 7
tiles, because the ammonite tile turns up for both dissection numbers
* the 4 IFSs for the 2nd cubic Pisot generate 4 tiles.
* the golden number does not have an qualifying order two; the golden
bee involves a reflection in one of the transforms.
Hence 6+7+4 = 17 tiles.
The other heuristic is what I call the grouped element technique, where t(i).r(i) is substituted for t(i). This only works under certain
restrictions, starting with the unit cell for the tiling have a symmetry
group consisting of more that the identity transform - the r(i) here
would be the members of the rotation group of the tile. For the order
two tiles, where this restriction applies, this just generates half-size
copies of the tile.
However one of the results of applying this technique to a 4 element
dissection of a twindragon is indeed a Harter-Heighway dragon. But as a
counter argument, the same technique can be applied to the tame
twindragon to generate an analogous tile, which does not have a 2
element dissection.
--
SRH
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