Re: Having trouble with substitution tile at higher iteration levels--> takes forever!
- To: mathgroup at smc.vnet.net
- Subject: [mg55903] Re: Having trouble with substitution tile at higher iteration levels--> takes forever!
- From: Roger Bagula <rlbagulatftn at yahoo.com>
- Date: Sat, 9 Apr 2005 03:56:01 -0400 (EDT)
- References: <d2tfkj$qnb$1@smc.vnet.net> <d32ul1$cv7$1@smc.vnet.net> <d356ou$p0k$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Dear Rolf Mertig,
Using you program I was able to get a better
version level 14 of the
{p,q,r}={1,1,1}
It turns out to be the border of Rauzy's first Pisot tile.
I had a version of that border tile from box counting experiments
and they are the same apparently.
So Dr. Richard Kenyon has invented a new border method
that is parallel to the older "body" tile method.
It is based on a six sign symmetrical substitution instead of a three
substitution method.
This pretty much makes Kenyon the most advanced scientist in the tiling
field at the moment. I think with his body of work he should be a
Field's medalist, since this seems to be a singular advance to the
tiling field of mathematics.
Roger Bagula wrote:
> Rolf Mertig wrote:
>
>>With a straightforward use of Split I can i=15
>>in about 16 minutes (on a fast Opteron machine).
>>I am sure there must be a better way, but this is at least
>>a slight improvement.
>>
>>Regards,
>>
>>Rolf Mertig
>>GluonVision GmbH
>>Berlin, Germany
>>
>>
>
>
> Dear Rolf Mertig,
> Your method does help , thanks.
> I've been busy and didn't see it this morning.
> There are iundocumented tiles ( until now):
> You just change the initial polynomial definitions.
> I just did: ( old program not to level 16)
> n=4, q=2,p=1,r=1
> n=3:
> q=3,p=1,r=1
> q=1,p=2,r=1
> q=1,p=1,r=2
> q=1,p=1,r=1 ( not as good, but is related to the Rauzy Pisot tile)
> They all gives tiles. q=3 and p=2 are really slow.
> You only really have to got to level 9 to see they are going to tile or not.
> I doubt Dr. Kenyon spent this much time on the examples section.
> There may be more by this method as well.
>
> Have you done any work on the French Rauzy substitution method in
> Mathematica? As searched everywhere and can't find anything om it.
> Having it Mathematica would be a real help.
> Specifically how to get the characteristic polynomial/ eigenvalues from
> a set of substitutions.
> Roger L. Bagula email: rlbagula at sbcglobal.net or
> rlbagulatftn at yahoo.com
> 11759 Waterhill Road,
> Lakeside, Ca. 92040 telephone: 619-561-0814}
>
--
Roger L. Bagula email: rlbagula at sbcglobal.net or
rlbagulatftn at yahoo.com
11759 Waterhill Road,
Lakeside, Ca. 92040 telephone: 619-561-0814}