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}