Re: Having trouble with substitution tile at higher iteration levels--> takes forever!
- To: mathgroup at smc.vnet.net
- Subject: [mg55877] Re: Having trouble with substitution tile at higher iteration levels--> takes forever!
- From: Roger Bagula <rlbagulatftn at yahoo.com>
- Date: Fri, 8 Apr 2005 01:37:11 -0400 (EDT)
- References: <email@example.com> <firstname.lastname@example.org>
- Sender: owner-wri-mathgroup at wolfram.com
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}