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- Newsgroups: sci.math
- Subject: Re: A Word-Problem
- Message-ID: <1992Aug13.113616.14692@husc3.harvard.edu>
- From: elkies@ramanujan.harvard.edu (Noam Elkies)
- Date: 13 Aug 92 11:36:15 EDT
- References: <Bstrq9.MKr@cs.psu.edu> <1992Aug11.132731.14626@husc3.harvard.edu> <1992Aug13.135704.2280@unibi.uni-bielefeld.de>
- Organization: Harvard Math Department
- Nntp-Posting-Host: ramanujan.harvard.edu
- Lines: 36
-
- In article <1992Aug13.135704.2280@unibi.uni-bielefeld.de>
- umatf071@unibi.uni-bielefeld.de (0105) writes:
- "This is the problem the word-problem comes from:
- "
- " 1
- " 1 1
- " 2 2 3
- " 3 2 3 3
- " 3 3 1 1 2
- " 1 2 2 1 2 2
- " 1 1 2 3 3 1 1
- " 2 3 3 1 3 2 1 3
- " 2 2 3 1 1 2 2 3 3
- " o
- "This figure shows that you can tile a triangle T9 with T2: o o
- "
- "QUESTION: which T(n) can be tiled with T2?
- [...]
-
- So it was a tiling problem, but not the one I had imagined...
-
- " Noam Elkies is right, that the homomorphism of David Sibley shows that
- " it is impossible to tile an odd rectangle with the L-triomino (only
- " 180-degree rotation allowed) too. In this case you have to show that
- " a^3 b <> b a^3.
-
- I omitted a step here: the commutators [a^3,b^2] and [b^2,a^3] vanish
- (aaabb=a(aabb)=a(baba)=(abab)a=(bbaa)a=bbaaa, and likewise aabbb=bbbaa;
- in ffect I'm tiling 2x3 and 3x2 rectangles with the allowed L-triominos),
- and aaabbb=bababa iff a^3 and b^3 commute since b(abab)a=b(bbaa)a;
- so the group-theory identity [a^m,b^n]=1 that would result from a tiling
- of any odd mxn rectangle by allowed L-triominos is equivalent to any of
- [aaa,b]=1, [aaa,bbb]=1, or aaabbb=bababa.
-
- --Noam D. Elkies (elkies@zariski.harvard.edu)
- Dept. of Mathematics, Harvard University
-