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- Newsgroups: sci.math
- Path: sparky!uunet!mcsun!Germany.EU.net!news.uni-bielefeld.de!umatf071
- From: umatf071@unibi.uni-bielefeld.de (0105)
- Subject: Re: A Word-Problem
- Message-ID: <1992Aug13.135704.2280@unibi.uni-bielefeld.de>
- Date: Thu, 13 Aug 92 13:57:04 GMT
- References: <1992Aug9.172411.26212@unibi.uni-bielefeld.de> <Bstrq9.MKr@cs.psu.edu> <1992Aug11.132731.14626@husc3.harvard.edu>
- Organization: Universitaet Bielefeld
- Lines: 67
-
- 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?
-
- SOLUTION:
- The number of points of T(n) is ( n+1 \choose 2 ) = t(n).
- 3 | t(n) iff n = 0, 2 (mod 3).
-
- CASE I: if n = 0, 2, 9, 11 (mod 12) then T(n) is tileable.
- From T9 you can get T11 and T12 adding:
-
- 1 2 2 1 2 2 1 2 2 1
- 1 1 2 1 1 2 1 1 2 1 1
-
- or
- 1 1 2 2 3 3 1 1 2 2
- 2 1 3 2 1 3 2 1 3 2 1
- 2 2 3 3 1 1 2 2 3 3 1 1
-
- With T11, T12, and k times
-
- 1 2 2 1 2 2 1 2 2 1 2 2
- 1 1 2 1 1 2 1 1 2 1 1 2
-
- you can construct T(n+12) from T(n).
-
- CASE II: if n = 3, 5, 6, 8 (mod 12) then T(n) is not tileable.
- This is the hard case. You can look at the equivalent triomino
- puzzle. In this case T5 would look like:
-
- o The allowed L-triominoes are:
- o o o o o
- o o o = T5 o o and o
- o o o o
- o o o o o (only 180-degree rotation)
-
- Now let's try group theory:
- You can read about this technique in the article:
- Dmitry V. Fomin, Getting it together with "polyominoes",
- quantum 2:2 (Nov/Dec 1991) 20-23, 61
- So you get the group: G = < a, b| aabb=baba, bbaa=abab >.
- If a^n b^n <> (ba)^n for n = 3, 5, 6, 8 (mod 12) then is T(n) not
- tileable. Is a^n b^n = (ba)^n for some n = 3, 5, 6, 8 (mod 12) then
- a^3 b^3 = (ba)^3. This you can see be the same technique as in CASE I.
- You have the additional possibility of reduction in the group. If you
- show a^3 b^3 <> (ba)^3 then CASE II is solved. David Sibley found a
- homomorphism, which shows this inequality. So CASE II is proofed.
-
- Annotation:
- Noam Elkies is right, that the homomorphis of David Sibley shows that
- it is implossible 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.
-
- Torsten Sillke
-