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- Newsgroups: sci.math
- Path: sparky!uunet!pipex!pavo.csi.cam.ac.uk!camcus!cet1
- From: cet1@cus.cam.ac.uk (C.E. Thompson)
- Subject: Re: Tiling problem
- Message-ID: <1992Dec20.000625.28213@infodev.cam.ac.uk>
- Sender: news@infodev.cam.ac.uk (USENET news)
- Nntp-Posting-Host: grus.cus.cam.ac.uk
- Organization: U of Cambridge, England
- References: <israel.723716857@unixg.ubc.ca> <israel.723837962@unixg.ubc.ca> <1gggutINN29q@access.usask.ca> <1gm373INNafn@access.usask.ca> <1992Dec18.002343.3944@infodev.cam.ac.uk> <1992Dec18.111210.2852@odin.diku.dk>
- Date: Sun, 20 Dec 1992 00:06:25 GMT
- Lines: 76
-
- In article <1992Dec18.111210.2852@odin.diku.dk>, torbenm@diku.dk
- (Torben AEgidius Mogensen) writes:
- |> cet1@cus.cam.ac.uk (C.E. Thompson) writes:
- |>
- |> >It occured to me that one could put more constraints on the problem by using
- |> >periodic boundary conditions. This brings the probabilisticly suggested largest
- |> >square down from about 17 to about 11, and so it might be more ammenable to
- |> >a brute-force-and-ignorance computational attack.
- |> >
- |> >This version could be stated as: colour the tiles of the infinite plane square
- |> >lattice black and white, periodically with period N in both directions, such
- |> >that the only (orthogonally oriented) squares with all four corners the same
- |> >colour have both sides divisible by N.
- |>
- |> Well, since someone already posted a 13x13 solution, an estimate of 11
- |> is a bit on the low side. But the suggestion sounds interesting.
-
- My first reaction to this was to point out that the "periodic boundary
- conditions" version is really more restrictive -- in term of the original
- problem one is demanding that i x (N-i) rectangles ("wrap-round squares")
- have their 4 corners not all the same, as well as the i x i squares, for
- 1 <= i <= N-1. Thus the N = 12 solution that Robert Israel posted:
-
- bb
- ||
- 000011011111
- 011010010101
- 001100110011
- 010110100110
- 010100101010
- 011001100101
- 101101001100
- 100101010110
- 110011010010
- a-> 101010011001 <-a
- a-> 110011001011 <-a
- 100110100101
- ||
- bb
-
- has several wrap-round squares (e.g. the 1x11 rectangles marked a and b)
- which violate the stronger condition.
-
- But I then noticed, to my considerable astonishment, that Robert Israel's
- later N = 13 solution:
-
- 0001101010110
- 0101011000011
- 1100001101110
- 0110101011000
- 0101100001101
- 0000110101011
- 1011101100001
- 0110000110111
- 0011010101100
- 1010110000110
- 1000011011101
- 1101110110000
- 1011000011010
-
- *does* satisfy the stronger condition, having no i x (13-i) rectangles with
- four corners the same. Unless this is a consequence of his search strategy,
- there is something very odd going on.
-
- The probabilistic argument I had in mind, following Torben Morgensen's
- original one, involves assuming that the probability of each sub-square
- violating the conditions is 7/8, independently of the others. For N=13 there
- are 650 regular sub-squares (giving Torben's estimate of 2^169*(7/8)^650 =
- 1.511e13 solutions) and another 364 wrap-round ones, giving an estimate of
- 2^169*(7/8)^1014 = 1.175e-8 solutions. So the undoubted existence of at
- least one solution suggests that the assumption of independence was not
- a very good one!
-
- Chris Thompson
- JANET: cet1@uk.ac.cam.phx
- Internet: cet1@phx.cam.ac.uk
-