I am almost finished with the section on the the transformation T, Binary Rule, and substitution rule (section 5 of our outline). The only thing that remains is to show that A_infty = union of T^(2k)C (k=0,1,...). I am not sure if we ever did prove this, and I'm not quite sure how to do it. I think what we did show is that this union of even iterates of T applied to the checkerboard satisfies the Binary Rule. However, we never proved, in turn, that the Binary Rule holds for the Exactly 1 Rule.
I think the approach we are taking in the writeup is:
1) Prove that U(T^(2k)C) = A_infty (not done)
2) Prove that U(T^(2k)C) satisfies the Binary Rule, and hence A_infty satisfies the Binary Rule (done)
3) Prove that the substitution rule satisfies the Binary Rule, and hence produces A_infty (done)
How do we prove item 1)?
I've uploaded my writeup here: PDF, LaTeX. I have done a lot of work on the other sections, so please read over that if you can (or we can scrutinize it in lab sometime).
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3 comments:
Charlie -
The idea is to show that the cloning recursion on dyadic blocks for Exactly 1 (rotated) applies in exactly the same way to generate
U(T^(2k)C)recursively. It's possible we didn't write down exactly this argument last term. Perhaps you want to reverse the order of (1) and (2) so you can simply check that the cloning scheme matches with the Binary Rule. Let me know if you still have questions...
Here's a start: Each dyadic box B_N is the union of finitely many T^(2k)C. If we can prove precisely how this works, then it should be a simple task to extend this to A_infty = U_{k=0}^infty (T^(2k)C) by induction.
I wrote some mathematica code to determine just how many T^(2k)C are needed to cover B_N. Here's what I found (this is supposed to be a table):
N, Size (=(2^N)-1), # of T^(2k)C required
2, 3, 2
3, 7, 4
4, 15, 6
5, 31, 8
6, 63, 10 (it took a whole 3 seconds to confirm this one!)
For example, B_2 = U(C, T^2C), B_3 = U(C, T^2C,T^4C,T^8C), etc.
pattern: for dyadic block of size 2^N-1, one requires (2N-2) of these T^(2k)C's.
That is, B_N = U_{k=0}^(2N-2) T^(2k)C.
This approach is too much work.
In (2) we have a self-contained argument that
U_{k=0}^infty (T^(2k)C)
satisfies the Binary Rule. So it suffices to show that the subset of Z^2 defined by the Binary Rule agrees with A_infty. I claim this is easy to check using the cloning construction of A_infty. Take an (x,y) in B_N and find its 3 clones in B_N+1 - B_N. Check that they satisfy the Binary Rule iff (x,y) does. That should do it.
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