Bill>NeilB wrote a sort of pixelated IFS that convinced us that this set contains no convex patch of positive area. I think I have to take this back; it actually does look like there are visible patches of area in the Levy C-curve where two or more octagonal-ish sections intersect. My original approach was based on the fact that all the vertices in every even level of the C-curve (counting a line as the 0th level) lie on a regular, axis-aligned square grid, and so correspond to coordinates on a suitably-sized bitmap. Thus, you can compute higher-resolution levels of the C-curve from lower ones using fast(ish) integer operations. The low-resolution bitmaps obviously missed the patches, but I've since rendered a set of higher resolution images where they do show up (such as the reasonably OK http://neilbickford.com/assets/levyc-20.png .) At level 24 (the 5MB, 16384x10240 http://neilbickford.com/assets/levyc-24.png ) the largest shape actually breaks up along a diagonal line - but that's almost certainly the last time it does. Finally, for anyone wondering, Wechsler/(Bailey and Kim)'s constructions around these patches look a bit like http://neilbickford.com/assets/levyc-tri20-4k-crop.png . Allan>Work remaining to do is to characterize all the different kinds of patches ... Eytan Alster's (unfortunately paywalled) paper "The Finite Number of Interior Compound Shapes of the Levy Dragon" might be relevant. --Neil Bickford On Fri, Mar 13, 2015 at 3:08 AM, Allan Wechsler <acwacw@gmail.com> wrote:
A quarter to six in the morning.
The area of the unit Levy C-curve is exactly 1/4.
Take the initializer to be an (open) isoceles right triangle with vertices 0, 1, and (1+i)/2. At level 0, the area is clearly 1/4.
I contend that with each level of recursion, area is neither created nor destroyed, but simply shuffled around.
It should be clear enough that area is not created. The recursion step splits each of the 2^n level-n modules, area 1/(4*2^n), into two level-(n+1) modules, area 1/(4*2^(n+1)). There are now 2^(n+1) of them. At best, if none of them overlap, the area is still 1/4.
The fact that area is not destroyed is harder to see. For this to be true, no two level-n modules can overlap. At each level, the complex plane may be completely divided into non-overlapping module sites of area 1/(4*2^n), each of which is either filled or empty. In order for two filled modules to overlap at all, they must coincide. But each module site has a unique parent at the previous level, so if two filled modules are superimposed, their parents must have been superimposed before them. At level 0, however, there is only a single filled module. The invidious superimposition can never get started.
Work remaining to do is to characterize all the different kinds of patches, determine how many of each kind there are at each size scale, and their locations.
On Thu, Mar 12, 2015 at 5:38 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Here is an intriguing little theorem. Let G be the topological interior of the C-curve (assuming, pace Gosper, that the object is well-defined and does not depend on the choice of initializer). Let L(z) = (1+i)z/2, and R(z) = ((1-i)z + 1+i)/2. The C-curve is defined as the forward limit of the ensemble {L, R}. Theorem: L(G) and R(G) are disjoint. Proof: If L(G) and R(G) intersected, then the area of L(G) U R(G) would be strictly less than the area of G. But G = L(G) U R(G) by construction.
Corrollary: No open component of the C-curve lies on its symmetry axis.
On Thu, Mar 12, 2015 at 4:52 PM, Allan Wechsler <acwacw@gmail.com> wrote:
@Dan, that looks very close to the right thing, but it doesn't look like it shows the sections very clearly. @RWG, I don't think I agree that the limit set depends on the initializer. I bet I can prove that any compact initializer will have the same result. I can also prove that the union of a particular constellation of 22 C-curve units has nonzero area, and I think I have found that constellation in the 14th iteration.
On Thu, Mar 12, 2015 at 4:41 PM, Dan Asimov <asimov@msri.org> wrote:
On Mar 12, 2015, at 1:38 PM, Allan Wechsler <acwacw@gmail.com> wrote:
If the endpoints of the curve are 0 and 1, I think that if you zoom
on
(3+5i)/8, there is an open patch just to its northwest (in the -1+i direction). It starts to crystallize at 16-18 levels. If this isn't enough information to find the open patch, I am going to have to write some code (unless somebody already has a pan-and-zoom C-curve explorer implemented).
Don't know if this is what you have in mind, but it looks to be at least in the ball park:
< http://www.mathlab.cornell.edu/~twk6/program.html >
--Dan
The patch I've identified looks tiny (area on the order of 2^(-15)?), but it may not be the largest patch, and there may be lots of them. I'm going to guess that the area of the unit C-curve is less than 2^(-8), but I'm now convinced that, pace Neil B., it is greater than 0.
On Thu, Mar 12, 2015 at 4:06 PM, Dan Asimov <asimov@msri.org> wrote:
Appears to be an interesting article about the C curve (aka Dragon curve) by S. Bailey, T. Kim, R. S. Strichartz: "Inside the Lévy dragon", Amer. Math. Monthly, 2002.
It mentions 16 distinct known shapes that these components of open subsets can take. (Apparently some Israeli mathematician showed there is actually a total of 21 different shapes.)
--Dan
On Mar 12, 2015, at 11:58 AM, Bill Gosper <billgosper@gmail.com> wrote: > > NeilB wrote a sort of pixelated IFS that convinced us that this set contains no convex patch of positive area. The article seems to say that the open sets follow from a proven dimension of 2.
I read it the other way round: that the existence of open subsets of the curve implies the Hausdorff dimension must be 2 (since it lies in R^2).
> Tentatively granting D = 2, does openness follow?
> Can someone show me an open subset of the (D = 2) boundary of the Mandelbrot set? > Exercise (NBickford): Must an open subset of R² have D = 2?
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