Re: [math-fun] spacefilling surface
On 2018-12-13 14:04, Andy Latto wrote:
Aren't there even simpler solutions? If f:R -> R^3 is space-filling, define
g:R^2 -> R^3
by
g(x,y) = f(x).
Holy cow, doesn't that cover the image a Continuum number of times!? Warren Smith offered:
just let F(t) be a plane filling curve and then the spacefilling surface mapping (s,t) -> (x,y,z) is (s, F(t)).
DAWK—Just "cylindrize" the planefill! Double DAWK: All along, gosper.org/base2funnel1.png was sitting in my Math folder. —rwg
On Thu, Dec 13, 2018 at 4:34 PM Bill Gosper <billgosper@gmail.com> wrote:
Rohan requested a continuous map from a 2D surface onto a non-porous
patch
of positive 3D volume. There's one under our noses, but maybe not where you think. We certainly have continuous maps from 1D onto 3D, so it is tempting to compose such a volume-filling function with some sort of (necessarily multivalued) 1D inverse of a planefilling function. But such inverses are wildly discontinuous, no matter which values you choose as the definition. Although 1/2+i/2 is "merely" triple point, the Hilbert "curve" is dense with quadruple points. The preimage of a sufficiently small neighborhood of such a point will comprise at least four disjoint pieces. Could the 1D to 3D map possibly smooth everything out? Maybe, but there's a much cleaner solution. SPOILER: gosper.org/3dsno.png gives a continuous mapping of an equilateral triangle onto a standard 3 simplex. —rwg
Here's a similar question, which I don't know the answer to: Let A = { a_1, ..., a_n } and B = { b_1, ..., b_m } be iterated function systems. Define an IFS-morphism f to be a function from {1, ..., n} to { 1, ... m } such that b_{f(i_1} b_{f(i_2)} ... = b_{f(j_1} b_{f(j_2)} ... whenever a_{i_1} a_{i_2} ... = a_{j_1} a_{j_2} ... for two infinite sequence of symbols {i_k} and {j_k}. Many classic space filling curves can be given as IFS-morphisms. For example, a Hilbert curve can be seen as the IFS-morphism f from {1/4x, 1/4x + 1/4, 1/4x + 1/2, 1/4x + 3/4} to {(iz + 1 + i)/2, z/2, (z+1)/2, (-iz + 1 + 2i)/2} with f(i) = i. The question is: are there any IFS-morphisms from A to B where A has a two-dimensional attractor and B has a three-dimensional attractor? -Thomas C On Sat, Dec 15, 2018 at 4:05 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2018-12-13 14:04, Andy Latto wrote:
Aren't there even simpler solutions? If f:R -> R^3 is space-filling, define
g:R^2 -> R^3
by
g(x,y) = f(x).
Holy cow, doesn't that cover the image a Continuum number of times!?
Warren Smith offered:
just let F(t) be a plane filling curve and then the spacefilling surface mapping (s,t) -> (x,y,z) is (s, F(t)).
DAWK—Just "cylindrize" the planefill! Double DAWK: All along, gosper.org/base2funnel1.png was sitting in my Math folder. —rwg
On Thu, Dec 13, 2018 at 4:34 PM Bill Gosper <billgosper@gmail.com>
wrote:
Rohan requested a continuous map from a 2D surface onto a non-porous
patch
of positive 3D volume. There's one under our noses, but maybe not where you think. We certainly have continuous maps from 1D onto 3D, so it is tempting to compose such a volume-filling function with some sort of (necessarily multivalued) 1D inverse of a planefilling function. But such inverses are wildly discontinuous, no matter which values you choose as the definition. Although 1/2+i/2 is "merely" triple point, the Hilbert "curve" is dense with quadruple points. The preimage of a sufficiently small neighborhood of such a point will comprise at least four disjoint pieces. Could the 1D to 3D map possibly smooth everything out? Maybe, but there's a much cleaner solution. SPOILER: gosper.org/3dsno.png gives a continuous mapping of an equilateral triangle onto a standard 3 simplex. —rwg
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