Curve Filling a Rectangle
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Curve Filling a Rectangle
The Hilbert curve completely fills a square. Can a modified version be used to fill a rectangle? My instinct says yes, but I wanted to check anyway.
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Re: Curve Filling a Rectangle
Sure, the Hilbert curve is defined with an x(s) and y(s), and if you take two bump functions and feed those in, you can get what you want.
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Re: Curve Filling a Rectangle
Given the lines are either horizontal or vertical, and you steadily fill the gaps between horizontals by vertical fractures and the gaps between the verticals by horizontal fractures, I'd say that taking a nonunitXunit ratio box and progressively filling it with similarly ratioed higherorder curves (applied as a transform in the same orientation as the box, i.e. complimentary ratios as you recurve around the corner of the bigger curve before it) would hit total horizontal filling by infinite widthless vertical linesegments at the same time as vertical filling by the similar stack of horizontal ones.
Or, by another way of looking at it, if ∞ = 4∞ (which it does, arguably, from various standard usages of alephnull) then ∞*(1/2)=∞*(2/1), so a 1:2 rectangle gets filled just as much at the absolute limit of spacefilling in both axes.
But I can also imagine counterinterpretations. Hilbert curves might not work, but Peany ones would?
PTW, then.
(Ninja says it more succinctly than me.)
Or, by another way of looking at it, if ∞ = 4∞ (which it does, arguably, from various standard usages of alephnull) then ∞*(1/2)=∞*(2/1), so a 1:2 rectangle gets filled just as much at the absolute limit of spacefilling in both axes.
But I can also imagine counterinterpretations. Hilbert curves might not work, but Peany ones would?
PTW, then.
(Ninja says it more succinctly than me.)
Re: Curve Filling a Rectangle
The square and the rectangle are homeomorphic.
Take the obvious homeomorphism between the square and the rectangle. Compose this with the curve. The result should be a curve that fills the rectangle.
Right?
Take the obvious homeomorphism between the square and the rectangle. Compose this with the curve. The result should be a curve that fills the rectangle.
Right?
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 Eebster the Great
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Re: Curve Filling a Rectangle
Couldn't you just substitute, say, x/2 for x? I'm missing the reason why you have to actually do anything at all.
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Re: Curve Filling a Rectangle
aka what madako said, yeah. It's a trivial mapping.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))
 MartianInvader
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Re: Curve Filling a Rectangle
Just wanted to add, the beauty of a spacefilling curve is that it's purely topological  which means you can compose it with any continuous function (well, any surjective continuous function) and you still have a spacefilling curve.
So you can start with the Hilbert curve and "stretch" it to fill a rectangle, a triangle, a circle, a star, or pretty much any other connected 2d region.
So you can start with the Hilbert curve and "stretch" it to fill a rectangle, a triangle, a circle, a star, or pretty much any other connected 2d region.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

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Re: Curve Filling a Rectangle
I did not know that. Thanks for the information.
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Re: Curve Filling a Rectangle
By "surjective continuous function," do you mean in one variable (i.e. ℝ→ℝ or [0,1]→ℝ)? If you mean ℝ→ℝ^{2}, then by surjectivity you already have a spacefilling curve.
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Re: Curve Filling a Rectangle
Presumably ℝ²→ℝ²; that's the only thing that typechecks when composed with the spacefilling curve, which maps [0,1]→ℝ²
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))
 Eebster the Great
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Re: Curve Filling a Rectangle
Yeah it wasn't clear to me in which order he was composing the functions, but in this case, again, that's just what surjective means.
 MartianInvader
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Re: Curve Filling a Rectangle
Eebster the Great wrote:Yeah it wasn't clear to me in which order he was composing the functions, but in this case, again, that's just what surjective means.
I meant from the square to a region of R^{2}. And being surjective isn't enough, as I said it also needs to be continuous. Otherwise, you don't have a spacefilling curve, just a spacefilling... function, I guess.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
 Eebster the Great
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Re: Curve Filling a Rectangle
Fair enough. Essentially, you can transform any curve filling any (simplyconnected open) subset of R^{2} to one filling any other by just composing it with any continuous function from the one onto the other. That such a function always exists follows from the Riemann Mapping Theorem, and for most practical subsets, it's really easy to find such a continuous function.
 MartianInvader
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Re: Curve Filling a Rectangle
You can even do a little better than what the Riemann mapping theorem gets you, since you only need a surjective mapping and not a fullon homeomorphism. You can wrap the square around a donut shape (annulus), for example (you don't need simple connectedness, just path connectedness I believe).
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
 Eebster the Great
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Re: Curve Filling a Rectangle
Right, the Riemann Mapping Theorem actually gives you a biholomorphic function. Are biholomorphic functions the isomorphisms of complex analysis?
What is a sufficient condition for the existence of a continuous surjection that is stricter than the hypothesis of the Riemann Mapping Theorem?
What is a sufficient condition for the existence of a continuous surjection that is stricter than the hypothesis of the Riemann Mapping Theorem?
 MartianInvader
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Re: Curve Filling a Rectangle
Eebster the Great wrote:What is a sufficient condition for the existence of a continuous surjection that is stricter than the hypothesis of the Riemann Mapping Theorem?
We're being a little fast and loose with our conditions anyway, since the Riemann Mapping theorem is about open sets, but anything covered by a path is going to be compact (and therefore closed). Anyways, I'm pretty sure that any compact pathconnected set can be surjected onto by the square, although the only way I can think to prove it is to essetially construct a spacefilling curve directly.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
 Soupspoon
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Re: Curve Filling a Rectangle
Have you tried logarithms?
 MartianInvader
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Re: Curve Filling a Rectangle
Oh my goodness, of course! Logarithms! I don't know why I didn't see it before! Soupspoon, you're brilliant! And definitely someone who knows what they're talking about and not at all faking it.
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Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
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Re: Curve Filling a Rectangle
Qapla'!
(vo' SuwomIy maDyar pagh vIpawtaH)
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Re: Curve Filling a Rectangle
Soupspoon wrote:Qapla'!
Please hold your Qapla's until after the lecture.
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