Curve Filling a Rectangle

[jewish_scientist]

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.

[doogly]

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.

[Soupspoon]

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 non-unitXunit ratio box and progressively filling it with similarly ratioed higher-order 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 line-segments 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 aleph-null) then ∞*(1/2)=∞*(2/1), so a 1:2 rectangle gets filled just as much at the absolute limit of space-filling in both axes.

But I can also imagine counter-interpretations. Hilbert curves might not work, but Peany ones would?

PTW, then.

(Ninja says it more succinctly than me.)

[madaco]

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?

[Eebster the Great]

Couldn't you just substitute, say, x/2 for x? I'm missing the reason why you have to actually do anything at all.

[Xanthir]

aka what madako said, yeah. It's a trivial mapping.

[MartianInvader]

Just wanted to add, the beauty of a space-filling 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 space-filling 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.

[jewish_scientist]

I did not know that. Thanks for the information.

[Eebster the Great]

By "surjective continuous function," do you mean in one variable (i.e. ℝ→ℝ or [0,1]→ℝ)? If you mean ℝ→ℝ², then by surjectivity you already have a space-filling curve.

[Xanthir]

Presumably ℝ²→ℝ²; that's the only thing that typechecks when composed with the space-filling curve, which maps [0,1]→ℝ²

[Eebster the Great]

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]

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 R². And being surjective isn't enough, as I said it also needs to be continuous. Otherwise, you don't have a space-filling curve, just a space-filling... function, I guess.

[Eebster the Great]

Fair enough. Essentially, you can transform any curve filling any (simply-connected open) subset of R² 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]

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 full-on homeomorphism. You can wrap the square around a donut shape (annulus), for example (you don't need simple connectedness, just path connectedness I believe).

[Eebster the Great]

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?

[MartianInvader]

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 path-connected set can be surjected onto by the square, although the only way I can think to prove it is to essetially construct a space-filling curve directly.

[Soupspoon]

Have you tried logarithms?

[MartianInvader]

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.

I'm going to have to move you way up in the rankings in my latest project - ranking people from best to worst.

[Soupspoon]

Qapla'!

(vo' SuwomIy maDyar pagh vIpawtaH)

[gd1]

Qapla'!

Please hold your Qapla's until after the lecture.

[mashnut]

Qapla'!

Please hold your Qapla's until after the lecture.

That's how I got kicked out of the last lecture though