I am trying to find the image of $[0,1/2)$ in Hilbert's Space Filling Curve.

What I thought is $[0,1/2]\times$ $[0,1]$ \ $[1/4,1/2]\times[1/2,3/4]$. Is my understanding correct?

The sets considered are as usual. From $[0,1]$ to $[0,1]^2$.

  • $\begingroup$ Why isn't it $[0,1/2]\times[0,1]\setminus\{(1/2,1/2)\}?$ $\endgroup$ – saulspatz Jul 14 at 3:03

What I thought is ...

From the definition of Hilbert Curve;
for each parameter $t \in I := [0,1] $ a sequence of nested intervals $$I \supset [a_1, b_1] \supset \dots \supset [a_n, b_n] \supset \dots $$
exists, such that each interval is obtained by splitting its predecessor into four parts of equal size.
Any such sequence of intervals can be mapped one by one to a sequence of nested 2D interval. These nested intervals will converge to a uniquely defined point $ h(t) \in Q:= [0,1]\times[0,1]$.

The point to point mapping is determined by the iteration of the curve.

  • Iteration 0 :

$$[0,1] \longmapsto [0,1]\times[0,1]$$

  • Iteration 1 :

$$ [0,\frac{1}{4}] \longmapsto [0,\frac{1}{2}]\times[0,\frac{1}{2}]\\ [\frac{1}{4},\frac{1}{2}] \longmapsto [0,\frac{1}{2}]\times[\frac{1}{2},1]\\ [\frac{1}{2},\frac{3}{4}] \longmapsto [\frac{1}{2},1]\times[0,\frac{1}{2}] \\ [\frac{3}{4},1] \longmapsto [\frac{1}{2},1]\times[\frac{1}{2},1] \\ $$

and so on. The following image shows the said iterations of the hilbert curve iteration of HC (src Why does the Hilbert curve fill the whole square?)

That answers your question.


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