1
$\begingroup$

i am trying to compute a space-filling curve for a given shape (an isosceles right triangle in this case) like following figure (call it figure_1)

enter image description here

wikipedia gives a formula

H(x,y) = (h(x), h(y))

i don't know how to use this formula.

is there a doable mathematical or programmatic way to do this?

$\endgroup$
  • $\begingroup$ The information given in your Figure 1 is not enough to specify a space filling curve. What you are missing is a careful description of the inductive step: you must describe a single method which when applied to Figure 1 produces a new Figure 2, and that when applied to Figure 2 produces a new Figure 3, and so on with an infinite sequence of Figures 1,2,3,4,5,6,... $\endgroup$ – Lee Mosher May 27 '19 at 14:39
  • $\begingroup$ Furthermore, your figure $1$ must be given by an explicit function $f_1$ from $[0,1]$ to your triangle, and your inductive step must produce an infinite sequence of continuous functions $f_1,f_2,f_3,f_4,f_5,f_6,...$ from $[0,1]$ to your triangle. Without that, your question does not make much sense. $\endgroup$ – Lee Mosher May 27 '19 at 14:39
1
$\begingroup$

I wouldn't advise trying to follow the "Outline of the construction" section in the Wikipedia article. It it pretty abstract and not particularly enlightening, as it punts most of the interesting parts to theorems that it merely asserts or assumes known -- and most importantly it does not correspond to the more intuitively accessible Peano or Hilbert solutions shown in the figures. (So you're bound to become extremely confused if you attempt to match them up. They don't match).

Instead, what you could do is just take one of the standard curves that fill the unit square, and compose it with something like $$ \langle x,y\rangle \mapsto \langle (1-y)x,y\rangle $$ that maps the unit square continuously onto your desired shape.

$\endgroup$
  • $\begingroup$ would you please elaborate the detail in the context of figure_1? such as the 1st step to do this. $\endgroup$ – shi95 May 27 '19 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.