# Fractal walking: well defined case or not?

Please consider the following recursive diagram: diagram

Each triangle is connected at the midpoint of a side to the corner of an inner triangle which is 1/4 times the size. The total line length of the figure is convergent.

The puzzle question that goes with the figure is: starting at the top, what is the shortest distance we have to walk to cover the entire figure?

What I would like to know is: is the answer to this question is well defined? We can use a recursive procedure, but to go the shortest distance, we must:

1. Go along the edges until we can cross to the inner triangle.
2. Do the same procedure for the inner triangle.
3. Cross to the outer triangle again and finish going around the edges.

Is this simply a kind of a Zeno's paradox? To me it seems that when the procedure is changed to first finish the outer triangle completely before moving inwards, there is no problem: we can keep going and the total distance is convergent. But by the first method, we first have to do an infinite number of things, and then finish up the outer triangle.

-
The first method does not seem to be well-defined – Ilya Apr 4 '12 at 8:46
Zeno's paradoxes aren't really paradoxes at all but rather common misconceptions about the "nature" of infinities. – example Apr 4 '12 at 9:18
It makes complete sense if you consider the solution not as a sort of supertask with an infinite number of steps, but rather as a parametrized curve (a continuous function from $[0,1]$ to the plane) which has finite arc length and whose trace (the range of the function) must equal the set of points comprising the fractal diagram. – Rahul Apr 4 '12 at 9:30
I think you are right, parametrizing the path reduces the problem to a much clearer case. – Mathijs Apr 5 '12 at 21:50

Let $L(s)$ be the length of the path you walk for such a "triangle" with edges of length $s$. And let $\alpha$ be the ratio between one side of the triangle and the connecting line between two triangles. The path thus has a length of
$$L(s) = \frac{3}{2}s + \alpha s + L\left(\frac{s}{4}\right) + \alpha s + \frac{3}{2}s$$ $$= (3+2\alpha)s + L\left(\frac{s}{4}\right)$$ $$= (3+2\alpha)s + (3+2\alpha)\frac{s}{4} + L\left(\frac{s}{16}\right)$$ and so on. What we get in the end is $$L(s) = (3+2\alpha)s\sum_{i=0}^\infty\frac{1}{4^i} = \frac{4}{3}(3+2\alpha)s$$