Binary fractal tree with equidistant leaves on a circle

[Ventrella]

Does there exist a binary fractal tree…

(reference: http://ecademy.agnesscott.edu/~lriddle/ifs/pythagorean/symbinarytree.htm )

…whose leaves (endpoints) lie on a circle and are equidistant?

Consider a binary fractal tree with branches decreasing in length by a scaling factor r (0 < r < 1) for each consecutive generation of branching. My first question is: do there exist two numbers representing the left and right branching angles θ1 and θ2 - being constant for every generation of branchings - such that the smallest branches converge on a circle at the limit? My initial hunch is that the answer is no, given the various shapes found by adjusting these angles, and some intuitions about fractals not being very good at imitating Euclidian geometry.

As a thought experiment: what if the lengths and angles could be anything? What if the tree could be drawn in a free-form manner (as if you were doodling on paper), with the only constraint being that it must be topologically equal to a binary fractal tree? In this case, it seems that the answer would be yes. But I want to preserve some of the classic tree’s beautiful self-similar nature by way of some elegant branching and length relationships, and to essentially come up with an expression that has as few factors as possible.

Specifically: can such a fractal tree exist if the scaling factor r is kept constant, and if the branch angles θ1 and θ2 change by some delta d per generation of branching? If so, I suspect that θ1 and θ2 would have to change in a nonlinear way in order to enable convergence to a circle. If the algorithm is sufficiently elegant and compact, then it could be used as a thought-provoking (if impractical) expression of a circle.

 

[jedishrfu]

Why not try writing a program to test it?

You could use the Processing IDE and Java to draw the fractal branches.

Www.processing.org

 

[Ventrella]

Hi jedishrfu,

I am planning on writing an algorithm, for sure :) I am familiar with Processing, but I would prefer to do it from scratch in javascript/canvas.

I just wanted to share the idea beforehand in case anyone had some insights about the math involved. I am wondering if it can be achieved without using any trigonometric functions.
-j

 

[jedishrfu]

I think you’ll need trig to accomplish this or you could build trig lookup tables to degree granularity or finer where finer means more memory but using the trig function would probably be just as fast in the end.

Processing has a JavaScript mode, I think it’s p5js

 

[willem2]

You can exactly compute some of the points after an infinite number of branchings, as long as the branching pattern is regular. You will need complex numbers.
Suppose we have a tree defined by the mappings L(z) = 1 + lz and R(z) 1+rz. abs(l) <1 and abs(r) < 1. for a symmetric tree $r = \bar{l}$.
if S is the set of points of the tree, S = {z∈S : L(z) } ∪ {z∈S : R(z) }
The limit points are what you get if you start with 0 and then apply L or R and infinite number of times in any order.
There's an endpoint for any infinite sequence of R and L mappings, so there is an uncountable number of limit points.

If you always take the left branch starting in the root at 0 you end up in 1 + l + l^2 + k^3 + ... = 1/(1-l).
if you take the left branch once, and than the right branch an infinite number of times, you end up in L(1/(1-r)) = 1 + r/(1-l)
To see that a circle is not possible we only need the limitpoints $1/(1-l), R(1/(1-l)), R(R((1/(1-l))$, etc. you get when you take the right branch n times and then the left branch an infinite number of times.
Because R is a rotation + a translation + a scaling, these numbers lie on a logarithmic spiral, and therefore not on a circle.