Re: name this golden ratio based fractal
- From: Roger Bagula <rlbagula@xxxxxxxxxxxxx>
- Date: Sat, 18 Aug 2007 16:16:32 GMT
jankrihau@xxxxxxxxxxx wrote:
I thought of a simple fractal and I just wonder if it has a name.Most of these fractal are types are based on polynomials:
Start with a point and draw three lines of length 1 from it so that
each angle is 120 degrees. The iteration step is to draw two new lines
from each "open" endpoint of the lines of the previous iteration, so
that each angle is still 120 degrees, but the lengths should be
(-1+sqrt(5))/2 (the golden ratio) of those in the previous step. This
ratio is optimal (I reckon) under the constraint that the lines do not
cross each other.
---
J K Haugland
http://home.no.net/zamunda
x^3-1
for 120 degrees
x^2-x-1
for the golden mean.
If you form the product:
(x^3-1)*(x^2-x-1 )
which is representationve of a substitution matrix:
M = {{0, 1, 0, 0, 0},
{0, 0, 1, 0, 0},
{1, 0, 0, 0, 0},
{0, 0, 0, 1, 1},
{0, 0, 0, 1, 0}}
1->2
2->3
3->1
4->4,5
5->4
which gives a vector matrix Markov:
v[0] = {0, 0, 0, 0, 1}
{0, 0, 0, 0, 1}
v[n_] := v[n] = M.v[n - 1]
a = Table[v[n][[4]], {n, 0, 20}]
{0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,
144, 233, 377, 610, 987, 1597, 2584, 4181, 6765}
b = Table[v[n][[5]], {n, 0, 20}]
{1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597,
2584, 4181}
That is an outward spiral with ratio of the golden mean.
The other terms are zero.
The dynamics aren't dominated by the
cyclotomic x^3-1 part , but by the Pisot x^2-x-1 part.If you use lengths 1 at
120 degress and two offshoots, you get a set of triangular shaped binary tree structures.
If you use 90 degrees you get the H fractal ( sometimes called the road map).
If you use a length shorter than one without scale reduction
they eventually cross, with a scale reduction you get a limiting tree that doesn't cross.
These kinds of fractals are delt with with L systems and
IFS usually. The name given this kind of structure is a pythagorean tree
when based on squares and triangles.If the right triangle is formed :
c^2=a^2+b^2
where c=1 and b=phi:
a=Sqrt[1-Phi^2)=Sqrt[1 - 0.6180339887498948^2]=0.786151
Where c=Phi and b=Phi^2
a=Sqrt[Phi^2-Phi^4]=0.485868
Clear[a]
Phi = 0.6180339887498948
a[1] = 0.786151
a[n_] := a[n] = a[n - 1]*Phi
Table[a[n], {n, 1, 20}]
{0.786151`, 0.4858680382897186`,
0.3002829617102814`, 0.18558507657943715`, 0.11469788513084422`,
0.0708871914485929`, 0.0438106936822513`, 0.02707649776634159`,
0.01673419591590971`, 0.010342301850431878`, 0.006391894065477831`,
0.003950407784954045`, 0.002441486280523785`, 0.0015089215044302594`,
0.0009325647760935253`, 0.0005763567283367339`, 0.00035620804775679115`,
0.00022014868057994264`, 0.00013605936717684845`, 0.00008408931340309415`}
which converges eventually to, zero. So the sum of the length comes to near 2.05817.
Table[Apply[Plus, Table[a[n], {n, 1, m}]], {m, 20, 200, 5}]
I hope that helps...
.
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