Draw a Golden Spiral!

Question

Given a number n, you must output a golden spiral with the number of quarter turns being n.

The spiral must be completely black and the background must be completely white. This must be output as a φ^⌊n/4⌋*4 by φ^⌊n/4⌋*4 PNG file (you may output a PPM and convert it later). The shortest code wins.

The radius of the smallest quarter turn should be φ^-3.

In geometry, a golden spiral is a logarithmic spiral whose growth factor is φ, the golden ratio.That is, a golden spiral gets wider (or further from its origin) by a factor of φ for every quarter turn it makes.

Answer 1

Mathematica, 169 bytes

All those annoying options. :(

f=Export[p=GoldenRatio;"a.png",ParametricPlot[p^(2t-3){Cos[a=t*Pi],Sin@a},{t,0,#/2},PlotRange->{r={m=-p^(#-3),-m},r},PlotStyle->Black,Axes->1<0,ImageSize->p^#~Floor~4]]&

This function saves the spiral to a.png in wherever your current working directory is.

The result of f[9]:

The result of f[12]:

I hope I understood all the bits about the scaling correctly.

There's not a whole lot to say about the code. I'm using ParametricPlot to draw curve as a parameterised function in t which advances by 1/2 for each quarter turn. The rest is done in the options:

• PlotRange->{r={m=-p^(#-3),-m},r} makes sure we have an square aspect ratio which just covers the outermost point of the spiral.
• PlotStyle->Black overwrites Mathematica's blue default colour.
• Axes->1<0 is just Axes->False (and turns off the axes, who knew!).
• ImageSize->p^#~Floor~4 sets the correct dimensions in pixels, by noticing that ⌊n/4⌋*4 just means "n rounded down to the nearest multiple of 4".

Answer 2

Python 3 - 290 269

This will get crushed by any kind of built-in parametrics, but I figured I'd post it as an example of a point-by-point solution.

Edit: Thanks for the golfing tips! If you have any more, please make them.

from math import *
from PIL import Image as I, ImageDraw as D
n=eval(input())
G=(1+5**.5)/2
w=int(G**(4*(n//4)))
i=I.new("RGB",(w,w),"white")
d=D.Draw(i)
k=pi/180
d.line([(G**(j/90)*cos(j*k)+w/2,G**(j/90)*sin(j*k)+w/2)for j in range(n*90)],fill=0)
i.save("s.png","PNG")