Implementing the Koch Curve?

Question

I was looking at the wikipedia page for the Koch Snowflake (here) and was bothered by the all the examples all being in the logo/turtle style. So i set out to make my own that returned a list or coordinates.

My implementation is in python and i basically ripped off the python turtle implementation but replaced the turtle specific stuff with basic trig. It resulted in some ugly code. My challenge for you is to either improve my code or come up with a more elligant solution of your own. It can be in python, or your favorite language.

My Code:

from math import sin, cos, radians

def grow(steps, length = 200, startPos = (0,0)):
    angle = 0
    try:
        jump = float(length) / (3 ** steps)
    except:
        jump = length

    set="F"
    for i in xrange(steps): set=set.replace("F", "FLFRFLF")

    coords = [startPos]
    for move in set:
        if move is "F": 
            coords.append(
              (coords[-1][0] + jump * cos(angle),
               coords[-1][1] + jump * sin(angle)))
        if move is "L":
            angle += radians(60)
        if move is "R":
            angle -= radians(120)

    return coords

EDIT: due to lazy copying, i forgot the import

Answer 1

I don't see it as particularly ugly and I'd only refactor it incrementally, e.g. as a first step (I've removed the try/except because I don't know what you're trying to ward against... if it needs to get back in it should be a bit more explicit, IMHO):

import math

angles = [math.radians(60*x) for x in range(6)]
sines = [math.sin(x) for x in angles]
cosin = [math.cos(x) for x in angles]

def L(angle, coords, jump):
    return (angle + 1) % 6
def R(angle, coords, jump):
    return (angle + 4) % 6
def F(angle, coords, jump):
    coords.append(
        (coords[-1][0] + jump * cosin[angle],
         coords[-1][1] + jump * sines[angle]))
    return angle

decode = dict(L=L, R=R, F=F)

def grow(steps, length=200, startPos=(0,0)):
    pathcodes="F"
    for i in xrange(steps):
        pathcodes = pathcodes.replace("F", "FLFRFLF")

    jump = float(length) / (3 ** steps)
    coords = [startPos]
    angle = 0

    for move in pathcodes:
        angle = decode[move](angle, coords, jump)

    return coords

If a second step was warranted I'd probably roll this functionality up into a class, but I'm not sure that would make things substantially better (or, better at all, in fact;-).

Answer 2

Mathematica is superior when it comes to math stuff:

points = {{0.0, 1.0}};
koch[pts_] := Join[
    pts/3,
    (RotationMatrix[60 Degree].#/3 + {1/3, 0}) & /@ pts,
    (RotationMatrix[-60 Degree].#/3 + {1/2, 1/Sqrt[12]}) & /@ pts,
    (#/3 + {2/3, 0}) & /@ pts
];
Graphics[Line[Nest[koch, points, 5]], PlotRange -> {{0, 1}, {0, 0.3}}] //Print