The implementation of the Koch curve?

Question

I looked at the wikipedia page for the Koch snowflake ( here ) and was worried about all the examples in the logo / turtle. So I decided to make my own, which returned a list or coordinates.

My implementation is in python, and I basically ripped off the python tortoise implementation, but replaced the specific turtle material with a basic trigger. This led to some ugly code. My task for you is to either improve my code, or come up with a more personal solution. It could be in python or your favorite language.

My code is:

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 copy I forgot about import

Answer 1

, , . ( try/except, , ... , , 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

, , , , , (, , ; -).

Answer 2

, , :

https://stackoverflow.com/questions/7420248

/, .

, StackOverflow!

Answer 3

Mathematica , :

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

Answer 4

- , , , , Python , , . begin_poly() end_poly() , , get_poly(), .

, , ( ) :

import turtle
from random import random, randrange

def koch_curve(turtle, steps, length):
    if steps == 0:
        turtle.forward(length)
    else:
        for angle in [60, -120, 60, 0]:
            koch_curve(turtle, steps - 1, length / 3)
            turtle.left(angle)

def koch_snowflake(turtle, steps, length):
    turtle.begin_poly()

    for _ in range(3):
        koch_curve(turtle, steps, length)
        turtle.right(120)

    turtle.end_poly()

    return turtle.get_poly()

turtle.speed("fastest")

turtle.register_shape("snowflake", koch_snowflake(turtle.getturtle(), 3, 100))

turtle.reset()

turtle.penup()

turtle.shape("snowflake")

width, height = turtle.window_width() / 2, turtle.window_height() / 2

for _ in range(24):
    turtle.color((random(), random(), random()), (random(), random(), random()))
    turtle.goto(randrange(-width, width), randrange(-height, height))
    turtle.stamp()

turtle.done()

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