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I got a challenge in a learning group today to draw this shape in python turtle library.
I cannot figure out a way to express the geometrical solution to find the angles to turn and the size of the line I need.
Can you please tell me how to draw the first polygon alone? I already know how to make the pattern.
I am in fifth grade. So please give me a solution that I can understand.
Here is the solution I came up with. It is based on this diagram:
Math background
My solution uses "trigonometry", which is a method for calculating the length of one side of a triangle from the length of another side and the angles of the triangle. This is advanced math which I would expect to be taught maybe in 9th or 10th grade. I do not expect someone in 5th grade to know trigonometry. Also I cannot explain every detail of trigonometry, because I would have to write a lot and I do not think I have the teaching skills to make it clear. I would recommend you to look at for example this video to learn about the method:
https://www.youtube.com/watch?v=5tp74g4N8EY
You could also ask your teacher for more information, or research about it on the internet on your own.
Step 1: Calculating the angles
We can do this without trigonometry.
First, we see there is a "pentagon" (5-sided polygon) in the middle. I want to know the inner angle of a corner in this "pentagon". I call this angle X:
How can we calculate the angle X? We first remember that the sum of the inner angles in a triangle is 180°. We see that we can divide a 5-sides polygon into 5-2 triangles like this:
The sum of the inner angle of each of these 5-2 triangles is 180°. So for the whole 5-sided polygon, the sum of the inner angles is 180° * (5-2). Since all angles have the same size, each angle is 180°*(5-2) / 5 = 108°. So we have X = 108°.
The angle on the other side is the same as X. This allows us the calculate the angle between the two X. I will call this angle Y:
Since a full circle is 360°, we know that 360° = 2*X + 2*Y. Therefore, Y = (360° - 2*X) / 2. We know that X = 108°, so we get Y = 72°.
Next, we see there is a triangle containing the Y angle. I want to know the angle Z at the other corner of the triangle:
The inner angles of a triangle sum up to 180°*(3-2) = 180°. Therefore, we know that 180° = 2*Y + Z, so Z = 180° - 2*Y. We know that Y = 72°, so we get Z = 36°.
We will use the angle Z a lot. You can see that every corner of the green star has angle Z. The blue star is the same as the green star except it is rotated, so all blue corners also have angle Z. The corners of the red star are twice as wide as the corners of the green and blue stars, so the corners of the red star have the angle 2*Z.
Step 2: Calculating the lengths
First, we observe that all outer corners are on a circle. We call the radius of this circle R. We do not have to calculate R. Instead, we can take any value we want for R. We will always get the same shape but in different sizes. We could call R a "parameter" of the shape.
Given some value for R, I want to know the following lengths:
Calculating A:
We start with A. We can see the following triangle:
The long side of the triangle is our radius R. The other side has length A/2 and we do not care about the third side. The angle in the right-most corner is Z/2 (with Z = 36° being the angle we calculated in the previous section). The angle S is a right-angle, so S = 90°. We can calculate the third angle T because we know that the inner angles of a triangle sum up to 180°. Therefore, 180° = S + Z/2 + T. Solving for T, we get T = 180° - S - Z/2 = 180° - 90° - 36°/2 = 72°.
Next, we use trigonometry to calculate A/2. Trigonometry teaches us that A/2 = R * sin(T). Putting in the formula for T, we get A/2 = R * sin(72°). Solving for A, we get A = 2*R*sin(72°).
If you pick some value for R, for example R = 100, you can now calculate A with this formula. You would need a calculator for sin(72°), because it would be extremely difficult to calculate this in your head. Putting sin(72) into my calculator gives me 0.951056516. So for our choice R = 100, we know that A = 2 * R * sin(72°) = 2 * 100 * 0.951056516 = 190.211303259.
Calculating B:
We use the same technique to find a formula for B. We see the following triangle:
So the bottom side is the length of our radius R. The right side is B/2. We do not care about the third side. The right-most angle is three times Z/2. The angle S is a right-angle, so we have S = 90°. We can calculate the remaining angle T with 180° = S + T + 3*Z/2. Solving for T, we get T = 180° - S - 3*Z/2 = 180° - 90° - 3*36°/2 = 36°. Ok so T = Z, we could have also seen this from the picture, but now we have calculated it anyways.
Using trigonometry, we know that B/2 = R * sin(T), so we get the formula B = 2 * R * sin(36°) to calculate B for some choice of R.
Calculating C:
We see the following triangle:
So the bottom side has length A/2 and the top side has length B. We already have formulas for both of these sides. The third side is C, for which we want to find a formula. The right-most angle is Z. The angle S is a right-angle, so S = 90°. The top-most angle is three times Z/2.
Using trigonometry, we get C = sin(Z) * B.
Calculating D:
We see the following triangle:
We already have a formula for C. We want to find a formula for D. The top-most angle is Z/2 (I could not fit the text into the triangle). The bottom-left angle S is a right-angle.
Using trigonometry, we know that D = tan(Z/2) * C. The tan function is similar to the sin from the previous formulas. You can again put it into your calculator to compute the value, so for Z = 36°, I can put tan(36/2) into my calculator and it gives me 0.324919696.
Calculating E:
Ok this is easy, E = 2*D.
Halfway done already!
Calculating F:
This is similar to A and B:
We want to find a formula for F. The top side has length F/2. The bottom side has the length of our radius R. The right-most corner has angle Z. S is a right-angle. We can calculate T = 180° - S - Z = 180° - 90° - Z = 90° - Z.
Using trigonometry, we get F/2 = R * sin(T). Putting in the formula for T gives us F/2 = R*sin(90° - Z). Solving for F gives us F = 2*R*sin(90°-Z).
Calculating G:
We see the following triangle:
The top side has length F, we already know a formula for it. The right side has length G, we want to find a formula for it. We do not care about the bottom side. The left-most corner has angle Z/2. The right-most corner has angle 2*Z. The bottom corner has angle S, which is a right-angle, so S = 90°. It was not immediately obvious to me that the red line and the green line are perfectly perpendicular to each other so that S really is a right-angle, but you can verify this by using the formula for the inner angles of a triangle, which gives you 180° = Z/2 + 2*Z + S. Solving for S gives us S = 180° - Z/2 - 2*Z. Using Z = 36°, we get S = 180° - 36°/2 - 2* 36° = 90°.
Using trigonometry, we get G = F * sin(Z/2).
Calculating H:
We see the following triangle:
The right side has length G, we already have formula for that. The bottom side has length H, we want to find a formula for that. We do not care about the third side. The top corner has angle Z, the bottom-right corner has angle S. We already know that S is a right-angle from the last section.
Using trigonometry, we get H = G * tan(Z).
Calculating I:
This is easy, I is on the same line as A. We can see that A can be divided into A = I + H + E + H + I. We can simplify this to A = 2*I + 2*H + E. Solving for I gives us I = (A - 2*H - E)/2.
Calculating J:
Again this is easy, J is on the same line as F. We can see that F can be divided into F = G + J + G. We can simplify that to F = 2*G + J. Solving for J gives us J = F - 2*G.
Writing the Python program
We now have formulas for all the lines we were interested in! We can now put these into a Python program to draw the picture.
Python gives you helper functions for computing sin and tan. They are contained in the math module. So you would add import math to the top of your program, and then you can use math.sin(...) and math.tan(...) in your program. However, there is one problem: These Python functions do not use degrees to measure angles. Instead they use a different unit called "radians". Fortunately, it is easy to convert between degrees and radians: In degrees a full circle is 360°. In radians, a full circle is 2*pi, where pi is a special constant that is approximately 3.14159265359.... Therefore, we can convert an angle that is measured in degrees into an angle that is measured in radians, by dividing the angle by 360° and then multiplying it by 2*pi. We can write the following helper functions in Python:
import math
def degree_to_radians(angle_in_degrees):
full_circle_in_degrees = 360
full_circle_in_radians = 2 * math.pi
angle_in_radians = angle_in_degrees / full_circle_in_degrees * full_circle_in_radians
return angle_in_radians
def sin_from_degrees(angle_in_degrees):
angle_in_radians = degree_to_radians(angle_in_degrees)
return math.sin(angle_in_radians)
def tan_from_degrees(angle_in_degrees):
angle_in_radians = degree_to_radians(angle_in_degrees)
return math.tan(angle_in_radians)
We can now use our functions sin_from_degrees and tan_from_degrees to compute sin and tan from angles measured in degrees.
Putting it all together:
from turtle import *
import math
# Functions to calculate sin and tan ###########################################
def degree_to_radians(angle_in_degrees):
full_circle_in_degrees = 360
full_circle_in_radians = 2 * math.pi
angle_in_radians = angle_in_degrees / full_circle_in_degrees * full_circle_in_radians
return angle_in_radians
def sin_from_degrees(angle_in_degrees):
angle_in_radians = degree_to_radians(angle_in_degrees)
return math.sin(angle_in_radians)
def tan_from_degrees(angle_in_degrees):
angle_in_radians = degree_to_radians(angle_in_degrees)
return math.tan(angle_in_radians)
# Functions to calculate the angles ############################################
def get_X():
num_corners = 5
return (num_corners-2)*180 / num_corners
def get_Y():
return (360 - 2*get_X()) / 2
def get_Z():
return 180 - 2*get_Y()
# Functions to calculate the lengths ###########################################
def get_A(radius):
Z = get_Z()
return 2 * radius * sin_from_degrees(90 - Z/2)
def get_B(radius):
Z = get_Z()
return 2 * radius * sin_from_degrees(90 - 3*Z/2)
def get_C(radius):
Z = get_Z()
return sin_from_degrees(Z) * get_B(radius)
def get_D(radius):
Z = get_Z()
return tan_from_degrees(Z/2) * get_C(radius)
def get_E(radius):
return 2 * get_D(radius)
def get_F(radius):
Z = get_Z()
return 2 * radius * sin_from_degrees(90 - Z)
def get_G(radius):
Z = get_Z()
return get_F(radius) * sin_from_degrees(Z/2)
def get_H(radius):
Z = get_Z()
return get_G(radius) * tan_from_degrees(Z)
def get_I(radius):
A = get_A(radius)
E = get_E(radius)
H = get_H(radius)
return (A - E - 2*H) / 2
def get_J(radius):
F = get_F(radius)
G = get_G(radius)
return F - 2*G
# Functions to draw the stars ##################################################
def back_to_center():
penup()
goto(0, 0)
setheading(0)
pendown()
def draw_small_star(radius):
penup()
forward(radius)
pendown()
Z = get_Z()
left(180)
right(Z/2)
E = get_E(radius)
H = get_H(radius)
I = get_I(radius)
for i in range(0,5):
penup()
forward(I)
pendown()
forward(H)
penup()
forward(E)
pendown()
forward(H)
penup()
forward(I)
left(180)
right(Z)
back_to_center()
def draw_green_star(radius):
pencolor('green')
draw_small_star(radius)
def draw_blue_star(radius):
pencolor('blue')
Z = get_Z()
left(Z)
draw_small_star(radius)
def draw_red_star(radius):
pencolor('red')
Z = get_Z()
penup()
forward(radius)
pendown()
left(180)
right(Z)
G = get_G(radius)
J = get_J(radius)
for i in range(0,10):
pendown()
forward(G)
penup()
forward(J)
pendown()
forward(G)
left(180)
right(2*Z)
back_to_center()
def draw_shape(radius):
draw_green_star(radius)
draw_blue_star(radius)
draw_red_star(radius)
radius = 400
draw_shape(radius)
done()
Output:
Here's a different solution. It's based on a kite polygon where the upper portion is a pair of 3-4-5 right triangles and the lower portion is a pair of 8-15-17 right triangles:
from turtle import Screen, Turtle
KITES = 10
RADIUS = 100
def kite(t):
t.right(37)
t.forward(100)
t.right(81)
t.forward(170)
t.right(124)
t.forward(170)
t.right(81)
t.forward(100)
t.right(37)
turtle = Turtle()
turtle.penup()
turtle.sety(-RADIUS)
for _ in range(KITES):
turtle.circle(RADIUS, extent=180/KITES)
turtle.pendown()
kite(turtle)
turtle.penup()
turtle.circle(RADIUS, extent=180/KITES)
turtle.hideturtle()
screen = Screen()
screen.exitonclick()
(Yes, I'm obsessed with this puzzle.) I was sufficiently impressed by the brevity of #AnnZen's solution, I decided to see if I could come up with an even shorter one. The only unique structure in this polygon is the side of the kite:
So the problem becomes drawing ten of these in a circular fashion, and then reversing the code to draw them again in the opposite direction:
from turtle import *
for _ in range(2):
for _ in range(10):
fd(105)
lt(90)
fd(76.5)
pu()
bk(153)
rt(54)
pd()
lt(72)
lt, rt = rt, lt
done()
My Short code:
from turtle import *
for f, t in [(0,-72),(71,108),(71,0)]*10+[(29,90),(73,72),(73,90),(29,72)]*10:fd(f),rt(t)
Here's the solution:
import turtle
#turtle.tracer(0)
a = turtle.Turtle()
for _ in range(10):
a.forward(100)
a.right(90)
a.forward(73)
a.right(72)
a.forward(73)
a.backward(73)
a.right(108)
a.forward(73)
a.right(90)
a.penup()
a.forward(100)
a.pendown()
a.forward(100)
a.right(108)
#turtle.update()
Let's look at a yet another approach to drawing this shape. We'll start with the same diagram as #f9c69e9781fa194211448473495534
Using a ruler (1" = 100px) and protractor on the OP's original image, we can approximate this diagram with very simple code:
from turtle import Screen, Turtle
turtle = Turtle()
turtle.hideturtle()
turtle.penup() # center on the screen
turtle.setposition(-170, -125)
turtle.pendown()
for _ in range(10):
turtle.forward(340)
turtle.left(126)
turtle.forward(400)
turtle.left(126)
screen = Screen()
screen.exitonclick()
This is equivalent to drawing with a pencil and not lifting it nor overdrawing (backing up) on any line.
To make the shape we want pop out, we divide the two lines that we draw above into segments that are thinner and thicker. The first line breaks into three segments and the second into five:
To do this, we simply break the forward() calls in our loop into wide and narrow segments:
for _ in range(10):
turtle.width(4)
turtle.forward(105)
turtle.width(1)
turtle.forward(130)
turtle.width(4)
turtle.forward(105)
turtle.left(126)
turtle.width(1)
turtle.forward(76.5)
turtle.width(4)
turtle.forward(76.5)
turtle.width(1)
turtle.forward(94)
turtle.width(4)
turtle.forward(76.5)
turtle.width(1)
turtle.forward(76.5)
turtle.left(126)
Finally, we replace the thickness changes with lifting and lowering the pen:
Which is simply a matter of replacing the width() calls with penup() and pendown():
for _ in range(10):
turtle.pendown()
turtle.forward(105)
turtle.penup()
turtle.forward(130)
turtle.pendown()
turtle.forward(105)
turtle.left(126)
turtle.penup()
turtle.forward(76.5)
turtle.pendown()
turtle.forward(76.5)
turtle.penup()
turtle.forward(94)
turtle.pendown()
turtle.forward(76.5)
turtle.penup()
turtle.forward(76.5)
turtle.left(126)
I basically got my angle vector from two points:
Point aa is going to Point bb.... (Window.width=800, Window.height=600)
aa = (50.0*Window.width/100.0, 50.0*Window.height/100.0)
bb = (10.0*Window.width/100.0, 70.0*Window.height/100.0)
Angle = Vector(bb)-Vector(aa)
print(Angle)
[-320, 120]
Is there a way to convert the vector into a single number that represents the angle...like 90.0, 45.0, 180, etc...? Keep in mind, we're going from aa to bb....that angle.
Another way to put it is, I want to convert the above "Angle" value into a single number.
There's no Z axis here. Only the 2d two points of aa and bb. The Window x-cord is defined going from left to right, with 0 starting at the left.
The Window y-cord is defined going from bottom to top, with 0 starting at the bottom.
Trying to be more clear here...
Vector point aa is like the middle of a circle. Degree zero will start at the the middle-right edge of the circle and going counter clockwise will increase the degrees, until you hit 360 which places you back at the middle-right edge.
Vector point aa can move but regardless of its position in the Window, I want to calculated the angle between aa and the point its moving towards (bb), with aa being the center of the circle and degrees going counter clockwise like I explained above.
Hope that sheds some more light.
Ok, I found a python function that works but not perfect.
def GetAngleOfLineBetweenTwoPoints(self, p1, p2):
xDiff = p2[0] - p1[0]
yDiff = p2[1] - p1[1]
return degrees(atan2(yDiff, xDiff))
What happens is,
I do get the correct degree but it's in a split screen format and what I mean by that is, the top half of my screen going from right to left is 0 to 180 positive. The bottom half of my screen going from right to left is 0 to -180.
See what's going on there?
How can I make that function return a value between 0 to 360, like a whole circle instead of the 0 to 180 +/- split like it's currently doing?
You can use some trigonometry :)
tan(angle) = opposite / adjacent
angle = arctan ( opposite / adjacent )
your opposite will be: bb.height - aa.height
your adjacent will be: bb.width - aa.width
I hope that helps.
trigonometry
This answer uses radians https://stackoverflow.com/a/2827475/4711754
# Copy of link
import math
def dotproduct(v1, v2):
return sum((a*b) for a, b in zip(v1, v2))
def length(v):
return math.sqrt(dotproduct(v, v))
def angle(v1, v2):
"""angle between two vectors"""
return math.acos(dotproduct(v1, v2) / (length(v1) * length(v2)))
To convert to degree divide by 2*math.pi and multiple by 360.
Given I believe you want angle to horizontal you'll want
def angle(v1, v2=(1,0)):
"""angle between two vectors"""
return math.acos(dotproduct(v1, v2) / (length(v1) * length(v2)))*(360/(2*math.pi))
Examples
>>> a=(0, 100)
>>> angle(a)
90.0
>>> a=(100, 0)
>>> angle(a)
0.0
>>> a=(40, 40)
>>> angle(a)
45.00000000000001
>>> a=(-40, 40)
>>> angle(a)
135.0
EDIT
Below is version is origin vector is center of window
def angle(v1):
"""angle around window"""
v2=(1, 0)
v1 = (v1[0] - Window.width/2, v1[1] - Window.height/2)
angle_between_radians = math.acos(dotproduct(v1, v2) / (length(v1) * length(v2)))
angle_between_degrees = angle_between_radians *(360/(2*math.pi))
if v1[1] >= 0:
return angle_between_degrees
else:
return 360 - angle_between_degrees
# or hardcoded this would be
def angle(v1):
"""angle around window"""
v2=(1, 0)
v1 = (v1[0] - 400, v1[1] - 300)
angle_between_radians = math.acos(dotproduct(v1, v2) / (length(v1) * length(v2)))
angle_between_degrees = angle_between_radians *(360/(2*math.pi))
if v1[1] >= 0:
return angle_between_degrees
else:
return 360 - angle_between_degrees
Examples
>>> angle((400, 400))
90.0
>>> angle((500, 300))
0.0
>>> angle((440, 340))
45.00000000000001
>>> angle((360, 340))
135.0
>>> angle((360, 260))
225.0
Got it!
from math import atan2, degrees, pi, cos, sin
def GetAngleOfLineBetweenTwoPoints(self, p1, p2):
xDiff = p2[0] - p1[0]
yDiff = p2[1] - p1[1]
val = degrees(atan2(yDiff, xDiff))
if str(val).find('-')!=-1:
val = float(360)-float(val)
return val
Yaaaaaayyyyyyy!!!!!!!
I'm new to this, but I'm trying to create a program that goes on a Random Walk in turtle and I have no idea what I'm doing.This is the program that I need to create Requirements, and this is what I have so far Code. I'm starting to get a little lost, and am wondering if I 'm headed towards the right direction or if i need to scrap it and start all over again
Thanks,
import turtle
import random
import math
def start():
myS = turtle.Screen()
myS.title("Random Walk")
border = 50
myWin.setworldcoordinates(-border, -border, border, border)
def go(heading, step_size):
turtle.setheading(heading)
turtle.forward(step_size)
def random_walk(step_size, steps):
angle = random.random() * 2 * math.pi
x = 0
y = 0
x = x + math.cos(angle)
y = y + math.sin(angle)
coord = (x, y)
for _ in range(steps):
go(random.choice(coord), step_size)
if __name__ == '__main__':
turtle.hideturtle()
turtle.speed('fastest')
random_walk(15, 1000)
Your random_walk function does way to many things; I think it grew over your head a little. What it should do for every step:
Calculate a random angle between 0 and 2*math.pi
Calculate x and y with the given formulas
Call turtle.goto(x,y) to go to these coordinates
Also there was a little typo in start.
import turtle
import random
import math
def start():
myS = turtle.Screen()
myS.title("Random Walk")
border = 50
myS.setworldcoordinates(-border, -border, border, border)
def go(heading, step_size):
turtle.setheading(heading)
turtle.forward(step_size)
def random_walk(steps):
x = 0
y = 0
for i in range(steps):
angle = random.random()*2*math.pi
x = x + math.cos(angle)
y = y + math.sin(angle)
turtle.goto(x,y)
if __name__=="__main__":
start()
random_walk(100)
Because this looks like a homework assignment I don't want to just fix your errors but instead give you pointers to the right solution. I will also hide hints behind spoilers...
if you run random_walk() with e.g. parameters random_walk(1, 1) your will notice that the direction of the walk will not appear random. Use some form of debugging to find out why that is
your go() functions expects an angle, look at what random.choice() produces
if you fix the issue number 1. you will see that the turtle will not draw a random line at all again
looks like the angle does not change every step. find a solution that updates the angle before each step!
units of angles: find out what unit setheading() expects: hint
your requirements mention the output of straight line distance and distance travelled. This somehow sounds like you need some form of tally variable (one or several). hint: if you read the turtle documentation carefully you might find a funtion that maybe makes this task easier
read the documentation for turtle.position().
I'm using Python Turtles to draw a circle using forward() and right().
I have a for loop counting from 0 to 359, and each time it triggers, it moves the turtle forward 1 and right 1.
But the problem is I need specific diameters. I am nearly 100% sure I'll need to use trig, but I've tried to no avail.
I can't figure out the math how to do it. We're supposed to use forward() and right(), NOT circle().
Thanks!
Here is a working example:
import turtle
import math
def circle(radius):
turtle.up()
# go to (0, radius)
turtle.goto(0,radius)
turtle.down()
turtle.color("black")
# number of times the y axis has been crossed
times_crossed_y = 0
x_sign = 1.0
while times_crossed_y <= 1:
# move by 1/360 circumference
turtle.forward(2*math.pi*radius/360.0)
# rotate by one degree (there will be
# approx. 360 such rotations)
turtle.right(1.0)
# we use the copysign function to get the sign
# of turtle's x coordinate
x_sign_new = math.copysign(1, turtle.xcor())
if(x_sign_new != x_sign):
times_crossed_y += 1
x_sign = x_sign_new
return
circle(100)
print('finished')
turtle.done()
Well, a complete circle is 360°, and you are planning on turning 360 times, so each turn should be:
right( 360 ° / 360 ), or
right(1)
The distance traveled will be one circumference, or π * diameter, so your forward might be:
forward( diameter * π / 360 )
I haven't tested this yet -- give it a try and see how it works.
This is one of the exercises in "Think Python," in chapter 4. It really is a horrible exercise to have this early in the book, especially with the "hint" given. I'm using forward and left here, but you can switch left with right.
You should have the polygon function:
def polygon(t, length, n):
for i in range(n):
bob.fd(length)
bob.lt(360 / n)
Then you create a circle function:
def circle(t):
polygon(t, 1, 360)
That will draw a circle, no radius needed. The turtle goes forward 1, then left 1 (360 / 360), 360 times.
Then, if you want to make the circle bigger, you calculate the circumference of the circle. The hint says:
Hint: figure out the circumference of the circle and make sure that
length * n = circumference.
Ok, so the formula for circumference = 2 * pi * radius. And the hint says length * n = circumference. n = 360 (number of sides/degrees). We have circumference, so we need to solve for length.
So:
def circle(t, r):
circumference = 2 * 3.14 * r
length = circumference / 360
polygon(t, length, 360)
Now, call the function with whatever radius you want:
circle(bob, 200)
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
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;-).
I liked your question so much that I posted an answer to it as a new question, so that other people could improve it:
https://stackoverflow.com/questions/7420248
I used no Logo/Turtle stuff, neither trigonometry.
Congrats for being the first one bringing this problem to StackOverflow!
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
Something to consider, if not for your implementation then for testing your implementation, is that Python turtle can record what it's doing and give you back the coordinates. You use begin_poly() and end_poly() around the code you want to record and then use get_poly() afterwards to get the points.
In this example, I'll draw the snowflake based on code from this site then register those coordinates back as a new turtle shape that I'll randomly (and quickly) stamp about the screen:
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()
You can have the pen up and turtle hidden during polygon generation if you don't want this step to be seen by the user.