Fill the five pointed stars with random colors with random colors. Refer to this project that draw uncolored five pointed star fractal.
Colored Five Pointed Star Fractal
Category Archives: math
Five Pointed Star Fractal with Python and Turtle (Source Code)
Draw a five pointed star fractal that have nested five pointed star as shown.
Hint: Create a function that draws five pointed star in any location, orientation, and size.
Source Code:
import turtle
import math
screen = turtle.Screen()
screen.title('Five Pointed Star Fractal - PythonTurtle.Academy')
screen.setup(1000,1000)
screen.setworldcoordinates(-1000,-1000,1000,1000)
turtle.speed(0)
turtle.hideturtle()
turtle.pensize(1)
def five_pointed_star(x,y,direction,r): #x,y is the center
turtle.up()
turtle.goto(x,y)
turtle.seth(direction)
turtle.fd(r)
turtle.right(180-18)
turtle.down()
length = r*math.sin(math.pi*2/5)/(1+math.sin(math.pi/10))
for _ in range(5):
turtle.fd(length)
turtle.left(72)
turtle.fd(length)
turtle.right(180-36)
def five_pointed_star_fractal(x,y,direction,r): #x,y is the center
five_pointed_star(x,y,direction,r)
if r < 20: return
five_pointed_star_fractal(x,y,180+direction,r*math.sin(math.pi/10)/math.cos(math.pi/5))
five_pointed_star_fractal(0,-40,90,1000)
Five Pointed Star with Python Turtle
Draw a five pointed star as shown preferably with a loop.
Pentagram Fractal with Python Turtle (Source Code)
Draw the following fractal made up of nesting pentagrams with either recursion or iteration.
Hint: Making a function that draws pentagram in any location, radius, and direction will help!
Source Code:
import turtle
import math
screen = turtle.Screen()
screen.title('Pentagram Fractal - PythonTurtle.Academy')
screen.setup(1000,1000)
screen.setworldcoordinates(-1000,-1000,1000,1000)
turtle.speed(0)
turtle.hideturtle()
def star(x,y,direction,r): #x,y is the center
turtle.up()
turtle.goto(x,y)
turtle.seth(direction)
turtle.fd(r)
turtle.right(180-18)
turtle.down()
length = 2*r*math.sin(math.pi*2/5)
for _ in range(5):
turtle.fd(length)
turtle.right(180-36)
def star_fractal(x,y,direction,r):
star(x,y,direction,r)
if r < 50: return
star_fractal(x,y,180+direction,r*math.sin(math.pi/10)/math.cos(math.pi/5))
star_fractal(0,0,90,1000)Mandelbrot Set with Different Paramaters (Source Code)
Using the center coordinate and zoom factor to draw the following Mandelbrot Set visualizations.
Source Code:
from PIL import Image
import colorsys
import math
px, py = -0.7746806106269039, -0.1374168856037867 #Tante Renate
px, py, zoom = -0.74384935657398, -0.13170134084746293, 5788441.443619884
px, py, zoom = 2.613577e-1, -2.018128e-3, 3.354786e+3
px, py, zoom = -0.59990625, -0.4290703125, 1024
px, py, zoom = -1.038650e-1, -9.584393e-1, 1.674667e+5
px, py, zoom = -0.761574, -0.0847596, 78125
px, py, zoom = -1.62917,-0.0203968, 3125
px, py, zoom = -0.75,0,1
R = 3
max_iteration = 512
w, h = 1250,1250
mfactor = 1
def Mandelbrot(x,y,max_iteration,minx,maxx,miny,maxy):
zx = 0
zy = 0
RX1, RX2, RY1, RY2 = px-R/2, px+R/2,py-R/2,py+R/2
cx = (x-minx)/(maxx-minx)*(RX2-RX1)+RX1
cy = (y-miny)/(maxy-miny)*(RY2-RY1)+RY1
i=0
while zx**2 + zy**2 <= 4 and i < max_iteration:
temp = zx**2 - zy**2
zy = 2*zx*zy + cy
zx = temp + cx
i += 1
return i
def gen_Mandelbrot_image():
bitmap = Image.new("RGB", (w, h), "white")
pix = bitmap.load()
for x in range(w):
for y in range(h):
c=Mandelbrot(x,y,max_iteration,0,w-1,0,h-1)
v = c**mfactor/max_iteration**mfactor
hv = 0.67-v*2
#if hv<0: hv+=1
r,g,b = colorsys.hsv_to_rgb(hv,1,1-(v-0.1)**2/0.9**2)
r = min(255,round(r*255))
g = min(255,round(g*255))
b = min(255,round(b*255))
pix[x,y] = int(r) + (int(g) << 8) + (int(b) << 16)
bitmap.save("Mandelbrot_"+str(px)+"_"+str(py)+"_"+str(zoom)+".jpg")
bitmap.show()
R=3/zoom
gen_Mandelbrot_image()Circle Inscribed Inside an Equilateral Triangle
Draw a circle that is perfectly inscribed inside an equilateral triangle as shown. You may want to use some math skills in solving this problem.
N Overlapping Circle with Python and Turtle (Source Code)
Generalize the three overlapping circles project by letting it draw any number of overlapping circles. The following is the overlapping circles from size 2 to size 7.
Solution
import turtle
import math
screen = turtle.Screen()
screen.title('N Overlapping Circles - PythonTurtle.Academy')
screen.setup(1000,1000)
turtle.hideturtle()
turtle.speed(0)
turtle.pensize(2)
def draw_circle(x,y,radius):
turtle.up()
turtle.goto(x,y-radius)
turtle.seth(0)
turtle.down()
turtle.circle(radius,steps=360)
r = 150
n = 3
r2 = r/2/math.sin(math.radians(180/n))
angle = 90
for _ in range(n):
draw_circle(r2*math.cos(math.radians(angle)),
r2*math.sin(math.radians(angle)),
r)
angle += 360/nThree Overlapping Circles with Python and Turtle (Tutorial)
Draw the following overlapping circles. Please note that each circle passes through the center of the other two circles.
Solution
We observe that the centers of the three circles form an equilateral triangle and the length of the equilateral triangle is the radius of the circle. So, the first step is to figure out the coordinates of the vertices of the equilateral triangle (with some math), and then draw three circles given the these three coordinates as the centers. So, it will be helpful to create a function that draws circle based on the center and radius. The following is the source code:
import turtle
screen = turtle.Screen()
screen.title('Three Circles - PythonTurtle.Academy')
screen.setup(1000,1000)
turtle.hideturtle()
turtle.speed(0)
def draw_circle(x,y,radius):
turtle.up()
turtle.goto(x,y-radius)
turtle.seth(0)
turtle.down()
turtle.circle(radius,steps=360)
r = 150
draw_circle(0,r,r*3**0.5)
draw_circle(-r/2*3**0.5,-r/2,r*3**0.5)
draw_circle(r/2*3**0.5,-r/2,r*3**0.5)What’s next?
Julia Set with Different Parameters
In this project, we are generate several Julia Sets by varying the constant c in the normal Julia Set function. Turtle will be too slow to do this work. Instead we are going to use the PIL library. Please check out Julia Set Animation project for source code.
Related Projects:
Julia Set Fractal Animation
In this project, we are going to animate the Julia Set by varying the constant c in the normal Julia Set function. Turtle will be too slow to do this work. Instead we are going to use the PIL library to generate 360 images. Then, a video editing software can be used to combine 360 images into a video file. The following is the source code:
from PIL import Image
import colorsys
import math
cx = 0.7885
cy = 0
R = (1+(1+4*(cx**2+cy**2)**0.5)**0.5)/2+0.001
max_iteration = 200
w, h, zoom = 1000,1000,1
def julia(cx,cy,R,max_iteration,x,y,minx,maxx,miny,maxy):
zx = (x-minx)/(maxx-minx)*2*R-R
zy = (y-miny)/(maxy-miny)*2*R-R
i=0
while zx**2 + zy**2 < R**2 and i < max_iteration:
temp = zx**2 - zy**2
zy = 2*zx*zy + cy
zx = temp + cx
i += 1
return i
def gen_julia_image(sequence,cx,cy):
bitmap = Image.new("RGB", (w, h), "white")
pix = bitmap.load()
for x in range(w):
for y in range(h):
c=julia(cx,cy,R,max_iteration,x,y,0,w-1,0,h-1)
r,g,b = colorsys.hsv_to_rgb(c/max_iteration,1,0.9)
r = min(255,round(r*255))
g = min(255,round(g*255))
b = min(255,round(b*255))
pix[x,y] = int(b) + (int(g) << 8) + (int(r) << 16)
bitmap.save("julia_"+str(sequence)+".jpg")
for i in range(360):
cx = 0.7885*math.cos(i*math.pi/180)
cy = 0.7885*math.sin(i*math.pi/180)
gen_julia_image(i,cx,cy)