PYGAME- “ Koch Snowflake ”

 

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             Fractals are one example of "algorithmic art" . Here discussing about a particular kind of fractal known as the Koch Snowflake. A fractal is an image that keeps on repeating itself every new time it gets smaller and smaller. The snowflake is a fractal of the Koch curve. Helge von Koch who is responsible for the snowflake. He wrote about the fractal snowflake in a paper written in 1906. The most interesting part about a Koch snowflake is that when you continue this creating process the curve is infinitely long but has an finite area. This means the the sides of the snowflake are never ending but at the same time the area it encloses is only going to be 1.6 times bigger then the original area of the original triangle. 

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Design Steps

Constructing a Koch snowflake is a 

 step by step process.

Step1

Start with an equilateral triangle.

Step2

Split each side into three equal segments.

step3

Create an smaller equilateral triangle in the middle of the three parts from step 2

step4

Remove the line segment that is the base of the triangle from step 3

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              If you continue repeating this procedure, the curve will never self-intersect, and in the limit you get a shape known as the Koch snowflake. Each time you repeat the process this is call an iteration. To find the number of sides all you need to know is how many iterations you have completed. 

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The Math Behind It

Number of sides (n)

            For each iteration, one side of the figure from the previous stage becomes four sides in the following stage. Since we begin with three sides, the formula for the number of sides in the Koch Snowflake is

                                   n = 3*4^a        in the a^th iteration.

For iterations 0, 1, 2 and 3, the number of sides are 3, 12, 48 and 192, respectively.

 Length of side (length)

     In every iteration, the length of a side is 1/3 the length of a side from the preceding stage. If we begin with an equilateral triangle with side length x, then the length of a side in iteration a is

                                              length = x*3^-a

For iterations 0 to 3, length = a, a/3, a/9 and a/27. 

Line equations 

  

    Some basic line equations are used in the construction of a snowflake, that are

1) Midpoint of a line is :-  consider a line with end points (x1,y1) and (x2,y2).

           it's mid point is (X,Y)            

                                                   X = (x1+x2)/2

                                                            

                                                   Y = (y1+y2)/2 

2) Distance  between two points :-  Consider two points (x1,y1) and (x2,y2) 

              distance = ( ( x1 − x 2 )² + ( y1 − y 2 )² )^1/2 

3) Find point on a line :-  consider a line having points (x1,y1) and (x2,y2).

       Distance  between these two point is D. To find an another point (X,Y) on the

       line, L distance away from the point (x1,y1) is

                                       X = x1 + k (x2 - x1)                    

                                       Y = y1 + k (y2 - y1)            where k = L/D

 

 
                                                                 

PROGRAM

#importing pygame
import pygame,sys
from pygame.locals import *

#initialise screeen
pygame.init()
screen= pygame.display.set_mode((600, 600))
pygame.display.set_caption('koch snowflake')
White=(255,255,255)
Black=(0,0,0)
Blue= (0,0,255)
screen.fill(White)
clock=pygame.time.Clock()

#divide line in to three and remove the middle section
def cutline(x,y,n):
    a=((x[0]+((y[0]-x[0])*.33333333333333333)))
    b=((x[1]+((y[1]-x[1])*.33333333333333333)))
    c=((x[0]+((y[0]-x[0])*.66666666666666666)))
    d=((x[1]+((y[1]-x[1])*.66666666666666666)))
    if n<=2:
       pygame.draw.line(screen,White,( a,b),(c,d),4)
    else:
       pygame.draw.line(screen,White,( a,b),(c,d),1)
    l=[(a,b),(c,d)]
    return l
                                                  #                                                       
#find midpoint of a line and draw two sides of triangle(like this  __/\__ )
def point(a,b,c):
    p=((a[0]+b[0])*.5)
    q=((a[1]+b[1])*.5)
    x=((c[0]+((p-c[0])*1.3333333333333333333)))
    y=((c[1]+((q-c[1])*1.3333333333333333333)))
    n=(x,y)
    pygame.draw.line(screen,Black,( b),(x,y),1)
    pygame.draw.line(screen,Black,( a),(x,y),1)
    return n

# divide a line and find a point 1/3 distance away from one point
def next(p,q):
    a=((p[0]+((q[0]-p[0])*.33333333333333333)))
    b=((p[1]+((q[1]-p[1])*.33333333333333333)))
    l=(a,b)
    return l

# main loop
while True:
   n=0
   l=[]
   l1=[]
   screen.fill(White)
   pygame.display.update()

# draw a equilateral triangle
   pygame.draw.line(screen,Black,( 150,450),(450, 450),1)
   topy=((((300**2) - (150**2))**.5))
   p=(450-topy)
   pygame.draw.line(screen,Black,( 150,450),(300, p),1)
   pygame.draw.line(screen,Black,( 300,p),(450, 450),1)
   l1=[(150,450),(450,450),(300,p)]
   pygame.display.update()
   l.append((l1))
   l1=[(150,450),(300,p),(450,450)]
   l.append((l1))
   l1=[(300,p),(450,450),(150,450)]
   l.append((l1))
 
# sub loop
   while n<=6:
     l2=l
     l=[]

# iteration process

     for i in range(len(l2)):
         a,b=cutline(l2[i][0],l2[i][1],n)
         c=point(a,b,l2[i][2])
         x=next(l2[i][0],l2[i][2])
         y=next(l2[i][1],l2[i][2])
         l1=[a,c,b]
         l.append((l1))
         l1=[b,c,a]
         l.append((l1))
         l1=[l2[i][0],a,x]
         l.append((l1))
         l1=[l2[i][1],b,y]
         l.append((l1))
   
   
     n+=1
     for event in pygame.event.get():
        if event.type==QUIT:
            pygame.quit()
            sys.exit()
     clock.tick(1)
     pygame.display.update()