Sierpinski triangle recursion using turtle graphics

Question:

I am trying to write a program that draws a sierpinski tree with python using turtle. Here is my idea:

import turtle
def draw_sierpinski(length,depth):
    window = turtle.Screen()
    t = turtle.Turtle()
    if depth==0:
        for i in range(0,3):
            t.fd(length)
            t.left(120)
    else:
        draw_sierpinski(length/2,depth-1)
        t.fd(length/2)
        draw_sierpinski(length/2,depth-1)
        t.bk(length/2)
        t.left(60)
        t.fd(length/2)
        t.right(60)
        draw_sierpinski(length/2,depth-1)
    window.exitonclick()


draw_sierpinski(500,1)

The program does not reach the 2nd line after the else statement and I don’t know why. Can anyone help me?

Asked By: AlexConfused

||

Answers:

I don’t think you should be creating the turtle or window object inside the function. Since draw_sierpinski gets called four times if you originally call it with depth 1, then you’ll create four separate windows with four separate turtles, each one drawing only a single triangle. Instead, I think you should have only one window and one turtle.

import turtle
def draw_sierpinski(length,depth):
    if depth==0:
        for i in range(0,3):
            t.fd(length)
            t.left(120)
    else:
        draw_sierpinski(length/2,depth-1)
        t.fd(length/2)
        draw_sierpinski(length/2,depth-1)
        t.bk(length/2)
        t.left(60)
        t.fd(length/2)
        t.right(60)
        draw_sierpinski(length/2,depth-1)


window = turtle.Screen()
t = turtle.Turtle()
draw_sierpinski(500,1)
window.exitonclick()

Result:

enter image description here


These results look pretty good for a depth 1 triangle, but what about when we call draw_sierpinski(100,2)?

enter image description here

Ooh, not so good. This occurs because the function should draw the shape, and then return the turtle to its original starting position and angle. But as is evident from the depth 1 image, the turtle doesn’t return to its starting position; it ends up halfway up the left slope. You need some additional logic to send it back home.

import turtle
def draw_sierpinski(length,depth):
    if depth==0:
        for i in range(0,3):
            t.fd(length)
            t.left(120)
    else:
        draw_sierpinski(length/2,depth-1)
        t.fd(length/2)
        draw_sierpinski(length/2,depth-1)
        t.bk(length/2)
        t.left(60)
        t.fd(length/2)
        t.right(60)
        draw_sierpinski(length/2,depth-1)
        t.left(60)
        t.bk(length/2)
        t.right(60)

window = turtle.Screen()
t = turtle.Turtle()
draw_sierpinski(100,2)
window.exitonclick()

Result:

enter image description here

Answered By: Kevin

Here you go.

import turtle

def sier(side, level):
    if level == 1:
        for i in range(3):
            turtle.fd(side)
            turtle.left(120)
    else:
        sier(side/2, level-1)
        turtle.fd(side/2)
        sier(side/2, level-1)
        turtle.bk(side/2)
        turtle.left(60)
        turtle.fd(side/2)
        turtle.right(60)
        sier(side/2, level-1)
        turtle.left(60)
        turtle.bk(side/2)
        turtle.right(60)
def main():
    sier(200, 4)

if __name__ == '__main__':
    main()
    turtle.mainloop()
Answered By: Navneet Sinha

This is the best code for sierpinski triangle

def sierpinski(a, n):
if n == 0:
    t.begin_fill()
    for i in range(3):
        t.fd(a)
        t.lt(120)
    t.end_fill()
    return
sierpinski(a / 2, n - 1)
t.pu()
t.fd(a / 2)
t.pd()
sierpinski(a / 2, n - 1)
t.pu()
t.lt(120)
t.fd(a / 2)
t.rt(120)
t.pd()
sierpinski(a / 2, n - 1)
#
# We should return home! This is important!
#
t.pu()
t.lt(60)
t.bk(a / 2)
t.rt(60)
t.pd()
Answered By: lisi
from turtle import *
import turtle
t = turtle.Turtle()
Window = turtle.Screen()

Window.bgcolor('white')

turtle.color('white')
goto(-200, -200)
def serp_tri(side, level):
    if level == 1:
        for i in range(3):
            turtle.color('black')
            turtle.ht()
            turtle.fd(side)
            turtle.left(120)
            turtle.speed(100000)

else:
    turtle.ht()
    serp_tri(side/2, level-1)
    turtle.fd(side/2)
    serp_tri(side/2, level-1)
    turtle.bk(side/2)
    turtle.left(60)
    turtle.fd(side/2)
    turtle.right(60)
    serp_tri(side/2, level-1)
    turtle.left(60)
    turtle.bk(side/2)
    turtle.right(60)
    turtle.speed(100000)

def main():
    serp_tri(400, 8)

if __name__ == '__main__':
    main()
    turtle.mainloop()

I looked at a similar program and wrote this using some of the same things. This will give you the biggest triangle that you can get. Hope this helps!

Answered By: HW00D

As a suggestion here is my solution. Any comments are very much appreciated as it seems like it is still not the most efficient algorithm.

import turtle

def sier(tur, order, size):
    """ Draw Sierpinski triangle """
    if order == 0:
        for _ in range(3):
            tur.forward(size)
            tur.left(120)
    else:
        step = size / 2
        for t1, m1, t2, m2 in [(0, step, 0, 0),
                               (120, step, -120, 0),
                               (-60, step, 60, -(step))]:
            sier(tur, order - 1, step)
            tur.left(t1)
            tur.forward(m1)
            tur.left(t2)
            tur.forward(m2)


if __name__ == '__main__':
    odr = int(input("Enter the order: "))
    sz = int(input("Enter the size: "))

    root = turtle.Screen()
    root.bgcolor("lightgreen")

    alex = turtle.Turtle()
    alex.color('blue')
    alex.speed(100)

    sier(alex, odr, sz)

    root.mainloop()
Answered By: Nicko
# PEP8 Verified
'''
The Sierpinski function relies heavily on the getMid function. getMid takes
as arguments two endpoints and returns the point halfway between them. In
addition, this program has a function that draws a filled triangle using
the begin_fill and end_fill turtle methods.
'''


import turtle


def drawTriangle(points, color, myTurtle):
    myTurtle.fillcolor(color)
    myTurtle.up()
    myTurtle.goto(points[0][0], points[0][1])
    myTurtle.down()
    myTurtle.begin_fill()
    myTurtle.goto(points[1][0], points[1][1])
    myTurtle.goto(points[2][0], points[2][1])
    myTurtle.goto(points[0][0], points[0][1])
    myTurtle.end_fill()


def getMid(p1, p2):
    return ((p1[0] + p2[0]) / 2, (p1[1] + p2[1]) / 2)


def sierpinski(points, degree, myTurtle):
    colormap = ['blue', 'red', 'green', 'white', 'yellow', 'violet', 'orange']
    drawTriangle(points, colormap[degree], myTurtle)
    if degree > 0:
        sierpinski([points[0],
                    getMid(points[0], points[1]),
                    getMid(points[0], points[2])],
                   degree-1, myTurtle)
        sierpinski([points[1],
                    getMid(points[0], points[1]),
                    getMid(points[1], points[2])],
                   degree-1, myTurtle)
        sierpinski([points[2],
                    getMid(points[2], points[1]),
                    getMid(points[0], points[2])],
                   degree-1, myTurtle)


def main():
    myTurtle = turtle.Turtle()
    myWin = turtle.Screen()
    myPoints = [[-100, -50], [0, 100], [100, -50]]
    sierpinski(myPoints, 3, myTurtle)
    myWin.exitonclick()

main()
Answered By: Kenneth Chang

starting from Navneet Sinha, i would suggest this:

def sierpinski(t,order,size):
try:
    order=int(order)
    size=float(size)
    if order==0:
        for i in range(0,3):
            t.pendown()             
            t.forward(size)         
            t.left(120)
            t.penup()               
    else:
        for (angle,move) in ([0,size],[60,-size],[-60,-size]):
            sierpinski(t,order-1,size/2)       
            t.right(angle)                              
            t.forward(move/2)                           
            t.left(angle)
except ValueError:
    None

def test_turtle():
    import turtle

    screen=turtle.Canvas()
    tess=turtle.Turtle()
    tess.shape("arrow")
    tess.shapesize(0.2)
    tess.speed(0)

    ords=input("define the order of the fractal: ")
    sz=input("define the size of the segment: ")

    tess.penup()
    tess.backward(float(sz)/2)
    tess.right(90)
    tess.forward(float(sz)/3)
    tess.left(90)
    tess.pendown()

    sierpinski(tess,ords,sz)

    screen.mainloop()

test_turtle()    
Answered By: andrea

The code is really simple:

def sierpinski(size:int,depth:int,up:bool = True):
    if depth==0:
        t.forward(size)
        return
    size = size/2
    depth = depth-1
    left = lambda deg:t.left(deg)
    right = lambda deg:t.right(deg)
    if not up:
        k = left
        left = right
        right = k
        del k
    left(60)
    sierpinski(size,depth,not up)
    right(60)
    sierpinski(size,depth,up)
    right(60)
    sierpinski(size,depth,not up)
    left(60)
Answered By: Mahek Shamsukha

A recursive way i found to draw what i think you were expecting. My goal was actually drawing it using thousands of dots but after breaking my head a little, i’ve settled for this solution:

import turtle
def sierpinski(length, level):
    if level == 0:
        for i in range(3):
            turtle.forward(length)
            turtle.left(120)
    else:
        sierpinski(length/2, level-1)
        turtle.forward(length/2)
        sierpinski(length/2, level-1)
        turtle.backward(length/2)
        turtle.left(60)
        turtle.forward(length/2)
        turtle.right(60)
        sierpinski(length/2, level-1)
        turtle.left(60)
        turtle.backward(length/2)
        turtle.right(60)
turtle.speed(0)
sierpinski(200, 4)
Answered By: FelixOakz