Using recursion to draw nested Triangles

Question

I'd like to draw a series of nested triangles using recursion.

My faulty code below:

def recursiveTri(x, y, shrink):

    tt.penup()
    tt.setx(x)
    tt.sety(y)

    if x > -10:
        return

    for element in range(3):
        tt.pendown()
        tt.forward(x)
        tt.right(120)

    recursiveTri(x + shrink, y - shrink, shrink)

def main():
    recursiveTri(-300,300,30)

main()

The current code produces the following:

Actual output

Here is what I mean by nested shapes, except that I would like to draw triangles instead of squares:

Nested squares

@Babydesta. import turtle as tt. It's heavily implied, and the code is pretty unambiguous — Mad Physicist
– Mad Physicist, Commented Apr 8, 2020 at 22:10

cdlane · Accepted Answer · 2020-04-09 02:30:08Z

3

And now, just for fun, the "better living through stamping" solution:

import turtle

CURSOR_SIZE = 20

def recursiveTri(side, shrink):
    if side > 10:
        turtle.shapesize(side / CURSOR_SIZE)

        turtle.stamp()

        recursiveTri(side - shrink, shrink)

turtle.hideturtle()
turtle.shape('triangle')
turtle.fillcolor('white')

recursiveTri(300, 30)

turtle.dot()
turtle.exitonclick()

This is the default orientation, you can turn it anyway you want before calling recursiveTri(). Stamping is an alternative to drawing that works best with simple geometric patterms like this.

answered Apr 9, 2020 at 2:30

cdlane

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Mad Physicist · Accepted Answer · 2020-04-08 22:44:49Z

The problem is with

tt.forward(x)

Remember that your x is always negative, and the length of a side is not x, but rather -2 * x, if you want to be symmetrical about zero. Since your triangles are nested, you can also compute the initial y from x, given that it is 1/3 of the way up the main bisector. -sqrt(3) / 3 * x therefore puts the center of the circles circumscribed around and inscribed in your triangle at 0, 0.

In fact, it is probably easier to just fix the length of a side, and compute x and y from that:

import turtle as tt
from math import sqrt

def recursiveTri(side, shrink):
    if side < 10: return

    tt.penup()
    tt.goto(-side / 2, sqrt(3) / 6 * side)

    tt.pendown()

    for _ in range(3):
        tt.forward(side)
        tt.right(120)

    recursiveTri(side - shrink, shrink)

tt.penup()
tt.home()
tt.dot()
recursiveTri(300, 30)

In this case, shrink is the total amount removed from each side. If you want it to be the amount you step forward, change the recursive call to recursiveTri(side - 2 * shrink, shrink).

The result (without multiplying shrink by 2) is

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