Using Decisions with Turtles¶

Here’s an example of conditional execution (using an if statement) with a turtle. We can make the turtle do some action when a condition is true. In this example if the turtle gets to right side of the space (drawing area), pick it up and move it back to the left side of the space further down so that it can draw more.

from turtle import *       # use the turtle library
from sys import *          # use the system library
setExecutionLimit(50000)   # let this take up to 50 seconds
space = Screen()           # create a turtle screen (space)

width = 400                # set the desired width
space.setup(width, width)  # set the space width and height
max_x = width // 2         # set the max x value to half the width

jaz = Turtle()             # create a turtle named jaz
jaz.shape('turtle')        # set the shape for jaz to turtle
jaz.penup()                # pick up the pen (don't draw)
jaz.goto(-1 * max_x, 100)  # go to the left side of the space
jaz.pendown()              # put the pen down to draw with
jaz.left(60)               # turn the turtle left 60 degrees

for x in range(10):        # repeat the body 10 times
    jaz.forward(100)           # go forward 100
    jaz.right(120)             # turn right 120 degrees
    jaz.forward(100)           # go forward 100
    jaz.left(120)              # turn left 120 degrees
    if (jaz.xcor() >= max_x):  # if at right edge of space
        jaz.penup()                # pick up the pen
        jaz.goto(-1 * max_x, jaz.ycor() - 100)  # move left & down
        jaz.pendown()              # put the pen down

This code calculates max_x as half the width since the drawing area uses the cartesian coordinate system with (0,0) as the center. Since the width is 400 the maxX is 200. We move the turtle to -1 * maxX which is -200. When the turtle’s x coordinate is greater than or equal to the maxX, which means it is at the right edge, then we move the turtle back to the left edge and down 100.

For more information on what this code does listen to the audio tour.

12
Using a range of 12 will complete the pattern on this line, but what should it be to finish the pattern in the space?
14
This will stop in the middle of the last line of the pattern. How many would fill the last row?
16
Each iteration of the loop draws one peak and there are 4 peaks per row. There is room for 4 rows so the answer is 4 * 4 = 16.
18
This would try to draw on the line below the end of the drawing area.

Modify the code above to draw a different pattern. Try turning changing the direction of every turn, what does that do?

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