Sierpiński Carpet

By the end of this lab you'll be able to:

• Write an 8-way recursive function (not 2 or 3 — eight!)
• Understand how a 9-subdivision with one center hole creates the Sierpiński Carpet
• Calculate how much of the original area remains at each depth level

A square divided into 9 equal sub-squares, the center removed, then each of the 8 remaining sub-squares subdivided again — to 4 levels. The lattice of holes becomes denser with each level.

Welcome to the Sierpiński Carpet!

The Sierpiński Triangle split triangles into 3. Now we split squares into 9 and remove the center — giving 8 recursive calls per level! Let's code it together!

How It Works

carpet(x, y, size, depth) draws a filled square. If depth > 0, divide the square into a 3×3 grid and recurse on the 8 non-center sub-squares. The center sub-square is left empty (white background).

The 8 positions: all 9 grid positions minus (1, 1) — the center.

Sample Code

import turtle
monty = turtle.Turtle()
monty.speed(0)
monty.hideturtle()

def draw_square(x, y, size, color):
    monty.penup()
    monty.goto(x, y)
    monty.setheading(0)
    monty.pendown()
    monty.color(color)
    monty.begin_fill()
    for _ in range(4):
        monty.forward(size)
        monty.left(90)
    monty.end_fill()

def carpet(x, y, size, depth):
    colors = ['royalblue', 'cornflowerblue', 'deepskyblue', 'lightblue', 'lightcyan']
    draw_square(x, y, size, colors[min(depth, len(colors) - 1)])
    if depth > 0:
        third = size / 3
        for row in range(3):
            for col in range(3):
                if row == 1 and col == 1:
                    continue
                carpet(x + col * third, y + row * third, third, depth - 1)

carpet(-160, -160, 320, 4)

What Do You Think Will Happen?

At depth 4, the remaining filled fraction is (8/9)^4 ≈ 62% of the original area. Will the carpet look mostly filled-in, or mostly holes? Make your guess — then click Run to find out!

Try It Now

▶ Run ↻ Reset

Mostly filled-in — (8/9)^4 ≈ 62% remains at depth 4, so more than half the area is still colored. But the holes are clearly visible at multiple scales. Were you right?

How It Works

for row in range(3): for col in range(3) generates all 9 sub-square positions. if row == 1 and col == 1: continue skips the center. The remaining 8 get recursive calls.

At depth n, the remaining area fraction is (8/9)^n. As depth → ∞, this approaches 0 — the Sierpiński Carpet has zero area but infinite boundary length.

Explanation Table

Line	What it does
third = size / 3	Divide the square into thirds
for row, col in range(3), range(3)	Generate all 9 sub-square positions
if row==1 and col==1: continue	Skip the center — the defining hole
carpet(x + col*third, y + row*third, ...)	Recurse on each of the 8 sub-squares

Learning Check

Your Turn — Try Depth 3

Change depth=4 to depth=3. The remaining area is (8/9)^3 ≈ 70%. Count the visible center holes at each level — how many are there at depth 3? Predict (1 + 8 + 64 = ?) then check!

▶ Run ↻ Reset

1 + 8 + 64 = 73 holes at depth 3 — one big center hole, 8 medium holes, 64 small holes.

Experiments

1. Use warm colors. Replace the blues with ['darkred', 'orangered', 'orange', 'gold', 'lightyellow']. You'll know it worked when the carpet turns orange/red.

2. Skip a different cell. Change if row == 1 and col == 1 to if row == 0 and col == 0 (skip the bottom-left). You'll know it worked when holes appear in the corners instead of centers.

3. Skip two cells. Add an extra or (row == 0 and col == 2) condition. You'll know it worked when two types of holes appear.

4. Try depth 5. Much slower but more detailed. You'll know it worked when you see tiny square holes at multiple scales.

Eight-Way Recursion!

You used 8 recursive calls — more than any previous fractal in this gallery! As depth → ∞, the carpet has zero area but is still "measurably" 2D. Up next: Barnsley Fern — a fractal drawn by random transformations.