Exploring Some Geometric Themes · Lesson 1 of 13
Fractals and the Sierpinski Carpet
“Discover self-similarity by constructing a square pattern and explaining how its pieces and holes grow.”
Recognise self-similar patterns in geometric constructions and nature. Construct the first stages of a Sierpinski Carpet. Count retained squares, newly removed squares, and total holes separately. Explain the repeated multiplication behind a general counting rule.
Look closely at a fern. One branch has a shape that resembles the whole leaf, and smaller branches repeat something of the same arrangement. A tree also divides into branches, which divide again. These patterns invite a different kind of geometry: instead of studying one finished triangle or square, we study a rule that keeps making new detail.
In this lesson, you will build such a pattern from a square. The drawing becomes more complicated at every step, but the instructions remain simple. Learning to describe the instructions will let you predict a stage that would take a long time to draw by hand.
Fractals and Self-Similarity
Imagine enlarging a small part of a picture and recognising the same arrangement that appeared in the whole picture. This resemblance across scales is called self-similarity. The scales may be different, but the structure is repeated.
A mathematical fractal can be approached by repeating a geometric rule indefinitely. Each picture that we actually draw is a finite stage of the construction. We can draw three or four stages clearly, while imagining what would happen if we continued the rule without stopping.
A pattern is self-similar when parts of it resemble the whole at a smaller scale.
Ferns, branching trees, clouds, and coastlines can show approximate repetition across scales. A real branch is not an exact reduced copy of a tree, and natural detail eventually stops at a physical scale. Our square construction is more precise: every retained square receives exactly the same operation.
Constructing the Sierpinski Carpet
Begin with a filled square and call it step 0. Dividing each side into three equal parts gives nine equal small squares. Remove only the middle one. The eight squares surrounding the gap form step 1.
For step 2, apply the same instruction inside each of those eight retained squares. Each one becomes a miniature version of step 1, so eight new holes appear. The original large hole stays empty. At step 3, repeat the operation inside each of the 64 smallest retained squares.
Always apply a step to every currently retained smallest square. Applying it to just one corner would make an unfinished stage. Also, do not subdivide a region that has already been removed. A hole is empty space, so it cannot produce retained squares at a later step.
Problem
Start with a square of side 27 cm. What are the sides of the smallest retained squares at steps 1, 2, and 3?
- 1.At every step a retained square is divided into three equal lengths in each direction. Divide its side by 3.
- 2.At step 1 the side is 27 ÷ 3 = 9 cm. At step 2 it is 9 ÷ 3 = 3 cm.
- 3.At step 3 the side is 3 ÷ 3 = 1 cm. The increasing number of squares therefore comes with decreasing individual size.
Counting the Retained Squares
Trying to count every tiny square in a complicated drawing is slow and easy to get wrong. Instead, follow what happens to one square. It produces eight retained squares. Since every square follows the same rule, the next count is eight times the previous count.
Let Rₙ mean the number of smallest retained squares after n steps. The subscript n tells us which stage we mean. Start at R₀ = 1; then repeatedly multiply by 8. The power 8ⁿ records n factors of 8, and 8⁰ = 1 agrees with the starting picture.
| Step | Retained squares | Side as fraction of original |
|---|---|---|
| 0 | 1 | 1 |
| 1 | 8 | 1/3 |
| 2 | 64 | 1/9 |
| 3 | 512 | 1/27 |
| 4 | 4096 | 1/81 |
Problem
There are 512 retained squares at step 3. How many will there be at step 5?
- 1.Moving from step 3 to step 5 requires two applications of the rule.
- 2.Step 4 has 512 × 8 = 4096 retained squares.
- 3.Step 5 has 4096 × 8 = 32768 retained squares. Equivalently, 8⁵ = 32768.
New Holes and Total Holes
The word “holes” needs a careful question. Do we want the holes created during one step, or all the holes visible after that step? These are different counts because earlier holes remain in the picture.
At step 1, one hole is created. At step 2, each of eight retained squares creates one new hole, giving eight new holes and nine holes altogether. At step 3, the 64 squares from step 2 create 64 new holes. Adding them to the nine older holes gives 73.
| Step | New holes at this step | Total holes after this step |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 1 | 1 |
| 2 | 8 | 9 |
| 3 | 64 | 73 |
| 4 | 512 | 585 |
If Hₙ is the total number of holes after n steps, the next total is the old total plus Rₙ new holes. A hole of one size counts as one hole, just as a hole of another size does. Counting holes does not tell us their total area; we will investigate area separately.
Problem
A student says, “At step 3 there are 64 holes.” Explain what is correct in this statement and repair it.
- 1.Step 2 contains 64 smallest retained squares, each of which creates a hole at step 3.
- 2.Therefore 64 is the number of new holes at step 3.
- 3.The earlier 1 + 8 = 9 holes remain. The total is 9 + 64 = 73 holes.
Do not mix the count of smallest retained squares with the count of holes. After step 2 there are 64 retained squares but only 9 holes. Name the object and the stage before choosing a calculation.
Make, Predict, and Check
A paper or grid drawing can connect the arithmetic to the actual construction. Work at a scale where you can still see the small squares clearly.
- Draw a 9 by 9 grid of equal cells and outline its nine 3 by 3 regions.
- Shade the central 3 by 3 region as removed; this represents the first hole.
- In each of the other eight regions, shade its central cell. Count newly shaded regions separately from the first one.
- Predict the next stage on a larger grid, and explain why the next number of new holes is 64.
The picture gives 64 retained unit cells and 9 separate removed regions at step 2. A repeated rule lets you reach the next answer without drawing every detail. A smaller drawing of the same stage changes lengths but does not change these counts.
Quiz
How many smallest retained squares are present after step 3?
At step 4, how many holes are newly created?
Which instruction correctly produces step 2?
Why does a fern illustrate the idea of self-similarity?
Which count describes step 2 correctly?
Practice Problems
- Find the retained-square counts at steps 4 and 5.
- Find the total number of holes after step 4.
- A carpet starts with side 81 cm. Find the smallest-square side at step 4.
- Explain why there are 4096 new holes at step 5, but more than 4096 holes altogether.
- Two students draw step 2 using different starting square sizes. Which counts must agree, and which measurements may differ?
1. Multiply 512 by 8 to obtain 4096 at step 4. 2. Multiply 4096 by 8 to obtain 32768 at step 5.
Key Takeaways
Self-similarity means resemblance across different scales. Each carpet step replaces every retained square by eight smaller retained squares. The retained-square count is 8ⁿ after n steps, starting from step 0. New holes and total holes are different quantities. Mathematical rules can be repeated exactly; natural examples are usually approximate.
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The Sierpinski Triangle and Fractal Area