---
jupyter:
  description: 'The Koch snowflake with planex2d: six steps from a triangle to 12288 corners, the area against
    A0 (1 + 3/5 (1 − (4/9)^n)) on its way to 8/5 of the triangle, the perimeter growing by 4/3 per step, and the whole
    snowflake inside the circle through the triangle''s corners.'
---

# The Koch snowflake

How can a shape that fits inside a circle have an endless boundary? The Koch snowflake starts from an equilateral
triangle. In every step, the middle third of each edge is replaced by two sides of a small equilateral triangle
pointing outward. Every edge becomes four edges a third as long, so the perimeter grows by $4/3$ per step, without
end. The area grows too, but by less each time, and tends to $8/5$ of the triangle's area $A_0$. After $n$ steps,

$$A_n = A_0 \left(1 + \tfrac{3}{5} \left(1 - \left(\tfrac{4}{9}\right)^n\right)\right), \qquad P_n = 3 s \left(\tfrac{4}{3}\right)^n.$$

The example builds six steps as polylines and lets planex2d compute the area and the perimeter of each, which are
checked against the formulas. It also checks that every step stays inside the circle through the triangle's corners.

This page is part of a collection of [geometry and design examples](geometry-and-design.md) using planex2d.

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`Open this example in the playground <https://planex2d.org/playground/>`__, the interactive code editor in the browser, to change the code and run it.
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![The Koch snowflake after five steps, the first steps drawn inside it](../_static/gallery/koch-snowflake.png){.result}

```python
import math

import numpy as np
import planex as px
```

```python nbsphinx="hidden"
from docs_figures import Row
```

## The steps

The triangle has the side $s = 100$ and runs counterclockwise, so outward is to the right of each edge. A step turns the
middle third of every edge by 60° to the right and adds the corner of the small triangle. NumPy does this for all edges
at once:

```python
s = 100
triangle = np.array([(0, 0), (s, 0), (s / 2, s * math.sqrt(3) / 2)])


def koch(points):
    """every edge's middle third replaced by two sides of an equilateral triangle, pointing outward"""
    third = (np.roll(points, -1, axis=0) - points) / 3
    c, si = math.cos(math.radians(-60)), math.sin(math.radians(-60))
    turned = np.column_stack([c * third[:, 0] - si * third[:, 1], si * third[:, 0] + c * third[:, 1]])
    return np.stack([points, points + third, points + third + turned, points + 2 * third], axis=1).reshape(-1, 2)
```

## Area and perimeter

Each step is a closed polyline, and planex2d's face inside it gives the area and the length of its outline (Figure 1):

```python
A0 = math.sqrt(3) / 4 * s**2
points = triangle
flakes = {}
for n in range(7):
    flakes[n] = px.FromPolyline(points, closed=True).Face()
    print("n", n, "| corners", len(points), "| area / A0", flakes[n].mass / A0, "formula",
          1 + 3 / 5 * (1 - (4 / 9)**n), "| perimeter", flakes[n].wires[0].mass, "3 s (4/3)^n", 3 * s * (4 / 3)**n)
    points = koch(points)
# px.Plot(flakes[6])
```

```python nbsphinx="hide-input"
Row(*[(flakes[n], f"After {n} {'step' if n == 1 else 'steps'}: {3 * 4**n} edges", {"lw": 1.0 if n > 3 else 1.8})
      for n in (1, 2, 3, 6)], caption="The Koch snowflake step by step")
```

## Inside a circle

The corners of the small triangles never reach past the circle through the triangle's corners, so the whole snowflake
stays inside it, and the Boolean difference of each step and the disc is empty. Its area stays below $8/5$ of the
triangle's while its perimeter grows without end, so the isoperimetric quotient $4 \pi A / P^2$, which is 1 for the
circle, falls toward zero:

```python
center = (s / 2, s * math.sqrt(3) / 6)
disc = px.Circle(center, s / math.sqrt(3)).Face()
for n, flake in flakes.items():
    print("n", n, "| outside the disc", (flake - disc).mass, "| 4πA/P²", 4 * math.pi * flake.mass / flake.wires[0].mass**2)
```

See [The isoperimetric inequality](isoperimetric-inequality.md) for the quotient and
[The coastline paradox](coastline-paradox.md) for a real boundary whose length depends on the ruler.

```python nbsphinx="hidden"
import sys
import matplotlib.pyplot as plt
sys.path.insert(0, "..")
from docs_figures import FACE, PLANEX
from gallery import save_figure

fig, ax = plt.subplots(figsize=(6, 4))
flakes[6].Plot(ax=ax, face_col=FACE, line_col=PLANEX, legend=False)
for artist in ax.lines + ax.collections + ax.patches:
    artist.set_linewidth(0.8)
ax.set_xlim(center[0] - 90, center[0] + 90)
ax.set_ylim(center[1] - 60, center[1] + 60)
ax.set_aspect("equal")
ax.axis("off")
save_figure(fig, "koch-snowflake")
plt.close(fig)
```