Larry Riddle, Agnes Scott College

Start with an equilateral triangle **T=S(0)**. Scale **T** by a
factor of 1/3 and place 3 copies along each of the three sides of
**T** as illustrated in the diagram below to form a new image
**S(1)**. Next scale **T** by a factor of 1/9 = (1/3)^{2} and
place 12=4×3 copies along the sides of **T(1)** as illustrated to
form the image **S(2)**. For the next iteration, take 48=4×12
copies of **T** scaled by a factor of 1/27=(1/3)^{3} and place them
around the sides of **S(2)** to form the image **S(3)**.
Continue this construction. The Koch Snowflake is the limiting image
of the construction.

The sequence of sets **S(0)**, **S(1)**, **S(2)**,
**S(3)**, .... that are constructed this way form a Cauchy
sequence in the Hausdorff metric, and the limit is the Koch
snowflake [Proof].

Function

System

IFS

Animation

(hexagons)

Here is one example of an iterated function system for the Koch snowflake. This is based on scaled hexagons [Details].

\({f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/2} & {- \sqrt 3 /6 } \\
{\sqrt 3 /6} & { 1/2} \\
\end{array}} \right]{\bf{x}} \) |
scale by \(1/ \sqrt{3} \), rotate by 30° |

\({f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/\sqrt 3 } \\
{1/3} \\
\end{array}} \right]\) |
scale by 1/3 |

\({f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{0} \\
{2/3} \\
\end{array}} \right]\) |
scale by 1/3 |

\({f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{-1/ \sqrt 3} \\
{1/3} \\
\end{array}} \right]\) |
scale by 1/3 |

\({f_5}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{- 1/\sqrt 3} \\
{- 1/3} \\
\end{array}} \right]\) |
scale by 1/3 |

\({f_6}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{0} \\
{- 2/3} \\
\end{array}} \right]\) |
scale by 1/3 |

\({f_7}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/ \sqrt 3} \\
{- 1/3} \\
\end{array}} \right]\) |
scale by 1/3 |

This produces the following attractor consisting of 6 smaller copies of the snowflake scaled by 1/3 and one copy in the middle scaled by \(1/\sqrt{3}\). The area of the middle section is 3 times the area of the surrounding copies.

IFS

Animation

(line segments)

IFS

Animation

(triangles)

Here is another iterated function system for the Koch snowflake. This is based on a dragon curve construction [Details].

\({f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ - 1/6} & {\sqrt 3 /6} \\
{ - \sqrt 3 /6} & { - 1/6} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/6} \\
{\sqrt 3 /6} \\
\end{array}} \right]\) |
scale by 1/3, rotate by −120° |

\({f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/6} & {- \sqrt 3 /6} \\
{ \sqrt 3 /6} & { 1/6} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/6} \\
{\sqrt 3 /6} \\
\end{array}} \right]\) |
scale by 1/3, rotate by 60° |

\({f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/3} \\
{\sqrt 3 /3} \\
\end{array}} \right]\) |
scale by 1/3 |

\({f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/6} & { \sqrt 3 /6} \\
{- \sqrt 3 /6} & { 1/6} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{2/3} \\
{\sqrt 3 /6} \\
\end{array}} \right]\) |
scale by 1/3, rotate by −60° |

\({f_5}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/2} & {- \sqrt 3 /6} \\
{ \sqrt 3 /6} & { 1/2} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/3} \\
{0} \\
\end{array}} \right]\) |
scale by \(1/\sqrt{3} \), rotate by 30° |

\({f_6}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ -1/3} & {0} \\
{ 0} & { -1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{2/3} \\
{0} \\
\end{array}} \right]\) |
scale by 1/3, rotate by 180° |

\({f_7}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0} \\
{ 0} & { 1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{2/3} \\
{0} \\
\end{array}} \right]\) |
scale by 1/3 |

L-system

Animation

The boundary of the snowflake consists of three copies of the Koch
curve placed around the three sides of the initial equilateral
triangle and facing outwards.

The boundary can be constructed by the following L-system:

Angle 60

Axiom F++F++F

F —> F−F++F−F

Dimension

As shown above, the Koch snowflake is self-similar with six pieces scaled by 1/3 and one piece scaled by \(1/\sqrt{3}\). Moran's equation tells us that the similarity dimension *d* must satisfy

\[1 = 6{\left( {\frac{1}{3}} \right)^d} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^d}\]

The unique solution to this equation is *d* = 2. Notice, however, that the boundary of the Koch snowflake consists of three copies of the Koch curve, which has a fractal dimension of 1.26186.

Properties

The Koch snowflake is also known as the Koch island.

The Koch snowflake along with six copies scaled by \(1/\sqrt 3\) and rotated by 30° can be used to tile the plane [Example].

The length of the boundary of **S(n)** at the **n**th
iteration of the construction is \(3{\left( {\frac{4}{3}} \right)^n} s\), where
*s* denotes the length of each side of the original equilateral
triangle. Therefore the Koch snowflake has a perimeter of infinite
length.

The area of **S(n)** is
\[\frac{{\sqrt 3 {s^2}}}{4}\left( {1 + \sum\limits_{k = 1}^n {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} } \right).\]
Letting **n** go to infinity shows that
the area of the Koch snowflake is \(\dfrac{2\sqrt 3}{5}{s^2}\) . Since the area of the original equilateral triangle is \(\dfrac{\sqrt 3}{4}{s^2}\), this means that the area of the snowflake is 8/5 times the area of the original equilateral triangle [Details].

Notice in particular that the snowflake has a finite area
bounded by a perimeter of infinite length! This apparent paradox plays a role in the novel *The Curve of the Snowflake* by William Grey Walter.

Start with an equilateral triangle **T** as before. This time, however, rather than add 3 scaled copies of the triangle to the outside, point the new triangles into the interior of **T** and **remove** those regions from **T**. This leaves three diamond shaped regions, each of which consists of two scaled copies of **T**, for a total of 6 new equilateral triangles. Now repeat the construction on each of those 6 triangles, and so on. This produces the figure on the left above. The outside boundary of this figure, shown on the right above, consists of three copies of a Koch curve placed around the three edges of **T** but pointed inside **T**. The region bounded by this curve is also known as the Koch anti-snowflake. Click on the link above for more details.

Animation

This variation on the Koch snowflake was created by William Gosper. Start with a pattern of seven regular hexagons. Eight vertices are connected as shown above to create the basic red motif. The red line segments are oriented by the arrows. Now replace each red line segment by a copy of the motif scaled by \(1/ \sqrt 7 \) such that the copy lies to the left of the oriented edge it is replacing. Notice that this will place the copy of the motif inside the hexagon containing that edge. This recursive procedure is continued and in the limit will produce the image of a space filling curve shown above, known as the flowsnake (a play on words of snowflake). This version of Gosper's flowsnake was first described by Martin Gardner in one of this Mathematical Games columns in Scientific American. Click on the link above for more details.

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