Larry Riddle, Agnes Scott College

Construction

Animation

Start with an equilateral triangle **T=S(0)** as before. This time, however, rather than adding 3 scaled copies of the triangle to the outside, point the new triangles into the interior of **T** and **remove** those regions from **T** to get **S(1)**. This leaves three diamond shaped regions, each of which consists of two copies of **T**, each scaled by a factor of 1/3, for a total of 6 new equilateral triangles. Now repeat the steps on each of those 6 triangles to get **S(2)** consisting of 36 copies of **T**, each scaled by a factor of 1/9 = (1/3)^{2} .

Continue this construction to get a decreasing sequence of sets \[ S(0) \supseteq S(1) \supseteq S(2) \supseteq S(3) \supseteq \cdots \] The Koch anti-snowflake is the intersection of all the sets in this sequence, that is, the set of points that remain after this construction is repeated infinitely often.

Iteration

Click the iterations to the left for another illustration of how the anti-snowflake is formed.

Function

System

Here is one possible IFS for the Koch anti-snowflake. It scales each triangle by 1/3 and rotates three of the triangles by 180°. Other rotations are possible because of the symmetry of an equilateral triangle. Changing the rotation will likely change the translation for that function.

\({f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 1/3} & {0 } \\
{0} & { 1/3} \\
\end{array}} \right]{\bf{x}} \) |
scale by 1/3 |

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

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

\({f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ -1/3} & {0 } \\
{0} & { -1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{1/2} \\
{\sqrt 3 /6} \\
\end{array}} \right]\) |
scale by 1/3, rotate by 180° |

\({f_5}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ -1/3} & {0 } \\
{0} & { -1/3} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
{5/6} \\
{\sqrt 3 /6} \\
\end{array}} \right]\) |
scale by 1/3, rotate by 180° |

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

Boundary

L-System

Animation

The outside boundary of the anti-snowflake consists of three copies of the Koch curve placed around the three sides of the initial equilateral triangle and facing inwards. The region bounded by this curve is also known as the Koch anti-snowflake.

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

Angle 60

Axiom F++F++F

F —> F+F−−F+F

Dimension

The Koch anti-snowflake is self-similar with 6 non-overlapping copies of itself, each scaled by the factor
**r** = 1/3. Therefore the similarity dimension, **d**,
of the attractor of the IFS is the solution to

\[\sum\limits_{k = 1}^6 {{r^d}} = 1\quad \Rightarrow \quad d = \frac{{\log (1/6)}}{{\log (r)}} = \frac{{\log (1/6)}}{{\log (1/3)}} = \frac{{\log (6)}}{{\log (3)}} = 1.63093\]

The outside boundary of the Koch anti-snowflake consists of three copies of the Koch curve which has a fractal dimension of 1.26186.Properties

If we only consider the outside boundary of the anti-snowflake, then the area bounded by this infinite Koch anti-snowflake curve is exactly 2/5 that of the original triangle [Details].