Larry Riddle, Agnes Scott College

## Koch Anti-Snowflake

#### Description

Construction
Video

The Koch snowflake can be formed by iterating the construction of the Koch curve along each of the three sides of an equilateral triangle, with the Koch curves facing outwards. The Koch anti-snowflake is formed by iterating the construction of the Koch curve around the three sides of the equilateral triangle but facing inwards. The region bounded by the three curves is also known as the Koch anti-snowflake. Click on the image below to toggle between the boundary curves and the filled anti-snowflake.

Iteration

Click the iterations to the left for an illustration of how the anti-snowflake is formed using the Koch curve construction.

#### L-System

Boundary
L-System
Animation

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

Angle 60
Axiom F++F++F
F —> F+F−−F+F

#### SpecialProperties

The area of the anti-snowflake bounded by the three Koch curves is exactly 2/5 that of the original triangle [Details].

#### Variation

Construction
Animation

Start with an equilateral triangle T=S(0) as with the Koch snowflake. This time, however, rather than adding 3 scaled copies of the triangle to the outside as with the snowflake, 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 intersection of all the sets in this sequence, that is, the set of points that remain after this construction is repeated infinitely often, lies inside the Koch anti-snowflake described above and its outer boundary is the same as that of the anti-snowflake.

#### IteratedFunctionSystem

IFS
Animation

Here is one possible IFS for this variation of the Koch anti-snowflake that is based on the following design. It scales the initial equilateral triangle of side length 1 by 1/3 and rotates three of the scaled triangles by 180°, as illustrated by the rotated letter L.

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°

#### SimilarityDimension

This variation of 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 consists of three copies of the Koch curve which has a fractal dimension of 1.26186.