Larry Riddle, Agnes Scott College

There are several ways to describe a fudgeflake.
The fudgeflake can be formed from three copies of the terdragon.

Alternatively, one can mimic the construction of the Koch snowflake by placing copies of the lower boundary of the terdragon around three sides of an equilateral triangle to form the boundary of the fudgeflake.

Construction

Animation

For the third method, start with a regular hexagon. Starting at a fixed vertex of the hexagon and going around clockwise, replace each of the six sides by the motif consisting of two sides of an isosceles triangle with an angle of 120° with the side of the hexagon forming the base for the triangle. Alternate directions so that the motif points outside the hexagon, then inside (this is the same motif as in the construction of the terdragon boundary). Repeat this construction ad infinitum, always starting at the top vertex and alternating between outside and inside. The figure below shows the first two iterations in this construction.

Function

System

Using the third construction above, the sides of the triangles form parts of three hexagons.

Each hexagon has been scaled by \(1 / \sqrt 3 \) and rotated by 30°. This yields the following IFS for the fudgeflake.

IFS

Animation

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

\({f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{1/2} & { - \sqrt 3 /6} \\
{\sqrt 3 / 6} & {1/2} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
1/2 \\
\sqrt 3 / 6 \\
\end{array}} \right]\) |
scale by \(1/ \sqrt 3\), rotate by 30° |

\({f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{1/2} & { - \sqrt 3 /6} \\
{\sqrt 3 / 6} & {1/2} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
1/2 \\
-\sqrt 3 / 6 \\
\end{array}} \right]\) |
scale by \(1/ \sqrt 3\), rotate by 30° |

The fudgeflake consists of three self-similar pieces corresponding to the three functions in the IFS.

L-system

Animation

The L-system for the boundary of the fludgeflake is based on starting with an equilateral triangle and constructing copies of the terdragon boundary along the three sides.

Angle 30

Axiom FX++++FX++++FX

F —> Z

X —> −FY++FX−

Y —> +FY−−FX+

Dimension

The fudgeflake is self-similar with 3 non-overlapping copies of itself, each scaled by the factor
\(r =1 / \sqrt 3\). Therefore the similarity dimension, **d**, of the
attractor of the IFS is the solution to

\[\sum\limits_{k = 1}^3 {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/3)}}{{\log (1/\sqrt 3 )}} = 2\]

Properties

Since the fudgeflake consists of three copies of the terdragon, it has area
\(3 \cdot \dfrac{1}{{2\sqrt 3 }} = \dfrac{{\sqrt 3 }}{2}\) [Details].
A fudgeflake built from an equilateral triangle with sides of length b has area \(\dfrac{{b^2 \sqrt 3 }}{2}\). A fudgeflake built from a regular hexagon with sides of length b has area \(\dfrac{{3b^2 \sqrt 3 }}{2}\). A problem in the third round of the 2002-2003 competition of the USA Mathematical Talent Search described a fudgeflake and asked for the area. You can see three student solutions here (under item 5/3/14 at the end of the document).

Tiling

Animation

Copies of the fudgeflake tile the plane. This is because regular hexagons will tile the plane. View the animation for a demonstration.

Boundary

Animation

The boundary of the fudgeflake has the same dimension log(4)/log(3) = 1.523627 as the boundary of the terdragon [Details]. Because of the self-similarity of the fudgeflake, repeated applications of the IFS to the image of the boundary of the fudgeflake will produce scaled versions that fill in the fudgeflake as shown in the animation.

- Mandelbrot, Benoit.
*The Fractal Geometry of Nature,*W.H. Freeman and Co. 1983. [Preview available at Google Books]