# Snowflake and Anti-Snowflake Curves

Fig. 1

The snowflake curve made its first appearence in a 1906 paper written by the Swedish mathematician Helge von Koch. It is a closed curve of infinite perimeter that encloses a finite area.

Start with a equallateral triangle of side length $a$ and area $A_0$, the snowflake curve is constructed iteratively (see Fig. 1). At each iteration, an outward equallateral triangle is created in the middle third of each side. The base of the newly created triangle is removed:

Fig. 2

Let $s_i$ be the number of sides the snowflak curve has at the end of $i^{th}$ iteration. From Fig. 2, we see that

$\begin{cases}s_i = 4s_{i-1}\\ s_0=3\end{cases}$.

Fig. 3

Solving for $s_i$ (see Fig. 3) gives

$s_i = 3\cdot 4^i.$

Suppose at the end of $i^{th}$ iteration, the length of each side is $a^*_i$, we have

$\begin{cases}a^*_i = \frac{a^*_{i-1}}{3}\\ a^*_0=a\end{cases}$,

therefore,

$a^*_i = (\frac{1}{3})^i a$

and, $p_i$, the perimeter of the snowflake curve at the end of $i^{th}$ iteration is

$p_i = s_{i-1} (4\cdot a^*_i) = (3\cdot4^{i-1})\cdot (4\cdot(\frac{1}{3})^ia)=3(\frac{4}{3})^i a$.

It follows that

$\lim\limits_{i\rightarrow \infty}p_i =\infty$

since $|\frac{4}{3}|>1$.

Let $A^*_{i-1}, A^*_i$ be the area of equallateral triangle created at the end of $(i-1)^{th}$ and $i^{th}$ iteration respectively. Namely,

$A^*_{i-1} = \boxed{\frac{1}{2}a^*_{i-1}\sqrt{(a^*_{i-1})^2-(\frac{a^*_{i-1}}{2})^2}}$

and

$A^*_{i} = \frac{1}{2}\frac{a^*_{i-1}}{3}\sqrt{(\frac{a^*_{i-1}}{3})^2-(\frac{1}{2}\frac{a^*_{i-1}}{3})^2}= \frac{1}{9}\cdot\boxed{\frac{1}{2}a^*_{i-1}\sqrt{(a^*_{i-1})^2-(\frac{a^*_{i-1}}{2})^2}}=\frac{1}{9}A^*_{i-1}$.

Since

$\begin{cases}A^*_i = \frac{1}{9} A^*_{i-1} \\ A^*_0=A_0\end{cases}$,

solving for $A^*_i$ gives

$A^*_i = (\frac{1}{9})^i A_0$.

Hence, the area added at the end of $i^{th}$ iteration

$\Delta A_i = s_{i-1} A^*_i = (3\cdot4^{i-1})\cdot(\frac{1}{9})^{i}A_0$.

After $n$ iterations, the total enclosed area

$A_n =A_0 + \sum\limits_{i=1}^{n}\Delta A_i = A_0 + \sum\limits_{i=1}^{n}(3\cdot4^{i-1})\cdot(\frac{1}{9})^i A_0=\frac{8}{5}A_0-\frac{3}{5}(\frac{4}{9})^{n}A_0$.

As the number of iterations tends to infinity,

$\lim\limits_{n\rightarrow \infty}A_n = \frac{8}{5}A_0$.

i.e., the area of the snowflake is $\frac{8}{5}$ of the area of the original triangle.

If at each iteration, the new triangles are pointed inward, the anti-snowflake is generated (see Fig. 4 ).

Fig. 4 First four iterations of anti-snowflake curve

Like the Snowflake curve, the perimeter of the anti-snowflake curve grows boundlessly, whereas its total enclosed area approaches a finite limit (see Exercise-1).

Exercise-1 Let $A_0$ be the area of the original triangle. Show that the area encloded by the anti-snowflake curve approaches $\frac{5}{2}A_0$.

Exercise-2 An “anti-square curve” may be generated in a manner similar to that of the anti-snowflake curve (see Fig. 5). Find:

(1) The perimeter at the end of $i^{th}$ iteration

(2) The enclosed area at the end of $i^{th}$ iteration

Fig. 5 First four iterations of anti-square curve

# A pair of non-identical twins

A complex number $x + i y$ can be plotted in a complex plain where the $x$ coordinate is the real axis and the $y$ coordinate the imaginary.

Let’s consider the following iteration:

$z_{n+1} = z_{n}^2 + c\quad\quad\quad(1)$

where $z, c$ are complex numbers.

If (1) are started at $z_0 = 0$ for various values of $c$ and plotted in c-space, we have the Mandelbrot set:

When $c$ is held fixed and points generated by (1) are plotted in z-space, the result is the Julia set:

# Chaos Esthétique

Drawing fractal by hand is tedious at best and close to impossible at worst. With the help of a CAS program, all you need is a few lines of code with a for-loop operation.

The fractal structure in Fig. 1 is generated from Gumnowski-Mira chaos model

$\begin{cases} F(x)=a x+{{2(1-a)x^2} \over { 1+x^2}} \\ x_{n+1}=y_{n}+F(x_{n}) \\ y_{n+1}=-x_{n} +F(x_{n+1}) \end{cases}$

Fig.1 $a=0.31, (x_0, y_0) = (12, 0)$

Replace the term $x_n$ by $-bx_n$ with b = 0.9998, the result is shown in Fig. 2.

Fig.2 $a=0.7, (x_0, y_0) = (15, 0)$