Larry Riddle, Agnes Scott College

## Koch Snowflake

#### Description

Construction
Animation

Construction
Video

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 $$\displaystyle S = \lim_{n \rightarrow \infty} S(n)$$ is the Koch snowflake [Proof].

Iteration

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

#### IteratedFunctionSystem

IFS
Animation
(hexagons)

IFS
Demonstration
Video

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

Koch converging
to his snowflake!

The Koch snowflake S is the unique set satisfying $$\displaystyle S = \bigcup_{k=1}^7 f_k(S)$$. The attractor consists 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 $${f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/6} & { \sqrt 3 /6} \\ {- \sqrt 3 /6} & { 1/6} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {2/3} \\ {\sqrt 3 /3} \\ \end{array}} \right]$$ 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

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.

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

Angle 60
Axiom F++F++F
F —> F−F++F−F

The boundary of the snowflake can also be considered to consist of six copies of the Koch curve placed around the six sides of a regular hexagon and facing inwards.

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

Angle 60
Axiom F+F+F+F+F+F
F —> F+F−−F+F

The six copies of the Koch curve around a hexagon consists of three pairs, each of which form a Koch curve around an equilateral triangle inscribed in the hexagon. Toggle between the two versions using the buttons below.

#### SimilarityDimension

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 multiple copies of the Koch curve, which has a fractal dimension of 1.26186.

#### SpecialProperties

There is no doubt that Helge Von Koch is responsible for the Koch curve which he constructed in 1904 as an example of a non-differentiable curve, that is, a continuous curve that does not have a tangent at any of its points. But is Koch also responsible for the first construction of the snowflake that now bares his name? The image of the snowflake is not present in his original articles. The mathematician Yann Demichel has investigated the question "Who Invented Von Kochâ€™s Snowflake Curve?" in a fascinating article about the history of the snowflake. He credits Professor Edward Kasner of Columbia University, at least, for the name "snowflake curve" which appeared in the book Mathematics and Imagination that Kasner and his student, James Newman, published in 1940 (although Koch's name never appears in the book!).

The Koch snowflake along with copies scaled by $$1/\sqrt 3$$ and rotated by 30° can be used to tile the plane [Example 1, Example 2], or copies of the Koch snowflake and anti-snowflake can also tile the plane [Example]. See details.

The length of the boundary of S(n) at the nth 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. It is also possible to compute the area of the snowflake using fractal similarity. [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.

#### Anti-Snowflake

Start with an equilateral triangle T as before. Place three copies of a Koch curve around the three edges of T but pointed inside T. This creates the Koch anti-snowflake as shown above on the left. The region bounded by this curve is also known as the Koch anti-snowflake. As a slightly different variation, rather than adding 3 scaled copies of the triangle T 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 right above. The outside boundary of this figure is the curve shown on the left. Click on the link above for more details. (The name "anti-snowflake curve" also appears in the 1940 book Mathematics and Imagination by Kasner and Newman referenced above.)

Animation

#### Flowsnake

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.

#### Snowflake Curve

This variation is from the book Curious Curves by Darst, Palagallo, and Price. The three scaled copies of the initial equilateral triangle are taken from the interior of the equilateral triangle after it has been subdivided into 9 equal parts. This recursive procedure is continued and the limit will produce the image of a snowflake. It is shown in the book that the snowflake is actually a curve, that is, the image of a nontrivial continuous map from [0,1] into the plane.

#### References

1. Barcellos, Anthony. "The Fractal Geometry of Mandelbrot," The College Mathematics Journal, Vol. 15, No. 2, (1984), 98-114.
2. Camp, Dane. "A Fractal Excursion," Mathematics Teacher, April 1991, 265-275.
3. Darst, Richard, with Judith Palagallo and Thomas Price. Curious Curves, World Scientific Publishing Company, 2010 [Preview available at Google Books]
4. Demichel, Yann. Who invented von Koch's Snowflake Curve?, arXiv:2308.15093 [math.HO].
5. Edgar, Gerald A. Measure, Topology, and Fractal Geometry, Springer-Verlag, 1990.
6. Gardner, Martin. "Mathematical Games," Scientific American, December 1976, 124-129. [Reprinted in Penrose Tiles to Trapdoor Ciphers, Freeman (1977)
7. Kasner, E. and J. R. Newman, Mathematics and Imagination, Simon & Schuster, New York, 1940.
8. Lauwerier, Hans. Fractals: Endlessly Repeated Geometrical Figures, translated by Sophia Gill-Hoffstadt, Princeton University Press, 1991.
9. Mandelbrot, Benoit. The Fractal Geometry of Nature, W.H. Freeman and Co. 1983. [Preview available at Google Books]
10. Peitgen, Heinz-Otto with Hartmut Jurgens and Dietmar Saupe. Fractals for the Classroom, Part One: Introduction to Fractals and Chaos, Springer-Verlag New York, Inc. 1990.
11. Prusinkiewicz, Przemyslaw and James Hanan. Lindenmayer Systems, Fractals, and Plants, Lecture Notes in Biomathematics #79, Springer-Verlag 1989.
12. Prusinkiewicz, Przemyslaw and Aristid Lindenmayer. The Algorithmic Beauty of Plants, Springer-Verlag, 1990. [Available from the Algorithmic Botany website]
13. Walter, William Grey. The Curve of the Snowflake, W. W. Norton & Co. Inc., NY, 1956 (science fiction novel). [Available at the Open Library or Internet Archive (login required for both)]