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# Koch snowflake

By Wikipedia Contributors

The Koch snowflake (also known as the Koch curve, Koch star, or Koch island) is a mathematical curve and one of the earliest fractal curves to have been described. It is based on the Koch curve, which appeared in a 1904 paper titled "On a continuous curve without tangents, constructible from elementary geometry" (original French title: Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire) by the Swedish mathematician Helge von Koch.

The progression for the area of the snowflake converges to 8/5 times the area of the original triangle, while the progression for the snowflake's perimeter diverges to infinity. Consequently, the snowflake has a finite area bounded by an infinitely long line.

## Construction

The Koch snowflake can be constructed by starting with an equilateral triangle, then recursively altering each line segment as follows:

1. divide the line segment into three segments of equal length.
2. draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
3. remove the line segment that is the base of the triangle from step 2.

After one iteration of this process, the resulting shape is the outline of a hexagram.

The Koch snowflake is the limit approached as the above steps are followed over and over again. The Koch curve originally described by Helge von Koch is constructed with only one of the three sides of the original triangle. In other words, three Koch curves make a Koch snowflake.

## Properties

### Perimeter of the Koch snowflake

After each iteration, the number of sides of the Koch snowflake increases by a factor of 4, so the number of sides after n iterations is given by:

${N}_{n}={N}_{n-1}\cdot 4=3\cdot {4}^{n}\phantom{\rule{thinmathspace}{0ex}}$ If the original equilateral triangle has sides of length s, the length of each side of the snowflake after n iterations is:

${S}_{n}=\frac{{S}_{n-1}}{3}=\frac{s}{{3}^{n}}\phantom{\rule{thinmathspace}{0ex}}$ the perimeter of the snowflake after n iterations is:

${P}_{n}={N}_{n}\cdot {S}_{n}=3\cdot s\cdot {\left(\frac{4}{3}\right)}^{n}\phantom{\rule{thinmathspace}{0ex}}$ The Koch curve has an infinite length because the total length of the curve increases by a factor of four thirds with each iteration. Each iteration creates four times as many line segments as in the previous iteration, with the length of each one being one-third the length of the segments in the previous stage. Hence the length of the curve after n iterations will be (4/3)n times the original triangle perimeter, which is unbounded as n tends to infinity.

#### Limit of perimeter

As the number of iterations tends to infinity, the limit of the perimeter is:

$\underset{n\to \infty }{lim}{P}_{n}=\underset{n\to \infty }{lim}3\cdot s\cdot {\left(\frac{4}{3}\right)}^{n}=\infty \phantom{\rule{thinmathspace}{0ex}}$ since |4/3| > 1.

A ln 4/ln 3-dimensional measure exists, but has not been calculated so far. Only upper and lower bounds have been invented.

### Area of the Koch snowflake

In each iteration a new triangle is added on each side of the previous iteration, so the number of new triangles added in iteration n is:

${T}_{n}={N}_{n-1}=3\cdot {4}^{n-1}=\frac{3}{4}\cdot {4}^{n}\phantom{\rule{thinmathspace}{0ex}}$ The area of each new triangle added in an iteration is one ninth of the area of each triangle added in the previous iteration, so the area of each triangle added in iteration n is:

${a}_{n}=\frac{{a}_{n-1}}{9}=\frac{{a}_{0}}{{9}^{n}}\phantom{\rule{thinmathspace}{0ex}}$ where a0 is the area of the original triangle. The total new area added in iteration n is therefore:

${b}_{n}={T}_{n}\cdot {a}_{n}=\frac{3}{4}\cdot {\left(\frac{4}{9}\right)}^{n}\cdot {a}_{0}$ The total area of the snowflake after n iterations is:

${A}_{n}={a}_{0}+\sum _{k=1}^{n}{b}_{k}={a}_{0}\left(1+\frac{3}{4}\sum _{k=1}^{n}{\left(\frac{4}{9}\right)}^{k}\right)={a}_{0}\left(1+\frac{1}{3}\sum _{k=0}^{n-1}{\left(\frac{4}{9}\right)}^{k}\right)\phantom{\rule{thinmathspace}{0ex}}$ Collapsing the geometric sum gives:

${A}_{n}={a}_{0}\left(1+\frac{3}{5}\left(1-{\left(\frac{4}{9}\right)}^{n}\right)\right)=\frac{{a}_{0}}{5}\left(8-3{\left(\frac{4}{9}\right)}^{n}\right)\phantom{\rule{thinmathspace}{0ex}}$ #### Limits of area

The limit of the area is:

$\underset{n\to \infty }{lim}{A}_{n}=\underset{n\to \infty }{lim}\frac{{a}_{0}}{5}\cdot \left(8-3{\left(\frac{4}{9}\right)}^{n}\right)=\frac{8}{5}\cdot {a}_{0}\phantom{\rule{thinmathspace}{0ex}}$ since |4/9| < 1.

So the area of the Koch snowflake is 8/5 of the area of the original triangle. Expressed in terms of the side length s of the original triangle this is:

$\frac{2{s}^{2}\sqrt{3}}{5}.$ ### Other properties

The Koch snowflake is self-replicating with six copies around a central point and one larger copy at the center. Hence it is an irreptile which is irrep-7.

The fractal dimension of the Koch curve is ln 4/ln 3 ≈ 1.26186. This is greater than the dimension of a line (1) but less than Peano's space-filling curve (2).

The Koch curve is continuous everywhere but differentiable nowhere.

## Tessellation of the plane

It is possible to tessellate the plane by copies of Koch snowflakes in two different sizes. However, such a tessellation is not possible using only snowflakes of one size. Since each Koch snowflake in the tessellation can be subdivided into seven smaller snowflakes of two different sizes, it is also possible to find tessellations that use more than two sizes at once. Koch snowflakes and Koch antisnowflakes of the same size may be used to tile the plane.

## Thue–Morse sequence and turtle graphics

A turtle graphic is the curve that is generated if an automaton is programmed with a sequence. If the Thue–Morse sequence members are used in order to select program states:

• If t(n) = 0, move ahead by one unit,
• If t(n) = 1, rotate counterclockwise by an angle of π/3,

the resulting curve converges to the Koch snowflake.

## Representation as Lindenmayer system

The Koch curve can be expressed by the following rewrite system (Lindenmayer system):

Alphabet : F
Constants : +, −
Axiom : F
Production rules:
F → F+F--F+F

Here, F means "draw forward", - means "turn right 60°", and + means "turn left 60°".

To create the Koch snowflake, one would use F--F--F (an equilateral triangle) as the axiom.

## Variants of the Koch curve

Following von Koch's concept, several variants of the Koch curve were designed, considering right angles (quadratic), other angles (Cesàro), circles and polyhedra and their extensions to higher dimensions (Sphereflake and Kochcube, respectively)

Variant Illustration Construction
1D, 85° angle
The Cesàro fractal is a variant of the Koch curve with an angle between 60° and 90° (here 85°).
1D, 90° angle
1D, 90° angle
Minkowski Sausage The first 2 iterations. Its fractal dimension equals 3/2 and is exactly half-way between dimension 1 and 2. It is therefore often chosen when studying the physical properties of non-integer fractal objects.
1D, ln 3/ln 5
1D, ln 3.33/ln 5
Another variation. Its fractal dimension equals ln 3.33/ln 5 = 1.49.
2D, triangles
2D, 90° angle
Extension of the quadratic type 1 curve. The illustration at left shows the fractal after the second iteration
.
3D
A three-dimensional fractal constructed from Koch curves. The shape can be considered a three-dimensional extension of the curve in the same sense that the Sierpiński pyramid and Menger sponge can be considered as extensions of the Sierpinski triangle and Sierpinski carpet. The version of the curve used for the shape uses 85° angles.

Squares can be used to generate similar fractal curves. Starting with a unit square and adding to each side at each iteration a square with dimension one third of the squares in the previous iteration, it can be shown that both the length of the perimeter and the total area are determined by geometric progressions. The progression for the area converges to 2 while the progression for the perimeter diverges to infinity, so as in the case of the Koch snowflake, we have a finite area bounded by an infinite fractal curve. The resulting area fills a square with the same center as the original, but twice the area, and rotated by π/4 radians, the perimeter touching but never overlapping itself.

The total area covered at the nth iteration is:

${A}_{n}=\frac{1}{5}+\frac{4}{5}\sum _{k=0}^{n}{\left(\frac{5}{9}\right)}^{k}\phantom{\rule{1em}{0ex}}$ while the total length of the perimeter is:

${P}_{n}=4{\left(\frac{5}{3}\right)}^{n}a\phantom{\rule{thinmathspace}{0ex}}$ which approaches infinity as n increases.