Koch snowflake

The Koch snowflake (also known as the Koch curve, Koch star, or Koch island) is a mathematical curve and one of the earliest fractals 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" by the Swedish mathematician Helge von Koch.

Koch antisnowflake
First four iterations
Sixth iteration

The Koch snowflake can be built up iteratively, in a sequence of stages. The first stage is an equilateral triangle, and each successive stage is formed from adding outward bends to each side of the previous stage, making smaller equilateral triangles. The areas enclosed by the successive stages in the construction of the snowflake converge to 8/5 times the area of the original triangle, while the perimeters of the successive stages increase without bound. Consequently, the snowflake encloses a finite area, but has an infinite perimeter.

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.

The first iteration of this process produces the outline of a hexagram.

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

A Koch curve–based representation of a nominally flat surface can similarly be created by repeatedly segmenting each line in a sawtooth pattern of segments with a given angle.

Properties

Perimeter of the Koch snowflake

Each iteration multiplies the number of sides in the Koch snowflake by four, so the number of sides after n iterations is given by:

$N_{n}=N_{n-1}\cdot 4=3\cdot 4^{n}\,.$ If the original equilateral triangle has sides of length s, the length of each side of the snowflake after n iterations is:

$S_{n}={\frac {S_{n-1}}{3}}={\frac {s}{3^{n}}}\,.$ The perimeter of the snowflake after n iterations is:

$P_{n}=N_{n}\cdot S_{n}=3\cdot s\cdot {\left({\frac {4}{3}}\right)}^{n}\,.$ The Koch curve has an infinite length, because the total length of the curve increases by a factor of 4/3 with each iteration. Each iteration creates four times as many line segments as in the previous iteration, with the length of each one being 1/3 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 and is unbounded, as n tends to infinity.

Limit of perimeter

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

$\lim _{n\rightarrow \infty }P_{n}=\lim _{n\rightarrow \infty }3\cdot s\cdot \left({\frac {4}{3}}\right)^{n}=\infty \,,$ since |4/3| > 1.

An 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:

$T_{n}=N_{n-1}=3\cdot 4^{n-1}={\frac {3}{4}}\cdot 4^{n}\,.$ The area of each new triangle added in an iteration is 1/9 of the area of each triangle added in the previous iteration, so the area of each triangle added in iteration n is:

$a_{n}={\frac {a_{n-1}}{9}}={\frac {a_{0}}{9^{n}}}\,.$ 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:

$A_{n}=a_{0}+\sum _{k=1}^{n}b_{k}=a_{0}\left(1+{\frac {3}{4}}\sum _{k=1}^{n}\left({\frac {4}{9}}\right)^{k}\right)=a_{0}\left(1+{\frac {1}{3}}\sum _{k=0}^{n-1}\left({\frac {4}{9}}\right)^{k}\right)\,.$ Collapsing the geometric sum gives:

$A_{n}=a_{0}\left(1+{\frac {3}{5}}\left(1-\left({\frac {4}{9}}\right)^{n}\right)\right)={\frac {a_{0}}{5}}\left(8-3\left({\frac {4}{9}}\right)^{n}\right)\,.$ Limits of area

The limit of the area is:

$\lim _{n\rightarrow \infty }A_{n}=\lim _{n\rightarrow \infty }{\frac {a_{0}}{5}}\cdot \left(8-3\left({\frac {4}{9}}\right)^{n}\right)={\frac {8}{5}}\cdot a_{0}\,,$ since |4/9| < 1.

Thus, 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:

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

The Koch snowflake is self-replicating with six smaller copies surrounding one larger copy at the center. Hence, it is an irrep-7 irrep-tile (see Rep-tile for discussion).

The fractal dimension of the Koch curve is ln 4/ln 3  1.26186. This is greater than that of a line (=1) but less than that of 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 (dimension, angle)IllustrationConstruction
≤1D, 60-90° angleThe Cesàro fractal is a variant of the Koch curve with an angle between 60° and 90° (here 85°).
≈1.46D, 90° angle
1.5D, 90° angleMinkowski Sausage
≤2D, 90° angleMinkowski Island
≈1.37D, 90° angle
≤2D, 90° angleAnticross-stitch curve, the quadratic flake type 1, with the curves facing inwards instead of outwards (Vicsek fractal)
≈1.49D, 90° angleAnother variation. Its fractal dimension equals ln 3.33/ln 5 = 1.49.
≤2D, 60° angle
≤2D, 90° angleExtension of the quadratic type 1 curve. The illustration at left shows the fractal after the second iteration .
≤3D, anyA 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 extensions of the Sierpinski triangle and Sierpinski carpet. The version of the curve used for this 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:

$A_{n}={\frac {1}{5}}+{\frac {4}{5}}\sum _{k=0}^{n}\left({\frac {5}{9}}\right)^{k}\quad {\mbox{giving}}\quad \lim _{n\rightarrow \infty }A_{n}=2\,,$ while the total length of the perimeter is:

$P_{n}=4\left({\frac {5}{3}}\right)^{n}a\,,$ which approaches infinity as n increases.