Arunn's Notebook

The Koch Curve

Arunn Narasimhan

The Koch Curve or the von Koch snowflake was discovered by Helge von Koch (1870-1924) in 1904. It is a closed fractal curve of infinite length within a finite region of space, enclosing a finite area. The construction of such a curve and some of its unique properties is explained in this note.

Take an equilateral triangle as shown below in Figure 1, with sides of, say, 1 unit each. This is the zeroth iteration of the Koch curve.

For the first iteration we split each side of the equilateral triangle into 3 equal parts, with the middle 1/3rd being replaced essentially by another smaller equilateral triangle made of side of 1/3 unit. The resulting figure is shown in Fig. 1b.

What we have done in the first iteration is to replace each side made of 1 unit stick by four, 1/3 unit sticks, arranged such that their end points remain fixed between the original 1 unit length of iteration zero. This procedure automatically fixes the angles of the protruding equilateral triangle in Figure 1b, and thus increases the original area confined by the equilateral triangle of Figure 1a, to that of Figure 1b.


PIC
Figure 1: Koch Curve: iteration zero to five in (a) to (f) respectively


By following the previous procedure, in the second iteration, we could now generate Figure 1c out of Figure 1b, with each of the sides of Figure 1b now being replaced in their middle thirds with even smaller equilateral triangles of sides made of 1/9 unit sticks.

Results of iteration three, four and five are shown in Figure 1d, e and f respectively.

The procedure could be repeated many times, to reach the von Koch Snowflake, when the iteration tends to infinity. There is an applet available at http://www.efg2.com/lab, for trying out more iterations and other Koch type of curves, starting from other basic geometrical figures instead of the equilateral triangle we used here.

Intuitively we could see that this resulting figure would have infinite length for its sides - if we could possibly draw it - while enclosing a finite area as shown in Fig. 1. How much would be the area when we actually do the infinite iterations?

To answer this let us put this procedure in a mathematical footing involving simple algebra.

The derivation for calculating the enclosed area done here follows the explanation given in [1]. Similar explanations are available in the Wikipedia page [2].

Let after n iterations (n 0)

  • Nn = number of sides
  • ln = length of each side
  • Ln = length of perimeter = Nn × ln

For instance, if l0 = 1, N0 = 3, then L0 = 3.

Since we increase in every iteration the number of sides by a factor of four (compare Figure 1a and 1b) and since N0 = 3, then it means

          n
Nn  = 3× 4 ,wheren  = 0,1,2,⋅⋅⋅

Further, for each iteration, the length of a side decreases by a factor of 3. for instance we used a stick of length unit 1 to form the side of Figure 1a, while in Figure 1b each side of Figure 1a is replaced by 4 sticks of length unit 1/3 of the original. Since l0 = 1, we could write then,

       (   )n
ln = 1⋅  1-   = -1,n = 0,1,2,⋅⋅⋅
         3      3n

The perimeter after the nth iteration becomes

                n  1--     4-n
Ln = Nnln = 3 ⋅4 ⋅ 3n = 3⋅(3 )

which, in other words, means

                      [    4  ]
limn→ ∞ [Ln ] = limn→ ∞ 3 ⋅(-)n  = ∞
                           3

So, the length of this closed curve does actually goes to infinity. Now, for the area it encloses.

We know from our high school trigonometry that the area of a triangle can be calculated using the Heron’s formula

∘ ----------------------  ∘ 3-----1-----1----1---  √3--
  s(s - ln)(s - ln)(s - ln) =   --ln × -ln × -ln ×--ln = ---(ln)2
                            2     2     2    2      4

where ln is the length of a side, with s = 12 × (ln + ln + ln) = 32 × ln.

Using the above formula, for Figure 1a, the zeroth iteration of the Koch curve, we could write

     √ --
A0 = --3
      4

By comparing Figure 1a and 1b we could see that we are increasing the area by adding 3 × 4(n - 1) equilateral triangles (a triangle for each side) with ln = 13n. For example, with n = 1, we increase from A0 to A1, by adding three equilateral triangles, each with l1 = 13.

The area of each of these triangles is obviously,

  --           --
√ 3 ( 1 )2   √ 3  1
-4-  3n-   = -4- ⋅9n-

and so for the figure after n iterations

                        √--
                  n-1   -3-  -1-
An  = An- 1 + 3 × 4  ×   4 × 9n

              1-   4-n- 1
= An- 1 + A0 × 3 × (9 )

For clarity, if we write explicitly for the first few n values

                        (  )0      [     (  )0 ]
n = 1 : A1 = A0 + A0 × 1  4-  = A0  1+  1- 4-
                      3   9             3  9

                        (  )             (  )      (  )
                      1- 4- 1     [    1-  4-0   1- 4-  ]
n = 2 : A2 = A1 + A0 × 3 9    = A0 1 + 3   9   + 3  9

                       1( 4)2      [    1( 4)0    1( 4)1    1( 4)2 ]
n = 3 : A3 = A2 + A0 × -- --  = A0  1+  -- --  +  -- --  +  -- --
                       3  9             3  9      3  9      3  9

Therefore, in general we could write

            [    1 ( 4)0   1 (4 )1   1 (4 )2     ]
 lim An  = A0 1 + --  --  + -- --   + -- --  +  ⋅⋅⋅
n→ ∞             3   9     3  9      3  9

                   (  )
    [    1-(    4-   4- 2     )]
= A0 1 + 3  1+  9 +  9   + ⋅⋅⋅

The expression inside the braces of the above equation can be seen to be in a geometric progression, which then could be simplified as

            [            ]
lim  An =  A0 1 + 1-⋅--1-- =  8A0
n→∞              3  1-  49    5

So, the area enclosed by the closed curve of infinite length is actually only 60 percent more than that of the original area of the equilateral triangle we started in Figure 1. A remarkable property indeed.

Reference

[1] When Least is Best by Paul J. Nahin. [Amazon Link]

[2] http://en.wikipedia.org/wiki/Koch_curve

© Arunn Narasimhan | Original version written ~ July 23, 2010 | Last revision on Apr 01, 2012

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