The "self-similarity dimension" of a shape
Suppose X is a bounded subset of R, R2, or R3. Suppose that s is a positive real
number. Let s X denote X "rescaled by s"--thus a shape geometrically similar to X but in
which all distances have been multiplied by s. For example, if X is a subset of R3 then
s X = { s (x,y,z) : (x,y,z) ! X}.
Note that all distances in sX are s times the corresponding distances in X.
Let m(X) = the "measure" or content of X. It is a number describing how much
"stuff" is in X. For a circular disk X with area A = "r2 the appropriate number is the area,
m(X) = "r2. For the circular boundary B of the disk X, the appropriate number is the
circumference or arclength, so m(B) = 2"r. For a cubical box Y in space with side a, the
appropriate number is the volume, m(Y) = a3.
In these three examples, the "dimension" of the shape is obvious. The circular disk
X with area has dimension 2; its circular perimeter B has dimension 1; the cubical box Y has
dimension 3. We wish to generalize this intuitive notion of dimension.
Definition: The self-similarity dimension D of a bounded subset X of Rn is the
number that satisfies
m (s X) = sD m(X)
for all s > 0, if it exists, assuming that 0 < m(X) < #.
The self-similarity dimension D agrees with our intuitive notion in the three
examples above. If X is the disk of radius r, then m(X) = " r2 and sX is the disk of radius
sr. Hence m(sX) = "(sr)2 = s2 " r2 = s2 m(X). The self-similarity dimension is 2.
If B is the circular boundary of X, then m(B) = 2"r. But sB is a circle of radius sr, so
m(sB) = 2"(sr) = s 2"r = s m(B). Hence the self-similarity dimension of B is 1. Finally, if
Y is the cubical box with m(Y) = a3, then sY is the box with side s a hence volume m(sY) =
(s a)3 = s3 a3 = s3 m(Y), so the self-similarity dimension of Y is 3.
Computation of D:
Suppose that X is composed of congruent copies of s1 X, s2 X, ...., sk X with negligible
overlaps. Then the self-similarity dimension D of X is the solution of the equation:
s1D + ... + skD = 1.
We shall sometimes omit the word "self-similarity" and just call D the dimension
of X or even the fractal dimension of X. But there are lots of uses of the word
"dimension" and other notions of the dimension of X. Informally, a "fractal" is a shape X
whose dimension is not an integer.
Why do we obtain this formula? Since the copies s1 X, s2 X, ...., sk X have
negligible overlap, we expect
m(s1 X) + m(s2 X,) + ... + m(sk X) = m(X).
Hence
s1D m(X) + s2D m(X) + .... + skD m(X) = m(X).
Assume that 0 < m(X) < #. Then we may divide by m(X) to obtain
s1D + ... + skD = 1.
Example. Consider the 3 by 3 square X (including its interior). Its area is
m(X) = 32 = 9. For a positive number s, the set sX is a square whose sides are 3s. Hence
m(sX) = (3s)2 = 9 s2.
Hence m(sX) = 9 s2 = s2 m(X)
Hence the dimension D = 2. The square is 2-dimensional (as we expect). It is not a fractal
since 2 is an integer.
Example. Consider the following shape, known as the Sierpinski triangle, which
here will be denoted S. Note that X is composed of three rescaled copies of X, each with
1/2 the linear measurements. (These are the upper portion, the lower right portion, and the
lower left portion.)
Hence X is the union of 3 copies congruent to (1/2)X. Overlaps are single points,
which are negligible. Hence
m(X) = m((1/2)X) + m((1/2)X) + m((1/2)X)
m(X) = (1/2)D m(X) + (1/2)D m(X) + (1/2)D m(X)
1 = 3 (1/2)D
2D = 3
D ln 2 = ln 3
D = ln 3 / ln 2 = 1.58496
Hence X is a fractal
Example. Consider the following shape X, called the Koch snowflake curve.
Note that X is made up of four parts, each congruent to (1/3) X, with negligible overlap.
We therefore expect 4 m ((1/3) X) = m(X)
4 (1/3)D m(X) = m(X)
4 (1/3) D = 1
3D = 4
D = ln 4 / ln 3 = 1.26186
Since D is not an integer, X is a fractal of fractal dimension ln 4 / ln 3.
For a final example consider the following shape X:
This shape X is like a distorted snowflake. But its four sections are not quite the same. The
lower left, middle left, and middle right are all intended to be congruent to (1/3)X, but the
right left is smaller, indeed only (1/6)X.
Hence the self-similarity dimension D satisfies
(1/3)D + (1/3)D + (1/3)D + (1/6)D = 1
Hence
3 (1/3)D + (1/6)D = 1
or multiply by 6D to obtain
3 2D + 1 = 6D .
This equation cannot be solved exactly, but a numerical method on a calculator yields
D = 1.12911, giving an approximation to its self-similarity dimension.