1
$\begingroup$

I have found that the fractal dimension of a self-similar object is: $$\text{fractal dimension} = \frac{\log(\text{number of self-similar pieces})}{\log(\text{magnification factor})} $$

See here details for the formula from above.

Therefore, using that formula can we conclude that the dimension of a Koch snowflake is $\frac{\log 6 }{\log 3}$?

When I searched online it's dimension appeared to be $\frac{\log 4}{\log 3}$. Why? Where do things fall apart with the first formula?

$\endgroup$
  • 1
    $\begingroup$ So you're saying that when you expand the snowflake (or some part of it) by a factor of 3, you can identify 6 pieces which each look like the original? What are they? $\endgroup$ – Nate Eldredge Oct 20 '19 at 16:54
  • 1
    $\begingroup$ If a set is self-similar then, subject to some other hypotheses, the dimension of that set (for certain notions of "dimension") can be computed as described in the question. For the von Koch snowflake, what six pieces are you identifying? Are you sure that you don't mean the von Koch curve, instead? Can you explain in more detail what you are talking about? $\endgroup$ – Xander Henderson Oct 20 '19 at 20:24
8
$\begingroup$

First off, the solid Koch Snowflake is, in fact, self-similar; it consists of seven copies of itself - six of which, shown in gray in figure below, are scaled by the factor $1/3$ and one of which, shown in red in the figure below, is scaled by the factor $1/\sqrt{3}$.

enter image description here

The formula that you mention,

$$ \text{dimension} = \frac{\log(\text{number of self-similar pieces})}{\log(\text{magnification factor})}, $$

works only for simpler sets, where all the pieces have the same scaling factor. More generally, a self-similar set can consist of $N$ copies of itself scaled by the factors $\{r_1,r_2,\ldots,r_N\}$. In this case, the similarity dimension of the set is defined to be the unique value of $s>0$ such that $$ r_1^s + r_2^s + \cdots + r_N^s = 1. $$ Note that if $r_1=r_2=\cdots=r_N = r$, then the equation simplifies to $N r^s=1$. In this case, you can solve for $s$ to get the simpler formula.

For the solid Koch curve, we expect the dimension to be 2. In fact, if we set $s=2$ and use the scaling factors for the solid Koch flake, we get.

$$ 6 (1/3)^2 + (1/\sqrt{3})^2 = 2/3+1/3 = 1, $$ as expected.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And what is the relationship between the similarity dimension $s=2$ and the fractal dimension of the Koch curve, $\log4/\log3$? $\endgroup$ – A. Goodier Nov 5 '19 at 21:41
  • 1
    $\begingroup$ @mvpq There is no direct relationship in that, if $E$ is a self-similar set in the plane with dimension 2, then there are many possible values for the dimension of the boundary of $E$. Often, however, the boundary may be described via a generalized type of IFS - see here or here. $\endgroup$ – Mark McClure Nov 5 '19 at 22:37
  • 3
    $\begingroup$ This is beautiful. It would be ridiculous if the post is deleted. FYI: I have voted to reopen the question. $\endgroup$ – Jack Nov 29 '19 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.