Measuring Fractal Dimension ?

Arnaud Delobelle arnodel at
Thu Jun 18 16:56:57 EDT 2009

David C. Ullrich <ullrich at> writes:

> On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson
> <dickinsm at> wrote:
>>On Jun 17, 3:46 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
>>> Mark Dickinson <dicki... at> writes:
>>> > It looks as though you're treating (a portion of?) the Koch curve as
>>> > the graph of a function f from R -> R and claiming that f is
>>> > uniformly continuous.  But the Koch curve isn't such a graph (it
>>> > fails the 'vertical line test',
>>> I think you treat it as a function f: R -> R**2 with the usual
>>> distance metric on R**2.
>>Right.  Or rather, you treat it as the image of such a function,
>>if you're being careful to distinguish the curve (a subset
>>of R^2) from its parametrization (a continuous function
>>R -> R**2).  It's the parametrization that's uniformly
>>continuous, not the curve,
> Again, it doesn't really matter, but since you use the phrase
> "if you're being careful": In fact what you say is exactly
> backwards - if you're being careful that subset of the plane
> is _not_ a curve (it's sometimes called the "trace" of the curve".

I think it is quite common to refer to call 'curve' the image of its
parametrization.  Anyway there is a representation theorem somewhere
that I believe says for subsets of R^2 something like:

    A subset of R^2 is the image of a continuous function [0,1] -> R^2
    iff it is compact, connected and locally connected.

(I might be a bit -or a lot- wrong here, I'm not a practising
mathematician) Which means that there is no need to find a
parametrization of a plane curve to know that it is a curve.

To add to this, the usual definition of the Koch curve is not as a
function [0,1] -> R^2, and I wonder how hard it is to find such a
function for it.  It doesn't seem that easy at all to me - but I've
never looked into fractals.


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