OpenEuphoria

Re: Dimension of sequences

• Posted by Fernando Bauer <fmbauer at ?o?mail.com> Sep 15, 2007
• 632 views

CChris wrote:

Hi CChris. Thanks for your reply.
> 
> Fernando Bauer wrote:
> > 
> > Hello All,
> > 
> > A basic question about sequences.
> > 
> > Suppose that a "retangular sequence" is a sequence generated by using
> > iteratively
> > the function repeat() beginning with an atom. Then, I think we can say that
> > the dimension of a retangular sequence is the number of calls to repeat()
> > function.
> > (atom=dimension 0, vector=dimension 1, matrix=dimension 2, ...).
> > 
> 
> They call it the dimension of a tensor in physics. Well there are two
> dimensions
> because of covariant and contravariant indexes, but Euphoria ignores this.
Thanks for the information, even if I don't understand what are these indices..

> 
> > Now, let's say we have a sequence like { 1, {1,1} } which is not retangular.
> > Then, a question arise:
> > 
> > What is the dimension of a non-retangular sequence ?
> > 
> > a) the maximum depth of the sequence.
> > b) an integer number.
> > c) a fractal number.
> > d) a sequence which depends on the structure.
> > e) the dimension concept does not apply.
> > f) I don't know.
> > g) other.
> > 
> 
> Depends on what you wish to do with the dimension concept.
Are you saying that there isn't an unique definition for dimension of sequences
?

> In keeping with how the dimension of a topological space is defined, I'd say
> "the maximum depth" (a/).

Then, suppose you have the following sequence:
{
 {1,1,1},
 {1},
 {1,1,1}
}

If I can view this sequence as a representation of the letter "U", using your
answer below, I can conclude that the dimension of this sequence is 1.
But in this case the maximum depth is 2 (using Ricardo Forno's MaxDepth()
definition in genfunc.e library). So, what is the dimension in this case ? 1 or
2?

> 
> > Trying to answer that question, others more basics and related to that arise
> > to me (sorry if they are stupid!):
> > 
> > What is the dimension of the circumference ?
> > a) 1 , because the area of the circumference is zero. It is a curved 1D
> > object.
> > b) 2 , because the circumference exists in a bidimensional space.
> > c) Both. It has 2 types of dimensions!
> > 
> 
> No, the dimension is clearly one, if only because of the scaling issue. If you
> multiply the radius of a circle by c, the length of any arc on this circle is
> multiplied by c, which is power(c,1), as well, so the dimension is 1. For a
> disc, areas are multiplied by c squared, so discs have dimension 2. That's how
> Haüsdorff's dimension is defined, because the relevant measure is the length
> on circles and the area on discs.
> 
> > Same question for a line not closed as, for example, the form of letter "U".
> 
> Same answer: 1. Because there is a length, but not an area.
See above its possible implication for sequences.

> 
> However, if you do the following iterative process:
> 1/start with a line segment
> 2/for each line segment in your objct, replace the middle third segment by the
> other two sides of an equiklateral triangle. Try making orientation
> consistent.
> Then the limit object has dimension... ln(4)/ln(3). It isn't either a curve
> any longer - even though any finite application of algorithm above results in
> a curve - nor a surface yet. Consider that you replace three segments ___ with
> four _/\_ .

Ok, it's the Koch curve. And if the process starts with an equilateral triangle,
it is denominated, as you know, Koch snowflake, which encloses a finite area with
an infinite perimeter. A beautiful image!

> > 
> > Thanks for your reply,
> > 
> > Regards,
> > Fernando Bauer
> 
> CChris

Regards,
  Fernando