Re: IFS transformations

From: Ville Ahonen (devnull_at_thesubmachine.com)
Date: 01/08/05 

Date: Sat, 08 Jan 2005 17:28:14 +0200

Ok, that last post was a mess. Here's what I actually meant. In one step,

(x', y') = ( x + e, y + f) ( s 0 ) ( s1 0 ) ( 1 s2 ) ( cos t sin t )
                            ( 0 s ) ( 0 1 ) ( 0 1 ) ( -sin t cos t )

, with the variables e for horisontal translation, f for vertical
translation, s for scale, s1 for strech, s2 for skewness and angle t. Is
this correct?

Ville Ahonen wrote:
> Thanks alot, Roger and Stewart. As I don't have a mathematical
> background, some (=most :-) ) of the stuff Roger wrote was a bit beyond
> my understanding. But I think I got the basic idea. I've summed up what
> I understood from your posts below. Please correct me if I'm wrong in my
> assumptions.
>
> I will first calculate
>
> (x', y') = (x, y) ( 1 0 ) + (e, f)
> ( 0 1 )
>
> , then take the (x', y') from above and scale it
>
> (x', y') = (x, y) ( s 0 )
> ( 0 s )
>
> , strech the above
>
> (x', y') = (x, y) ( s1 0 )
> ( 0 1 )
>
> , skew the above
>
> (x', y') = (x, y) ( 1 s2 )
> ( 0 1 )
>
> and finally rotate the above
>
> (x', y') = (x, y) ( cos t sin t )
> ( -sin t cos t )
>
> with the variables e for horisontal translation, f for vertical
> translation, s for scale, s1 for strech, s2 for skewness and angle t.
> After that I would add the resulting (x', y') to the old (x, y), plot it
> and reiterate n times to produce the picture.
>
> - Ville