Given a Pattern, find the fractal Is it possible, given a pattern or image, to calculate the equation of the fractal for that given pattern? 
For example, many plants express definite fractal patterns in their growth. Is there a formula or set of processes that give the fractal equation of their growth?
 A: For images, there is an automatic method, fractal compression.
A: As requested, I'm posting my comment as an answer here:
The Wikipedia link provided by Jim, with respect to "L-systems" lists, as an open problem: 

"Given a structure, find an L-system that can produce that structure." 

It is a great question, though! The same link provided by Jim (on L-systems), lists a book reference, including a link to the book in pdf format that you might be interested in: Przemyslaw Prusinkiewicz, Aristid Lindenmayer - Algorithmic Beauty of Plants  (for URL: algorithmicbotany.org/papers/#abop). There you can download any/all of the following pdfs: 
Chapter 1 - Graphical modeling using L-systems (2Mb; LQ, 1Mb)
Chapter 2 - Modeling of trees (4Mb; LQ, 300kb)
Chapter 3 - Developmental models of herbaceous plants (1.7Mb; LQ, 500kb)
Chapter 4 - Phyllotaxis (2.5Mb; LQ, 500kb)
Chapter 5 - Models of plant organs (1.2Mb; LQ, 300kb)
Chapter 6 - Animation of plant development (650kb; LQ, 160kb)
Chapter 7 - Modeling of cellular layers (3.7Mb; LQ, 800kb)
Chapter 8 - Fractal properties of plants (1.2Mb; LQ, 300kb)

There are also many additional links and resources that may be of interest, which you can access directly from the Wikipedia entry. 
A: There are two common ways to describe the shapes of simple fractals:


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*You can specify a generating set of self-similarities for the fractal.  These self-similarities form an iterated function system, which can be used to reconstruct the original fractal.  This is the method used to construct the Barnsley fern:





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*You can specify a Lindenmayer system (or L-system) for the fractal.  Wikipedia gives the following examples of "weeds" constructed using an L-system:



(Note: Both pictures are from Wikipedia.)
