Is there some known way to create the Mandelbrot set (the boundary),
with an iterated function system (IFS)?


Julia sets can be formed by iterating the two functions $z \mapsto \pm \sqrt{z-c},$
and this quickly converges to the Julia set. Formally, the two functions form a Hutchinson operator,
and the Julia set is the unique compact fixed set. 
That is, $J = f\_1(J) \cup f\_2(J)$ where $f\_1,f\_2$ are the two functions, and $J$ is the Julia set.


However, Mandelbrot is the iteration of many functions, with start value 0,
so there is no obvious way to define an IFS. Is there some non-obvious way?


That is, find maps $g\_1,\dots,g\_n$ s.t. the Mandelbrot $M$ can be expressed as
$M = \bigcup\_i g\_i(M),$ where each $g\_i$ maps $M$ to some subset of $M,$ but not the entire $M.$



Is there some known way to create the Mandelbrot set (the boundary), with an iterated function system?
I know very little about the Mandelbrot set or complex dynamics, but there are several features which strongly suggest that its boundary is not the attractor of an iterated function system, at least a reasonable one.


For example.


* Around [Misiurewicz point](https://en.wikipedia.org/wiki/Misiurewicz_point)s (which are only countable but dense), the Mandelbrot set looks locally like the corresponding Julia set, in particular it looks locally very different for each Misiurewicz point. Under some natural assumptions, sets invariant under IFS's look locally like the set itself.
* If an attractor of an IFS is connected, it is locally connected. The boundary of the Mandelbrot set is connected but it is a famous open problem whether it is locally connected.
* Under some weak contractivity assumptions on the generating maps, attractors of IFS's cannot have full dimension in the ambient space, while the boundary of the Mandelbrot set has dimension 2.


Also it is not quite true that all Julia sets are fixed sets for Hutchinson operators, at least under the usual definition of Hutchinson operator which requires strict contractivity of the maps (the maps $z\to\pm\sqrt{z-c}$ are in general not contractions, even on the Julia set itself)


