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# Minimum number of contractions needed to obtain a particular invarientinvariant set

Consider the Koch curve $G \subseteq \mathbb{R}^2$. Clearly $G$ is the invarient invariant set (IS) of the iterated function system (IFS) $\lbrace \phi_1, \phi_2, \phi_3, \phi_4 \rbrace$. Where (not wanting to jump between $\mathbb{R}^2$ and $\mathbb{C}$ but doing so for ease):

$\phi_1(x) = \frac{1}{3} x$, $\phi_2(x) = \frac{1}{3} (x \exp(\frac{i \pi}{3}) + 1)$, $\phi_3(x) = \frac{1}{3} (x \exp(-\frac{i \pi}{3}) + 1 + \exp(\frac{i \pi}{3}))$, $\phi_4(x) = \frac{1}{3} (x + 2)$

However can we do better? i.e. can we find an IFS consisting of fewer contractions such that its IS is $G$?

In this case, yes. The IFS $\lbrace \psi_1, \psi_2 \rbrace$ also has $G$ as its IS where:

$\psi_1(x) = \frac{1}{\sqrt{3}} x \exp(-\frac{5 i \pi}{6}) + \frac{1}{3} (1 + \exp(\frac{i \pi}{3}))$, $\psi_2(x) = \frac{1}{\sqrt{3}} x \exp(\frac{5 i \pi}{6}) + 1$

And as we know that an IFS consisting of a single contraction has a single point as its IS, we know that this is the best that we can do.

If $G \subseteq \mathbb{R}^n$ is the IS of the IFS $\lbrace \phi_1, \phi_2, \ldots, \phi_m \rbrace$ when can we tell if there exists an IFS with $G$ as its IS and consisting of strictly less than $m$ contractions?

As a specific example: how about the serpinski Sierpinski gasket / carpet? Can we do better that the obvious 3 / 8 constraction construction IFS?

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# Minimum number of contractions needed to obtain a particular invarient set

Consider the Koch curve $G \subseteq \mathbb{R}^2$. Clearly $G$ is the invarient set (IS) of the iterated function system (IFS) $\lbrace \phi_1, \phi_2, \phi_3, \phi_4 \rbrace$. Where (not wanting to jump between $\mathbb{R}^2$ and $\mathbb{C}$ but doing so for ease):

$\phi_1(x) = \frac{1}{3} x$, $\phi_2(x) = \frac{1}{3} (x \exp(\frac{i \pi}{3}) + 1)$, $\phi_3(x) = \frac{1}{3} (x \exp(-\frac{i \pi}{3}) + 1 + \exp(\frac{i \pi}{3}))$, $\phi_4(x) = \frac{1}{3} (x + 2)$

However can we do better? i.e. can we find an IFS consisting of fewer contractions such that its IS is $G$?

In this case, yes. The IFS $\lbrace \psi_1, \psi_2 \rbrace$ also has $G$ as its IS where:

$\psi_1(x) = \frac{1}{\sqrt{3}} x \exp(-\frac{5 i \pi}{6}) + \frac{1}{3} (1 + \exp(\frac{i \pi}{3}))$, $\psi_2(x) = \frac{1}{\sqrt{3}} x \exp(\frac{5 i \pi}{6}) + 1$

And as we know that an IFS consisting of a single contraction has a single point as its IS, we know that this is the best that we can do.

If $G \subseteq \mathbb{R}^n$ is the IS of the IFS $\lbrace \phi_1, \phi_2, \ldots, \phi_m \rbrace$ when can we tell if there exists an IFS with $G$ as its IS and consisting of strictly less than $m$ contractions?