Abstract
Let $$\lambda _{1},\ldots ,\lambda _{n}$$ be real numbers in $$(0,1)$$ and $$p_{1},\ldots ,p_{n}$$ be points in $$\mathbb {R}^{d}$$ . Consider the collection of maps $$f_{j}:\mathbb {R}^{d}\rightarrow \mathbb {R}^{d} $$ given by $$\begin{aligned} f_{j}(x)=\lambda _{j} x +\left( 1-\lambda _{j}\right) p_{j}. \end{aligned}$$ It is a well known result that there exists a unique nonempty compact set $$\Lambda \subset \mathbb {R}^{d}$$ satisfying $$\Lambda =\cup _{j=1}^{n} f_{j}(\Lambda ).$$ Each $$x\in \Lambda $$ has at least one coding, that is a sequence $$(\epsilon _{i})_{i=1}^{\infty }\in \{1,\ldots ,n\}^{\mathbb {N}}$$ that satisfies $$\lim _{N\rightarrow \infty }f_{\epsilon _{1}}\ldots f_{\epsilon _{N}} (0)=x.$$ We study the size and complexity of the set of codings of a generic $$x\in \Lambda $$ when $$\Lambda $$ has positive Lebesgue measure. In particular, we show that under certain natural conditions almost every $$x\in \Lambda $$ has a continuum of codings. We also show that almost every $$x\in \Lambda $$ has a universal coding. Our work makes no assumptions on the existence of holes in $$\Lambda $$ and improves upon existing results when it is assumed $$\Lambda $$ contains no holes.
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.