Abstract
We say that a language $L$ is \emph{constantly growing} if there is a constant $c$ such that for every word $u\in L$ there is a word $v\in L$ with $\vert u\vert<\vert v\vert\leq c+\vert u\vert$. We say that a language $L$ is \emph{geometrically growing} if there is a constant $c$ such that for every word $u\in L$ there is a word $v\in L$ with $\vert u\vert<\vert v\vert\leq c\vert u\vert$. Given two infinite languages $L_1,L_2$, we say that $L_1$ \emph{dissects} $L_2$ if $\vert L_2\setminus L_1\vert=\infty$ and $\vert L_1\cap L_2\vert=\infty$. In 2013, it was shown that for every constantly growing language $L$ there is a regular language $R$ such that $R$ dissects $L$. In the current article we show how to dissect a geometrically growing language by a homomorphic image of intersection of two context-free languages. Consider three alphabets $\Gamma$, $\Sigma$, and $\Theta$ such that $\vert \Sigma\vert=1$ and $\vert \Theta\vert=4$. We prove that there are context-free languages $M_1,M_2\subseteq \Theta^*$, an erasing alphabetical homomorphism $\pi:\Theta^*\rightarrow \Sigma^*$, and a nonerasing alphabetical homomorphism $\varphi : \Gamma^*\rightarrow \Sigma^*$ such that: If $L\subseteq \Gamma^*$ is a geometrically growing language then there is a regular language $R\subseteq \Theta^*$ such that $\varphi^{-1}\left(\pi\left(R\cap M_1\cap M_2\right)\right)$ dissects the language $L$.
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have