{"title":"Dissecting power of intersection of two context-free languages","authors":"Josef Rukavicka","doi":"10.46298/dmtcs.9063","DOIUrl":null,"url":null,"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 $L$.","PeriodicalId":55175,"journal":{"name":"Discrete Mathematics and Theoretical Computer Science","volume":null,"pages":null},"PeriodicalIF":0.7000,"publicationDate":"2023-10-02","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Discrete Mathematics and Theoretical Computer Science","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.46298/dmtcs.9063","RegionNum":4,"RegionCategory":"数学","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 1
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 $L$.
期刊介绍:
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Discrete Algorithms
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