diff --git a/main.tex b/main.tex index e3d0a9d85e78829e959577e244c2e068bc4615ee..0d271a059abf186071eaf83971022bf68ea6b65f 100644 --- a/main.tex +++ b/main.tex @@ -45,6 +45,7 @@ \newcommand{\push}{\mathsf{push}} \newcommand{\pop}{\mathsf{pop}} \newcommand{\auta}{\mathcal A} +\newcommand{\autb}{\mathcal B} \newtheorem{theorem}{Theorem} \newtheorem{corollary}{Corollary} @@ -114,7 +115,6 @@ A non-deterministic \msomi (\nmsomi) $S$ from $\ssign$-structures to $\tsign$-st We call string-to-string (monadic) interpretations the particular case when the two signatures have unary symbols and one binary symbol which is a linear order (this can be enforced syntactically). An \expreg function is a string-to-string monadic interpretation. -We denote by \textsf{ExpF} the class of \expreg functions. @@ -184,8 +184,23 @@ From the backward translation theorem. -\subsection{\Expreg functions of polynomial growth} +\subsection{Growth} +\subsubsection{Exponential growth} + +\begin{remark} +An \msomi of dimension $d$ has growth $O((2^n)^d)=O(2^{dn})=2^{O(n)}$. + +\end{remark} +\begin{corollary} +\begin{itemize}~ + +\item \Expreg functions are \emph{not} closed under composition. + +\item \Expreg functions are \emph{bot} closed under pre-composition by functions of superlinear growth, in particular polyregular functions. + +\end{itemize} +\end{corollary} \begin{theorem} An \msomi of growth $O(n^d)$ can be realized by an \msoi of dimension $d$. \end{theorem} @@ -214,13 +229,57 @@ From a linear bounded automaton, the next configuration relation can be defined Does the model $a_1\cdots a_n\mapsto f\circ f_{a_1}\circ \cdots \circ f_{a_n}(\epsilon)$ compute \dlin ? \end{question} + +\section{Nested invisible pebble transducer} + +An invisible-pebble automaton $\auta$ is given by: +\begin{itemize} +\item a finite set of letters $A$, and two extra letters $\set{\vdash,\dashv}$ +\item a finite set of states $Q$ +\item a transition function $\delta: A_{\vdash,\dashv}{\times} Q \rightarrow {\set{\mleft,\mright}}{\times}Q + {\set{\push}}{\times}Q^2 +\set{\pop}$ + + +\end{itemize} + +Let us explain how the automaton is run over a word $u\in A^*$. The automaton actually runs over the word $v={\vdash} u {\dashv}$. +A \emph{configuration} $c$ over the word $v$ of length $n$ is a word over $Q\times\set{1,\ldots,n}$. The configuration is actually a stack of reading heads over the input word. The head of the stack is the \emph{active head}. +The \emph{initial configuration} of $\auta$ over $v$ is the word $(q_0,1)$. Given a non-empty configuration $c(p,i)$ the \emph{successor configuration} $c'$ is defined according to $\delta(v[i],p)=\alpha$ in the following way: +\begin{itemize} +\item if $\alpha=(\mleft,q)$, and $i>1$ then $c'=c(q,i-1)$ +\item if $\alpha=(\mright,q)$, and $i<n$ then $c'=c(q,i+1)$ +\item if $\alpha=(\push,q_1,q_2)$ then $c'=c(q_1,i)(q_2,i)$ +\item if $\alpha= \pop$ then $c'=c$ +\item in all other cases, $c'$ is not defined +\end{itemize} + + +A \emph{$1$-nested} invisible-pebble automaton $\auta$ is simply an invisible pebble automaton. +A \emph{$d+1$-nested} invisible-pebble automaton $\auta$ is given by a pair $(\auta,\autb)$ + +\begin{theorem} + Any transduction defined by a nested marble transducer can be expressed as an \msomi. +\end{theorem} + +\begin{question} + +Does $\ipt\subseteq \mt\circ\msot$ hold? +\end{question} + + \section{Nested marble transducer} We define a model of transducers based on marble transducers. -pebble transducer model. candidate: - nested marble transducers. A finite nesting height. a marble of height k cannot go through another marble of height k. However it can go through marbles of height $<k$. - %TODO: find a better name than "marble" + A finite nesting height. a marble of height k cannot go through another marble of height k. However it can go through marbles of height $<k$. + +A marble automaton $\auta$ is given by: +\begin{itemize} +\item a finite set of letters $A$, and two extra letters $\set{\vdash,\dashv}$ +\item a finite set of states $Q$ +\item a transition function $\delta: A_{\vdash,\dashv}{\times} Q \rightarrow {\set{\mleft,\mright}}{\times}Q + {\set{\push}}{\times}Q^2 +\set{\pop}$ + + +\end{itemize} \begin{theorem} Any transduction defined by a nested marble transducer can be expressed as an \msomi. \end{theorem} @@ -254,32 +313,8 @@ The function $f$ above is not definable by a nested marble transducer. -\section{Nested invisible pebble transducer} - -An invisible-pebble automaton $\auta$ is given by: -\begin{itemize} -\item a finite set of letters $A$, and two extra letters $\set{\vdash,\dashv}$ -\item a finite set of states $Q$ -\item a transition function $\delta: A_{\vdash,\dashv}\times Q \rightarrow \set{\mleft,\mright,\push,\pop}\times Q$ - - -\end{itemize} - -Let us explain how the automaton is run over a word $u\in A^*$. The automaton actually runs over the word $v={\vdash} u {\dashv}$. -A \emph{configuration} $c$ over the word $v$ of length $n$ is a word over $Q\times\set{1,\ldots,n}$. The configuration is actually a stack of reading heads over the input word. The head of the stack is the \emph{active head}. -The \emph{initial configuration} of $\auta$ over $v$ is the word $(q_0,1)$. Given a non-empty configuration $c(p,i)$ the \emph{successor configuration} $c'$ is defined according to $\delta(v[i],p)=(t,q)$ in the following way: -\begin{itemize} -\item if $t=\mleft$, and $i>1$ then $c'=c(q,i-1)$ -\item if $t=\mright$ and $i<n$ then $c'=c(q,i+1)$ -\item if $t=\push$ then $c'=c(p,i)(q,i)$ -\item if $t=\pop$ and $c=d(r,j)$ then $c'=d(q,j)$ -\item in all other cases, $c'$ is not defined -\end{itemize} -\begin{question} -Does $\ipt\subseteq \mt\circ\msot$ hold? -\end{question} \section{Some remaining questions} \subsection{Expressiveness}