m

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}