main.tex

\documentclass[a4paper,UKenglish,cleveref, autoref, thm-restate]{article}

\usepackage{amssymb,amsthm,amsmath,stmaryrd}
\usepackage{xspace}

\newcommand{\expreg}{exp-regular\xspace}
\newcommand{\Expreg}{Exp-regular\xspace}


\newcommand{\mso}{\textsf{MSO}\xspace}
\newcommand{\fo}{\textsf{FO}\xspace}
\newcommand{\msot}{\textsf{MSO-T}\xspace}
\newcommand{\fot}{\textsf{FO-T}\xspace}
\newcommand{\msoi}{\textsf{MSO-I}\xspace}
\newcommand{\foi}{\textsf{FO-I}\xspace}
\newcommand{\msomi}{\textsf{MSO-MI}\xspace}
\newcommand{\fomi}{\textsf{FO-MI}\xspace}
\newcommand{\nmsomi}{\textsf{NMSO-MI}\xspace}


\newcommand{\sst}{\textsf{SST}\xspace}
\newcommand{\mt}{\textsf{MT}\xspace}
\newcommand{\ipt}{\textsf{IPT}\xspace}
\newcommand{\mtt}{\textsf{MTT}\xspace}

\newcommand{\rev}{\textsf{rev}\xspace}



\newcommand{\dlin}{\textsc{DLinSpace}\xspace}

\newcommand{\ssign}{\mathfrak{S}}
\newcommand{\tsign}{\mathfrak{T}}

\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\tuple}[1]{\left(#1\right)}
\newcommand{\sem}[1]{\left\llbracket #1\right\rrbracket}

\newcommand{\eg}{\textit{e.g.~}}
\newcommand{\ie}{\textit{i.e.~}}

\newcommand{\mleft}{\mathsf{left}}
\newcommand{\mright}{\mathsf{right}}
\newcommand{\push}{\mathsf{push}}
\newcommand{\pop}{\mathsf{pop}}
\newcommand{\auta}{\mathcal A}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{conjecture}{Conjecture}

\theoremstyle{definition}
\newtheorem{remark}{Remark}
\newtheorem{question}{Question}
\newtheorem{example}{Example}


\bibliographystyle{alpha}% the mandatory bibstyle

\title{\Expreg transductions} %TODO Please add
\author{}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}
\maketitle



\begin{abstract}

\end{abstract}


\section*{Introduction}
\label{sec:intro}

\section{Words, transducers and logic}
\label{sec:prelim}
\subsection{Words, languages and transductions}
\subsection{Relational structures and logical interpretations}

\subsubsection{Relational structures}
\paragraph{The orderned model of words}

\paragraph{The powerset model of words}~\\

Any \mso-formula can be seen as an \fo-formula over the powerset model
\subsubsection{Logical interpretation}
Here define \msoi and \foi.
\subsection{Monadic interpretation}

\footnote{We
could add copy to \msomi just as in Courcelle to avoid undesirable
cornercase behaviours with small words.}

Given two relational signatures $\ssign,\tsign$, an \mso monadic interpretation $T$ of dimension $d$ from structures over $\ssign$ to structures over $\tsign$ is given by:
\begin{itemize}
\item An \mso-formula $\phi_{U}(\bar X)$ defining the universe of the output structure
\item For each relation $R\in \tsign$ of arity $k$, a formula $\phi_R(\bar X _1,\ldots, \bar X_k )$

\end{itemize}
where $\bar X$ denotes a $d$-tuple of monadic variables.

Given a structure $A$ over $\ssign$, we define its image by $T$ by a structure $B=\sem T(B)$ over $\tsign$: the universe of $B$ is the set $U=\set{\bar S\mid\ A \models \phi_U(\bar S)}$, given $R\in \tsign$ of arity $k$, we define $R$ in $B$ as the set $\set{\tuple{\bar S_1,\ldots,\bar S_k}\in U^k\mid\ \phi_R(\bar S_1,\ldots,\bar S_k)}$.
%Given tuples $\bar X_1,\ldots,\bar X_l$, we extend $\sem T$ by $\sem T(A,\bar X_1,\ldots,\bar X_l)=B,\set{\bar X_1},\ldots,\set{\bar X_l}$.


A non-deterministic \msomi (\nmsomi) $S$ from $\ssign$-structures to $\tsign$-structures with $k$ parameters is given by an \msomi $T$ from $\ssign\uplus\set{X_1,\ldots,X_k}$-structures to $\tsign$-structures where $X_1,\ldots,X_k$ are additional unary symbols. Let $\pi$ denote the natural projection from $\ssign\uplus\set{X_1,\ldots,X_k}$-structures to $\ssign$-structures. We define $\sem S(A)=\set{\sem T(C)\mid\ \pi(C)=A}$.


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. 
As an example, consider the following function. Let $\Sigma=\{a,b\}$
and $\Gamma = \Sigma\times \{0,1\}\cup \#$. Let $u\in\Sigma^*$.
Given a
subset $U\subseteq Pos(u)$, we let $u_U\in \Gamma^*$ such that
$|u_U| = |u|$ and for all positions $p$, $u_U(p) = (u(p), p\in
U)$. Given another subset $V$, $U\leq_{lex} V$ if $U = V$ or the
smallest $x$ such that $x\in V$ iff $x\not\in U$ satisfies $x\in V$
and $x\not\in U$. Finally, we let $subsets(u) = \prod_{U\subseteq Pos(u) \text{ in
    lexicographic order}} u_U\#$. For example,
$$
subsets(ab) = (a,0)(b,0)\#(a,0)(b,1)\#(a,1)(b,0)\#(a,1)(b,1)\#
$$
It is easy to see that $subsets\in\textsf{ExpF}$. 




\section{Expressiveness of \expreg\ functions}

\subsection{Backward translation theorem}

\begin{theorem}[Backward translation theorem]
    Let $T$ be an \msomi of dimension $d$ from structures over $\ssign$ to
     structures over $\tsign$, and let $\phi(x_1,\dots,x_k)$ be an \fo-formula over
    $\tsign$.
    We can define an \mso-formula $\psi(\bar X_1,\ldots,\bar X_k)$  such that for any structure $A$ over $\ssign$, $A\models\psi(\bar S_1,\ldots,\bar S_k) $ if and only if $\sem T(A)\models \phi(\bar S_1,\ldots,\bar S_k)$.\footnote{Note that $\bar S_i$ denotes either a tuple of sets of positions of $A$, or a position of $\sem T(A)$}
\end{theorem}

\begin{proof}
    Idea: any first-order quantification $\exists x$ in $\phi$ is
    replaced by the monadic quantification $\exists \overline{X}$,
    the atoms $R(x_1,\dots,x_k)$ are replaced by
    $\phi_R(\bar X_1,\ldots,\bar X_k)$, etc. 
\end{proof}

\begin{corollary}
    The inverse image of any \fo-definable language by an \expreg
    function is regular. In particular, \expreg functions have
    regular domains. 
\end{corollary}

\begin{remark}
\label{rem:power}
An \msomi can also be seen as an \foi over the powerset model.
\end{remark}

\begin{question}
Does $\msot\circ \msomi \subseteq \msomi$ hold? Using Krohn-Rhodes theorem of Miko\l aj, it seems that the only difficult case
      is to post-compose an \msomi with a reversible Mealy machine. That is the problem can be rephrased as: does $\rev\circ \msomi \subseteq \msomi$ hold? 
\end{question}


\begin{question}
Given a (fixed) \msomi realizing $f$, can one compute the image of a word $u$ in time $O \big( |u|+|f(u)|\big)$ ?
\end{question}
\subsection{Closure properties}

\subsubsection{Postcomposition by \foi}
\begin{theorem}
    $\foi\ \circ \msomi\ = \msomi$
\end{theorem}

\begin{proof}
From the backward translation theorem.
    Use standard ideas for composition of logical transductions: formula
    substitutions. The first-order parameters of the \foi become
    monadic parameters. 
    
    Alternatively this comes from Remark~\ref{rem:power}, and the fact that \foi are closed under composition.
\end{proof}


\subsubsection{Precomposition by \msot}
\begin{theorem}
    $\msomi\ \circ \msot\ = \msomi$
\end{theorem}

\begin{proof}
    From the usual proof of closure under composition of \msot.
\end{proof}



\subsection{\Expreg functions of polynomial growth}

\begin{theorem}
    An \msomi of growth $O(n^d)$ can be realized by an \msoi of dimension $d$.
\end{theorem}








\section{The case of the successor}

\begin{theorem}
\msomi with successor is equivalent to \dlin reductions.
\end{theorem}

\begin{proof}
Given an \msomi, we can define a rational letter-to-letter relation which realizes the successor and from this obtain a linear bounded automaton.
From a linear bounded automaton, the next configuration relation can be defined in \mso.
\end{proof}

\begin{question}


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 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"
    
\begin{theorem}
   Any transduction defined by a nested marble transducer can be expressed as an \msomi. 
\end{theorem}
\begin{remark}
~
\begin{itemize}


\item Nested marble transducers subsume both pebble and marble transducers.

\item They are closed under postcomposition by polyregular functions. TODO check
\end{itemize}


\end{remark}

\begin{example}
Let $A=\set{a,b}$. Let us define an \expreg function $f:\left(A\cup \set{\sharp}\right)^*\rightarrow (A\times \set{0,1}\cup \set{\natural})^*$.
What the function $f$ does is list all subsets of positions (without $\sharp$) and output the corresponding subword. The subword are output in a specific order, and separated by $\natural$. Let us decribe the order in which these subsets of positions are listed.

Over a word $u$ without $\sharp$, $f$ outputs the list of all subwords (with multiplicities), in the lexicographic order over the subset of positions of $u$, \eg $f(aab)=\natural b\natural a\natural ab\natural a\natural ab\natural aa\natural aab$; each subword corresponding the the subsets $000,001$, $010$, $011$, $100$, $101$, $110,111$, in respective order.
Basically, the most significant figure inside a block between two $\sharp$ symbols is to the left. However, the most significant blocks are to the right, \eg: $f(a\sharp b)=\natural a \natural b\natural ab$.
\end{example}

\begin{conjecture}
The function $f$ above is not definable by a nested marble transducer.
\end{conjecture}



\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 $\ipt=\mt\circ\msot$ hold?
\end{question}
\section{Some remaining questions}

\subsection{Expressiveness}



   





\subsection{Computational models}

\begin{itemize}
    \item what about \mtt ? They have doubly-exponential growth, but do
      they capture \msomi  ?
    \item recursive programming language corresponding or being
      captured by \msomi ?
	\item Krohn-Rhodes like decomposition \eg $\msoi\circ \mathsf{exp} \circ \msot$, with $\mathsf{exp}$ being some simple class of exponential growth function.
\end{itemize}




\bibliography{biblio}

\end{document}