One-way definability of two-way word transducers

TL;DR

This paper investigates when two-way word transducers can be equivalently implemented by one-way transducers, providing new algorithms with specific complexity bounds and optimal constructions for certain cases.

Contribution

It introduces a new algorithm to decide one-way definability of two-way transducers with improved complexity bounds and constructs equivalent one-way transducers when possible.

Findings

Decidability of one-way definability with doubly exponential space complexity

Optimal construction of equivalent one-way transducers in the sweeping case

Complexity reduction for sweeping transducers compared to general two-way transducers

Abstract

Functional transductions realized by two-way transducers (or, equally, by streaming transducers or MSO transductions) are the natural and standard notion of "regular" mappings from words to words. It was shown in 2013 that it is decidable if such a transduction can be implemented by some one-way transducer, but the given algorithm has non-elementary complexity. We provide an algorithm of different flavor solving the above question, that has doubly exponential space complexity. In the special case of sweeping transducers the complexity is one exponential less. We also show how to construct an equivalent one-way transducer, whenever it exists, in doubly or triply exponential time, again depending on whether the input transducer is sweeping or two-way. In the sweeping case our construction is shown to be optimal.

Equations

$$\mathsf{pump}_{L}^{n+1}(u)~{}=~{}u[1,x_{1}]\cdot\big{(}u[x_{1}+1,x_{2}]\big{)}^{n+1}\cdot u[x_{2}+1,|u|].$$

$${\mathsf{pump}}_L^{n+1}(\rho )={\rho }_0\mathsf{tr}({\ell }_1{)}^n{\rho }_1\dots {\rho }_{k-1}\mathsf{tr}({\ell }_k{)}^n{\rho }_k$$

$$\mathsf{out}(\mathsf{tr}({\ell }_1))\mathsf{out}(\rho [{\ell }_1,{\ell }_2])\mathsf{out}(\mathsf{tr}({\ell }_2))$$

$$v_0^{(n_1,n_2)}{\mathbf{v}}_1^{{\mathbf{n}}_1}{v}_2^{(n_1,n_2)}{\mathbf{v}}_3^{{\mathbf{n}}_2}{v}_4^{(n_1,n_2)}=w_0{\mathbf{w}}_1^{{\mathbf{n}}_2}{w}_2{\mathbf{w}}_3^{{\mathbf{n}}_1}{w}_4$$

$${\mathbf{v}}_1{\mathbf{v}}_1^{{\mathbf{n}}_1}{v}_2^{(n_1,n_2)}{\mathbf{v}}_3^{{\mathbf{n}}_2}{\mathbf{v}}_3$$

$$\text{makebox[0.0pt][l]period ∣v1∣v1n1⋅v1 ⋅ v2(n1,n2) ⋅ v3}{\mathbf{v}}_1^{{\mathbf{n}}_1}\cdot \stackrel{\text{period }|v_3|}{\overbrace{{\mathbf{v}}_1\cdot v_2^{(n_1,n_2)}\cdot {\mathbf{v}}_3\cdot {\mathbf{v}}_3^{{\mathbf{n}}_2}}}.$$

$$\begin{array}[]{rcl}u^{(m_{1},m_{2})}&=&\mathsf{pump}_{L_{1}}^{m_{1}+1}(\mathsf{pump}_{L_{2}}^{m_{2}+1}(u))\\[4.30554pt] \rho^{(m_{1},m_{2})}&=&\mathsf{pump}_{L_{1}}^{m_{1}+1}(\mathsf{pump}_{L_{2}}^{m_{2}+1}(\rho)).\end{array}$$

$$\begin{array}[]{rcl}X_{2}^{(m_{1},m_{2})}&=&\big{\{}x_{3}+i(x_{4}-x_{3})~{}:~{}0\leq i\leq m_{2}+1\big{\}}\\[4.30554pt] X_{1}^{(m_{1},m_{2})}&=&\big{\{}x_{1}+j(x_{2}-x_{1})+m_{2}(x_{4}-x_{3})~{}:~{}0\leq j\leq m_{1}+1\big{\}}.\end{array}$$

$$\begin{array}[]{rcl}\mathsf{pump}_{L^{\prime}_{1}}^{n_{1}+1}(\mathsf{pump}_{L^{\prime}_{2}}^{n_{2}+1}(u^{(k_{0},k_{0})}))&=&u^{(f(n_{1}),g(n_{2}))}\\[4.30554pt] \mathsf{pump}_{L^{\prime}_{1}}^{n_{1}+1}(\mathsf{pump}_{L^{\prime}_{2}}^{n_{2}+1}(\rho^{(k_{0},k_{0})}))&=&\rho^{(f(n_{1}),g(n_{2}))}\\[4.30554pt] \mathsf{pump}_{L^{\prime}_{1}}^{n_{1}+1}(\mathsf{pump}_{L^{\prime}_{2}}^{n_{2}+1}(\lambda^{(k_{0},k_{0})}))&=&\lambda^{(f(n_{1}),g(n_{2}))}.\end{array}$$

$$w^{(f(n_1),g(n_2))}=w_0{\mathbf{w}}_1^{{\mathbf{n}}_2}{w}_2{\mathbf{w}}_3^{{\mathbf{n}}_1}{w}_4$$

$$w^{(f(n_1),g(n_2))}=v_0^{(n_1,n_2)}{\mathbf{v}}_1^{{\mathbf{n}}_1}{v}_2^{(n_1,n_2)}{\mathbf{v}}_3^{{\mathbf{n}}_2}{v}_4^{(n_1,n_2)}$$

$$v_0^{(n_1,n_2)}{\mathbf{v}}_1^{{\mathbf{n}}_1}{v}_2^{(n_1,n_2)}{\mathbf{v}}_3^{{\mathbf{n}}_2}{v}_4^{(n_1,n_2)}=w_0{\mathbf{w}}_1^{{\mathbf{n}}_2}{w}_2{\mathbf{w}}_3^{{\mathbf{n}}_1}{w}_4.$$

$$\big{(}\mathsf{out}(\mathsf{tr}(\ell_{1}))\big{)}~{}\big{(}\mathsf{out}(\mathsf{tr}(\ell_{1}))\big{)}^{k_{1}n_{1}}~{}v_{2}^{(n_{1},n_{2})}~{}\big{(}\mathsf{out}(\mathsf{tr}(\ell_{2}))\big{)}^{k_{2}n_{2}}~{}\big{(}\mathsf{out}(\mathsf{tr}(\ell_{2}))\big{)}$$

$$\big{(}\mathsf{out}(\mathsf{tr}(\ell_{1}))\big{)}~{}\big{(}\mathsf{out}(\mathsf{tr}(\ell_{1}))\big{)}^{k^{\prime}_{1}}~{}{v^{\prime}_{2}}^{(k^{\prime}_{1},k^{\prime}_{2})}~{}\big{(}\mathsf{out}(\mathsf{tr}(\ell_{2}))\big{)}^{k^{\prime}_{2}}~{}\big{(}\mathsf{out}(\mathsf{tr}(\ell_{2}))\big{)}$$

$$\mathsf{out}(\mathsf{tr}({\ell }_1))\mathsf{out}(\rho [{\ell }_1,{\ell }_2])\mathsf{out}(\mathsf{tr}({\ell }_2))$$

$$(L_0,{\ell }_0,L_1,{\ell }_1)(L_2,{\ell }_2,L_3,{\ell }_3)\dots \dots (L_{2k},{\ell }_{2k},L_{2k+1},{\ell }_{2k+1})$$

$$v_{2i}\mathsf{out}(\rho [{\ell }_{2i},{\ell }_{2i+1}])v_{2i+1}$$

$$\mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+3}]\big{)}~{}v_{2i+3}~{}=~{}\text{\makebox[0.0pt][l]{$\underbrace{\phantom{\mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+2}]\big{)}~{}\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+1}]\big{)}~{}}}_{\text{period }p_{2i+1}}$}}\mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+2}]\big{)}\overbrace{~{}\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+1}]\big{)}~{}\mathsf{out}\big{(}\rho[\ell_{2i+1},\ell_{2i+3}]\big{)}~{}v_{2i+3}}^{\text{periods }p_{2i+2}\text{ and }p_{2i+3}}.$$

$$w_{5}~{}=~{}\mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+2}]\big{)}~{}v_{2i+2}~{}\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+3}]\big{)}~{}v_{2i+3}$$

$${\mathsf{pump}}_L^{n+1}(\rho )={\rho }_0\mathsf{tr}({\ell }_1{)}^n{\rho }_1\cdots {\rho }_{k-1}\mathsf{tr}({\ell }_k{)}^n{\rho }_k$$

$$\mathsf{out}(\mathsf{tr}({\ell }_1))\mathsf{out}(\rho [{\ell }_1,{\ell }_2])\mathsf{out}(\mathsf{tr}({\ell }_2))$$

$$T{|}_D\subseteq T^{\prime }\subseteq T.$$

$$T^{\prime }\subseteq T\text{ and }\mathsf{dom}(T^{\prime })\supseteq D$$

$$T\text{ one-way definable}\text{if and only if}\mathsf{dom}(T)\cap D^{\complement }=\varnothing .$$

$$w_{\rho,\mathcal{I}}~{}=~{}\mathsf{out}\big{(}\mathsf{tr}(\ell_{1})\big{)}~{}\mathsf{out}\big{(}\rho[\ell_{1},\ell_{2}]\big{)}~{}\mathsf{out}\big{(}\mathsf{tr}(\ell_{2})\big{)}.$$

$$\left(\begin{array}[]{l}w\cdot u\\ w\cdot v\end{array}\right)\ .$$

$$\left(\begin{array}[]{l}w\cdot u\\ w\cdot\#^{|u|}\end{array}\right)\qquad\qquad\text{where }u=f(w)=g(w)\ .$$

$$m\ne |f(w)|$$

Full text

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. One can deplore that said figures are not always present to help understand some notation heavy proofs, but as it stands, the paper is nicely illustrated.size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. The fact that some long, intricate proofs do not benefit the same degree of structure and illustrations is the only blemish.size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. It is announced that those results are already published in two other papers [3,4]. If some results are new to this version, it should be made explicit. If not, disregard this comment. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. done, added paragraph in intro

1. Introduction

Since the early times of computer science, transducers have been identified as a fundamental computational model. Numerous fields of computer science are ultimately concerned with transformations, ranging from databases to image processing, and an important challenge is to perform transformations with low costs, whenever possible.

The most basic form of transformations is obtained using devices that process an input with finite memory and produce outputs during the processing. Such devices are called finite-state transducers. Word-to-word finite-state transducers were considered in very early work in formal language theory [29, 1, 16], and it was soon clear that they are much more challenging than finite-state word acceptors (classical finite-state automata). One essential difference between transducers and automata over words is that the capability to process the input in both directions strictly increases the expressive power in the case of transducers, whereas this does not for automata [28, 30]. In other words, two-way word transducers are strictly more expressive than one-way word transducers.

We consider in this paper functional transducers, that compute functions from words to words. Two-way word transducers capture very nicely the notion of regularity in this setting. Regular word functions, in other words, functions computed by functional two-way transducers111We know from [17, 15] that deterministic and non-deterministic functional two-way transducers have the same expressive power, though the non-deterministic variants are usually more succinct., inherit many of the characterizations and algorithmic properties of the robust class of regular languages. Engelfriet and Hoogeboom [17] showed that monadic second-order definable graph transductions, restricted to words, are equivalent to two-way transducers — this justifies the notation “regular” word functions, in the spirit of classical results in automata theory and logic by Büchi, Elgot, Rabin and others. Recently, Alur and Cerný [2] proposed an enhanced version of one-way transducers called streaming transducers, and showed that they are equivalent to the two previous models. A streaming transducer processes the input word from left to right, and stores (partial) output words in some given write-only registers, that are updated through concatenation and constant updates.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Few reasons are given for a specific focus towards two-way transducers as opposed to equivalent models. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. I answered to the editor that the main reason is that we are more familiar with it, so it seems natural. But we shouldn’t mention it here. size=,color=blue!30,inlinesize=,color=blue!30,inlinetodo: size=,color=blue!30,inlineA. I agree not to say it here, but I would not say this is the real reason - and I changed the answer. The real reason (for me) is that pumping SST is much less intuitive.

Two-way transducers raise challenging questions about resource requirements. One crucial resource is the number of times the transducer needs to re-process the input word. In particular, the case where the input can be processed in a single pass, from left to right, is very attractive as it corresponds to the setting of streaming, where the (possibly very large) inputs do not need to be stored in order to be processed. The one-way definability of a functional two-way transducer, that is, the question whether the transducer is equivalent to some one-way transducer, was considered quite recently: [19] shows that one-way definability of string transducers is a decidable property. However, the decision procedure of [19] has non-elementary complexity, which raises the question about the intrinsic complexity of this problem.

In this paper we provide an algorithm of elementary complexity solving the one-way definability problem. Our decision algorithm has single or doubly exponential space complexity, depending on whether the input transducer is sweeping (namely, it performs reversals only at the extremities of the input word) or genuinely two-way. We also describe an algorithm that constructs an equivalent one-way transducer, whenever it exists, in doubly or triply exponential time, again depending on whether the input transducer is sweeping or two-way. For the construction of an equivalent one-way transducer we obtain a doubly exponential lower bound, which is tight for sweeping transducers. Note that for the decision problem, the best lower bound known is only polynomial space [19]. size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. The subclass of sweeping transducers is central to the paper: not only is the algorithm of better complexity in this particular case, it is also used as an easy case to prepare for the general proof. However, beyond this property, it is unclear whether this class has other interesting properties, or has garnered interest in the part. This information should, ideally, be made available in the introduction. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. added a paragraph below.size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Were deterministic two-way word transducers considered at all (Introduction)? If not, do you plan to consider them (Conclusion)? size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. added the footnote in intro, and a paragraph in conclusionsize=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. I think the question is ”do you have a better complexity if the given transducer is deterministic?” like usually. We maybe also need to add something like : One can notice that if we want to decide definability by deterministic one-way transducers, an efficient procedure is known since Choffrut78 size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. you can complete the conclusion if you want. To me it is sufficient.

size=,color=green!30,size=,color=green!30,todo: size=,color=green!30,O. added this paragraphsize=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Changed a bit (without the reviewer comment the paragraph was a bit isolated)size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. changed a bit more

Our initial interest in sweeping transducers was the fact that they provide a simpler setting for characterizing one-way definability. Later it turned out that sweeping transducers enjoy interesting and useful connections with streaming transducers: they have the same expressiveness as streaming transducers where concatenation of registers is disallowed. The connection goes even further, since the number of sweeps corresponds exactly to the number of registers [5]. The results of this paper were refined in [5], and used to determine the minimal number of registers required by functional streaming transducers without register concatenation.

Related work.

Besides the papers mentioned above, there are several recent results around the expressivity and the resources of two-way transducers, or equivalently, streaming transducers. First-order definable transductions were shown to be equivalent to transductions defined by aperiodic streaming transducers [20] and to aperiodic two-way transducers [10]. An effective characterization of aperiodicity for one-way transducers was obtained in [18].

Register minimization for right-appending deterministic streaming transducers was shown to be decidable in [14]. An algebraic characterization of (not necessarily functional) two-way transducers over unary alphabets was provided in [12]. It was shown that in this case sweeping transducers have the same expressivity. The expressivity of non-deterministic input-unary or output-unary two-way transducers was investigated in [22].

In [32] a pumping lemma for two-way transducers is proposed, and used to investigate properties of the output languages of two-way transducers. In this paper we also rely on pumping arguments over runs of two-way transducers, but we require loops of a particular form, that allows to identify periodicities in the output. size=,color=green!30,size=,color=green!30,todo: size=,color=green!30,O. TODO ref Ismael to be improved before resubmission

The present paper unifies results on one-way definability obtained in [4] and [6]. Compared to the conference versions, some combinatorial proofs have size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. I have removed the reference to Saarelabeen simplified, and the complexity of the procedure presented in [4] has been improved by one exponential. size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. please check. I deleted the sentence on the under-approximation, because I don’t find it understandable at this point.size=,color=green!30,size=,color=green!30,todo: size=,color=green!30,O. added this paragraph

Overview.

Section 2 introduces basic notations for two-way and sweeping transducers, and Section 3 states the main result and provides a roadmap of the proofs. For better readability our paper is divided in two parts: in the first part we consider the easier case of sweeping transducers, and in the second part the general case. The two proofs are similar, but the general case is more involved since we need to deal with special loops (see Section 6). Since the high-level ideas of the proofs are the same and the sweeping case illustrates them in a simpler way, the proof in that setting is a preparation for the general case. Both proofs have the same structure: first we introduce some combinatorial arguments (see Section 4 for the sweeping case and Section 7 for the general case), then we provide the characterization of one-way definability (see Section 5 for the sweeping case and Section 8 for the general case). Finally, Section 9 establishes the complexity of our algorithms.

2. Preliminaries

We start with some basic notations and definitions for two-way automata and transducers. We assume that every input word $w=a_{1}\cdots a_{n}$ has two special delimiting symbols $a_{1}=\mathop{\vdash}$ and $a_{n}=\mathop{\dashv}$ that do not occur elsewhere: $a_{i}\notin\{\mathop{\vdash},\mathop{\dashv}\}$ for all $i=2,\dots,n-1$.

A two-way automaton is a tuple $A=(Q,\Sigma,\mathop{\vdash},\mathop{\dashv},\Delta,I,F)$, where

• •

$Q$ is a finite set of states,

• •

$\Sigma$ is a finite alphabet (including $\mathop{\vdash},\mathop{\dashv}$),

• •

$\Delta\subseteq Q\times\Sigma\times Q\times\{\mathsf{left},\mathsf{right}\}$ is a transition relation,

• •

$I,F\subseteq Q$ are sets of initial and final states, respectively.

By convention, left transitions on $\mathop{\vdash}$ are not allowed; on the other hand, right transitions on $\mathop{\dashv}$ are allowed, but, as we will see, they will necessarily appear as last transitions of successful runs. A configuration of $A$ has the form $u\,q\,v$, with $uv\in\mathop{\vdash}\,\Sigma^{*}\,\mathop{\dashv}$ and $q\in Q$. A configuration $u\,q\,v$ represents the situation where the current state of $A$ is $q$ and its head reads the first symbol of $v$ (on input $uv$). If $(q,a,q^{\prime},\mathsf{right})\in\Delta$, then there is a transition from any configuration of the form $u\,q\,av$ to the configuration $ua\,q^{\prime}\,v$; we denote such a transition by $u\,q\,av\trans{a,\mathsf{right}}ua\,q^{\prime}\,v$. Similarly, if $(q,a,q^{\prime},\mathsf{left})\in\Delta$, then there is a transition from any configuration of the form $ub\,q\,av$ to the configuration $u\,q^{\prime}\,bav$, denoted as $ub\,q\,av\trans{a,\mathsf{left}}u\,q^{\prime}\,bav$. A run on $w$ is a sequence of transitions. It is successful if it starts in an initial configuration $q\,w$, with $q\in I$, and ends in a final configuration $w\,q^{\prime}$, with $q^{\prime}\in F$ — note that this latter configuration does not allow additional transitions. The language of $A$ is the set of words that admit a successful run of $A$.

The definition of two-way transducers is similar to that of two-way automata, with the only difference that now there is an additional output alphabet $\Gamma$ and the transition relation is a finite subset of $Q\times\Sigma\times\Gamma^{*}\times Q\times\{\mathsf{left},\mathsf{right}\}$, which associates an output over $\Gamma$ with each transition of the underlying two-way automaton. For a two-way transducer $T=(Q,\Sigma,\mathop{\vdash},\mathop{\dashv},\Gamma,\Delta,I,F)$, we have a transition of the form $ub\,q\,av\trans{a,d\,\mid w}u^{\prime}\,q^{\prime}\,v^{\prime}$, outputting $w$, whenever $(q,a,w,q^{\prime},d)\in\Delta$ and either $u^{\prime}=uba~{}\wedge~{}v^{\prime}=v$ or $u^{\prime}=u~{}\wedge~{}v^{\prime}=bav$, depending on whether $d=\mathsf{right}$ or $d=\mathsf{left}$. The output associated with a run $\rho=u_{1}\,q_{1}\,v_{1}\trans{a_{1},d_{1}\mid w_{1}}<\qquad>\dots\trans{a_{n},d_{n}\mid w_{n}}<\qquad>u_{n+1}\,q_{n+1}\,v_{n+1}$ of $T$ is the word $\mathsf{out}(\rho)=w_{1}\cdots w_{n}$. A transducer $T$ defines a relation $\mathscr{L}(T)$ size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. I was wondering if we ever use this notation, but I found it on page 10… Maybe it is the only place…consisting of all pairs $(u,w)$ such that $w=\mathsf{out}(\rho)$, for some successful run $\rho$ on $u$. The domain of $T$, denoted $\mathsf{dom}(T)$, is the set of input words that have a successful run. For transducers $T,T^{\prime}$, we write $T^{\prime}\subseteq T$ to mean that $\mathsf{dom}(T^{\prime})\subseteq\mathsf{dom}(T)$ and the transductions computed by $T,T^{\prime}$ coincide on $\mathsf{dom}(T)$.

A transducer is called one-way if it does not contain transition rules of the form $(q,a,w,q^{\prime},\mathsf{left})$. It is called sweeping if it can perform reversals only at the borders of the input word. A transducer that is equivalent to some one-way (resp. sweeping) transducer is called one-way definable (resp. sweeping definable).

The size of a transducer takes into account both the state space and the transition relation, and thus includes the length of the output of each transition.

Crossing sequences.

The first notion that we use throughout the paper is that of crossing sequence. We follow the convenient presentation from [24], which appeals to a graphical representation of runs of a two-way transducer, where each configuration is seen as a point (location) in a two-dimensional space. Let $u=a_{1}\cdots a_{n}$ be an input word (recall that $a_{1}=\mathop{\vdash}$ and $a_{n}=\mathop{\dashv}$) and let $\rho$ be a run of a two-way automaton (or transducer) on $u$. The positions of $\rho$ are the numbers from [math] to $n$, corresponding to “cuts” between two consecutive letters of the input. For example, position [math] is just before the first letter $a_{1}$, position $n$ is just after the last letter $a_{n}$, and any other position $x$, with $1\leq x<n$, is between the letters $a_{x}$ and $a_{x+1}$. We will denote by $u[x_{1},x_{2}]$ the factor of $u$ between the positions $x_{1}$ and $x_{2}$ (both included). size=,color=green!30,size=,color=green!30,todo: size=,color=green!30,O. added

size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Please check this reformulation in terms of configurations

Each configuration $u\,q\,v$ of a two-way run $\rho$ has a specific position associated with it. For technical reasons we need to distinguish leftward and rightward transitions. If the configuration $u\,q\,v$ is the target of a rightward transition then the position associated with $u\,q\,v$ is $x=|u|$. The same definition also applies when $u\,q\,v$ is the initial configuration, for which we have $u=\varepsilon$ and $x=|u|=0$. Otherwise, if $u\,q\,v$ is the target of a leftward transition then the position associated with $u\,q\,v$ is $x=|u|+1$. Note that in both cases, the letter read by the transition leading to $u\,q\,v$ is $a_{x}$. size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. rewritten a bitA location of $\rho$ is any pair $(x,y)$, where $x$ is the position of some configuration of $\rho$ and $y$ is any non-negative integer for which there are at least $y+1$ configurations in $\rho$ with the same position $x$. The second component $y$ of a location is called level. For example, in Figure 1 we represent a possible run of a two-way automaton together with its locations $(0,0)$, $(1,0)$, $(2,0)$, $(2,1)$, etc. Each location is naturally associated with a configuration, and thus a state. Formally, we say that $q$ is the state at location $\ell=(x,y)$ in $\rho$, and we denote this by writing $\rho(\ell)=q$, if the $(y+1)$-th configuration of $\rho$ with position $x$ has state $q$. Finally, we define the crossing sequence of $\rho$ at position $x$ as the tuple $\rho|x=(q_{0},\dots,q_{h})$, where the $q_{y}$’s are all the states at locations of the form $(x,y)$, for $y=0,\dots,h$.

As shown in Figure 1, a two-way run can be represented as an path between locations annotated by the associated states. We observe in particular that if a location $(x,y)$ is the target of a rightward transition, then this transition has read the symbol $a_{x}$; similarly, if $(x,y)$ is the target of a leftward transition, then the transition has read the symbol $a_{x+1}$. We also observe that, in any successful run $\rho$, every crossing sequence has odd length and every rightward (resp. leftward) transition reaches a location with even (resp. odd) level. In particular, we can identify four types of transitions between locations, depending on the parities of the levels (the reader may refer again to Figure 1):

\phantom{\!+\!1}~{}(x,2y)$$(x\!+\!1,2y^{\prime})~{}\phantom{\!+\!1}$$(x,2y\!+\!1)$$(x\!+\!1,2y^{\prime}\!+\!1)$$\phantom{\!+\!1}~{}(x,2y)$$(x,2y\!+\!1)$$(x,2y\!+\!1)$$(x,2y\!+\!2)$$a_{x+1},\mathsf{right}$$a_{x+1},\mathsf{left}$$a_{x+1},\mathsf{left}$$a_{x},\mathsf{right}

Hereafter, we will identify runs with the corresponding annotated paths between locations.

It is also convenient to define a total order $\mathrel{\unlhd}$ on the locations of a run $\rho$ by letting $\ell_{1}\mathrel{\unlhd}\ell_{2}$ if $\ell_{2}$ is reachable from $\ell_{1}$ by following the path described by $\rho$ — the order $\mathrel{\unlhd}$ on locations is called run order. Given two locations $\ell_{1}\mathrel{\unlhd}\ell_{2}$ of a run $\rho$, we write $\rho[\ell_{1},\ell_{2}]$ for the factor of the run that starts in $\ell_{1}$ and ends in $\ell_{2}$. Note that the latter is also a run and hence the notation \mathsf{out}\big{(}\rho[\ell_{1},\ell_{2}]\big{)} is permitted. size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Parity of levels needs to be preservedWe will often reason with factors of runs up to isomorphism, that is, modulo shifting the coordinates of their locations while preserving the parity of the levels. Of course, when the last location of (a factor of) a run $\rho_{1}$ coincides with the first location of (another factor of) a run $\rho_{2}$, then $\rho_{1}$ and $\rho_{2}$ can be concatenated to form a longer run, denoted by $\rho_{1}\rho_{2}$. size=,color=green!30,size=,color=green!30,todo: size=,color=green!30,O. added, see review 2 belowsize=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. rephrased a bit, and I think corrected a problem with negative numbers. But do we really need to say this?This operation can be performed even if the two locations, say $(x_{1},y_{1})$ and $(x_{2},y_{2})$, are different, provided that $y_{1}=y_{2}\bmod 2$: in this case it suffices to shift the positions (resp. levels) of the locations of the first run $\rho_{1}$ by $x_{2}$ (resp. $y_{2}$) and, similarly, the positions (resp. levels) of the locations of the second run $\rho_{2}$ by $x_{1}$ (resp. $y_{1}$). size=,color=red!30,size=,color=red!30,todo: size=,color=red!30,F. it’s not really enough, with our current definition the first location of $\rho_{2}$ is $(x,0)$ is $x$ is the position of the cut. So we cannot even concatenate $\rho_{1}$ and $\rho_{2}$. I have an alternate paragraph where the locations of the subruns use the levels of the original run.size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. I don’t understand your comment Felix. Does it still apply?

size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. There seems to be some problems in you definition of crossing sequences. First, according to your definition, the state at location $(x,y)$ is the state reached by the $(y+1)$-th transition that crosses x. In order for this to be consistent with Figure 1, you should swap u and u’ in your definition of a position crossed by a transition. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. Correct, there was a mistake there! Check my fix above.Second, since, following your definition, the state at a given location always is the target of a transition, the first state of the run has no location (it should be added explicitly). size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. Correct, also fixedThird, not allowing more freedom in the second components of the locations seems to cause some problems while considering subruns. Let me explain what I mean with an example. Let us suppose that we split the run $\rho$ of your Figure 1 into two runs $\rho_{1}$ and $\rho_{2}$, where $\rho_{2}$ is composed of the last three transitions. Then the locations of the states of the run $\rho_{1}$ (considered as a whole run) are equal to the locations of the same states considered as part of $\rho$. However, the locations of the states of the run $\rho_{2}$ (considered as a whole run) are (1,0), (2,0), (3,0), (4,0), whereas in $\rho$, the locations of the corresponding states are (1,2), (2,2), (3,0), (4,0). As a consequence, $\rho_{1}$ and $\rho_{2}$ can not be concatenated to reform $\rho$, since to do so we need that “the last location of […] $\rho_{1}$ coincides with the first location of […] $\rho_{2}$” (p.5 l.15). size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. Added a sentence to explicitFinally (and this one is not a real problem), you introduce precise notions, but sometimes do not use them in the following parts of the paper: for example, p.4, “never uses a left transition from position x” corresponds to “no transition crosses position x”. Why not use the latter ? size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. “no left transition on $\mathop{\vdash}$” comes before the definition of “crosses” and is more explicit, so I would not change.size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. there is no “crosses” anymore, right?

Intercepted factors.

For simplicity, we will denote by $\omega$ the maximal position of the input word. We will consider intervals of positions of the form $I=[x_{1},x_{2}]$, with $0\leq x_{1}<x_{2}\leq\omega$. The containment relation $\subseteq$ on intervals is defined expected, as $[x_{3},x_{4}]\subseteq[x_{1},x_{2}]$ if $x_{1}\leq x_{3}<x_{4}\leq x_{2}$. A factor of a run $\rho$ is a contiguous subsequence of $\rho$. A factor of $\rho$ intercepted by an interval $I=[x_{1},x_{2}]$ is a maximal factor that visits only positions $x\in I$, and never uses a left transition from position $x_{1}$ or a right transition from position $x_{2}$. Figure 2 on the right shows the factors $\alpha,\beta,\gamma,\delta,\zeta$ intercepted by an interval $I$. The numbers that annotate the endpoints of the factors represent their levels.

Functionality.

We say that a transducer is functional (equivalently, one-valued, or single-valued) if for each input $u$, at most one output $w$ can be produced by any possible successful run on $u$. Of course, every deterministic transducer is functional, while the opposite implication fails in general. To the best of our knowledge determining the precise complexity of the determinization of a two-way transducer (whenever an equivalent deterministic one-way transducer exists) is still open. From classical bounds on determinization of finite automata, we only know that the size of a determinized transducer may be exponential in the worst-case. One solution to this question, which is probably not the most efficient one, is to check one-way definability: if an equivalent one-way transducer is constructed, one can check in PTime if it can be determinized [11, 33, 7].

The following result, proven in Section 9, is the reason to consider only functional transducers:

Proposition 1**.**

The one-way definability problem for non-functional sweeping transducers is undecidable.

Unless otherwise stated, hereafter we tacitly assume that all transducers are functional. Note that functionality is a decidable property, as shown below. The proof of this result is similar to the decidability proof for the equivalence problem of deterministic two-way transducers [23], as it reduces the functionality problem to the reachability problem of a $1$-counter automaton of exponential size. A matching PSpace lower bound follows by a reduction of the emptiness problem for the intersection of finite-state automata [26].

Proposition 2**.**

Functionality of two-way transducers can be decided in polynomial space. This problem is PSpace-hard already for sweeping transducers.

A (successful) run of a two-way transducer is called normalized if it never visits two locations with the same position, the same state, and both either at even or at odd level. It is easy to see that if a successful run $\rho$ of a functional transducer visits two locations $\ell_{1}=(x,y)$ and $\ell_{2}=(x,y^{\prime})$ with the same state $\rho(\ell_{1})=\rho(\ell_{2})$ and with $y=y^{\prime}\mod 2$, then the output produced by $\rho$ between $\ell_{1}$ and $\ell_{2}$ is empty: otherwise, by repeating the non-empty factor $\rho[\ell_{1},\ell_{2}]$, we would contradict functionality. So, by deleting the factor $\rho[\ell_{1},\ell_{2}]$ we obtain a successful run that produces the same output. Iterating this operation leads to an equivalent, normalized run.

Normalized runs are interesting because their crossing sequences have bounded length (at most ${\boldsymbol{H}}=2|Q|-1$). Throughout the paper we will implicitly assume that successful runs are normalized. The latter property can be easily checked on crossing sequences.

3. One-way definability: overview

In this section we state our main result, which is the existence of an elementary algorithm for checking whether a two-way transducer is equivalent to some one-way transducer. We call such transducers one-way definable. Before stating our result, we start with a few examples illustrating the reasons that may prevent a transducer to be one-way definable.

{exa}

We consider two-way transducers that accept any input $u$ from a given regular language $R$ and produce as output the word $u\,u$. We will argue how, depending on $R$, these transducers may or may not be one-way definable.

1. (1)

If $R=(a+b)^{*}$, then there is no equivalent one-way transducer, as the output language is not regular. If $R$ is finite, however, then the transduction mapping $u\in R$ to $u\,u$ can be implemented by a one-way transducer that stores the input $u$ (this requires at least as many states as the cardinality of $R$), and outputs $u\,u$ at the end of the computation. 2. (2)

A special case of transduction with finite domain is obtained from the language $R_{n}=\{a_{0}\,w_{0}\,\cdots a_{2^{n}-1}\,w_{2^{n}-1}\>:\>a_{i}\in\{a,b\}\}$, where $n$ is a fixed natural number, the input alphabet is $\{a,b,0,1\}$, and each $w_{i}$ is the binary encoding of the index $i=0,\dots,2^{n}-1$ (hence $w_{i}\in\{0,1\}^{n}$). According to Proposition 5 below, the transduction mapping $u\in R_{n}$ to $u\,u$ can be implemented by a two-way transducer of size $\mathcal{O}(n^{2})$, but every equivalent one-way transducer has size (at least) doubly exponential in $n$. 3. (3)

Consider now the periodic language $R=(abc)^{*}$. The function that maps $u\in R$ to $u\,u$ can be easily implemented by a one-way transducer: it suffices to output alternatively $ab$, $ca$, $bc$ for each input letter, while checking that the input is in $R$.

{exa}

We consider now a slightly more complicated transduction that is defined on input words of the form $u_{1}\>\#\>\cdots\>\#\>u_{n}$, where each factor $u_{i}$ is over the alphabet $\Sigma=\{a,b,c\}$. The associated output has the form $w_{1}\>\#\>\cdots\>\#\>w_{n}$, where each $w_{i}$ is either $u_{i}\>u_{i}$ or just $u_{i}$, depending on whether or not $u_{i}\in(abc)^{*}$ and $u_{i+1}$ has even length, with $u_{n+1}=\varepsilon$ by convention.

The natural way to implement this transduction is by means of a two-way transducer that performs multiple passes on the factors of the input: a first left-to-right pass is performed on $u_{i}\,\#\,u_{i+1}$ to produce the first copy of $u_{i}$ and to check whether $u_{i}\in(abc)^{*}$ and $|u_{i+1}|$ is even; if so, a second pass on $u_{i}$ is performed to produce another copy of $u_{i}$.

Observe however that the above transduction can also be implemented by a one-way transducer, using non-determinism: when entering a factor $u_{i}$, the transducer guesses whether or not $u_{i}\in(abc)^{*}$ and $|u_{i+1}|$ is even; depending on this it outputs either $(abc\,abc)^{\frac{|u_{i}|}{3}}$ or $u_{i}$, and checks that the guess is correct while proceeding to read the input.

The main result of our paper is an elementary algorithm that decides whether a functional transducer is one-way definable:

Theorem 3**.**

There is an algorithm that takes as input a functional two-way transducer $T$ and outputs in $3\textsc{ExpTime}$ a one-way transducer $T^{\prime}$ satisfying the following properties:

1. (1)

$T^{\prime}\subseteq T$, 2. (2)

$\mathsf{dom}(T^{\prime})=\mathsf{dom}(T)$* if and only if $T$ is one-way definable.* 3. (3)

$\mathsf{dom}(T^{\prime})=\mathsf{dom}(T)$* can be checked in $2\textsc{ExpSpace}$.*

Moreover, if $T$ is a sweeping transducer, then $T^{\prime}$ can be constructed in $2\textsc{ExpTime}$ and $\mathsf{dom}(T^{\prime})=\mathsf{dom}(T)$ is decidable in ExpSpace.

Remark 4**.**

The transducer $T^{\prime}$ constructed in the above theorem is in a certain sense maximal: for every $v\in\mathsf{dom}(T)~{}\setminus~{}\mathsf{dom}(T^{\prime})$ and every one-way transducer $T^{\prime\prime}$ with $\mathsf{dom}(T^{\prime})\subseteq\mathsf{dom}(T^{\prime\prime})\subseteq\mathsf{dom}(T)$ there exists some witness input $v^{\prime}$ obtained from $v$ such that $v^{\prime}\in\mathsf{dom}(T)~{}\setminus~{}\mathsf{dom}(T^{\prime\prime})$. We will make this more precise at the end of Section 8.

We also provide a two-exponential lower bound for the size of the equivalent transducer. As the lower bound is achieved by a sweeping transduction (even a deterministic one), this gives a tight lower bound on the size of any one-way transducer equivalent to some sweeping transducer.

Proposition 5**.**

size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. I have added the fact that the sweeping transducers are even deterministic

There is a family $(f_{n})_{n\in\mathbb{N}}$ of transductions such that

1. (1)

$f_{n}$* can be implemented by a deterministic sweeping transducer of size $\mathcal{O}(n^{2})$,* 2. (2)

$f_{n}$* can be implemented by a one-way transducer,* 3. (3)

every one-way transducer that implements $f_{n}$ has size $\Omega(2^{2^{n}})$.

Proof 3.1**.**

The family of transformations is precisely the one described in Example 3 (2), where $f_{n}$ maps inputs of the form $u=a_{0}\,w_{0}\,\cdots\,a_{2^{n}-1}~{}w_{2^{n}-1}$ to outputs of the form $u\,u$, where $a_{i}\in\{a,b\}$ and $w_{i}\in\{0,1\}^{n}$ is the binary encoding of $i$. A deterministic sweeping transducer implementing $f_{n}$ first checks that the binary encodings $w_{i}$, for $i=0,\dots,2^{n}-1$, are correct. This can be done with $n$ passes: the $j$-th pass uses $\mathcal{O}(n)$ states to check the correctness of the $j$-th bits of the binary encodings. Then, the sweeping transducer performs two additional passes to copy the input twice. Overall, the sweeping transducer has size $\mathcal{O}(n^{2})$.

As already mentioned, every one-way transducer that implements $f_{n}$ needs to remember input words $u$ of exponential length in order to output $u\,u$, which roughly requires doubly exponentially many states. A more formal argument providing a lower bound for the size of a one-way transducer implementing $f_{n}$ goes as follows.

First of all, one observes that given a one-way transducer $T$, the language of its outputs, i.e., $L^{\text{out}}_{T}=\{w\>:\>(u,w)\in\mathscr{L}(T)\text{ for some }u\}$ is regular. More precisely, if $T$ has size $N$, then the language $L^{\text{out}}_{T}$ is recognized by an automaton of size linear in $N$. Indeed, while parsing $w$, the automaton can guess an input word $u$ and a run on $u$, together with a factorization of $w$ in which the $i$-th factor corresponds to the output of the transition on the $i$-th letter of $u$. Basically, this requires storing as control states the transition rules of $T$ and the suffixes of outputs.

Now, suppose that the function $f_{n}$ is implemented by a one-way transducer $T$ of size $N$. The language $L^{\text{out}}_{T}=\{u\,u\>:\>u\in\mathsf{dom}(f_{n})\}$ is then recognized by an automaton of size $\mathcal{O}(N)$. Finally, we recall a result from [21], which shows that, given a sequence of pairs of words $(u_{i},v_{i})$, for $i=1,\dots,M$, every non-deterministic automaton that separates the language $\{u_{i}\,v_{i}\>:\>1\leq i\leq M\}$ from the language $\{u_{i}\,u_{j}\>:\>1\leq i\neq j\leq M\}$ must have at least $M$ states. By applying this result to our language $L^{\text{out}}_{T}$, where $u_{i}=v_{i}$ for all $i=1,\dots,M=2^{2^{n}}$, we get that $N$ must be at least linear in $M$, and hence $N\in\Omega(2^{2^{n}})$.

size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. improved a bit, but we may do better

The proof of Theorem 3 will be developed in the next sections. The main idea is to decompose a run of the two-way transducer $T$ into factors that can be easily simulated in a one-way manner. We defer the formal definition of such a decomposition to Section 5, while here we refer to it simply as a “$\boldsymbol{B}$-decomposition”, where $\boldsymbol{B}$ is a suitable number computed from $T$. The reader can refer to Figure 7 on page 7, which provides some intuitive account of a $\boldsymbol{B}$-decomposition for a sweeping run. Roughly speaking, each factor of a $\boldsymbol{B}$-decomposition either already looks like a run of a one-way transducer (e.g. the factors $D_{1}$ and $D_{2}$ of Figure 7), or it produces a periodic output, where the period is bounded by $\boldsymbol{B}$ (e.g. the factor between $\ell_{1}$ and $\ell_{2}$). Identifying factors that look like runs of one-way transducers is rather easy. On the other hand, to identify factors with periodic outputs we rely on a notion of “inversion” of a run. Again, we defer the formal definition and the important combinatorial properties of inversions to Section 4. The reader can refer to Figure 4 on page 4 for an example of an inversion of a run of a sweeping transducer. Intuitively, this is a portion of run that is potentially difficult to simulate in a one-way manner, due to existence of long factors of the output that are generated following the opposite order of the input. Finally, the complexity of the decision procedure in Theorem 3 is analyzed in Section 9.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. As neither inversions nor blocks are defined by Page 8, the presence of Theorem 3.6 is not as enlightening as one might hope. If the notion of inversion and their relevance is made quite clear at a glance of Page 11, the definition of run decomposition is not as intuitive. Hence, Theorem 3.6 cannot be appreciated before Page 17. It might be a good idea to give an intuition about inversion and diagonal/blocks before 3.6 (this solution being preferable, but made challenging by how involved the notion of decomposition seems to be in a first read). size=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. added the references to the corresponding figuressize=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. and Gabriele added some intuitions on decompositions

Roadmap.

In order to provide a roadmap of our proofs, we state below the equivalence between the key properties related to one-way definability, inversions of runs, and existence of decompositions:

Theorem 6**.**

Given a functional two-way transducer $T$, an integer $\boldsymbol{B}$ can be computed such that the following are equivalent:

• P1)

$T$* is one-way definable,*

• P2)

for every successful run of $T$ and every inversion in it, the output produced amid the inversion has period at most $\boldsymbol{B}$,

• P3)

every input has a successful run of $T$ that admits a $\boldsymbol{B}$-decomposition.

As the notions of inversion and $\boldsymbol{B}$-decomposition are simpler to formalize for sweeping transducers, we will first prove the theorem assuming that $T$ is a sweeping transducer; we will focus later on unrestricted two-way transducers. Specifically, in Section 4 we introduce the basic combinatorics on words and the key notion of inversion for a run of a sweeping transducer, and we prove the implication P1 $\Rightarrow$ P2. In Section 5 we define $\boldsymbol{B}$-decompositions of runs of sweeping transducers, prove the implication P2 $\Rightarrow$ P3, and sketch a proof of P3 $\Rightarrow$ P1 (as a matter of fact, this latter implication can be proved in a way that is independent of whether $T$ is sweeping or not, which explains why we only sketch the proof in the sweeping case). Section 6 lays down the appropriate definitions concerning loops of two-way transducers, and analyzes in detail the effect of pumping special idempotent loops. In Section 7 we further develop the combinatorial arguments that are used to prove the implication P1 $\Rightarrow$ P2 in the general case. Finally, in Section 8 we prove the implications P2 $\Rightarrow$ P3 $\Rightarrow$ P1 in the general setting, and show how to decide the condition $\mathsf{dom}(T^{\prime})=\mathsf{dom}(T)$ of Theorem 3.

4. Basic combinatorics for sweeping transducers

We fix for the rest of the section a functional sweeping transducer $T$, an input word $u$, and a (normalized) successful run $\rho$ of $T$ on $u$.

Pumping loops.

Loops turn out to be a basic concept for characterizing one-way definability. Formally, a loop of $\rho$ is an interval $L=[x_{1},x_{2}]$ such that $\rho|x_{1}=\rho|x_{2}$, namely, with the same crossing sequences at the extremities. The run $\rho$ can be pumped at any loop $L=[x_{1},x_{2}]$, and this gives rise to new runs with iterated factors. Below we study precisely the shape of these pumped runs.

{defi}

[anchor point, trace] Given a loop $L$ and a location $\ell$ of $\rho$, we say that $\ell$ is an anchor point in $L$ if $\ell$ is the first location of some factor of $\rho$ that is intercepted by $L$; this factor is then denoted222This is a slight abuse of notation, since the factor $\mathsf{tr}(\ell)$ is not determined by $\ell$ alone, but requires also the knowledge of the loop $L$, which is usually clear from the context. as $\mathsf{tr}(\ell)$ and called the trace of $\ell$.

Observe that a loop can have at most ${\boldsymbol{H}}=2|Q|-1$ anchor points, since we consider only normalized runs.

Given a loop $L$ of $\rho$ and a number $n\in\mathbb{N}$, we can replicate $n$ times the factor $u[x_{1},x_{2}]$ of the input, obtaining a new input of the form

[TABLE]

size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. The notation $u[x_{1},x_{2}]$ is not defined.size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. fixed

Similarly, we can replicate $n$ times the intercepted factors $\mathsf{tr}(\ell)$ of $\rho$, for all anchor points $\ell$ of $L$. In this way we obtain a successful run on $\mathsf{pump}_{L}^{n+1}(u)$ that is of the form

[TABLE]

where $\ell_{1}\mathrel{\unlhd}\dots\mathrel{\unlhd}\ell_{k}$ are all the anchor points in $L$ (listed according to the run order $\mathrel{\unlhd}$), $\rho_{0}$ is the prefix of $\rho$ ending at $\ell_{1}$, $\rho_{k}$ is the suffix of $\rho$ starting at $\ell_{k}$, and for all $i=1,\dots,k-1$, $\rho_{i}$ is the factor of $\rho$ between $\ell_{i}$ and $\ell_{i+1}$. Note that $\mathsf{pump}_{L}^{1}(\rho)$ coincides with the original run $\rho$. As a matter of fact, one could define in a similar way the run $\mathsf{pump}_{L}^{0}(\rho)$ obtained from removing the loop $L$ from $\rho$. However, we do not need this, and we will always parametrize the operation $\mathsf{pump}_{L}$ by a positive number $n+1$.

An example of a pumped run $\mathsf{pump}_{L_{1}}^{3}(\rho)$ is given in Figure 3, together with the indication of the anchor points $\ell_{i}$ and the intercepted factors $\mathsf{tr}(\ell_{i})$.

Output minimality.

We are interested into factors of the run $\rho$ that lie on a single level and that contribute to the final output, but in a minimal way, in the sense that is formalized by the following definition:

{defi}

Consider a factor $\alpha=\rho[\ell,\ell^{\prime}]$ of $\rho$. We say that $\alpha$ is output-minimal if $\ell=(x,y)$ and $\ell^{\prime}=(x^{\prime},y)$, and all loops $L\subsetneq[x,x^{\prime}]$ produce empty output at level $y$.

From now on, we set the constant $\boldsymbol{B}={\boldsymbol{C}}|Q|^{\boldsymbol{H}}+1$, where ${\boldsymbol{C}}$ is the capacity of the transducer, that is, the maximal length of an output produced on a single transition (recall that $|Q|^{\boldsymbol{H}}$ is the maximal number of crossing sequences). As shown below, $\boldsymbol{B}$ bounds the length of the output produced by an output-minimal factor:

Lemma 7**.**

For all output-minimal factors $\alpha$, $|\mathsf{out}(\alpha)|\leq\boldsymbol{B}$.

Proof 4.1**.**

Suppose by contradiction that $|\mathsf{out}(\alpha)|>\boldsymbol{B}$, with $\alpha=\rho[\ell,\ell^{\prime}]$, $\ell=(x,y)$ and $\ell=(x^{\prime},y)$.

Let $X$ be the set of all positions $x^{\prime\prime}$, with $\min(x,x^{\prime})<x^{\prime\prime}<\max(x,x^{\prime})$, that are sources of transitions of $\alpha$ that produce non-empty output. Clearly, the total number of letters produced by the transitions that depart from locations in $X\times\{y\}$ is strictly larger than $\boldsymbol{B}-1$. Moreover, since each transition emits at most ${\boldsymbol{C}}$ symbols, we have $|X|>\frac{\boldsymbol{B}-1}{{\boldsymbol{C}}}=|Q|^{\boldsymbol{H}}$. Now, recall that crossing sequences are sequences of states of length at most ${\boldsymbol{H}}$. Since $|X|$ is larger than the number of crossing sequences, $X$ contains two positions $x_{1}<x_{2}$ such that $\rho|x_{1}=\rho|x_{2}$. In particular, $L=[x_{1},x_{2}]$ is a loop strictly between $x,x^{\prime}$ with non-empty output on level $y$. This shows that $\rho[\ell,\ell^{\prime}]$ is not output-minimal.

Inversions and periodicity.

Next, we define the crucial notion of inversion. Intuitively, an inversion in a run identifies a part of the run that is potentially difficult to simulate in a one-way manner because the order of generating the output is reversed w.r.t. the input. Inversions arise naturally in transducers that reverse arbitrarily long portions of the input, as well as in transducers that produce copies of arbitrarily long portions of the input.

{defi}

An inversion of the run $\rho$ is a tuple $(L_{1},\ell_{1},L_{2},\ell_{2})$ such that

1. (1)

$L_{1},L_{2}$ are loops of $\rho$, 2. (2)

$\ell_{1}=(x_{1},y_{1})$ and $\ell_{2}=(x_{2},y_{2})$ are anchor points of $L_{1}$ and $L_{2}$, respectively, 3. (3)

$\ell_{1}\mathrel{{\lhd}}\ell_{2}$ and $x_{1}>x_{2}$

(namely, $\ell_{2}$ follows $\ell_{1}$ in the run, but the position of $\ell_{2}$ precedes the position of $\ell_{1}$), 4. (4)

for both $i=1$ and $i=2$, $\mathsf{out}(\mathsf{tr}(\ell_{i}))\neq\varepsilon$ and $\mathsf{tr}(\ell_{i})$ is output-minimal.

The left hand-side of Figure 4 gives an example of an inversion, assuming that the outputs $v_{1}=\mathsf{tr}(\ell_{1})$ and $v_{2}=\mathsf{tr}(\ell_{2})$ are non-empty and the intercepted factors are output-minimal.

The rest of the section is devoted to prove the implication P1 $\Rightarrow$ P2 of Theorem 6. We recall that a word $w=a_{1}\cdots a_{n}$ has period $p$ if for every $1\leq i\leq|w|-p$, we have $a_{i}=a_{i+p}$. For example, the word $abc\,abc\,ab$ has period $3$.

We remark that, thanks to Lemma 7, for every inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$, the outputs $\mathsf{out}(\mathsf{tr}(\ell_{1}))$ and $\mathsf{out}(\mathsf{tr}(\ell_{2}))$ have length at most $\boldsymbol{B}$. By pairing this with the assumption that the transducer $T$ is one-way definable, and by using some classical word combinatorics, we show that the output produced between the anchor points of every inversion has period that divides the lengths of $\mathsf{out}(\mathsf{tr}(\ell_{1}))$ and $\mathsf{out}(\mathsf{tr}(\ell_{2}))$. In particular, this period is at most $\boldsymbol{B}$. The proposition below shows a slightly stronger periodicity property, which refers to the output produced between the anchor points $\ell_{1},\ell_{2}$ of an inversion, but extended on both sides with the words $\mathsf{out}(\mathsf{tr}(\ell_{1}))$ and $\mathsf{out}(\mathsf{tr}(\ell_{2}))$. We will exploit this stronger periodicity property later, when dealing with overlapping portions of the run delimited by different inversions (cf. Lemma 13).

Proposition 8**.**

If $T$ is one-way definable, then the following property P2 holds:

For all inversions $(L_{1},\ell_{1},L_{2},\ell_{2})$ of $\rho$, the period $p$ of the word

[TABLE]

divides both $|\mathsf{out}(\mathsf{tr}(\ell_{1}))|$ and $|\mathsf{out}(\mathsf{tr}(\ell_{2}))|$. Moreover, $p\leq\boldsymbol{B}$.

The above proposition thus formalizes the implication P1 $\Rightarrow$ P2 of Theorem 6. Its proof relies on a few combinatorial results. The first one is Fine and Wilf’s theorem [27]. In short, this theorem says that, whenever two periodic words $w_{1},w_{2}$ share a sufficiently long factor, then they have the same period. Here we use a slightly stronger variant of Fine and Wilf’s theorem, which additionally shows how to align a common factor of the two words $w_{1},w_{2}$ so as to form a third word containing a prefix of $w_{1}$ and a suffix of $w_{2}$. This variant of Fine-Wilf’s theorem will be particularly useful in the proof of Lemma 13, while for all other applications the classical statement suffices. size=,color=green!30,size=,color=green!30,todo: size=,color=green!30,O. This last sentence is outdated, right?size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. This refers to the variant of Fine-Wilf, which still used in the referenced lemma. But I rephrased so as to make it clear. Let me know if it makes more sense now.

Theorem 9** (Fine-Wilf’s theorem).**

If $w_{1}=w^{\prime}_{1}\,w\>w^{\prime\prime}_{1}$ has period $p_{1}$, $w_{2}=w^{\prime}_{2}\,w\,w^{\prime\prime}_{2}$ has period $p_{2}$, and the common factor $w$ has length at least $p_{1}+p_{2}-\gcd(p_{1},p_{2})$, then $w_{1}$, $w_{2}$, and $w_{3}=w^{\prime}_{1}\,w\,w^{\prime\prime}_{2}$ have period $\gcd(p_{1},p_{2})$.

The second combinatorial result required in our proof concerns periods of words with iterated factors, like those that arise from considering outputs of pumped runs, and it is formalized precisely by the lemma below. To improve readability, we often highlight the important iterations of factors inside a word.

Lemma 10**.**

Assume that $v_{0}\>\boldsymbol{v_{1}^{n}}\>v_{2}\>\cdots\>v_{k-1}\>\boldsymbol{v_{k}^{n}}\>v_{k+1}$ has period $p$ for some $n>p$. Then $v_{0}\>\boldsymbol{{v_{1}}^{n_{1}}}\>v_{2}\>\cdots\>v_{k-1}\>\boldsymbol{{v_{k}}^{n_{k}}}\>v_{k+1}$ has period $p$ for all $n_{1},\ldots,n_{k}\in\mathbb{N}$.

Proof 4.2**.**

Assume that $w=v_{0}\>\boldsymbol{v_{1}^{n}}\>v_{2}\>\cdots\>v_{k-1}\>\boldsymbol{v_{k}^{n}}\>v_{k+1}$ has period $p$, and that $n>p$. Consider an arbitrary factor $v_{i}^{p}$ of $w$. Since $v_{i}^{p}$ has periods $p$ and $|v_{i}|$, it has also period $r=\gcd(p,|v_{i}|)$. By Fine-Wilf (Theorem 9), we know that $w$ has period $r$ as well. Moreover, since the length of $v_{i}$ is multiple of $r$, changing the number of repetitions of $v_{i}$ inside $w$ does not affect the period $r$ of $w$. Since $v_{i}$ was chosen arbitrarily, this means that, for all $n_{1},\dots,n_{k}\in\mathbb{N}$, $v_{0}\>\boldsymbol{{v_{1}}^{n_{1}}}\>v_{2}\>\cdots\>v_{k-1}\>\boldsymbol{{v_{k}}^{n_{k}}}\>v_{k+1}$ has period $r$, and hence period $p$ as well.

Recall that our goal is to show that the output produced amid every inversion has period bounded by $\boldsymbol{B}$. The general idea is to pump the loops of the inversion and compare the outputs of the two-way transducer $T$ with those of an equivalent one-way transducer $T^{\prime}$. The comparison leads to an equation between words with iterated factors, where the iterations are parametrized by two unknowns $n_{1},n_{2}$ that occur in opposite order in the left, respectively right hand-side of the equation. Our third and last combinatorial result considers a word equation of this precise form, and derives from it a periodicity property. size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. In Corollary 4.8, the definition of $v^{(n_{1},n_{2})}$ is hard to process, especially when introduced in the middle of a proof. The notion is used later in the part: it warrants a proper definition and/or clear example either in the preliminaries or at the beginning of Part 4. size=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. I don’t understand this remark size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. answered in review1-answers.txt: the best place is here.

size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. This reviewer commend has led to a number of corrections, and the removal of Saarela tool For the sake of brevity, we use the notation $v^{(n_{1},n_{2})}$ to represent words with factors iterated $n_{1}$ or $n_{2}$ times, namely, words of the form $v_{0}\>v_{1}^{n_{i_{1}}}\>v_{2}\>\cdots\>v_{k-1}\>v_{k}^{n_{i_{k}}}\>v_{k+1}$, where the $v_{0},v_{1},v_{2},\dots,v_{k-1},v_{k},v_{k+1}$ are fixed words (possibly empty) and each index among $i_{1},\dots,i_{k}$ is either $1$ or $2$.

Lemma 11**.**

Consider a word equation of the form

[TABLE]

where $n_{1},n_{2}$ are the unknowns and $v_{1},v_{3}$ are non-empty words. If the above equation holds for all $n_{1},n_{2}\in\mathbb{N}$, then

[TABLE]

has period $\gcd(|v_{1}|,|v_{3}|)$ for all $n_{1},n_{2}\in\mathbb{N}$.

Proof 4.3**.**

The idea of the proof is to let the parameters $n_{1},n_{2}$ of the equation grow independently, and apply Fine and Wilf’s theorem (Theorem 9) a certain number of times to establish periodicities in overlapping factors of the considered words.

We begin by fixing $n_{1}$ large enough so that the factor $\boldsymbol{v_{1}^{n_{1}}}$ of the left hand-side of the equation becomes longer than $|w_{0}|+|w_{1}|$ (this is possible because $v_{1}$ is non-empty). Now, if we let $n_{2}$ grow arbitrarily large, we see that the length of the periodic word $\boldsymbol{w_{1}^{n_{2}}}$ is almost equal to the length of the left hand-side term $v_{0}^{(n_{1},n_{2})}~{}\boldsymbol{v_{1}^{n_{1}}}~{}v_{2}^{(n_{1},n_{2})}~{}\boldsymbol{v_{3}^{n_{2}}}~{}v_{4}^{(n_{1},n_{2})}$: indeed, the difference in length is given by the constant $|w_{0}|+|w_{2}|+n_{1}\cdot|w_{3}|+|w_{4}|$. In particular, this implies that $\boldsymbol{w_{1}^{n_{2}}}$ covers arbitrarily long prefixes of $\boldsymbol{v_{1}}~{}v_{2}^{(n_{1},n_{2})}~{}\boldsymbol{v_{3}^{n_{2}+1}}$, which in its turn contains long repetitions of the word $v_{3}$. Hence, by Theorem 9, the word $\boldsymbol{v_{1}}~{}v_{2}^{(n_{1},n_{2})}~{}\boldsymbol{v_{3}^{n_{2}+1}}$ has period $|v_{3}|$.

We remark that the periodicity shown so far holds for a large enough $n_{1}$ and for all but finitely many $n_{2}$, where the threshold for $n_{2}$ depends on $n_{1}$: once $n_{1}$ is fixed, $n_{2}$ needs to be larger than $f(n_{1})$, for a suitable function $f$. In fact, by using size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Saarela has been replaced by the new lemma here Lemma 10, with $n_{1}$ fixed and $n=n_{2}$ large enough, we deduce that the periodicity holds for large enough $n_{1}\in\mathbb{N}$ and for all $n_{2}\in\mathbb{N}$.

We could also apply a symmetric reasoning: we choose $n_{2}$ large enough and let $n_{1}$ grow arbitrarily large. Doing so, we prove that for a large enough $n_{2}$ and for all but finitely many $n_{1}$, the word $\boldsymbol{v_{1}^{n_{1}+1}}~{}v_{2}^{(n_{1},n_{2})}~{}\boldsymbol{v_{3}}$ is periodic with period $|v_{1}|$. As before, with the help of size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. same here for Saarela… Lemma 10, this can be strengthened to hold for large enough $n_{2}\in\mathbb{N}$ and for all $n_{1}\in\mathbb{N}$.

Putting together the results proven so far, we get that for all but finitely many $n_{1},n_{2}$,

[TABLE]

Finally, we observe that the prefix $\boldsymbol{v_{1}^{n_{1}+1}}\cdot v_{2}^{(n_{1},n_{2})}\cdot\boldsymbol{v_{3}}$ and the suffix $\boldsymbol{v_{1}}\cdot v_{2}^{(n_{1},n_{2})}\cdot\boldsymbol{v_{3}^{n_{2}+1}}$ share a common factor of length at least $|v_{1}|+|v_{3}|$. By Theorem 9, we derive that $\boldsymbol{v_{1}^{n_{1}+1}}\cdot v_{2}^{(n_{1},n_{2})}\cdot\boldsymbol{v_{3}^{n_{2}+1}}$ has period $\gcd(|v_{1}|,|v_{3}|)$ for all but finitely many $n_{1},n_{2}$. Finally, by using again size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. …and here Lemma 10, we conclude that the periodicity holds for all $n_{1},n_{2}\in\mathbb{N}$.

We are now ready to prove the implication P1 $\Rightarrow$ P2:

Proof 4.4** (Proof of Proposition 8).**

Let $T^{\prime}$ be a one-way transducer equivalent to $T$, and consider an inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$ of the successful run $\rho$ of $T$ on input $u$. The reader may refer to Figure 4 to get basic intuition about the proof technique. For simplicity, we assume that the loops $L_{1}$ and $L_{2}$ are disjoint, as shown in the figure. If this were not the case, we would have at least $\max(L_{1})>\min(L_{2})$, since the anchor point $\ell_{1}$ is strictly to the right of the anchor point $\ell_{2}$. We could then consider the pumped run $\mathsf{pump}_{L_{1}}^{k}(\rho)$ for a large enough $k>1$ in such a way that the rightmost copy of $L_{1}$ turns out to be disjoint from and strictly to the right of $L_{2}$. We could thus reason as we do below, by replacing everywhere (except in the final part of the proof, cf. Transferring periodicity to the original run) the run $\rho$ with the pumped run $\mathsf{pump}_{L_{1}}^{k}(\rho)$, and the formal parameter $m_{1}$ with $m_{1}+k$.

Inducing loops in $T^{\prime}$.*  We begin by pumping the run $\rho$ and the underlying input $u$, on the loops $L_{1}$ and $L_{2}$, in order to induce new loops $L^{\prime}_{1}$ and $L^{\prime}_{2}$ that are also loops in a successful run of $T^{\prime}$. Assuming that $L_{1}$ is strictly to the right of $L_{2}$, we define for all numbers $m_{1},m_{2}\in\mathbb{N}$:*

[TABLE]

In the pumped run $\rho^{(m_{1},m_{2})}$, we identify the positions that mark the endpoints of the occurrences of $L_{1},L_{2}$. More precisely, if $L_{1}=[x_{1},x_{2}]$ and $L_{2}=[x_{3},x_{4}]$, with $x_{1}>x_{4}$, then the sets of these positions are

[TABLE]

Periodicity of outputs of pumped runs.*  We use now the fact that $T^{\prime}$ is a one-way transducer equivalent to $T$. size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. you fix $\lambda^{m_{1},m_{2}}$ to be a particular run of T’ on the input $u^{m_{1},m_{2}}$, and then, on line 10, you state that for some particular $m_{1}$ and $m_{2}$, this run can be obtained by pumping $\lambda^{k_{0},k_{0}}$. This does not seem true in general: since $T$ is not deterministic, we might just have chosen a different run.size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. CorrectedWe first recall (see, for instance, [16, 8]) that every functional one-way transducer can be made unambiguous, namely, can be transformed into an equivalent one-way transducer that admits at most one successful run on each input. This means that, without loss of generality, we can assume that $T^{\prime}$ too is unambiguous, and hence it admits exactly one successful run, say $\lambda^{(m_{1},m_{2})}$, on each input $u^{(m_{1},m_{2})}$.*

Since $T^{\prime}$ has finitely many states, we can find, for a large enough number $k_{0}$ two positions $x^{\prime}_{1}<x^{\prime}_{2}$, both in $X_{1}^{(k_{0},k_{0})}$, such that $L^{\prime}_{1}=[x^{\prime}_{1},x^{\prime}_{2}]$ is a loop of $\lambda^{(k_{0},k_{0})}$. Similarly, we can find two positions $x^{\prime}_{3}<x^{\prime}_{4}$, both in $X_{2}^{(k_{0},k_{0})}$, such that $L^{\prime}_{2}=[x^{\prime}_{3},x^{\prime}_{4}]$ is a loop of $\lambda^{(k_{0},k_{0})}$. By construction $L^{\prime}_{1}$ (resp. $L^{\prime}_{2}$) consists of $k_{1}\leq k_{0}$ (resp. $k_{2}\leq k_{0}$) copies of $L_{1}$ (resp. $L_{2}$), and hence $L^{\prime}_{1},L^{\prime}_{2}$ are also loops of $\rho^{(k_{0},k_{0})}$. In particular, this implies that for all $n_{1},n_{2}\in\mathbb{N}$:

[TABLE]

where $f(n_{1})=k_{1}n_{1}+k_{0}$ and $g(n_{2})=k_{2}n_{2}+k_{0}$.

Now recall that $\rho^{(f(n_{1}),g(n_{2}))}$ and $\lambda^{(f(n_{1}),g(n_{2}))}$ are runs of $T$ and $T^{\prime}$ on the same word $u^{(f(n_{1}),g(n_{2}))}$, and they produce the same output. Let us denote this output by $w^{(f(n_{1}),g(n_{2}))}$. Below, we show two possible factorizations of $w^{(f(n_{1}),g(n_{2}))}$ based on the shapes of the pumped runs $\lambda^{(f(n_{1}),g(n_{2}))}$ and $\rho^{(f(n_{1}),g(n_{2}))}$. For the first factorization, we recall that $L^{\prime}_{2}$ precedes $L^{\prime}_{1}$, according to the ordering of positions, and that the run $\lambda^{(f(n_{1}),g(n_{2}))}$ is one-way (in particular loops have only one anchor point and one trace). We thus obtain: size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. eq. 4.3 The exponents should be $n_{2}+1$ and $n_{1}+1$. (same for eq. 4.4) size=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. Yes but not sure it makes sense to keep the +1 everywhere then size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. We agreed to change the two itemized lists below and replace ‘right border’ by ‘left border’ everywhere. I have implemented this…

[TABLE]

where

• •

$w_{0}$* is the output produced by the prefix of $\lambda^{(k_{0},k_{0})}$ ending at the only anchor point of $L^{\prime}_{2}$,*

• •

$\boldsymbol{w_{1}}$* is the trace of $L^{\prime}_{2}$,*

• •

$w_{2}$* is the output produced by the factor of $\lambda^{(k_{0},k_{0})}$ between the anchor points of $L^{\prime}_{2}$ and $L^{\prime}_{1}$,*

• •

$\boldsymbol{w_{3}}$* is the trace of $L^{\prime}_{1}$,*

• •

$w_{4}$* is the output produced by the suffix of $\lambda^{(k_{0},k_{0})}$ starting at the anchor point of $L^{\prime}_{1}$. Hence, $w_{0}~{}w_{2}~{}w_{4}$ is the output of $\lambda^{(k_{0},k_{0})}$.*

size=,color=red!30,size=,color=red!30,todo: size=,color=red!30,F. added the last sentence to avoid the $+1$ or not confusion. In addition to the anchor point/trace usage that we talked about

For the second factorization, we consider $L^{\prime}_{1}$ and $L^{\prime}_{2}$ as loops of $\rho^{(k_{0},k_{0})}$. We recall that $\ell_{1},\ell_{2}$ are anchor points of the loops $L_{1},L_{2}$ of $\rho$, and that there are corresponding copies of these anchor points in the pumped run $\rho^{(f(n_{1}),g(n_{2}))}$. We define $\ell^{\prime}_{1}$ (resp. $\ell^{\prime}_{2}$) to be the first (resp. last) location in $\rho^{(f(n_{1}),g(n_{2}))}$ that corresponds to $\ell_{1}$ (resp. $\ell_{2}$) and that is an anchor point of a copy of $L^{\prime}_{1}$ (resp. $L^{\prime}_{2}$). For example, if $\ell_{1}=(x_{1},y_{1})$, with $y_{1}$ even, then \ell^{\prime}_{1}=\big{(}x_{1}+f(n_{2})(x_{4}-x_{3}),y_{1}\big{)}. Thanks to Equation 2 we know that the output produced by $\rho^{(f(n_{1}),g(n_{2}))}$ is of the form

[TABLE]

where

• •

$\boldsymbol{v_{1}}=\mathsf{out}(\mathsf{tr}(\ell^{\prime}_{1}))$, where $\ell^{\prime}_{1}$ is seen as an anchor point in a copy of $L^{\prime}_{1}$,

• •

$\boldsymbol{v_{3}}=\mathsf{out}(\mathsf{tr}(\ell^{\prime}_{2}))$, where $\ell^{\prime}_{2}$ is seen as an anchor point in a copy of $L^{\prime}_{2}$

(note that the words $v_{1},v_{3}$ depend on $k_{0}$, but not on $n_{1},n_{2}$),

• •

$v_{0}^{(n_{1},n_{2})}$* is the output produced by the prefix of $\rho^{(f(n_{1}),g(n_{2}))}$ that ends at $\ell_{1}^{\prime}$ (this word may depend on the parameters $n_{1},n_{2}$ since the loops $L^{\prime}_{1},L^{\prime}_{2}$ may be traversed several times before reaching the first occurrence of $\mathsf{tr}(\ell^{\prime}_{1})$),*

• •

$v_{2}^{(n_{1},n_{2})}$* is the output produced by the factor of $\rho^{(f(n_{1}),g(n_{2}))}$ that starts at $\ell^{\prime}_{1}$ and ends at $\ell^{\prime}_{2}$,*

• •

$v_{4}^{(n_{1},n_{2})}$* is the output produced by the suffix of $\rho^{(f(n_{1}),g(n_{2}))}$ that starts at $\ell^{\prime}_{2}$.*

Putting together Equations (3) and (4), we get

[TABLE]

Recall that the definition of inversions (Definition 4) states that the words $v_{1},v_{3}$ are non-empty. This allows us to apply Lemma 11, which shows that the word $\boldsymbol{v_{1}}~{}\boldsymbol{v_{1}^{n_{1}}}~{}v_{2}^{(n_{1},n_{2})}~{}\boldsymbol{v_{3}^{n_{2}}}~{}\boldsymbol{v_{3}}$ has period $p=\gcd(|v_{1}|,|v_{3}|)$, for all $n_{1},n_{2}\in\mathbb{N}$.

Note that the latter period $p$ still depends on $T^{\prime}$, since the words $v_{1},v_{3}$ were obtained from loops $L^{\prime}_{1},L^{\prime}_{2}$ on the run $\lambda^{(k_{0},k_{0})}$ of $T^{\prime}$. However, because each loop $L^{\prime}_{i}$ consists of $k_{i}$ copies of the original loop $L_{i}$, we also know that $v_{1}=(\mathsf{out}(\mathsf{tr}(\ell_{1})))^{k_{1}}$ and $v_{3}=(\mathsf{out}(\mathsf{tr}(\ell_{2})))^{k_{2}}$. By Theorem 9, this implies that for all $n_{1},n_{2}\in\mathbb{N}$, the word

[TABLE]

has a period that divides $|\mathsf{out}(\mathsf{tr}(\ell_{1}))|$ and $|\mathsf{out}(\mathsf{tr}(\ell_{2}))|$.

Transferring periodicity to the original run.*  The last part of the proof amounts at showing a similar periodicity property for the output produced by the original run $\rho$. By construction, the iterated factors inside $v_{2}^{(n_{1},n_{2})}$ in the previous word are all of the form $v^{k_{1}n_{1}+k_{0}}$ or $v^{k_{2}n_{2}+k_{0}}$, for some words $v$. By taking out the constant factors $v^{k_{0}}$ from the latter repetitions, we can write $v_{2}^{(n_{1},n_{2})}$ as a word with iterated factors of the form $v^{k_{1}n_{1}}$ or $v^{k_{2}n_{2}}$, namely, as ${v^{\prime}_{2}}^{(k_{1}n_{1},k_{2}n_{2})}$. So the word*

[TABLE]

is periodic, with period that divides $|\mathsf{out}(\mathsf{tr}(\ell_{1}))|$ and $|\mathsf{out}(\mathsf{tr}(\ell_{2}))|$, for all $k^{\prime}_{1}\in\{k_{1}n\>:\>n\in\mathbb{N}\}$ and all $k^{\prime}_{2}\in\{k_{2}n\>:\>n\in\mathbb{N}\}$. We now apply size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Also here Saarela has been replaced by the new lemma Lemma 10, once with $n=k^{\prime}_{1}$ and once with $n=k^{\prime}_{2}$, to conclude that the latter periodicity property holds also for $k^{\prime}_{1}=1$ and $k^{\prime}_{2}=1$. This shows that the word

[TABLE]

is periodic, with period that divides $|\mathsf{out}(\mathsf{tr}(\ell_{1}))|$ and $|\mathsf{out}(\mathsf{tr}(\ell_{2}))|$.

5. One-way definability in the sweeping case

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. From Part 5 onwards, a major point is that an output you can bound is an output you can guess, and a periodic output is an output you don’t have to guess, as you can produce it ”out of order”. On your Figures (and any you might care to add during potential revisions), having a specific visual representation (if possible coherent throughout the paper) to differentiate path sections with empty, bounded, or periodic output will make your point way more intelligible. size=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. so this is just about uniformizing the figure for sweeping and for two way ?

In the previous section we have shown the implication P1 $\Rightarrow$ P2 for a functional sweeping transducer $T$. Here we close the cycle by proving the implications P2 $\Rightarrow$ P3 and P3 $\Rightarrow$ P1. In particular, we show how to derive the existence of successful runs admitting a $\boldsymbol{B}$-decomposition and construct a one-way transducer $T^{\prime}$ that simulates $T$ on those runs. This will basically prove Theorem 6 in the sweeping case.

Run decomposition.

We begin by giving the definition of $\boldsymbol{B}$-decomposition for a run $\rho$ of $T$. Intuitively, a $\boldsymbol{B}$-decomposition of $\rho$ identifies factors of $\rho$ that can be easily simulated in a one-way manner. The definition below describes precisely the shape of these factors.

First we need to recall the notion of almost periodicity: a word $w$ is almost periodic with bound $p$ if $w=w_{0}~{}w_{1}~{}w_{2}$ for some words $w_{0},w_{2}$ of length at most $p$ and some word $w_{1}$ of period at most $p$.

We need to work with subsequences of the run $\rho$ that are induced by particular sets of locations, not necessarily consecutive. Recall that $\rho[\ell,\ell^{\prime}]$ denotes the factor of $\rho$ delimited by two locations $\ell\mathrel{\unlhd}\ell^{\prime}$. Similarly, given any set $Z$ of locations, we denote by $\rho|Z$ the subsequence of $\rho$ induced by $Z$. Note that, even though $\rho|Z$ might not be a valid run

of the transducer, we can still refer to the number of transitions in it and to the size of the produced output $\mathsf{out}(\rho|Z)$. Formally, a transition in $\rho|Z$ is a transition from some $\ell$ to $\ell^{\prime}$, where both $\ell,\ell^{\prime}$ belong to $Z$. The output $\mathsf{out}(\rho|Z)$ is the concatenation of the outputs of the transitions of $\rho|Z$ (according to the order given by $\rho$).

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Fig. 5 is moderately informative… Figure 17 and 18 are a more inspiring illustration of the intuition behind diagonal and blocks. Their ”restricted” counterpart could be simpler to represent.size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. done

{defi}

Consider a factor $\rho[\ell,\ell^{\prime}]$ of a run $\rho$ of $T$, where $\ell=(x,y)$, $\ell^{\prime}=(x^{\prime},y^{\prime})$ are two locations with $x\leq x^{\prime}$. We call $\rho[\ell,\ell^{\prime}]$

• •

a floor if $y=y^{\prime}$ is even, namely, if $\rho[\ell,\ell^{\prime}]$ lies entirely on the same level and is rightward oriented;

@afterheading

• •

a $\boldsymbol{B}$-diagonal if there is a sequence $\ell=\ell_{0}\mathrel{\unlhd}\ell_{1}\mathrel{\unlhd}\dots\mathrel{\unlhd}\ell_{2n+1}=\ell^{\prime}$, where each $\rho[\ell_{2i+1},\ell_{2i+2}]$ is a floor, each $\rho[\ell_{2i},\ell_{2i+1}]$ produces an output of length at most $2{\boldsymbol{H}}\boldsymbol{B}$, and the position of each $\ell_{2i}$ is to the left of the position of $\ell_{2i+1}$;

• •

a $\boldsymbol{B}$-block if the output produced by $\rho[\ell,\ell^{\prime}]$ is almost periodic with bound $2\boldsymbol{B}$, and the output produced by the subsequence $\rho|Z$, where Z=[\ell,\ell^{\prime}]~{}\setminus~{}\big{(}[x,x^{\prime}]\times[y,y^{\prime}]\big{)}, has length at most $2{\boldsymbol{H}}\boldsymbol{B}$.

Before continuing we give some intuition on the above definitions. The simplest concept is that of floor, which is a rightward oriented factor of a run. Diagonals are sequences of consecutive floors interleaved by factors that produce small (bounded) outputs. We see an example of a diagonal in Figure 5, where we marked the important locations and highlighted with thick arrows the two floors that form the diagonal. The factors of the diagonal that are not floors are represented instead by dotted arrows. The third notion is that of a block. An important constraint in the definition of a block is that the produce output must be almost periodic, with small enough bound. In Figure 6, the periodic output is represented by the thick arrows, either solid or dotted, that go from location $\ell$ to location $\ell^{\prime}$. In addition, the block must satisfy a constraint on the length of the output produced by the subsequence $\rho|Z$, where Z=[\ell,\ell^{\prime}]~{}\setminus~{}\big{(}[x,x^{\prime}]\times[y,y^{\prime}]\big{)}. The latter set $Z$ consists of location that are either to the left of the position of $\ell$ or to the right of the position of $\ell^{\prime}$. For example, in Figure 6 the set $Z$ coincides with the area outside the hatched rectangle. Accordingly, the portion of the subsequence $\rho|Z$ is represented by the dotted bold arrows.

Diagonals and blocks are used as key objects to derive a notion of decomposition for a run of a sweeping transducer. We formalize this notion below.

{defi}

A $\boldsymbol{B}$-decomposition of a run $\rho$ of $T$ is a factorization $\prod_{i}\,\rho[\ell_{i},\ell_{i+1}]$ of $\rho$ into $\boldsymbol{B}$-diagonals and $\boldsymbol{B}$-blocks.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Fig.6 does not look like a fully factorized run: what happens for example, between the end of $D_{1}$ and $l_{1}$? Can a concrete example be devised? This works rather well in Part 8.size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Improved the figure and the explanation

Figure 7 gives an example of such a decomposition. size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. dans Fig 7, les aretes de $B_{2}$ vont dans la mauvaise directionsize=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. I have corrected this. Now the figure is a bit more regular (perhaps too much), but at least conveys the correct information!Each factor is either a diagonal $D_{i}$ or a block $B_{i}$. Inside each factor we highlight by thick arrows the portions of the run that can be simulated by a one-way transducer, either because they are produced from left to right (thus forming diagonals) or because they are periodic (thus forming blocks). We also recall from Definition 5 that most of the output is produced inside the hatched rectangles, since the output produced by a diagonal or a block outside the corresponding blue or red hatched rectangle has length at most $2{{\boldsymbol{H}}^{2}}\boldsymbol{B}$. Finally, we observe that the locations delimiting the factors of the decomposition are arranged following the natural order of positions and levels. All together, this means that the output produced by a run that enjoys a decomposition can be simulated in a one-way manner.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Part 5: In ”From periodicity of inversions to existence of decompositions”, the general proof structure in general and transition between lemmas in particular could be made more visually and conceptually explicit, as the global idea of the proof is difficult to grasp at first. A mini ”road map” as made in Part 3 for the whole of Part 4 and 5 might be another way to help the reader through these rather intricate proofs.

From periodicity of inversions to existence of decompositions.

Now that we set up the definition of $\boldsymbol{B}$-decomposition, we turn towards proving the implication P2 $\Rightarrow$ P3 of Theorem 6. In fact, we will prove a slightly stronger result than P2 $\Rightarrow$ P3, which is stated further below. Formally, when we say that a run $\rho$ satisfies P2 we mean that for every inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$ of $\rho$, the word $\mathsf{out}(\mathsf{tr}(\ell_{1}))~{}\mathsf{out}(\rho[\ell_{1},\ell_{2}])~{}\mathsf{out}(\mathsf{tr}(\ell_{2}))$ has period $\gcd(|\mathsf{out}(\mathsf{tr}(\ell_{1}))|,|\mathsf{out}(\mathsf{tr}(\ell_{2}))|)\leq\boldsymbol{B}$. We aim at proving that every run that satisfies P2 enjoys a decomposition, independently of whether other runs do or do not satisfy P2:

Proposition 12**.**

If $\rho$ is a run of $T$ that satisfies P2, then $\rho$ admits a $\boldsymbol{B}$-decomposition.

size=,color=red!30,size=,color=red!30,todo: size=,color=red!30,F. a few lines for the roadmap asked by the reviewer, but I don’t think they’re useful at allsize=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. I think it is good

Let us fix a run $\rho$ of $T$ and assume that it satisfies P2. To show that $\rho$ admits a $\boldsymbol{B}$-decomposition, we will identify the blocks of the decomposition as equivalence classes of a suitable relation based on inversions (cf. Definition 5). Then, we will use combinatorial arguments (notably, Lemmas 7 and 13) to prove that the constructed blocks satisfy the desired properties. Finally, we will show how the resulting equivalence classes form all the necessary blocks of the decomposition, in the sense that the factors in between those classes are diagonals.

We begin by introducing the equivalence relation by means of which we can then identify the blocks of a decomposition of $\rho$.

{defi}

Let $\mathrel{\text{\sf S}}$ be the relation that pairs every two locations $\ell,\ell^{\prime}$ of $\rho$ whenever there is an inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$ of $\rho$ such that $\ell_{1}\mathrel{\unlhd}\ell,\ell^{\prime}\mathrel{\unlhd}\ell_{2}$, namely, whenever $\ell$ and $\ell^{\prime}$ occur within the same inversion. Let $\mathrel{\text{\sf S}}^{*}$ be the reflexive and transitive closure of $\mathrel{\text{\sf S}}$.

It is easy to see that every equivalence class of $\mathrel{\text{\sf S}}^{*}$ is a convex subset with respect to the run order on locations of $\rho$. Moreover, every non-singleton equivalence class of $\mathrel{\text{\sf S}}^{*}$ is a union of a series of inversions that are two-by-two overlapping. One can refer to Figure 8 for an intuitive account of what we mean by two-by-two overlapping: the thick arrows represent factors of the run that lie entire inside an $\mathrel{\text{\sf S}}^{*}$-equivalence class, each inversion is identified by a pair of consecutive anchor points with the same color. According to the run order, between every pair of anchor points with the same color, there is at least one anchor point of different color: this means that the inversions corresponding to the two colors are overlapping.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Fig.7 might benefit from some $L_{1}$, $L_{2}$ appearing, to illustrate the point made in Claim 5.6 size=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. I don’t agreesize=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. I also disagree. Argued in review1-answer.txt.

Formally, we say that an inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$ covers a location $\ell$ when $\ell_{1}\mathrel{\unlhd}\ell\mathrel{\unlhd}\ell_{2}$. We say that two inversions $(L_{1},\ell_{1},L_{2},\ell_{2})$ and $(L_{3},\ell_{3},L_{4},\ell_{4})$ are overlapping if $(L_{1},\ell_{1},L_{2},\ell_{2})$ covers $\ell_{3}$ and $(L_{3},\ell_{3},L_{4},\ell_{4})$ covers $\ell_{2}$ (or the other way around).

The next lemma uses the fact that $\rho$ satisfies P2 to deduce that the output produced inside every $\mathrel{\text{\sf S}}^{*}$-equivalence class has period at most $\boldsymbol{B}$. Note that the proof below does not exploit the fact that the transducer is sweeping.

Lemma 13**.**

If $\rho$ satisfies P2 and $\ell\mathrel{\unlhd}\ell^{\prime}$ are two locations of $\rho$ such that $\ell\mathrel{\text{\sf S}}^{*}\ell^{\prime}$, then the output $\mathsf{out}(\rho[\ell,\ell^{\prime}])$ produced between these locations has period at most $\boldsymbol{B}$.

Proof 5.1**.**

The claim for $\ell=\ell^{\prime}$ holds trivially, so we assume that $\ell\mathrel{{\lhd}}\ell^{\prime}$. Since the $\mathrel{\text{\sf S}}^{*}$-equivalence class that contains $\ell,\ell^{\prime}$ is non-singleton, we know that there is a series of inversions

[TABLE]

that are two-by-two overlapping and such that $\ell_{0}\mathrel{\unlhd}\ell\mathrel{{\lhd}}\ell^{\prime}\mathrel{\unlhd}\ell_{2k+1}$. Without loss of generality, we can assume that every inversion $(L_{2i},\ell_{2i},L_{2i+1},\ell_{2i+1})$ is maximal in the following sense: there is no other inversion $(\tilde{L},\tilde{\ell},\tilde{L}^{\prime},\tilde{\ell}^{\prime})\neq(L_{2i},\ell_{2i},L_{2i+1},\ell_{2i+1})$ such that $\tilde{\ell}\mathrel{\unlhd}\ell_{2i}\mathrel{\unlhd}\ell_{2i+1}\mathrel{\unlhd}\tilde{\ell}^{\prime}$.

For the sake of brevity, let $v_{i}=\mathsf{out}(\mathsf{tr}(\ell_{i}))$ and $p_{i}=|v_{i}|$. Since $\rho$ satisfies P2 (recall Proposition 8), we know that, for all $i=0,\dots,n$, the word

[TABLE]

has period that divides both $p_{2i}$ and $p_{2i+1}$ and is at most $\boldsymbol{B}$. In order to show that the period of $\mathsf{out}(\rho[\ell,\ell^{\prime}])$ is also bounded by $\boldsymbol{B}$, it suffices to prove the following claim by induction on $i$:

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Claim 5.6, end of page 18: making explicit what words you wish to use Fine-Wilf on might make for a more readable proof. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. answered in review1-answer.txt

{clm}

For all $i=0,\dots,k$, the word \mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+1}]\big{)}\>v_{2i+1} has period at most $\boldsymbol{B}$ that divides $p_{2i+1}$.

The base case $i=0$ follows immediately from our hypothesis, since $(L_{0},\ell_{0},L_{1},\ell_{1})$ is an inversion. For the inductive step, we assume that the claim holds for $i<k$, and we prove it for $i+1$. First of all, we factorize our word as follows:

[TABLE]

By the inductive hypothesis, the output produced between $\ell_{0}$ and $\ell_{2i+1}$, extended to the right with $v_{2i+1}$, has period that divides $p_{2i+1}$. Moreover, because $\rho$ satisfies P2 and $(L_{2i+2},\ell_{2i+2},L_{2i+3},\ell_{2i+3})$ is an inversion, the output produced between the locations $\ell_{2i+2}$ and $\ell_{2i+3}$, extended to the left with $v_{2i+2}$ and to the right with $v_{2i+3}$, has period that divides both $p_{2i+2}$ and $p_{2i+3}$. Note that this is not yet sufficient for applying Fine-Wilf’s theorem, since the common factor \mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+1}]\big{)} might be too short (possibly just equal to $v_{2i+2}$). The key argument here is to prove that the interval $[\ell_{2i+2},\ell_{2i+1}]$ is covered by an inversion which is different from those that we considered above, namely, i.e. $(L_{2i+2},\ell_{2i+2},L_{2i+1},\ell_{2i+1})$. For example, $[\ell_{2},\ell_{1}]$ in Figure 8 is covered by the inversion $(L_{2},\ell_{2},L_{1},\ell_{1})$.

For this, we have to prove that the anchor points $\ell_{2i+2}$ and $\ell_{2i+1}$ are correctly ordered w.r.t. $\mathrel{\unlhd}$ and the ordering of positions (recall Definition 4). First, we observe that $\ell_{2i+2}\mathrel{\unlhd}\ell_{2i+1}$, since $(L_{2i},\ell_{2i},L_{2i+1},\ell_{2i+1})$ and $(L_{2i+2},\ell_{2i+2},L_{2i+3},\ell_{2i+3})$ are overlapping inversions. Next, we prove that the position of $\ell_{2i+1}$ is strictly to the left of the position of $\ell_{2i+2}$. By way of contradiction, suppose that this is not the case, namely, $\ell_{2i+1}=(x_{2i+1},y_{2i+1})$, $\ell_{2i+2}=(x_{2i+2},y_{2i+2})$, and $x_{2i+1}\geq x_{2i+2}$. Because $(L_{2i},\ell_{2i},L_{2i+1},\ell_{2i+1})$ and $(L_{2i+2},\ell_{2i+2},L_{2i+3},\ell_{2i+3})$ are inversions, we know that $\ell_{2i+3}$ is strictly to the left of $\ell_{2i+2}$ and $\ell_{2i+1}$ is strictly to the left of $\ell_{2i}$. This implies that $\ell_{2i+3}$ is strictly to the left of $\ell_{2i}$, and hence $(L_{2i},\ell_{2i},L_{2i+3},\ell_{2i+3})$ is also an inversion. Moreover, recall that $\ell_{2i}\mathrel{\unlhd}\ell_{2i+2}\mathrel{\unlhd}\ell_{2i+1}\mathrel{\unlhd}\ell_{2i+3}$. This contradicts the maximality of $(L_{2i},\ell_{2i},L_{2i+1},\ell_{2i+1})$, which we assumed at the beginning of the proof. Therefore, we must conclude that $\ell_{2i+1}$ is strictly to the left of $\ell_{2i+2}$.

Now that we know that $\ell_{2i+2}\mathrel{\unlhd}\ell_{2i+1}$ and that $\ell_{2i+1}$ is to the left of $\ell_{2i+2}$, we derive the existence of the inversion $(L_{2i+2},\ell_{2i+2},L_{2i+1},\ell_{2i+1})$. Again, because $\rho$ satisfies P2, we know that the word $v_{2i+2}~{}\mathsf{out}(\rho[\ell_{2i+2},\ell_{2i+1}])~{}v_{2i+1}$ has period at most $\boldsymbol{B}$ that divides $p_{2i+2}$ and $p_{2i+1}$. Summing up, we have:

1. (1)

w_{1}~{}=~{}\mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+1}]\big{)}~{}v_{2i+1}* has period $p_{2i+1}$,* 2. (2)

w_{2}~{}=~{}v_{2i+2}~{}\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+1}]\big{)}~{}v_{2i+1}* has period $p=\gcd(p_{2i+2},p_{2i+1})$,* 3. (3)

w_{3}~{}=~{}v_{2i+2}~{}\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+3}]\big{)}~{}v_{2i+3}* has period $p^{\prime}=\gcd(p_{2i+2},p_{2i+3})$.*

We are now ready to exploit our stronger variant of Fine-Wilf’s theorem, that is, Theorem 9.

We begin with (1) and (2) above. Let w=\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+1}]\big{)}~{}v_{2i+1} be the common suffix of $w_{1}$ and $w_{2}$. First note that since $p$ divides $p_{2i+2}$, the word $w$ is also a prefix of $w_{2}$, thus we can write $w_{2}=w\,w^{\prime}_{2}$. Second, note that the length of $w$ is at least $|v_{2i+1}|=p_{2i+1}=p_{2i+1}+p-\gcd(p_{2i+1},p)$. We can apply now Theorem 9 to $w_{1}=w^{\prime}_{1}\,w$ and $w_{2}=w\,w^{\prime}_{2}$ and obtain:

1. (4)

w_{4}~{}=~{}w^{\prime}_{1}\>w\>w^{\prime}_{2}~{}=~{}\mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+2}]\big{)}~{}v_{2i+2}~{}\mathsf{out}\big{(}\rho[\ell_{2i+2},\ell_{2i+1}]\big{)}~{}v_{2i+1}* has period $p$.*

We apply next Theorem 9 to (2) and (3), namely, to the words $w_{2}$ and $w_{3}$ with $v_{2i+2}$ as common factor. It is not difficult to check that $|v_{2i+2}|=p_{2i+2}\geq p+p^{\prime}-p^{\prime\prime}$ with $p^{\prime\prime}=\gcd(p,p^{\prime})$, using the definitions of $p$ and $p^{\prime}$: we can write $p_{2i+2}=p^{\prime\prime}rq=p^{\prime\prime}r^{\prime}q^{\prime}$ with $p=p^{\prime\prime}r$ and $p^{\prime}=p^{\prime\prime}r^{\prime}$. It suffices to check that $p^{\prime\prime}rq+p^{\prime\prime}r^{\prime}q^{\prime}\geq 2(p^{\prime\prime}r+p^{\prime\prime}r^{\prime}-p^{\prime\prime})$, hence that $rq+r^{\prime}q^{\prime}\geq 2r+2r^{\prime}-2$. This is clear if $\min(q,q^{\prime})>1$. Otherwise the inequality $p_{2i+2}\geq p+p^{\prime}-p^{\prime\prime}$ follows easily because $p=p^{\prime\prime}$ or $p^{\prime}=p^{\prime\prime}$ holds. Hence we obtain that $w_{2}$ and $w_{3}$ have both period $p^{\prime\prime}$.

Applying once more Theorem 9 to $w_{3}$ and $w_{4}$ with $v_{2i+2}$ as common factor, yields period $p^{\prime\prime}$ for the word

[TABLE]

Finally, the periodicity is not affected when we remove factors of length multiple than the period. In particular, by removing the factor $v_{2i+2}$ from $w_{5}$, we obtain the desired word \mathsf{out}\big{(}\rho[\ell_{0},\ell_{2i+3}]\big{)}~{}v_{2i+3}, whose period $p^{\prime\prime}$ divides $p_{2i+3}$. This proves the claim for the inductive step, and completes the proof of the proposition.

The $\mathrel{\text{\sf S}}^{*}$-classes considered so far cannot be directly used to define the blocks in the desired decomposition of $\rho$, since the $x$-coordinates of their endpoints might not be in the appropriate order. The next definition takes care of this, by enlarging the $\mathrel{\text{\sf S}}^{*}$-classes according to $x$-coordinates of the anchor points in the equivalence class.

{defi}

Consider a non-singleton $\mathrel{\text{\sf S}}^{*}$-equivalence class $K=[\ell,\ell^{\prime}]$. Let $\mathsf{an}(K)$ be the restriction of $K$ to the anchor points occurring in some inversion, and $X_{\mathsf{an}(K)}=\{x\>:\>\exists y\>(x,y)\in\mathsf{an}(K)\}$ be the projection of $\mathsf{an}(K)$ on positions. We define $\mathsf{block}(K)=[\tilde{\ell},\tilde{\ell}^{\prime}]$, where

• •

$\tilde{\ell}$ is the latest location $(\tilde{x},\tilde{y})\mathrel{\unlhd}\ell$ such that $\tilde{x}=\min(X_{\mathsf{an}(K)})$,

• •

$\tilde{\ell}^{\prime}$ is the earliest location $(\tilde{x},\tilde{y})\mathrel{\unrhd}\ell^{\prime}$ such that $\tilde{x}=\max(X_{\mathsf{an}(K)})$

(note that the locations $\tilde{\ell},\tilde{\ell}^{\prime}$ exist since $\ell,\ell^{\prime}$ are anchor points in some inversion).

Lemma 14**.**

If $K$ is a non-singleton $\mathrel{\text{\sf S}}^{*}$-equivalence class, then $\rho|\mathsf{block}(K)$ is a $\boldsymbol{B}$-block.

Proof 5.2**.**

Consider a non-singleton $\mathrel{\text{\sf S}}^{*}$-class $K=[\ell,\ell^{\prime}]$ and let $\mathsf{an}(K)$, $X_{\mathsf{an}(K)}$, and $\mathsf{block}(K)=[\tilde{\ell},\tilde{\ell}^{\prime}]$ be as in Definition 5. The reader can refer to Figure 5 to quickly recall the notation. We need to verify that $\rho[\tilde{\ell},\tilde{\ell}^{\prime}]$ is a $\boldsymbol{B}$-block (cf. Definition 5), namely, that:

• •

$\tilde{\ell}=(\tilde{x},\tilde{y})$, $\tilde{\ell}^{\prime}=(\tilde{x}^{\prime},\tilde{y}^{\prime})$, with $\tilde{x}\leq\tilde{x}^{\prime}$,

• •

the output produced by $\rho[\tilde{\ell},\tilde{\ell}^{\prime}]$ is almost periodic with bound $2\boldsymbol{B}$,

• •

the output produced by the subsequence $\rho|Z$, where Z=[\tilde{\ell},\tilde{\ell}^{\prime}]~{}\setminus~{}\big{(}[\tilde{x},\tilde{x}^{\prime}]\times[\tilde{y},\tilde{y}^{\prime}]\big{)}, has length at most $2{\boldsymbol{H}}\boldsymbol{B}$.

The first condition $\tilde{x}\leq\tilde{x}^{\prime}$ follows immediately from the definition of $\tilde{x}$ and $\tilde{x}^{\prime}$ as $\min(X_{\mathsf{an}(K)})$ and $\max(X_{\mathsf{an}(K)})$, respectively.

Next, we prove that the output produced by the factor $\rho[\tilde{\ell},\tilde{\ell}^{\prime}]$ is almost periodic with bound $2\boldsymbol{B}$. By Definition 5, we have $\tilde{\ell}\mathrel{\unlhd}\ell\mathrel{{\lhd}}\ell^{\prime}\mathrel{\unlhd}\tilde{\ell}^{\prime}$, and by Lemma 13 we know that $\mathsf{out}(\rho[\ell,\ell^{\prime}])$ is periodic with period at most $\boldsymbol{B}$ ($\leq 2\boldsymbol{B}$). So it suffices to show that the lengths of the words $\mathsf{out}(\rho[\tilde{\ell},\ell])$ and $\mathsf{out}(\rho[\ell^{\prime},\tilde{\ell}^{\prime}])$ are at most $2\boldsymbol{B}$. We shall focus on the former word, as the arguments for the latter are similar.

First, we note that the factor $\rho[\tilde{\ell},\ell]$ lies entirely to the right of position $\tilde{x}$, and in particular, it starts at an even level $\tilde{y}$. This follows from the definition of $\tilde{\ell}$, and whether $\ell$ itself is at an odd/even level. In particular, the location $\ell$ is either at the same level as $\tilde{\ell}$, or just one level above.

Now, suppose, by way of contradiction, that $|\mathsf{out}(\rho[\tilde{\ell},\ell])|>2\boldsymbol{B}$. We head towards a contradiction by finding a location $\ell^{\prime\prime}\mathrel{{\lhd}}\ell$ that is $\mathrel{\text{\sf S}}^{*}$-equivalent to the first location $\ell$ of the $\mathrel{\text{\sf S}}^{*}$-equivalence class $K$. Since the location $\ell$ is either at the same level as $\tilde{\ell}$, or just above it, the factor $\rho[\tilde{\ell},\ell]$ is of the form $\alpha\,\beta$, where $\alpha$ is rightward factor lying on the same level as $\tilde{\ell}$ and $\beta$ is either empty or a leftward factor on the next level. Moreover, since $|\mathsf{out}(\rho[\tilde{\ell},\ell])|>2\boldsymbol{B}$, we know that either $|\mathsf{out}(\alpha)|>\boldsymbol{B}$ or $|\mathsf{out}(\beta)|>\boldsymbol{B}$. Thus, Lemma 7 says that one of the two factors $\alpha,\beta$ is not output-minimal. In particular, there is a loop $L_{1}$, strictly to the right of $\tilde{x}$, that intercepts a subfactor $\gamma$ of $\rho[\tilde{\ell},\ell]$, with $\mathsf{out}(\gamma)$ non-empty and output-minimal.

Let $\ell^{\prime\prime}$ be the first location of the factor $\gamma$. Clearly, $\ell^{\prime\prime}$ is an anchor point of $L$ and $\mathsf{out}(\mathsf{tr}(\ell^{\prime\prime}))\neq\varepsilon$. Further recall that $\tilde{x}=\min(X_{\mathsf{an}(K)})$ is the leftmost position of locations in the class $K=[\ell,\ell^{\prime}]$ that are also anchor points of inversions. In particular, there is a loop $L_{2}$ with some anchor point $\ell^{\prime\prime}_{2}=(\tilde{x},y^{\prime\prime}_{2})\in\mathsf{an}(K)$, and such that $\mathsf{tr}(\ell^{\prime\prime})$ is non-empty and output-minimal. Since $\ell^{\prime\prime}\mathrel{{\lhd}}\ell\mathrel{\unlhd}\ell^{\prime\prime}_{2}$ and the position of $\ell^{\prime\prime}$ is to the right of the position of $\ell^{\prime\prime}_{2}$, we know that $(L_{1},\ell^{\prime\prime},L_{2},\ell^{\prime\prime}_{2})$ is also an inversion, and hence $\ell^{\prime\prime}\mathrel{\text{\sf S}}^{*}\ell^{\prime\prime}_{2}\mathrel{\text{\sf S}}^{*}\ell$. But since $\ell^{\prime\prime}\mathrel{{\lhd}}\ell$, we get a contradiction with the assumption that $\ell$ is the first location of a $\mathrel{\text{\sf S}}^{*}$-class. In this way we have shown that $|\mathsf{out}(\rho[\ell_{1},\ell])|\leq 2\boldsymbol{B}$.

It remains to show that the output produced by the subsequence $\rho|Z$, where Z=[\tilde{\ell},\tilde{\ell}^{\prime}]~{}\setminus~{}\big{(}[\tilde{x},\tilde{x}^{\prime}]\times[\tilde{y},\tilde{y}^{\prime}]\big{)}, has length at most $2{\boldsymbol{H}}\boldsymbol{B}$. For this it suffices to prove that $|\mathsf{out}(\alpha)|\leq\boldsymbol{B}$ for every factor $\alpha$ of $\rho[\tilde{\ell},\tilde{\ell}^{\prime}]$ that lies at a single level and either to the left of $\tilde{x}$ or to the right of $\tilde{x}^{\prime}$. By symmetry, we consider only one of the two types of factors. Suppose, by way of contradiction, that there is a factor $\alpha$ at level $y^{\prime\prime}$, to the left of $\tilde{x}$, and such that $|\mathsf{out}(\alpha)|>\boldsymbol{B}$. By Lemma 7 we know that $\alpha$ is not output-minimal, so there is some loop $L_{2}$ strictly to the left of $\tilde{x}$ that intercepts an output-minimal subfactor $\beta$ of $\alpha$ with non-empty output. Let $\ell^{\prime\prime}$ be the first location of $\beta$. We know that $\tilde{\ell}\mathrel{{\lhd}}\ell^{\prime\prime}\mathrel{\unlhd}\tilde{\ell}^{\prime}$. Since the level $\tilde{y}$ is even, this means that the level of $\ell^{\prime\prime}$ is strictly greater than $\tilde{y}$. Since we also know that $\ell$ is an anchor point of some inversion, we can take a suitable loop $L_{1}$ with anchor point $\ell$ and obtain that $(L_{1},\ell,L_{2},\ell^{\prime\prime})$ is an inversion, so $\ell^{\prime\prime}\mathrel{\text{\sf S}}^{*}\ell$. But this contradicts the fact that $\tilde{x}$ is the leftmost position of $\mathsf{an}(K)$. We thus conclude that $|\mathsf{out}(\alpha)|\leq\boldsymbol{B}$, and this completes the proof that $\rho|\mathsf{block}(K)$ is a $\boldsymbol{B}$-block.

The next lemma shows that blocks do not overlap along the input axis:

Lemma 15**.**

Suppose that $K_{1}$ and $K_{2}$ are two different non-singleton $\mathrel{\text{\sf S}}^{*}$-classes such that $\ell\mathrel{{\lhd}}\ell^{\prime}$ for all $\ell\in K_{1}$ and $\ell^{\prime}\in K_{2}$. Let $\mathsf{block}(K_{1})=[\ell_{1},\ell_{2}]$ and $\mathsf{block}(K_{2})=[\ell_{3},\ell_{4}]$, with $\ell_{2}=(x_{2},y_{2})$ and $\ell_{3}=(x_{3},y_{3})$. Then $x_{2}<x_{3}$.

Proof 5.3**.**

Suppose by contradiction that $K_{1}$ and $K_{2}$ are as in the statement, but $x_{2}\geq x_{3}$. By Definition 5, $x_{2}=\max(X_{\mathsf{an}(K_{1})})$ and $x_{3}=\min(X_{\mathsf{an}(K_{2})})$. This implies the existence of some inversions $(L,\ell,L^{\prime},\ell^{\prime})$ and $(L^{\prime\prime},\ell^{\prime\prime},L^{\prime\prime\prime},\ell^{\prime\prime\prime})$ such that $\ell=(x_{2},y)$ and $\ell^{\prime\prime\prime}=(x_{3},y^{\prime\prime\prime})$. Moreover, since $\ell\mathrel{\unlhd}\ell^{\prime\prime\prime}$ and $x_{2}\geq x_{3}$, we know that $(L,\ell,L^{\prime\prime\prime},\ell^{\prime\prime\prime})$ is also an inversion, thus implying that $K_{1}=K_{2}$.

For the sake of brevity, we call $\mathrel{\text{\sf S}}^{*}$-block any factor of the form $\rho|\mathsf{block}(K)$ that is obtained by applying Definition 5 to a non-singleton $\mathrel{\text{\sf S}}^{*}$-class $K$. The results obtained so far imply that every location covered by an inversion is also covered by an $\mathrel{\text{\sf S}}^{*}$-block (Lemma 14), and that the order of occurrence of $\mathrel{\text{\sf S}}^{*}$-blocks is the same as the order of positions (Lemma 15). So the $\mathrel{\text{\sf S}}^{*}$-blocks can be used as factors for the $\boldsymbol{B}$-decomposition of $\rho$ we are looking for. Below, we show that the remaining factors of $\rho$, which do not overlap the $\mathrel{\text{\sf S}}^{*}$-blocks, are $\boldsymbol{B}$-diagonals. This will complete the construction of a $\boldsymbol{B}$-decomposition of $\rho$.

Formally, we say that a factor $\rho[\ell,\ell^{\prime}]$ overlaps another factor $\rho[\ell^{\prime\prime},\ell^{\prime\prime\prime}]$ if $[\ell,\ell^{\prime}]\>\cap\>[\ell^{\prime\prime},\ell^{\prime\prime\prime}]\neq\emptyset$, $\ell^{\prime}\neq\ell^{\prime\prime}$, and $\ell\neq\ell^{\prime\prime\prime}$.

Lemma 16**.**

Let $\rho[\ell,\ell^{\prime}]$ be a factor of $\rho$, with $\ell=(x,y)$, $\ell^{\prime}=(x^{\prime},y^{\prime})$, and $x\leq x^{\prime}$, that does not overlap any $\mathrel{\text{\sf S}}^{*}$-block. Then $\rho[\ell,\ell^{\prime}]$ is a $\boldsymbol{B}$-diagonal.

Proof 5.4**.**

Consider a factor $\rho[\ell,\ell^{\prime}]$, with $\ell=(x,y)$, $\ell^{\prime}=(x^{\prime},y^{\prime})$, and $x\leq x^{\prime}$, that does not overlap any $\mathrel{\text{\sf S}}^{*}$-block. We will focus on locations $\ell^{\prime\prime}$ with $\ell\mathrel{\unlhd}\ell^{\prime\prime}\mathrel{\unlhd}\ell^{\prime}$ that are anchor points of some loop with $\mathsf{out}(\mathsf{tr}(\ell^{\prime\prime}))\neq\varepsilon$. We denote by $A$ the set of all such locations.

First, we show that the locations in $A$ are monotonic size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Monotonic$\to$Monotonous? size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. we keep monotonicw.r.t. the position order. Formally, we prove that for all $\ell_{1},\ell_{2}\in A$, if $\ell_{1}=(x_{1},y_{1})\mathrel{\unlhd}\ell_{2}=(x_{2},y_{2})$, then $x_{1}\leq x_{2}$. Suppose that this were not the case, namely, that $A$ contained two anchor points $\ell_{1}=(x_{1},y_{1})$ and $\ell_{2}=(x_{2},y_{2})$ with $\ell_{1}\mathrel{{\lhd}}\ell_{2}$ and $x_{1}>x_{2}$. Let $L_{1},L_{2}$ be the loops of $\ell_{1},\ell_{2}$, respectively, and recall that $\mathsf{out}(\mathsf{tr}(\ell_{1})),\mathsf{out}(\mathsf{tr}(\ell_{2}))\neq\varepsilon$. This means that $(L_{1},\ell_{1},L_{2},\ell_{2})$ is an inversion, and hence $\ell_{1}\mathrel{\text{\sf S}}^{*}\ell_{2}$. But this contradicts the hypothesis that $\rho[\ell,\ell^{\prime}]$ does not overlap any $\mathrel{\text{\sf S}}^{*}$-block.

Next, we identify the floors of our diagonal. Let $y_{0},y_{1},\dots,y_{n-1}$ be all the even levels that have locations in $A$. For each $i=0,\dots,n-1$, let $\ell_{2i+1}$ (resp. $\ell_{2i+2}$) be the first (resp. last) anchor point of $A$ at level $y_{i}$. Further let $\ell_{0}=\ell$ and $\ell_{2n+1}=\ell^{\prime}$. Clearly, each factor $\rho[\ell_{2i+1},\ell_{2i+2}]$ is a floor. Moreover, thanks to the previous arguments, each location $\ell_{2i}$ is to the left of the location $\ell_{2i+1}$.

It remains to prove that each factor $\rho[\ell_{2i},\ell_{2i+1}]$ produces an output of length at most $2{\boldsymbol{H}}\boldsymbol{B}$. By construction, $A$ contains no anchor point at an even level and strictly between $\ell_{2i}$ and $\ell_{2i+1}$. By Lemma 7 this means that the outputs produced by subfactors of $\rho[\ell_{2i},\ell_{2i+1}]$ that lie entirely at an even level have length at most $\boldsymbol{B}$. Let us now consider the subfactors $\alpha$ of $\rho[\ell_{2i},\ell_{2i+1}]$ that lie entirely at an odd level, and let us prove that they produce outputs of length at most $2\boldsymbol{B}$. Suppose that this is not the case, namely, that $|\mathsf{out}(\alpha)|>2\boldsymbol{B}$. In this case we show that an inversion would exist at this level. Formally, we can find two locations $\ell^{\prime\prime}\mathrel{{\lhd}}\ell^{\prime\prime\prime}$ in $\alpha$ such that the prefix of $\alpha$ that ends at location $\ell^{\prime\prime}$ and the suffix of $\alpha$ that starts at location $\ell^{\prime\prime\prime}$ produce outputs of length greater than $\boldsymbol{B}$. By Lemma 7, those two factors would not be output-minimal, and hence $\alpha$ would contain disjoint loops $L_{1},L_{2}$ with anchor points $\ell^{\prime\prime}_{1},\ell^{\prime\prime}_{2}$ forming an inversion $(L_{1},\ell^{\prime\prime}_{1},L_{2},\ell^{\prime\prime}_{2})$. But this would imply that $\ell^{\prime\prime}_{1},\ell^{\prime\prime}_{2}$ belong to the same non-singleton $\mathrel{\text{\sf S}}^{*}$-equivalence class, which contradicts the hypothesis that $\rho[\ell,\ell^{\prime}]$ does not overlap any $\mathrel{\text{\sf S}}^{*}$-block. We must conclude that the subfactors of $\rho[\ell_{2i},\ell_{2i+1}]$ produce outputs of length at most $2\boldsymbol{B}$.

*Overall, this shows that the output produced by each factor $\rho[\ell_{2i},\ell_{2i+1}]$ has length at most $2{\boldsymbol{H}}\boldsymbol{B}$. *

We have just shown how to construct a $\boldsymbol{B}$-decomposition of the run $\rho$ that satisfies P2. This proves Proposition 12, as well as the implication P2 $\Rightarrow$ P3 of Theorem 6.

From existence of decompositions to an equivalent one-way transducer.

We now focus on the last implication P3 $\Rightarrow$ P1 of Theorem 6. More precisely, we show how to construct a one-way transducer $T^{\prime}$ that simulates the outputs produced by the successful runs of $T$ that admit $\boldsymbol{B}$-decompositions. In particular, $T^{\prime}$ turns out to be equivalent to $T$ when $T$ is one-way definable. Here we will only give a proof sketch of this construction (as there is no real difference between the sweeping and two-way cases) assuming that $T$ is a sweeping transducer; a fully detailed construction of $T^{\prime}$ from an arbitrary two-way transducer $T$ will be given in Section 8 (Proposition 28), together with a procedure for deciding one-way definability of $T$ (Proposition 31).

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. The structure of Prop 8.6 is all but indistinguishable visually/ This is made all the more aggravating by the fact that there is no corresponding proof in the sweeping case (merely the intuition sketched on Page 23. size=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. added a sentence size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. answered in review1-answer.txt

Proposition 17**.**

Given a functional sweeping transducer $T$ a one-way transducer $T^{\prime}$ can be constructed in $2\textsc{ExpTime}$ such that the following hold:

1. (1)

$T^{\prime}\subseteq T$, 2. (2)

$\mathsf{dom}(T^{\prime})$* contains all words that induce successful runs of $T$ admitting $\boldsymbol{B}$-decompositions.*

In particular, $T^{\prime}$ is equivalent to $T$ iff $T$ is one-way definable.

Proof 5.5** (Proof sketch.).**

Given an input word $u$, the one-way transducer $T^{\prime}$ needs to guess a successful run $\rho$ of $T$ on $u$ that admits a $\boldsymbol{B}$-decomposition. This can be done by guessing the crossing sequences of $\rho$ at each position, together with a sequence of locations $\ell_{i}$ that identify the factors of a $\boldsymbol{B}$-decomposition of $\rho$. To check the correctness of the decomposition, $T^{\prime}$ also needs to guess a bounded amount of information (words of bounded length) to reconstruct the outputs produced by the $\boldsymbol{B}$-diagonals and the $\boldsymbol{B}$-blocks. For example, while scanning a factor of the input underlying a diagonal, $T^{\prime}$ can easily reproduce the outputs of the floors and the guessed outputs of factors between them. In a similar way, while scanning a factor of the input underlying a block, $T^{\prime}$ can simulate the almost periodic output by guessing its repeating pattern and the bounded prefix and suffix of it, and by emitting the correct amount of letters, as it is done in Example 3. In particular, one can verify that the capacity of $T^{\prime}$ is linear in ${\boldsymbol{H}}\boldsymbol{B}$. Moreover, because the guessed objects are of size linear in ${\boldsymbol{H}}\boldsymbol{B}$ and ${\boldsymbol{H}}\boldsymbol{B}$ is a simple exponential in the size of $T$, the size of the one-way transducer $T^{\prime}$ has doubly exponential size in that of $T$.

6. The structure of two-way loops

While loop pumping in a sweeping transducer is rather simple, we need a much better understanding when it comes to pump loops of unrestricted two-way transducers. This section is precisely devoted to untangling the structure of two-way loops. We will focus on specific loops, called idempotent, that generate repetitions with a “nice shape”, very similar to loops of sweeping transducers.

We fix throughout this section a functional two-way transducer $T$, an input word $u$, and a (normalized) successful run $\rho$ of $T$ on $u$. As usual, ${\boldsymbol{H}}=2|Q|-1$ is the maximal length of a crossing sequence of $\rho$, and ${\boldsymbol{C}}$ is the maximal number of letters output by a single transition.

size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. I have some questions concerning the definition of the flows and effects. Initially you fix a transducer T, a word u and a run $\rho$ of T on u, and you define the flow corresponding to an interval I of $\rho$. On line 10, you say that “we consider effects that are not necessarily associated with intervals of specific runs”. Then you define the set of all flows, and all effects. If I understand correctly, a flow has to correspond to an interval I, whereas an effect is a triple containing a flow (corresponding to an interval I) and two crossing sequences (that might not correspond to I). My first question is: wouldn’t it be more natural to have either

1. also flows that do not correspond to an interval,
2. or only effects that correspond to intervals? I realise that in case 1, defining the flows in a way for Lemma 6.5 to still hold seems complicated, but I was wondering if you had considered it. Second, I was wondering if it wouldn’t be easier to fix the set of nodes of the flows to $\{0,\dots,2|Q|-2\}$. This would allow a smoother definition of $G\cdot G^{\prime}$ (l.25): as of now, the set of nodes of $G\cdot G^{\prime}$ is not stated explicitly. If I am not mistaken, it is the set of nodes that are part of G and admit an outgoing edge, or are part of $G^{\prime}$ and admit an incoming edge (which is not that complicated, but wouldn’t be needed at all if we always have all the nodes). size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. addressed in review2-answer.txt

Flows and effects.

We start by analyzing the shape of factors of $\rho$ intercepted by an interval $I=[x_{1},x_{2}]$. We identify four types of factors $\alpha$ intercepted by $I$ depending on the first location $(x,y)$ and the last location $(x^{\prime},y^{\prime})$:

• •

$\alpha$ is an ${\mathsf{LL}}$-factor if $x=x^{\prime}=x_{1}$,

• •

$\alpha$ is an ${\mathsf{RR}}$-factor if $x=x^{\prime}=x_{2}$,

• •

$\alpha$ is an ${\mathsf{LR}}$-factor if $x=x_{1}$ and $x^{\prime}=x_{2}$,

• •

$\alpha$ is an ${\mathsf{RL}}$-factor if $x=x_{2}$ and $x^{\prime}=x_{1}$.

In Figure 2 we see that $\alpha$ is an ${\mathsf{LL}}$-factor, $\beta,\delta$ are ${\mathsf{LR}}$-factors, $\zeta$ is an ${\mathsf{RR}}$-factor, and $\gamma$ is an ${\mathsf{RL}}$-factor.

{defi}

Let $I=[x_{1},x_{2}]$ be an interval of $\rho$ and $h_{i}$ the length of the crossing sequence $\rho|x_{i}$, for both $i=1$ and $i=2$.

The flow $F_{I}$ of $I$ is the directed graph with set of nodes $\{0,\dots,\max(h_{1},h_{2})-1\}$ and set of edges consisting of all $(y,y^{\prime})$ such that there is a factor of $\rho$ intercepted by $I$ that starts at location $(x_{i},y)$ and ends at location $(x_{j},y^{\prime})$, for $i,j\in\{1,2\}$.

The effect $E_{I}$ of $I$ is the triple $(F_{I},c_{1},c_{2})$, where $c_{i}=\rho|x_{i}$ is the crossing sequence at $x_{i}$.

For example, the interval $I$ of Figure 2 has the flow graph $0\mapsto 1\mapsto 3\mapsto 4\mapsto 2\mapsto 0$.

It is easy to see that every node of a flow $F_{I}$ has at most one incoming and at most one outgoing edge. More precisely, if $y<h_{1}$ is even, then it has one outgoing edge (corresponding to an ${\mathsf{LR}}$- or ${\mathsf{LL}}$-factor intercepted by $I$), and if it is odd it has one incoming edge (corresponding to an ${\mathsf{RL}}$- or ${\mathsf{LL}}$-factor intercepted by $I$). Similarly, if $y<h_{2}$ is even, then it has one incoming edge (corresponding to an ${\mathsf{LR}}$- or ${\mathsf{RR}}$-factor), and if it is odd it has one outgoing edge (corresponding to an ${\mathsf{RL}}$- or ${\mathsf{RR}}$-factor).

In the following we consider effects that are not necessarily associated with intervals of specific runs. The definition of such effects should be clear: they are triples consisting of a graph (called flow) and two crossing sequences of lengths $h_{1},h_{2}\leq{\boldsymbol{H}}$, with sets of nodes of the form $\{0,\ldots,\max(h_{1},h_{2})-1\}$, that satisfy the in/out-degree properties stated above. It is convenient to distinguish the edges in a flow based on the parity of the source and target nodes. Formally, we partition any flow $F$ into the following subgraphs:

• •

$F_{\mathsf{LR}}$ consists of all edges of $F$ between pairs of even nodes,

• •

$F_{\mathsf{RL}}$ consists of all edges of $F$ between pairs of odd nodes,

• •

$F_{\mathsf{LL}}$ consists of all edges of $F$ from an even node to an odd node,

• •

$F_{\mathsf{RR}}$ consists of all edges of $F$ from an odd node to an even node.

We denote by $\mathcal{F}$ (resp. $\mathcal{E}$) the set of all flows (resp. effects) augmented with a dummy element $\bot$. We equip both sets $\mathcal{F}$ and $\mathcal{E}$ with a semigroup structure, where the corresponding products $\circ$ and $\odot$ are defined below (similar definitions appear in [9]). Later we will use the semigroup structure to identify the idempotent loops, that play a crucial role in our characterization of one-way definability.

{defi}

For two graphs $G,G^{\prime}$, we denote by $G\cdot G^{\prime}$ the graph with edges of the form $(y,y^{\prime\prime})$ such that $(y,y^{\prime})$ is an edge of $G$ and $(y^{\prime},y^{\prime\prime})$ is an edge of $G^{\prime}$, for some node $y^{\prime}$ that belongs to both $G$ and $G^{\prime}$. Similarly, we denote by $G^{*}$ the graph with edges $(y,y^{\prime})$ such that there exists a (possibly empty) path in $G$ from $y$ to $y^{\prime}$.

The product of two flows $F,F^{\prime}$ is the unique flow $F\circ F^{\prime}$ (if it exists) such that:

• •

$(F\circ F^{\prime})_{\mathsf{LR}}=F_{\mathsf{LR}}\cdot(F^{\prime}_{\mathsf{LL}}\cdot F_{\mathsf{RR}})^{*}\cdot F^{\prime}_{\mathsf{LR}}$,

• •

$(F\circ F^{\prime})_{\mathsf{RL}}=F^{\prime}_{\mathsf{RL}}\cdot(F_{\mathsf{RR}}\cdot F^{\prime}_{\mathsf{LL}})^{*}\cdot F_{\mathsf{RL}}$,

• •

$(F\circ F^{\prime})_{\mathsf{LL}}=F_{\mathsf{LL}}~{}\cup~{}F_{\mathsf{LR}}\cdot(F^{\prime}_{\mathsf{LL}}\cdot F_{\mathsf{RR}})^{*}\cdot F^{\prime}_{\mathsf{LL}}\cdot F_{\mathsf{RL}}$,

• •

$(F\circ F^{\prime})_{\mathsf{RR}}=F^{\prime}_{\mathsf{RR}}~{}\cup~{}F^{\prime}_{\mathsf{RL}}\cdot(F_{\mathsf{RR}}\cdot F^{\prime}_{\mathsf{LL}})^{*}\cdot F_{\mathsf{RR}}\cdot F^{\prime}_{\mathsf{LR}}$.

If no flow $F\circ F^{\prime}$ exists with the above properties, then we let $F\circ F^{\prime}=\bot$.

The product of two effects $E=(F,c_{1},c_{2})$ and $E^{\prime}=(F^{\prime},c^{\prime}_{1},c^{\prime}_{2})$ is either the effect $E\odot E^{\prime}=(F\circ F^{\prime},c_{1},c^{\prime}_{2})$ or the dummy element $\bot$, depending on whether $F\circ F^{\prime}\neq\bot$ and $c_{2}=c^{\prime}_{1}$.

For example, let $F$ be the flow of interval $I$ in Figure 2. Then $(F\circ F)_{\mathsf{LL}}=\{0\mapsto 1,2\mapsto 3\}$, $(F\circ F)_{\mathsf{RR}}=\{1\mapsto 2,3\mapsto 4\}$, and $(F\circ F)_{\mathsf{LR}}=\{4\mapsto 0\}$ — one can quickly verify this with the help of Figure 10.

It is also easy to see that $(\mathcal{F},\circ)$ and $(\mathcal{E},\odot)$ are finite semigroups, and that for every run $\rho$ and every pair of consecutive intervals $I=[x_{1},x_{2}]$ and $J=[x_{2},x_{3}]$ of $\rho$, $F_{I\cup J}=F_{I}\circ F_{J}$ and $E_{I\cup J}=E_{I}\odot E_{J}$. In particular, the function $E$ that associates each interval $I$ of $\rho$ with the corresponding effect $E_{I}$ can be seen as a semigroup homomorphism.

Loops and components.

Recall that a loop is an interval $L=[x_{1},x_{2}]$ with the same crossing sequences at $x_{1}$ and $x_{2}$. We will follow techniques similar to those presented in Section 4 to show that the outputs generated in non left-to-right manner are essentially periodic. However, differently from the sweeping case, we will consider only special types of loops:

{defi}

A loop $L$ is idempotent if $E_{L}=E_{L}\odot E_{L}$ and $E_{L}\neq\bot$.

For example, the interval $I$ of Figure 2 is a loop, if one assumes that the crossing sequences at the borders of $I$ are the same. By comparing with Figure 10, it is easy to see that $I$ is not idempotent. On the other hand, the loop consisting of 2 copies of $I$ is idempotent.

As usual, given a loop $L=[x_{1},x_{2}]$ and a number $n\in\mathbb{N}$, we can introduce $n$ new copies of $L$ and connect the intercepted factors in the obvious way. This results in a new run $\mathsf{pump}_{L}^{n+1}(\rho)$ on the word $\mathsf{pump}_{L}^{n+1}(u)$. Figure 10 shows how to do this for $n=1$ and $n=2$. Below, we analyze in detail the shape of the pumped run $\mathsf{pump}_{L}^{n+1}(\rho)$ (and the produced output as well) when $L$ is an idempotent loop. We will focus on idempotent loops because pumping non-idempotent loops may induce permutations of factors that are difficult to handle. For example, if we consider again the non-idempotent loop $I$ to the left of Figure 10, the factor of the run between $\beta$ and $\gamma$ (to the right of $I$, highlighted in red) precedes the factor between $\gamma$ and $\delta$ (to the left of $I$, again in red), but this ordering is reversed when a new copy of $I$ is added.

When pumping a loop $L$, subsets of factors intercepted by $L$ are glued together to form factors intercepted by the replication of $L$. The notion of component introduced below identifies groups of factors that are glued together.

{defi}

A component of a loop $L$ is any strongly connected component of its flow $F_{L}$ (note that this is also a cycle, since every node in it has in/out-degree $1$).

Given a component $C$, we denote by $\min(C)$ (resp. $\max(C)$) the minimum (resp. maximum) node in $C$. We say that $C$ is left-to-right (resp. right-to-left) if $\min(C)$ is even (resp., odd).

An $(L,C)$-factor is a factor of the run that is intercepted by $L$ and that corresponds to an edge of $C$.

We will usually list the $(L,C)$-factors based on their order of occurrence in the run. For example, the loop $I$ of Figure 10 contains a single component $C=0\mapsto 1\mapsto 3\mapsto 4\mapsto 2\mapsto 0$ which is left-to-right. Another example is given in Figure 11, where the loop $L$ has three components $C_{1},C_{2},C_{3}$ (colored in blue, red, and green, respectively): $\alpha_{1},\alpha_{2},\alpha_{3}$ are the $(L,C_{1})$-factors, $\beta_{1},\beta_{2},\beta_{3}$ are the $(L,C_{2})$-factors, and $\gamma_{1}$ is the unique $(L,C_{3})$-factor.

Below, we show that the levels of each component of a loop (not necessarily idempotent) form an interval.

Lemma 18**.**

*Let $C$ be a component of a loop $L=[x_{1},x_{2}]$. The nodes of $C$ are precisely the levels in the interval $[\min(C),\max(C)]$. Moreover, if $C$ is left-to-right (resp. right-to-left), then $\max(C)$ is the smallest level $\geq\min(C)$ such that between $(x_{1},\min(C))$ and $(x_{2},\max(C))$ (resp. $(x_{2},\min(C))$ and $(x_{1},\max(C))$) there are equally many ${\mathsf{LL}}$-factors and ${\mathsf{RR}}$-factors intercepted by $L$. *

Proof 6.1** (Proof idea).**

The proof of this lemma is rather technical and deferred to Appendix A, since the lemma is not at the core of the proof of the main result. Let us first note that with the definition of $\max(C)$ stated in the lemma it is rather easy to see that the interval $[\min(C),\max(C)]$ is a union of cycles (i.e., components). This can be shown by arguing that every node in $[\min(C),\max(C)]$ has in-degree and out-degree one. What is much less obvious is that $[\min(C),\max(C)]$ is connected, thus consists of a single cycle.

The crux is thus to show that the nodes visited by every cycle of the flow (or, equally, every component) form an interval. For this, we use an induction based on portions of the cycle, namely, on paths of the flow. The difficulty underlying the formalization of the inductive invariant comes from the fact that, differently from cycles, paths of a flow may visit sets of levels that do not form intervals. An example is given in Figure 12, which represents some edges of a flow forming a path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$ and covering a non-convex set of nodes: note that there could be a large gap between the nodes $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ and $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$ due to the unbalanced numbers of ${\mathsf{LL}}$-factors and ${\mathsf{RR}}$-factors below $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$.

Essentially, the first part of the proof of the lemma amounts at identifying the sources $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ (resp. $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$) of the ${\mathsf{LR}}$-factors (resp. ${\mathsf{RL}}$-factors), and at showing that the latter factors are precisely of the form $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$ (resp. $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$). Once these nodes are identified, we show by induction on $i$ that every two consecutive nodes $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ and $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$ must be connected by a path whose intermediate nodes form the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}]$. Finally, we argue that every cycle (or component) $C$ visits all and only the nodes in the interval $[\min(C),\max(C)]$.

The next lemma describes the precise shape and order of the intercepted factors when the loop $L$ is idempotent.

Lemma 19**.**

If $C$ is a left-to-right (resp. right-to-left) component of an idempotent loop $L$, then the $(L,C)$-factors are in the following order: $k$ ${\mathsf{LL}}$-factors (resp. ${\mathsf{RR}}$-factors), followed by one ${\mathsf{LR}}$-factor (resp. ${\mathsf{RL}}$-factor), followed by $k$ ${\mathsf{RR}}$-factors (resp. ${\mathsf{LL}}$-factors), for some $k\geq 0$.

Proof 6.2**.**

Suppose that $C$ is a left-to-right component of $L$. We show by way of contradiction that $C$ has only one ${\mathsf{LR}}$-factor and no ${\mathsf{RL}}$-factor. By Lemma 18 this will yield the claimed shape. Figure 13 can be used as a reference example for the arguments that follow.

We begin by listing the $(L,C)$-factors. As usual, we order them based on their occurrences in the run $\rho$. Let $\gamma$ be the first $(L,C)$-factor that is not an ${\mathsf{LL}}$-factor, and let $\beta_{1},\dots,\beta_{k}$ be the $(L,C)$-factors that precede $\gamma$ (these are all ${\mathsf{LL}}$-factors). Because $\gamma$ starts at an even level, it must be an ${\mathsf{LR}}$-factor. Suppose that there is another $(L,C)$-factor, say $\zeta$, that comes after $\gamma$ and it is neither an ${\mathsf{RR}}$-factor nor an ${\mathsf{LL}}$-factor. Because $\zeta$ starts at an odd level, it must be an ${\mathsf{RL}}$-factor. Further let $\delta_{1},\dots,\delta_{k^{\prime}}$ be the intercepted ${\mathsf{RR}}$-factors that occur between $\gamma$ and $\zeta$. We claim that $k^{\prime}<k$, namely, that the number of ${\mathsf{RR}}$-factors between $\gamma$ and $\zeta$ is strictly less than the number of ${\mathsf{LL}}$-factors before $\gamma$. Indeed, if this were not the case, then, by Lemma 18, the level where $\zeta$ starts would not belong to the component $C$.

Now, consider the pumped run $\rho^{\prime}=\mathsf{pump}_{L}^{2}(\rho)$, obtained by adding a new copy of $L$. Let $L^{\prime}$ be the loop of $\rho^{\prime}$ obtained from the union of $L$ and its copy. Since $L$ is idempotent, the components of $L$ are isomorphic to the components of $L^{\prime}$. In particular, we can denote by $C^{\prime}$ the component of $L^{\prime}$ that is isomorphic to $C$. Let us consider the $(L^{\prime},C^{\prime})$-factors of $\rho^{\prime}$. The first $k$ factors are isomorphic to the $k$ ${\mathsf{LL}}$-factors $\beta_{1},\ldots,\beta_{k}$ from $\rho$. However, the $(k+1)$-th element has a different shape: it is isomorphic to $\gamma~{}\beta_{1}~{}\delta_{1}~{}\beta_{2}~{}\cdots~{}\delta_{k^{\prime}}~{}\beta_{k^{\prime}+1}~{}\zeta$, and in particular it is an ${\mathsf{LL}}$-factor. This implies that the $(k+1)$-th edge of $C^{\prime}$ is of the form $(y,y+1)$, while the $(k+1)$-th edge of $C$ is of the form $(y,y-2k)$. This contradiction comes from having assumed the existence of the ${\mathsf{RL}}$-factor $\zeta$, and is illustrated in Figure 13.

Remark 20**.**

Note that every loop in the sweeping case is idempotent. Moreover, the $(L,C)$-factors are precisely the factors intercepted by the loop $L$.

Pumping idempotent loops.

To describe in a formal way the run obtained by pumping an idempotent loop, we need to generalize the notion of anchor point in the two-way case (the reader may compare this with the analogous definitions in Section 4 for the sweeping case). Intuitively, the anchor point of a component $C$ of an idempotent loop $L$ is the source location of the unique ${\mathsf{LR}}$- or ${\mathsf{RL}}$-factor intercepted by $L$ that corresponds to an edge of $C$ (recall Lemma 19):

{defi}

Let $C$ be a component of an idempotent loop $L=[x_{1},x_{2}]$. The anchor point of $C$ inside $L$, denoted333In denoting the anchor point — and similarly the trace — of a component $C$ inside a loop $L$, we omit the annotation specifying $L$, since this is often understood from the context. $\mathsf{an}(C)$, is either the location $(x_{1},\max(C))$ or the location $(x_{2},\max(C))$, depending on whether $C$ is left-to-right or right-to-left.

We will usually depict anchor points by black circles (like, for instance, in Figure 11).

It is also convenient to redefine the notation $\mathsf{tr}(\ell)$ for representing an appropriate sequence of transitions associated with each anchor point $\ell$ of an idempotent loop:

{defi}

Let $C$ be a component of some idempotent loop $L$, let $\ell=\mathsf{an}(C)$ be the anchor point of $C$ inside $L$, and let $i_{0}\mapsto i_{1}\mapsto i_{2}\mapsto\dots\mapsto i_{k}\mapsto i_{k+1}$ be a cycle of $C$, where $i_{0}=i_{k+1}=\max(C)$. For every $j=0,\dots,k$, further let $\beta_{j}$ be the factor intercepted by $L$ that corresponds to the edge $i_{j}\mapsto i_{j+1}$ of $C$.

The trace of $\ell$ inside $L$ is the run $\mathsf{tr}(\ell)=\beta_{0}~{}\beta_{1}~{}\cdots~{}\beta_{k}$.

Note that $\mathsf{tr}(\ell)$ is not necessarily a factor of the original run $\rho$. However, $\mathsf{tr}(\ell)$ is indeed a run, since $L$ is a loop and the factors $\beta_{i}$ are concatenated according to the flow. As we will see below, $\mathsf{tr}(\ell)$ will appear as (iterated) factor of the pumped version of $\rho$, where the loop $L$ is iterated.

As an example, by referring again to the components $C_{1},C_{2},C_{3}$ of Figure 11, we have the following traces: $\mathsf{tr}(\mathsf{an}(C_{1}))=\alpha_{2}\>\alpha_{1}\>\alpha_{3}$, $\mathsf{tr}(\mathsf{an}(C_{2}))=\beta_{2}\>\beta_{1}\>\beta_{3}$, and $\mathsf{tr}(\mathsf{an}(C_{3}))=\gamma_{1}$.

The next proposition shows the effect of pumping idempotent loops. The reader can note the similarity with the sweeping case.

Proposition 21**.**

Let $L$ be an idempotent loop of $\rho$ with components $C_{1},\dots,C_{k}$, listed according to the order of their anchor points: $\ell_{1}=\mathsf{an}(C_{1})\mathrel{{\lhd}}\cdots\mathrel{{\lhd}}\ell_{k}=\mathsf{an}(C_{k})$. For all $n\in\mathbb{N}$, we have

[TABLE]

where

• •

$\rho_{0}$* is the prefix of $\rho$ that ends at the first anchor point $\ell_{1}$,*

• •

$\rho_{k}$* is the suffix of $\rho$ that starts at the last anchor point $\ell_{k}$,*

• •

$\rho_{i}$* is the factor $\rho[\ell_{i},\ell_{i+1}]$, for all $1\leq i<k$.*

Proof 6.3**.**

Along the proof we sometimes refer to Figure 11 to ease the intuition of some definitions and arguments. For example, in the left hand-side of Figure 11, the run $\rho_{0}$ goes until the first location marked by a black circle; the runs $\rho_{1}$ and $\rho_{2}$, resp., are between the first and the second black dot, and the second and third black dot; finally, $\rho_{3}$ is the suffix starting at the last black dot. The pumped run $\mathsf{pump}_{L}^{n+1}(\rho)$ for $n=2$ is depicted to the right of the figure.

Let $L=[x_{1},x_{2}]$ be an idempotent loop and, for all $i=0,\dots,n$, let $L^{\prime}_{i}=[x^{\prime}_{i},x^{\prime}_{i+1}]$ be the $i$-th copy of the loop $L$ in the pumped run $\rho^{\prime}=\mathsf{pump}_{L}^{n+1}(\rho)$, where $x^{\prime}_{i}=x_{1}+i\cdot(x_{2}-x_{1})$ (the “[math]-th copy of $L$” is the loop $L$ itself). Further let $L^{\prime}=L^{\prime}_{0}\cup\dots\cup L^{\prime}_{n}=[x^{\prime}_{0},x^{\prime}_{n+1}]$, that is, $L^{\prime}$ is the loop of $\rho^{\prime}$ that spans across the $n+1$ occurrences of $L$. As $L$ is idempotent, the loops $L^{\prime}_{0},\dots,L^{\prime}_{n}$ and $L^{\prime}$ have all the same effect as $L$. In particular, the components of $L^{\prime}_{0},\dots,L^{\prime}_{n}$, and $L^{\prime}$ are isomorphic to and in same order as those of $L$. We denote these components by $C_{1},\dots,C_{k}$.

We let $\ell_{j}=\mathsf{an}(C_{j})$ be the anchor point of each component $C_{j}$ inside the loop $L$ of $\rho$ (these locations are marked by black dots in the left hand-side of Figure 11). Similarly, we let $\ell^{\prime}_{i,j}$ (resp. $\ell^{\prime}_{j}$) be the anchor point of $C_{j}$ inside the loop $L^{\prime}_{i}$ (resp. $L^{\prime}$). From Definition 6, we have that either $\ell^{\prime}_{j}=\ell^{\prime}_{1,j}$ or $\ell^{\prime}_{j}=\ell^{\prime}_{n,j}$, depending on whether $C_{j}$ is left-to-right or right-to-left (or, equally, on whether $j$ is odd or even).

Now, let us consider the factorization of the pumped run $\rho^{\prime}$ induced by the locations $\ell^{\prime}_{i,j}$, for all $i=0,\dots,n$ and for $j=1,\dots,k$ (these locations are marked by black dots in the right hand-side of the figure). By construction, the prefix of $\rho^{\prime}$ that ends at location $\ell^{\prime}_{0,1}$ coincides with the prefix of $\rho$ that ends at $\ell_{1}$, i.e. $\rho_{0}$ in the statement of the proposition. Similarly, the suffix of $\rho^{\prime}$ that starts at location $\ell^{\prime}_{n,k}$ is isomorphic to the suffix of $\rho$ that starts at $\ell_{k}$, i.e. $\rho_{k}$ in the statement. Moreover, for all odd (resp. even) indices $j$, the factor $\rho^{\prime}[\ell^{\prime}_{n,j},\ell^{\prime}_{n,j+1}]$ (resp. $\rho^{\prime}[\ell_{0,j},\ell_{0,j+1}]$) is isomorphic to $\rho[\ell_{j},\ell_{j+1}]$, i.e. the $\rho_{j}$ of the statement.

The remaining factors of $\rho^{\prime}$ are those delimited by the pairs of locations $\ell^{\prime}_{i,j}$ and $\ell^{\prime}_{i+1,j}$, for all $i=0,\dots,n-1$ and all $j=1,\dots,k$. Consider one such factor $\rho^{\prime}[\ell^{\prime}_{i,j},\ell^{\prime}_{i+1,j}]$, and assume that the index $j$ is odd (the case of an even $j$ is similar). This factor can be seen as a concatenation of factors intercepted by $L$ that correspond to edges of $C_{j}$ inside $L^{\prime}_{i}$. More precisely, $\rho^{\prime}[\ell^{\prime}_{i,j},\ell^{\prime}_{i+1,j}]$ is obtained by concatenating the unique ${\mathsf{LR}}$-factor of $C_{j}$ — recall that by Lemma 19 there is exactly one such factor — with an interleaving of the ${\mathsf{LL}}$-factors and the ${\mathsf{RR}}$-factors of $C_{j}$. As the components are the same for all $L^{\prime}_{i}$’s and for $L$, this corresponds precisely to the trace $\mathsf{tr}(\ell_{j})$ (cf. Definition 6). Now that we know that $\rho^{\prime}[\ell^{\prime}_{i,j},\ell^{\prime}_{i+1,j}]$ is isomorphic to $\mathsf{tr}(\ell_{j})$, we can conclude that $\rho^{\prime}[\ell^{\prime}_{0,j},\ell^{\prime}_{n,j}]\>=\>\rho^{\prime}[\ell^{\prime}_{0,j},\ell^{\prime}_{1,j}]~{}\dots~{}\rho^{\prime}[\ell^{\prime}_{n-1,j},\ell^{\prime}_{n,j}]$ is isomorphic to $\mathsf{tr}(\ell_{j})^{n}$.

7. Combinatorics in the two-way case

In this section we develop the main combinatorial techniques required in the general case. In particular, we will show how to derive the existence of idempotent loops with bounded outputs using Ramsey-based arguments, and we will use this to derive periodicity properties for the outputs produced between inversions.

As usual, $\rho$ is a fixed successful run of $T$ on some input word $u$.

Ramsey-type arguments.

We start with a technique used for bounding the lengths of the outputs of certain factors, or subsequences of a two-way run. This technique is a Ramsey-type argument, more precisely it relies on Simon’s “factorization forest” theorem [31, 13], which is recalled below. The classical version of Ramsey theorem would yield a similar result, but without the tight bounds that we get here.

Let $X$ be a set of positions of $\rho$. A factorization forest for $X$ is an unranked tree, where the nodes are intervals $I$ with endpoints in $X$ and labeled with the corresponding effect $E_{I}$, the ancestor relation is given by the containment order on intervals, the leaves are the minimal intervals $[x_{1},x_{2}]$, with $x_{2}$ successor of $x_{1}$ in $X$, and for every internal node $I$ with children $J_{1},\dots,J_{k}$, we have:

• •

$I=J_{1}\cup\dots\cup J_{k}$,

• •

$E_{I}=E_{J_{1}}\odot\dots\odot E_{J_{k}}$,

• •

if $k>2$, then $E_{I}=E_{J_{1}}=\dots=E_{J_{k}}$ is an idempotent of the semigroup $(\mathcal{E},\odot)$.

Recall that in a normalized run there are at most $|Q|^{{\boldsymbol{H}}}$ distinct crossing sequences. Moreover, a flow contains at most ${\boldsymbol{H}}$ edges, and each edge has one of the 4 possible types ${\mathsf{LL}},{\mathsf{LR}},{\mathsf{RL}},{\mathsf{RR}}$, so they are at most $4^{{\boldsymbol{H}}}$ different flows. Hence, the effect semigroup $(\mathcal{E},\odot)$ has size at most ${\boldsymbol{E}}=4^{{\boldsymbol{H}}}\cdot|Q|^{2{\boldsymbol{H}}}\cdot|Q|^{2{\boldsymbol{H}}}=(2|Q|)^{2{\boldsymbol{H}}}$. size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. the way you bound the number of flows could be explained more precisely. Here you just state that there is at most H edges, and that each edge has one of the possible 4 types. Then, you bound the number of flows by $4^{H}$, which seems to correspond to any choice of types for the H edges. However, as you state, there is at most H edges, but there could be less.size=,color=red!30,size=,color=red!30,todo: size=,color=red!30,F. doneFurther recall that ${\boldsymbol{C}}$ is the maximum number of letters output by a single transition of $T$. Like we did in the sweeping case, we define the constant ${\boldsymbol{B={\boldsymbol{C}}\cdot{\boldsymbol{H}}\cdot(2^{3{\boldsymbol{E}}}+4)+4{\boldsymbol{C}}}}$ that will be used to bound the lengths of some outputs of $T$. Note that now $\boldsymbol{B}$ is doubly exponential with respect to the size of $T$, due to the size of the effect semigroup. size=,color=red!30,size=,color=red!30,todo: size=,color=red!30,F. added this explanation, it could be misleading otherwise

Theorem 22** (Factorization forest theorem [13, 31]).**

For every set $X$ of positions of $\rho$, there is a factorization forest for $X$ of height at most $3{\boldsymbol{E}}$.

The above theorem can be used to show that if $\rho$ produces an output longer than $\boldsymbol{B}$, then it contains an idempotent loop and a trace with non-empty output. Below, we present a result in the same spirit, but refined in a way that it can be used to find anchor points inside specific intervals. To formally state the result, we consider subsequences of $\rho$ induced by sets of locations that are not necessarily contiguous. Recall the notation $\rho|Z$ introduced on page 5: $\rho|Z$ is the subsequence of $\rho$ induced by the location set $Z$. For example, Figure 14 depicts a set $Z=[\ell_{1},\ell_{2}]\cap(I\times\mathbb{N})$ by a hatched area, together with the induced subrun $\rho|Z$, represented by thick arrows.

Theorem 23**.**

Let $I=[x_{1},x_{2}]$ be an interval of positions, $K=[\ell_{1},\ell_{2}]$ an interval of locations, and $Z=K\>\cap\>(I\times\mathbb{N})$. If $|\mathsf{out}(\rho|Z)|>\boldsymbol{B}$, then there exists an idempotent loop $L$ and an anchor point $\ell$ of $L$ such that

1. (1)

$x_{1}<\min(L)<\max(L)<x_{2}$* (in particular, $L\subsetneq I$),* 2. (2)

$\ell_{1}\mathrel{{\lhd}}\ell\mathrel{{\lhd}}\ell_{2}$* (in particular, $\ell\in K$),* 3. (3)

$\mathsf{out}(\mathsf{tr}(\ell))\neq\varepsilon$.

Proof 7.1**.**

Let $I$, $K$, $Z$ be as in the statement, and suppose that \big{|}\mathsf{out}(\rho|Z)\big{|}>\boldsymbol{B}. We define $Z^{\prime}=Z~{}\setminus~{}(\{\ell_{1},\ell_{2}\}\cup\{x_{1},x_{2}\}\times\mathbb{N})$ and we observe that there are at most $2{\boldsymbol{H}}+2$ locations that are missing from $Z^{\prime}$. This means that $\rho|Z^{\prime}$ contains all but $4{\boldsymbol{H}}+4$ transitions of $\rho|Z$, and because each transition outputs at most ${\boldsymbol{C}}$ letters, we have $|\mathsf{out}(\rho|Z^{\prime})|>\boldsymbol{B}-4{\boldsymbol{C}}\cdot{\boldsymbol{H}}-4{\boldsymbol{C}}={\boldsymbol{C}}\cdot{\boldsymbol{H}}\cdot 2^{3{\boldsymbol{E}}}$.

For every level $y$, let $X_{y}$ be the set of positions $x$ such that $(x,y)$ is the source location of some transition of $\rho|Z^{\prime}$ that produces non-empty output. For example, if we refer to Figure 15, the vertical dashed lines represent the positions of $X_{y}$ for a particular level $y$; accordingly, the circles in the figure represent the locations of the form $(x,y)$, for all $x\in X_{y}$. Since each transition outputs at most ${\boldsymbol{C}}$ letters, we have $\sum_{y}|X_{y}|>{\boldsymbol{H}}\cdot 2^{3{\boldsymbol{E}}}$. Moreover, since there are at most ${\boldsymbol{H}}$ levels, there is a level $y$ (which we fix hereafter) such that $|X_{y}|>2^{3{\boldsymbol{E}}}$.

We now prove the following:

{clm}

There are two consecutive loops $L_{1}=[x,x^{\prime}]$ and $L_{2}=[x^{\prime},x^{\prime\prime}]$ with endpoints $x,x^{\prime},x^{\prime\prime}\in X_{y}$ and such that $E_{L_{1}}=E_{L_{2}}=E_{L_{1}\cup L_{2}}$.

Proof 7.2**.**

By Theorem 22, there is a factorization forest for $X_{y}$ of height at most $3{\boldsymbol{E}}$. Since $\rho$ is a valid run, the dummy element $\bot$ of the effect semigroup does not appear in this factorization forest. Moreover, since $|X_{y}|>2^{3{\boldsymbol{E}}}$, we know that the factorization forest contains an internal node $L^{\prime}=[x^{\prime}_{1},x^{\prime}_{k+1}]$ with $k>2$ children, say $L_{1}=[x^{\prime}_{1},x^{\prime}_{2}],\dots,L_{k}=[x^{\prime}_{k},x^{\prime}_{k+1}]$. By definition of factorization forest, the effects $E_{L^{\prime}}$, $E_{L_{1}}$, …, $E_{L_{k}}$ are all equal and idempotent. In particular, the effect $E_{L^{\prime}}=E_{L_{1}}=\dots=E_{L_{k}}$ is a triple of the form $(F_{L^{\prime}},c_{1},c_{2})$, where $c_{i}=\rho|x_{i}$ is the crossing sequence at $x^{\prime}_{i}$. Finally, since $E_{L^{\prime}}$ is idempotent, we have that $c_{1}=c_{2}$ and this is equal to the crossing sequences of $\rho$ at the positions $x^{\prime}_{1},\dots,x^{\prime}_{k+1}$. This shows that $L_{1},L_{2}$ are idempotent loops.

Turning back to the proof of the theorem, we know from the above claim that there are two consecutive idempotent loops $L_{1}=[x,x^{\prime}]$ and $L_{2}=[x^{\prime},x^{\prime\prime}]$ with the same effect and with endpoints $x,x^{\prime},x^{\prime\prime}\in X_{y}\subseteq I\>\setminus\>\{x_{1},x_{2}\}$ (see again Figure 15).

Let $\tilde{\ell}_{1}=(x,y)$ and $\tilde{\ell}_{2}=(x^{\prime\prime},y)$, and observe that $\tilde{\ell}_{1},\tilde{\ell}_{2}\in Z^{\prime}$. In particular, $\tilde{\ell}_{1}$ and $\tilde{\ell}_{2}$ are strictly between $\ell_{1}$ and $\ell_{2}$. Suppose by symmetry that $\tilde{\ell}_{1}\mathrel{\unlhd}\tilde{\ell}_{2}$. Further let $C$ be the component of $L_{1}\cup L_{2}$ (or, equally, of $L_{1}$ or $L_{2}$) that contains the node $y$. Below, we focus on the factors of $\rho[\tilde{\ell}_{1},\tilde{\ell}_{2}]$ that are intercepted by $L_{1}\cup L_{2}$: these are represented in Figure 15 by the thick arrows. By Lemma 19 all these factors correspond to edges of the same component $C$, namely, they are $(L_{1}\cup L_{2},C)$-factors.

Let us fix an arbitrary factor $\alpha$ of $\rho[\tilde{\ell}_{1},\tilde{\ell}_{2}]$ that is intercepted by $L_{1}\cup L_{2}$, and assume that $\alpha=\beta_{1}\cdots\beta_{k}$, where $\beta_{1},\dots,\beta_{k}$ are the factors intercepted by either $L_{1}$ or $L_{2}$.

{clm}

If $\beta,\beta^{\prime}$ are two factors intercepted by $L_{1}=[x,x^{\prime}]$ and $L_{2}=[x^{\prime},x^{\prime\prime}]$, with $E_{L_{1}}=E_{L_{2}}=E_{L_{1}\cup L_{2}}$, and $\beta,\beta^{\prime}$ are adjacent in the run $\rho$ (namely, they share an endpoint at position $x^{\prime}$), then $\beta,\beta^{\prime}$ correspond to edges in the same component of $L_{1}$ (or, equally, $L_{2}$).

Proof 7.3**.**

*Let $C$ be the component of $L_{1}$ and $y_{1}\mapsto y_{2}$ the edge of $C$ that corresponds to the factor $\beta$ intercepted by $L_{1}$. Similarly, let $C^{\prime}$ be the component of $L_{2}$ and $y_{3}\mapsto y_{4}$ the edge of $C^{\prime}$ that corresponds to the factor $\beta^{\prime}$ intercepted by $L_{2}$. Since $\beta$ and $\beta^{\prime}$ share an endpoint at position $x^{\prime}$, we know that $y_{2}=y_{3}$. This shows that $C\cap C^{\prime}\neq\emptyset$, and hence $C=C^{\prime}$. *

The above claim shows that any two adjacent factors $\beta_{i},\beta_{i+1}$ correspond to edges in the same component of $L_{1}$ and $L_{2}$, respectively. Thus, by transitivity, all factors $\beta_{1},\dots,\beta_{k}$ correspond to edges in the same component, say $C^{\prime}$. We claim that $C^{\prime}=C$. Indeed, if $\beta_{1}$ is intercepted by $L_{1}$, then $C^{\prime}=C$ because $\alpha$ and $\beta_{1}$ start from the same location and hence they correspond to edges of the flow that depart from the same node. The other case is where $\beta_{1}$ is intercepted by $L_{2}$, for which a symmetric argument can be applied.

So far we have shown that every factor of $\rho[\tilde{\ell}_{1},\tilde{\ell}_{2}]$ intercepted by $L_{1}\cup L_{2}$ can be factorized into some $(L_{1},C)$-factors and some $(L_{2},C)$-factors. We conclude the proof with the following observations:

• •

By construction, both loops $L_{1},L_{2}$ are contained in the interval of positions $I=[x_{1},x_{2}]$, and have endpoints different from $x_{1},x_{2}$.

• •

Both anchor points of $C$ inside $L_{1},L_{2}$ belong to the interval of locations $K\>\setminus\>\{\ell_{1},\ell_{2}\}$. This holds because $\rho[\tilde{\ell}_{1},\tilde{\ell}_{2}]$ contains a factor $\alpha$ that is intercepted by $L_{1}\cup L_{2}$ and spans across all the positions from $x$ to $x^{\prime\prime}$, namely, an ${\mathsf{LR}}$-factor. This factor starts at the anchor point of $C$ inside $L_{1}$ and visits the anchor point of $C$ inside $L_{2}$. Moreover, by construction, $\alpha$ is also a factor of the subsequence $\rho|Z^{\prime}$. This shows that the anchor points of $C$ inside $L_{1}$ and $L_{2}$ belong to $Z^{\prime}$, and in particular to $K\>\setminus\>\{\ell_{1},\ell_{2}\}$.

• •

The first factor of $\rho[\tilde{\ell}_{1},\tilde{\ell}_{2}]$ that is intercepted by $L_{1}\cup L_{2}$ starts at $\tilde{\ell}_{1}=(x,y)$, which by construction is the source location of some transition producing non-empty output. By the previous arguments, this factor is a concatenation of $(L_{1},C)$-factors and $(L_{2},C)$-factors. This implies that the trace of the anchor point of $C$ inside $L_{1}$, or the trace of $C$ inside $L_{2}$ produces non-empty output.

Inversions and periodicity.

The first important notion that is used to characterize one-way definability is that of inversion. It turns out that the definition of inversion in the sweeping case (see page 4) can be reused almost verbatim in the two-way setting. The only difference is that here we require the loops to be idempotent and we do not enforce output-minimality (we will discuss this latter choice further below, with a formal definition of output-minimality at hand).

{defi}

An inversion of the run $\rho$ is a tuple $(L_{1},\ell_{1},L_{2},\ell_{2})$ such that

1. (1)

$L_{1},L_{2}$ are idempotent loops, 2. (2)

$\ell_{1}=(x_{1},y_{1})$ and $\ell_{2}=(x_{2},y_{2})$ are anchor points inside $L_{1}$ and $L_{2}$, respectively, 3. (3)

$\ell_{1}\mathrel{{\lhd}}\ell_{2}$ and $x_{1}>x_{2}$, 4. (4)

for both $i=1$ and $i=2$, $\mathsf{out}(\mathsf{tr}(\ell_{i}))\neq\varepsilon$.

Figure 16 gives an example of an inversion involving the idempotent loop $L_{1}$ with anchor point $\ell_{1}$, and the idempotent loop $L_{2}$ with anchor point $\ell_{2}$. The intercepted factors that form the corresponding traces are represented by thick arrows; the one highlighted in red are those that produce non-empty output.

The implication P1 $\Rightarrow$ P2 of Theorem 6 in the two-way case is formalized below exactly as in Proposition 8, and the proof is very similar to the sweeping case. More precisely, it can be checked that the proof of the first claim in Proposition 8 was shown independently of the sweeping assumption — one just needs to replace the use of Equation 2 with Proposition 21. The sweeping assumption was used only for deriving the notion of output-minimal factor, which was crucial to conclude that the period $p$ is bounded by the specific constant $\boldsymbol{B}$. In this respect, the proof of Proposition 24 requires a different argument for showing that $p\leq\boldsymbol{B}$:

Proposition 24**.**

If $T$ is one-way definable, then the following property P2 holds:

For all inversions $(L_{1},\ell_{1},L_{2},\ell_{2})$ of $\rho$, the period $p$ of the word

[TABLE]

divides both $|\mathsf{out}(\mathsf{tr}(\ell_{1}))|$ and $|\mathsf{out}(\mathsf{tr}(\ell_{2}))|$. Moreover, $p\leq\boldsymbol{B}$.

We only need to show here that $p\leq\boldsymbol{B}$. Recall that in the sweeping case we relied on the assumption that the factors $\mathsf{tr}(\ell_{1})$ and $\mathsf{tr}(\ell_{2})$ of an inversion are output-minimal, and on Lemma 7. In the general case we need to replace output-minimality by the following notion:

{defi}

Consider pairs $(L,C)$ consisting of an idempotent loop $L$ and a component $C$ of $L$.

1. (1)

On such pairs, define the relation $\sqsubset$ by $(L^{\prime},C^{\prime})\sqsubset(L,C)$ if $L^{\prime}\subsetneq L$ and at least one $(L^{\prime},C^{\prime})$-factor is contained in some $(L,C)$-factor. 2. (2)

A pair $(L,C)$ is output-minimal if $(L^{\prime},C^{\prime})\sqsubset(L,C)$ implies $\mathsf{out}(\mathsf{tr}(\mathsf{an}(C^{\prime})))=\varepsilon$.

Note that the relation $\sqsubset$ is not a partial order in general (it is however antisymmetric). Moreover, it is easy to see that the notion of output-minimal pair $(L,C)$ generalizes that of output-minimal factor introduced in the sweeping case: indeed, if $\ell$ is the anchor point of a loop $L$ of a sweeping transducer and $\mathsf{tr}(\ell)$ satisfies Definition 4, then the pair $(L,C)$ is output-minimal, where $C$ is the unique component whose edge corresponds to $\mathsf{tr}(\ell)$.

The following lemma bounds the length of the output trace $\mathsf{out}(\mathsf{tr}(\mathsf{an}(C)))$ for an output-minimal pair $(L,C)$:

Lemma 25**.**

For every output-minimal pair $(L,C)$, $|\mathsf{out}(\mathsf{tr}(\mathsf{an}(C)))|\leq\boldsymbol{B}$.

Proof 7.4**.**

Consider a pair $(L,C)$ consisting of an idempotent loop $L=[x_{1},x_{2}]$ and a component $C$ of $L$. Suppose by contradiction that $|\mathsf{out}(\mathsf{tr}(\mathsf{an}(C)))|>\boldsymbol{B}$. We will show that $(L,C)$ is not output-minimal.

Recall that $\mathsf{tr}(\mathsf{an}(C))$ is a concatenation of $(L,C)$-factors, say, $\mathsf{tr}(\mathsf{an}(C))=\beta_{1}\cdots\beta_{k}$. Let $\ell_{1}$ (resp. $\ell_{2}$) be the first (resp. last) location that is visited by these factors. Further let $K=[\ell_{1},\ell_{2}]$ and $Z=K\>\cap\>(L\times\mathbb{N})$. By construction, the subsequence $\rho|Z$ can be seen as a concatenation of the factors $\beta_{1},\dots,\beta_{k}$, possibly in a different order than that of $\mathsf{tr}(\mathsf{an}(C))$. This implies that $|\mathsf{out}(\rho|Z)|>\boldsymbol{B}$.

By Theorem 23, we know that there exist an idempotent loop $L^{\prime}\subsetneq L$ and a component $C^{\prime}$ of $L^{\prime}$ such that $\mathsf{an}(C^{\prime})\in K$ and $\mathsf{out}(\mathsf{tr}(\mathsf{an}(C^{\prime})))\neq\varepsilon$. Note that the $(L^{\prime},C^{\prime})$-factor that starts at the anchor point $\mathsf{an}(C^{\prime})$ (an ${\mathsf{LR}}$- or ${\mathsf{RL}}$-factor) is entirely contained in some $(L,C)$-factor. This implies that $(L^{\prime},C^{\prime})\sqsubset(L,C)$, and thus $(L,C)$ is not output-minimal.

We remark that the above lemma cannot be used directly to bound the period of the output produced amid an inversion. The reason is that we cannot restrict ourselves to inversions $(L_{1},\ell_{1},L_{2},\ell_{2})$ that induce output-minimal pairs $(L_{i},C_{i})$ for $i=1,2$, where $C_{i}$ is the unique component of the anchor point $\ell_{i}$. An example is given in Figure 16, assuming that the factors depicted in red are the only ones that produce non-empty output, and the lengths of these outputs exceed $\boldsymbol{B}$. On the one hand $(L_{1},\ell_{1},L_{2},\ell_{2})$ is an inversion, but $(L_{1},C_{1})$ is not output-minimal. On the other hand, it is possible that $\rho$ contains no other inversion than $(L_{1},\ell_{1},L_{2},\ell_{2})$: any loop strictly contained in the red factor in $L_{1}$ will have the anchor point after $\ell_{2}$.

We are now ready to show the second claim of Proposition 24.

Proof 7.5** (Proof of Proposition 24).**

The proof of the second claim requires a refinement of the arguments that involve pumping the run $\rho$ simultaneously on three different loops. As usual, we assume that the loops $L_{1},L_{2}$ of the inversion are disjoint (otherwise, we preliminarily pump one of the two loops a few times).

Recall that the word

[TABLE]

has period p=\gcd\big{(}|\mathsf{out}(\mathsf{tr}(\ell_{1}))|,|\mathsf{out}(\mathsf{tr}(\ell_{2}))|\big{)}, but that we cannot bound $p$ by assuming that $(L_{1},\ell_{1},L_{2},\ell_{2})$ is output-minimal. However, in the pumped run $\rho^{(2,1)}$ we do find inversions with output-minimal pairs. For example, as depicted in the right part of Figure 17, we can consider the left and right copy of $L_{1}$ in $\rho^{(2,1)}$, denoted by $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{L}}_{1}$ and $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1}$, respectively. Accordingly, we denote by $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{\ell}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{\ell}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{\ell}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{\ell}}_{1}$ and $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1}$ the left and right copy of $\ell_{1}$ in $\rho^{(2,1)}$. Now, let $(L_{0},C_{0})$ be any output-minimal pair such that $L_{0}$ is an idempotent loop, $\mathsf{out}(\mathsf{tr}(\mathsf{an}(C_{0})))\neq\varepsilon$, and either $(L_{0},C_{0})=(\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{L}}_{1},C_{1})$ or $(L_{0},C_{0})\sqsubset(\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{L}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{L}}_{1},C_{1})$. Such a loop $L_{0}$ is represented in Figure 17 by the red vertical stripe. Further let $\ell_{0}=\mathsf{an}(C_{0})$.

We claim that either $(L_{0},\ell_{0},L_{2},\ell_{2})$ or $(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1},L_{0},\ell_{0})$ is an inversion of the run $\rho^{(2,1)}$, depending on whether $\ell_{0}$ occurs before or after $\ell_{2}$. First, note that all the loops $L_{0}$, $L_{2}$, $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1}$ are idempotent and non-overlapping; more precisely, we have $\max(L_{2})\leq\min(L_{0})$ and $\max(L_{0})\leq\min(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1})$. Moreover, the outputs of the traces $\mathsf{tr}(\ell_{0})$, $\mathsf{tr}(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1})$, and $\mathsf{tr}(\ell_{2})$ are all non-empty. So it remains to distinguish two cases based on the ordering of the anchor points $\ell_{0}$, $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1}$, $\ell_{2}$. If $\ell_{0}\mathrel{{\lhd}}\ell_{2}$, then $(L_{0},\ell_{0},L_{2},\ell_{2})$ is an inversion. Otherwise, because $(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1},L_{2},\ell_{2})$ is an inversion, we know that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1}\mathrel{{\lhd}}\ell_{2}\mathrel{\unlhd}\ell_{0}$, and hence $(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1},L_{0},C_{0})$ is an inversion.

Now, we know that $\rho^{(2,1)}$ contains the inversion $(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1},L_{2},\ell_{2})$, but also an inversion involving the output-minimal pair $(L_{0},C_{0})$, with $L_{0}$ strictly between $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1}$ and $L_{2}$. For all $m_{0},m_{1},m_{2}$, we define $\rho^{(m_{0},m_{1},m_{2})}$ as the run obtained from $\rho^{(2,1)}$ by pumping $m_{0},m_{1},m_{2}$ times the loops $L_{0},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},L_{2}$, respectively. By reasoning as we did in the proof of Proposition 8 (cf. Periodicity of outputs of pumped runs), one can show that there are arbitrarily large output factors of $\rho^{(m_{0},m_{1},m_{2})}$ that are produced within the inversion on $\ell_{0}$ (i.e. either $(L_{0},\ell_{0},L_{2},\ell_{2})$ or $(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1},L_{0},\ell_{0})$) and that are periodic with period $p^{\prime}$ that divides $|\mathsf{out}(\mathsf{tr}(\ell_{0}))|$. In particular, by Lemma 25, we know that $p^{\prime}\leq\boldsymbol{B}$. Moreover, large portions of these factors are also produced within the inversion $(\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{L}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{L}}_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\ell}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\ell}}_{1},L_{2},\ell_{2})$, and hence by Theorem 9 they have period $\gcd(p,p^{\prime})$.

To conclude the proof we need to transfer the periodicity property from the pumped runs $\rho^{(m_{0},m_{1},m_{2})}$ to the original run $\rho$. This is done exactly like in Proposition 8 by relying on size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Usual replacement of Saarela… Lemma 10: we observe that the periodicity property holds for large enough parameters $m_{0},m_{1},m_{2}$, hence for all values of the parameters, and in particular for $m_{0}=m_{1}=m_{2}=1$. This shows that the word

[TABLE]

has period $\gcd(p,p^{\prime})\leq\boldsymbol{B}$.

So far we have shown that the output produced amid every inversion of a run of a one-way definable two-way transducer is periodic, with period bounded by $\boldsymbol{B}$ and dividing the lengths of the trace outputs of the inversion. This basically proves the implication P1 $\Rightarrow$ P2 of Theorem 6. In the next section we will follow a line of arguments similar to that of Section 5 to prove the remaining implications P2 $\Rightarrow$ P3 $\Rightarrow$ P1.

8. The characterization in the two-way case

In this section we generalize the characterization of one-way definability of sweeping transducers to the general two-way case. As usual, we fix through the rest of the section a successful run $\rho$ of $T$ on some input word $u$.

From periodicity of inversions to existence of decompositions.

We continue by proving the second implication P2 $\Rightarrow$ P3 of Theorem 6 in the two-way case. This requires showing the existence of a suitable decomposition of a run $\rho$ that satisfies property P2. Recall that P2 says that for every inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$, the period of the word $\mathsf{out}(\mathsf{tr}(\ell_{1}))~{}\mathsf{out}(\rho[\ell_{1},\ell_{2}])~{}\mathsf{out}(\mathsf{tr}(\ell_{2}))$ divides $\gcd(|\mathsf{out}(\mathsf{tr}(\ell_{1}))|,|\mathsf{out}(\mathsf{tr}(\ell_{2}))|)\leq\boldsymbol{B}$. The definitions underlying the decomposition of $\rho$ are similar to those given in the sweeping case:

{defi}

Let $\rho[\ell,\ell^{\prime}]$ be a factor of a run $\rho$ of $T$, where $\ell=(x,y)$, $\ell^{\prime}=(x^{\prime},y^{\prime})$, and $x\leq x^{\prime}$. We call $\rho[\ell,\ell^{\prime}]$

• •

a $\boldsymbol{B}$-diagonal if for all $z\in[x,x^{\prime}]$, there is a location $\ell_{z}$ at position $z$ such that $\ell\mathrel{\unlhd}\ell_{z}\mathrel{\unlhd}\ell^{\prime}$ and the words \mathsf{out}(\rho|Z_{\ell_{z}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}) and \mathsf{out}(\rho|Z_{\ell_{z}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}) have length at most $\boldsymbol{B}$, where Z_{\ell_{z}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}=[\ell_{z},\ell^{\prime}]\>\cap\>\big{(}[0,z]\times\mathbb{N}\big{)} and Z_{\ell_{z}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}=[\ell,\ell_{z}]\>\cap\>\big{(}[z,\omega]\times\mathbb{N}\big{)};

• •

a $\boldsymbol{B}$-block if the word $\mathsf{out}(\rho[\ell,\ell^{\prime}])$ is almost periodic with bound $\boldsymbol{B}$, and $\mathsf{out}(\rho|Z^{\shortleftarrow})$ and $\mathsf{out}(\rho|Z^{\shortrightarrow})$ have length at most $\boldsymbol{B}$, where Z^{\shortleftarrow}=[\ell,\ell^{\prime}]\>\cap\>\big{(}[0,x]\times\mathbb{N}\big{)} and Z^{\shortrightarrow}=[\ell,\ell^{\prime}]\>\cap\>\big{(}[x^{\prime},\omega]\times\mathbb{N}\big{)}.

The definition of $\boldsymbol{B}$-decomposition is copied verbatim from the sweeping case, but uses the new notions of $\boldsymbol{B}$-diagonal and $\boldsymbol{B}$-block:

{defi}

A $\boldsymbol{B}$-decomposition of a run $\rho$ of $T$ is a factorization $\prod_{i}\,\rho[\ell_{i},\ell_{i+1}]$ of $\rho$ into $\boldsymbol{B}$-diagonals and $\boldsymbol{B}$-blocks.

To provide further intuition, we consider the transduction of Example 3 and the two-way transducer $T$ that implements it in the most natural way. Figure 20 shows an example of a run of $T$ on an input of the form $u_{1}\>\#\>u_{2}\>\#\>u_{3}\>\#\>u_{4}$, where $u_{2},\,u_{4}\in(abc)^{*}$, $u_{1},\,u_{3}\not\in(abc)^{*}$, and $u_{3}$ has even length. The factors of the run that produce long outputs are highlighted by the bold arrows. The first and third factors of the decomposition, i.e. $\rho[\ell_{1},\ell_{2}]$ and $\rho[\ell_{3},\ell_{4}]$, are diagonals (represented by the blue hatched areas); the second and fourth factors $\rho[\ell_{2},\ell_{3}]$ and $\rho[\ell_{4},\ell_{5}]$ are blocks (represented by the red hatched areas).

To identify the blocks of a possible decomposition of $\rho$, we reuse the equivalence relation $\mathrel{\text{\sf S}}^{*}$ introduced in Definition 5. Recall that this is the reflexive and transitive closure of the relation $\mathrel{\text{\sf S}}$ that groups any two locations $\ell,\ell^{\prime}$ that occur between $\ell_{1},\ell_{2}$, for some inversion $(L_{1},\ell_{1},L_{2},\ell_{2})$.

The proof that the output produced inside each $\mathrel{\text{\sf S}}^{*}$-equivalence class is periodic, with period at most $\boldsymbol{B}$ (Lemma 13) carries over in the two-way case without modifications. Similarly, every $\mathrel{\text{\sf S}}^{*}$-equivalence class can be extended to the left and to the right by using Definition 5, which we report here verbatim for the sake of readability, together with an exemplifying figure.

{defi}

Consider a non-singleton $\mathrel{\text{\sf S}}^{*}$-equivalence class $K=[\ell,\ell^{\prime}]$. Let $\mathsf{an}(K)$ be the restriction of $K$ to the anchor points occurring in some inversion, and $X_{\mathsf{an}(K)}=\{x\>:\>\exists y\>(x,y)\in\mathsf{an}(K)\}$ be the projection of $\mathsf{an}(K)$ on positions. We define $\mathsf{block}(K)=[\tilde{\ell},\tilde{\ell}^{\prime}]$, where

• •

$\tilde{\ell}$ is the latest location $(\tilde{x},\tilde{y})\mathrel{\unlhd}\ell$ such that $\tilde{x}=\min(X_{\mathsf{an}(K)})$,

• •

$\tilde{\ell}^{\prime}$ is the earliest location $(\tilde{x},\tilde{y})\mathrel{\unrhd}\ell^{\prime}$ such that $\tilde{x}=\max(X_{\mathsf{an}(K)})$

(note that the locations $\tilde{\ell},\tilde{\ell}^{\prime}$ exist since $\ell,\ell^{\prime}$ are anchor points in some inversion).

As usual, we call $\mathrel{\text{\sf S}}^{*}$-block any factor of $\rho$ of the form $\rho|\mathsf{block}(K)$ that is obtained by applying the above definition to a non-singleton $\mathrel{\text{\sf S}}^{*}$-class $K$. Lemma 14, which shows that $\mathrel{\text{\sf S}}^{*}$-blocks can indeed be used as $\boldsymbol{B}$-blocks in a decomposition of $\rho$, generalizes easily to the two-way case:

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Most proofs are quite notation-heavy. A small figure to represent the positions and factors involved is a good thing to have. For example, Lemma 5.8 has Fig.5, but the corresponding (and slightly more involved) Lemme 8.4 hasn’t. This seems like a worthwile investment for any long proof, or any time a huge number of notations are involved. size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. I think the reviewer is referring to the lack of a figure for the previous definition, exactly like for Definition 5.7. I put it.size=,color=cyan!30,inlinesize=,color=cyan!30,inlinetodo: size=,color=cyan!30,inlineR2. -p.37 l.21-22 the $\ell$ should be replaced with $\tilde{\ell}$ size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. I didn’t find wheresize=,color=red!30,inlinesize=,color=red!30,inlinetodo: size=,color=red!30,inlineF. second and third line of the list of properties we will prove. I agree.

Lemma 26**.**

If $K$ is a non-singleton $\mathrel{\text{\sf S}}^{*}$-equivalence class, then $\rho|\mathsf{block}(K)$ is a $\boldsymbol{B}$-block.

Proof 8.1**.**

The proof is similar to that of Lemma 14. The main difference is that here we will bound the lengths of some outputs using a Ramsey-type argument (Theorem 23), instead of output-minimality of factors (Lemma 7). To follow the various constructions and arguments the reader can refer to Figure 8.

Let $K=[\ell,\ell^{\prime}]$, $\mathsf{an}(K)$, $X_{\mathsf{an}(K)}$, and $\mathsf{block}(K)=[\tilde{\ell},\tilde{\ell}^{\prime}]$ be as in Definition 8, where $\tilde{\ell}=(\tilde{x},\tilde{y})$, $\tilde{\ell}^{\prime}=(\tilde{x}^{\prime},\tilde{y}^{\prime})$, $\tilde{x}=\min(X_{\mathsf{an}(K)})$, and $\tilde{x}^{\prime}=\max(X_{\mathsf{an}(K)})$. We need to verify that $\rho|\mathsf{block}(K)$ is a $\boldsymbol{B}$-block, namely, that:

• •

$\tilde{x}\leq\tilde{x}^{\prime}$,

• •

$\mathsf{out}(\rho[\tilde{\ell},\tilde{\ell}^{\prime}])$* is almost periodic with bound $\boldsymbol{B}$,*

• •

$\mathsf{out}(\rho|Z^{\shortleftarrow})$* and $\mathsf{out}(\rho|Z^{\shortrightarrow})$ have length at most $\boldsymbol{B}$, where Z^{\shortleftarrow}=[\tilde{\ell},\tilde{\ell}^{\prime}]\>\cap\>\big{(}[0,x]\times\mathbb{N}\big{)} and Z^{\shortrightarrow}=[\tilde{\ell},\tilde{\ell}^{\prime}]\>\cap\>\big{(}[x^{\prime},\omega]\times\mathbb{N}\big{)}.*

The first condition $\tilde{x}\leq\tilde{x}^{\prime}$ follows immediately from $\tilde{x}=\min(X_{\mathsf{an}(K)})$ and $\tilde{x}^{\prime}=\max(X_{\mathsf{an}(K)})$.

Next, we prove that the output produced by the factor $\rho[\tilde{\ell},\tilde{\ell}^{\prime}]$ is almost periodic with bound $\boldsymbol{B}$. By Definition 8, we have $\tilde{\ell}\mathrel{\unlhd}\ell\mathrel{{\lhd}}\ell^{\prime}\mathrel{\unlhd}\tilde{\ell}^{\prime}$, and by Lemma 13 we know that $\mathsf{out}(\rho[\ell,\ell^{\prime}])$ is periodic with period at most $\boldsymbol{B}$. So it suffices to bound the length of the words $\mathsf{out}(\rho[\tilde{\ell},\ell])$ and $\mathsf{out}(\rho[\ell^{\prime},\tilde{\ell}^{\prime}])$. We shall focus on the former word, as the arguments for the latter are similar.

First, we show that the factor $\rho[\tilde{\ell},\ell]$ lies entirely to the right of position $\tilde{x}$ (in particular, it starts at an even level $\tilde{y}$). Indeed, if this were not the case, there would exist another location $\ell^{\prime\prime}=(\tilde{x},\tilde{y}+1)$, on the same position as $\tilde{\ell}$, but at a higher level, such that $\tilde{\ell}\mathrel{{\lhd}}\ell^{\prime\prime}\mathrel{\unlhd}\ell$. But this would contradict Definition 8 ($\tilde{\ell}$ is the latest location $(x,y)\mathrel{\unlhd}\ell$ such that $x=\min(X_{\mathsf{an}(K)})$).

Suppose now that the length of $|\mathsf{out}(\rho[\tilde{\ell},\ell])|>\boldsymbol{B}$. We head towards a contradiction by finding a location $\ell^{\prime\prime}\mathrel{{\lhd}}\ell$ that is $\mathrel{\text{\sf S}}^{*}$-equivalent to the first location $\ell$ of the $\mathrel{\text{\sf S}}^{*}$-equivalence class $K$. Since the factor $\rho[\tilde{\ell},\ell]$ lies entirely to the right of position $\tilde{x}$, it is intercepted by the interval $I=[\tilde{x},\omega]$. So $|\mathsf{out}(\rho[\tilde{\ell},\ell])|>\boldsymbol{B}$ is equivalent to saying $|\mathsf{out}(\rho|Z)|>\boldsymbol{B}$, where Z=[\tilde{\ell},\ell]\>\cap\>\big{(}[\tilde{x},\omega]\times\mathbb{N}\big{)}. Then, Theorem 23 implies the existence of an idempotent loop $L$ and an anchor point $\ell^{\prime\prime}$ of $L$ such that

• •

$\min(L)>\tilde{x}$,

• •

$\tilde{\ell}\mathrel{{\lhd}}\ell^{\prime\prime}\mathrel{{\lhd}}\ell$,

• •

$\mathsf{out}(\mathsf{tr}(\ell^{\prime\prime}))\neq\varepsilon$.

Further recall that $\tilde{x}=\min(X_{\mathsf{an}(K)})$ is the leftmost position of locations in the class $K=[\ell,\ell^{\prime}]$ that are also anchor points of inversions. In particular, there is an inversion $(L_{1},\ell^{\prime\prime}_{1},L_{2},\ell^{\prime\prime}_{2})$, with $\ell^{\prime\prime}_{2}=(\tilde{x},y^{\prime\prime}_{2})\in K$. Since $\ell^{\prime\prime}\mathrel{{\lhd}}\ell\mathrel{\unlhd}\ell^{\prime\prime}_{2}$ and the position of $\ell^{\prime\prime}$ is to the right of the position of $\ell^{\prime\prime}_{2}$, we know that $(L,\ell^{\prime\prime},L_{2},\ell^{\prime\prime}_{2})$ is also an inversion, and hence $\ell^{\prime\prime}\mathrel{\text{\sf S}}^{*}\ell^{\prime\prime}_{2}\mathrel{\text{\sf S}}^{*}\ell$. But since $\ell^{\prime\prime}\neq\ell$, we get a contradiction with the assumption that $\ell$ is the first location of the $\mathrel{\text{\sf S}}^{*}$-class $K$. In this way we have shown that $|\mathsf{out}(\rho[\tilde{\ell},\ell])|\leq\boldsymbol{B}$.

It remains to bound the lengths of the outputs produced by the subsequences $\rho|Z^{\shortleftarrow}$ and $\rho|Z^{\shortrightarrow}$, where Z^{\shortleftarrow}=[\tilde{\ell},\tilde{\ell}^{\prime}]\>\cap\>\big{(}[0,\tilde{x}]\times\mathbb{N}\big{)} and Z^{\shortrightarrow}=[\tilde{\ell},\tilde{\ell}^{\prime}]\>\cap\>\big{(}[\tilde{x}^{\prime},\omega]\times\mathbb{N}\big{)}. As usual, we consider only one of the two symmetric cases. Suppose, by way of contradiction, that $|\mathsf{out}(\rho|Z^{\shortleftarrow})|>\boldsymbol{B}$. By Theorem 23, there exist an idempotent loop $L$ and an anchor point $\ell^{\prime\prime}$ of $L$ such that

• •

$\max(L)<\tilde{x}$,

• •

$\tilde{\ell}\mathrel{{\lhd}}\ell^{\prime\prime}\mathrel{{\lhd}}\tilde{\ell}^{\prime}$,

• •

$\mathsf{out}(\mathsf{tr}(\ell^{\prime\prime}))\neq\varepsilon$.

By following the same line of reasoning as before, we recall that $\ell$ is the first location of the non-singleton class $K$. From this we derive the existence an inversion $(L_{1},\ell^{\prime\prime}_{1},L_{2},\ell^{\prime\prime}_{2})$ where $\ell^{\prime\prime}_{1}=\ell$. We claim that $\ell\mathrel{\unlhd}\ell^{\prime\prime}$. Indeed, if this were not the case, then, because $\ell^{\prime\prime}$ is strictly to the left of $\tilde{x}$ and $\ell$ is to the right of $\tilde{x}$, there would exist a location $\ell^{\prime\prime\prime}$ between $\ell^{\prime\prime}$ and $\ell$ that lies at position $\tilde{x}$. But $\tilde{\ell}\mathrel{{\lhd}}\ell^{\prime\prime}\mathrel{\unlhd}\ell^{\prime\prime\prime}\mathrel{\unlhd}\ell$ would contradict the fact that $\tilde{\ell}$ is the latest location before $\ell$ that lies at the position $\tilde{x}$. Now that we know that $\ell\mathrel{\unlhd}\ell^{\prime\prime}$ and that $\ell^{\prime\prime}$ is to the left of $\tilde{x}$, we observe that $(L_{1},\ell^{\prime\prime}_{1},L,\ell^{\prime\prime})$ is also an inversion, and hence $\ell^{\prime\prime}\in\mathsf{an}(K)$. Since $\ell^{\prime\prime}$ is strictly to the left of $\tilde{x}$, we get a contradiction with the definition of $\tilde{x}$ as leftmost position of the locations in $\mathsf{an}(K)$. So we conclude that $|\mathsf{out}(\rho|Z^{\shortleftarrow})|\leq\boldsymbol{B}$.

The proof of Lemma 15, which shows that $\mathrel{\text{\sf S}}^{*}$-blocks do not overlap along the input axis, carries over in the two-way case, again without modifications. Finally, we generalize Lemma 16 to the new definition of diagonal, which completes the construction of a $\boldsymbol{B}$-decomposition for the run $\rho$:

Lemma 27**.**

Let $\rho[\ell,\ell^{\prime}]$ be a factor of $\rho$, with $\ell=(x,y)$, $\ell^{\prime}=(x^{\prime},y^{\prime})$, and $x\leq x^{\prime}$, that does not overlap any $\mathrel{\text{\sf S}}^{*}$-block. Then $\rho[\ell,\ell^{\prime}]$ is a $\boldsymbol{B}$-diagonal.

Proof 8.2**.**

Suppose by way of contradiction that there is some $z\in[x,x^{\prime}]$ such that, for all locations $\ell^{\prime\prime}$ at position $z$ and between $\ell$ and $\ell^{\prime}$, one of the two conditions holds:

1. (1)

|\mathsf{out}(\rho|Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}})|>\boldsymbol{B}, where Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}=[\ell^{\prime\prime},\ell^{\prime}]\>\cap\>\big{(}[0,z]\times\mathbb{N}\big{)}, 2. (2)

|\mathsf{out}(\rho|Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}})|>\boldsymbol{B}, where Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}=[\ell,\ell^{\prime\prime}]\>\cap\>\big{(}[z,\omega]\times\mathbb{N}\big{)}.

First, we claim that each of the two conditions above are satisfied at some locations $\ell^{\prime\prime}\in[\ell,\ell^{\prime}]$ at position $z$. Consider the highest even level $y^{\prime\prime}$ such that $\ell^{\prime\prime}=(z,y^{\prime\prime})\in[\ell,\ell^{\prime}]$ (use Figure 18 as a reference). Since $z\leq x^{\prime}$, the outgoing transition at $\ell^{\prime\prime}$ is rightward oriented, and the set Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}} is empty. This means that condition (1) is trivially violated at $\ell^{\prime\prime}$, and hence condition (2) holds at $\ell^{\prime\prime}$ by the initial assumption. Symmetrically, condition (1) holds at the location $\ell^{\prime\prime}=(z,y^{\prime\prime})$, where $y^{\prime\prime}$ is the lowest even level with $\ell^{\prime\prime}\in[\ell,\ell^{\prime}]$.

Let us now compare the levels where the above conditions hold. Clearly, the lower the level of location $\ell^{\prime\prime}$, the easier it is to satisfy condition (1), and symmetrically for condition (2). So, let $\ell^{+}=(z,y^{+})$ (resp. $\ell^{-}=(z,y^{-})$) be the highest (resp. lowest) location in $[\ell,\ell^{\prime}]$ at position $z$ that satisfies condition (1) (resp. condition (2)).

We claim that $y^{+}\geq y^{-}$. For this, we first observe that $y^{+}\geq y^{-}-1$, since otherwise there would exist a location $\ell^{\prime\prime}=(z,y^{\prime\prime})$, with $y^{+}<y^{\prime\prime}<y^{-}$, that violates both conditions (1) and (2). Moreover, $y^{+}$ must be odd, otherwise the transition departing from $\ell^{+}=(z,y^{+})$ would be rightward oriented and the location $\ell^{\prime\prime}=(z,y^{+}+1)$ would still satisfy condition (1), contradicting the definition of highest location $\ell^{+}$. For similar reasons, $y^{-}$ must also be odd, otherwise there would be a location $\ell^{\prime\prime}=(z,y^{-}-1)$ below $\ell^{-}$ that satisfies condition (2). But since $y^{+}\geq y^{-}-1$ and both $y^{+}$ and $y^{-}$ are odd, we need to have $y^{+}\geq y^{-}$.

In fact, from the previous arguments we know that the location $\ell^{\prime\prime}=(z,y^{+})$ (or equally the location $(x,y^{-})$) satisfies both conditions (1) and (2). We can thus apply Theorem 23 to the sets Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}} and Z_{\ell^{\prime\prime}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}, deriving the existence of two idempotent loops $L_{1},L_{2}$ and two anchor points $\ell_{1},\ell_{2}$ of $L_{1},L_{2}$, respectively, such that

• •

$\max(L_{2})<z<\min(L_{1})$,

• •

$\ell\mathrel{{\lhd}}\ell_{1}\mathrel{{\lhd}}\ell^{\prime\prime}\mathrel{{\lhd}}\ell_{2}\mathrel{{\lhd}}\ell^{\prime}$,

• •

$\mathsf{out}(\mathsf{tr}(\ell_{1})),\mathsf{out}(\mathsf{tr}(\ell_{2}))\neq\varepsilon$.

In particular, since $\ell_{1}$ is to the right of $\ell_{2}$ w.r.t. the order of positions, we know that $(L_{1},\ell_{1},L_{2},\ell_{2})$ is an inversion, and hence $\ell_{1}\mathrel{\text{\sf S}}^{*}\ell_{2}$. But this contradicts the assumption that $\rho[\ell,\ell^{\prime}]$ does not overlap with any $\mathrel{\text{\sf S}}^{*}$-block.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Part 5, ”From existence of decompositions to an equivalent one-way transducer”: so far the paper did a great job hinting at the proof that remains. Here an intuition is given for the construction later detailed in 8.6, but no mention of an intuition for Proposition 9.2: if it is at all possible to do so, this might help the paper’s overall flow. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. answered in review1-answers.txt

From existence of decompositions to an equivalent one-way transducer.

It remains to prove the last implication P3 $\Rightarrow$ P1 of Theorem 6, which amounts to construct a one-way transducer $T^{\prime}$ equivalent to $T$.

Hereafter, we denote by $D$ the language of words $u\in\mathsf{dom}(T)$ such that all successful runs of $T$ on $u$ admit a $\boldsymbol{B}$-decomposition. So far, we know that if $T$ is one-way definable (P1), then $D=\mathsf{dom}(T)$ (P3). As a matter of fact, this reduces the one-way definability problem for $T$ to the containment problem $\mathsf{dom}(T)\subseteq D$.

We will see later (in Section 9) how the latter problem can be decided in double exponential space by further reducing it to checking the emptiness of the intersection of the languages $\mathsf{dom}(T)$ and $D^{\complement}$, where $D^{\complement}$ is the complement of $D$.

Below, we show how to construct a one-way transducer $T^{\prime}$ of triple exponential size such that $T^{\prime}\subseteq T$ and $\mathsf{dom}(T^{\prime})$ is the set of all input words that have some successful run admitting a $\boldsymbol{B}$-decomposition (hence $\mathsf{dom}(T^{\prime})\supseteq D$). In particular, we will have that

[TABLE]

Note that this will prove P3 to P1, as well as the second item of Theorem 3, since $D=\mathsf{dom}(T)$ if and only if $T$ is one-way definable. A sketch of the proof of this construction when $T$ is a sweeping transducer was given at the end of Section 5.

Proposition 28**.**

Given a functional two-way transducer $T$, a one-way transducer $T^{\prime}$ satisfying

[TABLE]

can be constructed in $3\textsc{ExpTime}$. Moreover, if $T$ is sweeping, then $T^{\prime}$ can be constructed in $2\textsc{ExpTime}$.

Proof 8.3**.**

Given an input word $u$, the transducer $T^{\prime}$ will guess (and check) a successful run $\rho$ of $T$ on $u$, together with a $\boldsymbol{B}$-decomposition $\prod_{i}\rho[\ell_{i},\ell_{i+1}]$. The latter decomposition will be used by $T^{\prime}$ to simulate the output of $\rho$ in left-to-right manner, thus proving that $T^{\prime}\subseteq T$. Moreover, $u\in D$ implies the existence of a successful run that can be decomposed, thus proving that $\mathsf{dom}(T^{\prime})\supseteq D$. We now provide the details of the construction of $T^{\prime}$.

Guessing the run $\rho$ is standard (see, for instance, [30, 24]): it amounts to guess the crossing sequences $\rho|x$ for each position $x$ of the input. Recall that this is a bounded amount of information for each position $x$, since the run is normalized. As concerns the decomposition of $\rho$, it can be encoded by the endpoints $\ell_{i}$ of its factors, that is, by annotating the position of each $\ell_{i}$ as the level of $\ell_{i}$. In a similar way $T^{\prime}$ guesses the information of whether each factor $\rho[\ell_{i},\ell_{i+1}]$ is a $\boldsymbol{B}$-diagonal or a $\boldsymbol{B}$-block.

Thanks to the definition of decomposition (see Definition 8 and Figure 20), every two distinct factors span across non-overlapping intervals of positions. This means that each position $x$ is covered by exactly one factor of the decomposition. We call this factor the active factor at position $x$. The mode of computation of the transducer will depend on the type of active factor: if the active factor is a diagonal (resp. a block), then we say that $T^{\prime}$ is in diagonal mode (resp. block mode). Below we describe the behaviour for these two modes of computation.

Diagonal mode.*  We recall the key condition satisfied by the diagonal $\rho[\ell,\ell^{\prime}]$ that is active at position $x$ (cf. Definition 8 and Figure 18): there is a location $\ell_{x}=(x,y_{x})$ between $\ell$ and $\ell^{\prime}$ such that the words \mathsf{out}(\rho|Z_{\ell_{x}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}) and \mathsf{out}(\rho|Z_{\ell_{x}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}) have length at most $\boldsymbol{B}$, where Z_{\ell_{x}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{135.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}=[\ell_{x},\ell^{\prime}]\>\cap\>\big{(}[0,x]\times\mathbb{N}\big{)} and Z_{\ell_{x}}^{\mspace{-2.0mu}\text{\rotatebox[origin={c}]{315.0}{\vrule height=0.0pt,width=0.0pt\shortrightarrow}}\mspace{-2.0mu}}=[\ell,\ell_{x}]\>\cap\>\big{(}[x,\omega]\times\mathbb{N}\big{)}.*

Besides the run $\rho$ and the decomposition, the transducer $T^{\prime}$ will also guess the locations $\ell_{x}=(x,y_{x})$, that is, will annotate each $x$ with the corresponding $y_{x}$. Without loss of generality, we can assume that the function that associates each position $x$ with the guessed location $\ell_{x}=(x,y_{x})$ is monotone, namely, $x\leq x^{\prime}$ implies $\ell_{x}\mathrel{\unlhd}\ell_{x^{\prime}}$. While the transducer $T^{\prime}$ is in diagonal mode, the goal is to preserve the following invariant:

After reaching a position $x$ covered by the active diagonal, $T^{\prime}$ must have produced the output of $\rho$ up to location $\ell_{x}$.

To preserve the above invariant when moving from $x$ to the next position $x+1$, the transducer should output the word $\mathsf{out}(\rho[\ell_{x},\ell_{x+1}])$. This word consists of the following parts:

1. (1)

The words produced by the single transitions of $\rho[\ell_{x},\ell_{x+1}]$ with endpoints in $\{x,x+1\}\times\mathbb{N}$. Note that there are at most ${\boldsymbol{H}}$ such words, each of them has length at most ${\boldsymbol{C}}$, and they can all be determined using the crossing sequences at $x$ and $x+1$ and the information about the levels of $\ell_{x}$ and $\ell_{x+1}$. We can thus assume that this information is readily available to the transducer. 2. (2)

The words produced by the factors of $\rho[\ell_{x},\ell_{x+1}]$ that are intercepted by the interval $[0,x]$. Thanks to the definition of diagonal, we know that the total length of these words is at most $\boldsymbol{B}$. These words cannot be determined from the information on $\rho|x$, $\rho|x+1$, $\ell_{x}$, and $\ell_{x+1}$ alone, so they need to be constructed while scanning the input. For this, some additional information needs to be stored.

More precisely, at each position $x$ of the input, the transducer stores all the outputs produced by the factors of $\rho$ that are intercepted by $[0,x]$ and that occur after a location of the form $\ell_{x^{\prime}}$, for any $x^{\prime}\geq x$ that is covered by a diagonal. This clearly includes the previous words when $x^{\prime}=x$, but also other words that might be used later for processing other diagonals. Moreover, by exploiting the properties of diagonals, one can prove that those words have length at most $\boldsymbol{B}$, so they can be stored with triply exponentially many states. Using classical techniques, the stored information can be maintained while scanning the input $u$ using the guessed crossing sequences of $\rho$. 3. (3)

The words produced by the factors of $\rho[\ell_{x},\ell_{x+1}]$ that are intercepted by the interval $[x+1,\omega]$. These words must be guessed, since they depend on a portion of the input that has not been processed yet. Accordingly, the guesses need to be stored into memory, in such a way that they can be checked later. For this, the transducer stores, for each position $x$, the guessed words that correspond to the outputs produced by the factors of $\rho$ intercepted by $[x,\omega]$ and occurring before a location of the form $\ell_{x^{\prime}}$, for any $x^{\prime}\leq x$ that is covered by a diagonal.

Block mode.*  Suppose that the active factor $\rho[\ell,\ell^{\prime}]$ is a $\boldsymbol{B}$-block. Let $I=[x,x^{\prime}]$ be the set of positions covered by this factor. Moreover, for each position $z\in I$, let Z^{\shortleftarrow}_{z}=[\ell,\ell^{\prime}]\>\cap\>\big{(}[0,z]\times\mathbb{N}\big{)} and Z^{\shortrightarrow}_{z}=[\ell,\ell^{\prime}]\>\cap\>\big{(}[z,\omega]\times\mathbb{N}\big{)}. We recall the key property of a block (cf. Definition 8 and Figure 19): the word $\mathsf{out}(\rho[\ell,\ell^{\prime}])$ is almost periodic with bound $\boldsymbol{B}$, and the words $\mathsf{out}(\rho|Z^{\shortleftarrow}_{x})$ and $\mathsf{out}(\rho|Z^{\shortrightarrow}_{x^{\prime}})$ have length at most $\boldsymbol{B}$.*

For the sake of brevity, suppose that $\mathsf{out}(\rho[\ell,\ell^{\prime}])=w_{1}\,w_{2}\,w_{3}$, where $w_{2}$ is periodic with period $\boldsymbol{B}$ and $w_{1},w_{3}$ size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. $w_{1},w_{3}$ I assume. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. fixed (was $w_{1},w_{2}$)have length at most $\boldsymbol{B}$. Similarly, let $w_{0}=\mathsf{out}(\rho|Z^{\shortleftarrow}_{x})$ and $w_{4}=\mathsf{out}(\rho|Z^{\shortrightarrow}_{x^{\prime}})$. The invariant preserved by $T^{\prime}$ in block mode is the following:

After reaching a position $z$ covered by the active block $\rho[\ell,\ell^{\prime}]$, $T^{\prime}$ must have produced the output of the prefix of $\rho$ up to location $\ell$, followed by a prefix of $\mathsf{out}(\rho[\ell,\ell^{\prime}])=w_{1}\,w_{2}\,w_{3}$ of the same length as $\mathsf{out}(\rho|Z^{\shortleftarrow}_{z})$.

The initialization of the invariant is done when reaching the left endpoint $x$. At this moment, it suffices that $T^{\prime}$ outputs a prefix of $w_{1}\,w_{2}\,w_{3}$ of the same length as $w_{0}=\mathsf{out}(\rho|Z^{\shortleftarrow}_{x})$, thus bounded by $\boldsymbol{B}$. Symmetrically, when reaching the right endpoint $x^{\prime}$, $T^{\prime}$ will have produced almost the entire word $\mathsf{out}(\rho[\ell,\ell^{\prime}])\,w_{1}\,w_{2}\,w_{3}$, but without the suffix $w_{4}=\mathsf{out}(\rho|Z^{\shortrightarrow}_{x^{\prime}})$ of length at most $\boldsymbol{B}$. Thus, before moving to the next factor of the decomposition, the transducer will produce the remaining suffix, so as to complete the output of $\rho$ up to location $\ell_{i_{x}+1}$.

It remains to describe how the above invariant can be maintained when moving from a position $z$ to the next position $z+1$ inside $I=[x,x^{\prime}]$. For this, it is convenient to succinctly represent the word $w_{2}$ by its repeating pattern, say $v$, of length at most $\boldsymbol{B}$. To determine the symbols that have to be output at each step, the transducer will maintain a pointer on either $w_{1}\,v$ or $w_{3}$. The pointer is increased in a deterministic way, and precisely by the amount $|\mathsf{out}(\rho|Z^{\shortleftarrow}_{z+1})|-|\mathsf{out}(\rho|Z^{\shortleftarrow}_{z})|$. The only exception is when the pointer lies in $w_{1}\,v$, but its increase would go over $w_{1}\,v$: in this case the transducer has the choice to either bring the pointer back to the beginning of $v$ (representing a periodic output inside $w_{2}$), or move it to $w_{3}$. Of course, this is a non-deterministic choice, but it can be validated when reaching the right endpoint of $I$. Concerning the number of symbols that need to be emitted at each step, this can be determined from the crossing sequences at $z$ and $z+1$, and from the knowledge of the lowest and highest levels of locations that are at position $z$ and between $\ell$ and $\ell^{\prime}$. We denote the latter levels by $y^{-}_{z}$ and $y^{+}_{z}$, respectively.

Overall, this shows how to maintain the invariant of the block mode, assuming that the levels $y^{-}_{z},y^{+}_{z}$ are known, as well as the words $w_{0},w_{1},v,w_{3},w_{4}$ of bounded length. Like the mapping $z\mapsto\ell_{z}=(z,y_{z})$ used in diagonal mode, the mapping $z\mapsto(y^{-}_{z},y^{+}_{z})$ can be guessed and checked using the crossing sequences. Similarly, the words $w_{1},v,w_{3}$ can be guessed just before entering the active block, and can be checked along the process. As concerns the words $w_{0},w_{4}$, these can be guessed and checked in a way similar to the words that we used in diagonal mode. More precisely, for each position $z$ of the input, the transducer stores the following additional information:

1. (1)

the outputs produced by the factors of $\rho$ that are intercepted by $[0,z]$ and that occur after the beginning $\ell^{\prime\prime}$ of some block, with $\ell^{\prime\prime}=(x^{\prime\prime},y^{\prime\prime})$ and $x^{\prime\prime}\geq z$; 2. (2)

the outputs produced by the factors of $\rho$ that are intercepted by $[z,\omega]$ and that occur before the ending $\ell^{\prime\prime\prime}$ of a block, where $\ell^{\prime\prime\prime}=(x^{\prime\prime\prime},y^{\prime\prime\prime})$ and $x^{\prime\prime\prime}\leq z$.

By the definition of blocks, the above words have length at most $\boldsymbol{B}$ and can be maintained while processing the input and the crossing sequences. Finally, we observe that the words, together with the information given by the lowest and highest levels $y^{-}_{z},y^{+}_{z}$, for both $z=x$ and $z=x^{\prime}$, are sufficient for determining the content of $w_{0}$ and $w_{4}$.

We have just shown how to construct a one-way transducer $T^{\prime}\subseteq T$ such that $\mathsf{dom}(T^{\prime})\supseteq D$. From the above construction it is easy to see that the number of states and transitions of $T^{\prime}$, as well as the number of letters emitted by each transition, are at most exponential in $\boldsymbol{B}$. Since $\boldsymbol{B}$ is doubly exponential in the size of $T$, this shows that $T^{\prime}$ can be constructed from $T$ in $3\textsc{ExpTime}$. Note that the triple exponential complexity comes from the lengths of the words that need to be guessed and stored in the control states, and these lengths are bounded by $\boldsymbol{B}$. However, if $T$ is a sweeping transducer, then, according to the results proved in Section 5, the bound $\boldsymbol{B}$ is simply exponential. In particular, in the sweeping case we can construct the one-way transducer $T^{\prime}$ in $2\textsc{ExpTime}$.

Generality of the construction.

We conclude the section with a discussion on the properties of the one-way transducer $T^{\prime}$ constructed from $T$. Roughly speaking, we would like to show that, even when $T$ is not one-way definable, $T^{\prime}$ is somehow the best one-way under-approximation of $T$. However, strictly speaking, the latter terminology is meaningless: if $T^{\prime}$ is a one-way transducer strictly contained in $T$, then one can always find a better one-way transducer $T^{\prime\prime}$ that satisfies $T^{\prime}\subsetneq T^{\prime\prime}\subsetneq T$, for instance by extending $T^{\prime}$ with a single input-output pair. Below, we formalize in an appropriate way the notion of “best one-way under-approximation”.

We are interested in comparing the domains of transducers, but only up to a certain amount. In particular, we are interested in languages that are preserved under pumping loops of runs of $T$. Formally, given a language $L$, we say that $L$ is $T$-pumpable if $L\subseteq\mathsf{dom}(T)$ and for all words $u\in L$, all successful runs $\rho$ of $T$ on $u$, all loops $L$ of $\rho$, and all positive numbers $n$, the word $\mathsf{pump}_{L}^{n}(u)$ also belongs to $L$. Clearly, the domain $\mathsf{dom}(T)$ of a transducer $T$ is a regular $T$-pumpable language.

Another noticeable example of $T$-pumpable regular language is the domain of the one-way transducer $T^{\prime}$, as defined in Proposition 28. Indeed, $\mathsf{dom}(T^{\prime})$ consists of words $u\in\mathsf{dom}(T)$ that induce successful runs with $\boldsymbol{B}$-decompositions, and the property of having a $\boldsymbol{B}$-decomposition is preserved under pumping.

The following result shows that $T^{\prime}$ is the best under-approximation of $T$ within the class of one-way transducers with $T$-pumpable domains:

Corollary 29**.**

Given a functional two-way transducer $T$, one can construct a one-way transducer $T^{\prime}$ such that

• •

$T^{\prime}\subseteq T$* and $\mathsf{dom}(T^{\prime})$ is $T$-pumpable,*

• •

for all one-way transducers $T^{\prime\prime}$, if $T^{\prime\prime}\subseteq T$ and $\mathsf{dom}(T^{\prime\prime})$ is $T$-pumpable, then $T^{\prime\prime}\subseteq T^{\prime}$.

Proof 8.4**.**

The transducer $T^{\prime}$ is precisely the one defined in Proposition 28. As already explained, its domain $\mathsf{dom}(T^{\prime})$ is a $T$-pumpable language. In particular, $T^{\prime}$ satisfies the conditions in the first item.

For the conditions in the second item, consider a one-way transducer $T^{\prime\prime}\subseteq T$ with a $T$-pumpable domain $L=\mathsf{dom}(T^{\prime\prime})$. Let $\tilde{T}$ be the transducer obtained from $T$ by restricting its domain to $L$. Clearly, $\tilde{T}$ is one-way definable, and one could apply Proposition 24 to $\tilde{T}$, using $T^{\prime\prime}$ as a witness of one-way definability. In particular, when it comes to comparing the outputs of the pumped runs of $\tilde{T}$ and $T^{\prime\prime}$, one could exploit the fact that the domain $L$ of $T^{\prime\prime}$, and hence the domain of $\tilde{T}$ as well, is $T$-pumpable. This permits to derive periodicities of inversions with the same bound $\boldsymbol{B}$ as before, but only restricted to the successful runs of $T$ on the input words that belong to $L$. As a consequence, one can define $\boldsymbol{B}$-decompositions of successful runs of $T$ on words in $L$, thus showing that $L\subseteq\mathsf{dom}(T^{\prime})$. This proves that $T^{\prime\prime}\subseteq T^{\prime}$.

9. Complexity of the one-way definability problem

In this section we analyze the complexity of the problem of deciding whether a transducer $T$ is one-way definable. We begin with the case of a functional two-way transducer. In this case, thanks to the results presented in Section 8 page 8, we know that $T$ is one-way definable if and only if $\mathsf{dom}(T)\subseteq D$, where $D$ is the language of words $u\in\mathsf{dom}(T)$ such that all successful runs of $T$ on $u$ admit a $\boldsymbol{B}$-decomposition. In particular, the one-way definability problem reduces to an emptiness problem for the intersection of two languages:

[TABLE]

The following lemma exploits the characterization of Theorem 6 to show that the language $D^{\complement}$ can be recognized by a non-deterministic finite automaton $A$ of triply exponential size w.r.t. $T$. In fact, this lemma shows that the automaton recognizing $D^{\complement}$ can be constructed using doubly exponential workspace. As before, we gain an exponent when restricting to sweeping transducers.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. Lemma 9.1: D is a poor choice of notation for the automaton given that there already is a language D. Might A (same font as current D) be an adequate choice? size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. Done. BTW. if anyone has changed the macros A, T, etc. mind that now they are rendered as simple capital letters instead as with mathcal. I like this, and it is more uniform than using caligraphic A for automata and normal T for transducers… But we should check we are consistent!

Lemma 30**.**

Given a functional two-way transducer $T$, an NFA $A$ recognizing $D^{\complement}$ can be constructed in $2\textsc{ExpSpace}$. Moreover, when $T$ is sweeping, the NFA $A$ can be constructed in ExpSpace.

Proof 9.1**.**

Consider an input word $u$. By Theorem 6 we know that $u\in D^{\complement}$ iff there exist a successful run $\rho$ of $T$ on $u$ and an inversion $\mathcal{I}=(L_{1},\ell_{1},L_{2},\ell_{2})$ of $\rho$ such that no positive number $p\leq\boldsymbol{B}$ is a period of the word

[TABLE]

The latter condition on $w_{\rho,\mathcal{I}}$ can be rephrased as follows: there is a function $f:\{1,\dots,\boldsymbol{B}\}\rightarrow\{1,\dots,|w_{\rho,\mathcal{I}}|\}$ such that w_{\rho,\mathcal{I}}\big{(}f(p)\big{)}\neq w_{\rho,\mathcal{I}}\big{(}f(p)+p\big{)} for all positive numbers $p\leq\boldsymbol{B}$. In particular, each of the images of the latter function $f$, that is, $f(1),\dots,f(\boldsymbol{B})$, can be encoded by a suitable marking of the crossing sequences of $\rho$. This shows that the run $\rho$, the inversion $\mathcal{I}$, and the function $f$ described above can all be guessed within space $\mathcal{O}(\boldsymbol{B})$: $\rho$ is guessed on-the-fly, the inversion is guessed by marking the anchor points, and for $f$ we only store two symbols and a counter $\leq\boldsymbol{B}$, for each $1\leq i\leq\boldsymbol{B}$. That is, any state of $A$ requires doubly exponential space, resp. simply exponential space, depending on whether $T$ is arbitrary two-way or sweeping.

As a consequence of the previous lemma, the emptiness problem for the language $\mathsf{dom}(T)\cap D^{\complement}$, and thus the one-way definability problem for $T$, can be decided in $2\textsc{ExpSpace}$ or ExpSpace, depending on whether $T$ is two-way or sweeping:

Proposition 31**.**

The problem of deciding whether a functional two-way transducer $T$ is one-way definable is in $2\textsc{ExpSpace}$. When $T$ is sweeping, the problem is in ExpSpace.

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. The transition from 9.2 to 9.3 is a tad abrupt. The kind of argument you make before Corollary 8.7 to explain that this makes your result more robust or tight than I expected in first approach could be useful here. size=,color=orange!30,inlinesize=,color=orange!30,inlinetodo: size=,color=orange!30,inlineG. Done, see below

The last result of the section shows that functional two-way transducers are close to be the largest class for which a characterization of one-way definability is feasible: as soon as we consider arbitrary transducers (including non-functional ones), the problem becomes undecidable.

Proposition 32**.**

The one-way definability problem for non-functional sweeping transducers is undecidable.

Proof 9.2**.**

The proof uses some ideas and variants of constructions provided in [25], concerning the proof of undecidability of the equivalence problem for one-way non-functional transducers.

We show a reduction from the Post Correspondence Problem (PCP). A PCP instance is described by two finite alphabets $\Sigma$ and $\Delta$ and two morphisms $f,g:\Sigma^{*}\rightarrow\Delta^{*}$. A solution of such an instance is any non-empty word $w\in\Sigma^{+}$ such that $f(w)=g(w)$. We recall that the problem of testing whether a PCP instance has a solution is undecidable.

Below, we fix a tuple $\tau=(\Sigma,\Delta,f,g)$ describing a PCP instance and we show how to reduce the problem of testing the non-existence of solutions of $\tau$ to the problem of deciding one-way definability of a relation computed by a sweeping transducer. Roughly, the idea is to construct a relation $B_{\tau}$ between words over a suitable alphabet $\Gamma$ that encodes all the non-solutions to the PCP instance $\tau$ (this is simpler than encoding solutions because the presence of errors can be easily checked). The goal is to have a relation $B_{\tau}$ that (i) can be computed by a sweeping transducer and (ii) coincides with a trivial one-way definable relation when $\tau$ has no solution.

We begin by describing the encodings for the solutions of the PCP instance. We assume that the two alphabets of the PCP instance, $\Sigma$ and $\Delta$, are disjoint and we use a fresh symbol $\#\not\in\Sigma\cup\Delta$. We define the new alphabet $\Gamma=\Sigma\cup\Delta\cup\{\#\}$ that will serve both as input alphabet and as output alphabet for the transduction. We call encoding any pair of words over $\Gamma$ of the form $(w\cdot u,w\cdot v)$, where $w\in\Sigma^{+}$, $u\in\Delta^{*}$, and $v\in\{\#\}^{*}$. We will write the encodings as vectors to improve readability, e.g., as

[TABLE]

We denote by $E_{\tau}$ the set of all encodings and we observe that $E_{\tau}$ is computable by a one-way transducer (note that this transducer needs $\varepsilon$-transitions). We then restrict our attention to the pairs in $E_{\tau}$ that are encodings of valid solutions of the PCP instance. Formally, we call good encodings the pairs in $E_{\tau}$ of the form

[TABLE]

All the other pairs in $E_{\tau}$ are called bad encodings. Of course, the relation that contains the good encodings is not computable by a transducer. On the other hand, we can show that the complement of this relation w.r.t. $E_{\tau}$ is computable by a sweeping transducer. Let $B_{\tau}$ be the set of all bad encodings. Consider $(w\cdot u,w\cdot\#^{m})\in E_{\tau}$, with $w\in\Sigma^{+}$, $u\in\Delta^{*}$, and $m\in\mathbb{N}$, and we observe that this pair belongs to $B_{\tau}$ if and only if one of the following conditions is satisfied:

1. (1)

$m<|u|$, 2. (2)

$m>|u|$, 3. (3)

$u\neq f(w)$, 4. (4)

$u\neq g(w)$.

We explain how to construct a sweeping transducer $\mathcal{S}_{\tau}$ that computes $B_{\tau}$. Essentially, $\mathcal{S}_{\tau}$ guesses which of the above conditions holds and processes the input accordingly. More precisely, if $\mathcal{S}_{\tau}$ guesses that the first condition holds, then it performs a single left-to-right pass, first copying the prefix $w$ to the output and then producing a block of occurrences of the symbol $\#$ that is shorter than the suffix $u$. This task can be easily performed while reading $u$: it suffices to emit at most one occurrence of $\#$ for each position in $u$, and at the same time guarantee that, for at least one such position, no occurrence of $\#$ is emitted. The second condition can be dealt with by a similar strategy: first copy the prefix $w$, then output a block of $\#$ that is longer than the suffix $u$. To deal with the third condition, the transducer $\mathcal{S}_{\tau}$ has to perform two left-to-right passes, interleaved by a backward pass that brings the head back to the initial position. During the first left-to-right pass, $\mathcal{S}_{\tau}$ copies the prefix $w$ to the output. During the second left-to-right pass, it reads again the prefix $w$, but this time he guesses a factorization of it of the form $w_{1}\>a\>w_{2}$. On reading $w_{1}$, $\mathcal{S}_{\tau}$ will output $\#^{|f(w_{1})|}$. After reading $w_{1}$, $\mathcal{S}_{\tau}$ will store the symbol $a$ and move to the position where the suffix $u$ begins. From there, it will guess a factorization of $u$ of the form $u_{1}\>u_{2}$, check that $u_{2}$ does not begin with $f(a)$, and emit one occurrence of $\#$ for each position in $u_{2}$. The number of occurrences of $\#$ produced in the output is thus $m=|f(w_{1})|+|u_{2}|$, and the fact that $u_{2}$ does not begin with $f(a)$ ensures that the factorizations of $w$ and $u$ do not match, i.e.

[TABLE]

Note that the described behaviour does not immediately guarantee that $u\neq f(w)$. Indeed, it may still happen that $u=f(w)$, but as a consequence $m\neq|u|$. This case is already covered by the first and second condition, so the computation is still correct in the sense that it produces only bad encodings. On the other hand, if $m$ happens to be the same as $|u|$, then $|u|=m\neq|f(w)|$ and thus $u\neq f(w)$. A similar behaviour can be used to deal with the fourth condition.

We have just shown that there is a sweeping non-functional transducer $\mathcal{S}_{\tau}$ that computes the relation $B_{\tau}$ containing all the bad encodings. Note that, if the PCP instance $\tau$ admits no solution, then all encodings are bad, i.e., $B_{\tau}=E_{\tau}$, and hence $B_{\tau}$ is one-way definable. It remains to show that when $\tau$ has a solution, $B_{\tau}$ is not one-way definable. Suppose that $\tau$ has solution $w\in\Sigma^{+}$ and let \big{(}w\cdot u,\>w\cdot\#^{|u|}\big{)} be the corresponding good encoding, where $u=f(w)=g(w)$. Note that every exact repetition of $w$ is also a solution, and hence the pairs \big{(}w^{n}\cdot u^{n},\>w^{n}\cdot\#^{n\cdot|u|}\big{)} are also good encodings, for all $n\geq 1$.

Suppose, by way of contradiction, that there is a one-way transducer $T$ that computes the relation $B_{\tau}$. For every $n,m\in\mathbb{N}$, we define the encoding

[TABLE]

and we observe that $\alpha_{n,m}\in B_{\tau}$ if and only if $n\neq m$ (recall that $w\neq\varepsilon$ is the solution of the PCP instance $\tau$ and $u=f(w)=g(w)$). Below, we consider bad encodings like the above ones, where the parameter $n$ is supposed to be large enough. Formally, we define the set $I$ of all pairs of indices $(n,m)\in\mathbb{N}^{2}$ such that (i) $n\neq m$ (this guarantees that $\alpha_{n,m}\in B_{\tau}$) and (ii) $n$ is larger than the number $|Q|$ of states of $T$.

We consider some pair $(n,m)\in I$ and we choose a successful run $\rho_{n,m}$ of $T$ that witnesses the membership of $\alpha_{n,m}$ in $B_{\tau}$, namely, that reads the input $w^{n}\cdot u^{m}$ and produces the output $w^{n}\cdot\#^{m\cdot|u|}$. We can split the run $\rho_{n,m}$ into a prefix $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{\rho}}_{n,m}$ and a suffix $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\rho}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\rho}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\rho}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\rho}}_{n,m}$ in such a way that $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{\rho}}_{n,m}$ consumes the prefix $w^{n}$ and $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{\rho}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{\rho}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{\rho}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{\rho}}_{n,m}$ consumes the remaining suffix $u^{m}$. Since $n$ is larger than the number of state of $T$, we can find a factor $\hat{\rho}_{n,m}$ of $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{\rho}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{\rho}}_{n,m}$ that starts and ends with the same state and consumes a non-empty exact repetition of $w$, say $w^{n_{1}}$, for some $1\leq n_{1}\leq|Q|$. We claim that the output produced by the factor $\hat{\rho}_{n,m}$ must coincide with the consumed part $w^{n_{1}}$ of the input. Indeed, if this were not the case, then deleting the factor $\hat{\rho}_{n,m}$ from $\rho_{n,m}$ would result in a new successful run that reads $w^{n-n_{1}}\cdot u^{m}$ and produces $w^{n-n_{2}}\cdot\#^{m\cdot|u|}$ as output, for some $n_{2}\neq n_{1}$. This however would contradict the fact that, by definition of encoding, the possible outputs produced by $T$ on input $w^{n-n_{1}}\cdot u^{m}$ must agree on the prefix $w^{n-n_{1}}$. We also remark that, even if we do not annotate this explicitly, the number $n_{1}$ depends on the choice of the pair $(n,m)\in I$. This number, however, range over the fixed finite set J=\big{[}1,|Q|\big{]}.

We can now pump the factor $\hat{\rho}_{n,m}$ of the run $\rho_{n,m}$ any arbitrary number of times. In this way, we obtain new successful runs of $T$ that consume inputs of the form $w^{n+k\cdot n_{1}}\cdot u^{m}$ and produce outputs of the form $w^{n+k\cdot n_{1}}\cdot\#^{m}$, for all $k\in\mathbb{N}$. In particular, we know that $B_{\tau}$ contains all pairs of the form $\alpha_{n+k\cdot n_{1},m}$. Summing up, we can claim the following:

{clm}

There is a function $h:I\rightarrow J$ such that, for all pairs $(n,m)\in I$,

[TABLE]

We can now head towards a contradiction. Let $\tilde{n}$ be the maximum common multiple of the numbers $h(n,m)$, for all $(n,m)\in I$. Let $m=n+\tilde{n}$ and observe that $n\neq m$, whence $(n,m)\in I$. Since $\tilde{n}$ is a multiple of $h(n,m)$, we derive from the above claim that the pair $(n+\tilde{n},m)=(m,m)$ also belongs to $I$. However, this contradicts the definition of $I$, since we observed earlier that $\alpha_{n,m}$ is a bad encoding if and only if $n\neq m$. We conclude that $B_{\tau}$ is not one-way definable when $\tau$ has a solution.

10. Conclusions

It was shown in [19] that it is decidable whether a given two-way transducer can be implemented by some one-way transducer. However, the provided algorithm has non-elementary complexity.

The main contribution of our paper is a new algorithm that solves the above question with elementary complexity, precisely in $2\textsc{ExpSpace}$. The algorithm is based on a characterization of those transductions, given as two-way transducers, that can be realized by one-way transducers. The flavor of our characterization is different from that of [19]. The approach of [19] is based on a variant of Rabin and Scott’s construction [28] of one-way automata, and on local modifications of the two-way run. Our approach instead relies on the global notion of inversion and more involved combinatorial arguments. The size of the one-way transducer that we obtain is triply exponential in the general case, and doubly exponential in the sweeping case, and the latter is optimal. size=,color=red!30,size=,color=red!30,todo: size=,color=red!30,F. add last sentence. Maybe we can do more on future work ? The characterization of sweeping in term of rational relations maybe ?size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. I did not like the sentence in the end, so I moved it here and rephrased (and it was not future work!)The approach described here was adapted to characterize functional two-way transducers that are equivalent to some sweeping transducer with either known or unknown number of passes (see [5], [3] for details).

size=,color=brown!30,inlinesize=,color=brown!30,inlinetodo: size=,color=brown!30,inlineR1. The conclusion gives no clue towards future works, either current, considered or at all possible. size=,color=green!30,inlinesize=,color=green!30,inlinetodo: size=,color=green!30,inlineO. added the following paragraph

Our procedure considers non-deterministic transducers, both for the initial two-way transducer, and for the equivalent one-way transducer, if it exists. size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Improved and mentioned that determinism does not bring much to our problemDeterministic two-way transducers are as expressive as non-deterministic functional ones. This means that starting from deterministic two-way transducers would address the same problem in terms of transduction classes, but could in principle yield algorithms with better complexity. size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. Added this tiny remarkWe also recall that Proposition 5 gives a tight lower bound for the size of a one-way transducer equivalent to a given deterministic sweeping transducer. This latter result basically shows that there is no advantage in considering one-way definability for deterministic variants of sweeping transducers, at least with respect to the size of the possible equivalent one-way transducers.

size=,color=orange!30,size=,color=orange!30,todo: size=,color=orange!30,G. slightly rephrased

A variant of the one-way definability problem asks whether a given two-way transducer is equivalent to some deterministic one-way transducer. A decision procedure for this latter problem is obtained by combining our characterization with the classical algorithm that determines whether a one-way transducer can be determinized [11, 7, 33]. In terms of complexity, it is conceivable that a better algorithm may exist for deciding definability by deterministic one-way transducers, since in this case one can rely on structural properties that characterize deterministic transducers. size=,color=blue!30,size=,color=blue!30,todo: size=,color=blue!30,A. I deleted the last phrase (“on the contrary, our approach etc”) because at this point, the reader should have remembered that we do only guessing.

Acknowledgment

The authors wish to acknowledge fruitful discussions with Emmanuel Filiot, Ismaël Jecker and Sylvain Salvati. We also thank the referees for their very careful reading and the suggestions for improvement.

Appendix A

Here we give a fully detailed proof of Lemma 18, for which we recall the statement below:

Lemma 18. *Let $C$ be a component of a loop $L=[x_{1},x_{2}]$. The nodes of $C$ are precisely the levels in the interval $[\min(C),\max(C)]$. Moreover, if $C$ is left-to-right (resp. right-to-left), then $\max(C)$ is the smallest level $\geq\min(C)$ such that between $(x_{1},\min(C))$ and $(x_{2},\max(C))$ (resp. $(x_{2},\min(C))$ and $(x_{1},\max(C))$) there are equally many ${\mathsf{LL}}$-factors and ${\mathsf{RR}}$-factors intercepted by $L$. *

Proof A.1**.**

To ease the understanding the reader may refer to Figure 12, that shows some factors intercepted by $L$ and the corresponding edges in the flow.

We begin the proof by partitioning the set of levels of the flow into suitable intervals as follows. We observe that every loop $L=[x_{1},x_{2}]$ intercepts equally many ${\mathsf{LL}}$-factors and ${\mathsf{RR}}$-factors. This is so because the crossing sequences at $x_{1},x_{2}$ have the same length $h$. We also observe that the sources of the factors intercepted by $L$ are either of the form $(x_{1},y)$, with $y$ even, or $(x_{2},y)$, with $y$ odd. For any location $\ell\in\{x_{1},x_{2}\}\times\mathbb{N}$ that is the source of an intercepted factor, we define $d_{\ell}$ to be the difference between the number of ${\mathsf{LL}}$-factors and the number of ${\mathsf{RR}}$-factors intercepted by $L$ that end at a location strictly before $\ell$. Intuitively, $d_{\ell}=0$ when the prefix of the run up to location $\ell$ has visited equally many times the position $x_{1}$ and the position $x_{2}$. For the sake of brevity, we let $d_{y}=d_{(x_{1},y)}$ for an even level $y$, and $d_{y}=d_{(x_{2},y)}$ for an odd level $y$. Note that $d_{0}=0$. We also let $d_{h+1}=0$, by convention.

We now consider the numbers $z$’s, with $0\leq z\leq h+1$, such that $d_{z}=0$, that is: $0=z_{0}<z_{1}<\dots<z_{k}=h+1$. Using a simple induction, we prove that for all $i\leq k$, the parity of $z_{i}$ is the same as the parity of its index $i$. The base case $i=0$ is trivial, since $z_{0}=0$. For the inductive case, suppose that $z_{i}$ is even (the case of $z_{i}$ odd is similar). We prove that $z_{i+1}$ is odd by a case distinction based on the type of factor intercepted by $L$ that starts at level $z_{i}$. If this factor is an ${\mathsf{LR}}$-factor, then it ends at the same level $z_{i}$, and hence $d_{z_{i}+1}=d_{z_{i}}=0$, which implies that $z_{i+1}=z_{i}+1$ is odd. Otherwise, if the factor is an ${\mathsf{LL}}$-factor, then for all levels $z$ strictly between $z_{i}$ and $z_{i+1}$, we have $d_{z}>0$, and since $d_{z_{i+1}}=0$, the last factor before $z_{i+1}$ must decrease $d_{z}$, that is, must be an ${\mathsf{RR}}$-factor. This implies that $(x_{2},z_{i+1})$ is the source of an intercepted factor, and thus $z_{i+1}$ is odd.

The levels $0=z_{0}<z_{1}<\dots<z_{k}=h+1$ induce a partition of the set of nodes of the flow into intervals of the form $Z_{i}=[z_{i},z_{i+1}-1]$. To prove the lemma, it is suffices to show that the subgraph of the flow induced by each interval $Z_{i}$ is connected. Indeed, because the union of the previous intervals covers all the nodes of the flow, and because each node has one incoming and one outgoing edge, this will imply that the intervals coincide with the components of the flow.

Now, let us fix an interval of the partition, which we denote by $Z$ to avoid clumsy notation. Hereafter, we will focus on the edges of subgraph of the flow induced by $Z$ (we call it subgraph of $Z$ for short). We prove a few basic properties of these edges. For the sake of brevity, we call ${\mathsf{LL}}$-edges the edges of the subgraph of $Z$ that correspond to the ${\mathsf{LL}}$-factors intercepted by $L$, and similarly for the ${\mathsf{RR}}$-edges, ${\mathsf{LR}}$-edges, and ${\mathsf{RL}}$-edges.

We make a series of assumption to simplify our reasoning. First, we assume that the edges are ordered based on the occurrences of the corresponding factors in the run. For instance, we may say the first, second, etc. ${\mathsf{LR}}$-edge (of the subgraph of $Z$) — from now on, we tacitly assume that the edges are inside the subgraph of $Z$. Second, we assume that the first edge of the subgraph of $Z$ starts at an even node, namely, it is an ${\mathsf{LL}}$-edge or an ${\mathsf{LR}}$-edge (if this were not the case, one could apply symmetric arguments to prove the lemma). From this it follows that the subgraph contains $n$ ${\mathsf{LR}}$-edges interleaved by $n-1$ ${\mathsf{RL}}$-edges, for some $n>0$. Third, we assume that $\min(Z)=0$, in order to avoid clumsy notations (otherwise, we need to add $\min(Z)$ to all the levels considered hereafter).

Now, we observe that, by definition of $Z$, there are equally many ${\mathsf{LL}}$-edges and ${\mathsf{RR}}$-edges: indeed, the difference between the number of ${\mathsf{LL}}$-edges and the number of ${\mathsf{RR}}$-edges at the beginning and at the end of $Z$ is the same, namely, $d_{z}=0$ for both $z=\min(Z)$ and $z=\max(Z)$. It is also easy to see that the ${\mathsf{LL}}$-edges and the ${\mathsf{RR}}$-edges are all of the form $y\rightarrow y+1$, for some level $y$. We call these edges incremental edges.

For the other edges, we denote by $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ (resp. $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$) the source level of the $i$-th ${\mathsf{LR}}$-edge (resp. the $i$-th ${\mathsf{RL}}$-edge). Clearly, each $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ is even, and each $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$ is odd, and $i\leq j$ implies $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}<\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}$ and $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}<\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j}$. Consider the location $(x_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i})$, which is the source of the $i$-th ${\mathsf{LR}}$-edge (e.g. the edge in blue in the figure). The latest location at position $x_{2}$ that precedes $(x_{1},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i})$ must be of the form $(x_{2},\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1})$, provided that $i>1$. This implies that, for all $1<i\leq n$, the $i$-th ${\mathsf{LR}}$-edge is of the form $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$. For $i=1$, we recall that $\min(Z)=0$ and observe that the first location at position $x_{2}$ that occurs after the location $(x_{1},0)$ is $(x_{2},0)$, and thus the first ${\mathsf{LR}}$-edge has a similar form: $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{1}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1$, where $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}=-1$ by convention.

Using symmetric arguments, we see that the $i$-th ${\mathsf{RL}}$-edge (e.g. the one in red in the figure) is of the form $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$. In particular, the last ${\mathsf{LR}}$-edge starts at the level $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}=\max(Z)$.

Summing up, we have just seen that the edges of the subgraph of $Z$ are of the following forms:

• •

$y\rightarrow y+1$*                        (incremental edges),*

• •

$\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$*                        ($i$-th ${\mathsf{LR}}$-edge, for $i=1,\dots,n$),*

• •

$\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$*                        ($i$-th ${\mathsf{RL}}$-edge, for $i=1,\dots,n-1$).*

In addition, we have $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1=\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}+2d_{\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}}$. Since $d_{z}>0$ for all $\min(Z)<z<\max(Z)$, this implies that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}>\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$.

The goal is to prove that the subgraph of $Z$ is strongly connected, namely, it contains a cycle that visits all its nodes. As a matter of fact, because components are also strongly connected subgraphs, and because every node in the flow has in-/out-degree $1$, this will imply that the considered subgraph coincides with a component $C$, thus implying that the nodes in $C$ form an interval. Towards this goal, we will prove a series of claims that aim at identifying suitable sets of nodes that are covered by paths in the subgraph of $Z$. Formally, we say that a path covers a set $Y$ if it visits all the nodes in $Y$, and possibly other nodes. As usual, when we talk of edges or paths, we tacitly understand that they occur inside the subgraph of $Z$. On the other hand, we do not need to assume $Y\subseteq Z$, since this would follow from the fact that $Y$ is covered by a path inside $Z$. For example, the right hand-side of Figure 12 shows a path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$ that covers the set $Y=\{\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i},\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1\}\cup[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}]$.

The covered sets will be intervals of the form

[TABLE]

Note that the sets $Y_{i}$ are well-defined for all $i=1,\dots,n-1$, but not for $i=n$ since $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n}$ is not defined either (the subgraph of $Z$ contains only $n-1$ ${\mathsf{RL}}$-edges).

{clm}

For all $i=1,\dots,n-1$, there is a path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$ that covers $Y_{i}$ (for short, we call it an incremental path).

Proof A.2**.**

We prove the claim by induction on $i$. The base case $i=1$ is rather easy. Indeed, we recall the convention that $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1=\min(Z)=0$. In particular, the node $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1$ is the target of the first ${\mathsf{LR}}$-edge of the subgraph of $Z$. Before this edge, according to the order induced by the run, we can only have ${\mathsf{LL}}$-edges of the form $y\rightarrow y+1$, with $y=0,2,\dots,\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{1}-2$. Similarly, after the ${\mathsf{LR}}$-edge we have ${\mathsf{RR}}$-edges of the form $y\rightarrow y+1$, with $y=1,3,\dots,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{1}-2$. Those incremental edges can be connected to form the path $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1\rightarrow^{*}\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{1}$ that covers the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{1}]$ . By prepending to this path the ${\mathsf{LR}}$-edge $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{1}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1$, and by appending the ${\mathsf{RL}}$-edge $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{1}\rightarrow\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{1}+1$, we get a path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{1}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{1}+1$ that covers the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{1}]$. The latter interval is precisely the set $Y_{1}$.

For the inductive step, we fix $1<i<n$ and we construct the desired path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$. The initial edge of this path is defined to be the ${\mathsf{LR}}$-edge $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$. Similarly, the final edge of the path will be the ${\mathsf{RL}}$-edge $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$, which exists since $i<n$. It remains to connect $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$ to $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$. For this, we consider the edges that depart from nodes strictly between $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}$ and $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$.

Let $y$ be an arbitrary node in $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}-1]$. Clearly, $y$ cannot be of the form $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j}$, for some $j$, because it is strictly between $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}$ and $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$. So $y$ cannot be the source of an ${\mathsf{RL}}$-edge. Moreover, recall that the ${\mathsf{LL}}$-edges and the ${\mathsf{RR}}$-edges are the of the form $y\rightarrow y+1$. As these incremental edges do not pose particular problems for the construction of the path, we focus mainly on the ${\mathsf{LR}}$-edges that depart from nodes inside $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}-1]$.

Let $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j-1}+1$ be such an ${\mathsf{LR}}$-edge, for some $j$ such that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}\in[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}-1]$. If we had $j\geq i$, then we would have $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}\geq\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}>\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$, but this would contradict the assumption that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}\in[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}-1]$. So we know that $j<i$. This enables the use of the inductive hypothesis, which implies the existence of an incremental path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j}+1$ that covers the interval $Y_{j}$.

Finally, by connecting the above paths using the incremental edges, and by adding the initial and final edges $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$ and $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$, we obtain a path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$. It is easy to see that this path covers the interval $Y_{i}$.

Next, we define

[TABLE]

We prove a claim similar to the previous one, but now aiming to cover $Y$ with a cycle. Towards the end of the proof we will argue that the set $Y$ coincides with the full interval $Z$, thus showing that there is a component $C$ whose set of notes is precisely $Z$.

{clm}

There is a cycle that covers $Y$.

Proof A.3**.**

It is convenient to construct our cycle starting from the last ${\mathsf{LR}}$-edge, that is, $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1$, since this will cover the upper node $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}=\max(Z)$. From there we continue to add edges and incremental paths, following an approach similar to the proof of the previous claim, until we reach the node $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}$ again. More precisely, we consider the edges that depart from nodes strictly between $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}$ and $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}$. As there are only $n-1$ ${\mathsf{RL}}$-edges, we know that every node in the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1,\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}-1]$ must be source of an ${\mathsf{LL}}$-edge, an ${\mathsf{RR}}$-edge, or an ${\mathsf{LR}}$-edge. As usual, incremental edges do not pose particular problems for the construction of the cycle, so we focus on the ${\mathsf{LR}}$-edges. Let $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1$ be such an ${\mathsf{LR}}$-edge, with $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\in[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1,\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}-1]$. Since $i<n$, we know from the previous claim that there is a path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}+1$ that covers $Y_{i}$. We can thus build a cycle $\pi$ by connecting the above paths using the incremental edges and the ${\mathsf{LR}}$-edge $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}\rightarrow\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1$.

By construction, the cycle $\pi$ covers the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1,\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}]$, and for every $i<n$, if $\pi$ visits $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$, then $\pi$ covers $Y_{i}$. So to complete the proof — namely, to show that $\pi$ covers the entire set $Y$ — it suffices to prove that $\pi$ visits each node $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$, with $i<n$.

Suppose, by way of contradiction, that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ is the node with the highest index $i<n$ that is not visited by $\pi$. Recall that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}>\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$. This shows that

[TABLE]

As we already proved that $\pi$ covers the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1,\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}]$, we know that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\in[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j+1}]$ for some $j$ with $i\leq j<n-1$. Now recall that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ is the highest node that is not visited by $\pi$. This means that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j+1}$ is visited by $\pi$. Moreover, since $j+1<n$, we know that $\pi$ uses the incremental path from $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j+1}$ to $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{j+1}+1$, which covers $Y_{j+1}=[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j+1}]$. But this contradicts the fact that $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}$ is not visited by $\pi$, since $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{i}\in[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{j+1}]$.

We know that the set $Y$ is covered by a cycle of the subgraph of $Z$, and that $Z$ is an interval whose endpoints are consecutive levels $z<z^{\prime}$, with $d_{z}=d_{z^{\prime}}=0$. For the homestretch, we prove that $Y=Z$. This will imply that the nodes of the cycle are precisely the nodes of the interval $Z$. Moreover, because the cycle must coincide with a component $C$ of the flow (recall that all the nodes have in-/out-degree $1$), this will show that the nodes of $C$ are precisely those of $Z$.

To prove $Y=Z$ it suffices to recall its definition as the union of the interval $[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n-1}+1,\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}]$ with the sets $Y_{i}$, for all $i=1,\dots,n-1$. Clearly, we have that $Y\subseteq Z$. For the converse inclusion, we also recall that $\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{0}+1=0=\min(Z)$ and $\mathchoice{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\displaystyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\textstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptstyle{y}}{\overarrow@\arrowfill@\relbar\relbar\shortrightarrow\scriptscriptstyle{y}}_{n}=\max(Z)$. Consider an arbitrary level $z\in Z$. Clearly, we have either $z\leq\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$, for some $1\leq i<n$, or $z>\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{n}$. In the former case, by choosing the smallest index $i$ such that $z\leq\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}$, we get $z\in[\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i-1}+1,\mathchoice{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\displaystyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\textstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptstyle{y}}{\overarrow@\arrowfill@\shortleftarrow\relbar\relbar\scriptscriptstyle{y}}_{i}]$, whence $z\in Y_{i}\subseteq Y$. In the latter case, we immediately have $z\in Y$, by construction.