Finite-state Independence and Normal Sequences

TL;DR

This paper explores finite-state independence of normal sequences, providing characterizations via automata and an algorithm to construct independent normal words.

Contribution

It introduces three new characterizations of finite-state independence for normal words and presents an algorithm for constructing such independent sequences.

Findings

Three characterizations of finite-state independence using different automata models

An algorithm to construct pairs of finite-state independent normal words

Enhanced understanding of the structure of normal sequences

Abstract

We consider the previously defined notion of finite-state independence and we focus specifically on normal words. We characterize finite-state independence of normal words in three different ways, using three different kinds of asynchronous deterministic finite automata with two input tapes containing infinite words. Based on one of the characterizations we give an algorithm to construct a pair of finite-state independent normal words.

Equations

$$\mathrm{01234567891011121314151617181920212223...}$$

$$|w{|}_u$$

$$||w|{|}_u$$

$$\underset{n\to \infty }{\lim }\frac{||x[\mathrm{1..}(n\ell )]|{|}_u}{n}=|A{|}^{-\ell }.$$

$$q_0\stackrel{u_{1,1},\dots ,u_{k,1}}{\to }{q}_1\stackrel{u_{1,2},\dots ,u_{k,2}}{\to }{q}_2\cdots q_{n-1}\stackrel{u_{1,n},\dots ,u_{k,n}}{\to }{q}_n.$$

$$q_0\stackrel{u_{1,1},\dots ,u_{k,1}}{\to }{q}_1\stackrel{u_{1,2},\dots ,u_{k,2}}{\to }{q}_2\stackrel{u_{1,3},\dots ,u_{k,3}}{\to }{q}_3\cdots$$

$$p\stackrel{u_1,\dots ,u_{\ell }|v_{\ell +1},\dots ,v_k}{\to }q$$

$${\rho }_{\mathcal{A}}(x)=\underset{n\to \infty }{\mathrm{liminf}}\frac{|v_1{v}_2\cdots v_n|}{|u_1{u}_2\cdots v_n|}.$$

$$\rho (x)=\inf \{{\rho }_{\mathcal{A}}(x):\text{A is a one-to-one 1-deterministic 2-automaton}\}$$

$$q_0\stackrel{u_1,v_1|w_1}{\to }{q}_1\stackrel{u_2,v_2|w_2}{\to }{q}_2\stackrel{u_3,v_3|w_3}{\to }{q}_3\cdots$$

$${\rho }_{\mathcal{A}}(x/y)=\underset{n\to \infty }{\mathrm{liminf}}\frac{|w_1{w}_2{w}_3\cdots |}{|u_1{u}_2{u}_3\cdots |}.$$

$$q_0\stackrel{{\bar{a}}_1,{\bar{b}}_1}{\to }{q}_1\stackrel{{\bar{a}}_2,{\bar{b}}_2}{\to }{q}_2\stackrel{{\bar{a}}_3,{\bar{b}}_3}{\to }{q}_3\cdots$$

$$\{i:0\le i\le n-1,q_i=q\}.$$

$$\{i:1\le i\le n,q_{i-1}\stackrel{{\bar{a}}_i,{\bar{b}}_i}{\to }{q}_i=\tau \}.$$

$$M=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 1& 0\end{array}\right)$$

$$\underset{n\to \infty }{\lim }\frac{|\gamma [\mathrm{1..}n]{|}_q}{n}=\pi (q).$$

$$\underset{n\to \infty }{\lim }\frac{|\gamma [\mathrm{1..}n]{|}_{\tau }}{n}=\underset{n\to \infty }{\lim }\frac{|\gamma [\mathrm{1..}n]{|}_{{\tau }^{\prime }}}{n}.$$

$$\hat{E}=\{{\tau }_0\stackrel{\bar{a},\bar{b}}{\to }\tau :q_0\in I,\tau =q_0\stackrel{\bar{a},\bar{b}}{\to }q\}\cup \{{\tau }^{\prime }\stackrel{\bar{a},\bar{b}}{\to }\tau :{\tau }^{\prime }=r\stackrel{\bar{c},\bar{d}}{\to }p,\tau =p\stackrel{\bar{a},\bar{b}}{\to }q\}.$$

$$q_0\stackrel{{\bar{a}}_1,{\bar{b}}_1}{\to }{q}_1\stackrel{{\bar{a}}_2,{\bar{b}}_2}{\to }{q}_2\stackrel{{\bar{a}}_3,{\bar{b}}_3}{\to }{q}_3\cdots$$

$${\tau }_0\stackrel{{\bar{a}}_1,{\bar{b}}_1}{\to }{\tau }_1\stackrel{{\bar{a}}_2,{\bar{b}}_2}{\to }{\tau }_2\stackrel{{\bar{a}}_3,{\bar{b}}_3}{\to }{\tau }_3\cdots$$

$$v_n(q)=|\gamma [\mathrm{1..}{k}_n]{|}_q/k_n$$

$$v_n^{\prime }(q)=|\gamma [\mathrm{2..}{k}_n+1]{|}_q/k_n.$$

$$(q,u,\lambda )\stackrel{a,\lambda }{\to }(q,ua,\lambda )$$

$$(q,u,v)\stackrel{\lambda ,b}{\to }(q,u,vb)$$

$$(q,a{u}^{\prime },v)\stackrel{a^{\prime },\lambda }{\to }(q,u^{\prime }{a}^{\prime },v)$$

$$(q,u,b{v}^{\prime })\stackrel{\lambda ,b^{\prime }}{\to }(q,u,v^{\prime }{b}^{\prime })$$

$$(q,u,v)\stackrel{uw{u}^{\prime \prime }{u}^{\prime },v{v}^{\prime \prime }{v}^{\prime }}{\to }(q^{\prime },u^{\prime },v^{\prime })$$

$$|\{a:p\stackrel{a,\lambda }{\to }q\}|/|A|\text{ or }|\{b:p\stackrel{\lambda ,b}{\to }q\}|/|B|$$

$$z=\overline{001}\underline{01}\overline{1}\underline{0001}\overline{01}\underline{1}\overline{0001}\underline{0001}\cdots$$

$$\underset{n\to \infty }{\lim }\frac{|\gamma [\mathrm{1..}n]{|}_{\sigma }}{n}\ne \underset{n\to \infty }{\lim }\frac{|\gamma [\mathrm{1..}n]{|}_{{\sigma }^{\prime }}}{n}.$$

$$\pi (\tau )=\{\begin{array}{ll}\underset{n\to \infty }{\lim }\frac{|\gamma [\mathrm{1..}{k}_n]{|}_{\tau }}{|\gamma [\mathrm{1..}{k}_n]{|}_q}& \text{if limn→∞∣γ[1..kn]∣q/kn=0}\\ \frac{1}{2}& \text{otherwise}\end{array}$$

$$\sum _{\text{τ starts at q}}\pi (\tau )=1.$$

Full text

1 Introduction and statement of results

As defined by Émile Borel [7], for an alphabet with at least two symbols, an infinite word $x$ is normal if all blocks of symbols of the same length occur in $x$ with the same limiting frequency. The most famous normal word was given by Champernowne in [10],

[TABLE]

Borel showed that almost all words are normal. In [4] we introduced the notion of finite-state independence for pairs of infinite words and we showed that almost all pairs of normal words are finite-state independent.

In this work we characterize the notion finite-state-independence specifically for normal words, in terms of computations in deterministic asynchronous finite automata with two input tapes. We give three characterizations.

For the first characterization we consider the notion of fairness of a run in a given finite automaton for a given pair of input words. A run is fair if the frequency of each state is determined by the stationary distribution associated with the automaton, hence not determined by the input words. This notion of fairness can also be phrased in terms of frequencies of edges leaving each state.

The second characterization considers selectors, which are finite automata with two input tapes and one output tape such that the symbols in the output tape are obtained by a selection of the symbols in the first input tape, while the symbols in the second input tape act as a consultative oracle. We require that the selector be oblivious which means that whether a symbol is selected or not does not depend on its value. This characterization of finite-state independence of normal words extends Agafonov’s [1] characterization of normality based on selection by finite automata.

The third characterization considers shufflers, which are finite automata with two input tapes and one output tape such that, after the run, the output tape contains all the symbols from the two normal words but shuffled. The output intercalates symbols from each of the input words, preserving the order in which they appear in the input words.

We can now state the first theorem.

Theorem 1** (Characterization Theorem).**

Let $x$ and $y$ be two normal words respectively on the alphabets $A$ and $B$. The following statements are equivalent.

The words $x$ and $y$ are finite-state independent. 2. 2.

For every deterministic two-tapes finite automaton $\mathcal{A}$, the run on $x$ and $y$ in $\mathcal{A}$ is fair. 3. 3.

For every oblivious selector $\mathcal{S}$, the results $\mathcal{S}(x,y)$ and $\mathcal{S}(y,x)$ are also normal.

Furthermore, if alphabets $A$ and $B$ are equal, the following statement is also equivalent.

For every shuffler $\mathcal{S}$, the result $\mathcal{S}(x,y)$ is also normal.

Based on the characterization of finite-state independence of normal words in terms of shufflers given in Theorem 1, we obtain the following.

Theorem 2**.**

For every alphabet $A$, there is an algorithm that computes a pair of finite-state independent normal words.

The proof exhibits an algorithm that outputs a pair of finite-state independent normal words $(x,y)$ by outputting, at each step, one new symbol extending either the currently computed prefix of $x$ or the currently computed prefix of $y$. Unfortunately, the computational complexity of this algorithm is doubly exponential, which means that to obtain the $n$-th symbol of the pair of finite-state independent normal words the algorithm performs a number of operations that is doubly exponential in $n$. Our construction of a pair of finite-state independent normal words has some similarity with the construction of sequences representing the fractional expansion of absolutely normal numbers (a number is absolutely normal if its fractional expansion in each integer base is a normal word). Our algorithm here has some similarity with Turing’s algorithm for computing absolutely normal numbers [20, 5], which also has doubly exponential computational complexity.

The paper is organized as follows. In Section 2 we present the primary definitions of finite automata, normality and finite-state independence. We devote Section 3 to the notions of fairness, selecting and shuffling. In Section 4 we give the proof of Theorem 1 (Characterization Theorem). Section 5 is devoted to Theorem 2, which gives the announced algorithm to compute a pair of finite-state independent normal words. Finally in section 6 we report some open problems.

2 Primary definitions

Let $A$ be finite set of symbols, that we refer as the alphabet. We write $A^{\omega}$ for the set of all infinite words in alphabet $A$, $A^{*}$ for the set of all finite words, $A^{\leq k}$ for the set of all words of length up to $k$, and $A^{k}$ for the set of words of length exactly $k$. The length of a finite word $w$ is denoted by $|w|$. The empty word is denoted by $\lambda$.

2.1 Normality

We start with some notation. The positions of finite and infinite words are numbered starting at $1$. To denote the symbol at position $i$ of a word $w$ we write $w[i]$ and to denote the substring of $w$ from position $i$ to $j$ we write $w[i..j]$.

Definition 3**.**

For $w$ and $u$ two words, the number $|w|_{u}$ of occurrences of $u$ in $w$ and the number $|\!|w|\!|_{u}$ of aligned occurrences of $u$ in $w$ are respectively given by

[TABLE]

For example, $|aaaaa|_{aa}=4$ and $|\!|aaaaa|\!|_{aa}=2$. Notice that the definition of aligned occurrences has the condition $i=1\mod|u|$ instead of $i=0\mod|u|$, because the positions are numbered starting at $1$. Of course, when a word $u$ is just a symbol, $|w|_{u}$ and $|\!|w|\!|_{u}$ coincide. Counting aligned occurrences of a word of length $r$ over alphabet $A$ is exactly the same as counting occurrences of the corresponding symbol over alphabet $A^{r}$. To be precise, consider alphabet $A$, a length $r$, and an alphabet $B$ with $|A|^{r}$ symbols. The set of words of length $r$ over alphabet $A$ and the set $B$ are isomorphic, as witnessed by the isomorphism $\pi:A^{r}\to B$ induced by the lexicographic order in the respective sets. Thus, for any $w\in A^{*}$ such that $|w|$ is a multiple of $r$, $\pi(w)$ has length $|w|/r$ and $\pi(u)$ has length $1$, as it is just a symbol in $B$. Then, for any $u\in A^{r}$, $|\!|w|\!|_{u}=|\pi(w)|_{\pi(u)}$.

We now present the definition of Borel normality [7] directly on infinite words. An infinite word $x$ is simply normal to word length $\ell$ if, for every $u\in A^{\ell}$,

[TABLE]

An infinite word $x$ is normal if it is simply normal to every word length. There are several other equivalent formulations of normality, they can be read from [2, 8, 13].

2.2 Automata

In this work we consider asynchronous finite automata running on a tuple of infinite words with no accepting condition. A thorough presentation of these automata is in the books [15, 17].

We consider $k$-tape automata, also known as $k$-tape transducers. In the rest of the paper, we use names such as compressors, selectors shufflers or splitters for some subclasses of these automata to emphasize their use. To simplify the presentation, we assume here that the same alphabet $A$ for the $k$ tapes. A $k$-automaton is a tuple ${\mathcal{A}}=\langle Q,A,\delta,I\rangle$, where $Q$ is the finite state set, $A$ is the alphabet, $\delta$ is the transition relation, $I$ the set of initial states. The set of transition relations is a finite subset of $Q\times(A^{*})^{k}\times Q$. A transition is thus a tuple $\langle p,u_{1},\ldots,u_{k},q\rangle$ where $p$ is its starting state, $\langle u_{1},\ldots,u_{k}\rangle$ is its label and $q$ is its ending state A transition is written p\mathchoice{\xrightarrow{u_{1},\ldots,u_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{1},\ldots,u_{k}}}}}{\text{Error}}{\text{Error}}q. As usual, two transitions are consecutive if the ending state of the first one is the starting state of the second one. A finite run is a finite sequence of consecutive transitions

[TABLE]

The label of the run is the component-wise concatenation of the labels of the transitions. More precisely, it is the tuple $\langle v_{1},\ldots,v_{k}\rangle$ where each $v_{j}$ for $1\leq j\leq k$ is equal to $u_{j,1}u_{j,2}\cdots u_{j,n}$. Such a run is written shortly as q_{0}\mathchoice{\xrightarrow{v_{1},\ldots,v_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle v_{1},\ldots,v_{k}}}}}{\text{Error}}{\text{Error}}q_{n}. An infinite run is an infinite sequence of consecutive transitions

[TABLE]

As for the finite case, the label of the infinite run is the component-wise concatenation of the labels of the transitions. More precisely, it is the tuple $\langle x_{1},\ldots,x_{k}\rangle$ where each $x_{j}$ for $1\leq j\leq k$ is equal to $u_{j,1}u_{j,2}u_{j,3}\cdots$. Note that some label $x_{j}$ might be finite although the run is infinite since some transitions may have empty labels. The run is accepting if its first state $q_{0}$ is initial and each word $x_{j}$ is infinite. Such an accepting run is written shortly q_{0}\mathchoice{\xrightarrow{x_{1},\ldots,x_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle x_{1},\ldots,x_{k}}}}}{\text{Error}}{\text{Error}}\infty. The tuple $\langle x_{1},\ldots,x_{k}\rangle$ is accepted if there exists at least one accepting run with label $\langle x_{1},\ldots,x_{k}\rangle$. Notice that there is no constraint on the states occurring infinitely often in an accepting run.

In this work we consider only deterministic $k$-automata whose transition function is determined by a subset of the $k$ tapes. We say that a $k$-automaton is $\ell$-deterministic, with $1\leq\ell\leq k$, if the following two conditions are fulfilled:

the set $I$ of initial states is a singleton set; 2. 2.

for each state $p$, there is an integer $1\leq i\leq\ell$ such that for each transition p\mathchoice{\xrightarrow{u_{1},\ldots,u_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{1},\ldots,u_{k}}}}}{\text{Error}}{\text{Error}}q starting from $p$, $u_{i}$ is a symbol and $u_{1},\ldots,u_{i-1},u_{i+1},\ldots,u_{\ell}$ are empty. Furthermore if p\mathchoice{\xrightarrow{u_{1},\ldots,u_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{1},\ldots,u_{k}}}}}{\text{Error}}{\text{Error}}q and p\mathchoice{\xrightarrow{u^{\prime}_{1},\ldots,u^{\prime}_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u^{\prime}{1},\ldots,u^{\prime}{k}}}}}{\text{Error}}{\text{Error}}q^{\prime} are two transitions starting from $p$, then $u_{i}\neq u^{\prime}_{i}$.

The $\ell$-deterministic automaton is called $\ell$-complete if for each state $p$ and each symbol $a$, there is an integer $i$ (depending only on $p$) and a transition p\mathchoice{\xrightarrow{u_{1},\ldots,u_{k}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{1},\ldots,u_{k}}}}}{\text{Error}}{\text{Error}}q starting from $p$ such that $1\leq i\leq\ell$ and $u_{i}=a$. The $\ell$-determinism guarantees that for each tuple $\langle x_{1},\ldots,x_{\ell}\rangle$ of infinite words, there exists at most one run such that the first $\ell$ components of its label are $\langle x_{1},\ldots,x_{\ell}\rangle$. Even if the automaton is $\ell$-complete, this run might be not accepting since one of its labels might be finite.

The $3$-automaton at the left of Figure 2 accepts a triple $\langle x,y,z\rangle$ of infinite words over the alphabet $\{0,1\}$ whenever $z$ is the join of $x$ and $y$; recall that the join of two infinite words $x=a_{1}a_{2}a_{3}\cdots$ and $y=b_{1}b_{2}b_{3}\cdots$ is the infinite word $z=a_{1}b_{1}a_{2}b_{2}a_{3}\cdots$. This automaton is $2$-deterministic. The $3$-automaton pictured at the right of Figure 2 accepts a triple $\langle x,y,z\rangle$ of infinite words over the alphabet $\{0,1\}$ whenever $z$ is a shuffle of the symbols in $x$ and $y$. This automaton is not $2$-deterministic. Indeed the first condition on transitions is not fulfilled by the two transitions q_{0}\mathchoice{\xrightarrow{0,\lambda,0}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle 0,\lambda,0}}}}{\text{Error}}{\text{Error}}q_{0} and q_{0}\mathchoice{\xrightarrow{\lambda,0,0}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,0,0}}}}{\text{Error}}{\text{Error}}q_{0}.

Let $\mathcal{A}$ be an $\ell$-deterministic $k$-automaton. For each tuple $\langle x_{1},\ldots,x_{\ell}\rangle$ of infinite words, there exists at most one tuple $\langle y_{\ell+1},\ldots,y_{k}\rangle$ of infinite words such that the $k$-tuple $\langle x_{1},\ldots,x_{\ell},y_{\ell+1},\ldots,y_{k}\rangle$ is accepted by $\mathcal{A}$. The automaton $\mathcal{A}$ realizes then a partial function from $(A^{\omega})^{\ell}$ to $(A^{\omega})^{k-\ell}$ and the tuple $\langle y_{\ell+1},\ldots,y_{k}\rangle$ is denoted by $\mathcal{A}(x_{1},\ldots,x_{\ell})$. The $1$-deterministic $2$-automata are also called sequential transducers in the literature. When a $k$-automaton is $\ell$-deterministic, each transition is written

[TABLE]

to emphasize that the first $\ell$ tapes are input tapes and that the $k-\ell$ remaining ones are output tapes.

Let $\mathcal{A}$ be a $1$-deterministic $2$-automaton. We say that $\mathcal{A}$ is a compressor if the (partial) function $x\mapsto\mathcal{A}(x)$ which maps $x$ to the output $\mathcal{A}(x)$ is one-to-one. The compression ratio of an infinite word $x$ for $\mathcal{A}$ is given by the unique accepting run q_{0}\mathchoice{\xrightarrow{u_{1}|v_{1}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{1}|v_{1}}}}}{\text{Error}}{\text{Error}}q_{1}\mathchoice{\xrightarrow{u_{2}|v_{2}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{2}|v_{2}}}}}{\text{Error}}{\text{Error}}q_{2}\mathchoice{\xrightarrow{u_{3}|v_{3}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u_{3}|v_{3}}}}}{\text{Error}}{\text{Error}}q_{3}\cdots where $x=u_{1}u_{2}u_{3}\cdots$ as

[TABLE]

This compression ratio for a given automaton $\mathcal{A}$ can have any non-negative real value. In particular, it can be greater than $1$. An infinite word $x$ is compressible by a $1$-deterministic $2$-automaton $\mathcal{A}$ if $\rho_{\mathcal{A}}(x)<1$. The compression ratio of a given word $x,$ $\rho(x)$, is the infimum of the compression ratios achievable by all one-to-one $1$-deterministic $2$-automata, namely,

[TABLE]

For every infinite word $x$, $\rho(x)$ is less than or equal to $1$, because there exists a compressor $\mathcal{A}_{0}$ which copies each symbol of the input to the output, so $\rho_{\mathcal{A}_{0}}(x)$ is equal to $1$. The compression ratio of the word $x=0^{\omega}$ is $\rho(x)=0$ because for each positive real number $\varepsilon$ there exists a compressor $\mathcal{A}$ such that $\rho_{\mathcal{A}}(x)<\varepsilon$. Notice that in this case the compression ratio equal to [math] is not achievable by any compressor $\mathcal{A}$. It follows from the results in [18, 11] that the words $x$ with compression ratio $\rho(x)$ equal to $1$ are the exactly the normal words. A direct proof of this result appears in [2, Characterization Theorem].

2.3 Finite-state independence

Roughly, two infinite words, possibly over different alphabets, are finite-state independent if none of them helps to compress the other using $3$-automata. In our setting, a compressor is a $2$-deterministic $3$-automata $\mathcal{A}$ such that for any fixed infinite word $y$, the function $x\mapsto\mathcal{A}(x,y)$ which maps $x$ to the output $\mathcal{A}(x,y)$ is one-to-one. This guarantees that if $y$ is known, $x$ can be recovered from $\mathcal{A}(x,y)$. Note that we do not require that the function $(x,y)\mapsto\mathcal{A}(x,y)$ be one-to-one, which would be a much stronger assumption. For example, the $2$-deterministic $3$-automaton $\mathcal{A}$ which maps the infinite words $x$ and $y$ to the infinite word $z$ satisfying $z[i]=x[i]+y[i]\mod|A|$ for each $i\geq 1$ is, indeed a compressor but the function $(x,y)\mapsto\mathcal{C}(x,y)$ is not one-to-one.

Definition 4** ([4]).**

Let $\mathcal{A}$ be a compressor. For simplicity in the presentation we assume just one alphabet. However, it is possible to have three different alphabets, one for each input tape and one for the output tape. The conditional compression ratio of an infinite word $x$ with respect to $y$ in $\mathcal{A}$ is given by the unique accepting run

[TABLE]

such that $x=u_{1}u_{2}u_{3}\cdots$ and $y=v_{1}v_{2}v_{3}\ldots$ as

[TABLE]

In case the input tape and the output tape have respective alphabets $A$ and $B$ of different sizes, the formula above should be multiplied by $\log|A|/\log|B|$. Notice that the number of symbols read from $y$, namely $|v_{1}v_{2}v_{3}\cdots|$, is not taken into account in the value of $\rho_{\mathcal{A}}(x/y)$.

The conditional compression ratio of an infinite word $x$ given an infinite word $y$, $\rho(x/y)$, is the infimum of the compression ratios $\rho_{\mathcal{A}}(x/y)$ of all compressors $\mathcal{A}$ with input $x$ and oracle $y$.

Definition 5** ([4]).**

Two infinite words $x$ and $y$, possibly over different alphabets, are finite-state independent if $\rho(x/y)=\rho(x)$, $\rho(y/x)=\rho(y)$ and the compression ratios of $x$ and $y$ are non-zero.

Notice that the compression ratios of $x$ and $y$ should not be zero. This means that a word $x$ such that $\rho(x)=0$ is finite-state independent of no word. Without this requirement, two words $x$ and $y$ such that $\rho(x)=\rho(y)=0$ would be finite-state independent. In particular, each word $x$ with $\rho(x)=0$ would be finite-state independent of itself. From the definition of finite-state independence follows that, if the infinite words $x$ and $y$ are finite-state independent, each suffix of $x$ is finite-state independent of each suffix of $y$.

Finite-state independence for a pair of normal words differs from the classical notion of normality for dimension $2$ also known as joint normality [13]. When two normal words are finite-state independent then they are also jointly normal, but the reverse implication fails. A witness for this appears in [4] with two normal words $x$ and $y$ such that $x$ is identical to the intercalation of the symbols of $x$ and $y$ (in our construction the sequence $x$ satisfies that for every position $n$, $x[n]=x[2n]$). Thus, from the normality of $x$ follows that $x$ and $y$ are jointly normal. However, given $x$ we can obtain $y$ as the subsequence of $x$ in the odd positions, hence $x$ and $y$ are not finite-state independent. This already suggests that the concept of finite-state independence can not be obtained with synchronous automata.

Although finite-state independence of normal words is more demanding than joint normality, it still holds that almost all pairs of normal words are finite-state independent. This is proved in [4, Theorem 5.1].

3 Fairness, selecting and shuffling

3.1 Fairness

We use the terminology of Markov chains for strongly connected components of an automaton. A strongly connected component of an automaton is called recurrent if any state reachable from it is still in it. It is called transient otherwise. By extension, a state is called recurrent (respectively, transient) whenever it belongs to a recurrent (respectively, transient) strongly connected component. Let $\mathcal{A}$ be a $2$-deterministic $2$-automaton and let $x$ and $y$ be two infinite words, possibly over different alphabets. Let $\gamma$ be the run of $\mathcal{A}$ on $x$ and $y$

[TABLE]

where each $\bar{a}_{i}$ and each $\bar{b}_{i}$ is either a symbol or the empty word and each q_{i-1}\mathchoice{\xrightarrow{\bar{a}_{i},\bar{b}_{i}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\bar{a}{i},\bar{b}{i}}}}}{\text{Error}}{\text{Error}}q_{i} is a transition of $\mathcal{A}$. With a slight abuse of notation let $|\gamma[1..n]|_{q}$ denote the number of occurrences of the state $q$ in the first $n$ states of $\gamma$. More precisely, this is the cardinality of the set

[TABLE]

Similarly, for each transition \tau=p\mathchoice{\xrightarrow{\bar{a},\bar{b}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\bar{a},\bar{b}}}}}{\text{Error}}{\text{Error}}q let $|\gamma[1..n]|_{\tau}$ denote the number of occurrences of $\tau$ in the first $n$ transitions of $\gamma$. More precisely, this is the cardinality of the set

[TABLE]

We first introduce the notion of fairness for states. It is based on on a notion of stationary distribution of an automaton which is now defined. We associate with $2$-deterministic and $2$-complete $2$-automaton $\mathcal{A}$ a Markov chain described by a stochastic matrix $M$. Let $A$ and $B$ be the alphabets for the first and second tape of $\mathcal{A}$. The state set of the Markov chain is the state set $Q$ of $\mathcal{A}$. The dimension of the matrix $M$ is thus the number $|Q|$ of states and its rows and columns are indexed by element of $Q$. For two states $p$ and $q$, the $(p,q)$-entry of $M$ is the sum of the weights of all transitions from $p$ to $q$ where the weights are as follows. The weight of a transition of the form p\mathchoice{\xrightarrow{a,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda}}}}{\text{Error}}{\text{Error}}q (respectively p\mathchoice{\xrightarrow{\lambda,b}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,b}}}}{\text{Error}}{\text{Error}}q) is $1/|A|$ (respectively $1/|B|$).

If the automaton $\mathcal{A}$ is strongly connected then the Markov chain is irreducible. By [19, Theorem 1.5], there exists a unique stationary distribution, that is, a line vector $\pi$ such that $\pi M=\pi$ and $\sum_{q\in Q}{\pi(q)}=1$. By definition, this vector is called the stationary distribution of the automaton $\mathcal{A}$. For example, the matrix of the associated Markov chain for the $2$-automaton in Figure 3 is the $2\times 2$-matrix $M$ given by

[TABLE]

and the stationary distribution is thus given by $\pi(q_{1})=2/3$ and $\pi(q_{2})=1/3$.

If the automaton $\mathcal{A}$ is not strongly connected, the stationary distribution $\pi$ of $\mathcal{A}$ is defined as follows. For each transient state $q$, $\pi(q)$ is equal to [math]. For any recurrent state $q$, $\pi(q)=\hat{\pi}(q)$ where $\hat{\pi}$ is the stationary distribution of the strongly connected component of $q$, considered as a whole automaton. This is well-defined because this stationary distribution only depends on the edges of the automaton and not on its initial and final states.

Let $\mathcal{A}$ be a $2$-deterministic $2$-automaton and let $x$ and $y$ be two infinite words, possibly over different alphabets. Let \gamma=q_{0}\mathchoice{\xrightarrow{\bar{a}_{1},\bar{b}_{1}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\bar{a}{1},\bar{b}{1}}}}}{\text{Error}}{\text{Error}}q_{1}\mathchoice{\xrightarrow{\bar{a}_{2},\bar{b}_{2}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\bar{a}{2},\bar{b}{2}}}}}{\text{Error}}{\text{Error}}q_{2}\cdots be the run of $\mathcal{A}$ on $x$ and $y$. This run is called fair for states if for any state $q$ which occurs in $\gamma$,

[TABLE]

The run $\gamma$ is called fair for edges if for any pair of transitions $\tau$ and $\tau^{\prime}$ starting from the state

[TABLE]

Let $\mathcal{A}$ be a $2$-deterministic $2$-automaton. By analogy with the line graph, the line automaton of $\mathcal{A}$ is the automaton $\hat{\mathcal{A}}$ whose states are the transitions of $\mathcal{A}$. More formally, its state set is $\hat{Q}=E\cup\{\tau_{0}\}$ where $E$ is the set of transitions of $\mathcal{A}$ and $\tau_{0}$ is a fresh element (it does not belong to $E$) being the initial state. Its set $\hat{E}$ of transitions is given by

[TABLE]

There is a tight correspondence between runs in $\mathcal{A}$ and runs in $\hat{\mathcal{A}}$. To each run in $\mathcal{A}$

[TABLE]

starting from the initial state $q_{0}$ of $\mathcal{A}$ corresponds the run

[TABLE]

where $\tau_{0}$ is the fresh initial state of $\hat{\mathcal{A}}$ and $\tau_{i}$ is the transition q_{i-1}\mathchoice{\xrightarrow{\bar{a}_{i},\bar{b}_{i}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\bar{a}{i},\bar{b}{i}}}}}{\text{Error}}{\text{Error}}q_{i} for each $i\geq 1$. Conversely, each run in $\hat{\mathcal{A}}$ starting from $\tau_{0}$ comes from a run in $\mathcal{A}$. The following lemma relates the stationary distribution of $\hat{\mathcal{A}}$ with the stationary distribution of $\mathcal{A}$.

Lemma 6**.**

The stationary distribution $\hat{\pi}$ of $\hat{\mathcal{A}}$ maps each transition \tau=p\mathchoice{\xrightarrow{\bar{a},\bar{b}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\bar{a},\bar{b}}}}}{\text{Error}}{\text{Error}}q to $\hat{\pi}(\tau)=\pi(p)/n_{p}$ where $\pi$ is the stationary distribution of $\mathcal{A}$ and $n_{p}$ is the number of transitions starting from state $p$ in $\mathcal{A}$.

Proof.

The proof of the Lemma 6 is routine. ∎

Next we relate the fairness for states and the fairness for edges: the two notions are equivalent as long as they hold for all automata. We start with an auxiliary lemma.

Lemma 7**.**

A run which is fair for edges ends in a recurrent strongly connected component.

Proof.

Let $P$ be the subset of states $\{q:\liminf_{n\to\infty}{\frac{|\gamma[1..k_{n}]|_{q}}{n}}>0\}$. The set $P$ cannot be empty because there a finitely many states. The hypothesis implies that every state reachable from a state in $P$ is also in $P$. Since a recurrent state is reachable from any state, $\gamma$ reaches a recurrent state as it was claimed. ∎

Consider the $2$-deterministic $2$-automaton $\mathcal{A}$ pictured in Figure 4 and the infinite words $x=y=0^{\omega}$. The run on $x$ an $y$ in $\mathcal{A}$ is fair for states because both states $q_{0}$ and $q_{1}$ have frequency $1/2$ but it is not fair for edges because the transition q_{0}\mathchoice{\xrightarrow{1,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle 1,\lambda}}}}{\text{Error}}{\text{Error}}q_{1} is never used.

Proposition 8**.**

Let $x$ and $y$ be two infinite words. The run on $x$ and $y$ is fair for states in any $2$-deterministic $2$-automaton if and only if it is fair for edges in any $2$-deterministic $2$-automaton.

Proof.

We first show that fairness for states implies fairness for edges. Let $\mathcal{A}$ be a $2$-deterministic $2$-automaton. This implication follows from the hypothesis applied to the line automaton $\hat{\mathcal{A}}$ and Lemma 6.

We now prove that fairness for states and fairness for edges coincide. Let $\mathcal{A}$ be a $2$-deterministic $2$-automaton and let $\gamma$ be the run on $x$ and $y$ in $\mathcal{A}$. By Lemma 7 the run $\gamma$ visits a recurrent state of $\mathcal{A}$. Therefore, we now assume that $\mathcal{A}$ is strongly connected. To prove the statement about frequencies of states, it is sufficient to show that for each increasing sequence of integers $(k_{n})_{n\geq 0}$ such that $\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{q}/k_{n}}$ exists, this limit is equal to $\pi(q)$. Let $(k_{n})_{n\geq 0}$ be such a sequence. Replace $(k_{n})_{n\geq 0}$ by one of its sub-sequences so that $\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{q}/k_{n}}$ exists for each state $q$. It has already been shown in the previous paragraph that these limits cannot be [math].

We introduce two sequences $(v_{n})_{n\geq 0}$ and $(v^{\prime}_{n})_{n\geq 0}$ of line vectors and a sequence $(M_{n})_{n\geq 0}$ of matrices. For each state $q$, the $q$-entries of the vectors $v_{n}$ and $v^{\prime}_{n}$ are given by

[TABLE]

and

[TABLE]

For each pair of states $p$ and $q$, the $(p,q)$-entry of $M_{n}$ is the sum over all transitions $\tau$ from $p$ to $q$ of the ratio $|\gamma[1..k_{n}]|_{\tau}/|\gamma[1..k_{n}]|_{p}$. A routine check yields that $v_{n}M_{n}=v^{\prime}_{n}$ holds for each integer $n\geq 1$. Both sequences $(v_{n})_{n\geq 0}$ and $(v^{\prime}_{n})_{n\geq 0}$ converge to the same line vector $v$ given by $v(q)=\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{q}/k_{n}}$. From the hypothesis, the sequence $(M_{n})_{n\geq 0}$ converges to the matrix $M$ of the Markov chain associated with $\mathcal{A}$. Taking limits gives that $vM=v$. By the uniqueness of the stationary distribution of $M$, $v(q)=\pi(q)$ holds for each state $q$. ∎

By Proposition 8, the two notions of fairness, fairness for states and fairness for edges are equivalent. This allows us to use the notion of fairness without mentioning which one is meant. The following lemmas on fairness are used in the proof of Theorem 1.

Let $\mathcal{A}$ be $2$-deterministic $2$-automaton and let $k$ and $\ell$ be two positive integers. We introduce a new automaton $\mathcal{A}_{k,\ell}$. Its state set is $Q\times A^{\leq k}\times\{\lambda\}\cup Q\times A^{k}\times B^{\leq\ell}$ and its transitions are defined as follows.

[TABLE]

Note that the states in $Q\times A^{\leq k}\times\{\lambda\}\cup Q\times A^{k}\times B^{<\ell}$ are obviously transient. The purpose of these states is to gather the first $k$ symbols of $x$ and the first $\ell$ symbols of $y$ to reach the state $(q_{0},u,v)$ where $q_{0}$ is the initial state of $\mathcal{A}$ and $u$ and $v$ are the prefixes of $x$ and $y$ of length $k$ and $\ell$ respectively.

Lemma 9**.**

If $\mathcal{A}$ is strongly connected, then the restriction of $\mathcal{A}_{k,\ell}$ to the set $Q\times A^{k}\times B^{\ell}$ is also strongly connected.

Proof.

Let $(q,u,v)$ and $(q^{\prime},u^{\prime},v^{\prime})$ be two states in $Q\times A^{k}\times B^{\ell}$. There exist a word $w$ in $A^{*}\cup B^{*}$ and states $r$ of $\mathcal{A}$ such that either q\mathchoice{\xrightarrow{uw,v}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle uw,v}}}}{\text{Error}}{\text{Error}}r or q\mathchoice{\xrightarrow{u,vw}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u,vw}}}}{\text{Error}}{\text{Error}}r is a finite run in $\mathcal{A}$. By symmetry, it can be assumed that q\mathchoice{\xrightarrow{uw,v}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle uw,v}}}}{\text{Error}}{\text{Error}}r is a finite run in $\mathcal{A}$. Since $\mathcal{A}$ is strongly connected, there exists a run r\mathchoice{\xrightarrow{u^{\prime\prime},v^{\prime\prime}}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle u^{\prime\prime},v^{\prime\prime}}}}}{\text{Error}}{\text{Error}}q^{\prime}. Then

[TABLE]

is a run in $\mathcal{A}_{k,\ell}$. ∎

Lemma 10**.**

If $\mathcal{A}$ is strongly connected and $\pi$ is its stationary distribution, then the stationary distribution of $\mathcal{A}_{k,\ell}$ is given by $\pi(q,u,v)=\pi(q)/|A|^{k}|B|^{\ell}$ for each state $(q,u,v)\in Q\times A^{k}\times B^{\ell}$.

Proof.

Le $M=(m_{p,q})$ be the $Q\times Q$-matrix of $\mathcal{A}$. Each entry $m_{p,q}$ is equal to either

[TABLE]

The vector $\pi$ is the unique vector satisfying $\pi M=\pi$ and $\sum_{q\in Q}{\pi(q)}=1$. Let $(q,u,v)$ be a fixed state. For each transition p\mathchoice{\xrightarrow{a,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda}}}}{\text{Error}}{\text{Error}}q, there is a transition (p,au^{\prime},v)\mathchoice{\xrightarrow{a^{\prime},\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a^{\prime},\lambda}}}}{\text{Error}}{\text{Error}}(q,u,v) where $u=u^{\prime}a^{\prime}$ ($u^{\prime}$ is the prefix of length $k-1$ of $u$ and $a$ is its last symbol). ∎

3.2 Selecting

We present the definition of a selector that we use to characterize finite-state independence of normal words, to be given in Theorem 1. Given a normal infinite word, the problem of selection is how to select symbols from an infinite word so that the word defined by the selected symbols satisfies a designated property. An early result of Wall [21] shows that selecting the symbols of a normal word in the positions given by an arithmetical progression yields again a normal word. Agafonov [1] extended Wall’s result and proved that any selection by finite automata preserves normality (a complete proof can be found in [3, Theorem 7.1]). The selections admitted by Agafonov must be performed by an oblivious $1$-deterministic $2$-automaton. Oblivious means that the choice of selecting or not the next symbol only depends on the current state and not on the next symbol.

Other forms of selection by finite-automata do not preserve normality. For instance [3, Theorem 7.3] shows that the two-sided selection rule “select symbols in between two zeroes” from $x$, does not preserve normality.

In order to characterize finite-state independence we consider selection by a finite automaton from an infinite word, conditioned to another infinite word that can be used in the selection process as an oracle.

Definition 11**.**

A selector is a $2$-deterministic $3$-automaton such that each of its transitions has one of the types p\mathchoice{\xrightarrow{a,\lambda|a}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda|a}}}}{\text{Error}}{\text{Error}}q (type I), p\mathchoice{\xrightarrow{a,\lambda|\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda|\lambda}}}}{\text{Error}}{\text{Error}}q (type II), or p\mathchoice{\xrightarrow{\lambda,b|\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,b|\lambda}}}}{\text{Error}}{\text{Error}}q (type III) for two symbols $a,b\in A$. It is oblivious if all transitions starting at a given state have the same type.

A transition of type p\mathchoice{\xrightarrow{a,\lambda|a}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda|a}}}}{\text{Error}}{\text{Error}}q (type I) copies a symbol from the first input $x$ to the output tape. A transition of the types p\mathchoice{\xrightarrow{a,\lambda|\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda|\lambda}}}}{\text{Error}}{\text{Error}}q (type II) or p\mathchoice{\xrightarrow{\lambda,b|\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,b|\lambda}}}}{\text{Error}}{\text{Error}}q (type III) skips a symbol from either the first input $x$ or the second input $y$. It follows then that the output word $z=\mathcal{S}(x,y)$ is obtained by selecting symbols from $x$. This justifies the terminology.

Since a selector is $2$-deterministic, all transitions starting at a given state either have type I and II or have type III. When it is oblivious it is not possible anymore that two transitions starting at the same state have types I and II. Whether or not a symbol is copied from the first input tape to the output tape only depends on the state and not on the symbol.

The automaton pictured in Figure 5 is an oblivious selector. It selects symbols from the first input $x$ which are at a position where there is a symbol $1$ in the second input $y$.

3.3 Shuffling

We present the definition of a shuffler we use to characterize finite-state independence of normal words in Theorem 1. A general presentation of shufflers can be read in [16]. An infinite word $z$ is the shuffle of $x$ and $y$ if it can be factorized as $z=u_{1}v_{1}u_{2}v_{2}u_{3}\cdots$ where the sequences of words $(u_{i})_{i\geq 1}$ and $(v_{i})_{i\geq 1}$ satisfy $x=u_{1}u_{2}u_{3}\cdots$ and $y=v_{1}v_{2}v_{3}\cdots$. We restrict to shuffles of words on the same alphabet, done by $2$-deterministic $3$-automata. We prove that if $x$ and $y$ are normal words, $x$ and $y$ are finite-state independent exactly when any shuffle of them is also normal. The interleaving of the symbols from $x$ and $y$ must be driven by a deterministic and oblivious automaton reading $x$ and $y$. Here oblivious means that the choice of inserting in the shuffled word $z$ a symbol either from $x$ or from $y$ is only made upon the current state of the automaton and not upon the current symbols read from $x$ and $y$.

Definition 12**.**

A shuffler is a $2$-deterministic $3$-automaton such that each of its transitions has either the type p\mathchoice{\xrightarrow{a,\lambda|a}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda|a}}}}{\text{Error}}{\text{Error}}q (type I) or the type p\mathchoice{\xrightarrow{\lambda,a|a}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,a|a}}}}{\text{Error}}{\text{Error}}q (type II).

Notice that the determinism of a shuffler $\mathcal{S}$ implies that for each of its states $p$, all the transitions leaving $p$ have the same type, either type I or type II. A transition of type I copies a symbol from the first input $x$ to the output and a transition of type II copies a symbol from the second input $y$ to the output. It follows then that the third word $z=\mathcal{S}(x,y)$ is obtained by shuffling $x$ and $y$. This justifies the terminology.

Consider infinite words $x=\overline{0011010001}\cdots$ and $y=\underline{01000110001}\cdots$ and let $\mathcal{S}$ be the shuffler pictured in Figure 6. Then, the infinite word $z=\mathcal{S}(x,y)$ has the form

[TABLE]

where the underlines and the overlines have been added to mark the origin of each symbol.

If two normal words $x$ and $y$ are on different alphabets then, in general, their shuffling $\mathcal{S}(x,y)$ is not normal. For instance, if $x$ and $y$ are words on different alphabets their join is not normal. Thus, we assume now a unique alphabet.

Exchanging the input and output tapes of a shuffler $\mathcal{S}$ gives a $1$-deterministic $3$-automaton that we call the splitter corresponding to $\mathcal{S}$. This is due to the very special form of the transitions of shufflers. If the output $z=\mathcal{S}(x,y)$ of the shuffler $\mathcal{S}$ on inputs $x$ and $y$ is fed to the corresponding splitter, the two outputs are $x$ and $y$. The fact that the corresponding splitter is $1$-deterministic yields the following lemma which really requires that the alphabets on the two tapes are equal.

Lemma 13**.**

Let $\mathcal{S}$ be a shuffler and $q$ one of its states. For each finite word $w$, there is exactly one run of length $|w|$ starting at $q$ and outputting $w$.

4 Proof of the Characterization Theorem

4.1 From independence to fairness and back

We first prove that finite-state independence implies fairness of the run in each $2$-deterministic $2$-automata.

Proof of Theorem 1, (1) implies (2).

For simplicity we assume that $A$ is the binary alphabet $\{0,1\}$ but the proof can easily be extended to the general case. We suppose by contradiction that there is a $2$-deterministic $2$-automaton $\mathcal{A}$ such that the run on $x$ and $y$ in $\mathcal{A}$ is not fair for edges and we claim that $x$ and $y$ are not finite-state independent.

By definition, each transition of $\mathcal{A}$ is of the form p\mathchoice{\xrightarrow{a,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda}}}}{\text{Error}}{\text{Error}}q or q\mathchoice{\xrightarrow{\lambda,b}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,b}}}}{\text{Error}}{\text{Error}}q for some symbols $a$ and $b$. For the rest of the proof, transitions of the form p\mathchoice{\xrightarrow{a,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle a,\lambda}}}}{\text{Error}}{\text{Error}}q are called of type I and transitions of the form p\mathchoice{\xrightarrow{\lambda,b}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,b}}}}{\text{Error}}{\text{Error}}q are called of type II. Since the automaton is deterministic, all the transitions starting at each state $q$ have the same type. A state $q$ is said to be of type I (respectively II) if all transitions starting at $q$ have type I (respectively II).

We suppose that there is exists a state $p$ and two transitions $\sigma$ and $\sigma^{\prime}$ starting from $p$ such that

[TABLE]

meaning that either at one of the two limits does not exist or that they both exist but they are not equal. By symmetry it can be assumed that all transitions starting at $p$, including $\sigma$ and $\sigma^{\prime}$, are of type I.

We show that $x$ can be compressed given $y$. There is a lack of symmetry between $x$ and $y$ because transitions $\sigma$ and $\sigma^{\prime}$ are of type I. By replacing $(k_{n})_{n\geq 0}$ by one of its subsequences, it can be assumed that, for each transition $\tau$, $\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{\tau}/k_{n}}$ exists and that $\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{\sigma}/k_{n}}\neq\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{\sigma^{\prime}}/k_{n}}$. Since the frequency of each state is equal to the sum of the frequencies of the transitions which start at it, the limit $\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{q}/k_{n}}$ exists for each state $q$. Denote this limit by $\pi(q)$.

For each transition $\tau$ starting at a state $q$, let $\pi(\tau)$ be defined as follows.

[TABLE]

Since $\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{\sigma}/k_{n}}\neq\lim_{n\to\infty}{|\gamma[1..k_{n}]|_{\sigma^{\prime}}/k_{n}}$, $\pi(\sigma)\neq\pi(\sigma^{\prime})$. Furthermore, the following equality holds for each state $q$.

[TABLE]

Since $x$ is normal it suffices to show that $\rho(x/y)<1$. Let $\ell$ be a block length to be fix later. Let $\gamma$ be a finite run of length $\ell$, so $\gamma$ is a sequence $\tau_{1}\tau_{2}\cdots\tau_{\ell}$ of $\ell$ consecutive transitions. Let $\pi(\gamma)$ be defined as follows.

[TABLE]

Let $q$ be a state and $\bar{v}$ be a word of length $\ell$. Let $\Gamma_{q,\bar{v}}$ be the set of runs of length $\ell$, starting at $q$ and reading a prefix of $\bar{v}$ on the second tape,

[TABLE]

Notice that the sets $\Gamma_{q,\bar{v}}$ are not always pairwise disjoint. The word $v$ read by the run $\gamma$ on the second tape can be the prefix of several words $\bar{v}$. If $v$ is the prefix of both $\bar{v}$ and $\bar{v}^{\prime}$, then the run $\gamma$ belongs to both $\Gamma_{q,\bar{v}}$ and $\Gamma_{q,\bar{v}^{\prime}}$.

We claim that for each state $q$ and each word $\bar{v}$,

[TABLE]

We prove it by induction on the length of the run, that we call $\ell$. If $\ell=0$, the only run $\gamma\in~{}\Gamma_{q,\bar{v}}$ is the empty run so $\pi(\gamma)=1$. Suppose now that $\ell\geq 1$. We distinguish two cases. First case: the transitions starting at $q$ are of type I. Suppose first that the transitions starting at $q$ are the two transitions \tau_{0}=q\mathchoice{\xrightarrow{0,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle 0,\lambda}}}}{\text{Error}}{\text{Error}}q_{0} and \tau_{1}=q\mathchoice{\xrightarrow{1,\lambda}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle 1,\lambda}}}}{\text{Error}}{\text{Error}}q_{1}. And suppose that $\bar{v}=\bar{v}^{\prime}a$ where $\bar{v}^{\prime}=\bar{v}[1..\ell-1]$ and $a$ is the last symbol of $\bar{v}$. The set $\Gamma_{q,\bar{v}}$ is then equal to the disjoint union $\Gamma_{q,\bar{v}}=\tau_{0}\Gamma_{q_{0},\bar{v}^{\prime}}\cup\tau_{1}\Gamma_{q_{1},\bar{v}^{\prime}}$. The result follows from the inductive hypothesis since $\pi(\Gamma_{q,\bar{v}})=\pi(\tau_{0})\pi(\Gamma_{q_{0},\bar{v}^{\prime}})+\pi(\tau_{1})\pi(\Gamma_{q_{1},\bar{v}^{\prime}})=\pi(\tau_{0})+\pi(\tau_{1})=1$. Second case: the transitions starting at $q$ have type II. Suppose that $\bar{v}=a\bar{v}^{\prime}$ where $a$ is the first symbol of $\bar{v}$ and $\bar{v}^{\prime}=\bar{v}[2..\ell]$. The transition \tau=q\mathchoice{\xrightarrow{\lambda,a}}{\xrightarrow{\smash{\lower 1.0pt\hbox{\scriptstyle\lambda,a}}}}{\text{Error}}{\text{Error}}q^{\prime} is the first transition of each run in $\Gamma_{q,\bar{v}}$ and $\Gamma_{q,\bar{v}}=\tau\Gamma_{q^{\prime},\bar{v}^{\prime}}$. The result follows from the inductive hypothesis since $\pi(\Gamma_{q,\bar{v}})=\pi(\Gamma_{q^{\prime},\bar{v}^{\prime}})=1$. Since $\sum_{\gamma\in\Gamma_{q,\bar{v}}}{\pi(\gamma)}=1$, there exists, for each state $q$ and each word $\bar{v}$, a prefix-free set $P_{q,\bar{v}}=\{w_{\gamma,\bar{v}}:\gamma\in\Gamma_{q,\bar{v}}\}$ such that $|w_{\gamma,\bar{v}}|\leq\lceil-\log\pi(\gamma)\rceil$ holds for each run $\gamma\in\Gamma_{q,\bar{v}}$. These words can be used to define a compressor $\mathcal{C}$ which runs as follows on two inputs. It simulates $\mathcal{A}$ and it has $\ell$ symbols of look ahead on the second tape. For each run $\gamma$ of length $\ell$, the compressor outputs $w_{\gamma,\bar{v}}$ on the third tape. The choice of $w_{\gamma,\bar{v}}$ depends on the look ahead $\bar{v}$.

We finally show that $\rho_{\mathcal{C}}(x/y)<1$. The run $\gamma$ of $\mathcal{A}$ on $x$ and $y$ can be factorized as $\gamma=\gamma_{1}\gamma_{2}\gamma_{3}\cdots$ where each run $\gamma_{i}$ has length $\ell$. The output of the compressor $\mathcal{C}$ is then $w_{\gamma_{1},\bar{v}_{1}}w_{\gamma_{2},\bar{v}_{2}}w_{\gamma_{3},\bar{v}_{3}}\cdots$ where the words $\bar{v}_{1},\bar{v}_{2},\bar{v}_{3},\ldots$ are the corresponding look ahead of $\ell$ symbols. Let $\varepsilon,\delta>0$ be two positive real numbers. Let $n$ be an integer large enough such that $|\gamma[1..k_{n}]|_{\tau}\leq(1+\delta)\pi(q)\pi(\tau)k_{n}$ for each transition $\tau$ starting at $q$. Then,

[TABLE]

Then, for each state $q$,

[TABLE]

and the relation is strict for $q=p$. Since $\pi(p)>0$, for $\varepsilon$ small enough, $\delta$ and $\ell$ can be chosen such that

[TABLE]

We obtain

[TABLE]

Since \sum_{\text{q of type I}}\pi(q) is the limit of the ratio between the number of symbols read from $x$ and the length of the run, we conclude $\rho_{\mathcal{C}}(x/y)<1$. ∎

We now prove that fairness of the run in each $2$-deterministic $2$-automata implies finite-state independence.

Proof of Theorem 1, (2) implies (1).

Let $x$ and $y$ be two normal words such that statement (2) of Theorem 1 holds. We show that $x$ and $y$ are finite-state independent. It is sufficient to show that $x$ cannot be compressed with the help of $y$, since the other incompressibility result is obtained by exchanging the roles of $x$ and $y$.

Let $\mathcal{C}$ be a $2$-deterministic $3$-automaton such that for each $y$, the function $x\mapsto\mathcal{C}(x,y)$ is one-to-one. By Lemma 7, the automaton $\mathcal{C}$ can be assumed to be strongly connected. Let $q_{0}$ be the initial state of $\mathcal{C}$. Let $\gamma$ be the run of $\mathcal{C}$ on $x$ and $y$ and let $z$ be the output of $\mathcal{C}$ along $\gamma$, that is, $z=\mathcal{C}(x,y)$. Let $\varepsilon>0$ be a positive real number. We claim that the compression ratio $\rho_{\mathcal{C}}(x/y)$ satisfies $\rho_{\mathcal{C}}(x/y)>1-\varepsilon$. Since this holds for each $\varepsilon>0$, this shows that $\rho_{\mathcal{C}}(x/y)\geq 1$.

Let $k$ be a positive integer to be fixed later. Since $y$ is normal, there exists a constant $K>0$ such that if $u\sqsubset x$, $v\sqsubset y$ and $w\sqsubset z$ ($u$, $v$ and $w$ are prefixes of $x$, $y$ and $z$ respectively) such that

[TABLE]

then $|v|\leq K|u|$, see [4, Lemma 5.3]. The run $\gamma$ is decomposed

[TABLE]

where $|u_{i}|=k$ for each integer $i\geq 1$. Note that the lengths of each word $v_{i}$ and each word $w_{i}$ are arbitrary. Our aim is to prove that for $N$ large enough $|w_{1}\cdots w_{N}|\geq(1-4\varepsilon)|u_{1}\cdots u_{N}|$.

Let $\ell$ be the integer $\lceil kK/\varepsilon\rceil$. By definition of $\ell$, the cardinality of the set $\{i\leq N:|v_{i}|>\ell\}$ is less than $\varepsilon N$. Otherwise we would have $|v_{1}\cdots v_{N}|>K|u_{1}\cdots u_{N}|$ which contradicts the definition of the constant $K$. The indices $i$ such that $|v_{i}|>\ell$ are ignored in the sequel.

Let $v^{\prime}_{i}$ be the prefix of length $\ell$ of the infinite word $v_{i}v_{i+1}v_{i+2}\cdots$. Unless $|v_{i}|>\ell$, $v_{i}$ is a prefix of $v^{\prime}_{i}$. Let $v^{\prime}\in B^{\ell}$ be a fixed word of length $\ell$. We claim that the cardinality of the set

[TABLE]

is bounded by $(\ell+1)|Q|^{2}|A|^{k(1-\varepsilon)}$. For each choice of $p$, $q$, $v$ and $w$, there is at most one possible $u$. Otherwise, the function $x\mapsto\mathcal{C}(x,y)$ would not be one-to-one. The terms $(\ell+1)$, $|Q|^{2}$ and $|A|^{k(1-\varepsilon)}$ account respectively for the number of choices for $v$, $p$ and $q$, and $w$. Note that the number of choices of $v$ is $\ell+1$ because $v$ is a prefix of the fixed word $v^{\prime}$ of length $\ell$. The integer $k$ is chosen such that $|A|^{k}-(\ell+1)|Q|^{2}|A|^{k(1-\varepsilon)}$ is greater than $(1-\varepsilon)|A|^{k}$. This is possible because $|Q|$ is constant and $\ell$ grows linearly with $k$.

By fairness and by Lemma 10, it follows that for $N$ great enough and for any words $u\in A^{k}$ and $v^{\prime}\in A^{\ell}$

[TABLE]

Summing up for all $u\notin X_{v^{\prime}}$ and all $v^{\prime}\in A^{\ell}$ gives that for $N$ great enough

[TABLE]

and subtracting the number of $i$ such that $|v_{i}|\geq\ell$ gives

[TABLE]

For each $i$ in the previous set, $w_{i}\geq(1-\varepsilon)k$. Therefore the length of the output $w_{1}\cdots w_{N}$ is at least $(1-3\varepsilon)(1-\varepsilon)kN\geq(1-4\varepsilon)kN$. This completes the proof since the length of the input $u_{1}\cdots u_{N}$ is $kN$. ∎

4.2 From independence/fairness to selecting and back

Proof of Theorem 1, (1) implies (3).

We need to prove that selection from a normal word $x$ with a finite-state independent normal oracle $y$ preserves normality. Mutatis mutandis this proof is the same as that given in [3, Theorem 7.1], but now one should consider $2$-deterministic $3$-automata, and the normal word $y$ as a consultative oracle. ∎

Proof of Theorem 1, (3) implies (2).

Suppose that fairness does not hold. By Proposition 8, there is a $2$-deterministic automaton $\mathcal{A}$ with the following property. Let $\gamma$ be the run of $\mathcal{A}$ on $x$ and $y$. There are in $\mathcal{A}$ and two transitions $\tau$ and $\tau^{\prime}$ starting from the same state $p$ and an increasing sequence $(k_{n})_{n\geq 0}$ of integers such that

[TABLE]

Since $\mathcal{A}$ is $2$-deterministic, all transitions starting at $q$ read symbols from the same tape. The automaton $\mathcal{A}$ can be turned into a selector $\mathcal{S}$ as follows. Transitions starting at $p$ select the digit they read but all other transitions do not select the digit they read. The previous inequality shows that the output of the selector $\mathcal{S}$ is not even simply normal. This is a contradiction with the hypothesis. ∎

We end this section with the following result that shows that the finite-state independence of two normal words implies the finite-state independence of one and a word that results from selection of the other.

Proposition 14**.**

Let $x$ and $y$ be normal and finite-state independent words. If $y^{\prime}$ is obtained by oblivious selection from $y$, then $x$ and $y^{\prime}$ are still finite-state independent.

Proof.

We show that if $x$ and $y^{\prime}$ are not finite-state independent, then $x$ and $y$ are also not finite-state independent. We suppose that $x$ and $y^{\prime}$ are not finite-state independent. This means either that $x$ can be compressed with the help of $y^{\prime}$ or that $y^{\prime}$ can be compressed with the help of $x$. Suppose first that $x$ can be compressed by a compressor $\mathcal{C}$ with the help of $y^{\prime}$. Combining this compressor with the selector $\mathcal{S}$ which selects $y^{\prime}$ from $y$ yields a compressor $\mathcal{C}^{\prime}$ which compresses $x$ with the help $y$. Indeed, this compressor $\mathcal{C}^{\prime}$ skips symbols from $y$ which are not selected by $\mathcal{S}$ and simulates $\mathcal{C}$ on those symbols which are selected by $\mathcal{S}$.

Suppose second that $y^{\prime}$ can be compressed by a compressor $\mathcal{C}$ with the help of $x$. We claim that $y$ can also be compressed with the help of $x$. The selector $\mathcal{S}$ which selects $y^{\prime}$ from $y$ is used as a splitter to split $y$ into $y^{\prime}$ made of the selected symbols and $y^{\prime\prime}$ made of the non selected symbols. Then, the compressor $\mathcal{C}$ is used to compress $y^{\prime}$ with the help of $x$ into a word $z$. Finally, words $z$ and $y^{\prime\prime}$ are merged into a word $z^{\prime}$ by blocks of the same length $m$. Each block of length $m$ contains either $m$ symbols from $z$ or $m$ symbols from $y^{\prime\prime}$ plus an extra symbol indicating whether the block contains symbols from $z$ or symbols from $y^{\prime\prime}$. The combination of all these automata yields a compressor which compresses $y$ with the help of $x$. ∎

4.3 From independence/fairness to shuffling and back

Proof of Theorem 1, (2) implies (4).

Suppose $x$ and $y$ are normal. Let $\gamma$ be the run of the shuffler $\mathcal{S}$ with inputs $x$ and $y$ and let $\ell$ be a given length. For each state $q$ of $\mathcal{S}$ and each word $w$ of length $\ell$, there exists by Lemma 13 a unique run $\rho_{q,w}$ starting at state $q$ and outputting $w$.

For each word $w$ of length $\ell$, the number of occurrences of $w$ in the prefix $z[1..n]$ of $z$ is given by

[TABLE]

By Lemma 7, the run $\gamma$ reaches a recurrent strongly connected component. Thus, it can be assumed without loss of generality that $\mathcal{S}$ is strongly connected. By Lemmas 9 and 10, for any two finite runs $\rho$ and $\rho^{\prime}$ of the same length and starting from the same state, the following equality holds.

[TABLE]

The result follows directly from this equality. ∎

Proof of Theorem 1, (4) implies (1).

Suppose that $x$ and $y$ are not finite-state independent and $x$ is compressible with the help of $y$. Let $\mathcal{A}$ be the compressor such that $\rho_{\mathcal{A}}(x/y)<\rho(x)$. Consider the shuffler $\mathcal{S}$ that mimics $\mathcal{A}$ and copies each digit of $x$ (respectively of $y$) as soon as it is read by ${\mathcal{A}}$. We claim that $\mathcal{S}(x,y)$ is compressible, hence not normal. For compressing $\mathcal{S}(x,y)$, first define a splitter $\mathcal{S}^{\prime}$ exchanging the inputs and outputs in the transition of $\mathcal{S}$. Thus, $\mathcal{S}^{\prime}(\mathcal{S}(x,y))=(x,y)$. By composing $\mathcal{S}^{\prime}$ with ${\mathcal{A}}$ we can compress $x$ using $y$ and obtain a compressed word $x^{\prime}$. Let $m$ be the block size used in this compression. Finally, words $y$ and $x^{\prime}$ are merged into a word $z$ interleaving a block of $m$ symbols from $x$ with a block of $m$ symbols from $y$. Since the hypothesis ensures $x$ is compressible, so is word $z$. From this word $z$ we can recover $(x^{\prime},y)$, from which we can recover $(x,y)$ and then obtain $S(x,y)$, as required. ∎

5 An algorithm for a pair of independent normal words

To prove Theorem 2 we give an explicit algorithm based on the characterization of finite-state independent normal words in terms of shufflers (Theorem 1 statement (4)). The algorithm we present here is an adaptation of Turing’s algorithm for computing an absolutely normal number [20, 5]. But instead of computing the expansion of a number that is normal in every integer base here we compute a pair of normal infinite words such that every shuffling of them produced by a finite-state shuffler is normal. We start with auxiliary definitions and some properties. We write $\log$ for the logarithm in base $e$ and $\log_{b}$ for any other base $b$.

Definition 15**.**

1.  For a shuffler ${\mathcal{S}}$, a real number $\varepsilon>0$, a finite word $\gamma\in A^{*}$ and a positive integer $n$, we define the set

[TABLE]

2. Assume an enumeration of shufflers ${\mathcal{S}}_{1},{\mathcal{S}}_{2},\ldots$ and define the set

[TABLE]

3. For each positive integer $n$, let $\ell_{n}=({\log_{|A|}n})/{3}$, $t_{n}=n$ and $\varepsilon_{n}=2\;\;\sqrt{(\log n\log_{|A|}n)/n}$.

[TABLE]

Lemma 16** (Lemma 8 in [20], adapted from Theorem 148 in [12]).**

Let $r$ and $n$ be positive integers. For every real $\varepsilon$ such that ${6}/{\lfloor n/r\rfloor}\leq\varepsilon\leq{1}/{|A|^{r}}$ and for every $\gamma\in A^{r}$, if $N(\gamma,i,n)=|\{w\in A^{n}:|w|_{\gamma}=i\}|$ then

[TABLE]

For a word $u\in A^{*}$ we denote by $[u]$ the set of infinite words that start with $u$, and we call it the cylinder determined by $u$,

[TABLE]

For the Cartesian product of two cylinders $[u]\times[v]$ we write $([u],[v])$, and we call the pair of cylinders determined by $(u,v)$.

Proposition 17**.**

For every shuffler ${\mathcal{S}}$, every $n,r,\varepsilon$ such that ${6}/{\lfloor n/r\rfloor}\leq\varepsilon\leq{1}/{|A|^{r}}$ and every $\gamma\in A^{r}$,

[TABLE]

Proof.

Consider the set

[TABLE]

Finally, Lemma 16 gives the needed upper bound for $|\overline{P}(\varepsilon,\gamma,n)|$. ∎

For any set $B\subseteq A^{\omega}\times A^{\omega}$ we write $\overline{B}$ to denote its complement, $(A^{\omega}\times A^{\omega})\setminus B$.

Proposition 18**.**

For any $\varepsilon$, $t$, $\ell$ and $n$, such that ${6}/{\lfloor n/\ell\rfloor}\leq\varepsilon\leq{1}/{|A|^{\ell}}$,

[TABLE]

Proof.

By Definition 15,

[TABLE]

The number of terms of this triple sum is bounded by

[TABLE]

From the lower bound given in Proposition 17 we obtain that for every shuffler ${\mathcal{S}}$ and for every word $\gamma\in A^{\leq\ell}$,

[TABLE]

Therefore,

[TABLE]

Recall the values given in Definition 15 $\ell_{n}=({\log_{|A|}n})/{3}$, $t_{n}=n$, $\varepsilon_{n}=2\sqrt{(\log n\log_{|A|}n)/n}$ and $F_{n}=F(\varepsilon_{n},t_{n},\ell_{n},n)$.

Proposition 19**.**

Let $n_{\text{\em start}}=\min\{n:\varepsilon_{n}\geq 6/\lfloor n/\ell_{n}\rfloor\}$. Then for every $n\geq n_{\text{\em start}}$, $\ell_{n},t_{n}\geq 1$,

[TABLE]

Proof.

To apply Proposition 18 it is required that ${6}/{\lfloor n/\ell_{n}\rfloor}\leq\varepsilon_{n}\leq{1}/{|A|^{\ell_{n}}}$. Then, for every $n\geq n_{\text{\em start}}$ the required inequality holds. So, application of Proposition 18 yields

[TABLE]

If $n_{start}$ is as determined by Proposition 19, then $\bigcap_{n\geq n_{start}}F_{n}$ is not empty and consists just of pairs of finite-state independent normal words. We can actually show that the intersection of a subsequence of $F_{n}$’s with $n$ growing at most exponentially, also consists just of pairs of finite-state independent normal words. The next definition fixes $n_{0}$ as $\log n_{\text{\em start}}$ and defines the sets $G_{n}$ which are used in the proof of Theorem 2.

Definition 20**.**

Let $n_{0}=\log_{|A|}\min\{n:\varepsilon_{n}\geq 6/\lfloor n/\ell_{n}\rfloor\}$. We define a sequence $(G_{n})_{n\geq 0}$ of finite sets of pairs of cylinders in $A^{\omega}\times A^{\omega}$, such that for every $n$, $G_{n+1}\subseteq G_{n}$ as

[TABLE]

Lemma 21**.**

The set $\bigcap_{n\geq 0}G_{n}$ consists exclusively of pairs of finite-state independent normal words.

Proof.

Fix $n_{0}$ as defined in Definition 20. Suppose $(u,v)\in\bigcap_{n\geq 0}G_{n}$. To show that $u$ and $v$ are finite-state independent we show that for any shuffler ${\mathcal{S}}$, ${\mathcal{S}}(u,v)$ is a normal sequence. Fix a finite word $w\in A^{*}$. Pick $m_{0}$ such that if $i$ is the index of ${\mathcal{S}}$ in the enumeration of shufflers, $t_{m_{0}}\geq i$, $\ell_{m_{0}}\geq|w|$, $m_{0}\geq n_{0}$ and $\varepsilon_{m_{0}}<1/|A|^{|w|}$.

Let’s see that for any $m$ greater than $m_{0}$ the following holds. Let $k$ be such that $|A|^{k}\leq m<|A|^{k+1}$. Then, using that $(u,v)\in F_{|A|^{k+1}}$,

[TABLE]

This implies that

[TABLE]

We conclude that ${\mathcal{S}}(u,v)$ is normal applying Theorem 4.6 in [8] which establishes that a word $x$ is normal if, and only if, there exists a positive number $C$ such that for every finite word $w$,

[TABLE]

Hence, taking $C$ equal to $2|A|$ we obtain that $S(u,v)$ is normal. Now we prove that both, $u$ and $v$, are normal too. Consider the selector ${\mathcal{S}}^{\prime}$ defined as the splitter that reverses ${\mathcal{S}}$ and then ignores the second output tape. That is, if ${\mathcal{S}}(u,v)=z$ then ${\mathcal{S}}^{\prime}(z)=u$. Since ${\mathcal{S}}(u,v)$ is normal, by Agafonov’s theorem $u$ is normal. A similar argument proves that $v$ is also normal. We proved that every $(u,v)\in\bigcap_{n\geq 0}G_{n}$ is a pair of normal words satisfying statement (4) of Theorem 1. Hence, $(u,v)$ is a pair of finite-state independent normal words. ∎

Proof of Theorem 2.

For clarity we present the proof for the alphabet $A=\{0,1\}$, hence $|A|=2$. It is straightforward transfer the proof to any alphabet of an arbitrary size. We prove that Algorithm 5.1 constructs of a pair of finite-state independent normal words. From the algorithm is immediate that the sequence $(I_{n})_{n\geq 0}$ is such that for every $n$, $I_{n+1}\subset I_{n}$, and $\mu(I_{n+1})={\mu(I_{n})}/{2}$. We show that for every $n$, $\mu(I_{n}\cap G_{n})>0$. We prove by induction that for every $n$,

[TABLE]

For the base case, $n=0$, $\mu(G_{0}\cap I_{0})=1>2^{-1}$. For the inductive step, $n+1$, since

[TABLE]

we have

[TABLE]

Then, at least one of $G_{n+1}\cap I_{n}^{0}$ and $G_{n+1}\cap I_{n}^{1}$ must have measure greater than $2^{-2(n+1)-1}$, as required. Since $(I_{n})_{n\geq 0}$ is a nested sequence of intervals of strictly decreasing but positive measure, and for every $n$, $\mu(G_{n}\cap I_{n})>0$, we conclude that

[TABLE]

contains a unique pair $(u,v)$. And by Lemma 21 all the elements in $\bigcap_{n\geq 0}G_{n}$ are pairs of finite-state independent normal words. This concludes the proof. ∎

5.1 Computational complexity

Algorithm 5.1 computes a sequence $(I_{n})_{n\geq 0}$ of pairs of cylinders in $\{0,1\}^{\omega}\times\{0,1\}^{\omega}$ such that $\bigcap_{i\geq 0}I_{n}$ contains a unique pair $(u,v)$ of finite-state independent words. We now establish its computational complexity.

Proposition 22**.**

Algorithm 5.1 has doubly exponential complexity: to output $n$ symbols of the finite-state independent normal words $u$ and $v$ the algorithm performs a number of mathematical operations that is doubly exponential in $n$.

Proof.

As in Turing’s original construction, the complexity of each step of our algorithm is dominated by the computation of the set $F_{n_{0}+2^{2^{n}+1}}$, which is doubly exponential. Notice that the measures of the inspected sets can be calculated in simply exponential time, and the rest of the computation takes constant time.

The construction works by taking a sequence of “good sets” $(G_{n})_{n\geq 0}$ and a sequence $(I_{n})_{n\geq 0}$ of pairs of cylinders in $\{0,1\}^{\omega}\times\{0,1\}^{\omega}$. For the initial step, $n=0$, $\mu(G_{0})=1,\mu(I_{0})=1$, and $\mu(G_{0}\cap I_{0})=1$. For subsequent steps, we refine $G_{n}$ into $G_{n+1}$ and choose one suitable half of $I_{n}$ to be $I_{n+1}$. We now find out the length $s_{n}$ of the shuffling that need to be inspected at step $n$ of the algorithm. At step $n$, $G_{n+1}=G_{n}\cap F_{s_{n}}$ and $\mu(G_{n+1})\geq\mu(G_{n})-\mu(\overline{F_{s_{n}}})$. The algorithm chooses the half of $I_{n}$ whose intersection with $G_{n+1}$ is at least $(\mu(G_{n})-\mu(\overline{F_{s_{n}}}))/2$. We need that for each $n$, this measure is positive:

[TABLE]

Therefore, we require $\sum_{n=1}^{\infty}2^{n}\mu(\overline{F_{s_{n-1}}})<1$ while Proposition 19 establishes that $\mu(\overline{F_{s_{n-1}}})<1/s_{n-1}^{2}$. Thus, we require $s_{n-1}\geq 2^{n}$, which shows the needed exponential growth in the index of the sets $F_{s_{n}}$. Notice that the algorithm fixes $s_{n}=2^{n+1}$ and the computation of the set $F_{s_{n}}$ requires the inspection of $2^{s_{n}}$ words of length $s_{n}$. Then at step $n$ the algorithm performs a number of operations that is doubly exponential in $n$. Finally notice that at step $n$ the algorithm outputs $n$ symbols in the form of two words $u_{n}$, $v_{n}$, such that $|u_{n}|+|v_{n}|=n$. ∎

6 Open problems

As a conclusion, we would like to mention a few open problems.

The characterization of finite-state independence of normal words given in Theorem 1 uses asynchronous deterministic finite automata with no extra memory (counters, stack). Determine if the same characterization holds for the non-deterministic version of the same finite automata. We have pursued this line of investigation in [3, 9] for the characterization of normality in terms of incompressibility by finite-automata and essentially we found that, without extra memory, non-determinism, two-way does not add compressibility power. 2. 2.

Give a purely combinatorial characterization of finite-state independence of normal words. We aim at a condition on the two sequences that is defined in combinatorial terms, without mentioning automata (in the same way that the definition of normality can be stated in terms of frequency of blocks). 3. 3.

There are efficient algorithms that compute absolutely normal numbers with nearly quadratic complexity as [6] or, as recently announced, in poly-logarithmic linear complexity [14]. It may be possible to adapt those algorithms to efficiently compute a pair of finite-state independent normal sequences. 4. 4.

Construct a normal word that is finite-state independent of some given normal word. That is, given a word that has been proved to be normal, as Champernowne’s word, we aim to construct another normal word that is finite-state independent of it.

Acknowledgements.

The authors are members of the Laboratoire International Associé INFINIS, CONICET/Universidad de Buenos Aires–CNRS/Université Paris Diderot and they are partially supported by the ECOS project PA17C04. Carton is partially funded by the DeLTA project (ANR-16-CE40-0007).