Re-pairing brackets

TL;DR

This paper studies a bracket re-pairing game, establishing bounds on the complexity of re-pairing strategies and applying these results to automata theory, notably deriving quasi-polynomial lower bounds for automata translation.

Contribution

It provides new lower bounds on re-pairing strategy width and applies these bounds to prove automata translation complexity results, answering an open question.

Findings

Upper and lower bounds on re-pairing strategy width.

Sub-exponential width strategies for complete binary tree sequences.

Quasi-polynomial lower bounds on automata translation complexity.

Abstract

Consider the following one-player game. Take a well-formed sequence of opening and closing brackets. As a move, the player can pair any opening bracket with any closing bracket to its right, erasing them. The goal is to re-pair (erase) the entire sequence, and the complexity of a strategy is measured by its width: the maximum number of nonempty segments of symbols (separated by blank space) seen during the play. For various initial sequences, we prove upper and lower bounds on the minimum width sufficient for re-pairing. (In particular, the sequence associated with the complete binary tree of height $n$ admits a strategy of width sub-exponential in $\log n$.) Our two key contributions are (1) lower bounds on the width and (2) their application in automata theory: quasi-polynomial lower bounds on the translation from one-counter automata to Parikh-equivalent nondeterministic finite automata. The latter result answers a question by Atig et al. (2016).

Equations

$$\mathrm{width}(Z(n))$$

$$\mathrm{width}(Y(\ell ))$$

$$n^{\Omega \left(\sqrt{\log n/\log \log n}\right)}$$

$$Z(1)=+-;Z(n+1)=+Z(n)Z(n)-$$

$$p=(p_1,\dots ,p_{N/2}),\text{where }{p}_i=({\ell }_i,r_i)$$

$$\underset{\text{k}}{\underbrace{++\dots ++}}\sigma \underset{\text{k}}{\underbrace{--\dots --}}.$$

$$k_{2^q\cdot s+1}=0,k_{2^q\cdot s+t}=\lceil {\log }_{2}t\rceil -1\text{for }1<t\le 2^q,0\le s.$$

$$\mathrm{width}(W_n)=\Omega (\sqrt{\log |W_n|/\log \log |W_n|}).$$

$$X(a_0)={+}^{a_0}{-}^{a_0},$$

$$X(a_0,\dots ,a_k)={+}^{a_k}X(a_0,\dots ,a_{k-1})X(a_0,\dots ,a_{k-1}){-}^{a_k}.$$

$$K=\Big{\{}(x_{0},\dots,x_{n-1})\colon\sum_{i=0}^{n-1}(-1)^{i}x_{i}=0;\ \sum_{i=0}^{k}(-1)^{i}x_{i}\geq 0,\ 0\leq k<n-1\Big{\}};$$

$$S\to \epsilon |+S-S.$$

$$T(+{\sigma }_1-{\sigma }_2)=(T(+{\sigma }_1-),T({\sigma }_2)),$$

$$\mathop{\mathrm{width}}\nolimits(I,E)=\max_{t}\big{|}T^{-1}(t)\big{|}.$$

$$\tau (I,E)=(\mathrm{root},v_1)\cdot \tau (I_1,E_1)\cdot (\mathrm{root},v_1)\cdot (\mathrm{root},v_2)\cdot \tau (I_2,E_2)\cdot (\mathrm{root},v_2),$$

$$\tau (I,E)=(\mathrm{root},v_1)\cdot \tau (I_1,E_1)\cdot (\mathrm{root},v_1),$$

$$\tau (I,E)=U_0{e}_1{D}_1{e}_1{U}_1{e}_2{D}_2{e}_2\dots U_{k-1}{e}_k{D}_k{e}_k{U}_k.$$

$$\tau (I,E):\{1,\dots ,2|E|\}\to E$$

$$\pi :\{1,\dots ,2|E|\}\to \{-1,+1\}.$$

$$\pi (S)=\pi (i_1)\cdot \pi (i_2)\cdot \dots \cdot \pi (i_{\ell }).$$

$$E_t=\{i:T(\tau (I,E)(i))=t\}.$$

$$\begin{array}[]{l|c|c|c|c|c|c|c|c|}i&1&2&3&4&5&6&7&8\\ \hline\cr\pi&+1&&-1&+1&&&-1&\end{array}$$

$$p_{\pi }=(({\ell }_{t_{\max }},r_{t_{\max }}),\dots ,({\ell }_{t_1},r_{t_1})),$$

$$[s_1,f_1]\subseteq [s_2,f_2]\text{and}{t}_1<t_2.$$

$$\mathop{\mathrm{width}}\nolimits(\sigma)\leq 1+\max_{1\leq i\leq t}\big{(}\mathop{\mathrm{width}}\nolimits(\sigma_{i})\,\big{)}.$$

$$\mathrm{width}(\sigma )\le \max (\mathrm{width}({\pi }_1),1+\mathrm{width}({\pi }_2)).$$

$${\sigma }^{(f)}=+{\sigma }_1^{(f)}\dots {\sigma }_{m(f)}^{(f)}-$$

$$\mathrm{width}(\sigma )\le 1+\mathrm{width}({\sigma }^{(1)})\le 1+1+1+\underset{|\tau |\le N/2}{\max }\mathrm{width}(\tau ).$$

$$k_t\le q-1\le \frac{n-1}{2}\le n-\frac{n+1}{2}\le n-q.$$

$$r_t=\max \{j:2^j\text{ divides }t\},\text{where }1\le t\le N-1.$$

$${+}^{k+1}Z(n)Z(n){-}^{k+1}.$$

Full text

For various initial sequences, we prove upper and lower bounds on the minimum width sufficient for re-pairing. (In particular, the sequence associated with the complete binary tree of height $n$ admits a strategy of width sub-exponential in $\log n$.) Our two key contributions are (1) lower bounds on the width and (2) their application in automata theory: quasi-polynomial lower bounds on the translation from one-counter automata to Parikh-equivalent nondeterministic finite automata. The latter result answers a question by Atig et al. (2016).

1 Introduction

Consider the following one-player game. Take a well-formed sequence of opening and closing brackets; that is, a word in the Dyck language. As a move, the player can pair any opening bracket with any closing bracket to its right, erasing them. The two brackets do not need to be adjacent or matched to each other in the word. The goal is to re-pair (erase) the entire word, and the complexity of a play is measured by its width: the maximum number of nonempty segments of symbols (‘islands’ separated by blank space) seen during the play. Here is an example:

\begin{array}[]{rcccccccc}\text{Initial word:}&\text{{(}}&\text{{(}}&\text{{)}}&\text{{(}}&\text{{)}}&\text{{)}}\\ \text{After move 1:}&\text{{(}}&\text{{\ }}&\text{{\ }}&\text{{(}}&\text{{)}}&\text{{)}}\\ \text{After move 2:}&\text{{(}}&\text{{\ }}&\text{{\ }}&\text{{\ }}&\text{{)}}&\text{{\ }}\\ \text{After move 3:}&\text{{\ }}&\text{{\ }}&\text{{\ }}&\text{{\ }}&\text{{\ }}&\text{{\ }}\end{array}

Note that move $2$ pairs up two brackets not matched to each other in the initial word; such moves are permitted without restrictions. At the beginning, there is a single segment, which splits into two after the first move. Both segments disappear simultaneously after the third move; the width of the play is equal to $2$. In this example, width $1$ is actually sufficient: a better strategy is to erase the endpoints first and then erase the two matched pairs in either order.

For a word $\sigma$, the width of $\sigma$ is the minimum width sufficient for re-pairing it. Is it true that all well-formed (Dyck) words, no matter how long, have bounded complexity, i.e., can be re-paired using width at most $c$, where $c$ is independent of the word? The answer to this simply formulated combinatorial question turns out to be negative, but there does not appear to be a simple proof for this: strategies, perhaps surprisingly, turn out quite intricate. In the present paper, we study this and related questions.

Motivation. First of all, we find the re-pairing problem interesting in its own right—as a curious combinatorial game, easily explained using just the very basic concepts in discrete mathematics. So our motivation is, in part, driven by the appeal of the problem itself.

We have first identified the re-pairing problem when studying an open question in automata theory, the complexity of translation of one-counter automata (OCA) on finite words into Parikh-equivalent nondeterministic finite automata (NFA) [ACH*+*16]. This translation arose in model checking, namely in the context of availability expressions [AAMS15, HMO10]. It is more generally motivated by recent lines of research on the classic Parikh theorem [Par66]: on its applications in verification (see, e.g., [GM12, EGP14, HL12]) and on its extensions and refinementsrequired for them [EGKL11, KT10, Kop15]. It has been unknown [ACH*+*16] whether the translation in question can be made polynomial, and in this paper we answer this question negatively: using our results on the re-pairing problem, we obtain a quasi-polynomial lower bound on the blowup in the translation.

The re-pairing problem is a curious case study in the theory of non-uniform models of computation. As it turns out, restricted strategies, where paired brackets are always matched to each other in the word, have a close connection with the black-and-white pebble game on binary trees, a classic setting in computational complexity (see, e.g., surveys by Nordström [Nor13, Nor15]). We show that unrestricted strategies in the re-pairing game make it significantly more complex than pebbling, strengthening this model of computation.

Our two key contributions are (i) lower bounds on the width (against this model) and (ii) the connection to automata theory: lower bounds on the translation from OCA to Parikh-equivalent NFA.

Our lower bounds on the width are obtained by bounding the set of (Dyck) words that can be re-paired using width $k$ (for each $k$). To put this into context, classic models of matrix grammars [Roz87] and deterministic two-way transducers [Roz85], as well as a more recent model of streaming string transducers [AC11, AC10], extend our model of computation with additional finite-state memory. (We refer the reader to surveys [FR16, Mus17, MP19] for more details.) In terms of transducers, our technique would correspond to determining the expressive power of machines of bounded size. Existing results of this kind (see [FR16, Mus17]) apply to variants of the model where concatenation is restricted: with [AR13] or without [DRT16] restrictions on output and in the presence of nondeterminism [BGMP16]. Our result is, to the best of our knowledge, the first lower bound against the unrestricted model.

In the application of the re-pairing problem to automata theory, our lower bounds on the size of NFA for the Parikh image can be viewed as lower bounds on the size of commutative NFA, i.e., nondeterministic automata over the free commutative monoid (cf. [Huy83, Huy85, Esp97, KT10, Kop15, HH16]). To the best of our knowledge, we are the first to develop lower bounds (on description size) for this simple non-uniform model of computation. It is well-known that, even for usual NFA, obtaining lower bounds on the size (number of states) is challenging and, in fact, provably (and notoriously) hard; and that the available toolbox of techniques is limited (see, e.g., [GH06, HK08] and [HPS09]). From the ‘NFA perspective’, we first develop a lower bound for a stronger model of computation and then import this result to NFA using combinatorial tools, which we thus bring to automata theory: the Birkhoff—von Neumann theorem on doubly stochastic matrices (see, e.g., [Sch03, p. 301]) and the Nisan—Wigderson construction of a family of sets with pairwise low intersection [NW94]. The obtained lower bounds point to a limitation of NFA that does not seem to have the form of the usual communication complexity bottleneck (cf. [Sha09, Theorem 3.11.4], [HPS09], and the book by Hromkovic [Hro97]); exploring and exploiting this further is a possible direction for future research.

Our contribution

In this paper, we define and study the re-pairing problem (game), as sketched above. Our main results are as follows:

We show that every well-formed (Dyck) word $\sigma$ has a re-pairing of width $O(\log|\sigma|)$, where $|\sigma|$ is the length of $\sigma$.

This re-pairing always pairs up brackets that are matched to each other in $\sigma$; we call re-pairings with this property simple. It is standard that well-formed words are associated with trees; for words $\sigma$ associated with binary trees, we show that the minimum width of a simple re-pairing is equal (up to a constant factor) to the minimum number of pebbles in the black-and-white pebble game on the associated tree, a quantity that has been studied in computational complexity [CS76, LT80, Nor13, Nor15] and captures the amount of space used by nondeterministic computation.

In particular, this means [Lou79, Mey81, LT80] that for the word $Z(n)$ associated with a complete binary tree of height $n$, the minimum width of simple re-pairings is $\Theta(n)$, which is logarithmic in the length of $Z(n)$. 2. 2.

For $Z(n)$, we show how to beat this bound, giving a (non-simple) recursive re-pairing strategy of width $2^{O(\sqrt{\log n})}$. This is a function sub-exponential in $\log n$; it grows faster than all $(\log n)^{k}$, but slower than all $n^{\varepsilon}$, $\varepsilon>0$. 3. 3.

For $Z(n)$ and for a certain ‘stretched’ version of it, $Y(\ell)$, we prove lower bounds on the width of re-pairings:

[TABLE] 4. 4.

As an application of our lower bounds, we prove that there is no polynomial-time translation from one-counter automata (OCA) on finite words to Parikh-equivalent nondeterministic finite automata (NFA). This shows that optimal translations must be quasi-polynomial, answering a question by Atig et al. [ACH*+*16].

To prove this result, we consider OCA from a specific complete family, $(\mathcal{H}_{n})_{n\geq 2}$, identified by Atig et al. [ACH*+*16]. (There is a polynomial translation from (any) OCA to Parikh-equivalent NFA if and only if these OCA $\mathcal{H}_{n}$ have Parikh-equivalent NFA of polynomial size.) We prove, for every Dyck word $\sigma_{n}$ of length $O(\sqrt{n})$, a lower bound of $n^{\Omega(\mathop{\mathrm{width}}\nolimits(\sigma_{n}))}$ on the minimum size of NFA accepting regular languages Parikh-equivalent to $L(\mathcal{H}_{n})$. Based on the words $Y(\ell)$, we get a lower bound of

[TABLE]

on the size of NFA. Note that this holds for NFA that accept not just a specific regular language, but any language Parikh-equivalent to the one-counter language $L(\mathcal{H}_{n})$ (there are infinitely many such languages for each $n$).

Background and related work

Parikh image of one-counter languages.

The problem of re-pairing brackets in well-formed words is linked to the following problem in automata theory.

The Parikh image (or commutative image) of a word $u$ over an alphabet $\Sigma$ is a vector of dimension $|\Sigma|$ in which the components specify how many times each letter from $\Sigma$ occurs in $u$. The Parikh image of a language $L\subseteq\Sigma^{*}$ is the set of Parikh images of all words $u\in L$. It is well-known [Par66] that for every context-free language $L$ there exists a regular language $R$ with the same Parikh image (Parikh-equivalent to $L$). If $L$ is generated by a context-free grammar of size $n$, then there is a nondeterministic finite automaton (NFA) of size exponential in $n$ that accepts such a regular language $R$ (see [EGKL11]); the exponential in this translation is necessary in the worst case.

When applying this translation to a language from a proper subclass of context-free languages, it is natural to ask whether this blowup in description size can be avoided. For languages recognized by one-counter automata (OCA; a fundamental subclass of pushdown automata), the exponential construction is suboptimal [ACH*+*16]. If an alphabet $\Sigma$ is fixed, then for every OCA with $n$ states over $\Sigma$ there exists a Parikh-equivalent NFA of polynomial size (the degree of this polynomial depends on $|\Sigma|$). And even in general, if the alphabet is not fixed, for every OCA with $n$ states over an alphabet of cardinality at most $n$ there exists a Parikh-equivalent NFA of size $n^{O(\log n)}$, quasi-polynomial in $n$. Whether this quasi-polynomial construction is optimal has been unknown, and we prove in the present paper a quasi-polynomial lower bound.

We note that the gap between NFA of polynomial and quasi-polynomial size grows to exponential when the translation is applied iteratively, as is the case in Abdulla et al. [AAMS15].

Matrix grammars of finite index and transducers.

The question of whether all well-formed (Dyck) words can be re-paired using bounded width can be linked to a question on matrix grammars, a model of computation studied since the 1960s [Abr65]. Matrix grammars are a generalization of context-free grammars in which productions are applied in ‘batches’ prescribed by the grammar. This formalism subsumes many classes of rewriting systems, including controlled grammars, L systems, etc. (see, e.g., [DPS97]).

The index of a derivation in a matrix grammar is the maximum number of nonterminals in a sentential form in this derivation (this definition applies to ordinary context-free grammars as well) [Bra67, GS68]. Bounding the index of derivations, i.e., restricting grammars to finite index is known to reduce the class of generated languages; this holds both for ordinary context-free [GS68, Sal69, Gru71] and matrix grammars [Bra67]. Languages generated by finite-index matrix grammars have many characterizations: as languages output by deterministic two-way transducers with one-way output tape [Raj72], or produced by EDT0L systems of finite index [Lat79, Proposition I.2]; images of monadic second-order logic (MSO) transductions [EH01]; and, most recently, output languages of streaming string transducers [AC11, AC10]. (See also the survey by Filiot and Reynier [FR16].)

Encoding the rules of our re-pairing problem in the matrix grammar formalism leads to a simple sequence of grammars with index $k=2,3,\ldots$ for subsets of the Dyck language $D_{1}$; the question of whether all Dyck words can be re-paired using bounded width is the same as asking if any of these grammars has in fact (bounded-index) derivations for all Dyck words. A 1987 paper by Rozoy [Roz87] is devoted to the proof that, in fact, no matrix grammar can generate all words in $D_{1}$ using bounded-index derivations without also generating some words outside $D_{1}$. This amounts to saying that no finite-index matrix grammar generates $D_{1}$; and a non-constant lower bound on the width in the re-pairing problem could be extracted from the proof.

Unfortunately, the proof in that paper seems to be flawed. Fixing the argument does not seem possible, and we are not aware of an alternative proof. (We discuss the details and compare the proof to our construction in Appendix A).

2 Basic definitions

The Dyck language.

We use non-standard notation for brackets in words from the Dyck language: the opening bracket is denoted by $+$ and the closing bracket by $-$; we call these symbols pluses and minuses, accordingly. Moreover, in some contexts it is convenient to interpret $+$ and $-$ as integers $+1$ and $-1$.

Let $N$ be an even integer. A word $\sigma=(\sigma(1),\dots,\sigma(N))$, $\sigma(i)\in\{+1,-1\}$, is a Dyck word (or a well-formed word) if it has an equal number of $+1$ and $-1$ and for every $1\leq k\leq N$ the inequality $\sum_{i=1}^{k}\sigma(i)\geq 0$ is satisfied. The height of a position $i$ in a well-formed word $\sigma$ is $h(i)=\sum_{j=1}^{i}\sigma(j)$. As usual, $|\sigma|$ denotes the length of the word $\sigma$ (the number of symbols in it).

Dyck words are naturally associated with ordered rooted forests (i.e., with sequences of ordered rooted trees). E.g., words $Z(n)$ defined by

[TABLE]

can be associated with complete binary trees of height $n-1$. Recall that the height of a rooted tree is the maximum length of a path (number of edges) from the root to a leaf.

Note that we described a re-pairing of the word $Z(2)$ in section 1.

Re-pairings and their width.

A re-pairing of a well-formed word $\sigma$ is a sequence of pairs

[TABLE]

and the following properties are satisfied:

1. (R1)

$\sigma(\ell_{i})=+1$, $\sigma(r_{i})=-1$, $\ell_{i}<r_{i}$ for all $i$; 2. (R2)

every number from the interval $[1,N]$ occurs in exactly one pair $p_{j}$.

(We use the word ‘interval’ to refer to a set of the form $[a,b]=\{x\in\mathbb{Z}:a\leq x\leq b\}$.)

The intuition is that the index $i$ corresponds to discrete time, and at time $i$ the two symbols $\sigma(\ell_{i})$ and $\sigma(r_{i})$ are (re-)paired (or erased). Denote by $B_{t}(p)=\{b\in[1,N]:(b=\ell_{i})\lor(b=r_{i}),\ i\leq t\}$ the set of points from $[1,N]$ that correspond to symbols erased at times $[1,t]$.

It is easy to see that re-pairings exist for every well-formed word. By induction on the length of the word one can prove a stronger statement: a word $(p_{1},\dots,p_{t})$ can be extended to a re-pairing iff all numbers in the pairs $p_{i}=(\ell_{i},r_{i})$ are different, the property (R1) is satisfied, and the remaining signs (those which have not been erased) constitute a well-formed word. We now define the following quantities:

• •

The width of a set $S$ of integers, $\mathop{\mathrm{width}}\nolimits(S)$, is the smallest number of intervals the union of which is equal to $S$.

• •

The width of a re-pairing $p$ at time $t$ is $\mathop{\mathrm{width}}\nolimits(B_{t}(p))$.

• •

The width of a re-pairing $p$ of a well-formed word $\sigma$, $\mathop{\mathrm{width}}\nolimits(p)$, is $\max_{t}\mathop{\mathrm{width}}\nolimits(B_{t}(p))$, i.e., the maximum of the width of this re-pairing over all time points.

• •

The width of a well-formed word $\sigma$, $\mathop{\mathrm{width}}\nolimits(\sigma)$, is $\min_{p}\mathop{\mathrm{width}}\nolimits(p)$, where the minimum is over all re-pairings of $\sigma$.

We will look into how big the width of a well-formed word of length $N$ can be, that is, we are interested in $\max_{\sigma}\mathop{\mathrm{width}}\nolimits(\sigma)$, where the maximum is over all well-formed words of length $N$.

Remark 1*.*

Section 1 discussed the minimization of the maximum number of the “surviving” (non-erased) intervals. This quantity cannot differ from the width defined above by more than $1$.

Remark 2*.*

A tree-based representation of re-pairings is described in Section 5 and, in more details, in Appendix, Section B.2.

3 Simple bounds and simple re-pairings

In this section we establish several basic facts on the width of well-formed words and re-pairings (proofs are provided in Appendix, Section C). A careful use of bisection leads to the following upper bound:

Theorem 1**.**

$\mathop{\mathrm{width}}\nolimits(\sigma)=O(\log|\sigma|)$* for all well-formed words $\sigma$.*

We call a re-pairing of a well-formed word $\sigma$ simple if at all times it pairs up two signs that are matching in the word $\sigma$. The re-pairing that the proof of Theorem 1 constructs is simple.

We now show a link between simple re-pairings and strategies in the following game. Let $G$ be an acyclic graph (in our specific case it will be a tree with edges directed from leaves to root). Define a black-and-white pebble game on $G$ (see, e.g., [LT80, Nor15]) as follows. There is only one player, and black and white pebbles are placed on the nodes of the graph. The following moves are possible:

1. (M1)

place a black pebble on a node, provided that all its immediate predecessors carry pebbles; 2. (M2)

remove a black pebble from any node; 3. (M3)

place a white pebble on any node; and 4. (M4)

remove a white pebble from a node, provided that all its immediate predecessors carry pebbles.

(In a tree, immediate predecessors are immediate descendants, i.e., children. Rules (M1) and (M4) are applicable to all sources, i.e., leaves of $G$.) At the beginning there are no pebbles on any nodes. A sequence of moves in the game is a strategy; it is successful if it achieves the goal: reaching a configuration in which all sinks of the graph carry pebbles and there are no white pebbles on any nodes. By $\mathrm{bw}(G)$ we will denote the minimum number of pebbles sufficient for a successful strategy in the black-and-white pebble game on $G$.

Theorem 2**.**

Suppose the tree $D$ associated with a well-formed word $\sigma$ is binary. Then the minimum width of a simple re-pairing for $\sigma$ is $\Theta(\mathrm{bw}(D))$.

Since $D$ is a tree, it follows from the results of the papers [Lou79, Mey81, LT80] (see also [Sav98, pp. 526–528]) that the value of $\mathrm{bw}(D)$ at most doubles if the strategies are not allowed any white pebbles. The optimal number of (black) pebbles in such strategies is determined by the so-called Strahler number (see, e.g., [ELS14] and [LT80]):

Corollary 1**.**

For binary trees, the following two quantities are within a constant factor from each other: the minimum width of a simple re-pairing for $\sigma$ and the maximum height of a complete binary tree which is a graph-theoretic minor of the tree $D$.

By Corollary 1, the upper bound in Theorem 2 has the same order of magnitude as (or lower than) the upper bound from Theorem 1. The latter gives a simple re-pairing too, but also holds for non-binary trees $D$.

The lower bound in Theorem 2 relies on the re-pairing being simple. For instance, for the word $Z(n)$ associated with a complete binary tree (see (2)), the minimum width of a simple re-pairing is $\Theta(n)$, but the (usual) width is $o(n^{\varepsilon})$ for all $\varepsilon>0$ (section 4).

4 Upper bound for complete binary trees

Recall the words $Z(n)$, defined by equation (2) on page 2.

Theorem 3**.**

$\mathop{\mathrm{width}}\nolimits(Z(n))=2^{O(\sqrt{\log n})}$.

The upper bound from the previous section gives $\mathop{\mathrm{width}}\nolimits(Z(n))=O(n)$, whilst the functions $f(n)=a\cdot 2^{b\sqrt{n}}$ for $a,b>0$ are such that $(\log n)^{k}=o(f(n))$ and $f(n)=o(n^{\varepsilon})$ for all $k,\varepsilon>0$.

To prove Theorem 3 we need a family of framed words $Z(n)^{(k)}$. Denote by $\sigma^{(k)}$ the word

[TABLE]

Using the brackets terminology, this is the word $\sigma$ which is enclosed by $k$ pairs of openings and closing brackets. We will call such words $k$-framed.

Remark 3*.*

If $k\geq|\sigma|/2$, then $\mathop{\mathrm{width}}\nolimits(\sigma^{(k)})\leq 2$, because a re-pairing can erase the signs of $\sigma$ from left to right, pairing each ${-}$ with a ${+}$ from the prefix and each ${+}$ with a ${-}$ from the suffix. This re-pairing is, of course, not simple.

We construct a family of re-pairings $p(q,n,k)$ of framed words $Z(n)^{(k)}$, where $k\leq n$ and $1\leq q\leq(n+1)/2$ is a parameter. The definition will be recursive, and $q$ will control the ‘granularity’ of the recursion.

Oveview.

On each step of the re-pairing $p(q,n,k)$ the leftmost remaining ${-}$ is erased. For $n\leq 2$, it is paired with the leftmost remaining ${+}$. For $n>2$, it is paired with the ${+}$ that we choose using the following recursive definition.

At each step of the re-pairing $p(q,n,k)$, we define an auxiliary subsequence of the word $Z(n)^{(k)}$ that forms a word $Z(q)^{(k^{\prime})}$. If the leftmost remaining minus is not in the subsequence, then we pair it with the leftmost non-erased plus. Otherwise we consider the re-pairing $p(q^{\prime},q,k^{\prime})$ of the word $Z(q)^{(k^{\prime})}$, where we pick $q^{\prime}$ and $k^{\prime}$ below, and pair the minus using this re-pairing (more details to follow).

Stages of the re-pairing $p(q,n,k)$.

The re-pairing $p(q,n,k)$ is divided into stages, indexed by $t=1,\ldots,2^{n-q}$. Denote by $Z_{t}$, $1\leq t\leq N=2^{n-q}$, the $t$th leftmost occurrence (factor) of $Z(q)$ in the word $Z(n)^{(k)}$. Stage $t$ begins at the moment when all minuses to the left of the start position $i_{t}$ of the factor $Z_{t}$ are erased, and ends when stage $t+1$ begins. Define an integer sequence $k_{t}$ as follows:

[TABLE]

At the beginning of stage $t$, the subsequence $Z^{\prime}_{t}$ of $Z(n)^{(k)}$ is formed by the $k^{\prime}=k_{t}$ rightmost non-erased pluses to the left of $i_{t}$; followed by the symbols of the factor $Z_{t}$; followed by the $k_{t}$ leftmost non-erased minuses to the right of the end position of $Z_{t}$. The symbols of $Z^{\prime}_{t}$ written together, form the word $Z(q)^{(k_{t})}$.

Choose $1\leq q^{\prime}\leq q/3$ such that the width of the re-pairing $p(q^{\prime},q,k_{t})$ is minimal. At the first part of stage $t$, the re-pairing $p(q,n,k)$ pairs the signs in $Z^{\prime}_{t}$ according to the re-pairing $p(q^{\prime},q,k_{t})$. The first part ends when either all minuses to the left of the factor $Z_{t+1}$ are erased or the sequence $Z^{\prime}_{t}$ is exhausted (whichever is earlier). In the latter case the final part of stage $t$ is started. At each step of this part, the leftmost non-erased minus is paired with the leftmost non-erased plus.

Claim 1**.**

Re-pairings $p(q,n,k)$ are well-defined.

Define $W_{n}=\min_{q}\max_{0\leq k\leq n}\mathop{\mathrm{width}}\nolimits(p(q,n,k))$, where the minimum is over $15\leq q\leq n/3$ for $n\geq 45$ and over $1\leq q\leq n/3$ for $3\leq n<45$.

Claim 2**.**

W_{n}\leq\min\limits_{15\leq q\leq n/3}\Big{(}\displaystyle\frac{2n}{q}+2W_{q}+3\Big{)}* for $n\geq 45$.*

Somewhat strangely, we have been unable to find solutions to recurrences of this form in the literature.

Claim 3**.**

$W_{n}=2^{O(\sqrt{\log n})}$.

Since $\mathop{\mathrm{width}}\nolimits(Z(n))\leq W_{n}$, this implies the upper bound of Theorem 3.

Proof idea for Claim 2.

(For complete proofs of these claims see Appendix, Section D.) Assume $15\leq q\leq n/3$. We notice that at each step at most two of the factors $Z_{t}$, $Z_{t+1}$ are partially erased. (All other factors $Z_{t^{\prime}}$ either have been erased completely ($t^{\prime}<t$) or are yet untouched ($t^{\prime}>t+1$).) Furthermore, non-erased signs to the left of partially erased factors form several intervals; each of them, except possibly the leftmost, has size at least $q$.

Note that, at each moment in time, the non-erased signs form a well-formed (Dyck) word, so the height of each position in $Z(n)$ with respect to these signs only is nonnegative. Since the height of positions in the word $Z(n)^{(k)}$ cannot exceed $n+k\leq 2n$, it follows that a partially erased factor $Z_{t}$ can be preceded by at most $2n/q+1$ non-erased intervals (runs of pluses). This leads to the recurrence of Claim 2.

5 Lower bounds

Theorem 4**.**

There exists a sequence of well-formed words $W_{n}$ such that

[TABLE]

The words in this sequence are similar to the words $Z(n)$ associated with complete binary trees. They are associated with a ‘stretched’ version of the complete binary tree, i.e., one in which every edge is subdivided into several edges. More precisely, let $a_{0}$, $a_{1}$, …, $a_{k}$ be a finite sequence of positive integers. Define the following sequence of well-formed words inductively:

[TABLE]

The words we use to prove Theorem 4 have the form $Y(m,\ell)=X(a_{0},\dots,a_{m\ell-1})$, where $a_{i}=2^{\lfloor i/\ell\rfloor}$, $m\geq 1$, and $\ell\geq 1$. In particular, Y(\ell)=Y\big{(}\lfloor\ell\cdot\log\ell\rfloor,\ell\big{)}. (Notice that $Z(n)=X(1,\ldots,1)=Y(1,n)$.) Our method applies both to $Y(\ell)$ and $Z(n)$, giving the bounds in equation (1) on page 1.

We give a proof overview below, details are provided in Appendix, Section E. We use a tree representation of re-pairings. Informally, this tree tracks the sequence of mergers of erased intervals. This sequence, indeed, is naturally depicted as an ordered rooted binary tree as shown in Fig. 1(a). This tree is essentially a derivation tree for a word in an appropriate matrix grammar. Edges of a rooted tree are divided into levels according to the distance to the root. We think of this distance as a moment of time in the derivation process. The derived word can be read off the tree by following the left-to-right depth-first traversal.

For formal definitions see Appendix, Section B.2.2.

Our proof is inductive, and one of the ideas is what the induction should be over. Observe that every factor $w$ of a Dyck word $W$ induces a connected subgraph, which we call a fragment; see Fig. 1(b). The width of a tree or a fragment is defined in natural way: it is the maximal number of edges at a level of the tree. E.g., the fragment shown in Fig. 1(b) has width 2.

Our inductive statement applies to fragments. Fix a well-formed word $W$; in the sequel we specialize the argument to $Z(n)$ and $Y(\ell)$. Denote by $L(W,k)$ the maximum length of a factor $w$ associated with a fragment of width at most $k$ in trees that derive the word $W$. Put differently, given $W$, consider all possible trees that derive $W$. Fragments of width at most $k$ in these trees are associated with factors of the word $W$, and $L(W,k)$ is the maximum length of such a factor. Note that in this definition the width of the (entire) trees is not restricted.

It is clear from the definition that the sequence of numbers $L(W,k)$ is non-decreasing: $L(W,1)\leq L(W,2)\leq\dots\leq L(W,k)$. We obtain upper bounds on the numbers $L(W,k)$ by induction. For $Z(n)$, we show that $L(Z(n),k)=L(Z(n),k-1)^{O(k)}$ for big enough $n$ and $k$. Here and below, implicit constants in the asymptotic notation do not depend on $n$ and $k$. From this we get $L(Z(n),k)=2^{2^{O(k)}\cdot k!}$. We observe that if $\mathop{\mathrm{width}}\nolimits(W)\leq k$, then $|W|\leq L(W,k+1)$. Since $|Z(n)|=\Theta(2^{n})$, it follows that every derivation tree of the word $Z(n)$ must have width $k$ satisfying $n=2^{O(k)}\cdot k!$, that is, $\mathop{\mathrm{width}}\nolimits(Z(n))=\Omega(\log n/\log\log n)$.

For $Y(\ell)$, we show a stronger inequality, $L(Y(\ell),k)\leq\mathop{\mathrm{poly}}(\ell,k)\cdot(ck)^{\ell}\cdot L(Y(\ell),k-1)$, which is sufficient for a lower bound $\mathop{\mathrm{width}}\nolimits(Y(\ell))=\Omega(\ell)$.

To prove the inductive upper bound on $L(W,k)$ we need to show that narrow fragments cannot be associated with long factors. For this purpose we use two ideas.

Combinatorial properties of increases and drops in $Z(n)$ and

$Y(\ell)$.

Denote by $\Delta(u)$ the difference $h(j)-h(i)$, where $i$ and $j$ are the start and end positions of a factor $u$. The value $\Delta(u)$ is the increase in height on the factor.

The first property is that every factor of $Z(n)$ of length $x$ contains a sub-factor ${-}^{d}$ with $d\geq\log x-O(1)$. The second combinatorial property of $Z(n)$ is as follows: for sufficiently large $x$ and every two factors $u$ and ${-}^{x}$ of the word $Z(n)$, if $\Delta(u)\geq x$ and $u$ is located to the left of ${-}^{x}$, then the distance between these factors is at least $2^{x}$. Here and below the distance between the factors is the length of the smallest factor of $W$ containing both of them.

For the word $Y(\ell)$, similar properties hold, but the functions $\log x$ and $2^{x}$ are replaced by the functions $\Omega(\ell\cdot(x/9)^{1/\ell})$ and $\Omega((x/2\ell)^{\ell})$, respectively.

Balance within a single time period.

Consider a factor $w$ of the word $W$ associated with a fragment of width at most $k$ (in a tree derivation that generates $W$). Denote this fragment $F$.

Notice that, in a Dyck word, every ${-}$ is matched by a ${+}$ somewhere to the left of it. Thus, for a factor ${-}^{d}$, there exists a factor $u$ to the left of ${-}^{d}$ with a matching height increase: $\Delta(u)\geq d$. We strengthen this balance observation to identify a pair of matching factors $-^{d}$, $u$ (with a slightly smaller height increase in $u$) which also satisfies the following conditions:

1. (a)

$d$ is large enough (of magnitude indicated by the first combinatorial property); 2. (b)

the factors $u$ and ${-}^{d}$ are derived during overlapping time intervals, 3. (c)

the factor $u$ sits to the left of ${-}^{d}$ and inside $w$, and 4. (d)

the sub-fragment associated with the factor between $u$ and ${-}^{d}$ has width strictly smaller than the width of the entire fragment $F$.

These conditions enable us to upper-bound the distance between $u$ and ${-}^{d}$ through a function of $L(W,k-1)$. On the other hand, this distance is lower-bounded by the second combinatorial property. Comparing the bounds shows how to bound $|w|$, and thus $L(W,k)$, from above by a function of $L(W,k-1)$.

6 An application: Lower bounds for commutative NFA

In this section we link the re-pairing problem for well-formed (Dyck) words to the descriptional complexity (number of states in NFA) of the Parikh image of languages recognized by one-counter automata (OCA).

We consider a slightly simplified version of complete languages $(\mathcal{H}_{n})_{n\geq 2}$ introduced by Atig et al. [ACH*+*16] (see Section 1). Each of them is over the alphabet $\{c_{ij}:0\leq i<j<n\}\cup\{a_{i}:0\leq i<n\}$ and can be recognized by an OCA with $n+2$ states. We will assume throughout that $n$ is even. In what follows, we need only the Parikh image $U_{n}$ of this language. We call $U_{n}$ the universal one-counter set (since it is provably the hardest case for translations from OCA to Parikh-equivalent NFA). This is the set of $n(n+1)/2$-dimensional vectors $(y_{ij},x_{i})=(y_{ij}\colon 0\leq i<j<n;\ x_{i}\colon 0\leq i<n)$ of nonnegative integers that satisfy the following conditions:

1. (U1)

$y_{ij}\in\{0,1\}$ and the directed graph on vertices $[n]=\{0,1,\dots,n-1\}$ with the set of edges $\{(i,j)\text{\ such that\ }y_{ij}=1\}$ is a monotone path from [math] to $n-1$ (i.e., one with $i<j$ in each edge); we call such paths chains; 2. (U2)

(balance) the vector $x=(x_{i})$ belongs to the cone $K$ of balanced vectors:

[TABLE] 3. (U3)

(compatibility) if $x_{j}>0$ for some $j>0$, then $y_{ij}=1$ for some $i$; if $x_{j}>0$ for some $j<n-1$, then $y_{jk}=1$ for some $k$.

We skip the standard definition of nondeterministic finite automata (NFA). The meaning of the numbers $x_{i}$ and $y_{ij}$ is that they specify the number of occurrences of letters $a_{i}$ and $c_{ij}$ on accepting paths in (the transition graph of) the NFA. An NFA recognizes a language with Parikh image $U_{n}$ iff for each vector from $U_{n}$ the NFA has an accepting path with these counts of occurrences, and for all other vectors no such path exists. There exists [ACH*+*16] an NFA with $n^{O(\log n)}$ states that recognizes a language with Parikh image $U_{n}$. Our goal is to prove that this superpolynomial dependency on $n$ is unavoidable.

Theorem 5**.**

Let $\sigma_{n}$ be an arbitrary Dyck word of length $O(\sqrt{n})$. Suppose an NFA ${\mathcal{A}}_{n}$ recognizes a language with Parikh image $U_{n}$. Then the number of states of ${\mathcal{A}}_{n}$ is at least $n^{\Omega(\mathop{\mathrm{width}}\nolimits(\sigma_{n}))}$.

Corollary 2**.**

If an NFA ${\mathcal{A}}_{n}$ recognizes a language with Parikh image $U_{n}$, then its number of states is $n^{\Omega(\sqrt{\log n/\log\log n})}$.

Corollary 2 follows from Theorems 4 and 5.

Since $U_{n}$ is the Parikh image of a language recognized by an OCA with $n+2$ states, it follows that there is no polynomial translation from OCA to Parikh-equivalent NFA.

Our proof of Theorem 5 makes three steps:

In the NFA for $U_{n}$ we find accepting paths $\pi_{F}$, parameterized by sets $F\subseteq\{0,1,\ldots,\frac{n}{2}-1\}$, $|F|=|\sigma_{n}|$, and extract re-pairings of $\sigma_{n}$ from them. Roughly speaking, the parameter $F$ determines the set of positions $j$ for which the path $\pi_{F}$ has $x_{j}>0$.

Intuitively, as $\pi_{F}$ goes through any strongly connected component (SCC) in the NFA, the re-pairing erases pairs $(2i,2j+1)$ such that a cycle in this SCC reads letters $a_{2i}$ and $a_{2j+1}$. To get a bijection between even and odd indices, we use the Birkhoff—von Neumann theorem on doubly stochastic matrices (see, e.g., [Sch03, p. 301]). 2. 2.

With every SCC $V$ in the NFA, we associate an auxiliary set $B(V)$. We show that each path $\pi_{F}$ visits an SCC $V_{F}$ for which $|B(V_{F})|\geq\mathop{\mathrm{width}}\nolimits(\sigma_{n})$. 3. 3.

By making $F$ range in a family $\mathcal{F}$ of sets with low intersection, we ensure that no other path $\pi_{F^{\prime}}$ can visit the SCC $V_{F}$. So the NFA has at least $|\mathcal{F}|$ SCCs, and therefore at least $|\mathcal{F}|$ states. The low intersection property means that $|F_{1}\cap F_{2}|\leq d$ for all $F_{1},F_{2}\in{\mathcal{F}}$. We choose $d=\mathop{\mathrm{width}}\nolimits(\sigma_{n})-1$; the family $\mathcal{F}$ of size $n^{\Omega(d)}$ can be obtained by the Nisan—Wigderson construction [NW94].

For the details of the proof see Appendix, Section F.

7 Open problems

Our work suggests several directions for future research. The first is computing the width of $Z(n)$ as well as of other words, closing the gap between the upper and lower bounds. Obtaining super-constant lower bounds (for infinite families of words, both constructively and non-constructively) seems particularly difficult. Our lower bound on the width of $Y(n)$ leaves a gap between $n^{\Omega(\sqrt{\log n/\log\log n})}$ and $n^{O(\log n)}$ for the size of blowup in an OCA to Parikh-equivalent NFA translation, and our second problem is to close this gap.

The third problem is to recover a proof of Rozoy’s statement that the Dyck language $D_{1}$ is not generated by any matrix grammar of finite index [Roz87], or equivalently by any two-way deterministic transducer with one-way output tape [Roz85]. We expect that our lower bound construction for the width can be extended appropriately.

Last but not least, our re-pairing game corresponds to the following family of deterministic two-way transducers $\mathcal{T}_{k}$ generating Dyck words. The input to a transducer $\mathcal{T}_{k}$ encodes a derivation tree of width $k$, in the sense defined in section 5. Symbols corresponds to layers of the tree; there are $O(k^{2})$ symbols in the alphabet that encode the branching and $O(k^{2})$ symbols that encode the positions of a pair of brackets ($+$ and $-$). The transducer $\mathcal{T}_{k}$ simulates a traversal of the tree and outputs the generated word; it has $O(k)$ states. All words of width at most $k$ are generated by $\mathcal{T}_{k}$.

Our final problem is to determine if there exist smaller transducers that generate all Dyck words of length $n$, $D_{1}\cap\{{+},{-}\}^{n}$, and do not generate any words outside $D_{1}$. Here $n$ is such that all words of length $n$ have width at most $k$.

Acknowledgment

We are grateful to Georg Zetzsche for the reference to Rozoy’s paper [Roz87].

This research has been supported by the Royal Society (IEC\R2\170123). The research of the second author has been funded by the Russian Academic Excellence Project ‘5-100’. Supported in part by RFBR grant 17–51-10005 and by the state assignment topic no. 0063-2016-0003.

Appendix A On matrix grammars of finite index

Rozoy’s 1987 paper [Roz87] is devoted to the proof that no matrix grammar can generate all words in the Dyck language $D_{1}$ using bounded-index derivations without also generating some words outside $D_{1}$. This amounts to saying that no finite-index matrix grammar can generate $D_{1}$.

Unfortunately, the proof in that paper seems to be flawed. The main tool in the proof is an inductive statement, relying on several technical claims. The quantifiers in the use of one of these claims are swapped when compared against the formulation (and proof) of the claim. (The earlier conference version [Roz85] of that paper uses an equivalent formalism of deterministic two-way transducers instead of matrix grammars; while the high-level structure of the results is present in the conference version, it does not contain the details of the proofs.)

In more detail, the main tool in Rozoy’s proof is a certain inductive statement, $P(h)$, which is part of the proof of Proposition 6.2.1. This statement $P(h)$, $h=1,2,\ldots,2k$, relies on several technical claims, and in particular on Corollary 6.1.2. This Corollary says, roughly, that, given a (part of a) derivation tree, a position $x$ in the derived word, and a family of disjoint time intervals $C_{1}$, …, $C_{p}$, there exists an index $i$ such that a certain quantity $\mathcal{O}(x|C_{i})$ enjoys what one can call an averaging upper bound. The quantity $\mathcal{O}(x|C_{i})$ is the height of the position $x$ computed relative to just the symbols generated at times $t\in C_{i}$, but not at other times.

Unfortunately, the use of this Corollary does not match its formulation. This occurs in the proofs of the base cases $P(1)$ and $P(2)$, and in the proof of the inductive step. In each of these three cases, time intervals $C_{j}$, $j\in J$, are chosen to correspond to occurrences of the factor $Z(m)$ in $Z(n)$, denoted by $Z(m)_{j}$, $j\in J$. But no position $x$ is chosen. Then the Corollary is invoked to find an index $i$ such that $\mathcal{O}(x|C_{i})$ enjoys an averaging upper bound, where $x$ denotes the beginning of $Z(m)_{i}$, the $i$th occurrence of $Z(m)$. It appears as if this use of Corollary would require a different formulation: instead of “for every position $x$ there exists an index $i$ such that $\mathcal{O}(x|C_{i})$ is upper-bounded”, it would need a “given positions $x_{j}$, $j\in J$, there exists an $i$ such that for the pair $C_{i},x_{i}$ the quantity $\mathcal{O}(x_{i}|C_{i})$ is upper-bounded”.

Given the existing formulation, it seems possible that for each of the positions $x_{j}$ (the start positions of the occurrences of $Z(m)$), there indeed exists a suitable $i$, but this $i$ is different from $j$. In other words, the quantities $\mathcal{O}(x_{j}|C_{i})$ may be bounded for $j\neq i$, but not for $j=i$. A formulation with the swapped quantifiers, “there exists an $i$ such that for all $x$”, would suffice to rule out this bad case, but it is not clear if it is possible to prove such a statement.

Rozoy’s construction has certainly influenced our lower bound construction (for the width of re-pairings). The family of statements $P(h)$ in her construction amounts to an inductive upper bound on the length of the factors $Z(m)$ that can occur in the derivation tree for $Z(n)$ if this tree has width at most $k$ (we are glossing over some technical details here). In our proofs, instead of factors of the form $Z(m)$ we consider arbitrary factors in Definition 1: the quantity $L(W,k)$ upper-bounds the length of all factors (of $W$) that can be associated with fragments of width at most $k$. We do not restrict the width of the entire tree either. The proof of the main inductive step (Lemma 3) then crucially depends on an upper bound for width $k-1$ being available for all words, not just for words of a particular form, $Z(m)$. We do not know if it is possible to use a weaker inductive statement, one mentioning words of the form $Z(m)$ only, for a lower bound on $\mathop{\mathrm{width}}\nolimits(Z(n))$.

Appendix B Trees, Dyck words, and re-pairings

In this section we review the various use of trees in our work. The interpretation of Dyck words as trees (subsection B.1) is required by our results on simple re-pairings and pebble games (Section C below and Section 3 in the main text). The tree representation of re-pairings (subsection B.2) is required for our lower bounds on the width of re-pairings (Section E below and Section 5 in the main text).

We will use ordered rooted trees. Let us recall the standard definitions.

A tree with an empty set of nodes (an empty tree) is denoted by $\varepsilon$. Suppose the set of nodes $I$ is nonempty; then it has a distinguished node $\mathop{\mathrm{root}}$ (the root of the tree). Consider a mapping $a\colon I\setminus\{\mathop{\mathrm{root}}\}\to I$. The node $a(v)$ is the immediate ancestor (parent) of a node $v$, and the node $v$ is the immediate descendant (child) of $a(v)$. The transitive closure of the parent (resp. child) relation is the ancestor (resp. descendant) relation. No node is its own descendant, and all non-root nodes are descendants of the root. These are exactly the defining properties of a rooted tree. The set of children of each node is linearly ordered. A leaf is a node without descendants.

A node $v$ and the set of its descendants form an ordered rooted tree in a natural way, which will be referred to as the subtree rooted at $v$.

If every node in a tree has at most two children, the tree is binary. If $a^{-1}(v)=(v_{1},v_{2})$, then the node $v_{1}$ is the left child and $v_{2}$ is the right child.

For a non-root node $v$, denote by $s(v)$ the sibling of $v$, that is, the node with the same parent different from $v$, if such a node exists in the tree.

B.1 Dyck words and trees

Well-formed words are naturally associated with ordered rooted forests (i.e., with sequences of ordered rooted trees) in the following way. Well-formed words are exactly those generated by the unambiguous grammar

[TABLE]

The rooted forest $T(\sigma)$ associated with the word $\sigma$ is defined recursively using the derivation in this grammar: $T(\varepsilon)=\varepsilon$, and

[TABLE]

where $T({+}\sigma_{1}{-})$ is the tree in which the children of the root are the roots of the trees from the rooted forest $T(\sigma_{1})$. The two symbols $+$ and $-$ in the word ${+}\sigma_{1}{-}$ are associated with the root of the tree $T({+}\sigma_{1}{-})$; these symbols are also said to be matched to each other.

This way all symbols in the word $\sigma$ are split into pairs of symbols matched to each other, and every pair is associated with a node in the rooted forest $T(\sigma)$.

B.2 Tree representation of re-pairings

To prove lower bounds on the width, we will need a different representation of re-pairings. Informally, this representation will track the sequence of mergers of erased intervals. This sequence is naturally depicted as an ordered rooted binary tree. The width of the tree will match the width of the re-pairing (in that the two numbers will differ by at most $1$). This tree is essentially a derivation tree for the word in an appropriate matrix grammar.

B.2.1 Trees and traversals

Here and below, all trees wil be binary and edges in trees will be directed from root to leaves. That is, in an ordered rooted binary tree, the set $E$ of edges consists of ordered pairs of the form $(a(v),v)$. A ranking function $T$ will be defined on this set, and we will call it the time function. For edges $e$ departing from the root, the time $T(e)$ is equal to $1$; for all pairs of edges $(x,y)$, $(y,z)$ with a common endpoint the time function satisfies $T(x,y)+1=T(y,z)$. Whenever $T(e)=t$, we will say that the edge $e$ exists at time $t$.

The width of a tree $(I,E)$ with the set of nodes $I$ and set of edges $E$ is now defined as the maximum (over all time points $t$) number of edges existing at time $t$, i.e.,

[TABLE]

The width of an arbitrary subset of edges, $\mathop{\mathrm{width}}\nolimits(I,E^{\prime})$ for $E^{\prime}\subseteq E$, is defined analogously.

Recall the definition of the (left-to-right depth-first) traversal of the tree (we denote this traversal by $\tau(I,E)$). This is a sequence of edges in which every edge occurs twice; it is defined recursively as follows. For a tree that consists of the root node only (and no edges), the traversal is the empty sequence. If $v_{1}$ and $v_{2}$ are the two children of the root, then

[TABLE]

where $\cdot$ denotes the concatenation of sequences, and $(I_{1},E_{1})$ and $(I_{2},E_{2})$ are the subtrees rooted at $v_{1}$ and $v_{2}$, respectively. Similarly, if there is only one child $v_{1}$, then

[TABLE]

where $(I_{1},E_{1})$ is the subtree rooted at $v_{1}$. Every tree has exactly one traversal.

An example of a traversal is shown in Fig. 2.

For every $t$, edges that exist at time $t$ cut the traversal as follows.

Claim 4**.**

*The edges $e_{1}$, $\dots$, $e_{k}$ of the tree $(I,E)$ that exist at time $t$ occur in the traversal in pairs: *

[TABLE]

All edges from $U_{i}$ exist at times before $t$, and all edges from $D_{i}$ at times after $t$.

Proof.

Assume that the edges $e_{i}$ are sorted according to the indices of their first occurrences in the traversal.

Let $v_{i}^{<},v_{i}^{>}$ be the two endpoints of the edge $e_{i}$, so that $v_{i}^{<}$ is closer to the root than $v_{i}^{>}$. It is clear from the definition of the traversal that the part of the traversal between the two occurrences of $e_{i}$—denote this part by $D_{i}$—is the traversal of the subtree rooted at $v_{i}^{>}$; therefore, all edges in that part exist at times strictly greater than (after) $t$.

Since an undirected tree has exactly one path between any two edges, we have $D_{i}\cap D_{j}=\varnothing$ whenever $i\neq j$.

For the same reason, the part $U_{i}$ of the traversal between the second occurrence of $e_{i}$ and the first occurrence of $e_{i+1}$ does not touch any edges from subtrees rooted at $v_{k}^{>}$. The same also holds for the initial and final parts of the traversal: $U_{0}$ from the beginning of the traversal up until $e_{1}$ and $U_{k}$ from the second occurrence of $e_{k}$ to the end of the traversal. ∎

B.2.2 Tree derivations

Tree derivations, or derivations, are similar to re-pairings defined in section 2. Informally, at any time point $t$, one $+$ and one $-$ are placed on edges that exist at time $t$; the plus must be placed to the left of the minus. A sign can be placed on an edge in two ways: on the left side of the edge or on the right side of it. This is shown in Fig. 3.

The formal definition is as follows. Let $(I,E)$ be a ordered rooted binary tree. The traversal $\tau(I,E)$ defines a function

[TABLE]

in natural way (any sequence is formally a function of this form).

Consider partial functions of the form

[TABLE]

These functions are partial words (or patterns) over the alphabet $\{-1,+1\}$. We define by $\mathop{\mathrm{Dom}}\nolimits\pi$ the domain of the partial function, i.e., the set of all $i$ such that $\pi(i)$ is defined.

For any subset $S\subseteq\{1,\dots,2|E|\}$ a partial word $\pi$ defines a word $\pi(S)$ over the alphabet $\{-1,+1\}$ by the rule: if $S\cap\mathop{\mathrm{Dom}}\nolimits\pi=\{i_{1},i_{2},\dots,i_{\ell}\}$, where $i_{1}<i_{2}<\dots<i_{\ell}$, then

[TABLE]

Let $E_{t}$ be the set of places in the traversal $\tau(I,E)$ occupied by edges existing at time $t$. Formally,

[TABLE]

For brevity we use notation $\pi_{t}=\pi(E_{t})$.

Given a tree $(I,E)$, a tree derivation (or a derivation) is a partial word $\pi$ such that, for all points in time $t$, either the word $\pi_{t}$ is empty, or it is ${+}{-}$.

Every edge $e$ occurs in the traversal exactly twice and thus may derive at most two signs in a tree derivation $\pi$. Let $\tau(I,E)^{-1}(e)=\{i,j\}$, where $i<j$. If $\pi(i)$ is defined, then the sign $\pi(i)$ is said to be placed on the left side of the edge. If $\pi(j)$ is defined, the sign $\pi(j)$ is said to be placed on the right side of the edge.

For an example of a derivation shown in Fig. 3, the corresponding partial function is as follows:

[TABLE]

Blank spaces in the table mean that the function is not defined.

Let $\pi$ be a derivation based on the tree $(I,E)$. We will say that the word $\sigma(\pi,I,E)=\pi(\{1,2,\dots,2|E|\})$ is generated by the derivation $\pi$, or, alternatively, that the word is derived by the tree. We will also sometimes say that a sign in the word is derived at time $t$ if it is derived by an edge that exists at time $t$.

B.2.3 Fragments

Consider an arbitrary factor of the tree traversal $\tau(I,E)$. The set of edges that occur in this factor at least once forms a subgraph in the tree $(I,E)$, which we will call a fragment of this tree. It is easy to check that all fragments are (weakly) connected subgraphs of the graph $(I,E)$. Thus, a fragment is a tree in graph theory terms.

Let $\pi$ be a derivation based on a tree $(I,E)$. An interval $[i,j]\subseteq\{1,2,\dots,2|E|\}$ defines a factor $w=\pi([i,j])$ of the word $\sigma(\pi,I,E)$ and each factor of $\sigma(\pi,I,E)$ can be represented in this form. Moreover, w.l.o.g. we will assume that $i,j\in\mathop{\mathrm{Dom}}\nolimits\pi$.

Suppose that $w=\pi([i,j])$ for a factor $w$ of the word $\sigma(\pi,I,E)$. The interval $[i,j]$ selects a factor in the tree traversal. We say that the fragment $F_{w}$ corresponding to this factor of the tree traversal is associated to the factor $w$.

Note that the only one fragment is associated to a factor of $\sigma(\pi,I,E)$ but the converse is not true. A fragment is a set of edges by definition. It may correspond to different factors of the tree traversal because each edge occurs twice in the traversal.

B.3 Derivations and re-pairings

We now show how to link re-pairings of words and derivations.

Let $\pi$ be a derivation based on a tree $(I,E)$. Consider the word $\sigma(\pi,I,E)$. Take any point in time $t$ for which this word has two (paired) signs; they are said to be derived at this time point. Denote by $\ell_{t}$ the index of the $+$ sign from this pair in the word $\sigma(\pi,I,E)$; and by $r_{t}$ the index of the $-$ sign. We obtain a sequence

[TABLE]

where $t_{\max}$ is the maximum point in time at which edges of the tree exist. The time in this sequence flows in direction opposite to that in the tree; indeed, as we already mentioned at the beginning of section B.2, the tree depicts a sequence of mergers of erased intervals.

Fig. 1(a) shows an example where a re-pairing of a word is obtained from a tree derivation.

Claim 5**.**

The word $\sigma(\pi,I,E)$ is well-formed, and the sequence $p_{\pi}$ is its re-pairing. The width of $p_{\pi}$ does not exceed the width of the tree $(I,E)$.

Proof.

It is easy to see that the word $\sigma(\pi,I,E)$ is well-formed and the sequence $p_{\pi}$ satisfy the requirements on re-pairings. Indeed, if $\pi_{t}={+}{-}$, then the plus is placed to the left of the minus by definition of words $\pi(S)$.

Consider a time point $t$. Suppose $k$ edges exist at this time, $e_{1}$, …, $e_{k}$ if read from left to right. It follows from Claim 4 that the re-pairing $p_{\pi}$ has, at this time point, at most $k$ erased intervals in the word $\sigma(\pi,I,E)$. Indeed, in terms of the factorization (5) from Claim 4, each interval is formed by the symbols of the word $\sigma(\pi,I,E)$ that correspond to the edges from the subsequence $D_{i}$ (this is the traversal of the subtree of descendants of the edge $e_{i}$).

Some of these intervals can be empty, however, they do in any case cover all signs that have been erased by time $t$. ∎

We now describe a link in the opposite direction. Let $p$ be a re-pairing of the word $\sigma$. We will construct a tree $(I,E)$ of width $\leq\mathop{\mathrm{width}}\nolimits(p)+1$ and a derivation $\pi$ based on this tree such that $p_{\pi}=p$.

Let us first construct a tree $(I^{\prime},E^{\prime})$ in which the nodes are pairs of the form

[TABLE]

The time here is specified according to the sequence $p$. The intervals are specified by their start and end positions. This way a node $([s,f],t)$ corresponds to a maximal interval $[s,f]$, between positions $s$ and $f$, which has been erased completely by time $t$.

The ancestor—descendant relation on the pairs of this form is the conjunction of the set inclusion for intervals and the ordering on time points: a pair $([s_{1},f_{1}],t_{1})$ is a descendant of a pair $([s_{2},f_{2}],t_{2})$ if and only if

[TABLE]

Here we have taken into account that the time on the tree and in the re-pairing flows in opposite directions.

All possible ways in which the erased intervals can evolve when a new pair $(\ell_{t},r_{t})$ of signs is erased in the re-pairing $p$ are shown in Fig. 4.

All intervals not adjacent to the freshly erased (paired) signs remain unchanged (case (1)). The node of the tree $(I^{\prime},E^{\prime})$ that corresponds to this interval at the time right after these two signs are paired has exactly one child. The cases (2)–(6) are similar: in these cases the interval evolves, but does not merge with other intervals. For this reason, the nodes corresponding to such intervals at this time have exactly one child too. As a special case, these scenarios can capture a new interval emerging (the dashed area in the picture is empty), that is, a new leaf appearing in the tree $(I^{\prime},E^{\prime})$.

In the case (7) one of the freshly erased (paired) signs is adjacent to two intervals. Dashed lines in the picture indicate positions of signs that may also be paired at this time. The node of the tree $(I^{\prime},E^{\prime})$ that corresponds to the new interval has two children in this case. Case (8) shows another way two intervals can merge, in which there are two signs between the intervals, and they are paired with one another at this point. There are no further cases of two-interval mergers: indeed, since just two signs are erased at a single time point, there may be at most two signs between the merging intervals.

Finally, case (9) shows the last possible scenario, one where three intervals are merged into a single one. The relative position of signs is determined uniquely in this case.

We will now transform the tree $(I^{\prime},E^{\prime})$ into a binary tree $(I,E)$ and a derivation $\pi$ based on the latter tree such that $p_{\pi}=p$. As seen from Fig. 4, the tree $(I^{\prime},E^{\prime})$ may have nodes with three children, in which case auxiliary nodes need to be inserted so that the tree $(I,E)$ would be binary.

The way the signs are positioned on the edges of the tree $(I,E)$ is based on Fig. 4 and shown in Fig. 5.

Note that due to the case (9) there is no bijection between the points in time in the tree and the points in time in the re-pairing. Still, the ordering of erased pairs in time in the tree is the reverse of the ordering in the re-pairing $p$. This ensures the equality $p_{\pi}=p$.

Another observation is that in the cases (5)–(6) the width of the tree becomes greater by $1$ than the width of the re-pairing $p$, because the signs are placed on (auxiliary) edges that lead to auxiliary leaves. Informally, one can think of this as of a new interval (of $+$ and $-$) being “born” first and merged with an old interval second. In the re-pairing $p$ these events occur simultaneously.

Combining all the arguments above, we obtain a new characterization of the width of a well-formed word.

Theorem 6**.**

The difference between the width of a well-formed word $\sigma$ and the minimum width of a tree $(I,E)$ for which there exists a derivation $\pi$ generating the word $\sigma$, i.e., $\sigma=\sigma(\pi,I,E)$, is at most $1$.

Appendix C Proofs for Section 3

C.1 Proof of the logarithmic upper bound

In this subsection we prove Theorem 1. We will use several observations.

Claim 6**.**

*Let $\sigma=\sigma_{1}\cdot\sigma_{2}\cdot\ldots\cdot\sigma_{t}$ be a factorization of a well-formed word into well-formed words. Then *

[TABLE]

Proof.

Denote by $p_{i}$ an optimal re-pairing of a word $\sigma_{i}$.

Consider a re-pairing $p$ of $\sigma$ which erases the words $\sigma_{1}$, $\sigma_{2}$, …, $\sigma_{t}$ consecutively and according to the optimal re-pairings $p_{1},\ldots,p_{t}$. The width of this re-pairing at all times when the word $\sigma_{i}$ is being re-paired cannot exceed the number of erased intervals in the re-pairing $p_{i}$ plus possibly an additional interval that consists of the fully erased word $\sigma_{1}\cdot\ldots\cdot\sigma_{i-1}$. ∎

Claim 7**.**

Let a well-formed word $\sigma$ factorize as $L\cdot\pi_{1}\cdot R$, where $\pi_{1}$ is a well-formed word. Then the word $\pi_{2}=LR$ is also well-formed and

[TABLE]

Proof.

The first statement is obvious.

Consider a re-pairing $p$ of the word $\sigma$ which first erases the word $\pi_{1}$ (optimally, using a re-pairing $p_{1}$) and then the word $\pi_{2}$ (also optimally, using a re-pairing $p_{2}$).

Before the beginning of $\pi_{2}$’s re-pairing, the width of $p$ does not exceed $\mathop{\mathrm{width}}\nolimits(p_{1})$. From that point on, the word $\pi_{1}$ is erased completely, so the width of $p_{2}$ can be increased by at most $1$ (which corresponds to the erased interval $\pi_{1}$). ∎

Proof of Theorem 1.

Use induction on the length of the well-formed word, $N=|\sigma|$. Construct a sequence of nested factors of $\sigma$ in the following way.

Any Dyck word is a concatenation of Dyck primes, i.e. words of the form ${+}w{-}$, where $w$ is a Dyck word.

If $\sigma=\sigma_{1}\ldots\sigma_{m}$, where $\sigma_{i}$ are Dyck primes, then pick the factor $\sigma^{(1)}=\sigma_{j}$ of maximum width.

Since $\sigma^{(1)}$ is a Dyck prime, $\sigma^{(1)}={+}\sigma_{1}^{(1)}\dots\sigma_{m(1)}^{(1)}{-}$, where $\sigma_{i}^{(1)}$ are Dyck primes.

If some $\sigma_{j}^{(1)}$ has length greater than $N/2$, set $\sigma^{(2)}=\sigma_{i}^{(1)}$. Note that $\sigma^{(1)}=L_{2}\sigma^{(2)}R_{2}$, and $|L_{2}R_{2}|<N/2$.

Repeating this procedure produces a sequence of Dyck primes $\sigma^{(i)}$: if $\sigma^{(i)}={+}\sigma_{1}^{(i)}\dots\sigma_{m(i)}^{(i)}{-}$ and $|\sigma_{j}^{(i)}|>N/2$ for some $1\leq j\leq m$, then $\sigma^{(i+1)}=\sigma_{j}^{(i)}$. It is easy to check that $\sigma^{(1)}=L_{i+1}\sigma^{(i+1)}R_{i+1}$ and $|L_{i+1}R_{i+1}|<N/2$.

The proccess stops on a Dyck prime $\sigma^{(f)}$ such that

[TABLE]

where $|\sigma_{j}^{\prime}|\leq N/2$ for all $1\leq j\leq m(f)$. We have $\sigma^{(1)}=L_{f}\sigma^{(f)}R_{f}$ and $|L_{f}R_{f}|<N/2$.

Applying Claim 6 to the factorization of $\sigma$, Claim 7 to the factorization $\sigma^{(1)}=L_{f}\sigma^{(f)}R_{f}$, and then again Claims 7 and 6 to the factorization of $\sigma^{(f)}$, obtain the inequality

[TABLE]

It is now easy to see that $\max\limits_{|\sigma|=N}\mathop{\mathrm{width}}\nolimits(\sigma)\leq 3\log_{2}N$. ∎

C.2 Simple re-pairings and pebble games: Proof of Theorem 2

In this subsection we prove Theorem 2.

We will rely on the interpretation of Dyck words as trees, discussed in Section B.1.

We will use slightly modified rules of the black-and-white pebble game, where instead of moves of the form (M4) we will only permit those of the form

1. (M4’)

replace a white pebble on a node with a black pebble, provided that all its immediate predecessors carry pebbles.

It is easy to see that replacing the rule (M4) with the rule (M4’) leaves the value of $\mathrm{bw}(G)$ unchanged.

Note that the variant of the game considered by Lengauer and Tarjan [LT80] also permits two types of moves where pebbles can be moved. Such a move can be replaced by two moves that our rules admit; the value of $\mathrm{bw}(G)$ either remains unchanged, or increases by $1$. Since we are only interested in asymptotic bounds, all results carry over between the two variants.

Consider a simple re-pairing $p$ of a word $\sigma$. Construct, based on $p$, a new re-pairing $p^{\prime}$ of the same word as follows. Each time $p$ erases a pair of signs associated with a non-root node $v$ (in the tree associated with $\sigma$), the new re-pairing $p^{\prime}$ will erase two pairs of signs associated with nodes $v$ and $s(v)$ (if these signs have not been erased previously and if, for $s(v)$, this node exists in the tree; see Fig. 6). It is easy to see that the re-pairing $p^{\prime}$ is simple.

Claim 8**.**

$\mathop{\mathrm{width}}\nolimits(p^{\prime})\leq 3\cdot\mathop{\mathrm{width}}\nolimits(p)$.

Proof.

Consider a point in time when $p^{\prime}$ erases two pairs of signs associated with nodes $v$ and $s(v)$. One of the signs associated with $s(v)$ is adjacent to one of the signs associated with $v$ and, therefore, does not increase the number of erased intervals at any point compared to $p$. The other sign associated with $s(v)$ can join an existing interval or form a new, separate interval. In the latter case we will say that this sign becomes special. It stops being special when the sign adjacent to it and associated with the node $a(v)$ is erased. The position of these signs is shown in Fig. 6.

It is easy to see that at all times the number of erased intervals in $p^{\prime}$ cannot be greater than the sum of the number of erased intervals in $p$ and the number of special signs. Indeed, every sign that has stopped being special has already joined another interval. (Note that the converse does not hold.)

We now bound the number of special signs from above. Consider an arbitrary point in time. Let the sign $\sigma(i)$ be special and associated, as previously, with the node $s(v)$. Then the signs associated with nodes $v$ and $s(v)$ are already erased, and the signs associated with $a(v)$ are not. Observe that in this case the position between the two adjacent signs associated with $v$ and $a(v)$ is an endpoint of an erased interval at this point in time. Moreover, this position is an endpoint of an erased interval in the re-pairing $p$ too. If we assign this position to the special sign $\sigma(i)$, then no position will be assigned to two or more signs, and so the number of special signs will be at most double the number of erased intervals in the re-pairing $p$. Combined with the argument above, this gives the inequality $\mathop{\mathrm{width}}\nolimits(p^{\prime})\leq 3\cdot\mathop{\mathrm{width}}\nolimits(p)$. ∎

Based on the re-pairing $p^{\prime}$, construct a strategy in the black-and-white pebble game on the associated tree, using the following rules:

1. (P1)

when a pair of signs is erased, place a white pebble on the associated node $v$; 2. (P2)

every time and everywhere, when a node $v$ carries a white pebble and all children of $v$ carry pebbles, replace the pebble on $v$ with a black one; 3. (P3)

every time and everywhere, when the tree has a pair of nodes $v$ and $w$ such that $v$ is a descendant of $w$, both carry black pebbles, and the path between them has no pebbles, remove the pebble from $v$.

Rules (P2) and (P3) are always applied until neither is applicable; in other words, rule (P1) is applicable only if rules (P2) and (P3) are not. We emphasize that our strategy must comply with this restriction.

Let $v$ be a node in the tree $D$. Consider the path from $v$ to the root of the tree. Let $u$ be the first node on this path that carries a pebble. We will use the following notation: $\chi(v)$ is $1$ if the pebble on $u$ is black and [math] if the pebble is white or if there is no such node $u$.

Claim 9**.**

At all points in time the following invariant holds: for all pairs of (matching) signs in the word $\sigma$, the pair has been erased if and only if the associated node $v$ carries a pebble or satisfies the equality $\chi(v)=1$.

Proof.

Use induction on time (the number of performed moves). At the beginning nothing is erased and $\chi(v)=0$ for all $v$. We will now consider an application of each of the three rules that define our construction of the strategy in the pebble game.

First suppose rule (P1) is applied. By the inductive hypothesis, we have $\chi(v)=0$, so placing a pebble onto $v$ does not change $\chi(v)$ for any node in the tree.

Now suppose rule (P2) is applied. Under the conditions of the replacement, the set of nodes that carry pebbles is unchanged, and for all other nodes the value of $\chi$ remains the same as previously.

Finally, suppose rule (P3) is applied. It is clear that the invariant will continue to hold for the node $v$; and for the nodes $u$ that had $\chi(u)=1$ due to the black pebble on $v$ this value will remain unchanged due to the black pebble on $w$. ∎

Claim 10**.**

After completion of the re-pairing, the root carries a black pebble and there are no white pebbles on the tree.

Proof.

Consider the point in time when the re-pairing is complete and rules (P2) and (P3) are not applicable. The root carries a pebble by the invariant, because it has no outgoing edges. Let us prove that there are no white pebbles on the tree. Assume for the sake of contradiction that $v$ is a node which carries a white pebble and the descendants of which carry no white pebbles. This node cannot be a leaf, because, if it were, rule (P2) would be applicable. On the other hand, all children of $v$ carry pebbles, since the associated pairs of signs are erased and the invariant holds. But then rule (P2) is applicable to $v$. This contradiction concludes the proof. ∎

Claim 10 means that the constructed strategy is successful.

Claim 11**.**

The number of pebbles used by the strategy does not exceed $4\cdot\mathop{\mathrm{width}}\nolimits(p^{\prime})+3$.

Proof.

First consider an arbitrary point in time when rules (P2) and (P3) are not applicable and, for every node $v$, if the associated pair of signs has been erased, then the pair of signs associated with $s(v)$ has also been erased (if this node exists in the tree). By our choice of the re-pairing $p^{\prime}$, the rule (P1) is applied at most twice between any two consecutive points in time with this property.

We now find two adjacent signs in the word $\sigma$, exactly one of which has been erased, (a) for each white pebble and (b) for each black pebble lying on a node which is not a root and which has no descendants that carry white pebbles.

First consider scenario (a). If a node $v$ carries a white pebble, then $v$ cannot be a leaf, because of rule (P2). By the same rule, at least one child $u$ of this node carries no pebble. Since the invariant holds, the two signs associated with the node $u$ have not been erased yet, but the signs associated with $v$ have. This gives us the two desired signs: the erased one is associated with the node $v$, and the surviving one with its child $u$.

Now consider scenario (b). Let $v$ be a black pebble satisfying the condition in question. The signs associated with $v$ are erased; by our choice of the re-pairing $p^{\prime}$, so are the signs associated with $s(v)$, if this node exists in the tree. Now observe that the signs associated with $a(v)$ have not been erased yet: otherwise, due to the invariant, either the node $v$ carries a black pebble (this pebble cannot be white because of rule (P2)), or the condition $\chi(a(v))=1$ is satisfied and there is a black pebble on one of the ancestors of $v$; but then in both cases rule (P3) is applicable. Therefore, in this scenario we have also found two adjacent signs, associated with nodes $v$ and $a(v)$, where the first has been erased and the second has not.

It is straightforward that all the pairs of adjacent signs that we have found are distinct. Indeed, in both scenarios these signs are associated with pairs of nodes in the tree, one of which is a child of the other. But the erased sign is associated with the parent node in scenario (a), and with the child node in scenario (b).

Therefore, the sum of the number of white pebbles and the number of black pebbles, except those that lie on the root or on a node which has a descendant carrying a white pebble, does not exceed double the number of the erased intervals. Now observe that for all excepted black pebbles (apart from one on the root—if there is such a pebble) the path to the descendant carrying a white pebble cannot meet a black pebble before a white one (due to rule (P3)). Charge this white pebble to our black pebble; clearly, each white pebble can be charged at most one black pebble. So the total number of pebbles cannot exceed by more than $1$ the number of erased intervals times $4$.

We have considered only a subset of (the set of) all time points. At all other times the number of pebbles cannot be greater by more than $2$, because of the form of rule (P1). This completes the proof. ∎

Combining Claims 8 and 11 (as well as Claim 10) now gives the lower bound in Theorem 2.

The upper bound is given by the following statement:

Claim 12**.**

For every successful strategy in the black-and-white pebble game on a binary tree there exists a re-pairing of the associated word with width at most twice the number of pebbles.

Proof.

Use the results of the papers [Lou79, Mey81, LT80] (see also [Sav98, p. 526–528]): if the graph $G$ is a tree, then for every successful strategy in the black-and-white pebble game on $G$ there exists another successful strategy that uses at most twice the original number of pebbles, but no white pebbles at all. Given this fact, it suffices to construct a re-pairing with the required width just for strategies using black pebbles only.

With no loss of generality, we consider only the strategies having an auxiliary property. Every time a black pebble is removed from a node $v$ using rule (M2), we will assume that the path from $v$ to the root has a black pebble. Indeed, if this is not the case, then the strategy in the pebble game can be modified as follows: instead of removing the pebble from $v$, we could have refrained from putting it on $v$ altogether; this pebble will necessarily be placed on $v$ in the future anyway with the goal of reaching the root. This modification yields a strategy that uses the same or lower number of pebbles and, at the same time, fewer moves. Thus, repeating modifications of this sort leads to a strategy satisfying the property stated above.

We now construct the required re-pairing in the following natural way: every time a node $v$ receives a pebble, we will erase the two signs associated with $v$ (unless they have been erased previously). We prove by induction on the number of moves that the number of erased intervals cannot exceed the number of used pebbles and, moreover, every node that has carried a pebble is associated with a pair of signs in the word $\sigma$, between which all signs have already been erased.

Indeed, this condition holds at the very beginning. When a black pebble is put on a node $v$ according to rule (M1), the children of that node (if they exist in the tree) already carry pebbles and the associated intervals have already been erased and become one erased interval. Erasing the pair of signs associated with $v$ can only extend this interval (and perhaps attach it to another interval). Finally, when a black pebble is removed from a node according to rule (M2), we know from the property above that the path from this node to the root has a black pebble. This means that the intervals associated with these two nodes are nested in one another, so the removal of the smaller will not violate the inequality in question. These two are the only rules in the game that use black pebbles only, so this completes the proof. ∎

This concludes the proof of Theorem 2.

Appendix D Proofs for Section 4

D.1 Correctness of the definition

The only possible issue with the definition of the re-pairing $p(q,n,k)$ is the choice of subsequence $Z^{\prime}_{t}$. All other actions are specified either by recursively defined re-pairings, or by the direct rule ‘pair the current minus with the leftmost non-erased plus’.

We will prove that the re-pairing $p(q,n,k)$, $q\leq(n+1)/2$, is well-defined by induction on $n$.

The base case, $n\leq 2$, is trivial because there are no stages at all in this case.

For the induction step, assume that for all $n<n_{0}$ the re-pairings $p(q,n,k)$, $q\leq(n+1)/2$, are well-defined.

We prove that the re-pairing $p(q,n,k)$ is well-defined by inner induction on stages. The base of the inner induction is obvious, since $k_{1}=0$ by definition.

For the inner induction step, assume that the re-pairing $p(q,n_{0},k)$ is well-defined for all stages $t^{\prime}<t$. At the beginning of the stage $t$, no sign in the factor $Z_{t}$ has been erased. Indeed, this is obvious for the minuses, because $p(q,n,k)$ picks them in a greedy way, i.e., in the same order in which they occur in the word. But the property in question also extends to the pluses, because they cannot be paired with minuses to the left of $Z_{t}$.

We now show that we can choose, from the sequence of non-erased signs, a subsequence $Z^{\prime}_{t}$, as specified in the definition of $p(q,n,k)$. Recall that at each moment in time the non-erased signs form a Dyck word. So, to pick up $k_{t}$ pluses to the left of the factor $Z_{t}$ (a Dyck word itself) it suffices to pick up $k_{t}$ minuses to the right of the factor.

By definition (4), $k_{t}\leq q-1$. On the other hand, the word $Z(n)^{(k)}$ has the suffix $-^{k+n-q}$ that lies to the right of any factor $Z(q)$. Taking into account the restriction on the parameter values $q\leq(n+1)/2$ we get

[TABLE]

This inequality guarantees that there are at least $k_{t}$ minuses to the right of the factor $Z_{t}$, for each $t$.

Remark 4*.*

Although the sequence $Z^{\prime}_{t}$ contains $k_{t}$ minuses to the right of the factor $Z_{t}$, only the minuses to the left of $Z_{t+1}$ are erased during stage $t$, by our definition of stages.

D.2 Properties of the re-pairing

Let $w_{t}$ be the factor between $Z_{t}$ and $Z_{t+1}$, i.e., the (unique) word such that $Z_{t}\cdot w_{t}\cdot Z_{t+1}$ is a factor of $Z(n)$.

Claim 13**.**

$w_{0}=+^{r_{0}}$, where $r_{0}=k+n-q$; $w_{N}=-^{r_{N}}$, $r_{N}=r_{0}$; and $w_{i}=-^{r_{i}}+^{r_{i}}$ for $1\leq i\leq N-1$. Here

[TABLE]

Proof.

Use induction on $n$. The base case, $n=q$, is trivial. For the inductive step, consider the factorization of $Z(n+1)^{(k)}$ into

[TABLE]

The arithmetic condition on $r_{0}=r_{N}$ is obvious. There are exactly $2^{n-q}$ occurrences of $Z(q)$ in $Z(n)$. Thus, we need to check only the word between the last occurrence of $Z(q)$ in the first factor $Z(n)$ and the first occurrence of $Z(q)$ in the second factor $Z(n)$. This word is the concatenation of the suffix $-^{n-q}$ from the first factor $Z(n)$ and the prefix $+^{n-q}$ from the second factor $Z(n)$. So it does have the right form, and $r_{2^{n-q}}=n-q$. This proves (6). ∎

Note that in the final part of a stage $t$, possibly empty, we pair minuses in the word $w_{t}$ with the leftmost non-erased pluses.

In the sequel, we say that a stage is a narrow-frame stage if the final part of it is non-empty. Otherwise, we call a stage wide-frame. This classification will be crucial in the following analysis.

From Claim 13 we conclude that a stage $t$ is narrow-frame iff $r_{t}>k_{t}$. Let us compare the values of $r_{t}$ and $k_{t}$. Note that the sequence $k_{t}$ is periodic. For $t=2^{q}\cdot s+a$, $0<a\leq 2^{q}$, we get

[TABLE]

Claim 14**.**

For $1\leq q\leq n/3$, $n\geq 15$, and $k\leq n$, the re-pairing $p(q,n,k)$ erases all pluses from the prefix $+^{k}$ of the word $Z(n)^{(k)}$ before the first minus from the suffix $-^{k}$.

Proof.

We prove that the pluses from the prefix $+^{k}$ are exhausted before the factors $Z(q)$ in the second half of the word start getting erased. It is clear that at this point in time no minus from the suffix has been paired.

It follows from (7) that there are exactly $q$ narrow-frame stages during a period $2^{q}\cdot s<t\leq 2^{q}\cdot(s+1)$. Thus, at least $q$ leftmost non-erased pluses in total are erased during these stages. The first $k$ pluses, therefore, will be exhausted after at most $\lceil k/q\rceil$ periods. There are $2^{n-1-q}$ factors $Z(q)$ in the first half of the word $Z(n)$. So it is sufficient to satisfy the inequality

[TABLE]

Note that for $q=1$ we have $\lceil\frac{k}{q}\rceil=k$. So for $n\geq 15$, $k\leq n$ we get

[TABLE]

For $q\geq 2$, we have $\lceil\frac{k}{q}\rceil\leq k/2+1$. So for $n\geq 12$, $k\leq n$ and $1<q\leq n/3$ we get

[TABLE]

Corollary 3**.**

For $15\leq q\leq(n+1)/2$ and $k\leq n$, all pluses in $Z^{\prime}_{t}$ from the prefix ${+}^{k_{t}}$ (i.e., those preceding the factor $Z_{t}$) are erased during the stage $t$ of the re-pairing $p(q,n,k)$.

Now we are going to prove that the re-pairing $p(q,n,k)$ maintains a specific shape of the sequence of non-erased pluses. We call a stage $t$ normal if, at the beginning of the stage, the following conditions hold:

1. (N1)

Non-erased pluses to the left of the factor $Z_{t}$ are divided, for some $s$, into (possibly empty) $s$ intervals (runs of ‘skipped’ pluses) $P_{1}$, …, $P_{s}$ of length $m_{1}$, …, $m_{s}$, respectively, and a group of $k_{t}$ pluses that form a prefix of $Z^{\prime}_{t}$. (The latter do not have to form a contiguous interval, i.e., a factor of $Z(n)$, and can be scattered inside $Z(n)$.) 2. (N2)

The last group lie in the factor $Z_{t-1}\cdot w_{t-1}$. 3. (N3)

Each $P_{i}$ is a subfactor of $w_{j_{i}}$, $j_{i}<t$. (Or, equivalently, all factors $Z_{i}$, $i<t-1$, have been completely erased.) 4. (N4)

$m_{i}\geq q$ for $i>1$.

Lemma 1**.**

For $15\leq q\leq(n+1)/2$ and $k\leq n$, each stage of the re-pairing $p(q,n,k)$ is normal.

Proof.

Use induction on stages. The base case, $t=1$, is trivial.

For the induction step, assume that a stage $t-1\geq 1$ is normal. The condition (N2) for it and Corollary 3 imply that at the beginning of stage $t$ all symbols in the factor $Z_{t-2}$, if any, have been erased. Thus, the condition (N3) holds for stage $t$. In addition, everywhere below we will also rely on the following consequence of Claim 14: the group of $k_{t-1}$ pluses from the beginning of $Z^{\prime}_{t-1}$ gets erased completely by the beginning of stage $t$.

The key intuition for the rest of the proof is based on the definition of the sequence $Z^{\prime}_{t}$. In particular, note that all minuses that can be erased on stage $t-1$ and earlier lie in the factors $Z_{t-1}$ and $w_{t-1}$. Since $w_{t-1}={-}^{r_{t-1}}{+}^{r_{t-1}}$, it follows that none of the pluses in ${+}^{r_{t-1}}$ can be erased by the beginning of stage $t$. These are the $r_{t-1}$ rightmost non-erased pluses to the left of $Z_{t}$, and recall that the definition of $Z^{\prime}_{t}$ requires $k_{t}$ such pluses. So for the rest of normality conditions, we need relations between $r_{t-1}$ and $k_{t}$, where $t=2^{q}\cdot s+a$, $1\leq a\leq 2^{q}$. They are as follows:

[TABLE]

If $t=2^{q}\cdot s+1$, $s\geq 1$, then the second line of (8) shows that stage $t-1$ is narrow-frame and, in fact, a group of $\geq q$ skipped pluses appears at the beginning of the stage $t$. It implies the conditions (N4) and (N1). The condition (N2) is trivial in this case, $k_{t}=0$.

If $t=2^{q}\cdot s$, $s\geq 1$, or $t=2^{q}\cdot s+a$, $2^{j}+1<a<2^{j+1}+1$, $1\leq j<q$, then the first and the last lines of (8) show that $k_{t}>r_{t-1}$. Note that $k_{t-1}=k_{t}$ in this case by equation (7). Applying Corollary 3 once more, we deduce that $k_{t-1}-r_{t-1}$ pluses remain non-erased in $Z_{t-1}$ and all of them lie in $Z^{\prime}_{t}$. It gives the conditions (N2) and (N1). The condition (N4) also holds because the runs of ‘skipped’ pluses do not change in this case.

If $t=2^{q}\cdot s+2$ or $2^{q}\cdot s+a$, $a=2^{j}+1$, $1\leq j<q$, then all signs in $Z_{t-1}$ are erased at the beginning of the stage $t$. The runs of ‘skipped’ pluses do not change. The $k_{t}=r_{t-1}$ pluses from $w_{t-1}$ are included into $Z^{\prime}_{t}$. Thus, the conditions (N2), (N1) and (N4) hold. ∎

Note that the properties mentioned in Section 4 follow from the conditions of normality. Thus, Lemma 1 implies them.

D.3 Proof of the recurrence for the width

We now prove Claim 2 from Section 4.

Claim 15**.**

For $15\leq q\leq(n+1)/2$, $1\leq q^{\prime}\leq q/3$, $k\leq n$,

[TABLE]

Proof.

To prove (9) we estimate the width of $p(q,n,k)$ at any moment in time. Recall that the erased signs constitute a well-formed word at any point in time as well as the non-erased signs. The height of any position in the entire word is the sum of the heights of this position with respect to the erased signs only and to the non-erased signs only.

In the re-paring $p(q,n,k)$, the skipped runs of pluses form a prefix of the word that consists of all non-erased signs. From the above argument we conclude that the height of this prefix, i.e. the number of signs in the skipped runs of pluses, is at most $n+k\leq 2n$ (the maximal height of a position in the framed word $Z(n)^{(k)}$ is $n+k$).

But the skipped runs of pluses $P_{1},\ldots,P_{s}$, except possibly the first one, have size at least $q$ due to normality condition (N4). This means that there are at most $2n/q+1$ of them. These skipped (non-erased) runs separate at most $2n/q+2$ erased intervals.

At any point in time at most two factors $Z(q)$, namely $Z_{t-1}$ and $Z_{t}$, are being erased, by normality condition (N3). Moreover, the only non-erased signs in $Z_{t-1}$ at this moment are pluses.

The case analysis in the proof of Lemma 1 shows that if both of these factors are partially erased at the moment, then $k_{t}=k_{t-1}$. It implies that in the re-pairings of $Z^{\prime}_{t-1}$ and $Z^{\prime}_{t}$ these pluses are erased in a consistent way. In the former they are erased from the left to the right during the last (narrow) stage of the re-pairing. In the latter they are located in the first run of ‘skipped’ pluses, and are thus erased from the left to the right.

This consistency of the re-pairings ensures that contribution of these factors to the number of erased intervals is at most

[TABLE]

(We ignore in this upper bound that $q^{\prime}$ is chosen optimally at each stage.)

Finally, one more erased interval separated from the ones listed above may occur inside $w_{t}$. Putting everything together, we obtain (9). ∎

Note that the right-hand side of (9) does not depend on $k$. To get the recurrence of Claim 2, observe that

[TABLE]

If $q<45$, then the inner minimum in this expression is, by definition, equal to $W_{q}$, and so

[TABLE]

Otherwise, if $q\geq 45$, notice that the minimum over a set decreases if a set becomes larger, so

[TABLE]

which gives us equation (10) again.

D.4 Proof of the upper bound on the width

It remains to obtain the upper bound from the recurrence (10), completing the proof of Theorem 3. We begin by simplifying the recurrence a little.

We will bound the sequence of $W_{n}$s from above by the sequence $x_{n}-3$, where $x_{n}$ is given by

[TABLE]

Claim 16**.**

$W_{n}\leq x_{n}-3$.

Proof.

For $n<45$, the equality $W_{n}=x_{n}-3$ holds. For greater values of $n$ the inequality is proved by induction, using the recurrences (10) and (11):

[TABLE]

To obtain asymptotic bounds on $x_{n}$, we need a different representation of these numbers. Denote by ${\mathcal{Z}}_{n}$ the set of sequences

[TABLE]

of positive integers satisfying the conditions (i) $z_{0}=n$, (ii) $15\leq z_{s}\leq 44$, and (iii) $z_{i}/z_{i+1}\geq 3$ for $i=0,\ldots,s-1$. For the sequence $z$ we denote by $s(z)$ its length minus $1$. Finally, let the function $\xi$ be defined on sequences by

[TABLE]

Claim 17**.**

$x_{n}=\min\limits_{z\in{\mathcal{Z}}_{n}}\xi(z)$* for all $n\geq 45$.*

Proof.

Use induction on $n$. The base case, $n=45$, is easily checked—the set ${\mathcal{Z}}_{45}$ contains a unique sequence $(45,15)$.

The induction step is given by the direct computation (which is actually Bellman’s principle of optimality) as follows:

[TABLE]

The second equality is the inductive hypothesis, and the last is the definition of the sequence $x_{n}$. ∎

It is easier to estimate the minimum of $\xi(z)$ over a bigger set ${\mathcal{R}}_{n}$, defined analogously to ${\mathcal{Z}}_{n}$ but in which the terms of the sequences are arbitrary real numbers (instead of positive integers only). More exactly,

[TABLE]

if (i) $z_{0}=n$, (ii) $15\leq z_{s}\leq 44$ and $z_{s}$ is integer, and (iii) $z_{i}/z_{i+1}\geq 3$ for $i=0,\ldots,s-1$.

Notice that the properties (i)–(iii) imply that for all sequences $z\in{\mathcal{R}}_{n}$ the length $s(z)$ does not exceed $\log(n/15)/\log 3$. Therefore, for every fixed $n$ the set ${\mathcal{R}}_{n}$ is a compact set, and in particular all functions continuous in $z_{i}$ attain a minimum on this set.

Since ${\mathcal{Z}}_{n}\subseteq{\mathcal{R}}_{n}$, we have $\min\limits_{z\in{\mathcal{R}}_{n}}\xi(z)\leq\min\limits_{z\in{\mathcal{Z}}_{n}}\xi(z)$. To obtain (a variant of) the converse inequality for the minima of $\xi(z)$ over the sets ${\mathcal{Z}}_{n}$ and ${\mathcal{R}}_{n}$, we will need two simple observations about real numbers.

Claim 18**.**

Let $x$ and $y$ be real numbers such that $y\geq 1$ and $x/y\geq 3$. Then $\lfloor x\rfloor/\lfloor y\rfloor\geq 3$.

Proof.

For the sake of contradiction, suppose $\lfloor x\rfloor<3\lfloor y\rfloor$, or equivalently $\lfloor x\rfloor\leq 3\lfloor y\rfloor-1$. Since $\lfloor x\rfloor+\{x\}\geq 3(\lfloor y\rfloor+\{y\})$, we obtain the inequality

[TABLE]

and hence $1\leq\{x\}-3\{y\}<1$. This is a contradiction. ∎

Claim 19**.**

If $x\geq 1$ and $y\geq 2$, then

[TABLE]

Proof.

The required inequality is implied by the chained inequality

[TABLE]

Claim 20**.**

$\min\limits_{z\in{\mathcal{Z}}_{n}}\xi(z)\leq\frac{3}{2}\min\limits_{z\in{\mathcal{R}}_{n}}\xi(z)$.

Proof.

Let $z^{*}=(z_{0}^{*},\dots,z_{s}^{*})$ be the sequence at which $\xi(z)$ attains its minimum on the set ${\mathcal{R}}_{n}$.

Consider the sequence $z^{\prime}=(z^{\prime}_{0},\dots,z^{\prime}_{s})$ with $z^{\prime}_{i}=\lfloor z_{i}^{*}\rfloor$. By Claim 18, we have $z^{\prime}\in{\mathcal{Z}}_{n}$. Applying the inequality from Claim 19 to each term in equation (12), we obtain the inequality $\xi(z^{\prime})\leq\frac{3}{2}\xi(z^{*})$. The desired inequality for the minima follows. ∎

Claim 21**.**

$x_{n}=2^{\Theta(\sqrt{\log n})}$.

Proof.

We obtain the upper bound by choosing an appropriate sequence from ${\mathcal{R}}_{n}$ and applying Claim 20. Let $s$ be the maximum integer $x$ satisfying the inequality

[TABLE]

i.e., $s=\Theta(\sqrt{\log n})$. Consider the sequence $z_{s-i}=2\cdot 2^{(i+1)^{2}/2}$, $2\leq i\leq s-1$. For $z_{0}=n$, the sequence $(z_{0},z_{1},\dots,z_{s-2},15)$ belongs to the set ${\mathcal{R}}_{n}$, since $z_{s-2}=2\cdot 2^{9/2}>3\cdot 15$ and for all $i\geq 2$ it is the case that

[TABLE]

But for this sequence we have

[TABLE]

By Claim 20, we have

[TABLE]

We now prove the lower bound. For $z\in{\mathcal{Z}}_{n}$ introduce the notation

[TABLE]

It follows from Claim 17 that

[TABLE]

Since $44\cdot r(z)^{s(z)}\geq n$, we have

[TABLE]

The minimum is attained at $s=\sqrt{\log(n/44)}$. We get the bound $x_{n}=\Omega(2^{\sqrt{\log n}})$, and the lower bound of the Claim follows. ∎

Claim 3 now follows from Claim 16 and Claim 21.

Appendix E Proofs of lower bounds on the width of re-pairings

In this section we prove lower bounds in equation (1). In particular, Theorem 4 will follow from the following equation:

[TABLE]

where the words $Y(\ell)$ are defined in Section 5. For the words $Z(n)$ associated with complete binary trees, the same method will give us the following bound:

[TABLE]

Recall that $Z(n)=X(1,\ldots,1)=Y(1,n)$.

We will rely, throughout the section, on the tree representation of re-pairings, discussed in Section B.2.

E.1 Proof overview

Fix a well-formed word $W$. We will restrict our attention to symmetric words, i.e., those that stay unchanged if they are reversed and all signs are flipped (inverted).

It is immediate, for instance, that all words $Y(m,\ell)$ are symmetric. We will rely on the fact that this symmetry transforms a re-pairing (resp. a derivation) into another re-pairing (resp. another derivation) of the same width.

We already introduced the function $L(W,k)$, somewhat informally, in Section 5. Now we give a more formal definition, based on the ones in Section B.2.2.

Let $\pi$ be a derivation of a word $W$ based on a tree $(I,E)$, i.e. $W=\sigma(\pi,I,E)$. In Section B.2.3 we associate a fragment $F_{w}$ to each factor $w$ of the word $\sigma(\pi,I,E)$. Note that $F_{w}$ depends on $\pi$, $I$, $E$ as well.

Definition 1**.**

In the above notation,

[TABLE]

Recall the informal meaning of the definition. Given a word $W$, consider all possible tree derivations that generate $W$. Fragments of width at most $k$ in the trees (on which the derivations are based) are associated with factors of the word $W$, and $L(W,k)$ is the maximum length of such a factor. Note that in this definition the width of the (entire) trees is not restricted.

It is clear from the definition that the sequence of numbers $L(W,k)$ is non-decreasing:

[TABLE]

Furthermore, if $\mathop{\mathrm{width}}\nolimits(W)\leq k$, then $|W|\leq L(W,k+1)$ by Theorem 6.

We will obtain upper bounds on the numbers $L(W,k)$ by induction. For $L(W,1)$ we will use a bound based on a simple fact characterizing derivations of well-formed words. We will say that a pair of symbols is separated by a position $i$ if the first symbol from the pair occurs to the left of $i$ and the second symbol to the right of $i$.

Claim 22**.**

Let $i$ be a position in a well-formed word $\sigma$. Then every re-pairing of $\sigma$ has exactly $h(i)$ pairs that are separated by this position.

Proof.

First consider all pluses to the left of $i$ that are not paired with minuses to the right of $i$. If the number of the pairs in the statement of the Claim exceeds $h(i)$, then to the left of $i$ there are more minuses than the considered pluses. But then these signs cannot be paired up among themselves, which is a contradiction.

Second, the number of pluses to the left of $i$ is greater by $h(i)$ than the number of minuses to the left of $i$. Therefore, some $h(i)$ among these pluses have to be paired with the minuses to the right of $i$. This completes the proof. ∎

By $H(u)$ we will denote the maximum height of a position in a word $u$.

Lemma 2**.**

For every symmetric word $W$, $L(W,1)\leq 4H(W)$.

Proof.

If a factor $w$ is associated with a fragment of width $1$, then this fragment is a simple path. Factorize the word $w$ into $w_{1}\cdot w_{2}$, where the symbols $w_{1}$ are generated by the left sides of the edges and the symbols $w_{2}$ by the right sides.

Assume without loss of generality that $|w_{1}|\geq|w_{2}|$. (If this is not the case, consider the symmetric re-pairing, obtained by ‘reflecting’ the indices of signs with respect to the middle of $W$ and by inverting all signs.)

Since the signs from the factor $w_{1}$ are generated by the left sides of the edges, they cannot be paired with one another in the re-pairing (or, equivalently, in the derivation) of the word $W$. So each minus from the factor $w_{1}$ is paired with a plus to the left of the start position of $w_{1}$; similarly, each plus from $w_{1}$ is paired with a minus to the right of the end position of $w_{1}$. In other words, every sign from the factor $w_{1}$ is paired with a sign outside $w_{1}$, and this pair of signs is separated by either the start or end position of $w_{1}$.

The start and end positions of $w_{1}$ have height not exceeding $H(W)$, the maximum height of a position in $W$. Therefore, these positions cannot separate more than $H(W)+H(W)$ signs, and so $|w_{1}|\leq 2H(W)$, i.e., $|w|\leq 4H(W)$. ∎

We use Claim 22 not only in the proof of Lemma 2, but also in the proof of the inductive upper bound on $L(W,k)$. Besides this Claim, we will need two ‘distance’ characteristics of the word $W$ and their properties.

Before we introduce them, let us fix some further notation. Given a word over the alphabet $\{+1,-1\}$, define $\Delta(i,j)=h(j)-h(i)$ the increase in height from position $i$ to position $j$; $i\leq j$. Similarly, for a factor $w$, we write $\Delta(w)=\Delta(i,j)$ where $i$ and $j$ are the start and end positions of $w$, respectively.111We assume that a position in a word points to a place between symbols. Formally, a position corresponds to a partitioning of the word $w$ into a prefix and suffix: $w=p\cdot s$. Thus, an $n$-symbol word has $n+1$ positions, including the start position (empty prefix) and the end position (empty suffix).

We will now introduce the following definitions. For a word $w\in\{{+},{-}\}$, we denote by $\mu(w)$ the maximum length of a factor of $w$ that consists of minuses only:

[TABLE]

Definition 2**.**

\displaystyle\varphi(W,x)=\min\{\,\mu(w)\colon\text{w$is a factor of$W and}\ |w|\geq x\,\}.

Informally, the value of $\varphi(W,x)$ shows how many minuses in a row are guaranteed to occur in factors of length $x$ or more.

Definition 3**.**

\psi(W,x)=\min\{\,|uv{-}^{x}|\colon uv{-}^{x}\ \text{is a factor of W and}\ \Delta(u)\geq x\,\}-1.

Informally, the value of $\psi(W,x)$ shows how far away to the left (from a run of minuses) one needs to jump in order to find a factor on which the height increase matches the height drop on this run of minuses. (More precisely, $\psi(W,x)$ determines the minimum size of the jump, over all factors ${-}^{x}$ of $W$.) Since the word $W$ is well-formed, such a factor has to exist somewhere on the left, if ${-}^{x}$ is indeed a factor of $W$.

In the proofs below we will also use another interpretation of $\psi(W,x)$, as the largest number $\ell$ satisfying the following property: every factor of $W$ of the form $u{-}^{x}$, where $|u{-}^{x}|\leq\ell$, contains no sub-factor $u^{\prime}$ such that $\Delta(u^{\prime})\geq x$.

To obtain an upper bound on $L(W,k)$, we will use a recurrence of the following form.

Lemma 3**.**

There exist constants $A$, $C$, $N$ such that for all symmetric words $W$ of maximum height at least $N$, for all $k\geq 2$, if $L(W,k-1)\geq C\cdot H(W)\cdot k$, then

[TABLE]

One can take $A=17$, $C=17^{2}$, and $N=18$.

Remark 5*.*

For the (sequences of) words $W$ that we consider, the functions $\varphi$ and $\psi$ are ‘approximate inverses’ of each other: for big enough values of parameters, the following bounds hold:

[TABLE]

Given these bounds, the left-hand side of the inequality in Lemma 3 is approximately bounded from below by the fraction $L(W,k)/(A\cdot H(W)\cdot k)$. The quality of this approximation is determined by the lower bounds on $\varphi$ and $\psi$ above and by the magnitude of the coefficient $1/6k$.

Lemma 3 will enable us to obtain an inductive upper bound on $L(W,k)$ in terms of$L(W,k-1)$, if the latter is not too small. A proof of the bounds (13) and (14) can then follow.

The proof of Lemma 3 is based on two auxiliary lemmas. Consider a factor $w$ of the word $W$ associated with a fragment of width at most $k$ (in a tree derivation that generates $W$). We show (see subsection E.2 below) that the existence of a long enough sub-factor of the form $-^{d}$ in the factor $w$ ensures the existence of another sub-factor $u$ which is derived during the same time period as $-^{d}$ and which has a big enough height increase (of at least $d/6k$). On the other hand, the distance (in the word $W$) between sub-factors of the factor $w$ that are derived during a single time period cannot be big (subsection E.3). We discuss the proof of Lemma 3 in subsection E.4.

In order to apply Lemma 3, we will need to estimate the value of the functions $\varphi$ and $\psi$. These estimates are given in subsection E.5. Finally, in subsection E.6 below, we discuss the proofs of the lower bounds (13) and (14) (see p. 13) and Theorem 4.

E.2 Increases and drops in height within a single time period

Consider a derivation $\pi$ based on a tree $(I,E)$. Restrict the tree to just the edges that exist during the time period between some $t_{1}$ and $t_{2}$. These edges form a forest, see Fig. 7.

The (weakly) connected components of this forest occur in the traversal of the tree $(I,E)$ consecutively, as the following factors: (in Fig. 7) from the root of the component at time $t_{1}$ to the leftmost edge at time $t_{2}$ (using the left side of the edges), then from the leftmost edge at time $t_{2}$ to the second leftmost edge at the same time (using the right side of the edges at first and then switching to the left side), and so on.

These factors of the traversal (a sequence of length $2|E|$) define a system of intervals $[i_{\alpha},j_{\alpha}]\subseteq\{1,\dots,2|E|\}$. We assume that the intervals are ordered: $j_{\alpha}<i_{\alpha+1}$. In the derivation $\pi$ based on the tree $(I,E)$, this system generates factors $u_{\alpha}=\pi([i_{\alpha},j_{\alpha}])$ of the word $\sigma(\pi,I,E)$. Denote by $S(t_{1},t_{2})$ the sequence formed by these factors, that is,

[TABLE]

This is shown in Fig. 7.

We will say that factors $u_{i}$ and $u_{j}$, $i<j$, in the sequence $S(t_{1},t_{2})$ are independent if no plus from $u_{i}$ is derived by $\pi$ at the same time point as a minus from $u_{j}$.

By the balance of paired signs in the derivation, the following condition is satisfied. (Recall that $\cdot$ denotes the concatenation of sequences.)

Claim 23**.**

Suppose $S(t_{1},t_{2})=S_{0}\cdot S_{1}\cdot S_{2}\cdot S_{3}$, where all factors in $S_{0}$ are independent from all factors in $S_{1}\cdot S_{2}$; and all factors in $S_{2}$ consist of minuses only. Then

[TABLE]

Proof.

Suppose $S_{1}$ and $S_{2}$ have the form

[TABLE]

Denote by $m_{i}$ the number of minuses in the factor $u_{i}$ that are paired (in the derivation) with symbols outside $u_{i}$; and by $p_{i}$ the analogous number of pluses. Then the height increase on the factor $u_{i}$ is $\Delta(u_{i})=p_{i}-m_{i}$, because the signs paired inside $u_{i}$ do not contribute to the height change.

Since $S_{0}$ is independent from $S_{1}$ and $S_{2}$, the concatenation of all factors in the sequence $S_{1}\cdot S_{2}$ is a prefix of a well-formed word. So any minus from $u_{j}$ can only be paired with a symbol from the words $u_{1}$, …, $u_{j}$. The factors $v_{i}$ consist of minuses only, and these minuses can only be paired with pluses from the factors $u_{1}$, $\dots$, $u_{\ell}$. Since the height of all positions in a well-formed word is nonnegative, we obtain the following ‘balance’ conditions:

[TABLE]

The last inequality is the desired one. ∎

This balance condition will give us the first ingredient in the proof of the inductive upper bound on $L(W,k)$.

Lemma 4**.**

Suppose a factor $-^{d}$ of the word $\sigma=\sigma(\pi,I,E)$ is derived during time period from $t_{1}$ to $t_{2}$.

Also suppose that $S(t_{1},t_{2})=S_{0}\cdot S_{1,2}\cdot S_{3}$, where all factors in $S_{0}$ are independent from all factors in $S_{1,2}$; and all symbols from the factor $-^{d}$ are contained in the factors from $S_{1,2}$.

Finally, suppose the width of the set of edges that derive (all symbols in) the factors in $S_{1,2}$ does not exceed $k$.

Then there exists a factor of the word $\sigma$ which is derived during the same time period as $-^{d}$, occurs to the left of $-^{d}$ in the word $\sigma$, is a sub-factor of one of the factors in $S_{1,2}$, and on which the height increase is at least $d/6k$.

Proof.

Denote by $t_{1}$ the earliest time point at which minuses from the factor $-^{d}$ are derived, and by $t_{2}$ the last such time point. Also denote by $t_{s}$ the time the first minus from $-^{d}$ is derived, and by $t_{f}$ the time the last minus from $-^{d}$ is derived. (See Fig. 8.)

The time period from $t_{1}$ to $t_{2}$ is divided by the time points $t_{s}$ and $t_{f}$ into at most three intervals. Each of these intervals, denote it $[t^{\prime},t^{\prime\prime}]$, induces a factorization $S_{0}^{\prime}\cdot S_{1}^{\prime}\cdot S_{2}^{\prime}\cdot S_{3}^{\prime}$; here the factors from $S_{2}^{\prime}$ are sub-factors of the factor $-^{d}$. (See Fig. 8.) Since the factors in $S_{0}^{\prime}$ are independent from the factors in $S_{1}^{\prime}\cdot S_{2}^{\prime}$, for each of these three time intervals Claim 23 is applicable. During at least one of these intervals, at least $d/3$ minuses from $-^{d}$ are derived; apply Claim 23 to this particular interval. We get

[TABLE]

where all the factors $u_{i}$ are derived during the time period $[t^{\prime},t^{\prime\prime}]\subseteq[t_{1},t_{2}]$ and are, by construction, sub-factors of the factors in $S_{1,2}$ (in the original factorization of $S(t_{1},t_{2}$); see Fig. 8.

The number of such factors is at most $2k$, because the number of their endpoints is at most the number of sides of edges that exist at times $t_{1}$ and $t_{2}$, i.e., $2k+2k=4k$; and each factor has exactly two endpoints.

Therefore, the height increase on at least one of these factors $u_{i}$ is at least $d/6k$. ∎

E.3 Distances within a single time period

Consider, as previously, a derivation $\pi$ based on a tree $(I,E)$, and a factor $w$ of the word $W=\sigma(\pi,I,E)$ associated with a fragment of width $k$. Denote this fragment $F$. Assume with no loss of generality that $w$ is the maximal factor associated with $F$.

Recall that edges of $F$ form a (weakly) connected subgraph, that is, a tree. The root of this tree is the node closest to the root of the entire tree $(I,E)$. Pick in $F$ any path of maximum length from the root and consider the point on this path furthest away from the root. This point corresponds to a position in the factor $w$ that splits it into $w=w_{1}\cdot w_{2}$. By symmetry, we can assume without loss of generality that $|w_{1}|\geq|w|/2$.

Our strategy is to prove an upper bound on the length of the factor $w$ (and, as a result, on the value of $L(W,k)$; see subsection E.4). In the present subsection we obtain an intermediate result: we identify a long enough factor $w_{1}$ inside $w$ and show that the distance between the symbols of this factor $w_{1}$ derived at a single time point is bounded from above (this bound is in terms of $L(W,k-1)$).

We will first identify some structure inside the tree of $F$. Specifically, we will need a ‘trunk’, spawning branches of smaller width.

Pick any path of maximum length from the root as this trunk. Removing the edges of the trunk would make our binary tree into a forest. Each (weakly) connected component of this forest is a rooted tree, and its root sits on the trunk. We will refer to these components as branches of the tree.

Branches are categorized as left or right, depending on their position relative to the trunk, see Fig. 9.

Left branches are ordered by the point of their attachment to the trunk. This matches the order of their occurrence in the tree traversal: whenever one branch starts before another, the tree traversal encounters the former before the latter.

Claim 24**.**

The width of each branch is strictly less than the width of the tree.

Proof.

This is obvious from Fig. 9: at each point in time when an edge of a branch exists, there also exists an edge of the trunk (it is a path of maximum length from the root). ∎

The point of the trunk that furthest away from the root (the lowest point) corresponds to a position in the factor $w$; this position splits this factor into $w=w_{1}\cdot w_{2}$. Here the symbols of $w_{1}$ are derived by the (sides of) edges that occur before this lowest point of the trunk in the traversal of the tree.

We will from now on focus on the word $w_{1}$, assuming without loss of generality that $|w_{1}|\geq|w_{2}|$. Recall that we only consider symmetric words $W$, so if this inequality fails (that is, if $|w_{2}|>|w_{1}|$), we can instead take a symmetric derivation of $W$. This derivation is obtained by reflecting the indices of symbols in each erased pair with respect to the middle of the word $W$ (and flipping the signs). In this derivation, the factor $w$ now corresponds to another factor, $\bar{w}^{R}$, obtained by flipping all the signs and reversing the word. This factor can be written as $\bar{w}^{R}=\bar{w}_{2}^{R}\cdot\bar{w}_{1}^{R}$, and in this factorization the length of the prefix is greater than the length of the suffix. Applying the argument below to the sub-factor $\bar{w}_{2}^{R}$ (instead of $w_{1}$), we will obtain the desired bound for $w_{2}$.

So, we have $|w_{1}|\geq|w_{2}|$. The left branches factorize the word $w_{1}$ as follows:

[TABLE]

The factor $b_{0}$ is derived by the edges of the trunk, which correspond to the prefix of $w_{1}$. Each factor $b_{i}$, $i>0$, is derived by the edges of a branch and the edges of the trunk preceding the next branch. Factorize $b_{i}=b^{\prime}_{i}\cdot b^{\prime\prime}_{i}$, according to which symbols are derived by the edges of the branch resp. the edges of the trunk; see Fig. 10. For $b_{0}$, the prefix $b^{\prime}_{0}$ in this factorization is the empty word.

Claim 25**.**

$|b^{\prime}_{i}|\leq L(W,k-1)$* and $|b^{\prime\prime}_{i}|\leq 3H(W)$ for all $i$.*

Proof.

First, the factors $b^{\prime}_{i}$ are derived by fragments of width strictly smaller than $k$, so the length of each of them is at most $L(W,k-1)$.

Second, observe that a plus from a factor $b^{\prime\prime}_{i}$ can only be paired with a minus outside $w_{1}$, because this minus occurs to the right of it (and is derived at the same time point), and therefore to the right of the lowest point of the trunk—which is where the factor $w_{1}$ ends. Applying Claim 22 to the lowest point of the trunk, we obtain that there are at most $H(W)$ such pluses On the other hand, the absolute value of the change in height on the factor $b^{\prime\prime}_{i}$ cannot be greater than $H(W)$. It follows that the length of $b^{\prime\prime}_{i}$ is at most $3H(W)$. ∎

Lemma 5**.**

Suppose the time periods during which the symbols of the factors $b_{i}$ and $b_{j}$, $i<j$, are derived, overlap. Then the distance between the start position of $b_{i}$ and end position of $b_{j}$ (in the word $W$) is at most $(k-1)(2L(W,k-1)+3H(W))$.

Proof.

Denote by $t^{(i)}_{\max}$ the last point in time when there exists some edge of the branch that derives the word $b^{\prime}_{i}$; see Fig. 10. The width of the entire fragment $F$ is at most $k$. Therefore, at the point in time $t^{(i)}_{\max}$, there exist edges of at most $k-1$ branches to the right of the $i$th branch (and including it) and to the left of the trunk. (Indeed, edges of all these branches and an edge from the trunk exist at time point $t^{(i)}_{\max}$.) If an edge of a branch exists at time point $t^{(i)}_{\max}$, then the branch will be referred to as long.

Suppose the long branch following the $i$th branch has index $\ell$. The factor

[TABLE]

is derived by the edges that belong to a fragment of width strictly smaller than $k$, because edges of the branch $b_{i}$ exist to the left of this fragment. So the length of this (small) fragment cannot exceed $L(W,k-1)$.

Therefore, every long branch, together with the factor up to the next long branch, have total length of at most

[TABLE]

Since there are at most $k-1$ such branches, the statement of the lemma follows. ∎

An immediate consequence of this lemma is the following statement.

Lemma 6**.**

If two factors in the word $w_{1}$ are derived during overlapping time intervals, then the shortest factor containing both has length at most $(k-1)(2L(W,k-1)+3H(W))$.

E.4 Proof of Lemma 3

Consider a derivation $\pi$ based on a tree $(I,E)$ and a factor $w$ of the word $W=\sigma(\pi,I,E)$ associated with a fragment of width $k$. Denote this fragment by $F$; as in the previous subsection assume with no loss of generality that $w$ is the maximal factor associated with this fragment $F$. Suppose the length of $w$ is $L(W,k)$.

In the previous subsection, we have identified a long enough factor $w_{1}$ inside $w$; more specifically, we have $|w_{1}|\geq|w|/2$.

The overall idea of the proof of the upper bound in Lemma 3 is to find, in our factor of the word $W$, a long enough sub-factor that consists of minuses only and apply Lemma 4 to this sub-factor. This lemma will yield another sub-factor $u$, one with big increase in height. On the other hand, Lemma 6 shows that the distance (in the word $W$) between the sub-factor $u$ and the sub-factor that consists of minuses only cannot be very big. This will give us the desired upper bound.

An issue with this plan is that a direct application of Lemma 4 will give us no information on whether the word $u$ is indeed a sub-factor of $w_{1}$. Indeed, the word $u$ may well sit to the left of $w_{1}$ inside the entire word $W$.

In order to avoid this issue, we will apply Claim 22 again.

Denote $H=H(W)\geq 1$. Split the word $w_{1}$ into $8kH+1$ factors with approximately equal lengths:

[TABLE]

More precisely, the length of each $w^{\prime}_{i}$ is chosen to be at least

[TABLE]

Note that if $H$ is sufficiently big, the inequality

[TABLE]

holds—indeed, one can check by direct calculation that it is sufficient that the following two inequalities hold: $kH\geq 36$ and $L(W,k)\geq L(W,k-1)\geq 17^{2}kH$. So we will use constants $N=18$ (restricting ourselves to $k\geq 2$), $A=17$, and $C=17^{2}$ in the statement of Lemma 3.

By Definition 2 of the function $\varphi$ (see p. 2), each of the factors $w^{\prime}_{i}$ contains a sub-factor $-^{d}$ with

[TABLE]

The start and end positions of the word $w_{1}$ separate at most $2H$ pairs of signs in the re-pairing $p_{\pi}$ (by Claim 22). At each time point when one of these ‘separated’ signs is derived (there are at most $4H$ of these time points), there exist at most $k$ edges in the left branches and trunk of the tree (in the fragment $F$) that may be deriving symbols from the words $w^{\prime}_{i}$. Each edge may be deriving symbols by its left or right side.

We now conclude that there is a word $w^{\prime\prime}=w^{\prime}_{j}$ with the following property: the signs of this word are not paired with symbols outside $w_{1}$. This word will be said to be isolated. Note, however, that the signs of $w^{\prime}_{j}$ may well be paired with signs outside $w^{\prime}_{j}$.

So, inside the isolated word $w^{\prime\prime}$ there is a factor $-^{d}$ but no signs paired with signs outside $w_{1}$. This enables us to apply Lemma 4. Indeed, let the factor $-^{d}$ be derived during time period $[t_{1},t_{2}]$; we have $[t_{1},t_{2}]\subseteq[\tau_{1},\tau_{2}]$, where $[\tau_{1},\tau_{2}]$ is the time interval when edges of the fragment $F$ exist. Now, since the bigger word $w$ was chosen to be maximal (among all words associated with the fragment $F$), each factor in the sequence $S(t_{1},t_{2})$ either is completely inside $w_{1}$ or sits completely before or after it. Then, in the statement of the lemma, let $S_{0}$ consist of all the factors that sit to the left of $w_{1}$, and let $S_{1,2}$ consist of all the factors that are sub-factors of $w_{1}$. It follows from Lemma 4 that the word $w_{1}$ contains a sub-factor $u$ which is derived within the same time interval $[t_{1},t_{2}]$ and on which the height increase is $\Delta(u)\geq d/6k$.

By Lemma 6, the shortest factor of the word $W$ containing both of these sub-factors $u$ and $-^{d}$ has length at most $(k-1)(2L(W,k-1)+3H)$. But by Definition 3 of the function $\psi$ (see p. 3), this length must be strictly greater than

[TABLE]

The function $\psi$ is easily seen to be non-decreasing; hence, we obtain the inequality

[TABLE]

the right-hand side of which does not exceed $3kL(W,k-1)$ under the conditions of the lemma (it suffices that the inequality $L(W,k-1)\geq 3H$ hold).

This completes the proof.

E.5 Combinatorial properties of the words $Z(n)$ and $Y(m,\ell)$

The inductive definition of the words

[TABLE]

where $a_{i}=2^{\lfloor i/\ell\rfloor}$, can be interpreted as follows. The signs of these words are grouped at the nodes of complete binary trees of height $n-1$ and $m\ell-1$, respectively. At the node of tree at distance $r\geq 0$ from the leaves, one pair of opposite signs is grouped in the case of $Z(n)$, and $a_{r}$ pairs in the case of $Y(m,\ell)$ (all the pluses in this group form a single run, and so do all the minuses).

Therefore, the length of the word $Z(n)$ is $2\cdot(2^{n}-1)$, double the number of nodes in a complete binary tree of height $n-1$.

We now estimate the length of the word $Y(m,\ell)$.

Claim 26**.**

$2^{m\ell}\leq|Y(m,\ell)|\leq 3\cdot 2^{m\ell}$* if $\ell>1$.*

Proof.

Note that $Y(1,\ell)=Z(\ell)$, so

[TABLE]

if $\ell>1$.

The set of nodes of the complete binary tree of height $m\ell-1$ is split into $m\ell$ levels. The root and the $\ell-1$ levels closest to the root, taken together, contribute

[TABLE]

signs to the length of $Y(m,\ell)$. All other nodes are divided into $2^{\ell}$ subtrees of height $(m-1)\ell$ each, one subtree corresponding to a factor $Y(m-1,\ell)$. This gives us a recurrence for the length of $Y(m,\ell)$:

[TABLE]

It is now easy to obtain the exact expression for $|Y(m,\ell)|$:

[TABLE]

The inequalities of the Claim now follow from this expression. Indeed, since $\ell>1$, we have $(2^{\ell}-1)/(2^{\ell-1}-1)\leq 3$, and the upper bound follows; the lower bound inequality

[TABLE]

also holds, since the conditions $\ell>1$ and $m\geq 1$ imply that

[TABLE]

To apply Lemma 3, we need bounds on the values of the functions $\varphi$ and $\psi$ for the words $Z(n)$ and $Y(m,\ell)$. The arguments in both cases will be similar to each other. We will rely on several simple facts, which we formulate for general words of the form $X(a_{0},\dots,a_{k-1})$ and will then use for $Z(n)$ and $Y(m,\ell)$.

We first generalize a property of $Z(n)$ formulated previously in section D, see Claim 13.

Fact 1**.**

For all $0\leq s\leq k-1$, the word $X(a_{0},\dots,a_{k-1})$ factorizes as

[TABLE]

where the factors $X_{i}$ are equal to $X(a_{0},\dots,a_{k-s-1})$, the factor $w_{0}=+^{r_{0}}$, $w_{i}=-^{r_{i}}+^{r_{i}}$ for $0<i<2^{s}$, $w_{2^{s}}=-^{r_{2^{s}}}=-^{r_{0}}$, and no number $r_{i}$ exceeds the maximum height of positions in the word $X(a_{0},\dots,a_{k-1})$.

Proof.

Applying the inductive definition $s$ times, we find $2^{s}$ factors of the form $X(a_{0},\dots,a_{k-s-1})$ in $X(a_{0},\dots,a_{k-1})$. Put differently, these occurrences of $X(a_{0},\dots,a_{k-s-1})$ in $X(a_{0},\dots,a_{k-1})$ correspond to the nodes of a complete binary tree at distance $s$ from the root.

One can see from here that these factors are separated from one another by words of the form $-^{r_{i}}+^{r_{i}}$, where each $r_{i}$ is at most the maximum height of positions in the word$X(a_{0},\dots,a_{k-1})$; the same holds for the prefix and suffix $w_{0}$ and $w_{2^{s}}$. ∎

Fact 2**.**

The height of all positions in the word $X(a_{0},\dots,a_{k-1})$ is at most

[TABLE]

and this bound is tight. The increase in height on factors that are not prefixes of $X(a_{0},\dots,a_{k-1})$ is strictly smaller than this bound.

Proof.

Use induction on $k$. The base case is $k=1$; it is obvious that that the maximum height of positions in the word $X(a)=+^{a}-^{a}$ is equal to $a$. It is as clear that every factor of $X(a)$, unless it is a prefix, has fewer than $a$ pluses, and so the increase in height on this factor is strictly smaller than $a$.

For the induction step, assume that for a word $X(a_{0},\dots,a_{k-1})$ the inductive statement holds. In the word

[TABLE]

the maximum height of positions does not exceed the height of positions inside either of the factors $X(a_{0},\dots,a_{k-1})$ increased by $a_{k}$. This bound is tight for those positions of the factors $X(a_{0},\dots,a_{k-1})$ that have maximum height inside these factors.

Suppose a non-empty factor $w$ is not a prefix of $X(a_{0},\dots,a_{k-1},a_{k})$. Consider the following three cases.

1. The factor $w$ starts in the prefix $+^{a_{k}}$, but not from the start position of the word. Then, using the same argument as above, the height increase on the factor $w$ is strictly smaller than the maximum height of $X(a_{0},\dots,a_{k-1})$ increased by $a_{k}$.

2. The factor starts inside either $X(a_{0},\dots,a_{k-1})$. Then the height increase on this factor cannot exceed the maximum height of positions in the word $X(a_{0},\dots,a_{k-1})$.

3. The factor starts inside the suffix $-^{a_{k}}$. Then the height increase on this factor is negative. ∎

We conclude from Fact 2 that the maximum height of positions in the words $Z(n)$ and $Y(m,\ell)$ satisfies the equalities

[TABLE]

Since the functions $\varphi$ and $\psi$ are defined using factors that consist of minuses only, we will need the following fact on the relative position of such factors inside $X(a_{0},\dots,a_{k-1})$.

Fact 3**.**

The suffix of $X(a_{0},\dots,a_{k-1})$ of length $a_{0}+\dots+a_{k-1}$ consists of minuses only.

Suppose a factor of $X(a_{0},\dots,a_{k-1})$ has the form $-^{x}$, and a number $s$ is such that

[TABLE]

Then this factor (i.e., this occurrence of $-^{x}$ in $X(a_{0},\dots,a_{k-1})$) can be extended leftwards to an occurrence of a word $X(a_{0},\dots,a_{s})-^{r}$ (for some $r\geq 0$).

Proof.

The first assertion is obvious from the inductive definition of $X(a_{0},\dots,a_{k-1})$.

For the second assertion, we will show by induction on $k$ that, for all $k>1$, (length-)maximal occurrences of factors that consist of minuses only (in the word $X(a_{0},\dots,a_{k-1})$) are suffixes of factors of the form $X(a_{0},\dots,a_{s})$.

The base case is $k=2$, i.e., words of the form

[TABLE]

Here the assertion is checked easily for the two maximal factors that consist of minuses only.

For the induction step, note that, in the word

[TABLE]

every (length-)maximal factor that consists of minuses only either is a suffix of the entire word $X(a_{0},\dots,a_{k-1},a_{k})$ (and then $s\in\{0,1,\ldots,k\}$), or ends inside either $X(a_{0},\dots,a_{k-1})$ and is bounded by the plus from the right— in which case we use the inductive assumption. ∎

We now establish the bounds on the functions $\varphi$ and $\psi$ for the words $Z(n)$ and $Y(m,\ell)$. They are proved under certain assumptions on the parameters and arguments of these functions; in all our cases these assumptions will hold.

Claim 27**.**

If $x\geq 6n$, then $\varphi(Z(n),x)\geq\log(x/12)$.

Proof.

We will find a condition on the length of a factor $u$ of the word $Z(n)$ which guarantees that the factor $u$ contains a sub-factor $Z(s)$ for big enough $s$. Fact 3 will then imply that this sub-factor contains $-^{s}$.

By Fact 1, every factor of length at least $2|Z(s)|+2n$ contains a sub-factor of the form $Z(s)$. Indeed, the distance between two consecutive occurrences of $Z(s)$ in $Z(n)$ cannot exceed $2n$. So for every factor of length $2|Z(s)|+2n$ one the following two conditions holds. Either this factor begins between the occurrences of $Z(s)$—and then its prefix of length $2n+|Z(s)|$ contains a sub-factor $Z(s)$ already; or the start position of this factor is inside an occurrence of $Z(s)$—in which case the prefix of this factor of length at most $2|Z(s)|+2n-1$ contains a sub-factor $Z(s)$.

Therefore, $\varphi(Z(n),x)\geq s$ if $x\geq 2|Z(s)|+2n$. Assuming $x\geq 6n$, we have

[TABLE]

and under this assumption it would suffice that the inequality $|Z(s)|\leq x/3$ hold—and this is equivalent to $2^{s+1}-2\leq x/3$, i.e., $s\leq\log(x/3+2)-1$. So we can pick $s=\lfloor\log(x/3+2)\rfloor-1>\log(x/3+2)-2$. We then obtain

[TABLE]

Claim 28**.**

If $x\geq 2$ and $n\geq 1$, then $\psi(Z(n),x)\geq 2^{x}$.

Proof.

From Facts 2 and 3 we conclude that a word with height increase at least $x$ cannot occur to the left of the factor $-^{x}$ earlier than the factor $Z(x)$ occurs. But the length of this factor is $2^{x+1}-2$. For $x\geq 2$ the inequality $(2^{x+1}-2)-1\geq 2^{x}$ holds, and the desired inequality on $\psi(Z(n),x)$ follows. ∎

We next obtain bounds on $\varphi(Y(m,\ell),x)$ and $\psi(Y(m,\ell),x)$.

Claim 29**.**

If $x\geq\max(6H(Y(m,\ell)),9\cdot 6^{\ell})$, then

[TABLE]

Proof.

Reasoning as in the proof of Claim 27 and using Facts 2 and 3, we conclude that a sub-factor which consists of $H(Y(s,\ell))$ minuses must occur in every factor of length at least $2|Y(s,\ell)|+2H(Y(m,\ell))$, and, moreover, for this inequality to hold under the given assumption on $x$, it would suffice that $|Y(s,\ell)|\leq x/3$.

The upper bound from Claim 26 means that the condition $3\cdot 2^{s\ell}\leq x/3$ is sufficient. This condition holds, if

[TABLE]

Applying equation (16) for the maximum height of positions in the word $Y(s,\ell)$ (see p. 16), we obtain

[TABLE]

The last inequality holds by the assumption that $x\geq 9\cdot 6^{\ell}$. ∎

Claim 30**.**

The following inequality holds:

[TABLE]

Proof.

The idea is the same as in the proof of Claim 28.

Let $s$ be such that $x>\ell\cdot(2^{s}-1)$. Then from Fact 3 and equation (16) for the maximum height of positions in the word $Y(s,\ell)$ (see p. 16) it follows that the factor $-^{x}$ can be extended leftwards to an occurrence of a word $Y(s,\ell)$; and then by Fact 2 this word has no sub-factor on which the height increases by $x$ or more.

The inequality $x>\ell\cdot(2^{s}-1)$ holds for

[TABLE]

Since the length of the word $Y(s,\ell)$ is at least 2^{s\ell}\geq\big{(}\frac{x}{2\ell})^{\ell}, this gives us the desired bound. ∎

E.6 Proof of Theorem 4

Apply Lemma 3 to the words $Y(\ell)=Y(m,\ell)$, where $m=\lfloor\ell\cdot\log\ell\rfloor$, for big enough $\ell$. We will rely on the bounds on the functions $\varphi(Y(\ell),x)$ and $\psi(Y(\ell),x)$ obtained in the previous subsection.

In this part of the present subsection, we will denote $L(k):=L(Y(\ell),k)$ and $H:=H(Y(\ell))$.

We obtain from Lemma 3 that, if

[TABLE]

then the following inequalities hold:

[TABLE]

These inequalities imply the following recurrence for the numbers $L(k)$:

[TABLE]

Apply this recurrence while it is still applicable, i.e., while the condition (17) holds. It suffices to use a rather crude approximation for the contribution of different factors. If $\ell\geq 2$, then the inequalities $\ell+2\leq 2\ell$ and

[TABLE]

hold. (Note that $L(1)\leq 4H$ by Lemma 2 (see p. 2), so for $k=2$ the inequality (17) fails and, therefore, for some $k^{\prime}\geq 2$ the following upper bound holds: $L(k^{\prime})<17k^{\prime}H\cdot\max(17H,9\cdot 6^{\ell})$. This implies, for instance, that the recurrence is applied at most $k-2$ times.)

Choose $k=\lfloor\ell/4\rfloor\leq\ell/4$. Then the first factor in (18) is at most

[TABLE]

The factor $H^{k}$ in (18) can be bounded, using the expression (16) for the maximum height of positions in the word $Y(m,\ell)$, by

[TABLE]

Now recall that if $L(k)<|Y(\ell)|$, then $\mathop{\mathrm{width}}\nolimits(Y(\ell))\geq k$. Compare the bound (18) with the lower bound on the length of the word $Y(\ell)$, taking into account the upper bounds for the individual factors above. Taking the logarithms of both sides, we get $\mathop{\mathrm{width}}\nolimits(Y(m,\ell))\geq\lfloor\ell/4\rfloor\geq\ell/4-1$, provided that

[TABLE]

(the right-hand side of this inequality does not exceed the logarithm of the length of $Y(\ell)$).

The main terms on both sides of this inequality are those with $\ell^{2}\cdot\log\ell$. On the left-hand side, this expression has coefficient $1/2+1/4=3/4$, and on the right-hand side, coefficient $1$. All other terms are $o(\ell^{2}\log\ell)$, so the inequality holds for all big enough $\ell$.

This gives us the upper bound $\mathop{\mathrm{width}}\nolimits(Y(\ell))\geq\ell/4-1$, which holds for big enough $\ell$. The inequality (13) (see p. 13) follows.

Let us now show that this inequality implies Theorem 4. From Claim 26 we obtain an upper bound on the length of the word $Y(\ell)$:

[TABLE]

It follows from this bound that

[TABLE]

which completes the proof of Theorem 4.

The lower bound of Theorem 4 is much higher than the upper bound on the width of the words associated with complete binary trees, $Z(n)$, given by Theorem 3. The same method gives a much weaker lower bound bound for $Z(n)$, namely the inequality (14).

Proof of inequality (14)

We apply Lemma 3 to the words $Z(n)$ for big enough $n$. We will rely on the bounds on the functions $\varphi(Z(n),x)$ and $\psi(Z(n),x)$ obtained in the previous subsection.

In this part of the present subsection, we will denote $L(k):=L(Z(n),k)$.

If the conditions

[TABLE]

are satisfied, then the following inequalities hold:

[TABLE]

The last inequality implies the following recurrence for the numbers $L(k)$:

[TABLE]

Taking the logarithms of both sides and coarsening slightly, we get

[TABLE]

provided that

[TABLE]

Condition (21) is stronger than the condition (19) of applicability of Lemma 3, but will suffice for us.

Apply this recurrence while it is still applicable, i.e., while the conditions (20) and (21) hold (notice that the former condition is equivalent to the inequality $L(k)\geq 12\cdot 17kn\cdot 2^{12k}$). This will give us the following upper bound on $\log L(k)$:

[TABLE]

Plug in $k=\lfloor 0.9\log n/\log\log n\rfloor$. For big enough $n$, we have

[TABLE]

Therefore, for big enough $n$ we have

[TABLE]

and then inequality (14) follows, i.e.,

[TABLE]

Appendix F Proofs of lower bounds for commutative NFA

Recall that a nondeterministic finite automaton (NFA) is a quintuple $(\Sigma,Q,q_{0},\delta,F)$ where $\Sigma$ is a (finite) input alphabet, $Q$ a (finite) set of states, $q_{0}\in Q$ an initial state, $\delta\subseteq Q\times\Sigma\times Q$ a transition relation, and $F\subseteq Q$ a set of final states. The transition graph of the NFA is a directed graph with vertices $Q$ that contains, for each $(q,a,q^{\prime})\in\delta$, an edge $(q,q^{\prime})$ labeled by the symbol $a$. The NFA accepts a word $\sigma\in\Sigma^{*}$ if the graph has a path from $q_{0}$ to some $q\in F$ such that the labels of the edges on the path form the word $\sigma$. The set of all accepted words is the language recognized by the NFA.

F.1 Vertices and paths in the graph of strongly connected components

Let ${\mathcal{A}}={\mathcal{A}}_{n}$ be an NFA which recognizes a language $L({\mathcal{A}})$ with Parikh image equal to $U_{n}$, the universal one-counter set (see Section 6).

Denote by $Q_{0}$, $Q_{1}$, …, $Q_{s}$ the (strongly connected) components in the transition graph of the automaton ${\mathcal{A}}$ which are reachable from the initial state and from which a final state is reachable. Assume without loss of generality that ${\mathcal{A}}$ has no other strongly connected components. Suppose the initial state of ${\mathcal{A}}$ belongs to the component $Q_{0}$.

We first make two observations about the structure of the transition graph.

The first observation is that every edge labeled $c_{ij}$ on a path from the initial state to a final state goes from one strongly connected component to another. Indeed, if this is not the case, then this edge belongs to a cycle in the graph. But then the Parikh image of $L({\mathcal{A}})$ violates condition (U1) in the definition of the universal one-counter set (by including vectors that have $y_{ij}>1$).

The second observation concerns the directed cycles in the transition graph of the automaton ${\mathcal{A}}$. Let $C$ be such a cycle. Denote by $x_{i}^{(C)}$ the number of edges in this cycle that are labeled by the symbol $a_{i}$.

Claim 31**.**

For the edges of every cycle $C$ in the transition graph of the automaton ${\mathcal{A}}$, the vector $(x_{i}^{(C)})$ belongs to the cone of balanced vectors.

Proof.

Consider paths from the initial state to a final state that differ in the number of traversals of the cycle $C$ only. If the cycle $C$ violates the balance condition, then, after sufficiently many traversals, the balance condition will be violated for the entire path too—but this contradicts the condition (U2) in the definition of the universal one-counter set. ∎

We construct, based on the transition graph of ${\mathcal{A}}$, another graph ${\mathcal{B}}$, the condensation of it (or the graph of strongly connected components). The vertices of this graph are strongly connected components in the transition graph of ${\mathcal{A}}$. The edges of ${\mathcal{B}}$ correspond to the edges of ${\mathcal{A}}$ between different components. An edge in ${\mathcal{B}}$ has label $(i,j)$ if the corresponding transition in ${\mathcal{A}}$ has label $c_{i,j}$; and label $\varepsilon$ if the transition in ${\mathcal{A}}$ has label $\varepsilon$ or $a_{i}$ for some $i$. By construction, the graph ${\mathcal{B}}$ may have parallel edges and is acyclic.

Vertices of ${\mathcal{B}}$ that contain initial (resp. final) state(s) of the NFA ${\mathcal{A}}$ are called initial and final, respectively. A path from the initial vertex to a final vertex is a complete path.

Recall that chains are defined on p. U1, when we describe condition (U1). Chains on $[n]$ are in bijection with sets of numbers from $[n]$ that they visit. We will denote the set of numbers visited by the edges of $\pi$ by $C(\pi)$. We have $\{0,n-1\}\subseteq C(\pi)\subseteq[n]$.

Claim 32**.**

Let $\pi$ be a complete path in the graph ${\mathcal{B}}$. Then the non-empty labels on (the edges of) this path induce a chain on the set $[n]$.

Proof.

Given $\pi$, construct an accepting path $\pi^{\prime}$ in ${\mathcal{A}}$ that consists of the edges that correspond to the edges of the path $\pi$, as well as edges inside the strongly connected components. Since all edges labeled $c_{ij}$ in the transition graph of ${\mathcal{A}}$ connect states that lie in different components, the set of non-empty labels on the edges of the path $\pi$ corresponds to a set of labels $c_{ij}$ of edges on the path $\pi^{\prime}$. Therefore, the claim follows from the property (U1) of the universal one-counter set. ∎

Let us now assign labels to vertices of ${\mathcal{B}}$. For each vertex $v$, we define four subsets of $[n]=\{0,1,\ldots,n-1\}$ that we denote $I(v)$, $L(v)$, $R(v)$, and $S(v)$. The intuition is as follows:

• •

$S(v)$ is the set of all numbers $i\in[n]$ such that the corresponding component in ${\mathcal{A}}$ contains a edge with label $a_{i}$;

• •

$I(v)$ specifies the (endpoints of) disjoint intervals in the set $[n]$ that are induced by (incomplete) paths from the initial vertex to $v$;

• •

$L(v)$ contains points from the interior of these intervals that must be visited by (all) these incomplete paths;

• •

$R(v)$ contains points from the exterior of these intervals that must be visited by the completion of (all) these paths (i.e., by paths from $v$ to final vertices).

Rigorous definitions of the sets $I(v)$, $L(v)$ and $R(v)$ are given below.

The following property links vertices to paths. Let $B(v)=S(v)\cup I(v)\cup L(v)\cup R(v)$.

Claim 33**.**

For every vertex $v$ in the graph ${\mathcal{B}}$ and every complete path $\pi$ that visits $v$, the inclusion $B(v)\subseteq C(\pi)$ holds.

We prove Claim 33, considering sets $S(v)$, $I(v)$, $L(v)$, $R(v)$ one by one.

Since a chain is a monotone path, it is uniquely determined by the set of numbers from the set $[n]$ that it visits (i.e., the set of vertices that are incident to at least one edge on the path). By Claim 32, the edges of a complete path $\pi$ induce a chain. We will denote the set of vertices visited by this chain by $C(\pi)$.

For every complete path, we have $\{0,n-1\}\subseteq C(\pi)$. The converse also holds:

Claim 34**.**

For all $\{0,n-1\}\subseteq S\subseteq[n]$ there exists a complete path $\pi$ such that $C(\pi)=S$.

Proof.

Let $S$ consist of numbers

[TABLE]

The vector of numbers

[TABLE]

belongs to the universal one-counter set. Therefore, the automaton ${\mathcal{A}}$ has an accepting path with this Parikh image. This path induces the required path in ${\mathcal{B}}$. ∎

Claim 35**.**

Let a vertex $v$ in the graph ${\mathcal{B}}$ belong to a complete path $\pi$. Then $S(v)\subseteq C(\pi)$.

Proof.

This is a consequence of the compatibility condition (U3) in the definition of the universal one-counter set. By definition, if $i\in S(v)$, then the component $Q_{v}$ in the transition graph of ${\mathcal{A}}$ contains an edge $e$ with label $a_{i}$ and there is an accepting path $\pi^{-1}$ in the transition graph of ${\mathcal{A}}$ such that $\pi^{-1}$ traverses this edge and corresponds to the path $\pi$ in the graph ${\mathcal{B}}$. The Parikh image of the word induced by this path $\pi^{-1}$ belongs to the universal one-counter set $U_{n}$, so it satisfies the compatibility condition. We have $x_{i}>0$; therefore, $\pi^{-1}$ traverses an edge with label $c_{i,x}$ or $c_{x,i}$ for some $x$, so the chain on $[n]$ induced by the path $\pi$ visits the number $i$. But then $i\in C(\pi)$. ∎

Consider, in the graph ${\mathcal{B}}$, a path from the initial vertex to a vertex $v$ that is a prefix of some complete path. Take the edges with labels of the form $(i,j)$ on this (incomplete) path. They form a subgraph of the chain and are therefore a disjoint union of non-intersecting paths: from the (vertex) number $\ell_{0}$ to $r_{0}$; from $\ell_{1}$ to $r_{1}$; …; from $\ell_{t}$ to $r_{t}$.

Claim 36**.**

Let $v$ be a vertex in the graph ${\mathcal{B}}$. Then for every path in ${\mathcal{B}}$ from the initial vertex to $v$ the set of numbers

[TABLE]

is the same.

Proof.

Observe that every path from $v$ to a final vertex is a completion of any path from the initial vertex to $v$.

Assume for the sake of contradiction that for paths $\pi$ and $\pi^{\prime}$ from the initial vertex to $v$ the sequences

[TABLE]

are different. Consider the first difference and suppose it is the left endpoint of a segment. Without loss of generality, $\ell_{i}<\ell^{\prime}_{i}$. Take some path $\pi^{\prime\prime}$ from $v$ to a final vertex. The unions $\pi\cup\pi^{\prime\prime}$ and $\pi^{\prime}\cup\pi^{\prime\prime}$ should induce chains on $[n]$; we will prove that this is impossible.

Let $\pi\cup\pi^{\prime\prime}$ and $\pi^{\prime}\cup\pi^{\prime\prime}$ induce two chains. Then the path $\pi^{\prime\prime}$ has no edges with labels of the form $(\ell_{i},x)$, because such a label already exists on the path $\pi$. Since we picked the first difference between the sequences (22), there are no edges with such labels on the path $\pi^{\prime}$ either. Therefore, there are no edges with such labels on the entire path $\pi^{\prime}\cup\pi^{\prime\prime}$. So $\ell_{i}\neq 0$ (since the chain starts from [math]).

But then the path $\pi^{\prime\prime}$ has an edge with label $(y,\ell_{i})$, because $\pi\cup\pi^{\prime\prime}$ induces a chain. It follows that $\pi^{\prime}\cup\pi^{\prime\prime}$ does not induce a chain, since the in-degree of $\ell_{i}$ is at least $1$, and the out-degree, as shown above, is [math]. This is a contradiction.

The case of the first difference in the right endpoint of a segment is analogous.

Previous arguments assumed implicitly that $i\leq t^{\prime}$. To complete the proof, we note that in the case $t^{\prime}<i$ (i.e., when the segment $[\ell_{i}^{\prime},r_{i}^{\prime}]$ is absent from the sequence for the path $\pi^{\prime}$) our analysis of the first difference in a left endpoint applies as well. ∎

By this claim, for every vertex $v$ in the graph ${\mathcal{B}}$, the set $I(v)$ is well-defined:

[TABLE]

The next claim is an immediate corollary of the previous one.

Claim 37**.**

$I(v)\subseteq C(\pi)$* for all complete paths that visit vertex $v$.*

The set $I(v)$ specifies a system of intervals inside $[0;n-1]$. The numbers in the interior of any of the intervals will be called internal for a vertex $v$, and the numbers outside the intervals will be called external. More precisely, a number $j$ is internal for a vertex $v$ if $\ell_{i}<j<r_{i}$ for some $i$. A number $j$ is external for a vertex $v$ if either $j<\ell_{0}$, or $r_{i}<j<\ell_{i+1}$ for some $i$, or $j>r_{t}$.

To each vertex $v$ we will associate two more subsets $L(v)$, $R(v)$ of the set $[n]$, defined as follows:

• •

$j\in L(v)$ iff (i) $j$ is internal for $v$ and (ii) every path $\pi$ in the graph ${\mathcal{B}}$ from the initial vertex to $v$ has at least one edge with label of the form $(x,j)$ or $(j,x)$.

• •

$j\in R(v)$ iff (i) $j$ is external for $v$ and (ii) every path $\pi$ in the graph ${\mathcal{B}}$ from $v$ to a final vertex has at least one edge with label of the form $(x,j)$ or $(j,x)$.

The following statement follows directly from these definitions.

Claim 38**.**

For every complete path $\pi$ in the graph ${\mathcal{B}}$ that visits a vertex $v$, the inclusions $L(v)\subseteq C(\pi)$ and $R(v)\subseteq C(\pi)$ hold.

Now Claim 33 follows from Claims 35, 37, and 38.

The sets $L(v)$ and $R(v)$ restrict possible elements of the sets $S(v^{\prime})$ for other vertices $v^{\prime}$ as follows:

Claim 39**.**

Let a vertex $v_{1}$ be visited by some complete path before another vertex $v_{2}$. Then the set $L(v_{1})$ contains all numbers from $S(v_{2})$ that are internal for $v_{1}$, and the set $R(v_{2})$ contains all numbers from $S(v_{1})$ that are external for $v_{2}$.

Proof.

Suppose $v_{1}$ occurs before $v_{2}$ on some complete path $\pi$; also suppose that the number $i$ is internal for $v_{1}$ and belongs to $S(v_{2})$.

We show by contradiction that $i\in L(v_{1})$. Assume that there is a path $\pi^{\prime}$ from the initial vertex to $v_{1}$ that has no edges with labels of the form $(x,i)$ and $(i,x)$. Denote by $\pi^{\prime\prime}$ the part of the path $\pi$ from $v_{1}$ to the final vertex. Then $\pi^{\prime}\cup\pi^{\prime\prime}$ is a complete path, the vertex $v_{2}$ is visited by this path, but $i\notin C(\pi^{\prime}\cup\pi^{\prime\prime})$. This contradicts Claim 35.

The second assertion is proved in a similar way. ∎

The strategy of the rest of the proof is as follows. In the following subsection, we describe a construction that produces paths $\pi$ with ‘predetermined’ $C(\pi)$. Our construction will have the following properties:

• •

Each path $\pi$ is originally chosen based (in a certain way) on a well-formed (Dyck) word $\sigma$; we show that from $\pi$ one can obtain a re-pairing of this word $\sigma$ (based on Claim 42 below).

• •

The width of this re-pairing will bound the cardinality of the $B(v)$ from below, for some vertex $u=u(\pi)$ on the path $\pi$ (Claim 43).

• •

Many (different) paths $\pi$ will be chosen and it will be ensured that the sets $C(\pi)$ have low pairwise intersection (subsection F.3).

Now suppose that the vertex $u=u(\pi)$ for a path $\pi$ is visited by another such path, say $\pi^{\prime}$. Then the intersection of the sets $C(\pi^{\prime})$ and $C(\pi)$ includes $B(u)$ by Claim 33. On the one hand, the cardinality $|B(u)|$ is greater than or equal to the width of the re-pairing; on the other hand, no two sets $C(\pi^{\prime})$ and $C(\pi)$ may overlap a lot. Therefore, with a careful choice of parameters, vertices $u=u(\pi)$ are not shared by the paths $\pi$, and so the NFA ${\mathcal{A}}$ should have at least as many strongly connected components as we can choose paths $\pi$.

F.2 From NFA and well-formed word to re-pairing

Let $\sigma$ be a well-formed (Dyck) word of length $s\leq\frac{n}{2}$. As everywhere in section 6, we assume that $n$ is even. Associate with the word $\sigma$ a set of vectors from the universal one-counter set. Each of these vectors will be determined by some $s$-element subset $F\supseteq\{0,\frac{n}{2}-1\}$ of the set $[\frac{n}{2}]$ and by a nonnegative integer $\lambda$; we will denote these vectors $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$. (These vectors depend on the word $\sigma$ as well, but our notation will not reflect this.)

Definition 4** (the set $C_{F}$ and vectors $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$).**

Given an $s$-element subset $F$ of the set $[\frac{n}{2}]$, define a subset $C_{F}$ of the set $[n]$ of the same size $s$ as follows. Sort the elements of $F$ in ascending order. If the $i$th least element is equal to $j$, then the set $C_{F}$ contains the number $2j$ if $\sigma(i)=+1$ and the number $2j+1$ otherwise. (Since every well-formed word begins with a $+1$ and ends with a $-1$, and $F\supseteq\{0,\frac{n}{2}-1\}$, we will always have $\{0,n-1\}\subseteq C_{F}$.)

The vectors $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$ are then defined by the following equations: $y_{ij}=1$ if $i$ and $j$ are two adjacent elements (in ascending order) in the set $C_{F}$, otherwise $y_{ij}=0$; $x_{i}=\lambda$ if $i\in C_{F}$, otherwise $x_{i}=0$.

Remark 6*.*

Notice that the correspondence between the elements of the set $C_{F}$ and the symbols of the word $\sigma$ defined in this way is a bijection. We will refer to symbols in the word $\sigma$ by specifying the corresponding numbers from the subset $C_{F}$.

Example 1**.**

Let $\sigma=Z(2)={+}{+}{-}{+}{-}{-}$; $s=|\sigma|=6$ and $n=16$; $s\leq\frac{n}{2}=8$. Consider the set $F=\{0,1,3,5,6,7\}\subseteq[8]=\{0,1,\ldots,7\}$. Then $C_{F}=\{0,2,7,10,13,15\}$, the vector $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$ satisfies the equations $y_{0,2}=y_{2,7}=y_{7,10}=y_{10,13}=y_{13,15}=1$ and $x_{0}=x_{2}=x_{7}=x_{10}=x_{13}=x_{15}=\lambda$, and all other components of this vector are equal to [math].

It is easy to see that the vectors $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$ defined in this way belong to the universal one-counter set. Conditions (U1) and (U3) hold by construction, and condition (U2) by the fact that the word $\sigma$ is well-formed.

Therefore, for every $\lambda$, the transition graph of the NFA ${\mathcal{A}}$ has at least one path $\tau(\lambda)$ from the initial state to a final state on which the labels form a word with Parikh image $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$, i.e., each symbol $c_{ij}$ occurs $y_{ij}$ times and each $a_{i}$ occurs $x_{i}$ times in this word. We can now decompose this path by separating (possibly empty) cycles $\gamma_{i}(\lambda)$:

[TABLE]

where all $q_{i}(\lambda)$ are distinct states of ${\mathcal{A}}$. It is clear that each cycle $\gamma_{i}(\lambda)$ lies in some strongly connected component of the transition graph of ${\mathcal{A}}$.

Since the set of possible values of the parameter $\lambda$ is infinite, there exists a sequence $\tau^{*}_{F}=(q_{0},q_{1},\dots,q_{T})$ of distinct states of the NFA ${\mathcal{A}}$ that coincides with infinitely many sequences $(q_{0},q_{1}(\lambda),\dots,q_{T}(\lambda))$ that occur in the decomposition (23).

The sequence $\tau^{*}_{F}$ corresponds to a complete path in the graph ${\mathcal{B}}$ up to the choice of parallel edges. Choose these edges, and thus a complete path $\pi_{F}$, in such a way that $\pi_{F}$ corresponds (in the graph ${\mathcal{B}}$) to infinitely many values of the parameter $\lambda$. Denote the vertices of this path

[TABLE]

The following claim holds by definition of the vectors $\boldsymbol{y}(\lambda)$.

Claim 40**.**

$C(\pi_{F})=C_{F}$.

In what follows, we will use a connection between accepting paths in the automaton ${\mathcal{A}}$ and re-pairings of the word $\sigma$. We construct such a re-pairing based on the sets $S(v_{i}^{(F)})$.

Also we need the following simple claim.

Claim 41**.**

The cone $K$ of balanced vectors is generated by the vectors

[TABLE]

Proof.

It is obvious that all vectors $\boldsymbol{e}_{i,j}$ belong to $K$.

Given a vector from $K$, we will show that it is a nonnegative linear combination of the vectors $\boldsymbol{e}_{i,j}$. The proof is by induction on $M$, the number of nonzero coordinates in the vector.

The base cases are $M=1$ and $M=2$. For $M=1$, the balance conditions cannot be satisfied, and for $M=2$, all balanced vectors are proportional to $\boldsymbol{e}_{i,j}$.

For the induction step, assume that the claim holds for all $M<M_{0}$. Given a vector $\boldsymbol{x}\in K$ with $M_{0}$ nonzero coordinates, consider the smallest nonzero coordinate $j$ with an odd index. The sum of the coordinates with even indices $i<j$ is at least $x_{j}$ (by the balance inequality), so by subtracting from $x$ an appropriate vector of the form

[TABLE]

we obtain a vector $\boldsymbol{x}^{\prime}\in K$ in which the number of nonzero coordinates is smaller than $M_{0}$. ∎

Claim 42**.**

The set $C_{F}$ can be decomposed as a union of non-overlapping pairs $\{\ell_{i},r_{i}\}_{i=1}^{s/2}$ such that for all $i$ there is a $k$ with $\{\ell_{i},r_{i}\}\subseteq S(v_{k}^{(F)})$. For each pair $(\ell_{i},r_{i})$, the corresponding pair of signs in $\sigma$ is formed by a plus and a minus, and the plus occurs to the left of the minus.

Proof.

It follows from the definition of the vectors $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$ that for each label $(i,j)$ on the edges of the path $\pi_{F}$ in the graph ${\mathcal{B}}$ the inclusion $\{i,j\}\subseteq C_{F}$ holds. By Claim 35, $S(v_{k}^{(F)})\subseteq C(\pi_{F})$, and $C(\pi_{F})=C_{F}$ by Claim 40.

Recall that, for every $\lambda$, the transition graph of the NFA ${\mathcal{A}}$ has at least one path $\tau(\lambda)$ from the initial state to a final state, on which the labels form a word with Parikh image $(\boldsymbol{y}_{F}(\lambda),\boldsymbol{x}_{F}(\lambda))$, i.e., each symbol $c_{ij}$ occurs $y_{ij}$ times and each $a_{i}$ occurs $x_{i}$ times in this word.

Decompose the path $\tau(\lambda)$ into cycles in a greedy way, by reading the sequence of states $q_{0},q_{1},\dots,q_{t}$ of the automaton ${\mathcal{A}}$ specified by the path $\tau(\lambda)$, from left to right. For each state $q_{i}$, we find the last occurrence of $q_{i}$ in the sequence: $q_{k}=q_{i}$ and $q_{j}\neq q_{i}$ for all $j>k$. The subsequence $q_{i},q_{i+1},\dots,q_{k}$ forms a cycle in the transition graph of the automaton ${\mathcal{A}}$; denote this cycle $\gamma_{i}(\lambda)$. After this, pick $q_{k+1}$ as the next state. Alternatively, if the state $q_{i}$ never occurs further on the path $\tau(\lambda)$, we set $k=i$ and let $\gamma_{i}(\lambda)$ be the empty cycle. We can now decompose this path by separating (possibly empty) cycles $\gamma_{i}(\lambda)$, as indicated in (23):

[TABLE]

Now each cycle $\gamma_{i}(\lambda)$ belongs to an SCC in the transition graph of ${\mathcal{A}}$, and the sequence of states $q_{0},q_{1}(\lambda),\ldots,q_{T}(\lambda)$ corresponds to a complete path through the (acyclic) graph ${\mathcal{B}}$. As previously in this section, we denote the vertices of this path $v_{0}^{(F)},v_{1}^{(F)},\ldots,v_{T}^{(F)}$. Note that the balance condition holds for every cycle in the transition graph of the automaton ${\mathcal{A}}$ (by Claim 31), and thus for each $\gamma_{i}(\lambda)$. Since the vector $\boldsymbol{x}_{F}(\lambda)$ belongs to the cone of balanced vectors as well, the balance condition also holds for the multiplicities of the labels $a_{i}$ on the other edges of the path $\tau(\lambda)$. Moreover, the number of these (other) edges cannot exceed $|Q({\mathcal{A}})|$, simply because our decomposition separated all cycles.

Therefore, if we take the expansion of the vector $\boldsymbol{x}_{F}(\lambda)$ as a nonnegative linear combination of the generators of the cone $K$, this expansion can be split into two terms as follows:

[TABLE]

Here the outer summation in the first term enumerates all vertices $v_{k}^{(F)}$ on the path, and the second term puts together the contribution of the edges not included in the cycles $\gamma_{i}(\lambda)$. We thus have $v_{ij}(\lambda)\leq|Q({\mathcal{A}})|$.

Denote by $Z^{\text{even}}_{F}$ the intersection of $C_{F}$ with the set of even integers and by $Z^{\text{odd}}_{F}$ the intersection of $C_{F}$ with the set of odd integers. Notice that if $u_{ij}^{(k)}(\lambda)>0$ or $v_{ij}(\lambda)>0$, then $i\in Z^{\text{even}}_{F}$ and $j\in Z^{\text{odd}}_{F}$, and that $|Z^{\text{even}}_{F}|=|Z^{\text{odd}}_{F}|=s/2$. Also denote

[TABLE]

then the coefficients of the expansion (24) satisfy the equations

[TABLE]

by definition of the vector $\boldsymbol{x}_{F}(\lambda)$. (Each equation in this system corresponds to one coordinate of $\boldsymbol{x}_{F}(\lambda)$.)

Since there are infinitely many possible values of $\lambda$, and the coefficients $v_{ij}(\lambda)$ are upper-bounded by the number of states of the automaton ${\mathcal{A}}$, it follows that the matrices $(u_{ij}(\lambda)/\lambda)$ of dimension $(s/2)\times(s/2)$ (in which the rows are indexed with even numbers from $C_{F}$, and the columns by odd numbers from $C_{F}$) have a limit point, $(u^{*}_{ij})$, as $\lambda\to\infty$. (Recall that we picked the path through ${\mathcal{B}}$ in such a way that it corresponds to infinitely many values of $\lambda$.) We see from the expansion (24) that if $u^{*}_{ij}>0$, then the set $\{i,j\}$ is contained in some $S(v_{k}^{(F)})$. Moreover, we have

[TABLE]

These conditions mean that $(u^{*}_{ij})$ is a doubly stochastic matrix of size $(s/2)\times(s/2)$. By the Birkhoff—von Neumann theorem (see, e.g., [Sch03, p. 301]), it is a convex combination of permutation matrices. Take some permutation matrix that occurs in this convex combination with a positive coefficient.

This permutation matrix specifies a bijection $\alpha\colon Z^{\text{even}}_{F}\to Z^{\text{odd}}_{F}$ between even and odd indices from $C_{F}$. The bijection has the following properties. If $\alpha(2i)=2j+1$, then $2j+1>2i$, since $u_{2i,2j+1}(\lambda)=0$ for $i>j$ due to the form of the generators of the cone of balanced vectors (Claim 41). Furthermore, every pair $\{2i,\alpha(2i)\}$ is included in some set $S(v_{k}^{(F)})$ by expansion (24). Since $Z^{\text{even}}_{F}\cup Z^{\text{odd}}_{F}=C_{F}$, we obtain the equality

[TABLE]

and the required partitioning of the set $C_{F}$ into pairs $(2i,\alpha(2i))$.

Now recall that even numbers in $C_{F}$ correspond to pluses in the word $\sigma$ and odd numbers to minuses. This correspondence is bijective by the construction of the set $C_{F}$. This proves the last assertion of the claim. ∎

Define a linear order on the pairs of signs identified by Claim 42. Sort the pairs $\{\ell_{i},r_{i}\}$ in the order in which they (or rather their $k$—the one for which $\{\ell_{i},r_{i}\}\subseteq S(v_{k}^{(F)})$) occur along the path $\pi_{F}$; pairs that belong to the same set $S(v_{k}^{(F)})$ can be ordered arbitrarily. This gives us a re-pairing $p_{F}$ of the word $\sigma$.

Claim 43**.**

\mathop{\mathrm{width}}\nolimits(p_{F})\leq\max_{i}\big{|}B(v_{i}^{(F)})\big{|}.

Proof.

In this argument, we will identify the signs of the word $\sigma$ with the corresponding numbers from the set $C_{F}$.

Recall that $B(v)=S(v)\cup I(v)\cup L(v)\cup R(v)$.

The idea behind the assertion is as follows. When the re-pairing has (just) erased all the pairs included in the sets $S(v_{1}^{(F)})$, $S(v_{2}^{(F)})$, …, $S(v_{t}^{(F)})$, the set of all erased signs is the union of intervals whose endpoints are specified by the set $I(v_{t}^{(F)})$, plus perhaps the signs that correspond to $R(v_{t}^{(F)})$, but except the signs that correspond to $L(v_{t}^{(F)})$. At all other points in time, the set of all erased signs is almost the same—except possibly for signs in one of the sets $S(v_{i}^{(F)})$. We will now make this precise and provide justification.

Consider a time point in the re-pairing when the pairs included in the sets $S(v_{1}^{(F)})$, $S(v_{2}^{(F)})$, …, $S(v_{t}^{(F)})$ have all been erased. The set $I(v_{i}^{(F)})$ defines k=\frac{1}{2}\big{|}I(v_{i}^{(F)})\big{|} intervals

[TABLE]

Suppose the number $i\in C_{F}$ belongs to one of these intervals and has not been erased yet. We then obtain from Claim 39 that $i\in L(v_{i}^{(F)})\cup I(v_{i}^{(F)})$. Similarly, if a $j\in C_{F}$ does not belong to these intervals and has been erased already, then $j\in R(v_{i}^{(F)})$ by the same Claim. Therefore, every sign in the word $\sigma$ that has been erased by this time either is covered by one of the $k/2$ intervals, or belongs to the set $R(v_{i}^{(F)})$; all the signs inside these $k/2$ intervals have been erased, except maybe the elements of the set $L(v_{i}^{(F)})$. But this means that the set of all erased signs is a union of at most \frac{1}{2}\big{|}I(v_{t}^{(F)})\big{|}+\big{|}L(v_{t}^{(F)})\big{|}+\big{|}R(v_{t}^{(F)})\big{|} intervals; so the width of the re-pairing at this time point does not exceed this quantity. (Whether the endpoints of the intervals are erased is irrelevant.)

Now consider an intermediate time point, when the pairs included in the sets $S(v_{1}^{(F)})$, $S(v_{2}^{(F)})$, …, $S(v_{t}^{(F)})$ have all been erased, and so have some of the pairs included in the set $S(v_{t+1}^{(F)})$. We claim that the width of the re-pairing at this time point cannot differ by more than \big{|}S(v_{t+1}^{(F)})\big{|} from the same width at time points (i) when the pairs included in the sets $S(v_{1}^{(F)})$, $S(v_{2}^{(F)})$, …, $S(v_{t}^{(F)})$ have all been erased or (ii) when the pairs included in the sets $S(v_{1}^{(F)})$, $S(v_{2}^{(F)})$, …, $S(v_{t+1}^{(F)})$ have all been erased. Indeed, when a pair is erased, this can change (increase or decrease) the number of intervals by at most $2$. It remains to observe that if one takes $\leq\ell$ steps of length at most $2$ to move from a point $X$ to point $Y$ on the line, then after each of these steps the distance from either $X$ or $Y$ to the current position does not exceed $\ell$. ∎

F.3 Proof of Theorem 5

We now obtain a lower bound on the number of states of the NFA ${\mathcal{A}}$ from the construction described in the previous subsections. We will pick in the set $[\frac{n}{2}]$ an appropriate big enough family ${\mathcal{F}}$ of subsets with low pairwise intersection: whenever $F_{1},F_{2}\in{\mathcal{F}}$, it should be the case that $|F_{1}\cap F_{2}|\leq d$. The value of the parameter $d$ will be chosen later.

There exist such families of cardinality $n^{\Omega(d)}$. We will use a construction by Nisan and Wigderson [NW94], modified slightly.

Pick an (odd) prime $p$ in the interval between $\sqrt{n/8-1/2}$ and $\sqrt{n/2-2}$ and set $m=p-1$. Also pick a subset $D$ of size $m$ in the finite field $\mathbb{F}_{p}$.

Embed $\mathbb{F}_{p}\times\mathbb{F}_{p}$ into $[\frac{n}{2}]\setminus\{0,\frac{n}{2}-1\}$. The members of the family ${\mathcal{F}}$ are obtained from the graphs of polynomials of degree (strictly) less than $d$ restricted to the subset $D$, by applying this embedding and adding the numbers [math] and $\frac{n}{2}-1$. There are $p^{d}$ such polynomials in total; if $d<m$, then different polynomials give rise to different subsets in the family ${\mathcal{F}}$, because the graphs of any two polynomials of degree $<d$ can have an overlap of size $<d$ only. This gives us a family ${\mathcal{F}}$ of $p^{d}=\Omega(n^{d/2})$ (sub)sets, each of size $\Omega(n^{1/2})$, with pairwise intersection of at most $d+1$.

For each subset $F\in{\mathcal{F}}$, apply the construction described in Subsection F.2, using the Dyck word $\sigma=\sigma_{n}$. For brevity, denote $w_{n}=\mathop{\mathrm{width}}\nolimits(\sigma_{n})$. This gives, for each $F\in{\mathcal{F}}$, a path $\pi_{F}$ in the graph ${\mathcal{B}}$. By Theorem 4 and Claim 43, this path satisfies the inequality

[TABLE]

Therefore, there is a vertex $v_{i}^{(F)}$ for which $|B(v_{i}^{(F)})|$ is at least $w_{n}$. By Claim 33, this set is included into $C(\pi_{F})=C_{F}$ (see Claim 40). Pick

[TABLE]

then the vertex $v_{i}^{(F)}$ cannot be visited by any other path $\pi_{F^{\prime}}\neq\pi_{F}$ due to the upper bound on the pairwise intersection of the subsets from ${\mathcal{F}}$.

We have thus identified for each path $\pi_{F}$, $F\in{\mathcal{F}}$, a unique vertex of the graph ${\mathcal{B}}$ that is visited by no other path from this family. So the number of vertices in the graph ${\mathcal{B}}$, i.e., the number of strongly connected components in the transition graph of the NFA ${\mathcal{A}}$, cannot be less than the number of members of the family ${\mathcal{F}}$, that is,

[TABLE]

The number of states of ${\mathcal{A}}$ cannot be less than that either, and this is exactly the assertion of Theorem 5.