Graph Reachability and Pebble Automata over Infinite Alphabets

Transcription

1 Graph Reachability and Pebble Automata over Infinite Alphabets Tony Tan Department of Computer Science Technion Israel Institute of Technology Haifa 32000, Israel Abstract We study the graph reachability problem as a language over an infinite alphabet. Namely, we view a word of even length a 0 b 0 a n b n over an infinite alphabet as a directed graph with the vertices a 0, b 0,..., a n, b n and the edges (a 0, b 0 ),..., (a n, b n ). We prove that for any positive integer k, k pebbles are sufficient for recognizing the existence of a path of length 2 k 1 from the vertex a 0 to the vertex b n, but are not sufficient for recognizing the existence of a path of length 2 k+1 2 from the vertex a 0 to the vertex b n. Based on this result, we establish a number of relations among some classes of languages over infinite alphabets. 1 Introduction In this paper we continue the study of pebble automata introduced in [4]. In short, pebble automata (PA) are finite state automata with a finite number of pebbles. The pebbles are placed on/lifted from the input word in the stack discipline first in last out and are intended to mark positions in the input word. One pebble can only mark one position and the last pebble placed serves as the head of the automaton. The automaton moves from one state to another depending on the current label and the equality tests among data values in the positions and the positions themselves currently marked by the pebbles. 1

2 It turns out that PA languages possess a very nice logical property: closure under all boolean operations. In fact, looking at the stack discipline imposed on the placement of the pebbles, one can rightly conclude that PA is a kind of extension of first order logic for word structure. Indeed, to simulate a first order sentence of quantifier rank k, a pebble automaton with k pebbles suffices: one pebble for each quantifier depth. In this paper we are trying to extend the study of PA to the realm of graphs. This is done by viewing a word of even length a 0 b 0 a n b n over an infinite alphabet as a directed graph with the symbols a 0, b 0,..., a n, b n as the vertices and the edges (a 0, b 0 ),..., (a n, b n ). With regards to reachability, we prove that PA behaves similarly to the classical result for first order logic: For any positive integer k, k pebbles are sufficient for recognizing the existence of a path of length 2 k 1 from the vertex a 0 to the vertex b n, but are not sufficient for recognizing the existence of a path of length 2 k+1 2 from the vertex a 0 to the vertex b n. This result can be easily translated to logic over graphs to establish a well known fact that to express a path of length 2 k in first order sentence, a quantifier rank k is necessary and sufficient. See [7]. Furthermore, this result, we believe, may contribute to the technical aspect of reasoning on languages over infinite alphabets as shown in this paper. For example, we establish the following relationships among the classes of languages over infinite alphabets which were previously unknown. 1. The strict hierarchy of the PA languages based on the number of pebbles. 2. The separation of monadic second order logic from the PA languages. 3. The separation of the one-way deterministic RA languages from the PA languages. Although the emptiness problem for PA, in general, is undecidable, we believe the technique introduced in this paper can be useful in general for the study of languages over infinite alphabets. For example, in [8] a similar technique is used to obtain definability result for Linear Temporal Logic augmented with one register freeze quantifier, a class of languages over infinite alphabets with decidable satisfaction problem. Related results. As we mention earlier, there is an analogy between our result with the well known quantifier lower bounds for (s, t)-reachability 2

3 which states that (s, t)-reachability in graphs is expressible in a first order sentence of quantifier rank k if and only if the distance from s to t is less than or equal to 2 k. See, for example, [9]. However, as far as the author can see, this classical result does not imply our result here. For one thing, PA is already strictly stronger than firstorder logic. See Proposition 4. For example, PA can easily check whether the length of the an input w is divisible by 4, thus, checking whether the number of edges is even. However, such property cannot be express in firstorder logic. In fact, as described in [7] every first order logic on graphs can be appropriately embedded into a first-order logic for languages over infinite alphabet. With a bit modification to the PA model, our result here can be translated to obtain the quantifier lower bound for (s, t)-reachability. Another result related to our result in this paper is the famous theorem of Fagin, Stockmeyer and Vardi that directed (s, t)-reachability is in monadic conp, but not in monadic NP [2]. To the best of our knowledge, this result has no connection with the result in this paper. One reason is that no bound of quantifier rank is established there. Furthermore, the logic introduced in [7] behaves differently from the standard logic for graphs. In [7] the domain is the edges, not the vertices. In such logic, (s, t)-reachability can be expressed in monadic NP when the domain is the edges. See Proposition 3 in [7]. This paper is organized as follows. In Section 2 we review two models of computations for words over infinite alphabet: monadic second-order logic MSO and pebble automata (PA). Section 3 is the core of the paper in which we present all our results (whose complete proofs can be found in Appendix A). Finally, in Section 4 we discuss how to adjust our results/proofs presented in Section 3 to both the stronger and weaker versions of PA: the alternating one-way PA and weak PA, respectively. The proofs of the results in this sections are presented in Appendices B and C. 2 Models of computations In Section 2.1 we recall the definition of pebble automata from [4], and in Section 2.2 we recall the logical framework for languages over infinite alphabets. We shall use the following notation: D is a fixed infinite alphabet not 3

4 containing the left-end marker or the right-end marker. The input word to the automaton is of the form w, where w D. Symbols of D are denoted by low case letters a, b, c, etc., possibly indexed, and words over D are denoted by low case letters u, v, w, etc., possibly indexed. The set of symbols occurring in w is denoted by [w]. 2.1 Pebble automata Definition 1 (See [4, Definition 2.3]) A non-deterministic (two-way) k-pebble automaton, (in short k-pa) over D is a system A = Q, q 0, F, µ whose components are defined as follows. Q, q 0 Q and F Q are a finite set of states, the initial state, and the set of final states, respectively; and µ is a finite set of transitions of the form α β such that α is of the form (i, a, P, V, q) or (i, P, V, q), where i {1,..., k}, a D {, }, P, V {i + 1,..., k}, and β is of the form (q, act), where q Q and act {left, right, place-pebble, lift-pebble}. A symbol a D which occurs in a transition (p, a, P, V ) β µ is called a constant symbol. We denote by Θ A the set of all constant symbols of A. Remark 2 Note the pebble numbering that differs from that in [4]. In the above definition we adopt the pebble numbering from [1] in which the pebbles placed on the input word are numbered from k to i and not from 1 to i as in. [4]. The reason for this reverse numbering is that it allows us to view the computation between placing and lifting pebble i as a computation of an (i 1) pebble automaton. This approach will be convenient in proofs by induction on the number of pebbles already placed on the input word. Given a word w = a 1 a n D, a configuration of A on w is a triple [i, q, θ], where i {1,..., k}, q Q and θ : {i, i+1,..., k} {0, 1,..., n, n+ 1}. The function θ defines the position of the pebbles and is called the pebble assignment. 1 The initial configuration is γ 0 = [k, q 0, θ 0 ], where θ 0 (k) = 0 is 1 The symbols in the positions 0 and n + 1 are and, respectively. 4

5 the initial pebble assignment. A configuration [i, q, θ] with q F is called an accepting configuration. A transition (i, a, P, V, p) β applies to a configuration [j, q, θ], if (1) i = j and p = q, (2) V = {l > i : a θ(l) = a θ(i) }, (3) P = {l > i : θ(l) = θ(i)}, and (4) a θ(i) = a. A transition (i, P, V, q) β applies to a configuration [j, q, θ], if conditions (1) (3) above hold and no transition of µ of the form (i, a, P, V, q) β applies to [j, q, θ]. We define the transition relation A as follows: [i, q, θ] A [i, q, θ ], if there is a transition α (p, act) µ that applies to [i, q, θ] such that q = p, for all j < i, θ (j) = θ(j), and - if act = left, then i = i and θ (i) = θ(i) 1, - if act = right, then i = i and θ (i) = θ(i) + 1, - if act = lift-pebble, then i = i + 1, - if act = place-pebble, then i = i 1, θ (i 1) = 0 and θ (i) = θ(i). As usual, we denote the transitive closure of A by A. When the automaton A is clear from the context, we shall omit the subscript A. For 1 i k, an i-configuration is a configuration of the form [i, q, θ], i.e., a configuration in in which the head pebble is pebble i. An i-run is a run from an i-configuration to an i-configuration in which pebble (i + 1) is never lifted. A word w D is accepted by A, if γ 0 γ, for some accepting configuration γ of A on the input w. The language L(A) consists of all words accepted by A. The automaton A is deterministic, if there is exactly one transition that applies to each configuration. If act {right, lift-pebble, place-pebble} for all transitions, then the automaton is one-way. It turns out that PA languages are quite robust. Namely, nondeterminism and two-wayness do not increase the expressive power, see Theorem 3 below. 5

6 Theorem 3 ([4, Theorem 4.6]) For each k 1, all versions of k-pebble automata: non-deterministic two-way, deterministic two-way, non-deterministic one-way and deterministic one-way, have the same recognition power. In view of Theorems 3, we shall always assume that pebble automata under consideration are deterministic and one way. Next, we define the hierarchy of languages accepted by PA. For k 1, PA k is the set of all languages accepted by k-pa, and PA is the set of all PA languages. 2 That is, PA = k 1 PA k. 2.2 Logic Formally, a word w = a 1 a n is represented by the logical structure with domain {1,..., n}; the natural ordering < on the domain with its induced successor +1; and the function val : {1,..., n} D, where val(i) = a i for each i = 1,..., n. The atomic formulas in this logic is of the form x < y, y = x + 1, val(x) = val(y), or val(x) = a for a D. The first-order logic FO is obtained by closing the atomic formulas under the propositional connectives and first-order quantification over {1,..., n}. The second-order formulas MSO are obtained by adding quantification over unary predicates on {1,..., n}. A sentence ϕ defines the set of words L(ϕ) = {w : w = ϕ}. If L = L(ϕ) for some sentence ϕ, then we say that the sentence ϕ expresses the language L. We use the same notation FO and MSO to denote the languages expressible by sentences in FO and MSO, respectively. That is, FO = {L(ϕ) : ϕ is an FO sentence} and MSO = {L(ϕ) : ϕ is an MSO sentence}. Proposition 4 FO D PA. More specifically, let ϕ FO D with quantifier rank k. Then, L(ϕ) PA k. 2 Note that we use the same notation, whose meaning, however, will be always clear from the context. 6

7 Proof. First, we note that PA k is closed under boolean operations. Let ϕ = Qx k ψ(x k ) where Q {, } and ψ(x k ) is of quantifier rank k 1. The proof is by straightforward induction on k. A k-pa A iterates pebble k through all possible positions in the input word. On each iteration, the automaton A recursively calls a (k 1)-PA A that accepts the language L(ψ(x k )), treating the position of pebble k as the assignment value for x k. - If Q =, then A accepts if and only if A accepts on all iterations. - If Q =, then A accepts if and only if A accepts on at least one iteration. We end this section with Theorem 5 below which states that all languages accepted by pebble automaton can be expressed by an MSO sentence. Theorem 5 ([4, Theorem 4.2]) PA MSO. Actually Theorem 4.2 in [4] states a stronger result: all languages accepted by alternating two-way PA can be expressed by an MSO sentence. However, we will postpone our discussion of alternating PA to Section 4. 3 Words of D as Graphs Let w = a 0 b 0 a n b n D be a word of even length. The word w induces a directed graph G w = (V w, E w ) whose the set of vertexes is V w = {a 0, b 0,..., a n, b n }, that is, the symbols that appear in w, and the set of edges is E w = {(a 0, b 0 ),..., (a n, b n )}. With such view, we will use the term graph when referring to an even length word. We also write s w = a 0 and t w = b n to denote the first and the last symbol in w, respectively. For convenience, we consider only the words w in which s w and t w occur only once. We need the following basic graph terminology. Let a and b be vertexes in a graph G. A path of length m from a to b is a sequence of m edges in G: (a i1, b i1 ),..., (a im, b im ) such that a i1 = a, b im = b and for each j = 1,..., m 1, b ij = a ij+1. The distance from a to b is the length of a minimal path from a to b in G. If there is no path from a to b in G, then we set the distance be. We define the following reachability languages. For m 1, R m = {w : the distance from s w to t w in G w is m} 7

8 and R = m=1,2,... R m Remark 6 When processing an input word w, an automaton A can remember by its state whether its head pebble is currently at an odd-number or even-number position in w. 3 Therefore, we shall always assume that the input word w is of even length (which also naturally fits the subject of this appear). In addition, unless indicated otherwise, we always denote an input word w by a 0 b 0 a n b n. That is, the odd-position symbols are denoted by a i s and the even-position symbols by b i s. Again, we remind the reader again that in k-pa the pebbles placed on the input word are numbered from k to i, see Remark 2. Proposition 7 For each k = 1, 2,..., R 2 k 1 PA k. Proof. The proof of this proposition is the standard implementation of Savitch s algorithm for (s-t)-reachability [5]. It is by induction on k. For the basis, we prove that R 3 PA 2. On input word w = a 0 b 0 a n b n, a 2-PA A 2 acts as follows. - Pebble 2 iterates through all a i. - On each iteration, the automaton uses pebbles 1 to check whether b 0 = a i. - If b 0 = a i, the automaton moves pebble 2 to the right to read b i, and check whether b i = a n. Now, we prove the induction step. On input word w = a 0 b 0 a n b n, a k-pa A k acts as follows. - Pebble k iterates through all a i. - On each iteration, the automaton recursively uses pebbles (k 1),..., 1 to check whether there is a path from a 0 to a i. That is, A k makes a recursive call of A k 1. 3 We count that the leftmost symbol of w is in position 1. 8

9 - If such path exists, the automaton lifts all pebbles 1,..., (k 1); moves pebble k to the right to read b i, and recursively uses pebbles k 1,..., 1) to check whether there is a path from b i to b n. That is, A k makes a recursive call of A k 1 one more time. A k accepts w if there is an index i such that there are a path from a 0 to a i and a path from b i to b n. It is straightforward to show that R 2 k 1 PA k. Lemma 8 below is the backbone of most of the results presented in this paper. Its proof is quite long. Therefore, we only sketch it in Section 3.1. A complete proof of the lemma can be found in Appendix A. Let n k = 2 k+1 2, k = 1, 2,.... Lemma 8 Let A be a k-pebble automaton such that L(A) R. Then R nk L(A). 4 Intuitively, the proof is as follows. We will show that for every k-pa A, there exists w R nk R nk 1 such that A cannot recognize a path from s w to t w. The proof is by induction on k. The basis, k = 1, is relatively trivial to demonstrate as 1 pebble is not even capable of comparing data values. Let A be a k-pa and let A k 1 be (k 1)-PA subautomaton of A. By the induction hypothesis, there exists w R nk R nk 1 such that A k 1 cannot recognize a path from s w to t w. The induction step is as follows. Let w be a data word isomorphic to w such that [w] [w ] =. We define the data word u such that where u = w }{{} a 1 a 2 }{{} a 2 a 3 }{{} w ; v 1 v 2 v 3 v 1, v 2, v 3 depend on the number of states of A; a 1 = t w ; a 3 = s w ; a 2, a 3 / [w] [w ]. 4 In fact, a similar result holds for alternating one-way k-pa, see Lemma 16. Therefore, Theorems 11 and 12 below extend onto alternating one-way pebble automata as well. 9

10 It is obvious that u R nk R nk 1. We claim that A cannot recognize a path from s u to t u. The proof is roughly as follows. When pebble k is above wv 1 a 1 a 2, by the induction hypothesis, the subautomaton A k 1 cannot recognize the path from s w to t w, thus, the path from s u to t u. When pebble k is above v 2 a 2 a 3 v 3 w, by the induction hypothesis, the subautomaton A k 1 cannot recognize the path from s w to t w, thus, the path from s u to t u. Therefore, the automaton cannot recognize the path from s u to t u. We want to notify the reader here that the stack discipline of the pebbles is very crucial in the application of the induction hypothesis. Pebble k is fixed on its current position once we start using the other (k 1) pebbles, and cannot be moved unless all the other (k 1) pebbles are lifted. Corollary 9 R nk / PA k. Proof. The corollary immediately follows from the lemma, because R k R. Corollary 10 R / PA. Proof. Assume to the contrary that R = L(A) for a k-pa A. Then, R nk L(A), which contradicts Lemma 8. The following theorem establishes the proper hierarchy of the PA languages. Theorem 11 For each k = 1, 2,..., PA k PA k+1. Proof. By Proposition 7, R 2 k+1 1 PA k+1. Suppose R 2 k+1 1 PA k. Let A be a k-pa that accepts R 2 1. k+1 Then, L(A) R and by Lemma 8, R 2 k+1 2 L(A). But this is a contradiction since L(A) = R 2 k+1 1 R 2 2. k+1 Another consequence of Lemma 8 is that the inclusion of PA in MSO provided by Theorem 5 is proper. 10

11 Theorem 12 PA MSO. Proof. Without loss of generality, we may assume that MSO contains two constant symbols, min and max, which denote minimum and the maximum elements of the domain, respectively. 5 Then the language R can be expressed in MSO as follows. There exist unary predicates S odd and P such that S odd is the set of all odd elements in the domain where min S odd and max S odd. sen- The predicate P satisfies the conjunction of the following FO tences: P S odd and min P and max 1 P, for all x P {max 1}, there exists exactly one y P such that val(x + 1) = val(y), and for all x P {min}, there exists exactly one y P such that val(y + 1) = val(x). Now, the theorem follows from Corollary 10. Next, we define a restricted version of the reachability languages. Recall that [w] denotes the set of symbols occurring in w. For a positive integer m 1, the language R + m consists of all words of the form where c 0 c 1 }{{} c 1 c 2 }{{} c 2 c 3 c m 3 c m 2 }{{} c m 2 c m 1 }{{} c m 1 c m u 1 u 2 u m 2 u m 1 c 0,..., c m D, and c i c i+1, for all i = 0,..., m 1; u 1,..., u m 1 D and each of u 1,..., u m 1 is of even length; for each i = 1,..., m 1, c i [u i ]. 5 For a word w = a 1 a n, the minimum and the maximum elements are 1 and n, respectively, whereas not 0 and n + 1 are reserved for the endmarkers and. 11

12 The language R + is defined as R + = m=1,2,... R + m. In some sense, we can compare the relation between R and R +, like the relation between nondeterministic and deterministic transitive closure logic. Remark 13 Actually, in the proof of Lemma 8 we show that for every k-pa A, there exist graphs G R + n k and G R + such that either A accepts both G and G, or rejects both of them are not. Therefore, R + PA. In fact, R + is not accepted by alternating one-way PA either. The following theorem answers the problem left open in [4, 6]: can twoway deterministic FMA be simulated by pebble automata. 6 Theorem 14 The language R + is accepted by one-way deterministic FMA, but is not accepted by pebble automata. Proof. Note that R + is accepted by a two register one-way deterministic FMA. On input word w = a 0 b 0 a n b n, the automaton stores b 0 in the first register and then moves right (using the second register to scan the input symbols) until it finds a vertex a i1 = b 0. If it finds one, then it stores b i1 in register 1 and moves right again until it finds another vertex a i2 = b i1. It repeats the process until either of the following holds. - a n is the same as the content of the register 1, or, - it cannot find a vertex currently stored in the first register. In the former case, th automaton accepts the input graph, and in the latter case does not. However, by Remark 13, R + is not a PA language. 7 6 See [3, Definition 1] for the formal definition of FMA. 7 As before, a similar separation result holds for alternating one-way PA as well. 12

13 3.1 Sketch of the Proof of Lemma 8 Recall that for i = 1, 2,..., n i = 2 i+1 2. An equivalent recursive definition is n 1 = 2, and n i+1 = 2n i + 2, for i 2. Let m > 1 be a positive integer. We define G i,m to be the set of graphs G of the following form: G = c 0 c 1 d {z} 0 d 1 c {z} 1 c 2 d {z} 1 d 2 c ni 2c ni 1 {z} v 1 w 1 v 2 v ni 1 d ni 2d ni 1 {z} c ni 1c ni (1) w ni 1 where the strings v 1, w 1,..., v ni 1, w ni 1 are defined as follows. For each j = 1,..., n i 1, - v j = e j,1 e j+1,1 e j,m 1 e j+1,m 1 ; - w j = f j,1 f j+1,1 f j,m 1 f j+1,m 1 ; and - the symbols {c 0,..., c ni, d 0,..., d ni 1} {e i,j, f i,j : 1 i n i 1, 1 j m 1} are pairwise different. Remark 15 Note that G i,m R ni. We just put u j = v j d j 1 d j w j, j = 1,..., n i 1. Figure 3.1 illustrates the real graph represented by an element of G k,m. Let G be the graph depicted in 1. We define the graph G as G = c 0 c 1 }{{} d 0 d 1 }{{} c 1 c 2 }{{} d 1 d 2 c ni 2c ni 1 }{{} v 1 w 1 v 2 v ni 1 d ni 2d ni 1 (2) That is, the graph G is G without the suffix w ni 1c ni 1c ni. Obviously, this graph is not in R. Let G i,m = {G : G i,m }. We claim that for each k-pa A, there exists an positive integer m such that A cannot distinguish between G and G, where G G k,m. 8 The proof is by induction on k and in the remainder of this section we will sketch it for the case of k = 3. 9 Note that in this case, n 3 = By cannot distinguish we mean that the automaton finishes its computations on both G and G in the same state. This notion will be precisely explained in Appendix A. 9 The complete proof is presented in Appendix A. 13

14 c 0 c 1 c 2 c 3 c nk 3 c nk 2 c nk 1 c nk e 1,1 e 2,1 e 3,1 e e 4,1 nk 3,1 e nk 2,1 e nk 1,1 e 1,2 e 2,2 e 3,2 e 4,2 e nk 3,2 e nk 2,2 e nk 1,2. e. 1,m 1 e 2,m 1 e 3,m 1 e 4,m 1 e nk 3,m 1 e nk 2,m 1 e nk 1,m 1 d 0 d 1 d 2 d 3 d nk 3 d nk 2 d nk 1 f 1,1 f 2,1 f 3,1 f f 4,1 nk 3,1 f nk 2,1 f nk 1,1 f 1,2 f 2,2 f 3,2 f 4,2 f nk 3,2 f nk 2,2 f nk 1,2. f. 1,m 1 f 2,m 1 f 3,m 1 f 4,m 1 f nk 3,m 1 f nk 2,m 1 f nk 1,m 1 The twisted arrow represents the order in which the edges are written in G G k,m. Figure 1: The graph induced by G G k,m. Let A = Q, q 0, µ, F be a 3-PA. We can assume that A is one-way and deterministic, and that µ does not contain transitions of the form (i, a, P, V, q) (q, act), i.e., Θ A =. Recall that, pebbles are placed on the input word in the order from 3 to 1, see Remark 2. We normalize the behavior of each pebble as follows. The automaton A can enters into a final state only after pebble 3 reaches the right-end marker and for each i = 2, 1 the following holds. Immediately after pebble (i + 1) moves right, pebble i is placed; pebble i is lifted only when it reaches the right end of the input; and 14

15 immediately after pebble i is lifted, pebble (i + 1) moves right. We shall need one more bit of notation. Let β i be defined by the following recursion: β 1 = Q and, for i 2, β i = Q 2 ((β i 1 )!) Q. Note that β j divides β i, whenever j i. Let m = β 4, 10 and let G and G be as follows. G = c 0 c 1 d 0 d 1 c 1 c 2 c 12 c 13 d 12 d 13 c 13 c 14 G = c 0 c 1 d 0 d 1 c 1 c 2 c 12 c 13 d 12 d 13 where G G 3,m. Note that for each i = 0,..., 8, we can partition G and G as follows, where +6 comes from i 2 = 6. G = u G {}}{{}}{{}}{ c 0 c 1 d 0 d 1 c i c i+1 d i d i+1 d i+4 d i+5 c i+5 c i+6 d 12 d 13 c 13 c 14 u G = c 0 c 1 d 0 d 1 c }{{} i c i+1 d i d i+1 d i+4 d i+5 c i+5c i+6 d 12d 13 }{{}}{{} u G u where G G 2,m, G G 2,m and u is isomorphic to u. Now (for the induction hypothesis) we assume that in the 2-runs of A on G and G, if (1) pebble 3 is not inside G and is not inside G, (2) both G and G do not contain the symbol under pebble 3, and (3) pebble 2 enters the patterns G and G in the same state, 11 then pebble 2 exits G and G in the same state; and prove (for the induction step) the following claim. Claim 1 The automaton A finishes the computation on both G and G in the same state. 10 Here the number 4 comes from the number of pebbles (3) plus 1, and in the general proof it will be β i Of course, by entering a pattern we mean arriving at its fist symbol. 15

16 Proof. (Sketch) We divide the graph G into three parts: A, B and C, and divide the graph G into two parts: D and E, as illustrated below. v 1 v 7 w 7 v 8 v 13 w 13 {}}{{}}{{}}{{}}{{}}{{}}{ G = c 0 c 1 d 0 d 1 c 6 c 7 d }{{} 6 d 7 c }{{} 7 c 8 d 7 d 8 c 12 c 13 d 12 d 13 c 13 c }{{ 14 } A B C v 1 v 7 w 7 v 8 v 13 {}}{{}}{{}}{{}}{{}}{ G = c 0 c 1 d 0 d 1 c 6 c 7 d }{{} 6 d 7 c 7 c 8 d 7 d 8 c 12 c 13 d 12 d }{{ 13 } D E The proof of the claim is divided into three stages according to the above division. Stage 1: We show that pebble 3 exits section A (in the run of A on G) in the same state as it exits section D (in the run of A on G). Recall that pebble 3 is the first pebble being placed on the input. First, note that there exist suffixes G 1 of G and G 1 of G, such that G 1 G 2,m and [A] [G 1 ] = [D] [G 1 ] =, as depicted in Figure 2 below. G = A G 1 {}}{{}}{ c 0 c 1 c 6 c 7 d 6 d 7 c 7 c 8 d 7 d 8 c 8 c 9 d 8 d 9 d 12 d 13 c 13 c 14 Pebble 3 is above A and D Since pebble 2 enters G 1 and G 1, it exits G 1 and G 1 in the same state G = c 0 c 1 c 6 c 7 d 6 d 7 c 7 c 8 d 7 d 8 c 8 c 9 d 8 d 9 d 12 d 13 }{{}}{{} D Figure 2: When pebble 3 is above A and D, we apply the induction hypothesis, according to which pebble 2 exits G 1 and G 1 in the same state. Since A starts the computation in the initial state q 0, pebble 3 reads the symbol c 0 in G in the same state (i.e., q 0 ) as it reads the symbol c 0 in G. Therefore, after pebble 2 is placed, it enters G 1 in the run of A on G in the same state as it enters G 1 in the run of A on G. Furthermore, since [A] [G 1 ] = [D] [G 1 ] =, the symbol under pebble 3 (i.e., c 0 ) is not in G 1 and G 1. By the induction hypothesis, pebble 2 exits G 1 and G 1 in the same state. Thus, when pebble 3 moves right, it reads c 1 in the run on G in the same 16 G 1

17 state as it reads c 1 in the run on G. This scenario repeats until pebble 3 exists sections A and D. Therefore, pebble 3 exits section A in the same state as it exits section D. Stage 2: We show that in the run of A on G pebble 3 enters and exits section B in the same state. At this stage we show that the behavior of pebble 3 when scanning v 7 d 6 d 7 w 7 c 7 c 8 is periodic. Namely, there exists an integer 1 p β 3 such that on every other 2p moves (to the right), pebble 3 enters the same state. We call such number p an interval of pebble 3. The proof of the existence of the interval p is by induction on the number of pebbles. In fact, we show that for i = 1, 2, 3, the number β i is the largest possible interval of pebble i and β i+1 is a multiple of all possible intervals of pebble i. Since the length of the section B = d 6 d 7 w 7 is 2m, pebble 3 enters and exists this section in the same state. The complete proof of this stage is presented in Appendix A.1. Stage 3: Pebble 3 exits section C (in the run of A on G) in the same state as it exits section E (in the run of A on G). First, note that there exist prefixes G 2 of G and G 2 of G such that G 2 G 2,m and [C] [G 2 ] = [E] [G 2 ] =, see Figure 3 below. Let G 3 be the suffix of G such that G = G 2 G 3 and G 4 be the suffix of G such that G = G 2 G 4. Then the graphs G 3 and G 4 are isomorphic. z } { G 2 C z } { z } { G = c 0 c 1 d 0 d 1 d 4 d 5 c 5 c 6 d 5 d 6 c 6 c 7 d 6 d 7 c 7 c 8 d 7 d 8 d 12 d 13 c 13 c 14 G 3 Since pebble 2 enters G 2 and G 2, it exits G 2 and G 2 in the same state Pebble 3 is above C and E G = c 0 c 1 d 0 d 1 d 4 d 5 c 5 c 6 d 5 d 6 c 6 c 7 d 6 d 7 c 7 c 8 d 7 d 8 d 12 d 13 {z } {z } G 2 {z E } G 4 Figure 3: When pebble 3 is above C and E, we apply the induction hypothesis, according to which pebble 2 exits G 2 and G 2 in the same state. By Stages 1 and 2, pebble 3 enters section C (in the run of A on G) in 17

18 the same state as it enters section E (in the run of A on G). Thus, pebble 3 reads the symbol c 7 in section C in the same state as it reads the symbol d 6 in section E. Therefore, pebble 2 enters G 2 and G 2 in the same state. In addition, since [C] [G 2 ] = [E] [G 2 ] =, the symbol under pebble 3 does not occur in either of G 2 or G 2. Thus, by the induction hypothesis, pebble 2 exits G 2 and G 2 in the same state. Next, the graphs G 3 and G 4 are isomorphic and the length of the pattern between G 2 and the first symbol of C (the position of pebble 3 in the run of A on G) equals to the length of the pattern between G 2 and the first symbol of E (the position of pebble 3 in the run of A on G). Thus, the computation of pebble 2 inside G 3 (in the run of A on G) is the same as its computation inside G 4 (in the run of A on G). Therefore, pebble 3 exits G 3 in the same state as it exits G 4. That is, pebble 3 moves right and reads c 8 in C in the same state as it moves right and reads d 7 in E. This scenario repeats until pebble 3 finishes scanning sections C and E. Thus, pebble 3 exits section C (in the run of A on G) in the same state as it exits section E (in the run of A on G). This completes the sketch of the proof of our claim that A finishes the computation on both G and G in the same state. 4 Alternating PA and weak PA Alternating PA As we have mentioned in the introduction, all the results of Section 3 extend onto the case of alternating one-way PA. (See Appendix B for the formal definition.) Let APA k = {L : L is accepted by an alternating one-way k-pa} and APA = k 1 APA k Lemma 16, whose proof is presented in Appendix B, is the extension of Lemma 8 onto alternating one-way PA. Lemma 16 Let A be an alternating one-way k-pebble automaton such that L(A) R. Then R nk L(A). 18

19 Each of the clauses of Theorem 17 below is easy consequence of the lemma. Theorem For each k = 1, 2,..., APA k APA k APA MSO. 3. The language R + is accepted by one-way deterministic FMA, but is not accepted by alternating one-way pebble automata. Weak PA There is another analogue of our results to another, but weaker, version of pebble automata. In the model defined in Section 2, the new pebble is placed in the beginning of the input word. An alternative would be to place the new pebble at the position of the most recent one. The model defined this way is usually referred as weak PA. Formally, it is defined by setting θ (i 1) = θ(i) (and keeping θ (i) = θ(i)) in the case of act = place-pebble of the definition of the transition relation (Definition 1). Obviously, twoway weak PA is just as expressive as two-way strong PA. However, one-way deterministic weak k-pa is weaker than strong k-pa. For example, R 2 k 1 is not a weak k-pa language, See Lemma 18 below. Furthermore, alternating, nondeterministic and deterministic weak k-pa have the same recognition power. See [8]. Let wpa k = {L : L is accepted by a weak k-pa} and wpa = k 1 wpa k Lemma 18 For each k = 1, 2,..., R k wpa k, but R k+1 / wpa k. The proof of the lemma is presented in Appendix C. Lemma 18 implies the strict hierarchy for wpa languages. Theorem 19 For each k = 1, 2,..., wpa k wpa k+1. 19

20 Acknowledgment The author would like to thank Michael Kaminski for all helpful discussions without which this paper would not exist, for pointing out that the original example of the author is, actually, the reachability problem in graphs, and for his comments on the first version of this paper. References [1] Mikolaj Bojanczyk, Mathias Samuelides, Thomas Schwentick, and Luc Segoufin. Expressive power of pebble automata. In Part I of the Proceedings of Automata, Languages and Programming, 33rd International Colloquium, ICALP 2006, Venice, Italy, July 10-14, 2006, volume 4051 of Lecture Notes in Computer Science, pages Springer, [2] Ronald Fagin, Larry J. Stockmeyer, and Moshe Y. Vardi. On monadic np vs. monadic co-np. Information and Computation, 120(1):78 92, [3] Michael Kaminski and Nissim Francez. Finite-memory automata. Theoretical Computer Science, 134(2): , [4] Frank Neven, Thomas Schwentick, and Victor Vianu. Finite state machines for strings over infinite alphabets. ACM Transactions on Computational Logic, 5(3): , [5] Walter J. Savitch. Relationships between nondeterministic and deterministic tape complexities. Journal of Computer and System Sciences, 4(2): , [6] Luc Segoufin. Automata and logics for words and trees over an infinite alphabet. In Proceedings of the 20th International Workshop/15th Annual Conference of the EACSL Computer Science Logic, CSL 2006, Szeged, Hungary September 25-29, 2006, pages 41 57, [7] Tony Tan. Logic for graph edges and pebble automata over infinite alphabets. Technical report, Department of Computer Science, Technion Israel Institute of Technology, Can be found in 20

21 [8] Tony Tan. On pebble automata for data languages with decidable emptiness problem. Technical report, Department of Computer Science, Technion Israel Institute of Technology, Can be found in [9] György Turán. On the definability of properties of finite graphs. Discrete Mathematics, 49(3): , A Proof of Lemma 8 In this appendix contains the proof of Lemma 8 that we outlined in Section 3.1. Namely, in Section A.1 we present the formalization of Stage 2 of the proof, whereas the proofs of Stages 1 and 3 are presented in Section A.2. Recall that A = Q, q 0, F, µ is a strong k-pa that accepts the language R nk. By Theorem 3, we may assume that A is deterministic and one-way. We also assume that Θ A =. Recall that, pebbles are placed on the input word in the order from k to 1, see Remark 2. By adding some extra states, we can normalize the behavior of each pebble i, i = k 1,..., 2, 1, as follows. After pebble (i + 1) moves right, pebble i is immediately placed (at the left endmarker ); pebble i is lifted only when it reaches the right-end marker of the input; and immediately after pebble i is lifted, pebble (i + 1) moves right. Furthermore, only pebble k can enter a final state and it may do so only after it reads the right-end marker of the input. Also, instead of writing the automaton A enters state q with pebble i as the head pebble, we shall write only pebble i enters the state q. That is, each pebble is actually viewed as an automaton itself. We start some terminology which we shall need for the proof. Recall that [w] denotes the set of symbols that occur in the word w. Let γ = [i, q, θ] be an i-configuration of A on the word w = c 1 c n. For j = i,..., k, we denote by val θ,w (j) = c θ(j), i.e. the symbol under pebble j with respect to the assignment θ on the word w. 21

22 Let γ = [i, q, θ] be an i-configuration on a word uvw. For i j k, we say that pebble j is outside the infix v, if it is not located above v and val θ,uvw (j) / [v]. The configuration γ is outside the infix v, if all pebbles from i to k are outside v. Let ρ : D D be a 1-1 mapping. The mapping ρ It induces an automorphism on D, also denoted by ρ, that is defined as follows. If w = c 1 c n D, then ρ(w) = ρ(c 1 ) ρ(c n ). Let w 1, w 2 D. We say that w 1 is isomorphic to w 2, denoted by w 1 = w2, if there exists a 1-1 mapping ρ : D D such that w 1 = ρ(w 2 ). Note that if w 1 = w2, then w 1 L(A) if and only if w 2 L(A). (Recall that we assume the automaton A does not have constant symbols in its description, i.e. Θ A =.) Definition 20 Let u 1, u 2 D. We say that u 1 and u 2 are A-equivalent, if A finishes the computation on both inputs u 1 and u 2 in the same state. Let uvw D and γ = [i, q, θ] be an i-configuration of A on uvw. We shall use the following notation. q(γ) = q and θ(γ) = θ. Let γ be outside v. The pebble assignment θ modulo v is a pebble assignment θ on uw, written as θ = θ mod v, where for all j = i,..., k, { θ(j) if 1 θ(j) u θ (j) = θ(j) v if u + v + 1 θ(j) u + v + w Recall that for 1 i k, an i-configuration is a configuration [i, q, θ] in which the head pebble is pebble i. An i-run is a run from an i-configuration to an i-configuration in which pebble (i+1) is never lifted. Let j < i. A j-run yielded from an i-configuration γ is a j-run that starts with an j-configuration γ where γ is the first j-configuration such that γ γ. Let w D and γ be an i-configuration of A on w. Let v be a prefix of w. We define R γ,j (w, v) to be the state of pebble j in which it exits v in the j-run of A on w yielded from γ. 22

23 A.1 Stage 2 of the Proof of Lemma 8 Recall that A = Q, q 0, µ, F is a strong k-pa, β 1 = Q, and for i 2, ) Q β i+1 = Q ((β 2 i )! Let m = β k+1 (fixed until the end of this appendix) and let G G k,m. Let i < k and let G i G i,m be an infix of G: for some w 1, w 2 D. Let G = w 1 G i w 2 (3) G i = G i 1 }{{} d n(i 1) 1d n(i 1) }{{} c c n(i 1) n (i 1)+1 v n(i 1) w n(i 1) }{{} v n(i 1) +1 d n(i 1) d n(i 1) +1 }{{} w n(i 1) +1 c n(i 1) +1c n(i 1) +2 }{{} v n(i 1) +2 where G i 1, G i 1 G i 1,m. We call the string v n(i 1) +1d n(i 1) d n(i 1) +1w n(i 1) +1 the middle section of the graph G i. For example, if G 1 = c 0 c 1 }{{} d 0 d 1 }{{} G 1,m v 1 w 1 d n(i 1) +1d n(i 1) +2 }{{} G i 1 w n(i 1) +1 then middle section of G 1 is v 1 d 0 d 1 w 1. Let γ be an (i + 1)-configuration on G outside G i. We are going to prove that in the i-run of A yielded from γ pebble i enters the symbols d n(i 1) d n(i 1) +1 and c n(i 1) +1c n(i 1) +2 in the same state. For what follows it will be convenient to rename the symbols in the middle section and c n(i 1) +1c n(i 1) +2 }{{} v n(i 1) +1 d n(i 1) d n(i 1) +1 }{{} c n(i 1) +1c n(i 1) +2 w n(i 1) +1 23

24 of G i as x 1 y 1 x m 1 y m 1 }{{} v n(i 1) +1 x m y m }{{} d n(i 1) d n(i 1) +1 x m+1 y m+1 x 2m 1 y 2m 1 }{{} w n(i 1) +1 x 2m y 2m }{{} c n(i 1) +1c n(i 1) +2 (4) where for j = 1,..., m 1, x j = e n(i 1) +1,j and y j = e n(i 1) +2,j; x m = d n(i 1) and y m = d n(i 1) +1; for j = m + 1,..., 2m 1, x j = f n(i 1) +1,j and y j = f n(i 1) +2,j; x 2m = c n(i 1) +1 and y 2m = c n(i 1) +2; see the notation in the beginning of Section 3.1. Recall that the symbols {x 1, y 1,..., x 2m 1, y 2m 1 } = [ v n(i 1) +1 d n(i 1) d n(i 1) +1 w n(i 1) +1 ] are pairwise different and neither of them occurs in either G i 1 or G i 1. The following proposition implies that pebble i enters the patterns d n(i 1) d n(i 1) +1 and c n(i 1) +1c n(i 1) +2 in the same state, as explained below. Proposition 21 Let G G k,m, G i G i,m, and let γ be an (i+1)-configuration outside G i. Let (q j 1, q j 2), j = 1,..., 2m, be a pair of states, where q j 1 is the state in which pebble i enters x j in the i-run of A yielded from γ, and q j 2 is the state in which pebble i enters y j in the i-run of A yielded from γ. Then, there exist two positive integers p, m 0 β i such that p β i+1 and for all j, j such that m 0 < j, j < 2m the following holds. If j j (mod p), then (q1, j q2) j = (q j 1, q j 2 ). This proposition states is that after every 2p moves to the right pebble i repeats its states. That is, (q1, j q2) j = (q j+p 1, q j+p 2 ). The repetition starts from an index m 0 < β i and the repetition period is p that divides β i+1 = m. Since the number of symbols between x m and x 2m is a factor of m, it follows that pebble i enters x m and x 2m in the same state. 24

25 We precede the proof of Proposition 21 with a brief note on the number of different repetition of states. Since γ is outside G i, the pebble assignment θ becomes irrelevant after pebbles i, (i 1),..., 1 enter G i. That is, the values p and m 0 depend only on the states q i, q i 1,..., q 1 in which pebbles i, i 1,..., 1 enter G i. Since the number of states of A is finite, there are only finitely many possible repetition of states. Each repetition of states has its own values of p and m 0, and we call the pair (p, m 0 ) a period of states. We will use the letter φ, possibly indexed, to denote a period of states. The components of φ = (p, m 0 ) are denoted by p(φ) = p and m 0 (φ) = m 0. The value p(φ) is called the interval of φ. Proof. (of Proposition 21) The proof is by induction on i. The base case: i = 1. Then, let x m+1 y m+1 x 2m 1 y 2m 1 c 1 c 2 u 2. G = u 1 G 1 u 2 = u 1 c 0 c 1 x 1 y 1 x m 1 y m 1 x m y }{{ m } d 0 d 1 for some u 1, u 2 D and G 1 G 1,m. Let γ be a 2-configuration outside G 1. By definition, none of the symbols seen by pebbles 2,..., k occur in G 1. Therefore, in the 1-run of A, starting at x 1 pebble 1 only moves (right) by the means of transitions of the form (1, q,, ) (q, right). Since A is deterministic and the number of states is finite, starting from some moment m 0 Q, the same sequence of transitions repeats itself. The length of such sequence p is Q, and, since β 2 = Q 2 (( Q )!) Q, it follows that have p β 2. That is, the desired period is (p, m 0 ). The repetition of states which pebble 1 eventually enters depends on the state of pebble 1 in which it enters G 1. Therefore, there are at most Q possible periods of states for pebble 1. The induction hypothesis: Let G i 1 G i 1,m be an infix of G where G = u 1 G i 1 u 2 for some u 1, u 2 D. Then, for every i-configuration γ outside G i 1, in the (i 1)-run of A on G yielded from γ pebble (i 1) enters into a period of 25

26 states when scanning the middle section of G i 1. Furthermore, there are at most Q such periods of states. The induction step: Let G i G i,m be an infix of G where for some u 1, u 2 D. Let G = u 1 G i u 2 (5) G i = G i 1 d n(i 1) 1d n(i 1) c n(i 1) c n(i 1) +1 d n(i 1) d n(i 1) +1 c n(i 1) +1c n(i 1) +2 d n(i 1) +1d n(i 1) +2 G i 1 where G i 1, G i 1 G i 1,m, and let x 1 y 1 x 2m 1 y 2m 1 be the middle section of G i, x 1y 1 x 2m 1y 2m 1 be the middle section of G i 1, and x 1y 1 x 2m 1y 2m 1 be the middle section of G i 1. By the induction hypothesis, when all pebbles i, i + 1,..., k are outside G i 1 and G i 1, pebble (i 1) enters into a period of states when scanning the middle sections x 1y 1 x 2my 2m and x 1y 1 x 2my 2m of G i 1 and G i 1, respectively. Let φ 1,..., φ l be all periods of states of pebble (i 1) in both x 1y 1 x 2my 2m and x 1y 1 x 2my 2m. We contend that there is a pair of positive integers (h, h ) such that 1. 1 h < h m, 2. h h mod p(φ 1 ),..., p(φ l ), and 3. (q h 1, q h 2 ) = (q h 1, q h 2 ). Indeed, let M be the least common multiple of p(φ 1 ),..., p(φ l ). Consider the sequence of pairs of states (q1 1+iM, q2 1+iM ), i = 0,..., Q 2. Since l Q ( Q, and 1 p(φ 1 ),..., p(φ l ) β i 1, M (β i 1 )!) implying Q 2 M β i. Therefore, for some i, j = 0,..., Q 2, i < j, (q 1+iM 1, q 1+iM 2 ) = (q 1+jM 1, q 1+jM 2 ), which proves our contention (with h = 1 + im and h = 1 + jm. Let (h, h ) be the pair satisfying conditions 1 3 above that is the smallest with respect to the lexicographic order of {1,..., 2m} {1,..., 2m}. 26

27 We claim that the desired period of states is (p, m 0 ), where p = h h and m 0 = h as stated below. Claim For all j, j m, if j j mod p, then (q1, j q2) j = (q j 1, q j 2 ). Obviously, it suffices to prove the claim for the case of j = j + p, which we shall do by induction on j. The base case j = h and j = h is immediate, because (q1 h, q2 h ) = (q1 h, q2 h ) by our choice of h and h, and for the induction step we assume and shall prove (q j 1 1, q j 1 2 ) = (q j 1+p 1, q j 1+p 2 ) (q j 1, q j 2) = (q j+p 1, q j+p 2 ). For the proof of q j 1 = q j+p 1 we shall need the following notation. In the (i 1)-run of A when pebble i is above y j 1, let (s 1, s 2, s 3, s 4, s 5 ) be the 5-tuple of states of pebble (i 1) that is defined as follows. s 1 is the state of pebble (i 1) in which it is placed on the input string (at the left-end marker symbol ). s 2 is the state of pebble (i 1) in which it arrives at the symbol x j 1. s 3 is the state of pebble (i 1) in which it arrives at the symbol x j 1. s 4 is the state of pebble (i 1) in which it exits symbol x 2m 1, i.e., in which it arrives at x 2m s 5 is the state of pebble (i 1) in which it arrives at the right-end marker. Similarly, in the (i 1)-run of A when pebble i is above y j+p 1, let (s 1, s 2, s 3, s 4, s 5) be the 5-tuple of states of pebble i 1 such that s 1 is the state of pebble (i 1) in which it is placed on the input string (at the left-end marker symbol ). s 2 is the state of pebble (i 1) in which it arrives at the symbol x j 1. s 3 is the state of pebble (i 1) in which it arrives at the symbol x j 1. s 4 is the state of pebble (i 1) in which it exits symbol x 2m 1, i.e., in which it arrives at x 2m 27

28 s 5 is the state of pebble (i 1) in which it arrives at the right-end marker. s 1 s 2 s 3 s 4 s 5 x j 1 y j 1 x j 1 x 2m 1 p p s 1 s 2 x j+p 1 y j+p 1 s 3 x j+p 1 s 4 x 2m 1 s 5 Figure 4: Positions of x j 1, y j 1, x j 1 and x j+p 1, y j+p 1, x j+p 1. Note the words x j 1... y j 1... x j 1 and x j+p 1... y j+p 1... x j+p 1 are isomorphic. Recall that p 0 mod p(φ 1 ), p(φ 2 ), where φ 1 is the period of pebble (i 1) on x 1y 1 x 2m 1y 2m 1, and φ 2 is the period of pebble (i 1) on x 1y 1 x 2m 1y 2m 1. For the proof of that q j 1 = q j+p 1 it suffices to show that (s 1, s 2, s 3, s 4, s 5 ) = (s 1, s 2, s 3, s 4, s 5), which is done as follows. Since q j 1 1 = q j+p 1 1, s 1 = s 1 is immediate. By the induction hypothesis, pebble (i 1) enters into a period of states when reading the middle section x 1y 1 x 2m 1y 2m 1 of G i 1. Since s 1 = s 1, pebble (i 1) enters the same period of states, say φ, in both cases: when pebble i is above y j 1 and when it is above y j+p 1. Since p(φ) divides p, s 2 = s 2 follows. The equality s 3 = s 3, will follow, if we show that for each l = 1,..., x j 1 y j 1 x j 1 (= x j+p 1 y j+p 1 x j+p 1 ), pebble (i 1) arrives at position l in x j 1 y j 1 x j 1 in the same state as it arrives at position l in x j+p 1 y j+p 1 x j+p 1. 28

29 By the induction hypothesis of Proposition 21, when pebbles 1,..., (i 2) run on the input word they enter periods of states in their respective middle sections in both G i 1 and G i 1. Note that each position in x j+p 1 y j+p 1 x j+p 1 is a shift to the right by p of the corresponding position in x j 1 y j 1 x j 1. Since p is a factor of all intervals of pebbles 1,..., (i 2), these pebbles enter the infix x j+p 1 y j+p 1 x j+p 1 (i.e., enter x j+p 1) in the same state as they enter the infix x j 1 y j 1 x j 1 (i.e., enter x j 1). In addition, since x j+p 1 y j+p 1 x j+p 1 and x j 1 y j 1 x j 1 are isomorphic, and γ is outside G i (see the statement of Proposition 21), all pebbles 1,..., (i 2) behave the same on x j+p 1 y j+p 1 x j+p 1 and x j 1 y j 1 x j 1. Thus, they all exit both strings in the same state. After this, each of the pebbles 1,..., (i 2) continues its computation and enter a period of states in their respective middle section of G i 1. Again, since p is a factor of all intervals of pebbles 1,..., (i 2), each of the pebbles exits its respective middle sections in G i 1 in the same state. Therefore, in both cases, all pebbles 1,..., (i 2) reach the right-end marker in the same state. Consequently, pebble (i 1) is in the same state in the respective positions in x j 1 y j 1 x j 1 and x j+p 1 y j+p 1 x j+p 1, s 3 = s 3 follows. By the induction hypothesis, pebble (i 1) enters a period of states when scanning the middle section x 1y 1 x 2m 1y 2m 1 of G i 1, and since s 3 = s 3, it enters the same period of states, say φ, in both cases when pebble i is above y j 1 and when it is above y j+p 1. Since p(φ) divides p, we have s 4 = s 4. Finally, s 5 = s 5 follows from s 4 = s 4. The proof of q j 2 = q j+p 2 is similar (and is omitted). This completes the proof of the claim and the whole proposition. A.2 Proof of Lemma 8 In this section we formalize the sketch of the proof of Lemma 8 from Section 3.1. Recall that A is a strong k-pa that accepts R nk. For m = β k+1 29

30 (defined in Section A.1) we are going to show that for all G G k,m, the graphs G and G are A-equivalent. We start with the following remark. Remark 22 Let G G k,m, where G = c 0 c 1 d 0 d 1 c 1 c 2 c nk 2c nk 1 d nk 2d nk 1 c nk 1c nk G = c 0 c 1 d 0 d 1 c 1 c 2 c nk 2c nk 1 d nk 2d nk 1 Then, for any i = 1,..., k, for any infix G i G i,m of G, there exists an infix G i of G such that G = G = u u z } { z } { c 0 c 1 d 0 d 1 c 1 c 2 G i c n c n +1 d n d n +1 c nk 1c nk u u z } { z } { c 0 c 1 d 0 d 1 c 1 c 2 G i d n 1d n c n c n +1 d nk 2d nk 1 and u = u. Also, if i > 1, then [u] [u ] = [u] [u ] =. We observe next that for any G i G i,m the following. a) There exists a prefix G i 1 and w 1, w 2 D such that of G i G i = G i 1 w 1 G i = G i 1 w 2 w 1 = w2, and [G i 1 ] [u ] = [G i 1 ] [u ] =. It is illustrated as follows. For some 0 j n k n i, G i = G i 1 {}}{ c j c j+1 d j d j+1 d j+ni 1 2d j+ni 1 1 c j+ni 1 1c j+ni 1 w 1 {}}{ d j+ni 1 1d j+ni 1 d j+ni 2d j+ni 1 c j+ni 1c j+ni G i = G i 1 {}}{ c j c j+1 d j d j+1 d j+ni 1 2d j+ni 1 1 w 2 {}}{ c j+ni 1 1c j+ni 1 d j+ni 1 1d j+ni 1 d j+ni 2d j+ni 1 30

31 b) There exists a suffix G i 1 and w D such that of G i G i = wg i 1 G i = wg i 1 and [G i 1] [u] = [G i 1] [u] =. It is illustrated as follows. For some 0 j n k n i, G i = w {}}{ c j c j+1 d j d j+1 d j+ni n i 1 1d j+ni n i 1 G i 1 {}}{ c j+ni n i 1 c j+ni n i 1 +1 d j+ni 2d j+ni 1 c j+ni 1c j+ni G i = w {}}{ c j c j+1 d j d j+1 d j+ni n i 1 1d j+ni n i 1 G i 1 {}}{ c j+ni n i 1 c j+ni n i 1+1 d j+ni 2d j+ni 1 The reason why we cut G i and G i into two, as shown in items (a) and (b), is that we can show, by induction on i, that i pebbles cannot distinguish between G i and G i. Lemma 8 follows from the Lemma 23 below. This lemma essentially states that, for each i = 1,..., k, if G i G i,m, then pebble i of A cannot distinguish between G i and G i. Recall that for w D, a prefix v of w, and an (i + 1)-configuration γ, R γ,i (w, v) denotes the state in which pebble i exits v in the i-run of A on w yielded from γ. Lemma 23 Let 1 i k 1. Let G i and G i be infixes of G and G, respectively, as described in Remark 22. Let γ 1 be an (i + 1)-configuration of A on G and γ 2 be an (i + 1)-configuration of A on G such that 1. γ 1 and γ 2 are outside G i and G i, respectively, 2. q(γ 1 ) = q(γ 2 ), 3. θ(γ 1 ) mod G i = θ(γ 2 ) mod G i, and 31