Efficient Construction of Semilinear Representations of Languages Accepted by Unary NFA

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1 Efficient Construction of Semilinear Representations of Languages Accepte by Unary NFA Zeněk Sawa Center for Applie Cybernetics, Department of Computer Science Technical University of Ostrava 17. listopau 15, Ostrava-Poruba, , Czech republic Abstract. Chrobak (1986) prove that a language accepte by a given noneterministic finite automaton with one-letter alphabet, i.e., a unary NFA, with n states can be represente as the union of O(n 2 ) arithmetic progressions, an Martinez (2002) has shown how to compute these progressions in polynomial time. To (2009) has pointe out recently that Chrobak s construction an Martinez s algorithm, which is base on it, contain a subtle error an has shown how they can be correcte. In this paper, anewsimpler anmore efficientalgorithm for thesame problem is presente. The running time of the presente algorithm is O(n 2 (n+m)), where n is the number of states an m the number of transitions of a given unary NFA. 1 Introuction It is well known that Parikh images of regular (an even contet-free) languages are semilinear sets [7, 4]. In unary languages, i.e., languages over a one-letter alphabet, wors can be ientifie with their lengths (i.e., a n can be ientifie with n), so the Parikh image of a unary language is just the set of lengths of wors of the language, an it can be ientifie with the language itself. It can be easily shown that each regular unary language can be represente as the union of a finite number of arithmetic progressions of the form {c+i i N} where c an are constants specifying the offset an the perio of a progression. A unary noneterministic finite automaton (a unary NFA) is an NFA with a one-letter alphabet. Given a unary NFA A, a set of arithmetic progressions representing the language accepte by A can be compute by eterminization of A; however, this straightforwar approach can prouce an eponential number of progressions. Chrobak [1] has shown that this eponential blowup is avoiable an that a language accepte by a unary NFA with n states can be represente as the union of O(n 2 ) progressions of the form {c+i i N} where c < p(n) for some p(n) O(n 2 ) an 0 n. The computational compleity of the Supporte by the Czech Ministry of Eucation, Grant No. 1M0567.

2 2 construction of these progressions was not analyze in [1], but it can be easily seen that a naive straightforwar implementation woul require eponential time. Later, Martinez [5, 6] has shown how the construction escribe in [1] can be realize in polynomial time. The eact compleity of Martinez s algorithm is O(kn 4 ) where n is the number of states of the automaton an k the number of strongly connecte components of its graph. The result was recently use for eample in [3,2] to obtain more efficient algorithms for some problems in automata theory an the verification of one-counter processes. In [8], To pointe out that Chrobak s construction an Martinez s algorithm (whose correctness relies on correctness of Chrobak s construction) contain a subtle error, an he has shown moifications that correct this error. In this paper, we give a simpler an more efficient algorithm for the same problem, i.e., for computing of a corresponing set of arithmetic progressions for a given unary NFA. The time compleity of the algorithm is O(n 2 (n+m)) an its space compleity O(n 2 ), where n is the number of states an m number of transitions of the unary NFA. Section 2 gives basic efinitions an formulates the main result, Section 3 escribes the algorithm an proofs of its correctness, an Section 4 contains a escription of an efficient implementation of the algorithm an an analysis of its compleity. 2 Definitions an Main Result The set of natural numbers {0,1,2,...} is enote by N. For i,j N such that i j, [i,j] enotes the set {i,i+1,...,j}, an [i,j) enotes the (possibly empty) set {i,i+1,...,j 1}. Given c, N, an arithmetic progression is the set {c+ i i N}, enote c+n, where c is calle the offset an the perio of the progression. The following efinitions are stanar (see e.g. [4]), ecept that they are specialize to the case where a one-letter alphabet is use. In such an alphabet, wors can be ientifie with their lengths. A unary noneterministic finite automaton (a unary NFA) is a tuple A = (Q,δ,I,F) where Q is a finite set of states, δ Q Q is a transition relation, an I,F Q are sets of initial an final states respectively. A path of length k from q to q, where q,q Q, is a sequence of states q 0,q 1,...,q k from Q where q = q 0, q k = q k, an (q i 1,q i ) δ for each i [1,k]. We use q q to enote that there eists a path of length k from q to q. A wor N is accepte by A if q 0 q f for some q 0 I an q f F. The language L(A) accepte by a unary NFA A is the set of all wors accepte by A. We consier the following problem: Problem: UNFA-Arith-Progressions Input: A unary NFA A. Output: A set {(c 1, 1 ), (c 2, 2 ),..., (c k, k )} of pairs of natural numbers such that L(A) = k i=1 (c i + i N).

3 3 The main result presente in this paper is: Theorem 1. There is an algorithm solving UNFA-Arith-Progressions with time compleity O(n 2 (n+m)) an space compleity O(n 2 ) where n is the number of states an m the number of transitions of a given unary NFA. The algorithm constructs O(n 2 ) pairs of numbers an each constructe pair (c i, i ) satisfies c i < 2n 2 +n an i n. 3 L(A) as Union of Arithmetic Progressions In this section, we escribe the algorithm for UNFA-Arith-Progressions an prove its correctness. In the rest of the section, we assume a fie unary NFA A = (Q,δ,I,F) with Q = n. 3.1 The algorithm The algorithm works as follows. It computes the resulting set R of pairs of numbers that represent arithmetic progressions as the union of the following sets R 1 an R 2 where: R 1 is the set of all of pairs (,0) where L(A) an [0,2n 2 +n), an R 2 is the set of all of pairs (c,) where [1,n], c [2n 2,2n 2 ), an n where for some q 0 I, q Q, an q f F we have q 0 q, q q, an q c n q f (note that c n). To compute R 1, it is sufficient to test for each [0,2n 2 +n)if L(A), an to compute R 2, it is sufficient to test for each of O(n 2 ) pairs (c,), where [1,n] an c [2n 2,2n 2 ), if the require conitions are satisfie. All these tests can be easily one in polynomial time an we can also see that R O(n 2 ). An efficient implementation of the algorithm, which avois some recomputations by precomputing certain sets of states, is escribe in Section 4 together with a more etaile analysis of its compleity. The correctness of the algorithm is ensure by the following crucial lemma an its corollary; the proof of the lemma is postpone to the net subsection. Lemma 2. Let 2n 2 +n. If L(A) then c+n for some (c,) R 2. Corollary 3. Let N. Then L(A) iff c+n for some (c,) R. Proof. ( ) Assume L(A). Either < 2n 2 + n an then (,0) R 1 an (+0N) = {}, or 2n 2 +n an then c+n for some (c,) R 2 by Lemma 2. ( ) It can be easily checke that c + N L(A) for each (c,) R. For (c,) R 1 this follows from the efinition, an for (c,) R 2 from the observation that if q 0 q, q q, an q c n q f for some q 0 I, q Q, an q f F n (where c n), then A accepts each wor from c+n.

4 4 3.2 Proof of Lemma 2 The rest of this section is evote to the proof of Lemma 2, which is one by the following sequence of simple propositions. The basic iea of the proof is that there eists a polynomial p(n) O(n 2 ) such that if q 1 q 2 for some q 1,q 2 Q an p(n) then there is a path α of length from q 1 to q 2 of the following form: α goes from q 1 to some state q by c 1 steps, then goes through a cycle of length [1,n] several times, an then goes from q to q 2 by c 2 steps. Obviously = c 1 +k +c 2 for some k N, an it will be also ensure that c 1 +c 2 < p(n). Every path α of length from q 1 to q 2 can be transforme into the escribe form by the following construction: we can ecompose α into elementary cycles, i.e., cycles where no state is repeate, an a simple path, i.e., a path where no state is repeate, from q 1 to q 2. We can o this by repeately removing elementary cycles from α. Using this ecomposition, we can construct a path of the require form by selecting one elementary cycle of some length, say, an by repeately cutting-out some subsets of the remaining elementary cycles, such that the sums of lengths of cycles in these subsets are multiples of, which means that they can be replace with iterations of the selecte cycle of length. However, when we cut-out cycles, we must be careful, because by cuttingout some cycles, some other cycles can become unreachable. An error of this kin was mae by Chrobak in [1] as pointe out by To in [8]. To ensure that none of the cycles becomes unreachable, we ivie elementary cycles into two categories removable an unremovable. Only removable cycles will be cut-out, an it will be ensure that it is safe to remove any subset of removable cycles. We say a sequence β 0,β 1,...,β r, where β 0 is a simple path from q 1 to q 2 an where β 1,β 2,...,β r are elementary cycles, is goo if for each i [1,r] there is some j [0,i), such that β i an β j share at least one state q. Note that from such goo sequence we can construct a path from q 1 to q 2, whose length is the sum of lengths of all β i, by starting with β 0 an repeately pasting-in β 1,β 2,...,β r (in this orer). Each cycle β i can be paste-in since it shares some state q with some β j where j < i (β i can be paste in by splitting it in q). Note that a ecomposition β 0,β 1,...,β r of an original path α, where β 0 is a simple path from q 1 to q 2 an where β 1,β 2,...,β r are elementary cycles in the reverse orer, in which they were remove from α (i.e., β r was remove first an β 1 last), is goo. We say a cycle β i, where i [1,r], is removable if for each state q of β i there is some j [0,i) such that β j contains q. Cycle β i that is not removable is unremovable. It can be easily checke that a sequence obtaine from β 0,β 1,...,β r by removing some arbitrary subset of removable cycles is also goo. The following proposition is the main tool that allows us to fin a subset of removable cycles such that the sum of lengths of cycles in this subset is a multiple of.

5 5 Proposition 4. Let 1. Every sequence 1, 2,..., r of natural numbers, where r, contains a non-empty subsequence i, i+1,..., j (where 1 i j r) such that ( i + i j ) 0 (mo ). Proof. Consierasequences 0,s 1,...,s r wheres i = i fori [0,r]. Thereareatmostifferentvaluesofs i moulo.sincer,bythepigeonhole principle we have s i s j (mo ) for some i,j such that 0 i < j r. The nonempty sequence i+1, i+2,..., j has the require property ( i+1 + i j ) 0 (mo ), since s j s i 0 (mo ). y Proposition 5. Let q 1,q 2 Q, N, an [1,n]. If q 1 q 2 then q 1 q 2 for some y [0,2n 2 n) such that y an y (mo ). Proof. Let us assume q 1 q 2 an let y N be the smallest number such that y y (mo ) an q 1 q 2 (such y eists, since y = satisfies these properties). Let β 0,β 1,...,β r be a goo ecomposition of a path of length y from q 1 to q 2 (β 0 is a simple path from q 1 to q 2 an β i for i [1,r] are elementary cycles). Let us assume that there are at least removable cycles in this ecomposition. Then, by Proposition 4, there is a nonempty subset of these removable cycles such that the sum of lengths of the cycles in this subset is a multiple of. By removing the cycles in this subset we obtain a goo sequence, from which we can construct a path from q 1 to q 2 of length y < y where y y (mo ). y So q 1 q 2 an y (mo ), which is a contraiction, since we have assume that y is the smallest such number. This implies that in the sequence β 0,β 1,...,β r there are at most 1 removable cycles. A cycle β i is unremovable iff it contains a state q that oes not belong to any β j with j < i, which implies that there are at most n 1 unremovable cycles (note that there is at least one state in β 0 ). The length of β 0 is at most n 1 an a length of each elementary cycle is at most n, which implies since n. y (n 1)+(n 1+ 1) n < 2n 2 n, Corollary 6. Let q 1 q 2 for some q 1,q 2 Q an N. If n then there eist q Q, c 1 [0,n), [1,n], an c 2 [0,2n 2 c n) such that q 1 1 q, q q, q c2 q 2, an (c 1 +c 2 )+N. Proof. By the pigeonhole principle, some q Q must be visite twice in the first n steps of a path from q 1 to q 2 of length n, an so for some c 1 [0,n), c 1 [1,n], an c 2 N we have q 1 q, q q, q c 2 q 2, an = c 1 ++c 2. By Proposition 5, there is some c 2 [0,2n 2 n) satisfying c 2 c 2, q c2 q 2, an c 2 c 2 (mo ). So c 2 = c 2 +k for some k N, an = c 1 ++c 2 = (c 1 +c 2 )+(k +1), which means that (c 1 +c 2 )+N. Proposition 7. Let q 1 q 2 for some q 1,q 2 Q an N. If 2n 2 +n then there eist q Q, c [0,2n 2 n n), an [1,n], such that q 1 q, c q q, q q 2, an (n+c)+n.

6 6 Proof. Assume q 1 q 2 where 2n 2 + n. By Corollary 6, there are some q Q, c 1 [0,n), [1,n], c 2 [0,2n 2 c n), an k N such that q 1 1 q, q, q c 2 q2, an = (c 1 +c 2 )+k. Let α be a path of length from q 1 to q 2 that goes from q 1 to q by c 1 steps, then goes k times through a cycle β of length, an then goes from q to q 2 by c 2 steps, an let q be the state reache after the first n steps of α. Note that since (c 1 +c 2 )+k = 2n 2 +n an c 1 +c 2 < 2n 2 (because c 1 < n an c 2 < 2n 2 n), we have k n. Togetherwith n c 1 < n this ensures that the state q is on the cycle β, which implies q 1 q, q q, an q n q 2. By Proposition 5, there is some c [0,2n 2 n) such that c c n, q q 2, an c n (mo ). This means that n+c (mo ), an since c n implies n+c, we have (n+c)+n. q Now we can prove Lemma 2. Proof (of Lemma 2). Assume that 2n 2 + n an L(A), so there are some q 0 I an q f F such that q 0 q f. By Lemma 7, there eist q Q, c [0,2n 2 n c n), an [1,n], such that q 0 q q q, q q f, an (n+c )+N. This means that for each c (n+c )+N, such that c, we have q c n q f an c+n. In particular, there is one such c in the interval [2n 2,2n 2 ), since n+c [n,2n 2 ). 4 Efficient Implementation To avoi recomputations, the algorithm precomputes some sets. For i N we i i efine S i = {q Q q 0 I : q 0 q} an T i = {q Q q f F : q q f }, an for q Q we efine Perios(q) = { [1,n] q q}. In particular, the algorithm precomputes the sets S n, T i for i [2n 2 2n,2n 2 n), an Perios(q) n for q S n. To test for a given q if q 0 q for some q 0 I, the algorithm tests if q S n, to test if q c n q f for some q f F, it tests if q T c n, an to test if q q, it tests if Perios(q). All these sets can be implemente as bit arrays, so operations like aing an element to a set, testing if an element is member of a set, an so on, can be performe in a constant time. It is also obvious that for Q Q, the sets Succ(Q ) = {q Q q Q : (q,q) δ} an Pre(Q ) = {q Q q Q : (q,q ) δ} can be compute in time O(n + m) where m is the number of transitions (i.e., δ = m). Using subroutines for computing Pre an Succ, the precomputation of all necessary sets can be one in time O(n 2 (n+m)). For eample, S n can be precompute by computing sequence S 0,S 1,...,S n where S 0 = I, an S i+1 = Succ(S i ) for i 0, T i can be compute by T 0 = F, an T i+1 = Pre(T i ) for i 0, etc. Also all < 2n 2 + n such that L(A) can be foun in time O(n 2 (n+m)) by computing the sequence S 0,S 1,...,S 2n2 +n 1 an checking if S F for [0,2n 2 +n). There are O(n 2 ) pairs (c,) such that [1,n] an c [2n 2,2n 2 ), an for each of them, at most n states are teste. Since the corresponing tests for

7 7 one triple c,,q can be one in a constant time as escribe above,all triples can be teste in time O(n 3 ). We see that the overall running time of the algorithm is O(n 2 (n+m)). Duringthe computation,onlythevaluesofs n,t i fori [2n 2 n,2n 2 n), an Perios(q) for q S n nee to be store. Obviously, O(n 2 ) bits are sufficient to store these values. Other values are use only temporarily, can be iscare after their use, an o not take more than O(n 2 ) bits, so the overall space compleity of the algorithm is O(n 2 ). References 1. Chrobak, M.: Finite automata an unary languages. Theoretical Computer Science 47(2), (1986) 2. Göller, S., Mayr, R., To, A.W.: On the computational compleity of verifying one-counter processes. In: LICS 09. pp IEEE Computer Society (2009), 3. Gruber, H., Holzer, M.: Computational compleity of NFA minimization for finite an unary languages. In: LATA 08. Lecture Notes in Computer Science, vol. 5196, pp Springer (2008) 4. Kozen, D.C.: Automata an Computability. Springer-Verlag (1997) 5. Martinez, A.: Efficient computation of regular epressions from unary nfas. In: Descriptional Compleity of Formal Systems (DFCS) (2002) 6. Martinez, A.: Topics in Formal Languages: String Enumeration, Unary NFAs an State Compleity. Master s thesis, University of Waterloo (2002) 7. Parikh, R.J.: On contet-free languages. J. ACM 13(4), (1966) 8. To, A.W.: Unary finite automata vs. arithmetic progressions. Information Processing Letterss 109(17), (2009),