Residual Finite States Automata

Transcription

1 Residual Finite States Automata François Denis LIF, CMI, Université de Provence, Marseille Aurélien Lemay, Alain Terlutte Ý GRAPPA-LIFL, Université de Lille I January 7, 2002 Abstract We define a new variety of Non Deterministic Automata : a Residual Finite State Automata is a NFA all the states of which define residual languages of the language it recognizes. We prove that every regular language is recognized by a unique (canonical) RFSA which has a minimal number of states and a maximal number of transitions. Canonical RFSA are based on the notion of prime residual languages, i.e. that are not the union of other residual languages. We provide an algorithmic construction of the canonical RFSA from a given NFA. We study the size of canonical RFSA and the complexity of our constructions. 1 Introduction Regular languages are among the most studied objects in formal language theory. Deterministic finite automata (DFA) and non deterministic finite automata (NFA) are two basic types of representation of regular languages. DFA have many good properties: most classical constructions are polynomial and there exists a unique minimal element for a given regular language. Moreover, the Myhill-Nerode theorem shows that the states of a DFA correspond to natural components of the fdenis@cmi.univ-mrs.fr Ý lemay, terlutte@lifl.fr 1

2 language it recognizes: its residual languages or Brzozowski derivatives (here, we shall use the first name). NFA are a generalization of DFA which lose most properties of DFA but gain concision. For example, the minimal DFA which recognizes the language ¼ Ò has ¾ Ò ½ states while a minimal equivalent NFA has only Ò ¾ states. However, there may exist several non isomorphic equivalent minimal NFA and states of NFA may correspond to no natural component of the language they recognize. In this paper, we consider NFA all the states of which define residual languages of the language it recognizes. We call Residual Finite State Automata (RFSA) such automata. RFSA have been introduced in [DLT01c]. Clearly, all DFA are RFSA but the converse is false. We show that we can naturally associate with every regular language Ä a RFSA which has a minimal number of states and which we call the canonical RFSA of Ä: each of its states is associated with a prime residual language, i.e. a residual language which is not the union of other residual languages. We provide an algorithmic construction of the canonical RFSA equivalent to a given NFA which stems from the classical subset construction used to build the minimal DFA. We give some results on the size of RFSA: for example, there are canonical RFSA exponentially larger (resp. smaller) than equivalent minimal NFA (resp. DFA). Then, we study RFSA over an unary alphabet and we show that the gap between the sizes of minimal DFA and canonical RFSA is quadratic in the worst case: remind that the gap between the sizes of minimal DFA and NFA can be superpolynomial. Finally, we give some complexity results which show that natural constructions and decision problems are PSPACE-complete. In Section 2, we recall classical definitions and notations about regular languages and automata. We define RFSA in Section 3 and we study their properties in Section 4. In particular, we introduce the notion of canonical RFSA. The construction of the canonical RFSA using the subset method is given in Section 5. In Section 6, we study some particular (and pathological) RFSA. RFSA over unary alphabet are studied in Section 7. Section 8 is devoted to the study of the complexity of decision and construction problems on RFSA. We conclude in Section 9. 2 Preliminaries In this section, we recall some definitions on finite states automata. For more information, we invite the reader to consult [HU79, Yu97]. 2

3 2.1 Automata and languages Let be a finite alphabet, and let be the set of words on. We note the empty word and Ù the length of a word Ù in. A language is a subset of. A non deterministic finite automaton (NFA) is a quintuple É É ¼ Æ where É is a finite set of states, É ¼ É is the set of initial states, É is the set of final states, Æ is the transition function of the automaton defined from a subset of É to ¾ É. We also denote by Æ the extended transition function defined from a subset of ¾ É to ¾ É by Æ Õ µ Õ, Æ Õ Üµ Æ Õ Üµ, Æ É ¼ Ùµ Æ Õ ÙµÕ ¾ É ¼, Æ Õ Ùܵ Æ Æ Õ Ùµ ܵ where É ¼ É Ü ¾ Õ ¾ É and Ù ¾. A NFA is deterministic (DFA) if É ¼ contains only one element Õ ¼ and if Õ ¾ É, Ü ¾, Ö Æ Õ Üµµ ½. A NFA is trimmed if and only if Õ ¾ É, Û ½ ¾, Õ ¾ Æ É ¼ Û ½ µ and Û ¾ ¾, Æ Õ Û ¾ µ. A state Õ is reachable by the word Ù if Õ ¾ Æ É ¼ Ùµ. A word Ù ¾ is recognized by a NFA if Æ É ¼ Ùµ and the language recognized by is Ä ÙÆ É ¼ Ùµ. We denote by Ê µ the class of recognizable languages. It can be proved that every recognizable language can be recognized by a DFA. There exists a unique minimal DFA that recognizes a given recognizable language (minimal with regard to the number of states and unique up to an isomorphism). Finally, the Kleene theorem [Kle56] proves that the class of regular languages Ê µ is identical to Ê µ. The reversal Ù Ê of a word Ù is defined inductively by Ê and ÜÚµ Ê Ú Ê Ü for Ü ¾ and Ú ¾ The reversal Ä Ê of a language Ä is defined by Ä Ê Ù Ê Ù ¾ Ä. The reversal Ê of a NFA É É ¼ Æ is defined by Ê É É ¼ Æ Ê with Õ ¾ Æ Ê Õ ¼ ܵ if and only if Õ ¼ ¾ Æ Õ Üµ. It is clear that Ä µ Ê Ä Ê. Let Ä be a regular language. Let É É ¼ Æ be a NFA that recognizes Ä and let É ¼ É. We note Ä É ¼ the language defined by Ä É ¼ ÚÆ É ¼ Úµ. So, Ä Ä É¼. When É ¼ contains exactly one state Õ, we simply denote Ä É ¼ by Ä Õ. 3

4 2.2 Residual Languages Let Ä be a language over and let Ù ¾. The residual language of Ä with regard to Ù is defined by Ù ½ Ä Ú ¾ ÙÚ ¾ Ä. The Myhill-Nerode theorem [Myh57, Ner58] proves that the set of distinct residual languages of a language Ä is finite if and only if Ä is regular. Automata and residual languages are linked by the following properties. Let É É ¼ Æ be a NFA. For any state Õ ¾ É and any word Ù ¾, if Õ ¾ Æ É ¼ Ùµ then Ä Õ Ù ½ Ä Let É Õ ¼ Æ be a trimmed DFA. For every non empty residual language Ù ½ Ä, there exists a state Õ ¾ É such that Ä Õ Ù ½ Ä, For every state Õ ¾ É, there exists a residual language Ù ½ Ä such that Ù ½ Ä Ä Õ. Furthermore, if is the minimal DFA, the correspondence between states of and non empty residual languages of Ä is bijective. 3 Definition of Residual Finite States Automaton Definition 1. A Residual Finite State Automaton (RFSA) is a NFA É É ¼ Æ such that, for each state Õ ¾ É, Ä Õ is a residual language of Ä. More formally, Õ ¾ É, Ù ¾ such that Ä Õ Ù ½ Ä Trimmed DFA are RFSA. So, every regular language is recognized by a RFSA. Example 1. Let Ä ¼ where ¼ ½. This language is recognized by the following automata ½, ¾ and (Fig. 1, 2, 3): ½ is a NFA which is neither a DFA, nor a RFSA. Languages associated with states are: Ä Õ½ ¼, Ä Õ¾, Ä Õ. As for every Ù in, we have ÙÄ Ä and so, Ä Ù ½ Ä, neither Ä Õ¾ nor Ä Õ are residual languages. ¾ is the minimal DFA that recognizes Ä. ¾ is also a RFSA. We have Ä Õ½ ¼ ½ Ä, Ä Õ¾ ¼ ¼ ½ Ä, Ä Õ ¼ ¼¼ ½ Ä, Ä Õ ¼ ¼½ ½ Ä. 4

5 0,1» 0 Õ 0,1 ¹ Õ ½ Õ ¾ Õ Figure 1: ½ automaton is neither a DFA nor a RFSA Õ ¹ 1 Õ ½» 0 1 Þ Õ ¾ Õ 0 1 Õ Ý 1 Figure 2: ¾ is a minimal DFA. 0 0» 0,1 0» 0 Þ ¹ Õ ½ Õ ¾ Õ Ý 0,1» 0,1 Ý 0 Þ 0,1 Figure 3: is a canonical RFSA. 5

6 is a RFSA. Indeed, we have Ä Õ½ ½ Ä, Ä Õ¾ ¼ ½ Ä, Ä Õ ¼½ ½ Ä. One can notice that is not a DFA. Definition 2. Let É É ¼ Æ be a RFSA, and let Õ be a state of. The word Ù is a characterizing word for Õ if Ä Õ Ù ½ Ä The automaton is consistent if each state Õ is reachable by a characterizing word for Õ. We say that is strongly consistent if each state Õ is reachable by every characterizing word for Õ. Examples of not consistent RFSA and not strongly consistent RFSA are shown in Fig. 4 and Fig. 5. ¹ Õ ¼ ¹ 0 Õ ½ ¹ Õ ¾ Figure 4: A RFSA which recognizes Ä ¼. It is not consistent as the only word which reaches Õ ¼ is but Ä Õ¼ ½ Ä. ¹ Õ ¼ Õ ½ 0 0,1 ¹ Õ ¾ Figure 5: A RFSA which is consistent but not strongly consistent. If, for every state Õ of a NFA, there exists a word Ù Õ that only leads to Õ (that is such that Æ É ¼ Ù Õ µ Õ) then is a RFSA since Ä Õ ÙÕ ½ Ä. However, next example shows that the converse property is false. Example 2. Let Ä and consider the automaton described in Fig. 6. This automaton is a strongly consistent RFSA. But we can observe that there exists no word Ù such that Æ É ¼ Ùµ Õ ½. 6

7 » ¹ Õ ½ Õ ¾ ¹ Õ»» Õ» Figure 6: A RFSA recognizing the language 4 Properties of Residual Finite States Automata 4.1 General Properties Definition 3. Let Ä be a regular language. A residual language Ù ½ Ä is prime if it is not equal to the union of the residual languages it strictly contains: Ù ½ Ä is prime if Ú ½ Ä Ú ½ Ä Ù ½ Ä Ù ½ Ä A residual language is composed if it is not prime. Note that a prime residual language is not empty and that the set of distinct prime residual languages of a regular language is finite. Proposition 1. Let É É ¼ Æ be a RFSA. For each prime residual language Ù ½ Ä, there exists a state Õ ¾ Æ É ¼ Ùµ such that Ä Õ Ù ½ Ä. Proof. As Ù ½ Ä is prime, Æ É ¼ Ùµ is not empty. Let Æ É ¼ Ùµ Õ ½ Õ and let Ú ½ Ú be words such that Ä Õ Ú ½ Ä for every ½. We have Ù ½ Ä ½ Ä Õ ½ Ú ½ Ä As Ù ½ Ä is prime, there exists Ú such that Ù ½ Ä Ú ½ Ä Ä Õ. As a corollary, a RFSA has at least as many states as the number of prime residual languages of Ä. 4.2 Saturation operator We define a saturation operator which may add transitions and initial states to a NFA without modifying the language it recognizes. 7

8 Definition 4. Let É É ¼ Æ a NFA. The saturated of is the automaton É É ¼ Æ where É ¼ Õ ¾ É Ä Õ Ä and Æ Õ Üµ Õ ¼ ¾ É ÜÄ Õ ¼ Ä Õ. We say that an automaton is saturated if. Lemma 1. Let ½ and ¾ be two NFA sharing the same set of states É. If Ä ½ Ä ¾ and if for every state Õ ¾ É, Ä ½Õ Ä ¾Õ (Ä ½Õ and Ä ¾Õ being the languages corresponding to Õ in both automata), then ½ ¾. Proof. The state Õ is an initial state of if and only if ½ Ä ½Õ Ä ½, i.e. if and only if Õ is an initial state of ¾. In the same way, Õ ¼ ¾ Æ ½ Õ Üµ in if and only if ½ ÜÄ ½Õ Ä ½Õ, i.e. if and only ¼ if Õ ¼ ¾ Æ ¾ Õ Üµ in. ¾ We denote by Ä Õ the set: Ä Õ Ù Æ Õ Ùµ. Proposition 2. Let É É ¼ Æ be a NFA and let É É ¼ Æ be the saturated of. For each Õ in É, we have Ä Õ Ä Õ. Proof. Clearly, Ä Õ Ä Õ as the saturated of an automaton is obtained by adding transitions and initial states. To prove the converse inclusion, we prove by induction that for every integer Ò and every state Õ Ä Õ Ò Ä Õ If Ò ¼, the property is true as and have the same final states. Let Ù ÜÚ ¾ Ä Õ Ò with Ò ½ and let Õ ¼ ¾ Æ Õ Üµ such that Ú ¾ Ä Õ¼. Because of the induction hypothesis, Ú ¾ Ä Õ ¼. As Õ ¼ ¾ Æ Õ Üµ, we have ÜÄ Õ ¼ Ä Õ and therefore ÜÚ ¾ Ä Õ. Corollary 1. Let be a NFA and be its saturated. Then and recognize the same language and µ. Proof. We have Ä Ä Õ Õ ¾ É ¼ Ä Õ Õ ¾ É ¼ Ä ÕÕ ¾ É ¼ Ä Due to the previous point and to the Proposition 2, Lemma 1 can be applied on and to prove that µ ; the saturated of a saturated automaton is itself. 8

9 Corollary 2. If is a RFSA then is also a RFSA. Proof. The saturated of a RFSA is a RFSA as the saturation does not change the languages associated with the states. It could have seemed better to define the saturation operator another way: in order to saturate a NFA, add initial states and transitions as long as this operation does not modify the language recognized by the automaton. Unfortunately, this procedure does not lead to a unique NFA in the general case: in the automaton defined in Fig. 4, it is possible to add a transition from Õ ½ to Õ ¼ and from Õ ¼ to Õ ¾ without modifying the language but not simultaneously. However, we have the following lemma: Lemma 2. Let be a consistent RFSA, let Õ Õ ¼ be two states of and let Ü be a letter such that adding the transition Õ Ü Õ ¼ µ to does not modify the language recognized by. Then, ÜÄ Õ ¼ Ä Õ. Proof. Let Ú ¾ Ä Õ ¼ and let Ù such that Ä Õ Ù ½ Ä and Õ is reachable by Ù. As adding the transition Õ Ü Õ ¼ µ to does not modify the language recognized by, the word ÙÜÚ ¾ Ä. Then, ÜÚ ¾ Ù ½ Ä and ÜÚ ¾ Ä Õ. Therefore, ÜÄ Õ ¼ Ä Õ. Corollary 3. In order to compute the saturated of a consistent RFSA, the following procedure can be applied: add as much initial states and transitions as possible as long as the recognized language is not modified. Proof. A state Õ of can become an initial state without changing the recognized language if and only if Ä Õ Ä. The result directly comes from this remark and previous lemma. There exist saturated RFSA which are not consistent (see Fig. 7). Moreover, there exists consistent saturated RFSA which are not strongly consistent (see Fig. 8). However, we get the following result. Proposition 3. If is a saturated RFSA and if Õ is a prime state of (i.e. such that Ä Õ is a prime residual language of Ä ) then for every word Ù such that Ä Õ Ù ½ Ä, Õ is reachable by Ù. Proof. If Ù, Ä Õ Ä and as is saturated, Õ is initial. Suppose now that Ä Õ ÙÜ ½ Ä, where Ü ¾. As Ä Õ is a prime residual language, there exists Õ ¼¼ ¾ Æ É ¼ Ùܵ such that Ä Õ ¼¼ ÙÜ ½ Ä. Let Õ ¼ be such that Õ ¼ ¾ Æ É ¼ Ùµ and Õ ¼¼ ¾ Æ Õ ¼ ܵ. Let Ú ¾ Ä Õ. As Ä Õ Ä Õ ¼¼, ÜÚ ¾ Ä Õ ¼. That is, ÜÄ Õ Ä Õ ¼ and as is saturated, Õ ¾ Æ Õ ¼ ܵ. Therefore, Õ is reachable by ÙÜ in. 9

10 Ø ¹ Õ ½ ¹ Õ ¼ Ü ½ Ø ½ Õ ¾ Ø Ø ¹ Õ ¹ Õ Û ¹ Õ ½ Õ Ü ¾ Õ Õ ¹ Þ Õ Figure 7: A saturated inconsistent RFSA: the state Õ defines the residual language of but it is not reachable by. Ü ½ ½ Õ ½ ¹ ½ Õ ¹ Õ ¼ ¹ Õ ¾ ¹ Õ ¹ Õ ½ Õ Ü ¾ Õ Õ ¹ Õ Õ Figure 8: A saturated consistent RFSA which is not strongly consistent; and define the same residual language but the state Õ is not reachable by. 10

11 4.3 Reduction operator We define a reduction operator which may delete states in a NFA without changing the language it recognizes. Definition 5. Let É É ¼ Æ be a NFA, and let Õ be a state of É. We note Ê Õµ Ë Õ ¼ ¾ É Ò Õ Ä Õ ¼ Ä Õ. We say that Õ is erasable in if Ä Õ Ä Õ ¼ Ò Õ ¼ ¾ Ê Õµ. If Õ is erasable, we define Õµ ¼ É ¼ É ¼ ¼ ¼ Æ ¼ where: É ¼ ÉÒÕ, É ¼ ¼ É ¼ if Õ ¾ É ¼, and É ¼ ¼ É ¼ Ò Õµ Ê Õµ otherwise, ¼ Ì É ¼, for every Õ ¼ ¾ É ¼ and every Ü ¾ Æ ¼ Õ ¼ ܵ Æ Õ ¼ ܵ if Õ ¾ Æ Õ ¼ ܵ Æ Õ ¼ ܵ Ò Õµ Ê Õµ otherwise. If Õ is not erasable, we define Õµ. Note that if is a saturated NFA and if Õ is an erasable state in, then Õµ is obtained from by deleting Õ from all components of. Definition 6. Let be a NFA. If there is no erasable state in A, we say that is reduced. Proposition 4. Let be a NFA, Õ a state of and ¼ Õµ. Let Õ ¼ ¾ ÉÒÕ. We denote by Ä Õ ¼ (resp. Ä ¼ Õ ¼) the language generated from Õ¼ in the automaton (resp. ¼ ). For every state Õ ¼ Õ, Ä Õ ¼ Ä ¼ Õ ¼. As a consequence, Ä Ä ¼. Proof. If Õ is not an erasable state, the proposition is straightforward. Suppose now that Õ is an erasable state. Let É É ¼ Æ and Õµ ¼ É ¼ É ¼ ¼ ¼ Æ ¼. We first prove by induction on Ò that, for every state Õ ¼ Õ and every integer Ò, Ä Õ ¼ Ò Ä ¼ Õ Ò ¼ For Ò ¼, if ¾ Ä Õ ¼ then Õ ¼ ¾ and as Õ ¼ Õ, Õ ¼ ¾ ¼ and ¾ Ä ¼ Õ ¼. Let Ù ÜÙ ¼ ¾ Ä Õ ¼ where Ü ¾ and let Õ ¼ ½ ¾ É such that Õ ¼ ½ ¾ Æ Õ ¼ ܵ and Æ Õ ¼ ½ Ù ¼ µ. 11

12 If Õ ¼ ½ ¾ É ¼, we can apply the induction hypothesis: Ù ¼ ¾ Ä ¼ Õ½ ¼ Æ ¼ Õ ¼ ܵ, we have Ù ¾ Ä ¼ Õ ¼. and as Õ ¼ ½ ¾ Otherwise, Õ ¼ ½ Õ and there exists a state Õ¼¼ ½ Ù ¼ ¾ Ä. Using the induction again, we have Õ ¼¼ ½ Ù¼ ¾ Ä ¼ Õ½ ¼¼ we have Ù ¾ Ä ¼ Õ ¼. Õ such that Ä Õ ¼¼ ½ and as Õ ¼¼ Ä Õ and ½ ¾ Æ ¼ Õ ¼ ܵ, So, we have shown that for every state Õ ¼ Õ, Ä Õ ¼ Ä ¼ Õ ¼. We prove now by induction that for every state Õ ¼ Õ and every integer Ò, Ä ¼ Õ ¼ Ò Ä Õ ¼ Ò For Ò ¼, if ¾ Ä ¼ Õ ¼ then Õ¼ ¾ ¼ and then ¾ Ä Õ ¼. Let Ù ÜÙ ¼ ¾ Ä ¼ Õ ¼ where Ü ¾ and let Õ¼ ½ ¾ É ¼ such that Õ ¼ ½ ¾ Æ ¼ Õ ¼ ܵ and Æ ¼ Õ ¼ ½ Ù ¼ µ ¼. We can apply the induction hypothesis: Ù ¼ ¾ Ä Õ ¼ ½. If Õ ¼ ½ ¾ Æ Õ ¼ ܵ, we directly have Ù ¾ Ä Õ ¼. Otherwise, we have Ä Õ ¼ ½ Ä Õ and Õ ¾ Æ Õ ¼ ܵ. We also have Ù ¾ Ä Õ ¼. So, we have shown that for every state Õ ¼ Õ, Ä ¼ Õ ¼ Ä Õ ¼. We show that and ¼ recognize the same languages by studying the two following cases. If Õ ¾ É ¼, we have Ä Ë Ë Ë Ä Õ¼ Ä ¼ Õ ¼ Õ ¼ ¾É ¼ Õ ¼ ¾É ¼ Õ ¼ ¾É ¼ ¼ Ä ¼ Õ ¼ Ä ¼. If Õ ¾ É ¼, we have Ä Õ ¼ ¾É ¼ Ä Õ¼ Õ ¼ ¾É ¼ Õ ¼ Õ Õ ¼ ¾É ¼ Õ ¼ Õ Õ ¼ ¾É ¼ Õ ¼ Õ Ä ¼ Õ ¼ Ä Õ¼ µ Ä Õ Ä Õ¼ µ Ä ¼ Õ ¼ µ Õ ¼ ¾Ê Õµ Õ ¼ ¾Ê Õµ Ä Õ ¼µ Ä ¼ Õ ¼µ Õ ¼ ¾É ¼ ¼ Ä ¼ 12

13 The reduction operator does not change the language recognized by the automaton. Corollary 4. operator is an internal operator in the class of RFSA. Proof. Straightforward as from Proposition 4, the residual languages are unchanged by. We shall now show that saturation and reduction operators can be swapped. Lemma 3. Let É É ¼ Æ be a NFA and let Õ be a state of É. Then the automaton Õµ is saturated. Proof. Let us denote by Ä ¼ (resp. by Õ Ä ¼ Õ¼) the language associated with a state Õ¼ in Õµ (resp. in ), Æ ¼ (resp. Æ) the transition function of Õµ (resp. in ) and Ä the language recognized by automata, and Õµ. If Ä ¼ Õ ¼ Ä then Ä Õ ¼ Ä and so Õ¼ is initial in and in Õµ. If ÜÄ ¼ Õ ¼ ļ Õ ¼¼ then ÜÄ Õ ¼ Ä Õ ¼¼ and so Õ¼ ¾ Æ Õ ¼¼ ܵ and Õ ¼ ¾ Æ ¼ Õ ¼¼ ܵ. Proposition 5. Let É É ¼ Æ be a NFA recognizing a regular language Ä and Õ a state of É. We have Õµ Õµ Proof. We can observe that Õµ and Õµ have the same set of states. Furthermore, languages associated with every state Õ ¼ are identical in the two automata because of Propositions 2 and 4. Because of Lemma 1, the saturated of these automata are isomorphic, therefore Õµ Õµ. But as Õµ is a saturated automaton, the proposition is proved. 13

14 4.4 Canonical RFSA Definition 7. Let Ä be a regular language. We define the canonical RFSA of Ä the following way: É É ¼ Æ where is the alphabet of Ä, É is the set of prime residual languages of Ä, so É Ù ½ Ä Ù ½ Ä is prime, its initial states are prime residual languages included in Ä, so É ¼ Ù ½ Ä ¾ É Ù ½ Ä Ä, its final states are prime residual languages containing the empty word, so Ù ½ Ä ¾ É ¾ Ù ½ Ä, its transition function is defined by Æ Ù ½ Ä Üµ Ú ½ Ä ¾ É Ú ½ Ä Ùܵ ½ Ä. Example 3. An example of canonical RFSA is shown in Fig. 3. This definition assumes that the canonical RFSA is a RFSA, we shall prove this presumption below. We have showed that the operator transforms a RFSA into a RFSA, and that it can be swapped with the saturation operator. We shall now show that, if is a saturated RFSA, the reduction operator converges and that the resulting automaton is the canonical RFSA of the language recognized by. Proposition 6. Let Ä be a regular language. If É É ¼ Æ is a reduced saturated RFSA recognizing Ä, then is (isomorphic to) the canonical RFSA of Ä. Proof. As is a RFSA, every prime residual language Ù ½ Ä of Ä can be defined as a language Ä Õ associated with some state Õ ¾ É (Proposition 1). As there are no erasable states in, for every state Õ, Ä Õ is a prime residual language and distinct states define distinct languages. As is saturated, prime residual languages contained in Ä correspond to initial states of É ¼. For the same reason, we can verify that the transition function is the same as the one defined in the canonical RFSA. 14

15 Let ¼ Ò be a sequence of NFA such that for every index ½ Ò, there exists a state Õ of ½ such that ½ Õ µ. Proposition 5 and 6 show that if ¼ is a saturated RFSA and if Ò is reduced, then Ò is the canonical RFSA of the language recognized by ¼. Theorem 1. The canonical RFSA of a regular language Ä is a strongly consistent RFSA which recognizes Ä and is minimal regarding to the number of states. Moreover, the canonical RFSA possesses a maximal number of transitions. Proof. It is a RFSA that recognizes Ä since it can be obtained from any RFSA that recognizes Ä using saturation and reduction operators, and since these two operators do not change the language recognized by the automaton nor the fact that the automaton is a RFSA. It possesses a minimal number of states because of Proposition 1 and it is strongly consistent from Proposition 3. It has a maximal number of transitions from Corollary 3. It would have been possible to give an alternative definition of canonical RFSA. Definition 8. Let Ä be a regular language. For any set of distinct residual languages Ê of Ä, define ÑÜ Êµ Ä ¼ ¾ ÊÄ ¼¼ ¾ Ê Ä ¼ Ä ¼¼ µ Ä ¼ Ä ¼¼. Let É É ¼ Æ be the canonical RFSA which recognizes Ä. The simplified canonical RFSA of Ä is the automaton ¼ É É ¼ ¼ Ƽ where É ¼ ¼ ÑÜ É ¼ µ and Æ ¼ Õ Üµ ÑÜ Æ Õ Üµµ. It can be shown that the simplified canonical RFSA of Ä is a RFSA which recognizes Ä. Example 4. The simplified canonical RFSA for ¼ has transitions less than the canonical RFSA and is the minimal RFSA with respect to number of transitions. However, it may happen that several non isomorphic RFSA have as many states but less transitions than the simplified canonical RFSA. Example 5. Consider the language Ä. The minimal RFSA of Ä has 6 states. The equivalent canonical RFSA has 17 transitions. The simplified canonical RFSA has 14 transitions as for example, ½ Ä ½ Ä ½ Ä and ÑÜ ½ Ä ½ Ä µ ½ Ä. There exist three non isomorphic RFSA with 13 transitions as ½ Ä ½ Ä ½ Ä ½ Ä ½ Ä ½ Ä ½ Ä and none with less transitions. 15

16 1 0» 0 Þ» 0,1 ¹ Õ ½ Õ ¾ Õ Ý 0 Þ 1 Figure 9: The simplified canonical RFSA for ¼. 5 Construction of the canonical RFSA using the subset method In the previous section, we provided a way to build the canonical RFSA from a given DFA using saturation and reduction operators. This method requires to check whether a residual language is included in another one and to check whether a residual language is composed or not. These checks can be very expensive, even for simple automata. We present in this section another method which stems from a classical construction of the minimal DFA of a language and which is easier to implement. The subset construction is a classical method used to build a DFA equivalent to a given NFA. Let É É ¼ Æ be a NFA. The method consists in building the set of reachable sets of states of. We note É Ê µ Ô ¾ ¾ É Ù ¾ s. t. Æ É ¼ Ùµ Ô and we define the subset automaton µ É É ¼ Æ with É É Ê µ É ¼ É ¼ Ô ¾ É Ô Æ Ô Üµ Æ Ô Üµ if Æ Ô Üµ and otherwise. The automaton µ is a deterministic trimmed automaton that recognizes the same language as. We remind that Ä Ê (resp. Ê ) denotes the reversal of a language Ä (resp. of an automaton ). The following result provides a method to build the minimal DFA of Ä. Theorem 2. [Brz62] Let Ä be a regular language and an automaton such that Ê is a DFA that recognizes Ä Ê. Then µ is the minimal DFA recognizing Ä. 16

17 We can deduce from this theorem that Ê µ Ê µ is the minimal DFA recognizing the language Ä. We adapt the subset construction technique to deal with inclusions of set of states. We say that Ô ¾ É Ê µ is coverable if there exist states Ô ½ Ô Ð ¾ É Ê µ Ò Ô, such that Ô Ð ½ Ô. We define the automaton µ É É ¼ Æ with É Ô ¾ É Ê µ Ô is not coverable É ¼ Ô ¾ É Ô É ¼ Ô ¾ É Ô Æ Ô Üµ Ô ¼ ¾ É Ô ¼ Æ Ô Üµ Lemma 4. Let be a NFA. µ is a RFSA recognizing Ä whose all states are reachable. Proof. µ can be obtained from the DFA µ by using the reduction operator defined in Section 4.3. Then Corollary 4 can be applied. Theorem 3. Let Ä be a regular language and let be a NFA such that Ê is a RFSA recognizing Ä Ê whose all states are reachable. Then µ is the canonical RFSA recognizing Ä. In order to prove this theorem, we introduce some lemmas. Lemma 5. Let É É ¼ Æ be a NFA such that Ê is a trimmed RFSA and let Ä be the language recognized by. Let Õ ¾ É, let Ä Õ be the language associated with Õ in and Ä Ê Õ be the language associated with Õ in Ê. Let Ú ¾ be such that Ä Ê Õ ÚÊ µ ½ Ä Ê. Let Ô ¾ É Ê µ be a reachable set of states. Then Ú ¾ Ä Ô if and only if Õ ¾ Ô. Proof. Let Ù be such that Ô Æ É ¼ Ùµ; thus Ä Ô Ù ½ Ä. We have Õ ¾ Ô Õ ¾ Æ É ¼ Ùµ Ù Ê ¾ Ä Ê Õ ÚÊ µ ½ Ä Ê Ú Ê Ù Ê ¾ Ä Ê ÙÚ ¾ Ä Ú ¾ Ù ½ Ä Ä Ô 17

18 Lemma 6. Let É É ¼ Æ be a NFA such that Ê is a trimmed RFSA and let Ä be the language recognized by. For every Ô, Ô ¼ ¾ É Ê µ, Ä Ô Ä Ô ¼ if and only if Ô Ô ¼. Proof. If Ô Ô ¼ then Ä Ô Õ¾Ô Ä Õ Õ¾Ô ¼Ä Õ Ä Ô ¼. Conversely, let Õ ¾ Ô. As Ê is a RFSA, there exists a word Ú such that Ä Ê Õ ÚÊ µ ½ Ä Ê. From Lemma 5, we know that Ú is in Ä Ô. As Ä Ô Ä Ô ¼, Ú also belongs to Ä Ô ¼ and, using Lemma 5, we have Õ ¾ Ô ¼. Lemma 7. Let be a NFA such that Ê is a trimmed RFSA and let Ä the language recognized by. For every Ô, Ô ½, Ô ¾ Ô Ò ¾ É Ê µ, Ä Ô ½Ò Ä Ô is equivalent to Ô ½Ò Ô. Proof. It is obvious that Ô ½Ò Ô implies Ä Ô ½Ò Ä Ô. If Ä Ô ½Ò Ä Ô then, Ä Ô Ä Ô for every ½ Ò. Using the previous lemma, we have, Ô Ô for all and so, ½Ò Ô Ô. Suppose that Ä Ô ½Ò Ä Ô and let Õ ¾ Ô. Since Ê is a trimmed RFSA, there exists a word Ú such that Ä Ê Õ ÚÊ µ ½ Ä Ê. From Lemma 5, we have Ú ¾ Ä Ô. As Ä Ô ½Ò Ä Ô, there exists an index such that Ú ¾ Ä Ô. From Lemma 5 again, we have Õ ¾ Ô. So Ô ½Ò Ô. Proof of the theorem 3. Reachable sets of states correspond to residual languages and Lemma 7 shows that composed states correspond to coverable states. So, Ô ¾ É if and only if Ä Ô is a prime residual language of Ä and É can naturally be identified with the set of states of the canonical RFSA. É ¼ Ô ¾ É Ô É ¼. Lemma 6 tells us that Ô É ¼ is equivalent to Ä Ô Ä É¼ Ä. So É ¼ Ô ¾ É Ä Ô Ä corresponds to the set of initial states of the canonical RFSA. Ô ¾ É Ô. We have ¾ Ä Ô if and only if there exists Õ ¾ Ô. So Ô ¾ É ¾ Ä Ô is the set of final states of the canonical RFSA. Æ Ô Üµ Ô ¼ ¾ É Ô ¼ Æ Ô Üµ. As states correspond to residual languages, for each state Ô ¾ É, there exists Ù Ô ¾ such that Ô Æ É ¼ Ù Ô µ and Ä Ô ÙÔ ½ Ä. 18

19 From Lemma 6, Ô ¼ Æ Ô Üµ Æ É ¼ Ù Ô Üµ is equivalent to Ä Ô ¼ Ä Æ É¼ Ù Ôܵ Ù Ô Üµ ½ Ä. So, Æ Ô Üµ Ô ¼ ¾ É Ô ¼ Ù Ô Üµ ½ Ä and Æ can be identified with the transition function of the canonical RFSA. We can deduce from Lemma 4 and Theorem 3 that Ê µ Ê µ is the canonical RFSA of Ä. We can also define a subset construction for the simplified canonical automaton by ¼ µ É É ¼ ¼ Æ ¼ with É Ô ¾ É Ê µ Ô is not coverable É ¼ ¼ Ô ¾ É Ô É ¼ Ò Ô ¼ ¾ É Ø Ô Ô ¼ É ¼ Ô ¾ É Ô Æ Ô Üµ Ô ¼ ¾ É Ô ¼ Æ Ô Üµ Ò Ô ¼¼ ¾ É Ø Ô ¼ Ô ¼¼ Æ Ô Üµ The simplified canonical is obtained by ¼ ¼ Ê µ Ê µ. We can observe that, if we are interested in covering without saturation (if a state is covered, we delete it and we relead its transitions to covering states), we obtain a RFSA which has the same number of states (non-coverable states) and fewer transitions. Nevertheless, we have to remind that there may exist several non isomorphic RFSA minimal both with regard to the number of states and the number of transitions. Note that, as in the deterministic case, this construction may produce cumbersome intermediate automata; indeed, it is possible to find examples for which Ê µ has an exponential number of states with regard to the number of states of or Ê µ Ê µ. We can observe this situation with the reversal of the automaton used in Proposition 8. 6 Results on size of RFSA We classically take the number of states of an automaton as a measure of its size. The size of a canonical RFSA is bounded by the size of the equivalent minimal DFA and by the size of one of its equivalent minimal NFA. We show that both bounds can be reached even if there exists an exponential gap between them. Proposition 7. There exist languages for which the minimal DFA has a size exponentially bigger than the size of the canonical RFSA, and for which the canonical RFSA has the same size as minimal NFA. 19

20 Proof. Consider the languages ¼ Ò languages, where Ò is an integer and ¼ ½. Residual languages of Ä ¼ Ò are languages of the form Ä ËÔ¾È Ô µ where È ¼ Ò. One can observe that there exist ¾ Ò ½ distinct residual languages. Therefore, the minimal DFA recognizing ¼ Ò has ¾ Ò ½ states. There exist only Ò ¾ prime residual languages: Ä, Ä ¼,..., Ä Ò, so the canonical RFSA of Ä has Ò ¾ states. It can be proved that minimal NFA have also Ò ¾ states. Proposition 8. The size of the canonical RFSA of a language Ä can be exponential with regard to the size of a smallest NFA recognizing Ä. Proof. We can verify this proposition on automata Ò É É ¼ Æ defined by ¼ ½, É Õ ¼ Ò ½, É ¼ Õ ¼ Ò¾, Õ ¼. Æ Õ ¼µ Õ ½ for ¼ Ò ½, Æ Õ Ò ½ ¼µ Õ ¼, Æ Õ ¼ ½µ Õ ¼, Æ Õ ½µ Õ ½ (for ½ Ò) and Æ Õ ½ ½µ Õ Ò ½. The automaton is represented in Figure 10. The reversal of automaton Ò is trimmed and deterministic, thus we can apply Theorem 3. The automaton Ò µ is the canonical RFSA. The initial state in the subset construction has Ò¾ elements. Moreover the reachable sets of states are all sets of states with Ò¾ elements. So, none of them is coverable. The canonical RFSA Ò µ has a size exponentially bigger than the initial NFA. Every non empty residual language of a regular language Ä can be described by a word whose length is bounded by the number of states of the minimal DFA. Next proposition shows that this is no longer true if we consider RFSA. 20

21 1 0 ¹ Õ ½ Õ ¼ Õ «1 0 Õ ¾ Figure 10: Automaton. Each pair of states is an uncoverable set of states in the subset construction. Proposition 9. There exist regular languages for which the smallest characterizing word for some residual language is longer than any polynomial in the number of states of the canonical RFSA. Proof. Let È Ô ½ Ô Ò be a set of Ò distincts prime number. Let us define the automaton È É É ¼ Æ by: Ô Ô ¾ È is the alphabet of the language É Õ Ô Ô ¾ È ¼ Ô É ¼ Õ Ô ¼ Ô ¾ È É ¼ and Æ Õ Ô µ ÕÔ ½µÑÓ for ¼ Ô Ô ¾ È Ô Æ Õ Ô Ô ¼µ ÕÔ ÕÔ ½ for ¼ Ô ½ Ô Ô ¼ ¾ È Æ Õ Ô Ô ½ Ô ¼µ ÕÔ¼ ¼ for Ô Ô ¼ ¾ È See Figure 11 for È ¾. Let Æ Ô ½ Ô Ò and let Ù Æ ½ for ½ Ò and ¼ Ô. We can check that Æ É ¼ Ù µ Õ Ô. Therefore, Ä Õ Ô Ù ½ Ä and È is a RFSA. 21

22 ¾ ¾ Þ ¹ Õ ¾ ¼ Ý Õ ¾ ½ ¾ ¾ ¾ ¾» ¾ Þ ¹ Õ ¼ Õ ½ Õ ¾ ¾ Figure 11: Automaton È for È ¾ 22

23 Let ½ Ò, ¼ Ô and ½ Ò, ¼ Ð Ô. If Ô Ô Ð, then Ô ¾ Ä Ô Õ Ò Ä Ô Õ Ð If Ô Ô Ð, then ¾Ô ¾ Ä Õ Ô Ò Ä Ô Õ Ð are different, none of them is included Therefore, all residual languages Ä Ô Õ in another one and È has the same number of states than the canonical RFSA. We can check that for every Ù ¾ Æ, Æ É ¼ Ùµ ½. So, the smallest word Ù Õ such that Ä Õ Ù ½ Õ Ä has a length Ù Õ Æ. Now, let È be some polynomial. We can choose different prime numbers Ô ½ Ô Ò such that Ô ½ Ô Ò È Ô ½ Ô Ò µ È È µ. 7 RFSA over a one-letter alphabet Here, we consider the case where the underlying alphabet possesses only one letter and we compare the state complexity of DFA and RFSA. Such a study has already been done for DFA and NFA in [Chr86] where it has been shown that the minimal DFA equivalent to a given NFA with Ò states could have Ò ÐÓ Ç Ô Òµ states. Here we show that the number of states of the minimal DFA over a one-letter alphabet is at most quadratic relatively to the number of states of the equivalent canonical RFSA. Let Ü, let Ä be a non empty regular language over, let É Õ ¼ Æ be the trimmed minimal DFA which recognizes Ä and let Ò É ½. Let us define for ½ Ò, Õ Æ Õ ¼ Ü µ and Ä Ä Õ. A DFA over a one letter alphabet is always of the following form, for some Ò ½ ¾ ¼ Ò (final states are omitted) : Ü ¹ Õ ¼ Ü ¹ Õ ½ Ü ¹ Ü ¹ Õ Ò½ Ü ¹ ܹ Õ Ò 7.1 Inclusions relations between residual languages of an infinite regular language over a one-letter alphabet We suppose here that Ä is infinite. Hence, for every integer, Æ Õ ¼ Ü µ. Previous definition can be extended: for all integers, let Õ Æ Õ ¼ Ü µ and Ä Ä Õ. Let Ò ½ be the smallest index such that Õ Ò½ Æ Õ Ò Üµ and let Ò Ò ½ ½. Note that ¼ and that for all Ò ½ and every non negative integer, Õ Õ. 23

24 Lemma 8. For all integers, Ä Ä µ Ä Ä Proof. Let Ù ¾ Ä. We have Ü Ù ¾ Ä, i.e. Ü Ù ¾ Ä, then Ü Ù ¾ Ä and Ü Ù ¾ Ä. Therefore Ù ¾ Ä. Lemma 9. For all integers, Moreover, if Ä Ä, ÑÒ µ Ò ½. Ä Ä µ divides µ Proof. Let Ñ be an integer such that Ñ ¼ and Ò ½ Ñ ¼. Let ½ Ò ½ Ñ µ and ½ Ò ½ Ñ µ. We have Ä ½ Ä Ò½ Ñ Ä Ò½ Ä ½ and ½ Ò ½. Applying Lemma 8 with ½ Ò ½ we obtain, Ä Ò½ Ä ½ Ä ¾½ Ò ½ Ä ½ Ò ½ µ Ò ½ Ä Ò½ As is minimal, this implies that Õ ½ Õ Ò½, i.e. that divides ½ Ò ½ and also. ÑÜ µ ÑÒ µ As ÑÜ µ ÑÒ µ, we must have ÑÒ µ Ò ½. Proposition 10. We have the following cases 1. If is a loop, i.e. if Ò ½ ¼ then there are no inclusion relations between residual languages. 2. If there are no inclusion relations between Ä Ò½ ½ and Ä Ò then there are no inclusions between residual languages. 3. If Ä Ò½ ½ Ä Ò then Ä Ä µ. 4. If Ä Ò Ä Ò½ ½ then Ä Ä µ. Proof. The first case is clear from Lemma 9. If there exist some such that Ä Ä, we have ÑÒ µ Ò ½ ½. Let Ò ½ ½ ÑÒ µ, Ä ÑÒ µ Ä Ò½ ½ and Ä ÑÜ µ Ä Ò. Now, using the Lemma 8, the three last points are clear. 24

25 7.2 Prime and composed residual languages Proposition 11. Every non empty residual language of a finite regular language over a one-letter alphabet is prime. Proof. Let Ä be a finite regular language over and let be its trimmed minimal DFA. We have Õ Ò ¾ and Æ Õ Ò Üµ. For any ¼ Ò, Ü Ò ¾ Ä. Let. If Ü Ò ¾ Ä then and Ü Ò ¾ Ä, i.e. Ä Ä. Hence, Ä is prime. Lemma 10. Let Ä be an infinite regular language over and let be its trimmed minimal DFA. If some residual language of Ä is composed then is not a loop, i.e. Ò ½ ¼, and Ä Ò½ ½ Ä Ò. Proof. Suppose that some residual language of Ä is composed. From Proposition 10, Ò ½ ¼ and it must exist an inclusion relation between Ä Ò½ ½ and Ä Ò. Suppose that Ä Ò Ä Ò½ ½. Let Õ be a composed state. From Proposition 10 and Lemma 9, Ò ½. Let Ù ¾ Ä Ò½ ½. We have Ü Ò ½ ½ Ù ¾ Ä. Let such that Ä Ä and Ü Ò ½ ½ Ù ¾ Ä. We have Ù ¾ Ä Ò½ ½. From Proposition 10, we have and from Lemma 9, we have Ñ with Ñ ¼. So, Ù ¾ Ä Ò½ ½ Ä Ò Ñ ½µ Ä Ò. Therefore, Ä Ò½ ½ Ä Ò which is contradictory. Proposition 12. Let Ä be a regular language over a one-letter alphabet and let be its trimmed minimal DFA. If some state Õ is composed then all the states which follow Õ are composed too. Proof. From previous lemmas, Ä is infinite, is not a loop, Ä Ò½ ½ Ä Ò and Ä Ä implies. Let be the first index such that Õ is composed and let Ê Ä Ä. We have Ä Ä ¾ Ê. It is easy to verify that for every Ò, Ä Ä ¾ Ê and Ä Ä for every ¾ Ê. 7.3 Ratio between the size of a minimal DFA and the equivalent canonical RFSA for a one-letter alphabet regular language If all states of a minimal DFA are prime, the canonical RFSA has the same size than the minimal DFA. We consider here the case where Ä Ò½ ½ Ä Ò. Let É È be the set of prime states of É and Ò Ô É È. Due to Lemma 12, Ò Ô implies that Õ is composed. Let Ë É È Õ ¼ É È Æ Ë be the trimmed NFA 25

26 built from Õ ÒÔ µ where is the reduction operator defined in Section 4.3: Ë is a RFSA recognizing Ä that has the same number of states than the canonical RFSA. Let Ò ¼ be the smallest index such that Õ Ò¼ ¾ Æ Ë Õ ÒÔ ½ ܵ. Lemma 11. For every ¼, Æ Ë Õ ¼ Ü µ Æ Ë Õ ¼ Ü ½ µ. Moreover, if Æ Ë Õ ¼ Ü µ Æ Ë Õ ¼ Ü ÒÔ Ò¼ µ for some Ò ¼ then Æ Ë Õ ¼ Ü µ Æ Ë Õ ¼ Ü ÒÔ Ò¼ µ. Proof. If Ò Ô, Æ Ë Õ ¼ Ü µ Õ. Now let Ò Ô. Due to the way Æ Ë is built, Õ ¾ Æ Ë Õ ¼ Ü µ µ Ò ¼ Ò Ô. We have Æ Ë Õ ¼ Ü ½ µ Æ Ë Õ Üµ Õ ¾ Æ Ë Õ ¼ Ü µ. If Ò Ô ½ then Æ Ë Õ Üµ Õ ½ and if Ò Ô ½ then Õ Ò¼ ¾ Æ Ë Õ Üµ. And as Õ Ò¼ ¾ Æ Ë Õ Üµ Ò Ô ½, we have Æ Ë Õ ¼ Ü ½ µ Æ Ë Õ ¼ Ü µ, which proves the first point. For every Ò ¼, we have Õ ¾ Æ Ë Õ Ü ÒÔ Ò¼ µ. So, for every Ò ¼, Æ Ë Õ ¼ Ü µ Æ Ë Õ ¼ Ü ÒÔ Ò¼ µ. This proves the second point. Proposition 13. Let Ä be a regular language over a one-letter alphabet, and let and Ê be respectively the minimal DFA and the canonical RFSA of Ä. Let Ò (resp. Ò Ê ) be the number of states of (resp. Ê ). We have Ò Ò Ê ½µ ¾ ¾. Proof. Let us define the function by µ Æ Ë Õ ¼ Ü µ: has Ò Ê prime states; For every Ò such that ½ Ò Ò Ê, we can show using Lemma 11 that either there are Ò Ê Ò ¼ states Õ such that µ Ò but in this case there are no index such that µ Ò, otherwise there are at most Ò Ê Ò ¼ ½ states Õ such that µ Ò; There is at most one state Õ such that µ Ò Ê Ò ¼. From this, we can calculate that has no more than Ò Ê Ò Ê ½µ Ò Ê ¾µ ½ Ò Ê ½µ ¾ ¾ states.this upper bound is reached in the automaton described in Figure 12. Ü Ê Ü Ê Ü Ê Ü Ü... Ü Ê ¹ Õ ½ Õ ¾ Õ Õ Ò Figure 12: The minimal DFA corresponding to this canonical RFSA has Ò ½µ ¾ ¾ states 26

27 8 Complexity results We have defined notions of RFSA, saturated automata, canonical RFSA ; in this section, we evaluate the complexity of constructions and decision problems linked to them. We shall mainly use two classical results of complexity concerning finite automata (quoted from [HU79]). Proposition 14. Deciding whether two NFA recognize the same language is a È ËÈ ÓÑÔÐØ problem. As an immediate corollary: given two NFA and ¼, deciding whether Ä Ä ¼ is a È ËÈ ÓÑÔÐØ problem. Deciding whether the intersection of two DFA is empty can be done in quadratic time. On the other hand: Proposition 15. Deciding whether the intersection of Ò DFA is empty or not is a È ËÈ ÓÑÔÐØ problem. The first notion that we defined is saturation. Clearly, deciding whether a DFA is saturated is a polynomial problem. For NFA, we have the following result. Proposition 16. Deciding whether a NFA is saturated is a È ËÈ ÓÑÔÐØ problem. Proof. Given an oracle which decides whether the language recognized by a given NFA is included in another one, we can build the saturated of a given NFA within polynomial time. Given an oracle which builds the saturated of a given NFA, we can say whether a given NFA is saturated within polynomial time. It remains to prove that the inclusion problem between two languages represented by NFA polynomially reduces to the problem of deciding whether a NFA is saturated. Let É É ¼ Æ and ¼ É ¼ É ¼ ¼ ¼ Æ ¼ be two NFA. We can suppose that and ¼ are trimmed and that they have a unique initial state which cannot be reached from other states. Let É ¼ Õ ¼, É Õ ¼ Õ ½ Õ Ð, É ¼ ¼ Õ¼ ¼ and É ¼ Õ ¼ ¼ Õ¼ ½ Õ¼ Ð ¼. We complete the alphabet by adding Рм ¾ new letters Ü ½ Ü Ð Ü ¼ ½ ܼ Ð ¼ Þ Ø: let ¼ be the new alphabet. We consider two new states Õ and Õ and let ¼ ÉÉ ¼ Õ Õ Õ¼ ¼ ¼ Õ Æ ¼¼ where Æ ¼¼ contains the transitions of Æ Æ ¼ and the transitions defined below: 27

28 Õ ¾ Æ Õ Ü µ for ½ Ð and Õ ¾ Æ Õ ¼ ܼ µ for ½ м Õ ¾ Æ Õ ¼ ܵ Æ Õ ¼ ¼ ܵ for Ü ¾ Õ ¾ Æ Õ Üµ for Ü ¾ Õ ¾ Æ Õ Ü µ for ½ Ð and Õ ¾ Æ Õ Ü ¼ µ for ½ м Õ ¾ Æ Õ Þµ and Õ ¾ Æ Õ Øµ (see Figure 13). Now, it is easy to verify that is saturated if and only if Ä Ä ¼. Õ ¹ Õ ¼ Õ ¼ ¼ Ê Õ Ü Ü, Ü ¼, Ø Þ Ü ¼ Õ Õ ¼ Figure 13: The automaton is saturated iff Ä Õ¼ Ä Õ ¼ ¼ Next proposition shows that deciding whether a given NFA is a RFSA is also difficult. Proposition 17. Deciding whether a NFA is a RFSA is a È ËÈ ÓÑÔÐØ problem. Proof. First, we prove that the problem of deciding whether the union of Ò regular languages described by DFA is equal to can be polynomially reduced to the problem of deciding whether a NFA is a RFSA. We can consider Ò DFA ½ É ½ Õ ½ É ½ ƽ Ò É Ò Õ Ò É Ò ÆÒ. Let Ü ½ Ü Ò Ý ½ Ý Ò be ¾Ò ¾ new letters. We can build the NFA É É Á É Æ where 28

29 É ½Ò É Õ ½ Õ Ò Õ Õ Õ where Õ ½ Õ Ò Õ Õ Õ are new states, É Á Õ ½ Õ Ò Õ Õ É Ë ½Ò É Õ Æ Ë½Ò Æ µ Ë½Ò Õ Ü Õ µ Õ Ý Õ µ Õ Õ µµ Õ Õ µ Õ Õ µ Õ Õ µ Õ Ü Õ µü ¾ (see Figure 8). Every states, except maybe Õ, defines a residual language of the described language: states Õ correspond to residual languages of words Ý, states Õ of sets É correspond to residual languages of Ü Ù where Æ Õ Ùµ Õ, Õ corresponds to the residual language of and Õ corresponds to the residual language of. State Õ defines a residual language of the language if and only if the union of recognized languages by automaton is equal to. That is, is an RFSA if and only if the union of languages described by automaton is equal to It remains to show that we can decide whether a given NFA is a RFSA within polynomial space. Consider the subset construction defined in Section 5. The reachable set of states of can be enumerated within polynomial space: therefore, for each state Õ of and for each reachable set of states, decide whether they define equal regular languages. The NFA is a RFSA if and only if all its states are equivalent to some reachable set of states. Using similar techniques, it can easily be shown that the problem of building the canonical RFSA equivalent to a given DFA is a PSPACE-complete problem. 9 Conclusion Ideas developped in this paper come from a work done in the domain of Grammatical Inference. A main problem in this field is to infer efficiently (a representation 29

30 ݽ ½ Ò ÝÒ Õ½ Õ ½ ÕÒ Õ... Ò Ü½ ÜÒ Õ a a a Õ b Õ b Figure 14: is an RFSA iff the union of languages described by the is equal to. of) a regular language from a finite set of examples of this language. Some positive results can be proved when regular languages are represented by Deterministic Finite Automata (DFA). For example, it has been shown that Regular Languages represented by DFA can be infered from given data ([Gol78, Hig97]). In this framework, classical inference algorithms such as RPNI [OG92] need a polynomial number of examples relatively to the size of the minimal DFA that recognizes the language to be infered. So, regular languages as simple as ¼ Ò cannot be infered efficiently using these algorithms since their minimal DFA have an exponential number of states. Hence, it is a natural idea to think of using other kind of representations for regular languages, such as Non deterministic Finite Automata (NFA). Unfortunately, it has been shown that Regular Languages represented by NFA cannot be efficiently infered from given data ([Hig97]). The main difficulty that arises when one try to build a NFA from examples comes from the fact that states do not correspond to natural components of the associated language. So, we defined RFSA in order to obtain an automata representation of regular languages for which states correspond to residual languages. We used RFSA for grammatical inference in [DLT00], where we designed an inference algorithm that computes the canonical RFSA of a target regular language from given data and some other papers [DLT01a, DLT01b]. Consequently, languages such as ¼ Ò become efficiently learnable. In conclusion, the class of RFSA has a very simple definition, several nice 30

31 properties and can be useful for applications. References [Brz62] [Chr86] [DLT00] J. A. Brzozowski. Canonical regular expressions and minimal state graphs for definite events. In Mathematical Theory of Automata, volume 12 of MRI Symposia Series, pages M. Chrobak. Finite automata and unary languages. Theorical Computer Science, pages 47(2): , F. Denis, A. Lemay, and A. Terlutte. Learning regular languages using non deterministic finite automata. In ICGI 2000, 5th International Colloquium on Grammatical Inference, volume 1891 of Lecture Notes in Artificial Intelligence, pages Springer Verlag, [DLT01a] F. Denis, A. Lemay, and A. Terlutte. Inférence de langages réguliers représentés par des afer. In Gilles Bisson, editor, CAP 2001, pages 43 58, [DLT01b] F. Denis, A. Lemay, and A. Terlutte. Learning regular languages using rfsa. In ALT Springer Verlag, [DLT01c] F. Denis, A. Lemay, and A. Terlutte. Residual finite state automata. In 18th Annual Symposium on Theoretical Aspects of Computer Science, volume 2010 of Lecture Notes in Computer Science, pages , [Gol78] [Hig97] [HU79] [Kle56] E.M. Gold. Complexity of automaton identification from given data. Inform. Control, 37: , Colin De La Higuera. Characteristic sets for polynomial grammatical inference. Machine Learning, 27: , J.E. Hopcroft and J.D. Ullman. Introduction to Automata Theory, Languages, and Computation. Addison-Wesley, S. C. Kleene. Representation of events in nerve nets and finite automata. In C. Shannon and J. McCarthy, editors, Automata Studies, Annals of Math. Studies 34. New Jersey,

32 [Myh57] J. Myhill. Finite automata and the representation of events. Technical Report , WADC, [Ner58] A. Nerode. Linear automaton transformation. In Proc. American Mathematical Society, volume 9, pages , [OG92] J. Oncina and P. Garcia. Inferring regular languages in polynomial update time. In Pattern Recognition and Image Analysis, pages 49 61, [Yu97] Sheng Yu. Handbook of Formal Languages, Regular Languages, volume 1, chapter 2, pages Springer Verlag,