Computer Science & Engineering Department. IIT Kharagpur. Theory of Computation: CS41001 Lecture I

Transcription

1 Computer Science & Engineering Department IIT Kharagpur Theory of Computation: CS4 Lecture I Instructor: Goutam Biswas Autumn Semester Büchi Automata. Finite and Infinite Sequences An alphabet Σ is afinite set ofsymbols. The collectionoffinite length sequences of symbols from Σ is called Σ. Σ = {σ σ k : σ i Σ, k N } = k N Σ k, where N = {,,2, } and Σ k = {σ σ k : σ i Σ} for all k N. It is not difficult to prove that Σ is a countably infinite set. A Σ-language (or a language over Σ) L is a subset of Σ. So the set of Σ- languagesis uncountable 2 and most ofthese languagescannot have anyeffective finite description. We may view x Σ k, as a map from {,2,,k} Σ, where x(i) Σ is the i th symbol of x from the left. We shall write x i for x(i). The set Σ k is the collection of all possible maps from {,2,,k} Σ. If there are d symbols in Σ, the size of Σ k is d k. The null string ε is the map from Σ. An infinite string α is a map from N Σ, where N = {,2,3, }. As usual α(i) is the i th symbol of α, and we write α i. The collection of all infinite strings over Σ is called Σ ω. It is also not difficult to show that Σ ω is uncountable 3 if Σ >. An ω-language over Σ is a subset of Σ ω. We call Σ = Σ ω Σ. Example. It is clear that is a subset of both Σ as well as Σ ω. If L Σ and α Σ ω, then Lα = {xα : x L} is an ω-language over Σ..2 Finite State Recognisers We look into languages both over Σ and also over Σ ω, that can be recognised by machine models having finite number of states. It has finite memory, so it can remember finite amount of history for deciding the future course of action..2. Languages over Σ It is known that most of the Σ -languages cannot have effective descriptions. But some of them (countably many) can be recognised by a mathematical machine model known as nondeterministic finite automaton (nfa). An nfa N, over Σ is specified by four data, N = (Q,I,F,T), where Q is a finite set of states, I Q is the set of initial states, F Q is the set of final states, and T Q Σ Q, is the transition relation. The transition relation gives a directed graph with set of states as vertices. Edges of the graph are labeled with the symbols of Σ. If (p,σ,q) T, then there is an edge from p to q labeled with σ. The reading is that the machine from the state p will go to state q on input (on event) σ. A run or computation of the machine on an input x = σ σ 2 σ k Σ, is a finite sequence of states s,s 2,,s k+, such that. s I, and 2. (s i,σ i,s i+ ) T, for all i, i k. If Σ = d, then any element of Σ may be viewed as a numeral in radix-(d+) numeral. So there is an injective map from Σ N, the set of natural numbers. 2 There is no surjective map from Σ PΣ (Cantor). 3 If Σ has two symbols, then a map from N Σ may be viewed as the characteristic function of a subset of N. So the collection of all such functions is bijective with the power set of N. And it is known (Cantor) that power set cannot be equinumerous to the original set.

2 An nfa N accepts x = σ σ 2 σ k Σ, if there is a run of the machine on x such that the last state s k+ F. The language of the automaton L(N) is the collection of all strings for which there are accepting runs or computations of the automaton. A language L Σ is is called regular, if there is an nfa N such that L = L(N). An nfa is deterministic (dfa) if following two conditions are satisfied.. I =, 2. for each p Q and σ Σ, there is at most one q Q so that (p,σ,q) T. An nfa or dfa is complete if for each p Q and σ Σ, there is at least one q Q so that (p,σ,q) T. It can be shown that the model of dfa and nfa are of equivalent in terms of recognising languages of Σ i.e. if L Σ is recognised by an nfa, then there is a dfa for L. But an nfa is a more succinct model. In the worst case the equivalent dfa model of an n state nfa can have O(2 n ) states. Ex. Show that the collection of all regular languages over Σ is a countable Boolean algebra..2.2 Languages over Σ ω In case of languages over Σ ω we have a model similar to nfa, known as Büchi automaton 4. A Büchi automatonis alsospecified by four data, B = (Q,I,F,T), with the symbols having their same meaning. But a run or computation of this model on an infinite sequence of elements from Σ is an infinite sequence of states (non-terminating). Let the sequence of events or inputs be α = α α 2 Σ ω. The corresponding infinite run or computation is the sequence of states s,s 2, Q ω, such that. s I, and 2. (s i,α i,s i+ ) T, for all i N. The obvious question is what can be the an accepting or successful run? A successful or accepting run of a Büchi automaton is the one that starts from an initial state, and at least one of the final states is present infinitely often in it. Given a run r : N Q we define a subset of states Inf(r) = {p Q : r (p) is infinite}, where r (p) is the set of pre-images of the state p Q. So an infinite sequence α = α α 2 is accepted by a Büchi automaton B = (Q,I,F,T), ifr = s,s 2, is arun 5 ofthe automatononαandf Inf(r). The language of the Büchi automaton B, L(B) = {α Σ ω : B has a successful run on α}. A Büchi automaton is said to be incomplete if there is a some s Q and σ Σ, there is no state t Q such that (s,σ,t) T. An incomplete Büchi automaton can be completed by introducing a new state g Q and the following set of transitions.. T = {(s,σ,g) : (s,σ,t) T for any t Q}. 2. T 2 = {(g,σ,g) : σ Σ}. The equivalent complete Büchi automaton, B = (Q {g},i,f,t T T 2 ). Example 2. Following are a few examples of Büchi automaton. Let Σ = {,} in all these cases.. Each string has either exactly one or exactly one. 4 In 962 it was introduced by J R Büchi in his paper On a decision method in restricted second order arithmetic. 5 There may be more than one run of B on α, as the machine is non-deterministic. 2

3 s s2 s3 s4 The machine has two start states, I = {s,s 3 } and two final states, F = {s 2,s 4 }. It is incompletely specified. 2. Each should be followed by two s. s s2 s3 s4 3. There are finitely many s in each sequence., s, s2 4. After every there must be either even number of consecutive s or there must be infinite number of consecutive s. s5 s s2 s3 s4 5. Each sequence is of the form xα, where x = () k and α Σ ω, where k. s s2 s3 s4, 3

4 .2.3 Closure Properties The class of languages over Σ ω recognised by Büchi automaton is closed under (i) union, (ii) intersection and (iii) projection. We present this fact in form of following lemmas. Lemma. If L and L 2, languages over Σ ω, are recognised by Büchi automata B = (Q,I,F,T ) and B 2 = (Q 2,I 2,F 2,T 2 ) respectively, then L L 2 is also recognised by a Büchi automaton. Proof: We construct the recogniser for L = L L 2. Without any loss of generality we assume that Q Q 2 =. Our recogniser for L is B = (Q Q 2,I I 2,F F 2,T T 2 ). The claim that any α L L 2 is accepted by B is straight forward. If α has an successful run on B, then it also has a successful run on B. Similar is the case if α has a successful run on B 2. In the other direction, if α has a successful run on B, then it must have a successful run on B or B 2 component of B. By our construction these two components are disjoint. Example 3. Consider the following two languages: L : each string has either exactly one or exactly one. L 2 : there are finitely many s in each sequence. The automaton for L = L L 2 is s s2 s3 s4, s5, s6 Lemma 2. If L and L 2, languages over Σ ω, are recognised by Büchi automata B = (Q,I,F,T ) and B 2 = (Q 2,I 2,F 2,T 2 ) respectively, then L L 2 is also recognised by a Büchi automaton. Proof: The construction of intersection machine requires two things. Given an input α Σ ω,. we should simulate computations of both B and B 2 simultaneously, and 2. keep track of visiting final states of both the machines. The first condition calls for the set of states of the intersection machine as the Cartesian product of Q and Q 2. For a successful run on an input α on the intersection machines, the final states of each of the component machines must appear infinitely often in the run. We use a switch in the states of the intersection machine to remember that the final states of B and B 2 are visited alternately. This will not create any problem as on any successful run on an input α L L 2, of the intersection machine, as after every occurrence of a final state of either B or B 2, there are infinitely many occurrences of final states of both B and B 2. 4

5 We describe the product or intersection machine as follows: B = (Q Q 2 {,2,3},I I 2 {},Q Q 2 {3},T). where transition table T is defined as follows:. ((s,s 2,),σ,(s,s 2,)), where (s,σ,s ) T, (s 2,σ,s 2) T 2, and s F. Expecting a final state of B. 2. ((s,s 2,),σ,(s,s 2,2)), where (s,σ,s ) T, (s 2,σ,s 2 ) T 2, and s F. Expecting a final state of B ((s,s 2,2),σ,(s,s 2,2)), where (s,σ,s ) T, (s 2,σ,s 2) T 2, and s 2 F 2. Expecting a final state of B ((s,s 2,2),σ,(s,s 2,3)), where (s,σ,s ) T, (s 2,σ,s 2) T 2, and s 2 F 2. Visited both the final states. 5. ((s,s 2,3),σ,(s,s 2,)), where (s,σ,s ) T, (s 2,σ,s 2 ) T 2, and s F. Expecting a final state of B. 6. ((s,s 2,3),σ,(s,s 2,2)), where (s,σ,s ) T, (s 2,σ,s 2 ) T 2, and s F. Expecting a final state of B 2. It is possible to work with two-valued switch i.e. Q = Q Q 2 {,2}, where F = F F 2 {2} etc. Definition : Let Σ = Σ Σ 2, and α = (α,α 2 )(α 2,α 22 )(α 3,α 32 ). The Σ -projectionofα isπ (α) = α = α α 2 α 3 Σ ω. Similarly,Σ 2-projection of α is π 2 (α) = α 2 = α 2 α 22 α 32 Σ ω 2. If L Σ ω = (Σ Σ 2 ) ω, then the projection languages of L are π (L) = {π (α) : α L} and π 2 (L), defined similarly. Lemma 3. If L Σ ω, where Σ = Σ Σ 2, is recognised by a Büchi automaton B = (Q,I,F,T), then π (L) and π 2 (L) are also Büchi recognisable. Proof: We construct the machine B = (Q,I,F,T ) to recognise π (L), where for all σ Σ, (s,σ,t) T, if there exists some σ 2 Σ 2, such that (s,(σ,σ 2 ),t) T. Let α = (α,α 2 )(α 2,α 22 )(α 3,α 32 ) L Σ ω = (Σ Σ 2 ) ω. So there isasuccessfulrunr = s s 2 ofb onα, wheres I and(s i,(α i,α i2 ),s i+ ) T, for all i =,2,. But then according to the construction there is a successful run of B on π (α) = α α 2. The other direction can also be proved similarly. We can construct recogniser for π 2 (L) in a similar manner..2.4 Büchi Characterisation Theorem First we show that for any regular language L over Σ, the language of infinite words L ω is recognised by some Büchi automaton, where L ω = {x x 2 : where all x i L, i and x i ε }. We start with a regularlanguageloverσ. There is an nfa N = (Q,I,F,T) thatrecognisesl. Itiswellknownthatthenfacanbeconvertedtoanequivalent dfa by subset construction. Let M = (Q,s,F,T) be the equivalent dfa. The following lemma claims that there is an equivalent dfa with a start state having no incoming transition. Lemma 4. If a regular language L is recognised by a dfa M = (Q,s,F,T), then there is an equivalent dfa M = (Q,s,F,T ) such that L = L(M ) and there is no incoming transition to the initial state s of M. Proof: The construction of M is simple. If there is no incoming transition to s, then M = M. If there are incoming transitions to s, we introduce a new state s Q. The incoming transitions to s are removed and are directed towards s. The state s has all the outgoing transitions of s.. Q = Q {s }. 2. T = (T \ {(t,σ,s ) : t Q,σ Σ}) {(t,σ,s ) : if (t,σ,s ) T} {(s,σ,t) : if (s,σ,t) T}. 5

6 3. F = F if s F, else F = F {s }. We can prove by induction on the length of x = σ σ k Σ, that there is a run r = s s s k of M on x if and only if there is a run r = s s s k of M on x, where s i = s i if s i s and s i = s otherwise, for all i =,,k. Example 4. Let L = {x {,} : x has a substring }. Following dfa accepts L, s s2 s3 s4 We wish to remove incoming transition to the initial state s. s, s s2 s3 s4 Given a dfa M = (Q,s,F,T) for the language L Σ where there is no incoming transition to any initial state, the construction of the Büchi automaton for L ω is simple. Lemma 5. If L Σ is recognised by a dfa M = (Q,s,F,T), there there is a Büchi automaton for L ω. Proof: Without any loss of generality we assume that no initial state of M has any incoming transition. We copy all in-coming transitions to the final states of M to the initial state (this makes it non-deterministic). So the machine after consuming a string of L(N) Σ will start afresh from the initial state and this continues ad infinitum. Formally the Büchi automaton B = (Q b,i b,f b,t b ) is. Q b = Q, I b = {s }, F b = I b. 2. T b = T {(t,σ,s ) : if (t,σ,f) T, where f F}. A successful run is as follows: s,q,,q k,s q 2,,q 2k,s, The strings consumed in the portion s,q j,,q jkj,s belongs to L because any x Σ that drives the machine from the initial state s to the state q jkj and then to s again, also drives M form s to a final state of M. Example 5. The Büchi automaton corresponding to L ω of Example..2.4 is s, s s2 s3 s4, 6

7 Lemma 6. If L and L 2 are regular languages over Σ, then L L ω 2 is Büchi recognisable. Proof: We have two dfa s M = (Q,s,F,T ) and M 2 = (Q 2,s 2,F 2,T 2 ) for L and L 2 respectively. We assume that Q Q 2 = and the initial state of Q 2 does not have any in-coming transition. We know how to construct a Büchi automaton corresponding to L ω 2. Let it be B 3 = (Q 2 {s 2 },{s 2},{s 2 },T 3 ), where s 2 Q Q 2, and T 3 is defined as earlier. Now we can construct a Büchi automaton B = (Q,I,F,T) for L L ω 2 as follows:. Q = Q Q 2 {s 2 }, 2. I = {s }, F = {s 2 }, 3. T T 3 {(t,σ,s 2 ) : if (t,σ,f) T and f F }. We add transitions (t,σ,s 2 ) where (t,σ,f) T and f F. An infinite string α Σ ω drives B on a successful run if its initial segment is a string of L, and then there is an infinite repetition of strings of L 2. Theorem 7. (Büchi) An ω-language L Σ ω is Büchi recognisable if and only if L = k i= L iki ω, where L i,k i Σ are regular languages. Proof: We already have done the ground work for the proof of the if ( ) part. We know that L i Ki ω is Büchi recognisable and Büchi recognisablelanguages are closed under union. To prove the only if ( ) part we proceed as follows. Let B = (Q,I,F,T) be a Büchi automaton such that L = L(B). For every p,q Q, we define an nfa M p,q = (S,p,q,T) and L p,q = L(M p,q ) Σ. As Q is finite, there are finite number of such languages. We define the following ω-language using these regular languages. L = L p,q L ω q,q. p I,q F We claim that L = L. Suppose α L, then there is some p I and q F such that α L p,q L ω q,q. So this α drives B starting from one of its start state p to a final state q and then in the run the state q appears infinitely often. So α L i.e. L L. Conversely, suppose α L. So there is a successful run of the machine B on α. The machine starts from a start state p, and in the infinite sequence of states, at least one of the final states q appears infinitely often. So the run may be viewed as p q q q. The initial segment of α that drives B from state p to state q, so the segment belongs L p,q and each portion of α that drives themachinefromstateq toq belongstol q,q. Soα L p,q L ω q,q i.e. L L..2.5 Complement of Büchi Language We know that Σ \ L is a regular language if and only if L Σ is a regular language. The collection of regular languages over Σ is closed under complementation. The question is whether the collection of Büchi recognisable languages are closed under complementation. The answer is positive but the proof is more involved than the case of regular languages. The main reason is that as a model deterministic Büchi automaton is strictly less powerful than its non-deterministic counterpart. Let α Σ ω. We define α(m,n) or α m,n as the string α(m)α(m+) α(n ) Σ, wheremandnaretwopositiveintegersandm < n. Similarly, α(m, ) is the ω-string α(m)α(m+) Σ ω. Let L Σ. We define L Σ ω as follows: L = {α Σ ω : α(,n) L for infinitely many n N}. So any string α L has infinite number of prefixes from L. This may be viewed as some kind of limit operator. 7

8 .2.6 Deterministic Büchi Automata We have the following theorem that characterises the language recognised by a deterministic automaton. Theorem 8. An ω-language L over Σ is recognisable by a deterministic Büchi automaton if and only if there is a regular language L such that L = L. Proof: Let B = (Q,s,F,T) be adeterministic Büchi automatonand L ω (B) be the ω-language accepted by B. We may view B as an dfa accepting L(B) Σ. We shall show that L(B) = L ω (B). Let α = α α 2 L ω (B) and the corresponding successful run be r = s s. In r there are infinitely many occurrences of final states of B. So there are infinite number of prefixes of α belonging to L(B) i.e. α L(B). So L ω (B) L(B). Let α = α α L(B). As the machine is deterministic there is only one run and there are infinite number of prefixes of α belonging to L(B). So some final state is visited infinitely often implies that α L ω (B) i.e. L(B) L ω (B). Determinism is important in the second part of the proof. If the machine is non-deterministic, it is possible to have an α with infinite number of prefixes in the regular language, but no run of the corresponding Büchi automaton on α have infinitely many final states. Next example makes it clear. Example 6. Consider the following non-deterministic automaton N., s s2 An α = () ω has infinitely many prefixes,,,, from the regular language of N. But there is no successful run of N as a Büchi automaton on this α. Theorem 9. There is a Büchi recognisable language that cannot be accepted by a deterministic Büchi automaton. Proof: We fix the alphabet Σ = {,}. Define the ω-language L = {α Σ ω : there are finite number of occurrences of in α}. L = Σ {} ω and is recognised by the non-deterministic Büchi automaton of example (.2.6). We prove by contradiction that L cannot be recognised by a deterministic Büchi automaton. Assume that L is recognised by a deterministic Büchi automaton. So there is a regular language L Σ such that L = L (deterministic Büchi language). It is clear that ω L, so there is some positive integer n, such that n L ( L = L and there are infinite number of such prefixes). But the infinite string n ω L as it has finite number of s. So again there is some positive integer n 2 such that n n2 L. By a similar argument we have n n2 ω L and there is a positive integer n 3 such that n n2 n3 L. Continuing this way we construct an ω-string α = n n2 n3 ni, where each n i, such that α has infinite number of prefixes belonging to L. So α L = L. But that is contradictory as it has infinite number of s. Corollary. Thereis a deterministic Büchi automaton B such that Σ ω \L ω (B) is not recognised by any deterministic Büchi automaton. Proof: We define L = {w : w Σ, Σ} {,}. It is not difficult to construct a dfa that recognises L. An element α ω-language L has infinitely many prefixes of in L i.e. α has infinite number of s in it. The language L = Σ ω \ L = {α Σ ω : only finitely many prefixes of α are in L. So there are finite number of s in α. But we have already proved in theorem (.2.6) that there is no deterministic Büchi automaton for L. 8

9 .2.7 Equivalences Classes on NFA Definition 2: Let B = (Q,I,F,T)be a Büchi automaton and p,q Q. We define L p,q Σ as the collection of all strings that drives the machine from state p to state q. This is the language L(M p,q ), where M p,q = (Q,{p},{q},T). Similarly for A Q, we define L A p,q Σ as the collection of all strings that drives the machine (i) from state p to state q, and (ii) one of the states in the run is in A. L A p,q may be viewed as r A (L p,rl r,q ). So both these languages are regular. We define a binary relation B on Σ in the following way. Let x,y Σ, then x B y if. for all pairs of states p,q Q, the input x drives the machine from p to q (p x q) if and only if y drives the machine from p to q (p y q), and also 2. for all pairs of states p,q Q, the input x drives the machine from p to q through a final state (p F x q) if and only if y drives the machine from p to q through a final state (p F y q). It is not difficult to prove that B is an equivalence relation. i Afterward we shall characterise the equivalence classes of Σ under B. Definition 3: An equivalence relation on Σ is said to be a congruence relation if for all x,y,z Σ, x B y implies that xz B yz. Lemma. Given a Büchi automata B = (Q,I,F,T), the equivalence relation B is a congruence relation. Proof: Suppose x,y,z Σ and x B y i.e. for all p,q Q, (i) p x q if and only if p y q, and (ii) p F x q if and only if p F y q. p xz r if and only if there is some q Q, so that p x q and q z r if and only if p y q and q z r i.e. p yz r. If p F xz r, then there are two possibilities, (i) there is some q Q, so that p F x q and q z r, or (ii) there is some q Q, so that p x q and q F z r. In the first case we have p F y q and q z r i.e. p F yz r. And in the second case we have p y q and q F z r i.e. p F yz r. Lemma 2. Given a Büchi automata, the congruence relation B partitions Σ in finite number of equivalence classes i.e B has finite index. Proof: We define two functions e,e 2 : Σ 2 Q Q in the following way: and e (x) = {(p,q) Q Q : p x q}, e 2 (x) = {(p,q) Q Q : p F x q}. Let A Q Q. The preimage of A, e (A) = L A Σ, if non-empty, is such that for any two x,y L A, and for any p,q Q, p x q if and only if p y q. So the preimages of two subsets of Q Q are disjoint subsets of Σ. Number of subsets of Q Q are finite, so the partitions are also finite. Similarly the preimage of A Q Q under e 2, e 2 (A) = L A Σ, if nonempty, is such that for any two x,y L A, and for any p,q Q, p F x q if and only if p F y q. This also divides Σ in finite number of partitions. Refinement of these two partitions gives the equivalence classes of B. So the number of equivalence classes are finite. It is well known (Mihill-Nerode theorem) that if there is an equivalence relation of finite index over Σ, and the relation is a congruence relation, then the equivalence classes are regular languages. Example 7. Consider the following Büchi automaton B = (Q,I,F,T) over the alphabet Σ = {,}., s2 s s3 9

10 The map e : Σ 2 Q Q is as follows: e ({ε}) = {(,),(2,2),(3,3)}, e ({ n : n }) = {(,),(,2),(2,2)}, e ({ n : n }) = {(,),(,3),(3,3)}, e ({x : x Σ Σ }) = {(,),(,2)}, e ({x : x Σ Σ }) = {(,),(,3)}, So the map e induces an equivalence relation of finite index on Σ. The classes are, E = {ε}, A = { n : n }, B = { n : n }, C = {x : x Σ Σ }, D = {x : x Σ Σ }. Similarly, map e 2 : Σ 2 Q Q is as follows: e 2 ({ε}) = {(2,2),(3,3)}, e 2 ({ n : n }) = {(,2),(2,2)}, e 2 ({ n : n }) = {(,3),(3,3)}, e 2 ({x : x Σ Σ }) = {(,2)}, e 2 ({x : x Σ Σ }) = {(,3)}, This map also induces the same equivalence relation. So both the map together induces the same equivalence relation of finite index and the equivalence classes are A,B,C,D,E. Lemma 3. Given a Büchi automaton B = (Q,I,F,T), the equivalence class [x] B ([x] in short) is the intersection of all languages of the form L p,q, L F p,q, Σ \L p,q, Σ \L F p,q which contain x. Proof: Let T x be the intersection of all the sets of the form L p,q, L F p,q, Σ \L p,q and Σ \L F p,q, containing x. If y [x], then y B x and y belongs to all sets of the above form containing x 6. So y is in the intersection T x. Suppose y T x, the intersection of all the above mentioned sets. If p x q, then x L p,q. By our definition of T x L p,q. So y L p,q, implies that p y q. If p x q, then x Σ \L p,q. Again by definition y T x Σ \L p,q i.e. p y q. Similarly, if p F x q, then p F y q. And if p F x q then p F y q. It is clear that given a Büchi automaton B = (Q,I,F,T), the corresponding nfa induces an equivalence (congruence) relation B on Σ, where each equivalence class is regular..2.8 Complementation Theorem The plan for the construction of a Büchi automaton that recognises the complement of a Büchi recognisable language is as follows: Let B be a Büchi automaton. We have already shown how a congruence relation of finite index over Σ can be constructed using B as an nfa. We also know that these equivalence classes are regular languages. So for any two equivalence classes L and L 2, the language L L ω 2 is Büchi recognisable. We shall prove that if L L ω 2 L ω (B), then L L ω 2 L ω(b). We also prove that each α L ω (B) belongs to some L L ω 2, where L and L 2 are two equivalence classes. So the complement language Σ ω \ L ω (B) is the union of all L L ω 2 such that L L ω 2 L ω(b) =, where L and L 2 are equivalence classes. Example 8. In the example (.2.7), the equivalence classes are E = {ε}, A = { n : n }, B = { n : n }, C = {x : x Σ Σ }, D = {x : x Σ Σ }. The Büchi recognisable language L ω (B) = A ω B ω CA ω DA ω CB ω DB ω. The complement language consists of AC ω, BC ω etc. Definition 4: AcongruencerelationE onσ saturatesanω-languagelifforany pair of equivalence classes L and L 2, whenever L L ω 2 L, then L L ω 2 L. Lemma 4. Let B = (Q,I,F,T) be a Büchi automaton and L = L ω (B). The congruence relation B defined on Σ saturates L. In other words, for any pair 6 Given a p,q Q, x L p,q if and only if y L p,q. Similarly x L F p,q y L F p,q. if and only if

11 of equivalence classes L and L 2, if there an α L L ω 2 that has a successful run on B, then all strings of L L ω 2 have successful runs on B i.e. L L ω 2 L. Proof: LetL andl 2 be equivalenceclassesandα = uv v 2 L L ω 2 L ω(b), where u L and v i L 2, i =,2,. So there is a successful run r = s s s 2 of B on α, where s I and Inf(r) F. Given α and its successful run r, we get the following infinite sequence of states: s q q 2, such that. u drives the machine from the start state s to q, s u q, and 2. every v i drives the machine from q i to q i+, i.e. q i vi q i+, for i =,2,. As r is a successful run, there are infinitely many final states. So there are infinitely many v i s such that v i drives the machine from q i to q i+ through some final state, q i F v i q i+. Take any arbitrary word β = u v v 2 L L ω 2, where u L and v i L 2. Moreover, u B u and v i B v i, as L and L 2 are equivalence classes. So s u q and q i vi q i+ for i =,2,. Also if q i F v i q i+ then so is q i F v q i i+. We conclude that β has a successful run on B. Nowwegoto provethe secondpart, whereweclaim that everyinfinite string α belongs to some L L ω 2, where L and L 2 are equivalence classes. This will enable us to claim that L ω (B) = L L ω 2, where the union is over all pairs of equivalence classes L and L 2 such that L ω (B) L L ω 2. So the complement language Σ ω \L ω (B) = L L ω 2, where the union is over all pairs of equivalence classesl and L 2 such that L ω (B) L L ω 2 =. Hence the complement language is also Büchi recognisable. Let B = (Q,I,F,T) be a Büchi automaton and B be the equivalence relation defined earlier. Let α = α α 2 Σ ω. The symbol at the i th -position of α is α i. Let k,l and m be positions in α so that k,l < m. We say that positions k and l merge at m if α k,m B α l,m i.e. the string α k α m and α l α m belong to the same equivalence class of B. It is also clear that if α k,m B α l,m, then α k,m B α l,m, where m m i.e. if the positions k and l merge at m, then they merge at all positions after this as B is a congruence relation. Definition 5: Given a Büchi automata B = (Q,I,F,T) and an ω-string α we define a binary relation on the positions in the string, or on the set of positive integers N as follows: k α l, if there is a position m > k,l, such that α k,m B α l,m. We may also write k α l(at m) or simply k α l(m). The binary relation is clearly reflexive and symmetric. We prove that it is transitive. Let j,k,l N and j α k(m), where m is the smallest merge position and k α l(n), where n is the smallest merge position. But we know that j α k at any merge position m m and k α l at any merge position n n. Let p = max(m,n), then α j,p B α k,p and α k,p B α l,p. So we have α j,p B α l,p i.e. j α l at p. So the binary relation on N is an equivalence relation. Lemma 5. Given a Büchi automaton B = (Q,I,F,T) and an ω-string α, the equivalence relation α has finite index i.e. it partitions N in finite number of equivalence classes. Proof: The proof is by contradiction. Assume that α creates infinitely many equivalence classes. So there are infinitely many positions k < k 2 < such that for every k i and k j (i j), k i α k j i.e. there is no position m > k i,k j, such that α ki,m B α kj,m. But we know that B is of finite index. Let the number of equivalence classes of Σ under B be n. Consider positions k < k 2 < < k n+. For all position m > k,,k n+, by our assumption none of α k,m,,α kn+,m are equivalent. But that is impossible as there are only n equivalence classes of Σ under B. The is a contradiction and hence the index of α is finite. Now we know that given a Büchi automaton B and an ω-word α, the equivalence relation α partitions N (positions in α) in finite number of equivalence classes. So there must be an equivalence class with infinite number of positions

12 (positive integers) in it. Let points (positions) in such an equivalence class be k < k < where we may assume that < k. Each k i,k j in this sequence of positions are α equivalent with respect to some position m > k i,k j. We consider the following sequence of finite words: α k,k,α k,k 2,. Again due to finite number of equivalence classes of Σ under B, there will be an infinite subsequence of α k,k,α k,k 2,, whose strings are B equivalent. We pick-up those k j s (positions where α k,k j s are B equivalent and belong to an equivalence class of infinite size). So we have an infinite sequence of positions, k = l,l,l 2, such that l i α l j for all i,j and α l,l i B α l,l j. We claim now that there is an infinite sequence of positions p,p,p 2, such that for all i < j, p i merges with p j at p j+. The sequence is constructed as follows:. Basis: p = l,p = l. 2. Inductive step: If we have already identified the positions p,p,,p n, where p i α p j at position p j+, j =,,n. There is a position m such that p i α p j (m), for all i,j, i,j n. We choose the smallest such m, and take l k as p n+ such that m l k. So we have the infinite sequence of positions, p,p,p 2,, satisfying the following conditions:. < p, 2. α p,p i B α p,p j for all i,j. 3. For all positions p i and p j, i < j, p i α p j, merges at p j+. We define two subsets of Σ as follows: L = {x Σ : x B α,p }, L 2 = {x Σ : x B α p,p }. Clearly L and L 2 are two equivalence classes under B. Lemma 6. The ω-string α L L ω 2. Proof: We consider the position sequence p < p < generated earlier. We know that α pi,p i+ B α p,p i+ (condition (3)). We also have α p,p B α p,p i b α p,p i+ (condition (2)). So, α p,p B α pi,p i+ i.e. the string α pi,p i+ L 2. Hence the infinite string α can be written as α = α,p α p,p α p,p 2, where α,p L and α pi,p i+ L 2 i.e. α L L ω 2. Theorem 7. Let B be a Büchi automaton over Σ. The language Σ ω \L ω (B) is Büchi recognisable. Proof: We have already proved that L L ω 2 L ω (B), where L and L 2 are equivalence classes generated by B, and L L ω 2 L ω(b). Again from the previous lemma we know that if α L ω (B), then α L L ω 2 for some equivalence classes. So we conclude that L L ω 2 = L ω (B), where the union is over all pairs of equivalence classes L and L 2 such that L L ω 2 L ω(b). In the previous theorem we proved that any ω-string is a member of some L L ω 2. So, the complement language, Σω \L ω (B) = L L ω 2, where L and L 2 are equivalence classes generated by B, and the union is over all pairs of such L and L 2 such that L L ω 2 L ω(b) =. So the conclusion is that the collection of Büchi recognisable languages are closed under complementation. References [BKAN] Automata Theory and its Applications, by Bakhadyr Khoussainov and Anil Nerode, Birkhäuser, Springer Int. Ed., 2, ISBN

13 [DPJEP] Infinite Words Automat, Semigroups, Logic and Games, by Dominique Perrin and Jean-Éric Pin, Elsevier, Academic Press, 24, ISBN