Finite State Automata - PDF Free Download

Trento 2005 p. 1/4 Finite State Automata Automata: Theory and Practice Paritosh K. Pandya (TIFR, Mumbai, India) Unversity of Trento 10-24 May 2005

Trento 2005 p. 2/4 Finite Word Langauges Alphabet Σ is collection of symbols (letters). E.g. Σ = {a,b}. Finite word is a finite sequence of letters. E.g. aabb. Set of all finite words is denoted by Σ. Langauge U is a set of words, i.e. U Σ. Example: (a) Words over Σ = {a,b} with equal number of a and b. E.g. aabb or abba. Language recognition problem: To determine whether a word belongs to a language. Most computational problems can be encoded as language recognition problem. Automata are computational devices to solve langauge recognition problems.

Trento 2005 p. 3/4 Finite State Automata Basic model of computational systems with finite memory. Widely applicable Embedded System Controllers. Languages: Esterel, Lustre, Verilog, SMV. Synchronous Circuits. Regular Expression Pattern Matching Grep, Lex, Perl, Awk, Emacs. Protocols Network Protocols Architecture: Bus, Cache Coherence... Mobile Telephony. Good decision procedures

Trento 2005 p. 4/4 Topics DFA, NFA and Equivalence Closure Properties and Decision Problems Regular Expressions and McNaughton Yamada Lemma Homomorphisms DFA minization Pumping Lemma Myhill Nerode Theorem Bisimulation and collapsing nondeterministic automta Textbook: Dexter Kozen, Automata and Computability. Springer, 1997.

Trento 2005 p. 5/4 Notation a,b Σ finite alphabet. u,v,w Σ finite words. ǫ empty word. u.v catenation. u i = u.u..u repeated i-times. U,V Σ Finite word languages. For a set S we use 2 S to denote all its subsets. 2 S def = {X X S}.

Trento 2005 p. 6/4 Deterministic Finite State Automaton Example: DFA A 1 over Σ = {a,b}. a b b s 1 s 2 a

Trento 2005 p. 6/4 Deterministic Finite State Automaton Example: DFA A 1 over Σ = {a,b}. a b b s 1 s 1 s 2 a The run of A 1 over the word abba is : a b b a s 1 s 2 s 2 s 1. For a given word the run is unique.

Trento 2005 p. 7/4 DFA Definition DFA is (Q, Σ,δ,I,F) Q Finite set of states. Σ is the finite alphabet. q 0 Q the initial states. F Q set of final states. δ : Q Σ Q, transition function (total). Notation: We use q a q to denote δ(q,a) = q. A run of DFA A on u = a 0,a 1,...,a n 1 is a sequence of a states q 0,q 1,...,q n s.t. q i i qi+1 for 0 i < n. For a given word w the DFA has a unique run. A run accepting if last state q n F.

Trento 2005 p. 8/4 Regular Langauges Language accepted by A L(A) = {u Σ A has an accepting run on u}. A langauge accepted by a DFA is called a regular language.

Trento 2005 p. 8/4 Regular Langauges Language accepted by A L(A) = {u Σ A has an accepting run on u}. A langauge accepted by a DFA is called a regular language. Define ˆδ(q,u) inductively over u Σ. ˆδ(q,ǫ) = q, ˆδ(q,ua) = δ(ˆδ(q,u),a). For X Q define ˆδ(X,u) def = q X ˆδ(q,u).

Trento 2005 p. 9/4 Nondeterministic FSA NFA A 2. a,b b s 1 s b 2

Trento 2005 p. 9/4 Nondeterministic FSA NFA A 2. a,b b s 1 s b 2 A run of A 2 over the word abbb is: a b b b s 1 s 1 s 2 s 2. s 1 Another run over the same word abbb is: a b b b s 1 s 1 s 1 s 1. s 1 A 2 has no accepting run over the word aba as: a b a s 1 s 2 s 1 NFA can have 0, 1 or more than one runs on a given word.

Trento 2005 p. 9/4 Nondeterministic FSA NFA A 2 recognises words which end in b. a,b b s 1 s b 2 A run of A 2 over the word abbb is: a b b b s 1 s 1 s 2 s 2. s 1 Another run over the same word abbb is: a b b b s 1 s 1 s 1 s 1. s 1 A 2 has no accepting run over the word aba as: a b a s 1 s 2 s 1 NFA can have 0, 1 or more than one runs on a given word.

Trento 2005 p. 10/4 NFA Definition Nodeterministic Finite State Automaton: NFA is (Q, Σ,δ,I,F) Q Finite set of states. I Q set of initial states. F Q set of final states. δ Q Σ Q transition relation (edges). a We use q q to denote (q,a,q ) δ. Alternate Definition of δ δ : Q Σ 2 Q. Exercise: Given that Q = n and Σ = k, give the number of syntactically distinct NFAs.

Trento 2005 p. 11/4 NFA Runs A run of NFA A on u = a 0,a 1,...,a n 1 is a sequence of a states q 0,q 1,...,q n s.t. q i i qi+1 for 0 i < n and q 0 I. For a given word w the NFA can have many runs. A run accepting if last state q n F. Language accepted by A is L(A) = {u Σ A has an accepting run on u}. Define ˆδ(S,w) inductively on w as follows. ˆδ(S,a) = s S δ(s,a) ˆδ(S,ǫ) = S, ˆδ(S,wa) = ˆδ(ˆδ(S,w),a). Claim x L(A) ˆδ(I,x) F

Trento 2005 p. 12/4 Subset Construction Theorem (determinisation) Given NFA A = (Q, Σ,δ,I,F) we can construct a DFA A s.t. L(A) = L(A ).

Trento 2005 p. 12/4 Subset Construction Theorem (determinisation) Given NFA A = (Q, Σ,δ,I,F) we can construct a DFA A s.t. L(A) = L(A ). Subset construction: 0,1 p q 1 0,1 Consider ˆδ(I,w) the set of states reached after word w. We define (S, a). 0 1 {p} {p} {p,q} {p,q} {p,r} {p,q,r} {p,r} {p} {p,q} {p,q,r} {p,r} {p,q,r} r

Trento 2005 p. 13/4 Subset Construction (cont) Let A = (2 Q, Σ,,I,F ) where (S,a) = {q q δ(p,a) p S} F = {S 2 Q S F }. Claim: L(A) = L(A ). Proof Method: Show that ˆδ(S,w) = ˆ (S,w). Complexity A 2 A. Lower Bound Example: Consider NFA recognising words over {0, 1} such that the n-th last symbol is 1. Claim: Equivalent DFA will have at least 2 n states. (Home Exercise: Prove this.)

Trento 2005 p. 14/4 Closure Properties Theorem (boolean closure) Given NFA A 1,A 2 over Σ we can construct NFA A over Σ s.t. L(A) = L(A 1 ) (Complement). Size: A = 2 A 1. L(A) = L(A 1 ) L(A 2 ) (union). Size: A = A 1 + A 2. L(A) = L(A 1 ) L(A 2 ) (intersection). A = A 1 A 2.

Trento 2005 p. 15/4 Synchronous Product: Example s 0 a b b b b a a a s 1 a t 2 t 0 b t 1 A 1 recognises words with even number of b. A 2 recognises words with number of a mod 3 = 0. The Product Automaton A 1 A 2 with F = {s 0,t 0 }. s 0 a a a t s 0 t 1 2 s 0 t 0 b b b b s 1 t 2 s 1 t 0 a a a b b s 1 t 1

Trento 2005 p. 16/4 Synchronous Product Construction Let A 1 = (Q 1, Σ,δ 1,I 1,F 1 ) and A 2 = (Q 2, Σ,δ 2,I 2,F 2 ). Then, A 1 A 2 = (Q, Σ,δ,I,F) where Q = Q 1 Q 2. I = I 1 I 2. F = F 1 F 2. < p,q > a < p,q > iff p a p and q Theorem L(A 1 A 2 ) = L(A 1 ) L(A 2 ). a q. Semantics of many concurrent systems E.g. Esterel, Statecharts, Lustre, Synchronous circuits, SMV.

Trento 2005 p. 17/4 Synchronised Product Construction Let A 1 = (Q 1, Σ 1,δ 1,I 1,F 1 ) and A 2 = (Q 2, Σ 2,δ 2,I 2,F 2 ). Then, A 1 A 2 = (Q, Σ,δ,I,F) where Q = Q 1 Q 2. Σ = Σ 1 Σ 2. I = I 1 I 2. F = F 1 F 2. < p,q > < a p,q > if a Σ 1 Σ 2 and p a p and a q q. < p,q > a < p,q > if a Σ 1, a Σ 2 and p a p. < p,q > a < p,q > if a Σ 1, a Σ 2 and q a q.

Trento 2005 p. 18/4 Asynchronous Product Construction Let A 1 = (Q 1, Σ 1,δ 1,I 1,F 1 ) and A 2 = (Q 2, Σ 2,δ 2,I 2,F 2 ). Then, A 1 A A 2 = (Q, Σ,δ,I,F) where Q = Q 1 Q 2. Σ = Σ 1 Σ 2. I = I 1 I 2. F = F 1 F 2. < p,q > a < p,q > if a Σ 1 and p a p. < p,q > a < p,q > if a Σ 2 and q a q.

Trento 2005 p. 19/4 Decision Problems Theorem (Emptiness) Given NFA A we can decide whether L(A) =. Method Forward/Backward Reachability in Automaton graph using say Depth First Search. Complexity is O( Q + δ ). In practice, symbolic techniques are used and often more effective. Theorem (Language Containment) Given NFA A 1 and A 2 we can decide whether L(A 1 ) L(A 2 ). Method: L(A 1 ) L(A 2 ) iff L(A 1 ) L(A 2 ) =. Complexity is O( A 1 2 A 2 ).

Trento 2005 p. 20/4 Kleene Closure Let U,V Σ. (Catenation) Let U V def = {x y x U y V }. (Iteration) Let U i def = {x 1 x 2... x i x k U, 1 k < i}. By convention U 0 def = {ǫ}. (Kleene Closure) Let U def = 0 i U i. Theorem Regular langauges are closed under the operations U V and U V and U. Proof method: Using ǫ-nfa.

Trento 2005 p. 21/4 Regular Expressions Syntax: ǫ a reg 1.reg 2 reg 1 + reg 2 reg.

Trento 2005 p. 21/4 Regular Expressions Syntax: ǫ a reg 1.reg 2 reg 1 + reg 2 reg. Language L(reg) is defined inductively: L( ) =, L(a) = {a}, L(ǫ) = {ǫ}. L(re 1 re 2 ) = L(re 1 ) L(re 2 ), L(re 1 + re 2 ) = L(re 1 ) L(re 2 ), L(re ) = (L(re 1 )) Example: a.(b + bb).a. Words over a,b having either 1 b or 2 consecutive b. Equivalence of RegExp and FSA (McNaughton, Yamada) Theorem: For every regular expression reg we can construct a language equivalent ǫ-nfa of size O( reg ).

Trento 2005 p. 22/4 Automata to RegExp Theorem: For every NFA A = (Q, Σ,δ,I,F) we can construct a langauge equivalent regular expression reg(a).

Trento 2005 p. 22/4 Automata to RegExp Theorem: For every NFA A = (Q, Σ,δ,I,F) we can construct a langauge equivalent regular expression reg(a). Construction Define regular expression α X p,q for X Q and p,q Q. This denotes set of words w such that A has a run on w from p to q where all intermediate states are from X.

Trento 2005 p. 23/4 Automata to RegExp (2) α X p,r can be recursively defined using the following rules: α p,r = a 1 +... + a k where r δ(p,a i ) provided p r

Trento 2005 p. 23/4 Automata to RegExp (2) α X p,r can be recursively defined using the following rules: α p,r α p,p = a 1 +... + a k where r δ(p,a i ) = a 1 +... + a k + ǫ where p δ(p,a i ) provided p r

Trento 2005 p. 23/4 Automata to RegExp (2) α X p,r can be recursively defined using the following rules: α p,r α p,p = a 1 +... + a k where r δ(p,a i ) = a 1 +... + a k + ǫ where p δ(p,a i ) For any q X, α X p,r = α (X {q}) p,r + α (X {q}) p,q provided p r ( α (X {q}) q,q ) α (X {q}) q,r

Trento 2005 p. 24/4 Example Compute α p,q,r p,p α p,q,r p,p as It is easy to see that αp,p p,r = 0, αq,p p,r = 00(0 ), 0 p 1 0 q 1 0 = α p,r p,p + α p,r p,q (α p,r q,q) α p,r q,p. α p,r r p,q = 0 1 αq,q p,r = ǫ + 01 + 00(0 )1 Hence α p,q,r p,p = 0 + 0 1 (ǫ + 01 + 00(0 )1) 00(0 )

Trento 2005 p. 25/4 Kleene Algebra Axiomatizing equivalence of regular expression. Not covered in this course. (See Kozen.)

Trento 2005 p. 26/4 Homomorphism Homomorphism is a map h : Σ Γ with the property u,v Σ. h(u v) = h(u) h(v). Consequences: h(ǫ) = ǫ. Giving h(a) for a Σ uniquely determines h.

Trento 2005 p. 26/4 Homomorphism Homomorphism is a map h : Σ Γ with the property u,v Σ. h(u v) = h(u) h(v). Consequences: h(ǫ) = ǫ. Giving h(a) for a Σ uniquely determines h. Theorem If h : Σ Γ is a homomorphism and B Γ is regular then h 1 (B) = {x Σ h(x) B} is also regular. Proof method: Given DFA BB = (Q, Γ,δ,q 0,F) for B construct DFA AA for h 1 (B) as follows. AA def = (Q, Σ,δ,q 0,F) with δ (q,a) = ˆδ(q,h(a)).

Trento 2005 p. 27/4 Homomorphism (2) Theorem If h : Σ Γ is a homomorphism and A Σ is regular then h(a) = {h(x) x A} is also regular.

Trento 2005 p. 27/4 Homomorphism (2) Theorem If h : Σ Γ is a homomorphism and A Σ is regular then h(a) = {h(x) x A} is also regular. Proof method: Given regular epxression re for A, obtain h(re) by substituting every letter a by h(a). Then L(h(re)) = h(a). Theorem (projection) Given NFA(A 1 ) over Σ and surjection h : Σ Γ, we can construct NFA(A 2 ) over Γ s.t. L(A 2 ) = h(l(a 1 )). Proof Method: Substitute label a by h(a) in each transition of A 1 to get A 2. Note that this can make DFA into NFA. We have A 2 = A 1.

Trento 2005 p. 28/4 DFA Minimization 0 a b 1 2 a b b a 3 4 a,b a,b 5 a,b

Trento 2005 p. 28/4 DFA Minimization 0 a b 1 2 a b b a 3 4 a,b a,b 5 a,b p q def = x Σ. (ˆδ(p,x) F ˆδ(q,x) F)

Trento 2005 p. 28/4 DFA Minimization 0 a b 1 2 a b b a 3 4 a,b a,b 5 a,b p q def = x Σ. (ˆδ(p,x) F ˆδ(q,x) F) Equivalent Automaton a,b 6 7 a,b 8 a,b

Trento 2005 p. 29/4 DFA Minimization (2) p q def = x Σ. (ˆδ(p,x) F ˆδ(q,x) F)

Trento 2005 p. 29/4 DFA Minimization (2) p q def = x Σ. (ˆδ(p,x) F ˆδ(q,x) F) Proposition is an equivalence relation, i.e. (a) p. p p, (b) p,q. p q q p (c) p,q,r. p q q r p r (Exercise: Check that above properties are true.) partitions Q into equivalence classes. Let [p] denote the equivalence class of p. Example The classes are {0}, {1, 2} and {3, 4, 5}.

Trento 2005 p. 30/4 Quotient Automaton Given DFA A = (Q, Σ,δ, [q 0 ],F) and as before, the Quotient automaton is A/ def = (Q, Σ,δ, [q 0 ],F ), where Q = {[p] p Q} δ ([p],a) = [δ(p,a)] F = {[f] f F } Well-formedness Theorem(Congruence) p q a Σ. δ(p,a) δ(q,a). Now we can show that quotient automaton recognises the same language. Theorem ˆδ ([p],w) = [ˆδ(p,w)]. Corollary L(A/ ) = L(A).

Trento 2005 p. 31/4 Minimization algorithm 1. Make pairs table with (p,q) Q Q and p < q. 2. Mark (p,q) if p F q / F or vice versa. 3. Repeat following steps until no change occurs. (a) Pick unmarked state (p, q). (b) If (δ(p,a),δ(q,a)) is marked for some a Σ then mark (p,q) Termination: In each pass at least one new pair must get marked. Lemma (p,q) is marked iff x Σ.ˆδ(p,x) F ˆδ(q,x) / F or vice versa.

Trento 2005 p. 32/4 Pumping Lemma Idea: Consider a long word recognized by a DFA with n states. Consider substring y s.t. y n. Some state during y part of the run must repeat. The string between repeating states can be pumped. Lemma For every regular language L n > 0 such that x,y,z with x y z L and y n u,v,w s.t. y = u v w with v > 0 and x u v i w z L for all 0 i.

Trento 2005 p. 33/4 Proving non-regularity Game with Demon Demon claims L is regular and chooses n. You choose a string xyz L with y n. Demon partitions y = uvw with v 0. You show that xu(v i )wz / L for some i for your choice.

Trento 2005 p. 34/4 Myhill-Nerode Relations Let be an equivalence relation over Σ. is right congruent if x,y Σ, a Σ we have x y x a y a

Trento 2005 p. 34/4 Myhill-Nerode Relations Let be an equivalence relation over Σ. is right congruent if x,y Σ, a Σ we have x y x a y a Let R Σ (not necessarily regular). refines R if x,y Σ x y (x R y R) is of finite index if partitions the Σ into only finitely many equivalence classes.

Trento 2005 p. 35/4 Machine Equivalence Given a DFA A = (Q, Σ,δ,q 0,F) define the induced equivalence A over Σ as follows: def x A y = ˆδ(q 0,x) = ˆδ(q 0,y).

Trento 2005 p. 35/4 Machine Equivalence Given a DFA A = (Q, Σ,δ,q 0,F) define the induced equivalence A over Σ as follows: def x A y = ˆδ(q 0,x) = ˆδ(q 0,y). Proposition A is a Myhill-Nerode relation refining L(A). Proof Method Check the following: (a) A is an equivalence relation over Σ. (b) x A y a. xa A ya. (c) x A y x L(A) y L(A) (d) A is of finite index.

Trento 2005 p. 36/4 From Equivalence to DFA Let be Myhill-Nerode refining R. Define DFA A def = (Q, Σ,δ,q 0,F) as follows: Q def = {[x] x Σ } def q 0 = [ǫ] F def = {[x] x R} δ([x],a]) def = [xa]. (Check well-formedness.)

Trento 2005 p. 37/4 Correspondence The A and A are inverses of each other. Theorem A =. Theorem If A is automaton without unreachable states then A A is isomorphic to A.

Trento 2005 p. 38/4 Language Induced Equivalence An equivalence relation 1 refines equivalence relation 2 provided x 1 y x 2 y (Set theoretically, 1 2.) Intuitively, 1 makes finer partitions than 2.

Trento 2005 p. 38/4 Language Induced Equivalence An equivalence relation 1 refines equivalence relation 2 provided x 1 y x 2 y (Set theoretically, 1 2.) Intuitively, 1 makes finer partitions than 2. Definition (Language Induced Equivalence) Given R Σ, def x R y = z Σ. (xz R yz R). Proposition R is right congruent. Proposition R refines R.

Trento 2005 p. 39/4 Myhill-Nerode Theorem Let R Σ. The following statements are equivalent. 1. R is regular 2. There exists a Myhill-Nerode relation refinining R. 3. The relation R is of finite index. The automaton A R gives the minimal DFA for R.

Trento 2005 p. 40/4 State Minimization is optimal Let A recognise R. Consider the quotient automaton M = A/ as before, and assume that it has no inaccessible states. Lemma x R y x M y

Trento 2005 p. 41/4 Bisimulation Postponed to the End of course. See Kozen Supplementary Lecture B.