Chap. 2 Finite Automata

Transcription

1 Chap. 2 Finite Automata 2.1 An Informal Picture of Finite Automata A man with a wolf, goat, and cabbage is on the left bank of a river A boat carries one man and only one of the other three. The wolf eats the goat without the man. The goat eats the cabbage without the man. Is it possible to cross the river without the goat or cabbage being eaten? states: occupants of each bank after a crossing 16 subsets of the man(m), wolf(w), goat(g), and cabbage(c). 16 states: mwgc-, WC-mG,, -mwgc input symbol: four ways of crossing the river man cross alone(m), with the wolf(w), the goat(g), or cabbage(c) input string: a sequence of crossings(input symbols) state transition: From a state to another state by a input symbol 9/13/12 Kwang-Moo Choe 1

2 mcgw- m,c,w,g -mcgw g g m,c,w m-cgw mc-gw mw-cg m,c,w g g CW-mG m m mg-cw m mcw-g m c c w w W-mCG mgw-c w w m m m G-mCW C-mGW g g g g m mcg-w c c 9/13/12 Kwang-Moo Choe 2

3 2.2 Deterministic Finite Automata A deterministic finite automaton consists of: 1. A finite set of states, denoted Q, 2. A finite set of input symbols, denoted Σ, 3. A transition function, δ: Q Σ Q Let p, q Q, and a Σ, if we write δ(q, a) = p, meaning that when current state is in q and input symbol is a, the current state is changed to p, and input symbol is changed to the next(right) one. 4. A start(initial) state, q 0 Q, and 5. A set of final or accepting states, F Q. A DFA A = (Q, Σ, δ, q 0, F) five-tuple notation 9/13/12 Kwang-Moo Choe 3

4 2.2.2 How a DFA Processes Strings How a DFA decides whether or not accept a input string(sequence of input symbols) Suppose a 1 a 2 a n is a sequence of input symbols 1. We start out the DFA with in its start state q We consult the transition function, δ, with current state and current input symbol, asumme in the first trial δ(q 0, a 1 ) = q 1, 3. Repeat this step with the next state q 1 and the next input symbol a 2. Assume δ(q 1, a 2 ) = q 2, δ(q 2, a 3 ) = q 3,, δ(q n-1, a n ) = q n. 4. Finally, check if δ(q n-1, a n ) = q n F or not. If q n F, accept a 1 a 2 a n. If q n F does not accept a 1 a 2 a n. δ(δ( δ(δ(q 0, a 1 ), a 2 ), ), a n-1 ), a n ) = q n where δ(q i-1, a i ) = q i for 1 i n. 9/13/12 Kwang-Moo Choe 4

5 Example 2.1 Let Σ = {0, 1} and L = {x01y Σ * x, y Σ * }. State q 0 : initial state If it sees 0, go to a new state q 1. If it sees 1 stay in the state q 0. It has never seen 0. State q 1 : The last symbol it has seen is 0. If it sees 1, go to a new state q 2. If it sees 0, stay in the state q 1. State q 2 : It has ever seen is 01. final state Whether it sees 0 or 1, stay in the state q 2. A = (Q, Σ, δ, q 0, F) Q = {q 0, q 1, q 2 } Σ = {0, 1} δ = {δ(q 0, 0)=q 1, δ(q 0, 1)=q 0, δ(q 1, 0)=q 1, δ(q 1, 1)=q 2, δ(q 2, 0)=q 2, δ(q 2, 1)=q 2 } F = {q 2 } 9/13/12 Kwang-Moo Choe 5

6 2.2.3 Simple notations for DFA s Transition diagram 1 0 0, 1 0 q 0 q 1 1 q 2 Transition table 0 1 q 0 q 1 q 0 q 1 q 1 q 2 * q 2 q 2 q 2 9/13/12 Kwang-Moo Choe 6

7 2.2.4 Extend the domain of transition function from symbols to strings δ: Q Σ Q δ^: Q Σ * Q δ^(q, ε) = B q δ^(q, wa) = R δ(δ^(q, w), a) basis q Q, w Σ *, a Σ, wa Σ +. recursion δ^(q 0, a 1 a 2 a n ) = R δ(δ^(q 0, a 1 a 2 a n-1 ), a n ) 1st recursion = R δ(δ(δ^(q 0, a 1 a 2 a n-2 ), a n-1 ), a n ) 2nd recursion = R = R δ(δ( δ(δ^(q 0, a 1 ), a 2 ), ), a n-1 ), a n ) (n-1)-th recursion = R δ(δ( δ(δ(δ^(q 0, ε), a 1 ), a 2 ), ), a n-1 ), a n ) n-th recursion = B δ(δ( δ(δ(q 0, a 1 ), a 2 ), ), a n-1 ), a n ) basis 9/13/12 Kwang-Moo Choe 7

8 If δ^(q, w) = p, for some q, p Q and w Σ *, w is the path from the state q to the state p. Since δ^ δ, we may use δ instead of δ^. Example q 0 1 q even numbers of 0 s and even numbers of 1 s. 0 q q 3 9/13/12 Kwang-Moo Choe 8

9 2.2.5 The Language of DFA We define the language of a DFA A = (Q, Σ, δ, q 0, F), L(A), L(A) = {w Σ * δ^(q 0, w) F} or {w Σ * δ(q 0, w) F}, for short. L(A) Σ * or L(A) 2 Σ*. Definition: Regular languages Let L be a language. If L = L(A) for some DFA A, then we say L is a regular language. a class of languages type-3 languages in Chomsky s language hierarchy Definition: Equivalence of Automata Let M 1 and M 2 be two automata. We say two automata M 1 and M 2 are equivalent, if L(M 1 ) = L(M 2 ). 9/13/12 Kwang-Moo Choe 9

10 The state of automata summarizes the information concerning past inputs that is needed to determine the behavior of the automata on subsequent inputs. Aho and Ullman Let M = (Q, Σ, δ, q 0, F) be a DFA and q Q. We define accessible input strings from initial state to the state q. L(q) = {w Σ * δ^(q 0, w) = q} L(M) = q F L(q). Consider a binary relation R M on Σ * and x R M y, if δ^(q 0, x) = δ^(q 0, y). Then R M is a quivalent relation and the equivalent class [x] RM = L(q), if δ^(q 0, x) = q. Since R A is equivalent, [x] RM is a partion of Σ *. 9/13/12 Kwang-Moo Choe 10

11 2.3 Nondeterministic Finite Automaton A = (Q, Σ, δ, q 0, F) is a nondeterministic finite automaton(nfa), if 1. Q is a finite set of states, 2. Σ is a finite set of input symbols, open denoted, 3. δ is a transition function, δ: Q Σ 2 Q. If q Q, a Σ, and {p 1, p 2,, p n } Q(or 1 i n, p i Q) then we write δ(q, a) = {p 1, p 2,, p n }, meaning that when current state is in q and input symbol is a, the state may be changed to any one of p 1, p 2,, p n, and input symbol is changed to the next(right) one. 4. q 0 Q is a start state, and 5. F Q is a set of final or accepting states. Fig. 2.9, 2.10, /13/12 Kwang-Moo Choe 11

12 DFA: δ: Q Σ Q q Q, a Σ, δ(q, a) Q. Otherwise (total) function partial function If A = (Q, Σ, δ, q 0, F) be a FA with δ: Q Σ \ Q partial function, L(A) = L(B) for DFA B = (Q {d}, Σ, δ, q 0, F) where d Q, δ = δ {δ(q, a) = d δ(q, a) Q} {δ(d, a) = d a Σ} d: dead state B is the DFA(with total function) DFA: δ: Q Σ Q. DFA but partial function: δ: Q Σ Q { }. NFA: δ: Q Σ 2 Q. 9/13/12 Kwang-Moo Choe 12

13 2.3.3 The Extended Transition Function DFA δ: Q Σ Q δ^: Q Σ * Q NFA δ: Q Σ 2 Q δ : 2 Q Σ 2 Q in this lecture δ ^: 2 Q Σ * 2 Q in this lecture δ^: Q Σ * 2 Q in the text Two step extension 1. Set extension: δ δ We extend the 1st domain of δ to set of states. δ : 2 Q Σ 2 Q. Let P Q, we define δ (P, a) = {q Q p P, q δ(p, a)} 9/13/12 Kwang-Moo Choe 13

14 or = p P δ(p, a) We may write δ (p, a) instead of δ ({p}, a) when needed. 2. String extension : δ δ ^(same as δ δ^) δ ^: 2 Q Σ * 2 Q. δ ^(P, ε) = B P basis δ ^(P, wa) = R δ (δ ^(P, w), a) recursion P Q(or P 2 Q ), w Σ *, a Σ, wa Σ +. We may use δ instead of δ or δ ^, since δ δ, δ ^. Let N = (Q, Σ, δ, q 0, F) be an NFA. Then the language defined by the NFA N, denoted as L(N) is, L(N) = {w Σ * δ ^({q 0 }, w) F }. L(N) Σ * or L(N) 2 Σ*. 9/13/12 Kwang-Moo Choe 14

15 2.3.5 Equivalence of DFA and NFA Example 2.10(Fig. 2.12) Subset construction(nfa DFA) Given an NFA N = (Q N, Σ, δ N, q 0N, F N ), construct DFA D = (Q D, Σ, δ D, q 0D, F D ). 1. Q D = {P P Q N } = 2 Q N. 3. δ D (P, a) = {q Q N p P Q N, q δ N (p, a)} or = p P δ N (p, a) or = δ N (P, a) set extension of δ N. δ D (P, a) = δ N (P, a) or δ D = δ N in short, where δ D : Q D Σ Q D, δ N : 2 Q N Σ 2 Q N, and Q D = 2 Q N. 4. q 0D = {q 0N }. 5. F D = {F Q D F F N } 9/13/12 Kwang-Moo Choe 15

16 Theorem 2.11 L(D) = L(N) in the above subset construction Proof δ^ D (q 0D, w) = δ ^N ({q 0N }, w) for all w Σ*. δ^d : Q D Σ* Q D = 2 Q N Σ * 2 Q N. δ ^N : 2Q N Σ * 2 Q N. We may use q 0N instead of {q 0N }. Induction on w. basis: Let w = 0, δ^ D (q 0D, ε) = δ^ D ({q 0N }, ε) = {q 0N }= δ ^N ({q 0N }, ε). induction: Let w = xa where a Σ, x Σ *, and w Σ +. ( w = x + 1) δ^ D (q 0D, xa) = δ D (δ^ D ({q 0N }, x), a) by definition of δ^d. = δ N (δ^d ({q 0N }, x), a) by subset construction = δ N (δ ^N ({q 0N }, x), a) by induction hypothesis = δ ^ N ({q 0N }, xa) by definition of δ ^N. 9/13/12 Kwang-Moo Choe 16

17 δ ^ D ({q 0 }, x) F D, if and only if, δ ^N (q 0, x) F N. Q.E.D. Theorem 2.12 A language L is defined by some DFA, if and only if L is defined by some NFA.(NFA DFA) Proof: (If) subset construction(nfa DFA) (Only if) NFA can easily simulate DFA(DFA NFA). Let a DFA D = (Q, Σ, δ D, q 0, F). We define an NFA(Q, Σ, δ N, q 0, F) δ N (q, a) = {p δ D (q, a) = p}. Proof for w Σ *, if and only if, δ^ D (q, w) = p, then δ^d (q, w) = {p}. Then L(D) = L(N). L is regular, iff L = L(N) for some NFA N. 9/13/12 Kwang-Moo Choe 17

18 2.5 Finite Automata with Epsilon-Transitions A = (Q, Σ, δ, q 0, F) is a fa with epsilon transition(ε-nfa), if Q, Σ, q 0, and F are same as NFA, but 3. δ is a transition function, δ: Q (Σ {ε}) 2 Q. If q Q, and {p 1, p 2,, p n } Q, then we write δ(q, ε) = {p 1, p 2,, p n }, meaning that when current state is in q and regardless of input symbol the state may be changed to any one of p 1, p 2,, p n, and next input symbol is not changed. Ex. 2.16, Fig /13/12 Kwang-Moo Choe 18

19 2.5.3 Epsilon-closure basis: q ε * (q) (same as ECLOSE(q) in the text) recursion: If p ε * (q), and r δ(p, ε), then r ε * (q). Example 2.18 Why ε * (q) instead of ECLOSE(q) Let ε = {(p, q) q δ(p, ε)}. Then ε Q Q(a binary relation on Q) and ECLOSE(q) = ε * (q) reflexive-transitive closure of ε. 9/13/12 Kwang-Moo Choe 19

20 2.5.4 The Extended Transition and Language for ε-nfa s NFA: δ: Q Σ 2 Q. ε-nfa: δ: Q ({ε} Σ) 2 Q. Let s divide δ of ε-nfa into δ 1 and δ 0. δ 1 : Q Σ 2 Q. δ 0 : Q {ε} 2 Q. δ = δ 1 δ 0. δ 1 : is same as δ in NFA, we use δ instead of δ 1, since δ 1 δ. δ 0 f ε what is the bijection f? We use ε instead of δ 0. ε = {(p, q) q δ(p, ε)} Q Q binary relation on Q. ε * = i N ε i 0 Q Q binary relation on Q. ε, ε * : Q 2 Q. function from Q to 2 Q. 9/13/12 Kwang-Moo Choe 20

21 Let s extend the domain of δ(=δ 1 ), ε, and ε * from state to set of states. Let P Q, we define δ : 2 Q Σ 2 Q. δ (P, a) = {q Q p P, δ(p, a) = q} or = p P δ(p, a) We may write δ (p, a) instead of δ ({p}, a) when needed. We may write δ(p, a) instead of δ (P, a) when needed. ε : 2 Q 2 Q. ε (P) = {q Q p P, ε(p) = q} or = p P ε(p) ε * : 2 Q 2 Q. ε * (P) = {q Q p P, ε * (p) = q} = p P ε * (p) We write ε(p) instead of ε (P) and ε * (P) instead of ε * (P). 9/13/12 Kwang-Moo Choe 21

22 Let s extend the of domain of δ from symbol to strings. δ^: 2 Q Σ * 2 Q. δ^(p, ε) = ε * (P) basis δ^(p, wa) = ε * (δ(δ^(p, w), a)) recursion P Q(or P 2 Q ), w Σ *, a Σ, wa Σ +. δ^(q 0, a 1 a 2 a n ) = ε * (δ(δ^(q 0, a 1 a 2 a n-1 ), a n )) 1st recursion = ε * (δ(ε * (δ(δ^(q 0, a 1 a 2 a n-2 ), a n-1 )), a n )) 2nd recursion = = ε * (δ(ε * (δ( ε * (δ(δ^(q 0, a 1 ), a 2 )), ), a n-1 )), a n ))(n-1)th recursion = ε * (δ(ε * (δ( ε * (δ(ε * (δ(δ^(q 0, ε), a 1 )), a 2 )), )), a n-1 )), a n )) n-th = ε * (δ(ε * (δ( ε * (δ(ε * (δ(ε * (q 0 ), a 1 )), a 2 )), )), a n-1 )), a n )) basis δ n = ε * δ ε * δ ε * = ε * (δ ε * ) n. 9/13/12 Kwang-Moo Choe 22

23 We may write δ * instead of δ^, since δ^ = i N 0 δ i = δ *. Let E = (Q, Σ, δ, q 0, F) be an ε-nfa. Then the language defined by the ε-nfa E, denoted as L(E) is, L(E) = {w Σ * δ^(q 0, w) F }. L(E) Σ * or L(E) 2 Σ* 9/13/12 Kwang-Moo Choe 23

24 2.5.5 Eliminating ε-transitions Given an ε-nfa E = (Q E, Σ, δ E, q 0, F E ), construct the equivalent DFA D = (Q D, Σ, δ D, q 0D, F D ) such that L(D) = L(N). 1. Q D = {P P Q E } = 2 Q E. 3. q 0D = ε * ({q 0 }) 4. δ D (P, a) = ε * (δ E (P, a)) set extension of δ E and ε-closure where P Q E (P Q D ), a Σ, and δ D (P, a) Q E (or δ D (P, a) Q D ) 5. F D = {F Q D F F E } Theorem 2.22 A language L is accepted by some ε-nfa if and only if L is accepted by some DFA. Proof: (If) ε-nfa can easily simulate DFA(DFA ε-nfa). (Only if) Eliminating ε-transitions(ε-nfa DFA) 9/13/12 Kwang-Moo Choe 24

25 Let E = (Q E, Σ, δ E, q 0, F E ) be a ε-nfa and DFA D = (Q D, Σ, δ D, q 0D, F D ) is the one in We show δ E^(q0, w) = δ D^(q0D, w) by induction on w. basis: δ E^(q0, ε) = ε * (q 0 ) by basis definition of δ E^. = q 0D = by construction of δ D. = δ D^(q0D, ε) by basis definition of δ D^. Induction: Let w = xa Σ +, a Σ, and x Σ *. δ E^(q0, xa) = δ E (ε * (δ E^(q0, x), a) by recursive definition of δ E^. = δ E (ε * (δ D^(q0D, x), a)) by induction hypothesis. = δ D (δ D^(q0D, x), a) by construction of δ D. = δ D^(q0D, xa) by recursive definition of δ D^. 9/13/12 Kwang-Moo Choe 25

26 2.A Extended Finite Automata(XFA) X = (Q, Σ, q 0, δ, F) is an extended finite state automata(xfa), if δ: Q Σ * 2 Q. Given an XFA X = (Q X, Σ, q 0, δ X, F X ), construct an equivalent ε-nfa E = (Q E, Σ, q 0, δ E, F E ) such that L(X) = L(E). Construction algorithm Q E := Q X ; F E := F X ; δ E := ; for (δ X (q, x) = P) δ X where x = a 1 a 2 a n do if 0 n 1 δ E (q, x) := P where x Σ {ε} n 2 δ E (q, a 1 ) = δ E (q, a 1 ) {p 1 }; od fi for 2 i n-1 do δ E (p i-1, a i ) := {p i } od; δ E (p n-1, a n ) = P; Q E := Q E {p 1, p 2,, p n-1 } where Q E {p 1, p 2,, p n-1 } 9/13/12 Kwang-Moo Choe 26

27 The following statements are equivalent, 1. A language L is regular. 2. L = L(D) for some DFA D. δ : Q Σ Q. 3. L = L(P) for some FA P with partial function δ. δ : Q Σ Q { }. 4. L = L(N) for some NFA N. δ : Q Σ 2 Q. 5. L = L(E) for some ε-nfa E. δ : Q ({ε} Σ) 2 Q. 6. L = L(X) for some XFA X. δ : Q Σ * 2 Q. 9/13/12 Kwang-Moo Choe 27