Finite Automata and Regular languages

Transcription

1 Finite Automata and Regular languages Huan Long Shanghai Jiao Tong University

2 Acknowledgements Part of the slides comes from a similar course in Fudan University given by Prof. Yijia Chen. chen/ chen/teaching/toc/ Textbook Introduction to the theory of computation Michael Sipser, MIT Third edition, 2012

3 Outline Finite automata and regular language Nondeterminism automata Equivalence of DFA and NFA Regular expression Pumping lemma for regular languages Some decision problems related to FA

4 Finite Automata Definition A deterministic finite automata (DFA) is a 5-tuple (Q, Σ, δ, q 0, F ) where 1. Q is a finite set called the states, 2. Σ is a finite set called the alphabet, 3. δ : Q Σ Q is the transition function, 4. q 0 Q is the start state, 5. F Q is the set of accept states.

5 Computation by DFA Let M = (Q, Σ, δ, q 0, F ) be a DFA and let w = w 1 w 2 w n be a string with w i Σ for all i [n]. Then M accepts w if there exists a sequence of states r 0, r 1,..., r n in Q such that: 1. r 0 = q 0, 2. δ(r i, w i+1 ) = r i+1 for i = 0,..., n 1, 3. r n F. For a set A, we say that M recognizes A if A = {l M accepts l}

6 Regular languages Definition A language is called regular if some finite automata recognizes it.

7 The regular operators Definition Let A and B be languages. We define the following three regular operations: Union: A B = {x x A x B} Concatenation: A B = {xy x A y B} Kleene star: A = {x 1 x 2... x k k 0 x i A}

8 Nondeterminism Definition A nondeterministic finite automata (NFA) is a 5-tuple (Q, Σ, δ, q 0, F ) where 1. Q is a finite set called the states, 2. Σ is a finite set called the alphabet, 3. δ : Q Σ ɛ P(Q) is the transition function, where Σ ɛ = Σ {ɛ}, 4. q 0 Q is the start state, 5. F Q is the set of accept states.

9 Computation by NFA Let N = (Q, Σ, δ, q 0, F ) be a NFA and let w = y 1 y 2 y m be a string with y i Σ ɛ for all i [m]. Then N accepts w if there exists a sequence of states r 0, r 1,..., r n in Q such that: 1. r 0 = q 0, 2. r i+1 δ(r i, y i+1 ) for i = 0,..., m 1, 3. r n F.

10 Equivalence of NFAs and DFAs Theorem Every NFA has an equivalent DFA, i.e., they recognize the same language.

11 Proof (1) NFA: N = (Q, Σ, δ, q 0, F ) Main idea: view a NFA as occupying a set of states at any moment. Step 1: For any state q Q, compute its silently reachable class E(q): initially set E(q) = {q}; repeat E (q) = E(q) x E(q), if y δ(x, ɛ) y E(q), E(q) = E(q) {y} until E(q) = E (q) return E(q).

12 Proof (2) Step 2: build the equivalent DFA NFA: N = (Q, Σ, δ, q 0, F ) DFA: M = (Q, Σ, δ, q 0, F ) 1. Q = P(Q); 2. Let R Q and a Σ, define δ (R, a) = {E(q) q Q ( r R)(q σ(r, a))} ; 3. q 0 = {q 0}; 4. F = {R Q R F }.

13 Corollary A language is regular if and only if some NFA recognizes it.

14 Recall: regular operators Definition Let A and B be languages. We define the following three regular operations: Union: A B = {x x A x B} Concatenation: A B = {xy x A y B} Kleene star: A = {x 1 x 2... x k k 0 x i A}

15 Closure under the regular operators Theorem The class of regular languages is closed under the,, operations. Proof. Let N 1 = (Q 1, Σ 1, δ 1, q 1, F 1 ) recognize A 1, N 2 = (Q 2, Σ 2, δ 2, q 2, F 2 ) recognize A 2 ; We will build NFAs which recognize A 1 A 2, A 1 A 2, A 1 respectively.

16 I. Closure under union : A 1 A 2 is regular N 1 = (Q 1, Σ 1, δ 1, q 1, F 1 ) recognize A 1, N 2 = (Q 2, Σ 2, δ 2, q 2, F 2 ) recognize A 2 ; Define the NFA as: 1. Q = {q 0 } Q 1 Q 2 ; 2. q 0 is the new start state; 3. F = F 1 F 2 ; 4. For any q Q and any a Σ ɛ {q 1, q 2 } q = q 0 a = ɛ q = q δ(q, a) = 0 a ɛ δ 1 (q, a) q Q 1 δ 2 (q, a) q Q 2

17 II. Closure under concatenation : A 1 A 2 is regular N 1 = (Q 1, Σ 1, δ 1, q 1, F 1 ) recognize A 1, N 2 = (Q 2, Σ 2, δ 2, q 2, F 2 ) recognize A 2 ; Define the NFA as: 1. Q = Q 1 Q 2 ; 2. the start state is q 1 ; 3. the set of accept states is F 2 ; 4. For any q Q and any a Σ ɛ δ 1 (q, a) q Q 1 F 1 δ δ(q, a) = 1 (q, a) q F 1 a ɛ δ 1 (q, a) {q 2 } q F 1 a = ɛ δ 2 (q, a) q Q 2

18 III. Closure under Kleene star : A 1 is regular N 1 = (Q 1, Σ, δ 1, q 1, F 1 ) recognize A 1 ; Define the NFA as: 1. Q = {q 0 } Q 1 ; 2. the new start state is q 0 ; 3. F = {q 0 } F 1 ; 4. For any q Q and any a Σ ɛ δ 1 (q, a) q Q 1 F 1 δ 1 (q, a) q F 1 a ɛ δ(q, a) = δ 1 (q, a) {q 1 } q F 1 a = ɛ {q 1 } q = q 0 a = ɛ q = q 0 a ɛ

19 Regular expression Given alphabet Σ, we say that R is a regular expression if R is 1. a for some a Σ, 2. ɛ, 3., 4. (R 1 R 2 ), where R 1 and R 2 are regular expressions, 5. (R 1 R 2 ), where R 1 and R 2 are regular expressions, 6. (R 1 ), where R 1 is a regular expression. Sometimes, we use R 1 R 2 instead of (R 1 R 2 ) if no confusion arises.

20 Other closure property Given N = (Q, Σ, δ, q, F ) the set of language recognized by N is A, then Complement: A = Σ A Intersection: A B = {x x A x B} Lemma The class of regular languages is closed under complementation and intersection. Proof. w.l.o.g, N is a DFA, then N = (Q, Σ, δ, q, Q F ) will recognize A. A B = A B.

21 Language defined by regular expressions regular expression R language L(R) a {a} ɛ {ɛ} (R 1 R 2 ) L(R 1 ) L(R 2 ) (R 1 R 2 ) L(R 1 ) L(R 2 ) (R1 ) L(R 1)

22 Equivalence with finite automata Theorem A language is regular if and only if some regular expression describes it. Proof. Only if: build the NFAs; (relatively easy) If: Automata = regular expressions. Sketch: (Dynamic programming) R(i, j, k) = R(i, j, k 1) R(i, k, k 1)R(k, k, k 1) R(k, j, k 1) L(M) = {R(1, j, n) q j F }.

23 Languages need counting L 1 = {l {0, 1} l has an equal number of 0s and 1s}. L 2 = {l {0, 1} l has an equal number of occurrences of 01 and 10 as substrings}. L 2 is regular; L 1 is or is not regular? It is not regular!

24 The pumping lemma for regular languages Lemma If A is a regular language, then there is a number p (i.e., the pumping length where if s is any string in A of length at least p, then s my be divided into three pieces, s = xyz, satisfying the following conditions: 1. y > 0, 2. xy p, 3. for each i 0, we have xy i z A. Any string xyz in A can be pumped along y.

25 Proof Pigeonhole principle Let M = (Q, Σ, δ, q 1, F ) be a DFA recognizing A and p = Q. Let s = s 1 s 2 s n be a string in A with n p. Let r 1,, r n+1 be the sequence of states that M enters while processing s, i.e., r i+1 = δ(r i, s i ) for i [n]. Among the first p + 1 states in the sequence, two must be the same, say r j and r k with j < k p + 1. Define x = s 1 s j 1, y = s j s k 1, z = s k s n.

26 Example (1) The language L = {0 n 1 n n 0} is not regular. Proof. If it is regular, choose p be the pumping length and consider s = 0 p 1 p.by the Pumping lemma, s = xyz with xy i z L for all i 0. As xy p and y > 0, y = 0 i for some i > 0. But then xz = a n i b n L. Contradicting the lemma.

27 Example (2) The language L = {w w has an equal number of 0s and 1s} is not regular. Proof. If it is regular, then L 0 1 would also be regular. However, this latter language is precisely the language in Example (1), which is not regular.

28 Problems from formal language theory Decision Problems Acceptance: does a given string belong to a given language? Emptiness: is a given language empty? Equality: are given two languages equal?

29 Language Problems concerning FA Theorem The following three problems: Acceptance: Given a DFA (NFA) A and a string w, does A accept w? Emptiness: Given a DFA (NFA) A is the language L(A) empty? Equality: Given two DFA (NFA) A and B is L(A) equal to L(B)? The corresponding decision problems are all decidable. Proof.