Computational Models - Lecture 1 1

Transcription

1 Computational Models - Lecture 1 1 Handout Mode Ronitt Rubinfeld and Iftach Haitner. Tel Aviv University. February 29/ March 02, Based on frames by Benny Chor, Tel Aviv University, modifying frames by Maurice Herlihy, Brown University. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

2 Talk Outline Languages, words and alphabets Finite automata and regular languages Regular operations Sipser s book, chapter 1.1 Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

3 Part I Languages, words and alphabets Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

4 Languages, words and alphabets Definition 1 An alphabet Σ is a finite set of letters. Σ = {a, b, c,..., z} the English alphabet. Σ = {α, β, γ,..., ζ} the Greek alphabet. Σ = {0, 1} the binary alphabet. Σ = {0, 1,..., 9} the digital alphabet. Definition 2 A word (i.e., string) over Σ, is a finite sequence of letters from Σ. The collection of all strings over Σ is denoted by Σ. For {0, 1}, the binary alphabet, ε, 1, 0, , are all members of Σ. Definition 3 A language over Σ is a (possibly infinite) subset of Σ. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

5 Language Examples Modern English. Ancient Greek. All prime numbers, written using digits. A = {w {0, 1} : w has at most seventeen 0 s}. B = {0 n 1 n : n 0}. C = {w {0, 1} : w has an equal number of 0 s and 1 s}. Make sure you understand what the above notions stand for... Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

6 Part II Finite Automata Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

7 Example: A One-Way Automatic Door front pad rear pad door open when person approaches hold open until person clears don t open when someone standing behind door Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

8 The Automatic Door as DFA REAR BOTH NEITHER closed FRONT open FRONT REAR BOTH NEITHER States: OPEN CLOSED Sensor: FRONT: someone on front pad REAR: someone on rear pad BOTH: someone(s) on both pads NEITHER no one on either pad. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

9 The Automatic Door as DFA A DFA is Deterministic Finite Automata REAR BOTH NEITHER closed FRONT open FRONT REAR BOTH NEITHER neither front rear both closed closed open closed closed open closed open open open Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

10 DFA: Informal definition The machine M 1 : q 1 q q 2 3 States: q 1, q 2, and q 3. Start state: q 1 (arrow from outside ). Accept state: q 2 (double circle). State transitions: arrows tagged with letters. 0,1 Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

11 DFA: Informal definition (cont.) q 1 q q 2 3 0,1 On an input string DFA begins in start state q1 after reading each symbol, DFA makes state transition with matching label. After reading last symbol, DFA produces" output: accept if DFA is an accepting state. reject otherwise. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

12 DFA: Informal definition (cont..) q 1 q q 2 3 0,1 What happens on the following input strings: In general?! Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

13 DFA: Informal definition (cont...) q 1 q q 2 3 This DFA accepts All input strings that end with a 1 All input strings that contain at least one 1, and end with an even number of 0 s No other strings Proof:? 0,1 Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

14 DFA - Formal Definition Definition 4 A deterministic finite automaton (DFA) is a 5-tuple (Q, Σ, δ, q 0, F), where Q is a finite set called the states Σ is a finite set called the alphabet δ : Q Σ Q is the transition function q 0 Q is the start state F Q is the set of accept states Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

15 Back to M q 1 q q 2 3 0,1 M 1 = (Q, Σ, δ, q 1, F) where Q = {q 1, q 2, q 3 }, Σ = {0, 1}, the transition function δ is q 1 is the start state F = {q 2 }. 0 1 q 1 q 1 q 2 q 2 q 3 q 2 q 3 q 2 q 2 Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

16 Another Example a s b a q 1 r 1 b b a a b b q 2 r 2 a Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

17 Formal Model of Computation Definition 5 M = (Q, Σ, δ, q 0, F) accepts w Σ if δ M (q 0, w) F. Definition 6 ( δ) δ M : Q Σ Q is defined by δ M (q, w) = { δ( δ(q, w1,...,n 1 ), w n ), n = w 1 q, w = ε.. w 1,...,k stands for the word w 1,..., w k the k-letter prefix of w (w 1,...,0 = w = ε) Note that δ M (q, σ) = δ(q, σ) for σ Σ. We write δ when M is clear from the context. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

18 An equivalent definition Definition 7 (Equivalent definition) M = (Q, Σ, δ, q 0, F) accepts w = w 1 w 2... w n, if r 0,..., r n Q s.t., r 0 = q 0. δ(r i, w i+1 ) = r i+1, for all 0 i < n. r n F. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

19 The language of a DFA Definition 8 L(M), the language of a DFA M, is the set of strings that M accepts. M may accept many strings M accepts only one language. What language does M accept if it accepts no strings? Definition 9 A language is called regular, if some deterministic finite automaton accepts it. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

20 Example: DFA M 2 Q = {q 1, q 2 }, Σ = {0, 1}, F = {q 2 }, δ =? What is L(M 2 ) := {w {0, 1} : δ(q 1, w) = q 2 }? Theorem 10 L(M 2 ) = {w Σ : # 1 (w) is odd}. # 1 (w) number of ones in w. Proof by induction on the word length What does it mean? The j th assumption: w {0, 1} j is in L(M) iff # 1 (w) is odd Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

21 Proving L(M 2 ) = L := {w Σ : # 1 (w) is odd} Basis (length 0): ε / L and δ(q 1, ɛ) = q 1 = ɛ / L(M 2 ). Step: assume hypothesis holds for words of length j 0. Let x = yσ {0, 1} j+1 for σ {0, 1} If # 1 (y) is even. By assumption δ(q 1, y) = q 1. σ = 1 = 1. # 1 (x) is odd = x L 2. δ(q1, x) := δ( δ(q 1, y), σ) = δ(q 1, 1) = q 2 = x L(M 2 ). σ = 0 = 1. # 1 (x) is even = x / L 2. δ(q1, x) := δ( δ(q 1, y), σ) = δ(q 1, 0) = q 1 = x / L(M 2 ). If # 1 (y) is odd... We proved for x {0, 1} j+1 : x L(M 2 ) x L (hence, y = j). Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

22 The language of M q 1 q q 2 3 Theorem 11 L(M 1 ) = {w10 2k : k 0, w {0, 1} } 0,1 Proof: Claim 12 (implies the theorem) Let L i = {x {0, 1} : δ(q 1, x) = q i } and let L 1 = {0 k : k 0} L 2 = {w10 2k : k 0, w {0, 1} } L 3 = {w10 2k+1 : k 0, w {0, 1} } Then, L i = L i for every i {1, 2, 3} Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

23 Proving Claim 12 We need to prove that i {1, 2, 3}: x L i x L i. Recall that x L i δ(x) = q i Proof by induction on word length. Induction basis: Easy to see that hypothesis holds for ɛ. Induction step: Assume hypothesis holds for words of length j 0. Let x = yσ {0, 1} j+1 for σ {0, 1}. We prove the hypothesis for x, separately for each i {1, 2, 3} Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

24 L 1 = {0 k : k 0} Recall x = yσ {0, 1} j+1 for σ {0, 1}. Proving: x L 1 = δ(q 1, x) = q 1 ( = x L 1 ). x = 0 j+1, y = 0 j and σ = 0. Since y L 1, by i.h. δ(q 1, y) = q 1 Therefore, δ(q 1, x) = δ( δ(q, y), σ) = δ(q 1, 0) = q 1. Proving: δ(q 1, x) = q 1 = x L 1. Let q y = δ(q 1, y) (hence, δ(q 1, x) = δ(q y, σ) = q 1 ) q y = q 1 and σ = 0. (?) By i.h. y = 0 j. Hence, x = yσ = 0 j 0 = 0 j+1 L 1. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

25 L 2 = {w10 2k : k 0, w {0, 1} } Recall x = yσ {0, 1} j+1 for σ {0, 1}. Proving x L 2 = δ(q 1, x) = q 2. Assume σ = 1 Since δ(q i, 1) = q 2 for any i = δ(q 1, x) = q 2. Assume x = w10 2k for k > 0 ( = y = w10 2k 1 and σ = 0) Hence, y L 3. By i.h. δ(q 1, y) = q 3 Thus, δ(q 1, x) = δ(q 3, 0) = q 2. Proving δ(q 1, x) = q 2 = x L 2. Assume σ = 1. = x L 2. (?) Assume σ = 0 q y := δ(q 1, y) = q 3 By i.h. y = w10 2k+1 for some k 0 Therefore x = yσ = w10 2k+1 0 L 2. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

26 L 3 = {w10 2k+1 : k 0, w {0, 1} } Recall x = yσ {0, 1} j+1 for σ {0, 1}. Proving x L 3 = δ(q 1, x) = q 3. x = w10 2k+1, y = w10 2k and σ = 0 y L 2 By i.h. δ(q 1, y) = q 2. Therefore, δ(q 1, x) = δ(q 2, 0) = q 3. Proving δ(q 1, x) = q 3 = x L 3. Let q y = δ(q 1, y) Hence, q y = q 2 and σ = 0 (?) By i.h. y = w10 2k Therefore, x = yσ = w10 2k 0 L 3. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

27 Part III Regular Operations Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

28 Additional examples of regular languages Let Σ = {0, 1}. {w {0, 1} : # 1 (w) 0 mod 7}. Sequence of 0 followed by sequence of 1, i.e., {0 m 1 n : m, n 0}. Any finite language. All the above languages are regular Is there a simple proof? Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

29 The regular operations Let A and B be languages. The union operation: A B = {x : x A x B} The concatenation operation: A B = {xy : x A y B} The star operation: A = {x 1 x 2... x k : k 0 and each x i A} Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

30 The regular operations Examples Let A= {good, bad} and B = {boy, girl}. Union Concatenation A B = {good, bad, boy, girl} A B = {goodboy, goodgirl, badboy, badgirl} Star A = {ε, good, bad, goodgood, goodbad, badbad, badgood,...} Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

31 Closure under union Theorem 13 If L 1 and L 2 are regular languages, then so is L 1 L 2. Approach to Proof: Some DFA M 1 accepts L 1 Some DFA M 2 accepts L 2 Construct DFA M that accepts L 1 L 2. Attempted Proof Idea: first emulate M 1, and if M 1 doesn t accept, then emulate M 2. What s wrong with this? Fix: Emulate both machines simultaneously. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

32 Closure Under Union: Correct Proof Suppose M 1 = (Q 1, Σ, δ 1, q 1, F 1 ) accepts L 1, M 2 = (Q 2, Σ, δ 2, q 2, F 2 ) accepts L 2. Define M as follows (M will accept L 1 L 2 ): Q = Q 1 Q 2. Σ is the same. For each (r 1, r 2 ) Q and a Σ, δ((r 1, r 2 ), a) = (δ 1 (r 1, a), δ 2 (r 2, a)) q 0 = (q 1, q 2 ) F = {(r 1, r 2 ): r 1 F 1 or r 2 F 2 }. Formal proof (next slide) (hey, why not choose F = F 1 F 2?) Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

33 Correctness of the construction Claim 14 L(M) = L(M 1 ) L(M 2 ). Follows by the next claim. Claim 15 δ M ((q 1, q 2 ), x) = ( δ 1 (q 1, x), δ 2 (q 2, x)). Proof: By induction on word length. DIY... Proving Claim 14: x L(M 1 ) = δ 1 (q 1, x) = r 1 F 1. (similar if x L(M 2 ).) By Claim 15, δ M ((q 1, q 2 ), x) = (r 1, ) F = x L(M). x L(M) = δ M ((q 1, q 2 ), x) = (r 1, r 2 ) F. By Claim 15, (r 1, r 2 ) = ( δ 1 (q 1, x), δ 2 (q 2, x)). Hence, either r 1 F 1 or r 2 F 2 = x L(M 1 ) L(M 2 ). Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

34 What about concatenation? Theorem 16 If L 1, L 2 are regular languages, then so is L 1 L 2. Example: L 1 = {good, bad} and L 2 = {boy, girl}. L 1 L 2 = {goodboy, goodgirl, badboy, badgirl} This is much harder to prove. Idea: Simulate M 1 for a while, then switch to M 2. Problem: But when do you switch? This leads us into non-determinism, wait for next class... Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

35 Part IV Non-deterministic Finite Automata Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

36 NFA non-deterministic Finite Automata 0,1 0,1 0 0 q q 1 q q 4 May have more than one transition labeled with the same symbol, May have no transitions labeled with a certain symbol, May have transitions labeled with ε, the symbol of the empty string. Will deal with this latter Every DFA is also an NFA. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

37 Non-deterministic computation 0,1 0,1 0 0 q q 1 q q 4 What happens when more than one transition is possible? The machine splits into multiple copies Each branch follows one possibility Together, branches follow all possibilities. If the input doesn t appear, that branch dies. Automaton accepts if some branch accepts. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

38 Computation on ,1 0,1 0 0 q q 1 q q 4 symbol 1 q 1 0 q 1 0 q 1 q 2 1 q q 1 q 2 3 q 1 q 4 Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

39 Why non-determinism? Theorem 17 (Informal, to be proved soon) Deterministic and non-deterministic finite automata, accept exactly the same set of languages. Q.: So why do we need NFA s? Design a finite automaton for the language L all binary strings with a 1 in their third-to-the-last position? Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

40 NFA for L 0,1 1 0,1 q q 1 q 2 3 0,1 q 4 Guesses which symbol is third from the last, and checks that indeed it is a 1. If guess is premature, that branch dies, and no harm occurs. Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41

41 DFA for L Have 8 states, encoding the last three observed letters. A state for each string in {0, 1} 3. Add transitions on modifying the suffix, give the new letter. Mark as accepting, the strings q q 100 q 010 q q 001 q q q DFA has few bugs... Ronitt Rubinfeld and Iftach Haitner (TAU) Computational Models, Lecture 1 February 29/ March 02, / 41