Great Theoretical Ideas in Computer Science. Lecture 4: Deterministic Finite Automaton (DFA), Part 2

Transcription

1 5-25 Great Theoretical Ideas in Computer Science Lecture 4: Deterministic Finite Automaton (DFA), Part 2 January 26th, 27

2 Formal definition: DFA A deterministic finite automaton (DFA) M =(Q,,,q,F) M is a 5-tuple Q = {q,q,q 2,q 3 } ={, } : Q! Q q q q q q 2 q 2 q 2 q 3 q 2 q 3 q q 2 q is the start state F = {q,q 2 }

3 Formal definition: DFA accepting a string Let M =(Q,,,q,F) be a DFA. For q 2 Q, w 2, (q, w) = state we end up in when we start at and read w. q M accepts w if (q,w) 2 F. Otherwise M rejects w.

4 Definition: Regular languages Let M be a DFA. We let L(M) denote the set of strings that M accepts. Definition: A language L = L(M) L is called regular if for some DFA M.

5 Non-regular languages Theorem: The language L = { n n : n 2 N} is not regular. Theorem: The language L = {a 2n : n 2 N} is not regular.

6 The big picture All languages P( ) Regular languages { n n : n 2 N} {a 2n : n 2 N}..

7 Regular languages Questions:. Are all languages regular? (Are all decision problems computable by a DFA?) 2. Are there other ways to tell if a language is regular?

8 Closure properties of regular languages

9 Closed under complementation Proposition: Let be some finite alphabet. If L is regular, then so is L = \L. Proof: If L is regular, then there is a DFA recognizing L. M =(Q,,,q,F) Then M =(Q,,,q,Q\F ) recognizes L. So L is regular.

10 Proof: Closed under complementation Closure properties can be used to show languages are not regular. Corollary: If L is non-regular, then so is L. If L L = L By contrapositive: is regular, then by the previous Proposition is regular. Examples: {, } \{ n n : n 2 N} {a} \{a 2n : n 2 N} are non-regular.

11 Theorem: Let be some finite alphabet. If L and L 2 are regular, then so is L [ L 2. Proof: Let M =(Q,,,q,F) be a DFA deciding and M =(Q,,,q,F ) be a DFA deciding L 2. We construct a DFA M =(Q,,,q,F ) that decides L [ L 2, as follows:. L

12 The mindset Imagine yourself as a DFA. Rules: ) Can only scan the input once, from left to right. 2) Can only remember constant amount of information. should not change based on input length

13 Step : Imagining ourselves as a DFA

14 Example M L = strings with even number of s L 2 = strings with length divisible by 3. M 2, p p p 2,,

15 Input: M M 2, p p p 2,,

16 Input: M M 2, p p p 2,,

17 Input: M M 2, p p p 2,,

18 Input: M M 2, p p p 2,,

19 Input: M M 2, p p p 2,,

20 Input: M M 2, p p p 2,,

21 Input: M M 2, p p p 2,,

22 Input: M M 2, p p p 2,,

23 Input: M Accept M 2, p p p 2,,

24 Main idea: Construct a DFA that keeps track of both at once. M M 2, p p p 2,,

25 Main idea: Construct a DFA that keeps track of both at once. p p p 2 p p p 2

26 Main idea: Construct a DFA that keeps track of both at once. p p p 2? p p p 2

27 Main idea: Construct a DFA that keeps track of both at once. p p p 2 p p p 2

28 Main idea: Construct a DFA that keeps track of both at once. p p p 2? p p p 2

29 Main idea: Construct a DFA that keeps track of both at once. p p p 2 p p p 2

30 Main idea: Construct a DFA that keeps track of both at once. p p p 2? p p p 2

31 Main idea: Construct a DFA that keeps track of both at once. p p p 2 p p p 2

32 Main idea: Construct a DFA that keeps track of both at once. p p p 2? p p p 2

33 Main idea: Construct a DFA that keeps track of both at once. p p p 2 p p p 2

34 Closed under union Main idea: Construct a DFA that keeps track of both at once. p p p 2 p p p 2

35 Input: p p p 2 p p p 2

36 Input: p p p 2 p p p 2

37 Input: p p p 2 p p p 2

38 Input: p p p 2 p p p 2

39 Input: p p p 2 p p p 2

40 Input: p p p 2 p p p 2

41 Input: p p p 2 p p p 2

42 Input: p p p 2 p p p 2

43 Input: p p p 2 p p p 2

44 Input: p p p 2 p p p 2

45 Input: p p p 2 p p p 2

46 Input: p p p 2 p p p 2

47 Input: p p p 2 p p p 2

48 Input: Decision: Accept p p p 2 p p p 2

49 Step 2: Formally defining the DFA

50 Proof: Let M =(Q,,,q,F) be a DFA deciding L and M =(Q,,,q,F ) be a DFA deciding L 2. We construct a DFA M =(Q,,,q,F ) that decides L [ L 2, as follows: - Q = Q Q = {(q, q ):q 2 Q, q 2 Q } - ((q, q ),a)=( (q, a), (q,a)) - q =(q,q ) - F = {(q, q ):q 2 F or q 2 F } It remains to show that. L(M )=L [ L 2 L(M ) L [ L 2 :... L [ L 2 L(M ):...

51 Closed under intersection Corollary: Let be some finite alphabet. If L and L 2 are regular, then so is L \ L 2. Proof: Follows from: - L \ L 2 = L [ L 2 - regular languages are closed under complementation - regular languages are closed under union

52 Closed under intersection Closure properties can be used to show languages are not regular. Example: Let L {, } be the language consisting of all words with an equal number of s and s. We claim L is not regular. Suppose it was regular. { n m : n, m 2 N} \ L = { n n : n 2 N} regular regular regular contradiction

53 More closure properties Closed under union: L,L 2 regular =) L [ L 2 regular. L [ L 2 = {x 2 : x 2 L or x 2 L 2 } Closed under concatenation: L,L 2 regular =) L L 2 regular. L L 2 = {xy : x 2 L,y 2 L 2 } Closed under star: L regular =) L regular. L = {x x 2 x k : k, 8i x i 2 L}

54 awesome vs regular What is the relationship between awesome and regular? awesome regular In fact: awesome = regular

55 awesome = regular Theorem: Can define regular languages recursively as follows: - ; is regular. - For every a 2, {a} is regular. - L,L 2 regular =) L [ L 2 regular. - L,L 2 regular =) L L 2 regular. - regular regular. =) L L

56 Regular expressions Definition: A regular expression is defined recursively as follows: - ; is a regular expression. - is a regular expression. - For every a 2, a is a regular expression. - R,R 2 regular expr. =) (R [ R 2 ) regular expr. - R,R 2 regular expr. =) (R R 2 ) regular expr. - R regular expr. =) (R ) regular expr.

57 Regular expressions Examples: ((( [ ) )( [ ) ) = {w 2 {, } : w has at least one } {w 2 {, } : w has exactly one } [ [ [ {w 2 {, } : w starts and ends with same symbol}

58 Closed under concatenation Theorem: Let be some finite alphabet. If L and L 2 are regular, then so is L L 2.

59 The mindset Imagine yourself as a DFA. Rules: ) Can only scan the input once, from left to right. 2) Can only remember constant amount of information. should not change based on input length

60 Step : Imagining ourselves as a DFA

61 M q q q 2 q 3 q 4 M 2 q q q 2 Given w 2, we need to decide if w = uv for u 2 L,v2 L 2. Problem: don t know where u ends, v begins. When do you stop simulating M and start simulating M 2?

62 M q q q 2 q 3 q 4 M 2 q q q 2 Suppose God tells you u ends at w 3. w w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 w q q q q 3 q q2 q2 q2 q q2 q2 q thread: a simulation of M and then M 2 that corresponds to breaking up w into uv where u 2 L.

63 M q q q 2 q 3 q 4 automatic teleportation M 2 q q q 2 w w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 w w q q q q 3 q 2 q 4 q q q 3 q q q 3 q 2 thread q 2 q 2 q 2 q q 2 q 2 q q thread2 q q q q q 2 q q thread3 q q q 2 q q thread4 q q 2

64 M q q q 2 q 3 q 4 automatic teleportation M 2 q q q 2 w w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 w w q q q q 3 q 2 q 4 q q 3 q q q 3 q 2 thread q q2 q2 q2 q q2 q2 q q thread2 q q q q q2 q q thread3 q q q2 q q thread4 q q 2

65 w w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 w w q q q q 3 q 2 q 4 q q 3 q q q 3 q 2 q q 2 q 2 q 2 q q 2 q 2 q q q q q q q 2 q q q q q 2 q q q q 2 Q Q Q Q Q Q Q Q Q Q Q This keeps track of every possible thread. At any point, need to remember: - an element of Q constant amount of - a subset of Q information

66 Step 2: Formally defining the DFA

67 M =(Q,,,q,F) M 2 =(Q,,,q,F ) Q = Q P(Q ) : Q P(Q )! Q P(Q ) for q 2 Q, S 2 P(Q ), a 2 (q, S, a) q = (q, ;) if q 62 F q = (q,{q}) otherwise ( (q, a), { (s, a) :s 2 S}) if (q, a) 62 F (q, S, a) ( (q, a), { (s, a) :s 2 S} [{q} ) otherwise F = {(q, ):q 2 Q, S S 2 P(Q ), S \ F 6= ;}

68 Next Time