CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 4

1. CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 4 School of Innovation, Design and Engineering Mälardalen University 2012 1

2. Content Regular Expressions and Regular LanguagesNFADFA Subset Construction (Delmängdskonstruktion)FA  RE State Elimination (Tillståndseliminering)Minimizing DFA by Set Partitioning (Särskiljandealgoritmen)Grammar: Lineargrammar. Regulargrammar. Application: Compiler. LexicalAnalysis.

3. Two Basic Theoremsabout Finite State Automata Based on C Busch, RPI, Models of Computation

4. 1: KleeneTheoremNFADFA Let M be NFA accepting L. Then there exists DFA M´ that accepts L as well.For each NFA there exists corresponding DFA that accepts the same language.

5. II: Finite Language TheoremAny finite language is regular.Any finite language is FSA-acceptable. Proof: every finite language is a union of single strings which are regular because described by regular expressions Example L = {abba, abb, abab} L is expressed by the regular expression “abba|abb|abab“ FSA = Finite State Automaton = DFA/NFA

6. Some Properties of Regular Languages, Summary

7. Union: Concatenation: Are Regular Languages Star: Properties For regular languages and we will prove that:

8. Union: Concatenation: Star: Regular languages are closed under

9. Regular language Regular language NFA NFA Single final state Single final state

10. Example

11. Union (Thompson’s construction) • NFA for

12. NFA for Example

13. Concatenation (Thompson’s construction) • NFA for

14. Example • NFA for

15. Star Operation (Thompson’s construction) • NFA for

16. Example • NFA for

17. Reverse of a Regular Language

18. Theorem The reverse of a regular language is a regular language Proof idea Construct NFA that accepts : invert the transitions of the NFA that accepts

19. Proof Since is regular, there is NFA that accepts Example:

20. Invert Transitions

21. Make old initial state a final state and vice versa

22. Add a new initial state

23. is regular Resulting machine accepts

24. Regular ExpressionsandRegular Languages

25. Theorem: Regular expressions are representation of regular languages and they are equivalent. Regular expressions describe regular languages in formal language theory. They have the same expressive power as regular grammars.

26. 1. For any regular expression the language is regular Theorem - Part 1 Languages Generated by Regular Expressions Regular Languages

27. 2. For any regular language there is a regular expression with Theorem - Part 2 Languages Generated by Regular Expressions Regular Languages

28. Proof - Part 1 For any regular expression the language is regular Proof by induction on the size of

29. NFAs Induction Basis • Primitive Regular Expressions: (where ) Regular Languages

30. Assume for regular expressions and Inductive Hypothesis that and are regular languages

31. Inductive Step • We will prove: regular languages

32. By definition of regular expressions:

33. By inductive hypothesis we know: and are regular languages Regular languages are closed under union concatenation star

34. Therefore regular languages is a regular language And trivially

35. Proof – Part 2 For any regular language there is a regular expression with Proof by construction of regular expression

36. Since is regular take the NFA that accepts it Single final state

37. From construct the equivalent • Generalized Transition Graph • transition labels are regular expressions Example QED

38. Summary: Operations on Regular Expressions • RE Regular language description • a+b {a,b} • (a+b)(a+b) {aa, ab, ba, bb} • a* {, a, aa, aaa, …} • a*b {b,ab, aab, aaab, …} • (a+b)* {, a, b, aa, ab, ba, aaa, bbb…}

39. Algebraic Properties

40. Operator Precedence • 1. Kleene star • 2. Concatenation • 3. Union • allows dropping unnecessary parentheses.

41. Converting Regular Expression to aDFA

42. 6 Example: From a Regular Expression to an NFA Example : (a+b)*abb step by step construction (a+b) a  3 2  1   4 5 b

43. 7 Example: From a Regular Expression to an NFA Example : (a+b)*abb (a+b)*  a 3 2     0 1 6   4 5 b 

44. Example: From a Regular Expression to an NFA Example : (a+b)*abb (a+b)*abb  a 3 2    0  1 6 a 7 8   4 5 b b 9  b 10

45. Converting FA to a Regular ExpressionState Eliminationhttp://www.math.uu.se/~salling/Movies/FA_to_RegExpression.mov

46. , 2 1 3 Example • Add a new start (s) and a new final (f) state: • From s to the ex-starting state (1) • From each ex-final state (1,3) to f

47. s ; ; , ; , ; , 2 1 The original: 3 f Let’s remove state 1! Each combination of input/output to 1 will generate a new path once state 1 is gone

48. s (  ) , ; When state 1 is gone we must be able to make all those transitions! , 2 1 Previous: 3 , f

49. s (  ) ; , 2 1 3 f

50. s ( (   ) ) ; , , 2 1 3 f