1. Costas Busch - RPI1 Standard Representations of Regular Languages Regular Languages DFAs NFAs Regular Expressions Regular Grammars

2. Costas Busch - RPI2 When we say: We are given a Regular Language We mean:Language is in a standard representation

3. Costas Busch - RPI3 Elementary Questions about Regular Languages

4. Costas Busch - RPI4 Membership Question Question:Given regular language and string how can we check if ? Answer:Take the DFA that accepts and check if is accepted

5. Costas Busch - RPI5 DFA

6. Costas Busch - RPI6 Given regular language how can we check if is empty: ? Take the DFA that accepts Check if there is any path from the initial state to a final state Question: Answer:

7. Costas Busch - RPI7 DFA

8. Costas Busch - RPI8 Given regular language how can we check if is finite? Take the DFA that accepts Check if there is a walk with cycle from the initial state to a final state Question: Answer:

9. Costas Busch - RPI9 DFA is infinite DFA is finite

10. Costas Busch - RPI10 Given regular languages and how can we check if ? Question: Find if Answer:

11. Costas Busch - RPI11 and

12. Costas Busch - RPI12 or

13. Costas Busch - RPI13 Non-regular languages

14. Costas Busch - RPI14 Regular languages Non-regular languages

15. Costas Busch - RPI15 How can we prove that a language is not regular? Prove that there is no DFA that accepts Problem: this is not easy to prove Solution: the Pumping Lemma !!!

16. Costas Busch - RPI16 The Pigeonhole Principle

17. Costas Busch - RPI17 pigeons pigeonholes

18. Costas Busch - RPI18 A pigeonhole must contain at least two pigeons

19. Costas Busch - RPI19........... pigeons pigeonholes

20. Costas Busch - RPI20 The Pigeonhole Principle........... pigeons pigeonholes There is a pigeonhole with at least 2 pigeons

21. Costas Busch - RPI21 The Pigeonhole Principle and DFAs

22. Costas Busch - RPI22 DFA with states

23. Costas Busch - RPI23 In walks of strings:no state is repeated

24. Costas Busch - RPI24 In walks of strings:a state is repeated

25. Costas Busch - RPI25 If string has length : Thus, a state must be repeated Then the transitions of string are more than the states of the DFA

26. Costas Busch - RPI26 In general, for any DFA: String has length number of states A state must be repeated in the walk of...... walk of Repeated state

27. Costas Busch - RPI27 In other words for a string : transitions are pigeons states are pigeonholes...... walk of Repeated state

28. Costas Busch - RPI28 The Pumping Lemma

29. Costas Busch - RPI29 Take an infinite regular language There exists a DFA that accepts states

30. Costas Busch - RPI30 Take string with There is a walk with label :......... walk

31. Costas Busch - RPI31 If string has length (number of states of DFA) then, from the pigeonhole principle: a state is repeated in the walk...... walk

32. Costas Busch - RPI32...... walk Let be the first state repeated in the walk of

33. Costas Busch - RPI33 Write......

34. Costas Busch - RPI34...... Observations:lengthnumber of states of DFA length

35. Costas Busch - RPI35 The string is accepted Observation:......

36. Costas Busch - RPI36 The string is accepted Observation:......

37. Costas Busch - RPI37 The string is accepted Observation:......

38. Costas Busch - RPI38 The string is accepted In General:......

39. Costas Busch - RPI39 In General:...... Language accepted by the DFA

40. Costas Busch - RPI40 In other words, we described: The Pumping Lemma !!!

41. Costas Busch - RPI41 The Pumping Lemma: Given a infinite regular language there exists an integer for any string with length we can write with and such that:

42. Costas Busch - RPI42 Applications of the Pumping Lemma

43. Costas Busch - RPI43 Theorem: The language is not regular Proof: Use the Pumping Lemma

44. Costas Busch - RPI44 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

45. Costas Busch - RPI45 Let be the integer in the Pumping Lemma Pick a string such that: length We pick

46. Costas Busch - RPI46 it must be that length From the Pumping Lemma Write: Thus:

47. Costas Busch - RPI47 From the Pumping Lemma: Thus:

48. Costas Busch - RPI48 From the Pumping Lemma: Thus:

49. Costas Busch - RPI49 BUT: CONTRADICTION!!!

50. Costas Busch - RPI50 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:

51. Costas Busch - RPI51 Regular languages Non-regular languages