1. 1 More Applications of the Pumping Lemma

2. 2 The Pumping Lemma: Given a infinite regular language there exists an integer for any string with length we can write with and such that:

3. 3 Regular languages Non-regular languages

4. 4 Theorem: The language is not regular Proof: Use the Pumping Lemma

5. 5 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

6. 6 pick Let be the integer in the Pumping Lemma Pick a string such that: length

7. 7 Write it must be that length From the Pumping Lemma

8. 8 From the Pumping Lemma: Thus: We have:

9. 9 Therefore: BUT: CONTRADICTION!!!

10. 10 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:

11. 11 Regular languages Non-regular languages

12. 12 Theorem: The language is not regular Proof: Use the Pumping Lemma

13. 13 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

14. 14 pick Let be the integer in the Pumping Lemma Pick a string such that: length

15. 15 Write it must be that length From the Pumping Lemma

16. 16 From the Pumping Lemma: Thus: We have:

17. 17 Therefore: BUT: CONTRADICTION!!!

18. 18 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:

19. 19 Regular languages Non-regular languages

20. 20 Theorem: The language is not regular Proof: Use the Pumping Lemma

21. 21 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

22. 22 pick Let be the integer in the Pumping Lemma Pick a string such that: length

23. 23 Write it must be that length From the Pumping Lemma

24. 24 From the Pumping Lemma: Thus: We have:

25. 25 Therefore: And since: There is :

26. 26 However:for for any

27. 27 Therefore: BUT: CONTRADICTION!!! and

28. 28 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:

29. 29 Lex

30. 30 Lex: a lexical analyzer A Lex program recognizes strings For each kind of string found the lex program takes an action

31. 31 Var = 12 + 9; if (test > 20) temp = 0; else while (a < 20) temp++; Lex program Identifier: Var Operand: = Integer: 12 Operand: + Integer: 9 Semicolumn: ; Keyword: if Parenthesis: ( Identifier: test.... Input Output

32. 32 In Lex strings are described with regular expressions “if” “then” “+” “-” “=“ /* operators */ /* keywords */ Lex program Regular expressions

33. 33 (0|1|2|3|4|5|6|7|8|9)+ /* integers */ /* identifiers */ Regular expressions (a|b|..|z|A|B|...|Z)+ Lex program

34. 34 integers [0-9]+(0|1|2|3|4|5|6|7|8|9)+

35. 35 (a|b|..|z|A|B|...|Z)+ [a-zA-Z]+ identifiers

36. 36 Each regular expression has an associated action (in C code) Examples: \n Regular expressionAction linenum++; [a-zA-Z]+ printf(“identifier”); [0-9]+ prinf(“integer”);

37. 37 Default action: ECHO; Prints the string identified to the output

38. 38 A small program % [a-zA-Z]+printf(“Identifier\n”); [0-9]+printf(“Integer\n”); [ \t\n] ; /*skip spaces*/

39. 39 1234 test var 566 78 9800 Input Output Integer Identifier Integer

40. 40 % [a-zA-Z]+ printf(“Identifier\n”); [0-9]+ prinf(“Integer\n”); [ \t] ; /*skip spaces*/. printf(“Error in line: %d\n”, linenum); Another program %{ int linenum = 1; %} \nlinenum++;

41. 41 1234 test var 566 78 9800 + temp Input Output Integer Identifier Integer Error in line: 3 Identifier

42. 42 Lex matches the longest input string “if” “ifend” Regular Expressions Input: ifend if ifn Matches: “ifend” “if” nomatch Example:

43. 43 Internal Structure of Lex Lex Regular expressions NFADFA Minimal DFA The final states of the DFA are associated with actions