1. 1 Linear Bounded Automata LBAs

2. 2 Linear Bounded Automata are like Turing Machines with a restriction: The working space of the tape is the space of the input string

3. 3 Left-end marker Input string Right-end marker Working space of tape All computation is done between end markers

4. 4 We define LBA’s to be NonDeterministic Open Problem: NonDeterministic LBA’s have same power with Deterministic LBA’s ?

5. 5 Example languages accepted by LBAs: LBA’s have more power than NPDA’s

6. 6 Later in class we will prove: LBA’s have less power than Turing Machines

7. 7 A Universal Turing Machine

8. 8 Turing Machines are “hardwired” They execute only one program Limitation of Turing Machines: Real Computers are reprogrammable

9. 9 Solution: Universal Turing Machine is a reprogrammable machine simulates any other Turing Machine

10. 10 Universal Turing machine simulates any Turing machine Input of Universal Machine: Description of transitions of Initial tape contents of

11. 11 Universal Turing Machine Description of Tape Contents of State of

12. 12 Alphabet Encoding Symbols: Encoding:

13. 13 State Encoding States: Encoding: Head Move Encoding Move: Encoding:

14. 14 Transition Encoding Transition: Encoding: separator

15. 15 Machine Encoding Transitions: Encoding: separator

16. 16 Input of Universal Turing Machine: encoding of the simulated machine

17. 17 A Turing Machine is described with a string of 0’s and 1’s The set of Turing machines form a language: each string of the language is the encoding of a Turing Machine

18. 18 Countable Sets

19. 19 Infinite sets are either: Countable Uncountable

20. 20 Countable set: There is a one to one correspondence between elements of the set and positive integers

21. 21 Example: Even integers: The set of even integers is countable Positive integers: Correspondence: corresponds to

22. 22 Example:The set of rational numbers is countable Rational numbers:

23. 23 Naive Approach Rational numbers: Positive integers: Correspondence: Doesn’t work: we will never count numbers with nominator 2

24. 24 Better Approach

25. 25

26. 26

27. 27

28. 28

29. 29

30. 30 Rational Numbers: Correspondence: Positive Integers:

31. 31 We proved: the set of rational numbers is countable by giving an enumeration procedure

32. 32 Definition An enumeration procedure for is a Turing Machine that generates any string of in finite number of steps Let be a set of strings

33. 33 Enumeration Machine for Finite time: strings output

34. 34 Enumeration Machine Configuration Time 0 Time

35. 35 Time

36. 36 A set is countable if there is an enumeration procedure for it

37. 37 Example: The set of all strings is countable We will describe the enumeration procedure

38. 38 Naive procedure: Produce the strings in lexicographic order: Doesn’t work: strings starting with will never be produced

39. 39 Better procedure: Produce all strings of length 1 Produce all strings of length 2.......... Proper Order

40. 40 Produce strings: Length 2 Length 3 Length 1 Proper Order

41. 41 Theorem: The set of all Turing Machines is countable

42. 42 Theorem: The set of all Turing Machines is countable Proof: Find an enumeration procedure for the set of Turing Machine strings Any Turing Machine is encoded with a string of 0’s and 1’s

43. 43 1. Generate the next string of 0’s and 1’s in proper order 2. Check if the string defines a Turing Machine if YES: print string on output if NO: ignore string Enumeration Procedure: Repeat

44. 44 Uncountable Sets

45. 45 A set is uncountable if it is not countable Definition:

46. 46 Theorem: Let be an infinite countable set. The powerset of is uncountable

47. 47 Proof: Since is countable, we can write Element of

48. 48 Elements of the powerset have the form:

49. 49 We encode each element of the power set with a string of 0’s and 1’s Powerset element Encoding

50. 50 Let’s assume for contradiction that the powerset is countable. We can enumerate the elements of the powerset

51. 51 Powerset element Encoding

52. 52 Take the powerset element whose bits are the complements if the diagonal

53. 53 New element: (Diagonal complement)

54. 54 The new element must be some This is impossible: The i-th bit must be the complement of itself

55. 55 We have contradiction! Therefore the powerset is uncountable