1. Fall 2006Costas Busch - RPI1 Reductions

2. Fall 2006Costas Busch - RPI2 Problem is reduced to problem If we can solve problem then we can solve problem

3. Fall 2006Costas Busch - RPI3 Language is reduced to language There is a computable function (reduction) such that: Definition:

4. Fall 2006Costas Busch - RPI4 Computable function : which for any string computes There is a deterministic Turing machine Recall:

5. Fall 2006Costas Busch - RPI5 If: a: Language is reduced to b: Language is decidable Then: is decidable Theorem: Proof: Basic idea: Build the decider for using the decider for

6. Fall 2006Costas Busch - RPI6 Decider for Decider for compute accept reject accept reject (halt) Input string END OF PROOF Reduction YES NO

7. Fall 2006Costas Busch - RPI7 Example: is reduced to:

8. Fall 2006Costas Busch - RPI8 Turing Machine for reduction DFA We only need to construct:

9. Fall 2006Costas Busch - RPI9 Let be the language of DFA construct DFA by combining and so that: DFA Turing Machine for reduction

10. Fall 2006Costas Busch - RPI10

11. Fall 2006Costas Busch - RPI11 Decider Decider for compute Input string YES NO Reduction

12. Fall 2006Costas Busch - RPI12 If: a: Language is reduced to b: Language is undecidable Then: is undecidable Theorem (version 1): (this is the negation of the previous theorem) Proof: Using the decider for build the decider for Suppose is decidable Contradiction!

13. Fall 2006Costas Busch - RPI13 Decider for Decider for compute accept reject accept reject (halt) Input string Reduction END OF PROOF If is decidable then we can build: CONTRADICTION! YES NO

14. Fall 2006Costas Busch - RPI14 Observation: In order to prove that some language is undecidable we only need to reduce a known undecidable language to

15. Fall 2006Costas Busch - RPI15 State-entry problem Input: Turing Machine State Question:Does String enter state while processing input string ? Corresponding language:

16. Fall 2006Costas Busch - RPI16 Theorem: (state-entry problem is unsolvable) Proof: Reduce (halting problem) to (state-entry problem) is undecidable

17. Fall 2006Costas Busch - RPI17 Decider for YES NO state-entry problem decider DeciderCompute Reduction YES NO Given the reduction, if is decidable, then is decidable A contradiction! since is undecidable Halting Problem Decider

18. Fall 2006Costas Busch - RPI18 Compute Reduction We only need to build the reduction: So that:

19. Fall 2006Costas Busch - RPI19 halting states special halt state Construct from : A transition for every unused tape symbol of

20. Fall 2006Costas Busch - RPI20 halts on statehalts halting states special halt state

21. Fall 2006Costas Busch - RPI21 halts on state on input halts on inputTherefore: Equivalently: END OF PROOF

22. Fall 2006Costas Busch - RPI22 Blank-tape halting problem Input:Turing Machine Question:Doeshalt when started with a blank tape? Corresponding language:

23. Fall 2006Costas Busch - RPI23 Theorem: (blank-tape halting problem is unsolvable) Proof: Reduce (halting problem) to (blank-tape problem) is undecidable

24. Fall 2006Costas Busch - RPI24 Decider for YES NO blank-tape problem decider DeciderCompute Reduction YES NO Given the reduction, If is decidable, then is decidable A contradiction! since is undecidable Halting Problem Decider

25. Fall 2006Costas Busch - RPI25 Compute Reduction We only need to build the reduction: So that:

26. Fall 2006Costas Busch - RPI26 no yes Write on tape Tape is blank? Run with input Construct from : If halts then halt

27. Fall 2006Costas Busch - RPI27 halts when started on blank tape halts on input no yes Write on tape Tape is blank? Run with input

28. Fall 2006Costas Busch - RPI28 END OF PROOF halts when started on blank tape halts on input Equivalently:

29. Fall 2006Costas Busch - RPI29 If: a: Language is reduced to b: Language is undecidable Then: is undecidable Theorem (version 2): Proof: Using the decider for build the decider for Suppose is decidable Contradiction! Then is decidable

30. Fall 2006Costas Busch - RPI30 Suppose is decidable Decider for accept reject (halt)

31. Fall 2006Costas Busch - RPI31 Suppose is decidable Decider for accept reject (halt) Then is decidable (we have proven this in previous class) reject accept (halt) Decider for NO YES NO

32. Fall 2006Costas Busch - RPI32 Decider for Decider for compute accept reject accept reject (halt) Input string Reduction If is decidable then we can build: CONTRADICTION! YES NO

33. Fall 2006Costas Busch - RPI33 Decider for Decider for compute accept reject accept reject (halt) Input string Reduction END OF PROOF CONTRADICTION! Alternatively: NO YES NO

34. Fall 2006Costas Busch - RPI34 Observation: In order to prove that some language is undecidable we only need to reduce some known undecidable language to or to (theorem version 1) (theorem version 2)

35. Fall 2006Costas Busch - RPI35 Undecidable Problems for Turing Recognizable languages is empty? is regular? has size 2? Let be a Turing-acceptable language All these are undecidable problems

36. Fall 2006Costas Busch - RPI36 is empty? is regular? has size 2? Let be a Turing-acceptable language

37. Fall 2006Costas Busch - RPI37 Empty language problem Input: Turing Machine Question:Isempty? Corresponding language:

38. Fall 2006Costas Busch - RPI38 Theorem: (empty-language problem is unsolvable) is undecidable Proof: Reduce (membership problem) to (empty language problem)

39. Fall 2006Costas Busch - RPI39 Decider for YES NO empty problem decider DeciderCompute Reduction YES NO Given the reduction, if is decidable, then is decidable membership problem decider A contradiction! since is undecidable

40. Fall 2006Costas Busch - RPI40 Compute Reduction We only need to build the reduction: So that:

41. Fall 2006Costas Busch - RPI41 skip input string write on tape run on input yes then accept If Tape of input string accepts ? Construct from :

42. Fall 2006Costas Busch - RPI42 skip input string write on tape run on input yes then accept If accepts ? Tape of

43. Fall 2006Costas Busch - RPI43 skip input string write on tape run on input yes then accept If accepts ? Tape of

44. Fall 2006Costas Busch - RPI44 skip input string write on tape run on input yes then accept If accepts ? During this phase this area is not touched working area of

45. Fall 2006Costas Busch - RPI45 skip input string write on tape run on input yes then accept If accepts ? altered Simply check if entered an accept state

46. Fall 2006Costas Busch - RPI46 skip input string write on tape run on input yes then accept If accepts ? Now check input string

47. Fall 2006Costas Busch - RPI47 skip input string write on tape run on input yes then accept If accepts ? The only possible accepted string t r o y

48. Fall 2006Costas Busch - RPI48 skip input string write on tape run on input yes then accept If accepts ? accepts does not accept

49. Fall 2006Costas Busch - RPI49 Therefore: accepts Equivalently: END OF PROOF

50. Fall 2006Costas Busch - RPI50 is empty? is regular? has size 2? Let be a Turing-acceptable language

51. Fall 2006Costas Busch - RPI51 Regular language problem Input: Turing Machine Question:Isa regular language? Corresponding language:

52. Fall 2006Costas Busch - RPI52 Theorem: (regular language problem is unsolvable) is undecidable Proof: Reduce (membership problem) to (regular language problem)

53. Fall 2006Costas Busch - RPI53 Decider for YES NO regular problem decider DeciderCompute Reduction YES NO Given the reduction, If is decidable, then is decidable membership problem decider A contradiction! since is undecidable

54. Fall 2006Costas Busch - RPI54 Compute Reduction We only need to build the reduction: So that:

55. Fall 2006Costas Busch - RPI55 skip input string write on tape run on input yes then accept If has the form Tape of input string accepts ? Construct from :

56. Fall 2006Costas Busch - RPI56 skip input string write on tape run on input yes accepts ? accepts does not accept then accept If has the form not regular regular

57. Fall 2006Costas Busch - RPI57 Therefore: accepts Equivalently: END OF PROOF is not regular

58. Fall 2006Costas Busch - RPI58 is empty? is regular? has size 2? Let be a Turing-acceptable language

59. Fall 2006Costas Busch - RPI59 Does have size 2? Size2 language problem Input: Turing Machine Question: Corresponding language: (accepts exactly two strings?)

60. Fall 2006Costas Busch - RPI60 Theorem: (regular language problem is unsolvable) is undecidable Proof: Reduce (membership problem) to (size 2 language problem)

61. Fall 2006Costas Busch - RPI61 Decider for YES NO size2 problem decider DeciderCompute Reduction YES NO Given the reduction, If is decidable, then is decidable membership problem decider A contradiction! since is undecidable

62. Fall 2006Costas Busch - RPI62 Compute Reduction We only need to build the reduction: So that:

63. Fall 2006Costas Busch - RPI63 skip input string write on tape run on input yes then accept If Tape of input string accepts ? Construct from :

64. Fall 2006Costas Busch - RPI64 skip input string write on tape run on input yes accepts ? accepts does not accept 2 strings 0 strings then accept If

65. Fall 2006Costas Busch - RPI65 Therefore: accepts Equivalently: END OF PROOF has size 2

66. Fall 2006Costas Busch - RPI66 RICE’s Theorem is empty? is regular? has size 2? Undecidable problems: This can be generalized to all non-trivial properties of Turing-acceptable languages

67. Fall 2006Costas Busch - RPI67 Non-trivial property: A property possessed by some Turing-acceptable languages but not all : is empty? Example: YES NO

68. Fall 2006Costas Busch - RPI68 : is regular? More examples of non-trivial properties: YES NO YES : has size 2? NO YES NO

69. Fall 2006Costas Busch - RPI69 Trivial property: A property possessed by ALL Turing-acceptable languages : has size at least 0?Examples: True for all languages : is accepted by some Turing machine? True for all Turing-acceptable languages

70. Fall 2006Costas Busch - RPI70 We can describe a property as the set of languages that possess the property : is empty? Example: YES NO If language has property then

71. Fall 2006Costas Busch - RPI71 : has size 1? NO YES Example:Suppose alphabet is NO

72. Fall 2006Costas Busch - RPI72 Non-trivial property problem Does have the non-trivial property ? Input: Turing Machine Question: Corresponding language:

73. Fall 2006Costas Busch - RPI73 Rice’s Theorem: is undecidable (the non-trivial property problem is unsolvable) Proof: Reduce (membership problem) to or

74. Fall 2006Costas Busch - RPI74 We examine two cases: Case 1: Case 2: Examples: : is empty? : is regular? : has size 2? Example:

75. Fall 2006Costas Busch - RPI75 Let be the Turing machine that accepts Case 1: Since is non-trivial, there is a Turing-acceptable language such that:

76. Fall 2006Costas Busch - RPI76 Reduce (membership problem) to

77. Fall 2006Costas Busch - RPI77 Decider for YES NO Non-trivial property problem decider DeciderCompute Reduction YES NO Given the reduction, if is decidable, then is decidable membership problem decider A contradiction! since is undecidable

78. Fall 2006Costas Busch - RPI78 Compute Reduction We only need to build the reduction: So that:

79. Fall 2006Costas Busch - RPI79 skip input string write on tape run on input yes then accept If Tape of input string accepts ? Construct from :

80. Fall 2006Costas Busch - RPI80 skip input string write on tape run on input yes then accept If accepts ? For this phase we can run machine that accepts, with input string

81. Fall 2006Costas Busch - RPI81 skip input string write on tape run on input yes accepts ? accepts does not accept then accept If

82. Fall 2006Costas Busch - RPI82 Therefore: accepts Equivalently:

83. Fall 2006Costas Busch - RPI83 Let be the Turing machine that accepts Case 2: Since is non-trivial, there is a Turing-acceptable language such that:

84. Fall 2006Costas Busch - RPI84 Reduce (membership problem) to

85. Fall 2006Costas Busch - RPI85 Decider for YES NO Non-trivial property problem decider DeciderCompute Reduction YES NO Given the reduction, if is decidable, then is decidable membership problem decider A contradiction! since is undecidable

86. Fall 2006Costas Busch - RPI86 Compute Reduction We only need to build the reduction: So that:

87. Fall 2006Costas Busch - RPI87 skip input string write on tape run on input yes then accept If Tape of input string accepts ? Construct from :

88. Fall 2006Costas Busch - RPI88 skip input string write on tape run on input yes accepts ? accepts does not accept then accept If

89. Fall 2006Costas Busch - RPI89 Therefore: accepts Equivalently: END OF PROOF