1. Reductions

2. Problem is reduced to problem
If we can solve problem
then we can solve problem

3. Definition:
Language
is reduced to
language
There is a computable
function (reduction) such that:

4. Recall:
Computable function :
There is a deterministic Turing machine
which for any string computes

5. Theorem:
If: a: Language is reduced to
b: Language is decidable
Then: is decidable
Proof:
Basic idea:
Build the decider for
using the decider for

6. Decider for Reduction compute Decider for END OF PROOF Input string
YES
Input
string
YES
accept
accept
compute
Decider
for
(halt)
(halt) NO NO
reject
reject
(halt)
(halt)
END OF PROOF

7. Example:
is reduced to:

8. We only need to construct:
Turing Machine
for reduction
DFA

9. Let be the language of DFA
Turing Machine
for reduction
DFA
construct DFA
by combining and so that:

11. Decider for
Reduction
Input
string
YES
compute
YES
Decider NO NO

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

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

14. Observation:
In order to prove
that some language is undecidable
we only need to reduce a
known undecidable language to

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

16. Theorem:
is undecidable
(state-entry problem is unsolvable)
Proof:
Reduce
(halting problem) to
(state-entry problem)

17. Decider for Reduction Compute Decider Given the reduction,
Halting Problem Decider
Decider for
state-entry problem
decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
if is decidable,
then is decidable
A contradiction!
since
is undecidable

18. We only need to build the reduction:
Compute
So that:

19. Construct from :
special
halt state
halting
states
A transition for every unused
tape symbol of

20. special
halt state
halting
states
halts
halts on state

21. Therefore: halts on input halts on state on input Equivalently:
END OF PROOF

22. Blank-tape halting problem
Input:
Turing Machine
Question:
Does
halt when started with
a blank tape?
Corresponding language:

23. Theorem: is undecidable Proof: Reduce (halting problem) to
(blank-tape halting problem is unsolvable)
Proof:
Reduce
(halting problem) to
(blank-tape problem)

24. Decider for Reduction Compute Decider Given the reduction,
Halting Problem Decider
Decider for
blank-tape problem
decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
If is decidable,
then is decidable
A contradiction!
since
is undecidable

25. We only need to build the reduction:
Compute
So that:

26. Construct from : no yes Run with input If halts then halt
Tape is blank?
yes
Run
Write on tape
with input
If halts then halt

27. halts when started on blank tapeno
Tape is blank?
yes
Run
Write on tape
with input
halts on input
halts when started on blank tape

28. halts when started on blank tape
halts on input
halts when started on blank tape
Equivalently:
END OF PROOF

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

30. Suppose is decidable
reject
Decider
for
(halt)
accept
(halt)

31. Suppose is decidable Then is decidable Decider for Decider for
(we have proven this in previous class)
Decider for NO
YES
reject
accept
Decider
for
(halt)
(halt)
YES NO
accept
reject
(halt)
(halt)

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

33. Alternatively: Decider for Reduction compute Decider forNO
Input
string
YES
reject
accept
compute
Decider
for
(halt)
(halt)
YES NO
accept
reject
(halt)
(halt)
CONTRADICTION!
END OF PROOF

34. that some language is undecidable we only need to reduce some
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. Undecidable Problems for Turing Recognizable languages
Let be a Turing-acceptable language
is empty?
is regular?
has size 2?
All these are undecidable problems

36. Let be a Turing-acceptable language
is empty?
is regular?
has size 2?

37. Empty language problem
Input:
Turing Machine
Question: Is
empty?
Corresponding language:

38. (empty language problem)
Theorem:
is undecidable
(empty-language problem is unsolvable)
Proof:
Reduce
(membership problem) to
(empty language problem)

39. Decider for Reduction Compute Decider Given the reduction,
membership problem decider
Decider for
empty problem
decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
if is decidable,
then is decidable
A contradiction!
since
is undecidable

40. We only need to build the reduction:
Compute
So that:

41. Construct from : Tape of input string yes write skip input run on tape
on input
yes If
accepts ?
then accept

42. Tape of yes write skip input run on tape string on input If accepts ?
then accept

43. Tape of yes write skip input run on tape string on input If accepts ?
then accept

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

45. Simply check if entered an accept state
altered
skip input
string
write
on tape
run
on input
yes If
accepts ?
then accept

46. Now check input string yes write skip input run on tape string
on input
yes If
accepts ?
then accept

47. The only possible accepted string
t r o y
skip input
string
write
on tape
run
on input
yes If
accepts ?
then accept

48. yes accepts does not accept write skip input run on tape string
on input
yes If
accepts ?
then accept

49. Therefore:
accepts
Equivalently:
END OF PROOF

50. Let be a Turing-acceptable language
is empty?
is regular?
has size 2?

51. Regular language problem
Input:
Turing Machine
Question: Is
a regular language?
Corresponding language:

52. (regular language problem)
Theorem:
is undecidable
(regular language problem is unsolvable)
Proof:
Reduce
(membership problem) to
(regular language problem)

53. Decider for Reduction Compute Decider Given the reduction,
membership problem decider
Decider for
regular problem
decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
If is decidable,
then is decidable
A contradiction!
since
is undecidable

54. We only need to build the reduction:
Compute
So that:

55. Construct from : Tape of input string yes write skip input run on tape
on input
yes
If has the form
accepts ?
then accept

56. yes not regular accepts does not accept regular write skip input run
string
write
on tape
run
on input
yes
If has the form
accepts ?
then accept

57. Therefore:
accepts
is not regular
Equivalently:
END OF PROOF

58. Let be a Turing-acceptable language
is empty?
is regular?
has size 2?

59. Corresponding language:
Size2 language problem
Input:
Turing Machine
Question:
Does have size 2?
(accepts exactly two strings?)
Corresponding language:

60. (size 2 language problem)
Theorem:
is undecidable
(regular language problem is unsolvable)
Proof:
Reduce
(membership problem) to
(size 2 language problem)

61. Decider for Reduction Compute Decider Given the reduction,
membership problem decider
Decider for
size2 problem
decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
If is decidable,
then is decidable
A contradiction!
since
is undecidable

62. We only need to build the reduction:
Compute
So that:

63. Construct from : Tape of input string yes write skip input run on tape
on input
yes If
accepts ?
then accept

64. yes 2 strings accepts does not accept 0 strings write skip input run
on tape
run
on input
yes If
accepts ?
then accept

65. Therefore:
accepts
has size 2
Equivalently:
END OF PROOF

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

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

68. More examples of non-trivial properties:
: is regular?
YES
YES NO
: has size 2? NO NO
YES

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

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

71. Example:
Suppose alphabet is
: has size 1? NO
YES NO NO

72. Non-trivial property problem
Input:
Turing Machine
Question:
Does have the non-trivial
property ?
Corresponding language:

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

74. We examine two cases:
Case 1:
Examples:
: is empty?
: is regular?
Case 2:
Example:
: has size 2?

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

76. Reduce
(membership problem) to

77. Decider for Reduction Compute Decider Given the reduction,
membership problem decider
Decider for
Non-trivial property
problem decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
if is decidable,
then is decidable
A contradiction!
since
is undecidable

78. We only need to build the reduction:
Compute
So that:

79. Construct from : Tape of input string yes write skip input run on tape
on input If
yes
accepts ?
then accept

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

81. yes accepts does not accept write skip input run on tape string
on input If
yes
accepts ?
then accept

82. Therefore:
accepts
Equivalently:

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

84. Reduce
(membership problem) to

85. Decider for Reduction Compute Decider Given the reduction,
membership problem decider
Decider for
Non-trivial property
problem decider
Reduction
YES
YES
Compute
Decider NO NO
Given the reduction,
if is decidable,
then is decidable
A contradiction!
since
is undecidable

86. We only need to build the reduction:
Compute
So that:

87. Construct from : Tape of input string yes write skip input run on tape
on input If
yes
accepts ?
then accept

88. yes accepts does not accept write skip input run on tape string
on input If
yes
accepts ?
then accept

89. Therefore:
accepts
Equivalently:
END OF PROOF

90. This Summary is an Online Content from this Book: Michael Sipser, Introduction to the Theory of Computation, 2ndEdition It is edited for Computation Theory Course by: T.Mariah Sami Khayat Teacher Adam University College For Contacting:
Kingdom of Saudi Arabia
Ministry of Education
Umm AlQura University
Adam University College
Computer Science Department
المملكة العربية السعودية
وزارة التعليم
جامعة أم القرى
الكلية الجامعية أضم
قسم الحاسب الآلي