Automata and Formal Languages - CM0081Undecidable Problems About Turing Machines
AndrΓ©s Sicard-RamΓrez
Universidad EAFIT
Semester 2018-1
Reductions
DefinitionLet π1 and π2 be two problems. A reduction from π1 to π2 is a Turingmachine that takes an instance of π1 written on its tape and halts with aninstance of π2 that have the same answer (i.e. a reduction is an algorithm).
yes
no
yes
no
π1 π2
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Reductions
Theorem (9.7.a)If there is a reduction from π1 to π2 then if π1 is undecidable then so π2.(Hint: Suppose the π2 is decidable and find a contradiction).
Theorem (9.7.b)If there is a reduction from π1 to π2 then if π1 is not recursively enumerablethen so π2. (Hint: Suppose the π2 is recursively enumerable and find acontradiction).
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Reductions
Theorem (9.7.a)If there is a reduction from π1 to π2 then if π1 is undecidable then so π2.(Hint: Suppose the π2 is decidable and find a contradiction).
Theorem (9.7.b)If there is a reduction from π1 to π2 then if π1 is not recursively enumerablethen so π2. (Hint: Suppose the π2 is recursively enumerable and find acontradiction).
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Turing Machines that Accept the Empty Language
NotationHenceforth, weβll regard strings as the Turing machines they represent.
Two languagesLe = {π β£ πΏ(π) = β },
Lne = {π β£ πΏ(π) β β }.
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Turing Machines that Accept the Empty Language
Theorem (9.8)Lne is recursively enumerable,
ProofConstruction of a non-determinist TM to accept Lne:β
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.8].Undecidable Problems About Turing Machines 6/28
Turing Machines that Accept the Empty Language
Theorem (9.8)Lne is recursively enumerable,
ProofConstruction of a non-determinist TM to accept Lne:β
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.8].Undecidable Problems About Turing Machines 7/28
Turing Machines that Accept the Empty Language
Theorem (9.9)Lne is not recursive.
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Turing Machines that Accept the Empty Language
Proof.1. Reduction from Lu to Lne where the pair (π, π€) is converted in π β²,
such that π€ β πΏ(π), if only if, πΏ(π β²) β β .
2. The key is to have π β² ignore its input.β
3. Lne is not recursive by Theorem 9.7.a.
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.9].Undecidable Problems About Turing Machines 9/28
Turing Machines that Accept the Empty Language
Proof.1. Reduction from Lu to Lne where the pair (π, π€) is converted in π β²,
such that π€ β πΏ(π), if only if, πΏ(π β²) β β .2. The key is to have π β² ignore its input.β
3. Lne is not recursive by Theorem 9.7.a.
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.9].Undecidable Problems About Turing Machines 10/28
Turing Machines that Accept the Empty Language
Proof.1. Reduction from Lu to Lne where the pair (π, π€) is converted in π β²,
such that π€ β πΏ(π), if only if, πΏ(π β²) β β .2. The key is to have π β² ignore its input.β
3. Lne is not recursive by Theorem 9.7.a.
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.9].Undecidable Problems About Turing Machines 11/28
Turing Machines that Accept the Empty Language
Theorem (9.10)Le is not recursively enumerable. (Hint: Le is the complement of Lne).
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Riceβs Theorem
Set of the recursively enumerable languagesREL = {πΏ β£ πΏ is recursively enumerable }.
Properties (subsets) of the recursively enumerable languagesProperty π of REL: π β REL.
Example (Trivial properties)π (πΏ) = β or π(πΏ) = REL.
Exampleπ (πΏ): πΏ is a language regular.
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Riceβs Theorem
Set of the recursively enumerable languagesREL = {πΏ β£ πΏ is recursively enumerable }.
Properties (subsets) of the recursively enumerable languagesProperty π of REL: π β REL.
Example (Trivial properties)π (πΏ) = β or π(πΏ) = REL.
Exampleπ (πΏ): πΏ is a language regular.
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Riceβs Theorem
Set of the recursively enumerable languagesREL = {πΏ β£ πΏ is recursively enumerable }.
Properties (subsets) of the recursively enumerable languagesProperty π of REL: π β REL.
Example (Trivial properties)π (πΏ) = β or π(πΏ) = REL.
Exampleπ (πΏ): πΏ is a language regular.
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Riceβs Theorem
Set of the recursively enumerable languagesREL = {πΏ β£ πΏ is recursively enumerable }.
Properties (subsets) of the recursively enumerable languagesProperty π of REL: π β REL.
Example (Trivial properties)π (πΏ) = β or π(πΏ) = REL.
Exampleπ (πΏ): πΏ is a language regular.
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Riceβs Theorem
Theorem (Riceβs theorem, 9.11)Every non-trivial property of REL is undecidable.
How to prove Riceβs theorem?We identify a property π by the Turing machines π such that πΏ(π) β π .
Theorem (Riceβs theorem (second version))If π β REL is a non-trivial property then
πΏπ = {π β£ πΏ(π) β π}
is undecidable.
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Riceβs Theorem
Theorem (Riceβs theorem, 9.11)Every non-trivial property of REL is undecidable.
How to prove Riceβs theorem?We identify a property π by the Turing machines π such that πΏ(π) β π .
Theorem (Riceβs theorem (second version))If π β REL is a non-trivial property then
πΏπ = {π β£ πΏ(π) β π}
is undecidable.
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Riceβs Theorem
Theorem (Riceβs theorem, 9.11)Every non-trivial property of REL is undecidable.
How to prove Riceβs theorem?We identify a property π by the Turing machines π such that πΏ(π) β π .
Theorem (Riceβs theorem (second version))If π β REL is a non-trivial property then
πΏπ = {π β£ πΏ(π) β π}
is undecidable.
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Riceβs Theorem
Proof: Case β β π .1. Let πΏ be a language and ππΏ be a Turing machine such πΏ β β ,
πΏ β π and πΏ = πΏ(ππΏ). Reduction from Lu to πΏπ where the pair(π, π€) is converted in π β² such that:β
i) πΏ(π β²) = β (i.e. π β² β πΏπ ) if π€ β πΏ(π) andii) πΏ(π β²) = πΏ (i.e. π β² β πΏπ ) if π€ β πΏ(π).
2. πΏπ is not recursive by Theorem 9.7.a.
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.10].Undecidable Problems About Turing Machines 20/28
Riceβs Theorem
Proof: Case β β π .1. Let πΏ be a language and ππΏ be a Turing machine such πΏ β β ,
πΏ β π and πΏ = πΏ(ππΏ). Reduction from Lu to πΏπ where the pair(π, π€) is converted in π β² such that:β
i) πΏ(π β²) = β (i.e. π β² β πΏπ ) if π€ β πΏ(π) andii) πΏ(π β²) = πΏ (i.e. π β² β πΏπ ) if π€ β πΏ(π).
2. πΏπ is not recursive by Theorem 9.7.a.
β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.10].Undecidable Problems About Turing Machines 21/28
Riceβs Theorem
Proof: Case β β π .1. Let πΏ be a language and ππΏ be a Turing machine such πΏ β β ,
πΏ β π and πΏ = πΏ(ππΏ). Reduction from Lu to πΏπ where the pair(π, π€) is converted in π β² such that:β
i) πΏ(π β²) = β (i.e. π β² β πΏπ ) if π€ β πΏ(π) andii) πΏ(π β²) = πΏ (i.e. π β² β πΏπ ) if π€ β πΏ(π).
2. πΏπ is not recursive by Theorem 9.7.a.β Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.10].
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Riceβs Theorem
Proof: Case β β π .1. By the previous case, π is undecidable, i.e. πΏπ is undecidable.2. πΏπ = πΏπ .3. Suppose πΏπ is decidable then πΏπ would be also decidable
(contradiction).4. Therefore, πΏπ is undecidable.
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Riceβs Theorem
All the problems about Turing machines that involve only the languagethat the TM accepts are undecidable.
ExamplesIs the language accepted by the TM empty?Is the language accepted by the TM finite?Is the language accepted by the TM regular?Is the language accepted by the TM context-free language?Does the language accepted by the TM contain the string βhelloworldβ?Does the language accepted by the TM contain all the even numbers?
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Riceβs Theorem
All the problems about Turing machines that involve only the languagethat the TM accepts are undecidable.
ExamplesIs the language accepted by the TM empty?Is the language accepted by the TM finite?Is the language accepted by the TM regular?Is the language accepted by the TM context-free language?Does the language accepted by the TM contain the string βhelloworldβ?Does the language accepted by the TM contain all the even numbers?
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Riceβs Theorem
Riceβs theorem does not imply that everything about Turing machines isundecidable.
ExampleIt is decidable if a Turing machine has five states.
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Riceβs Theorem
Riceβs theorem does not imply that everything about Turing machines isundecidable.ExampleIt is decidable if a Turing machine has five states.
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References
Hopcroft, J. E., Motwani, R. and Ullman, J. D. (2007). Introduction toAutomata theory, Languages, and Computation. 3rd ed. Pearson Education.
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