cm0081 undecidable problems about turing machines

Automata and Formal Languages - CM0081Undecidable Problems About Turing Machines

Andrés Sicard-Ramírez

Universidad EAFIT

Semester 2018-1

Reductions

DefinitionLet 𝑃1 and 𝑃2 be two problems. A reduction from 𝑃1 to 𝑃2 is a Turingmachine that takes an instance of 𝑃1 written on its tape and halts with aninstance of 𝑃2 that have the same answer (i.e. a reduction is an algorithm).

yes

no

yes

no

𝑃1 𝑃2

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Reductions

Theorem (9.7.a)If there is a reduction from 𝑃1 to 𝑃2 then if 𝑃1 is undecidable then so 𝑃2.(Hint: Suppose the 𝑃2 is decidable and find a contradiction).

Theorem (9.7.b)If there is a reduction from 𝑃1 to 𝑃2 then if 𝑃1 is not recursively enumerablethen so 𝑃2. (Hint: Suppose the 𝑃2 is recursively enumerable and find acontradiction).

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Reductions

Theorem (9.7.a)If there is a reduction from 𝑃1 to 𝑃2 then if 𝑃1 is undecidable then so 𝑃2.(Hint: Suppose the 𝑃2 is decidable and find a contradiction).

Theorem (9.7.b)If there is a reduction from 𝑃1 to 𝑃2 then if 𝑃1 is not recursively enumerablethen so 𝑃2. (Hint: Suppose the 𝑃2 is recursively enumerable and find acontradiction).

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Turing Machines that Accept the Empty Language

NotationHenceforth, we’ll regard strings as the Turing machines they represent.

Two languagesLe = {𝑀 ∣ 𝐿(𝑀) = ∅},

Lne = {𝑀 ∣ 𝐿(𝑀) ≠ ∅}.

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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:†

†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:†

†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 (𝑀, 𝑤) is converted in 𝑀 ′,

such that 𝑤 ∈ 𝐿(𝑀), if only if, 𝐿(𝑀 ′) ≠ ∅.

2. The key is to have 𝑀 ′ ignore its input.†

3. Lne is not recursive by Theorem 9.7.a.

†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 (𝑀, 𝑤) is converted in 𝑀 ′,

such that 𝑤 ∈ 𝐿(𝑀), if only if, 𝐿(𝑀 ′) ≠ ∅.2. The key is to have 𝑀 ′ ignore its input.†

3. Lne is not recursive by Theorem 9.7.a.

†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 (𝑀, 𝑤) is converted in 𝑀 ′,

such that 𝑤 ∈ 𝐿(𝑀), if only if, 𝐿(𝑀 ′) ≠ ∅.2. The key is to have 𝑀 ′ ignore its input.†

3. Lne is not recursive by Theorem 9.7.a.

†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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Rice’s Theorem

Set of the recursively enumerable languagesREL = {𝐿 ∣ 𝐿 is recursively enumerable }.

Properties (subsets) of the recursively enumerable languagesProperty 𝑃 of REL: 𝑃 ⊆ REL.

Example (Trivial properties)𝑃 (𝐿) = ∅ or 𝑃(𝐿) = REL.

Example𝑃 (𝐿): 𝐿 is a language regular.

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Rice’s Theorem

Set of the recursively enumerable languagesREL = {𝐿 ∣ 𝐿 is recursively enumerable }.

Properties (subsets) of the recursively enumerable languagesProperty 𝑃 of REL: 𝑃 ⊆ REL.

Example (Trivial properties)𝑃 (𝐿) = ∅ or 𝑃(𝐿) = REL.

Example𝑃 (𝐿): 𝐿 is a language regular.

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Rice’s Theorem

Set of the recursively enumerable languagesREL = {𝐿 ∣ 𝐿 is recursively enumerable }.

Properties (subsets) of the recursively enumerable languagesProperty 𝑃 of REL: 𝑃 ⊆ REL.

Example (Trivial properties)𝑃 (𝐿) = ∅ or 𝑃(𝐿) = REL.

Example𝑃 (𝐿): 𝐿 is a language regular.

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Rice’s Theorem

Set of the recursively enumerable languagesREL = {𝐿 ∣ 𝐿 is recursively enumerable }.

Properties (subsets) of the recursively enumerable languagesProperty 𝑃 of REL: 𝑃 ⊆ REL.

Example (Trivial properties)𝑃 (𝐿) = ∅ or 𝑃(𝐿) = REL.

Example𝑃 (𝐿): 𝐿 is a language regular.

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Rice’s Theorem

Theorem (Rice’s theorem, 9.11)Every non-trivial property of REL is undecidable.

How to prove Rice’s theorem?We identify a property 𝑃 by the Turing machines 𝑀 such that 𝐿(𝑀) ∈ 𝑃 .

Theorem (Rice’s theorem (second version))If 𝑃 ⊆ REL is a non-trivial property then

𝐿𝑃 = {𝑀 ∣ 𝐿(𝑀) ∈ 𝑃}

is undecidable.

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Rice’s Theorem

Theorem (Rice’s theorem, 9.11)Every non-trivial property of REL is undecidable.

How to prove Rice’s theorem?We identify a property 𝑃 by the Turing machines 𝑀 such that 𝐿(𝑀) ∈ 𝑃 .

Theorem (Rice’s theorem (second version))If 𝑃 ⊆ REL is a non-trivial property then

𝐿𝑃 = {𝑀 ∣ 𝐿(𝑀) ∈ 𝑃}

is undecidable.

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Rice’s Theorem

Theorem (Rice’s theorem, 9.11)Every non-trivial property of REL is undecidable.

How to prove Rice’s theorem?We identify a property 𝑃 by the Turing machines 𝑀 such that 𝐿(𝑀) ∈ 𝑃 .

Theorem (Rice’s theorem (second version))If 𝑃 ⊆ REL is a non-trivial property then

𝐿𝑃 = {𝑀 ∣ 𝐿(𝑀) ∈ 𝑃}

is undecidable.

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Rice’s Theorem

Proof: Case ∅ ∉ 𝑃 .1. Let 𝐿 be a language and 𝑀𝐿 be a Turing machine such 𝐿 ≠ ∅,

𝐿 ∈ 𝑃 and 𝐿 = 𝐿(𝑀𝐿). Reduction from Lu to 𝐿𝑃 where the pair(𝑀, 𝑤) is converted in 𝑀 ′ such that:†

i) 𝐿(𝑀 ′) = ∅ (i.e. 𝑀 ′ ∉ 𝐿𝑃 ) if 𝑤 ∉ 𝐿(𝑀) andii) 𝐿(𝑀 ′) = 𝐿 (i.e. 𝑀 ′ ∈ 𝐿𝑃 ) if 𝑤 ∈ 𝐿(𝑀).

2. 𝐿𝑃 is not recursive by Theorem 9.7.a.

†Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.10].Undecidable Problems About Turing Machines 20/28

Rice’s Theorem

Proof: Case ∅ ∉ 𝑃 .1. Let 𝐿 be a language and 𝑀𝐿 be a Turing machine such 𝐿 ≠ ∅,

𝐿 ∈ 𝑃 and 𝐿 = 𝐿(𝑀𝐿). Reduction from Lu to 𝐿𝑃 where the pair(𝑀, 𝑤) is converted in 𝑀 ′ such that:†

i) 𝐿(𝑀 ′) = ∅ (i.e. 𝑀 ′ ∉ 𝐿𝑃 ) if 𝑤 ∉ 𝐿(𝑀) andii) 𝐿(𝑀 ′) = 𝐿 (i.e. 𝑀 ′ ∈ 𝐿𝑃 ) if 𝑤 ∈ 𝐿(𝑀).

2. 𝐿𝑃 is not recursive by Theorem 9.7.a.

†Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.10].Undecidable Problems About Turing Machines 21/28

Rice’s Theorem

Proof: Case ∅ ∉ 𝑃 .1. Let 𝐿 be a language and 𝑀𝐿 be a Turing machine such 𝐿 ≠ ∅,

𝐿 ∈ 𝑃 and 𝐿 = 𝐿(𝑀𝐿). Reduction from Lu to 𝐿𝑃 where the pair(𝑀, 𝑤) is converted in 𝑀 ′ such that:†

i) 𝐿(𝑀 ′) = ∅ (i.e. 𝑀 ′ ∉ 𝐿𝑃 ) if 𝑤 ∉ 𝐿(𝑀) andii) 𝐿(𝑀 ′) = 𝐿 (i.e. 𝑀 ′ ∈ 𝐿𝑃 ) if 𝑤 ∈ 𝐿(𝑀).

2. 𝐿𝑃 is not recursive by Theorem 9.7.a.†Figure from Hopcroft, Motwani and Ullman [2007, Fig. 9.10].

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Rice’s Theorem

Proof: Case ∅ ∈ 𝑃 .1. By the previous case, 𝑃 is undecidable, i.e. 𝐿𝑃 is undecidable.2. 𝐿𝑃 = 𝐿𝑃 .3. Suppose 𝐿𝑃 is decidable then 𝐿𝑃 would be also decidable

(contradiction).4. Therefore, 𝐿𝑃 is undecidable.

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Rice’s 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 “helloworld”?Does the language accepted by the TM contain all the even numbers?

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Rice’s 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 “helloworld”?Does the language accepted by the TM contain all the even numbers?

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Rice’s Theorem

Rice’s theorem does not imply that everything about Turing machines isundecidable.

ExampleIt is decidable if a Turing machine has five states.

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Rice’s Theorem

Rice’s 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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