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CSC3130Formal Languages and

Automata Theory

Tutorial 9Undecidable Problem

KN Hung

SHB 1026

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Agenda

• Recap– Decidibility, Undecidibility, Turing-

Recognizable …

• Reduction

• Undecidable Problem

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Recap

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Recap: Recognize

• Given a language L and a machine M “M recognizes L” iff

“For any w ∈ L, M halts and goes to an accepting state”

• “M is a recognizer for L”

• However, for w ∉ L, M may– Reject– Loop forever

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Recap: Decide

• Given a language L and a machine M “M decides L” iff

“For any w ∈ L, M halts and accepts AND

Otherwise, M rejects”

• “M is a decider for L”

• Meaning: M can determine whether a given string is in L or not

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Recognize VS Decide

• Recognize = Halts for accepting input

• Decide = Halts for all inputs

• Decide => Recognize

• Make sure that you can distinguish these two concepts!

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Recap: ~able / ~ability

• L is Turing-recognizable iffthere is a TM that recognizes L

• L is decidable iffthere is a TM that decides L

• L is undecidable iffthere is no TM that decides L

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Recognizability VS Decidibility

Turing-Recognizable

Decidable

For a language, L, located here,there is no machine that halts for all input string

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Some Decidable Problems

• The following problems are decidable– ADFA = {<A, w> | A is a DFA that accepts w}

– ANFA = {<A, w> | A is a NFA that accepts w}

– EDFA = {<A, w> | A is a DFA and L(A) = Ø}

– EQDFA = {<A, B> | A, B are DFAs, L(A) = L(B)}

– ACFG = {<G, w> | G is a CFG that generates w}

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Some Undecidable Problems

• The following problems are undecidable– ATM = {<M, w> | M is a TM that accepts w}

• To prove undecidability of a problem using reduction, ATM is used very frequently (nearly in all situations)

• Used in Example 1 and 2 (Later slides)

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Reduction

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Reduction

• Occurs in daily life…• Problem A: Travel to Japan• Problem B: Buying a air ticket to Japan• Problem A reduces to Problem B

– Meaning: Settling B automatically solves A– Put in logical statement:

B is solvable A is solvable

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Reduction

Buy an air ticket

Travel to Japan

Reduce toReduce to

The problem “To buy an air ticket” enclosesThe problem “To travel to Japan”

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Reduction

B

A

Reduce toReduce to

Problem B “encloses” ASo, B is solved => A is solvedThus, B is no harder than A!

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Reduction

• A reduces to B• A is reducible to B

– If B is solved, then A is solved– i.e., Whenever there is a solution to B,

then there is a solution to A

• If A cannot be solved, then B cannot be solved (Contrapositive)

• Thus, B is no harder than A

B

AReduced toReduced to

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Undecidable Problems

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Example 1

• Show that REGULARTM is undecidable

• REGULARTM = {<M> | M is a TM and L(M) is a regular language}

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REGULARTM is undecidable?

• Idea (By Contradiction):– Assume there is a TM R that decides

REGULARTM

– Construct a TM S that decides ATM

– Contradiction!

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Construction of S

• On input <M, w>– where M is a TM and w is a string:

• Construct M’ by M and w

• Run R on <M’>

• Output the answer

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Idea of M’

• Strategy: Convert M into another TM M’: M’ recognizes a regular languageiff M accepts w

• If M accepts w, let M’ recognizes {0, 1}*• Otherwise, let M’ recognizes {0n1n}• With the above properties

– When we run R on <M’>,the output is “accept” iff M accepts w

– Why…? [See next few slides…]

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Construction of M’

• M’ = “On input x:1. If x has the form 0n1n, accept

2. If x does not have this form, run M on input w and output answer”

• {0n1n} is recognized by M’ anyway

• How about the other strings?

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Construction of M’

• For strings of form 0n1n, M’ accepts• For all the other strings, listing all cases…

– If M accepts w• they are all accepted by M’

– If M rejects w• they are all rejected by M’

– If M loops on w• they all loop on M’

M’ = “On input x:

1. If x has the form 0n1n, accept

2. If x does not have this form, run M on input w and output answer”

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L(M’) in relation to S

• If M accepts w– M’ recognizes {0n1n}∪{0,1}*– {0,1}* is regular– R will accept M’

• If M does not accept w– M’ recognizes {0n1n}∪Ø– {0n1n} is not regular– R will reject M’

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REGULARTM is undecidable

• Assume “R that decides REGULARTM” exists

• We construct S to decide ATM

“On input <M, w>, where M is a TM and w is a string:1. Construct the following TM M’

M’ = “On input x:

1. If x has the form 0n1n, accept

2. If x does not have this form, run M on input w and accept if M accepts w”

2. Run R on input <M’>

3. Output the output of R”

• Now, S decides ATM… Contradiction!

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Example 2

• L = {<M, w> | M accepts all strings shorter than w}

• Show that L is undecidable

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L is undecidable?

• L = {<M, w> | M accepts all strings shorter than w}

• Again, we try to use the undecidability of ATM to achieve a contradiction

• Idea (By Contradiction):– Assume there is a TM R that decides L

– Construct a TM (by R), S that decides ATM

– Contradiction!

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Construction of S

• On input <M, w>– where M is a TM and w is a string:

• Construct M’ by M and w

• Run R on <M’, ?>– This time, we have a TM that takes 2

arguments – a TM, and a string – as inputs– Think what string to put as the 2nd input

• Output the answer

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Idea of M’

• Strategy: Convert M into another TM M’: M’ accepts all strings shorter than ziff M accepts w

• We want to achieve the following:– When we run R on <M’, z>,

the output is “accept” iff M accepts w

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Idea of M’

• Strategy: Convert M into another TM M’: M’ accepts all strings shorter than “1”iff M accepts w

• We want to achieve the following:– When we run R on <M’, “1”>,

the output is “accept” iff M accepts w

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Construction of M’

• M’ = “On input x:1. If x =ε

Run M on w

Output the answer

2. If x ≠ε Reject

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Construction of M’

• For M’– If M accepts w

• the only possibility is input isε AND M’ accepts ε

– If M rejects w, • Case 1: x =ε

– M’ rejects

• Case 2: x ≠ε– M’ rejects

– If M loops on w• Case 1: x =ε

– M’ loops

• Case 2: x ≠ε– M’ rejects

M’ = “On input x: If x =ε run M on w and output answer If x ≠ε reject”

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Construction of M’

• For M’– If M accepts w

• M’ accepts only ε

– Otherwise• M’ does not accepts any string

– including εM’ = “On input x:

If x =ε run M on w and output answer If x ≠ε reject”

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L(M’) in relation to S

• If M accepts w– L(M’) = {ε}– {ε} is exactly the language that contains all

string shorter than “1”– S accepts <M’, “1”>

• Otherwise (i.e., M loops on or rejects w)– L(M’) = Ø– S rejects <M’, “1”>

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L is undecidable

• Assume “R that decides REGULARTM” exists• We construct S to decide ATM

“On input <M, w>, where M is a TM and w is a string:1.Construct the following TM M’

M’ = “On input x:1. If x =ε

Run M on xOutput the answer

2. If x ≠εReject

2. Run R on input <M’, “1”>3. Output the output of R”

• Now, S decides ATM… Contradiction!

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Conclusion

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Pf: Undecidability by Reduction

R• Given an undecidable language, L• Assume such a decider TM R, exists

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• Given an undecidable language, L• Assume such a decider TM R, exists• Construct a TM, M’, by M and w

– M is from input

Pf: Undecidability by Reduction

R

M M’Modify

Input

w

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• Given an undecidable language, L• Assume such a decider TM R, exists• Construct a TM, M’, by M and w

– M is from input

• Feed M’ to R

Pf: Undecidability by Reduction

R

M M’Modify

Input

w

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• Given an undecidable language, L• Assume such a decider TM R, exists• Construct a TM, M’, by M and w

– M is from input

• Feed M’ to R• If you are right

– R will accept M’ iff M accepts w– R will reject M’ iff M does not accept w

• Thus, you obtain a decider for ATM

– Contradiction!!

Pf: Undecidability by Reduction

R

M M’Modify

Input

“M acc w” then ACC

Otherwise,REJ

w

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Q & A

• Thanks for coming!

• I understand that this topic is confusing…

• Should have some questions?