CSC3130 Formal Languages and Automata Theory Tutorial 9 Undecidable Problem

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CSC3130Formal Languages andAutomata
TheoryTutorial 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 gt 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 ltA, wgt A is a DFA that accepts w
• ANFA ltA, wgt A is a NFA that accepts w
• EDFA ltA, wgt A is a DFA and L(A) Ø
• EQDFA ltA, Bgt A, B are DFAs, L(A) L(B)
• ACFG ltG, wgt G is a CFG that generates w

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Some Undecidable Problems

• The following problems are undecidable
• ATM ltM, wgt 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
Reduce to
Travel to Japan
The problem To buy an air ticket
enclosesThe problem To travel to Japan
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Reduction
B
Reduce to
A
Problem B encloses ASo, B is solved gt A is
solvedThus, B is no harder than A!
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Reduction
B

• 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

Reduced to
A
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Undecidable Problems
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Example 1

• Show that REGULARTM is undecidable
• REGULARTM ltMgt 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 ltM, wgt
• where M is a TM and w is a string
• Construct M by M and w
• Run R on ltMgt
• 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 ltMgt,
• the output is accept iff M accepts w
• Why? See next few slides

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Construction of M

• M On input x
• If x has the form 0n1n, accept
• 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
• If x has the form 0n1n, accept
• 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 ltM, wgt, where M is a TM and w is a
  string
• Construct the following TM M
• M On input x
• If x has the form 0n1n, accept
• If x does not have this form, run M on input w
  and accept if M accepts w
• 2. Run R on input ltMgt
• 3. Output the output of R
• Now, S decides ATM Contradiction!

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Example 2

• L ltM, wgt M accepts all strings shorter than
  w
• Show that L is undecidable

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L is undecidable?

• L ltM, wgt 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 ltM, wgt
• where M is a TM and w is a string
• Construct M by M and w
• Run R on ltM, ?gt
• 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 ltM, zgt,
• 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 1iff M
  accepts w
• We want to achieve the following
• When we run R on ltM, 1gt,
• the output is accept iff M accepts w

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Construction of M

• M On input x
• If x e
• Run M on w
• Output the answer
• If x ?e
• Reject

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Construction of M

• For M
• If M accepts w
• the only possibility is input ise AND M
  accepts e
• If M rejects w,
• Case 1 x e
• M rejects
• Case 2 x ?e
• M rejects
• If M loops on w
• Case 1 x e
• M loops
• Case 2 x ?e
• M rejects

M On input x If x e run M on
w and output answer If x ?e
reject
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Construction of M

• For M
• If M accepts w
• M accepts only e
• Otherwise
• M does not accepts any string
• including e

M On input x If x e run M on
w and output answer If x ?e
reject
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L(M) in relation to S

• If M accepts w
• L(M) e
• e is exactly the language that contains all
  string shorter than 1
• S accepts ltM, 1gt
• Otherwise (i.e., M loops on or rejects w)
• L(M) Ø
• S rejects ltM, 1gt

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L is undecidable

• Assume R that decides REGULARTM exists
• We construct S to decide ATM
• On input ltM, wgt, where M is a TM and w is a
  string
• Construct the following TM MM On input x
• If x e
• Run M on x
• Output the answer
• If x ?e
• Reject
• 2. Run R on input ltM, 1gt
• 3. Output the output of R
• Now, S decides ATM Contradiction!

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Conclusion
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Pf Undecidability by Reduction

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

R
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Pf Undecidability by Reduction

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

R
Input
Modify
M
M
w
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Pf Undecidability by Reduction

• 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

R
Input
Modify
M
M
w
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Pf Undecidability by Reduction
M acc w then ACC
Otherwise,REJ

• 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!!

R
Input
Modify
M
M
w
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Q A

• Thanks for coming!
• I understand that this topic is confusing
• Should have some questions?