1. Intro to Theory of Computation
LECTURE 15
Last time
Countable/uncountable sets.
Diagonalization
Undecidable/unrecognizable languages (ATM is undecidable)
Today
ATM is unrecognizable
Reductions CS
332
Sofya Raskhodnikova
10/7/2019

2. Classes of languages recognizable decidable CFL regular ATM
{0 𝑛 1 𝑛 0 𝑛 ∣𝑛≥0}
{0 𝑛 1 𝑛 ∣𝑛≥0}
regular 1*
ATM = {〈M,w〉| M is a TM, w is a string, and M accepts w }
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3. Exercise The fact that ATM is undecidable means that
if we are given input <M,w>, then M is not a decider
there is no TM S such that L(S)= ATM
there is no TM S that accepts on strings in ATM and halts and rejects on strings on strings not in ATM .
Both B and C are correct.
None of the above.
{madam, dad, but, tub, book}
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4. Classes of languages recognizable decidable CFL regular ATM ATM
{0 𝑛 1 𝑛 0 𝑛 ∣𝑛≥0}
{0 𝑛 1 𝑛 ∣𝑛≥0}
regular 1*
ATM = {〈M,w〉| M is a TM, w is a string, and M accepts w }
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5. Theorem. Language L is decidable iff L and L are Turing-recognizable
Proof: 1)
L is decidable ⇒ 𝑳 and 𝑳 are Turing-recognizable. 2)
𝑳 and 𝑳 are Turing-recognizable ⇒ L is decidable.
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6. Prove that the following languages are Turing-recognizable
ATM = { 𝑴, 𝒘 ∣ 𝑴 is a TM that accepts string 𝒘}
HALTTM = { 𝑴, 𝒘 ∣ 𝑴 is a TM that halts on string 𝒘 }
Corollary is not Turing-recognizable.
ATM
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7. Classes of languages recognizable decidable CFL regular ATM ATM
{0 𝑛 1 𝑛 0 𝑛 ∣𝑛≥0}
{0 𝑛 1 𝑛 ∣𝑛≥0}
regular 1*
ATM = {〈M,w〉| M is a TM, w is a string, and M accepts w }
10/7/2019

8. Problems in language theory
ADFA decidable
ACFG decidable
EDFA decidable
ECFG decidable
EQDFA decidable
ATM
undecidable
ETM ?
EQCFG ?
EQTM ?
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9. Reductions Now that we have an undecidable language, can we get more?
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10. What is a reduction?
A reduction from problem A to problem B is an algorithm for problem A that uses a subroutine for problem B.
Many of the deciders we constructed use reductions to one of these problems: ADFA, EDFA, EQDFA, ACFG, ECFG.
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11. Old Theorem. EQDFA is decidable.
EQDFA = { ⟨ 𝑫 𝟏 , 𝑫 𝟐 ⟩| 𝑫 𝟏 , 𝑫 𝟐 are DFAs & 𝑳 𝑫 𝟏 =𝑳( 𝑫 𝟐 )}
Proof: The following TM M decides EQDFA.
M = `` On input 〈 𝑫 𝟏 , 𝑫 𝟐 〉, where 𝑫 𝟏 , 𝑫 𝟐 are DFAs:
Construct a DFA D that recognizes
the set difference of 𝑳 𝑫 𝟏 and 𝑳( 𝑫 𝟐 ).
Run the decider for EDFA on <D>.
If it accepts, accept. O.w. reject.’’
That’s a reduction from EQDFA to EDFA.
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12. How to tell a difference between a CS student and a CE student?
Put an empty kettle in the middle of the kitchen floor and ask your subjects to boil some water.
Both subjects will fill the kettle with water, turn on the stove and turn the flame on.
Put the kettle full of water on the stove and ask the subjects to boil the water.
CE student will turn the flame on.
CS student will empty the kettle and put it in the middle of the kitchen floor… thereby reducing the problem to the one that has already been solved!
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13. Using reductions to prove undecidability
We want to prove that language L is undecidable.
Idea: Use a proof by contradiction.
Suppose to the contrary that L is decidable.
Use a decider for L as a subroutine to construct a decider for ATM.
But ATM is undecidable. Contradiction!
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14. Prove that HALTTM is undecidable
HALTTM= { 𝑴, 𝒘 ∣ 𝑴 is a TM that halts on string 𝒘 }
Proof: Suppose to the contrary that HALTTM is decidable,
and let R be a TM that decides it.
We construct TM S that decides ATM.
S = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Run TM R on input <M,w>.
If it rejects, reject.
If R accepts, simulate M on w.
If it accepts, accept. O.w. reject.’’
M must halt on w,
since R accepted.
That’s a reduction from ATM to HALTTM.
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15. Prove that ETM is undecidable
ETM = { 𝑴 ∣ 𝑴 is a TM and 𝑳 𝑴 =∅}
Proof: Suppose to the contrary that ETM is decidable,
and let R be a TM that decides it.
We construct TM S that decides ATM.
S = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Run TM R on input ???
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16. Prove that ETM is undecidable
ETM = { 𝑴 ∣ 𝑴 is a TM and 𝑳 𝑴 =∅}
Proof: Suppose to the contrary that ETM is decidable,
and let R be a TM that decides it.
We construct TM S that decides ATM.
S = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Construct TM 𝑀′.
Run TM R on input <𝑴′>.
If , accept. O.w. reject.’’
𝑀′ = `` On input x,
Ignore the input.
Run TM M on input w.
If it accepts, accept.’’
it rejects
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17. Prove that CFLTM is undecidable
CFLTM = { 𝑴 ∣ 𝑴 is a TM and 𝑳 𝑴 is context-free}
Proof: Suppose to the contrary that CFLTM is decidable,
and let R be a TM that decides it.
We construct TM S that decides ATM.
S = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Construct TM 𝑀′.
Run TM R on input <𝑴′>.
If , accept. O.w. reject.’’
𝑀′ = `` On input x,
If x is not of the form 𝟎 𝒏 𝟏 𝒏 𝟐 𝒏 , reject.
Run TM M on input w.
If it accepts, accept.’’
it rejects
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