1. Intro to Theory of Computation
LECTURE 16
Last time
ATM is unrecognizable
Reductions
Today
Mapping reductions CS
332
Sofya Raskhodnikova
3/17/2016

2. Problems in language theory
ADFA decidable
ACFG decidable
EDFA decidable
ECFG decidable
EQDFA decidable
ATM
undecidable
ETM
undecidable
EQCFG ?
EQTM ?
3/17/2016

3. 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!
3/17/2016

4. Exercise To prove that ETM is undecidable
we assumed ETM had a decider and used it to construct a decider for ATM
we assumed ATM had a decider and used it to construct a decider for ETM
we constructed a TM 𝑺 that on input <𝑴,𝒘> decides whether 𝑴 accepts 𝒘, assuming the existence of a TM 𝑹 that decides on input <𝑴′> whether the language of <𝑴′> is empty
There is more than one correct answer.
None of the above.
{madam, dad, but, tub, book}
3/17/2016

5. Prove that EQTM is undecidable
EQTM = { 𝑴 𝟏 , 𝑴 𝟐 ∣ 𝑴 𝟏 , 𝑴 𝟐 are TMs, 𝑳 𝑴 𝟏 =𝑳( 𝑴 𝟐 )}
Proof: Suppose to the contrary that EQTM 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 TMs 𝑴 ′ ,𝑴′′.
Run TM R on input < 𝑴 ′ ,𝑴′′>.
If , accept. O.w. reject.’’
𝑀′ = `` On input x,
Ignore the input.
Run TM M on input w.
If it accepts, accept.’’
𝑀′′ = ``Accept.’’
it accepts
3/17/2016

6. Proof 2 that EQTM is undecidable
EQTM = { 𝑴 𝟏 , 𝑴 𝟐 ∣ 𝑴 𝟏 , 𝑴 𝟐 are TMs, 𝑳 𝑴 𝟏 =𝑳( 𝑴 𝟐 )}
Proof: Suppose to the contrary that EQTM is decidable,
and let R be a TM that decides it.
We construct TM S that decides ETM. What do we change?
S = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string: 𝑴
Construct TMs 𝑴 ′ ,𝑴′′.
Run TM R on input < 𝑴 ′ ,𝑴′′>.
If , accept. O.w. reject.’’
TM 𝑴 ′
𝑀′ = `` On input x,
Ignore the input.
Run TM M on input w.
If it accepts, accept.’’
𝑀′ = ``Reject.’’
𝑀′′ = ``Accept.’’
< 𝑴, 𝑴 ′ >
it accepts
3/17/2016

7. Problems in language theory
ADFA decidable
ACFG decidable
EDFA decidable
ECFG decidable
EQDFA decidable
ATM
undecidable
ETM
undecidable
EQCFG ?
EQTM
undecidable
3/17/2016

8. Mapping Reductions Proving undecidability and unrecognizability
3/17/2016

9. Computable functions
A function 𝒇: 𝚺 ∗ → 𝚺 ∗ is computable if some TM M, on every input 𝒘, halts with only 𝒇(𝒘) on its tape.
Example 1: 𝒇( 𝒙,𝒚 )=𝒙+𝒚.
Example 2: 𝒇 𝑴,𝒘 = 𝑴′ , where 𝑴 is a TM and 𝒘 is a string, and 𝑴′ is a TM that ignores its input and runs 𝑴 on 𝒘.
3/17/2016

10. if there is a computable function 𝒇, such that for all strings 𝒘,
Mapping reductions
Given languages A and B,
A ≤ 𝒎 B
if there is a computable function 𝒇,
such that for all strings 𝒘,
𝒘∈ 𝑨 iff 𝒇(𝒘)∈𝑩. A B 𝒇
3/17/2016

11. Exercise If 𝐀 ≤ 𝒎 𝑩 , we can conclude that 𝑨 ≤ 𝒎 𝑩 𝑩 ≤ 𝒎 A 𝐀 ≤ 𝒎 𝑩
𝑨 ≤ 𝒎 𝑩
𝑩 ≤ 𝒎 A
𝐀 ≤ 𝒎 𝑩
𝐁 ≤ 𝒎 𝑨
None of the above.
{madam, dad, but, tub, book}
3/17/2016

12. Mapping reductions: decidability
Theorem. If A ≤ 𝒎 B and B is decidable, then A is decidable.
Proof: 𝐋𝐞𝐭 𝑴 be a decider for 𝑩 and
𝒇 be a mapping reduction from A to B.
Construct a decider for A:
``On input w:
Compute f(w).
Run M on f(w).
If it accepts, accept. O.w. reject.”
3/17/2016

13. Using mapping reductions to prove undecidability
Theorem. If A ≤ 𝒎 B and B is decidable, then A is decidable.
Corollary. If A ≤ 𝒎 B and A is undecidable, then B is undecidable.
Example: If ATM≤ 𝒎 B, then B is undecidable.
3/17/2016

14. Mapping reductions: recognizability
Theorem. If A ≤ 𝒎 B and B is Turing-recognizable, then A is Turing-recognizable.
Proof: 𝐋𝐞𝐭 𝑴 be a TM that recognizes 𝑩 and
𝒇 be a mapping reduction from A to B.
Construct a TM that recognizes A:
``On input w:
Compute f(w).
Run M on f(w).
If it accepts, accept. O.w. reject.”
3/17/2016

15. Using mapping reductions to prove unrecognizability
Theorem. If A ≤ 𝒎 B and B is Turing-recognizable, then A is Turing-recognizable.
Corollary. If A ≤ 𝒎 B and A is unrecognizable, then B is unrecognizable.
Example: If ATM≤ 𝒎 B, then B is unrecognizable.
3/17/2016

16. Old proof that EQTM is undecidable
EQTM = { 𝑴 𝟏 , 𝑴 𝟐 ∣ 𝑴 𝟏 , 𝑴 𝟐 are TMs, 𝑳 𝑴 𝟏 =𝑳( 𝑴 𝟐 )}
Proof: Suppose to the contrary that EQTM 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 TMs 𝑴 ′ ,𝑴′′.
Run TM R on input < 𝑴 ′ ,𝑴′′>.
If , accept. O.w. reject.’’
𝑀′ = `` On input x,
Ignore the input.
Run TM M on input w.
If it accepts, accept.’’
𝑀′′ = ``Accept.’’
it accepts
3/17/2016

17. ATM ≤ 𝒎 EQTM Proof: The following TM computes the reduction:
F = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Construct TMs 𝑴 ′ ,𝑴′′.
Output < 𝑴 ′ ,𝑴′′>.”
𝑀′ = `` On input x,
Ignore the input.
Run TM M on input w.
If it accepts, accept.’’
𝑀′′ = ``Accept.’’
ATM
EQTM 𝒇
3/17/2016

18. Conclusions from ATM ≤ 𝒎 EQTM
Since ATM is undecidable, so is EQTM
ATM ≤ 𝒎 EQTM
Since ATM is unrecognizable, so is EQTM
3/17/2016

19. Prove that EQTM is unrecognizable
Proof: We give a mapping reduction ATM ≤ 𝒎 EQTM
The following TM computes the reduction:
F = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Construct TMs 𝑴 ′ ,𝑴′′.
Output < 𝑴 ′ ,𝑴′′>.”
𝑀′ = `` On input x,
Ignore the input.
Run TM M on input w.
If it accepts, accept.’’
𝑀′′ = ``Reject.’’
ATM
EQTM 𝒇
3/17/2016

20. Problems in language theory
ADFA decidable
ACFG decidable
EDFA decidable
ECFG decidable
EQDFA decidable
ATM
undecidable
ETM
undecidable
EQCFG ?
EQTM
undecidable
3/17/2016