1. Intro to Theory of Computation
LECTURE 19
Last time
Reductions
Mapping reductions
Today
Computation history method CS
464
Sofya Raskhodnikova
3/17/2016

2. Mapping Reductions Proving undecidability and unrecognizability
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3. 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
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4. Using mapping reductions to prove 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

5. 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

6. Using mapping reductions to prove 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

7. 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

8. I-clicker question (frequency: AC)
If 𝐀 ≤ 𝒎 𝑩 , we can conclude that
𝑨 ≤ 𝒎 𝑩
𝑩 ≤ 𝒎 A
𝐀 ≤ 𝒎 𝑩
𝐁 ≤ 𝒎 𝑨
None of the above.
{madam, dad, but, tub, book}
3/17/2016

9. 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

10. 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

11. Conclusions from ATM ≤ 𝒎 EQTM
Since ATM is undecidable, so is EQTM
ATM ≤ 𝒎 EQTM
Since ATM is unrecognizable, so is EQTM
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12. 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

13. Problems in language theory
ADFA decidable
ACFG decidable
EDFA decidable
ECFG decidable
EQDFA decidable
ATM
undecidable
ETM
undecidable
EQCFG ?
EQTM
undecidable
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14. Computation history method
Proving undecidability for languages that do not involve TM descriptions
Computation history method
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15. A linear bounded automaton (LBA)
A linear bounded automaton (LBA) is a TM variant that has bounded tape, with the number of tape squares equal to the size of the input.
FINITE STATE CONTROL
Linear memory (by increasing the alphabet – can use ``tracks’’ on the tape). Every CFL can be decided by an LBA. 1 1 1
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16. Configurations
A configuration of an LBA is a setting of state, head position and tape contents.
110〈𝑞7〉
Snapshot in the middle of computation q7 1
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17. I-clicker question (frequency: AC)
How many distinct configurations does an LBA have if it has 𝒒 states, 𝒈 symbols in tape alphabet, and the tape of length 𝒏?
𝒒𝒈𝒏
𝒒+𝒈+𝒏
𝒒 𝒈 𝒏
𝒒𝒏 𝒈 𝒏
None of the above.
{madam, dad, but, tub, book}
3/17/2016

18. Prove that ALBA is decidable
ALBA= { 𝑩, 𝒘 ∣ 𝑩 is an LBA that accepts string 𝒘 }
Idea: Given 𝑩, 𝒘 , simulate B on w.
If it halts, we know the answer.
If it loops, we can detect because B repeats a configuration.
S = `` On input 〈𝑩,𝒘〉, where 𝑩 is an LBA and w is a string:
Simulate B on w for 𝒒𝒏 𝒈 𝒏 steps.
If it accepts, accept.
If it rejects or does not halt, reject.’’
3/17/2016

19. starting configuration accepting configuration
Computation history
An accepting computation history for a TM M on input w is a sequence of configurations entered by M on input w:
A log
𝑪 𝟎 # 𝑪 𝟏 #…# 𝑪 ℓ
starting configuration
accepting configuration
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20. Starting configurationq0 1
𝑪 𝟎 =〈 𝑞 𝟎 〉𝐰
〈 𝑞 𝟎 〉𝐰
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21. Accepting configuration
qacc 1
𝑪 ℓ =〈 𝑞 𝟎 〉𝐰
… 𝑞 𝒂𝒄𝒄 …
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22. Each 𝑪 𝒊+𝟏 legally follows from 𝑪 𝒊q7 q5 1 2 1 1
𝑪 𝒊 = 〈𝑞7〉
𝑪 𝒊+𝟏 = 〈 𝑞 𝟎 〉𝐰
𝟐 〈𝑞𝟓〉 1 1 0
3/17/2016

23. Given a TM M and a string w,
LBAs can check computation histories of TMs
Given a TM M and a string w,
we can construct an LBA that checks whether its input is the accepting computation history of M on w.
3/17/2016

24. Prove that ELBA is undecidable
ELBA = { 𝑩 ∣ 𝑩 is a LBA and 𝑳 𝑩 =∅ }
Proof: Suppose to the contrary that TM R decides ELBA .
We construct TM S that decides ATM.
S = `` On input 〈𝑴,𝒘〉, where 𝑴 is a TM and w is a string:
Construct an LBA 𝑩 from 𝑴 and 𝒘:
Run TM R on input <𝑩>.
If , accept. O.w. reject.’’
𝑩 = `` On input x,
Accept if x = 𝑪 𝟎 #…# 𝑪 ℓ is the accepting configfaddur computation history of M on w:
𝑪 𝟎 is the starting configuration of 𝑴 on 𝑤
Each 𝑪 𝒊+𝟏 legally follows from 𝑪 𝒊
𝑪 ℓ is an accepting configuration for 𝑴 ’’
it rejects
3/17/2016