CSE 105 THEORY OF COMPUTATION Fall 2016 - - PowerPoint PPT Presentation

CSE 105

THEORY OF COMPUTATION

Fall 2016 http://cseweb.ucsd.edu/classes/fa16/cse105-abc/

Language of a TM

• L(M) = {w | M accepts w}
• M may reject or loop on strings not in L(M)
• A language X is recognizable if X=L(M) for some TM M
• A TM M is a decider if M(w) halts on every input w
• A language X is decidable is X=L(M) for some decider M

Summary

• Theorem: A language L is decidable if and only if it is

both recognizable and co-recognizable: D = RE ∩ coRE

RE co-RE Context-free Regular Decidable

Closure properties (D)

• Decidable languages are closed under

–

Union

–

Intersection

–

Set Complement

–

Set Difference

–

….

• Proof: similar to the proof for union

Closure properties (D)

• Theorem: If A and B are decidable, then AUB is decidable
• Proof:

–

Let M, M’ be deciders such that L(M)=A, L(M’)=B

–

We build a decider for AUB, M’’(w) =

1) Run M(w). If M(w) accepts, then accept 2) Run M’(w). If M’(w) accepts, then accept 3) Otherwise, reject

Closure properties for RE

• Theorem: If A and B are in RE, then AUB is also in RE
• Proof:

–

Let M, M’ be TMs such that L(M)=A, L(M’)=B

–

We need to build a TM such that L(M’’)=AUB. M’’(w) =

1) Run M(w). If M(w) accepts, then accept 2) Run M’(w). If M’(w) accepts, then accept 3) Otherwise, reject

Closure properties

• Theorem: If A and B are recognizable, then AUB is

recognizable

• Proof:

–

Let M, M’ be TMs such that L(M)=A, L(M’)=B

–

We build a TM for AUB, M’’(w) =

1) Run M(w). If M(w) accepts, then accept 2) Run M’(w). If M’(w) accepts, then accept 3) Otherwise, reject

Question: Is M’’ a decider? A) Yes, because it always terminate B) No, beause it may loop in step 1 C) No, because it may loop in step 2 D) I can’t decide, I am not a decider

Closure properties

• Theorem: If A and B are recognizable, then AUB is

recognizable

• Proof:

–

Let M, M’ be TMs such that L(M)=A, L(M’)=B

–

We build a TM for AUB, M’’(w) =

1) Run M(w). If M(w) accepts, then accept 2) Run M’(w). If M’(w) accepts, then accept 3) Otherwise, reject

Question: What property best describes the language of M’’? A) L(M’’) = A U B B) A ⊆ L(M’’) ⊆ AUB C) A∩B ⊆ L(M’’) ⊆ A U B D) L(M’’) = A – B E) None of the above

Closure Properties (RE)

• Can we fix the proof, and show that RE is closed under

union?

• Is RE closed under

–

Union?

–

Intersection?

–

Complement?

–

Set Difference?

Closure of RE under union

• Theorem: If A and B are recognizable, then AUB is

recognizable

• Proof:

–

Let M, M’ be TMs such that L(M)=A, L(M’)=B

–

We build a TM for AUB, M’’(w) =

• 1. For t=1,2,3,….

1) Run M(w) for t steps. If M(w) accepts, then accept 2) Run M’(w) for t steps. If M’(w) accepts, then accept 3) Otherwise, continue to next t

Closure of RE under union

• Theorem: If A and B are recognizable, then AUB is

recognizable

• Proof:

–

Let M, M’ be TMs such that L(M)=A, L(M’)=B

–

We build a TM for AUB, M’’(w) =

• 1. For t=1,2,3,….

1) Run M(w) for t steps. If M(w) accepts, then accept 2) Run M’(w) for t steps. If M’(w) accepts, then accept 3) Otherwise, continue to next t

Question: Is M’’ a decider? A) Yes, it always terminate B) No, beause it may loop in step 1 C) No, because it may loop in step 2 D) No, because of infinite loop “for t=1,2,3 ...” E) I don’t know

Summary

• Theorem: A language L is decidable if and only if it is

both recognizable and co-recognizable: D = RE ∩ coRE

RE co-RE Context-free Regular Decidable ??? ??? ??? Are there languages

• utside of RE U coRE

Are there languages in RE - Dec? Are there languages in coRE - Dec?

Undecidable languages

• Are there languages that are not decidable?

–

E.g., HALT = {<M,w> | M is a TM such that M(w) terminates}

• Claim: HALT is recognizable because it is the language of

the following TM MHALT(<M,w>) =

• 1. Run M on input w.
• 2. Accepts.

Undecidable languages

• Are there languages that are not decidable?

–

E.g., HALT = {<M,w> | M is a TM such that M(w) terminates}

• Claim: HALT is recognizable because it is the language of

the following TM MHALT(<M,w>) =

• 1. Run M on input w.
• 2. Accepts.

Question: Is the claim correct? A) Yes, because L(MHALT)=HALT B) No, because MHALT may loop in step 1 C) No, because MHALT never rejects D) It depends on the input w

Undecidable languages

• Are there languages that are not decidable?

–

E.g., HALT = {<M,w> | M is a TM such that M(w) terminates}

• Claim: HALT is recognizable because it is the language of

the following TM MHALT(<M,w>) =

• 1. Run M on input w.
• 2. Accepts.

Question: does MHALT decide HALT? A) Yes, because L(MHALT)=HALT B) No, because MHALT may loop in step 1 C) No, because MHALT never rejects D) It depends on the input w

Summary

• Theorem: A language L is decidable if and only if it is

both recognizable and co-recognizable: D = RE ∩ coRE

RE co-RE Context-free Regular Decidable ??? ??? ??? Are there languages

• utside of RE U coRE

Is HALT in RE - Dec?

Countable Sets

• You can make an infjnite list of all natural numbers

N: 0,1,2,3,4,….

• You can make an infjnite list containing all integers

Z: 0,1,-1,2,-2,3,-3,4,-4,…

• Defjnition: a set S is countable if there is a

bijection f: N → S

–

Informally: you can list the elements of S: f(0),f(1),f(2),….

–

N and Z are countable

Countable Sets

• You can make an infjnite list of all natural numbers

N: 0,1,2,3,4,….

• You can make an infjnite list containing all integers

Z: 0,1,-1,2,-2,3,-3,4,-4,…

• Defjnition: a set S is countable if there is a

bijection f: N → S

–

Informally: you can list the elements of S: f(0),f(1),f(2),….

–

N and Z are countable Question: Is any of these sets countable? Q (rational numbers), R (real numbers), C (complex numbers) A) None of them is countable B) Q is countable, but R and C are not C) Q and R are countable, but C is not because D) They are all countable

Diagonalization

• Theorem: R is not countable.
• Proof: assume for contradiction it is countable
• List its element [0,1)

(1)0.0100010… (2)0.1101000… (3)0.0110001… (4)0.1010101… (5)0.1111000…

• Flip the diagonal: r=0.10011…
• The real number r is not on the list: Contradiction!

There are non-recognizable languages

• The set of all strings {0,1}* is countable

–

Just list them in lexicographic order: [w[i]]i = ε,0,1,00,01,10,11,000,…

• The set of all Turing machines is countable because TM can be

represented by strings.

• Languages can be represented by real numbers:

–

r=0.10010101011100…. ↔ {w[i] | the ith digit of r is 1}

• The set of all languages P({0,1}*) is not countable
• RE is strictly contained in P({0,1}*) because RE is countable and

P({0,1}*) is not.

Next Time

• Can we find specific (and possibly interesting) non-

recognizable languages?

• Reading: Sipser Chapters 3 and 4
• Friday Nov 11: Veterans’ day – No class
• HW6 due on Nov 15
• Haskell 4 due Nov 21
• Exam 3: Nov 18