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 Decidable Context-free RE co-RE Regular
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 Question: Is M’’ a decider? A) Yes, because it always terminate ● Theorem: If A and B are recognizable, then AUB is B) No, beause it may loop in step 1 recognizable C) No, because it may loop in step 2 D) I can’t decide, I am not a decider ● 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
Closure properties Question: What property best describes the language of M’’? A) L(M’’) = A U B ● Theorem: If A and B are recognizable, then AUB is B) A ⊆ L(M’’) ⊆ AUB recognizable C) A∩B ⊆ L(M’’) ⊆ A U B D) L(M’’) = A – B ● Proof: E) None of the above 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
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 Question: Is M’’ a decider? A) Yes, it always terminate ● Theorem: If A and B are recognizable, then AUB is B) No, beause it may loop in step 1 C) No, because it may loop in step 2 recognizable D) No, because of infinite loop “for t=1,2,3 ...” ● Proof: E) I don’t know 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
Summary Are there languages outside of RE U coRE Are there languages ● Theorem: A language L is decidable if and only if it is in RE - Dec? both recognizable and co-recognizable: D = RE ∩ coRE ??? Decidable ??? ??? Context-free RE co-RE Regular 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 M HALT (<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 Question: Is the claim correct? M HALT (<M,w>) = A) Yes, because L(M HALT )=HALT B) No, because M HALT may loop in step 1 1. Run M on input w. C) No, because M HALT never rejects 2. Accepts. 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 Question: does M HALT decide HALT? M HALT (<M,w>) = A) Yes, because L(M HALT )=HALT B) No, because M HALT may loop in step 1 1. Run M on input w. C) No, because M HALT never rejects 2. Accepts. D) It depends on the input w
Summary Are there languages outside of RE U coRE Is HALT ● Theorem: A language L is decidable if and only if it is in RE - Dec? both recognizable and co-recognizable: D = RE ∩ coRE ??? Decidable ??? ??? Context-free RE co-RE Regular
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 Question: Is any of these sets countable? Q (rational numbers), R (real numbers), C (complex numbers) ● You can make an infjnite list of all natural numbers A) None of them is countable N: 0,1,2,3,4,…. B) Q is countable, but R and C are not ● You can make an infjnite list containing all integers C) Q and R are countable, but C is not because Z: 0,1,-1,2,-2,3,-3,4,-4,… D) They are all countable ● 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 –
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 i th 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
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