cse 105
play

CSE 105 THEORY OF COMPUTATION Fall 2016 - PowerPoint PPT Presentation

Sep 06, 2022 •307 likes •554 views

CSE 105 THEORY OF COMPUTATION Fall 2016 http://cseweb.ucsd.edu/classes/fa16/cse105-abc/ T odays lecture Countable and uncountable sets Diagonalization There are languages not in RE or coRE An interesting undecidable language

  1. CSE 105 THEORY OF COMPUTATION Fall 2016 http://cseweb.ucsd.edu/classes/fa16/cse105-abc/

  2. T oday’s lecture ● Countable and uncountable sets ● Diagonalization ● There are languages not in RE or coRE ● An interesting undecidable language

  3. 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 –

  4. 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 –

  5. Diagonalization ● Theorem: R is not countable. Even [0,1) is uncountable! ● 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!

  6. Some countable sets ● The set of all strings {0,1}* is countable Just list them in lexicographic order: ε,0,1,00,01,10,11,000,… – Lexicographic order: fjrst sort by length, then alphabetically. – ● The set TM={<M> | M is a T uring Machine} ● RE: The set of all T uring Recognizable languages L(M 1 ), L(M2),...L(Mk),… – ● coRE: L(M 1 ), L(M 2 ),...L(M k ),… ● RE U coRE

  7. Some uncountable sets ● Recall: {0,1}*= {w[i] | i in N} is countable ● R is uncountable ● [0,1] is uncountable ● The set of all languages P({0,1}*) is not countable Languages can be represented by real numbers: – r=0.10010101011100…. ↔ {w[i] | the i th digit of r is 1} – Difgerent reals are mapped to difgerent languages – There are at least as many languages as r in [0,1] –

  8. Summary uncountable L ● RE U coRE is countable ● P({0,1}*) is uncountable ● There is a language L in D P({0,1}*) - (RE U coRE)! Context-free RE coRE Regular countable

  9. Undecidable Lanugages ● There are uncountably many undecidable languages! ● In fact, there are uncountably many languages that are not even in RE (or coRE)! ● Questions: Can we find a specific language not in RE or coRE? – Can we find interesting languages not in RE or coRE? – Is HALT TM undecidable? –

  10. Diagonalization ● [0,1) is uncountable because given any list of r in [0,1) ● 0.0100010… ● 0.1101000… ● 0.0110001… ● 0.1010101… ● 0.1111000… we can build an r=0.10011… that is not in the list ● We can make a list of all L in RE: L(M 1 ), L(M 2 ),L(M 3 )… ● Can we build a language not in this list?

  11. A language not in RE ● We want a language L that is difgerent from L(M 1 ) – difgerent from L(M 2 ) – difgerent from L(M 3 ) – ….. – difgerent from L(M k ) – ….. –

  12. A language not in RE Two numbers are different if they ● We want a language L that is differ at some digit difgerent from L(M 1 ) – Two languages are different if they differ at some string w: difgerent from L(M 2 ) – Either w in L(M) but not in L, difgerent from L(M 3 ) Or w in L but not in L(M) – ….. – difgerent from L(M k ) – W? ….. – W?

  13. A language not in RE ● We want a language L that is difgerent from L(M 1 ) at <M 1 > – difgerent from L(M 2 ) at <M 2 > – difgerent from L(M 3 ) at <M 3 > – ….. – difgerent from L(M k ) at <M k > – ….. – ● L = {<M> | M it a TM such that <M> is not is L(M) }

  14. A “diagonal” language ● Diag={<M> | M is a TM s.t. <M> is not in L(M) } ● Why is Diag difgerent from L(M k )? We need a string w that belongs to one but not the other – Let w = <M k > – If <M k > is in L(M k ), then – Not (Not “<M k > is in L(M k )”) ● “<M k > is not in L(M k )” is false ● <M k > is not in Diag ● If w is not in Diag, then w is in L(M k ) –

  15. A “diagonal” language ● Diag={<M> | M is a TM s.t. <M> is not in L(M) } ● Why is Diag difgerent from L(M k )? Question: What can you tell about Diag? We need a string w that belongs to one but not the other – Let w = <M k > – A) Diag is in RE, but not in coRE If <M k > is in L(M k ), then – B) Diag is coRE, but not in RE Not (Not “<M k > is in L(M k )”) ● C) Diag is neither in RE nor in coRE “<M k > is not in L(M k )” is false ● D) Diag is decidable <M k > is not in Diag ● E) I don’t know If w is not in Diag, then w is in L(M k ) –

  16. Diag is in coRE ● Diag={<M> | M is a TM s.t. <M> is not in L(M) } ● Here is a recognizer for Diag M diag (w) = 1) Check if w=<M> for some TM M. If not, accept. 2) Parse w as <M> for some TM M 3) Run M on input w 4) If M(w) accepts, then accept, else reject.

  17. Is M diag a decider? Diag is in coRE A) Yes, because L(M diag )=Diag ● Diag={<M> | M is a TM s.t. <M> is not in L(M) } B) No, because it can loop at step 1 C) No, because it can loop at step 2 ● Here is a recognizer for Diag D) No, because it can loop at step 3 M diag (w) = E) I don’t know 1) Check if w=<M> for some TM M. If not, accept. 2) Parse w as <M> for some TM M 3) Run M on input w 4) If M(w) accepts, then accept, else reject.

  18. What can you say about Diag Diag is in coRE A) Diag is decidable B) Diag is in RE, but not coRE ● Diag={<M> | M is a TM s.t. <M> is not in L(M) } C) Diag is in coRE, but not RE ● Here is a recognizer for Diag D) Diag is neither in RE nor coRE E) I don’t know M diag (w) = 1) Check if w=<M> for some TM M. If not, accept. 2) Parse w as <M> for some TM M 3) Run M on input w 4) If M(w) accepts, then accept, else reject.

  19. Summary ● Diag is in coRE because L(M diag )=Diag ● Diag is not in RE because Diag ≠ L(M k ) for all TM M k ● Diag is undecidable ● Diag is in coRE, but not RE ● So far: We have found specific languages not in RE or coRE. ● Can we find more interesting examples? What about HALT TM = {<M,w> | M(w) terminates}? – What about A TM = {<M,w> | M(w) accepts}? –

  20. A TM is undecidable ● Assume for contradiction A TM is decidable ● A TM is also decidable ● MM = {<M,<M>> | M is a TM} is decidable ● DD = MM ∩ A TM is decidable ● Notice: DD = { <M,<M>> | M is a TM such that M is not in L(M) } –

  21. What can you say about DD? A TM is undecidable A) DD is decidable ● Assume for contradiction A TM is decidable B) DD is undecidable because DD = Diag ● A TM is also decidable C) DD is undecidable but ● MM = {<M,<M>> | M is a TM} is decidable I am not sure why D) DD is both decidable and ● DD = MM ∩ A TM is decidable undecidable ● Notice: E) I don’t know DD = { <M,<M>> | M is a TM such that M is not in L(M) } –

  22. A TM is undecidable ● Assume for contradiction A TM is decidable ● DD = { <M,<M>> | M is a TM such that M is not in L(M) } is decidable. Let M DD be a decider for DD. ● M’ diag (<M>) = Run M DD (<M,<M>>) L(M’ diag )=Diag – M’ diag is a decider (M’ diag always terminates) – ● This proves that Diag is decidable: contradiction!

  23. For next Time ● Try to prove that HALT TM is undecidable ● Reading: Sipser Chapters 3 and 4 ● HW6 due on Nov 15 ● Haskell 4 due Nov 21 ● Exam 3 : Nov 18

Recommend

More recommend