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Theory of Computation 1 Theory of Computation Peer Instruction Lecture Slides by Dr. Cynthia Lee, UCSD are licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Based on a work at www.peerinstruction4cs.org.Dr. Cynthia Lee, UCSDCreative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported Licensewww.peerinstruction4cs.org

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INFINITE LOOPS “ARE WE THERE YET?” Flashback from CSE 8A/B: 2

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If there exists a TM M that accepts every string in L, and rejects all other strings, then L is: a)Turing decidable b)Turing recognizable, but not decidable c)Not enough information 3

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If there exists a TM M that accepts every string in L, and loops for all strings not in L, then L is: a)Turing decidable b)Turing recognizable, but not decidable c)Not enough information 4

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If there exists a for all TM M such that M recognizes L, M is not a decider, then L is: a)Turing decidable b)Turing recognizable, but not decidable c)Not enough information 5

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DECIDABILITY IMPORTANT DECIDABLE LANGUAGES More Proofs 6

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String Encoding Notation Turing Machines always take a single input string of finite length – M(w) But sometimes we want them to take as input several strings, or an object, or several objects We use to denote “encode this as a single string” – Several strings can be combined into one string by using a delimiter character not in the main alphabet – Object(s) can be encoded as a finite string A video V can be encoded as a (very long) string of 0’s and 1’s The diagram for TM M can be encoded as a string Three TM’s M 1, M 2 and M 3 7

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String Encoding Notation Since we already know these encodings can be done, we just put triangle brackets around them to denote “encode this as a string” – For standard objects (TMs, DFAs, etc), no further explanation is necessary—use book as your guide NOTE! Languages CANNOT generally be encoded as a string because if there are infinitely many strings, there is no way to concatenate all the strings together into a single finite string – DFAs, NFAs, REs, PDAs, CFGs and TMs can all be encoded as strings ( etc) – However their languages cannot

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Prove that the class of Turing-decidable languages is closed under Union Given: Two decidable languages A and B, and TMs that decide them, M A and M B. Want to Show: A TM M U that decides A U B. Construction: – M U (w) = //w is a string 1.Simulate running M A (w) – If it accepts, accept. If it rejects, go to step 2: 2.Simulate running M B (w) – If it accepts, accept. If it rejects, reject. Correctness: //this is for you to debate! Conclusion: M U is a TM that decides AUB, therefore AUB is decidable, and decidable languages are closed under union. QED. This proof is: (a)Correct (b)Incorrect

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Prove that the class of Turing-recognizable languages is closed under Union Given: Two Turing-recognizable languages A and B, and TMs that recognize them, M A and M B. Want to Show: A TM M U that recognizes A U B. Construction: – M U (w) = //w is a string 1.Simulate running M A (w) – If it accepts, accept. If it rejects, go to step 2: 2.Simulate running M B (w) – If it accepts, accept. If it rejects, reject. Correctness: //this is for you to debate! Conclusion: M U is a TM that recognizes AUB, therefore AUB is Turing-recognizable, and Turing-recognizable languages are closed under union. QED. This proof is: (a)Correct (b)Incorrect

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Prove that the class of Turing-recognizable languages is closed under Intersection Given: Two Turing-recognizable languages A and B, and TMs that recognize them, M A and M B. Want to Show: A TM M I that recognizes A intersect B. Construction: – M I (w) = //w is a string 1.Simulate running M A (w) – If it rejects, reject. If it accepts, go to step 2: 2.Simulate running M B (w) – If it accepts, accept. If it rejects, reject. Correctness: //this is for you to debate! Conclusion: M I is a TM that recognizes A intersect B, therefore A intersect B is Turing-recognizable, and Turing- recognizable languages are closed under intersection. QED. This proof is: (a)Correct (b)Incorrect

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Prove that the class of Turing-decidable languages is closed under Intersection Given: Two decidable languages A and B, and TMs that decide them, M A and M B. Want to Show: A TM M I that decides A intersect B. Construction: – M I (w) = //w is a string 1.Simulate running M A (w) – If it rejects, reject. If it accepts, go to step 2: 2.Simulate running M B (w) – If it accepts, accept. If it rejects, reject. Correctness: //this is for you to debate! Conclusion: M I is a TM that decides A intersect B, therefore A intersect B is decidable, and decidable languages are closed under intersection. QED. This proof is: (a)Correct (b)Incorrect

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Prove that the class of Turing-decidable languages is closed under Complement Given: A decidable language A, and a TM that decides it, M A. Want to Show: A TM M C that decides the complement of A. Construction: – M C (w) = //w is a string 1.Simulate running M A (w) – If it accepts, reject. If it rejects, accept. Correctness: //this is for you to debate! Conclusion: M C is a TM that decides the complement of A, therefore the complement of A is decidable, and decidable languages are closed under complement. QED. This proof is: (a)Correct (b)Incorrect

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Prove that the class of Turing-recognizable languages is closed under Complement Given: A Turing-recognizable language A, and a TM that recognizes it, M A. Want to Show: A TM M C that recognizes the complement of A. Construction: – M C (w) = //w is a string 1.Simulate running M A (w) – If it accepts, reject. If it rejects, accept. Correctness: //this is for you to debate! Conclusion: M C is a TM that recognizes the complement of A, therefore the complement of A is Turing- recognizable, and Turing-recognizable languages are closed under complement. QED. This proof is: (a)Correct (b)Incorrect

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Co-Turing-Recognizable Languages Our current classes of langauges: – Regular – Context-free – Turing-decidable (AKA just “decidable”) – Turing-recognizable (or r.e.) – co-Turing-recognizable (or co-r.e.) We now have the new class “co-Turing-recognizable” (or “co-r.e.” for short) – Language A is co-r.e. if the complement of A is r.e. 15

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CARDINALITY INFINITY AND INFINITIES To infinity, and beyond! (really) 16

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Set Theory and Sizes of Sets How can we say that two sets are the same size? Easy for finite sets--what about infinite sets? Georg Cantor (1845-1918), who invented Set Theory, proposed a way of comparing the sizes of two sets that does not involve counting how many things are in each – Works for both finite and infinite SET SIZE EQUALITY: – Two sets are the same size if there is a bijective (one-to-one and onto) function mapping from one to the other – Intuition: neither set has any element “left over” in the mapping 17

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One-to-one and Onto f is one-to-one but NOT onto. Does this prove that |N| ≠ |N|? Draw a function that is onto but not one-to-one. 18

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One-to-one and Onto f is: a)One-to-One b)Onto c)Correspondence (both (a) and (b)) d)Neither 19

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One-to-one and Onto f is: a)One-to-One b)Onto c)Correspondence (both (a) and (b)) d)Neither 20

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It gets even weirder: Rational Numbers 21 1/11/21/31/41/51/6… 2/12/22/32/42/52/6… 3/13/23/33/43/53/6… 4/14/24/34/44/54/6… 5/15/25/35/45/55/6... 6/16/26/36/46/56/6 ………………… Q = {m/n | m,n are in N}

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Sizes of Infinite Sets The number of Natural Numbers is equal to the number of positive Even Numbers, even though one is a proper subset of the other! – |N| = |E+|, not |N| = 2|E+| The number of Rational Numbers is equal to the number of Natural Numbers – |N| = |Q|, not |Q| ≈ |N| 2 But it gets even weirder than that: It might seem like Cantor’s definition of “same size” for sets is so overly broad, that any two sets of infinite size could be proven to be the “same size” – Not so!!!! 22

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Thm. |R| != |N| Proof by contradiction: Assume |R| = |N|, so a correspondence f exists between N and R. 23 Want to show: no matter how f is designed (we don’t know how it is designed so we can’t assume anything about that), it cannot work correctly. Specifically, we will show a number z in R that can never be f(n) for any n, no matter how f is designed. Therefore f is not onto, a contradiction.

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Thm. |R| != |N| Proof by contradiction: Assume a correspondence f exists between N and R. 24 nf(n) 1.100000… 2.333333… 3.314159… …… We construct z as follows: z’s n th digit is the n th digit of f(n), PLUS ONE* (*wrap to 1 if the digit is 9) Below is an example f What is z in this example? a).244… b).134… c).031… d).245…

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Thm. |R| != |N| Proof by contradiction: Assume a correspondence f exists between N and R. 25 nf(n) 1.d 1 1 d 1 2 d 1 3 d 1 4 … 2.d 2 1 d 2 2 d 2 3 d 2 4 … 3.d 3 1 d 3 2 d 3 3 d 3 4 … …… We construct z as follows: z’s n th digit is the n th digit of f(n), PLUS ONE* (*wrap to 1 if the digit is 9) Below generalized version of f(n) What is z? a).d 1 1 d 1 2 d 1 3 … b).d 1 1 d 2 2 d 3 3 … c).[d 1 1 +1] [d 2 2 +1] [d 3 3 +1] … d).[d 1 1 +1] [d 2 1 +1] [d 3 1 +1] …

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Thm. |R| != |N| Proof by contradiction: Assume a correspondence f exists between N and R. 26 nf(n) 1.d 1 1 d 1 2 d 1 3 d 1 4 … 2.d 2 1 d 2 2 d 2 3 d 2 4 … 3.d 3 1 d 3 2 d 3 3 d 3 4 … …… How do we reach a contradiction? Must show that z cannot be f(n) for any n How do we know that z ≠ f(n) for any n? a)We can’t know if z = f(n) without knowing what f is and what n is b)Because z’s n th digit differs from n‘s n th digit c)Because z’s n th digit differs from f(i)’s n th digit

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Thm. |R| != |N| 27 Proof by contradiction: Assume |R| = |N|, so a correspondence f exists between N and R. Want to show: f cannot work correctly. Let z = [z’s n th digit = (n th digit of f(n)) + 1]. Note that z is in R, but for all n in N, z != f(n). Therefore f is not onto, a contradiction. So |R| ≠ |N| |R| > |N|

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Diagonalization 28 nf(n) 1.d 1 1 d 1 2 d 1 3 d 1 4 d 1 5 d 1 6 d 1 7 d 1 8 d 1 9 … 2.d 2 1 d 2 2 d 2 3 d 2 4 d 2 5 d 2 6 d 2 7 d 2 8 d 2 9 … 3.d 3 1 d 3 2 d 3 3 d 3 4 d 3 5 d 3 6 d 3 7 d 3 8 d 3 9 … 4.d 4 1 d 4 2 d 4 3 d 4 4 d 4 5 d 4 6 d 4 7 d 4 8 d 4 9 … 5.d 5 1 d 5 2 d 5 3 d 5 4 d 5 5 d 5 6 d 5 7 d 5 8 d 5 9 … 6.d 6 1 d 6 2 d 6 3 d 6 4 d 6 5 d 6 6 d 6 7 d 6 8 d 6 9 … 7.d 7 1 d 7 2 d 7 3 d 7 4 d 7 5 d 7 6 d 7 7 d 7 8 d 7 9 … 8.d 8 1 d 8 2 d 8 3 d 8 4 d 8 5 d 8 6 d 8 7 d 8 8 d 8 9 … 9.d 9 1 d 9 2 d 9 3 d 9 4 d 9 5 d 9 6 d 9 7 d 9 8 d 9 9 … ……

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Some infinities are more infinite than other infinities 29 Natural numbers are called countable Any set that can be put in correspondence with N is called countable (ex: E+, Q) Real numbers are called uncountable Any set that can be put in correspondence with R is called “uncountable” But it gets even weirder… There are more than two categories!

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Some infinities are more infinite than other infinities 30 |N| is called א 0 |E+| = |Q| = א 0 |R| is maybe א 1 Although we just proved that |N| < |R|, and nobody has ever found a different infinity between |N| and |R|, mathematicians haven’t proved that there are not other infinities between |N| and |R|, making |R| = א 2 or greater Sets exist whose size is א 0, א 1, א 2, א 3 … An infinite number of aleph numbers! An infinite number of different infinities

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Famous People: Georg Cantor (1845-1918) His theory of set size, in particular transfinite numbers (different infinities) was so strange that many of his contemporaries hated it – Just like many CSE 105 students! “scientific charlatan” “renegade” “corrupter of youth” “utter nonsense” “laughable” “wrong” “disease” “I see it, but I don't believe it!” –Georg Cantor 31 “The finest product of mathematical genius and one of the supreme achievements of purely intellectual human activity.” –David Hilbert

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What does this have to do with computer science? Or Turing Machines? The set of all possible Turing Machines is countable The set of all possible languages is uncountable Therefore…. a)The language NATURAL = { | w is a natural number} is not Turing decidable b)Some languages are not Turing decidable c)Some languages are not Turing recognizable d)Some Turing Machines don’t recognize any language e)None or more than one of these That’s a start, but more diagonalization gets us a more specific result 32

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PARADOXXODARAP Preparing for DIAGONALIZATION! 33 XODARAPPARADOX

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Warm-up You can prove a language is Regular by drawing a DFA that recognizes the language. a)TRUE b)FALSE c)Not enough information to decide between (a) and (b) d)Other Dr. Lee wrote down number 5. a)TRUE b)FALSE c)Not enough information to decide between (a) and (b) d)Other 34

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This sentence is false. a)TRUE b)FALSE c)Not enough information to decide between (a) and (b) d)Other 35

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Liar’s Paradox “This sentence is false.” – This has been perplexing people since at least the Greeks in 4 th century BCE (2300 years!) What are some key features of this that make it a paradox? 36

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Two more examples This sentence is 36 characters long. a)TRUE b)FALSE c)Not enough information to decide between (a) and (b) d)Other 37 This sentence is in French. a)TRUE b)FALSE c)Not enough info d)Other

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The Barber A certain town has only one barber (a man). Every man in the town is clean-shaven. For each man m in the town, the barber shaves m if and only if m does not shave himself. Question: Does the barber shave himself? a)YES b)NO c)Not enough information d)Other 38

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Making Lists Suppose you have many, many lists. So many, in fact, that some of your lists are lists of lists (to help you organize your lists), and some of them even include themselves. 39

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List Organization Question Since some of your lists include themselves, and you know that self-reference is dangerous business, you make a list of all lists that do not include themselves. Call this list NON-DANGER-LIST. Question: – Should NON-DANGER-LIST include itself? a)YES b)NO c)Not enough information d)Other 40

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Grandparent Paradox (Time Travel Paradox) You travel back in time and prevent one pair of your biological grandparents from ever meeting each other (assume this prevents your birth). Pop culture version: 41

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CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative.

CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative.

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