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Remaining Discussions from Previous Class

Remaining Discussions from Previous Class. Please be precise in your writing Specially because some of the proofs are written in plain English Queue automata are equivalent to Turing Machines Transitions for the 2-tape Turing machine The notion of algorithm

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Remaining Discussions from Previous Class

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  1. Remaining Discussions from Previous Class • Please be precise in your writing • Specially because some of the proofs are written in plain English • Queue automata are equivalent to Turing Machines • Transitions for the 2-tape Turing machine • The notion of algorithm • Hilbert’s 10th problem (1900): “process”, “finite number of operations” • Algorithm = Turing machines • Church-Turing Thesis, 1936 • Matijaseviĉ solution to Hilbert’s 10th problem (1970) • Not decidable

  2. Turing-Enumerable Héctor Muñoz-Avila

  3. Riddle: How can we tell if two sets have the same number of elements without counting their elements? B A a b c d e f 1 2 3 4 5 6

  4. Comparing Sets Size without Counting 2 sets A and B have the same size if there is a function f: A B such that: • For every x  A there is one and only y  B such that f(x) = y (i.e., has to be a function) • Every y  B has one x  A such that f(x) = y In such situations f is said to be a bijective function

  5. f(x) = x/2 is a bijective from E to N N = {f(2), f(4), f(6) ,…} Example (2) E = {n : n is an even natural number} N = {n : n is a natural number} Does E and N have the same size? Yes:

  6. f(x) = 1/x is a bijective from R1 to (0,1] 1 same size! 1 Example (3) R1 = {r : r is a real positive number greater than 1} (0,1] = {r : r is a real number between 0 and 1} Does R1 and (0,1] have the same size? Yes:

  7. Example (4) How about: R+ = {r : r is a real non-negative number} N = {n : n is a natural number} Every attempt fails: f(x) = x (leaves numbers like 0.5 out) f(x) = floor(x) (assigns the same value for numbers like 1.2 and 1.3) How can we know for sure that there is no bijective function from R+ to N?

  8. Enumerability • We know that there is an enumeration for all the natural numbers: 1, 2, 3, 4, … , 101000, … The point is that for any natural, say 101000, it will eventually be listed! • There is no such enumeration for [0,1), the set of all the real numbers between 0 and 1 (i.e., 0, 0.01, 0.1003, 3/) • Thus there can’t be any bijective function f: N [0,1), otherwise: {f(0), f(1), f(2), …} would be an enumeration for [0,1) • Surprisingly there is a enumeration for the rational numbers (the irrational numbers are the ones that are non enumerable!)

  9. The Rational Numbers are Enumerable • The set of all rational numbers: {p/q : p, q are natural numbers} is enumerable: Note: you could easily write a program in C++ that prints this enumeration (and runs forever) 1 2 3 4 5 … q … 1 1/1 1/2 1/3 1/4 1/5 … 1/q … 2 2/1 2/2 2/3 2/4 2/5 … 2/q … 3 3/1 … 4 4/1 … 5 5/1 … 5/q … … p p/1 … p/q … … Enumeration: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, …, p/q, …

  10. [0,1) Is Not Enumerable By contradiction: suppose that there is an enumeration for all the real numbers between 0 and 1: # 1: 0.012304565 ... # 2: 0.10002344345 ... # 3: 0.865732546789 … … #23: 0.434555…6… …

  11. The 23-rd digit [0,1) Is Not Enumerable (II) We construct a number  as follows: for each number n in the enumeration, we look at the n-th digit in n: #1: 0.012304565 ... #2: 0.10002344345 ... #3: 0.865732546789 … … #23: 0.434555…6… …  = 0.005…6… Obviously  is a real number between 0 and 1

  12.  = 0.120…7… Obviously  is a real number between 0 and 1 Question: is  = #1? or  = #2? or … or  = #23? or … …  … [0,1) Is Not Enumerable(III) We construct a number  as follows: we change each digit in  for a different digit:  = 0.005…6…  #1: 0.012304565 ... #2: 0.10002344345 ... #3: 0.865732546789 … … #23: 0.434555…6… … Thus, it is not possible to enumerate all the real numbers between 0 and 1!  

  13. Summary of Enumerability • Two sets have the same cardinality (read: size) if there is a bijective function from one into the other one • The set of the natural numbers is enumerable • The set of all rational numbers are enumerable • Therefore, the set of natural numbers has the same “cardinality” = as the set of rational numbers • The set of real numbers is not enumerable • Therefore, the cardinality of the real numbers is larger than the cardinality of the natural numbers

  14. Consequences • This means that even though the natural and the real numbers are both infinite, the size of the set of the real numbers is “bigger” than the size of the set of the natural numbers. • This has been known for mathematicians for quite a long time • This is all nice and beautiful, but what the %$%%^# does this has to do with Turing machines? • Astonishingly, this result is relevant for Turing machines!

  15. Enumerability and Turing Machines Definition: A language L is Turing-enumerable if there is a Turing machine that enumerates all words in L in its tape (may run forever): w1 w2 w3… • Our book does not define Turing-enumerability • Rather it says that there is an Enumerator Turing machine that enumerates all words in L • These two notions are equivalent

  16. 2 2 1 a1 a2 1 2 compute successor word in second tape copy word from second tape into first tape * is Turing-Enumerable Lemma. If  is finite then * is enumerable M(*): (for  = {a,b}) (the first tape acts as the printer)

  17. T’: • Uses 3 tapes • Machine deciding L runs in Tape 2 1 If w is in L T Run M(*) stepwise in Tape 3 0 If w is not in L T Tape 3: w1w2… Decidability implies Turing-enumerability Theorem 1. If a language L is decidable then the language is Turing-enumerable • Given: T is the Turing machine that decides L • Construct: T’ a Turing machine that enumerates L • Copy latest word w output in Tape 3 into Tape 2, if w is in L, append it to the end of Tape 1 and add a blank afterwards

  18. We are recognizing if w is in the language T’: T • Uses 2 tapes • Put word w in Tape 1 Tape: w1w2… Run T stepwise in Tape 2 T Tape 2: w1w2… Turing-Enumerability Implies Semidecidability Theorem 2. If a language L is Turing-enumerablethen the language is Turing-recognizable • Given: T is the Turing machine that enumerates L • Construct: T’ a Turing machine that recognizes L • If the latest word output in Tape 2 is equal to word in Tape 1 then halt

  19. Other Enumerability Results Theorem 4. If a language L is Turing-recognizable then L is Turing-enumerable Theorem 5. The set of all Turing Machines is enumerable Theorem 6. The set of all functions f: N  N is not enumerable Theorem 7. There exists f: N N that is not Turing-recognizable

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