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 10 th problem (1900): “process”, “finite number of operations” –Algorithm = Turing machines Church-Turing Thesis, 1936 –Matijaseviĉ solution to Hilbert’s 10 th problem (1970) Not decidable

Turing-Enumerable Héctor Muñoz-Avila

Riddle: How can we tell if two sets have the same number of elements without counting their elements? abcdefabcdef A B

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

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: f(x) = x/2 is a bijective from E to N N = {f(2), f(4), f(6),…}

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

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?

Enumerability We know that there is an enumeration for all the natural numbers: 1, 2, 3, 4, … The point is that for any natural, say , 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, , 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!)

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

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

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

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

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

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 Astonishingly, this result is relevant for Turing machines! This is all nice and beautiful, but what the %$%^# does this has to do with Turing machines?

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):  w 1 w 2 w 3 … 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

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

Decidability implies Turing- enumerability Theorem 1. If a language L is decidable then the language is Turing-enumerable T 1If w is in L 0 If w is not in L Given: T is the Turing machine that decides L Construct: T’ a Turing machine that enumerates L T w1w2…w1w2… Tape 3: Run M(  *) stepwise in Tape 3 T’: Uses 3 tapes  Machine deciding L runs in Tape 2  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

Turing-Enumerability Implies Semidecidability Theorem 2. If a language L is Turing-enumerable then the language is Turing-recognizable If the latest word output in Tape 2 is equal to word in Tape 1 then halt Given: T is the Turing machine that enumerates L Construct: T’ a Turing machine that recognizes L T’: Uses 2 tapes  Put word w in Tape 1  T w1w2…w1w2… Tape: We are recognizing if w is in the language T w1w2…w1w2… Tape 2: Run T stepwise in Tape 2

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