Busch Complexity Lectures: A Universal Turing Machine. Costas Busch - LSU 1

Transcription

1 Busch Complexity Lectures: A Universal Turing Machine Costas Busch - LU

2 A limitation of Turing Machines: Turing Machines are hardwired they execute only one program Real Computers are re-programmable Costas Busch - LU

3 olution: Universal Turing Machine Attributes: Reprogrammable machine imulates any other Turing Machine Costas Busch - LU 3

4 Universal Turing Machine simulates any Turing Machine M Input of Universal Turing Machine: Description of transitions of M Input string of M Costas Busch - LU 4

5 Three tapes Tape Description of M Universal Turing Machine Tape Tape Contents of M Tape 3 tate of M Costas Busch - LU 5

6 Tape Description of M We describe Turing machine as a string of symbols: M We encode M as a string of symbols Costas Busch - LU 6

7 Alphabet Encoding ymbols: a b c d Encoding: Costas Busch - LU 7

8 tate Encoding tates: q q q3 q 4 Encoding: Head Move Encoding Move: L R Encoding: Costas Busch - LU 8

9 Transition Encoding Transition: δ ( q, a) = ( q, b, L) Encoding: 0000 separator Costas Busch - LU 9

10 Turing Machine Encoding Transitions: δ ( q, a) = ( q, b, L) δ ( 3 q, b) = ( q, c, R) Encoding: separator Costas Busch - LU 0

11 Tape contents of Universal Turing Machine: binary encoding of the simulated machine M Tape Costas Busch - LU

12 A Turing Machine is described with a binary string of 0 s and s Therefore: The set of Turing machines forms a language: each string of this language is the binary encoding of a Turing Machine Costas Busch - LU

13 Language of Turing Machines L = { 00000, , , (Turing Machine ) (Turing Machine ) } Costas Busch - LU 3

14 Countable ets Costas Busch - LU 4

15 Infinite sets are either: Countable or Uncountable Costas Busch - LU 5

16 Countable set: There is a one to one correspondence (injection) of elements of the set to Positive integers (,,3, ) Every element of the set is mapped to a positive number such that no two elements are mapped to same number Costas Busch - LU 6

17 Example: The set of even integers is countable Even integers: (positive) 0,, 4, 6, Correspondence: Positive integers:,, 3, 4, n corresponds to n + Costas Busch - LU 7

18 Example: The set of rational numbers is countable Rational numbers:, 3, 4 7 8, Costas Busch - LU 8

19 Naïve Approach Rational numbers: Nominator,,, 3 Correspondence: Positive integers:,, 3, Doesn t work: we will never count numbers with nominator :,, 3, Costas Busch - LU 9

20 Better Approach Costas Busch - LU 0

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26 Rational Numbers:,,,, 3, Correspondence: Positive Integers:,, 3, 4, 5, Costas Busch - LU 6

27 We proved: the set of rational numbers is countable by describing an enumeration procedure (enumerator) for the correspondence to natural numbers Costas Busch - LU 7

28 Let Definition be a set of strings (Language) An enumerator for that generates (prints on tape) all the strings of and is a Turing Machine one by one each string is generated in finite time Costas Busch - LU 8

29 strings s, s, s3, Enumerator Machine for output (on tape) s, s, s3, Finite time: t, t, t3, Costas Busch - LU 9

30 Enumerator Machine Time 0 Configuration q 0 Time t x # s q s prints s Costas Busch - LU 30

31 Time t prints s x # s q s Time t 3 prints s 3 x3 # s 3 q s Costas Busch - LU 3

32 Observation: If for a set then the set is countable there is an enumerator, The enumerator describes the correspondence of to natural numbers Costas Busch - LU 3

33 Example: The set of strings is countable = { a, b, c } + Approach: We will describe an enumerator for Costas Busch - LU 33

34 Naive enumerator: Produce the strings in lexicographic order: s s = = a aa aaa aaaa... Doesn t work: strings starting with will never be produced b Costas Busch - LU 34

35 Better procedure: Proper Order (Canonical Order). Produce all strings of length. Produce all strings of length 3. Produce all strings of length 3 4. Produce all strings of length 4... Costas Busch - LU 35

36 Produce strings in Proper Order: s s = = a b c aa ab ac ba bb bc ca cb cc aaa aab aac... length length length 3 Costas Busch - LU 36

37 Theorem: The set of all Turing Machines is countable Proof: Any Turing Machine can be encoded with a binary string of 0 s and s Find an enumeration procedure for the set of Turing Machine strings Costas Busch - LU 37

38 Enumerator: Repeat. Generate the next binary string of 0 s and s in proper order. Check if the string describes a Turing Machine if YE: print string on output tape if NO: ignore string Costas Busch - LU 38

39 Binary strings Turing Machines s s End of Proof Costas Busch - LU 39

40 Uncountable ets Costas Busch - LU 40

41 We will prove that there is a language Lʹ which is not accepted by any Turing machine Technique: Turing machines are countable Languages are uncountable (there are more languages than Turing Machines) Costas Busch - LU 4

42 Theorem: If is an infinite countable set, then the powerset of is uncountable. The powerset is the set whose elements are all possible sets made from the elements of Example: = { a, b} = {,{ a},{ b},{ a, b}} Costas Busch - LU 4

43 Proof: ince is countable, we can write = { s, s, s3, } Element of Costas Busch - LU 43

44 Elements of the powerset have the form: { s, s 3} { s5, s7, s9, s 0 } Costas Busch - LU 44

45 We encode each element of the powerset with a binary string of 0 s and s Powerset element (in arbitrary order) Binary encoding s s s 3 s 4 { s } { s, s 3} 0 0 { s, s3, s 4 } 0 Costas Busch - LU 45

46 Observation: Every infinite binary string corresponds to an element of the powerset: Example: Corresponds to: { s, s, s, s, } Costas Busch - LU 46

47 Let s assume (for contradiction) that the powerset is countable Then: we can enumerate the elements of the powerset = 3 { t, t, t, } Costas Busch - LU 47

48 Powerset element suppose that this is the respective Binary encoding t t t t Costas Busch - LU 48

49 Take the binary string whose bits are the complement of the diagonal t t t 3 t Binary string: t = 00 (birary complement of diagonal) Costas Busch - LU 49

50 The binary string t =00 corresponds to an element of the powerset : t = s, s, } { 3 4 Costas Busch - LU 50

51 Thus, t must be equal to some ti t =t i However, the i-th bit in the encoding of the complement of the i-th bit of, thus: t is ti t t i Contradiction Costas Busch - LU 5

52 ince we have a contradiction: The powerset of is uncountable End of proof Costas Busch - LU 5

53 The set of all strings: = { a, b} * An Application: Languages Consider Alphabet : A = { a, b} = { λ, a, b, aa, ab, ba, bb, aaa, aab, } infinite and countable because we can enumerate the strings in proper order Costas Busch - LU 53

54 Consider Alphabet : A = { a, b} The set of all strings: = { a, b} * = { λ, a, b, aa, ab, ba, bb, aaa, aab, } infinite and countable Any language is a subset of : L = { aa, ab, aab} Costas Busch - LU 54

55 Consider Alphabet : A = { a, b} The set of all trings: = A * = { a, b} * = { λ, a, b, aa, ab, ba, bb, aaa, aab, } infinite and countable The powerset of contains all languages: = {,{ λ},{ a},{ a, b},{ aa, b},...,{ aa, ab, aab}, } uncountable Costas Busch - LU 55

56 Consider Alphabet : A = { a, b} Turing machines: Languages accepted accepts By Turing Machines: countable M M M 3 L countable L L 3 Denote: X = { L, L, L 3, } countable Note: X * ( = { a, b} ) Costas Busch - LU 56

57 Languages accepted by Turing machines: X countable All possible languages: uncountable Therefore: X ( since X, we get X ) Costas Busch - LU 57

58 Conclusion: There is a language Lʹ by any Turing Machine: not accepted X L ʹ and Lʹ X (Language Lʹ by any algorithm) cannot be described Costas Busch - LU 58

59 Non Turing-Acceptable Languages Lʹ Turing-Acceptable Languages Costas Busch - LU 59

60 Note that: X = { L, L, L, } 3 is a multi-set (elements may repeat) since a language may be accepted by more than one Turing machine However, if we remove the repeated elements, the resulting set is again countable since every element still corresponds to a positive integer Costas Busch - LU 60