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On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM COT 6421 Paresh Gupta by Alan Mathison Turing.

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Presentation on theme: "On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM COT 6421 Paresh Gupta by Alan Mathison Turing."— Presentation transcript:

1 On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM COT 6421 Paresh Gupta by Alan Mathison Turing

2 2 ENTSCHEIDUNGSPROBLEM? Decision Problem “ Does there exists a fully automatic procedure or mechanical method which decides whether given mathematical statements from first order logic are universally true or false? ”

3 3 Posed by In 17 th Century Gottfried Leibnitz  Mechanical machine to decide mathematical statements 1928 - David Hilbert and Wilhelm Ackermann

4 4 Index of Turing’s Paper(1/2) 1.Computing Machines 2.Definitions Automatic Machines Computing Machines Circle and circle-free Machines Computable sequences and numbers 3.Examples of computable machines 4.Abbreviated tables Further examples

5 5 5.Enumeration of computable sequences 6.The universal computing machine 7.Detailed description of universal machine 8.Application of diagonal process 9.The extent of the computable numbers 10.Examples of large classes of numbers which are computable 11.Application to the Entsheidungsproblem Index of Turing’s Paper(2/2)

6 6 Computable Numbers Real numbers whose decimal representations are calculable by finite means Computable numbers does not mean computable functions of real or integral variable, or computable predicates Class of computable numbers is enumerable

7 7 Computing Machines Analogous to humans machine is capable of finitely many conditions to compute a number “ m-configurations ” –Example- q 1, q 2, …, q R Supplied with “ tape ”, divided into sections (called squares) each capable of bearing a symbol At any moment there is just one symbol S(r) in the r- th square is in the machine Pair (q R, S(r)) form a configuration

8 8 Automatic machines Motion of the machine is completely determined by the configurations Denoted as a-machine c-machine are choice machines Computing machines a-machine which prints figures ( 0 or 1) on its tape and some other tape symbols Binary symbols left on the tape is called the sequence computed by the machine

9 9 Circular and circle-free machines Circular –A computing machine that never writes more than a finite number of symbols ( 0’s or 1’s) –If it reaches a configuration from which there is no possible move, or moves and keeps printing same set symbols Computable sequence and numbers – A sequence is computable if it can be computed by circle-free machine –A number is computable if it differs by an integer from the number computed by a circle-free machine

10 10 Enumeration of Computable Sequences (1/3) Transition Table Standard Description ( S.D) –Concatenation of the expressions q i S j S k Lq m separated by semi-colons. –Each q i is replaced by the letter “D” followed by letter “A” repeated i times – S j is replaced by “D” followed by “C” repeated j times –Entirely made up of letters “A”, “C”, “D”, “L”, “R”, “N”, and “;” –E.g. DADDCRDAA;DAADDRDAAA;DAAADDCCRD;

11 11 Description Number (D.N) –Replacing letters “A”, “C”, “D”, “L”, “R”, “N”, and “;” by “1”, “2”, “3”,…, “7” respectively gives a description of the machine in Arabic numeral The D.N uniquely determines the structure and S.D of a machine The machine whose D.N is n is described as M(n) E.g. 31173113353111731113322531111731111335317 Enumeration of Computable Sequences (2/3)

12 12 Enumeration of Computable Sequences (3/3) To each computable sequence there corresponds at least one D.N, while to no D.N does there correspond more than one computable sequence Computable sequence and numbers are therefore enumerable Computable Sequences Description Number (D.N) Relationship between computable sequences and D.N

13 13 The Universal Computing Machine (1/2) F-squares –The symbols on F-squares form a continuous sequence E-squares –The symbols on E-squares are liable to be erased It is possible to invent a single machine which can compute any computable sequence How does U work? –If this machine U, is supplied with a tape on the beginning of which is written the S.D of the machine M, then U will compute the same sequence as M –Note that S.D of M is nothing but the rules of operation of M

14 14 The Universal Computing Machine (2/2) U bears on its tape the symbols е on an F-square and again an е on the next F-square; followed by the S.D of the machine M and a double colon “::” The instructions of S.D are separated by semi-colons Each instructions consists of five parts i.“D” followed by a sequence of letters “A”. This describes the relevant m-configuration. ii.“D” followed by a sequence of letters “C”. This describes the scanned symbol. iii.“D” followed by another sequence of letters “C”. This describes the symbol into which the scanned symbol is to be changed. iv. “L”, “R”, “N”, describing whether the machine is to move to left, right, or not at all. v.“D” followed by a sequence of letters “A”. This describes the final m-configuration.

15 15 Application of diagonal process (1/3) Turing brings out the significance of circle-free machines by application of diagonal process to prove that a computing machine might never halt Implies that it is impossible to decide whether a given number is the D.N of a circle-free machine in finite number of steps Argument that real numbers are not enumerable (by Hobson, Theory of functions of a real variable Theory of functions of a real variable in 1907 ) cannot be used to prove that computable numbers are not enumerable

16 16 Application of diagonal process (2/3) Idea is that “ If the computable sequence is enumerable, let α n be the n- th computable sequence, and let Φ n (m) be the m-th figure in α n. Let β be the sequence with 1- Φ n (n) as its n-th figure. Since β is computable, there exists a number K such that 1- Φ n (n)=Φ k (n) for all n. Putting n=k, we have 1 = 2 Φ k (K), i.e 1 is even. This is impossible therefore computable sequences are not enumerable ” This argument is incorrect because of the assumption that β is computable. It would be true if we could enumerate computable sequences by finite means

17 17 Application of diagonal process (3/3) 12m..n α0α0 α1α1 αnαn Φn(m).. β1- Φn(n)..

18 18 Halting problem (1/4) Suppose we can invent a machine D which when supplied with the S.D of a machine M will test S.D and if M is circular flag it with the symbol “u” and if it is circle free mark it with “s” Using D with U we can compute a sequence β 1 with Φn(n) as its n-th figure

19 19 Halting Problem (2/4) We can construct a Machine H, which has its motion divided into sections In the first N-1 sections integers 1,2, …,N-1 have been written down and tested by machine D In the N-th section machine D tests N if N-1 is the D.N of a circle-free machine If N is satisfactory, then H has found the first N figures of the sequence of which a D.N is N are calculated If not then H goes on to the (N+1)-th section of its motion

20 20 Halting problem (3/4) Note that H is circle-free from its construction –Because each section comes to an end after a finite number of steps ( as per our assumption D takes finite amount of time) Contradiction – H is circular –Let K be the D.N of H –In the k-th section H tests if K is satisfactory –Since H is circle-free verdict of D for K cannot be circular (“u”) –On the contrary it also cannot be circle-free (“s”) because K- th section would have to compute first R(K-1) figures of the sequence and then R(k)-th figure –But R(k)-th figure would never be found

21 21 Halting problem (4/4) Therefore H is circular and D cannot exist which can test if given the S.D of a machine M, if M is circular or circle- free in finite amount of time

22 22 Halting problem- version 2 There can be no machine ξ, which when applied with the S.D of a machine M, will determine whether M ever prints a symbol ( 0 say ) If there exists such a machine ξ, then there is a general process for determining whether machine M prints 0 infinitely often Let M1 be a machine which prints the same sequence as M, except that it prints Õ where first 0 was printed by M M2 has prints two symbols 0 replaced by Õ, and so on M3, M4,.. can be created

23 23 Halting problem- version 2 Let F be a machine which, when supplied with the S.D of M will write down successively the S.D of M,M1, M2,… Combining F with machine ξ gives us a machine G G first writes S.D of M using M and uses ξ to test if it has any 0’s in it Symbol :0: is written if it is found that M never prints a 0 Similar actions are taken on M1, M2,… If G is tested with ξ and if G never prints :0: then M prints 0 infinitely many often Similarly we can design a general process for determining if M prints 1 infinitely often By combining all of these machines we have a way of determining whether M prints infinity of figures This implies we have a machine for determining whether M is circle-free, but that is impossible Therefore machine ξ cannot exist

24 24 Application to the Entsheidungsproblem (1/4) In 1931 Hilbert and Ackermann suggested that there exists a mechanical solution to the Entsheidungsproblem In 1936 Turing proposed that there can be no general process for determining whether a given formula Ŭ of functional calculus K is provable in a finite amount of time

25 25 Application to the Entsheidungsproblem (2/4) Different from Gödel’s Incompleteness Theorem which states that there propositions Ŭ such that neither Ŭ or – Ŭ is provable Note that if negation of Gödel's Theorem is proved then we have an immediate solution to Decision problem –i.e. For propositions Ŭ, either Ŭ or – Ŭ is provable

26 26 Application to the Entsheidungsproblem (3/4) Proof a)If symbol S 1 ( i.e. 0 ) appears on the tape in some complete configuration of M, then Un(M) is provable b)If Un(M) is provable, then S 1 appears on the tape in some complete configuration of M

27 27 Application to the Entsheidungsproblem (4/4) If we assume that Entsheidungsproblem can be solved Then we can determine if Un(M) is provable This implies that there is a process for determining that M ever prints a symbol 0. But this is impossible Hence Entsheidungsproblem cannot be solved

28 28 ?’s


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