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

On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM

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On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM. COT 6421 Paresh Gupta. by Alan Mathison Turing. ENTSCHEIDUNGSPROBLEM?. Decision Problem - PowerPoint PPT Presentation

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Page 1: On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM

On computable numbers, with an application to the

ENTSCHEIDUNGSPROBLEM

COT 6421Paresh Gupta

byAlan Mathison Turing

Page 2: On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM

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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? ”

Page 3: On computable numbers, with an application to the ENTSCHEIDUNGSPROBLEM

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Posed by

• In 17th Century Gottfried Leibnitz Mechanical machine to decide

mathematical statements

• 1928 - David Hilbert and Wilhelm Ackermann

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Index of Turing’s Paper(1/2)

1. Computing Machines2. Definitions

Automatic MachinesComputing MachinesCircle and circle-free MachinesComputable sequences and numbers

3. Examples of computable machines

4. Abbreviated tablesFurther examples

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5. Enumeration of computable sequences6. The universal computing machine7. Detailed description of universal

machine8. Application of diagonal process9. The extent of the computable numbers10.Examples of large classes of numbers

which are computable11.Application to the

Entsheidungsproblem

Index of Turing’s Paper(2/2)

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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

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Computing Machines

• Analogous to humans machine is capable of finitely many conditions to compute a number

• “ m-configurations ” – Example- q1, q2, …, qR

• 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 (qR , S(r)) form a configuration

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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

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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

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Enumeration of Computable Sequences (1/3)

Transition Table

• Standard Description ( S.D)– Concatenation of the expressions qiSjSkLqm separated

by semi-colons.– Each qi is replaced by the letter “D” followed by letter

“A” repeated i times–

Sj 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;

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• 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)

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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

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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

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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 partsi. “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.

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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

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Application of diagonal process (2/3)

• Idea is that “ If the computable sequence is enumerable, let αn

be the n-th computable sequence, and let Φn(m) be the m-th figure in αn. Let β be the sequence with 1- Φn(n) as its n-th figure. Since β is computable, there exists a number K such that 1- Φn(n)=Φk(n) for all n. Putting n=k, we have 1 = 2 Φk(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

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Application of diagonal process (3/3)

1 2 m .. n ..

α0

α1

..

αn Φn(m)

..

β 1- Φn(n)

..

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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 β1 with Φn(n) as its n-th figure

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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

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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

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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

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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

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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

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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

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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

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Application to the Entsheidungsproblem (3/4)

• Proofa) If symbol S1( 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 S1

appears on the tape in some complete configuration of M

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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

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