while programs

8/19/2019 While Programs

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WHILE PROGRAMS

Computability


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Computability. Grado en Ingeniería Informática del Software. Universidad de Oviedo 17/11/14

! Understand the concept of Computation Model! Design While Programs and Turing Machines.!

Use Church Thesis.

Goals

What can computers do in principle; and, more important,what is it that they in principle cannot?

1


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! What is the nature of computation?

! What can be computed?!

What can be computed efficiently?! How can we build computing devices?

! What is an algorithm?! What is a function?

Questions addressed by Computability

A function is computable if it isdetermined by some algorithm


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Etymology of the word Algorithm

Muhammad ibn Musa abu Djafar Al'Khwarizmi.(a.k.a Al'Khorezmi).

He was born around 780 aC in Khorezm, (south of

Aral sea, known nowadays as Khiva, Uzbekistán).He dead in Bagdad around 850 aC.

•

Algoritmi de numero Indorum.• Kitab al-jabr wa'l-muqabala (The art of solving equations).


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! It is the problem of determining, given an arbitrary computerprogram and an input, whether the program will finish runningor continue to run forever.

! Turing proved that a general algorithm to solve the haltingproblem for all possible program-input pairs cannot exist.

! It is a decision problem about properties of computer programson a fixed model of computation

! It is an abstract framework, there are no resource limitations onthe amount of memory or time required for the program'sexecution

Motivation: The Halting Problem


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! Context for computation.! Basic operations.! Combination rules.

Models of Computation

! While Programs.!

Turing Machines.


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! While Programs.

! Turing Machines.



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!

Variable names: An Arbitrary finite string of upper case letters and decimalnumbers provided it also starts with an upper case letter

E.g. X1, X2, TEMP, LIST1

! Operator symbols:

" To denote basic operators

" succ ( successor function ) , pred ( predecessor function ), 0 ( constant zero

function )

! Relation symbol “!” for the inequality which compares the values of

pairs of variables

! Program Symbols := ( assigment ), ; ( semi-colon ), begin, end , while, do,

( , ),

WHILE Programs: Basic Components


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! AssigmentX:=0

X:=succ(Y)X:=pred(Y)

! While statementwhile X ! Y do !

! may be an arbitrary statement! Compound statement

begin ! 1; ! 2; … ; ! n end

WHILE Programs: Statements

X ! Y is the test

! is the body

i are arbitrarystatements

and n ! 0

A while program is none other

than a compound statement we

choose to identify as a program,

but which can also be used as a

compound statement in someother larger program


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WHILE Programs: Example

!

How to write a while-program P that adds X to Y and leavethe resulting value in Z ?

Assign the value of X to Z

Add the value of Y to Z

There is no single step in our

programming language that will

carry it out

begin Z:=succ(X); Z:=pred(Z) endZ:=X


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!

How to write a while-program P that adds X to Y and leavethe resulting value in Z ?

Add the value of Y to ZIt may be decomposed further using andauxiliary variable U and

Set U to 0

Add 1 to Z and add 1 to U as many

times as required to make U=Y

begin

U:=0;while U!Y do begin

Z:= succ(Z);U:= succ(U)

end

end

Z:=Z+Y



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!

How to write a while-program P that adds X to Y

and leave the resulting value in Z ?

begin

Z:=X

Z:=Z+Y

end

! Whenever we use the macro

statement Z:=X+U in awhile-program, whe should

u n d e r s t a n d t h a t i t

abbreviates the code

begin

Z:=succ(X); Z:=pred(Z)

begin U:=0;

while U !Y do

begin

Z:= succ(Z);U:= succ(U)

end

end

end

2



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Z:= X Y Z:=X*Y

Z:= X div Y Z:= X mod Y

Z:= X**Y Z:=f(X 1 , … , X m )

!!Z:= XZ:=X+Y

Z:=n (n>0)

!

Macro Statement: An abbreviation for a multi-instruction statement ,a short-hand notation for its macro definition

WHILE Programs: Macro-Statements


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Result stored in X1

Macro +

Macro :=

begin X3:=0;

X4:=0;

while X3!X2 do

begin

X3:=succ(X3);

X4:=X4 + X1;

end

X1:=X4;

end


!

Provide with a while program for computing f(x,y) = x * y. ! Allowed macros for + and :=

‘x’ is stored in X1 and ‘y’ in X2


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begin

X3:=0;

X4:=0;

while X3!

X2 dobegin

X3:=succ(X3);

X5:=0;

while X5!X1 do

begin

X5:=succ(X5);X4:=succ(X4);

end

end

X1:=succ(X4);

X1:=pred(X1);

end

Replace macro + by

basic operations

Replacemacro := bybasic operations


!

Provide with a while program for computing f(x,y) = x * y. ! Without macros.


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beginX1:=succ(X1);

X3:=0;

X4:=0;

while X1!X4 do

beginX3:=succ(X3);

X1:=X1 X2;

end

X1:=pred(X3);

end

‘x’ is stored in X1 and ‘y’ in X2

Result stored in X1

!!

Macro for difference


!

Provide with a while program for computing f(x,y) = x div y. ! Using a macro for difference.


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!

Provide with a while program for computing f(x,y) = x div y. ! Without macros.

Replace macro for

difference by basic

sentences

begin

X1:=succ(X1);

X3:=0;

X4:=0;while X1!X4 do

begin

X3:=succ(X3);

X5:=0;

while X5!X2 do

begin X1:=pred(X1);

X5:=succ(X5);

end

end

X1:=pred(X3);

end



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!

Macro Test: Macro sentences defined inductively as follows ! Basic: X and Y are either natural numbers or variables, then

! X=Y, X


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!

Proposition: A statement of the form

while T do !

where T is a test and ! is an arbitrary statement, is a macro sentence in

the language of while programs

Without loss of generality we may assume that the test T does not involve naturalnumbers

Given any test T, we can write an arithmetical expression ET in terms of the variablesused in T and the arithmetic operators succ, pred, + and such that

!!

!"#

=

otherwise0

trueisTif 1

T E

begin

U:=ET ;V:=0 ;while U!V do

begin ! ;

U:= ET

endend



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!


while T do !

where T is a test and ! is an arbitrary statement, is a macro sentence in the

language of while programs

But….How to find E T for any arbitrary test T ?

HINT: Use the inductive definition of tests

Take tests of the form X < Y

Classify all other tests to be compositeT= X < Y

ET= (Y X) pred(Y X) !!!! !!



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!


while T do !

where T is a test and ! is an arbitrary statement, is a macro sentence in the

language of while programs

Assuming we have found expressions ET1 and ET2 for test T1 and T2

T= T1 ^T2 ET= pred( ET1+ ET2 )

T= T1 T2

T= ~T1

ET= ( ET1+ ET2 ) pred( ET1+ ET2 ) !!

ET= 1 ET1 !!



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Example: Construct ET for T = (Z"Y) ^ (Z>X)

E(Z"Y)^(Z>X) = pred(EZ"Y + EZ>X) =

= pred((1-EZ


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! Proposition: A statement of the formIf T then ! 1

If T then ! 1 else ! 2

Repeat ! until Twhere T is a test and ! 1 and ! 2 are arbitrary

statements, are all macro sentences in the language ofwhile programs



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Definition

A sentence ! is a k-variable statement, then the variables it uses are (asubset of) {X 1, X 2, … ,X k} variables

Definition

A state of computation for a k-variable while program is k-dimensionalvector over the natural numbers

State Vector: â=(a1, … ,ak )" Nk , where ai is the content of X i

Definition

Let P be a k-variable while program. A computation by P is a sequence,possible infinite, of the form

â0A1 â1 A2 â2 … ân-1 An ân …

Initial state vector Instructions

Size of sequence

WHILE Computable Functions


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Compute a computation sequence,given as initial state vector (4,2)

begin

X1:=0;

while X1!X2 do X1:= succ(X1)

end

â0=(4,2)

A1=X1:=0;

â1=(0,2)

A2

=X1"X2

â2=(0,2)

A3=X1=succ(X1)

â3=(1,2)

A4=X1"X2

â4=(1,2) A5 =X1=succ(X1)

â5 =(2,2)

A6=X1"X2 â6=(2,2)

Computation sequence. Example


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Let âi-1=(a1, … , ak ) be a non initial state vector

a) If Ai is a test, then âi= âi-1b) If Ai is a setting X n:=g(X m ), then:

âi =(a1 , … , an-1 , g(am ), an+1 , … , ak )

being g(.) either succ or pred or 0

begin

X1:=succ(0);

while X1!

X2 do X1:= pred(X1)end

â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X1!X2

â2

=(1,0)

A3=X1=pred(X1)

â3=(0,0)

A4=X1!X2

â4=(0,0)

Consistency Conditions


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Let P # begin 1 ; 2 ; … ; n end be

The first instruction of sequence is

a) If ! 1 is a setting, A1= ! 1b)

If ! 1 # while T do ! , A1=Tâ0A1 â1 A2 â2 … ân-1 An ân

begin

X1:=succ(0);while X1!X2 do X1:= pred(X1)

X2:=succ(X1);

end

â0=(0,0)

A1=X 1:=succ(0);

â1=(1,0)

A2=X1!X2

â2=(1,0) A3=X1=pred(X1)

â3=(0,0)

A4=X1!X2â4=(0,0)



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Next instruction: Let Ai be an intermediate step in computation1. If Ai # T # Xn!X m associated to

! j # while T do !

a) If an! am then Ai+1 # 1st instruction within the body of the corresponding while

statement

b) If an= a

m then

i. If ! j is the last instruction of the body of a while sentence, Ai+1 # while Test

ii. otherwise Ai+1 # 1st instruction of ! j+1

â0A1 â1 A2 â2 … ân-1 An ân

â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X 1! X 2â2=(1,0)

A3=X1=pred(X1)

â3=(0,0)

A4=X1!X2â4=(0,0)


begin

X1:=succ(0);

while X1!X2 do X1:= pred(X1)X2:=succ(X1);

end


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â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X1!X2â2=(1,0)

A3=X 1=pred(X 1 )

â3=(0,0)

A4=X 1! X 2

â4=(0,0)


! j # while T do !

a) If an! am then Ai+1 # 1st instruction within the body of the corresponding while

statement

b) If an

= am

then




begin

X1:=succ(0);


end


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â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X1!X2â2=(1,0)

A3=X1=pred(X1)

â3=(0,0)

A4=X 1! X 2

â4=(0,0)


! j # while T do !

a) If an! am then Ai+1 # 1st instruction within the body of the corresponding whilestatement

b) If an

= am

then




begin

X1:=succ(0);


end


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! j # while T do !

a) If an! am then Ai+1 # 1st instruction within the body of the corresponding whilestatement

b) If an

= am

then


ii. otherwise Ai+1 # 1st instruction of j+1


â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X1!X2

â2=(1,0) A3=X1=pred(X1)

â3=(0,0)

A4=X 1! X 2

â4=(0,0)

A5=X 2=succ(X 1 )


begin

X1:=succ(0);


end


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Next instruction: Let Ai be an intermediate step in computation2. If Ai is a setting,

i. If ! j is the last instruction of the body of a while sentence,

Ai+1 # while Test


begin

X1:=succ(0);


end

â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X1!X2â2=(1,0)

A3=X 1=pred(X 1 )

â3=(0,0) A4=X 1! X 2

â4=(0,0)

A5=X 2=succ(X 1 )

â0=(0,0)

A1=X1:=succ(0);

â1=(1,0)

A2=X1!X2â2=(1,0)

A3=X1=pred(X1)

â3=(0,0) A4=X 1! X 2

â4=(0,0)

A5=X 2=succ(X 1 )



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Next instruction: Let Ai be an intermediate step in computation2. If Ai is an assigment,

i. If ! j is the last instruction of the body of a while sentence,

Ai+1 # while Test



begin

X1:=succ(0);


end

â0=(0,0) A1=X 1:=succ(0);

â1=(1,0)

A2=X1!X2â2=(1,0)

A3

=X 1

=pred(X 1

)

â3=(0,0)

A4=X 1! X 2

â4=(0,0)

A5=X 2=succ(X 1 )

â0=(0,0) A1=X 1:=succ(0);

â1=(1,0)

A2=X 1! X 2

â2=(1,0)

A3

=X 1

=pred(X 1

)

â3=(0,0)

A4=X 1! X 2

â4=(0,0)

A5=X 2=succ(X 1 )



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! Given P, compute its semantic function with arity 1.

begin

X3:=0;

while X1!X3 do

begin

X2:=pred(X2);

X1:=pred(X1);

endX1:=X2;

end

! j=1 !

As k"

j, P is executed using (x,0,0) as initial statevector.! After executing X1stores ?

P uses 3 variables # k=3%

(1)(x) = 0

Semantic Functions


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begin

X3:=0;

while X1!X3 do

begin

X2:=pred(X2);

X1:=pred(X1);

endX1:=X2;

endP uses 3 variables # k=3

Semantic Functions

! J=3 ! As k"j, P is executed using (x,y,z) as initial state

vector.!

After executing X1stores ?

%

(3)(x,y,z) = y-x


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begin

X3:=0;

while X1!X3 do

begin

X2:=pred(X2);

X1:=pred(X1);

endX1:=X2;

endP uses 3 variables # k=3

Semantic Functions

! J=5 ! As k


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â0=(3,0,0,0,0)

X4:=X1+X2;

â1=(3,0,0,3,0)

X4>X5â2=(3,0,0,3,0)

X5:=succ(X5);

â3=(3,0,0,3,1)

X4:=pred(X4);

â3=(3,0,0,2,1)

X4>X5â4=(3,0,0,2,1)

X5:=succ(X5);

â5=(3,0,0,2,2)

X4:=pred(X4);

BeginX4:=X1+X2;While (X4>X5) do

BeginX5:=succ(X5);

X4:=pred(X4);End

X1:=X4;

While (X3&0) doBegin

X1:=succ(X1);X3:=pred(X3);

End

End

Assume j=1 !P(1)(a1)

â0=(4,0,0,0,0)

X4:=X1+X2;

â1=(4,0,0,4,0)

X4>X5â2=(4,0,0,4,0)

X5:=succ(X5);

â3=(4,0,0,4,1)

X4:=pred(X4);

â3=(4,0,0,3,1)

X4>X5â4=(4,0,0,3,1)

X5:=succ(X5);

â5=(4,0,0,3,2)

X4:=pred(X4);

%

(1)(x) = x/2

Semantic Functions

â6=(3,0,0,1,2)

X4>X5

â7 =(3,0,0,1,2)

X1:=X4;â8=(1,0,0,1,2)

â6=(4,0,0,2,2)

X4>X5

â7 =(4,0,0,2,2)

X1:=X4;â8=(2,0,0,2,2)

X3&0

â9=(2,0,0,2,2)


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3


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The j-ary semantic function, !P(j): Nj $ N, for a k variableprogram P is defined as follows.

Given an input vector (a1, …, aj), !P(j)(a1, …, aj) is evaluated

according to the following rules, with two cases arising:

1. k " j. Then !P(j)(a1, …, aj) is evaluated by applying P to theinitial state â0=(a1, …, aj,0, .. K-j .., 0)

2. k < j. Then !P(j)(a1, …, aj) is evaluated by applying P to the

initial state â0=(a1, …, ak).

In both cases, If and when P halts on this vector, the final value isstored on X1

In either case, if P fails to halt on the initial state vector, then thevalue of !P

(j)(a1, …, aj)= '

3

Semantic Functions


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! Every algorithm or effective procedure is computable.

Church’s Thesis


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! While Programs.

! Turing Machines.



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Alan Turing

1912 - 1954

Alan Turing was one of the foundingfathers of CS.

! His computer model was premonitionof the electronic computer that cametwo decades later

! Was instrumental in cracking the NaziEnigma cryptosystem in WWII

!

Invented the “Turing Test” used in AI

Alan Turing


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!

He was instrumental in cracking the Nazi Enigmacryptosystem in WWII.

! 1918 Arthur Scherbius built the Enigma" Advantage Enigma: cypher produced was very

difficult" Polish were good at cracking codes

" The French bought keys, couldn’t do anything with it

" Poland foresaw its invasion by Germany: gave allknowledge to England and France, destroyed itafterwards (1939)

" http://www.enigmaco.de/

A short story during the second world war


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! 1939 Turing was asked to help to crack the Enigma! Built with a team the Colussus, the first programmable

computer

!

Based on:" his own 1936 concept of the universal machine" the potential speed and reliability of electronic technology" the inefficiency in designing different machines for different

logical processes

! Cyphercode could be decrypted from 1943! All computers were destroyed, ordered by Churchill



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! Because of the construction of theColussus Turing thought it could bepossible to construct a computer with

the mind of a human being! “Turing was convinced that if a

computer could do all mathematicaloperations, it could also do anythinga person can do“



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! “Computing machinery andintelligence” (1950)

! Operating definition ofintelligence

! Loebner Prize! CAPTCHA

Turing Test


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. . .

Finite State Control

A finite-state control equipped with

an external storage device whichcan be extended in both directions

May be a blank or may hold

any one symbol from as p e c i f i e d f i n i t e t a p e

alphabet

! Two-way, infinite tape, broken into cells, each containing one symbol.! Two-way, read/write tape head.! Finite control: a program containing the position of the read head, current symbol being scanned,

and the current state.! An input string is placed on the tape, padded to the left and right infinitely with blanks, read/write

head is positioned at the left end of input string.

!

In one move, depending on the current state and the current symbol being scanned, the TM 1)

changes state, 2) prints a symbol over the cell being scanned, and 3) moves its’ tape head one cell

left or right.

Turing Machine


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!

First Instruction: Initial State. Tape head scans the first non blank symbolof the tape.

! Next Instruction:

Depending on the pair (State, symbol):

!

Write a symbol on the scanned square.

! Enter a new state of the finite-state control

! One of the options listed below:

– Shift the head left one square (L)

–

Shift the head right one square (R)–

Not shift it at all (N)

– Halt (H)

Turing Machine: Formal description


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! Halting Test:

! The machine halts when there is no instructions.

! Result:! If it halts in a final state, the result is coded in the tape.

! If it halts in a non final state, the result is undefined



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Definition

A Turing Machine is specified by a quintuple (X, Q, T, i, F) where:

! Q is a finite set of states.

!

i(Q is the initial state.

! F ) Q is the halting state set.

! X denotes the set of symbols that can be written on tape, including the

blank symbol (B).

! T :Q x X :$ X x {L,R,H,N} x Q



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An instantaneous description (ID) is a triple a1qa2, where :

a1= y1 … yn and a2= x1 … xm! q: the current state

! x1: The symbol the tape head is currently scanning

!

a2 = x2 … xm: Symbols from the tape head up to the rightmostnon-blank symbol

! a1 = y1 … ym: Symbols from the leftmost non-blank symbol upto the tape head

!

At the beginning of a computation a1= i x1 … xm

Instantaneous Description


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!

A transition provides us with the description!

y1 … yn q x1 … xm

! It computes T(q, x1)

!

If T(q, x1)=(x’1, q’, D):! y1 … yn q x1 … xm ->y1 … yn x’1 q’ x2 … xm

! If T(q, x1)=(x’1, q’, I)!

y1 … yn q x1 … xm ->y1 … yn-1 q’yn x’1 x2 … xm

!

If T(q, x1)=(x’1, q’, N)! y1 … yn q x1 … xm ->y1 … yn q’x’1 x2 … xm

Turing Machine: Transitions


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!

If T(q, x1)=(x’1, q’, H)

! y1 … yn q x1 … xm ->y1 … yn q’x’1 x2 … xm

! Special Cases:

!

With (x’1, q’, I): q x1 … xm -> q’B x’1 x2 … xm

! With (x’1, q’, I): y1 … yn q B -> y1 … q’yn x’1!

With (x’1, q’, D): y1 … yn q x1 -> y1 … yn x’1q’B

!

With (x’1, q

’, D): q x1 -> x

’1q’B

! With (x’1, q’, D): q B -> x’1q’B

Turing Machine: Transitions


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!

Interpret TM as Computers of Number-Theoretic partialfunctions

! Input Codification: Unary Notation (X = {0,1})

!

Given n(N, it is represented by a block of (n+1)consecutive1’s.

! Given a vector (n1, …, nk), it will coded as

1 .. (n1+1) .. 101.. (n2+1) .. 10 … 01.. (nk+1) .. 1

Turing Machine: Semantics


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Semantic Function

Given a TM, M=(X, Q, T, i, F) the k-ary semantic function of M is

defined as:

%

M(k) : Nk $ N

! To compute %M

(k)(x1,…,xk), start M using the initial configuration.

! If and when M halts, count the total number of 1’s (consecutive or

non) on the tape, interpret this number in unary notation and take

this value to be the output value

Turing Machine: Semantics


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Definition

A function f: Nk $ N, is Turing computable if there exists a TM

M, such that:

%M(k)(x1,…,xk) = f(x1,…,xk) * (x1,…,xk) ( N

k

Turing Machine: Computation


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0 1 1 1 0 1 1 0 0

First Step: Input Codification (Unary Notation)

Given n(N, it is represented by a block of (n+1) consecutive1’s.

Given a vector (n1, …, nk), it will be coded as

1 .. (n1+1) .. 101.. (n2+1) .. 10 … 01.. (nk+1) .. 1

INPUT VECTOR (x,y)

x y

X+Y


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0 1 1 1 0 1 1 0 0

Third Step: Transition function, how does the machine work?

T :Q x X :$ X x {L,R,N,H} x Q

T(q0, 1)=(0, q1, D)

Initial state

X+Y


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0 0 1 1 0 1 1 0 0


T(q1, 1)=(1, q1, D)

X+Y



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0 0 1 1 0 1 1 0 0

T(q1, 1)=(1, q1, D)



X+Y


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0 0 1 1 0 1 1 0 0


T(q1, 0)=(0, q2, D)


X+Y


65/106

65/106


0 0 1 1 0 1 1 0 0


T(q2, 1)=(0, q2, H)


X+Y


66/106

66/106


0 0 1 1 0 0 1 0 0

Third Step: Transition function, how does the machine

work?


T(q2, 1)=(0, q2, H)

Final state

X+Y


67/106

67/106



work?


T(q0, 1)=(0, q1, D)

T(q1, 1)=(1, q1, D)

T(q1

, 0)=(0, q2

, D)

T(q2, 1)=(0, q2, H)

(q0, 1 , 0, q1, D)

(q1, 1 , 1, q1, D)

(q1

, 0 , 0, q2

, D)

(q2, 1 , 0, q2, H)

X+Y


68/106

68/106


0 1 1 1 0 1 1 0 0


work?


T(q0, 1)=(0, q1, D)

T(q1, 1)=(0, q1, H)

q0 initial state q1 final state

X+Y: Second try


69/106

69/106


0 1 1 1 0 1 1 0 0


work?


T(q0, 1)=(0, q1, D)

Initial state

X+Y: Third try


70/106

70/106


0 0 1 1 0 1 1 0 0


work?


T(q1, 1)=(1, q1, D)

X+Y


71/106

71/106


0 0 1 1 0 1 1 0 0

T(q1, 1)=(1, q1, D)


work?


X+Y

X Y


72/106

72/106


0 0 1 1 0 1 1 0 0


work?


T(q1, 0)=(0, q0, D)

X+Y


73/106

X+Y


74/106

74/106


0 0 1 1 0 0 1 0 0

T(q1, 1)=(1, q1, D)


work?


X+Y

X+Y


75/106

75/106


0 0 1 1 0 0 1 0 0


work?


T(q1, 0)=(0, q0, D)

X+Y

X+Y


76/106

76/106


0 0 1 1 0 0 1 0 0


work?


T(q0, 0)=(0, q0, H)

X+Y

X+Y


77/106

77/106


(q0, 1 , 0, q1, D)

(q1, 1 , 1, q1, D)

(q1, 0 , 0, q0, D)

(q0, 0 , 0, q0, H)

(q0, 1 , 0, q1, D)

(q1, 1 , 1, q1, D)

(q1, 0 , 0, q2, D)

(q2, 1 , 0, q2, H)

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q1, H)

Which one is the best?

X+Y

2


78/106

78/106


0 1 1 1 1 0 0 0 0

x

2x

T(q0, 1)=(0, q1, D)

Initial state

# Delete the 1 due to codification

2X


79/106

79/106


T(q1, 1)=(0, q2, D) # Mark the 1 to duplicate

0 0 1 1 1 0 0 0 0

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

2X

2X


80/106

80/106


T(q2, 1)=(1, q2, D) # Inside the codification

0 0 0 1 1 0 0 0 0

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

2X

2X


81/106

81/106


0 0 0 1 1 0 0 0 0

T(q2, 1)=(1, q2, D) # Inside the codification

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

2X

2X


82/106

82/106


0 0 0 1 1 0 0 0 0

T(q2, 0)=(0, q3, D) # End of codification, newsituation, new state

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

2X

2X


83/106

83/106


0 0 0 1 1 0 0 0 0

T(q3, 0)=(1, q4, I) # looking for a cell to write 1.1 is duplicated(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

new situation,new state

2X

2X


84/106

84/106


0 0 0 1 1 0 1 0 0

T(q4, 0)=(0, q5, I) # separator found(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q4, 0 , 0, q5, I)

2X


2X


85/106

85/106


0 0 0 1 1 0 1 0 0

# inside the codification

2X


T(q5, 1)=(1, q5, I)

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)(q3, 0 , 1, q4, I)

(q4, 0 , 0, q5, I)(q5, 1 , 1, q5, I)

2X


86/106

86/106


0 0 0 1 1 0 1 0 0

# inside the codification

2X

T(q5, 1)=(1, q5, I)

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)(q3, 0 , 1, q4, I)

(q4, 0 , 0, q5, I)(q5, 1 , 1, q5, I)

2X


87/106

87/106


0 0 0 1 1 0 1 0 0

# mark found

2X

new situation,

new state

T(q5, 0)=(1, q1, D)

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)(q3, 0 , 1, q4, I)

(q4, 0 , 0, q5, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)


88/106

2X


89/106

89/106


0 0 1 0 1 0 1 0 0

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)(q4, 0 , 0, q5, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X

2X


90/106

90/106


0 0 1 0 1 0 1 0 0

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)(q4, 0 , 0, q5, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X

2X


91/106

91/106


0 0 1 0 1 0 1 0 0

T(q3, 1)=(1, q3, D)

# looking for a blank

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3

, 1 , 1, q3

, D)(q4, 0 , 0, q5, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


92/106

2X


93/106

93/106


0 0 1 0 1 0 1 1 0

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)(q4, 1 , 1, q4, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

T(q4, 1)=(1, q4, I)

# looking for a blank

2X

2X


94/106

94/106


0 0 1 0 1 0 1 1 0

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


95/106

2X


96/106

96/106


0 0 1 0 1 0 1 1 0

(q0, 1 , 0, q1, D)

(q1

, 1 , 0, q2

, D)

(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X

2X


97/106

97/106


0 0 1 1 1 0 1 1 0

2X

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)(q4, 1 , 1, q4, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


98/106

98/106


0 0 1 1 0 0 1 1 0

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X

2X


99/106

99/106


0 0 1 1 0 0 1 1 0

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X

2X


100/106

100/106


0 0 1 1 0 0 1 1 0

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q5, 1 , 1, q5, I)(q5, 0 , 1, q1, D)

2X

2X


101/106

101/106


0 0 1 1 0 0 1 1 0

(q0, 1 , 0, q1, D)(q

1

, 1 , 0, q2

, D)

(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


102/106

102/106


0 0 1 1 0 0 1 1 1

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


103/106

103/106


0 0 1 1 0 0 1 1 1

(q0, 1 , 0, q1, D)(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


104/106

104/106


0 0 1 1 0 0 1 1 1

(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)(q2, 1 , 1, q2, D)

(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


105/106

105/106


0 0 1 1 0 0 1 1 1(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

2X


106/106

106/106

0 0 1 1 1 0 1 1 1(q0, 1 , 0, q1, D)

(q1, 1 , 0, q2, D)

(q2, 1 , 1, q2, D)(q2, 0 , 0, q3, D)

(q3, 0 , 1, q4, I)

(q3, 1 , 1, q3, D)

(q4, 0 , 0, q5, I)

(q4, 1 , 1, q4, I)(q1, 0 , 1, f, H)

(q5, 1 , 1, q5, I)

(q5, 0 , 1, q1, D)

while programs

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