ch3_turingmachine

8/10/2019 Ch3_TuringMachine

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thuyt tnh ton -- Ch.3 - My Turing

L thuyL thuyt tt tnh tonh tonn(Theory of Computation)(Theory of Computation)

PGS.TS. Phan Huy KhPGS.TS. Phan Huy Khnhnh

[email protected]@dut.udn.vn,, [email protected]@gmail.com

Chng 3Chng 3

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CHNGCHNG 33

M HNH MM HNH MY TURINGY TURING

(Turing Machines Model)(Turing Machines Model)

Alan Mathison TuringAlan Mathison Turing ((23/06/191223/06/1912 07/06/195407/06/1954))nhnh toton hn hc,logic hc,logic hcc vv mmt m ht m hcc ngngiiAnhAnh

ththng ng c xem lc xem lcha cha cca nga ngnhnh Khoa hKhoa hc Mc My ty tnhnh

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TheThe ChomskyChomskys Languages LanguageHierarchyHierarchy

Context-Free LanguagesnnbaPDA Rww

Context-Sensitive Languages

anbncn ? ww ?

*a

Regular Languages

**ba

DFA

Languages accepted by

Turing Machines

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Handicapped machines

DFA (Deterministic Finite Automaton) limitationsTape head moves only one direction

2-way DFA has equivalent power

Tape is read-only

Tape length is a constant

PDA (Push-Down Deterministic Finite Automaton) limitationsTape head moves only one direction

2-way PDA has a little more power (can accept { ww | w* }

Tape is read-only, but stack is writable

Stack has only LIFO (last-in, first-out) access

Tape length is constant, but stack is not bounded

What about:Writable, 2-way tape?

Random-accessstack?

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Computability

Computible (there is an algorithm)Equivalence of DFAs

String membership in a regular language

Determining valid executions of a PDA

UncomputibleEquivalence of CFGs

Emptiness of the complement of a CFL(emptiness of CFL is computible)

Whether a C program ever terminates

How to prove it?

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Introduction to Turing MachinesIntroduction to Turing Machines ((TMTM))TM: a formal model of a (digital) computerTM: a formal model of a (digital) computer

running a particular programrunning a particular program

The TM is the invention ofThe TM is the invention of Alan TuringAlan Turing, 1936, 1936

Other formalisms equivalent :Other formalisms equivalent :

Random Access MachinesRandom Access Machines

Partial recursive functions (Partial recursive functions (KleeneKleene, 1952), 1952)

--calculuscalculus ((ChurchChurch, 1936), 1936)

Post systems (Post systems (PostPost, 1943), 1943)

etc...etc...

All these were preceded by the work ofAll these were preceded by the work of Kurt GodelKurt Godel, 1931,, 1931,which in his incompleteness theorem showed that there was nowhich in his incompleteness theorem showed that there was noway for a computer to answer all mathematical problemsway for a computer to answer all mathematical problems


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M tM t ttng qung qut mt my Turingy Turing

MMt bng v ht bng v hn cn c hai phhai pha tra tri vi v phphii

MMttuucc--ghighicc hayhay ghighi lnln bngbng,, theotheo haihai chichiuu

MMtt bb nhnh phph hhu hu hn xn xc c nh trnh trng thng thii

MMt cu vt cu vo o c c t trn bngt trn bng

MMt chng trt chng trnh lnh lm hom hot t ng mng my Turingy Turing

Blank symbol

...... a b ca

Input string

Read-Write head

Tape

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A Turing MachineA Turing Machine

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M tM t chi tichi tit ct cc thc thnh phnh phn TMn TM

Bng lBng lm vim vic :c :c chia thc chia thnh cnh cc lin tic lin tipp

MMi chi cha ma mt k tt k tnno o aa SS

S cS c chcha k ta k t trtrngng (blank symbol),, hohocc ##

PhPhn cu cn cu c chch (usable part)(usable part) wwS*S*t git giaa t nht nhtt hai k thai k t##NgNgi ta vii ta vitt #w#w##

u u ccghi :ghi :c hay ghi (lc hay ghi (lm thay m thay i) ni) ni dung ti dung tng cng ca bnga bng

Di chuyDi chuyn mn mt vt v trtrqua phqua phi, hoi, hoc trc tri,i, tngtng ng vng vi mi mt t

TTi mi mi thi thi ii im, chm, ch duy nhduy nht mt mt t c c c, hoc, hoc c c ghic ghi

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M tM t chi tichi tit ct cc thc thnh phnh phn (2)n (2)

MMt bt b nhnh phph hhu hu hn :n :

TiTip cp cn n c nc ni dung ti dung ti mi mi thi thi ii imm

CC thththay thay i i c nc ni dungi dung

BB nhnh phphc lic lit k lt k lc c nh nghnh nghaa

MMi phi phn tn tcca ba b nhnh phph xxc c nh mnh mt trt trng thng thii

Chng trChng trnh :nh :

TTp hp hp hp hu hu hn cn cc quy tc quy tcc

LLm thay m thay i ni ni dung trn bng vi dung trn bng v nni dung bi dung b nhnh phph

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A Turing MachineA Turing Machine

. . .. . .

Input tape

Read-Write head

Control Unit / Transition Diagram

Input String w *

a ba

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The TapeThe Tape

......

Read-Write head

The head moves Left or Right

No boundaries -- infinite length

The head at each time step:

1. Reads a symbol

2. Writes a symbol

3. Moves Left or Right


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Example Moves Left :

Time 0

Time 1

...... a a cb

...... a b k c

1. Reads

2. Writes

a

k

3. Moves Left14/189

Time 1

...... a b k c

Time 2

...... a k cf

1. Reads

2. Writes

b

f

3. Moves Right

Example Moves Right :

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The Input StringThe Input String

Blank symbol

head

...... a b ca

Head starts at the leftmost position

of the input string

Input string

Remark: the input string is never empty16/189

States & TransitionsStates & Transitions

1q 2qLba ,

Read Write Move Left

1q 2qRba ,

Move Right

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Example Moves Right :

1q 2qRba ,

...... a b caTime 1

1qcurrent state

...... a b cb

Time 2

2q

b

a

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1q 2qLba ,

...... a b cTime 1

1qcurrent state

...... a b cb

Time 2

2q

Example Moves Left :

b

a


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AcceptanceAcceptance

Accept Input If machine halts

in a final state

Reject Input

If machine haltsin a non-final state

or

If machine enters

an infinite loop

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Turing MachineTuring Machine Example (1)Example (1)

A Turing machine that accepts the language:

0q

Raa ,

L,1q

aa* = a+

(This Turing machine mimic a DFA)

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Time 0 aa

0q

a

aaTime 1

0q

a


0q L,

1q

a a, R

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aaTime 2

0q

a

0q L,

1q

a a, R


aaTime 3

0q

a

a a, R

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aaTime 4

1q

a

Halt & Accept


0q

Raa ,

L,1q

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Rejection Example (1)

baTime 0

0q

a

0q

Raa ,

L,1q


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baTime 1

0q

a

No possible Transition

Halt & Reject

Rejection Example (2)

0q

Raa ,

L,1q

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Infinite LoopInfinite Loop Example (1)Example (1)

A Turing machine

for language *)(* babaa ++

0q

Raa ,

L,1q

Lbb ,

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baTime 0

0q

a


0q

Raa ,

L,1q

Lbb ,

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0q

baTime 1 a


0q

Raa ,

L,1q

Lbb ,

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baTime 2

0q

a

0q

Raa ,

L,1q

Lbb ,


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baTime 2

0q

a

baTime 30q

a

baTime 4

0q

a

baTime 5

0q

a



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Because of the infinite loop:

The final state cannot be reached

The machine never halts

The input is not accepted

InfiniteInfinite Loop ResultLoop Result

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T= Mt cch khc m t ngn gn TM = {1, #}, hoc = {1} Nu khng mun chk ttrng ca bngQ = {q1, q2, q3}P= { q1, 1 1, R, q2,

q2, 1 1, R, q1,q1, # #, L, q3,q3, 1 1, L, q3,q3, 1 1, R, q3,q3, 1 #, R, q1 }

Another TM ExampleAnother TM Example

1|1, R

1q

3q 1|1, L

2q1|1, R

1|1, R

1|#, R #|#, L

c mt s chn s 1, v li q1

Tha nhn cu c s chn s 1Bi tp : M t hot ngon nhn cu 111111 ?

Bi tp : M t hot ngon nhn cu 111111 ?

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TM for the language aTM for the language annbbnn (1)(1)

0q

1q 2q

3q

Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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Time 0 ba

0q

a b


0q

1q 2q

3q

Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,

X, Y

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b

1q

a b Time 1


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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bx

1q

a b Time 2


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,


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y

2q

a b Time 3


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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yx

2q

a b Time 4


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,

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0q

a b Time 5


0q

1q 2q

3q

Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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yx

1q

x b Time 6


0q

1q 2q

3q

Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,

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y

1q

b Time 7


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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yx x y

2q

Time 8


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,


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y y

2q

Time 9


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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yx

0q

x y Time 10


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,

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y

3q

y Time 11


0q

1q 2q

3q

Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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yx

3q

x y Time 12


0q

1q 2q

3q

Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx,

Ryy ,

Ryy , 4q

L,

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Halt & Accept

y

4q

y Time 13


0q 1q 2q3q Ra ,

Raa ,

Ryy ,

Lyb ,

Laa ,

Lyy ,

Rx ,

Ryy ,

Ryy , 4q

L,

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If we modify the TM

for the Context-Free Language

we can easily construct a TMfor the Context-Sensitive Language

Observation:

}{ nnba

}{ nnn

cba


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TMTMAcceptingAccepting aannbbnnccnn,, nn11

input abc:q0abcxq1bc xyq2c xq3yz q3xyz xq0yz xyq4z xyzq5B xyzBq6B

input a2b2c2:q0 a

2b2c2xaq1b2c2 xayq2bc2 xaybq2c2 xayq3bzc xaq3ybzc xq3aybzc q3xaybzcxq0aybzc xxq1ybzc xxyyq2zc xxyyq3zz xq3xyyzz xxq0yyzz xxyyq4zz xxyyzzq5B xxyyzzBq6B

z/z, L ; b/b, Ly/y, L ; a/a, L

Idea: Finding matchinga,b,c and replacing by x,y,z

Idea: Finding matchinga,b,c and replacing by x,y,z

accepting matched x,y,z

Algorithm

q0

q4

q1 q2 q3

q6q5

a/x, R b/y, R c/z, L

y/y, R

z/z, R #/#, L

a/a, R ; y/y, R

y/y, R z/z, R

b/b, R ; z/z, R

x/x, R

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AlgorithmAlgorithm forforAcceptingAccepting aannbbnnccnn,, nn

11

BeginBeginWhileWhile Cha gCha gp xp xDoDo

GGp a, ghi xp a, ghi x;; DoDoqua Rqua RWhileWhileCha gCha gp bp bOdOdGGp b, ghi yp b, ghi y;; DoDoqua Rqua RWhileWhileCha gCha gp cp cOdOdGGp c, ghi zp c, ghi z;; DoDoqua Lqua LWhileWhileCha gCha gp xp xOdOd

Qua RQua R

OdOdDoDoQua RQua RWhileWhileGGp yp yOdOdDoDoQua RQua RWhileWhileGGp zp zOdOdIfIf GGp #p #ThenThen Qua L,Qua L,YesYesElseElse NoNoFiFi

EndEnd

q0

q4

q1 q2 q3

q6q5

a/x, R b/y, R c/z, L

y/y, R

z/z, R #/#, L

a/a, R,y/y, R

y/y, R z/z, R

b/b, R,z/z, R

x/x, R

z/z, L,b/b, Ly/y, L,a/a, L

Formal DefinitionsFormal Definitionsforfor

Turing MachinesTuring Machines

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Transition Function

1q 2qRba ,

),,(),( 21 Rbqaq =

A Move to right:

1q 2qLdc ,

),,(),( 21 Ldqcq =

A Move to left:

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Turing Machine:

),,,,,,( 0 FqQ =

States

Input

alphabetTape

alphabet

Transitionfunction

Initialstate blank

Set ofFinal states

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ConfigurationConfiguration

ba

1q

a

Instantaneous description:

c

baqca 1


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yx

2q

a b

Time i

yx

0q

a b

Time i+1

A Move:

TransitionTransition

aybqx 0xaybq2

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x

2q

a b

Time 1

xaybq2

yx

1q

x b

Time 4

bqxxy 1

yx

1q

x b

Time 3

ybqxx 1

yx

0q

a b

Time 2

aybqx 0

Transition processTransition process

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Equivalent notation:

Transition processTransition process

bqxxyybqxxaybqxxaybq 1102

bqxxyxaybq 12

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Initial configuration:

ba

0q

a b

wInput string

NotationNotation

wq0

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The Accepted LanguageThe Accepted Language

For any Turing Machine (TM) M:

Initial state Final state

} |{)( 210 xqxwqwML f=

qfF

L(M) is the Language Accepted by TM M

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TM computes palindromes

a|a,Rb|b, R

q2 q3

q7

# | #, L

a|#, R

q1

a|a, Rb|b, R

q5 q6#|#, L

b|#,R

q4

a|#, L

b|#, L

#|#, R

#|#, R

#|#, R

Odd palindrome

Odd palindrome

#|#, R

Even palindrome

a|a, Lb|b, L

Ex1: w=abbbba?Ex2: Draw M: L(M)=w#w for = {a, b, c} ?

nh du con amt tri nht

Vt qua phiGp # li li

Gp # qua phi

nh du con amt phi nht

Vt qua tri


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DefinitionsDefinitions

The languages accepted by TM

are called Recursively Enumerable (lit k quy)

Recursion is related to recursion in programs

Enumerable means there is some TM that can enumerate the strings in the

language

A language is Recursiveif it is accepted by some TM

that halts on all inputs

We'll usually assume that

TM halt if they enter a final state

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Standard Turing MachineStandard Turing Machine

Deterministic

Infinite tape in both directions

Tape is the input/output file (online)

The machine we described is the standard:

Computing FunctionsComputing Functionswithwith

Turing MachinesTuring Machines

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Domain: Result Region:

A function )(wf has:

D

NotationNotation

R

Dw)(wf

Rwf )(

fcan be TotaL or Partial Function

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A function may have many parameters

The number of parameters is called arity

yxyxf +=),(

Example: Addition function

with 2-arity

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Integer Domain

Unary:

Binary:

Decimal:

11111

101

5

We prefer unary representation:

easier to manipulate with TMs, first

Data representationData representation

n phn

Nh phn

Thp phn


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Definition:

A function is computable if

there is a Turing Machine such that:

f

Initial configuration Final configuration

Dw Domain

0q

w

fq

)(wf

final stateinitial state

For all

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Initial

Configuration

Final

Configuration

A function is computable if

there is a Turing Machine such that:

f

In other words:

Dw DomainFor all

)(0 wfqwq f

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ExampleExample

The function yyf +=),( is computable

Turing Machine:

Input string: yx0 unary

Output string: 0xy unary

yx, are integers

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0

0q

1 1 1 1

x y

1L L

Start

initial state

The 0 is the delimiter that

separates the two numbers

Input representationInput representation

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Finish

Start

Computing FunctionComputing Function

0q

1 1 1 1

x y

1L L

initial state

0

fq

1 1

yx +

L 11

final state

0

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0

fq

1 1

yx +

L 11Finish

final state

The 0 helps when we use

the result for other operations

Computing FunctionComputing Function


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Turing machine for function

yxyxf +=),(

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11

Exercice:Draw TM for inc(x), dec(x), double(x), x*y?Exercice:Draw TM for inc(x), dec(x), double(x), x*y? 80/189

Execution Example (1)

0

0q

1 1 1 1

Time 0

x y

Final Result

0

4q

1 1 1 1

yx +

11= (2)

11=y (2)

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0

0q

1 1Time 0 1 1

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11

0q

01 11 1Time 1


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0

0q

1 1 1 1Time 2

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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1q

1 11 11Time 3

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11



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1q

1 1 1 11Time 4

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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1q

1 11 11Time 5

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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2q

1 1 1 11Time 6

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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3q

1 11 01Time 7

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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3q

1 1 1 01Time 8

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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3q

1 11 01Time 9

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11



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3q

1 1 1 01Time 10

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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3q

1 11 01Time 11

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11


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4q

1 1 1 01Time 12

0q 1q 2q 3qL, L,01

L,11

R,

R,10

R,11

4q

R,11

HALT & accept


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AnotherAnother Example:Example:f(x) = 2xf(x) = 2x (1)(1)

The function f 2)( = is computable

Turing Machine:

Input string: x unary

Output string: xx unary

is integer

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1

fq

1 1

x2

L 11Finish

final state

0q

1 1

x

1LStart

initial state

AnotherAnother Example:Example:f(x) = 2xf(x) = 2x (2)(2)

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TM Pseudocode for f(x) = 2x

Replace every 1 with $

Repeat: Find rightmost $, replace it with 1

Go to right end, insert 1

Find rightmost $, replace it with 1

Go to right end, insert 1

Until no more $ remain

A Tip!A Tip!


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ExampleTM for f(x) = 2x

0q 1q 2q

3q

R,1$

L,1

L,

R$,1 L,11 R,11

R,

0q

1 1

Start

3q

1 11 1

Finish

1. Replace every 1 with $2. Repeat:

Find rightmost $,replace it with 1

Go to right end, insert 13.Until no more $ remain

1. Replace every 1 with $2. Repeat:

Find rightmost $,replace it with 1

Go to right end, insert 13.Until no more $ remain

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T= ; = { 0, 1, # } ; Q = {q1, q2, q3}

P= { q1, 1 1, R, q1,q2, 0 1, L, q3,q1, 0 0, R, q1,q2, 1 0 L, q2,

q1, # #, L, q2,q2, # 1, L, q3 }

TM compute succ(n)TM compute succ(n)

1q 3q

0|0, R

2q

1|1, R1|0, L

#|1, L

#|#, L

0|1, L

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An example: Unary to Binary ConversionAn example: Unary to Binary Conversion

Input string: #aaaaaa#

Output string?

#110###

2q

X|#, L

#|#, L

#|1, R

1|0, L

0|1, R

a|X, L1q0q

1|1, R

0|0, R

Find a

Increment 1

Erase X3q

0|0, R1|1, RX|X, R

X|X, L

Exercice: Draw TM for Inverse?Exercice: Draw TM for Inverse?100/189

Another ExampleAnother ExampleAnother Example

The function iscomputable =),( yf

0

1 yx>

yx

if

if

Turing Machine for

Input: yx0

Output: 1 0or

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Turing Machine Pseudocode:

Match a 1 from with a 1 fromx y

Repeat

Until all of or is matchedy

If a 1 from is not matched

erase tape, write 1

else

erase tape, write 0

x

)( yx>

)( yx

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TM move RWH to the right at nth = 2

q2q1

$ | $, Rq3

x | x, R

y | y, L x | x, Lq4

x|x, Rx \

$ x

Assume that in the Input Tape, we have Input String w1$w2$$wkwi , $

Draw a TM in order to move RWH to the right at w3:

P = { q1, x x, R, q1, vi x

q1, $ $, R, q2, vi $ q2, x x, R, q2, vi x q2, $ $, R, q3, vi y q3, x x, L, q4, vi x }

$

x \

Tng qut : Di RWH v v tr n,vi n t trc cu von$w1$w2$$wk

Tng qut : Di RWH v v tr n,vi n t trc cu von$w1$w2$$wk


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103/189

TM move RWH one case to the right

P= { q1, 0 #, R, q(0),q1, 1 #, R, q(1),q(0), 0 0, R, q(0),q(0), 1 0, R, q(1),q(1), 0 1, R, q(0),q(1), 1 1, R, q(1),

q(0), # 0, R, q(#),q(1), # 1, R, q(#),q(#), 0 #, R, q(0),q(#), 1 #, R, q(1)}

q(0)

q1

0 | #, R

0 | 0, R

q(1)1 | #, R

0 | 1, R 1 | 0, R q#

# | 1, R

# | 1, R

1 | 0, R# | 0, R

0 | #, R

1 | 1, R

Remember 0

Remember 1

Restore 0,

to the right

Restore 1

0# # * # # 0 #

1# # * # # 1 #

Or

104/189

CopyG: G* (G*)2 oror ww w#w

0 | 0, R

q2

b | b, R

1 | 1, R 0 | 0, L

q3

b | b, L

1 | 1, L

q4

1 | 1, R

q5

0 | 0, L

q1

q6

# | 1, L

# | #, L

1 | 1, L

0 | 0, R

b | b, R# | 0, L

b | b, L

1 | a, R

0 | b, R

a | 1, R

b | 0 , R

w w#wR?ww w#wR?

Combining Turing MachinesCombining Turing Machines

106/189

Block Diagram

Turing

Machineinput output

107/189

Example:

=),( y

yx + y>if

0 yif

Comparer

Adder

Eraser

yx,

yx,

y>

y

yx +

0

TuringTurings Thesiss Thesis


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109/189

Turings thesis*:

Any computation carried outby mechanical means

can be performed by a Turing Machine

(1930)

Thesis: A Accepted Statement but not DemontrationThesis: A Accepted Statement but not Demontration

110/189

Computer Science Law:

A computation is mechanical

if and only if

it can be performed by a Turing Machine

There is no known model of computation

more powerful than Turing Machines

111/189

Definition of Algorithm:

An algorithm for function

is a

Turing Machine which computes

)(wf

)(wf

112/189

When we say:

There exists an algorithm

Algorithms are Turing Machines

We mean:

There exists a Turing Machine

that executes the algorithm

VariationsVariationsofof the Turingthe Turing MachineMachine

114/189

Read-Write Head

Control Unit

Deterministic

The Standard ModelThe Standard Model

Infinite Tape

(Left or Right)

a a c ba cb b a a


20/32


115/189

Variations of the Standard ModelVariations of the Standard Model

1. Stay-Option

2. Semi-Infinite Tape3. Off-Line

4. Multitape

5. Multidimensional

6. Nondeterministic

Turing machines with:

116/189

We want to prove:

Each Class has the same powerwith the Standard Model

The variations form different

Turing Machine Classes

Variations of the Standard ModelVariations of the Standard Model

What is the Same Power of two classes?

(or Equivalence)

117/189

For any machine of first class1there is a machine of second class2

such that: )()( 21 LL =

And vice-versa

Same Power of two classes means:

Both classes of Turing machines accept

the same languages

NotionNotion

118/189

a technique to prove same powerSimulation:

Simulate the machine of one class

with a machine of the other class

First Class

Original Machine

1 1

2

Second Class

Simulation Machine

DemonstrationDemonstration

119/189

Configurations in the Original Machine

correspond to configurations

in the Simulation MachineOriginal Machine:

Simulation Machine:

nddd L10

nddd L10

Correspondence of theCorrespondence of the

configurationsconfigurations

120/189

The Simulation Machine

and the Original Machine

accept the same language

fdOriginal Machine:

Simulation Machine:fd

Final Configuration

Correspondence of theCorrespondence of the

configurationsconfigurations


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121/189

Turing Machines with StayTuring Machines with Stay--OptionOption

The head can stay in the same position

a a c ba cb b a a

Left, Right, Stay

L,R,S: moves

122/189

Example:

1q 2q

a a c ba cb b a a

Time 1

1q

Sba ,

ba c ba cb b a a

Time 2

2q

123/189

Stay-Option Machines

have the same power with

Standard Turing machines

Theorem:

124/189

Proof of Part 1 :

Part 1: Stay-Option Machines

are at least as powerful as

Standard machines

Proof: a Standard machine is also

a Stay-Option machine

(that never uses the S move)

125/189

Part 2: Standard Machines

are at least as powerful as

Stay-Option machines

Proof: a standard machine can simulate

a Stay-Option machine

Proof of Part 2 :

126/189

1q 2q

Lba ,

Stay-Option Machine

Simulation in Standard Machine

Simulationfor Left moves

Similar for Right moves

1q 2qLba ,


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127/189

1q 2qSba ,

Stay-Option Machine

1q 2qLba ,

3qRx,

Simulation in Standard Machine

For every symbol x

Simulationfor Stay

128/189

Example

Stay-Option Machine:

aab a

1q

1 ba b a

2q

21q 2q

Sba ,

Simulation in Standard Machine:

aab a

1q

1 ba b a

2q

2 ba b a

3q

3

129/189

Standard Machine--Multiple Track Tape

b

d

a

bba

a

c

track 1

track 2

one symbol

Turing Machines withTuring Machines with Multiple Track TapeMultiple Track Tape

130/189

1q 2qLdcab ),,(),(

b

d

a

bba

a

c

track 1

track 2

1q

Turing Machines withTuring Machines with Double Track TapeDouble Track Tape

b

d

a

b

c

d

a

c

track 1

track 2

2q

131/189

Turing Machines withTuring Machines with SemiSemi--InfiniteInfinite TapeTape

...# a b a c

132/189


simulate

Semi-infinite tape machines:

Trivial

StandardStandard SemiSemi--InfiniteInfinite TapeTape


23/32


133/189

Semi-infinite tape machines simulate

Standard Turing machines:Standard machine

.........

Semi-infinite tape machine

..................

SemiSemi--Infinite TapeInfinite Tape Standard (1)Standard (1)

134/189

Semi-infinite tape machine with two tracks

Standard machine

..................

reference point

a b c d e

.........#

#

Right part

Left part ac b

d e


135/189

1q2q

Standard machine

Rq2Lq1

Lq2 R

q1

Left part Right part



136/189

1q 2qRa ,

Standard machine

Lq1

Lq2

Lgxax ),,(),(

Rq1Rq2

Rxxa ),,(),(


Left part

Right part

For all symbols


137/189

.........


#

#

Right part

Left part ac b

d e

Lq1

Standard machine.................. a b c d e

1q

Time 1


138/189


Standard machine.................. b c d e

Time 2

2q

.........


#

#

Right part

Left part cd e

b

Lq2


24/32


139/189

At the border:

Lq1Rq1

R),#,(#)#,(#


Left part

Rq1Lq1

R),#,(#)#,(# Right part

140/189

.........


#

#

Right part

Left part c b

d e

Lq1

.........#

#

Right part

Left part c b

d e

Rq1

Time 1

Time 2

141/189

Theorem:

Semi-infinite tape machines

have the same power with


142/189

The OffThe Off--LineLine Machine (MMachine (My vy vi bng ti bng tch rch ri)i)

Control Unit

Input File

Tape

read-only

a b c

d eg

read-write

143/189

Off-line machines simulate

Standard Turing Machines:

Off-line machine:

1. Copy input file to tape

2. Continue computation as in

Standard Turing machine

144/189

1. Copy input file to tape

Input Filea b c

a b c

Standard machine

Off-line machine

Tape

a b c


25/32


26/32


151/189

a c e d

2q 2q

Time 2

RLdgfb ,),,(),( 1q 2q

a b c e

1q

1

q

Time 1Tape 1 Tape 2

MultitapeMultitape TMs (2)TMs (2)

152/189

Multitape machines simulate

Standard Machines:

Use just one tape

MultitapeMultitape StandardStandard

153/189

Standard machines simulate

Multitape machines:

Use a multi-track tape

A tape of the Multiple tape machine

corresponds to a pair of tracks

Standard machine:

StandardStandard MultitapeMultitape

154/189

a b c h e g

Multitape Machine

Tape 1 Tape 2

Standard machine with four track tape

a b c

e f0 0

0 0

1

1

Tape 1

head position

Tape 2

head position

h

0

155/189

Repeat for each state transition:

Return to reference point

Find current symbol in Tape 1

Find current symbol in Tape 2

Make transition

a b c

e f g0 0

0 0

1

1

Tape 1

head position

Tape 2

head position

h

0

##

#

#

Reference point

156/189

Theorem:

Multi-tape machineshave the same power with

Standard Turing Machines


27/32


157/189

Same power doesnt imply same speed:

Language }{ nnbaL=

Acceptance Time

Standard machine

Two-tape machine

2n

n

158/189

}{ nnbaL=

Standard machine:

Go back and forth times2n

Two-tape machine:

Copy to tape 2nb

Leave on tape 1na

Compare tape 1 and tape 2

n( steps)

n( steps)

n( steps)

159/189

MultiDimensionalMultiDimensional Turing MachinesTuring Machines

Two-dimensional tape

Position: +2, -1

MOVES: L,R,U,D

U: up D: down

x

y

a

b

c

HEAD

160/189

Multidimensional machines simulate

Standard machines:

Use one dimension

161/189

Standard machines simulate

Multidimensional machines:

Standard machine:

Use a two track tape

Store symbols in track 1

Store coordinates in track 2

162/189

Two-dimensional machine

x

y

a

b

c

1q

1q

Standard Machine

a

1b

#

symbols

coordinates1 # 2# 1c

#


28/32


163/189

Repeat for each transition

Update current symbol Compute coordinates of next position

Go to new position

Standard machine:

164/189

MultiDimensional Machines

have the same powerwith Standard Turing Machines

Theorem:

165/189

NonDeterministicNonDeterministic Turing MachinesTuring Machines

Lba ,

Rca ,

1q

2q

3q

Non Deterministic Choice

166/189

Lba ,

Rca ,

1q

2q

3q

a b c

1q

Time 0

Time 1

b b c

2q

Choice 1

c b c

3q

Choice 2

NondeterministicNondeterministic choiceschoices

167/189

Input string is accepted if

this a possible computationw

Initial configuration Final Configuration

Final state

NotionNotion

yqxwq0 f

168/189

NonDeterministic Machines simulate

Standard (deterministic) Machines:

Every deterministic machine

is also a nondeterministic machine


29/32


169/189

Deterministic machines simulate

NonDeterministic machines:

Keeps track of all possible computations

Deterministic machine:

170/189

Non-Deterministic Choices

Computation 1

1q

2q

4q

3q

5q

6q 7q

Computation 2

1q

2q

4q

3q

5q

6q 7q

171/189

Keeps track of all possible computations

Deterministic machine:

Simulation

Stores computations in a

two-dimensional tape

172/189

Lba ,

Rca ,

1q

2q

3q

a b c

1q

Time 0

NonDeterministic machine

Deterministic machine

a b c

1q

# # # # ####

# # #

##

# #

Computation 1

173/189

NonDeterministic machine

b b c

2q

# # # # #### #

#

# #

Computation 1

c b c3q ## Computation 2

Deterministic machine

b b c

2q

Choice 1

c b c

3qChoice 2

Time 1Lba ,

Rca ,

1q

2q

3q

174/189

Repeat

Execute a step in each computation:

If there are two or more choices

in current computation:

1. Replicate configuration

2. Change the state in the replica


30/32


175/189

Theorem: NonDeterministic Machines

have the same power withDeterministic machines

176/189

Remark:

The simulation

in the Deterministic machinetakes time exponential time compared

to the NonDeterministic machine

A Universal Turing MachineA Universal Turing Machine

178/189

Turing Machines are hardwired

they execute

only one program

A limitation of Turing Machines:

Real Computers are re-programmable

Solution: Universal Turing Machine(My Turing vn nng, hay phqut)

Reprogrammable machine Simulates any other Turing Machine

Attributes:

179/189

Universal Turing Machine

Simulates any other Turing Machine M

Input of Universal Turing Machine:

Description of transitions of M

Initial tape contents of M

180/189

UniversalTuring

Machine

Three tapes

Description of M

Tape 1

Tape contents of M

Tape 2

State of M

Tape 3



31/32


181/189

We describe TM Mas a string of symbols: We encode Mas a string of symbols

Alphabet Encoding

Description of M

Tape 1

Symbols: a b c d K

Encoding: 1 11 111 1111

Encoding in Tape 1

182/189

State Encoding

States: 1q 2q 3q 4q K

Encoding: 1 11 111 1111

Encoding in Tape 1

Description of M

Tape 1

183/189

Head Move Encoding

Move:

Encoding:

L R

1 11

Encoding in Tape 1

Description of M

Tape 1

184/189

Encoding in Tape 1

Transition Encoding

Transition: ),,(),( 21 Lbqaq =

Encoding: 10110110101

Using 0 for Separator

Description of M

Tape 1

185/189

Machine Encoding

Transitions:

),,(),( 21 Lbqaq =

Encoding:

),,(),( 32 Rcqbq =

10110110101 110111011110101100

Encoding in Tape 1

Description of M

Tape 1

Using 00 for Transition Separator186/189


Tape 1 contents of Universal TM M:

encoding of the simulated machineas a binary string of 0s and 1s

A TM is describedwith a binary string of 0s and 1s


32/32

187/189

Language of Turing Machines

L = { 010100101, 00100100101111, 111010011110010101, }

TM 1, TM 2, . . .

The set of Turing machines forms a language,

therefore :

each string of the language

is the binary encoding of a TM

188/189

Question: How powerful is RAM?

Can RAMs compute functions/languages that TMscannot?

Can RAMs compute functions/languages significantlyfaster then TMs?

Theorem:A TMA TM can simulate a RAM, provided that the elementary RAMcan simulate a RAM, provided that the elementary RAMinstructions can themselves be simulated by a TMinstructions can themselves be simulated by a TM

189/189

ProofProof

The idea is to use a multitape TM MThe idea is to use a multitape TM Mto perform simulation:to perform simulation:

Tape 1:Tape 1:keeps programkeeps program instructionsinstructions vvii in binary, numbered, and separated byin binary, numbered, and separated by

## andand **, i.e.,, i.e.,

#0*v#0*v00#1*v#1*v11#10*v#10*v22###i*v#i*vii##

Tape 2:Tape 2: keeps data registers:keeps data registers:#0*r#0*r00#1*r#1*r11#10*r#10*r22###i*r#i*rii##

Tape 3:Tape 3: keeps program counterkeeps program counter

Tape 4:Tape 4: keeps data address registerkeeps data address register

ch3_turingmachine

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