chapter 8 introduction to turing machines (part a) (1)

7/27/2019 Chapter 8 Introduction to Turing Machines (Part a) (1)

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Chapter 8 Introduction toTuring Machines (part a)

Rothenberg, Germany


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Outline

Problems that Computers Cannot Solve

The Turing Machine (TM)

(in part b) Programming Techniques for TMs

Extensions to the Basic TM

Restricted TMs

TMs and Computers


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8.0 Introduction

Concepts to be taught ---

Studying questions about what languages can

be defined be any computational device??There are specific problems that cannot be

solved by computers! --- undecidable!

Studying the Turing machine which seemssimple, but can be recognized as an accurate

model for what any physical computing

device is capable of doing.


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8.1 Problems That Computers

Cannot Solve

Purpose of this section ---

to provide an informal proof , C-

programming-based introduction to proof of aspecific problem that computers cannot solve.

The problem:

whether the first thing a C program prints ishello, world.

We will give the intuition behind the formal

proof.


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Cannot Solve

8.1.1 Programs that print Hello, World

A C program that prints Hello, World :

main(){

print(hello, world\n);

}

Define hello, world problemto be:

determine whether a given C program, with a given

input, prints hello, worldas the first 12 characters

that it prints.


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Cannot Solve

8.1.1 Programs that print Hello, World

The problem described alternatively using symbols:

Is there a programHthat could examine any

program P and input I for P, and tell whether P,

run with I as its input, would print hello, world?(A

programHmeans an algorithmin concept here.)

The answer is: undecidable! That is, there exists no

suchprogram

H. We will prove this by contradiction.


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Cannot Solve

8.1.2 Hypothetical Hello, World Tester

1ststep: assumeHexists in the following form:

2ndstep: transformHto another formH2in simple

ways which can be done by C programs

3rdstep: proveH2 not existing. SoHalso not existing.

Hello-world

tester

H

I

P

yes if P, with input I , pr in tshello world

no if not.Figure 8.3


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Cannot Solve 8.1.2 Hypothetical Hello, World Tester

2ndstep:

(1) transformHtoH1in the following way -

(2) transformH1toH

2in the following way

Use P both as input and program!

Hello-world

tester

H1

I

P

yes

hel lo, wo rld

Figure 8.4

Hello-world

tester

H2

P

yes

hel lo, wo rld

Figure 8.5


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Cannot Solve 8.1.2 Hypothetical Hello, World Tester

2ndstep (contd):

The function ofH2is:

given any program P as input,if P prints hello world as first output, thenH2makes

output yes;

if P does not prints hello worldas first output, thenH2

prints hello world.

Hello-world

tester

H2

P

yes

hel lo, wo rld

Figure 8.5


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Cannot Solve

8.1.2 Hypothetical Hello, World Tester3rdstep:

ProveH2does not exist as follows

LetPforH2in Fig. 8.5 (last figure) beH2itself, as

follows:

Hello-world

tester

H2

H2

yes

hel lo, wo rld

Figure 8.6


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Cannot Solve

8.1.2 Hypothetical Hello, World Tester3rdstep (contd): ProveH2does not exist by contradiction as follows (contd)

Now,

(1) if the boxH2, given itself as input, makes outputyes, then it

means thatthe inputH2, given itself as input, prints hello worldas first

output.

But this is contradictory because we just suppose thatH2, given

itself as input, makes outputyes.

Hello-world

tester

H2

H2

yes

hel lo, wo rld

Figure 8.6


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Cannot Solve

8.1.2 Hypothetical Hello, World Tester3rdstep (contd): ProveH2does not exist as follows (contd)

The above contradiction means the other alternative must betrue (since it must be one or the other), that is ---

(2) the boxH2, given itself as input, printshello, world. Thisthen means that

suchH2, when taken as input to the boxH2(itself), will makethe boxH2to make outputyes.

Contradiction again because we just say that the boxH2,

given itself as input, printshello, world.

Hello-world

testerH2

H2

yes

Hel lo, world

Figure 8.6


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Cannot Solve


3rdstep (contd):

ProveH2does not exist as follows (contd)

Since both cases lead to contradiction, we conclude that the

assumption thatH2exists is wrong by the principle of

contradiction for proof.

Hello-world

tester

H2

H2

yes

Hel lo, world

Figure 8.6


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Cannot Solve


3rdstep (contd):

H2does not exist

H1does not exist (otherwise,H2must exist)

Hdoes not exist (otherwise,H1must exist)

The above self-contradictory technique, similar to the

diagonalization technique (to be introduced later), was

used by Alan Turing for proving undecidable problems.


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Cannot Solve

8.1.3 Reducing One Problem to Another

Now we have an undecidable problem, which can be

used to prove other undecidable problems by a

technique of problem reduction.That is, if we knowP1is undecidable, then we may

reduce P1to a new problem P2, so that we may prove

P2

undecidable by contradiction in the following way:

If P2is decidable, then P1is decidable.

But P1 is known undecidable. So, contradiction!

Consequently, P2is undecidable.


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Cannot Solve


Example 8.1

We want to prove a new problemP2

(called calls-foo

problem):

does program Q, given input y, ever call function foo?

to be undecidable.

Solution: regard Q in P1;

reduceP1: the hello-world problem toP2: the calls-

foo problem in the following way:

(continued in the next page)


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Cannot Solve

8.1.3 Reducing One Problem to AnotherExample 8.1 (contd)

Solution: reduceP1toP2in the following way:

If Qhas a function called foo, renameitand all calls to that

function a new program Q1doing the same as Q. ()

Add to Q1a function foo doing nothing & notcalled a new Q2

Modify Q2to remember the first 12 characters that it prints,

storing them in a global arrayAQ3 Modify Q3so that whenever it executes any output statement, it

checksAto see if it has written 12 characters or more, and if so,

whetherhello, world are the first characters. In that case, call the

new function foo Rwith inputz=y.


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Cannot Solve


Example 8.1 (contd)

Solution (contd): Now,

(1) if Qwith input y prints hello, worldas its first output, thenRwill call foo;

(2) if Qwith input y does not print hello, world, thenRwill

never call foo.

(3) That is,R,with inputy,calls fooif and only if Q,with input

y,prints hello, world.

(4) So, if we can decide whetherR,with inputz,calls foo, then

we can decide whether Q, with inputy, prints hello, world.


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Cannot Solve

8.1.3 Reducing One Problem to AnotherThe above example illustrates how to reduce a

problem to another.

An illustration of this idea is:

ConstructP1

instance

P2

instanceDecide yes

no


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8.2 The Turing Machine

Concepts to be taught:

The study of decidability provides guidance to programmers

about what they might or might not be able to accomplish

through programming. Previous problems are dealt with programs. But not all

problems can be solved by programs.

We need a simple model to deal with other decision problems

(like grammar ambiguity problems) The Turing machine is one of such models, whose

configuration is easy to describe, but whose function is the

most versatile: all computations done by a modern computer

can be done by a Turing machine.


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8.2.1 The Quest to Decide All Mathematical

Questions

At the turn of 20thcentury, D, Hilbert asked:

whether it was possible to find an algorithm for

determining the truth or falsehood of any

mathematical proposition.(in particular, he asked if there was a way to decide

whether any formula in the 1st-order predicate

calculus, applied to integer, was true)


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8.2.1 The Quest to Decide All MathematicalQuestions

In 1931, K. Gdel published his incompleteness

theorem:

A certain formula in the predicate calculus applied

to integers could not be neither proved nor

disproved within the predicate calculus.

The proof technique is diagonalization,resembling the self-contradictory technique

used previously.


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Natures of computational model

Predicate calculus --- declarative

Partial-recursive functions --- computational (a

programming-language-like notion)

Turing machine --- computational (computer-like)

(invented by Alan Turing several years before true

computers were invented)


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8.2.1 The Quest to Decide All Mathematical

Questions

Equivalence of maximalcomputational model

They all compute the same functions or recognize

the same languages, having the same power ofcomputation


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Unprovable Church-Turing hypothesis (or

thesis)

Any general way to compute will allow us to

compute only the partial-recursive functions (orequivalently, what the Turing machine or modern-

day computers can compute).


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8.2.2 Notion for the Turing Machine

A model for Turing machine:

BX2

Finite

control

B B X1 Xi Xn B


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A move of Turing machine includes:

change state;

write a tape symbol in the cell scanned; move the tape head left or right.

Formal definition:

A Turing machine (TM) is a 7-tupleM= (Q, S, G, d,q0,B,F) where

Q: a finite set of states of the finite control;

S: a finite set of input symbols;

G: a set of tape symbols, with Sbeing a subset;


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Formal definition (contd):

d: a transition function d(q,X) = (p, Y,D) where

q: the current state, in Q;

X: a tape symbol being scanned;

p: the next state, in Q;

Y: the tape symbol written on the cell being scanned, used

to replaceX;

D: eitherL(left) orR (right) telling the move direction of

the tape head;


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Formal definition (contd):

q0: the start state, in Q;

B: the blank symbol in G,notin S (should not

be an input symbol);

F: the set of final or accepting states.

A TM is a deterministicautomaton with a two-

way infinitetape which can be read and writtenin

either direction.


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8.2.3 Instantaneous Descriptions for Turing

Machine

The instantaneous description(ID) of a TM isrepresented by

X1X2Xi1qXiXi+1Xnin which

qis the current state;The tape head is scanning the ith symbol from the left;

X1X2Xnis the portion of the tape between the

leftmost and the rightmost nonblank symbols.


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8.2.3 Instantaneous Descriptions for Turing

Machine

Moves of a TM Mdenoted by or as follows:If d(q,Xi) = (p, Y,L) (a leftward move), then we

write the following to describe the left move:

X1X2Xi1qXiXi+1Xn X1X2Xi2pXi1YXi+1Xn

Right moves are defined similarly.

_

|

_

|M

_|M


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8.2.3 Instantaneous Descriptions for TuringMachine

Example 8.2--- Design a TM to accept the languageL=

{0n1n| n1} as follows.

Starting at the left end of the input.

Change 0 to anX.

Move to the right over 0s and Ys until a 1.

Change 1 to Y.

Move left over Ys and 0s until anX.

Look for a 0 immediately to the right.

If a 0 is found, change it toXand repeat the above process.


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8.2.3 Instantaneous Descriptions for TuringMachine

Example 8.2--- Design a TM to accept the

languageL= {0n1n| n1} (contd).

M= ({q0~q4}, {0, 1}, {0, 1,X, Y,B}, d, q0,B, {q4})

Transition table: in the next page.


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symbol

state 0 1 X Y B

q0

(q1, X, R)1 - - (q3, Y, R)8 -q1

(q1, 0, R)2 (q2, Y, L )4 - (q1, Y, R)3 -

q2 (q2, 0, L )5 - (q0, X, R)7 (q2, Y, L )6 -q3 - - - (q

3, Y, R)

9 (q

4, B, R)

10

q4

- - - - -

8.2 The Turing Machine 8.2.3 Instantaneous Descriptions for TM

Example 8.2---


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8.2.3 Instantaneous Descriptions for TM

Example 8.2(contd)

To accept 0011 ---

(use instead of )

q0

0011 Xq1

011 X0q1

11 Xq2

0Y1 q2

X0Y1

Xq00Y1 XXq1Y1 XXYq11 XXq2YYXq2XYY

XXq0YYXXYq3YXXYYq3BXXYYBq4B

_|


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8.2.4 Transition Diagrams for TMs

If d(q,X) = (p, Y,L), we use labelX/Yon the arc.

If d(q,X) = (p, Y,R), we use labelX/Yon the arc.

Example 8.3 --- Transition diagram for Example 8.2.

See the textbook, p. 331.

Example 8.4 --- TM as a function-computingmachine.No final state is needed. For details, see thetextbook and part b.


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8.2.5 The Language of a TM

LetM= (Q, S, G, d, q0,B,F) be a TM. The language

accepted byMisL(M) = {w| wS*and q0w apbwithpF}

The set of languages accepted by a TM is often

called the recursively enumerable language or RElanguage.

The term RE came from computational formalism

that predate the TM.

_*|M


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8.2.6 TMs and Halting

Another notion for accepting strings by TMs

--- acceptance by halting.

We say a TM halts if it enters a state q

scanning a tape symbolX, and there is no

move in this situation, i.e., d(q,X) is

undefined.


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Acceptance by halting may be used for a TMs

functions other than accepting languages like Example8.4 and Example 8.5.

We assume that a TM always halts when it is in an

accepting state.It is not always possible to require that a TM halts even

it does not accept.


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Languages with TMs that do halt eventually,

regardless whether or not they accept, are calledrecursive(considered in Sec. 9.2.1)

TMs that always halt, regardless of whether or not

they accept, are a good model of an algorithm. So TMs that always halt can be used for studies of

decidability (see Chapter 9).

chapter 8 introduction to turing machines (part a) (1)

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