Turing Machines Lecture 14-16 Ref. handout page 57 -65.

Turing Machines

Lecture 14-16

Ref. handout page 57 -65

Alan Turing 1912-1954

Great mathematician The father of

the modern computer

Alan Turing’s Statueat Bletchley Park

Bletchley Park is near Milton Keynes Half hour from London

Historical Notes

For a long time it was believed that any mathematical problem could, at least in principle, be proved from the basic axioms

1931, Kurt Gödel proposed the theorem of undecidability – there exist theorems which can neither be proved or disproved

Later, Alonso Church and Alan Turing also found other problems which had no algorithmic solution

Alan Turing was interested in whether there was a way to define which

problems were/were not decidable (computable)?

can we create a machine to simulate the human brain so that those computable problems can be solved automatically?

1935-36, Turing was working on a paper, “computable numbers”. The Turing machine in this paper turned out to be the simplest prototype of all computers!

Turing’s Idea

a+b*x/y

thinks

State of mind changes

reads

makes notes

The Turing Machines

(Infinitely long) tape, one symbol per square

control unit

Currentstate

Read/Writehead

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

A Turing Machine Conceptualization

0

00

-

-- 0

start

reset

-

-

Actions of a Turing Machine

Move left/right one square

Change

state

Write a new symbol

onto the current tape square

depending on current state

and

current tape symbol

State Transitions

read write move

A BX,Y / L

State Transitions – comparing with FA and PDA

read write move for TM

input value on stack/ op of stack for PDA

A BX,Y / L

A State Transition Table

state read write move new

state

1 a b L 2

2 b b L 2

2 a b R 3

3 a b L 2

3 b c R 2

1

2

3

The Infinite Tape

a b aca

Initial position of read/writehead (the left most)

Initial data (finite)

Blank tape (infinite)

A Turing Machine Example

Test for a palindrome

aabbabbaaa □ □ □ □ □ □ □ □ □ □

If ‘a’ then replace with spacego to right hand endcheck for ‘a’if not found halt -> errorelse go to left hand end

do the samefor ‘b’

Cont. Transition Table for checking ‘a’

state read write move New state

start a □ findA

findA a

findA b

findA □ deleteA

deleteA a

deleteA b

deleteA □

Answer (cont. TM for detecting palindromes)

startreturn no

delB

yes

findB

delAfindAa,□/R

b,□/R

a,a/Rb,b/R

a,a/Rb,b/R

a,a/Lb,b/L

□,□/R

b,□/L

a,□/L

b,b/L

a,a/L

□,□/L

□,□/L

□,□/L

□,□/L

□,□/L

Memo for In-class test 14 [ /5]

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Turing Machine Tricks

Write as many y’ as there are x’s

before

□ □ □ □ x x x x □ □ □ □

after

□ □ □ □ x x x x □ y y y y □ □ □ □

Answer

Unary Numbers

0 x1 x x2 x x x3 x x x x 4 x x x x x5 x x x x x x 6 ......

x x x x + x x x = x x x x x x ? + ? = ?

The Limitations of TMs

control unit

Control Unitdoes one thing only

‘hardware’specific

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

The 2-Tape Turing Machine

control unit

data tape

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

program tape

program tape stores the description of a 1-tape TM’s transition table

The 2-tape TM – The Universal TM

control unit

Data tape

Program tape

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

2-Tape Transitions

state read write move read write move new

state

tape 1 tape 2

Includes ‘no move’

The Emulator

control unit

data tape

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

program tape

extras, e.g. the current state

An Example of UTM (Universal Turing Machine)

Adding one to a binary numbere.g.before □ □ □ □ 1 0 1 0 □ □ □ □after □ □ □ □ 1 0 1 1 □ □ □ □

before □ □ □ □ 1 0 1 1 □ □ □ □after □ □ □ □ 1 1 0 0 □ □ □ □

before □ □ □ □ 1 1 1 1 □ □ □ □after □ □ □ 1 0 0 0 0 □ □ □ □

An Example of UTM Cont.

How to go about writing a TM for adding one (a single-tape first)?

Idea (algorithm):

1. Move the head to the right most position

2. If the right most is 0, replace it by 1,stop

3. move to left one space

4. If the current value is ‘0’ or a blank replace

it by 1, stop

1. Go to step 3

An Example of UTM Cont.

start check

stop carry

1,1/R0,0/R

0,1,_□,1, _

□,□ /L

0,1,_

1,0,L

1,0,L

A TM which adds one in binary

state read write mov

new

state

start

start

1

0

1

0

R

R

start

start

start

check

□

0

□

1

L

_

check

stop

check

carry

1

1

0

0

L

L

carry

carry

carry

carry

0

□

1

1

_

_

stop

stop

Cont. change to 3-tape TM

S 1 1 R S S 0 0 R S S □ □ L C

Program tape

1 1 0 1

S

Data tape

Extra tape

What’s so Special?

UTM

•The lever – many lifting jobs

•The wheel – lots of uses

•UTM –anything you can

‘program’

Memo for In-class test 15 [ /5]

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A Real Computer

Control unit: processor

Data tape: user memory

Program tape:

program memory

Extra tape:

system memoryI/O facilities added

TMs and ‘Real’ Computer - 2

Write a TM which simulates a PC

Anything a real computer can do,a TM can do

TM

TMs and ‘Real’ Computer - 1

Write a Java program which simulates a TM

Anything a TM can do, a real computer can do

TM

The Church-Turing Thesis

Anything which can be computed can be computed by a Turing Machine

Anything which can’t be done by a TM can’t be done by any computer

corollary

So ......

If there is anything a TM can’t doit probably can’t be done

Is there anything which TMs can’t do???

But may be a TM can do anything we can imagine (?)

Decision Problems

It is harder to say what we will never be able to do rather than what we can do (computable).

Consider only simpler problems – decision problems where the answer is ‘yea’ or ‘no’.

Consider only TMs with two halt states ‘yes’ and ‘no’.

A problem is decidable if we can have a TM for it which eventually enters either the ‘yes’ or ‘no’ state depending on its input.

The Halting Problem

...... In searching of an undecidable problem.

The best-known such problem is called the halting problem.

Given an arbitrary program with an arbitrary input, can we make a Turing Machine to test whether the program stop or loop forever?

The Barber Paradox (by Bertrand Russell )

Once upon a time, a village barber put a notice outside his shop

I shave all and only thosemen in the village who do not shave themselves.

Q. Does the barber shave himself?

The Barber Paradox

U = { every man in the village }S = { men shave themselves }U \ S = { men don’t shave themselves}

Where does the barber belong to, S or U \ S ?

The Barber Paradox – trouble in either ways

Assume: the barber does shave himselfAny man in this village who shaves himself is not shaved by barber.Therefore, the barber does not shave himself.

Assume: the barber doesn’t shave himselfAny man in this village is shaved by the barber if and only if he is not shaved by himself. Therefore, the barber does shaves himself.

Why is the Barber Paradox important?

A paradox with importance to mathematical logic and set theory.

It was constructed to demonstrate the self-contradictory nature of the elementary set theory

It underlies the proof of Alan Turing’s proof of the undesirability of the halting problem.

Program Testing – (another way to look at the halting problem)

A Java program is stored in a file

The title is called ‘abc.java’

• I have tested ‘abc’ with different data• It always works• It’s never got into an infinite loop• Can I be sure it never will for any input?

Automated Program Testing

A file containing

a program

A file of test data

Prog Test

MyTest

infinite loop /halt

The Java Prog Test Class

public class Prog Test

{

Public boolean halt(String prog, String data)

{

if prog halts when given data as input

return true;

else

return false;

}

}

Using Prog Test

import Prog Textpublic class MyTest{

static public void main(String[] args){ String s = args[0];

if(ProgTest.halt(s,s)) // loop foreverwhile(true) continue;

}}

Running MyTest

java MyTest MyTest.java

command to run java

MyTest(programTo test)

with itselfas data

Suppose It Loops

static public void main(String[] args)

{ String s = args[0]; s = MyTest.java

if(ProgTest.halt(s,s))

// loop forever

while(true) continue;

}

ProgTest.halt(“MyTest.java, “MyTest.java”)

The Dilemma

If it halts it loops forever ...

If it loops forever it halts ...

The Solution

The result is nonsense –

MyTest must halt or loop

We got into this mess by assuming that

ProgTest.halts could be written

So ProgTest.halts doesn’t (can’t) exist.

Conclusion

ProgTest.halts can’t be written

No program exists which can tell whether any given program halts for any given data

There are some things computer can’t do

Memo for In-class test 16 [ /5]

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