[PDF] - W e m ust argue that the TM can do exactly what a PDF Document

SLIDE 1 Outline

f

T uring Mac hines and Complexit y 1. T uring mac hine (TM) = formal mo del

f

a computer running a particular program.

✦

W e m ust argue that the TM can do exactly what a computer can do, alb eit slo w er. 2. W e use the simplicit y

f

the TM mo del to pro v e formally that there are sp ecic problems (=languages) that the TM cannot solv e.

✦

Tw

classes:

\recursiv ely en umerable" = TM can accept the strings in the language but cannot tell for certain that a string is not in the language; \non-RE" = no TM can ev en recognize the mem b ers

f

the language in the RE sense. 3. W e then lo

k

at problems (languages) that do ha v e TM's that accept them and alw a ys halt; i.e., they not

nly

recognize the strings in the language, but they tell us when they are sure the string is not in the language.

✦

The classes P and N P are those languages recognizable b y deterministic (resp., nondeterministic) TM's that halt within a time that is some p

lynomial

in the input.

✦

P

lynomial

is as close as w e can get, b ecause real computers and dieren t mo dels

f

(deterministic) TM's can dier in their running time b y a p

lynomial

function, e.g., a problem migh t tak e O (n 2 ) time

n

a real computer and O (n 6 )

n

a TM. 4. NP-complete problems: Since w e don't kno w whether P = N P , but it app ears that at least some problems in N P tak e exp

nen

tial time, the b est w e can do is sho w that a certain problem is \NP-complete," = if this problem is in P , then all

f

N P is in P . 5. Some sp ecic problems that are NP-complete: satisabilit y

f

b

lean

(prop

sitional

logic) form ulas, tra v eling salesman, etc. In tuiti v e Argumen t Ab

ut

an Undecidabl e Problem Giv en a C program, do es it prin t hello, world. as the rst 13 c haracters

f
utput?

1

SLIDE 2

W

e pro v e there is no C program to solv e that problem b y supp

sing

that there w ere suc h a program H , the \hello-w

rld-tester."

✦

H tak es as input a C program P and an input le I for that program, and tells whether P , with input I , \prin ts hello w

rld"

(b y whic h w e mean it do es so as the rst 13 c haracters).

Mo

dify H to a new program H 1 that acts lik e H , but when H prin ts no, H 1 prin ts hello, world.

✦

Requires some though t: w e need to nd where no is prin ted and c hange the printf statemen t.

Mo

dify H 1 to H 2 . This program tak es

nly
ne

input, P , and acts lik e H 1 with b

th

its program and data inputs equal to P .

✦

I.e., H 2 (P ) = H 1 (P ; P ).

✦

Requires more though t: H 2 m ust buer its input so it can b e used as b

th

the P and I inputs to H 1 .

H

2 cannot exist. If it did, what w

uld

H 2 (H 2 ) do?

✦

If H 2 (H 2 ) = yes, then H 2 giv en H 2 as input eviden tly do es not prin t hello, world. But H 2 (H 2 ) = H 1 (H 2 ; H 2 ) = H (H 2 ; H 2 ), and H 1 prin ts yes if and

nly

if its rst input, giv en its second input as data, prin ts hello, world. Th us, H 2 (H 2 ) = yes implies H 2 (H 2 ) = hello, world.

✦

But if H 2 (H 2 ) = hello, world. then H 1 (H 2 ; H 2 ) = hello, world. and H (H 2 ; H 2 ) = no. Th us, H 2 (H 2 ) = hello, world. implies H 2 (H 2 ) 6= hello, world. The TM

Finite-state

con trol, lik e PD A.

One

read-write tap e serv es as b

th

input and un b

unded

storage device.

✦

T ap e divided in to c el ls.

✦

Eac h tap e holds

ne

sym b

l

from the tap e alphab et.

✦

T ap e is \semi-innite"; it ends

nly

at the left. 2

SLIDE 3

T

ap e he ad marks the \curren t" cell, whic h is the

nly

cell that can inuence the mo v e

f

the TM.

Initially

, tap e holds a 1 a 2

a

n B B

where

a 1 a 2

a

n is the input, c hosen from an input alphab et (subset

f

the tap e alphab et) and B is the blank. F

rmal

TM M = (Q; ; ;

;

q ; B ; F ), where:

Q

= n te set

f

states.

=

tap e alphab et;

=

input alphab et.

B

in

=

blank.

q

in Q = start sym b

l;

F

Q

= accepting states.

tak

es a state and tap e sym b

l,

returns a new state, replacemen t sym b

l

(either migh t not c hange) and a direction L=R for head motion. Example Non trivial examples are hard to come b y . Here's a TM that c hec ks its third sym b

l

is 0, accepts if so, and runs forev er, if not. M = (fp; q ; r ; s; tg; f0; 1g ; f0 ; 1; B g; p; B ; fsg) 1.

(p;

X ) = (q ; X ; R) for X = 0; 1. 2.

(q

; X ) = (r ; X ; R) for X = 0; 1. 3.

(r

; 0) = (s; 0; L). 4.

(r

; 1) = (t; 1; R). 5.

(t;

X ) = (t; X ; R) for X = 0; 1; B . ID's

f

a T uring Mac hine The ID (instan taneous description) captures what is going

n

at an y momen t: the curren t state, the con ten ts

f

the tap e, and the p

sition
f

the tap e head.

Keep

things nite b y dropping all sym b

ls

to the righ t

f

the head and to the righ t

f

the righ tmost non blank.

✦

Subtle p

in

t: although there is no limit

n

ho w far righ t the head ma y mo v e and write non blanks, at an y nite time, the TM has visited

nly

a nite prex

f

the innite tap e. 3

SLIDE 4

Notation:

q

sa

ys:

✦

is

the tap e con ten ts to the left

f

the head.

✦

The state is q .

✦

is

the non blank tap e con ten ts at

r

to the righ t

f

the tap e head.

One

mo v e indicated b y ` ; zero,

ne,
r

more mo v es represen ted b y ` * .

✦

Chec k the reader for the detailed denition

f

`. Example With input 0101, the sequence

f

ID's

f

the TM is: p0101 ` 0q 101 ` 01r 01 ` 0s101.

A

t that p

in

t it halts, since state s has no mo v e when the head is scanning 1. With input 0111 the sequence is: p0111 ` 0q 111 ` 01r 11 ` 011t1 ` 0111t ` 0111B t `

.
The

TM nev er halts, but con tin ues to mo v e righ t. Acceptance b y Final State and b y Halting One w a y to dene the language

f

a TM is b y the set

f

input strings that cause it to reac h an accepting state.

L(M

) = fw j q w ` * p for some p in F and an y

and
in
g.

Another w a y is to dene the set

f

strings that cause the TM to halt = ha v e no next mo v e.

H

(M ) = fw j q w ` * pX

,

and

(p;

X ) is not denedg.

✦

Subtle p

in

t: a TM can app e ar to halt if the next mo v e w

uld

tak e the head

the

left end

f

the tap e.

✦

Giv en an y TM, w e can mark the left end so that nev er happ ens; i.e., w e pro duce a mo died TM that accepts the same language and halts rather than fall

the

left end. Example

The

TM M

f
ur

previous example has L(M ) equal to those strings in the language

f

RE (0 + 1)(0 + 1)0(0 + 1)

.

4

SLIDE 5

H

(M ) is the language

f
+

+ 1 + (0 + 1)(0 + 1) + (0 + 1)(0 + 1)0(0 + 1)

.

Equiv alence

f

Acceptance b y Final State and Halting W e need to sho w L is L(M 1 ) for some TM M 1 if and

nly

if L is H (M 2 ) for some TM M 2 . If Mo dify M 2 as follo ws: 1. In tro duce

ne

accepting state r . 2. Whenev er there is no transition for M 2

n

state q and sym b

l

X , add a transition to state r , mo ving righ t (so w e can't p

ssibly

fall

the

left end) and lea ving sym b

l

X . Only-If Roughly , w e let M 2 sim ulate M 1 , but if M 1 en ters an accepting state, M 2 has no next mo v e and so halts.

Ma

jor problem: M 1 could halt without accepting.

✦

T

a

v

id

this problem, in tro duce state r that mo v es righ t

n

ev ery sym b

l,

sta ying in state r and lea ving the tap e sym b

ls

unc hanged.

✦

Giv e M 2 a transition to r (mo ving righ t)

n

ev ery state-sym b

l

com bination that do es not ha v e a rule.

Also,

remo v e all transitions where the state is an accepting state

f

M 1 , so M 2 will halt in those situations. F alling O the Left End

f

T ap e The reader talks ab

ut

the funn y situation where the TM w

uld

halt but falls

the

left end

f

tap e.

This

situation is not halting.

Neither

do es a TM accept if it tries to en ter an accepting state as it falls

the

left end.

W

e can prev en t falling

the

left end, b y marking the leftmost cell, as in the reader.

But

it app ears w e do not need to do so in

rder

to pro v e the equiv alence

f

halting/accepting, since neither

ccurs

when the TM falls

the

left end. 5