The rst index has to b e 1 b ecause only the pair 10 - - PDF document

SLIDE 1 P

• st's

Corresp

• ndence

Problem

• An

undecidable, but RE, problem that app ears not to ha v e an ything to do with TM's.

• Giv

en t w

• lists
• f

\corresp

• nding"

strings (w 1 ; w 2 ; : : : ; w n ) and x 1 ; x 2 ; : : : ; x n ), do es there exist a nonempt y sequence

• f

in tegers i 1 ; i 2 ; : : : ; i k suc h that w i 1 w i 2

• w

i+k = x i 1 x i 2

• x

i k ?

• In

tuition: w e can try all lists i 1 ; i 2 ; : : : ; i k in

• rder
• f

k . If w e nd a solution, the answ er is \y es." But if w e nev er nd a solution, ho w can w e b e sure no longer solution exists, so w e can nev er sa y \no." Example (1; 0; 010; 11) and (10; 10; 01; 1).

• A

solution is 1; 2; 1; 3; 3 ; 4.

• The

constructed string from b

• th

lists is 10101001011. Another Example F rom the course reader: (10; 011; 101) and (101; 11; 011).

• Another

argumen t wh y this instance

• f

PCP has no solution:

✦

The rst index has to b e 1 b ecause

• nly

the pair 10 and 101 b egin with the same sym b

• l.

✦

Then, whatev er indexes w e c ho

• se

to con tin ue, there will b e more 1's in the string constructed from the rst list than the second (b ecause in eac h corresp

• nding

pair there are at least as man y 1's in the second list). Th us, the t w

• strings

cannot b e equal. Plan to Sho w PCP is Undecidable 1. In tro duce MPCP , where the rst pair m ust b e tak en rst in a solution. 2. Sho w ho w to reduce MPCP to PCP . 3. Sho w ho w to reduce L u to MPCP .

✦

This is the

• nly

reason for MPCP: it mak es the reduction from L u easier. 4. Conclude that if PCP is decidable, so is MPCP , and so is L u (whic h w e kno w is false). 1

SLIDE 2 Reduction

• f

MPCP to PCP T ric k: giv en a MPCP instance, in tro duce a new sym b

• l

*.

• In

the rst list, * app ears after ev ery sym b

• l,

but in the second list, the * app ears b efor e ev ery sym b

• l.

✦

Example: the pair 10 and 011 b ecomes 1*0* and *0*1*1.

✦

Notice that no suc h pair can ev er b e the rst in a solution.

• T

ak e the rst pair w 1 and x 1 from the MPCP instance (whic h m ust b e c hosen rst in a MPCP solution) and add to the PCP instance another pair in whic h the *'s are as alw a ys, but w 1 also gets an extra * at the b eginning.

✦

Referred to as \pair 0."

✦

Example: if 10 and 011 is the rst pair, also add to the PCP instance the pair *1*0* and *0*1*1.

• Finally

, since the strings from the rst list will ha v e an extra * at the end, add to the PCP instance the pair $ and *$.

✦

$ is a new sym b

• l,

so this pair can b e used

• nly

to complete a matc h.

✦

Referred to as the \nal pair." Pro

• f

the Reduction is Correct

• If

the MPCP instance has a solution 1 follo w ed b y i 1 ; i 2 ; : : : ; i k , Then the PCP instance has a solution, whic h is the same, using the rst pair in place

• f

pair 1, and terminating the list with the nal pair.

• If

the PCP instance has a solution, then it m ust b egin with the \rst pair," b ecause no

• ther

pair b egins with the same sym b

• l.

Th us, remo ving the *'s and deleting the last pair giv es a solution to the MPCP instance. Reduction

• f

L u to MPCP

• In

tuition: The equal strings represen t a computation

• f

a TM M

• n

input w .

✦

Sequence

• f

ID's separated b y a sp ecial mark er #.

• First

pair is # and #q w #. 2

SLIDE 3

• String

from rst list is alw a ys

• ne

ID b ehind, unless an accepting state is reac hed, in whic h case the rst string can \catc h up."

• Some

example pairs: 1. X and X for ev ery tap e sym b

• l

X . Allo ws cop ying

• f

sym b

• ls

that don't c hange from

• ne

ID to the next. 2. If

• (q

; X ) = (P ; Y ; R), then q X and Y p is a pair. Sim ulates a mo v e for the next ID. 3. If q is an accepting state, then X q Y and q is a pair for all X and Y . Allo ws ID to \shrink to nothing" when an accepting state is reac hed. Undecidable Problems Ab

• ut

CFL's W e are applying the theory

• f

undecidabilit y in a useful w a y when w e sho w a real problem not to b e solv able b y computer.

• Example:

Y

• u

ma y think the CS154 pro ject w as hard, but at least there is an algorithm to con v ert RE's to DF A's, so it is at least p

• ssible

for y

• u

to succeed.

• Supp
• se

next spring's CS154 pro ject is to tak e a CF G and tell whether it is am biguous. Y

• u

can't do it b ecause the problem is undecidable! Con v erting PCP to CF G's F

• r

eac h list A = (w 1 ; w 2 ; : : : ; w n ) w e can construct a gramma r and a language that represen ts all sequences

• f

in tegers i 1 ; i 2 ; : : : ; i k and the strings w i 1 w i 2

• w

i k that are constructed from those lists

• f

in tegers.

• Use

a 1 ; a 2 ; : : : ; a n as new sym b

• ls

(not in the alphab et

• f

the PCP instance) represen ting the in tegers.

• The

\grammar for list A": S ! w 1 S a 1 j w 2 S a 2 j

• j

w n S a n j .

✦

Yields all concatenations

• f

w 's follo w ed b y the rev erse

• f

their index sequence. Reduction

• f

PCP to CF G Am biguit y Problem Giv en lists A and B , construct grammar as follo ws:

• S

! A j B .

• A

is the start sym b

• l

for a grammar from list A; B is the same for list B . 3

SLIDE 4

• If

there is a solution to the PCP instance, then the same string can b e deriv ed starting S ) lm A and S ) lm B .

✦

Con v ersely , the

• nly

w a y a string can ha v e t w

• leftmost

deriv ations is if they b egin in these t w

• w

a ys, b ecause the grammar

• f
• ne

list is unam biguous. Example Use the lists

• f
• ur

rst example: (1; 0; 010; 11) and (10; 10; 01; 1 ). Let a; b; c; d stand for the four index in tegers. The grammar is: S ! A j B A ! 1Aa j 0Ab j 010Ac j 11Ad j

• B

! 10B a j 10B b j 01B c j 1B d j

• A

string with t w

• leftmost

deriv ations: 10101001011dccaba.

✦

S ) lm 1Aa ) lm 10Aba ) lm 101Aaba ) lm 101010Acaba ) lm 101010010Accaba ) lm 10101001011Adccaba ) lm 10101001011dccaba.

✦

S ) lm 10B a ) lm 1010B ba ) lm 101010B aba ) lm 10101001B caba ) lm 1010100101B ccaba ) lm 10101001011B dccaba ) lm 10101001011dccaba. Undecidable Problem: Is the In tersection

• f

Tw

• CFL's

Empt y? Consider the t w

• list

languages from a PCP instance. They ha v e an empt y in tersection if and

• nly

if the PCP instance has a solution. Complemen ts

• f

List Languages W e can get

• ther

undecidabilit y results ab

• ut

CFL's if w e rst establish that the complemen t

• f

a list language is a CFL.

• PD

A is easier approac h.

• Accept

all ill-formed input (not a sequence

• f

sym b

• ls

follo w ed b y indexes) using the state.

• F
• r

inputs that b egin with sym b

• ls

from the alphab et

• f

the PCP instance, store them

• n

the stac k, accepting as w e go. 4

SLIDE 5

• When

index sym b

• ls

start, p

• p

the stac k, making sure that the righ t strings w ere found

• n

top

• f

the stac k; again, k eep accepting un til

• When

w e exp

• se

the b

• ttom-of-stac

k mark er, w e ha v e found a sequence

• f

strings from the PCP list and their matc hing indexes. This string is not in the complemen t

• f

the list language, so don't accept.

• If

more index sym b

• ls

come in, then w e ha v e a misma tc h, so start accepting again and k eep

• n

accepting. Undecidable Problem: Is a CFL Equal to

• ?
• T

ak e an instance

• f

PCP , sa y lists A and B .

• The

union

• f

the complemen ts

• f

their t w

• list

languages is

• if

the instance has no solution, and something less if there is a solution. Undecidable Problem: Is the In tersection

• f

Tw

• CFL's

Regular? Key idea: the in tersection

• f

list languages is regular if and

• nly

if it is empt y . Th us, PCP reduces to regularit y

• f

in tersection for CFL's.

• Ob

viously , if empt y , it is regular.

• Supp
• se

the in tersection

• f

t w

• list

languages, for A and B , L A \ L B , is nonempt y . Then there is a solution to this instance

• f

PCP , sa y string w and string

• f

index sym b

• ls

i.

✦

Example: for the running PCP instance, w = 10101001011 and i = abaccd.

• Then

i k is an index sequence that yields solution w k for all k .

✦

General principle: concatenation

• f

PCP solutions is a solution.

• Consider

homomorphism h(0) = w and h(1) = i R .

• h

1 (L A \ L B ) is f0 n 1 n j n

• 1g.
• Since

regular languages are closed under in v erse homom

• rphism

, if the in tersection w ere regular, so w

• uld

h 1 (L A \ L B ) b e.

• Since

w e kno w this language is not regular, w e conclude that L A \ L B is not regular. 5