[PDF] - W e can alw a ys rename v ariables whatev er w e lik PDF Document

SLIDE 1 Closure Prop erties

f

CFL's | Substitut i

n

If a substitution s assigns a CFL to ev ery sym b

l

in the alphab et

f

a CFL L, then s(L) is a CFL. Pro

f
T

ak e a grammar for L and a gramma r for eac h language L a = s(a).

Mak

e sure all the v ariables

f

all these grammars are dieren t.

✦

W e can alw a ys rename v ariables whatev er w e lik e, so this step is easy .

Replace

eac h terminal a in the pro ductions for L b y S a , the start sym b

l
f

the grammar for L a .

A

pro

f

that this construction w

rks

is in the reader.

✦

In tuition: this replacemen t allo ws an y string in L a to tak e the place

f

an y

ccurrence
f

a in an y string

f

L. Example

L

= f0 n 1 n j n

1g,

generated b y the grammar S ! 0S 1 j 01.

s(0)

= fa n b m j m

ng,

generated b y the grammar S ! aS b j A; A ! aA j ab.

s(1)

= fab; abcg, generated b y the grammar S ! abA; A ! c j . 1. Rename second and third S 's to S and S 1 , resp ectiv ely . Rename second A to B . Resulting grammars are: S ! 0S 1 j 01 S ! aS b j A; A ! aA j ab S 1 ! abB ; B ! c j

2.

In the rst grammar, replace b y S and 1 b y S 1 . The com bined grammar: S ! S S S 1 j S S 1 S ! aS b j A; A ! aA j ab S 1 ! abB ; B ! c j

Consequences
f

Closure Under Substitut i

n

1. Closed under union, concatenation, star.

✦

Pro

fs

are the same as for regular languages, e.g. for concatenation

f

CFL's L 1 , L 2 , use L = fabg, s(a) = L 1 , and s(b) = L 2 . 1

SLIDE 2 2. Closure

f

CFL's under homomorphi sm . Nonclosure Under In tersection

The

reader sho ws the follo wing language L = f0 i 1 j 2 k 3 l j i = k and j = l g not to b e a CFL.

✦

In tuitiv ely , y

u

need a v ariable and pro ductions lik e A ! 0A2 j 02 to generate the matc hing 0's and 2's, while y

u

need another v ariable to generate matc hing 1's and 3's. But these v ariables w

uld

ha v e to generate strings that did not in terlea v e.

Ho

w ev er, the simpler language f0 i 1 j 2 k 3 l j i = k g is a CFL.

✦

A grammar: S ! S 3 j A A ! 0A2 j B B ! 1B j

Lik

ewise the CFL f0 i 1 j 2 k 3 l j j = l g.

Their

in tersection is L. Nonclosure

f

CFL's Under Complemen t

Pro
f

1: Since CFL's are closed under union, if they w ere also closed under complemen t, they w

uld

b e closed under in tersection b y DeMorgan's la w.

Pro
f

2: The complemen t

f

L ab

v

e is a CFL. Here is a PD A P recognizing it:

✦

Guess whether to c hec k i 6= k

r

j 6= l . Sa y w e w an t to c hec k i 6= k .

✦

As long as 0's come in, coun t them

n

the stac k.

✦

Ignore 1's.

✦

P

p

the stac k for eac h 2.

✦

As long as w e ha v e not just exp

sed

the b

ttom-of-stac

k mark er when the rst 3 comes in, accept, and k eep accepting as long as 3's come in.

✦

But w e also ha v e to accept, and k eep accepting, as so

n

as w e see that the input is not in L(0

1
2
3
).

Closure

f

CFL's Under Rev ersal Just rev erse the b

dy
f

ev ery pro duction. 2

SLIDE 3 Closure

f

CFL's Under In v erse Homomorphism PD A-based construction.

Keep

a \buer" in whic h w e place h(a) for some input sym b

l

a.

Read

inputs from the fron t

f

the buer ( OK).

When

the buer is empt y , it ma y b e reloaded with h(b) for the next input sym b

l

b,

r

w e ma y con tin ue making

mo

v es. T esting Emptiness

f

a CFL As for regular languages, w e really tak e a represen tation

f

some language and ask whether it represen ts ;.

In

this case, the represen tation can b e a CF G

r

PD A.

✦

Our c hoice, since there are algorithms to con v ert

ne

to the

ther.
The

test: Use a CF G; c hec k if the start sym b

l

is useless? T esting Finiteness

f

a CFL

Let

L b e a CFL. Then there is some pumping- lemma constan t n for L.

T

est all strings

f

length b et w een n and 2n

1

for mem b ership (as in next section).

If

there is an y suc h string, it can b e pump ed, and the language is innite.

If

there is no suc h string, then n

1

is an upp er limit

n

the length

f

strings, so the language is nite.

✦

T ric k: If there w ere a string z = uv w xy

f

length 2n

r

longer, y

u

can nd a shorter string uw y in L, but it's at most n shorter. Th us, if there are an y strings

f

length 2n

r

more, y

u

can rep eatedly cut

ut

v x to get, ev en tually , a string whose length is in the range n to 2n

1.

T esting Mem b ership

f

a String in a CFL Sim ulating a PD A for L

n

string w do esn't quite w

rk,

b ecause the PD A can gro w its stac k indenitely

n
input,

and w e nev er nish, ev en if the PD A is deterministic. 3

SLIDE 4

There

is an O (n 3 ) algorithm (n = length

f

w ) that uses a \dynamic programmi ng" tec hnique.

✦

Called Co c k e-Y

unger-Kasami

(CYK) algorithm.

Start

with a CNF grammar for L.

Build

a t w

-dimensional

table:

✦

Ro w = length

f

a substring

f

w .

✦

Column = b eginning p

sition
f

the substring.

✦

En try in ro w i and column j = set

f

v ariables that generate the substring

f

w b eginning at p

sition

j and extending for i p

sitions.

✦

In reader, these en tries are denoted X j;i+j 1 , i.e., the subscripts are the rst and last p

sitions
f

the string represen ted, so the rst ro w is X 11 ; X 22 ; : : : ; X nn , the second ro w is X 12 ; X 23 ; : : : ; X n1;n , and so

n.

Basis : (ro w 1) X ii = the set

f

v ariables A suc h that A ! a is a pro duction, and a is the sym b

l

at p

sition

i

f

w . Induction : Assume the ro ws for substrings

f

length up to m

1

ha v e b een computed, and compute the ro w for substrings

f

length m.

W

e can deriv e a i a i+1

a

j from A if there is a pro duction A ! B C , B deriv es an y prex

f

a i a i+1

a

j , and C deriv es the rest.

Th

us, w e m ust ask if there is an y v alue

f

k suc h that

✦

i

k

< j .

✦

B is in X ik .

✦

C is in X k +1;j . Example In class, w e'll w

rk

the table for the grammar: S ! AS j S B j AB A ! a B ! b and the string aabb. 4