W ell sort of; y ou lose the abilit y to generate as a - - PDF document

SLIDE 1 Cleaning Up Grammars W e can \simplify" grammars to a great exten t. Some

• f

the things w e can do are: 1. Get rid

• f

useless sym b

• ls

| those that do not participate in an y deriv ation

• f

a terminal string. 2. Get rid

• f
• pr
• ductions

| those

• f

the form v ariable ! .

✦

W ell sort

• f;

y

• u

lose the abilit y to generate

• as

a string in the language. 3. Get rid

• f

unit pr

• ductions

| those

• f

the form v ariable ! v ariable. 4. Chomsky normal form |

• nly

pro duction forms are v ariable ! t w

• v

ariables and v ariable ! terminal. Useless Sym b

• ls

In

• rder

for a sym b

• l

X to b e useful, it m ust: 1. Deriv e some terminal string (p

• ssibly

X is a terminal). 2. Be r e achable from the start sym b

• l;

i.e., S ) * X

• .
• Note

that X w

• uldn't

really b e useful if

• r
• included

a sym b

• l

that didn't satisfy (1), so it is imp

• rtan

t that (1) b e tested rst, and sym b

• ls

that don't deriv e terminal strings b e eliminated b efor e testing (2). Finding Sym b

• ls

That Don't Deriv e An y T erminal String Recursiv e construction: Basis : A terminal surely deriv es a terminal string. Induction : If A is the head

• f

a pro duction whose b

• dy

is X 1 X 2

• X

k , and eac h X i is kno wn to deriv e a terminal string, then surely A deriv es a terminal string.

• Keep

going un til no more sym b

• ls

that deriv e terminal strings are disco v ered. Example S ! AB j C ; A ! 0B j C ; B ! 1 j A0; C ! AC j C 1.

• Round

1: and 1 are \in."

• Round

2: B ! 1 sa ys B is in. 1

SLIDE 2

• Round

3: A ! 0B sa ys A is in.

• Round

4: S ! AB sa ys S is in.

• Round

5: Nothing more can b e added.

• Th

us, C can b e eliminated, along with an y pro duction that men tions it, lea ving S ! AB ; A ! 0B ; B ! 1 j A0. Finding Sym b

• ls

That Cannot Be Deriv ed F rom the Start Sym b

• l

Another recursiv e algorithm: Basis : S is \in." Induction : If v ariable A is in, then so is ev ery sym b

• l

in the pro duction b

• dies

for A.

• Keep

going un til no more sym b

• ls

deriv able from S can b e found. Example S ! AB ; A ! 0B ; B ! 1 j A0.

• Round

1: S is in.

• Round

2: A and B are in.

• Round

3: and 1 are in.

• Round

4: Nothing can b e added.

• In

this case, all sym b

• ls

are deriv able from S , so no c hange to gramma r.

• Reader

has an example where not

• nly

are there sym b

• ls

not deriv able from S , but y

• u

must eliminate rst the sym b

• ls

that don't deriv e terminal strings,

• r

y

• u

get the wrong grammar. Eliminati ng

• Pro

duction s A v ariable A is nul lable if A ) * . Find them b y a recursiv e algorithm: Basis : If A !

• is

a pro duction, then A is n ullable. Induction : If A is the head

• f

a pro duction whose b

• dy

consists

• f
• nly

n ullable sym b

• ls,

then A is n ullable.

• Once

w e ha v e the n ullable sym b

• ls,

w e can add additional pro ductions and then thro w a w a y the pro ductions

• f

the form A !

• for

an y A. 2

SLIDE 3

• If

A ! X 1 X 2

• X

k is a pro duction, add all pro ductions that can b e formed b y eliminating some

• r

all

• f

those X i 's that are n ullable.

✦

But, don't eliminate all k if they are all n ullable. Example If A ! B C is a pro duction, and b

• th

B and C are n ullable, add A ! B j C . Eliminati ng Unit Pro ductions 1. Eliminate useless sym b

• ls

and

• pro

ductions. 2. Disco v er those pairs

• f

v ariables (A; B ) suc h that A ) * B .

✦

Because there are no

• pro

ductions, this deriv ation can

• nly

use unit pro ductions.

✦

Th us, w e can nd the pairs b y computing reac hablit y in a graph where no des = v ariables, and arcs = unit pro ductions. 3. Replace eac h com bination where A ) * B )

• and
• is
• ther

than a single v ariable b y A ! .

✦

I.e., \short circuit" sequences

• f

unit pro ductions, whic h m ust ev en tually b e follo w ed b y some

• ther

kind

• f

pro duction. Remo v e all unit pro ductions. Chomsky Normal F

• rm

0. Get rid

• f

useless sym b

• ls,
• pro

ductions, and unit pro ductions (already done). 1. Get rid

• f

pro ductions whose b

• dies

are mixes

• f

terminals and v ariables,

• r

consist

• f

more than

• ne

terminal. 2. Break up pro duction b

• dies

longer than 2.

• Result:

All pro ductions are

• f

the form A ! B C

• r

A ! a. No Mixed Bo dies 1. F

• r

eac h terminal a, in tro duce a new v ariable A a , with

• ne

pro duction A a ! a. 2. Replace a in an y b

• dy

where it is not the en tire b

• dy

b y A a .

✦

No w, ev ery b

• dy

is either a single terminal

• r

it consists

• nly
• f

v ariables. 3

SLIDE 4 Example A ! 0B 1 b ecomes A ! 0; A 1 ! 1; A ! A B A 1 . Making Bo dies Short If w e ha v e a pro duction lik e A ! B C D E , w e can in tro duce some new v ariables that allo w the v ariables

• f

the b

• dy

to b e in tro duced

• ne

at a time.

• A

b

• dy
• f

length k requires k

• 2

new v ariables.

• Example:

In tro duce F and G; replace A ! B C D E b y A ! B F ; F ! C G; G ! D E . Summary Theorem If L is an y CFL, there is a gramma r G that generates L

• fg,

for whic h eac h pro duction is

• f

the form A ! B C

• r

A ! a, and there are no useless sym b

• ls.

CFL Pumping Lemma Similar to regular-language PL, but y

• u

ha v e to pump t w

• strings

in the middle

• f

the string, in tandem (i.e., the same n um b er

• f

copies

• f

eac h). F

• rmally:
• 8

CFL L

• 9

in teger n

• 8

z in L, with jz j

• n
• 9

uv w xy = z suc h that jv w xj

• n

and jv xj >

• 8

i

• 0,

uv i w x i y is in L. Outline

• f

Pro

• f
• f

PL

• Let

there b e a Chomsky-normal

• fo

rm CF G for L with m v ariables. Pic k n = 2 m .

• Because

CNF grammars ha v e b

• dies
• f

no more than 2 sym b

• ls,

a string z

• f

length

• n

m ust ha v e some path with at least m + 1 v ariables.

• Th

us, some v ariable m ust app ear t wice

• n

the path.

✦

Compare with the DF A argumen t ab

• ut

a path longer than the n um b er

• f

states. 4

SLIDE 5

• F
• cus
• n

some path that is as long as an y path in the tree. In this path, w e can nd a duplication

• f

some v ariable A among the b

• ttom

m + 1 v ariables

• n

the path.

✦

Let the lo w er A deriv e w and the upp er A deriv e v w x.

• CNF

guaran tees us that jv w xj

• n

and v x 6= .

• By

rep eatedly replacing the lo w er A's tree b y the upp er A's tree, w e see uv i w x i y has a parse tree for all i > 1.

✦

And replacing the upp er b y the lo w er sho ws the case i = 0; i.e., uw y is in L. Example L = f0 k 2 j k is an y in tegerg is not a CFL.

• Supp
• se

it w ere. Then let n b e the PL constan t for L.

• Consider

z = n 2 . W e can write z = uv w xy , with jv w xj

• n

and jv xj > 0.

• Then

uv v w xxy is in L. But n 2 < juv v w xxy j

• n

2 + n < (n + 1) 2 , so there is no p erfect square that juv v w xxy j could b e.

• By

\pro

• f

b y con tradiction," L is not a CFL. 5