Th us, suc h a grammar is unlik ely to b e useful for - - PDF document

SLIDE 1 Am biguous Grammars A CF G is ambiguous if

• ne
• r

more terminal strings ha v e m ultiple leftmost deriv ations from the start sym b

• l.
• Equiv

alen tly: m ultiple righ tmost deriv ations,

• r

m ultiple parse trees. Example Consider S ! AS j ; A ! A1 j 0A1 j 01. The string 00111 has the follo wing t w

• leftmost

deriviations from S : 1. S ) lm AS ) lm 0A1S ) lm 0A11S ) lm 00111S ) lm 00111 2. S ) lm AS ) lm A1S ) lm 0A11S ) lm 00111S ) lm 00111

• In

tuitiv ely , w e can use A ! A1 rst

• r

second to generate the extra 1. Inheren tly Am biguous Languages A CFL L is inher ently ambiguous if every CF G for L is am biguous.

• Suc

h things exist; see course reader. Example The language

• f
• ur

example grammar is not inheren tly am biguous, ev en though the grammar is am biguous.

• Change

the grammar to force the extra 1's to b e generated last. S ! AS j

• A

! 0A1 j B B ! B 1 j 01 Wh y Care?

• Am

biguit y

• f

the grammar implies that at least some strings in its language ha v e dieren t structures (parse trees).

✦

Th us, suc h a grammar is unlik ely to b e useful for a programmi ng language, b ecause t w

• structures

for the same string (program) implies t w

• dieren

t meanings (executable equiv alen t programs) for this program.

✦

Common example: the easiest grammars for arithmetic expressions are am biguous and need to b e replaced b y more complex, 1

SLIDE 2 unam biguous grammars (see course reader).

• An

inheren tly am biguous language w

• uld

b e absolutely unsuitable as a programming language, b ecause w e w

• uld

not ha v e an y w a y

• f

xing a unique structure for all its programs. Pushdo wn Automata

• Add

a stac k to a F A.

• T

ypically nondeterministic.

• An

automaton equiv alen t to CF G's. Example Notation for \transition diagrams": a; Z =X 1 X + 2

• X

k = \on input a, with Z

• n

top

• f

the stac k, consume the a, mak e this state transition, and replace the Z

• n

top

• f

the stac k b y X 1 X 2

• X

k (with X 1 at the top). p q r Start 0; Z =X Z 0; X=X X 1; X= 1; X= 1; Z = Z ; Z = Z

• p

= starting to see a group

• f

0's and 1's; q = reading 0's and pushing X 's

• n

to the stac k; r = reading 1's and p

• pping

X 's un til the X 's are all p

• pp

ed.

• W

e can start a new group (transition from r to p)

• nly

when all X 's (whic h coun t the 0's) ha v e b een matc hed against 1's. F

• rmal

PD A P = (Q; ; ;

• ;

q ; Z ; F ), where Q, , q , and F ha v e their meanings from F A.

• =

stac k alphab et.

• Z

in

• =

start sym b

• l

= the

• ne

sym b

• l
• n

the stac k initially .

• =

transition function tak es a state, an input sym b

• l

(or ), and a stac k sym b

• l

and giv es y

• u

a nite n um b er

• f

c hoices

• f:

2

SLIDE 3 1. A new state (p

• ssibly

the same). 2. A string

• f

stac k sym b

• ls

to replace the top stac k sym b

• l.

Instan taneous Descripti

• ns

(ID's) F

• r

a F A, the

• nly

thing

• f

in terest ab

• ut

the F A is its state. F

• r

a PD A, w e w an t to kno w its state and the en tire con ten t

• f

its stac k.

• It

is also con v enien t to main tain a ction that there is an input string w aiting to b e read.

• Represen

ted b y an ID (q ; w ; ), where q = state, w = w aiting input, and

• =

stac k, top left. Mo v es

• f

the PD A If

• (q

; a; X ) con tains (p; ), then (q ; aw ; X

• )

` (p; w ;

• ).
• Extend

to ` * to represen t 0, 1,

• r

man y mo v es.

• Subscript

b y name

• f

the PD A, if necessary .

• Input

string w is accepted if (q ; w ; Z ) ` * (p; ;

• )

for an y accepting state p and an y stac k string

• .
• L(P

) = set

• f

strings accepted b y P . Example (p; 0110011; Z ) ` (q ; 110011; X Z ) ` (r ; 10011; Z ) ` (r ; 0011; Z ) ` (p; 0011; Z ) ` (q ; 011; X Z ) ` (q ; 11; X X Z ) ` (r ; 1; X Z ) ` (r ; ; Z ) ` (p; ; Z ) Acceptance b y Empt y Stac k Another

• ne
• f

those tec hnical con v eniences: when w e pro v e that PD A's and CF G's accept the same languages, it helps to assume that the stac k is empt y whenev er acceptance

• ccurs.
• N

(P ) = set

• f

strings w suc h that (q ; w ; Z ) ` * (p; ; ) for some state p.

✦

Note p need not b e in F .

✦

In fact, if w e talk ab

• ut

N (P )

• nly

, then w e need not ev en sp ecify a set

• f

accepting states. Example F

• r
• ur

previous example, to accept b y empt y stac k: 3

SLIDE 4 1. Add a new transition

• (p;

; Z ) = f(p; )g.

✦

That is, when starting to lo

• k

for a new 0-1 blo c k, the PD A has the

• ption

to p

• p

the last sym b

• l
• the

stac k instead. 2. p is no longer an accepting state; in fact, there ar e no accepting states. Equiv alence

• f

Acceptance b y Final State and Empt y Stac k A language is L(P 1 ) for some PD A P 1 if and

• nly

if it is N (P 2 ) for some PD A P 2 .

• Giv

en P 1 = (Q; ; ;

• ;

q ; Z ; F ), construct P 2 : 1. In tro duce new start state p and new b

• ttom-of-stac

k mark er X . 2. First mo v e

• f

P 2 : replace X b y Z X and go to state q . The presence

• f

X prev en ts P 2 from \acciden tally" empt ying its stac k and accepting when P 1 did not accept. 3. Then, P 2 sim ulates P 1 ; i.e., giv e P 2 all the transitions

• f

P 1 . 4. In tro duce a new state r that k eeps p

• pping

the stac k

• f

P 2 un til it is empt y . 5. If (the sim ulated) P 1 is in an accepting state, giv e P 2 the additional c hoice

• f

going to state r

• n
• input,

and th us empt ying its stac k without reading an y more input.

• Giv

en P 2 = (Q; ; ;

• ;

q ; Z ; F ), construct P 1 : 1. In tro duce new start state p and new b

• ttom-of-stac

k mark er X . 2. First mo v e

• f

P 1 : replace X b y Z X and go to state q . 3. In tro duce new state r for P 1 ; it is the

• nly

accepting state. 4. P 1 sim ulates P 2 . 5. If (the sim ulated) P 1 ev er sees X , it kno ws P 2 accepts, so P 1 go es to state r

• n
• input.

4