cfg-pda.ppt

8/11/2019 CFG-PDA.ppt

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Equivalence of CFG's and

PDA'sA language is context free if and only

if some pushdown automatonrecognizes it

As usual with if and only if theorems,

there are two directions to prove If a language is context free, then some pushdownautomaton recognizes it

If a pushdown automaton recognizes somelangauge, then it is context free


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Only If (CFG to PDA) Let L = L(G) for some CFG G = ( V, , P,

S ) Idea : have PDA A simulate leftmost

derivations in G , where a left-sententialform (LSF) is represented by:

1. The sequence of input symbols that A hasconsumed from its input, followed by

2. A 's stack, top left-most

Example: If ( q , abc d , S ) * (q, cd , A B C ),

then the LSF represented is abABC


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Moves of A If a terminal a is on top of the stack ,

then there had better be an a waiting on

the input. A consumes a from theinput and pops it from the stack The LSF represented doesn't change!

If a variableB

is on top of the stack ,then PDA A has a choice of replacingB on the stack by the body of anyproduction with head B


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Defining the PDA Define PDA A as follows:

Q contains a single state, q contains the terminal symbols of the grammar contains all terminal and non-terminal symbols from the

grammar F is the empty set ( A terminates by empty stack) Start stack symbol is the distinguished symbol of the

grammar is defined as follows:

For each production X in the grammar, create a move(q, ,X ) = (q, )

For each terminal symbol a in the grammar, create a move(q,a,a ) = (q, )


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Example

S a | aS | b SS | SSb | Sb S

PDA A = ({q},{a,b },{S,a,b }, ,q, ,S )

is defined as(q, ,S ) = { (q,a ),

(q,aS ),(q,bSS ),(q,SSb ),(q,SbS ) }

(q,a,a ) = (q, )(q,b,b ) = (q, )


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Processing of baa state input stack move

q baa S (q, ,S ) = (q,bSS )

q baa bSS (q,b,b ) = (q, )

q aa SS (q, ,S ) = (q,a )

q aa aS (q,a,a ) = (q, )q a S (q, ,S ) = (q,a )

q a a (q,a,a ) = (q, )

q - - - accept -

Generate bSS

Match b

Generate a

Match a

Generate a

Match a

S

b S S

a a

b a a

m a

t c h

m a

t c h

m a

t c h


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Converting from PDA to CFG

A PDA consumes a character A CFG generates a character We want to relate these two What happens when a PDA consumes

a character? It may change state It may change the stack


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Converting from PDA to CFGcontinued

Suppose X is on the stack and a is read What can happen to X ?

It can be popped It may replaced by one or more other stacksymbols

And so on The stack grows and shrinks and grows and shrinks

Eventually, as more input is consumed, X must bepopped (or well never reach an empty stack )

And the state may change many times We must track all of this!


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PDA to CFG

STRATEGY 1 Assume L = N (P ), where P = (Q, , , ,q 0,Z 0,F ), Fis empty (accept by empty stack)

Key idea : units of PDA action have the net effect of

popping one symbol from the stack, consumingsome input, and making a state change. The triple [ qZp ] is a CFG variable that generates exactly

those strings w such that P can read w from the input,pop Z (net effect), and go from state q to state p . More precisely, ( q,w,Z ) * ( p, , ) As a consequence of above, ( q,wx,Z ) * ( p, x, ) for any x and

.


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It's a Zen thing [qZp ] is at once a triple involvingstates and symbols of P , and yetto the CFG we construct it is asingle, indivisible object.

(OK, I know that's not a Zen thing, but you getthe point)


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Strategy A popping rule, e.g., ( p, ) in (q,a,Z ).

[qZp ] aPop Z , consume a

A rule that replaces one symbol and state by others,e.g., ( p,Y ) in (q,a,Z ).

For all states r in Q : [qZr ] a [ pYr ]Pop Z , consume a, move to state p, push Y A rule that replaces one stack symbol by two, e.g.,

( p,XY ) in (q,a ,Z ).

For all states r and s in Q: [qZs ] a [ pXr ][rYs]Pop Z , consume a, move to state p, push X, move to someother state, push Y , move to s

There may be some states r that cannot be reached from p whilepopping X. True, but does not affect grammar since the resultingvariables are useless and do not affect the language accepted by

the grammar


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(q,a ,Z ) = ( p,Y )

q

p

q

p

?

a , Z Y

a , Z Y

processY

consume apop Zpush Ymove to state p

Y not yet processed

consume apop Zpush Ymove to statep

Since we dont know which statethe PDA will be in afterprocessing Y, define a

production [ qZr n] a [ pYr n] thatends in each possible state r n


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ExamplePDA with transitions :

(q 1,0 ,Z 0) = {( q 1,XZ 0)}(q 2, ,Z 0) = {( q 3, )}(q 1,0,X ) = {( q 1,XX )}(q 1,1,X ) = {( q 2, )}(q 2,1,X ) = {( q 2, )}

S [q 1Z0q 3][q 1Z0q 3] 0 [q 1Xq 1] [q 1Z0q 3][q 1Z0q 3] 0 [q 1Xq 2] [q 2Z0q 3][q

1Z

0q

3] 0 [q

1Xq

3] [q

3Z

0q

3]

[q 1Xq 1] 0 [q 1Xq 1] [q 1Xq 1][q 1Xq 1] 0 [q 1Xq 2] [q 2Xq 1][q 1Xq 1] 0 [q 1Xq 3] [q 3Xq 1] [q 1Xq 2] 0 [q 1Xq 1] [q 1Xq 2]

[q 1Xq 2] 0 [q 1Xq 2] [q 2Xq 2][q 1Xq 2] 0 [q 1Xq 3] [q 3Xq 2] [q 1Xq 2] 1[q 2Xq 2] 1

[q 2Z0q 3]

S[q 1Z0q 3]0 [q 1Xq 2] [q 2Z0q 3]0 0 [q 1Xq 1] [q 1Xq 2] [q 2Z0q 3]0 0 0 [q 1Xq 2] [q 2Xq 2] [q 2Xq 2] [q 2Z0q 3]0 0 0 1 [q 2Xq 2] [q 2Xq 2] [q 2Z0q 3]0 0 0 1 1 [q 2Xq 2] [q 2Z0q 3]0 0 0 1 1 1 [q 2Z0q 3]

0 0 0 1 1 1

Impossible configurations involving q 3 are dimmed

Derivation of 000111 :


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Example

PDA withtransitions(q 0,a,Z 0) = {( q 0,A Z )}(q 0,b,Z 0) = {( q 0,B Z )}(q 0,a,A ) = {( q 0,A A)}(q 0,a,A ) = {( q 0,A A)}(q 0,b,A ) = {( q 0,B A )}(q 0,a,B ) = {( q 0,A B )}(q 0,b,B ) = {( q 0,B B )}(q

0,c,Z

0) = {( q

1,Z

0)}

(q 0,c,A ) = {( q 1,A )}(q 0,c,B ) = {( q 1,B )}(q 1,a,A ) = {( q 1, )}(q 1,b,B ) = {( q 1, )}

(q 1, ,Z 0) = {( q 1, ) }

S [q 0Z 0q ]

[q 0Z 0q ] a [q 0A p ] [pZ 0q ][q 0Z 0q ] b [q 0B p ] [pZ 0q ][q 0A q ] a [q 0A p ] [pA q ][q 0A q ] b [q 0B p ] [pA q ][q 0B q ] a [q 0A p ] [pB q ][q 0B q ] b [q 0B p ] [pB q ][q 0Z 0q ] c [q 1Z 0q ][q 0A q ] c [q 1A q ][q 0B q ] c [q 1B q ]

[q 1A q 1] a[q 1B q 1] b

[q 1Z 0q 1]

In the above, q and p can each

be either q0 or q1


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The Full StoryS [q0Z 0q0]S [q0Z 0q1][q0Z 0q0] a [q0 Aq 0] [q0Z 0q0][q0Z 0q0] a [q0 Aq 1] [q1Z 0q0][q0Z 0q1] a [q0 Aq 0] [q0Z 0q1][q0Z 0q1] a [q0 Aq 1] [q1Z 0q1]

[q0Z 0q0] b [q0Bq0] [q0Z 0q0][q0Z 0q0] b [q0Bq1] [q1Z 0q0][q0Z 0q1] b [q0Bq0] [q0Z 0q1][q0Z 0q1] b [q0Bq1] [q1Z 0q1][q0 Aq 0] a [q0 Aq 0] [q0 Aq 0][q0 Aq 0] a [q0 Aq 1] [q1 Aq 0][q0 Aq 1] a [q0 Aq 0] [q0 Aq 1][q0 Aq 1] a [q0 Aq 1] [q1 Aq 1][q0 Aq 0] b [q0Bq0] [q0 Aq 0][q0 Aq 0] b [q0Bq1] [q1 Aq 0][q0 Aq 1] b [q0Bq0] [q0 Aq 1]

[q0 Aq 1] b [q0Bq1] [q1 Aq 1]

[q0Bq0] a [q0 Aq 0] [q0Bq0][q0Bq0] a [q0 Aq 1] [q1Bq0][q0Bq1] a [q0 Aq 0] [q0Bq1][q0Bq1] a [q0 Aq 1] [q1Bq1][q0Bq0] b [q0Bq0] [q0Bq0][q0Bq0] b [q0Bq1] [q1Bq0][q0Bq1] b [q0Bq0] [q0Bq1][q0Bq1] b [q0Bq1] [q1Bq1][q0Z 0q0] c [q1Z 0q0][q0Z 0q1] c [q1Z 0q1][q0 Aq 0] c [q1 Aq 0]

[q0 Aq 1] c [q1 Aq 1][q0Bq0] c [q1Bq0][q0Bq1] c [q1Bq1][q1 Aq 1] a[q1Bq1] b[q

1Z

0q

1]

If we specify every non-terminal in terms of all possible states, the expansion wouldcontain all of the states below


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Deriving bacab

S [q0Z 0q1][q0Z 0q1] a [q0 Aq 1] [q1Z 0q1][q0Z 0q1] b [q0Bq1] [q1Z 0q1][q0 Aq 1] a [q0 Aq 1] [q1 Aq 1][q0 Aq 1] b [q0Bq1] [q1 Aq 1]

[q0Bq1] a [q0 Aq 1] [q1Bq1][q0Bq1] b [q0Bq1] [q1Bq1][q0Z 0q1] c [q1Z 0q1][q0 Aq 1] c [q1 Aq 1][q0Bq1] c [q1Bq1]

[q1 Aq 1] a[q1Bq1] b[q1Z 0q1]

Only some of the productions in the generated grammar will allow for a derivation; therest are unnecessary

PDA moves (q0, bacab , Z 0) |- ( q0, acab , BZ 0)

|- (q0, cab , ABZ 0)|- (q1, ab , ABZ 0)

|- (q1, b, BZ 0)|- (q1, , Z 0)

|- (q1, , )

Corresponding leftmost derivation S [q0,Z 0,q1]

b [q0,B,q1] [q1,Z 0,q1] ba [q0, A,q1] [q1,B,q1][q1,Z 0,q1] bac [q1, A,q1] [q1,B,q1][q1,Z 0,q1] baca [q1,B,q1][q1,Z 0,q1] bacab [q1,Z 0,q1] bacab


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PDA to CFG

STRATEGY 2 Convert PDA P into CFG G Modify P to be a normalized PDA N so that:

It has a single accept state, qaccept Create -transitions from old accept states to this new accept state

It empties the stack before accepting Push a special character $ on the stack in the start state (introducing a

new start state in the process) Introduce a new temporary state q

temp that replaces q

accept , which has

transitions popping all characters from the stack (except $) Introduce transition:

qtemp qaccept ,$


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Continued Each transition either pushes a symbol onto the stack

or pops one off the stack, but it does not do both atthe same time Replace a simultaneous pop/push move with a 2-transition

rule that goes through a new state E.g.

( read a from input, pop b from stack, push c ) Introduce special state q temp plus 2 transitions, one doing pop

and one doing push:

Replace a transition that neither pops nor pushes with twotransitions that push and then immediately pop some newly-created dummy stack symbol

qi q ja,b c

qia,b q jq temp

, c

qi q ja, qi

a, Xq jq temp

, X


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Normalizing the PDAEXAMPLE

, $

b , X

a, X Y

, $

a,

L (N) = (ba + b aa)*b) +


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Pure Push PopMake sure the stack is always active by replacing inactive stack movesby a push followed by immediate pop of a dummy symbol

, $

b , X

a, X Y

, $

a,

, $

b , X

a, X Y

, $

a, D

D


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Pure Push Pop Any move that replaces the top letter on the stack

should be changed into a pop followed by a push

, $

b , X

a, X Y

, $

a, D

, D

, $

b , X

a, X

, $

a, D

, D

Y


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Unique Accept StateTurn off original accept states and connect to a new

accept state

, $

b , X

a, X

, $

a, D

, D

, Y

, $

b , X

a, X

, $

a, D

, D

, Y

D D

D

Rememb er: eachmo ve must e i therpush o r pop f rom

the stack)


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Empty StackMake sure the stack empties its content by adding a

new dummy empty stack symbol and new start/acceptstates

, $

b , X

a, X

, $

a, D

, D

, Y

, D

D $ X

Y , D

,

, D


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PDA to CFGINTUITIVE DESCRIPTION

Consider normalized PDA N = (Q, , , ,q init ,qaccept ) Starts in q init with an empty stack Ends in qaccept with an empty stack

In general, can define the language L pq , for any two states p,q Q which isthe language of all strings that start in p with an empty stack, and end in q with an empty stack

For each pair of states p and q, define a symbol S pq inthe CFG for the language L

pq

Language of N is

Lq init q accept


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Steps to process w L pq

Two possibilities:1. During the processing of w the stack

becomes empty at some intermediatestate r This means a word of L pq can be formed by concatenating

a word of L pr (which brought N from state p to state r withan empty stack) and a word of Lrq (that took N from r to q )

2. Stack is never empty in the middle of N stransit from p to q in processing w The first transition (from, say, p to p 1) must have been a

push, and the last transition (from, say, q1 to q) must havebeen a pop, and the pop popped exactly the symbolpushed by the first transition from p to p 1


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In other words, if the PDA read a from input

as it moved from p to p 1, and read b as itmoved from q1 to q, then w =ayb , where y isan input that causes the PDA N to start from

p 1 with an empty stack and end in q1 with an

empty stack i.e., y Lp 1q 1 Formally, if there is a push transition (pushing

X onto the stack) from p to p 1 (reading a ) anda pop transition from q

1 to q (popping X and

reading b), then a word in L pq can beconstructed from the expression aL p 1q 1b

Note that either or both of a or b could be


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The Construction For every state p , introduce the rule

A pp Empty string can always be considered as getting you from p to

p without doing any thing to the stack, since nothing was read

CONCATENATION RULE: For the case where thestack empties in the middle of transition from

p to q, introduce, for all states p, q, r of N, therule

A pq A pr Arq


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RECURSION RULE: Case where stack neverempty: for any given states p, p 1, q 1, r of N ,such that there is a push transition from p to

p 1 and a pop transition from q 1to r (that pushand pop the same symbol), introduce anappropriate rule

Formally, for p, p 1, q 1, r of N with the form

introduce the rule A pr a A p1q1b

p1 p a, X

r q1b, X

push X pop X


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Formal DefinitionFROM S IPSER

P = (Q, , , ,q0,qaccept ) Non-terminals of G are { A pq | p,q Q}

Rules: For each p, q, r, s Q, t , and a, b , if ( p,a, )

contains ( r,t ) and (s,b,t ) contains ( q, ), put the rule A pq aA rsb in G

For each p, q, r Q, put the rule A pq

A pr

Arq

in G. For each p Q, put the rule A pp in G.


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A pq A pr Arq

Stackheight

Input string

Generated by A pr Generated by Arq

Generated by A pq

p r q

CONCATENATION RULE


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A pq aA rsb

Stackheight

Input string

Generated by Ars

Generated by A pq

q pr s

a b

RECURSION RULE


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PDA CFG

If paths for strings that are accepted by thePDA start and end with an empty stack, it ispossible to consider any such path, betweenany two states and recursively generate allsuch paths

This recursive relationship between paths willgive rise to the recursion at the heart of therepresentative context free grammar


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The Grammar The rules for generating paths give a

grammar to generate all labels of suchpaths

The grammar has non-terminals Aqr whichwill generate all strings x that areprocessed when passing from state q tostate r

Q: Under this assumption, what should theproduction body (right hand side) forthe start variable S be?


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The Grammar SymbolsA: S = Aq init q accept , where q ini t is the start

state and q accept is the final state In addition to this start variable, the other

variables are all Aqr for which there is apath going from q to r that starts andends with an empty stack

Note that Sipser doesnt require the extra condition that there be a path from q to r which starts and ends with an emptystack his method generates all possible combinations. However, those pairs q,r for which no such path exists willcreate useless variables A qr which end up cluttering the grammar and making the construction extremely ugly, even onthe simplest PDAs . On the other hand, it is not obvious how one would determine a priori which of the pairs dont havesuch paths, which probably explains why Sipser didnt include this condition.


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Grammar Rules

1. B ASIS RULE : Add a production Aqq foreach state q in the PDA

2. C ONCATENATION RULE : Add a production A pr A pq Aqr for all p,q,r when A pr , A pq and Aqr are all in V .

3. R ECURSION RULE : Add a production A ps aA

qr b for all p,s,q,r when

A ps and Aqr are in V Transitions ( q,X ) ( p,a, ), (s, ) (r,b,X ) for the same

stack symbol X exist in the PDA


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Example

PDA in the normalized form:

Q: What is the accepted language?

r s , $q , $

(, X), X


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A: CNP = correctly nested parentheses,

including sets of pairs [e.g., ()(())]. Thenumber of Xs on the stack reflects howdeep the current nesting is.

Q: What are the variables for theequivalent grammar? What is thestart variable?

r s , $q , $

(, X

), X


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A: V = {Aq s , Aq q , A r r , A s s }, S = Aq s

We dont need Arq , Asq , Asr becausethe paths go in the wrong direction

We dont need Aqr or Ars because cantadd or remove $ while at r I.e., not a transition where you both begin and end

with an empty stack

r s , $q , $

(, X

), X


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Productions from the Base

Rule Empty string can always be considered as getting

you from p to p without doing any thing to the stack,since nothing was read

r s , $q , $

(, X), X

Aqq , Arr , Ass


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Productions from the

Concatenation ruleIf you can get from some state p to another state p 1 starting and ending with the stack empty (regardlessof stack activity in the processing of moving from p to

p 1), and from q1 to q under the same conditions, thencombine paths to get a path from p to q.

r s , $q , $

(, X), X Aqs Aqq Aqs | Aqs Ass

Aqq Aqq Aqq Arr Arr Arr Ass Ass Ass State pairs q and r and r and s do not satisfy

the condition that the stack is empty in bothstates


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Productions from the Recursion RuleFor any given states p, p 1, q 1, q of N , such that there is a pushtransition from p to p1 and a pop transition from q1to q (that pushand pop the same symbol), i.e., there exist transitions ( p,a, )contains( p 1,X ) and (q1,b,X ) contains ( q, ), put the rule A pq aA

p1q1b

r s , $q , $

(, X), X

Aqs Arr = A rr Arr (Arr )

(q, , ) contains ( r,$ ) and (r , , X) contains (s , )

(r,(, ) contains (r , X ) and (r , ) , X) contains (r , )


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Showing that the grammarworks

Aqs Arr | Aqs Arr | Arr Arr | ( Arr )

Rename variables to get:S T | S

T | TT | (T )

S isnt needed as its whole purpose is to getyou to T

So the final (cleaned up) grammar isT | TT | ( T )


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Another Example

Consider the language L = {wcw R | w {a,b }*}. A non-normalized PDA for this

language isa, a

b, b

a, a

b, b

s fc,


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Convert to Normalized Form

a, a

b, b

a, a

b, b

s fc , D

s a

1. Create new start and accepting states2. All transitions either pop or push except c, ;

change to 2 transitions that push and pop adummy symbol

q tD

, $ ,$


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Generate Grammar

1. Add start symbol and a production Aqq for

each state q in the PDAS Asa , Ass Ass Aqq Aff Aaa

a, a

b, b

a, a

b, b

s fc, D

s aq , D


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Generate Grammar

2. Add a production A pr A pq Aqr for all p,q,r

when A pr , A pq and Aqr are all in V

a, a

b, b

a, a

b, b

s fc, D

s aq , D , $ ,$

A sa A ss A sa | A sa Aaa


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Generate Grammar

3. Add a production A ps aAqr b for all p,s,q,r when A ps and Aqr arein V and transitions ( q,X ) ( p,a, ), (s, ) (r,b,X ) for the samestack symbol X exist in the PDA

a, a

b, b

a,a

b,b

s fc, D

s aq , D , $ ,$

A sa $A s f $ | A s f c A qqA s f b A s f b |aA s f a


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Final Grammar

S A sa A ss A ss

Aqq A ff A aa A sa A ss A sa A sa A sa A aaA sa $A sf $A sf cA qq A sf bA sf bA sf aA sf a

S TR U

V W X T RT T TX T $Z$Z cVZ bZbZ aZa

T $Z$Z c

Z bZbZ aZa

More readableFinal grammar Simplified

R,U,V,W,X contribute only so can

be eliminated T RT and T TX then become T T , which is obviously unnecessary

S is superfluous because it only getsyou to T


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Deterministic PDAs Intuitively: never a choice of move

(q, a, Z) has at most one member for any q, a, Z (including a = ).

If (q, , Z ) is nonempty, then (q, a, Z ) must beempty for all input symbols a .

Why Care? Parsers, as in YACC, are really DPDA's. Thus, the question of what languages a DPDA can

accept is really the question of what programminglanguage syntax can be parsed conveniently.


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Some LanguageRelationships

Acceptance by empty stack is hardfor a DPDA Once it accepts, it dies and cannot accept any

continuation. Thus, N (P ) has the prefix property: if w is in N (P ),

then wx is NOT in N (P ) for any x .

However, parsers do accept byemptying their stack Trick: they really process strings followed by a

unique endmarker (typically $) e.g., if they acceptw $, they consider w to be a correct program.


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If L is a regular language, then L is

a DPDA language A DPDA can simulate a DFA, without using itsstack (acceptance by final state).

If L is a DPDA language, then L is aCFL that is n o t inherentlyambiguous

A DPDA yields an unambiguous grammar inthe standard construction Interesting fact: The class of languages

accepted by NPDAs is larger than thoseaccepted by DPDAs!


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Languages accepted bynondeterministic PDA

Languages accepted bydeterministic PDA

Languagesaccepted by FAor NFA

PDA more powerful than FA


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Cleaning Up Grammars

We can "simplify" grammars to a greatextent, e.g.:1. Get rid of useless symbols -- those that do not

participate in any derivation of a terminal string.

2. Get rid of -productions --those of the form variable .

But you lose the ability to generate as a string in the language.3. Get rid of unit productions -- those of the form variable

variable. Any CFG can be converted via these and

other methods to Chomsky Normal Form only production forms are variable two variables and

variable terminal .


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Getting Rid of the Empty

String Empty string is a nuisance with grammarsand languages in general

We will look at languages that do not contain

No loss of generality:For language L, let G = (V,T,S,P ) be a CFG that generates L- { }

Modify grammar by adding a new start variable S 0 and addproductions S 0 S | This grammar generates LTherefore any non-trivial conclusion we make for L - { }should transfer to L


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Useless Symbols

In order for a symbol X to be useful, itmust:1. Derive some terminal string (possibly X is a

terminal).2. Be reachable from the start symbol; i.e., S

X . Note that X wouldn't really be useful if or

included a symbol that didn't satisfy (1), so it isimportant that (1) be tested first, and symbols thatdon't derive terminal strings be eliminated before testing (2).

*


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Finding Symbols That Don'tDerive Any Terminal String

Recursive construction :

Basis : A terminal surely derives a terminalstring. Induction : If A is the head of a production

whose body is X 1 X 2 X k , and each X i isknown to derive a terminal string, thensurely A derives a terminal string.

Keep going until no more symbols that deriveterminal strings are discovered.

E l


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ExampleS AB | C

A 0B | CB 1 | A0C A C | C1

Round 1 : 0 and 1 are "in." Round 2 : B 1 says B is in. Round 3 : A 0B says A is in. Round 4 : S AB says S is in.

Round 5 : Nothing more can be added.

Thus, C can be eliminated, along with any productionthat mentions it, leaving S A B ; A 0 B ; B 1 | A0 .


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Finding Symbols That Can't BeDerived From the Start Symbol

Another recursive algorithm: Basis : S is "in." Induction : If variable A is in, then so is

every symbol in the production bodiesfor A.

Keep going until no more symbolsderivable from S can be found.


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ExampleS A BA 0 B B 1 | A0

Round 1 : S is in.

Round 2 : A and B are in. Round 3 : 0 and 1 are in. Round 4 : Nothing can be added. In this case, all symbols are derivable from S, so nochange to grammar.

Book has an example where not only are there symbolsnot derivable from S, but you must eliminate first thesymbols that don't derive terminal strings, or you get the

wrong grammar.


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Eliminating -Productions

A variable A is nullable if A . Find them bya recursive algorithm: Basis : If A is a production, then A is

nullable. Induction : If A is the head of a production

whose body consists of only nullable symbols,then A is nullable.

Once we have the nullable symbols, we canadd additional productions and then throwaway the productions of the form A forany A.

*


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If A X 1 X 2 X k is a production, add

all productions that can be formed byeliminating some or all of those X i 'sthat are nullable.

But, don't eliminate all k if they are allnullable.

Example If A BC is a production, and both B and C

are nullable, add A B | C


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Eliminating Unit Productions

1. Eliminate useless symbols and -productions.2. Discover those pairs of variables ( A, B ) such that

A B.

Because there are no -productions, this derivation can only useunit productions. Thus, we can find the pairs by computing reachablity in a graph

where nodes = variables, and arcs = unit productions.

3. Replace each combination where A B and is other than a single variable by A

I.e., "short circuit" sequences of unit productions, which musteventually be followed by some other kind of production.

4. Remove all unit productions.

*

* *


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Chomsky Normal Form

1. Get rid of useless symbols, -productions, andunit productions (already done).

2. Get rid of productions whose bodies are mixesof terminals and variables, or consist of morethan one terminal.

3. Break up production bodies longer than 2.

ResultAll productions are of the form A B C or A

a


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No Mixed Bodies

1. For each terminal a , introduce a newvariable Aa , with one production Aa a .

2. Replace a in any body where it is notthe entire body by Aa .

Now, every body is either a single terminal or itconsists only of variables.

Example A 0B1 becomes A0 0; A1 1; A

A0BA1


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Making Bodies Short If we have a production like A BCDE ,

we can introduce some new variables thatallow the variables of the body to beintroduced one at a time. A body of length k requires k - 2 new variables.

Example Introduce F and G; replace A BCDE by A

BF; F CG; G DE .


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Summary TheoremIf L is any CFL, there is agrammar G that generates L -{ }, for which each production isof the form A B C or A a ,

and there are no uselesssymbols.

cfg-pda.ppt

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