Automata, Grammars and Languages

C SC 473 Automata, Grammars & Languages

Automata, Grammars and Languages

Discourse 04

Context-Free Grammarsand

Pushdown Automata

C SC 473 Automata, Grammars & Languages 2

Backus-Naur Form Grammars (CFGs)• Algol 60, Algol 68—first “block-structured” languages

• Ex: • CF Grammar

<program> ::= <block><statement> ::= s | <block><block> ::= begin <list> end<list> ::= <statement> ; <list> | <statement>

begin s ; begin s;s;s end ;s end

P BSS BB LL S LL S

→→→→→→

s

b e;

:G { ;}Σ = s, b, e,

Nonter

minals

=varia

bles

rules=productions

terminals

V

Start variable “S”

R


Grammars are “Generators”

“yields” or “derives in one step” Apply one production to one variable in the string

nondeterministic

P B L⇑

⇒ ⇒ b e

S B L⇒ ⇒b e b e bb ee⇑bse S ⇒

⇑bb ee L

;S L⇓

⇒bb ee L

bbsee

; ;B L L L⇒ ⇒b e bb e e L ; L ⇒bs e L

;S L⇓

⇒b e ;S S ⇒b e L

; ;S S L⇓

⇒b e L


• One possible derivation. Variable being rewritten at each stage is underscored

• two choices at each derivation step: Which variable (nonterminal) to be rewritten? Which rule with that variable as LHS to be applied?

• All possible terminal strings obtainable in this way make up L(G)

; ; ;; ; ; ; ; ;; ; ; ; ; ; ;; ; ; ; ; ; ; ; ; ; ;; ; ; ; ; ; ; ;

; ; ; ; ( )

P B L S L S S LS S S S B S S L S

L S L S LL S L L

SL G

⇒ ⇒ ⇒ ⇒⇒ ⇒ ⇒⇒ ⇒ ⇒⇒ ⇒ ⇒⇒ ⇒∴ ∈

b e b e b eb e b e b b e ebs b e e bs b e se bs b e sebs bs e se bs bs e se bs bs s e sebs bs s e se bs bs s se se

bs bs s se se

A Particular Derivation


Why CFGs?• Most natural or artificial (e.g. programming) languages are not regular

• We know that the latter language is not regular, so …

• Ex: C programs

( ) { | 1}n nL G n∗ ∗∩ = ≥b seb se

main(){}, main(){{}}, main(){{{}}},

main(){} main(){} | 1n n n∗ ∗∩ = ≥C

K

{ }


Derivation (Parse) TreePB

b L eS L;s S ; L

SBsb eL

S L;s S ; L

s Ss; ; ; ;bs bs s se se

yield/frontier/terminal string =


Derivation (Parse) TreePB

S L;s S ; L

B Ss

S L;s S ; L

s Ss

Lb e

b eL


Derivation (Parse) Tree (cont’d)

PB

b L eS L;s S ; L

SBsb eL

S L;s S ; L

s Ss

1

2

3

456

7

8

9

10

11 12

13 14

15


Context-Free Grammar• Defn 2.2: A context-free grammar G is a 4-

tuple

• is a finite set, the variables (nonterminals)• is a finite set disjoint from V, the terminals• is a finite set of rules, of the form

• is the start variable

• Ex: strings with balanced parentheses. Formally:

• Ex: informally• Variables = upper case• Terminals = lower case

( , , ),G V R S= ΣV

RΣ

S V∈

0 ( , , ),G V R S= Σ S B={(,)} { }V BΣ = ={ , , ( )}R B B BB B Bε= → → →

, , ( )A w A V w V ∗→ ∈ ∈ ∪ Σ

0 : | | ( )G B B BB B Bε→ → →

tech

nicall

y, an

ordere

d

pair

(A, w)


Yields & Derives Relations• Defn. The relation yields (derives in 1 step)

is defined as follows: if is a rule in R, then

• Defn: derives in k steps:• Defn: derives:

• In other words:

• Defn: A derivation (of n steps) from is any sequence of strings satisfying:

1 2 1

1 2 1

iff or ( )( ) ( )( )

k

k

u v u v u u uu u u u v

∗−

−

⇒ = ∃ ∃ ∃⇒ ⇒ ⇒ ⇒

LL

, ( )u v V uAv uwv∗∀ ∈ ∪ Σ ⇒

( ) ( )V V∗ ∗⇒ ⊆ ∪ Σ × ∪ Σ

A w→

( )k⇒( )∗⇒

0 1 2 ku u u u⇒ ⇒ ⇒L

0u


Language Generated• Defn. The language generated by G is the set

of all terminal strings derived from S:

A partial derivation is one that starts with S and ends in a non-terminal string containing variables in V

• Ex:

Partial: Terminal or terminated:

( ) { | }L G w w S w∗ ∗= ∈ Σ ∧ ⇒

|| |

S aAS aA SbA SS ba

→→

S aAS aSbAS⇒ ⇒

S aAS aSbAS aabAS

aabbaS aabbaa

⇒ ⇒ ⇒⇒ ⇒


Derivations and Parse Trees• Ex: 0 : | | ( )G B B BB B Bε→ → →

( ) (( )) (())(())( ) (())()

( ) () ( )()(( ))() (())()

B BB B B B B BBB BB B B B B

B

⇒ ⇒ ⇒ ⇒⇒ ⇒

⇒ ⇒ ⇒ ⇒⇒ ⇒

leftmost :

rightmost :

B BBB

BB

( BB)

( B )

BB

( BB)

( B )

BB

( BB)

ε( B )

BB

( BB)

ε

( B )

( B )

BB

( BB)

ε

( B )ε Notice: completed (terminated) parse

tree is the same for both derivations—thoughthe sequence “grows” differently


Derivation Parse Tree• Proposition 1: For every (terminated or partial) derivation

there is an unique parse tree T with frontier constructible from D.

• Proposition 2: For every parse tree T in G and any traversal order that is top-down (visits parents before children), there is an unique derivation for the frontier of T from S, and it is constructible from T.

• Corollary 3: For every parse tree T in G there is an unique leftmost derivation constructible from T.

Pf: Pre-order traverse T, expanding variables as their nodes are visited.

1 2: nD S u u u⇒ ⇒ ⇒L

nu


Ex: Leftmost DerivationET∗ F

( )T

( )

F

EE + T

E

E + T

F

x

T

∗ FTFx

x

TFx

Fx|

|

( )|

E E T T

T T F F

F E x

→ +→ ∗→


Ex: Leftmost DerivationET∗ F

( )T

( )

F

EE + T

E

E + T

F

x

T

∗ FTFx

x

TFx

Fx

1

3

4

5

6

7

8

2

9

10

1112

1314

151617

1718

Preorder traversal


Syntactic Ambiguity

• 2 distinct parse trees for same terminal string

• 2 distinct leftmost derivations for same terminal string

• Leftmost derivation parse tree 1-to-1

• A CFG is unambiguous wL(G) w has an unique parse tree (unique leftmost derivation)

| |E E E E E a→ + ∗

E

+ E

E E

E

a ∗a a

E

∗ E

a

E

E E+a a

a a a+ ∗

E E E a E

a E E a a E

a a a

⇒ + ⇒ + ⇒+ ∗ ⇒ + ∗ ⇒+ ∗

terminal string =

E E E E E E

a E E a a E

a a a

⇒ ∗ ⇒ + ∗ ⇒+ ∗ ⇒ + ∗ ⇒+ ∗


Ex: Ambiguous Grammar--English

<Sent><NP><VP><NP><N>|<Adj><N><VP><V><Obj>|<V><AdvP><AdvP><Adv>|<AdvP><AdvP><Prep><Obj><Obj><Adj><N><N>fruit | flies | ……


“Fruit flies like a banana”<Sent>

<NP> <VP>

<Adj> <N> <V> <Obj>

fruit flies like<Adj> <N>

a banana<Sent>

<NP> <VP>

<N> <V> <AdvP>

fruit flies<Prep> <Obj>

like <Adj> <N>

a banana

<Sent><NP><VP><NP><N>|<Adj><N><VP><V><Obj>|<V><AdvP><AdvP><Adv>|<AdvP><AdvP><Prep><Obj><Obj><Adj><N><N>fruit | flies | ……


Right Linear Grammars & Regular Languages

• Defn: A CFG is right-linear iff each rule is of one the forms AwB or Aw where A, B are variables and w Σ*

Chomsky (1958) called these “Type 3”

• Thm: L is a regular language iff L=L(G) for some right-linear grammar G. There are algorithms for converting from finite automata to right-linear grammars, and conversely.

DFA M

NFA NReg. Expr

ERight-linear

Grammar

G

= conversion algorithm


Right-Linear & Regular (cont’d)• Pf: () Assume L=L(M) where is a DFA. Construct with R having rule if in and rule

if is a final state. Claim: Pf: easy induction on n The proof direction follows since

• Pf: () Assume L=L(G) where is right-linear. Construct NFA where

is a new symbol. has the transition

if in R and transition if

( , , , , )M Q s F= Σ( , , , )G Q R s= Σ

p aq→ p qap ε→

p

(p, a1 L an) êMn (q, ε) i f f p ⇒ G

n a1 L anq.

( , , , )G V R S= Σ( , , , ,{ })N V S f= Σ

f V∉ A BwA wB→ A fw A w→


Right-Linear & Regular (cont’d)• Claim: Pf: easy induction on n The proof direction follows since

A0 ⇒ G w1A1 ⇒ G w1w2A2 ⇒ G L ⇒ G w1 L wnAni f f(A0, w1 L wn) êM (A1, w2 L wn) êM L êM (An, ε)

S ⇒ G∗ wA ⇒ wx i f f

(s, w) êM∗ (A, x) êM (f , ε) W


Ex: Right-Linear FA• Ex:

• Ex:

: ||

G S aA bA aB dSB dA

→→→

f A BS ab

d

a

d

1q 2qab b

0q:M a ,a b

0 0 1

1 0 2

2 2 2

( ): | || ||

G M q aq bqq aq bqq aq bq

εε

→→→

( ):N G

“useless” rules—canbe eliminated


Pushdown Automaton• Defn 2.12: A pushdown automaton M is a 6-

tuple

• is a finite set, the states• is a finite, the input alphabet• is a finite set, the stack alphabet•

is the transition function • is the start state• is the set of accept (final)

states

( , , ), , ,M Q s F= Σ ΓQ

ΓΣ

: ( )Q Qε ε ε × Σ ×Γ → ×ΓPs Q∈F Q⊆

{ }{ }

ε

ε

εε

Σ = Σ ∪Γ = Γ ∪


PushDown Automaton

input Σ*

Finite Control

p

1 2 3A A A K

1 2 1iaa a −Kseen to come

1i i na a a+ K

current input symbol

stack Γ*

Top Bottom (no end-marker supplied)

1 2 3 1 2 3( , , )p aaa A A AK K

configuration:

(state, rest of input,Stack )

ia ∈ Σ

iA ∈ Γ


PDA (cont’d)

Finite Control

s

ε

w

( , , )s w εconfiguration:

Initially: start state


PDA (cont’d)

Finite Control p

Xα

ax

(p, ax, Xα) ê (q, x, Yα)

configurations:

Transition:

( , ) ( , , )q Y p a X∈

Finite Control q

Yα

x


PDA (cont’d)• Can have ε-move: consume no input

• Pop-move: erase top stack symbol

• Push-only move: ignore stack

• Any combination is possible

( , ) ( , , )q Y p X ε∈

, ,a X Yε ε ε= = =

(p, ax, Xα) ê (q, ax, Yα)

( , ) ( , , )q p a Xε ∈ (p, ax, Xα) ê (q, x, α)

( , ) ( , , )q Y p a ε∈ (p, ax, Xα) ê (q, x, YXα)

( , ) ( , , )q pε ε ε∈ (p, ax, Xα) ê (q, ax, Xα)


Finite Control f

α

f F∈

Finally:

configuration:( , , )f ε α

PDA (cont’d)

•Defn: recognizes iff for some , and some

•Defn:

(s, w, ε)êM∗ (f , ε, α)

f F∈M wα ∗∈Γ

( ) { : accepts }L M w M w=


Example: PDA• Recognizer for

( , , ), , ,M Q s F= Σ Γ { , , , }Q s p q f={ , }A BΓ =

{ , , }a b cΣ =

(s, abbcbba, ε) ê ( p, abbcbba, $) ê ( p, bbcbba, A$)

ê L ( p, cbba, BBA$) ê (q, bba, BBA$) ê (q, ba, BA$)

ê L (q, a, A$) ê (q, ε, $) ê (f , ε, ε) ñ

s

{ | { , }}RL wcw w a b ∗= ∈

{ }F f= , $ε ε → ,a Aε →,b Bε →

,c ε ε→,a A ε→,b B ε→

p

qf,$ε ε→

accepts

(s, acb, ε) ê ( p, acb, $) ê ( p, cb, A$) ê (q, b, A$) ñdoes not accept

(blocked)


Example: PDA w/ nondeterminism

• Last example (palindromes with center-mark) was a deterministic PDA (DPDA)

• NPDA for

(s, aa, ε) ê ( p, aa, $) ê ( p, a, A$)ê ( p, ε, AA$)

ê (q, ε, AA$) ñ

(s, aa, ε) ê ( p, aa, $) ê ( p, a, A$)ê (q, a, A$)ê (q, ε, $) ê (f , ε, ε) ñ

s{ | { , }}

{ | { , } is a palindrome}

RL ww w a b

x x a b

∗

∗

=

=

∈

∈

, $ε ε → ,a Aε →,b Bε →

,ε ε ε→,a A ε→,b B ε→

p

qf,$ε ε→

does not accept(blocked) Nondeterministic

“guess”


Example: PDA • Recall well-nested parentheses (()) (()())

(p, w, $) ê∗ ( p, ε, $) ⇔

∀ pr ef i xes x of w, |x|( ≥|x|) ∧ |w|( =|w|)

p

0 : | | ( )G B B BB B Bε→ → →

// if ( then +1(, Aε →s, $ε ε →

f,$ε ε→

// if ) then -1),A ε→

DPDA!


2{ | 0} { | 0}n n m mL a b n a b m= ≥ ∪ ≥

Example: PDA

,b A ε→

, $ε ε →

, $ε ε →

,a Aε →

,b A ε→

,$ε ε→

,a Aε →

//push a 2nd A, Aε ε →

,b A ε→

• “guesses” which pattern• “checks” whether guess is correct• accepts iff correct guess that checks

s

(s, ε, ε)ê0(s, ε, ε) ∧

s ∈F ⇒ ε i s r ecogni zed


CFG PDA• Thm 2.20: A language is CF a PDA recognizes it. There are algorithms for converting a grammar to an equivalent automaton, and conversely.

• Lemma 2.21: There is an algorithm for constructing, from any CFG G, a PDA M such that L(G) = L(M).

Pf: In constructing a PDA, we can permit, without losing generality, “multi- push” moves such as

where For we may break a multi-push into a sequence of single-push moves by introducing new states:

Henceforth we will allow multi-push moves in our PDAs.

1 2, ta A X X X→ L

, iA X ∈ Γ

, taA X→1, tXε ε −→ 1, Xε ε→

tn 2n1tn −• • •


CFG PDA• Idea: use nondeterminism. Given G, construct PDA

P to Load S on stack & simulate a leftmost derivation on the stack:

When a variable symbol A comes to stack top, “guess” a grammar rule Aα , pop A and push α

When a terminal character comes to stack top, compare to next input symbol.

If they match, pop the top and advance the input (“check off”)

If they fail to match, jam (not an accepting computation)

If the input holds a word in L(G) and P guesses the correct leftmost derivation (rules to apply), then all the input characters will be checked off against those at the top of the stack and the stack will empty as the last input is checked off. Otherwise at some point the PDA will jam


CFG PDA (cont’d)• Given construct

States: Input alphabet: Σ Stack alphabet: Start state: Accept states: Transition function:

Initialize stack: Simulate rules: Check off terminals: Detect null stack & accept:

start( , , , , , ):P Q q F= Σ Γstart loop{ , , }acceptQ q q q=

startq

( , , , )G V R S= Σ

accept{ }F q=

{$}V ∪ Σ ∪

start loop( , , ) {( , $)}q q S ε ε =

loop loop( ) ( , , ) {( , )}A R q A qα ε α∀ → ∈ =

loop loop( ) ( , , ) {( , )}a q a a q ε∀ ∈ Σ =

loop accept( , ,$) {( , )}q q ε ε=

,Aε α→,a a ε→

startq loopq acceptq, $Sε ε → ,$ε ε→

( )a∀ ∈ Σ ( )A Rα∀ → ∈


CFG PDA (cont’d)• Ex:

startq loopq acceptq, $Sε ε → ,$ε ε→

0 : | |G S S SS S sSdε→ → →

,Sε ε→,S SSε →,S sSdε →

,s s ε→,d d ε→

S ⇒ L

∗ xAα ⇔ (ql oop, xu, S$) ê∗ (ql oop, u, Aα$)

∴ S ⇒ L∗ w ⇔ (ql oop, w, S$) ê

∗ (ql oop, ε, ε$)ê(qaccept , ε, ε)


CFG PDA (cont’d)• G • P

(qst ar t, ssddsd, ε) ê

LS ⇒ (ql oop, ssddsd, S$) ê

LSS ⇒ (ql oop, ssddsd, SS$) ê

LsSdS ⇒ (ql oop, ssddsd, sSdS$) ê

(ql oop, sddsd, SdS$) ê

LssSddS ⇒ (ql oop, sddsd, sSddS$) ê

(ql oop, ddsd, SddS$) ê

LssddS ⇒ (ql oop, ddsd, ddS$) ê

(ql oop, dsd, dS$) ê

(ql oop, sd, S$) ê

LssddsSd ⇒ (ql oop, sd, sSd$) ê


CFG PDA (cont’d)• G • P

(ql oop, d, Sd) ê

ssddsd (ql oop, d, d$) ê

(ql oop, ε, ε$) ê

( )ssddsd L G∴ ∈ ( )ssddsd LP∴ ∈

(qst ar t, ssddsd, ε) ê∗

(qaccept , ε, ε)LS ssddsd∗⇒

CFG leftmost derivation PDA computation


PDA CFGLemma 2.27: There is an algorithm for constructing, from any PDA P, a CFG G such that L(G) = L(P).

Pf: Given a PDA we can convert it into a PDA with the following simplified structure:• it has only one accept state:

• add ε-transitions from multiple accept states

• it empties its stack just before entering the accept state:

•Loop on a state that just pops:

• each PDA transition is either a “pure push” or a “pure pop

- introduce new intermediate states

{ }acceptF q=,

acceptf qε ε ε→

→,

,pop pop

X

q q Xε ε→

∈ Γ→,a Xε →

,a X ε→

0( , , , , , )P Q q F= Σ Γ


PDA CFG (cont’d)

becomes becomes

• Idea of proof: construct G with variables for each p and q in the set of states Q. Arrange that if

generates terminal string x, then PDA P started in state p with an empty stack on input string x has a computation that reaches state q with an empty stack. And conversely, if P started in state p with an empty stack has a computation on input string x that reaches state q with an empty stack, then

How does P, when started on an empty stack in state p, operate on an input string x, ending with an empty stack in state q ? First move must be a push Last move must be a pop

,a X Y→ , ,a X Yε ε ε→ →,a ε ε→ , ,a X Xε ε ε→ →

pqApqA

.pq GA x⇒

,a Xε →,b X ε→


PDA CFG (cont’d)

Trace computation of P on x starting in state p with empty stack, and ending in state q with empty stack: (1) stack never empties

pq G rsA aA b⇒

p qa b

rs GA y⇒1 4 4 4 442 4 4 4 4 43

input

Stack height

r s

pq GA x⇒1 4 4 4 442 4 4 4 4 43

push X← pop X →

Fig. 1


PDA CFG (cont’d)

Trace computation of P on x starting in state p with empty stack, and ending in state q with empty stack: (2) stack empties somewhere

pq G pr rqA A A⇒

p q

r q GA z⇒1 4 4 4 442 4 4 4 4 43

input

Stack height

r

pr GA y⇒1 4 4 4 442 4 4 4 4 43

Fig. 2


PDA CFG (cont’d)

Construction. Given PDA construct with the following rules in R:

If

then

( , )( ) pq p qp q Q Q A A A∀ ∈ ∀ ∈ → r rr

&

pq rsA aA b→

0( , , , , ,{ })acceptP Q q q= Σ Γ

0,( , , , )

acceptq qG V R A= Σ

( ) ppp Q A ε∀ ∈ →( , , , )( )( , )p q r s Q X a b ε∀ ∈ ∀ ∈ Γ ∀ ∈ Σ

,a Xε →p r,b X ε→s q


PDA CFG (cont’d)

Claim 2.30: If then Pf: by induction on a derivation in G length k.

Base: k=1. The only derivations of length 1 are and we haveStep: Assume (IH) true for derivations of k steps.

WantClaim true for derivations of k+1 steps. Suppose that . The first derivation

step is either ofthe form orCase . Then

with So IH By construction, since

is a rule of G,

pq G p qA A A⇒ r r

pq G rsA aA b⇒

pq GA x⇒ (p, x, ε) êP∗ (q, ε, ε)

kpq GA x⇒

1pp GA ε⇒

(p, ε, ε) êP∗ ( p, ε, ε)

1kpq GA x+⇒

x ayb=.k

rs GA y⇒ (r , y, ε) êP∗ (s, ε, ε) .

pq rsA aA b→

pq G rsA aA b⇒


PDA CFG (cont’d)

Case . Then with

So IH

Putting these together:

pq G p qA A A⇒ r r

∴ ( p, x, ε) êP (r , yb, X) êP∗(s, b, X) êP (q, ε, ε)

x yz=& .k k

pr G rq GA y A z≤ ≤⇒ ⇒

(p, y, ε) êP∗ (r , ε, ε) &(r , z, ε) êP

∗ (q, ε, ε)

(p, yz, ε) êP∗ (r , z, ε) =(r , z, ε) êP

∗ (q, ε, ε) W

( ) ( , ) ( , , )&( , ) ( , , ).X r X p a q s b X ε ε ∀ ∈ ∈


PDA CFG (cont’d)

Claim 2.31: If then

Pf: by induction on a computation in P of length k:

Base: k=0. The only computations of length 0 are where x = ε. By

construction

Step: Assume (IH) true for computations of k steps. Want

Claim true for computations of k+1 steps. Suppose that . Two

cases: either the stackdoes not empty in midst of this computation (Fig. 1) or

itBecomes empty during the computation (Fig. 2). Call

theseCase 1 and Case 2.

.pq GA x∗⇒ (p, x, ε) êP

∗ (q, ε, ε)

1 .pp GA ε⇒ (p, x, ε) êP

0 ( p, ε, ε)

(p, x, ε) êPk (q, ε, ε)

(p, x, ε) êPk +1 (q, ε, ε) .


PDA CFG (cont’d)

Case 1: See Fig.1. The symbol X pushed in the 1st moveIs the same as that popped in the last move. Let the

1st

and last moves be governed by the push/pop transitions:

By construction, there is a rule in G

Let x = ayb. Since then

we must have By IH

Then Using we conclude

.pq rsA aA b→

.pq GA ayb x∗⇒ =.rs GA y∗⇒

(r , y, X) êPk−1 (s, ε, X)

( , ) ( , , )&( , ) ( , , ).r X p a q s b X ε ε ∈ ∈

(r , y, ε) êPk−1 (s, ε, ε) .

1pq G rsA aA b⇒


PDA CFG (cont’d)

Case 2: See Fig.2. Let r be the intermediate state where the stack becomes empty. Then

By the IH, and

Since by construction there is a rule in G of the form

then

pq pr rqA A A→

pr GA y∗⇒

( , )y z x yz∃ =

(p, y, ε) êP≤k (r , ε, ε) &(r , z, ε) êP

≤k (q, ε, ε)

.rq GA z∗⇒

.pq G pr rq GA A A yz x∗⇒ ⇒ = W


PDA CFG (cont’d)

Ex:

Rules of G:(1) push-pop pairs (1st kind): # #sf qqA A→

(, Aε →

s#, $ε →

f#,$ ε→

),A ε→

is a well-balanced string of parentheses{ {(, )}( ) }∈ *L P = #w# | w

q

s#, $ε →

q

q

(, Aε →

f

q

#,$ ε→

q

q),A ε→

q

:{ , , }

PQ s q f=

{ ,$}{#,(,)}AΓ =

Σ =

( )qq qqA A→


PDA CFG (cont’d)Note: If

(p´ unreachable) then (abbreviated ).

Such variables are useless; all rules involving them on left

or right sides can be eliminated as useless productions. For

this grammar(2) Rules of the 2nd Kind (with useless rules

removed—only 10/27 survive) in the order s,q,f:

(p, −, −) ñP∗ ( ′p , −, −)

ss ss ss

sq ss sq

sq sq qq

sf ss sf

sf sq qf

sf sf ff

A A AA A AA A AA A AA A AA A A

→→→→→→

{ | }pp Gx A x∗′ ⇒ = ∅ ppA ′ = ∅

, ,fq qs fsA A A= ∅ = ∅ = ∅

qq qq qq

qf qq qf

qf qf ff

ff ff ff

A A AA A AA A AA A A

→→→→


PDA CFG (cont’d)(2) Rules of the 3rd Kind:

Combining all rules with same LHS:

ss

qq

ff

AAA

εεε

→→→

||

# # | | |( )| |

||

ss ss ss

sq ss sq sq qq

sf qq ss sf sq qf sf ff

qq qq qq qq

qf qq qf qf ff

ff ff ff

A A AA A A A AA A A A A A A AA A A AA A A A AA A A

ε

ε

ε

→→→→→→


PDA CFG (cont’d)Simplify: easy to see that

Substituting this into rules:

Eliminate useless rules like

ss

ff

AA

εε

==

X X→

# # |( )| |

sq sq qq

sf qq sq qf

qq qq qq qq

qf qq qf

A A AA A A AA A A AA A A

ε

→→→→


PDA CFG (cont’d)Another kind of useless rule:

generate no terminal strings. Eliminate these variables

any and rules mentioning them. Final simplified grammar

is:

Note: chose to use endmarkers # for clarity, but these could have been ε, (input symbols can be anything in ) leading to the familiar grammar

# #( )| |

sf qq

qq qq qq qq

A AA A A A ε

→→

,sq qfA A

( )| |sf qq

qq qq qq qq

A AA A A A ε

→→

εΣ


Closure Properties

Regular Ops. The CFLs are closed under , , Pf: Homework

Intersection. The CFLs are not closed under intersection.

Example: Consider the two CFLs

Then We will later see (CF Pumping Lemma) that this last is not a CFL. �

However, if is regular and is CF, then

is CF.

1 2{ | 0}, { | 0}n n n nL a b c n L a b n c∗ ∗= ⋅ ≥ = ≥ ⋅

1 2 { | 0}.n n nL L a b c n∩ = ≥

1L2L 1 2L L∩

W


Closure Properties (cont’d)• Thm: The class of CFLs is closed under intersection with regular languages.

Pf: Assume and

Construction. Construct a “cross-product pda” M as follows:

where the transition function is defined by:

provided and

Machine M simulates the two given machines “in parallel”,

keeping each machine state in one component of the compound state [ , ].

1 1 1 1 1 1 1( ), ( , , , , , )L LP P Q s F= = Σ Γ

2 2 2 2 2 2 2( ), ( , , , , )R L M M Q s F= = Σ

1 2 1 2 1, 2 1 2( , , , ,[ ], )M Q Q s s F F= × Σ ∪ Σ Γ ×

1 2 1 2([ , ], ) ([ , ], , )p p Y q q a X∈

1 1 1( , ) ( , , )p Y q a X∈ 2 2 2( , )p q a=


What is Not Context-Free?• PDA have a limited computing ability. They cannot, for example, recognize repeated strings like w#w or strings that “count” in more than 2 places, such as .

• We will show that some languages are not CF using a CF Pumping Lemma, which gives a property that all CFLs must have. Then, to show that a language L is not CF, we somehow argue that it lacks this pumping property.

• Closure properties of CFLs can sometimes be used to simplify non-CFLs and make a pumping argument easier.

n n na b c


CF Pumping Lemma• Thm [Pumping Lemma for CFLs]. Suppose that

L is an infinite CF language. Then

• For comparison, here is the Regular P.L.:

( )( )[

( , , )(

( 0) )]i

p w w L w p

x y z w xyz y

xy p i xy z L

ε∃ ∀ ∈ ∧ ≥ ⇒∃ = ∧ ≠

∧ ≤ ∧ ∀ ≥ ∈


CF Pumping Lemma (cont’d)• Pf: Let where CFG G is a CFG in Chomsky Normal Form (Text, Theorem 2.9), i.e. a CFG in which all rules are of the (schematic) forms ABC or Aa (a ε). If is “sufficiently long”, then any derivation tree T for w must contain a “long” path—more precisely:

• Claim 1: If the derivation tree T for has no path longer than h then

Pf: Induction on h. Base: h = 1. Only possible tree is

and Step: Assume Let T have all paths and be

of form (in CNF)

( )w L G∈

1| | 2 .hw −≤

1.h > h≤

{ } ( )L L Gε− =

w

0| | 2.a =

S

a


CF Pumping Lemma (cont’d)

• Then have all paths of length By IH,

which implies .

Conversely, if a generated string is at least long,

then its parse tree must be at least high.

G has variables. Choose If

and then Claim 1 any parse tree T for w has

a path of length at least Such a path has at least nodes. ∴ some variable appears twice

on the path (note the leaf node is a terminal).

2 2| | 2 ,| | 2h hs t− −≤ ≤

S

1T 2TT =

s t w st=1 2,T T 1.h≤ −

1| | 2hw −≤2h

1h+

V 12 .Vp += w L∈w p≥

2.V +3V +


CF Pumping Lemma (cont’d)• Picture:

12Vw p +≥ =

T =R

R

height nodes

variables repeat

2 3

2

V V

V

≥ + ⇒ ≥ +

⇒ ≥ + ⇒

R

RR

u v x y zw =

aht 2V≤ +

12V p+≤ =

ChooseBottom

variables

1V +

vxy p∴ ≤

1T

2T


CF Pumping Lemma (cont’d)(1) Center portion is not too long:(2) Pumped portion not empty:

cannot both = ε.

1T

2T

vxy p≤,v y

=R

B C

2T

v x yx ε≠

R

B C

2T

v x yxε≠

or

In CNF, no variablegenerates ε


CF Pumping Lemma (cont’d)(3) Pumped strings in L : the following are

all parse trees

R

R

( 0) i ii uv xy z L∴ ∀ ≥ ∈ W

R

R

R

R

R

R

R

u v x y z

u vv x yy z

u vvv x yyy z

R

u x z

and:


CF Pumping: Applications• Ex: is not a CFL.

Pf: Suppose it is CF. Then the Pumping Lemma p

wL, |w|p uvxyz =w & vy ε & i0 u vi x yi z L.

Pick p as the constant guaranteed and choose n p/3 and

Where is

Cases: Assume first that

{ | 0}n n nL a b c n= ≥

, .n n nw a b c uvxyz vy ε= = ≠ ?vxy.v ε≠

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

a ab bc c

u v x y z

u v x y z

u v xyz

u v x y z

u v xyz

u v xyz

K K K K K K K K K


CF Pumping: Applications• In cases 1-3 has an imbalance. In case 4 it has

a b before a. In case 5 it has a c before b. In case 6 it has an a after a c. In any case, there is a contradiction to the pumped

word being in L.

The case where is symmetric. Contradiction.

Cor. The CFLs are not closed under complementation. Pf: is a CFL.

But is not a CFL.

Therefore cannot be CF. Ex: is not CF. Proof similar to

regular case. Ex: is not CF.

y ε≠

2 2uv xy z

{ | }p q rL a b c p q q r p r= ≠ ∨ ≠ ∨ ≠{ | 0}i i iL a b c a b c i∗ ∗ ∗∩ = ≥

L{ | prime}iL a i=

2

{ | 0}iL a i= ≥


Pumping: Applications (cont’d) Ex: is not CF.

Pf: Intersection with is not a CFL.

Therefore cannot be CF. Ex: is not CF.

Pf: By pumping on the word Similar to

Text, Example 2.38. Ex: is not CF.

Pf: Pump on the latter language in a way similar to the previous example to show it is not CF.

L

a b c{ { , , } | = = }L w a b c w w w∗= ∈

a b c∗ ∗ ∗

{ |i 1,j 1}i j i jL a b c d= ≥ ≥n n n nw a b c d=

{ | {0,1}}L ww w= ∈10 10 110 10 1 {10 10 110 10 1| , 1}n m n mL m n+ + + +∩ = ≥

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