CSCI 2670Introduction to Theory of

Computing

September 23, 2004

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Agenda• Yesterday

– Pushdown automata• Today

– Quiz– More on pushdown automata– Pumping lemma for CFG’s

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Example• Find for the PDA that accepts all

strings in {0,1}* with the same number of 0’s and 1’s

1. Need to keep track of “equilibrium point” with a $ on the stack

2. If stack top is not $, it contains the symbol currently dominating in the string

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Example• Find for the PDA that accepts all

strings in {0,1}* with the same number of 0’s and 1’s

3. Push a symbol on the stack as its read if1. It matches the top of the stack, or2. The top of stack is $

4. Pop the symbol off the top of the stack if you read a 0 and the top of stack is 1 or vice versa

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Example

ε,ε$

0,$0$0,0 000,11 1 0,1$ $

1,$1$1,1 111,00 0 1,0$ $

ε,$ ε

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Example

ε,ε$

0,$0$0,0 000,1 ε

1,$1$1,1 111,0 ε

ε,$ ε

This PDA is equivalent to the one on the previous slide

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Equivalence of PDA’s and CFG’s

Theorem: A language is context free if and only if some pushdown automaton recognizes it

Proved in two lemmas – one for the “if” direction and one for the “only if” direction

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CFG’s are recognized by PDA’s

Lemma: If a language is context free, then some pushdown automaton recognizes it

Proof idea: Construct a PDA following CFG rules

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Constructing the PDA• You can read any symbol in when

that symbol is at the top of the stack– Transitions of the form a,aε

• The rules will be pushed onto the stack – when a variable A is on top of the stack and there is a rule Aw, you pop A and push w

• You can go to the accept state only if the stack is empty

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Idea of PDA construction for AxBz

State control

a b

At

State control

a b

xBzt

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Actual construction for AxBz

ε,Az ε, ε B

ε, ε x

In an abuse of notation, we say (q,ε,A)=(q,xBz)

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Constructing the PDA• Q = {qstart, qloop, qaccept}E, where E

is the set of states used for replacement rules onto the stack

(the PDA alphabet) is the set of terminals in the CFG

(the stack alphabet) is the union of the terminals and the variables and {$} (or some suitable placeholder)

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Constructing the PDA is comprised of several rules

1. (qstart,ε,ε)=(qloop,S$)- Start with placeholder on the stack and

with the start variable2. (qloop,a,a)=(qloop,ε) for every a

- Terminals may be read off the top of the stack

3. (qloop,ε,A)=(qloop,w) for every rule Aw- Implement replacement rules

4. (qloop,ε,$)=(qaccept,ε)- Accept when the stack is empty

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Example• S SS | (S) | ()

qstart qloop qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S() S

$qloop

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

(S)$

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

S)$

(

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

SS)$

(

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

()S)$

(

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

)S)$

((

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

S)$

(()

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

())$

(()

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

))$

(()(

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

)$

(()()

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Example• Read (()())

qstart qacceptε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

$

(()())

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Example• Read (()())

qstartε, ε S$ ε, $ ε

(,(ε),)ε

ε,SSSε,S(S)ε,S()

qloop

(()())

qaccept

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The pumping lemma for RE’s• The pumping lemma for RE’s

depends on the structure of the DFA and the fact that a state must be revisited– Only a finite number of states

x

y

z

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The pumping lemma for CFG’s

• What might be repeated in a CFG?– The variables

TR

R

u v x y z

v & y will be repeated simultaneously

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The pumping lemma for CFG’s

TRR

u v x y z

TR

uv y

zR

x yvR

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The pumping lemma for CFG’s

TRR

u v x y z

TR

ux

z

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The pumping lemma for CFL’s

Theorem: If A is a context-free language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into five pieces s=uvxyz satisfying the conditions:

1. For each i 0, uvixyiz A2. |vy| > 03. |vxy| p

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Finding the pumping length of a CFL

• Let b equal the longest right-hand side of any rule (assume b > 1)– Each node in the parse tree has at

most b children– At most bh nodes are h steps from the

start node• Let p equal b|V|+2, where |V| is the

number of variables (could be huge!)– Tree height is at least |V|+2

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Proving the pumping lemma for CFG’s

• Let s be any string in A with length at least p = b|V|+2 and let be a parse tree for s with the fewest possible nodes– The height of is at least |V|+2– The longest path in has a length of

at least |V|+2• Leaf is a terminal and at least |V|+1

variables– Some variable appears twice in the

path

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Proving the pumping lemma for CFG’s

• Let R be the last variable to appear twice on this path– The last two occurrences of R are

both at most |V|+1 steps from the leaf T

RR

u v x y z}At most |V|

+1 steps

Question: What do we know about uvixyiz?Answer: It is accepted by the PDA

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Proving the pumping lemma for CFG’s

• Let R be the last variable to appear twice on this path– The last two occurrences of R are

both at most |V|+1 steps from the leaf T

RR

u v x y z}At most |V|

+1 steps

Question: What do we know about |vy|?Answer: It is non-empty (because tree has fewest possible nodes)

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Proving the pumping lemma for CFG’s

• Let R be the last variable to appear twice on this path– The last two occurrences of R are

both at most |V|+1 steps from the leaf T

RR

u v x y z}At most |V|

+1 steps

Question: What do we know about |vxy|?Answer: It has at most p symbols (because at most |V|+1 steps)

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Example• Show A is not context free, where

A={an|n is prime}Proof: Assume A is context-free and

let p be the pumping length of A. Let w=an for any np. By the pumping lemma, w=uvxyz such that |vxy|p, |vy|>0, and uvixyizA for all i=0,2,1….

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Example (cont.)• Show A is not context free, where

A={an|n is prime}Clearly, vy=ak for some kConsider the string uvn+1xyn+1z

This string add n copies of ak to w – i.e., this is an+nk

Since the exponent is n(1+k) this in not in A, which contradicts the pumping lemma. Therefore, A is not context free.

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