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# Equivalence of Pushdown Automata and Context-Free Grammar PowerPoint PPT Presentation

Equivalence of Pushdown Automata and Context-Free Grammar. Prof. H é ctor Mu ñ oz-Avila. Two Crucial Concepts. Nondeterministic computation Give us flexibility for constructing devices and understanding the power/limitations of these devices Induction

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Equivalence of Pushdown Automata and Context-Free Grammar

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## Equivalence of Pushdown Automata and Context-Free Grammar

Prof. Héctor Muñoz-Avila

### Two Crucial Concepts

• Nondeterministic computation

• Give us flexibility for constructing devices and understanding the power/limitations of these devices

• Induction

• Allow us to prove statements that otherwise would be hard to see why they are true

• In this class(es), we will illustrate these two powerful concepts once more

,   

q

q’

### Class Convention regarding Transitions in Pushdown Automata

• Strictly speaking, Transitions have the form:

• :(Q × (  {e}) × (  {e})) (Q × (  {e}))

• (q’,))  ((q, ,))

• We are going to allow to pop and push words in (  {e})*

• Can we represent a transition that pops/pushes words in (  {e})* with transitions that pop/pushes characters in (  {e}) ?

• Careful: order of pushing/popping individual characters matter!

• To avoid confusion, view the stack as a word and push/pop as adding/removing strings prefix from that word

### My Solution of the Homework

• Construct a pushdown automaton for words in {a,b} such that the number of a’s is twice the number of b’s

• 3 states

• Pushing marker for bottom of stack

• ….

(for each rule C  w in R)

(this rule needs to be expanded)

e, C  w

e, e 

e, e  S

e, e

,  e

(for each rule terminal  in )

### Equivalence of Pushdown Automata and Context-Free Grammars (part I)

Theorem. (Lemma 2.21) Given a context-free grammar CG = (,V,R,S) , then there is a pushdown automaton PA = (Q,,, ,s,F) such that L(CG) = L(PA)

Construction:

q

S  1T11

where 1, 1 are in * and T1 is in (  V)*

 12T2 21

 12… n n …1

where 2 is in in * and T2 is in (  V)*

(always taking the leftmost non terminal)

(s, 12…n n 1, e)

(q, 12…n n 1, S)

(q, 12… n n 1, 1T11)

*

*

*

(q, 2…n n 1, T11)

(Excluded

for simplicity)

(q, 2…n n 1, 2T2 21)

(q, 3… n n 2 1,T2 21)

### Sketch of the Proof

(taking the leftmost non terminal in T1)

(q, e,e)

### Equivalence of Pushdown Automata and Context-Free Grammars (Part II)

Theorem. (Lemma 2.27) Given a pushdown automata PA = (Q,,, ,s,F) then, there exists a context-free grammar CG = (,V,R,S) such that L(PA) = L(CG)

• Assumptions:

• PA has only one accepting state

• Stack is empty when accepting a word

• Each transitions pops XOR pushes one element in the stack

### Constructing the Grammar CG (1)

• CG will contain the variables: Apq for every two states p and q in PA

• Apq generates a word w PA empty-process w from p to q

• PA empty-process w from p to q

• w is given as input starting on state p with empty stack

• then PA will nondeterministically process all characters in w

• ending with the empty word in state q and the empty stack

a, e  t

b, t  e

p

s

r

q

### Constructing the Grammar CG (2)

• We are going to construct three kinds of rules for CG:

• Apq aArsb

for all p, q, r, s in Q, all a, b in (  {e}), andall t in  such that the following two transitions occur in the PA:

• Apq AprArq

for all p, q, r in Q

• App e

for all p in Q

That’s it!

Can you see it?

### Proving that CG is equivalent to PA (1)

If Apq generates a word w then PA empty-process w from p to q

• Proof by induction on # steps in Apq* w

• Basis: # steps is 1

• Induction: holds for k # steps, need to prove for k+1 # steps

• Two sub-cases depending of which rule was applied first:

• Apq aArsb or

• Apq AprArq

### Proving that CG is equivalent to PA (2)

If PA empty-process w from p to q then Apq generates a word w

• Proof by induction on # steps in processing w

• Basis: # steps is 0

• Induction: holds for k # steps, need to prove for k+1 # steps

• Two sub-cases depending on the following:

• Stack is empty only at the beginning and at the end of process or

• Stack gets empty somewhere in-between

### Corollary

• Let s be the start state in PA

• Let f be the accepting state in PA

• Therefore, Asf is the start variable in CG

• We just proved that:

Asf generates a word w if and only if PA accepts w

### Homework

• Show that if L1 and L2 are context-free languages then:

• L1 L2 is a context-free language

• L1L2 is a context-free language

(hint: if L1 and L2 are context free, then there is two grammars G1 generating L1 and G2 generating L2. How can you combine G1 and G2 to generate the union and concatenation?)

• 2.19

• 2.23

• 2.27