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

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

q q’,

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

– ….

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:

e, e e, e S

(for each rule terminal in ) , e

e, C w

(for each rule C w in R)(this rule needs to be expanded)

e, e q

Sketch of the ProofS 1T11 where 1, 1 are in * and T1 is in ( V)*

1 2T2 2 1 where 2 is in in * and T2 is in ( V)*

(taking the leftmost non terminal in T1)

…(always taking the leftmost non terminal)

1 2… n n …1

(s, 1 2… n n 1, e)

(q, 1 2… n n 1, S)

(q, 1 2… n n 1, 1T11)

*

(q, 2… n n 1, T11)

(q, 2… n n 1, 2T2 2 1)

*(q, 3… n n 2 1,T2 2 1)

* (q, e,e) (Excluded for simplicity)

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: 1. PA has only one accepting state2. Stack is empty when accepting a word3. 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

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}), and all t in such that the following two transitions occur in the PA:

– Apq Apr Arq

for all p, q, r in Q

– App e for all p in Q

p ra, e t

s qb, t e

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 Apr Arq

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