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Equivalence of Pushdown Automata and Context-Free Grammar

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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  1. Equivalence of Pushdown Automata and Context-Free Grammar Prof. Héctor Muñoz-Avila

  2. 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

  3. ,    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

  4. 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 • ….

  5. (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

  6. 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)

  7. 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

  8. 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

  9. 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?

  10. 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

  11. 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

  12. 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

  13. 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

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