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. 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 qq’ ,  

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. 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

6. Sketch of the Proof S   1 T 1  1 where  1,  1 are in  * and T 1 is in (   V)*   1  2 T 2  2  1 where  2 is in in  * and T 2 is in (   V)* (taking the leftmost non terminal in T 1 ) … (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,  1 T 1  1 )  * (q,  2 …  n  n  1, T 1  1 )  (q,  2 …  n  n  1,  2 T 2  2  1 )  * (q,  3 …  n  n  2  1,T 2  2  1 )  * (q, e,e) (Excluded for simplicity)

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: 1.PA has only one accepting state 2.Stack is empty when accepting a word 3.Each transitions pops XOR pushes one element in the stack

8. Constructing the Grammar CG (1) CG will contain the variables: A pq for every two states p and q in PA A pq 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. Constructing the Grammar CG (2) We are going to construct three kinds of rules for CG: –A pq  aA rs b 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: –A pq  A pr A rq for all p, q, r in Q – A pp  e for all p in Q pr a, e  t sq b, t  e That’s it! Can you see it?

10. Proving that CG is equivalent to PA (1) If A pq generates a word w then PA empty-process w from p to q Proof by induction on # steps in A pq  * 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: A pq  aA rs b or A pq  A pr A rq

11. Proving that CG is equivalent to PA (2) If PA empty-process w from p to q then A pq 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, A sf is the start variable in CG We just proved that: A sf generates a word w if and only if PA accepts w

13. Homework Show that if L 1 and L 2 are context-free languages then: –L 1  L 2 is a context-free language –L 1 L 2 is a context-free language (hint: if L 1 and L 2 are context free, then there is two grammars G 1 generating L 1 and G 2 generating L 2. How can you combine G 1 and G 2 to generate the union and concatenation?) 2.19 2.23 2.27