CSCI 2670 Introduction to Theory of Computing

1. CSCI 2670Introduction to Theory of Computing September 22, 2005

2. Agenda • Yesterday • Pushdown automata • Today • Equivalence of pushdown automata and CFG’s • Pumping lemma for CFG’s September 22, 2005

3. Announcements • Matrix Reloaded tonight! • 6:30 in Boyd 328 • Free popcorn • 25 cent Coke • Pizza will be ordered at cost if people want September 22, 2005

4. 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 We will only do the “only if” step – i.e., show that every context-free language has an associated PDA September 22, 2005

5. 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 September 22, 2005

6. Have a transition for each rule replacing the variable with its right hand side ε,Aw Finish only if the stack is empty ε, ε S$ ε,$ε qstart qaccept qloop Have a transition that allows us to read each alphabet symbol if it is at the top of the stack Start by pushing the start variable and stack bottom marker a,aε CFG’s are recognized by PDA’s Format of the new PDA September 22, 2005

7. Constructing the PDA • You can read any symbol in  when that symbol is at the top of the stack • Transitions of the form a,aε • The rules will be pushed onto the stack – when a variable A is on top of the stack and there is a rule Aw, you pop A and push w • You can go to the accept state only if the stack is empty September 22, 2005

8. State control State control a b a b x B z t A t Idea of PDA construction for AxBz September 22, 2005

9. ε, εB ε,Az ε, εx Actual construction for AxBz In an abuse of notation, we say (q,ε,A)=(q,xBz) September 22, 2005

10. 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) September 22, 2005

11. Constructing the PDA •  is comprised of several rules • (qstart,ε,ε)=(qloop,S$) • Start with placeholder on the stack and with the start variable • (qloop,a,a)=(qloop,ε) for every a • Terminals may be read off the top of the stack • (qloop,ε,A)=(qloop,w) for every rule Aw • Implement replacement rules • (qloop,ε,$)=(qaccept,ε) • Accept when the stack is empty September 22, 2005

12. ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε Example • S  SS | (S) | () September 22, 2005

13. Example • Read (()()) ε,SSS ε,S(S) ε,S() S $ ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε September 22, 2005

14. Example • Read (()()) ( S ) $ ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε September 22, 2005

15. Example • Read (()()) S ) $ ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε ( September 22, 2005

16. Example SS ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε ( September 22, 2005

17. Example ( ) S ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε ( September 22, 2005

18. Example ) S ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (( September 22, 2005

19. Example S ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (() September 22, 2005

20. ( ) ) $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (() September 22, 2005

21. ) ) $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()( September 22, 2005

22. ) $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()() September 22, 2005

23. $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()()) September 22, 2005

24. Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()()) September 22, 2005

25. y x z 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 September 22, 2005

26. T R R u v x y z v & y will be repeated simultaneously The pumping lemma for CFG’s • What might be repeated in a CFG? • The variables September 22, 2005

27. T R T R R R R u v x y z z u y y v v x The pumping lemma for CFG’s September 22, 2005

28. T R T R R x u v x y z u z The pumping lemma for CFG’s September 22, 2005

29. 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: • For each i  0, uvixyiz  A • |vy| > 0 • |vxy|  p September 22, 2005

30. 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 September 22, 2005

31. 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 np. By the pumping lemma, w=uvxyz such that |vxy|p, |vy|>0, and uvixyizA for all i=0,2,1…. September 22, 2005

32. Example (cont.) • Show A is not context free, where A={an|n is prime} Clearly, vy=ak for some k Consider 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. September 22, 2005

33. Have a fun weekend! September 22, 2005

34. Let's work on the practice test! September 22, 2005