CSCI 2670 Introduction to Theory of Computing

1. CSCI 2670Introduction to Theory of Computing September 23, 2004

2. Agenda • Yesterday • Pushdown automata • Today • Quiz • More on pushdown automata • Pumping lemma for CFG’s September 23, 2004

3. Example • Find  for the PDA that accepts all strings in {0,1}* with the same number of 0’s and 1’s • Need to keep track of “equilibrium point” with a $ on the stack • If stack top is not $, it contains the symbol currently dominating in the string September 23, 2004

4. Example • Find  for the PDA that accepts all strings in {0,1}* with the same number of 0’s and 1’s • Push a symbol on the stack as its read if • It matches the top of the stack, or • The top of stack is $ • Pop the symbol off the top of the stack if you read a 0 and the top of stack is 1 or vice versa September 23, 2004

5. 1,$1$ 1,1 11 1,00 0 1,0$ $ ε,$ ε ε,ε$ 0,$0$ 0,0 00 0,11 1 0,1$ $ Example September 23, 2004

6. 1,$1$ 1,1 11 1,0 ε ε,$ ε ε,ε$ 0,$0$ 0,0 00 0,1 ε Example This PDA is equivalent to the one on the previous slide September 23, 2004

7. 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 September 23, 2004

8. 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 23, 2004

9. 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 23, 2004

10. State control State control a b a b x B z t A t Idea of PDA construction for AxBz September 23, 2004

11. 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 23, 2004

12. 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 23, 2004

13. 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 23, 2004

14. ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε Example • S  SS | (S) | () September 23, 2004

15. Example • Read (()()) ε,SSS ε,S(S) ε,S() S $ ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε September 23, 2004

16. Example • Read (()()) ( S ) $ ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε September 23, 2004

17. Example • Read (()()) S ) $ ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε ( September 23, 2004

18. Example SS ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε ( September 23, 2004

19. Example ( ) S ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε ( September 23, 2004

20. Example ) S ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (( September 23, 2004

21. Example S ) $ • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (() September 23, 2004

22. ( ) ) $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (() September 23, 2004

23. ) ) $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()( September 23, 2004

24. ) $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()() September 23, 2004

25. $ Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()()) September 23, 2004

26. Example • Read (()()) ε,SSS ε,S(S) ε,S() ε, ε S$ ε,$ε qstart qloop qaccept (,(ε ),)ε (()()) September 23, 2004

27. 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 23, 2004

28. 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 23, 2004

29. 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 23, 2004

30. T R T R R x u v x y z u z The pumping lemma for CFG’s September 23, 2004

31. 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 23, 2004

32. 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 23, 2004

33. Proving the pumping lemma for CFG’s • Let s be any string in A with length at least p = b|V|+2 and let  be a parse tree for s with the fewest possible nodes • The height of  is at least |V|+2 • The longest path in  has a length of at least |V|+2 • Leaf is a terminal and at least |V|+1 variables • Some variable appears twice in the path September 23, 2004

34. T R R u v x y z } At most |V|+1 steps Proving the pumping lemma for CFG’s • Let R be the last variable to appear twice on this path • The last two occurrences of R are both at most |V|+1 steps from the leaf Question: What do we know about uvixyiz? Answer: It is accepted by the PDA September 23, 2004

35. T R R u v x y z } At most |V|+1 steps Proving the pumping lemma for CFG’s • Let R be the last variable to appear twice on this path • The last two occurrences of R are both at most |V|+1 steps from the leaf Question: What do we know about |vy|? Answer: It is non-empty (because tree has fewest possible nodes) September 23, 2004

36. T R R u v x y z } At most |V|+1 steps Proving the pumping lemma for CFG’s • Let R be the last variable to appear twice on this path • The last two occurrences of R are both at most |V|+1 steps from the leaf Question: What do we know about |vxy|? Answer: It has at most p symbols (because at most |V|+1 steps) September 23, 2004

37. 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 23, 2004

38. 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 23, 2004

39. Have a fun weekend! September 23, 2004