FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

1. 15-453 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

2. Some Homework 1 Statistics: Each question was graded out of 20 points. Max: 100 Mean: 78 Median: 79.02 Std Dev: 15.65

3. NOTE : Your 1st Project Report is Due Tuesday! Turn in your Homework! (New Homework on the Web)

4. PDAs ARE EQUIVALENT TO CFGs THURSDAY Jan 31

5. string pop push ε,ε → $ 0,ε→ 0 1,0→ε ε,$ → ε 1,0→ε

6. CONTEXT-FREE GRAMMARS Production rules terminals variable A → 0A1 A → ε  (VΣ)*  000A111 A  0A1  00A11  000111 (yields) A *000111 (derives)

7. A Language L is generated by a CFG  L is recognized by a PDA

8. Suppose L is generated by a CFG G = (V, Σ, R, S) Construct P = (Q, Σ, Γ, , q, F) that recognizes L A Language L is generated by a CFG  L is recognized by a PDA

9. Suppose L is generated by a CFG G = (V, Σ, R, S) Construct P = (Q, Σ, Γ, , q, F) that recognizes L ε,ε → S$ For each rule 'A → w’R: ε,A → wR For each terminal aΣ: a,a → ε ε,$ → ε

10. S → aTb T → Ta | ε ε,ε → $ ε,ε → T ε,S → b ε,ε → T ε,ε → S ε,T → a ε,$ → ε ε,ε → a ε,T → ε a,a → ε b,b → ε

11. S → aTb S*ab (derives) T → Ta | ε ε,ε → $ ε,ε → T ε,S → b ε,ε → T ε,ε → S ε,T → a ε,$ → ε ε,ε → a ε,T → ε a,a → ε b,b → ε

12. Suppose L is generated by a CFG G = (V, Σ, R, S) Describe P = (Q, Σ, Γ, , q, F) that recognizes L : (1) Push $ and then S on the stack (2) Repeat the following steps forever: (a) Pop the stack, call the result X. (b) If X is a variable A, guess a rule that matches A and push result into the stack (c) If X is a terminal, read next symbol from input and compare it to terminal. If they’re different, reject. (d) If X is $: then accept iff no more input

13. A Language L is generated by a CFG  L is recognized by a PDA

14. A Language L is generated by a CFG  L is recognized by a PDA Given PDA P = (Q, Σ, Γ, , q, F) Construct a CFG G = (V, Σ, R, S) such that L(G) = L(P) First, simplify P to have the following form: (1) It has a unique accept state, qacc (2) It empties the stack before accepting (3) Each transition either pushes a symbol or pops a symbol, but not both at the same time

15. SIMPLIFY ε,ε → $ 0,ε→ 0 q1 q0 1,0→ε ε,$ → ε q2 q3 1,0→ε

16. SIMPLIFY ε,ε → E ε,ε → $ 0,ε→ 0 Q q1 q0 1,0→ε ε,ε → ε ε,$ → ε q2 q3 1,0→ε ε,ε → ε q4 ε,E → ε q5 ε,σ → ε

17. SIMPLIFY ε,ε → E ε,ε → $ 0,ε→ 0 Q q1 q0 ε,ε → D 1,0→ε q’0 ε,$ → ε q2 q3 1,0→ε ε,D→ ε ε,ε → D q’3 ε,D → ε q4 ε,E → ε q5 ε,σ → ε

18. Idea For Our Grammar G: For every pair of states p and q in PDA P, G will have a variable Apq whose production rules will generate all strings x that can take: P from p with an empty stack to q with an empty stack V = {Apq | p,qQ } S = Aq0qacc

19. ε,ε → E ε,ε → $ 0,ε→ 0 Q q1 q0 ε,ε → D 1,0→ε q’0 ε,$ → ε q2 q3 1,0→ε ε,ε → D q’3  Aq0q1 generates? ε,D → ε Aq1q2 generates? q4 {0n1n | n > 0} ε,E → ε q5 Aq1q3 generates?  ε,σ → ε WANT: Apq generates all strings that take p with an empty stack to q with empty stack

20. WANT: Apq generates all strings that take p with an empty stack to q with empty stack Let x be such a string • P’s first move on x must be a push • P’s last move on x must be a pop Two possibilities: 1. The symbol popped at the end is exactly the one pushed at the beginning 2. The symbol popped at the end is not the one pushed at the beginning

21. x = ayb takes p with empty stack to q with empty stack 1. The symbol t popped at the end is exactly the one pushed at the beginning stack height r s a b input string pop t push t p q ─ ─ ─ ─ x ─ ─ ─ ─ δ(p, a, ε) → (r, t) Apq→ aArsb δ(s, b, t) → (q, ε)

22. 2. The symbol popped at the end is not the one pushed at the beginning stack height input string p r q Apq→ AprArq

23. Formally: V = {Apq | p, qQ } S = Aq0qacc For every p, q, r, s  Q, t  Γand a, b  Σε If (r, t)  (p, a, ε) and (q, ε)  (s, b, t) Then add the rule Apq→ aArsb For every p, q, r  Q, add the rule Apq→ AprArq For every p  Q, add the rule App→ ε

24. For all x, Apqgenerates x  xcan bring P from p with an empty stack to q with an empty stack  Proof (by induction on the number of steps in the derivation of x from Apq): Inductive Step: Base Case: The derivation has 1 step: Appε Assume true for derivations of length ≤ k and prove true for derivations of length k+1: Apq* x in k+1 steps Apq → AprArq Apq → aArsb (r,t)  (p,a,ε) and (q, ε)  (s,b,t) state push state alphabet pop

25. For all x, Apq generates x  x can bring P from p with an empty stack to q with an empty stack  Proof (by induction on the number of steps in the computation of P from p to q with empty stacks, on input x): Base Case: The computation has 0 steps So it starts and end in the same state: We must show that App* x But it must be that x = ε, so we are done

26. Inductive Step: Assume true for computations of length ≤ k, we’ll prove true for computations of length k+1 Suppose that P has a computation where x brings p to q with empty stacks in k+1 steps Two cases: 1. The stack is empty only at the beginning and the end of this computation Write x as ayb. Ars* y by I.H., use Apq →aArsb 2. The stack is empty somewhere in the middle of the computation Write x as yz. Apr* y, Arq* z by I.H., Apq → AprArq

27. A Language L is generated by a CFG  L is recognized by a PDA

28. Corollary: Every regular language is context-free

29. WWW.FLAC.WS