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### Lect14

1. 1. Lecture Pushdown Automata
3. 3. a l p h a b e tThe tape is divided into finitely many cells.Each cell contains a symbol in an alphabetΣ.
4. 4. The stack head always scans the topsymbol of the stack. It performs twobasic operations:Push: add a new symbol at the top.Pop: read and remove the top symbol. Alphabet of stack symbols: Γ
5. 5. a• The head scans at a cell on the tape and can read a symbol on the cell. In each move, the head can move to the right cell.
6. 6. • The finite control has finitely many states which form a set Q. For each move, the state is changed according to the evaluation of a transition function Q x (Γ U {ε}) δ : Q x (Σ U {ε}) x (Γ U {ε}) → 2 .
7. 7. a a q v p u• (p, u) ∈ δ(q, a, v) means that if the tape head reads a, the stack head read v, and the finite control is in the state q, then one of possible moves is that the next state is p, v is replaced by u at stack, and the tape head moves one cell to the right.
8. 8. a a q v p u• (p, u) ∈ δ(q, ε, v) means that this a ε-move.
9. 9. a a q p u• (p, u) ∈ δ(q, a, ε) means that a push operation performs at stack.
10. 10. a a q v p• (p, ε) ∈ δ(q, a, v) means that a pop operation performs at stack
11. 11. s• There are some special states: an initial state s and a final set F of final states.• Initially, the PDA is in the initial state s and the head scans the leftmost cell. The tape holds an input string. The stack is empty.
12. 12. x f• When the head gets off the tape, the PDA stops. An input string x is accepted by the PDA if the PDA stops at a final state and the stack is empty.• Otherwise, the input string is rejected.
13. 13. • The PDA can be represented by M = (Q, Σ, Γ, δ, s, F) where Σ is the alphabet of input symbols and Γ is the alphabet of stack symbols.• The set of all strings accepted by a PDA M is denoted by L(M). We also say that the language L(M) is accepted by M.
14. 14. • The transition diagram of a PDA is an alternative way to represent the PDA.• For M = (Q, Σ, Γ, δ, s, F), the transition diagram of M is an edge-labeled digraph G=(V, E) satisfying the following: V = Q (s = ,f= for f ∈ F) a, v/u E={q p | (p,u) ∈ δ(q, a, v) }.
15. 15. Example 1. Construct PDA to accept n n L= {0 1 | n > 0}Solution 1. 1, 0/ε 0, ε/0 1, 0/ε
16. 16. Solution 2. Consider a CFG G = ({S}, {0,1}, {S → ε | 0S1}, S). ε, S/1 ε, ε/S ε, ε/S ε, ε/0 ε, S/ε 0, 0/ε 1, 1/ε
17. 17. Theorem Every CFL can be accepted by a PDA.Proof. Consider a CFL L = L(G) for a CFG G = (V, Σ, R, S). ε, A/xn A → x1 x2 …xn in R ε, ε/S ε, ε/x1 a, a/ε for a in Σ TheoremA language L is CFL ⇔ it can be accepted by a PDA.
18. 18. Sometimes, constructing the PDA is easier thanconstructing CFG.
19. 19. Example 2Show that {x ∈ {a, b} | #b ( x) ≤ # a ( x) ≤ 2#b ( x)} is a CFL Construct a PDA or a CFG? PDA!!! Sometimes, constructing the PDA is easier than constructing CFG.
20. 20. IdeaTo check if #b ( x) ≤ # a ( x) ≤ 2#b ( x), we need to cancel a with b.For each b, we need to cancel sometimes one a and sometimestwo a. Do we need to make a deterministic choice at eachcancellation? No, the concept of nondeterministic computationsolves this trouble : As long as a correct choice exists, the inputstring x would be accepted!
21. 21. a a s p s• (p, s) ∈ δ(s, ε, ε) ε ,ε / s
22. 22. a a p p a s s(q, s) ∈ δ(p, a, s)(p, a) ∈ δ(q, ε, ε) a, s / s ε ,ε / a
23. 23. a a p p a a a ∈(q1, a) δ(p, a, a)(p, a) ∈ δ(q1, ε, ε) a, a / a ε ,ε / a
24. 24. a ap p b (p, ε) ∈ δ(p, a, b) a,b / ε
25. 25. a ap p b’ b (p, b’)∈ δ(p, a, b) a, b / b
26. 26. a ap p b’ (p, ε) ∈ δ(p, a, b’) a,b / ε
27. 27. b b p p b s s(r1, s)∈ δ(p, b, s)(p, b) ∈ δ(r1, ε, ε) b, s / s ε ,ε / b
28. 28. b b p p b b b(r2, b)∈ δ(p, b, b)(p, b) ∈ δ(r2, ε, ε) b, b / b ε ,ε / b
29. 29. b b p p b b’ b’(r3, b’) ∈ δ(p, b, b’)(p, b) ∈ δ(r3, ε, ε) b, b / b ε ,ε / b
30. 30. b bp p a (p, ε) ∈ δ(p, b, a) b, a / ε
31. 31. b b p p a a(t, ε) ∈ δ(p, b, a) (p, ε) ∈ δ(t, ε, a) b, a / ε ε,a /ε
32. 32. b b p p b’ a z z z≠a(t, ε) ∈ δ(p, b, a) (u, x) ∈ δ(t, ε, x) (p, b’) ∈ δ(u, ε, ε) b, a / ε ε, z / z ε ,b / ε z≠a
33. 33. p p s (p, ε) ∈ δ(p, ε, s) ε,s /ε
34. 34. a,b / ε a, b / b a,b / ε ε ,ε / s a, z / z z ≠ b,b ε ,ε / a ε,s /ε b, a / ε b, a / ε b, z / z z≠a ε,a /εε, z / z ε ,ε / b z≠a ε , ε / b
35. 35. Example 3Construct PDA accepting {x ∈ {a, b} | #b ( x) <# a ( x) ≤ 2#b ( x)}. IdeaTo have #b ( x) <# a ( x), we must have a b which cancel two a s.Let the first b do the job.
36. 36. a,b / εSolution a, b / b a, z / z a,b / ε 1 z ≠ b,b ε ,ε / s a, b / b a, z / z 11 z ≠ b,b ε ,ε / a b, a / ε ε,a /ε ε , ε / b ε ,ε / a ε, z / z z≠a ε,s /ε b, a / ε b, a / ε b, z / z z≠a b, z / z ε,a /ε 1 ε, z / z ε ,ε / b z≠a z≠a ε , ε / b ε ,ε / b
37. 37. Example 4Construct PDA accepting {x ∈ {a, b} | #b ( x) ≤# a ( x) < 2#b ( x)}. IdeaTo have # a ( x) < 2#b ( x), we must have a b which cancel one a.Let the first b do the job.
38. 38. a,b / εSolution a, b / b a, z / z a,b / ε 1 z ≠ b,b ε ,ε / s a,b / ε a, z / z 1 1 z ≠ b,b ε ,ε / a b, a / ε ε ,ε / a ε,s /ε b, a / ε b, a / ε b, z / z z≠a b, z / z ε,a /ε 1 ε, z / z ε ,ε / b z≠a z≠a ε , ε / b ε ,ε / b
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