CS Pushdown Automata

Transcription

1 Chap. 6 Pushdown Automata 6.1 Definition of Pushdown Automata Example 6.2 L ww R = {ww R w (0+1) * } Palindromes over {0, 1}. A cfg P 0 1 0P0 1P1. Consider a FA with a stack(= a Pushdown automaton; PDA). q 0 : Push input symbol onto stack, and stay in q 0 (gathering mode) or Go to state q 1 (matching mode) nondeterministically( -moves) Guess that it is the center of the palrindrome, now. q 1 : If the input symbol is same as the top of the stack, then pop the top of the stack and stay in q 1.(matching mode) If the input symbol is not same as the top of the stack, then reject. If no more input symbol and empty stack in state q 1, then go to the final state q 2 and accept, else reject. 10/24/17 Kwang-Moo Choe 1

2 6.1.2 The Formal Definition of Pushdown Automata A pushdown automaton(pda) P = (Q, T,,, q 0, Z 0, F) is 1. Q is a finite set of state alphabet, 2. T is a finite set of input alphabet, 3. is a finite stack alphabet, 4. is a transition function. : Q (T { }) 2 Q *. If (p, ) (q, a, X) for q, p Q, a T { }, *. pop stack top X and push stack string * onto stack. i) (p, ) (q, a, X) pop X ii) (p, X) (q, a, X) no stack change iii) (p, YX) (q, a, X) push Y 5. q 0 Q is an initial state, 6. Z 0 * is an initial stack content, 7. F Q is a set of final states. 10/24/17 Kwang-Moo Choe 2

3 Example 6.2 P ww R = ({q 0, q 1, q 2 }, {0, 1}, {0, 1, Z 0 },, q 0, Z 0, {q 2 }) 1. X, a T: (q 0, a, X) = {(q 0, ax)}. gathering mode(push) State q 0 : If see 0, then push 0 and if see 1, then push 1; stay in q (q 0,, X) = {(q 1, X)}. goto matching mode Go to the state q 1 on input(nondeterministic): guess center of Pal. 3. a T: (q 1, a, a) = {(q 1, )}. matching mode(pop) State q 1 : match input symbol against stack top symbol and pop it. else error! (q 1, 0, 1) = (q 1, 1, 0) = (q 1, 0, Z 0 )= (q 1, 1, Z 0 ) =. 4. (q 1,, Z 0 ) = {(q 2, Z 0 )}. end of matching! Accept! L(P ww R) = L ww R but P ww R is not deterministic! Ex. 6.3 Fig. 6.2 in p 230. q 0 0, X/0X 1, X/1X X={0,1,Z 0 }, X/X q 1 0, 0/ 1, 1/, Z 0 /Z 0 q 2 10/24/17 Kwang-Moo Choe 3

4 Configuration(Instantaneous description) of PDA (current state, remained input string, current stack contents) (q, x, ) Q T * *. P (Q T * * ) (Q T * * ) (q, ax X ) P (p, x, ), if (p, ) (q, a, X) (q, x, X ) P (p, x, ), if (p, ) (q,, X) We may use instead of P if P is understood. is a binary relation on (Q T * * ). * is a reflexive and transitive closure of. Recursive definition of *. I, J, K Q T * *, I B * I. If I J and J * K, I R * K. 10/24/17 Kwang-Moo Choe 4

5 Let P = (Q, T,,, q 0, Z 0, F) is a PDA. Then Thm. 6.5 If (q, x, ) * (p, y, ) for q, p Q, x, y T * and, *. Then (q, xw, ) * (p, yw, ) for any w T * and *. Adding lookahead input strings and lookback stack strings. Thm. 6.6 If (q, xw, ) * (p, yw, ) for q,p Q, x,y,w T * and, *. Then (q, x, ) * (p, y, ). Remooving lookahead input strings. 6.2 The language of a PDA Acceptance by Final State L(P) = {w T * (q 0, w, Z 0 ) * (f,, ), f F} Acceptance by Null Stack N(P) = {w T * (q 0, w, Z 0 ) * (f,, )} 10/24/17 Kwang-Moo Choe 5

6 6.2.3 From Empty Stack to Final State Thm. 6.9 If L = N(P N ) for some PDA P N = (Q N, T, N, N, q N 0, Z N 0, ). Then there is a PDA P F such that L = L(P F ). P F = (Q N {q 0 F, q F }, T, N {Z 0 F }, F, q 0 F, Z 0 F, {q F }) where q 0 F,q F Q N, Z 0 F N. F : 1. F (q F 0,, Z F 0 ) = {(q N 0, Z N 0 Z F 0 )}. push old stack bottom Z N F N, simulate P N with N. 3. q Q N, F (q,, Z 0 F ) = {(q F, (or Z 0 F or any N * ))}. If stack is empty(z 0 F : stack top), go to the final final state q F. See Fig. 6.4 in p p 0 = q F 0, X 0 = Z F 0, Z 0 = Z N 0, q 0 / = q N 0 /any. 10/24/17 Kwang-Moo Choe 6

7 6.2.4 From Final State to Empty Stack Thm If L = L(P F ) for some PDA P F = (Q F, T, F, F, q F 0, Z F 0, F). Then there is a PDA P N such that L = N(P N ). P N = (Q F {q 0 N, q E }, T, N, N, q 0 E, Z 0 N, ) where q 0 E, q E Q F, Z 0 N F. and N = F {Z 0 N }. N : 1. N (q E 0,, Z N 0 ) = {(q 0, Z F 0 Z N 0 )} push old stack bottom Z F N F, simulate P F with F. 3. f F, Z N, N (f,, Z) {(q E, )}. If final state, pop a stack symbol and go to the empty state q E. 4. Z N, N (q E,, Z) = {(q E, )} Pop all of the stack symbols in the empty state q E. See Fig. 6.7 in p. 240 p 0 = q 0 E, X 0 = Z 0 N, Z 0 = Z 0 F, q 0 = q 0 F, and... any/ = Z/ in the state p = q E. 10/24/17 Kwang-Moo Choe 7

8 6.3 Equivalence of PDA s and CFG s From Grammars to Pushdown Automata Theorem 6.13 If G = (N, T, P, S) is a cfg. Then PDA P. L(G) = N(P). Construct P = ({q}, T, N T,, q, S, ) guess and verify parser A N, (q,, A) = {(q, ) A P} guess A as (A P). a, (q, a, a) = {(q, a) a T} verify a. Proof A * lm x if and only if (q, x, A) * (q,, ), x *, (N ) *. (If) If (q, x, A) i (q,, ), then A lm * x for i 0. basis i = 0: x =, and A =. (q,, A) 0 (q,, A). A * lm A. induction Let i 1, and consider the next-to-last step. i) (q, x, A) i-1 (q,, B ) B guess (q,, ) = (q,, ) A lm * xb by IH and B P by construction of guess. A lm * xb lm x = x. 10/24/17 Kwang-Moo Choe 8

9 ii) (q, x, A) = (q, ya, A) i-1 (q, a, a ) verify (q,, ) (q, y, A) i-1 (q,, a ) A * lm ya x by IH (Thm 6.5; (q,, a )) (Only if) If A lm i x, then (q, x, A) * (q,, ) for i 0. basis i = 0, A lm 0 A. x = and A =. (q,, A) 0 (q,, A). induction Let i 1, and consider the next-to-last step. A lm i-1 yb lm y = yy = xa where =y, y *, (N ) *. (q, y, A) * (q,, B ) by IH. (q, yy, A) * (q, y, B ) (by T.6.5) B G (q, y, ) = (q, y, y ) y V (q,, ) = (q,, ) A lm * x if and only if (q, x, A) * (q,, ). If A = S, =, then S lm * x if and only if (q, x, S) * (q,, ) L(G) = N(P). 10/24/17 Kwang-Moo Choe 9

10 6.3.2 From PDA s to Grammars Theorem 6.14 If a PDA P = (Q, T,,, q 0, Z 0, ). Then there is a CFG G such that L(G) = N(P). null stack proof G = (Q Q {S}, T, P, S) P = {S [q 0 Z 0 q] q q} use [q 0 Z 0 q] instead of (q 0, Z 0, q) {[qap m ] a [py 1 p 1 ][p 1 Y 2 p 2 ] [p m-1 Y m p m ] (p, Y 1 Y 2 Y m ) (q, a, A), a T { }, q Q, 1 i m: p i Q, Y i } (if m = 0, [qap] a P, a T { }) To proove that [qap] lm * x T *, if and only if PDA Configurations(ID s) (q, x, A) * (p,, ) Q T * *. Nonterminal [qap] derives terminal string x if and only if x causes PDA P to pop A from stack starting in the state q and ending in the state p. 10/24/17 Kwang-Moo Choe 10

11 1) If (q, x, A) i (p,, ), then [qap] * lm x for i 1. basis i = 1, (q, x, A) (p,, ). (p, ) (q, x, A), x { }. [qap] x P where x T { }. [qap] * lm x. induction (q, x, A) = (q, ay, A) (p 1, y, Y 1 Y m ) i-1 (p,, ). p 2,, p m, p q and assume y = y 1 y m T *. (p 1,y 1 y m,y 1 Y m ) * (p 2,y 2 y m,y 2 Y m ) * (p m,y m,y m ) (p,, ). 1 i m, (p i, y i, Y i ) * (p i+1,, ). (Thm 6.5 and y i depends on Y i only) [p i Y i p i+1 ] lm * y i by IH. [p 1 Y 1 p 2 ][p 2 Y 2 p 3 ] [p m Y m p] lm * y 1 y 2 y m = y Since (p 1, Y 1 Y m ) (q, a, A), (q, ay, A) (p 1, y, Y 1 Y m ). [qap] a [p 1 Y 1 p 2 ][p 2 Y 2 p 3 ] [p m Y m p] P.( p i Q) [qap] lm a [p 1 Y 1 p 2 ][p 2 Y 2 p 3 ] [p m Y m p] lm * ay = x. 10/24/17 Kwang-Moo Choe 11

12 2) If [qap] i lm x, then (q, x, A) * (p,, ) for i 1. basis i = 1, [qap] x P, (p, ) (q, x, A) where x T { }. induction [qap] lm a [p 1 Y 1 p 2 ][p 2 Y 2 p 3 ] [p m Y m p] i-1 lm x *. x = ay 1 y m where 1 i m, [p i Y i p i+1 ] lm i y i where p m+1 =p. (p i, y i, Y i ) * (p i+1,, ) by IH. Since [qap] a [p 1 Y 1 p 2 ][p 2 Y 2 p 3 ] [p m Y mp] P, a [p 1 Y 1 p 2 ][p 2 Y 2 p 3 ] [p m Y m p] (q, a, A) where a T { }. (q, ay 1 y m, A) (p 1, y 1 y m, Y 1 Y m ) * * ( p,, ). [qap] lm * x, q, p q if and only if (q, x, A) * (p,, ). [q 0 Z 0 p] lm * x, q, p q if and only if (q 0, x, Z 0 ) * (p,, ). S lm [q 0 Z 0 p] lm * x, q, p q if and only if (q 0, x, Z 0 ) * (p,, ). 10/24/17 Kwang-Moo Choe 12

13 6.4 Deterministic Pushdown Automata Definition of a Deterministic PDA A PDA P = (Q, T,,, q 0, Z 0, F) with : Q (T { }) 2 Q * is deterministic, if 1. q Q, a T { }: (q, a, X) 1 or : q (T { }) \ Q *. 2. If (q, a, X), then (q,, X) or If (q,, X), then a : (q, a, X). Example 6.16 Palindromes over {0, 1} with center marker c. L wcw R = {wcw R w (0+1) * } P wcw R = ({q 0, q 1, q 2 }, {0, 1, c}, {0, 1, Z 0 },, q 0, Z 0, {q 2 }) L wcw R = L(P wcw R). 10/24/17 Kwang-Moo Choe 13

14 P wcw R = (Q, T,,, q 0, Z 0, F) is a deterministic PDA! 1. X, a {0, 1}: (q 0, a, X) = {(q 0, ax)}. gathering mode(push) 2. (q 0, c, X) = {(q 1, X)}. goto matching mode 3. a {0, 1}: (q 1, a, a) = {(q 1, )}. matching mode(pop) 4. (q 1,, Z 0 ) = {(q 2, )}. end of matching! q 0 0, X/0X 1, X/1X X={0,1,Z 0 } c, X/X q 1 0, 0/ 1, 1/ Stronger version of deterministic PDA : Q T \ Q *, Z 0 /Z 0 q Regular Languages and Deterministic PDA 10/24/17 Kwang-Moo Choe 14

15 Theorem 6.17 If L is regular, the L = L(P) for some PDA. proof Let A = (Q, T, A, q 0, F) is a DFA L = L(A). Then P = (Q, T, {Z 0 }, P, q 0, Z 0, F) with P (q, a, Z 0 ) = {(p, Z 0 ) A (q, a) = p} is a deterministic PDA. DFA is a DPDA and FA is a PDA! DPDA s and Context-free Languages P wcw R is a DPDA. But L wcw R is not regular but context-free. L wcw R = L(P wcw R). L ww R is not regular but context-free. But there is no DPDA P L ww R = L(P) Regular Languages L(DPDA) Context-free Languages DPDA s and Ambiguous Grammars 10/24/17 Kwang-Moo Choe 15

16 Theorem 6.20 If L = L(P) for some DPDA A, Then L has an unambiguous context-free grammar. proof If a PDA P is DPDA, then the CFG in Ex is unambiguous. Theorem 6.21 If L = L(P) for some DPDA A, Then L has an ambiguous grammar. proof end marker($) vs. But not important! ( {$} vs { }) 10/24/17 Kwang-Moo Choe 16

17 Context-free Languages = Context-free grammars = Pushdown Automata Regular Languages = Regular Grammars = Finite Automata = Regular Expressions Automata PDA FA m-dfa Languages context-free regular Expressions regular Grammars context-free CNF regular 10/24/17 Kwang-Moo Choe 17