1 pushdown automata there are context-free languages that are not regular. finite automata cannot...

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Pushdown Automata

• There are context-free languages that are not regular.

• Finite automata cannot recognize all context-free languages.

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Pushdown Automata• {a, b}* is regular.

• {akbk | k is a constant} is regular.

• {anbn | n 0} is not regular.

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Finite Automata

Control unitq0

Input file

yes/no

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Pushdown Automata

Control unitq0

Input file

yes/no

Stack

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Non-deterministic Pushdown Automata (NPDA)

M = (Q, , , , q0, z, F)Q: finite set of internal states: finite set of symbols - input alphabet: finite set of symbols - stack alphabet: Q ({}) finite subsets of Q * transition functionq0 Q: initial statez : stack start symbolF Q: set of final states

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Non-deterministic Pushdown Automata (NPDA)

: Q ({}) finite subsets of Q *

stack top

stack top replacement

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Example(q1, a, b) = {(q2, cd), (q3, )}

b

dc

q1

q2

q3

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ExampleM = (Q, , , , q0, z, F)

Q = {q0, q1, q2, q3} (q0, a, 0) = {(q1, 10), (q3, )} = {a, b} (q0, , 0) = {(q3, )} = {0, 1} (q1, a, 1) = {(q1, 11)}z = 0 (q1, b, 1) = {(q2, )}F = {q3} (q2, b, 1) = {(q2, )}(q2, , 0) = {(q3, )}

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Instantaneous Description

(q, w, u)

current state unread part of stack contents

input string

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move: (q1, aw, bx) (q2, w, yx)

iff (q2, y) (q1, a, b)

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(q1, x, y) (q2, u, v)

(q1, x, y) (q2, u, v)

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Language accepted by NPDA

Let M = (Q, , , , q0, z, F) be an NPDA.

L(M) = {w * | (q0, w, z) (qf, , u), qf F, u

*}

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ExampleL = {w {a, b}* | na(w) = nb(w)}

M = (Q, , , , q0, z, F) ?

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ExampleL = {w {a, b}* | na(w) = nb(w)}

M = (Q, , , , q0, z, F)Q = {q0, qf} (q0, , z) = {(qf, z)} = {a, b} (q0, a, z) = {(q0, 0z)} = {0, 1, z} (q0, b, z) = {(q0, 1z)}F = {qf} (q0, a, 0) = {(q0, 00)}(q0, b, 0) = {(q0, )}(q0, a, 1) = {(q0, )}(q0, b, 1) = {(q0, 11)}

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ExampleL = {wwR | w {a, b}+}

M = (Q, , , , q0, z, F) ?

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ExampleL = {wwR | w {a, b}+}

M = (Q, , , , q0, z, F)Q = {q0, q1, q2} = {a, b} = {a, b, z} F = {q2} (q0, a, a) = {(q0, aa)} (q0, , a) = {(q1, a)}(q0, b, a) = {(q0, ba)} (q0, , b) = {(q1, b)}(q0, a, b) = {(q0, ab)}(q0, b, b) = {(q0, bb)} (q1, a, a) = {(q1, )}(q0, a, z) = {(q0, az)} (q1, b, b) = {(q1, )}(q0, b, z) = {(q0, bz)} (q1, , z) = {(q2, z)}

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NPDA and Context-Free Languages

Greibach NF

(q0, , z) = {(q1, Sz)}S aSA | a (q1, a, S) = {(q1, SA), (q0, )}A bB (q1, b, A) = {(q1, B)}B b (q1, b, B) = {(q1, )}(q1, , z) = {(q2, )}

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TheoremFor any context-free language L not containing , there exists an NPDA M such that L = L(M).

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TheoremProof: G = (V, T, S, P)M = ({q0, q1, qf}, T, V{z}, , q0, z, {qf}) z V

(q0, , z) = {(q1, Sz)}(q1, u) (q1, a, A) iff A au P(q1, , z) = {(qf, z)}

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Example Greibach NF

(q0, , z) = {(q1, Sz)}S aA (q1, a, S) = {(q1, A)}A aABC | bB | a (q1, a, A) = {(q1, ABC), (q1, )}(q1, b, A) = {(q1, B)}B b (q1, b, B) = {(q1, )}C c (q1, c, C) = {(q1, )}(q1, , z) = {(q2, )}

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Context-Free Grammars for NPDA

M = (Q, , , , q0, z, F)

G = (V, T, S, P)

L(G) = L(M)

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M = (Q, , , , q0, z, F)

M^: Single final state Final state entered iff the stack is empty (qi, a, A) = {c1, c2, ..., cn}

ci = (qj, ) ci = (qj, BC)

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(qi, a, A) = {(qj, ), ... } ?

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(qi, a, A) = {(qj, ), ... } ?

At qi erase A and move to qj if receiving a

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(qi, a, A) = {(qj, BC), ... } ?

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(qi, a, A) = {(qj, BC), ... } ?

At qi erase A and move to qk if receiving a and

at qj erase BC and move to qk

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(qi, a, A) = {(qj, BC), ... } ?

At qi erase A and move to qk if receiving a andat qj erase BC and move to qk

At qi erase A and move to qk if receiving a andat qj erase B and move to qm and

at qm erase C and move to qk

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(qi, a, A) = {(qj, ), ... }

At qi erase A and move to qj if receiving a

(qiAqj) a

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(qi, a, A) = {(qj, BC), ... }

At qi erase A and move to qk if receiving a andat qj erase B and move to qm and

at qm erase C and move to qk

(qiAqk) a(qjBqm)(qmCqk)

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Start symbol: (q0zqf)

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Q = {q0, q1, q2} = {a} = {A, z} F = {q2}

(q0, a, z) = {(q0, Az)}(q0, a, A) = {(q0, A)}(q0, b, A) = {(q1, )}(q1, , z) = {(q2, )}

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Q = {q0, q1, q2 , q3} = {a} = {A, z} F = {q2}

(q0, a, z) = {(q0, Az)} (q0, a, z) = {(q0, Az)}(q0, a, A) = {(q0, A)} (q0, a, A) = {(q3, )}(q0, b, A) = {(q1, )} (q3, , z) = {(q3, Az)}(q1, , z) = {(q2, )} (q0, b, A) = {(q1, )} (q1, , z) = {(q2, )}

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Q = {q0, q1, q2 , q3} = {a} = {A, z} F = {q2}

(q0, a, z) = {(q0, Az)}(q0, a, A) = {(q3, )} (q0Aq3) a(q3, , z) = {(q3, Az)}(q0, b, A) = {(q1, )} (q0Aq1) b(q1, , z) = {(q2, )} (q1zq2)

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Q = {q0, q1, q2 , q3} = {a} = {A, z} F = {q2}

(q0, a, z) = {(q0, Az)} (q0zq0) ...(q0, a, A) = {(q3, )} (q0zq1) ...(q3, , z) = {(q3, Az)} (q0zq2) ...(q0, b, A) = {(q1, )} (q0zq3) ...(q1, , z) = {(q2, )}

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Q = {q0, q1, q2 , q3} = {a} = {A, z} F = {q2}

(q0, a, z) = {(q0, Az)} (q0zq0) a(q0Aq0)(q0zq0) |(q0, a, A) = {(q3, )} a(q0Aq1)(q1zq0) |(q3, , z) = {(q3, Az)} a(q0Aq2)(q2zq0)(q0, b, A) = {(q1, )} a(q0Aq3)(q3zq0)(q1, , z) = {(q2, )} ...

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TheoremIf L = L(M) for some NPDA M, then L is a context-free language.

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TheoremProof: M = (Q, , , , q0, z, {qf}) G = (V, T, S, P)

T = V = {(qiAqj) | A } S = (q0zqf)P:(qiAqj) a iff (qj, ) (qi, a, A)(qiAqm) a(qiBqm)(qmCqj) iff (qj, BC) (qi, a, A)

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Deterministic Pushdown Automata

A DPDA is a pushdown automaton that never has a

choice in its move:1. (q, a, b) contains at most one element.2. if (q, , b) is not empty, then (q, a, b) must be empty

for every a .

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Deterministic Context-Free Language

A language L is said to be a DCFL iff there exists a

DPDA M such that L = L(M).

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Q = {q0, q1, q2} (q0, a, 0) = {(q1, 10)} = {a, b} (q1, a, 1) = {(q1, 11)} = {0, 1} (q1, b, 1) = {(q2, )}z = 0 (q2, b, 1) = {(q2, )}F = {q0} (q2, , 0) = {(q0, )}

L = {anbn | n 0}

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Homework• Exercises: 5, 10, 13 of Section 7.1 - Linz’s book.

• Exercises: 1, 2, 4, 5, 8, 12 of Section 7.2 - Linz’s book.

• Exercises: 1, 2, 3, 6, 7 of Section 7.2 - Linz’s book.

• Presentations: Section 6.3 and Section 7.4.

1 pushdown automata there are context-free languages that are not regular. finite automata cannot...

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