Cs3102: Theory of Computation Class 6: Pushdown Automata Spring 2010 University of Virginia David Evans TexPoint fonts used in EMF. Read the TexPoint manual.

cs3102: Theory of Computation

Class 6: Pushdown Automata

Spring 2010University of VirginiaDavid Evans

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• Revisiting and Reversing the Pumping Lemma• Recognizing Non-Regular Languages• Pushdown Automata

Revisiting the Pumping Lemma

q0

qz

x

y

z

If input string is longer than |Q|, some state must repeat.

qi

If A is a regular language, then there is some number p (the pumping length) where for any string sA and |s| p, s may be divided into three pieces, s = xyz, such that |y|> 0, |xy| p, and for any i 0, xyizA.

Pumping Game for Language A1. Player 1: picks p2. Player 2: picks sA and |s| p3. Player 1: picks x, y, z, |y|> 0, |xy| p, s = xyz4. Player 2: picks i 0

Player 1 wins if xyiz A, Player 2 wins if xyizAIf Player 1 can always win: A is regularIf Player 2 can always win: A is not regular

q0

qz

x

y

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qi

What if Player 2 picks an sA?

Pump-Priming Game for Language A1. Player 1: picks p2. Player 2: picks sA and |s| p3. Player 1: picks x, y, z, |y|> 0, |xy| p, s = xyz4. Player 2: picks i 0

Player 1 wins if xyiz A, Player 2 wins if xyiz AIf Player 1 can always win: A is regularIf Player 2 can always win: A is not regular

q0

qz

x

y

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qi

A = { w | w has more 0s than 1s }

Pump-Priming Game for Language A1. Player 1: picks p2. Player 2: picks sA and |s| p3. Player 1: picks x, y, z, |y|> 0, |xy| p, s = xyz4. Player 2: picks i 0

Player 1 wins if xyiz A, Player 2 wins if xyiz AIf Player 1 can always win: A is regularIf Player 2 can always win: A is not regular

A Complementary View

• Regular Languages are closed under complement: if A is regular, then Ā is regularProof sketch:

DFA M = (Q, , , q0, F) recognizes A.

DFA M = (Q, , , q0, Q F) recognizes Ā.

• Thus, Player 2 wins by showing either A or Ā is non-regular.

s

All Languages

RegularLanguages

Can be recognized by some DFA

Finite Languages

anbn

wwR

What do we need to add to a DFA to recognize some non-regular languages?

DFA + Counter?A = anbn

DFA + Counter?

count = 0

A = anbn

a, count = count + 1 b, count = count 1

b, count = count 1

Accept if count == 0.

DFA + Stack = Deterministic Pushdown Automaton

A = anbn

DFA + Stack = Deterministic Pushdown Automaton

a, push a b, pop a

b, pop a

Accept if stack is empty.

A = anbn

Formalizing DPDA

(Q;§ ;¡ ;±;q0;F )(Q;§ ;¡ ;±;q0;F )

Note: Sipser only defines Nondeterministic PDA and calls it PDA.

DPDA Transitions

Inputs: state, alphabet symbol or ε, popped stack symbol or εOutputs: state, stack symbol to push or ε

It is a deterministic machine: must always follow possible ε-input transition.

a, ε Aq1 q2

Deterministic Pushdown Automaton

q1 q2

a, ε +

b, + ε

A = anbn

Bottom of stack marker

q0 b, + ε

ε, ε $

q3

ε, $ ε

This looks like nondeterminism, but is not: must always take possible ε-transitions a state with an (ε, h) transitions cannot have any other (a, h) transitions; a state with a (a, ε) transition, cannot have any other (a, h) transitions

A = { w | w has more 0s than 1s }

Computing Model for DPDA

First, we need to model the stack!

$

+

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push +

$

+

+

+

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pop +

$

+

+

+

Modeling the Stack

Modeling the Stack

This would be a normalway to define a stack, but not what we use!

DPDA Computation Model

±¤ : Q £ § ¤ £ ¡ ¤² ! Q £ ¡ ¤²

±¤(q;²;°) = (q;°)

8q2 Q;8a 2 § ;x 2 § ¤;° 2 ¡ ¤ :

±¤ : Q £ § ¤ £ ¡ ¤² ! Q £ ¡ ¤²8q2 Q;8a 2 § ;x 2 § ¤;° 2 ¡ ¤;h 2 ¡ :

±(q;a;²) ! (qt; ²) ) ±¤(q;ax;°) = ±¤(qt;x;°)

±(q;a;²) ! (qt;hp) ) ±¤(q;ax;°) = ±¤(qt;x;hp°)

±(q;a;ht) ! (qt;hp) ) ±¤(q;ax;ht°) =±¤(qt;x;hp°)

±(q;a;ht) ! (qt;²) ) ±¤(q;ax;ht°) =±¤(qt;x;°)

No push or pop:

Push only:

Pop only:

Push and pop:

This rule actually covers all four cases if we make h 2 ¡ ²

Dealing with -transitionsRemember the NFA DFA proof?

DPDA Computing Model

±(q;a;ht) ! (qt;hp) ) ±¤(q;ax;ht°) =±¤(qr ;x;°r )

±¤(q;²;°) = E (q;°)

Where E is the forced-follow -transitions function defined by:

E : Q £ ¡ ¤ ! Q £ ¡ ¤

E (q;°) = (q;°)±(q;²;°) = ; :

E (q;ht°) = E (qt;hp°)

±(q;²;ht°) = (qt;hp°) :

(qr ;°r ) = E (qt;hp°)where

Acceptance: PDA accepts w when:Accepting State Model

±¤(q0;w;²) ! (qf ;s) ^qf 2 F

Empty Stack Model

±¤(q0;w;²) ! (q;²)Is the set of languages accepted by DPDAs with each model the same?

A good, original proof is worth a challenge bonus. (Finding a published proof is not.)

Power of DPDAsL(DPDA) L(DFA) ?

1. Prove there is some DPDA that recognizes every regular language.

2. Prove there is some language that can be recognized by a DPDA that cannot be recognized by any PDA.

Construct DPDA from DFA by addingempty stack transitions:

±DP DA (q;a;²) = (±D F A (q;a);²)

A = fanbn jn ¸ 0g

Henceforth: assume DPDA is Accepting-State DPDA

s

All Languages

RegularLanguages

Can be recognized by some DFA

Finite Languages

anbn

DPDA Languages(Accepting State)

Charge

• Thursday:– Non-equivalence of NPDA and DPDA– Context-Free Grammars

• Next Tuesday:– PS2 Due– Languages that cannot be recognized by NPDA