7. Pushdown Automata

8/13/2019 7. Pushdown Automata

1/26

Pushdown Automata

Wen-Guey Tzeng

Department of Computer ScienceNational Chiao Tung University

1


2/26

Nondeterministic Pushdown Automata

2


3/26

Definition of NPDA

A nondeterministic pushdown acceptor (npda)is defined by M=(Q, , , , q0, z, F), where Q: a finite set of internal states

: the input alphabet: the stack alphabet

: Q({})a subset of Q*

q0Q: the initial state z: the stack start symbol

FQ: the set of final states

3


4/26

Example

M=(Q, , , , q0, z, F) Q={q0, q1, q2, q3}

={a, b}

={0, 1}

z=0

F={q3}

(q0, a, 0)={(q1, 10), (q3, )}(q0, , 0)={(q3, )}

(q1, a, 1)={(q1, 11)}(q1, b, 1)={(q2, )}(q2, b, 0)={(q2, )}(q2, , 0)={(q3, )}

4


5/26

Transition graph

5


6/26

Instantaneous description (q, w, u)

q: the current state

w: the current un-read input

u: the symbols in the current stack

Initial id (q0, w, z)

Transition: (q, aw, bx)(p, w, yx)

Example

(q0, ab, 0)(q1, b, 10)(q2, , 0)(q3, , )

6


7/26

Language accepted by npda

The language accepted by M is

L(M)={w *: (q0, w, z)* (p, , u), pF, u *}.

7


8/26

Example

Design an npda to accept

L={w{a,b}* : na(w)=nb(w)}

Idea: the stack holds extra as (or bs) up to

now

8


9/26

9


10/26

Example


L={w{a,b}* : na(w) nb(w)}

10


11/26

Example


L={wcwR: w{a,b}+}

Idea

The stack holds w first by pushing them in.

After seeing c, M switches to the matching

phase.

The stack matches wRwith the w in the stack.

11


12/26

Q={q0, q1, q2},

={a, b},

={a, b, z}, F={q2} Pushing w into the stack:

(q0, a, a)={ (q0, aa)}

(q0, b, a)={ (q0, ba)}

(q0, a, b)={ (q0, ab)}

(q0, b, b)={ (q0, bb)}

(q0

, a, z)={ (q0

, az)}

(q0, b, z)={ (q0, bz)}

12


13/26

After seeing c, switch to q1for matching

(q0, c, a)={ (q1, a)}

(q0, c, b)={ (q1, b)}

Matching w

R

with w in the stack(q1, a, a)={ (q1, )}

(q1, b, b)={ (q1, )}

In the end(q1, , z)={ (q2, z)}

13


14/26

Example


L={wwR: w{a,b}+}

Idea

The stack holds w first by pushing them in.

M switches to the matching phase.

The stack matches wRwith the w in the stack.

Difficulty: How does M know the end of w?

14


15/26

Q={q0, q

1, q

2}, ={a, b}, ={a, b, z}, F={q

2}

Pushing w into the stack:

(q0, a, a)={ (q0, aa)}

(q0, b, a)={ (q0, ba)}

(q0, a, b)={ (q0, ab)}

(q0, b, b)={ (q0, bb)}

(q0

, a, z)={ (q0

, az)}

(q0, b, z)={ (q0, bz)}

15


16/26

Switching from q0to q

1for matching

(nondeterministic)

(q0, , a)={ (q1, a)}

(q0, , b)={ (q1, b)}

Matching wRwith w in the stack

(q1, a, a)={ (q1, )}

(q1, b, b)={ (q1, )}

In the end

(q1, , z)={ (q2, z)}

16


17/26

Run M on the input abba:

17


18/26

Equivalence of ndpa and cfg

For any npda M, there is a cfg G such that L(G)=L(M)

For any cfg G, there is an npda M such that L(M)=L(G)

18


19/26

Convert cfg to pda

Example SaSbb|a

CFG in Greibach normal form

Productions: SaSA|a, AbB, Bb

Construct M Idea: use the stack to hold the derivation process, and match the input with

the derivation

Initial: (q0, , z)={ (q1, Sz) }

Process the input: (q1, a, S)={ (q1, SA), (q1, )}

(q1, b, A)={ (q1, B) } (q1, b, B)={ (q1, )}

In the end: (q1, , z)={ (q2, )}

19


20/26

Run M on the input aaabbbb

See the relation between *(q0, aaabbbb, z) and

S* aaabbbb

20


21/26

Convert pda to cfg

Idea: the grammar simulates the move of pda

Assumptions for an npda Only one final state qf

All transitions must have form, a{},

(qi, a, A)={ c1, c2, , cn}, where ci=(qj, ) or ci=(qi, BC)

Simulation each variable is of form (qiAqj) -- generating string w

Starting at state qi

Top stack symbol is A Read in string w

Ending at state qj

Stack symbol is popped out.

21


22/26

The construction

Start variable: S=(q0zqf)

For transitions a{},

(qi, a, A)={ c1, c2, , cn}, where ci=(qj, ) or ci=(qi, BC)

If ci=(qj, ), add production (qiAqj)a If ci=(qj, BC), add production (qiAqk)a(qjBql)(qlCqk),

for all qk, qlQ

22


23/26

Example

(q0, a, z)={ (q0, Az) }

(q0, a, A)={ (q0, A) }

(q0, b, A)={ (q1, ) }

(q1, , z)={ (q2, ) } Converted to satisfy the requirements

(q0, a, z)={ (q0, Az) }

(q3, , z)={ (q0, Az) }

(q0, a, A)={ (q3, ) }

(q0, b, A)={ (q1, ) }

(q1, , z)={ (q2, ) }

23


24/26

Work

24


25/26

The final result

25


26/26

Comments on dpda and dcfl

For efficient parsing, we need deterministic

cfl.

The corresponding deterministic pda

(q, a, b) contains one element

If (q, , b) is not empty, (q0, c, b) is empty for

c

26

Documents

7. Pushdown Automata