Pushdown Automata

1
Pushdown Automata

• Juan Carlos Guzmán
• CS 6413 Theory of Computation
• Southern Polytechnic State University

2
Summary

• Definitions
• Languages recognized
• Equivalence to Context-Free Grammars
• Deterministic PDAs

3
Pushdown Automaton

• A pushdown automaton is
• e-NFA
• States
• Transitions
• Tape
• Read-only
• Read-once
• Finite
• Plus a stack
• LIFO

4
Pushdown Automaton

• A pushdown automaton P is a seven-element tuple
• P (Q,S,G,d,q0,Z0,F )
• where
• Q is the set of states
• S is the alphabet of the input tape
• G is the alphabet of the stack
• d is the transition function
• q0 is the initial state
• Z0 is the initial stack symbol
• F is the set of final states

5
Pushdown Automaton

• L 0n1n
• P (q0,q1,q2,0,1,Z0,0,1,d,q0,Z0,q2),
  where
• d(q0,0,Z0) (q0,0Z0)
• d(q0,0,0) (q0,00)
• d(q0,1,0) (q1,e)
• d(q0,e,Z0) (q2,e)
• d(q1,1,0) (q1,e)
• d(q1,e,Z0) (q2,e)

6
Instantaneous Descriptions

• The instantaneous description of a PDA is the
  triple (q, w, ?), where
• q is a state
• w is the remaining input
• ? is the stack contents

7
Movement Relation

• Establishes the relation between different
  instantaneous descriptions
• (q,aw,X?) (p,w,??)
• if (p,?) ? d(q,a,X )

8
Languages of a PDA

• Remember that languages are sets of strings that
  satisfy a property
• For a phrase, the automaton, started in its
  initial state, lands in a final state with the
  input tape completely consumed
• Acceptance by final state
• L(P ) w (q0,w,Z0) (q,e,?), q?F
• For a phrase, the automaton, started in its
  initial state, consumes all stack symbols as the
  input tape completely consumed
• Acceptance by empty stack
• N(P ) w (q0,w,Z0) (q,e,e)

9
L(P ) vs. N(P ) vs. CFL

• For a particular PDA P, L(P) and N(P) may be
  different
• Take the families of languages implied by L(P)
  and N(P) over all PDAs P
• Do they refer to the same family?, or
• Is there a language in one of these families and
  not in the other?
• Is one of them more powerful than the other?
• How do these families relate to the Context-Free
  Languages?

10
Equivalences

• These families of languages are equal
• Context-Free Languages
• Languages accepted by final state of a PDA
• Languages accepted by empty stack of a PDA

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From Acceptance by Empty Stack to Final State

• For any L, if there exists PN such that LN(PN),
  then there is PF such that LL(PF)
• If a language is accepted by empty stack by a
  PDA, then construct a new PDA with an augmented
  stack
• A new initial stack symbol (X0)
• A new initial state (p0)
• A new final state (pf)
• Augment the transitions as follows
• The initial state should insert Z0 (old initial
  stack symbol into the stack)
• All old states should have a transition to move
  to the new final state if X0 is visible

12
From Acceptance by Empty Stack to Final State

• PN (Q,S,G,d,q0,Z0,?)
• PF (Q ?p0,pf,S,G?X0,d,p0,X0,pf), where
• d(p0,e,X0) (q0,Z0X0)
• d(q,e,X0) (pf , e), for q ?Q
• d(q,a,X) d(q,a,X), otherwise

13
From Acceptance by Final State to Empty Stack

• For any L, if there exists PF such that LL(PF),
  then there is PN such that LN(PN)

14
From Acceptance by Final State to Empty Stack

• PF (Q,S,G,d,q0,Z0,F)
• PF (Q ?p0,p,S,G?X0,d,p0,X0,?), where
• d(p0,e,X0) (q0,Z0X0)
• d(q,e,X) (p,e), for all q ?F ?p, all X ?
  GUX0
• d(q,a,X) d(q,a,X), otherwise

15
From Grammars to PDAs

• Given G, construct a PDA P that simulates
  leftmost derivations of G
• A sentential form can be characterized as xA?,
  where A is the leftmost variable
• We will call A? as its tail
• We want to make sure that
• S ? xA? ? xy, iff
• (q,xy,S) (q,y,A?) (q,e,e)

16
From Grammars to PDA

• Let G (V,T,Q,S )
• P (q,T,V ?T, d,q,S ), where
• d(q,e,A) (q,ß) A ? ß ? Q (derive)
• d(q,a,a) (q,e) (match)
• Note that there is not a one-to-one
  correspondence between the leftmost derivation in
  the grammar and the movements of the automaton
• Nice automaton but, does it recognize the same
  language as the grammar? Prove by induction (on
  the number of derivations taken to obtain the
  phrase

17
Example

• Let G (A,C,0,1,R,A), where
• R A ? 0 C e,
• C ? 0 C 1 1
• The PDA is P (q,0,1,0,1,A,C,d,q,A), where
• d(q,e,A) (q,0C ),(q,e)
• d(q,e,C) (q,0C 1),(q,1)
• d(q,0,0) (q,e)
• d(q,1,1) (q,e)

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Grammar vs. PDA

• (q,000111,A)
• (q,000111,0C)
• (q,00111,C)
• (q,00111,0C 1)
• (q,0111,C 1)
• (q,0111,0C 11)
• (q,111,C 11)
• (q,111,111)
• (q,11,11)
• (q,1,1)
• (q,e,e)

A
0
C
C
0
1
C
0
1
1
19
Correctness

• L(G) L(P)
• Vw, S gtw iff (q,w,S) -(q,eps,eps)
• By induction on the number of derivations

20
From PDAs to Grammars

• This is a more complex construction, since we
  have to deal with different states

21
From PDAs to Grammars

• Let P (Q, S,G,d,q0,Z0) be a PDA
• Let G (V, S,R,S ), such that
• S?V
• For all p,q ? Q, X ? G, pXq ? V
• For all p ? Q, S?q0Z0p ? R
• If (r,Y1Y2Yk) in d(q,a,X)
• then, for all states ri
• qXrk ? arY1r1r1Y2r2rk-1Ykrk

22
Example

• Recall sample automaton

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Grammar Variables

• S

• q01q0
• q01q1
• q01q2
• q11q0
• q11q1
• q11q2
• q21q0
• q21q1
• q21q2

• q00q0
• q00q1
• q00q2
• q10q0
• q10q1
• q10q2
• q20q0
• q20q1
• q20q2

• q0Z0q0
• q0Z0q1
• q0Z0q2
• q1Z0q0
• q1Z0q1
• q1Z0q2
• q2Z0q0
• q2Z0q1
• q2Z0q2

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Grammar Productions

• S
• S ? q0Z0q0
• S ? q0Z0q1
• S ? q0Z0q2
• d(q0,0,Z0) (q0,0Z0)
• q0Z0q0 ? 0q00q0q0Z0q0
• q0Z0q0 ? 0q00q1q1Z0q0
• q0Z0q0 ? 0q00q2q2Z0q0
• q0Z0q1 ? 0q00q0q0Z0q1
• q0Z0q1 ? 0q00q1q1Z0q1
• q0Z0q1 ? 0q00q2q2Z0q1
• q0Z0q2 ? 0q00q0q0Z0q2
• q0Z0q2 ? 0q00q1q1Z0q2
• q0Z0q2 ? 0q00q2q2Z0q2

• d(q0,0,0) (q0,00)
• q00q0 ? 0q00q0q00q0
• q00q0 ? 0q00q1q10q0
• q00q0 ? 0q00q2q20q0
• q00q1 ? 0q00q0q00q1
• q00q1 ? 0q00q1q10q1
• q00q1 ? 0q00q2q20q1
• q00q2 ? 0q00q0q00q2
• q00q2 ? 0q00q1q10q2
• q00q2 ? 0q00q2q20q2

25
Grammar Productions

• d(q0,1,0) (q1,e)
• q00q1 ? 1
• d(q0,e,Z0) (q2,e)
• q0Z0q2 ? e
• d(q1,1,0) (q1,e)
• q10q1 ? 1
• d(q1,e,Z0) (q2,e)
• q1Z0q2 ? e

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Grammar Notes

• The generated grammar has
• 28 variables
• 25 productions
• However, most of them are useless
• The highlighted variables and productions are not
  useless

• Variables
• q0Z0q2
• q1Z0q2
• q00q1
• q10q1
• Productions
• S?q0Z0q2
• q0Z0q2 ? 0q00q1q1Z0q2
• q00q1 ? 0q00q1q10q1
• q00q1 ? 1
• q0Z0q2 ? e
• q10q1 ? 1
• q1Z0q2 ? e

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PDA vs. Grammar
S

• (q0,000111,Z0)
• (q0,00111,0Z0)
• (q0,0111,00Z0)
• (q0,111,000Z0)
• (q1,11,00Z0)
• (q1,1,0Z0)
• (q1,e,Z0)
• (q2,e,e)

q0Z0q2
0
q00q1
q1Z0q2
q00q1
0
e
q10q1
q00q1
0
q10q1
1
1
1
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PDA vs. Grammar
8 1
S

• (q0,000111,Z0)
• (q0,00111,0Z0)
• (q0,0111,00Z0)
• (q0,111,000Z0)
• (q1,11,00Z0)
• (q1,1,0Z0)
• (q1,e,Z0)
• (q2,e,e)

8 1
q0Z0q2
7 2
8 7
0
q00q1
q1Z0q2
6 3
7 6
q00q1
0
e
q10q1
5 4
6 5
q00q1
0
q10q1
1
1
1
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Grammar Notes
S

• Lets rename the variables
• q0Z0q2 ? A
• q1Z0q2 ? B
• q00q1 ? C
• q10q1 ? D

• Productions
• S ? A
• A ? 0 C B e
• C ? 0 C D 1
• D ? 1
• B ? e
• Equivalent Grammar
• A ? 0 C e
• C ? 0 C 1 1

A
0
C
B
C
0
e
D
C
0
D
1
1
1
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Deterministic PDA

• There is never a choice of movement in any
  situation
• d(q,a,X) does not contain more than one entry
• If some d(q,a,X) is nonempty, then d(q,e,X) must
  be empty

31
Deterministic PDA

• Recognize some CFLs
• Recognize some languages that are not regular
• Do not recognize all CFLs

PDAs
DPDAs
FAs