Transcript No Slide Title

Pushdown Automata
CPSC 388
Ellen Walker
Hiram College
PDA = FA + Stack
input tape
control mechanism
state indicator
7
6
1
2
5 4 3
stack
A PDA Transaction
•
•
•
•
Read the next symbol from the input
Pop a symbol from the stack
Push a symbol on the stack
Enter a new state
Epsilon in PDA
• Read epsilon
– No character is read, input location stays
the same
• Push or pop epsilon
– The stack remains unchanged
Format of a Transition
• Items before ; are “input”, items after ;
are “action”
( Current state,
Input symbol,
Symbol popped ;
New state,
Symbol pushed )
PDA for anbn
• Assume initial stack contains “Z”
(delimiter for bottom of stack)
• S1, a,  ; S2, a - leave Z on bottom
• S2, a, ; S2, a - push a for each a
• S2, b, a ; S3,  - pop a for each b
• S3, b, a ; S3, 
• S3, , Z ; S4, Z
- Final state if Z
reached
PDA Diagram
• For each state: item read, popped,
pushed
• State transitions using arrows as for FA
• To check the item on top of the stack,
then push an additional item, JFLAP
allows a double-push
(a, Z, aZ)
PDA Diagram for anbn
This PDA requires n≥1. (How can we make it accept n=0?)
Formal Definition of PDA
• A PDA is of the form (S, , , , , F) where
– S is a finite collection of states
  is the machine's alphabet
  is the finite collection of stack symbols
  is the finite collection of transitions of the form
(p,s,x; q,y)

(an element of S) is the initial state
– F (a subset of S) is the set of accept states
Formal Definition of Example
• States: {q0, q1, q2, q3}
• Alphabet: {a, b}
• Stack symbols: {Z, a}
• Transitions (see slide 6)
• Initial state: q2
• Accept states: {q3}
All Strings with Equal # a’s and
b’s
• General idea:
– If you see an a
• If there is a b on the stack, pop it
• Otherwise, push an a
– If you see a b
• If there is an a on the stack, pop it
• Otherwise, push a b
– At the end of the string, if Z is on top, then
accept
Build a PDA for each:
• All strings of the form ambn, n>m
• All strings consisting of a’s and b’s with
at least 3 more a’s than b’s (Hint: use
epsilon-transitions)
• All palindromes over the alphabet
{a,b,c} (Hint: it needs to be
nondeterministic)
Every RL has a PDA to accept it
• Given the RL, construct an FA for it.
• Make a PDA with the same FA that
reads the input but ignores the stack
– For each FA transition (state1, char; state2)
create a pda transition (state1, char, ;
state2, )
PDA’s are more powerful than
FA’s
• Every FA has an equivalent PDA
– Proof by construction on previous slide
• At least one PDA does not have an
equivalent FA
– Proof by example: anbn
• Therefore the class of languages
accepted by PDA’s is a superset of the
class of languages accepted by FA’s
Deterministic vs.
Nondeterministic PDA’s
• Deterministic PDA’s
– Every state/stack combination has exactly one
transition
– No epsilon-transitions (epsilon for both input and
stack symbol)
• Nondeterministic PDA’s are more powerful
than deterministic PDA’s
– Example: strings from a*b* that have either twice
as many a’s as b’s or 3x as many a’s as b’s
Not All Languages are Context
Free
• There is a pumping lemma for Context
free languages
• Similar to RL pumping lemma, but you
have to pump in two places
• Non CFL: anbncn
• (Details are beyond the scope of this
course)