Pushdown Automaton _PDA_

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```					        Pushdown Automaton
(PDA)

A Pushdown Automaton is a nondeterministic
finite state automaton (NFA) that permits
ε-transitions and a stack.
Pushdown Automaton (PDA)
P  Q, , ,  ( qi , a, m )  ( qk , n ),... , q0 ,  0 , F 
Q: A finite set of states.
 : A finite set of input symbols.
: A finite stack alphabet.
: The transition function with input:
qi is a state in Q.
a is a symbol in  or a = e (the empty string).
m is a stack symbol, m  .
and the output is a finite set of pairs:
qk the new state.
n is the string of stack symbols that replaces m at the top of the stack.
If n = e, then the stack is popped.
q0: The start state.
0 : Initially, the PDA’s stack consists this symbol and nothing else.
F : The set of accepting states.
PDA Example: Lwwr  {ww | w  (0  1) }
R              *

The language, Lwwr, is the even-length
palindromes over alphabet {0,1}.

Lwwr is a Context-Free Language (CFL)
generated by the grammar:

S  0S 0 | 1S1 | e

One PDA for Lwwr is given on the following
slide...
PDA for Lwwr                                    P  Q, , ,  , q0 ,  0 , F 

Q  {q0 , q1 , q2 , q3}   {0,1}   {0,1, 0 } F  {q3}

1)    ( q0 ,0, 0 )  ( q0 ,0 0 )  ( q0 ,1, 0 )  ( q0 ,1 0 )

2)    (q0 ,0,0)  (q0 ,00)  (q0 ,0,1)  (q0 ,01)
 (q0 ,1,0)  ( q0 ,10)  (q0 ,1,1)  (q0 ,11)

3)     ( q0 , e , 0 )  ( q1 , 0 )  ( q0 , e ,0)  ( q1 ,0)  ( q0 , e ,1)  ( q1 ,1)

4)     ( q1,0,0)  ( q1, e )  ( q1,1,1)  ( q1, e )

5)     ( q1 , e , 0 )  ( q2 , 0 )

6)     ( q2 , Read_Past_ End_Of_Inp ut, 0 )  ( q3 , 0 )
A Graphical Notation for PDA’s
1.   The nodes correspond to the states of the PDA.
2.   An arrow labeled Start indicates the unique start
state.
3.   Doubly circled states are accepting states.
4.   Edges correspond to transitions in the PDA as
follows:
5.   An edge labeled (ai, m)/n from state q to state p
means that (q, ai, m) contains the pair (p, n),
perhaps among other pairs.
Graphical Notation for PDA of Lwwr
(0, 0)/00
(0, 1)/01              (0,0)/ε
(0, 0)/00        (1, 0)/10              (1,1)/ε
(1, 0)/10        (1, 1)/11

start                                               (ε,0) / 0
q0                         q1                        q2
(ε, 0) / 0
(ε, 0) / 0
(ε, 1) / 1
(EOF,0) / 0
All possibilities that do not have explicit
edges, have implicit edges that go to an
implicit reject state.
q3
•    This is a nondeterministic machine.
•    Think of the machine as following all possible paths.
•    Kill a path if it leads to a reject state.
•    If any path leads to an accept state, then the machine accepts.
Exercise 1
Design a PDA that recognizes legal sequences
of ‘if’ and ‘else’ statements in a C program.

In the PDA, let ‘i’ stands for ‘if’ and ‘e’ stands for
‘else’.

Hint: There is a problem whenever the number
of ‘else’ statements in any prefix exceeds the
number of ‘if’ statements in that prefix.
Exercise 2
Design a PDA to accept the language:

{a b c | i  j or j  k}
i   j k

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