Transcript Pushdown Automata PDAs Fall 2006 Costas Busch - RPI Pushdown Automaton -- PDA Input String Stack States Fall 2006 Costas Busch - RPI.

Pushdown Automata
PDAs
Fall 2006
Costas Busch - RPI
1
Pushdown Automaton -- PDA
Input String
Stack
States
Fall 2006
Costas Busch - RPI
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Initial Stack Symbol
stack
head
Stack
Stack
$
z
top
bottom special symbol
Appears at time 0
Fall 2006
Costas Busch - RPI
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The States
Input
symbol
Pop
symbol
Push
symbol
a,
b

c
q1
q2
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Costas Busch - RPI
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q1
a, b  c
q2
input

a


a

stack
b
h
e
$
Fall 2006
top
Replace
Costas Busch - RPI
c
h
e
$
5
q1
a,   c
q2
input

a


stack
b
h
e
$
Fall 2006
top
Push
Costas Busch - RPI
a

c
b
h
e
$
6
q1
a, b  
q2
input

a


a

stack
b
h
e
$
Fall 2006
top
Pop
Costas Busch - RPI
h
e
$
7
q1
a,   
q2
input

a


a

stack
b
h
e
$
Fall 2006
top
No Change
Costas Busch - RPI
b
h
e
$
8
Pop from Empty Stack
q1
input

a
a, b  c
q2

Pop
stack
Automaton halts!
top
If the automaton attempts to pop from
empty stack then it halts and rejects input
Fall 2006
Costas Busch - RPI
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Non-Determinism
PDAs are non-deterministic
Allowed non-deterministic transitions
a, b  c
q2
q1
a, b  c
Fall 2006
q1
, b  c
q2
  transition
q3
Costas Busch - RPI
10
Example PDA
PDA
M:
L(M )  {a b : n  0}
n n
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
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L(M )  {a b : n  0}
n n
Basic Idea:
1. Push the a’s
on the stack
2. Match the b’s on input
with a’s on stack
a,   a
b, a  
3. Match
found

,



b,
a



,
$

$
q3
q2
q0
q1
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Execution Example: Time 0
Input
a a a b b b
$
Stack
current
state
a,   a
b, a  

,    q b, a   q , $  $ q
q0
3
2
1
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Time 1
Input
a a a b b b
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 2
Input
a a a b b b
a
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 3
Input
a
a
$
a a a b b b
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 4
Input
a a a b b b
a
a
a
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 5
Input
a a a b b b
a
a
a
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 6
Input
a
a
$
a a a b b b
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 7
Input
a
$
a a a b b b
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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Time 8
Input
a a a b b b
$
Stack
a,   a
b, a  
accept

,



b,
a



,
$

$
q3
q2
q0
q1
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A string is accepted if there is
a computation such that:
All the input is consumed
AND
The last state is an accepting state
we do not care about the stack contents
at the end of the accepting computation
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Rejection Example:
Time 0
Input
a a b
$
Stack
current
state
a,   a
b, a  

,    q b, a   q , $  $ q
q0
3
2
1
Fall 2006
Costas Busch - RPI
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Rejection Example:
Time 1
Input
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
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Rejection Example:
Time 2
Input
a
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
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Rejection Example:
Time 3
Input
a
a
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
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Costas Busch - RPI
26
Rejection Example:
Time 4
Input
a
a
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
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Costas Busch - RPI
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Rejection Example:
Time 4
Input
a
a
a a b
$
reject
current
state
a,   a
Stack
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
Fall 2006
Costas Busch - RPI
28
There is no accepting computation for
The string
aab
aab
is rejected by the PDA
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
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Another PDA example
PDA
M:
a,   a
b,   b
q0
Fall 2006

L(M )  {vv : v {a, b} }
R
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
q2
30
Basic Idea:
1. Push v
on stack
a,   a
b,   b
q0
Fall 2006

L(M )  {vv : v {a, b} }
R
3. Match v R on input
with v on stack
2. Guess
middle
of input
,   
a, a  
b, b  
q1
Costas Busch - RPI
4. Match
found
, $  $
q2
31
Execution Example:
Time 0
Input
a b b a
$
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
32
Time 1
Input
a b b a
a,   a
b,   b
q0
Fall 2006
,   
a
$
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
33
Time 2
Input
b
a
$
a b b a
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
34
Time 3
Input
a b b a
Guess the middle
of string
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
b
a
$
Stack
q2
35
Time 4
Input
b
a
$
a b b a
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
36
Time 5
Input
a b b a
a,   a
b,   b
q0
Fall 2006
,   
a
$
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
37
Time 6
Input
a b b a
$
a,   a
b,   b
a, a  
b, b  
Stack
accept
q0
Fall 2006
,   
q1
Costas Busch - RPI
, $  $
q2
38
Rejection Example:
Time 0
Input
a b b b
$
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
39
Time 1
Input
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a
$
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
40
Time 2
Input
b
a
$
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
41
Time 3
Input
a b b b
Guess the middle
of string
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
b
a
$
Stack
q2
42
Time 4
Input
b
a
$
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
43
Time 5
There is no possible transition.
Input
a b b b
a,   a
b,   b
q0
Fall 2006
,   
Input is not
consumed
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
a
$
Stack
q2
44
Another computation on same string:
Input
Time 0
a b b b
$
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
45
Time 1
Input
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a
$
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
46
Time 2
Input
b
a
$
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
47
Time 3
b
b
a
$
Input
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
48
Time 4
b
b
b
a
$
Input
a b b b
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
Stack
q2
49
Time 5
Input
a b b b
a,   a
b,   b
q0
Fall 2006
,   
No accept state
is reached
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
b
b
b
a
$
Stack
q2
50
There is no computation
that accepts string abbb
abbb L(M )
a,   a
b,   b
q0
Fall 2006
,   
a, a  
b, b  
q1
Costas Busch - RPI
, $  $
q2
51
Pushing & Popping Strings
Pop
string
Input
symbol
q1
Fall 2006
a , w1  w 2
Costas Busch - RPI
Push
string
q2
52
Example:
q1
a , eb  cdf
q2
input
a
a
stack
pop
string
Fall 2006
e
b
h
e
$
top
top
Replace
Costas Busch - RPI
c
d
f
h
e
$
push
string
53
q1
a , eb  cdf
Equivalent
transitions
pop
a, e  
q1
a, b  
,   
a,   f
Fall 2006
q2
a,   d
Costas Busch - RPI
push
a,   c q
2
54
Another PDA example
L(M )  {w {a, b} : na (w)  nb (w)}
*
PDA
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
M
, $  $
Costas Busch - RPI
q2
55
Execution Example:
Time 0
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
current
state
Fall 2006
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
Costas Busch - RPI
$
Stack
q2
56
Time 1
Input
a b
b b
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
a a
, $  $
Costas Busch - RPI
0
$
Stack
q2
57
Time 3
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
, $  $
Costas Busch - RPI
0
$
Stack
q2
58
Time 4
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
, $  $
Costas Busch - RPI
1
$
Stack
q2
59
Time 5
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
, $  $
Costas Busch - RPI
1
1
$
Stack
q2
60
Time 6
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
, $  $
Costas Busch - RPI
1
1
$
Stack
q2
61
Time 7
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
, $  $
Costas Busch - RPI
1
$
Stack
q2
62
Time 8
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
Fall 2006
, $  $
Costas Busch - RPI
$
Stack
accept
q2
63
Formalities for PDAs
Fall 2006
Costas Busch - RPI
64
q1
a , w1  w 2
q2
Transition function:
 (q1, a ,w1 )  {(q2,w2 )}
Fall 2006
Costas Busch - RPI
65
a , w1  w2
q2
q1
a , w1  w3
q3
Transition function:
 (q1, a ,w1 )  {(q2,w2 ), (q3,w3 )}
Fall 2006
Costas Busch - RPI
66
Formal Definition
Pushdown Automaton (PDA)
M  (Q, Σ, Γ, δ, q0 , z, F )
Accept
states
Input
alphabet
Transition Initial
Stack
function
state
alphabet
Stack
start
symbol
States
Fall 2006
Costas Busch - RPI
67
Instantaneous Description
( q, u , s )
Current
state
Fall 2006
Remaining
input
Costas Busch - RPI
Current
stack
contents
68
Example:
Instantaneous Description
(q1, bbb, aaa$)
Time 4:
Input
a a a b b b
a,   a
b, a  
a
a
a
$
Stack

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
69
Example:
Instantaneous Description
(q2 , bb, aa$)
Time 5:
Input
a a a b b b
a,   a
b, a  
a
a
a
$
Stack

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
70
We write:
(q1, bbb, aaa$)  (q2 , bb, aa$)
Time 4
Fall 2006
Time 5
Costas Busch - RPI
71
A computation:
(q0 , aaabbb,$)  (q1, aaabbb,$) 
(q1, aabbb, a$)  (q1, abbb, aa$)  (q1, bbb, aaa$) 
(q2 , bb, aa$)  (q2 , b, a$)  (q2 ,  ,$)  (q3 ,  ,$)
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
72
(q0 , aaabbb,$)  (q1, aaabbb,$) 
(q1, aabbb, a$)  (q1, abbb, aa$)  (q1, bbb, aaa$) 
(q2 , bb, aa$)  (q2 , b, a$)  (q2 ,  ,$)  (q3 ,  ,$)
For convenience we write:

(q0 , aaabbb,$)  (q3 ,  ,$)
Fall 2006
Costas Busch - RPI
73
Language of PDA
Language
L(M )
accepted by PDA M :

L(M )  {w : (q0 ,w , z )  (qf , , s )}
Initial state
Fall 2006
Costas Busch - RPI
Accept state
74
Example:

(q0 , aaabbb,$)  (q3 ,  ,$)
aaabbb L(M )
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
75

(q0 , a b ,$)  (q3 ,  ,$)
n n
a b  L(M )
n n
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
76
Therefore:
L( M )  {a b : n  0}
n n
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Fall 2006
Costas Busch - RPI
77