Transcript Languages and Finite Automata

Pushdown Automata
PDAs
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Pushdown Automaton -- PDA
Input String
Stack
States
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Initial Stack Symbol
stack
head
Stack
Stack
$
z
top
bottom special symbol
Appears at time 0
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The States
Input
symbol
Pop
symbol
Push
symbol
a,
b

c
q1
q2
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q1
a, b  c
q2
input

a


a

stack
b
h
e
$
top
Replace
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c
h
e
$
5
q1
a,   c
q2
input

a


stack
b
h
e
$
top
Push
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a

c
b
h
e
$
6
q1
a, b  
q2
input

a


a

stack
b
h
e
$
top
Pop
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h
e
$
7
q1
a,   
q2
input

a


a

stack
b
h
e
$
top
No Change
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b
h
e
$
8
Non-Determinism
PDAs are non-deterministic
Allowed non-deterministic transitions
a, b  c
q2
q1
a, b  c
q1
, b  c
q2
  transition
q3
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Example PDA
PDA
M:
L(M )  {a b : n  0}
n n
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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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
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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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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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Rejection Example:
Time 4
Input
a
a
a a b
$
reject
current
state
a,   a
Stack
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
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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
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Another PDA example
PDA
M:
a,   a
b,   b
q0

L(M )  {vv : v {a, b} }
R
,   
a, a  
b, b  
q1
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, $  $
q2
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Basic Idea:
1. Push v
on stack
a,   a
b,   b
q0

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
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4. Match
found
, $  $
q2
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Execution Example:
Time 0
Input
a b b a
$
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
31
Time 1
Input
a b b a
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
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, $  $
Stack
q2
32
Time 2
Input
b
a
$
a b b a
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
33
Time 3
Input
a b b a
Guess the middle
of string
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
b
a
$
Stack
q2
34
Time 4
Input
b
a
$
a b b a
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
35
Time 5
Input
a b b a
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
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, $  $
Stack
q2
36
Time 6
Input
a b b a
$
a,   a
b,   b
a, a  
b, b  
Stack
accept
q0
,   
q1
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, $  $
q2
37
Rejection Example:
Time 0
Input
a b b b
$
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
38
Time 1
Input
a b b b
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
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, $  $
Stack
q2
39
Time 2
Input
b
a
$
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
40
Time 3
Input
a b b b
Guess the middle
of string
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
b
a
$
Stack
q2
41
Time 4
Input
b
a
$
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
42
Time 5
There is no possible transition.
Input
a b b b
a,   a
b,   b
q0
,   
Input is not
consumed
a, a  
b, b  
q1
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, $  $
a
$
Stack
q2
43
Another computation on same string:
Input
Time 0
a b b b
$
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
44
Time 1
Input
a b b b
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
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, $  $
Stack
q2
45
Time 2
Input
b
a
$
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
46
Time 3
b
b
a
$
Input
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
47
Time 4
b
b
b
a
$
Input
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
Stack
q2
48
Time 5
Input
a b b b
a,   a
b,   b
q0
,   
No accept state
is reached
a, a  
b, b  
q1
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, $  $
b
b
b
a
$
Stack
q2
49
There is no computation
that accepts string abbb
abbb L(M )
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
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, $  $
q2
50
Pushing & Popping Strings
Pop
string
Input
symbol
q1
a , w1  w 2
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Push
string
q2
51
Example:
q1
a , eb  cdf
q2
input
a
a
stack
pop
string
e
b
h
e
$
top
top
Replace
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c
d
f
h
e
$
push
string
52
q1
a , eb  cdf
Equivalent
transitions
pop
q1
a, e  
, b  
,   
,   f
q2
,   d
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push
,   c q
2
53
Another PDA example
L(M )  {w {a, b} : na (w)  nb (w)}
*
PDA
a, $  0$
a, 0  00
a, 1 
M
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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q2
54
Execution Example:
Time 0
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
current
state
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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$
Stack
q2
55
Time 1
Input
a b
b b
a, $  0$
a, 0  00
a, 1 
a a
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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0
$
Stack
q2
56
Time 3
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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0
$
Stack
q2
57
Time 4
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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1
$
Stack
q2
58
Time 5
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
Prof. Busch - LSU
1
1
$
Stack
q2
59
Time 6
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
Prof. Busch - LSU
1
1
$
Stack
q2
60
Time 7
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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1
$
Stack
q2
61
Time 8
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
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$
Stack
accept
q2
62
Formalities for PDAs
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63
q1
a , w1  w 2
q2
Transition function:
 (q1, a ,w1 )  {(q2,w2 )}
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64
a , w1  w2
q2
q1
a , w1  w3
q3
Transition function:
 (q1, a ,w1 )  {(q2,w2 ), (q3,w3 )}
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65
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
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66
Instantaneous Description
( q, u , s )
Current
state
Remaining
input
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Current
stack
contents
67
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
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68
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
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69
We write:
(q1, bbb, aaa$)  (q2 , bb, aa$)
Time 4
Time 5
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70
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
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71
(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 ,  ,$)
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72
Language of PDA
Language
L(M )
accepted by PDA M :

L(M )  {w : (q0 ,w , z )  (qf , , s )}
Initial state
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Accept state
73
Example:

(q0 , aaabbb,$)  (q3 ,  ,$)
aaabbb L(M )
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Prof. Busch - LSU
74

(q0 , a b ,$)  (q3 ,  ,$)
n n
a b  L(M )
n n
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
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75
Therefore:
L( M )  {a b : n  0}
n n
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
Prof. Busch - LSU
76