Transcript Languages and Finite Automata

Pushdown Automata
PDAs
1
Pushdown Automaton -- PDA
Input String
Stack
States
2
Initial Stack Symbol
stack
head
Stack
Stack
$
z
top
bottom special symbol
Appears at time 0
3
The States
Input
symbol
Pop
symbol
Push
symbol
a,
b

c
q1
q2
4
q1
a, b  c
q2
input

a


a

stack
b
h
e
$
top
Replace
c
h
e
$
5
q1
a,   c
q2
input

a


stack
b
h
e
$
top
Push
a

c
b
h
e
$
6
q1
a, b  
q2
input

a


a

stack
b
h
e
$
top
Pop
h
e
$
7
q1
a,   
q2
input

a


a

stack
b
h
e
$
top
No Change
b
h
e
$
8
Empty Stack
q1
input

stack
$
a
a, $  

top
q2

Pop
a

empty
The automaton HALTS
No possible transition after q2
9
A Possible Transition
q1
input

stack
$
a
a, $  b

top
q2

Pop
a

b
10
Non-Determinism
PDAs are non-deterministic
Allowed non-deterministic transitions
a, b  c
q2
q1
a, b  c
q1
, b  c
q2
  transition
q3
11
Example PDA
PDA
M
L(M )  {a b : n  0}
n n
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
12
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
13
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
14
Time 1
Input
a a a b b b
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
15
Time 2
Input
a a a b b b
a
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
16
Time 3
Input
a
a
$
a a a b b b
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
17
Time 4
Input
a a a b b b
a
a
a
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
18
Time 5
Input
a a a b b b
a
a
a
$
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
19
Time 6
Input
a
a
$
a a a b b b
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
20
Time 7
Input
a
$
a a a b b b
Stack
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
21
Time 8
Input
a a a b b b
$
Stack
a,   a
b, a  
accept

,



b,
a



,
$

$
q3
q2
q0
q1
22
A string is accepted if there is
a computation such that:
All the input is consumed
AND
The last state is an accepting state
At the end of the computation,
we do not care about the stack contents
(the stack can be empty at the last state)
23
The input string aaabbb
is accepted by the PDA:
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
24
In general,
L  {a b : n  0}
n n
is the language accepted by the PDA:
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
25
Rejection Example:
Time 0
Input
a a b
$
Stack
current
state
a,   a
b, a  

,    q b, a   q , $  $ q
q0
3
2
1
26
Rejection Example:
Time 1
Input
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
27
Rejection Example:
Time 2
Input
a
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
28
Rejection Example:
Time 3
Input
a
a
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
29
Rejection Example:
Time 4
Input
a
a
a a b
$
Stack
current
state
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
30
Rejection Example:
Time 4
Input
a
a
a a b
$
reject
current
state
a,   a
Stack
b, a  

,



b,
a



,
$

$
q3
q2
q1
q0
31
The input string aab
is rejected by the PDA:
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
32
A string is rejected if there is
no computation such that:
All the input is consumed
AND
The last state is an accept state
At the end of the computation,
we do not care about the stack contents
33
Another PDA example

L(M )  {vv : v {a, b} }
R
PDA
a,   a
b,   b
q0
,   
M
a, a  
b, b  
q1
, $  $
q2
34
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
4. Match
found
, $  $
q2
35
Execution Example:
Time 0
Input
a b b a
$
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
36
Time 1
Input
a b b a
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
, $  $
Stack
q2
37
Time 2
Input
b
a
$
a b b a
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
38
Time 3
Input
a b b a
Guess the middle
of string
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
b
a
$
Stack
q2
39
Time 4
Input
b
a
$
a b b a
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
40
Time 5
Input
a b b a
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
, $  $
Stack
q2
41
Time 6
Input
a b b a
$
a,   a
b,   b
a, a  
b, b  
Stack
accept
q0
,   
q1
, $  $
q2
42
Rejection Example:
Time 0
Input
a b b b
$
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
43
Time 1
Input
a b b b
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
, $  $
Stack
q2
44
Time 2
Input
b
a
$
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
45
Time 3
Input
a b b b
Guess the middle
of string
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
b
a
$
Stack
q2
46
Time 4
Input
b
a
$
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
47
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
, $  $
a
$
Stack
q2
48
Another computation on same string:
Input
Time 0
a b b b
$
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
49
Time 1
Input
a b b b
a,   a
b,   b
q0
,   
a
$
a, a  
b, b  
q1
, $  $
Stack
q2
50
Time 2
Input
b
a
$
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
51
Time 3
b
b
a
$
Input
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
52
Time 4
b
b
b
a
$
Input
a b b b
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
Stack
q2
53
Time 5
Input
a b b b
a,   a
b,   b
q0
,   
No final state
is reached
a, a  
b, b  
q1
, $  $
b
b
b
a
$
Stack
q2
54
There is no computation
that accepts string abbb
abbb L(M )
a,   a
b,   b
q0
,   
a, a  
b, b  
q1
, $  $
q2
55
Another PDA example
L(M )  {w {a, b} :
in everyprefixv, na (v)  nb (v)}
*
a,   a
b, a  
PDA
M
q0
56
Execution Example:
Time 0
Input
a a b
a,   a
$
b, a  
Stack
b, $  
q0
57
Time 1
Input
a a b
a,   a
a
$
b, a  
Stack
b, $  
q0
58
Time 2
Input
a,   a
a
a
$
b, a  
Stack
a a b
b, $  
q0
59
Time 3
Input
a a b
a,   a
a
$
b, a  
Stack
b, $  
accept
q0
60
Rejection example:
Time 0
Input
a b b b
$
Stack
q0
61
Time 1
Input
a b b b
a,   a
a
$
b, a  
Stack
b, $  
q0
62
Time 2
Input
a b b b
a,   a
$
b, a  
Stack
b, $  
q0
63
Time 3
Input
a b b b
a,   a
b, a  
Stack
b, $  
q0
64
Time 4
Input
a b b b
a,   a
b, a  
Stack
b, $  
Halt and Reject
q0
65
Pushing Strings
Input
symbol
Pop
symbol
Push
string
a,
b

w
q1
q2
66
Example:
q1
a, b  cdf
q2
input
a
a
stack
b
h
e
$
top
Push
c
d
f
h
e
$
pushed
string
67
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
, $  $
q2
68
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
, $  $
$
Stack
q2
69
Time 1
Input
a b
b b
a, $  0$
a, 0  00
a, 1 
a a
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
0
$
Stack
q2
70
Time 3
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
0
$
Stack
q2
71
Time 4
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
1
$
Stack
q2
72
Time 5
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
1
1
$
Stack
q2
73
Time 6
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
1
1
$
Stack
q2
74
Time 7
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
1
$
Stack
q2
75
Time 8
Input
a b
b b a a
a, $  0$
a, 0  00
a, 1 
b, $ 1$
b, 1 11
b, 0  
q1
, $  $
$
Stack
accept
q2
76
Formalities for PDAs
77
a,
b

w
q1
q2
Transition function:
 (q1, a, b)  {(q2 , w)}
78
a, b  w
q2
q1
a, b  w
q3
Transition function:
 (q1, a, b)  {(q2 , w), (q3 , w)}
79
Formal Definition
Pushdown Automaton (PDA)
M  (Q, Σ, Γ, δ, q0 , z, F )
States
Input
alphabet
Transition Initial
Stack
function
state
alphabet
Final
states
Stack
start
symbol
80
Instantaneous Description
( q, u , s )
Current
state
Remaining
input
Current
stack
contents
81
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
82
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
83
We write:
(q1, bbb, aaa$)  (q2 , bb, aa$)
Time 4
Time 5
84
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
85
(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 ,  ,$)
86
Formal Definition
Language
L(M )
of PDA M :

L( M )  {w : (q0 , w, s)  (q f ,  , s' )}
Initial state
Final state
87
Example:

(q0 , aaabbb,$)  (q3 ,  ,$)
aaabbb L(M )
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
88

(q0 , a b ,$)  (q3 ,  ,$)
n n
a b  L(M )
n n
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
89
Therefore:
L( M )  {a b : n  0}
n n
PDA M :
a,   a
b, a  

,



b,
a



,
$

$
q3
q2
q0
q1
90