1. Pushdown
Automata
PDAs

2. Pushdown Automaton --
PDA Input String
Stack
States

3. Initial Stack Symbol
Stack Stack
$ z
bottom
special symbol

4. The States
Input Pop Push
symbol symbol symbol
a, b c
q1 q2

5. a, b c
q1 q2
input
 a   a 
stack
b top c
h Replace h
e e
$ $

6. a, c
q1 q2
input
 a   a 
stack c
b top b
h Push h
e e
$ $

7. a, b
q1 q2
input
 a   a 
stack
b top
h Pop h
e e
$ $

8. a,
q1 q2
input
 a   a 
stack
b top b
h No Change h
e e
$ $

9. Non-Determinism
a, b c q2
q1 , b c
q1 q2
a, b c transition
q3
These are allowed transitions in
a Non-deterministic PDA (NPDA)

10. NPDA: Non-Deterministic
PDA
Example:
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

11. Execution Example: Time 0
Input
a a a b b b
$
Stack
current a, a b, a
state
, b, a ,$ $ q
q0 q1 q2 3

12. Time 1
Input
a a a b b b
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

13. Time 2
Input
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

14. Time 3
Input a
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

15. Time 4
a
Input
a
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

16. Time 5
a
Input
a
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

17. Time 6
Input a
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

18. Time 7
Input
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

19. Time 8
Input
a a a b b b
$
Stack
a, a b, a
accept
q0 , q1 b, a q2 ,$ $ q
3

20. The Language of PDA
A string is accepted if there is
a computation such that:
• All the input is consumed
• The last state is a final state
At the end of the computation,
we do not care about the stack contents

21. Formal Definition Automaton
Non-Deterministic Pushdown
NPDA
M (Q, Σ, Γ, δ, q0 , z, F ) Final
States states
Input Stack
alphabet Transition Initial start
Stack
function state symbol
alphabet

22. Instantaneous Description
( q, u , s )
Current Current
Remaining
state stack
input
contents

23. Acceptance by Final State
L(M)= set of all strings w such that starting
from initial ID machine consumes w from
input and enters an accepting state.
The contents of the stack at that time is
irrelevant.
{w | (q0, w, Z) * (q, , )
For some final state q of F and any stack
staring .

24. Acceptance by empty
stack
N(M) = (w | (q0,w,Z) * (q, , )
That is N(M) is the set of all strings w such
that M can consume w and at the same
time empties its stack.Here we do not
care whether q is a final state or not.

25. The input string aaabbb
is accepted by the NPDA:
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

26. In general,
n n
L {a b : n 0}
is the language accepted by the NPDA:
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

27. Another NPDA example
NPDA M
R
L( M ) {ww }
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

28. Execution Example: Time 0
Input
a b b a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

29. Time 1
Input
a b b a
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

30. Time 2
Input
b
a b b a
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

31. Time 3
Input
b
a b b a
Guess the middle a
of string $
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

32. Time 4
Input
b
a b b a
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

33. Time 5
Input
a b b a
a
$
Stack
a, a a, a
b, b b, b
q0 , ,$ $ q2
q1

34. Time 6
Input
a b b a
$
Stack
a, a a, a
b, b b, b
accept
q0 , q1 ,$ $ q2

35. Rejection Example: Time 0
Input
a b b b
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

36. Time 1
Input
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

37. Time 2
Input
b
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

38. Time 3
Input
b
a b b b
Guess the middle a
of string $
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

39. Time 4
Input
b
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

40. Time 5
Input There is no possible transition.
a b b b Input is not a
consumed
$
Stack
a, a a, a
b, b b, b
q0 , ,$ $ q2
q1

41. Another computation on same string:
Input Time 0
a b b b
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

42. Time 1
Input
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

43. Time 2
Input
b
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

44. Time 3
Input b
b
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

45. Time 4 b
Input b
b
a b b b
a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

46. Time 5 b
Input b
No final state b
a b b b is reached a
$
Stack
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

47. There is no computation
that accepts string abbb
abbb L(M )
a, a a, a
b, b b, b
q0 , q1 ,$ $ q2

48. Pushing Strings
Input Pop Push
symbol symbol string
a, b w
q1 q2

49. Example:
a, b cdf
q1 q2
input

 a 
 
 a 

c pushed
stack d
top string
b f
h Push h
e e
$ $

50. Another NPDA example
NPDA M
L( M ) {w : na nb }
a, $ 0$ b, $ 1$
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

51. Execution Example: Time 0
Input
a b b a a b
$
a, $ 0$ b, $ 1$
Stack
a, 0 00 b, 1 11
a, 1 b, 0
current
state ,$ $
q1 q2

52. Time 1
Input
a b b a a b
0
$
a, $ 0$ b, $ 1$ Stack
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

53. Time 3
Input
a b b b a a
0
$
a, $ 0$ b, $ 1$ Stack
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

54. Time 4
Input
a b b b a a
1
$
a, $ 0$ b, $ 1$ Stack
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

55. Time 5
Input
a b b b a a 1
1
$
a, $ 0$ b, $ 1$ Stack
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

56. Time 6
Input
a b b b a a 1
1
$
a, $ 0$ b, $ 1$ Stack
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

57. Time 7
Input
a b b b a a
1
$
a, $ 0$ b, $ 1$ Stack
a, 0 00 b, 1 11
a, 1 b, 0
,$ $ q2
q1

58. Time 8
Input
a b b b a a
$
a, $ 0$ b, $ 1$
Stack
a, 0 00 b, 1 11
a, 1 b, 0
accept
,$ $ q2
q1

59. Formalities for
NPDAs

60. a, b w
q1 q2
Transition function:
(q1, a, b) {(q2 , w)}

61. a, b w q2
q1
a, b w q3
Transition function:
(q1, a, b) {(q2 , w), (q3 , w)}

62. Formal Definition Automaton
Non-Deterministic Pushdown
NPDA
M (Q, Σ, Γ, δ, q0 , z, F ) Final
States states
Input Stack
alphabet Transition Initial start
Stack
function state symbol
alphabet

63. Instantaneous Description
( q, u , s )
Current Current
Remaining
state stack
input
contents

64. Example: Instantaneous Description
(q1, bbb, aaa$)
a
Time 4: Input a
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

65. Example: Instantaneous Description
(q2 , bb, aa$)
a
Time 5: Input a
a a a b b b a
$
Stack
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

66. We write:
(q1, bbb, aaa$)  (q2 , bb, aa$)
Time 4 Time 5

67. A computation:
(q0 , aaabbb,$)  (q1, aaabbb,$) 
(q1, aabbb, a$)  (q1, abbb, aa$)  (q1, bbb, aaa$) 
(q2 , bb, aa$)  (q2 , b, a$)  (q2 , ,$)  (q3 , ,$)
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

68. (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 , ,$)

69. Formal Definition
Language of NPDA M :
L( M ) {w : (q0 , w, s )  (q f , , s ' )}
Initial state Final state

70. Example:
(q0 , aaabbb,$)  (q3 , ,$)
aaabbb L(M )
NPDA M :
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

71. n n
(q0 , a b ,$)  (q3 , ,$)
n n
a b L(M )
NPDA M :
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

72. Therefore: n n
L( M ) {a b : n 0}
NPDA M :
a, a b, a
q0 , q1 b, a q2 ,$ $ q
3

73. NPDAs Accept
Context-Free
Languages

74. Theorem:
Context-Free Languages
Languages Accepted by
(Grammars) NPDAs

75. Proof - Step 1:
Context-Free Languages
Languages Accepted by
(Grammars) NPDAs
Convert any context-free grammar G
to a NPDA M with: L (G ) L( M )

76. Proof - Step 2:
Context-Free Languages
Languages Accepted by
(Grammars) NPDAs
Convert any NPDA M to a context-free
grammar G with: L (G ) L( M )

77. Converting
Context-Free
Grammars
to
NPDAs

78. An example grammar: S aSTb
S b
T Ta
T
What is the equivalent NPDA?

79. Grammar:
S aSTb
S b NPDA:
T Ta
,S aSTb
T
,S b
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

80. The NPDA simulates
leftmost derivations of the grammar
L(Grammar) = L(NPDA)

81. Grammar: S aSTb
S b
T Ta
T
A leftmost derivation:
S aSTb abTb abTab abab

82. NPDA execution: Time 0
Input a b a b
,S aSTb $
,S b Stack
,T Ta a, a
current ,T b, b
state
, S ,$ $
q0 q1 q2

83. Time 1
Input a b a b
S
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

84. Time 2 a
S
Input a b a b
T
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

85. Time 3 a
S
Input a b a b
T
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

86. Time 4
b
Input a b a b
T
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

87. Time 5
b
Input a b a b
T
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

88. Time 6
T
Input a b a b a
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

89. Time 7
T
Input a b a b a
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

90. Time 8
Input a b a b a
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

91. Time 9
Input a b a b
b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
q0 , S ,$ $
q1 q2

92. Time 10
Input a b a b
,S aSTb $
,S b Stack
,T Ta a, a
,T b, b
accept
q0 , S ,$ $
q1 q2

93. In general:
Given any grammar G
We can construct a NPDA M
With L(G ) L( M )

94. Constructing NPDA M from grammar G :
For any production For any terminal
A w a
,A w a, a
q0 , S ,$ $
q1 q2

95. Grammar G generates string w
if and only if
NPDA M accepts w
L(G) L( M )

96. Therefore:
For any context-free language
there is an NPDA
that accepts the same language

97. Converting
NPDAs
to
Context-Free
Grammars

98. For any NPDA M
we will construct
a context-free grammar G with
L( M ) L(G )

99. Intuition: The grammar simulates the machine
A derivation in Grammar G:
S  abc ABC   abc
Current configuration in NPDA M

100. A derivation in Grammar G:
terminals variables
S  abc ABC   abc
Input processed Stack contents
in NPDA M

101. Some Necessary
Modifications
First, we modify the NPDA:
• It has a single final state q f
• It empties the stack
when it accepts the input
Original NPDA Empty Stack
qf

102. Second, we modify the NPDA transitions:
all transitions will have form
a, B qj
qi
or
qi a, B CD q
j
B, C , D : stack symbols

103. Example of a NPDA in correct form:
L( M ) {w : na nb }
$ : initial stack symbol
a, $ 0$ b, $ 1$
a, 0 00 b, 1 11
a, 1 b, 0
,$ qf
q0

104. The Grammar
Construction
In grammar G :
Stack symbol
Variables: (qi Bq j )
states
Terminals:
Input symbols of NPDA

105. For each transition
a, B qj
qi
We add production (qi Bq j ) a

106. For each transition qi a, B CD q
j
We add production (qi Bqk ) a(q j Cql )(ql Dqk )
For all states qk , ql

107. Stack bottom symbol
Start Variable: (qo $q f )
Start state final state

108. Example:
a, $ 0$ b, $ 1$
a, 0 00 b, 1 11
a, 1 b, 0
,$ qf
q0
Grammar production: (q01q0 ) a

109. Example:
a, $ 0$ b, $ 1$
a, 0 00 b, 1 11
a, 1 b, 0
,$ qf
q0
Grammar productions:
(q0 $q0 ) b(q01q0 )(q0 $q0 ) | b(q01q f )(q f $q0 )
(q0 $q f ) b(q01q0 )(q0 $q f ) | b(q01q f )(q f $q f )

110. Example:
a, $ 0$ b, $ 1$
a, 0 00 b, 1 11
a, 1 b, 0
,$ qf
q0
Grammar production: (q0 $q f )

111. Resulting Grammar: (q0 $q f ) : start variable
(q0 $q0 ) b(q01q0 )(q0 $q0 ) | b(q01q f )(q f $q0 )
(q0 $q f ) b(q01q0 )(q0 $q f ) | b(q01q f )(q f $q f )
(q01q0 ) b(q01q0 )(q01q0 ) | b(q01q f )(q f 1q0 )
(q01q f ) b(q01q0 )(q01q f ) | b(q01q f )(q f 1q f )
(q0 $q0 ) a (q0 0q0 )(q0 $q0 ) | a (q0 0q f )(q f $q0 )
(q0 $q f ) a (q0 0q0 )(q0 $q f ) | a (q0 0q f )(q f $q f )

112. (q0 0q0 ) a (q0 0q0 )(q0 0q0 ) | a (q0 0q f )(q f 0q0 )
(q0 0q f ) a (q0 0q0 )(q0 0q f ) | a (q0 0q f )(q f 0q f )
(q01q0 ) a
(q0 0q0 ) b
(q0 $q f )

113. Derivation of string abba
(q0 $q f ) a(q0 0q0 )(q0 $q f )
ab(q0 $q f )
abb(q01q0 )(q0 $q f )
abba(q0 $q f ) abba

114. In general, in Grammar:
(q0 $q f ) w
if and only if
w is accepted by the NPDA

115. Explanation:
By construction of Grammar:
(qi Aq j ) w
if and only if
in the NPDA going from qi to q j
the stack doesn’t change below A
and A is removed from stack