Class8

Pushdown Automata
(PDA)

Pushdown Automata
PDAs
Homework (p.183) 2, 3abc, 4a~h, 5, 6, 9~11

1

Pushdown Automata
(PDA)

Pushdown Automaton -- PDA
Input String*
Stack with
symbol from 

States

 Gamma  Sigma 2

Pushdown Automata
(PDA)

Initial Stack Symbol

Stack

z

bottom
special symbol
3

Pushdown Automata

The States
(PDA)

Symbol on
Input symbol top of the The string
may be  stack that replaces

q1
a,b, x q
2

a   {} b x *

Input alphabet Stack alphabet

b & x are all for stack processing 4

Pushdown Automata
(PDA)

q1 a,b, ? q2
?   (q1, a, b)
input

… a …

stack
b top
h For the execution of the stack, there
e are 4 kinds of operations: replace,
z push, pop, and no change.
5

Pushdown Automata
(PDA)

q1 a,b, ? q2
(q2, c)   (q1, a, b): when it is on the state q1 and
input a is read, then b is replace by c and moves to q2

… a … … a …

stack
b top c
h Replace h
e e
z z 6

Pushdown Automata
(PDA)

q1 a,b, c q2
(q2, c)   (q1, a, b): when it is on the state q1 and
input a is read, then b is replace by c and moves to q2

… a … … a …

stack
b top c
h Replace h
e e
z z 7

Pushdown Automata
(PDA)

q1 a,b, ? q2
(q2, cb)   (q1, a, b): when it is on the state q1 and a is read,
input
then c is pushed, i.e. b is replace by cb, and moves to q2

… a … … a …
stack c
b top b
h Push h
e e
z In Def 7.1 (p.177), it is written as (q2, cb)  (q1, a, b)
for pushing.
z
For the execution of the stack, there are 4 kinds of
operations: replace, push, pop, and no change.
8

Pushdown Automata
(PDA)

q1 a,b, cb q2
(q2, cb)   (q1, a, b): when it is on the state q1 and a is read,
input
then c is pushed, i.e. b is replace by cb, and moves to q2

… a … … a …

stack c
b top b
h Push h
e e
z z
9

Pushdown Automata
(PDA)

q1 a,b, ? q2
(q2, )   (q1, a, b): when it is on the state q1 and a is read,
input
then b is popped, i.e. b is replace by , and moves to q2

… a … … a …
stack
b top Pop h
h e
e Beware, when we push a symbol into stack—we do z
not really care about what is the top symbol of the
z stack.
But if we want to pop a symbol out of a stack, we
need to know what is the top stack symbol (will be
popped up). 10

Pushdown Automata

a,b, 
(PDA)

q1 q2
(q2, )   (q1, a, b): when it is on the state q1 and a is read,
input
then b is popped, i.e. b is replace by , and moves to q2

… a … … a …

stack
b top
h Pop h
e e
z z 11

Pushdown Automata
(PDA)

q1 a,b, ? q2
(q2, b)   (q1, a, b): when it is on the state q1 and a is read,
input
nothing will be changed on the stack and moves to q2

… a … … a …

stack
b top b
h No Change h
e e
z z
12

Pushdown Automata
(PDA)

q1 a,b, b q2
(q2, b)   (q1, a, b): when it is on the state q1 and a is read,
input
nothing will be changed on the stack and moves to q2

… a … … a …

stack
b top b
h No Change h
e e
z z
13

Pushdown Automata
(PDA)

Every FA is a PDA

Consider regular language L={ a2nbm : n, m  0}
with FA as:
b , z, z
a , z, z
a , z, z

 , z, z

It becomes a PDA !

14

Pushdown Automata

Every FA is a PDA
(PDA)

LM   L(a * b)
a, z, z
a , z, z a b
b,,z, z

a, z, z
b, z, z b, , b z
a z,
q0 q1 q2

It becomes a PDA !
Do Hw#3 on p. 183
15

Pushdown Automata
(PDA)

While it is on state q1 and reads a Non-Determinism
If top of stack symbol is b: (q2, c)   (q1, a, b)
q2   transition
a,b, c
q1  ,b, c q2
q1 If top of stack symbol is b: moves to q2 without
consuming any input symbol, i.e. (q2,c) (q1,, b)
a,d , ed
q3
If top of stack symbol is d: (q3, ed)   (q1, a, d)
These are allowed transitions in
a Non-deterministic PDA (NPDA) 16

Formal Definition
Pushdown Automata
(PDA)

Non-Deterministic Pushdown Automaton
NPDA
M  (Q, Σ, Γ, δ, q0 , z, F ) Final
States states

Input Stack
alphabet Transition Initial start
Stack
function state symbol
alphabet
17

NPDA: Non-Deterministic
Pushdown Automata
(PDA)

PDA

Example:
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3

According to the def. 7.1, , should be
replaced by ,$$
transitions on q1 should be { a,$a$, a, aaa}
18

Execution Example: Time 0 (initial configuration)
Pushdown Automata
(PDA)

a a a b b b
z
Stack

a, z, az
current a, a, aa b, a, 
state

q0  , z, z q1 b,a,  q2
 , z, z q
3
19

Execution Example: Time 1
Pushdown Automata
(PDA)

a a a b b b
z
Stack

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q  ,z, z q
2 3
20

Pushdown Automata
(PDA)

a a a b b b
z
Stack

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 ,z, z q
3
21

Pushdown Automata
(PDA)

a a a b b b a
z z
Stack
a, Z , aZ
a, a, aa b, a, 

q0  ,Z, Z q b,a,  q2
 ,z, z q
1 3

演變過程請看投影片動畫
22

Pushdown Automata
(PDA)

a
a a a b b b a a
z z
Stack

a, Z , aZ
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 ,z, z q
3
演變過程請看投影片動畫 23

Pushdown Automata
(PDA)

a
a a a b b b a a
a a
z z
Stack
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 ,z, z q
3

Pushdown Automata
(PDA)

a
a a a b b b a
a
z
Stack
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 ,z, z q
3

Execution Example: End of input tape
Pushdown Automata
(PDA)

a
a a a b b b a
a
z
Stack
a, z, az
a, a, aa b, a, 
accept

q0  , z, z q1 b,a,  q2
 ,z, z q
3

Pushdown Automata
(PDA)

A string is accepted if there is
a computation such that:

• All the input is consumed
and
• The last state is a final state

At the end of the computation,
we do not care about the stack contents

30

Pushdown Automata
(PDA)

Therefore:
The input string aaabbb
is accepted by the NPDA:

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
31

Pushdown Automata
(PDA)

L( M ) = ? { anbn : n  0 }
is the language accepted by the NPDA:

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
32

Pushdown Automata
(PDA)

L( M ) = { anbn : n  0 }
Can we find a machine with 3 states?
Can we find a machine with 2 states?
What is L(M) if both q0 & q1 are final states?

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
33

Pushdown Automata
(PDA)

Pushing Strings
Input Top stack Push
symbol symbol string

q1
a, b, w q2

The top stack symbol b is
replaced by the string w
34

Example:
Pushdown Automata
(PDA)

q1 a, b, cdf q2
input top


 a 
 
 a 


c pushed
stack d
top string
b f
h Push h
e e
Z Z
35

Example:
Pushdown Automata
(PDA)

input
Beware: we can push a string into a stack
but we do not pop a string from a stack---
only a 
  
 pop one symbol or nothing at a a 
 time! 
c pushed
stack d
top string
b f
h Push h
e e
Z Z
In fact, to read from input tape or pop up from stack, there is at most one symbol at
a time. The main reason is we can only see one symbol at a time.
As to push into stack is like to record an information so we can write a string, or
push a string into a stack. 36

Pushdown Automata

Examples for NPDA
(PDA)

Construct an NPDA for

LM   {a b
n 2n
: n  0}
LM   {a b : n  0}
2n n

LM   {a b c
n m nm
: n, m  0}

37

Formal Def. for NPDA M
Pushdown Automata
(PDA)

L ={ anbn : n  0 }

M  ({q0 , q1 , q2 , q3},   {a, b},   {a, b, z},
 , q0 , z, F  {q0 , q3})

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
38

Pushdown Automata

Transition function: (PDA)

q2
a, b, w
q1
a, b, w
q3

 (q1, a, b)  {(q2 , w), (q3 , w)}

 : Q  ({}  )    finite subsets of Q   *
39

M  ({q0 , q1 , q2 , q3},   {a, b},   {a, b, z},
Pushdown Automata
(PDA)

 , q0 , z, F  {q0 , q3})
Transition Functions  ?

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
40

Pushdown Automata
(PDA)

Transition Functions 
(q1, a, z) = {(q1, az)}
(q1, a, a) = {(q1, aa)} (q2, b, a) = {(q2, )}
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
(q0, , z) = {(q1, z)} (q1, b, a) = {(q2, )} (q2, , z) = {(q3, z)}
41

Pushdown Automata
(PDA)

Instantaneous Description

( q, u , s )

Current Current
Remaining
state stack
input
contents

42

Pushdown Automata
(PDA)

Time 4
a
a a a b b b a
a
z
Instantaneous ( q1, bbb, aaaz)
Description Stack
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 ,z, z q
3
43

Pushdown Automata
(PDA)

a
a a a b b b a
a
z
Instantaneous ( q2, bb, aaz)
Description Stack
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 ,z, z q
3
44

Pushdown Automata
(PDA)

We write:

(q1 , bbb, aaaZ ) ├ (q2 , bb, aaZ )
Time 4 Time 5

45

Pushdown Automata
(PDA)
A computation for w= aaabbb:
(q0 , aaabbb, z )├ (q1 , aaabbb, z ) ├
(q1 , aabbb, az)├ (q1 , abbb, aaz) ├ (q1 , bbb, aaaz) ├

(q2 , bb, aaz)├ (q2 , b, az) ├ (q2 ,  , z ) ├(q3 ,  , z )

a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
46

Pushdown Automata
(PDA)

(q0 , aaabbb, z ) (q1 , aaabbb, z )├
├

(q1 , aabbb, az)├ (q1 , abbb, aaz) ├ (q1 , bbb, aaaz) ├
(q2 , bb, aaz) (q2 , b, az)├ (q2 ,  , z )├ (q3 ,  , z )
├

For convenience we write:

(q0 , aaabbb, z ) ├ (q3 ,  , z )
*

47

Pushdown Automata
(PDA)

Formal Definition
Language of NPDA M :
*

L( M )  {w : (q0 , w, z )├ (q f ,  , u )}

Initial state Final state empty

48

Example:
Pushdown Automata
(PDA)


(q0 , aaabbb, z ) ├ (q3 ,  , z )

aaabbb L(M )

NPDA M :
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3

49

Pushdown Automata
(PDA)


(q0 , a b , z ) ├ (q3 ,  , z )
n n

a b  L(M )
n n

NPDA M :
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
50

Pushdown Automata
(PDA)

Therefore: L( M )  {a b : n  0}
n n

NPDA M :
a, z, az
a, a, aa b, a, 

q0  , z, z q1 b,a,  q2
 , z, z q
3
51

Pushdown Automata
(PDA)

Another NPDA example
NPDA M for
L( M )  {ww : w {a, b}*}
R

52

Pushdown Automata

NPDA M for (PDA)

L( M )  {ww : w {a, b}*}
R

a, Z , aZ b, Z , bZ
a, a, 
a, a, aa b, a, ba
b, b, 
a, b, ab b, b, bb
, Z , Z
 , a, a
 ,Z, Z q2
q0  , b, b q1
Move to Move to
Push w next state Pop wR final state
53

NPDA M for L( M )  {ww : w  {a, b}*}
R Pushdown Automata
(PDA)

Try strings like: , abba, baa.
If it is accepted, write the computation.

a , z , az b, z , bz
a , a , aa b, a, ba a, a, 
a , b, ab b, b, bb b, b, 
 , z, z
 , a, a  ,z, z
q0 q1 q2
 , b, b 54

NPDA M for L( M )  {ww : w  {a, b}*}
R Pushdown Automata
(PDA)

M  ({q0 , q1 , q2 },   {a, b},   {a, b, z},
 , q0 , z, F  {q2 })
How about for L( M )  {wcw : w  {a, b}*}
R

a , z , az b, z , bz
a , a , aa b, a, ba a, a, 
a , b, ab b, b, bb b, b, 
 , z, z
 , a, a  ,z, z
q0 q1 q2
 , b, b 55

Pushdown Automata
(PDA)

Another NPDA example
NPDA M for
L( M )  {w : na  nb }

How to use the stack to recognize
the strings like :
abbbaa babbaaba 

56

M for L( M )  {w : na  nb }
Pushdown Automata
NPDA (PDA)

M  ({q1 , q2},   {a, b},   {0,1, z},  , q1 , Z , F  {q2})

Write a computation for abbbaa.

a,z,0 z b,z,1z
a,0,00 b,1,11
a,1,  b,0, 

q1
 , z, z q2
57

M for L( M )  {w : na  nb }
Pushdown Automata
NPDA (PDA)

How about NPDA for {w: na = nb+1 } ?
How about NPDA for {w: na = 2nb } ?
a,z,0 z b,z,1z
a,0,00 b,1,11
a,1,  b,0, 

q1
 , z, z q2
{w: na = 2nb }: one b and push 2 bs, or push one b and pop up one a, or pop up two
as, i.e., b, z, bbz; b, b, bbb in q1, but b, a, (pop up and go to another state, say q’,
such that ,a,  (pop up one more time); or , z, bz; or , b, bb; and all go back to
state q1.
58

M for L( M )  {w : na  2nb }
Pushdown Automata
NPDA (PDA)

a,Z ,0Z b,Z ,1Z b,Z ,11Z
a,0,00 b,1,11 b,1,111
a,1,  b,0, 
q3
 ,Z ,1Z
 ,0, 
q1 q2
 , Z, Z
59

Pushdown Automata

More Example on designing:
(PDA)

LM   {a b : n  m  2n}
n m

How about NPDA for {anbm: n < m } ?

How about NPDA for {anbm: n > m } ?

How about NPDA for {anbm: n  m } ?

60

Class8

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