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PushDown Automata
What is a stack?
• A stack is a Last In First Out data structure
where I only have access to the last element
inserted in the stack.
• In order to access other elements I have to
remove those that are on top one by one.
Stack
Stack
Stack
Stack
Stack
Stack
Stack
Stack
What is a PDA?
• A PDA is an enhanced finite automaton that
also contains an infinite stack.
• The transitions in a PDA are of the form
a, x ⟶ y
meaning that if you see an a in the input string
and the stack contains the symbol x on top
then you remove the x and add a y.
• The stack gives us extra power to recognize
non-regular languages.
Transitions
• Transitions of the form a, x ⟶ y require that
the next input symbol should be a and the top
stack symbol should be x.
q
a, x ⟶ y
x
w
Stack
q’
...abb...
Input
q
a, x ⟶ y
q’
y
w
...abb...
Stack
Input
Transitions
• Transitions of the form ε, x ⟶ y require that
the top stack symbol is x.
q
ε, x ⟶ y
x
w
Stack
q’
...abb...
Input
q
ε, x ⟶ y
q’
y
w
...abb...
Stack
Input
Transitions
• Transitions of the form a, ε ⟶ y require that
the next input symbol is a.
q
a, ε ⟶ y
x
w
Stack
q’
...abb...
Input
q
a, ε ⟶ y
q’
y
x
w
...abb...
Stack
Input
Transitions
• Transitions of the form ε, ε ⟶ y can be
followed without restrictions.
q
ε, ε ⟶ y
x
w
Stack
q’
...abb...
Input
q
ε, ε ⟶ y
q’
y
x
w
...abb...
Stack
Input
PDA Accept – Reject Status
• The PDA accepts when there exists a
computation path such that:
– The computation path ends in an accept state
– All the input is consumed
– (no requirement for the stack)
• The PDA rejects when all the paths:
– Either end in a non-accepting state
– Or are incomplete (meaning that at some point
there is no possible transition under the current
input and stack symbols)
A PDA for {anbn : n ≥ 0}
• We usually use the stack for counting.
• For this language for example, you first insert
all the as in the stack until you start seeing bs .
• When you see the first b start removing as
from the stack.
• When you have consumed the whole string
you check the stack: if it’s empty then this
means that the number of as equals the
number of bs.
Is the stack empty?
How can you check if the stack is empty?
• What we usually do is to place a special
symbol (for example a $) at the bottom of the
stack.
• Whenever we find the $ again we know that
we reached the end of the stack.
• In order to accept a string there is no need
for the stack to be empty.
Stack push and pop in PDA
• a, ε ⟶ t
when you see an a in the input push t on the
stack
• a, b ⟶ ε
when you see an a in the input and b is on the
top of the stack, pop b out.
A PDA for {anbn : n ≥ 0}
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
$
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
$
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
a
$
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
a
a
$
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
a
$
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
$
Visualization of {anbn:n ≥ 0}
aaabbb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
$
Visualization of {anbn:n ≥ 0}
aaabbb 
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
Visualization of {anbn:n ≥ 0}
aab
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
Visualization of {anbn:n ≥ 0}
aab
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
$
Visualization of {anbn:n ≥ 0}
aab
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
$
Visualization of {anbn:n ≥ 0}
aab
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
a
$
Visualization of {anbn:n ≥ 0}
aab
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
$
Visualization of {anbn:n ≥ 0}
aab 
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
$
Visualization of {anbn:n ≥ 0}
abb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
Visualization of {anbn:n ≥ 0}
abb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
$
Visualization of {anbn:n ≥ 0}
abb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
a
$
Visualization of {anbn:n ≥ 0}
abb
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
$
Visualization of {anbn:n ≥ 0}
abb 
q0
ε, ε ⟶ $
q1
b, a ⟶ ε
q3
ε, $ ⟶ ε
q2
$
PDA formally
• A PDA is a sextuple (Q, Σ, Γ, δ, q0, F), where:
– Q is the set of states
– Σ is the input alphabet
– Γ is the alphabet for the stack
– δ is the transition function
– q0 is the start state
– F is the set of accepting states
About Γ: The stack alphabet can contain any
symbol you want. It can be completely disjoint
from Σ.
L() : proper opening and closing
parenthesis
ε, ε ⟶ $
q0
(, ε ⟶ *
q1
ε, $ ⟶ ε
), * ⟶ ε
Try it yourself
• Create a PDA for the language:
L= = {w : w contains an equal number of 0s and 1s}
L= : equal number of 0s and 1s
q0
ε, ε ⟶ $
q2
0, ε ⟶ *
1, * ⟶ ε
q3
1, ε ⟶ *
0, * ⟶ ε
q1
L= : equal number of 0s and 1s
NPDA for this language
ε, ε ⟶ $
q0
q1
ε, $ ⟶ ε
0, ε ⟶ 0
0, 1 ⟶ ε
1, ε ⟶ 1
1, 0 ⟶ ε
PDA and Regular Languages
• Regular languages can be recognized by PDA:
– For every regular language there is an NFAε
recognizing it.
– Simply add ε⟶ε in every transition for the stack
(i.e just don’t use it at all).
• The languages recognized by PDA is a superset
of regular languages.
– As we saw the language L = {anbn : n≥0} is
recognized by some PDA.
– L is not regular.
Non-Determinism
• Non- determinism means that we can have
more than one choice.
• Non-Deterministic:
q2
q1
q3
Non-Determinism
• Non- determinism means that we can have
more than one choice.
• Non-Deterministic:
q2
q1
q3
Non-Determinism
• Non- determinism means that we can have
more than one choice.
• Non-Deterministic:
0, a ⟶ 0
0, a ⟶ 1
q1
q2
Non-Determinism
• Non- determinism means that we can have
more than one choice.
• Non-Deterministic:
q2
q1
q3
Non-Determinism
• Non- determinism means that we can have
more than one choices.
• Non-Deterministic:
q2
q1
q3
Non-Determinism
• Non- determinism means that we can have
more than one choices.
• Deterministic:
q2
q1
q3
Non-Determinism
• Non- determinism means that we can have
more than one choices.
• Deterministic:
q1
ε, ε ⟶ 0
q2
No other possible transitions
Definition of DPDA
(deterministic push down automata)
δ : Q x Σε x Γε  (Q x Γε) U {φ}
A DPDA has exactly one legal move in every
situation where its stack is non empty
Given any state q, any letter a, any stack letter x
Only one of the following is allowed to be non
empty
δ(q,a,x)
δ(q,a, ε)
δ(q, ε, x)
δ(q, ε, ε)
DPDA vs NPDA
• Although non-deterministic and deterministic FA
are equivalent this is not the case with PDA.
Non-determinism helps us recognize more
languages.
• Intuition:
LR = { wwR : w in {0,1}* }
An NPDA for this language pushes the first half
of the string in the stack and pops the second
half.
It has to guess where the middle of the string is.
L#R = { w#wR : w in {0,1}* }
q0
ε, ε ⟶ $
q1
0, ε ⟶ 0
1, ε ⟶ 1
#, ε ⟶ ε
q3
ε, $ ⟶ ε
q2
0, 0 ⟶ ε
1, 1 ⟶ ε
LR = { wwR : w in {0,1}* }
• Compare the previous DPDA with this NPDA
q0
ε, ε ⟶ $
q1
0, ε ⟶ 0
1, ε ⟶ 1
ε, ε ⟶ ε
q3
ε, $ ⟶ ε
q2
0, 0 ⟶ ε
1, 1 ⟶ ε
NPDA and CF languages
• It can be shown that non-deterministic PDA
are equivalent with context free grammars.
• NPDA accept exactly the set of CF languages.
• In order to prove that a language is CF you can
– Construct a CF grammar that generates it
– Construct a NPDA that recognizes it.

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