push down automata- theoretical foundations of computations

Course – TFC
Course Instructor
Dr. Umadevi V
Department of CSE, BMSCE
Webpage:https://sites.google.com/site/drvumadevi/
1 July 2021 CSE, BMSCE 1

Unit-4
Pushdown Automata
Definition of the Pushdown Automaton, The Languages of a PDA,
Equivalence of PDA’s and CFG’s , Deterministic Pushdown
Automata,
The Pumping Lemma for Context Free Languages, Closure
Properties of Context Free Languages

Push Down Automata (PDA)
Pushdown Automata is a finite automata with
extra memory called stack which helps Pushdown
automata to recognize Context Free Languages.

Schematic Representation of PDA
Rough Slide

Rough Slide
□ Example: L={an
bn
, n>=1}
□ Valid String “aabb”
a a b b 𝞮
z0
Input
Stack

Rough Slide
□ Example: L={an
bn
, n>=1}
□ Valid String “aabb” Invalid String “aabbb”
a a b b 𝞮
a
a
z0
Input
Stack
a a b b b 𝞮
z0
Input
Stack

□ Example: L={an
bn
, n>=1}
□ Valid String “aabb” Invalid String “aabbb”
a a b b 𝞮
a
a
z0
Input
Stack
a a b b b 𝞮
a
a
z0
Input
Stack

Formal Definition of PDA
P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
□ Q is the set of states
□ ∑ is the set of input alphabet
□ Γ is the set of symbols called stack alphabet
□ q0
∈ Q , is the initial state
□ Z0
∈ Γ, is the initial symbol on the stack
□ F is the set of final states
□ δ is a transition function which maps Q x {Σ ∪ ε} x Γ into Q x Γ*. In a given
state, PDA will read input symbol and stack symbol (top of the stack) and
move to a new state and change the symbol of stack.
δ(State, Input Symbol, Stack Symbol)=(Next State, Stack Symbol)

Classification of Push Down Automata
1. Deterministic Push Down Automata (DPDA)
2. Non-Deterministic Push Down Automata (NPDA)
Note: Unlike Finite Automata, DPDA and NPDA are not equivalent. For example,
the language
L={WWR
, W ∈(a+b)*) } is accepted by NPDA but not by any DPDA

Two ways a PDA accept the language
There are two ways a PDA can accept the language:
a. Empty Stack: If a sentence in a language causes PDA to empty its stack
then the sentence is accepted.
b. Final State: If a sentence in a language causes PDA to move to a state
which is member of a set of final states, then the sentence is accepted.

Question
Construct PDA ( or DPDA) to recognize the language L={an
bn
, n>=1} by empty stack
method. Show the moves made by PDA for the string aabb and aaabb

bn
, n>=1} by empty stack method.
Rough Slide
bn
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
a
q1
z0
a

bn
, n>=1} by empty
stack method.
bn
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
Push a
q0
X X
a Push a
q0
Pop a
q1
X
q1
z0
X X Pop Z0
q1
a
X Pop a
q1
X

bn
method.
q0
a, z0
| az0
q1
Representing PDA as a generalized
Transition Diagram
a, a | aa
b, a | 𝞮
𝞮, Z0
| 𝞮
b, a | 𝞮

bn
method.
Formal definition of PDA
P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
}
∑ = {a, b}
Γ ={a, Z0
}
q0
= q0
Z0
= Z0
F = { }
δ(q0
, a, z0
) =(q0
, az0
)
δ(q0
, a, a) =(q0
, aa)
δ(q0
, b, a) =(q1
, 𝞮)
δ(q1
, b, a) =(q1
, 𝞮)
δ(q1
, 𝞮, z0
) =(q1
, 𝞮)
Push Operation
Pop Operation

Instantaneous Description of a PDA
Rough Slide
□ The current configuration of PDA at any given instant can be described by Instantaneous
Description (ID).
□ An ID gives current state of PDA, the remaining string to be processed and the entire contents of
stack.
Instantaneous Description for the string "aabb"
(q0
, aabb, z0
)

□ The current configuration of PDA at any given instant can be described by Instantaneous
Description (ID).
□ An ID gives current state of PDA, the remaining string to be processed and the entire contents of
stack.
Instantaneous Description for the string "aabb"
(q0
, aabb, z0
) ˫ (q0
, abb, az0
)
˫ (q0
, bb, aaz0
)
˫ (q1
, b, az0
)
˫ (q1
, 𝞮, z0
)
˫ (q1
, 𝞮, 𝞮)
All the Input symbols
are read and Stack is empty,
therefore the string “aabb” is
Accepted.

Rough Slide
Instantaneous Description for the string "aaabb"
(q0
, aaabb, z0
)

Instantaneous Description for the string "aaabb"
(q0
, aaabb, z0
) ˫ (q0
, aabb, az0
)
˫ (q0
, abb, aaz0
)
˫ (q1
, bb, aaaz0
)
˫ (q1
, b, aaz0
)
˫ (q1
, 𝞮, az0
)
are read but the Stack is
not empty, therefore the string
“aaabb” is not accepted.

Question
b2n
method. Give the graphical representation of the PDA obtained. Show the moves made by
PDA for the string aabbbb and aabbb

b2n
Rough Slide
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
a
q1
z0
a

b2n
, n>=1} by empty
stack method.
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
Push aa
q0
X X
a Push aa
q0
Pop a
q1
X
q1
z0
X X Pop Z0
q1
a
X Pop a
q1
X
q0
a, z0
| aaz0
q1
Representing PDA as a generalized
Transition Diagram
a, a | aaa
b, a | 𝞮
𝞮, Z0
| 𝞮
b, a | 𝞮

b2n
method.
P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
}
∑ = {a, b}
Γ ={a, Z0
}
q0
= q0
Z0
= Z0
F = { }
δ(q0
, a, z0
) =(q0
,aaz0
)
δ(q0
, a, a) =(q0
,aaa)
δ(q0
, b, a) =(q1
, 𝞮)
δ(q1
, b, a) =(q1
, 𝞮)
δ(q1
, 𝞮, z0
) =(q1
, 𝞮)
Push Operation
Pop Operation

b2n
Rough Slide
Instantaneous Description for the string "aabbbb"
(q0
, aabbbb, z0
)

b2n
, n>=1} by empty
stack method.
Instantaneous Description for the string "aabbbb"
(q0
, aabbbb, z0
) ˫ (q0
, abbbb, aaz0
)
˫ (q0
, bbbb, aaaaz0
)
˫ (q1
, bbb, aaaz0
)
˫ (q1
, bb, aaz0
)
˫ (q1
, b, az0
)
˫ (q1
, 𝞮, z0
)
˫ (q1
, 𝞮, 𝞮)
are read and Stack is
empty, therefore the string
“aabbbb” is accepted.

b2n
, n>=1} by empty
stack method.
Instantaneous Description for the string "aabbb"
(q0
, aabbb, z0
) ˫ (q0
, abbb, aaz0
)
˫ (q0
, bbb, aaaaz0
)
˫ (q1
, bb, aaaz0
)
˫ (q1
, b, aaz0
)
˫ (q1
, 𝞮, az0
)
are read but Stack not is
“aabbb” is not accepted.

Question
The push down automata indicate the acceptance
of input string in terms of
a) Finial state
b) Empty stack
c) Both (a) and (b)
d) None of these

Question
a) Finial state
b) Empty stack
c) Both (a) and (b)
d) None of these

Question
The instantaneous description is PDA shows
a) Present state
b) Stack symbol
c) String to be processed
d) All of these

Question
a) Present state
b) Stack symbol
d) All of these

Question
Construct PDA ( or DPDA) to recognize the language L={a2n
bn
PDA for the string aaaabb and aabbb

Question
bn
PDA for the string aaaabb and aabbb
Solution:
Logic
a a a a b b 𝞮
Initial
State q0
Go To
State q1
Push a
Go To
State q0
Go To
State q1
Push a
Go To
State q0
Pop a
Go To
State q2
Pop a
Be in
State q2
Pop Z0
Be in
State q2

bn
Rough Slide
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
a
q1
z0
a
q2
z0
a

bn
, n>=1} by empty
stack method.
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
q1
X X
a q1
Pop a
q2
X
q1
z0
Push a
q0
X X
a Push a
q0
X X
q2
z0
X X Pop Z0
q2
a
X Pop a
q2
X

bn
, n>=1} by empty
stack method.
q0
q1
Representing PDA as a generalized Transition Diagram
𝞮, Z0
| 𝞮
b, a | 𝞮
q2
b, a | 𝞮
a, z0
| z0
a, a | a
a, z0
| az0
a, a | aa

bn
, n>=1} by empty
stack method.

Question
a) Finial state
b) Empty store
c) Both (a) and (b)
d) None of these

Question
a) Present state
b) Stack symbol
d) All of these

Question
Design Deterministic Push Down Automata for the language
L={WcWR
, W ∈ (a+b)*} by empty stack. Show instantaneous
description for the string aabcbaa
Example Strings:
c , aca , bcb , aacaa , abcba , abbcbba
Logic: Push w on to stack, when c is seen switch to different state, for wr
do the
Popping operation

Design DPDA for the language L={WcWR
, W ∈ (a+b)*} by empty stack.
Rough Slide
State Stack
Top
Symbol
Input Symbols
a b c 𝞮
q0
z0
a
b
q1
Z0
a
b

State Stack
Top
Symbol
Input Symbols
a b c 𝞮
q0
z0
Push a
q0
Push b
q0
q1
X
a Push a
q0
Push b
q0
q1
X
b Push a
q0
Push b
q0
q1
X
q1
Z0
X X X Pop Z0
q1
a Pop a
q1
X X X
b
X Pop b
q1
X X
q0
q1
Transition Diagram
a, z0
| az0
b, z0
| bz0
a, a | aa
a, b | ab
b, b | bb
b, a | ba
a, a | 𝞮
b, b | 𝞮
𝞮, Z0
| 𝞮
c, z0
| z0
c, a | a
c, b | b

P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
}
∑ = {a, b, c}
Γ ={a, b, Z0
}
q0
= q0
Z0
= Z0
F = { }
δ(q0
, a, z0
) =(q0
,az0
)
δ(q0
, b, z0
) =(q0
,bz0
)
δ(q0
, a, a) =(q0
,aa)
δ(q0
, a, b) =(q0
,ab)
δ(q0
, b, b) =(q0
,bb)
δ(q0
, b, a) =(q0
,ba)
δ(q0
, c, z0
) =(q1
,z0
)
δ(q0
, c, a) =(q1
,a)
δ(q0
, c, b) =(q1
,b)
δ(q1
, a, a) =(q1
, 𝞮)
δ(q1
, b, b) =(q1
, 𝞮)
δ(q1
, 𝞮, z0
) =(q1
, 𝞮)
Push Operation
Pop Operation
No action

Rough Slide
Instantaneous Description for the string "aabcbaa"
(q0
, aabcbaa, z0
)

Instantaneous Description for the string "aabcbaa"
(q0
, aabcbaa, z0
) ˫ (q0
, abcbaa, az0
)
˫ (q0
, bcbaa, aaz0
)
˫ (q0
, cbaa, baaz0
)
˫ (q1
, baa, baaz0
)
˫ (q1
, aa, aaz0
)
˫ (q1
, a, az0
)
˫ (q1
, 𝞮, z0
)
˫ (q1
, 𝞮, 𝞮)
“aabcbaa” is accepted.

Question
Design Deterministic Push Down Automata for the language L={W, W ∈ (a+b)*
and na
(w)==nb
(w)} by empty stack. Show instantaneous description for the
string aababb
Example Strings:
ab , ba , baba , aabb , baab , aababb
Logic:
Input Stack Top Operation
a Z0
Push a
b Z0
Push b
a a Push a
b a Pop a
a b Pop b
b b Push b

Design DPDA for the language L={W, W ∈ (a+b)* and na
(w)==nb
(w)} by empty stack.
Rough Slide
State Stack
Top
Symbol
Input Symbols
a b 𝞮
q0
z0
a
b
q1
Z0
a
b

(w)==nb
(w)} by empty
stack.
State Stack
Top
Symbol
Input Symbols
a b 𝞮
q0
z0
Push a
q0
Push b
q0
q1
a Push a
q0
Pop a
q0
X
b Pop b
q0
Push b
q0
X
q1
Z0
X X Pop Z0
q1
a
X X X
b
X X X

(w)==nb
(w)} by empty
stack.
Instantaneous Description for the string "aababb"
(q0
, aababb, z0
) ˫ (q0
, ababb, az0
)
˫ (q0
, babb, aaz0
)
˫ (q0
, abb, az0
)
˫ (q0
, bb, aaz0
)
˫ (q0
, b, az0
)
˫ (q0
, 𝞮, z0
)
˫ (q1
, 𝞮, z0
)
˫ (q1
, 𝞮, 𝞮)
“aababb” is accepted.

Question
and na
(w) < nb
(w)} by final state. Show instantaneous description for the string
abbab
Example Strings:
b , bb , babab , abbb , bbbaa , abbab
Logic:
Input Stack Top Operation
a Z0
Push a
b Z0
Push b
a a Push a
b a Pop a
a b Pop b
b b Push b
𝞮 b Move to Final State

and na
(w) < nb
(w)} by final state.
State Stack
Top
Symbol
Input Symbols
a b 𝞮
q0
z0
Push a
q0
Push b
q0
X
a Push a
q0
Pop a
q0
X
b Pop b
q0
Push b
q0
q1
q1
Z0
X X X
a
X X X
b
X X q1
q0
q1
Transition Diagram
a, z0
| az0
b, z0
| bz0
a, a | aa
a, b | 𝞮
b, b | bb
b, a | 𝞮
𝞮, b | b
𝞮, b | b

and na
(w) < nb
P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
}
∑ = {a, b}
Γ ={a, b, Z0
}
q0
= q0
Z0
= Z0
F = {q1
}
δ(q0
, a, z0
) =(q0
,az0
)
δ(q0
, b, z0
) =(q0
,bz0
)
δ(q0
, a, a) =(q0
,aa)
δ(q0
, b, b) =(q0
,bb)
δ(q0
, a, b) =(q0
, 𝞮)
δ(q0
, b, a) =(q0
, 𝞮)
δ(q0
, 𝞮, b) =(q1
,b)
δ(q1
, 𝞮, b) =(q1
,b)
Push Operation
Pop Operation
No action
Note: for na
(w) > nb
(w)
δ(q0
, 𝞮, a) =(q1
,a)
δ(q1
, 𝞮, a) =(q1
,a)

and na
(w) < nb
Instantaneous Description for the string "abbab"
(q0
, abbab, z0
) ˫ (q0
, bbab, az0
)
˫ (q0
, bab, z0
)
˫ (q0
, ab, bz0
)
˫ (q0
, b, z0
)
˫ (q0
, 𝞮, bz0
)
˫ (q1
, 𝞮, bz0
)
˫ (q1
, 𝞮, bz0
)
are read and reached to
Final state, therefore the string
“abbab” is accepted.

Question
Design Deterministic Push Down Automata for the language L={ai
bJ
ck
, J=i+k
,i,k≥0} by empty stack. Show instantaneous description for the string abbbcc
Example Strings:
i k J=i+k Strings
ai
bJ
ck
0 0 0 𝞮
1 1 2 abbc
1 0 1 ab
0 1 1 bc
2 2 4 aabbbbcc
2 1 3 aabbbc
1 2 3 abbbcc

Question
bJ
ck
, J=i+k
,i,k≥0} by empty stack. Show instantaneous description for the string abbbcc
Example Strings:
𝞮 , abbc , ab , bc , aabbbbcc , aabbbc , abbbcc
Logic:
ai
bi
bk
ck 𝞮
Push a Pop a Push b Pop Z0
Pop b
ai
bJ
ck
ai
bi
bk
ck

bJ
ck
, J=i+k ,i,k≥0} by empty
stack.
Rough Slide
State Stack Top
Symbol
Input Symbols
a b c 𝞮
q0
z0
a
b
q1
Z0
a
b
q2 Z0
a
b
q3 Z0
a
b

bJ
ck
, J=i+k ,i,k≥0} by
empty stack.
State Stack
Top
Symbol
Input Symbols
a b c 𝞮
q0
z0
Push a , q0
Push b, q1
X q3
a Push a , q0
Pop a, q1
X X
b X X X X
q1
Z0
X Push b, q1
X X
a X Pop a, q1
X X
b X Push b, q1
Pop b, q2
X
q2 Z0
X X X q3
a X X X X
b X X Pop b, q2
X
q3 Z0
X X X Pop z0
, q3
a X X X X
b X X X X

bJ
ck
, J=i+k
,i,k≥0} by empty stack.
Instantaneous Description for the string "abbbcc"
(q0
, abbbcc, z0
) ˫ (q0
, bbbcc, az0
)
˫ (q0
, bbcc, z0
)
˫ (q1
, bcc, bz0
)
˫ (q1
, cc, bbz0
)
˫ (q2
, c, bz0
)
˫ (q2
, 𝞮, z0
)
˫ (q3
, 𝞮, z0
)
˫ (q3
, 𝞮, 𝞮)
are read and stack is
“abbbcc” is accepted.

Question
bn
, n>=1} by final state
method.

bn
, n>=1} by final state method.
Rough Slide
States Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
a
q1
z0
a
q2
z0
a

Question
Design Non-deterministic Push Down Automata for the language L={WWR
, W ∈
(a+b)*} by empty stack. Show instantaneous description for the string abba
or
Design Non-deterministic Push Down Automata for the language L={x, x ∈
(a+b)* and x is a palindrome} by empty stack. Show instantaneous description
for the string abba

Question
, W ∈
Example Strings:
𝞮 , aa , bb , aaaa , abaaba , bbaabb , abbbba

Question
, W ∈
Example Strings:
𝞮 , aa , bb , aaaa , abaaba , bbaabb
Logic:
Input Stack
Top
Operation
a Z0
Push a, q0
b Z0
Push b, q0
a a Push a, q0
; Pop a , q1
b a Push b, q0
a b Push a, q0
b b Push b, q0
; Pop b, q1

Design NPDA for the language L={WWR
Rough Slide
Input Stack
Top
Operation
a Z0
Push a, q0
b Z0
Push b, q0
a a Push a, q0
; Pop a , q1
b a Push b, q0
a b Push a, q0
b b Push b, q0
; Pop b, q1
𝞮 Z0
Change state to q2
𝞮 Z0
Pop Z0
, q2
(q0
, abba, z0
)
Instantaneous Description for the string "abba"
Logic

Input Stack
Top
Operation
a Z0
Push a, q0
b Z0
Push b, q0
a a Push a, q0
; Pop a , q1
b a Push b, q0
a b Push a, q0
b b Push b, q0
; Pop b, q1
(q0
, abba, z0
) ˫ (q0
, bba, az0
)
˫ (q0
, ba,baz0
)
Instantaneous Description for the string "abba"
(q0
, a, bbaz0
) (q1
, a, az0
)
Logic
(q0
, 𝞮, abbaz0
)
(q1
, 𝞮, z0
)
(q1
, 𝞮, 𝞮)
Rejected
Accepted

Rough Slide
State Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
a
b
q1
Z0
a
b
q2
Z0
a
b

State Stack Top
Symbol
Input Symbols
a b 𝞮
q0
z0
Push a , q0
Push b , q0
q2
a Push a , q0
Push b , q0 X
Pop a , q1
b Push a , q0
Push b , q0 X
Pop b , q1
q1
Z0 X X q2
a Pop a , q1 X X
b X Pop b , q1 X
q2
Z0 X X Pop Z0
, q2
a X X X
b X X X

P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
, q2
}
∑ = {a, b}
Γ ={a, b, Z0
}
q0
= q0
Z0
= Z0
F = { }
δ(q0
, a, z0
) =(q0
,az0
)
δ(q0
, b, z0
) =(q0
,bz0
)
δ(q0
, a, b) ={(q0
,ab)
δ(q0
, b, a) ={(q0
,ba)
δ(q0
, a, a) ={(q0
, aa), (q1
, 𝞮)}
δ(q0
, b, b) ={(q0
,bb), (q1
, 𝞮)}
δ(q1
, a, a) =(q1
, 𝞮)
δ(q1
, b, b) =(q1
, 𝞮)
δ(q1
, 𝞮, z0
) =(q2
, Z0
)
δ(q2
, 𝞮, z0
) =(q2
, 𝞮)
q0
q1
Transition Diagram
a, z0
| az0
b, z0
| bz0
a, a | aa
a, b | ab
b, b | bb
b, a | ba
a, a | 𝞮
b, b | 𝞮
a, a | 𝞮
b, b | 𝞮
q2
𝞮, Z0
| z0
𝞮, Z0
| 𝞮

Equivalence of PDA and CFG
- Converting CFG to PDA
- Converting PDA to CFG

Procedure for converting PDA to CFG
1. The input symbols of PDA will be the terminals of CFG
2. If PDA moves from state qi
to qj
on consuming the input a ∈ Σ when Z is top of the
stack, then the non-terminals of CFG are the triplets of the form (qi
Zqj
)
3. If q0
is the start state and qf
is the final state then (q0
Zqf
) is the start symbol of the
CFG
4. The productions of CFG can be obtained from the transitions of PDA as shown below:
a. For each transition of the form δ(qi
, a, Z) = (qj
, AB) introduce the
productions of the form
(qi
Zqk
) -> a(qj
Aql
)(ql
Bqk
) where qk
and ql
will take all possible values from Q
a. For each transition of the form δ(qi
, a, Z) = (qj
, ε ) introduce the production
(qi
Zqj
) -> a
Note: Using this procedure, we may introduce lot of useless symbols, which in any way can be
eliminated.

Question
Obtain CFG that generates the language accepted by
the PDA
P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
}
∑ = {a, b}
Γ ={A, Z}
q0
= q0
Z0
= Z
F = {q1
}
δ(q0
, a, Z)=(q0
, AZ)
δ(q0
, b, A)=(q0
, AA)
δ(q0
, a, A)=(q1
, 𝞮) q0
q1
a, z | AZ
b, A | AA
a, A | 𝞮

Question
Obtain CFG that generates the language accepted by the PDA M=({q0
, q1
}, {a,
b}, {A, Z}, q0
, Z , {q1
} ,
{ δ(q0
, a, Z)=(q0
, AZ),
δ(q0
, b, A)=(q0
, AA),
δ(q0
, a, A)=(q1
, 𝞮))
Solution:
For δ of the form
δ(qi
, a, Z) = (qj
, ε )
Resulting Production
(qi
Zqj
) → a
δ(q0
, a, A)=(q1
, ε) (q0
Aq1
) → a

, q1
}, {a, b}, {A, Z}, q0
, Z , {q1
} ,
{δ(q0
, a, Z)=(q0
, AZ), δ(q0
, b, A)=(q0
, AA), δ(q0
, a, A)=(q1
, 𝞮))
1 July 2021 CSE, BMSCE
75
Solution (Contd…):
For δ of the form
δ(qi
, a, Z) = (qj
, AB)
(qi
Zqk
) → a(qj
Aql
)(ql
Bqk
)
where qk
and ql
will take all possible
values from Q
δ(q0
, a, Z)=(q0
, AZ) (q0
Zq0
) → a(q0
Aq0
)(q0
Zq0
) | a(q0
Aq1
)(q1
Zq0
)
(q0
Zq1
) → a(q0
Aq0
)(q0
Zq1
) | a(q0
Aq1
)(q1
Zq1
)

, q1
}, {a, b}, {A, Z}, q0
, Z , {q1
} ,
{δ(q0
, a, Z)=(q0
, AZ), δ(q0
, b, A)=(q0
, AA), δ(q0
, a, A)=(q1
, 𝞮))
1 July 2021 CSE, BMSCE
76
For δ of the form
δ(qi
, a, Z) = (qj
, AB)
(qi
Zqk
) → a(qj
Aql
)(ql
Bqk
)
where qk
and ql
values from Q
δ(q0
, a, Z)=(q0
, AZ) (q0
Zq0
) → a(q0
Aq0
)(q0
Zq0
) | a(q0
Aq1
)(q1
Zq0
)
(q0
Zq1
) → a(q0
Aq0
)(q0
Zq1
) | a(q0
Aq1
)(q1
Zq1
)
δ(q0
, b, A)=(q0
, AA) (q0
Aq0
) → b(q0
Aq0
)(q0
Aq0
) | b(q0
Aq1
)(q1
Aq0
)
(q0
Aq1
) → b(q0
Aq0
)(q0
Aq1
) | b(q0
Aq1
)(q1
Aq1
)
Start Symbol of the grammar is (q0
Zq1
)

, q1
}, {a, b}, {A, Z}, q0
, Z , {q1
} ,
{δ(q0
, a, Z)=(q0
, AZ), δ(q0
, b, A)=(q0
, AA), δ(q0
, a, A)=(q1
, 𝞮))
G=(V, T, P, S)
V= {(q0
Zq1
) , (q0
Aq1
) , (q0
Zq0
), (q0
Aq0
) ,
(q1
Zq0
) , (q1
Zq1
) , (q1
Aq0
) , (q1
Aq1
) }
T={a, b}
P={
(q0
Aq1
) → a
(q0
Zq0
) → a(q0
Aq0
)(q0
Zq0
) | a(q0
Aq1
)(q1
Zq0
)
(q0
Zq1
) → a(q0
Aq0
)(q0
Zq1
) | a(q0
Aq1
)(q1
Zq1
)
(q0
Aq0
) → b(q0
Aq0
)(q0
Aq0
) | b(q0
Aq1
)(q1
Aq0
)
(q0
Aq1
) → b(q0
Aq0
)(q0
Aq1
) | b(q0
Aq1
)(q1
Aq1
)
}
S=(q0
Zq1
)

Question
Obtain CFG that generates the language accepted by
the PDA
P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1 ,
q2
}
∑ = {a, b}
Γ ={A, Z}
q0
= q0
Z0
= Z
F = {q2
}
δ(q0
, a, Z)=(q0
, AZ)
δ(q0
, a, A)=(q0
, A)
δ(q0
, b, A)=(q1
, 𝞮)
δ(q1
, 𝞮, Z)=(q2
, 𝞮)

Converting PDA to CFG: M=({ q0
, q1 ,
q2
}, {a, b} , {A, Z}, q0 ,
Z, {q2
}, {
δ(q0
, a, Z)=(q0
, AZ), δ(q0
, a, A)=(q0
, A), δ(q0
, b, A)=(q1
, 𝞮), δ(q0
, 𝞮, Z)=(q2
, 𝞮)}
Solution:
To obtain CFG from the PDA, all transitions should be of the form
δ(qi
, a, Z) = (qj
, AB)
or
δ(qi
, a, Z) = (qj
, ε )
In the given transitions except the second transition, all transitions are in the required form. So, let
us take the second transition
δ(q0
, a, A)=(q0
, A)
This transition indicates that when input symbol a is encountered and top stack symbol is A, the PDA
remains in the state q0
and contents of the stack are not altered. This can be interpreted as delete A
from stack and insert A onto the stack. So once A is deleted from the stack we enter into a new state
q3
. But in state q3
without consuming any input we add A onto the stack. The corresponding
transitions are
δ(q0
, a, A)=(q3
, 𝞮)
δ(q3
, 𝞮, Z)=(q0
, AZ)

, q1 ,
q2
}, {a, b} , {A, Z}, q0 ,
Z, {q2
}, {
δ(q0
, a, Z)=(q0
, AZ), δ(q0
, a, A)=(q0
, A), δ(q0
, b, A)=(q1
, 𝞮), δ(q1
, 𝞮, Z)=(q2
, 𝞮)}
So the given PDA can be written using the following transitions
δ(q0
, a, Z)=(q0
, AZ)
δ(q0
, a, A)=(q3
, 𝞮)
δ(q3
, 𝞮, Z)=(q0
, AZ)
δ(q0
, b, A)=(q1
, 𝞮)
δ(q1
, 𝞮, Z)=(q2
, 𝞮)
For δ of the form
δ(qi
, a, Z) = (qj
, ε )
(qi
Zqj
) → a
δ(q0
, a, A)=(q3
, 𝞮) (q0
Aq3
)→ a
δ(q0
, b, A)=(q1
, 𝞮) (q0
Aq1
) → b
δ(q1
, 𝞮, Z)=(q2
, 𝞮) (q1
Zq2
) → 𝞮

, q1 ,
q2
}, {a, b} , {A, Z}, q0 ,
Z, {q2
}, {
δ(q0
, a, Z)=(q0
, AZ), δ(q0
, a, A)=(q0
, A), δ(q0
, b, A)=(q1
, 𝞮), δ(q1
, 𝞮, Z)=(q2
, 𝞮)}
δ(q0
, a, Z)=(q0
, AZ) , δ(q0
, a, A)=(q3
, 𝞮), δ(q3
, 𝞮, Z)=(q0
, AZ)
δ(q0
, b, A)=(q1
, 𝞮) , δ(q1
, 𝞮, Z)=(q2
, 𝞮)
For δ of the form
δ(qi
, a, Z) = (qj
, AB)
(qi
Zqk
) → a(qj
Aql
)(ql
Bqk
)
where qk
and ql
values from Q
δ(q0
, a, Z)=(q0
, AZ) (q0
Zq0
) → a(q0
Aq0
)(q0
Zq0
) | a(q0
Aq1
)(q1
Zq0
) |
a(q0
Aq2
)(q2
Zq0
) | a(q0
Aq3
)(q3
Zq0
)
(q0
Zq1
) → a(q0
Aq0
)(q0
Zq1
) | a(q0
Aq1
)(q1
Zq1
) |
a(q0
Aq2
)(q2
Zq1
) | a(q0
Aq3
)(q3
Zq1
)
(q0
Zq2
) → a(q0
Aq0
)(q0
Zq2
) | a(q0
Aq1
)(q1
Zq2
) |
a(q0
Aq2
)(q2
Zq2
) | a(q0
Aq3
)(q3
Zq2
)
(q0
Zq3
) → a(q0
Aq0
)(q0
Zq3
) | a(q0
Aq1
)(q1
Zq3
) |
a(q0
Aq2
)(q2
Zq3
) | a(q0
Aq3
)(q3
Zq3
)

, q1 ,
q2
}, {a, b} , {A, Z}, q0 ,
Z, {q2
}, {
δ(q0
, a, Z)=(q0
, AZ), δ(q0
, a, A)=(q0
, A), δ(q0
, b, A)=(q1
, 𝞮), δ(q1
, 𝞮, Z)=(q2
, 𝞮)}
δ(q0
, a, Z)=(q0
, AZ) , δ(q0
, a, A)=(q3
, 𝞮), δ(q3
, 𝞮, Z)=(q0
, AZ)
δ(q0
, b, A)=(q1
, 𝞮) , δ(q1
, 𝞮, Z)=(q2
, 𝞮)
For δ of the form
δ(qi
, a, Z) = (qj
, AB)
(qi
Zqk
) → a(qj
Aql
)(ql
Bqk
)
where qk
and ql
will take all possible values from Q
δ(q3
, 𝞮, Z)=(q0
, AZ) (q3
Zq0
) → (q0
Aq0
)(q0
Zq0
) | (q0
Aq1
)(q1
Zq0
) |
(q0
Aq2
)(q2
Zq0
) | (q0
Aq3
)(q3
Zq0
)
(q3
Zq1
) → (q0
Aq0
)(q0
Zq1
) | (q0
Aq1
)(q1
Zq1
) |
(q0
Aq2
)(q2
Zq1
) | (q0
Aq3
)(q3
Zq1
)
(q3
Zq2
) → (q0
Aq0
)(q0
Zq2
) | (q0
Aq1
)(q1
Zq2
) |
(q0
Aq2
)(q2
Zq2
) | (q0
Aq3
)(q3
Zq2
)
(q3
Zq3
) → (q0
Aq0
)(q0
Zq3
) | (q0
Aq1
)(q1
Zq3
) |
(q0
Aq2
)(q2
Zq3
) | (q0
Aq3
)(q3
Zq3
)
Start Symbol of the grammar is (q0
Zq2
)

Greibach Normal Form (GNF)
In Greibach Normal Form (GNF), there is a restriction on the terminals and
variables appearing on the right hand side of the production.
A CFG G = (V, T, P, S) is said to be in GNF if every production is of the form
A → a⍺
where a ∈ T and ⍺ ∈ V ∗ , i.e., ⍺ is a string of zero or more variables.
The first symbol on the right hand side of the production must be a terminal and
it can be followed by zero or more Variables.
If Grammar is in GNF, it can be easily converted into Push Down Automata (PDA),
which accepts Context Free Languages

Procedure for converting CFG to PDA
Given any grammar, first obtain the grammar in GNF (Greibach Normal Form) and then
obtain PDA.
The steps to be followed to convert a grammar G=(V,T,P,S) to its equivalent PDA are
1. Convert the grammar to GNF
2. Let q0
be the start state and z0
is initial symbol on stack. Without consuming any input,
push the start symbol S onto the stack and change state to q1
The transition for this can be δ(q0
, ε, z0
)=(q1
,Sz0
)
3. For each production of the form A -> aβ introduce the transition
δ(q1
, a, A)=(q1
, β)
4. Finally, in state q1
, without consuming any input change the state to qf
which is an
accepting state. The transition can be of the form δ(q1
, ε, z0
)=(qf
, z0
)

Question
Convert following Grammar to PDA. Show the instantaneous description for the
string “aabba”
S → aABC
A → aB | a
B → bA | b
C → a

Convert following Grammar to PDA.
S → aABC ; A → aB | a ; B → bA | b ; C → a
Solution:
Step1: Convert the grammar to GNF
The given grammar is in GNF
Step2: Let q0
be the start state and z0
is initial symbol on stack. Without consuming any
input, push the start symbol S onto the stack and change state to q1
The transition for this can be δ(q0
, ε, z0
)=(q1
,Sz0
)

Rough Slide
Step3: For each production of the form A -> aβ introduce the transition δ(q1
, a, A)=(q1
,
β)
Production Transition
S → aABC
A → aB
A → a
B → bA
B → b
C → a

Step3: For each production of the form A -> aβ introduce the transition δ(q1
, a, A)=(q1
,
β)
Step4:Finally, in state q1
, without consuming any input change the state to qf
which is an
accepting state. The transition can be of the form δ(q1
, ε, z0
)=(qf
, z0
)
Production Transition
S → aABC δ(q1
, a, S)=(q1
, ABC)
A → aB δ(q1
, a, A)=(q1
, B)
A → a δ(q1
, a, A)=(q1
, 𝞮)
B → bA δ(q1
, b, B)=(q1
, A)
B → b δ(q1
, b, B)=(q1
, 𝞮)
C → a δ(q1
, a, C)=(q1
, 𝞮)

P=(Q, ∑ , Γ , q0
, Z0
, δ, F)
Q = { q0
, q1
, qf
}
∑ = {a, b}
Γ ={S, A, B, C, Z0
}
q0
= q0
Z0
= Z0
F = {qf
}
δ(q0
, ε, z0
)=(q1
,Sz0
)
δ(q1
, a, S)=(q1
, ABC)
δ(q1
, a, A)=(q1
, B)
δ(q1
, a, A)=(q1
, 𝞮)
δ(q1
, b, B)=(q1
, A)
δ(q1
, b, B)=(q1
, 𝞮)
δ(q1
, a, C)=(q1
, 𝞮)
δ(q1
, ε, z0
)=(qf
, z0
)

Rough Slide
Transitions of PDA
δ(q0
, ε, z0
)=(q1
,Sz0
)
δ(q1
, a, S)=(q1
, ABC)
δ(q1
, a, A)=(q1
, B)
δ(q1
, a, A)=(q1
, 𝞮)
δ(q1
, b, B)=(q1
, A)
δ(q1
, b, B)=(q1
, 𝞮)
δ(q1
, a, C)=(q1
, 𝞮)
δ(q1
, ε, z0
)=(qf
, z0
)
Instantaneous Description for the string "aabba"

Transitions of PDA
δ(q0
, ε, z0
)=(q1
,Sz0
)
δ(q1
, a, S)=(q1
, ABC)
δ(q1
, a, A)=(q1
, B)
δ(q1
, a, A)=(q1
, 𝞮)
δ(q1
, b, B)=(q1
, A)
δ(q1
, b, B)=(q1
, 𝞮)
δ(q1
, a, C)=(q1
, 𝞮)
δ(q1
, ε, z0
)=(qf
, z0
)
(q0
, aabba, z0
) ˫ (q1
, aabba, Sz0
)
˫ (q1
, abba, ABCz0
)
Instantaneous Description for the string "aabba"
(q1
, bba, BBCz0
) (q1
, bba, BCz0
)
(q1
, ba, BCz0
)
Accepted
All Input Symbols are read and the final state is reached,
therefore the string “aabba” is accepted
(q1
, ba, ABCz0
)
(q1
, a, Cz0
) (q1
, a, ACz0
)
(q1
, 𝞮, z0
)
(qf
, 𝞮, z0
)

Question
S → aABB | aAA
A → aBB | a
B → bBB | A
C → a

Pumping Lemma for CFL

Pumping Lemma for Context Free Language (CFL)
Let L be context free language and is infinite. Let z is sufficiently
long string and z ∈ L ,
so that |z| >=n where n is some positive integer. If the string z can
be decomposed into combination of strings
z=uvwxy
Such that |vwx| <= n , |vx| >=1 ,
then uvi
wxi
y ∈ L for i=0,1,2,3,……

Using Pumping Lemma to show certain languages are not
CFL
The general strategy used to prove that a given language is not CFL
is shown below:
1. Assume that the language L is infinite and it is context free
2. Select string say z and break it into sub strings u, v, w, x and y
such that |z|>=n,
z=uvwxy where |vwx| <= n , |vx| >=1
3. Find any i such that uvi
wxi
y ∉ L . According to pumping lemma
uvi
wxi
y ∈ L . So, the result is a contradiction to the assumption
that the language is context free. Therefore, the given language L
is not context free.

Question
Show that the language L={an
bn
cn
, n≥1} is not Context Free Language (CFL).

Question
bn
cn
Solution:
Step1: Let L be CFL and let z= an
bn
cn
∈ L
Step2: Note that |z|=| an
bn
cn
|=3n >= n
Split z into uvwxy such that |vwx| ≤ n and |vx| ≥ 1

Question
bn
cn
Solution:
Step1: Let L be CFL and let z= an
bn
cn
∈ L
bn
cn
|=3n >= n
|vx| =|aj
al
|= (j+ l) ≥ 1
|vwx| =|aj
ak
al
|=(j+k+ l) ≤ n

Question
bn
cn
Step3: For all i, uvi
wxi
y ∈ L.
Hence L={an
bn
cn
, n≥0} is not CFL

Question
Show that the language L={WW , W ∈ (a+b)* }is not Context Free Language (CFL).

Show that the language L={WW , W ∈ (a+b)*} is not Context Free Language (CFL).
Rough Slide
Solution:
Step1: Let L be CFL and Let z= an
bn
an
bn
∈ L
bn
an
bn
|=4n >= n

Question
Show that the language L={W , W ∈ (0+1)* , |w| is a perfect square }is not Context Free
Language (CFL).

Show that the language L={W , W ∈ (0+1)* , |w| is a perfect square }is not Context Free Language
(CFL).
Rough Slide
Solution:
Step1: Let L be CFL and Let z= 0n2
∈ L
Step2: Note that |z|=| 0n2
|=n2
≥ n
Step 3: For all i, uvi
wxi
y ∈ L.

Show that the language L={W , W ∈ (0+1)* , |w| is a perfect square }is not Context Free Language
(CFL).
Rough Slide
Solution:
Step1: Let L be CFL and Let z= 0n2
∈ L
Step2: Note that |z|=| 0n2
|=n2
≥ n
Step 3: For all i, uvi
wxi
y ∈ L.
Let |vx| = m , m ≤ n.
But for i=2,
n2
< |uv2
wx2
y| = n2
+m ≤ (n2
+ n) < (n+1)2
And hence the length is nit a perfect square and therefore string is not in L. This
contradiction allows us to conclude that our assumption was wrong and L is not CFL.

Question
Show that the language L={an!
Solution:
Step1: Let L be CFL and it is infinite, Let z= an!
∈ L
Step2: Note that |z|=| an!
|=n! > n
|vwx| ≤ n, |aj
+ak
+al
| =(j+k+ l ) ≤ n
v=aj
, w=ak
, x=al
an!
= an!-(j+k+l )
aj
ak
al
|vx| ≥ 1, |aj
+al
| =(j+ l )≥ 1

Question
Show that the language L={an!
, n≥0 is not Context Free Language (CFL).
Solution (Contd..):
Step3: For all i, uvi
wxi
y ∈ L.
But for i=2,
an!-(j+k+l)
a2j
ak
a2l
= an!+j+l
|an!+j+l
|= n!+j+l ≤ n!+n
< n!+n!(n)
< n!(n+1)
n! < (n!+j+l) < (n+1)!
Since, (n!+j+l) lies in between n! and (n+1)! ,
the string generated by uv2
wx2
y = an!+j+l
∉ L
which is contradiction, hence L is not CFL

Question
Pumping lemma for context free language breaks
the strings into
a) Two parts
b) Three parts
c) Five parts
d) Six parts

Answer
Pumping lemma for context free language breaks
the strings into
a) Two parts
b) Three parts
c) Five parts
d) Six parts

Question
In pumping lemma for context free language
a) We start by assuming the given language is context free and
then we get contradict
b) We first convert the given language into regular language and
then apply steps on
c) Both (a) and (b)
d) None of these

Answer
In pumping lemma for context free language
a) We start by assuming the given language is context free
and then we get contradict
b) We first convert the given language into regular language and
then apply steps on
c) Both (a) and (b)
d) None of these

Closure properties of CFL

119
Closure properties of CFL
□ Closure properties consider operations on CFL that are
guaranteed to produce a CFL
□ The CFL’s are closed under union, concatenation, closure (star),
reversal, homomorphism, inverse homomorphism and
substitution.
□ CFL’s are not closed under intersection and complementation

120
CFLs are closed under Union Operation
□ Let L1
and L2
be two CFL’s generated by CFGs G1
=(V1
, T1
, S1
, P1
) and G2
=(V2
, T2
, S2
, P2
)

121
□ Let L1
and L2
=(V1
, T1
, S1
, P1
) and G2
=(V2
, T2
, S2
, P2
)
□ To get grammar for L1
∪ L2
?
■ Add new start symbol S and rules S → S1
| S2
■ We get grammar G = (V, T, P, S) where
V = V1
∪ V2
∪ { S }, where S ∉ V1
∪ V2
P = P1
∪ P2
∪ { S → S1
| S2
}

122
□ Let L1
and L2
=(V1
, T1
, S1
, P1
) and G2
=(V2
, T2
, S2
, P2
)
∪ L2
?
■ Add new start symbol S and rules S → S1
| S2
■ We get grammar G = (V, T, P, S) where
V = V1
∪ V2
∪ { S }, where S ∉ V1
∪ V2
P = P1
∪ P2
∪ { S → S1
| S2
}
□ Example:
■ L1
= { an
bn
| n ≥ 0 } , L2
= { bn
an
| n ≥ 0 }
■ G1
: S1
→ aS1
b | ε G2
: S2
→ bS2
a | ε
■ L1 ∪ L2 is G = ({S1
, S2
, S}, {a, b}, P, S) where
P = {P1 ∪ P2 ∪ {S → S1
| S2
}}
□ Union of two CFLs results in CFL.

123
CFLs are closed under Concatenation Operation
□ Let L1
and L2
=(V1
, T1
, S1
, P1
) and G2
=(V2
, T2
, S2
, P2
)
L2
?
■ Add new start symbol S and rule S → S1
S2
■ We get G = (V, T, P, S) where
V = V1
∪ V2
∪ { S }, where S ∉ V1
∪ V2
P = P1
∪ P2
∪ { S → S1
S2
}
□ Example:
■ L1
= { an
bn
| n ≥ 0 } with G1
: S1
→ aS1
b | ε
■ L2
= { bm
am
| m ≥ 0 } with G2
: S2
→ bS2
a | ε
■ L1
L2
= { an
b{n+m}
am
| n, m ≥ 0 } with G = ({S, S1
, S2
}, {a, b},
{S → S1
S2
, S1
→ aS1
b | ε, S2
→ bS2
a}, S)
□ Concatenation of Two CFLs results in a CFL

124
CFLs are closed under Kleene’s-Closure or
Star-Closure Operation
□ Let L1
be the CFL generated by CFG G1
=(V1
, T1
, S1
, P1
)
□ Kleene Star Closure L(G)=(L(G1
))*
□ Example:
■ L1
= {an
bn
| n ≥ 0}
(L1
)*= { a{n1}
b{n1}
... a{nk}
b{nk}
| k ≥ 0 and ni ≥ 0 for all i }
□ To get the grammar for (L1
)*
■ Add new start symbol S and rules S → SS1
| ε.
■ We get G = (V, T, P, S) where
V = V1
∪ { S }, where S ∉ V1
P = P1
∪ { S → SS1
| ε}

Context-free languages are not closed under Intersection
□ Consider two CFL’s
L1
={an
bn
cm
, n ≥ 0, m ≥ 0} and
L2
={an
bm
cm
, n ≥ 0, m ≥ 0}
□ But L1 ∩ L2 ={an
bn
cn
, n ≥ 0}, which is not CFL
□ Thus family of CFL’s is not closed under Intersection operation.

126
Intersection
□ The CFL’s are not closed under intersection
□ Example:
■ L = {0n
1n
2n
|n ≥ 1} is not context-free.
■ L1 = {0n
1n
2i
|n ≥ 1,i ≥1 }, L2 = {0i
1n
2n
|n ≥ 1,i ≥1 } are CFL’s with
corresponding grammars for L1: S->AB; A->0A1 | 01; B->2B | 2 ,
and for L2: S ->AB; A->0A | 0; B->1B2 | 12.
■ However, L = L1
∩ L2
■ Thus intersection of CFL’s is not CFL

Context-free languages are not closed under Complement
□ If the family of CFL’s were closed under Complementation then the right side
of the above expression would be a CFL for any L1
and L2
. But this contradicts
what we have just shown in previous slide that the intersection of two CFLs is
not necessarily Context free.
□ Consequently, the family of CFL is not closed under Complementation

128
Set Difference
□ L1
and L2
are CFLs. L1
- L2
is not necessarily a CFL
Proof:
■ L1 = ∑* - L
■ ∑* is regular and is also CFL
■ But ∑* - L = LC
■ If CFLs were closed under set difference, then ∑* - L = LC
would
always be a CFL.
■ But CFL’s are not closed under complementation

129
Reversal
□ The CFL’s are closed under reversal
□ This means then if L is a CFL, so LR
is a CFL
□ It is enough to reverse each production of a CFL for L, i.e.,
substitute A→α by A→αR
□ Example:
■ L = { an
bn
| n ≥ 0 } with P : S → aSb | ε
■ LR
= {bn
an
| n ≥ 0 } with PR
: S → bSa | ε

130
Homomorphism
□ Closure under homomorphism of CFL L for every a∈∑
□ Suppose L is a CFL over alphabet ∑ and h is a homomorphism
on ∑.
□ Let s be a substitution that replaces every a ∈ ∑, by h(a). i.e.,
s(a) = {h(a)}.
□ Then h(L) = s(L).
□ h(L) ={h(a1
)…h(ak
) | k ≥ 0} where h(ak
) is a homomorphism for
every ak
∈ ∑.

131
Inverse homomorphism
□ To recall: If h is a homomorphism, and L is any language, then
h-1
(L), called an inverse homomorphism, is the set of all strings
w such that h(w)∈L
□ The CFL’s are closed under inverse homomorphism.
□ Theorem: If L is a CFL and h is a homomorphism, then h-1
(L) is
a CFL

132
Inverse homomorphism – proof
Stack
PDA
h
Accept/
Reject
Input
a h(a)
Buffer

133
Proof Contd...
□ After input a is read, h(a) is placed in a buffer.
□ Symbols of h(a) are used one at a time and fed to PDA being
simulated.
□ Only when the buffer is empty does the PDA read another of its
input symbol and apply homomorphism to it.

134
Proof Contd...
□ Suppose h applies to symbols of alphabet Σ and produces
strings in T*.
□ Let PDA P = (Q, T, Γ, δ, q0
, Z0
, F) that accept CFL L by
final state.
□ Construct a new PDA P′ = (Q′, Σ, Γ, δ′, (q0
, ε), Z0
, F X {ε})
to simulate language of h-1
(L), where
■ Q′ is the set of pairs (q, x) such that
□ q is a state in Q
□ x is a suffix of some string h(a) for some input string a in Σ

135
Proof Contd...
■ δ′ is defined by
□ δ′((q, ε), a, X) = {((q, h(a)),a,X)}
□ If δ(q, b, X) = {(p, γ)} where b∈T or b = ε then δ′((q, bx), ε, X) = {((p, x), γ)}
■ The start state of P’ is (q0, ε)
■ The accepting state of P′ is (q, ε), where q is an accepting state of P.
■ (q0
,h(w),Z0
)|-*P (p,ε,γ) iff ((q0
,ε),w,Z0
) |-*P′ ((p, ε), ε, γ)
■ P accepts h(w) if and only if P′ accepts w, because of the way the accepting states of
P′ are defined.
■ Thus L(P′)=h-1
(L(P))

136
Intersection with RL
□ Theorem: If L is CFL and R is a regular language, then L ∩
R is a CFL.
Stack
PDA
FA
AN
D
Accept/
Reject

137
Intersection with RL Proof
□ P=(QP
, ∑, Γ, δP
, qP
, Z0
, FP
) be PDA to accept CFL by final
state
□ A=(QA
, ∑, δA
, qA
, FA
) be a DFA for RL
□ Construct PDA P′ = (Q, ∑, Γ, δ, qo
, Z0
, F) where
■ Q = Qp
X QA
■ qo
= (qp
, qA
)
■ F = (FP
X FA
)
■ δ is in the form δ((q, p), a, X) = ((r, s), γ) such that
1. s = δA
(p, a)
2. (r, γ) is in δP
(q, a, X)

138
Proof Contd…
□ For each move of PDA P, we make the same move in PDA P′ and
also we carry along the state of DFA A in a second component of
P′.
□ P′ accepts a string w iff both P and A accept w.
□ w is in L ∩ R.
□ The moves ((qp
, qA
), w, Z) |-*P′ ((q, p), ε, γ) are possible iff (qp
,
w, Z) |-*P (q, ε, γ) moves and p = δ*(qA
, w) transitions are
possible.

139
Set Difference with RL
□ For a CFL’s L, and a regular language R.
L - R is a CFL.
Proof:
■ R is regular and RC
is also regular
■ L - R = L ∩ RC
■ Complement of Regular Language (RC
) is regular
■ Intersection of a CFL and a regular language is CFL
■ CFL and Regular language are closed under intersection.

140
Complementation
□ LC
is not necessarily a CFL
□ Proof:
■ Assume that CFLs were closed under complement.
■ If L is a CFL then LC
is a CFL
■ Since CFLs are closed under union, L1
C
∪ L2
C
is a CFL
■ And by our assumption (L1
C
∪ L2
C
) C
is a CFL
■ But (L1
C
∪ L2
C
) C
= L1
∩ L2
which we just showed isn’t
necessarily a CFL.
■ Contradiction!

141
Substitution
□ Each symbol in the strings of one language is replaced by
an entire CFL language
□ Useful in proving some other closure properties of CFL
□ Example: S(0) = {an
bn
| n ≥1}, S(1) = {aa, bb} is a
substitution on alphabet Σ ={0, 1}.

142
Substitution
■ Theorem: If a substitution s assigns a CFL to every symbol in the alphabet
of a CFL L, then s(L) is a CFL.
■ Proof
□ Let G = (V, ∑, P, S) be grammar for L
□ Let Ga
= (Va
, Ta
, Pa
, Sa
) be the grammar for each a ∈ ∑ with V∩Va
=φ
□ G′= (V′, T′, P′, S) for s(L) where
■ V′ = V ∪ Va
■ T′ = union of Ta
for all a ∈ ∑
■ P′ consists of
▪ All productions in any Pa
for a ∈ ∑
▪ In productions of P, each terminal a is replaced by Sa
■ A detailed proof that this construction works is in the reader.
■ Intuition: this replacement allows any string in La to take the place
of any occurrence of a in any string of L.

143
Example (1)
□ L = {0n
1n
| n ≥1}, generated by the grammar
S→0S1|01,
□ s(0) = {an
bm
|m≤ n}, generated by the grammar S→aSb|A;
A→aA| ab,
□ s(1)={ab, abc}, generated by the grammar S → abA, A → c
|ε
□ Rename second and third S’s to S0
and S1
respectively.
Rename second A to B. Resulting grammars are:
S→0S1 | 01
S0
→aS0
b | A; A→aA | ab
S1
→abB; B→c | ε

144
Example(1) Contd...
□ In the first grammar replace 0 by S0
and 1 by S1
. The combined
grammar:
G′ = ({S, S0
, S1
, A, B}, {a, b}, P′, S),
where P′ = {S → S0
SS1
| S0
S1
, S0
→ aS0
b | A, A →aA | ab,
S1
→abB, B→ c | ε}

145
Application of Substitution
□ Closure under union of CFL’s L1
and L2
□ Closure under concatenation of CFL’s L1
and L2
□ Closure under Kleene’s star (closure * and positive
closure +
) of CFL’s L1
□ Closure under homomorphism of CFL Li
for every
ai
∈ ∑

Thanks for Listening

Closure properties of CFG’s

Question
The context free languages are closed under:
a) Intersection
b) Complement
c) Kleene’s-Closure
d) None of the mentioned

Question
Context free languages are not closed under:
a) Intersection
b) Complement
c) All of the mentioned

Applications of Context Free Grammars (CFG)
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