06_PumpingLemma compiler design of chapter 4.ppt

TK Prasad Pumping Lemma 1
Nonregularity Proofs

Regular Languages: Grand Unification
)
(
)
(
)
(
DFAs
L
NFAs
L
s
NFA
L


 
)
(
)
( RE
L
FA
L 
(Parallel Simulation)
(Rabin and Scott’s work)
(Collapsing graphs;
Structural Induction)
(S. Kleene’s work)
)
(
)
( RG
L
FA
L  (Construction)
(Solving linear equations)
)
(
)
( RE
L
RG
L 

Role of various representations
for Regular Languages
• Closure under complemention. (DFAs)
• Closure under union, concatenation, and Kleene
star. (NFA-s, Regular expression.)
• Consequence:
Closure under intersection by De Morgan’s Laws.
• Relationship to context-free languages. (Regular
Grammars.)
• Ease of specification. (Regular expression.)
• Building tokenizers/lexical analyzers. (DFAs)

regular.
not
is
}
0
|
{
Show 
 i
b
a
L i
i
Consider pairs of strings:
...
...
:
'
...
...
:
'
n
s
i
n
s
i
b
bb
b
v
a
aa
a
u
j
i
F
b
a
q
F
b
a
q
j
i
M
i
i
M



if
)
,
(
)
,
(
0
*
0
*


If L were regular, then there exists a DFA M
accepting L with the following property:
j
i
b
a
q
b
a
q j
i
M
i
i
M 
 if
)
,
(
)
,
( 0
*
0
*



j
i
a
q
a
q j
M
i
M 
 if
)
,
(
)
,
( 0
*
0
*


j
i
b
a
q
b
a
q i
j
M
i
i
M 
 for
)
,
(
)
,
( 0
*
0
*


JUSTIFICATION: Otherwise, from the definition of DFA,
CLAIM:
which contradicts the earlier conclusion.
In order to satisfy
j
i
a
q
a
q j
M
i
M 
 if
)
,
(
)
,
( 0
*
0
*


the machine M must have a unique state for every i.
Thus, M must have infinite number of states, if L
is regular. This violates the definition of DFA.
So, L must be non-regular.

Using Closure Properties
• Regular languages are closed under set-intersection.
• Note that regularity is a property
of a collection, and not a property
of an individual string in the
collection.
2
1
2
1
2
1
L
L
L
L
L
L
L






L1=bit strings with even parity
L2=bit strings with number of 1’s divisible by 3
L=bit strings with number of 1’s a multiple of 6

• Show that
is not regular.
• Proof: If L were regular, ought to be
regular. However, is known to be
non-regular. Hence, L cannot be regular.
• If R is a regular language and C is context-free,
then may not be regular.
• Proof:
C
R
C
C
R
i
b
a
C
b
a
R
i
i





}
0
|
{
*
*
}
in
s
'
#
in
'
#
|
*
}
,
{
{ 

 b
s
a
b
a
L 


R
L 
C
R
L 


Prelude to Pumping Lemma
• Is 46551 divisible by 46?
Necessary vs sufficient condition

Pumping Lemma for Regular Languages
• It is a necessary condition.
– Every regular language satisfies it.
– If a language violates it, it is not regular.
• RL => PL not PL => not RL
• It is not a sufficient condition.
– Not every non-regular language violates it.
• not RL =>? PL or not PL (no conclusion)

Basic Idea:
q0
b
a
a
a,b
b
b
a
q2 q3
q1
)
(M
L
ababbaaab
3
1
0
2
3
1
2
3
1
0 q
q
q
q
q
q
q
q
q
q
b
a
a
a
b
b
a
b
a










3
1
0
2
3
1
2
3
1
0 q
q
q
q
q
q
q
q
q
q
b
a
a
a
b
b
a
b
a









Note,
)
(M
L
ababb
So,
)
(M
L
abaaab
3
1
0
2
3
1
2
3
1
0 q
q
q
q
q
q
q
q
q
q
b
a
a
a
b
b
a
b
a









)
(
)
(
)
(
:
, M
L
aaab
abb
ab
j
i j
i





Fundamental Observation
• Given a “sufficiently” long string, the states
of a DFA must repeat in an accepting
computation. These cycles can then be used
to predict (generate) infinitely many other
strings in (of) the language.
Pigeon-Hole Principle

Pumping Lemma
• Let L be a regular language that is accepted
by a DFA M with k states. Let z be any
string in L with . Then z can
be decomposed as uvw with
L
w
uv
i
v
length
k
uv
length
i





:
0
and
,
0
)
(
,
)
(
k
z
length 
)
(

)
0
:
(
0)
|
|
(
)
|
|
(
)
(
:
|
|
:
L
w
uv
i
i
v
k
uv
s
uvw
u,v,w
k
s
L
s
i















For all sufficiently long strings (z)
There exists non-null prefix (uv)
and substring (v)
For all repetitions of the substring (v),
we get strings in the language.

Proving non-regularity
• If there exists an arbitrarily long string s L, and
for each decomposition s = uvw, there exists an i
such that , then L is non-regular.
)
0
:
(
0)
|
|
(
)
|
|
(
)
(
:
|
|
:
L
w
uv
i
i
v
k
uv
s
uvw
u,v,w
k
s
L
s
i















Negation of the necessary condition:

L
w
uvi


Examples
Applying Pumping Lemma

Proof by contradiction:
• Let be accepted by a k-state DFA.
• Choose
• For all prefixes of length
• show there exists such that
• i.e.,
regular.
not
is
}
number
prime
a
is
|
{ p
a
L p
p 
p
L
k
n
a
s n

 prime
a
is
where
,
,
k
j 
,
j
i p
j
n
i
j
L
a
a j


)
(
number.
composite
a
is
)
( j
n
i
j j 



• Choose
(For this specific problem happens to be
independent of j, but that need not always be the
case.)
• is non-regular because it violates the
necessary condition.
1

 n
ij
j
i
number!
composite
)
1
(
*
)
1
(







j
n
n
n*j
j
n
n
j*
p
L
,...
)
(
, 2
1
2
1
1
p
n
n
p
n
n
L
a
a
L
a
a 
 




Proof : (For this example, choice of initial string is crucial.)
• For this choice of s, the pumping lemma cannot
generate a contradiction!
• However, let instead.
}
|
{ m
n
b
a
L m
n
p 

DFA
of
states
of
number
where 
 n
a
s n
1 n
n
b
a
s 

:
String
Pumped
:
String
Original
1
*
1
n
j
n
j
i
n
j
n
j
b
a
a
b
a
a
s






• For
• Thus, by pumping the substring containing
a’s 0 times (effectively deleting it), the
number of a’s can be made smaller than the
number of b’s.
• So, by pumping lemma, L is non-regular.
n
j
n
j
n
i








1
1
0

• Proof by contradiction:
– If is regular, then so is , the complement of
– But which is known to be non-regular.
– So, cannot be regular.
regular.
not
is
}
number
composite
a
is
|
{ c
a
L c
c 
p
c L
L
a 


c
L
a 

c
L .
c
L
c
L

Summary: Proof Techniques
• Counter Examples
• Constructions/Simulations
• Induction Proofs
• Impossibility Proofs
• Proofs by Contradiction
• Reduction Proofs : Closure Properties

06_PumpingLemma compiler design of chapter 4.ppt

More Related Content

Similar to 06_PumpingLemma compiler design of chapter 4.ppt (20)

More from ranjan317165 (18)

Recently uploaded (20)

06_PumpingLemma compiler design of chapter 4.ppt