Examples
Applying Pumping Lemma
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Lp  {a p | p is a prime number }
is not regular.
Proof by contradiction:
• Let L p be accepted by a k-state DFA.
• Choose s  a n , where n is a prime  k
• For all prefixes of length j  k ,
j i
n j
(
a
)
a
 Lp
• show there exists i j , such that
• i.e., ( j  i  n  j ) is a composite number.
j
j
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• Choose
ij  n 1
(For this specific problem i j happens to be
independent of j, but that need not always be the
case.)
j*( n  1)  n  j  n*j  n
 n * ( j  1)
composite number!
a n1a n1  Lp , (a 2 )n1 a n2  Lp ,...
• L p is non-regular because it violates the
necessary condition.
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Lp  {a b | n  m}
n m
Proof : (For this example, choice of initial string is crucial.)
s  a n where n  number of states of DFA
• For this choice of s, the pumping lemma cannot
generate a contradiction!
• However, let s  a n 1b n instead.
Original String :
Pumped String :
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s  a j a n 1 j bn
i* j
a a
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n 1 j n
b
4
• For
i  0  n  j 1

n 1 j  n
• 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.
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Lc  {a c | c is a composite number }
is not regular.
• Proof by contradiction:

a
 Lc , the
– If Lc is regular, then so is
complement of Lc .

a
 Lc  Lp which is known to be non– But
regular.
– So, Lc cannot be regular.
• Proving Lc to be non-regular using pumping
lemma may be difficult/impossible.
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Source of the problem?
Regular (ultimately periodic)
…}
{
Prime (sparse)
…}
{
Composite(dense)
…}
{
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Summary of Proof Techniques Employed
• Counter Examples
• Constructions/Simulations
• Induction Proofs
• Impossibility Proofs
• Proofs by Contradiction
• Reduction Proofs : Closure Properties
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