15 Regular Pumping Lemma Examples

More Applications of the Pumping Lemma
This ppt is the work by Dr. Costas Busch, used with permission, and available from http://csc.lsu.edu/~busch/courses/theorycomp/fall2008/ 1

The Pumping Lemma:
• Given a infinite regular language
L
• there exists an integerm

| w | m with length

• for any string w L
• we can write w  x
• with

|x y|  m

• such that:
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(critical length

yz and |

i

xy z  L
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y | 1

i 0, 1, 2, ...
2

Non-regular languages

R

L {vv : v  *}

Regular languages

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Theorem:The language
R

L {vv : v  *}

 {a, b}

is not regular

Proof:

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Use the Pumping Lemma

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R

L {vv : v  *}
Assume for contradiction that L is a regular language

Since L is infinite we can apply the Pumping Lemma
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R

L {vv : v  *}
Let

m

be the critical length for L

Pick a string w

such that: w

 L

and length

We pick
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| w | m

m m m m

w a b b a
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From the Pumping Lemma: m m

m

we can write: w a b b a with lengths:

m

x y z

| x y | m, | y |1

m

m m m

w  xyz a...aa...a...ab...bb...ba...a x Thus:

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y

z

k

y a , 1 k m
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m m m m

x y z a b b a

k

y a , 1 k m

From the Pumping Lemma: x

i

y z  L

i 0, 1, 2, ...

Thus:
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2

xy z  L
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m m m m

x y z a b b a

k

y a , 1 k m

From the Pumping Lemma:x

2

y z  L

m m m

m+k
2

xy z = a...aa...aa...a...ab...bb...ba...a ∈ L

x
Thus:
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y a y m k m m m

b b a

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z

L
9

a

BUT:

m k m m m

b b a

L

k 1

R

L {vv : v  *}

a

m k m m m

b b a

L

CONTRADICTION!!!
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Therefore:

Our assumption thatL is a regular language is not true

Conclusion:L

is not a regular language

END OF PROOF
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Non-regular languages n l n l

L {a b c

: n, l 0}

Regular languages