Languages and Finite Automata
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Transcript Languages and Finite Automata
More Applications
of
the Pumping Lemma
Prof. Busch - LSU
1
The Pumping Lemma:
• Given a infinite regular language
• there exists an integer
• for any string
• we can write
• with
L
m (critical length)
w L with length | w | m
w x y z
| x y | m and | y | 1
• such that:
xy z L
i
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i 0, 1, 2, ...
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Non-regular languages
L {vv : v *}
R
Regular languages
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Theorem: The language
L {vv : v *}
R
{a, b}
is not regular
Proof:
Use the Pumping Lemma
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L {vv : v *}
R
Assume for contradiction
that L is a regular language
Since L is infinite
we can apply the Pumping Lemma
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L {vv : v *}
R
Let
m be the critical length for L
Pick a string
w such that: w L
and length
We pick
| w| m
wa b b a
m m m m
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From the Pumping Lemma:
we can write: w a b b a
m
with lengths:
m
m
m
x y z
| x y | m, | y | 1
m
m m m
w xyz a...aa...a...ab...bb...ba...a
x
Thus:
y
z
y a , 1k m
k
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x y za b b a
m m m m
y a , 1k m
k
From the Pumping Lemma:
xy z L
i
i 0, 1, 2, ...
Thus:
xy z L
2
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x y za b b a
m m m m
y a , 1k m
k
From the Pumping Lemma:
xy z L
2
m m m
m+k
2
xy z = a...aa...aa...a...ab...bb...ba...a ∈L
x
Thus:
y
a
y
z
m k m m m
b b a
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L
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a
BUT:
m k m m m
b b a
L
k 1
L {vv : v *}
R
a
m k m m m
b b a
L
CONTRADICTION!!!
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Therefore:
Our assumption that L
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
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Theorem: The language
n l n l
L {a b c
: n, l 0}
is not regular
Proof:
Use the Pumping Lemma
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n l n l
L {a b c
: n, l 0}
Assume for contradiction
that L is a regular language
Since L is infinite
we can apply the Pumping Lemma
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n l n l
L {a b c
Let
: n, l 0}
m be the critical length of L
Pick a string
w such that: w L
length
We pick
and
| w| m
wa b c
m m 2m
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From the Pumping Lemma:
We can write
w a b c
With lengths
| x y | m, | y | 1
m
m 2m
m
x y z
m
2m
w xyz a...aa...aa...ab...bc...cc...c
x
Thus:
y
z
y a , 1k m
k
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x y za b c
m m 2m
y a , 1k m
k
From the Pumping Lemma:
xy z L
i
i 0, 1, 2, ...
Thus:
0
x y z = xz ∈ L
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x y za b c
m m 2m
y a , 1k m
k
xz L
From the Pumping Lemma:
mk
m
2m
xz a...aa...ab...bc...cc...c L
x
Thus:
z
a
mk m 2m
b c
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L
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a
BUT:
mk m 2m
n l n l
L {a b c
a
L
b c
k 1
: n, l 0}
mk m 2m
b c
L
CONTRADICTION!!!
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Therefore:
Our assumption that L
is a regular language is not true
Conclusion: L is not a regular language
END OF PROOF
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Non-regular languages
L {a : n 0}
n!
Regular languages
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Theorem: The language
L {a : n 0}
n!
is not regular
n! 1 2 (n 1) n
Proof:
Use the Pumping Lemma
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L {a : n 0}
n!
Assume for contradiction
that L is a regular language
Since L is infinite
we can apply the Pumping Lemma
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L {a : n 0}
n!
Let
m be the critical length of
Pick a string
L
w such that: w L
length
We pick
wa
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| w| m
m!
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From the Pumping Lemma:
We can write
w a
With lengths
| x y | m, | y | 1
m!
x y z
m
w xyz a
m!
m!m
a...aa...aa...aa...aa...a
x
Thus:
y
z
y a , 1 k m
k
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x y za
y a , 1 k m
m!
k
From the Pumping Lemma:
xy z L
i
i 0, 1, 2, ...
Thus:
xy z L
2
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x y za
y a , 1 k m
m!
k
From the Pumping Lemma:
mk
xy z L
2
m!m
xy z a...aa...aa...aa...aa...aa...a L
2
x
Thus:
y
y
a
m! k
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z
L
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a
Since:
m! k
L
1 k m
L {a : n 0}
n!
There must exist
p such that:
m! k p!
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However:
m! k m! m
m! m!
for
m 1
m!m m!
m!(m 1)
(m 1)!
m! k (m 1)!
m! k p!
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for any
p
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a
BUT:
m! k
L
1 k m
L {a : n 0}
n!
a
m! k
L
CONTRADICTION!!!
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Therefore:
Our assumption that L
is a regular language is not true
Conclusion: L is not a regular language
END OF PROOF
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