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Formal Languages
Non-Regular LanguagesSlides based on RPI CSCI 2400
Thanks to Petros Drineas
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Evert: Session 3Posix notation
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Evert: Session 1Vollstaendige Induktion
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Regular languagesa* cb*
.et
(acb
Non-regular languages:{ nbann
,{:{
avvvR
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How can we prove that a language
is not regular?L
Prove that there is no DFA that acceptsL
Problem: this is not easy to prove
Solution: the Pumping Lemma !!!
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The Pigeonhole Principle
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pigeons
pigeonholes
4
3
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A pigeonhole mustcontain at least two pigeons
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...........
...........
pigeons
pigeonholes
n
m n
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The Pigeonhole Principle
...........
pigeonspigeonholes
nm
n
There is a pigeonhole
with at least 2 pigeons
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The Pigeonhole Principle
and
DFAs
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DFA with states4
1q 2q 3qa
b
4q
b
b b
b
a a
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1q 2q 3qa
b
4q
b
b
b
aa
a
In walks of strings:
aab
aa
a no state
is repeated
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In walks of strings:
1q 2q 3qa
b
4q
b
b
b
aa
a
...abbbabbabb
abbabb
bbaa
aabb a state
is repeated
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If string has length :
1q 2q 3qa
b
4q
b
b
b
aa
a
w ||w
Thus, a state must be repeated
Then the transitions of stringare more than the states of the DFAw
l
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In general, for any DFA:
String has length number of statesw
A state must be repeated in the walk ofw
q
q...... ......
walk of w
Repeated state
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In other words for a string :
transitions are pigeons
states are pigeonholesq
a
w
q...... ......
walk of w
Repeated state
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The Pumping Lemma
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Take an infinite regular language L
There exists a DFA that accepts L
mstates
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Take string withw w
There is a walk with label :w
.........
walk w
If t i h l th ||
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If string has lengthw w|| (numberof states
of DFA)
then, from the pigeonhole principle:
a state is repeated in the walk w
q...... ......
walk w
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q
q...... ......
walk w
Let be the first state repeated in the
walk of w
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Write xw
q...... ......
x
y
z
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q...... ......
x
y
z
Observations: yx||length number
of states
of DFA||ylength
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The string
is accepted
xObservation:
q...... ......
x
y
z
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The string
is accepted
yxObservation:
q...... ......
x
y
z
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The string
is accepted
yyObservation:
q...... ......
x
y
z
i
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The string
is accepted
yxi
In General:
,,1,0i
q...... ......
x
y
z
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zyxiIn General: ,,1,0
i
q...... ......
x
y
z
Language accepted by the DFA
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In other words, we described:
The Pumping Lemma !!!
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The Pumping Lemma:
Given an infinite regular languageL
there exists an integer m
for any string with lengthw w||
we can write xw
with andyx|| ||y
such that:zyxi ,,1,0i
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Applications
of
the Pumping Lemma
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Theorem: The language :{ nbaLnn
is not regular
Proof: Use the Pumping Lemma
{bLnn
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Assume for contradiction
that is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
:{ nbaLnn
{bLnn
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Let be the integer in the Pumping Lemma
Pick a string such that:w w
w||length
awWe pick
m
:{ nbaLnn
bmm
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it must be that length
From the Pumping Lemma||,||yx
aabayzmm......
, kayk
x y z
m m
Write: xbamm
Thus:
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kk
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From the Pumping Lemma:
aaaazy ...........2
x y
m m
Thus:
zyx2
azy , kayk
y
ba
km
b
km k
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bakm
:{ nbaLnn
BUT:
bamkm
CONTRADICTION!!!
k
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Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
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Regular languages
Non-regular languages :{ nbann