Non-regular languages
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{a b : n  0}
n n
Non-regular languages
{vv : v {a, b}*}
R
Regular languages
a *b
b*c  a
b  c ( a  b) *
etc...
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How can we prove that a language
is not regular?
L
Prove that there is no DFA that accepts
L
Problem: this is not easy to prove
Solution: the Pumping Lemma !!!
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The Pigeonhole Principle
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4 pigeons
3 pigeonholes
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A pigeonhole must
contain at least two pigeons
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n pigeons
...........
m
pigeonholes
nm
...........
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The Pigeonhole Principle
n pigeons
m
pigeonholes
nm
There is a pigeonhole
with at least 2 pigeons
...........
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The Pigeonhole Principle
and
DFAs
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DFA with
b
q1
4
states
b
b
a
q2
b
q3
a
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b
q4
a
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In walks of strings:
no state
is repeated
a
aa
aab
b
q1
b
b
a
q2
a
q3
a
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b
q4
a
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In walks of strings:
a state
is repeated
aabb
bbaa
abbabb
abbbabbabb...
b
q1
b
b
a
q2
a
q3
a
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b
q4
a
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If string
w has length | w |  4 :
Then the transitions of string w
are more than the states of the DFA
Thus, a state must be repeated
b
q1
b
b
a
q2
a
q3
a
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b
q4
a
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In general, for any DFA:
String
A state
w has length  number of states
q
walk of
must be repeated in the walk of
w
w
......
q
......
Repeated state
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In other words for a string
w:
a
transitions are pigeons
q
states are pigeonholes
walk of
w
......
q
......
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
m
states
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Take string
w with w L
There is a walk with label
w:
.........
walk
w
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If string
w has length | w |  m
(number
of states
of DFA)
then, from the pigeonhole principle:
a state is repeated in the walk
......
walk
q
w
......
w
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Let q be the first state repeated in the
walk of w
......
walk
q
......
w
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Write
w x y z
y
......
x
q
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......
z
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Observations:
length
length
| x y |  m number
of states
of DFA
| y | 1
y
......
x
q
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......
z
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Observation:
The string
is accepted
xz
y
......
x
q
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......
z
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Observation:
The string
is accepted
xyyz
y
......
x
q
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......
z
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Observation:
The string
is accepted
xyyyz
y
......
x
q
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......
z
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In General:
The string
is accepted
i
xy z
i  0, 1, 2, ...
y
......
x
q
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......
z
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In General:
i
x y z ∈L
i  0, 1, 2, ...
Language accepted by the DFA
y
......
x
q
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......
z
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In other words, we described:
The Pumping Lemma !!!
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The Pumping Lemma:
• Given a infinite regular language
• there exists an integer
• for any string
• we can write
• with
w L
L
m
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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Applications
of
the Pumping Lemma
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Theorem: The language L  {a nb n : n  0}
is not regular
Proof:
Use the Pumping Lemma
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L  {a b : n  0}
n 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 b : n  0}
n n
Let
m
be the integer in the Pumping Lemma
Pick a string
w such that: w  L
length
We pick
| w| m
wa b
m m
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Write:
a b xyz
m m
From the Pumping Lemma
it must be that length | x
y |  m, | y | 1
m
xyz  a b
m m
 a...aa...aa...ab...b
x
Thus:
m
y
z
y  a , k 1
k
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x y za b
y  a , k 1
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 za b
y  a , k 1
m m
k
From the Pumping Lemma:
xy z  L
2
mk
m
xy z  a...aa...aa...aa...ab...b  L
2
x
Thus:
a
y
y
z
m k m
b L
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a
BUT:
m k m
b L
k ≥1
L  {a b : n  0}
n n
a
m k m
b 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
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Non-regular languages
{a b : n  0}
n n
Regular languages
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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 integer in the Pumping Lemma
Pick a string
w such that: w  L
length
We pick
and
| w| m
wa b b a
m m m m
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Write
a b b a xyz
m m m m
From the Pumping Lemma
it must be that length | x
m
y |  m, | y | 1
m m m
xyz  a...aa...a...ab...bb...ba...a
x
Thus:
y
z
y  a , k 1
k
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x y za b b a
y  a , k 1
m m 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 za b b a
y  a , k 1
m m m 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
m k m m m
b b a
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z
L
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a
BUT:
m k m m m
L
b b a
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
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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
m
: n, l  0}
be the integer in the Pumping Lemma
Pick a string
w such that: w  L
length
We pick
and
| w| m
wa b c
m m 2m
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Write
m m 2m
a b c
xyz
From the Pumping Lemma
it must be that length | x
y |  m, | y | 1
m
m
2m
xyz  a...aa...aa...ab...bc...cc...c
x
Thus:
y
z
y  a , k 1
k
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x y za b c
m m 2m
y  a , k 1
From the Pumping Lemma:
k
xy z  L
i
i  0, 1, 2, ...
Thus:
0
x y z = xz ∈ L
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x y za b c
y  a , k 1
m m 2m
k
xz  L
From the Pumping Lemma:
mk
m
2m
xz  a...aa...ab...bc...cc...c  L
x
Thus:
z
a
mk m 2 m
b c
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L
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a
BUT:
mk m 2 m
b c
n l n l
L  {a b c
a
mk m 2 m
b c
L
k 1
: n, l  0}
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
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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 integer in the Pumping Lemma
Pick a string
w such that: w  L
length
We pick
wa
| w| m
m!
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Write
a
m!
xyz
From the Pumping Lemma
it must be that length | x
y |  m, | y | 1
m
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 za
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 za
y  a , 1 k  m
m!
k
From the Pumping Lemma:
xy z  L
2
m!m
mk
xy z  a...aa...aa...aa...aa...aa...a  L
2
x
Thus:
y
y
a
m! k
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
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Summary
Showing regular
construct DFA, NFA
construct regular expression
show L is the union, concatenation, intersection
(regular operations) of regular languages.
Showing non-regular
pumping lemma
assume regular, apply closure properties of
regular languages and obtain a known non-regular
language.
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