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...
2
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
pigeonholes
33 pigeonholes
5
A pigeonhole must
contain at least two pigeons
6
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
a
q3
b
q4
a
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In walks of strings:
a
{q1, q2}
aa {q1, q2, q3}
aab {q1, q2 , q3 , q4}
no state
is repeated
b
q1
b
a
q2
a
a
q3
b
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
a
q3
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
a
q3
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
17
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
......
z
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Observations:
length
length
| x y |  m number
of states
of DFA
| y | 1
y
......
x
q
......
z
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Observation:
xz
The string
is accepted
y
......
x
q
......
z
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Observation:
xyyz
The string
is accepted
y
......
x
q
......
z
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Observation:
The string
is accepted
xyyyz
y
......
x
q
......
z
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In General:
i
The string
xy z
i  0, 1, 2, ... is accepted
y
......
x
q
......
z
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In General:
i
x y z ∈L
i  0, 1, 2, ...
Language accepted by the DFA
y
......
x
q
......
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
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!!!
37
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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