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Formal Languages
Non-Regular Languages
AUST Zahle
Eng. Charbel Boustany

2
Regular languages
b
a* a
c
b 
*
...
etc
*
)
( b
a
c
b 

Non-regular languages
}
0
:
{ 
n
b
a n
n
}*}
,
{
:
{ b
a
v
vvR


3
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 !!!

6
A pigeonhole must
contain at least two pigeons

7
...........
...........
pigeons
pigeonholes
n
m m
n 

8
The Pigeonhole Principle
...........
pigeons
pigeonholes
n
m
m
n 
There is a pigeonhole
with at least 2 pigeons

9
The Pigeonhole Principle
and
DFAs

10
DFA with states
4
1
q 2
q 3
q
a
b
4
q
b
b b
b
a a

11
1
q 2
q 3
q
a
b
4
q
b
b
b
a a
a
In walks of strings:
aab
aa
a no state
is repeated

12
In walks of strings:
1
q 2
q 3
q
a
b
4
q
b
b
b
a a
a
...
abbbabbabb
abbabb
bbaa
aabb a state
is repeated

13
If string has length :
1
q 2
q 3
q
a
b
4
q
b
b
b
a a
a
w 4
|
| 
w
Thus, a state must be repeated
Then the transitions of string
are more than the states of the DFA
w

14
In general, for any DFA:
String has length number of states
w 
A state must be repeated in the walk of w
q
q
...... ......
walk of w
Repeated state

15
In other words for a string :
transitions are pigeons
states are pigeonholes
q
a
w
q
...... ......
walk of w
Repeated state

17
Take an infinite regular language L
There exists a DFA that accepts L
m
states

18
Take string with
w L
w
There is a walk with label :
w
.........
walk w

19
If string has length
w m
w 
|
| (number
of states
of DFA)
then, from the pigeonhole principle:
a state is repeated in the walk w
q
...... ......
walk w

20
q
q
...... ......
walk w
Let be the first state repeated in the
walk of w

21
Write z
y
x
w 
q
...... ......
x
y
z

22
q
...... ......
x
y
z
Observations: m
y
x 
|
|
length number
of states
of DFA
1
|
| 
y
length

23
The string
is accepted
z
x
Observation:
q
...... ......
x
y
z

24
The string
is accepted
z
y
y
x
Observation:
q
...... ......
x
y
z

25
The string
is accepted
z
y
y
y
x
Observation:
q
...... ......
x
y
z

26
The string
is accepted
z
y
x i
In General:
...
,
2
,
1
,
0

i
q
...... ......
x
y
z

27
L
z
y
x i
∈
In General: ...
,
2
,
1
,
0

i
q
...... ......
x
y
z
Language accepted by the DFA

28
In other words, we described:
The Pumping Lemma !!!

29
The Pumping Lemma:
• Given an infinite regular languageL
• there exists an integer m
• for any string with length
L
w m
w 
|
|
• we can write z
y
x
w 
• with and
m
y
x 
|
| 1
|
| 
y
• such that: L
z
y
x i
 ...
,
2
,
1
,
0

i

30
Applications
of
the Pumping Lemma

31
Theorem: The language }
0
:
{ 
 n
b
a
L n
n
is not regular
Proof: Use the Pumping Lemma

32
Assume for contradiction
that is a regular language
L
Since is infinite
we can apply the Pumping Lemma
L
}
0
:
{ 
 n
b
a
L n
n

33
Let be the integer in the Pumping Lemma
Pick a string such that:
w L
w 
m
w 
|
|
length
m
m
b
a
w 
We pick
m
}
0
:
{ 
 n
b
a
L n
n

34
it must be that length
From the Pumping Lemma
1
|
|
,
|
| 
 y
m
y
x
b
ab
aa
aa
a
b
a
xyz m
m
...
...
...
...


1
, 
 k
a
y k
x y z
m m
Write: z
y
x
b
a m
m

Thus:

35
From the Pumping Lemma: L
z
y
x i

...
,
2
,
1
,
0

i
Thus:
m
m
b
a
z
y
x 
L
z
y
x 
2
1
, 
 k
a
y k

36
From the Pumping Lemma:
L
b
ab
aa
aa
aa
a
z
xy 
 ...
...
...
...
...
2
x y z
k
m  m
Thus:
L
z
y
x 
2
m
m
b
a
z
y
x  1
, 
 k
a
y k
y
L
b
a m
k
m



37
L
b
a m
k
m


}
0
:
{ 
 n
b
a
L n
n
BUT:
L
b
a m
k
m


CONTRADICTION!!!
1
≥
k

38
Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:

39
Regular languages
Non-regular languages }
0
:
{ 
n
b
a n
n

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