1 more properties of regular languages. 2 we have proven regular languages are closed under: union...
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1
More Properties of
Regular Languages
2
We have proven
Regular languages are closed under:
Union
Concatenation
Star operation
Reverse
3
Namely, for regular languages and :1L 2L
21 LL
21LL
1L
Union
Concatenation
Star operation
Reverse RL1
RegularLanguages
4
We will prove
Regular languages are closed under:
Complement
Intersection
5
Namely, for regular languages and :1L 2L
1L
21 LL
Complement
Intersection
RegularLanguages
6
Complement
Theorem:For regular language the complement is regular L
L
Proof: Take DFA that accepts and make • nonfinal states final• final states nonfinal
Resulting DFA accepts
L
L
7
Example:a
b ba,
ba,
0q 1q 2q
)*( baLL
a
b ba,
ba,
0q 1q 2q
)*))((**( bababaaLL
8
Intersection
Theorem:For regular languages and the intersection is regular 21 LL
1L 2L
Proof: Apply DeMorgan’s Law:
2121 LLLL
9
21 , LL regular
21 , LL regular
21 LL regular
21 LL regular
21 LL regular
10
Standard Representations of
Regular Languages
11
Standard Representations of Regular Languages
Regular Languages
DFAs
NFAsRegularExpressions
RegularGrammars
12
When we say: We are given a Regular Language
We mean:
L
Language is in a standard representation
L
13
Elementary Questions
about
Regular Languages
14
Membership Question
Question: Given regular languageand string how can we check if ?
L
Lw w
Answer: Take the DFA that acceptsand check if is accepted
Lw
15
DFA
Lw
DFA
Lw
w
w
16
Given regular languagehow can we checkif is empty: ?
L
L
Take the DFA that accepts
Check if there is a path from the initial state to a final state
L
)( L
Question:
Answer:
17
DFA
L
DFA
L
18
Given regular languagehow can we checkif is finite?
L
L
Take the DFA that accepts
Check if there is a walk with cyclefrom the initial state to a final state
L
Question:
Answer:
19
DFA
L is infinite
DFA
L is finite
20
Given regular languages and how can we check if ?
1L 2L
21 LL Question:
)()( 2121 LLLLFind ifAnswer:
21
)()( 2121 LLLL
21 LL 21 LLand
21 LL
1L 2L 1L2L
21 LL 12 LL 2L 1L
22
)()( 2121 LLLL
21 LL 21 LLor
1L 2L 1L2L
21 LL 12 LL
21 LL
23
Non-regular languages
24
Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages}0:{ nba nn
}*},{:{ bawwwR
25
How can we prove that a languageis not regular?
L
Prove that there is no DFA that accepts L
Problem: this is not easy to prove
Solution: the Pumping Lemma !!!
26
The Pigeonhole Principle
27
pigeons
pigeonholes
4
3
28
A pigeonhole mustcontain at least two pigeons
29
...........
...........
pigeons
pigeonholes
n
m mn
30
The Pigeonhole Principle
...........
pigeons
pigeonholes
n
m
mn There is a pigeonhole with at least 2 pigeons
31
The Pigeonhole Principle
and
DFAs
32
DFA with states 4
1q 2q 3qa
b
4q
b
b b
b
a a
33
1q 2q 3qa
b
4q
b
b
b
a a
a
In walks of strings:
aab
aa
a no stateis repeated
34
In walks of strings:
1q 2q 3qa
b
4q
b
b
b
a a
a
...abbbabbabb
abbabb
bbaa
aabb a stateis repeated
35
If the walk of string has length
1q 2q 3qa
b
4q
b
b
b
a a
a
w 4|| w
then a state is repeated
36
If in a walk of a string transitions states of DFAthen a state is repeated
1q 2q 3qa
b
4q
b
b
b
a a
a
Pigeonhole principle for any DFA:
w
37
In other words for a string : transitions are pigeons
states are pigeonholes
1q 2q 3qa
b
4q
b
b
b
a a
a
q
a
w
38
In general:
A string has length number of states w
A state must be repeated in the walk wq
q...... ......
walk of w
39
The Pumping Lemma
40
Take an infinite regular languageL
DFA that accepts L
mstates
41
Take string with w Lw
There is a walk with label :w
.........
walk w
42
If string has length w mw || number of states
then, from the pigeonhole principle: a state is repeated in the walkq w
q...... ......
walk w
43
Write zyxw
q...... ......
x
y
z
44
q...... ......
x
y
z
Observations: myx ||length numberof states
1|| ylength
45
The string is accepted zxObservation:
q...... ......
x
y
z
46
The string is accepted
zyyxObservation:
q...... ......
x
y
z
47
The string is accepted
zyyyxObservation:
q...... ......
x
y
z
48
The string is accepted
zyx iIn General:
...,2,1,0i
q...... ......
x
y
z
49
In other words, we described:
The Pumping Lemma !!!
50
The Pumping Lemma:
• Given a infinite regular language L
• there exists an integer m
• for any string with length Lw mw ||
• we can write zyxw
• with andmyx || 1|| y
• such that: Lzyx i ...,2,1,0i
51
Applications
of
the Pumping Lemma
52
Theorem: The language }0:{ nbaL nn
is not regular
Proof: Use the Pumping Lemma
53
Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
}0:{ nbaL nn
54
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
Example: mmbawpick
m
}0:{ nbaL nn
55
Write: zyxba mm
it must be that: length
From the Pumping Lemma 1||,|| ymyx
Therefore: babaaaaba mm ............
1, kay kx y z
m m
56
From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmbazyx
Lbazyyxzyx mkm 2
Lzyx 2
We have: 1, kay k
57
Lba mkm Therefore:
}0:{ nbaL nnBUT:
Lba mkm
CONTRADICTION!!!
58
Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
59
Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages }0:{ nba nn
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