more properties of regular languages
DESCRIPTION
More Properties of Regular Languages. We have proven. Regular languages are closed under:. Union Concatenation Star operation Reverse. Namely, for regular languages and :. Union Concatenation Star operation Reverse. Regular Languages. We will prove. - PowerPoint PPT PresentationTRANSCRIPT
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More Properties of
Regular Languages
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We have proven
Regular languages are closed under:
Union
Concatenation
Star operation
Reverse
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Namely, for regular languages and :1L 2L
21 LL
21LL
1L
Union
Concatenation
Star operation
Reverse RL1
RegularLanguages
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We will prove
Regular languages are closed under:
Complement
Intersection
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Namely, for regular languages and :1L 2L
1L
21 LL
Complement
Intersection
RegularLanguages
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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
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Example:a
b ba,
ba,
0q 1q 2q
)*( baLL
a
b ba,
ba,
0q 1q 2q
)*))((**( bababaaLL
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Intersection
Theorem:For regular languages and the intersection is regular 21 LL
1L 2L
Proof: Apply DeMorgan’s Law:
2121 LLLL
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21 , LL regular
21 , LL regular
21 LL regular
21 LL regular
21 LL regular
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Standard Representations of
Regular Languages
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Standard Representations of Regular Languages
Regular Languages
DFAs
NFAsRegularExpressions
RegularGrammars
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When we say: We are given a Regular Language
We mean:
L
Language is in a standard representation
L
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Elementary Questions
about
Regular Languages
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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
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DFA
Lw
DFA
Lw
w
w
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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:
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DFA
L
DFA
L
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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:
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DFA
L is infinite
DFA
L is finite
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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
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)()( 2121 LLLL
21 LL 21 LLor
1L 2L 1L2L
21 LL 12 LL
21 LL
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Non-regular languages
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Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages}0:{ nba nn
}*},{:{ bawwwR
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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 !!!
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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 mn
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The Pigeonhole Principle
...........
pigeons
pigeonholes
n
m
mn There is a pigeonhole with at least 2 pigeons
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The Pigeonhole Principle
and
DFAs
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DFA with states 4
1q 2q 3qa
b
4q
b
b b
b
a a
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1q 2q 3qa
b
4q
b
b
b
a a
a
In walks of strings:
aab
aa
a no stateis repeated
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In walks of strings:
1q 2q 3qa
b
4q
b
b
b
a a
a
...abbbabbabb
abbabb
bbaa
aabb a stateis repeated
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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
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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
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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
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In general:
A string has length number of states w
A state must be repeated in the walk wq
q...... ......
walk of w
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The Pumping Lemma
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Take an infinite regular languageL
DFA that accepts L
mstates
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Take string with w Lw
There is a walk with label :w
.........
walk w
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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
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Write zyxw
q...... ......
x
y
z
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q...... ......
x
y
z
Observations: myx ||length numberof states
1|| ylength
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The string is accepted zxObservation:
q...... ......
x
y
z
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The string is accepted
zyyxObservation:
q...... ......
x
y
z
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The string is accepted
zyyyxObservation:
q...... ......
x
y
z
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The string is accepted
zyx iIn General:
...,2,1,0i
q...... ......
x
y
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 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
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Applications
of
the Pumping Lemma
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Theorem: The language }0:{ nbaL nn
is not regular
Proof: Use the Pumping Lemma
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Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
}0:{ nbaL nn
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Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
Example: mmbawpick
m
}0:{ nbaL nn
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Write: zyxba mm
it must be that: length
From the Pumping Lemma 1||,|| ymyx
Therefore: babaaaaba mm ............
1, kay kx y z
m m
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From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmbazyx
Lbazyyxzyx mkm 2
Lzyx 2
We have: 1, kay k
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Lba mkm Therefore:
}0:{ nbaL nnBUT:
Lba mkm
CONTRADICTION!!!
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Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
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Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages }0:{ nba nn