Download - CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University
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CD5560
FABER
Formal Languages, Automata and Models of Computation
Lecture 4
Mälardalen University
2005
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Content
- More Properties of Regular Languages (RL)- Standard Representations of RL- Elementary Questions about RL- Non-Regular Languages- The Pigeonhole Principle- The Pumping Lemma- Applications of the Pumping Lemma
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More Properties of
Regular Languages
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We have shown
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
Regular
Languages
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We will show
Regular languages are also closed under
Complement
Intersection
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Namely, for regular languages and :1L 2L
1L
21 LL
Complement
Intersection
Regular
Languages
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Complement
Theorem For regular language
the complement is regular LL
Proof Take DFA that accepts and change:L
• non-final states final states
• non-accepting states accepting states
LResulting DFA accepts
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Examplea
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
DFAs
NFAs
Regular
Expressions
Regular
Grammars
Regular
Languages
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Elementary Questionsabout
Regular Languages
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Membership Question
Question: Given regular language
and string
how can we check if ?
L
Lw w
Answer: Take the DFA that accepts
and check if is accepted
Lw
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Lw
DFAw
Lw
DFAw
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Take the DFA that accepts
Check if there is a path from
the initial state to a final state
L
Given regular language
how can we check
if is empty: ?
L
L )( L
Question:
Answer:
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DFA
L
L
DFA
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Given regular language
how can we check
if is finite?
L
L
Take the DFA that accepts
Check if there is a walk with cycle
from 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:
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)()( 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???
Chomsky’s Language Hierarchy
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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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The Pigeonhole Principle
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pigeons
pigeonholes
4
3
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A pigeonhole must
contain at least two pigeons
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...........
...........
pigeonsn
pigeonholesm 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 Principleand
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
ano state
is 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 state
is 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 DFA
then a state is repeated
Pigeonhole principle for any DFA:
w
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In other words for a string
transitions are pigeons
states are pigeonholesq
a
w
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A string has length number of states w
A state must be repeated in the walk of wq
In general
...... ......
walk of w
q
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The Pumping Lemmafor Regular Languages
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Take an infinite regular language L
DFA that accepts L
nstates
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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 w1|| nmw
then, from the pigeonhole principle:
a state is repeated in the walkq w
...... ......
walk w
( number of states)
q
n
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Write zyxw
...... ......
x
y
z
q
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myx ||
Lengths:
1|| y
...... ......
x
y
z
q
(from pigeon principle, as q is the first repetition in sequence)
(there is a walk in the graph)
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The string is accepted zxObservation
...... ......
x
y
z
q
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The string
is accepted
zyyxObservation
...... ......
x
y
z
q
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The string
is accepted
zyyyxObservation
...... ......
x
y
z
q
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The string
zyx iGenerally
...,2,1,0i
...... ......
x
y
z
q
is accepted
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The Pumping Lemma
Given an 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 to the contrary,
that is a regular languageL
Since is infinite
we 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
mmbawe.g. pick
m
}0:{ nbaL nn
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Write: zyxba mm
it must be that: length
From the Pumping Lemma
1||,|| ymyx
Therefore:
1, kay k
babaaaaba mm ............
x y z
m m
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From the Pumping Lemma: Lzyx i ...,2,1,0i
We can choose
mmbazyx
0i
We have:
1, kay k
Lbaw mkm
CONTRADICTION!
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Therefore: Our assumption that
is a regular language is not true
L
Conclusion
L is not a regular language
END OF PROOF
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Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages
}0:{ nba nn
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Theorem The language
Proof
Use the Pumping Lemma!
is not regular
*}:{ wwwL R },{ ba
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*}:{ wwwL R
Assume to the contrary,
that is a regular language
Since is infinite
we can apply the Pumping Lemma
L
L
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mmmm abbawpick
Pick a string such that: w Lw
mw ||length
Let be the integer in the Pumping Lemmam
*}:{ wwwL R
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Write zyxabba mmmm
it must be that length
From the Pumping Lemma
ababbabaaaaabba mmmm ..................
x y z
m m m m
1||,|| ymyx
1, kay k
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ababbabaaaaabba mmmm ..................
x y z
m m m m
1, kay k
We can choose 0i
CONTRADICTION!
So we get a’s on the left,
while is on the right:
km m
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Conclusion
L is not a regular language
Our assumption that
is a regular language is not true
Therefore:
END OF PROOF
L
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Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages
}0:{ nba nn *}:{ wwwR
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Theorem The language
is not regular
Proof
Use the Pumping Lemma
}0,:{ lncbaL lnln
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Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0,:{ lncbaL lnln
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mmm cbaw 2Pick
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
m
}0,:{ lncbaL lnln
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Write zyxcba mmm 2
cccbcabaaaaacba mmm ..................2
x y z
m m m2
it must be that length
From the Pumping Lemma
1||,|| ymyx
1, kay k
![Page 71: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/71.jpg)
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From the Pumping Lemma
Lzyx i ...,2,1,0i
Thus:
Lcbazxzyx mmkm 20
Lzyx 0
mmm cbazyx 2We have:
1, kay k
![Page 72: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/72.jpg)
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Lcba mmkm 2Therefore:
Lcba mmkm 2
BUT:
CONTRADICTION!
}0,:{ lncbaL lnln
![Page 73: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/73.jpg)
73
Conclusion
L is not a regular language
LTherefore: Our assumption that
is a regular language is not true
END OF PROOF
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74
Regular languages
Non-regular languages
}0,:{ lncba lnln
*}:{ wwwR }0:{ nba nn
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75
Theorem
The language
is not regular
ProofUse the Pumping Lemma
}0:{ ! naL n
nnn )1(21!
![Page 76: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/76.jpg)
76
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0:{ ! naL n
![Page 77: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/77.jpg)
77
!mawPick
Pick a string such that: w
Lw mw ||length
Let be the integer in the Pumping Lemmam
}0:{ ! naL n
![Page 78: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/78.jpg)
78
Write zyxam !
From the Pumping Lemma
aaaaaaaaaaam ...............!
x y z
m mm !
it must be that length 1||,|| ymyx
mkay k 1,
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79
From the Pumping Lemma:
Lzyx i ...,2,1,0i
Thus:
!mazyx
Lazyyxzyx km !2
Lzyx 2
We have:
mkay k 1,
![Page 80: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/80.jpg)
80
La km !Therefore:
!! pkm
}0:{ ! naL nAnd since:
mk 1
mk 1There is p
![Page 81: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/81.jpg)
81
However
)!1( m)1(! mm!! mmm
!! mm! mmkm !
for 1m
)!1(! mkm
!! pkm for any p
![Page 82: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/82.jpg)
82
La km !Therefore:
La km !
BUT:
CONTRADICTION!
}0:{ ! naL n mk 1and
![Page 83: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/83.jpg)
83
Conclusion
L is not a regular language
Our assumption that
is a regular language is not true
LTherefore:
END OF PROOF
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84
Regular languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln
*}:{ wwwR }0:{ nba nn
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85
Theorem
The language
is not regular
ProofUse the Pumping Lemma
}:{ primeiaL i
![Page 86: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/86.jpg)
86
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}:{ primeiaL i
![Page 87: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/87.jpg)
87
From the Pumping Lemma
it must be that length 1||,|| ymyx
From the Pumping Lemma:
Lzyx i ...,2,1,0i
The length of zxyw k 1
must be prime for each string of
mk
Lw mw ||length
L
![Page 88: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/88.jpg)
88
Thus:
But
which is not prime!
CONTRADICTION!
))(1(
))((
)()(
)()( 1
ylengthk
ylengthkk
ylengthxyzlength
zxyylengthzxylengthk
kk
)( mk
Lw
![Page 89: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/89.jpg)
89
Conclusion
L is not a regular language
Our assumption that
is a regular language is not true
LTherefore:
END OF PROOF
![Page 90: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/90.jpg)
90
Regular languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln
*}:{ wwwR }0:{ nba nn
}:{ primeiaL i
![Page 91: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/91.jpg)
91
Theorem
The language}0:{ nbaL nn
is not regular
Proof
Use the Pumping Lemma!
An alternative variant of proof from slide 53.
![Page 92: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/92.jpg)
92
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0:{ nbaL nn
![Page 93: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/93.jpg)
93
Pick a string such that: w Lw
mw ||length
mmbawPick
Let be the integer in the Pumping Lemmam
}0:{ nbaL nn
![Page 94: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/94.jpg)
94
Write: zyxba mm
it must be that: length
From the Pumping Lemma
1||,|| ymyx
Therefore: babaaaaba mm ............
1, kay kx y z
m m
![Page 95: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/95.jpg)
95
From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmbazyx
Lbazyyxzyx mkm 2
Lzyx 2
We have: 1, kay k
![Page 96: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/96.jpg)
96
Lba mkm Therefore:
}0:{ nbaL nnBUT:
Lba mkm
CONTRADICTION
![Page 97: CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 4 Mälardalen University](https://reader036.vdocuments.mx/reader036/viewer/2022070411/56814749550346895db4891e/html5/thumbnails/97.jpg)
97
Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF