cd5560 faber formal languages, automata and models of computation lecture 4 mälardalen university
Post on 14-Jan-2016
31 Views
Preview:
DESCRIPTION
TRANSCRIPT
1
CD5560
FABER
Formal Languages, Automata and Models of Computation
Lecture 4
Mälardalen University
2005
2
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
3
More Properties of
Regular Languages
4
We have shown
Regular languages are closed under
Union
Concatenation
Star operation
Reverse
5
Namely, for regular languages and :1L 2L
21 LL
21LL
1L
Union
Concatenation
Star operation
Reverse RL1
Regular
Languages
6
We will show
Regular languages are also closed under
Complement
Intersection
7
Namely, for regular languages and :1L 2L
1L
21 LL
Complement
Intersection
Regular
Languages
8
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
9
Examplea
b ba,
ba,
0q 1q 2q
)*( baLL
a
b ba,
ba,
0q 1q 2q
)))((( bababaaLL
10
Intersection
Theorem For regular languages and
the intersection is regular 21 LL 1L 2L
Proof Apply DeMorgan’s Law:
2121 LLLL
11
21 , LL regular
21 , LL regular
21 LL regular
21 LL regular
21 LL regular
12
Standard Representations of
Regular Languages
13
Standard Representations of Regular Languages
DFAs
NFAs
Regular
Expressions
Regular
Grammars
Regular
Languages
14
Elementary Questionsabout
Regular Languages
15
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
16
Lw
DFAw
Lw
DFAw
17
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:
18
DFA
L
L
DFA
19
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:
20
DFA
L is infinite
DFA
L is finite
21
Given regular languages and
how can we check if ? 1L 2L
21 LL Question:
)()( 2121 LLLLFind ifAnswer:
22
)()( 2121 LLLL
21 LL 21 LLand
21 LL
1L 2L 1L2L
21 LL 12 LL 2L 1L
23
)()( 2121 LLLL
21 LL 21 LLor
1L 2L 1L2L
21 LL 12 LL
21 LL
24
Non-Regular Languages
25
Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages???
Chomsky’s Language Hierarchy
26
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 !
27
The Pigeonhole Principle
28
The Pigeonhole Principle
29
pigeons
pigeonholes
4
3
30
A pigeonhole must
contain at least two pigeons
31
...........
...........
pigeonsn
pigeonholesm mn
32
The Pigeonhole Principle
...........
pigeons
pigeonholes
n
m
mn There is a pigeonhole
with at least 2 pigeons
33
The Pigeonhole Principleand
DFAs
34
DFA with states 4
1q 2q 3qa
b
4q
b
b b
b
a a
35
1q 2q 3qa
b
4q
b
b
b
a a
a
In walks of strings:
aab
aa
ano state
is repeated
36
In walks of strings:
1q 2q 3qa
b
4q
b
b
b
a a
a
...abbbabbabb
abbabb
bbaa
aabb a state
is repeated
37
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
38
If in a walk of a string
transitions states of DFA
then a state is repeated
Pigeonhole principle for any DFA:
w
39
In other words for a string
transitions are pigeons
states are pigeonholesq
a
w
40
A string has length number of states w
A state must be repeated in the walk of wq
In general
...... ......
walk of w
q
41
The Pumping Lemmafor Regular Languages
42
Take an infinite regular language L
DFA that accepts L
nstates
43
Take string with w Lw
There is a walk with label w
.........
walk w
44
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
45
Write zyxw
...... ......
x
y
z
q
46
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)
47
The string is accepted zxObservation
...... ......
x
y
z
q
48
The string
is accepted
zyyxObservation
...... ......
x
y
z
q
49
The string
is accepted
zyyyxObservation
...... ......
x
y
z
q
50
The string
zyx iGenerally
...,2,1,0i
...... ......
x
y
z
q
is accepted
51
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
52
Applications of
the Pumping Lemma
53
Theorem
The language}0:{ nbaL nn
is not regular
Proof
Use the Pumping Lemma!
54
Assume to the contrary,
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0:{ nbaL nn
55
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
mmbawe.g. pick
m
}0:{ nbaL nn
56
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
57
From the Pumping Lemma: Lzyx i ...,2,1,0i
We can choose
mmbazyx
0i
We have:
1, kay k
Lbaw mkm
CONTRADICTION!
58
Therefore: Our assumption that
is a regular language is not true
L
Conclusion
L is not a regular language
END OF PROOF
59
Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages
}0:{ nba nn
60
Theorem The language
Proof
Use the Pumping Lemma!
is not regular
*}:{ wwwL R },{ ba
61
*}:{ wwwL R
Assume to the contrary,
that is a regular language
Since is infinite
we can apply the Pumping Lemma
L
L
62
mmmm abbawpick
Pick a string such that: w Lw
mw ||length
Let be the integer in the Pumping Lemmam
*}:{ wwwL R
63
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
64
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
65
Conclusion
L is not a regular language
Our assumption that
is a regular language is not true
Therefore:
END OF PROOF
L
66
Regular languages
ba* acb *
...etc
*)( bacb
Non-regular languages
}0:{ nba nn *}:{ wwwR
67
Theorem The language
is not regular
Proof
Use the Pumping Lemma
}0,:{ lncbaL lnln
68
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0,:{ lncbaL lnln
69
mmm cbaw 2Pick
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
m
}0,:{ lncbaL lnln
70
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
71
From the Pumping Lemma
Lzyx i ...,2,1,0i
Thus:
Lcbazxzyx mmkm 20
Lzyx 0
mmm cbazyx 2We have:
1, kay k
72
Lcba mmkm 2Therefore:
Lcba mmkm 2
BUT:
CONTRADICTION!
}0,:{ lncbaL lnln
73
Conclusion
L is not a regular language
LTherefore: Our assumption that
is a regular language is not true
END OF PROOF
74
Regular languages
Non-regular languages
}0,:{ lncba lnln
*}:{ wwwR }0:{ nba nn
75
Theorem
The language
is not regular
ProofUse the Pumping Lemma
}0:{ ! naL n
nnn )1(21!
76
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0:{ ! naL n
77
!mawPick
Pick a string such that: w
Lw mw ||length
Let be the integer in the Pumping Lemmam
}0:{ ! naL n
78
Write zyxam !
From the Pumping Lemma
aaaaaaaaaaam ...............!
x y z
m mm !
it must be that length 1||,|| ymyx
mkay k 1,
79
From the Pumping Lemma:
Lzyx i ...,2,1,0i
Thus:
!mazyx
Lazyyxzyx km !2
Lzyx 2
We have:
mkay k 1,
80
La km !Therefore:
!! pkm
}0:{ ! naL nAnd since:
mk 1
mk 1There is p
81
However
)!1( m)1(! mm!! mmm
!! mm! mmkm !
for 1m
)!1(! mkm
!! pkm for any p
82
La km !Therefore:
La km !
BUT:
CONTRADICTION!
}0:{ ! naL n mk 1and
83
Conclusion
L is not a regular language
Our assumption that
is a regular language is not true
LTherefore:
END OF PROOF
84
Regular languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln
*}:{ wwwR }0:{ nba nn
85
Theorem
The language
is not regular
ProofUse the Pumping Lemma
}:{ primeiaL i
86
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}:{ primeiaL i
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
88
Thus:
But
which is not prime!
CONTRADICTION!
))(1(
))((
)()(
)()( 1
ylengthk
ylengthkk
ylengthxyzlength
zxyylengthzxylengthk
kk
)( mk
Lw
89
Conclusion
L is not a regular language
Our assumption that
is a regular language is not true
LTherefore:
END OF PROOF
90
Regular languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln
*}:{ wwwR }0:{ nba nn
}:{ primeiaL i
91
Theorem
The language}0:{ nbaL nn
is not regular
Proof
Use the Pumping Lemma!
An alternative variant of proof from slide 53.
92
Assume to the contrary
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0:{ nbaL nn
93
Pick a string such that: w Lw
mw ||length
mmbawPick
Let be the integer in the Pumping Lemmam
}0:{ nbaL nn
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
95
From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmbazyx
Lbazyyxzyx mkm 2
Lzyx 2
We have: 1, kay k
96
Lba mkm Therefore:
}0:{ nbaL nnBUT:
Lba mkm
CONTRADICTION
97
Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF
top related