regular properties
TRANSCRIPT
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 1
Single Accepting
State for NFAs
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 2
Any NFA can be converted
to an equivalent NFA
with a single accepting state
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 3
a
b
b
a NFA
Equivalent NFA
PP
a
b
b
a
Example
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 4
NFAIn General
Equivalent NFA
Singleaccepting
state
PPP
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 5
Extreme Case
NFA without accepting state
Add an accepting state
without transitions
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 6
Properties of
Regular Languages
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 7
1L 2L
21LLConcatenation:
*1LStar:
21 LL Union:
Are regular
Languages
For regular languages and
we will prove that:
1L
21 LL
Complement:
Intersection:
R
L1Reversal:
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 8
We say: Regular languages are closed under
21LLConcatenation:
*1LStar:
21 LL Union:
1L
21 LL
Complement:
Intersection:
R
L1Reversal:
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 9
1LRegular language
11 LML !
1M
Single accepting state
NFA 2M
2L
Single accepting state
22 LML !
Regular language
NFA
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 10
Example
}{1 baLn!
a
b
1M
_ abaL !2ab
2M
0un
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 11
Union
NFA for
1M
2M
21
LL
P
P
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 12
Example
a
b
ab
P
P
}{1 baLn
!
}{2 baL !
}{}{21baba
LL
n
!
NFA for
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 13
Concatenation
NFA for 21LL
1 2
P P
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 14
Example
NFA for
a
b ab
}{1 baLn!
}{2 baL !
}{}}{{21 bbaababaLLnn !!
P P
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 15
Star Operation
NFA for *1L
1M
P
P
*1LP
P P
1
21
Lw
wwww
i
k
! .
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 16
Example
NFA for *}{*1 baLn!
a
b
}{1 baLn
!
P
P
P
P
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 17
Reverse
R
L1
1
NFA for
d1M
1. Reverse all transitions
2. Make initial state accepting state
and vice versa
1L
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 18
Example
}{1 baLn!
a
b
1M
}{1nR
baL !a
b
d1M
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 19
Complement
1. Take the DFA that accepts 1L
1M1Ld
1M1L
2. Make final states non-final,
and vice-versa
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 20
Example
}{1 baLn
!
a
b
1M
ba,
ba,
}{*},{1 babaLn! a
b
d1M
ba,
ba,
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Fall 2005 Costas Busch - RPI 21
Intersection
1L regular
2L regularWe show 21 LL
regular
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Fall 2005 Costas Busch - RPI 22
DeMorgans Law: 2121 LLLL !
21 , LL regular
21 , LL regular
21 LL regular
21 LL regular
21 LL regular
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 23
Example
}{1 baLn
!
},{2 baabL !
regular
regular
}{21
abLL !
regular
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 24
1Lfor for 2LFA
1M
FA
2M
Construct a new FA that accepts
Machine Machine
M 21 LL
M simulates in parallel and1M 2M
Another Proof for Intersection Closure
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 25
States in M
ji pq ,
1M
2MState in State in
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Fall 2005 Costas Busch - RPI 26
1M 2M
1q 2qa
transition
1p 2pa
transition
FA FA
11, pq a
New transition
MFA
22, pq
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 27
0q
initial state
0p
initial state
New initial state
00 , pq
1M 2MFA FA
MFA
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Fall 2005 Costas Busch - RPI 28
iq
accept state
jp
accept states
New accept states
ji pq ,
kp
ki pq ,
1M 2MFA FA
MFA
Both constituents must be accepting states
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 29
Example:
}{1 baLn!
a
b
1M
0un
}{2m
abL !
b
b
2M
0q 1q 0p 1p
0um
2q 2pa
a
ba, ba,
ba,
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Fall 2005 Costas Busch - RPI 30
00, pq
Automaton for intersection
}{}{}{ ababbaL nn !!
10, pqa
21, pq
b
ab11, pq
20, pq
a
12, pq
22, pq
b
ba,
a
b
ba,
b
a
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8/9/2019 Regular Properties
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Fall 2005 Costas Busch - RPI 31
M simulates in parallel and1M 2M
M accepts string w if and only if
accepts string andw1Maccepts string w2M
)()()( 21 MLMLML !