-
7/27/2019 5- Regular Properties
1/29
Prof. Busch - LSU 1
Properties of
Regular Languages
-
7/27/2019 5- Regular Properties
2/29
Prof. Busch - LSU 2
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:
-
7/27/2019 5- Regular Properties
3/29
Prof. Busch - LSU 3
We say: Regular languages are closed under
21LLConcatenation:
*1LStar:
21 LL Union:
1L
21 LL
Complement:
Intersection:
R
L1Reversal:
-
7/27/2019 5- Regular Properties
4/29
Prof. Busch - LSU 4
a
b
b
aNFA
Equivalent
NFA
a
b
b
a
A useful transformation: use one accept state
2 accept states
1 accept state
-
7/27/2019 5- Regular Properties
5/29
Prof. Busch - LSU 5
NFA
Equivalent NFA
Singleaccepting
state
In General
-
7/27/2019 5- Regular Properties
6/29
Prof. Busch - LSU 6
NFA without accepting state
Add an accepting state
without transitions
Extreme case
-
7/27/2019 5- Regular Properties
7/29Prof. Busch - LSU 7
1LRegular language
11 LML
1
Single accepting state
NFA 2
2L
Single accepting state
22 LML
Regular language
NFA
Take two languages
-
7/27/2019 5- Regular Properties
8/29Prof. Busch - LSU 8
}{1 baLn
a
b
1
baL 2ab
2
0n
Example
-
7/27/2019 5- Regular Properties
9/29Prof. Busch - LSU 9
Union
NFA for
1
2
21 LL
-
7/27/2019 5- Regular Properties
10/29Prof. Busch - LSU 10
a
b
ab
}{1 baLn
}{2 baL
}{}{21 babaLLn
NFA for
Example
-
7/27/2019 5- Regular Properties
11/29Prof. Busch - LSU 11
Concatenation
NFA for 21LL
1 2
-
7/27/2019 5- Regular Properties
12/29Prof. Busch - LSU 12
NFA for
a
b ab
}{1 baLn
}{2 baL
}{}}{{21 bbaababaLL nn
Example
-
7/27/2019 5- Regular Properties
13/29Prof. Busch - LSU 13
Star Operation
NFA for *1L
1
*1L
1
21
Lw
wwww
i
k
-
7/27/2019 5- Regular Properties
14/29Prof. Busch - LSU 14
NFA for *}{*1 baL n
a
b
}{1 baLn
Example
-
7/27/2019 5- Regular Properties
15/29Prof. Busch - LSU 15
Reverse
R
L1
1
NFA for1
1. Reverse all transitions
2. Make initial state accepting state
and vice versa
1L
E l
-
7/27/2019 5- Regular Properties
16/29Prof. Busch - LSU 16
}{1 baLn
a
b
1
}{1nR baL
ab
1
Example
-
7/27/2019 5- Regular Properties
17/29Prof. Busch - LSU 17
Complement
1. Take the DFAthat accepts 1L
11L
11L
2. Make accepting states non-final,
and vice-versa
-
7/27/2019 5- Regular Properties
18/29
-
7/27/2019 5- Regular Properties
19/29Prof. Busch - LSU 19
Intersection
1L regular
2L regular
We show21
LL
regular
-
7/27/2019 5- Regular Properties
20/29Prof. Busch - LSU 20
DeMorgans Law: 2121 LLLL
21 , LL regular
21 , LL regular
21 LL regular
21 LL regular
21 LL regular
-
7/27/2019 5- Regular Properties
21/29Prof. Busch - LSU 21
Example
}{1 baLn
},{2 baabL
regular
regular
}{21 abLL
regular
-
7/27/2019 5- Regular Properties
22/29
Prof. Busch - LSU 22
1Lfor for 2LDFA1
DFA
2
Construct a new DFA that accepts
Machine Machine
21 LL
simulates in parallel and1 2
Another Proof for Intersection Closure
-
7/27/2019 5- Regular Properties
23/29
Prof. Busch - LSU 23
States in
ji
pq ,
1 2State in State in
DFA DFA
-
7/27/2019 5- Regular Properties
24/29
Prof. Busch - LSU 24
1 2
1q 2qa
transition1p 2p
a
transition
DFA DFA
11, pq a
New transition
DFA
22, pq
DFA DFA
-
7/27/2019 5- Regular Properties
25/29
Prof. Busch - LSU 25
0qinitial state
0p
initial state
New initial state
00 , pq
1 2DFA DFA
DFA
DFA DFA
-
7/27/2019 5- Regular Properties
26/29
Prof. Busch - LSU 26
iq
accept state
jp
accept states
New accept states
ji pq ,
kp
ki pq ,
1 2DFA DFA
DFA
Both constituents must be accepting states
E l
-
7/27/2019 5- Regular Properties
27/29
Prof. Busch - LSU 27
Example:
}{1 baLn
a
b
1
0
n}{2
mabL
b
b
2
0q 1q 0p 1p
0
m
2q 2pa
a
ba, ba,
ba,
f
-
7/27/2019 5- Regular Properties
28/29
Prof. Busch - LSU 28
00, pq
Automaton for intersection
}{}{}{ ababbaL mn
10, pqa
21, pq
b
ab11, pq
20, pq
a
12, pq
22, pq
b
ba,
a
b
ba,
b
a
-
7/27/2019 5- Regular Properties
29/29
simulates in parallel and1 2
accepts string w if and only if:
accepts string w1
and accepts string w2
)()()( 21 MLMLML
